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https://mathoverflow.net/questions/11699 | 12 | I am posting my comment from [this question](https://mathoverflow.net/questions/8853/what-assumptions-and-methodology-do-metaproofs-of-logic-theorems-use-and-employ) as a separate question, as was recommended to me.
(EDIT: I'm sorry if it ended up being too similar a question, I just wanted to phrase it in the terminology I think of it in)
First, let me mention that I agree with Hilbert's [formalism](http://en.wikipedia.org/wiki/Formalism_%28mathematics%29) - when we make the claim "X is true", it is really a *shorthand* for: "If we label certain strings of symbols 'true', and we label certain rules for symbol manipulation 'truth-preserving', then we can manipulate our initial strings using our rules to reach X". If someone disagrees with our choice of initial strings, or our choice of manipulation rules (which, together, I'll call a "system"), then they will obviously come to some different conclusions. Now, mathematicians happen to have preferences about which systems to study, and those preferences are often motivated by apparent *analogies* between those systems and the real world, but no system is "right" or "wrong".
I realize that this is not everyone's view of mathematics, but I wanted to explain my reasoning before asking my question:
In what system do logicians usually "prove" metamathematical theorems - i.e., claims about systems? As I mentioned, all theorems, metamathematical or otherwise, are really of the form "In the system Y, X is true." The potential problem I feel with some results like Godel's theorems is that X is a claim about systems which are not necessarily consistent with Y - how does this make sense?
| https://mathoverflow.net/users/1916 | Where are we working when we prove metamathematical theorems? | There are many flavors of "meta" in logic. Most make very minimal use of the metatheory. For example, the Montague Reflection Principle in Set Theory says the following:
>
> **Metatheorem.** For every formula $\phi(x)$ of the language of Set Theory, the following is provable: If $\phi(a)$ is true then there is an ordinal $\alpha$ such that $a \in V\_\alpha$ and $V\_\alpha \vDash \phi(a)$.
>
>
>
This is in fact an infinite collection of theorems, one for each formula $\phi(x)$. Gödel's theorem prevents ZFC from proving all of these instances at the same time. The proof of this metatheorem is by induction on the length of $\phi(x)$. The requirements on the metatheory are very minimal, all you need is enough combinatorics to talk about formulas, proofs, and enough induction to make sense of it all. This is by far the most common flavor of "meta" in logic. There really is not much to it. Most logicians don't worry about the metatheory in this context. Indeed, it is often irrelevant and, like most mathematicians, logicians usually believe in the natural numbers with full induction.
However, sometimes there is need for a much stronger metatheory. For example, the metatheory of Model Theory is usually taken to be ZFC. Gödel's Completeness Theorem is a nice contrasting example.
>
> **Completeness Theorem.** Every consistent theory is satisfiable.
>
>
>
Recall that a theory is consistent if one cannot derive a contradiction form it, while a theory is satisfiable if it has a model. The consistency of a theory is a purely syntactic notion. When the theory is effective, consistency can be expressed in simple arithmetical terms in a manner similar to the one discussed above. However, satisfiability is not at all of this form since the models of a theory are typically infinite structures. This is why this theorem is usually stated in ZFC. For the Completeness Theorem, one can often get by with substantially less. For example, the system WKL0 of second-order arithmetic suffices to prove the Completeness Theorem for countable theories (that exist in the metaworld in question), but such a weak metatheory would be nothing more than a major inconvenience for model theorists.
Sometimes there are multiple choices of metatheory and no consensus on which is most appropriate. This happens in the case of Forcing in Set Theory, which has three common points of view:
* Forcing is a way to extend a countable transitive model of ZFC' to a larger model of ZFC' with different properties. Here ZFC' is a finite approximation to ZFC, which makes this statement non-vacuous in ZFC because of the Reflection Principle above.
* Forcing is a way to talk about truth in an alternate universe of sets. Here, the alternate universe is often taken to be a Boolean-valued model, formalized within the original universe.
* Forcing is an effective way to transform a contradiction some type of extensions of ZFC into contradictions within ZFC itself. Here, ZFC could be replaced by an extension of ZFC.
The last point of view, which is the least common, is essentially arithmetical since it only talks about proofs. The second point of view is well suited for hard-core platonists who believe in one true universe of sets. The first point of view, which is probably most popular, essentially takes ZFC as a metatheory much like model theorists do.
Since there is no consensus, set theorists will often talk as if both $V$ and its forcing extension $V[G]$ are absolutely real universes of sets. Although this point of view is hard to justify formally, it is possible to makes sense of it using any one of the three viewpoints above and it has the advantage that it makes it easier to express ideas that go into forcing constructions, which is what set theorists really want to talk about.
Finally, the idea of analyzing which systems prove the consistency of other systems is very common in logic. In Set Theory, these systems often take the form of large cardinal axioms. In Second-Order Arithmetic these systems often take the form of comprehension principles. In First-Order Arithmetic these systems often take the form of induction principles. Together these form an incredibly long consistency strength hierarchy stretching from extremely weak basic arithmetical facts to incredibly deep large cardinal axioms. A very significant part of logic deals with studying this hierarchy and, as Andrej Bauer commented, logicians are usually very aware of where they are sitting in this hierarchy when proving metatheorems.
| 11 | https://mathoverflow.net/users/2000 | 11756 | 7,968 |
https://mathoverflow.net/questions/11761 | 13 | Is there any special reason that we use the sines and cosines functions in the Fourier Series, while we know that if we chose any maximal orthonormal system in L2, we would get the same result?
Is it something historical or what?
Thanks in advance.
| https://mathoverflow.net/users/3124 | Why sin and cos in the Fourier Series? | $1$. Mathematical reason.
There is one reason which makes the basis of complex exponentials look very natural, and the reason is from complex analysis. Let $f(z)$ be a complex analytic function in the complex plane, with period $1$.
Then write the substitution $q = e^{2\pi i z}$. This way the analytic function $f$ actually becomes a meromorphic function of $q$ around zero, and $z = i \infty$ corresponds to $q = 0$. The Fourier expansion of $f(z)$ is then nothing but the Laurent expansion of $f(q)$ at $q = 0$.
Thus we have made use of a very natural function in complex analysis, the exponential function, to see the periodic function in another domain. And in that domain, the Fourier expansion is nothing but the Laurent expansion, which is a most natural thing to consider in complex analysis.
Here you can make suitable modifications when $f$ is periodic in some domain which is not the whole complex plane. In that case in the $q$-domain, $f$ will be analytic in some circle around $0$, and you can use that to get a Laurent expansion. The modular forms for instance are defined only in the upper-half plane, and what we get here is called the $q$-expansion.
However from the point of view of Real analysis, $L^p$-spaces etc., any other base would do just as fine as the complex exponentials. The complex exponentials are special because of complex analytic reasons.
$2$. Physical reason.
There are historical reasons also. For instance, in electrical engineering or theory of waves, it is very useful to decompose a function into its frequency components and this is the reason for the great importance of Fourier analysis in electrical engineering or in electrical communication theory. The impedance offered by circuits depends on the frequency of the signal that is being fed in, and a circuit consisting of capacitors, inductors etc. react differently to different frequencies, and thus the sine/cosine wave decomposition is very natural from a physical point of view. And it was from this context, and also the theory of heat conduction, that Fourier analysis developed up.
| 14 | https://mathoverflow.net/users/2938 | 11763 | 7,973 |
https://mathoverflow.net/questions/11768 | 7 | Let $X$ be a (edit: nonsingular) projective complex algebraic curve. The genus of $X$ can be defined as the dimension of the space of holomorphic $1$-forms on $X$, which in turn can be defined either analytically or algebraically in terms of Kahler differentials. It can also be defined as the topological genus of $X$ considered as a surface, which in turn can be defined either topologically as the number of tori in a connected sum decomposition of $X$ or homologically in terms of the Betti numbers of $X$. Does anyone know of a reasonably self-contained reference where some or all of these equivalences are proven?
(There is a [related question](https://mathoverflow.net/questions/152/how-do-you-see-the-genus-of-a-curve-just-looking-at-its-function-field) about computing the genus of a curve from its function field as well as a nice [post by Danny Calegari](http://lamington.wordpress.com/2009/09/23/how-to-see-the-genus/) explaining the relationship to the Newton polygon, but I am mostly interested in the algebraic-to-topological step of going from Kahler differentials to the number of tori in a connected sum decomposition.)
| https://mathoverflow.net/users/290 | Reference for equivalent definitions of the genus | For $\mathrm{dim} H^0(X, \Omega^1\_X) = \dim H^1(X, \mathbb{Q})$ see <http://en.wikipedia.org/wiki/Hodge_theory>. For $\dim H^1(X, \mathbb{Q}) =$ number of tori use induction and the Mayer-Vietoris sequence.
(And for $\mathrm{dim} H^0(X, \Omega^1\_X) = \mathrm{dim} H^1(X, \mathcal{O}\_X)$ see <http://en.wikipedia.org/wiki/Serre_duality>.)
| 7 | https://mathoverflow.net/users/nan | 11769 | 7,976 |
https://mathoverflow.net/questions/11684 | 6 | Take $W = S\_n$ for simplicity, though other Weyl groups work too. Let $r\_i$ denote the $i$th simple reflection acting on ${\mathbb A}^n$, and $\partial\_i = 1/(x\_i - x\_{i+1}) ({\operatorname{Id}} - r\_i)$ denote the corresponding divided difference operator.
It's easy to show that the operators $r\_i + c \partial\_i$ satisfy the Coxeter relations. I know I saw this in a Lascoux article, but there are so many that I'm hoping MathOverflow can tell me which one so I don't have to pore over the French, or can suggest some other canonical reference, the older the better.
Separately, I'd like to know if any author explicitly discusses these in the context of the Steinberg variety, where the $c$ should be the equivariant cohomology parameter corresponding to dilation of the cotangent bundle, I guess.
| https://mathoverflow.net/users/391 | Reference for representation of Weyl group using r_ + c∂_ | The difference operators (as you presumably know) are defined in a paper of Bernstein–Gelfand–Gelfand on Schubert cells etc. which is probably roughly as old as you can get. The fact that the $r\_i + c\partial\_i$ satisfy the Coxeter relations is implied by (equivalent to, pretty much) the fact that the graded/degenerate affine Hecke algebra is isomorphic as a vector space to $\mathbb C[W]\otimes \mathbb C[x\_1,x\_2,\dotsc,x\_n]$ (the operators give the action of the $\mathbb C[W]$ subalgebra on the polynomial representation of the degenerate affine Hecke algebra). The first references for the degenerate affine Hecke algebra are Drinfeld's paper [Degenerate affine Hecke algebras and Yangians](https://doi.org/10.1007/BF01077318) (for type A) and Lusztig's paper [Cuspidal local systems and graded Hecke algebras, I](http://archive.numdam.org/item/PMIHES_1988__67__145_0/), and the Lusztig paper that Stephen references, which is pure algebra. I think it also arises in some form in Cherednik's paper [A new interpretation of Gelʹfand–Tzetlin bases](https://doi.org/10.1215/S0012-7094-87-05423-8).
The connection to the equivariant cohomology of the Steinberg is examined in Lusztig's cuspidal local systems papers (it is the easiest "Springer theory" case, i.e. where you don't have to worry about cuspidal local systems).
| 4 | https://mathoverflow.net/users/1878 | 11770 | 7,977 |
https://mathoverflow.net/questions/11758 | 10 | Is there an algorithm for solving the following problem: let $g\_1,\ldots,g\_n$ be permutations in some (large) symmetric group, and $g$ be a permutation that is known to be in the subgroup generated by $g\_1,\ldots,g\_n$, can we write $g$ explicitly as a product of the $g\_i$'s?
My motivation is that I'm TAing an intro abstract algebra course, and would like to use the Rubik's cube to motivate a lot of things for my students, and would, in particular, like to show them an algorithm to solve it using group theory. (That is, I can write down what permutation of the cubes I have, and want to decompose it into basic rotations, which I then invert and do in the opposite order to get back to the solved state.) Though I'm interested in the more general case, not just for the Rubik(n) groups, if a solution works out.
Note: I don't really know what keywords to use for solving this problem, if someone can point me to the right search terms to google to get the results I'm looking for, I'll gladly close this.
| https://mathoverflow.net/users/622 | Algorithm for decomposing permutations | Yes. The general rule of thumb is that groups described by permutations are computationally easy, groups described by generators and relations have computational problems that are generally undecidable, and matrix groups are somewhere in between.
There's a whole book "Permutation group algorithms" by Seress, Cambridge University Press, 2003.
The main technique for permutation groups is called the Schreier–Sims algorithm; there's a survey [here](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.90.3009&rep=rep1&type=pdf), for instance. The rough idea is to stabilize the permuted elements one at a time. As Mitch already said, though, this doesn't find the shortest word in the generators that produces any particular group element, which is a more difficult problem.
| 11 | https://mathoverflow.net/users/440 | 11783 | 7,983 |
https://mathoverflow.net/questions/11781 | 4 | Is it possible to compute complex powers in finite fields? Given a $\in \mathbb{F}\_p$ ($p$ prime), how can one compute $a^i$ per example?
| https://mathoverflow.net/users/3264 | Complex powers in finite fields | I certainly don't think that for arbitrary complex numbers, you'll be able to define a notion of exponentiation (take $\pi$ for example). However, I think if you restrict to algebraic numbers, you'll have more of a chance. Here's something I cooked up for (sometimes) handling an $n$th root of unity $\omega$:
I assume we would like exponentiation to $\omega$ to be a (group) automorphism of $\mathbb{F}\_q^\times\simeq\mathbb{Z}/(q-1)\mathbb{Z}$ - note that $\sigma(a)=a^\omega$ would have the property that $\sigma^n=$ id, so if there is an element of $Aut(\mathbb{F}\_q^\times)$ of order $n$, then you could just define exponentiation to $\omega$ (or any of its conjugates) to be that (group) automorphism of $\mathbb{F}\_q^\times$ (and set $0^\omega=0$). Note that the group of automorphisms $Aut(\mathbb{F}\_q^\times)\simeq Aut(\mathbb{Z}/(q-1)\mathbb{Z})$ is simply $(\mathbb{Z}/(q-1)\mathbb{Z})^\times$, which has order $\phi(q-1)$, so if $n\nmid\phi(q-1)$, this wouldn't work.
| 16 | https://mathoverflow.net/users/1916 | 11788 | 7,987 |
https://mathoverflow.net/questions/11798 | 28 | I would like to know what mathematics topics are the most important to learn before actually studying the theory on neural networks.
I ask that because I will start to learn about neural networks and machine learning on my own to help in the analysis I am doing on my PhD about patterns of genome evolution.
Thank you in advance.
| https://mathoverflow.net/users/1776 | Mathematics for machine learning | For basic neural networks (i.e. if you just need to build and train one), I think basic calculus is sufficient, maybe things like gradient descent and more advanced optimization algorithms. For more advanced topics in NNs (convergence analysis, links between NNs and SVMs, etc.), somewhat more advanced calculus may be needed.
For machine learning, mostly you need to know probability/statistics, things like Bayes theorem, etc.
Since you are a biologist, I don't know whether you studied linear algebra. Some basic ideas from there are definitely extremely useful. Specifically, linear transformations, diagonalization, SVD (that's related to PCA, which is a pretty basic method for dimensionality reduction).
The book by Duda/Hart/Stork has several appendices which describe the basic math needed to understand the rest of the book.
| 11 | https://mathoverflow.net/users/3035 | 11807 | 8,002 |
https://mathoverflow.net/questions/11774 | 35 | For the definitions of the equivalence relations on algebraic cycles see <http://en.wikipedia.org/wiki/Adequate_equivalence_relation>.
I want to know how far away from each other the equivalence relations on algebraic cycles are and what the intuition is for them.
My impression is that rational equivalence gives much bigger Chow groups than algebraic equivalence, and that algebraic equivalence, homological equivalence and numerical equivalence are quite tight together.
Take for example an elliptic curve. We have $CH^1(E) = \mathbb{Z} \times E(K)$, algebraic equivalence (take $C = E$) $\mathbb{Z}$ = numerical equivalence.
| https://mathoverflow.net/users/nan | difference between equivalence relations on algebraic cycles | I will focus on complex projective varieties.
Codimension one
---------------
The situation in codimension one is considerably simpler than in higher codimensions.
Codimension one rational equivalence classes are parametrized by $Pic(X)= H^1(X,\mathcal O\_X^{\ast})$ while algebraic equivalence classes are parametrized by the Neron-Severi group of $X$, which can be defined as the image of the Chern class map from $Pic(X)$ to $H^2(X,\mathbb Z)$. It follows that in codimension one
* the group of **rational** equivalence classes is a countable union of abelian varieties;
* the groups of **algebraic** equivalence classes and **homological** equivalence classes coincide, and are equal to $NS(X)$ a subgroup of $H^2(X,\mathbb Z)$;
* the group of **numerical** equivalence classes is the quotient of $NS(X)$ by its torsion subgroup.
Higher codimension
------------------
The higher codimension case, as pointed out by Tony Pantev, is considerably more complicate and algebraic and homological equivalence no longer coincide.
Concerning rational equivalence, Mumford proved that the Chow group of zero cycles of surfaces admitting non-zero holomorphic $2$-forms are **infinite dimensional**, contradicting a conjecture by Severi. The paper is Mumford, D. Rational equivalence of $0$-cycles on surfaces. J. Math. Kyoto Univ. 9 1968.
Warning
-------
The definitions of rational and algebraic equivalence at [wikipedia](http://en.wikipedia.org/wiki/Adequate_equivalence_relation) are not correct.
I will commment below on the algebraic equivalence.
There one can find the following definition.
>
> $Z ∼\_{alg} Z'$ if there exists a curve $C$ and a
> cycle $V$ on $X × C$ flat over C, such
> that $$V \cap \left( X \times\lbrace c\rbrace \right) = Z \quad \text{ and }
> \quad V \cap \left( X \times\lbrace c\rbrace \right) = Z' $$
> for two points $c$ and $d$ on the
> curve.
>
>
>
This is not correct. The correct definition is
>
> $Z ∼\_{alg} Z'$ if there exists a curve $C$ and a
> cycle $V$ on $X × C$ flat over C, such
> that $$V \cap \left( X \times\lbrace c\rbrace \right) - V \cap \left( X \times\lbrace d\rbrace \right) = Z - Z' $$
> for two points $c$ and $d$ on the
> curve.
>
>
>
To construct an example of two algebraically equivalent divisors which do not satisfy the wikipedia definition let $X$ be a projective variety with $H^1(X,\mathcal O\_X) \neq 0$ and
take a non-trivial line-bundle $\mathcal L$ over $X$ with zero Chern class.
If $Y = \mathbb P ( \mathcal O\_X \oplus \mathcal L)$ then $Y$ contains two copies $X\_0$ and $X\_{\infty}$ of $X$ ( one for each factor of $\mathcal O\_X \oplus \mathcal L$ ) which are algebraically equivalent but can't be deformed because their normal bundles are $\mathcal L$ and $\mathcal L^{\ast}$. This does not contradict the second definition because for sufficiently ample divisors $H$ it is clear $X\_0 + H$ can be deformed into $X\_{\infty} + H$.
| 37 | https://mathoverflow.net/users/605 | 11813 | 8,005 |
https://mathoverflow.net/questions/3269 | 50 | Serre's *A Course in Arithmetic* gives essentially the following proof of the three-squares theorem, which says that an integer $a$ is the sum of three squares if and only if it is not of the form $4^m (8n + 7)$ : first one shows that the condition is necessary, which is straightforward. To show it is sufficient, a lemma of Davenport and Cassels, using Hasse-Minkowski, shows that $a$ is the sum of three *rational* squares. Then something magical happens:
Let $C$ denote the circle $x^2 + y^2 + z^2 = a$. We are given a rational point $p$ on this circle. Round the coordinates of $p$ to the closest integer point $q$, then draw the line through $p$ and $q$, which intersects $C$ at a rational point $p'$. Round the coordinates of $p'$ to the closest integer point $q'$, and repeat this process. A straightforward calculation shows that the least common multiple of the denominators of the points $p'$, $p''$, ... are strictly decreasing, so this process terminates at an integer point on $C$.
Bjorn Poonen, after presenting this proof in class, remarked that he had no intuition for why this should work. Does anyone have a reply?
Edit: Let me suggest a possible reformulation of the question as follows. Complete the analogy: Hensel's lemma is to Newton's method as this technique is to \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.
| https://mathoverflow.net/users/290 | Intuition for the last step in Serre's proof of the three-squares theorem | The intuition for this method of passing from a rational solution to an integral solution seems pretty simple to me: passing from a rational solution to a nearby integral point (not necessarily a solution) is passing to a point whose denominators are 1, so you can anticipate that when you intersect the line through your rational solution and the nearby integral point with whatever curve or surface contains your solutions, the second intersection point on that line will have *denominators that have moved closer to 1*. That is, connecting a rational solution with some integral point will spit out a new solution whose denominators are somewhere between the denominators of your solution and the denominators of the integral point you used to produce the line.
Of course intuition is one thing and checking the details is another: you choose the integral point nearby and the math has to work out to show the denominators really get smaller in the second solution you produce. For instance, this method of proving the 3-square theorem goes through without a problem for a similar 2-square theorem (if an integer is a sum of two rational squares than it's a sum of two integral squares by the same method, replacing the sphere $x^2 + y^2 + z^2 = a$ with the circle $x^2 + y^2 = a$). But this intuitive way of creating an integral solution from a rational solution breaks down if you apply it to the 4-square theorem: the inequalities in the proof just barely fail to work (sort of like doing division with remainder and finding the remainder is as big as the divisor instead of smaller).
The intuition also breaks down if you slightly change the expression $x^2 + y^2$ (sticking to two variables). Consider $x^2 + 82y^2 = 2$ and the rational solution $(4/7,1/7)$. Its nearest integral point in the plane is $(1,0)$, and the line through these intersects the ellipse in $(16/13,-1/13)$, so the denominator has gone up. There actually are no integral solutions to $x^2 + 82y^2 = 2$. Or if we take $x^3 + y^3 = 13$ and the rational solution $(2/3,7/3)$, its nearest integral point in the plane is $(1,2)$, the line through these meets the curve again in $(7/3,2/3)$, whose nearest integral point in the plane is $(2,1)$, the line through them meets the curve in $(2/3,7/3),\ldots$
A few years ago when I was giving some lectures on the method of descent, I worked out some examples of this geometric "three-square" theorem (start with an equation $a = x^2 + y^2 + z^2$ where $a$ is an integer and $x, y,$ and $z$ are rational and produce in a few steps an equation where $x, y,$ and $z$ are integral) and I noticed in my initial examples that the denominators in each new step did not merely drop, but dropped as factors, e.g., if the common denominator at first was 15 then at the next step it was 5 and then 1. Maybe the denominators always decrease through factors like this? Nope, eventually I found a case where they don't: if you start with
$$
13 = (18/11)^2 + (15/11)^2 + (32/11)^2
$$
then the integral point nearest $(18/11,15/11,32/11)$ is $(2,1,3)$ and the line through these two points meets the sphere $13 = x^2 + y^2 + z^2$ in the new point $(2/3,7/3,8/3)$, so the denominator has fallen from 11 to 3, which is not a factor. (At the next step you will terminate in the integral solution $(0,3,2)$.)
| 34 | https://mathoverflow.net/users/3272 | 11822 | 8,009 |
https://mathoverflow.net/questions/11821 | 2 | Can anyone give me an explicit isomorphism between $SU(2)$ and the three sphere?
What about for higher spheres? This question [link text](https://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotient) seems to indicate that there exists a homeomorphism from $SU(n)/SU(n-1)$ to the $(2n-1)$-sphere.
| https://mathoverflow.net/users/1977 | $SU(2)$ and the three sphere | Elements of $SU(2)$ look like this:
$$ x = \begin{pmatrix} a & - \overline{b} \\ b & \overline{a} \end{pmatrix},$$
where $|a|^2 + |b|^2 = 1$. This follows easily from $x^\* = x^{-1}$. So you map that matrix to the point $(a,b)$ in $\mathbb{C}^2$, and this is your diffeomorphism.
| 8 | https://mathoverflow.net/users/703 | 11823 | 8,010 |
https://mathoverflow.net/questions/11364 | 20 | This question is inspired by [Cohomology of fibrations over the circle](https://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle) Moreover, it can be considered a subquestion of the above, but somehow it seems to me that some of the more interesting points were not addressed there. So I decided to ask a bit more specific question to emphasize some of those points, but I would not mind at all if someone merges this question with the above.
Let $E\to S^1$ be a fiber bundle with fiber $F$, and assume we know $H^{\bullet}(F,\mathbf{Q})$ as a ring and the monodromy action on it. Notice that since the base is a circle, the Leray spectral sequence degenerates in the second term for dimension reasons. So we have an exact sequence $0\to I\to H^{\bullet}(E,\mathbf{Q})\to Q\to 0$ where $I$ is the kernel and $Q$ is the image of $H^\*(E,\mathbf{Q})\to H^{\bullet}(F,\mathbf{Q})$. In this sequence we know $Q,I$ and the action of $Q$ on $I$ from the Leray spectral sequence.
1. Does this suffice to determine the rational cohomology of $E$ as a ring, up to isomorphism? My guess is that probably not, but I can't find a counter-example off hand.
2. If not, would the higher Massey products on $H^\*(F,\mathbf{Q})$ allow one to compute the cup product on $H^{\bullet}(E,\mathbf{Q})$?
3. If not, would a rational homotopy model $A$ of the fiber suffice, together with an automorphism $A\to A$ that covers up to homotopy the "monodromy" automorphism of the differential forms on $F$? My guess is that probably yes, but notice that computing models of fibrations with non-simply connected bases can be tricky in general: take for example the space $X$ of all (ordered) couples of distinct points of the real projective plane and the projection on the first factor: the fiber is a M\"obius band, which contracts to the circle and the monodromy action changes the sign of the generator in degree 1; but we have $H^i(X,\mathbf{Q})=\mathbf{Q}$ if $i=0,3$ and zero otherwise.
| https://mathoverflow.net/users/2349 | Cohomology of fibrations over the circle: how to compute the ring structure? | This is a continuation of Ryan's answer above, but it has become too large for a comment. I wanted to work out the details of Ryan's example explicitly, so that we can see explicitly where your conditions fail to determine the cohomology; perhaps this can help you to pin down precisely what conditions you want. It doesn't seem that we actually need Kitano here, just Johnson's classic results.
Let $S\_g\to M^3\to S^1$ be a mapping torus of an element of the Torelli group, i.e. a diffeomorphism $S\_g\to S\_g$ acting trivially on homology. Such a bundle admits cohomology classes satisfying the [Leray-Hirsch condition](http://en.wikipedia.org/wiki/Leray-Hirsch_theorem) [this is a fun exercise], implying that as $H^{\ast}(S^1)$-modules, $H^\ast(M^3) = H^\ast(S\_g) \otimes H^\ast(S^1)$. Thus the following do not depend on the monodromy:
* $Q = H^\ast(S\_g)$,
* $I$, which is $Q$ with grading shifted by 1 (if $H^\ast(S^1) = \mathbb{Z}[t]/t^2$, this is $tQ$)
* the action of $Q$ on $I$ (just the action of $Q$ on $tQ$),
* and the Massey products on $Q = H^\ast(S\_g)$ [although perhaps I misunderstand what you mean here].
However, Johnson's work implies that your 3-manifold has the same cohomology ring as the product $S\_g \times S^1$ iff the monodromy lies in the kernel of a certain homomorphism called the Johnson homomorphism; in particular, the ring $H^\ast(E)$ depends on the monodromy. It seems this shows that the answers to 1) and 2) are both "No".
Now we can compare this with your conditions to see exactly what information we're missing; it turns out to be exactly the "Johnson homomorphism". The exact sequence above $0\to I\to H^\ast(E)\to Q\to 0$ has a splitting as abelian groups $H^\ast(E) = Q\oplus tQ$ coming from the Leray-Hirsch theorem as above. The only information we don't know automatically is the cup product on $Q$ in this splitting with itself. We know when we project back to the $Q$ factor we recover the cup product there, which means that the missing information is the projection onto the $tQ$ factor. Letting e.g. $Q(1)$ denote the degree 1 part, the cup product is a map $Q(1) \wedge Q(1) \to H^2(E)$. Projecting onto the $tQ$ factor, we have $Q(1) \wedge Q(1) \to tQ(2)$. But both $Q(1)$ and $tQ(2)$ are isomorphic to $H^1(S\_g)$, so this projection of cup product is a map $\bigwedge^2 H^1(S\_g) \to H^1(S\_g)$. This exactly encodes the data that is not determined by your conditions; Johnson's beautiful result is that this map is exactly the Johnson homomorphism, originally defined from the algebraic properties of the monodromy. In particular he showed that this missing data could be zero or nonzero, and in fact can be anything in the subspace $\bigwedge^3 H^1<\textrm{Hom}(\bigwedge^2 H^1,H^1)$.
This was first laid out in Johnson's survey "A survey of the Torelli group" ([MR0718141](http://www.ams.org/mathscinet-getitem?mr=718141)), and the details are worked out carefully in Hain, "Torelli groups and geometry of moduli spaces of curves" ([MR1397061](http://www.ams.org/mathscinet-getitem?mr=1397061)). What Kitano is doing is different, or rather a generalization of this: showing that just as the cup product on $H^\ast(E)$ detects the Johnson homomorphism, the higher Massey products on $H^\ast(E)$ detect "higher Johnson homomorphisms" measuring deeper algebraic invariants. (If any of this is useful, please consider it a partial repayment for your beautiful summary of Hodge theory in [this answer](https://mathoverflow.net/questions/11393/why-the-similarity-between-hodge-theory-for-compact-riemannian-and-complex-manifo/11407#11407).)
| 15 | https://mathoverflow.net/users/250 | 11826 | 8,013 |
https://mathoverflow.net/questions/11828 | 16 | Recently I've been reading J.P. May's [A Concise Course in Algebraic Topology](http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf). In the section on the classification of covering groupoids, he mentions that sometimes a group G may have two conjugate subgroups H and H' such that H is properly contained in H' (on pp. 26-27, according to his numbering). This seems bizarre to me, and I'm pretty sure I've seen an example before, but I'm having trouble coming up with one now.
Anyways, he continues by saying that it is possible to have an endomorphism of a covering groupoid which is not an isomorphism. I'd like to come up with an example of this, and I'm pretty sure that for me obstruction lies in failing to completely grasp the group-theoretic statement above.
(Of course, when I think of a covering of groupoids I'm secretly thinking about a covering space, partly because this is his motivation for introducing groupoids and partly because it's just easier for me, so ideally but not necessarily the example would really just be a map of covering spaces over the same base space.)
| https://mathoverflow.net/users/303 | two conjugate subgroups and one is a proper subset of the other? plus, a covering space interpretation. | The subgroup $H=\mathbb Z^\mathbb N$ of $G\_1=\mathbb Z^\mathbb Z$ is mapped to a proper subgroup by translation. By considering the semidirect product $G\_1\rtimes\mathbb Z$, you can make translation on $G\_1$ an inner automorphism.
The first theorem on p.26 tells you that if $G=\pi(B,b)$, $H=p(\pi(E,e))$, $H\_1=p'(\pi(E',e'))$, the unique map $g\colon E\to E'$ satisfying $g(e)=e'$ is not an isomorphism. However, by the first proposition on p.23 there is an $e'\_1\in E'$ such that $p(\pi(E',e'\_1))=p(\pi(E,e))$, and the corresponding $E\to E'$ is an isomorphism.
| 10 | https://mathoverflow.net/users/2035 | 11834 | 8,018 |
https://mathoverflow.net/questions/11827 | 5 | Looked around a bit and couldn't seem to find a similar question. (either that or it was worded with vocabulary above the multivariable calculus I've taken. :))
Roughly worded: I would like to develop an algorithm (either in the form of "action to take each discrete time step" or "do these actions at exactly these times") for navigating a rigid body vehicle time-optimally in a frictionless environment from point A to point B.
This specific instance happens to be a Spaceship, whom I want to get from point A to point B utilizing Forward thrusters, reverse thrusters (for braking), torque thrusters, and an omnidirectional thruster for minor position/velocity corrections. The environment is 2 dimensional, though if someone knows preexisting work in 3 dimensions I can extrapolate a simpler solution from that. I've looked around the web a bit, and was unable to find anything other than some work on steering behaviors, which always assume point particles and thus do not factor torque into the equations.
I've worked for a couple of days on this problem using standard Newtonian equations, (p(t) = at^2/2 + vt + p(0)), but the entire problem is polluted by the torque calculations, such that torque is applied to turn towards and then slow down and stop at an angle that changes based on the objects velocity and the time it would take to turn to that angle. :-S 9 pages of scribbling and several frustrated nights later, a friend Reccomend I ask here.
A generic solution (which incorporates initial velocity and angle) with a high degree of accuracy would be preferred, though in the end if I have to just "fake it" to look nice (space based RTS), that would be fine too.
| https://mathoverflow.net/users/3275 | Navigation solution for frictionless vehicles. | The optimal solution is given by the proportional navigation guidance law, see for example the Wikipedia page: [on proportional navigation](http://en.wikipedia.org/wiki/Proportional_navigation). This solution applies for the general case where the target point B is moving. In our case it is fixed which doesn't change the character of the solution. The principle of the solution is as follows: If the velocity vector happens to be in the direction of the line of sight that is in the direction connecting the current position with the point B, then the vehicle should proceed in the same direction, otherwise, the line of sight has a rate of rotation which can be computed using using the instantaneous geometry; in this case the navigation law requires accelarting the vehicle perpendicular to the line of sight by a force proportional to rate of rotation of the line of sight. This solution leads to a minimum time to reach the point B.
| 1 | https://mathoverflow.net/users/1059 | 11847 | 8,028 |
https://mathoverflow.net/questions/11816 | 6 | The title says it all. In one of his answers to the question "Convex hull in CAT(0)" (I don't have the points to post a link, if someone doesn't mind link-ifying this that would be cool), Greg Kuperberg said that GL(n,C)/U(n) is a CAT(0) space. I was wondering why this is true, or if there's a reference for this.
| https://mathoverflow.net/users/2669 | Why is GL(n,C)/U(n) a CAT(0) space? | It also suffices to check that the sectional curvature of this space, using the Riemannian metric induced by the Killing form, is nonpositive. I recommend that you both figure out how to do the explicit calculation of the sectional curvature for this particular example and learn the general theory referred to in Andy's answer. They are, of course, essentially the same answer.
| 2 | https://mathoverflow.net/users/613 | 11862 | 8,041 |
https://mathoverflow.net/questions/11621 | 16 | Let $X$ be a smooth projective variety over $F$, and $E/F$ - finite Galois extension.
There is an extension of scalars map $CH^\\*(X) \to CH^\\*(X\_E)$. The image lands in the Galois invariant part of $CH^\\*(X\_E)$, and in the case of rational coefficients, all Galois-invariant cycles are in the image (EDIT: this follows from taking the trace argument).
With integer coefficients Galois-invariant classes don't have to descend.
For example, for $CH^1(X) = Pic(X)$ there is an exact sequence:
$$
0 \to Pic(X) \to Pic(X\_E)^{Gal(E/F)} \to Br(F),
$$
so we can say that the obstruction to descend a cycle lies in a Brauer group.
Are there any known obstructions to descend elements of higher groups $CH^i$ with integer coefficients?
In my case I have a cycle in $CH^\*(X\_E)^{Gal(E/F)}$ and I want to find out whether or not it's coming from $CH^\*(X)$.
(The actual cycle is described in [here](https://mathoverflow.net/questions/9200/geometry-of-the-multilagrangian-grassmannian).)
| https://mathoverflow.net/users/2260 | Obstructions to descend Galois invariant cycles | Let us keep the notations from above, and let's write $G:=\mathrm{Gal}(E/F)$. Let me quickly recall the origin of the Brauer obstruction: it really comes from the Hochschild-Serre spectral sequence
$$H^p(G,E^q(X\_E,\mathbf{G}\_m))\Longrightarrow E^{p+q}(X,\mathbf{G}\_m)$$
(I'm writing $E^{\ast}=H^{\ast}\_{\mathrm{et}}$ for étale cohomology here, because the system doesn't seem to like too many subscripts.) If we analyze this in low degrees, this gives us the following classical exact sequence (for any $F$-variety $X$):
$$0\to H^1(G,E^0(X\_E,\mathbf{G}\_m))\to\mathrm{Pic}(X)\to H^0(G,\mathrm{Pic}(X\_E))\to H^2(G,E^0(X\_E,\mathbf{G}\_m))\to\ker\left[\mathrm{Br}(X)\to\mathrm{Br}(X\_E)\right]$$
$$\to H^1(G,\mathrm{Pic}(X\_E))\to H^3(G,E^0(X\_E,\mathbf{G}\_m))$$
So we'd like to generalize the sequence above to the situation of $\mathrm{CH}^n(X)=H^n(X,\mathcal{K}\_n)$ (where $\mathcal{K}\_n$ is the Zariski-sheafification of the presheaf $K\_n$). Assume $X$ geometrically regular here. The Gersten resolution of $\mathcal{K}\_n$ on $X\_E$ is the complex
$$C^{\bullet}(X\_E)\colon\quad K\_nk(X\_E)\to\bigoplus\_{x\in X\_E^1}K\_{n-1}k(x)\to\bigoplus\_{x\in X\_E^2}K\_{n-2}k(x)\to\cdots\to\bigoplus\_{x\in X\_E^{n-1}}K\_1k(x)\to\bigoplus\_{x\in X\_E^n}K\_0k(x)$$
There's a similar complex $C^{\bullet}(X)$ giving the Gersten resolution of $\mathcal{K}\_n$ on $X$. We regard the complex $C^{\bullet}(X\_E)$ as a $G$-complex; write $\sigma$ for the map
$$C^{n-1}(X\_E)=\bigoplus\_{x\in X\_E^{n-1}}k(x)^{\times}=\bigoplus\_{x\in X\_E^{n-1}}K\_1k(x)\to\bigoplus\_{x\in X\_E^n}K\_0k(x)=Z^n(X\_E)$$
of $G$-modules. I want to argue that the kernel of this map is playing the role of $E^0(X\_E,\mathbf{G}\_m)$ for higher $n$. (When $n=1$, this kernel coincides with $E^0(X\_E,\mathbf{G}\_m)$.)
Now we might hope for a spectral sequence
$$H^p(G,H^q(C^{\bullet}(X\_E)[n]))\Longrightarrow H^{p+q}(C^{\bullet}(X)[n])$$
but of course $K$-theory doesn't quite satisfy Galois descent, so we don't have this convergence ($C^{\bullet}(X)[n]$ is not the homotopy fixed-point complex of $C^{\bullet}(X\_E)[n]$). But we're only trying to analyze a very small piece of this spectral sequence — the piece involving $\sigma$. For that, Hilbert Theorem 90 does the work, and we get the following exact sequence:
$$0\to H^1(G,\ker\sigma)\to\mathrm{CH}^n(X)\to H^0(G,\mathrm{CH}^n(X\_E))\to H^2(G,\ker\sigma)\to\ker\left[H^2\left(G,C^{n-1}(X\_E)\right)\to H^2(G,Z^n(X\_E))\right]$$
$$\to H^1(G,\mathrm{CH}^n(X\_E))\to H^3(G,\ker\sigma)$$
So we find an obstruction to descending cycles of codimension $n$ in $H^2(G,\ker\sigma)$. Is this the sort of thing you had in mind?
| 11 | https://mathoverflow.net/users/3049 | 11863 | 8,042 |
https://mathoverflow.net/questions/11845 | 3 | Automata theory is mainly concerned with Turing machines and all its relatives-in-spirit. $\lambda$-calculus is rather rarely mentioned in textbooks on automata theory.
What's the common name of the theory mainly concerned with $\lambda$-calculus and its relatives? (I think, "mathematical logic", "computability theory", "programming language theory" and "recursion theory" are too general, compared to "automata theory". But there should be an "$\lambda$-theory", shouldn't it?)
| https://mathoverflow.net/users/2672 | Theory mainly concerned with $\lambda$-calculus? | ["combinatory logic"](http://plato.stanford.edu/entries/logic-combinatory/)
| 4 | https://mathoverflow.net/users/2361 | 11876 | 8,050 |
https://mathoverflow.net/questions/11904 | 8 |
>
> **Question:** Is there a polynomial map from $\Bbb R^n$ to $\Bbb R^n$ under which the image of the positive orthant (the set of points with all coordinates positive) is all of $\Bbb R^n$ ?
>
>
>
Some observations:
My intuition is that the answer must be 'no'... but I confess my intuition for this sort of geometric problem is not very well-developed.
Of course it is relatively easy to show that the answer is 'no' when $n=1$. (In fact it seems like a nice homework problem for some calculus students.) But I can't seem to get any traction for $n>1$.
This feels like the sort of thing that should have an easy proof, but then I remember feeling that way the first time I saw the Jacobian conjecture... now I'm wary of statements about polynomial maps of $\Bbb R^n$ !
| https://mathoverflow.net/users/2502 | A polynomial map from $\Bbb R^n$ to $\Bbb R^n$ mapping the positive orthant onto $\Bbb R^n$? | The map $z\in\mathbb C\mapsto z^4\in\mathbb C$, when written out in coordinates, is a polynomial map which sends the *closed* first quadrant to the whole of $\mathbb R^2$---and by considering cartesian products you get the same for $\mathbb R^{2n}=\mathbb C^n$.
**Later:** as observed in a comment by [Charles](https://mathoverflow.net/users/622/charles-siegel), this can be turned into a solution for the *open* quadrant by composing with a translation, as in $z\in\mathbb C\mapsto (z-z\_0)^4\in\mathbb C$ with $z\_0$ in the open first quadrant.
| 16 | https://mathoverflow.net/users/1409 | 11905 | 8,071 |
https://mathoverflow.net/questions/11908 | 1 | This question is motivated by my most spectacular [answer](https://mathoverflow.net/questions/10239/is-it-true-that-as-z-modules-the-polynomial-ring-and-the-power-series-ring-over/10240#10240) on MO (:
Let $A$ be a module over $\mathbb Z$. $A$ is said to be torsion-free if $na=0$ implies $n=0$ or $a=0$ for any $n\in \mathbb Z, a\in A$. $A$ is torsionless if the map $\phi: A \to A^{\*\*}$ is injective (here ${}^\*$ means $\text{Hom}\_{\mathbb Z}(A,\mathbb Z)$).
If $A$ is finite, then torsion-free and torsionless are equivalent. In general, it is not hard to see that being torsionless implies torsion-free. On the other hand, $A=\mathbb Q$ is torsion-free but not torsionless since $A^\*=0$. But the question and answers quoted above (which shows that for $A=\mathbb Z[x]$, $\phi$ is an isomorphism) raised the following:
Question: If $A$ is a finite $\mathbb Z[x\_1,...,x\_d]$-module, are being torsion-free and torsionless equivalent?
| https://mathoverflow.net/users/2083 | Torsion-free and torsionless abelian groups | If by finite you mean finitely presented, then the answer is no. For instance, Let $A = \mathbb{Z}[x]/(2x-1) = \mathbb{Z}[\frac12]$. Then, like $\mathbb{Q}$, $\text{Hom}(A,\mathbb{Z}) = 0$.
| 5 | https://mathoverflow.net/users/1450 | 11910 | 8,073 |
https://mathoverflow.net/questions/11899 | 6 | In [wikipedia](http://en.wikipedia.org/wiki/Sheaf_%28mathematics%29#Sheaves), sheaves were first defined in the case of concrete categories (with usual identity and gluing axioms), then in the general case. (writing it as an "exact" sequence)
Do these two definitions agree? I find the definition for concrete categories case very strange, for if we consider a topological space of two points a,b with discrete topology, and let us consider a sheaf of topological spaces on it that assigns A to a, B to b. According to the "concrete-category-case" definition, we need the global sections to look like $A \times B$ such that the projection maps are continuous and nothing else. But if we look at the "equalizer" definition, we would require the global sections to carry the product topology as well.
So is wikipedia wrong? Or am I misunderstanding something? Thanks!
Edit: There is another not-quite-related question. In wikipedia, for the "equalizer" definition they require the category, where the sheaf's taking values in, to have products. Is this really necessary? In EGA Chapter 0 p.23 for example, the product is just "splitted", and we consider the large family of maps all together. It seems that these two approaches are just the same. Or am I wrong?
| https://mathoverflow.net/users/nan | Definition of sheaves in wikipedia | I think you're quite right; Wikipedia's "concrete definition" is only correct for concrete categories whose underlying-set functor is (not just faithful but) conservative, i.e. such that any morphism which is a bijection on underlying sets is an isomorphism in the category. The page does say that the concrete definition "applies to the most common examples such as sheaves of sets, abelian groups and rings," all of which have this property, but it ought to be fixed to make clear in exactly what situations this definition applies.
Secondly, I observe that the "normalisation" condition in the Wikipedia concrete definition is also odd. Since the empty set is covered by the empty family, the "local identity" and "gluing" conditions already imply that the underlying set of $F(\emptyset)$ is terminal. Saying that in addition, $F(\emptyset)$ itself is terminal is an additional condition, which is in fact a special case of the second, more generally applicable, definition.
Thirdly, I think you're also right that for the correct general definition, the category doesn't need to have any limits a priori; you can just assert that $F(U)$ is the limit of the appropriate diagram of the $F(U\_i)$ and $F(U\_i\cap U\_j)$.
Finally, let me go out on a limb and say that it seems to me that defining "sheaves with values in an arbitrary category" is often a misguided thing to do. More often, it seems like rather than "a sheaf with values in the category of X," the important notion is "an internal X in the category of sheaves of sets." For familiar cases such as groups, abelian groups, rings, small categories—in fact, for any finite limit theory—the two are the same, which may be what leads to the confusion. But the good notion of "sheaf of local rings," for instance, is not a sheaf with values in the category of local rings, but rather a sheaf of rings whose *stalks* are local (at least, when there are enough points), and that's the same as an *internal* local ring in the category of sheaves of sets. The situation is similar, I think, for "sheaves of topological spaces" (or locales). I'd be happy for people to point out where I'm wrong about this, though.
| 10 | https://mathoverflow.net/users/49 | 11911 | 8,074 |
https://mathoverflow.net/questions/11026 | 27 | Epstein's (et al.) "Word Processing in Groups" is a quite comprehensive monograph on automatic groups, finite automata in geometric group theory, specific examples like braid groups, fundamental groups of 3-dim manifolds etc. However, the book is from 1992, so much of the material summarizes research done by Cannon, Thurston, Holt etc. back in the '80s. I'm interested in how the theory of automatic groups (and, more generally, applications of formal languages in group theory) has progressed since then - have there been any significant new results, open problems, novel ideas, examples?
| https://mathoverflow.net/users/2192 | Automatic groups - recent progress | I'm not an expert in the area, but here's a few highlights:
Bridson [distinguished automatic and combable groups.](https://www.ams.org/mathscinet-getitem?mr=2016694)
Burger and Mozes found examples of [biautomatic simple groups.](https://www.ams.org/mathscinet-getitem?mr=1839489)
Mapping class groups were originally shown to be automatic by Mosher. Recently, Hamenstadt has shown that [they are biautomatic](https://arxiv.org/pdf/0912.0137.pdf).
See part 3 of McCammond's survey for an [update on open problems](http://aimath.org/pggt/Decision_Problems).
A well-known open problem is whether automatic groups have solvable conjugacy problem.
| 14 | https://mathoverflow.net/users/1345 | 11915 | 8,078 |
https://mathoverflow.net/questions/11907 | 2 | I am fitting a linear regression line to my data and computing the confidence interval for some predicted $y$ (independent variable) (<http://people.stfx.ca/bliengme/ExcelTips/RegressionAnalysisConfidence2.htm>).
Now I want to do the inverse. I need a way to measure the confidence for some predicted $x$ (dependent variable) using the same regression. How can I achieve this?
Thank you!
| https://mathoverflow.net/users/3285 | Linear Regression Confidence Interval | Usually you predict a dependent variable y and calculate a confidence interval, i.e. given x0, you calculate [y-, y+] where y will probably lie in.
For the reverse, if you have a y0 and want to find [x-, x+] for whatever reasons, regression will not help.
The appropriate tool for this kind of analysis could be structural equations modeling [1]
[1] <http://en.wikipedia.org/wiki/Structural_equation_modeling>
| 1 | https://mathoverflow.net/users/1313 | 11921 | 8,083 |
https://mathoverflow.net/questions/11737 | 0 | I designed a kernel function (to be used within SVM) which has the expression $tr(AB)$ in it. For efficient implementation of this, I was wondering if I could write $tr(AB)$ as an inner product: $\phi(A)^T \phi(B)$? What is the function $\phi()$?
| https://mathoverflow.net/users/3244 | Get rid of tr() in SVM kernel trick | Matus is right. But if the matrices $A$, and $B$ have certain properties like being symmetric, or diagonal, then simply just vectorizing the matrices and taking their inner product would be equal to the $tr(AB)$.
| 0 | https://mathoverflow.net/users/3287 | 11924 | 8,084 |
https://mathoverflow.net/questions/11934 | 15 | I recently stumbled across this number, and then (foolishly, most likely) decided to try to describe it in a blog post
<http://frothygirlz.com/2010/01/14/big-numbers-part-2/>
Q - Are there any comparisons of Graham's Number, hell, even G1, to other well known "big" numbers, such as googolplex?
I'd just like to have some way, however abstract, to be able to pretend that I have some sort of idea of the magnitude of this number.
Any help and/or tips would be much appreciated.
| https://mathoverflow.net/users/3290 | Magnitude of Graham's Number? | I think all that can really be said is that a googolplex is much, much smaller.
This isn't a direct comparison of a googolplex with Graham's number, but maybe it will help give some perspective:
Some back-of-the-envelope/Mathematica calculations tell me that
$10^{(10^{100})}\approx 3^{(3^{(3^{4.86})})}$
and so a googolplex is between $3\uparrow\uparrow 4=3^{(3^{(3^{3})})}$ and $3\uparrow\uparrow 5=3^{(3^{(3^{27})})}$ (much closer to $3\uparrow\uparrow 4$, obviously).
| 17 | https://mathoverflow.net/users/1916 | 11936 | 8,090 |
https://mathoverflow.net/questions/11932 | 18 | I find the following description of Artin motives in [Wikipedia](http://en.wikipedia.org/wiki/Motive_%28algebraic_geometry%29%23Tannakian_formalism_and_motivic_Galois_group). Since these seem to be quite related to number theory, I am interested to learn more in that context. I request the experts available in MO to provide me a reference which can be more useful to me than this terse description. Any comments and explanations also will be helpful. I apologize for asking a seemingly basic question; but I find it impossible to wade through the numerous references available on motives.
Fix a field $k$ and consider the functor
finite separable extensions $K$ of $k$ → finite sets with a (continuous) action of the absolute Galois group of $k$
which maps $K$ to the (finite) set of embeddings of $K$ into an algebraic closure of $k$. In Galois theory this functor is shown to be an equivalence of categories. Notice that fields are $0$-dimensional. Motives of this kind are called Artin motives. By $\mathbb Q$-linearizing the above objects, another way of expressing the above is to say that Artin motives are equivalent to finite $\mathbb Q$-vector spaces together with an action of the Galois group.
I have some idea of what are pure motives and mixed motives, in the context of algebraic varieties. What I exactly have in mind is to understand the modern statement of equivariant Tamagawa number conjecture. This would appear to be the simplest instance to keep in mind, if I go ahead.
| https://mathoverflow.net/users/2938 | References for Artin motives | André's book is the main reference for the "yoga" of motives. You'll find a description of Artin motives in the Voevodsky formalism in
Beilinson and Vologodsky - <http://www.math.uiuc.edu/K-theory/0832/>
Wildehaus - <http://www.math.uiuc.edu/K-theory/0918/>
From the tannakian view point, Artin motives are just representation of the usual Galois group. So, as motives, Artin motives are not that interesting. It's just the usual Galois theory of fields.
| 6 | https://mathoverflow.net/users/1985 | 11938 | 8,092 |
https://mathoverflow.net/questions/11939 | 4 | Given a finite graph $\Gamma$, one has the
[right-angled Artin group](http://en.wikipedia.org/wiki/Right-angled_Artin_group#Right-angled_Artin_groups) $A(\Gamma )$. Its generators $s\_1, \dots s\_n$ bijectively correspond to vertices of $\Gamma$ and the relators are $s\_is\_j=s\_js\_i$ provided the corresponding vertices are joined by an edge.
Let $A\_i(\Gamma)$ be the group obtained from $A(\Gamma)$ by setting $s\_i=1$; this corresponds to removing the vertex $i$ from $\Gamma$.
I know very little of these matters but it seems plausible that any nontrivial element of $A(\Gamma)$ projects to a nontrivial element of $A\_i(\Gamma)$ for some $i$; is this correct?
| https://mathoverflow.net/users/1573 | "Remove a vertex" map for right-angled Artin groups | No. Let $\Gamma$ be the graph with two vertices and no edges - the non-abelian free group of rank two - and let $g$ be the commutator of the two generators $s\_1$ and $s\_2$. Then $g$ is certainly non-trivial, but $g$ dies whenever you kill $s\_1$ or $s\_2$.
UPDATE:
For an example with a connected graph, let's take $\Gamma$ to be the straight-line graph with four vertices $a,b,c,d$ (so $[a,b]=[b,c]=[c,d]=1$). Now consider $g=[[c,a],[b,d]]$. Clearly this dies when you kill any generator. On the other hand,
$g=cac^{-1}a^{-1}bdb^{-1}d^{-1}aca^{-1}c^{-1}dbd^{-1}b^{-1}$
and a well-known solution to the word problem in right-angled Artin groups tells you that $g$ is non-trivial.
| 4 | https://mathoverflow.net/users/1463 | 11940 | 8,093 |
https://mathoverflow.net/questions/11929 | 6 | I was hoping that someone could help clarify a source of confusion for me, I must be doing and saying something wrong but I just don't know what:
Let $E$ be an elliptic curve over $\mathbb{Q}$ and let
$$L(s,E)=\sum\_{n=1}^{\infty}a\_n(E)n^{-s}$$
be the Hasse-Weil $L$-function of $E$. Finally, let $\tilde{E}$ be the reduction of $E$ mod $p$ and assume that $p$ is a prime for which $E$ has good reduction.
Then
$$a\_p(E)=p+1-|\tilde{E}(\mathbb{F}(p))|$$
and setting $a\_1(E)=1$ the $p$ power coefficients are given by
$$a\_{p^e}(E)=a\_p(E)a\_{p^{e-1}}(E)-pa\_{p^{e-2}}(E).$$
Now looking at Diamond and Shurman, for instance, I find that also we can write
$$a\_{p^e}(E)=p^e+1-|\tilde{E}(\mathbb{F}(p^e))|$$
but when I use this expression as a "definition" of $a\_{p^e}(E)$ and do some explicit calculations I don't get the right recursion, for instance I seem to get in practice
$$a\_{p^2}(E)=a\_p(E)^2 - 2p$$
instead of
$$a\_{p^2}(E)=a\_p(E)^2-p.$$
I must be misunderstanding something, but I can't figure out what. Any help?
| https://mathoverflow.net/users/3289 | Alternate expresion of L-series coefficients | There are two different recursions involved here, one for the points of $E$ over ${\mathbb F}\\_{p^n}$, and the other for the coefficients of the $L$-function.
If we write $a\_p = \alpha + \beta,$ where $\alpha\beta = p$ (so $\alpha$ and $\beta$ are
the two roots of the char. poly. of Frobenius), then
$$1 + p^n - E({\mathbb F}\\_{p^n}) = \alpha^n + \beta^n.$$
On the other hand, the Euler factor at $p$ for the $L$-function of $E$ is
$$(1 - \alpha p^{-s})^{-1}(1-\beta p^{-s})^{-1}$$
$$= (1 + \alpha p^{-s} + \alpha^2 p^{-2s} + \cdots )(1 + \beta p^{-s} + \beta^2 p^{-2s} +
\cdots )$$
$$= 1 + (\alpha + \beta) p^{-s} + (\alpha^2 + \alpha\beta + \beta^2) p^{-2s} +
\cdots ,$$
and so we conclude that $a\_{p^n}$ (the coefficient of $p^{-ns}$ in the $L$-function)
equals
$$\alpha^n + \alpha^{n-1} \beta + \cdots + \alpha\beta^{n-1} + \beta^n.$$
These formulas are simply different, as soon as $n > 1.$ The recursion given in
the question describes the second, and not the first.
| 4 | https://mathoverflow.net/users/2874 | 11944 | 8,096 |
https://mathoverflow.net/questions/11916 | 12 | In [Theory mainly concerned with lambda-calculus?](https://mathoverflow.net/questions/11845/theory-mainly-concerned-with-lambda-calculus/11861#11861), F. G. Dorais wrote, of the idea that the lambda-calulus defines a domain of mathematics:
>
> That would never stick unless there's another good reason. Besides, the schism between cs and math is very recent, I would contend that "functional programming" is actually a math term, historically speaking. More importantly, it would be wrong to use a term different than those who use it most, namely theoretical computer scientists, who are very competent mathematicians by the way.
>
>
>
The idea, I think, is that the overlap between the kind of constructive mathematics that follows the formulae-as-types correspondence, and pure functional programming is so substantial that the core of the two topics is essentially the same subject.
Is this true?
| https://mathoverflow.net/users/3154 | Is functional programming a branch of mathematics? | So, I'm a computer scientist working in this area, and my sense is the following:
You cannot do good work on functional programming if you are ignorant of the logical connection, period. However, while "proofs-as-programs" is a wonderful slogan, which captures a vitally important fact about the large-scale structure of the subject, it doesn't capture *all* the facts about the programming aspects.
The reason is that when we look at a lambda-term as a program, we additionally care about its *intensional* properties (such as space usage and runtime), whereas in mathematics we only care about the *extensional* properties (i.e., the relation it computes).[\*] Bubble sort and merge sort are the same function under the lens of $\beta\eta$ equality, but no computer scientist can believe these are the same algorithm.
I hope, of course, that one day these intensional factors will gain logical significance in the same way that the extensional properties already have. But we're not there yet.
[\*] FYI: Here I'm using "intensional" and "extensional" in a difference sense than is used in Martin-Lof type theory.
| 10 | https://mathoverflow.net/users/1610 | 11976 | 8,115 |
https://mathoverflow.net/questions/11977 | 2 | Let $G$ be a group and $X$ a set equipped with a *transitive* right $G$-action. Further, let $c: X\to X$ be a $G$-equivariant map. Is it true that $\text{Stab}(x) = \text{Stab}(c(x))$ for all $x\in X$?
This doesn't seem to be an interesting mathoverflow question on its own, but the reason I ask is the following: In Hovey's book on Model Categories Hovey proves some sort of five lemma for pointed model categories (Thm.6.5.3). In the end of the proof, he seems to conclude from $\alpha\text{Stab}(x)\alpha^{-1}\subset\text{Stab}(x)$ that equality holds; I don't understand how this can be done. The above statement comes from a try to fix this gap.
| https://mathoverflow.net/users/3108 | Equivariant map preserves stabilizer | That's essentially the same as [this question](https://mathoverflow.net/questions/11828). So the answer is negative: if $X=H\backslash G$, the stabilizer of $Ha\in X$ is
$a^{-1}Ha$; if $b$ is any element such that $bHb^{-1}\subset H$, the map
$$Ha\mapsto Hba$$ is $G$-equivariant; if $bHb^{-1}\ne H$, we get a counterexample.
| 7 | https://mathoverflow.net/users/2653 | 11982 | 8,117 |
https://mathoverflow.net/questions/11974 | 25 | According to Chang and Keisler's "Model Theory", Model Theory = Universal Algebra + Logic. Model theory generalized Universal Algebra in the sense that we allow relations while in Universal Algebra we only allow functions.
Also, we know that Category Theory generalized Universal Algebra. From wikipedia:
>
> Blockquote Given a list of operations and axioms in universal algebra, the corresponding algebras and homomorphisms are the objects and morphisms of a category. Category theory applies to many situations where universal algebra does not, extending the reach of the theorems. Conversely, many theorems that hold in universal algebra do not generalise all the way to category theory.
>
>
>
So this suggest there might be some overlapping between Model Theory and Category Theory. I hope some one can elaborate about the relation (if there is)?
| https://mathoverflow.net/users/2701 | Is there a relationship between model theory and category theory? | Between model theory and category theory broadly conceived: not anything really compelling, because a category, on its own, does not stand as an interpretation for anything.
Between model theory and categorical logic, however: yes, I think the overlap is large.
A spot of history: the man most deserving, in my opinion, of being called the father of model theory is Alfred Tarski, who came from a Polish school of logic that, I understand, was very much within the algebraic school. His model theory was more in the vein of a reworking of the Polish-style algebraic logic (this is not, in anyway, to talk down his achievement).
Blackburn *et al* (2001, pp 40-41) talk of a might-have-been for the Jónsson-Tarski representation theorem:
>
> ...while modal algebras were useful tools, they *seemed* of little help in guiding logical intuitions. The [theorem] should have swept this apparent shortcoming away for good, for in essence they showed how to represent modal algebras as the structures we now call models! In fact, they did a lot more that this. Their representation technique is essentially a model building technique, hence their work gave the technical tools needed to prove the completeness result that dominated [work on modal logic before Kripke].
>
>
>
They go on to present a nice anecdote showing how Tarski did not seem to think this algebraic approach provided a semantics for modal logic, even after Kripke stressed how important it was to Kripke semantics. It seems that sometimes algebraic logic and model theory are more similar than they appear.
Like model theory, categorical logic can seem to be a special way of doing algebraic logic. And with some theories, model theory and algebraic logic sometimes seem to differ only in trivialities; with categorical logic I am more hesitant in making sweeping judgements, but it sometimes feels that way to me too.
Ref: [Blackburn, de Rijke, & Venema (2001) *Modal Logic*, CUP.](https://www.cambridge.org/core/books/modal-logic/F7CDB0A265026BF05EAD1091A47FCF5B)
| 7 | https://mathoverflow.net/users/3154 | 11985 | 8,119 |
https://mathoverflow.net/questions/8930 | 3 | In my quest to understand all things Spearman, consider the following problem:
Given random variable $x$ with known variance, $\sigma^2$, and $p \in [-1,1]$, one can construct a random variable $y$ such that the Pearson correlation coefficient of the variables $x$ and $y$ is exactly $p$. (Let $y = p x + \sqrt{(1-p^2)} z,$ where $z$ is a random variable independent of $x$ with variance $\sigma^2$.)
I am wondering if there is an analogue for the Spearman Rank Correlation Coefficient. I am defining the *population* Spearman Correlation Coefficient of the jointly distributed $(x,y)$ as
$$E[sign((x\_i - x\_j)(y\_i - y\_j))].$$
(this is the subject of another of my questions here on M.O.)
It seems like I would need to know more than just the variance of $x$ to construct $y$, perhaps the entire CDF.
| https://mathoverflow.net/users/2570 | construct random variable with a fixed level of Spearman Coefficient to another | You don't even need $\sigma^2$ to construct such a variable.
Let $Z$ be $+1$ with probability $q$, and $-1$ with probability $1-q$, and independent of $X$.
Let $Y = e^XZ$. This squashes X to the positive reals preserving order, and then may change the sign.
$y\_1$ and $y\_2$ have the same ordering as $x\_1$ and $x\_2$ when the greater value isn't negated. That is, if $x\_1 \gt x\_2$, then $sign((x\_1 - x\_2)(y\_1-y\_2)) = sign(y\_1)$. If $x\_1\lt x\_2$, then $sign((x\_1 - x\_2)(y\_1-y\_2)) = sign(y\_2)$. So, $E[sign((x\_1 - x\_2)(y\_1-y\_2))] = E[Z].$
Choose $Z$ to have average value $p$ (set $q=\frac{(p+1)}2$), and then $X$ and $Y$ have Spearman Rank Correlation Coefficient $p$.
Actually, there is a little ambiguity (to me) about whether you allow $x\_1 = x\_2$, which I ignored above. If under your definition, the rank correlation of $X$ with itself is $\alpha$, then the rank correlation of $X$ with $Y$ is $\alpha p$, and you can get any value in $[-\alpha,\alpha]$.
| 3 | https://mathoverflow.net/users/2954 | 11995 | 8,127 |
https://mathoverflow.net/questions/11752 | 2 | In this post I want to look at an issue I was in doubt when looking at the comment of F. G. Dorais in the post [In model theory, does compactness easily imply completeness?](https://mathoverflow.net/questions/9309/in-model-theory-does-compactness-easily-imply-completeness)
F. G. Dorais remark was:
>
> Blockquote
> ...The first, which comes through rather clearly, is that Model Theory could ultimately be done without any formal syntax and deduction rules...
>
>
>
I think F. G. Dorais was talking about Fraisse's development of model theory via back and forth. It is, however, not clear to me that this is more free from syntax and deduction rule than the traditional one in a meaningful way.
I think Fraissen view does show that Model Theory could be done without a specific choice of syntax. But it seems unreasonable to think that the traditional model theorists would believe that a specific choice of syntax does matter.
The main question I want to ask:
Is there any difference between Fraissean point of view to traditional point of view BEYOND switching from formal syntax and deduction rule to its informal counter part?
It is not immediately clear to me that the Fraissean did anything more than doing so.
If this is the case then there is nothing genuine new about Fraissean point of view than the traditional one. For example, there seems to be no fundamental difference between writing $ N \vDash S0+S0=SS0 $ and speaking out loud "in $N$, one plus one is equal to two". In both cases, we use language which are ultimately meaningless. There is no more meaning in the utterance of "one" than writing $S0$. (If a parrot shout “one plus one is equal to two”, his statement would have no meaning).
In both cases meaning is given by the interpretations. The only difference is the interpretation in the case of "one" is more familiar than in the case of $S0$. This difference has nothing to deal with the subject matter of mathematics. Likewise, there is no meaningful difference between " there exist .. in M" and $M \vDash \exists ...$.
| https://mathoverflow.net/users/2701 | In what sense Fraissean view point shows Model Theory can be done without any formal syntax and deduction rule? | After your answer, I think I understand better where you see a problem. I don't think you fully appreciate the way of interpreting formulas from Fraïssé's point of view. For simplicity, I will follow your lead and stick with the case of a language with just one relation symbol. It's not hard to generalize, but that would introduce some unnecessary tedium.
First, think about how you would actually define a formula in the Fraïssean style. What you have are the $n$-types (${\sim\_\omega}$-equivalence classes of $n$-tuples). There is a (Hausdorff, in fact, zero-dimensional) topology on the space $S\_n$ of $n$-types which is induced by the ${\sim\_p}$-equivalences. A good way to think about this topology is to think of the set $S\_n$ of $n$-types as the inverse limit of the ${\sim\_p}$-quotients, so the basic clopen sets of the topology on $S\_n$ correspond to ${\sim\_p}$-equivalence classes for some $p < \omega$.
Formulas can then be viewed as clopen sets in $S\_n$. The meaning of ${\land}$, ${\lor}$, ${\lnot}$, is clear since clopen sets form a Boolean algebra. Before thinking about quantifiers, let's see what it means to satisfy a formula $\phi$, i.e. a clopen set in $S\_n$. Let's take a structure $(M,R)$ and pick an $n$-tuple $\bar{a}$ from $M$. Then $(M,R,\bar{a})$ has a specific type, which may or may not belong to the clopen set $\phi$ of $S\_n$. To (ab)use classical symbolism, we can write $(M,R) \vDash \phi(\bar{a})$ when the type of $(M,R,\bar{a})$ does belong to $\phi$. This gives the usual interpretation of ${\land}$, ${\lor}$, ${\lnot}$.
Returning to quantifiers, the existential quantifier is, in a certain sense, the projection $\exists x\_{n+1}:S\_{n+1} \to S\_n$ which simply forgets the last coordinate (suggested by the dummy variable symbol $x\_{n+1}$). More precisely, if $\psi$ is a clopen set in $S\_{n+1}$, then the set $\exists x\_{n+1}\psi$ of $n$-types that extend to an $(n+1)$-type in $\psi$ is a well-defined subset of $S\_n$. The fact that this is a clopen set is however not immediately obvious, nor is the fact that this is correct. (It's easier to think about this when $\psi$ is a basic clopen set, i.e. a ${\sim\_p}$-equivalence class for some $p < \omega$. Working through the definitions, you see that $\exists x\_{n+1}\psi$ is easy to understand in terms of ${\sim\_{p+1}}$-equivalence. It's also easier to see that this is indeed correct.)
Now that we understand how to view formulas from Fraïssé's point of view. How does compactness enter? The Compactness Theorem, in Fraïssé's view, simply says that the spaces $S\_n$ are all compact. (Or, in a more restricted sense, that $S\_0$ is compact.) In our case, the fact that the spaces $S\_n$ are compact is obvious, since they are inverse limits of finite spaces. However, this fact uses the cheat that we're only considering a language with only one relation symbol. For the general case, the ultrafilter construction reduces the problem to the case where the language is finite. (This works well in a relational language with constants, to handle functions you need some magic tricks.)
The point here is that you prove that the spaces $S\_n$ are compact *directly*, you don't need to know that clopen sets are actually formulas. The Classical Compactness Theorem then follows from the simple observation that formulas are closed sets. The other big theorems follow in the same way. For example, the Omitting Types Theorem follows from the fact that the Baire Category Theorem holds for compact spaces, again without explicit mention formulas.
What about the Completeness Theorem? Here, you definitely need formulas (or at least sentences), but we know how to interpret those so it's not a big deal. The Compactness Theorem tells us that any inconsistent set of sentences has a finite inconsistent subset. As a collection of deduction rules, we can take all rules
$\phi\_1,\dots,\phi\_k \vdash \psi$
where $\{\phi\_1,\dots,\phi\_k,\lnot\psi\}$ is an inconsistent set of sentences. This is a horrible system, but it's finitary, sound and complete for semantical consequence. (You can do something similar if you want deduction rules for formulas, but there's really no point to any of this.) This is a completely useless completeness theorem since there it does not give a useful description of this set of deduction rules. You would have a very hard time proving the Gödel Incompleteness Theorems from this...
| 3 | https://mathoverflow.net/users/2000 | 11998 | 8,129 |
https://mathoverflow.net/questions/215 | 19 | In the cohomological incarnation, the Riemann hypothesis part of the Weil conjectures for a smooth proper scheme of finite type over a finite field with $q$ elements says that: the eigenvalues of Frobenius acting on the $H^i\_\mathrm{et}(X, Q\_\ell)$ are algebraic integers with complex absolute value $q^{i/2}$.
For smooth proper curve $ C$ over $F\_q$ of genus g, the Riemann hypothesis is often stated in another way as $|N- q - 1| \leq 2g \sqrt(q)$ where $N$ is the number of rational points on $ C$.
Why are these two statements equivalent? Is (are) there any corresponding inequality(ies) which are equivalent to the Riemann hypothesis in higher dimensions?
| https://mathoverflow.net/users/81 | Equivalent statements of the Riemann hypothesis in the Weil conjectures | The reason why that inequality is equivalent to RH (for curves) is the *functional equation*.
The polynomial $L(z)$ I speak about below would arise in practice as the numerator of the zeta-function of the curve, with $z = q^{-s}$ and the usual version of the Riemann hypothesis for $L(q^{-s})$ is equivalent to the statement that the reciprocal roots of $L(z)$ all have absolute value $\sqrt{q}$. That's the form of the Riemann hypothesis I will be referring to in what follows.
Suppose we have a polynomial $L(z)$ over the complex numbers with constant term 1 and degree $d$, factored over its reciprocal roots:
$$
L(z) = (1 - \alpha\_1z)\cdots(1-\alpha\_dz), \ \ \ \alpha\_j \not= 0.
$$
Let $L^\*(z)$ be the polynomial with complex-conjugate coefficients to those of $L(z)$, so
$$
L^\*(z) = (1 - \overline{\alpha\_1}z)\cdots(1-\overline{\alpha\_d}z).
$$
Assume $L(z)$ and $L^\*(z)$ are connected by the functional equation
$$
L(1/qz) = \frac{W}{z^d}L^\*(z)
$$
for some constant $W$. If you compare coefficients of the same powers of $z$ on both sides, this functional equation implies the mapping $\alpha \mapsto q/\overline{\alpha}$ sends reciprocal roots of $L(z)$ to reciprocal roots of $L^\*(z)$ (and $W$ has absolute value $q^{d/2}$).
Lemma 1. Granting the functional equation above, the following conditions are equivalent:
i$)$ the reciprocal roots of $L(z)$ have absolute value $\sqrt{q}$ (RH for $L(z)$),
ii$)$ the reciprocal roots of $L(z)$ have absolute value $\leq \sqrt{q}$.
Proof. We only need to show ii implies i.
Assuming ii, let $\alpha$ be any reciprocal root of $L(z)$,
so $|\alpha| \leq \sqrt{q}$. By the functional
equation, $q/\overline{\alpha}$ is a reciprocal
root of $L^\*(z)$, so $q/\overline{\alpha} = \overline{\beta}$
for some reciprocal root $\beta$ of $L(z)$.
Then $|q/\overline{\alpha}| = |\overline{\beta}| = |\beta| \leq \sqrt{q}$ and thus $\sqrt{q} \leq |\alpha|$.
Therefore $|\alpha| = \sqrt{q}$ and i follows. QED
This lemma reduces the proof of the Riemann hypothesis for $L(z)$ from the equality $|\alpha\_j| = \sqrt{q}$ for all $j$ to the upper bound $|\alpha\_j| \leq \sqrt{q}$ for all $j$. Of course the functional equation was crucial in explaining why the superficially weaker inequality implies the equality.
Next we want to show the upper bound on the $|\alpha\_j|$'s in part ii of Lemma 1 is equivalent to a $O$-estimate on sums of powers of the $\alpha\_j$'s which superficially seems weaker.
We will be interested in the sums
$$
\alpha\_1^n + \cdots + \alpha\_d^n,
$$
which arise from the theory of zeta-functions as coefficients in an exponential generating function: since $L(z)$ has constant term 1, we can write (as formal power series over the complex numbers) $$L(z) = \exp\left(\sum\_{n \geq 1}N\_n z^n/n\right)$$
and then logarithmic differentiation shows
$$
N\_n = -(\alpha\_1^n + \dots + \alpha\_d^n)
$$
for all $n \geq 1$.
Lemma 2.
For nonzero complex numbers $\alpha\_1,\dots,\alpha\_d$ and a
constant $B > 0$, the following are equivalent:
i$)$
For some $A > 0$, $|\alpha\_1^n + \dots + \alpha\_d^n| \leq AB^n$ for
all $n \geq 1$.
ii$)$
For some $A > 0$ and positive integer $m$,
$|\alpha\_1^n + \dots + \alpha\_d^n| \leq AB^n$ for
all $n \geq 1$ with $n \equiv 0 \bmod m$.
iii$)$ $|\alpha\_j| \leq B$ for all $j$.
Part ii is saying you only need to show part i when $n$ runs through the (positive) multiples of any particular positive integer to know it is true for all positive integers $n$. It is a convenient technicality in the proof of the Riemann hypothesis for curves, but the heart of things is the connection between parts i and iii. (We'd be interested in part iii with $B = \sqrt{q}$.) You could set $m = 1$ to make the proof below that ii implies iii into a proof that i implies iii. The passage from i to iii is what Dave is referring to in his answer when he cites the book by Iwaniec and Kowalski.
Proof.
Easily i implies ii and (since $|\alpha\_j| = |\overline{\alpha\_j}|$)
iii implies i.
To show ii implies iii, we use a cute analytic trick. Assuming ii, the series
$$
\sum\_{n \equiv 0 \bmod m} (\alpha\_1^n + \dots + \alpha\_d^n)z^n
$$
is absolutely convergent for $|z| < 1/B$, so the series defines a holomorphic function on this disc. (The sum is over positive multiples of $m$, of course.)
When $|z| < 1/|\alpha\_j|$ for
all $j$, the series can be computed to be
$$
\sum\_{j=1}^{d} \frac{\alpha\_j^mz^m}{1-\alpha\_j^mz^m} =
\sum\_{j=1}^{d}\frac{1}{1-\alpha\_j^mz^m} - d,
$$
so the rational function $\sum\_{j=1}^{d} 1/(1-\alpha\_j^mz^m)$ is holomorphic
on the disc $|z| < 1/B$. Therefore the poles of this rational function must have absolute value $\geq 1/B$. Each $1/\alpha\_j$ is a pole, so $|\alpha\_j| \leq B$ for all $j$. QED
Theorem.
The following are equivalent:
i$)$ $L(z)$ satisfies the Riemann hypothesis ($|\alpha\_j| = \sqrt{q}$ for all $j$),
ii$)$ $N\_n = O(q^{n/2})$ as $n \rightarrow \infty$,
iii$)$ for some $m \geq 1$, $N\_n = O(q^{n/2})$ as $n \rightarrow
\infty$ through the multiples of $m$.
Proof.
Easily i implies ii and ii implies iii.
Assuming iii, we get $|\alpha\_j| \leq \sqrt{q}$ for all $j$ by
Lemma 2, and this inequality over all $j$ is equivalent to i
by Lemma 1. QED
Brandon asked, after Rebecca's answer, if the inequality implies the Weil conjectures (for curves) and Dave also referred in his answer to the Weil conjectures following from the inequality. In this context at least, you should not say "Weil conjectures" when you mean "Riemann hypothesis" since we used the functional equation in the argument and that is itself part of the Weil conjectures. The inequality does not imply the Weil conjectures, but only the Riemann hypothesis (after the functional equation is established).
That the inequality is logically equivalent to RH, and not just a consequence of it, has some mathematical interest since this is one of the routes to a proof of the Weil conjectures for curves.
P.S. Brandon, if you have other questions about the Weil conjectures for curves, ask your thesis advisor if you could look at his senior thesis. You'll find the above arguments in there, along with applications to coding theory. :)
| 20 | https://mathoverflow.net/users/3272 | 12001 | 8,131 |
https://mathoverflow.net/questions/12003 | 5 | In going from Riemann surface theory to the theory of algebraic curves over fields $k$ that are not necessarily $\mathbb{C}$, I would like to understand more about how the notion of a covering map carries over.
If I have a compact, connected Riemann surface $M$, a cover of $M$ by another such Riemann surface, say $N$, then I am taking this to mean a holomorphic map $f:N\rightarrow M$ of finite degree $m>0$ (that is, the generic fiber of $f$ consists of $m$ points). At a $p\in M$ that is a regular value of $f$ (i.e. $p$ is not a branch point), then there is a open neighborhood $U$ of $p$ such that $f^{-1}(U)$ is the disjoint union of $m$ copies of $U$, and 'open' refers to the topology determined by the complex analytic structure on $M$.
If I have $M$ and $N$ nonsingular algebraic curves over a field $k$, then what can be said about $f^{-1}(U)$ when $f:N\rightarrow M$ is a finite regular map? What I mean by this is when the topology is the Zariski topology, I assume that the statement "$f^{-1}(U)$ is the disjoint union of $m$ copies of $U$" translates to open sets in that topology. When we regard a compact Riemann surface as a nonsingular curve over $\mathbb{C}$, then will these notions coincide?
Sorry if this is a trivial/ill-posed question. (My experience so far is more with differential geometry and complex analytic geometry....)
Many thanks.
| https://mathoverflow.net/users/3310 | Covering maps of Riemann surfaces vs covering maps of $k$-algebraic curves | No, the property of having small neighbourhoods whose preimage is a disjoint union
of $n$ homeomorphic open sets does not hold in the Zariski topology (once $n > 1$, i.e.
the cover is non-trivial). The reason is
that non-empty Zariski open sets are always very big; in the case of a curve,
their complement is always just a finite number of points. In particular, two different
non-empty Zariski opens are never disjoint.
There are two ways that one rescues the situation: the first is to use the differential
topology view-point on covers: they are proper submersions between manifolds of the same dimension.
In the context of compact Riemann surfaces, both source and target have the same dimension,
and maps are automatically proper, so it is just the submersion property that is left
to think about. It is a property about how tangent spaces map, which can be translated into
the algebraic context (e.g. using the notion of Zariski tangent spaces).
So if $f: X \rightarrow Y$ is a regular morphism (regular morphism is the algebraic geometry terminology for an everywhere defined map given locally by rational functions) of projective curves, we can say that
$f$ is unramified at a point $p \in X$ if $f$ induces an isomorphism from the Zariski tangent
space of $X$ at $p$ to the Zariksi tangent space to $Y$ at $f(P)$.
For historical reasons, if $f$ is unramified at every point in its domain, we say that
$f$ is etale (rather than a cover), but this corresponds precisely to the notion of a covering map when we pass to Riemann surfaces.
This leads to the more sophisticated rescue: one considers all the etale maps from (not necessarily projective or connected) curves $X$ to $Y$, and considers them as forming a topology on $Y$, the so-called etale topology of $Y$. This leads to many important notions
and results, since it allows one to transport many topological notions (in particular, fundamental groups and cohomology) to the algebraic context.
| 10 | https://mathoverflow.net/users/2874 | 12004 | 8,133 |
https://mathoverflow.net/questions/11948 | 9 | Say a ring $R$ is an isolated hypersurface singularity if $R = k[x\_1, \ldots, x\_n]\_{(x\_1, \ldots, x\_n)}/(W)$, where $k$ is a field and $W \in k[x\_1, \ldots, x\_n]$ is such that the ideal $(\partial\_1 W, \ldots, \partial\_n W)$ is $(x\_1, \ldots, x\_n)$-primary. For finitely generated $R$-modules $M$ and $N$ define the function $\theta$ to be $\theta(M, N) = \lambda( \operatorname{Tor}^R\_{2i} (M,N)) - \lambda( \operatorname{Tor}^R\_{2i-1}(M,N))$ for any $i \gg 0$. Here $\lambda$ denotes the length of a module. The definition makes sense because all modules over $R$ have eventually 2-periodic resolutions (since $R$ is a hypersurface) and the $\operatorname{Tor}$'s have finite length for $i \gg 0$ (since $R$ is an "isolated singularity"). Hochster made this definition in his 1981 paper "The dimension of an intersection in an ambient hypersurface."
I'm looking for examples (or preferably a family) of isolated hypersurface singularity rings with $n \geq 4$ and modules $M$ over such rings with $\theta(M, - )$ non-zero. See Hailong's answer below for an equivalent formulation of this in terms of certain Chow groups when $k = \mathbb{C}$ and $W$ is homogeneous. I would prefer if $W$ were not homogeneous but am interested in all cases.
Of course any insight as to when $\theta$ is non-zero would be great but in general this is hard. For instance in the paper above Hochster showed that the direct summand conjecture is true if $\theta$ is non-zero for an explicit family of modules and rings. It is conjectured that $\theta$ is zero when $n$ is odd (this is known when $W$ is homogeneous, see <http://arxiv.org/abs/0910.1289v1>), and it is known that when $n=4$ the function $\theta$ is nonzero if and only if the class group of $R$ is nonzero.
EDIT: There is a physical interpretation of the above in the spirit of this post: [Matrix factorizations and physics](https://mathoverflow.net/questions/9733/matrix-factorizations-and-physics). My knowledge of physics is limited so I apologize in advance for any mistakes.
D-branes in a B-twisted topological Landau-Ginzburg models with potential $W$ are given by matrix factorizations of $W$. We only care about values of $\theta$ on maximal Cohen-Macaulay (MCM) $R$-modules, and all such modules are given by matrix factorizations of $W$. Thus MCM modules over $R$ can be thought of as D-branes. Now physicists talk about the BRST-cohomology of two branes $M,N$ (which I don't understand) but it seems that it is given by $\operatorname{Ext}\_R^2(M \oplus \Omega M, N \oplus \Omega N)$ (or equivalently the stable homomorphisms between these modules) where $\Omega( - )$ denotes the first syzygy; see for instance <http://arxiv.org/abs/0802.1624>. It is not hard to see, viewing the modules as matrix factorizations, that for two MCM modules $M$ and $N$ we have
$\theta(M, N) = \lambda( \operatorname{Ext}^1\_R( M^\*, N) ) - \lambda( \operatorname{Ext}^2\_R(M^\*, N) )$,
where $M^\*$ is the MCM module given by $\operatorname{Hom}\_R(M, R)$. Thus to find an example of modules with nonzero $\theta$ is equivalent to finding branes whose "even" and "odd" BRST cohomology have different dimensions over $k$.
| https://mathoverflow.net/users/3293 | Isolated hypersurface singularities, Chow groups and D-branes | Assume $k= \mathbf C$ and $W$ homogeneous. Let $X=Proj (k[x\_1,\cdots,x\_n]/(W))$. $X$ is then a smooth hypersuraface in $\mathbb P\_{n-1}$.
Assume $n=2d$ is even. Corollary 3.10 of the paper you quoted says that $\theta=0$ for all pairs iff the homological Chow group $CH^{d-1}\_{hom}$ modulo $[h]^{d-1}$ is not torsion (here $[h]$ is the class of the hyperplane section). So your question, in this case, is equivalent to
($l=d-1$):
**Examples of smooth hypersurfaces of dimension $2l$ such that $CH^{l}\_{hom}/([h]^{l})$ is not torsion** ?
(By the way, I think if you phrased your question this way, it probably would become more popular, consider how many geometry-inclined people visit this site! So if you want more and better answers, consider changing the title.)
Now, a cheap way to get examples you want is to take $W= x\_1x\_{d+1} + \cdots + x\_dx\_{2d}$. Then the cycle defined by $(x\_1,...,x\_d)$ will not be a multiple of a power of the hyperplane section. Why? Because, I am waving my hand a bit here, if it is then the intersection with the cycle $(x\_{d+1},\cdots, x\_{2d})$ would be positive. But they are disjointed in $X$!
The same trick works for generalized quadrics, i.e. if $W = f\_1g\_1 +\cdots +f\_dg\_d$.
EDIT: Let me give more details here. In this situation you can easily make $W$ non-homogeneous as you desire. But the trouble is you can't use my argument above as there is no longer a projective variety $X$. But one can get around this. Let $S=k[x\_1,\cdots,x\_{2d}]\_{m}$
here $m$ is the irrelevant ideal. Suppose $W = f\_1g\_1 +\cdots +f\_dg\_d$ and assume that $(f\_1,\cdots, f\_d, g\_1,\cdots, g\_d)$ is a full system of parameters in $S$. Let $R=S/(W)$, $P=(f\_1,\cdots,f\_d)$ and $Q=(g\_1,\cdots,g\_d)$. I claim that $\theta^R(R/P,R/Q) \neq 0$.
The reason is that $\theta^R(R/P,R/Q) = \chi^S(S/P,S/Q)$, the Serre's intersection multiplicity (see Hochster's original paper). Because $dim S/P + dim S/Q = d+d =dim S$, we must have $\chi^S(S/P,S/Q)>0$ by [Positivity](http://en.wikipedia.org/wiki/Serre%27s_multiplicity_conjectures), which is known in this case.
More exotic examples should be abound, and I am sure people who know more intersection theory can provide some, once they are aware of what this question is about. I would be interested in hearing more answers along that line.
| 8 | https://mathoverflow.net/users/2083 | 12010 | 8,137 |
https://mathoverflow.net/questions/12045 | 66 | The obvious ones are 0 and $e^{-x^2}$ (with annoying factors), and someone I know suggested hyperbolic secant. What other fixed points (or even eigenfunctions) of the Fourier transform are there?
| https://mathoverflow.net/users/3319 | What are fixed points of the Fourier Transform | The following is discussed in a little more detail on pages 337-339 of Frank Jones's book "Lebesgue Integration on Euclidean Space" (and many other places as well).
Normalize the Fourier transform so that it is a unitary operator $T$ on $L^2(\mathbb{R})$. One can then check that $T^4=1$. The eigenvalues are thus $1$, $i$, $-1$, and $-i$. For $a$ one of these eigenvalues, denote by $M\_a$ the corresponding eigenspace. It turns out then that $L^2(\mathbb{R})$ is the direct sum of these $4$ eigenspaces!
In fact, this is easy linear algebra. Consider $f \in L^2(\mathbb{R})$. We want to find $f\_a \in M\_a$ for each of the eigenvalues such that $f = f\_1 + f\_{-1} + f\_{i} + f\_{-i}$. Using the fact that $T^4 = 1$, we obtain the following 4 equations in 4 unknowns:
$f = f\_1 + f\_{-1} + f\_{i} + f\_{-i}$
$T(f) = f\_1 - f\_{-1} +i f\_{i} -i f\_{-i}$
$T^2(f) = f\_1 + f\_{-1} - f\_{i} - f\_{-i}$
$T^3(f) = f\_1 - f\_{-1} -i f\_{i} +i f\_{-i}$
Solving these four equations yields the corresponding projection operators. As an example, for $f \in L^2(\mathbb{R})$, we get that $\frac{1}{4}(f + T(f) + T^2(f) + T^3(f))$ is a fixed point for $T$.
| 86 | https://mathoverflow.net/users/317 | 12047 | 8,164 |
https://mathoverflow.net/questions/12009 | 70 | One the proofs that I've never felt very happy with is the classification of finitely generated abelian groups (which says an abelian group is basically uniquely the sum of cyclic groups of orders $a\_i$ where $a\_i|a\_{i+1}$ and a free abelian group).
The proof that I know, and am not entirely happy with goes as follows: your group is finitely presented, so take a surjective map from a free abelian group. The kernel is itself finitely generated (this takes a little argument in and of itself; note that adding a new generator to a subgroup of free abelian group either increases dimension after tensoring with $\mathbb{Q}$ or descreases the size of the torsion of the quotient), so our group is the cokernel of a map between finite rank free groups. Now, (and here's the part I dislike) look at the matrix for this map, and remember that it has a [Smith normal form](http://en.wikipedia.org/wiki/Smith_normal_form). Thus, our group is the quotient of a free group by a diagonal matrix where the non-zero entries are $a\_i$ as above.
I really do not think I should have to algorithmically reduce to Smith normal form or anything like that, but know of no proof that doesn't do that.
By the way, if you're tempted to say "classification of finitely generated modules for PIDs!" make sure you know a proof of that that doesn't use Smith normal form first.
| https://mathoverflow.net/users/66 | Is there a slick proof of the classification of finitely generated abelian groups? | I reject the premise of the question. :-)
It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A\_p$ and a stable image $A\_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A\_p$. A similar argument appears in Matthew Emerton's proof. As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.
The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give next.
**Theorem.** If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.
**Proof.** By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )
The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.
The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □
Compare the complexity of this argument to the other arguments supplied so far.
Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column. Thus, each modulus $x\_k$ that you produce divides the next modulus $x\_{k+1}$.
Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.
The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem. And it only needs minor changes to apply to general PIDs.
| 49 | https://mathoverflow.net/users/1450 | 12053 | 8,169 |
https://mathoverflow.net/questions/11978 | 43 | I was very surprised when I first encountered the [Mertens conjecture](http://en.wikipedia.org/wiki/Mertens_conjecture). Define
$$ M(n) = \sum\_{k=1}^n \mu(k) $$
The Mertens conjecture was that $|M(n)| < \sqrt{n}$ for $n>1$, in contrast to the Riemann Hypothesis, which is equivalent to $M(n) = O(n^{\frac12 + \epsilon})$ .
The reason I found this conjecture surprising is that it fails heuristically if you assume the Mobius function is randomly $\pm1$ or $0$. The analogue fails with probability $1$ for a random $-1,0,1$ sequence where the nonzero terms have positive density. The law of the iterated logarithm suggests that counterexamples are large but occur with probability 1. So, it doesn't seem surprising that it's false, and that the first counterexamples are uncomfortably large.
There are many heuristics you can use to conjecture that the digits of $\pi$, the distribution of primes, zeros of $\zeta$ etc. seem random. I believe random matrix theory in physics started when people asked whether the properties of particular high-dimensional matrices were special or just what you would expect of random matrices. Sometimes the right random model isn't obvious, and it's not clear to me when to say that an heuristic is reasonable.
On the other hand, if you conjecture that all naturally arising transcendentals have simple continued fractions which appear random, then you would be wrong, since $e = [2;1,2,1,1,4,1,1,6,...,1,1,2n,...]$, and a few numbers algebraically related to $e$ have similar simple continued fraction expansions.
What other plausible conjectures or proven results can be framed as heuristically false according to a reasonable probability model?
| https://mathoverflow.net/users/2954 | Heuristically false conjectures | I think this example fits, in 1985 H. Maier disproved a very reasonable conjecture on the distribution of prime numbers in short intervals. The probabilistic approach had been thoroughly examined by Harald Cramer. Nice paper by Andrew Granville including this episode in (mathematical) detail, page 23 (or 13 out of 18 in the pdf):
www.dms.umontreal.ca/~andrew/PDF/cramer.pdf
| 19 | https://mathoverflow.net/users/3324 | 12063 | 8,176 |
https://mathoverflow.net/questions/7074 | 11 | Let R be a local, normal domain of dimension 2. Suppose that R contains a finite field. I am interested in knowing when the class group of R is torsion. In characteristic 0, this is known to be related to R being a rational singularity.
Lipman showed that if X is a desingularization of Spec(R), then one has an exact sequence:
$0 \to Pic^{0}(X) \to Cl(R) \to H $
Here $Pic^{0}(X)$ is the numerically trivial part of the Picard group of $X$, and $H$ is a finite group. Thus the second one is torsion if and only if the first one is. I do not have much understanding of the first group, unfortunately.
Does anyone know an answer or reference to this? Does anyone know an example in positive characteristic such that $Cl(R)$ is not torsion? Thanks a lot.
| https://mathoverflow.net/users/2083 | Class groups of normal domains over finite fields | As requested in the comments, here's an example of a local, normal $2$-dimensional domain R in positive characteristic such that $\mathrm{Cl}(R)$ is not torsion: choose an elliptic curve $E \subset \mathbf{P}^2$ over a field $k$ such that $E(k)$ is not torsion, and take R to be the local ring at the origin of the affine cone on $E$ (i.e., $R = k[x,y,z]/(f)\_{(x,y,z)}$ where $f$ is a homoegenous cubic defining $E$). This can be done over $k = \overline{\mathbf{F}\_p(t)}$.
*Proof*: The normality follows from the fact that R is a hypersurface singularity (hence even Gorenstein) and isolated and $2$-dimensional (hence regular in codim 1). Blowing up at the origin defines a map $f:X \to \mathrm{Spec}(R)$. One can then show the following: $X$ is smooth, and $X$ can be identified with the Zariski localisation along the zero section of the total space of the line bundle $L = \mathcal{O}\_{\mathbf{P^2}}(-1)|\_E$ (these are general facts about cones). By Lipman's theorem, it suffices to show that $\mathrm{Pic}^0(X)$ contains non-torsion elements. As $X$ is fibered over $E$ with a section, the pullback $\mathrm{Pic}^0(E) \to \mathrm{Pic}^0(X)$ is a direct summand. As $\mathrm{Pic}^0(E) \simeq E(k)$ has non-torsion elements by assumption, so does $\mathrm{Pic}^0(X)$.
Also, an additional comment: In general, Lipman's theorem tells you that $\mathrm{Cl}(R)$ is torsion if and only if $\mathrm{Pic}^0(X)$ is torsion. Now $\mathrm{Pic}(X) \simeq \lim\_n \mathrm{Pic}(X\_n)$ where $X\_n$ is the $n$-th order thickening of the exceptional fibre $E$. Because we are blowing up a point, the sheaf of ideals $I$ defining $E$ is ample on $E$. The kernel and cokernel of $\mathrm{Pic}(X\_n) \to \mathrm{Pic}(X\_{n-1})$ are identified with $H^1(E,I|\_E^{\otimes n+1})$ and $H^2(E,I|\_E^{\otimes n+1})$. As $I|\_E$ is ample, it follows that the system "$\lim\_n \mathrm{Pic}(X\_n)$" is eventually stable. Thus, $\mathrm{Pic}(X) \simeq \mathrm{Pic}(X\_n)$ for $n$ sufficiently big. As $X\_n$ is a proper variety, it follows that if we are working over a finite field (resp. an algebraic closure of a finite field), then $\mathrm{Pic}^0(X)$ is finite (resp. ind-finite).
| 26 | https://mathoverflow.net/users/986 | 12073 | 8,181 |
https://mathoverflow.net/questions/12065 | 0 | Hi, I want to be able to solve a linear program that has constraints that are either zero or a range. An example below in LP\_Solve-like syntax shows what I want to do. This doesnt work. In general all the decision variables Qx can be 0 or a <= Qx <= b where a > 0 and b > a. All decision variables must be integers.
`max: 0.2 Q1 + 0.4 Q2 + 0.6 Q3 + 0.8 Q4 + 0.4 Q5 + 0.6 Q6 + 0.6 Q7 + 0.7 Q8;`
Q1 = 0 or Q1 >= 5;
Q2 = 0 or Q2 >= 1;
Q3 = 0 or Q3 >= 1;
Q4 = 0 or Q4 >= 1;
Q5 = 0 or Q5 >= 1;
Q6 = 0 or Q6 >= 1;
Q7 = 0 or Q7 >= 9;
Q8 = 0 or Q8 >= 1;
Q4 <= 30;
Q1 + Q2 + Q3 + Q4 + Q5 + Q6 + Q7 + Q8 <= 50;
How would I rewrite this to work? Is there a particular solver that should be used for this kind of task?
| https://mathoverflow.net/users/3323 | linear program with zeros | You can use the following modelling trick to transform you problem in a integer linear program: for each constraint of the type
$$ Q\_i = 0 \text{ or } L\_i \le Q\_i \le M\_i$$
(on an integer variable $Q\_i$) introduce a new binary variable $B\_i$ and write
$$ L\_i B\_i \le Q\_i \le M\_i B\_i $$
If, as in your example, you have no explicit value for $M\_i$, simply use a large upper bound on the values of $Q\_i$ (for example $50$ in your situation). Also, the constraint $Q\_i = 0 \text{ or } Q\_i \ge 1$ is redundant.
You can then solve the resulting program with any integer programming solver, such as lp\_solve or glpk.
By the way, the solution to your example is easily seen to be obtained with $Q\_4=30$ and $Q\_8=20$.
| 1 | https://mathoverflow.net/users/1184 | 12075 | 8,182 |
https://mathoverflow.net/questions/12095 | 4 | Let $k|\mathbb{F}\_q$ be a field extension. An $\mathbb{F}\_q$-structure on a $k$-algebra $A$ is an $\mathbb{F}\_q$-subalgebra $A \_0$ of $A$ such that $A \_0 \otimes \_{\mathbb{F}\_q} k \cong A$ via the *canonical* morphism $a \otimes \lambda \mapsto a \lambda$.
Now, my question is if this notion can be properly globalized to $k$-schemes? I saw a definition like: an $\mathbb{F}\_q$-structure on a $k$-scheme $X$ is an $\mathbb{F}\_q$-scheme $X \_0$ such that $X \cong X \_0 \times \_{\mathrm{Spec}(\mathbb{F}\_q)} \mathrm{Spec}(k)$ as $k$-schemes (see for example "Representations of finite groups of Lie type" by Digne and Michel, where $\cong$ is even replaced by $=$). But my problem is that here the particular choice of the canonical morphism as above does not appear so that on affines this definition is not the same as above. Is this a problem?
(The reason why I care about this is that I want to defined the (geometric) Frobenius on a $k$-Scheme with $\mathbb{F}\_q$-structure as the "base change" of the canonical Frobenius (raising to the $q$-th power) on the $\mathbb{F}\_q$-structure $X \_0$.)
| https://mathoverflow.net/users/717 | F_q-structures on schemes | I think that notion you cite from Digne and Michel is not a good one as you will not get a well-defined Frobenius. I suggest replacing $X\_0$ by a pair $(X\_0,p)$, where $p:X\to X\_0$ is a morphism of $\mathbb{F}\_q$ schemes such that $X$ is a product $X\_0\times\_{\mathbb{F}\_q} \mathrm{Spec}(k)$ via the structure morphism $X\to \mathrm{Spec}(k)$ and via $p$. The Frobenius on $X$ will then be the unique map $F:X\to X$ of $k$-schemes such that $F\circ p= F\_0\circ p$, where $F\_0:X\_0\to X\_0$ denotes the canonical Frobenius.
To compare this to the definition for $k$-algebras, note that we could define two $\mathbb{F}\_q$-structures $(X\_0,p)$ and $(X\_0',p')$ to be equivalent if there is an isomorphism $f:X\_0\to X\_0'$ such that $fp=p'$. Two equivalent $\mathbb{F}\_q$-structures on $X$ will then give rise to the same(!) geometric Frobenius. In the case of a $k$-algebra $A$ every $\mathbb{F}\_q$-structure will have a unique representative of the form $(\mathrm{Spec}(A\_0),p)$, where $A\_0\subset A$ and $p$ is induced by this inclusion.
| 7 | https://mathoverflow.net/users/2308 | 12101 | 8,193 |
https://mathoverflow.net/questions/12066 | 10 | Is there some condition on a complex algebraic surface that implies it has a smooth canonical divisor? I am searching for the sharpest possible condition, but sufficient criteria would be nice as well.
For example, the question is quite simple for surfaces that can be embedded in $P^3$. Roughly, just notice that in this case the linear system of the canonical line bundle is base point free and hence, by Bertini's theorem, we can find a smooth canonical divisor.
When we cannot avoid having a singular canonical divisor, then we are left with some singular curves. I am interested in computing the Euler characteristic of their symmetric product. I will post a question about this in a follow up enquire.
| https://mathoverflow.net/users/3314 | When is the canonical divisor of an algebraic surface smooth? | Any smooth projective surface with nonempty $K\_X$ is obtained by blowing up finitely many points on its unique minimal model. From the formula $K\_X=f^\*K\_Y+E$ for the blowup, you see that the exceptional divisors of the blowup are always in the base locus of $|K\_X|$. Thus, the problem is reduced to the minimal model.
Then you have to go through the classification of the minimal models of surfaces, that's been known for a hundred years now. (Using a book such as van de Ven "Complex surfaces", or Shafarevich et al, or Beauville...)
For a minimal surface $Y$ Kodaira dimension 0 for example, $12K\_Y=0$. So either $K\_Y\ne 0$ and then $|K\_X|=\emptyset$, or $K\_Y=0$ and then any divisor in $|K\_X|$ is $\sum a\_i E\_i$, where $E\_i$ are the exceptional divisors of the blowups.
For a minimal surface $Y$ of Kodaira dimension 2, the question is still somewhat tricky. If looking at higher multiples $|mK\_Y|$ suffices, then by a well known theorem (Bombierri? certainly I. Reider gave a very nice proof), $|5K\_Y|$ is free, so a general element is smooth (in characteristic 0). For $|K\_Y|$ I don't think the answer is known but why not search mathscinet.
Finally, for Kodaira dimension 1, an elliptic surface $\pi:Y\to C$, there is a well-known Kodaira's formula for the canonical class $K\_Y=\pi^\*K\_C + R$ with explicit rational coefficients in $R$. I'd play with that. Again, for higher multiples I think $|12K\_Y|$ works.
Of course, to your example of a hypersurface in $\mathbb P^3$ you can add the case of complete intersections, and other surfaces for which $K\_X$ is either zero or $\pm K\_X$ is very ample.
| 8 | https://mathoverflow.net/users/1784 | 12103 | 8,194 |
https://mathoverflow.net/questions/12089 | 3 | Is there an upper bound on the genus of a graph that has a book embedding on say k pages, or can the genus be arbitrarily large? If not a general bound is known, what happens for k=3?
| https://mathoverflow.net/users/3327 | Upper bound on the genus of a k-page graph | Every $n$-vertex genus-$g$ graph has at most $3n + 6g - 6$ edges, by Euler's formula. Now consider the three-page graph in which $n-3$ vertices are connected in a path (in one of the pages, it doesn't matter which one), followed by three vertices that are each connected (in separate pages) to everything in the path. It has $(n-4)+3(n-3)=4n-13$ edges, so as $n$ grows its genus must be unbounded, and grows linearly in the number of vertices.
In the other direction, a graph with genus g must have bounded pagenumber $O(\sqrt g)$; see Malitz, FOCS 1988, [doi:10.1109/SFCS.1988.21962](http://doi.ieeecomputersociety.org/10.1109/SFCS.1988.21962).
| 3 | https://mathoverflow.net/users/440 | 12106 | 8,197 |
https://mathoverflow.net/questions/12100 | 29 | In many formulation of Class Field theory, the Weil group is favored as compared to the Absolute Galois group. May I asked why it is so? I know that Weil group can be generalized better to Langlands program but is there a more natural answer?
Also we know that the abelian Weil Group is the isomorphic image of the reciprocity map of the multiplicative group (in the local case) and of the idele-class group (in the global case). Is there any sense in which the "right" direction of the arrow is the inverse of the reciprocity map?
Please feel free to edit the question into a form that you think might be better.
| https://mathoverflow.net/users/2701 | Why Weil group and not Absolute Galois group? | One reason we prefer the Weil group over the Galois group (at least in the local case) is that the Weil group is locally compact, thus it has "more" representations (over $\bf C$). In fact, all $\bf C$-valued characters of $Gal(\bar{\bf Q\_p} / \bf Q\_p)$ have finite image, where as that of $W\_{\bf Q\_p}$ can very well have infinite image. The same goes for general representations of these groups (recall that $\bf{GL}\_n(\bf C)$ has no small subgroups.)
The global Weil group (which is much more complicated than the local one), on the other hand, is a rather mysterious object that is pretty much untouched in modern number theory as far as I can tell. Supposedly the global Langlands group used in the global Langlands correspondence should be the extension of the global Weil group by a compact group, but this is still largely conjectural.
The standard reference is Tate's "Number Theory Background" in the Corvallis volumes (available for free at ams.org). Also Brooks Roberts has notes on Weil representations available at his website.
| 15 | https://mathoverflow.net/users/3247 | 12111 | 8,201 |
https://mathoverflow.net/questions/12109 | 25 | In *The Geometry of Schemes* by Eisenbud and Harris, Exercise I-32 asks one to show that a scheme $X$ is reduced if and only if every local ring $\mathcal{O}\_{X,p}$ is reduced for closed points $p \in X$. However, this does not seem to work in general, since $X$ may not have enough closed points. What additional hypotheses on $X$ do I need for such an assertion to hold?
| https://mathoverflow.net/users/3333 | Reduced scheme and closed points | There do exist schemes without a closed point, yes. (Liu, exercises 3.3.26/27)
But under some very reasonable additional conditions - I think quasi-compactness will be sufficient, if you are happy with using Zorn's lemma - the result holds. Use/prove the existence of a closed point, and the fact that localizing a reduced ring still gives you a reduced ring.
| 13 | https://mathoverflow.net/users/1107 | 12112 | 8,202 |
https://mathoverflow.net/questions/12092 | 11 | It is well known that Beck's theorem for Comonad is equivalent to Grothendieck flat descent theory on scheme.
There are several version of derived noncommutative geometry. I wonder whether someone developed the triangulated version of Beck's theorem. And What does it mean,if exists?
| https://mathoverflow.net/users/3156 | Is there triangulated category version of Barr-Beck's theorem? | There isn't a descent theory for derived categories per se - one can't glue objects in the derived category of a cover together to define an object in the base. (Trying to apply the usual Barr-Beck to the underlying plain category doesn't help.)
But I think the right answer to your question is to use an enriched version of triangulated categories (differential graded or $A\_\infty$ or stable $\infty$-categories), for which there is a beautiful Barr-Beck and descent theory, due to Jacob Lurie. (This is discussed at length in the n-lab I believe, and came up recently on the n-category cafe (where I wrote basically the same comment [here](http://golem.ph.utexas.edu/category/2010/01/quasicoherent_stacks.html#c031134)..)
This is proved in [DAG II: Noncommutative algebra](http://arxiv.org/abs/math/0702299). In the comonadic form it goes like this. Given an adjunction between $\infty$-categories (let's call the functors pullback and pushforward, to mimic descent), if we have
1. pullback is conservative (it respects isomorphisms), and
2. pullback respects certain limits (namely totalizations of cosimplicial objects,
which are split after pullback)
then the $\infty$-category downstairs is equivalent to comodules over the comonad
(pullback of pushforward). (There's an opposite monadic form as well)
This can be verified in the usual settings where you expect descent to hold.
In other words if you think of derived categories as being refined to $\infty$-categories (which have the derived category as their homotopy category), then everything you might want to hold does.
So while derived categories don't form a sheaf (stack), their refinements do:
you can recover a complex (up to quasiisomorphism) from a collection of complexes on a cover, identification on overlaps, coherences on double overlaps, coherences of coherences on triple overlaps etc.
More formally: define a sheaf as a presheaf $F$ which has the property that
for an open cover $U\to X$, defining a Cech simplicial object $U\_\bullet=\{U\times\_X U\times\_X U\cdots\times\_X U\}$, then $F(X)$ is the totalization of the cosimplicial object $F(U\_\bullet)$. Then enhanced derived categories form sheaves (in appropriate topologies) as you would expect. This is of course essential to having a good theory of noncommutative algebraic geometry!
| 23 | https://mathoverflow.net/users/582 | 12115 | 8,204 |
https://mathoverflow.net/questions/12119 | 26 | **Background**
Exercise 2.1.16b in Hartshorne (homework!) asks you to prove that if $0 \rightarrow F \rightarrow G \rightarrow H \rightarrow 0$ is an exact sequence of sheaves, and F is flasque, then $0 \rightarrow F(U) \rightarrow G(U) \rightarrow H(U) \rightarrow 0$ is exact for any open set $U$. My solution to this involved the axiom of choice in (what seems to be) an essential way.
Essentially, you are asking to $G(U) \rightarrow H(U)$ to be surjective when you only know that $G \rightarrow H$ is locally surjective. Ordinarily, you might not be able to glue the local preimages of sections in $H(U)$ together into a section of $G(U)$, but since $F$ is flasque, you can extend the difference on overlaps to a global section. This observation deals with gluing finitely many local preimages together. Zorn's lemma enters in to show that you can actually glue things together even if the open cover of $U$ is infinite.
Now, I have not really studied sheaf cohomology, but the idea I have is that it detects the failure of the global sections functor to be right exact. So if you can't even show sheaf cohomology vanishes for flasque sheaves without the axiom of choice, it seems like a lot of the machinery of cohomology would go out the window.
Now, just on the set theoretic level, it seems like there is something interesting going on here. Essentially the axiom of choice is a local-global statement (although I had never thought of it this way before this problem), namely that if $f:X \rightarrow Y$ is a surjection you can find a way to glue the preimages $f^{-1}(\{y\})$ of a surjection together to form a section of the map $f$.
This brings me to my
**Questions**
Can the above mentioned exercise in Hartshorne be proven without the axiom of choice?
How much homological machinery depends on choice?
Have any reverse mathematicians taken a look at sheaf cohomology as a subject to be "deconstructed"?
Have any constructive set theorists thought about using cohomological technology to talk about the extent to which choice fails in their brand of intuitionistic set theory? (it seems like topos models of such set theories might make the connection to sheaves and their cohomology very strong!)
My google-fu is quite weak, but searches for "reverse mathematics cohomology" didn't seem to bring anything up.
| https://mathoverflow.net/users/1106 | Reverse mathematics of (co)homology? | I don't have Hartshorne, so I can't address the specifics of this case. However, there is a very interesting paper by Andreas Blass *Cohomology detects failures of the Axiom of Choice* (TAMS 279, 1983, 257-269), which addresses questions of this type and should at least put you on the right track.
| 19 | https://mathoverflow.net/users/2000 | 12126 | 8,211 |
https://mathoverflow.net/questions/12129 | 5 | Hello, I'm writting something about Malcev categories and monadicity. The fact is that I need to know if Graph is or not complete (have all finite limits). It seems easy but I would like a real answer (not my feelings saying that it is) and I don't find that information anywhere.
Thank you for your answers.
| https://mathoverflow.net/users/3338 | Category of graphs. | First let me just mention that *complete* usually means "has all small limits". If you want to say "has all finite limits", you could use the term *finitely complete* or just *has finite limits*.
You did not explain which particular category of graphs you are talking about, but luckily almost any reasonable choice will have an affirmative answer: yes, the category of graphs has finite limits. I can tell you why if you tell me what objects and morphisms you have in mind.
| 5 | https://mathoverflow.net/users/1176 | 12131 | 8,214 |
https://mathoverflow.net/questions/11868 | 15 | *Formally étale* means that the infinitesimal lifting property is *uniquely* satisfied. If the map is also locally of finite presentation, then it is called *étale*. One of many characterizations (see EGA 4.5.17) of *étale* is flat and unramified. So my question is whether the weaker condition of *formally étale* still implies flatness?
| https://mathoverflow.net/users/917 | Does formally etale imply flat? | It seems that Anton Geraschenkos answer to a previous question [Is there an example of a formally smooth morphism that is not smooth](https://mathoverflow.net/questions/195/is-there-an-example-of-a-formally-smooth-morphism-which-is-not-smooth) does the trick here as well. His example of a formally smooth map that is not flat is indeed formally etale. So, formally etale does **not** imply flat.
| 14 | https://mathoverflow.net/users/917 | 12139 | 8,219 |
https://mathoverflow.net/questions/12118 | 15 | Let $R$ be a (noncommutative) ring. (For me, the words "ring" and "algebra" are isomorphic, and all rings are associative with unit, and usually noncommutative.) Then I think I know what "linear algebra in characteristic $R$" should be: it should be the study of the category $R\text{-bimod}$ of $(R,R)$-bimodules. For example, an $R$-algebra on the one hand is a ring $S$ with a ring map $R \to S$. But this is the same as a ring object in the $R\text{-bimod}$. When $R$ is a field, we recover the usual linear algebra over $R$; in particular, when $R = \mathbb Z/p$, we recover linear algebra in characteristic $p$.
Suppose that $G$ is an algebraic group (or perhaps I mean "group scheme", and maybe I should say "over $\mathbb Z$"); then my understanding is that for any *commutative* ring $R$ we have a notion of $G(R)$, which is the group $G$ with coefficients in $R$. (Probably there are some subtleties and modifications to what I just said.)
>
> **My question:** What is the right notion of an algebraic group "in characteristic $R$"?
>
>
>
It's certainly a bit funny. For example, it's reasonable to want $GL(1,R)$ to consist of all invertible elements in $R$. On the other hand, in $R\text{-bimod}$, the group $\text{Aut}(R,R)$ consists of invertible elements in the *center* $Z(R)$.
Incidentally, I'm much more interested in how the definitions must be modified to accommodate noncommutativity than in how they must be modified to accommodate non-invertibility. So I'm happy to set $R = \mathbb H$, the skew field of quaternions. Or $R = \mathbb K[[x,y]]$, where $\mathbb K$ is a field and $x,y$ are noncommuting formal variables.
| https://mathoverflow.net/users/78 | What is an algebraic group over a noncommutative ring? | It seems that you want some notions on noncommutative group scheme,right?
In fact, A.Rosenberg has introduced **noncommutative group scheme** in his work with Kontsevich
["noncommutative grassmannian and related constructions"](http://www.mpim-bonn.mpg.de/preblob/3621) (2008). Actually, this work gave a systematically treatment to the noncommutative grassmannian type space introduced in their early paper [noncommutative smooth space](https://ui.adsabs.harvard.edu/abs/1998math.....12158K/abstract) and the work of Rosenberg himself on [noncommutative spaces and schemes](http://www.mpim-bonn.mpg.de/preblob/950).
More comments: It seems that you want to know the linear algebra over noncommutative ring. I think you need to look at the paper by Gelfand and Retakh on [Quasideterminants, I](https://doi.org/10.1007/s000290050019). And the main motivation for the "noncommutative grassmannian and related constructions" is to give a geometric explanation to the work of Gelfand and Retakh.
All of these work are based on functor of point of view.
| 5 | https://mathoverflow.net/users/1851 | 12141 | 8,220 |
https://mathoverflow.net/questions/12140 | 2 | Apologies for this elementary question; but I was unable to find a reference otherwise.
Let $A, B, C$ be square matrices of the same dimension. Then,
$$\begin{vmatrix} A & C \\\ 0 & B \end{vmatrix} = |A||B|.$$
The above statement is quite easy to prove using linearity etc properties of the determinant. However this statement is supposed to be a special case of a general theorem of Lagrange(the existence of which I read about, in Artin's Galois theory book).
So what is this general theorem of Langrange, and could someone please provide a reference? Most stuff is usually in Lang's Algebra, but not this one.
| https://mathoverflow.net/users/2938 | Statement of Lagrange's theorem on determinants(elementary question). | Let $A\_{S,T}$ denote the submatrix of $A$ with rows indexed by the elements of $S$ and columns
by the elements of $T$; let $A'\_{S,T}$ denote the submatrix of $A$ with rows indexed by the
elements not in $S$ and columns by the elements not in $T$. Then we have an expansion
$$
\det(A) = \sum\_T (-1)^{\omega(S,T)} \det(A\_{S,T})\det(A'\_{S,T})
$$
where $\omega(S,T)=|S|+|T|$ and $S$ is a fixed $k$-subset of the rows of $A$ and $T$ runs over
all $k$-subsets of the columns.
As you can see this makes short work of your identity. If you can prove the usual row expansion using exterior algebra, you can prove this generalization. I do not know a reference, but this expansion will be in any of the classical books on determinants
(possibly ascribed to Laplace). It can be used to provide a nice proof of the Pluecker relations, satisfied by the $n\times n$ minors of an $n\times 2n$ matrix.
| 5 | https://mathoverflow.net/users/1266 | 12145 | 8,223 |
https://mathoverflow.net/questions/2703 | 22 | Continuing an amazingly interesting chain of answers about [motivic cohomology](https://mathoverflow.net/questions/2146/whats-the-yoga-of-motives), I thought I should learn about the Beilinson conjectures, referred there.
I have found some references, and they seem to present the conjectures from different sides, e.g. there's the statement about [**vanishing**](http://wwwmath.uni-muenster.de/u/pschnei/publ/beilinson-volume/Schneider.pdf) but then there are also connections to [**motivic polylogarithms**](http://www.math.uiuc.edu/K-theory/0152/).
What I miss from these articles in a general picture that would allow us to start somewhere natural. So,
>
> how would you describe an introduction into **Beilinson conjectures in motivic homotopy**?
>
>
>
Sorry for such a loaded question — I really don't know how to make it fit MathOverflow format better. One could theoreticlly post lost of specific questions on the topic, but to ask the right questions in this case you might need to know more than I do. Also, I know there are some technical developments, e.g. the language of derived stacks, and my hope would be that somebody could make a connection to these conjectures using some clear and suitable language.
| https://mathoverflow.net/users/65 | Beilinson conjectures | Let me talk about Beilinson's conjectures by beginning with $\zeta$-functions of number fields and $K$-theory. Space is limited, but let me see if I can tell a coherent story.
### The Dedekind zeta function and the Dirichlet regulator
Suppose $F$ a number field, with
$$[F:\mathbf{Q}]=n=r\_1+2r\_2,$$
where $r\_1$ is the number of real embeddings, and $r\_2$ is the number of complex embeddings. Write $\mathcal{O}$ for the ring of integers of $F$.
Here's the power series for the *Dedekind zeta function*:
$$\zeta\_F(s)=\sum|(\mathcal{O}/I)|^{-s},$$
where the sum is taken over nonzero ideals $I$ of $\mathcal{O}$.
Here are a few key analytical facts about this power series:
1. This power series converges absolutely for $\Re(s)>1$.
2. The function $\zeta\_F(s)$ can be analytically continued to a meromorphic function on $\mathbf{C}$ with a simple pole at $s=1$.
3. There is the *Euler product expansion*:
$$\zeta\_F(s)=\prod\_{0\neq p\in\mathrm{Spec}(\mathcal{O}\_F)}\frac{1}{1-|(\mathcal{O}\_F/p)|^{-s}}.$$
4. The Dedekind zeta function satisfies a *functional equation* relating $\zeta\_F(1-s)$ and $\zeta\_F(s).$
5. If $m$ is a positive integer, $\zeta\_F(s)$ has a (possible) zero at $s=1-m$ of order
$$d\_m=\begin{cases}r\_1+r\_2-1&\textrm{if }m=1;\\
r\_1+r\_2&\textrm{if }m>1\textrm{ is odd};\\
r\_2&\textrm{if }m>1\textrm{ is even},
\end{cases}$$
and its *special value* at $s=1-m$ is
$$\zeta\_F^{\star}(1-m)=\lim\_{s\to 1-m}(s+m-1)^{-d\_m}\zeta\_F(s),$$
the first nonzero coefficient of the Taylor expansion around $1-m$.
Our interest is in these special values of $\zeta\_F(s)$ at $s=1-m$. At the end of the 19th century, Dirichlet discovered an arithmetic interpretation of the special value $\zeta\_F^{\star}(0)$. Recall that the *Dirichlet regulator map* is the logarithmic embedding
$$\rho\_F^D:\mathcal{O}\_F^{\times}/\mu\_F\to\mathbf{R}^{r\_1+r\_2-1},$$
where $\mu\_F$ is the group of roots of unity of $F$. The covolume of the image lattice is the the *Dirichlet regulator* $R^D\_F$. With this, we have the
*Dirichlet Analytic Class Number Formula*. The order of vanishing of $\zeta\_F(s)$ at $s=0$ is $\operatorname{rank}\_\mathbf{Z}\mathcal{O}\_F^\times$, and the special value of $\zeta\_F(s)$ at $s=0$ is given by the formula
$$\zeta\_F^{\star}(0)=-\frac{|\mathrm{Pic}(\mathcal{O}\_F)|}{|\mu\_F|}R^D\_F.$$
Now, using what we know about the lower $K$-theory, we have:
$$K\_0(\mathcal{O})\cong\mathbf{Z}\oplus\mathrm{Pic}(\mathcal{O})$$
and
$$K\_1(\mathcal{O}\_F)\cong\mathcal{O}\_F^{\times}.$$
So the Dirichlet Analytic Class Number Formula reads:
$$\zeta\_F^{\star}(0)=-\frac{|{}^{\tau}K\_0(\mathcal{O})|}{|{}^{\tau}K\_1(\mathcal{O})|}R^D\_F,$$
where ${}^{\tau}A$ denotes the torsion subgroup of the abelian group $A$.
### The Borel regulator and the Lichtenbaum conjectures
Let us keep the notations from the previous section.
*Theorem* [Borel]. If $m>0$ is even, then $K\_m(\mathcal{O})$ is finite.
In the early 1970s, A. Borel constructed the *Borel regulator maps*, using the structure of the homology of $SL\_n(\mathcal{O})$. These are homomorphisms
$$\rho\_{F,m}^B:K\_{2m-1}(\mathcal{O})\to\mathbf{R}^{d\_m},$$
one for every integer $m>0$, generalizing the Dirichlet regulator (which *is* the Borel regulator when $m=1$). Borel showed that for any integer $m>0$ the kernel of $\rho\_{F,m}^B$ is finite, and that the induced map
$$\rho\_{F,m}^B\otimes\mathbf{R}:K\_{2m-1}(\mathcal{O})\otimes\mathbf{R}\to\mathbf{R}^{d\_m}$$
is an isomorphism. That is, the rank of $K\_{2m-1}(\mathcal{O})$ is equal to the order of vanishing $d\_m$ of the Dedekind zeta function $\zeta\_F(s)$ at $s=1-m$. Hence the image of $\rho\_{F,m}^B$ is a lattice in $\mathbf{R}^{d\_m}$; its covolume is called the *Borel regulator* $R\_{F,m}^B$.
Borel showed that the special value of $\zeta\_F(s)$ at $s=1-m$ is a rational multiple of the Borel regulator $R\_{F,m}^B$, *viz*.:
$$\zeta\_F^{\star}(1-m)=Q\_{F,m}R\_{F,m}^B.$$
Lichtenbaum was led to give the following conjecture in around 1971, which gives a conjectural description of $Q\_{F,m}$.
*Conjecture* [Lichtenbaum]. For any integer $m>0$, one has
$$|\zeta\_F^{\star}(1-m)|"="\frac{|{}^{\tau}K\_{2m-2}(\mathcal{O})|}{|{}^{\tau}K\_{2m-1}(\mathcal{O})|}R\_{F,m}^B.$$
(Here the notation $"="$ indicates that one has equality up to a power of $2$.)
### Beilinson's conjectures
Suppose now that $X$ is a smooth proper variety of dimension $n$ over $F$; for simplicity, let's assume that $X$ has good reduction at all primes. The question we might ask is, what could be an analogue for the Lichtenbaum conjectures that might provide us with an interpretation of the special values of $L$-functions of $X$? It turns out that since number fields have motivic cohomological dimension $1$, special values of their $\zeta$-functions can be formulated using only $K$-theory, but life is not so easy if we have higher-dimensional varieties; for this, we must use the weight filtration on $K$-theory in detail; this leads us to motivic cohomology.
Write $\overline{X}:=X\otimes\_F\overline{F}$. Now for every nonzero prime $p\in\mathrm{Spec}(\mathcal{O})$, we may choose a prime $q\in\mathrm{Spec}(\overline{\mathcal{O}})$ lying over $p$, and we can contemplate the decomposition subgroup $D\_{q}\subset G\_F$ and the inertia subgroup $I\_{q}\subset D\_{q}$.
Now if $\ell$ is a prime over which $p$ does not lie and $0\leq i\leq 2n$, then the inverse $\phi\_{q}^{-1}$ of the arithmetic Frobenius $\phi\_{q}\in D\_{q}/I\_{q}$ acts on the $I\_{q}$-invariant subspace $H\_{\ell}^i(\overline{X})^{I\_{q}}$ of the $\ell$-adic cohomology $H\_{\ell}^i(\overline{X})$. We can contemplate the characteristic polynomial of this action:
$$P\_{p}(i,x):=\det(1-x\phi\_{q}^{-1}).$$
One sees that $P\_{p}(i,x)$ does not depend on the particular choice of $q$, and it is a consequence of Deligne's proof of the Weil conjectures that the polynomial $P\_{p}(i,x)$ has integer coefficients that are independent of $\ell$. (If there are primes of bad reduction, this is expected by a conjecture of Serre.)
This permits us to define the *local $L$-factor* at the corresponding finite place $\nu(p)$:
$$L\_{\nu(p)}(X,i,s):=\frac{1}{P\_{p}(i,p^{-s})}$$
We can also define local $L$-factors at infinite places as well. For the sake of brevity, let me skip over this for now. (I can fill in the details later if you like.)
With these local $L$-factors, we define the *$L$-function of $X$* via the Euler product expansion
$$L(X,i,s):=\prod\_{0\neq p\in\mathrm{Spec}(\mathcal{O})}L\_{\nu(p)}(X,i,s);$$
this product converges absolutely for $\Re(s)\gg 0$. We also define the *$L$-function at the infinite prime*
$$L\_{\infty}(X,i,s):=\prod\_{\nu|\infty}L\_{\nu}(X,i,s)$$
and the *full $L$-function*
$$\Lambda(X,i,s)=L\_{\infty}(X,i,s)L(X,i,s).$$
Here are the expected analytical properties of the $L$-function of $X$.
1. The Euler product converges absolutely for $\Re(s)>\frac{i}{2}+1$.
2. $L(X,i,s)$ admits a meromorphic continuation to the complex plane, and the only possible pole occurs at $s=\frac{i}{2}+1$ for $i$ even.
3. $L\left(X,i,\frac{i}{2}+1\right)\neq 0$.
4. There is a *functional equation* relating $\Lambda(X,i,s)$ and $\Lambda(X,i,i+1-s).$
Beilinson constructs the *Beilinson regulator* $\rho$ from the part $H^{i+1}\_{\mu}(\mathcal{X},\mathbf{Q}(r))$ of rational motivic cohomology of $X$ coming from a smooth and proper model $\mathcal{X}$ of $X$ (conjectured to be an invariant of the choice of $\mathcal{X}$) to Deligne-Beilinson cohomology $D^{i+1}(X,\mathbf{R}(r))$. This has already been discussed [here](https://mathoverflow.net/questions/2132/what-is-the-beilinson-regulator). It's nice to know that we now have a precise relationship between the Beilinson regulator and the Borel regulator. (They agree up to exactly the fudge factor power of $2$ that appears in the statement of the Lichtenbaum conjecture above.)
Let's now assume $r<\frac{i}{2}$.
*Conjecture* [Beilinson]. The Beilinson regulator $\rho$ induces an isomorphism
$$H^{i+1}\_{\mu}(\mathcal{X},\mathbf{Q}(r))\otimes\mathbf{R}\cong D^{i+1}(X,\mathbf{R}(r)),$$
and if $c\_X(r)\in\mathbf{R}^{\times}/\mathbf{Q}^{\times}$ is the isomorphism above calculated in rational bases, then
$$L^{\star}(X,i,r)\equiv c\_X(r)\mod\mathbf{Q}^{\times}.$$
| 66 | https://mathoverflow.net/users/3049 | 12148 | 8,224 |
https://mathoverflow.net/questions/12136 | 3 | I’ve couldn’t find any information about the free category built up from that Freyd cover. Where can I find more about the Freyd cover of a category (not a topos!)?
**Edit:** The definition has been given in Lambek and Scott's "Higher order categorical logic". I think (according to L. Román) it is initial among all categories endowed with products and a weak nno.
**Edit:** (Added by Tom Leinster) Here's the definition of Freyd cover, taken from Lambek and Scott (22.1). Let $\mathcal{T}$ be a category with terminal object. Its **Freyd cover** $\hat{\mathcal{T}}$ is the comma category whose objects are the triples $(X, \xi, U)$ where:
* $X$ is a set
* $U$ is an object of $\mathcal{T}$
* $\xi: X \to \mathcal{T}(1, U)$ is a function.
Lambek and Scott emphasize that $\hat{\mathcal{T}}$ has a terminal object and that it comes equipped with a terminal-object-preserving functor $G: \hat{\mathcal{T}} \to \mathcal{T}$. Strictly speaking, the Freyd cover is the pair $(\hat{\mathcal{T}}, G)$, not just the category $\hat{\mathcal{T}}$ itself.
| https://mathoverflow.net/users/3338 | Freyd cover of a category. | I don't know anything about it myself, but here are some other phrases you might try looking up.
The Freyd cover of a category is sometimes known as the **Sierpinski cone**, or "**scone**". It's also a special case of Artin gluing. Given a category $\mathcal{T}$ and a functor $F: \mathcal{T} \to \mathbf{Set}$, the **Artin gluing** of $F$ is the comma category $\mathbf{Set}\downarrow F$ whose objects are triples $(X, \xi, U)$ where:
* $X$ is a set
* $T$ is an object of $\mathcal{T}$
* $\xi$ is a function $X \to F(U)$.
So the Freyd cover is the special case $F = \mathcal{T}(1, -)$.
You can find more on Artin gluing in this important (and nice) paper:
>
> Aurelio Carboni, Peter Johnstone, Connected limits, familial representability and Artin glueing, *Mathematical Structures in Computer Science* 5 (1995), 441--459
>
>
>
plus
>
> Aurelio Carboni, Peter Johnstone, Corrigenda to 'Connected limits...', *Mathematical Structures in Computer Science* 14 (2004), 185--187.
>
>
>
(Incidentally, my Oxford English Dictionary tells me that the correct spelling is 'gluing', but some people, such as these authors, use 'glueing'. I'm sure Peter Johnstone has a reason.)
| 6 | https://mathoverflow.net/users/586 | 12149 | 8,225 |
https://mathoverflow.net/questions/12154 | 7 | For p a constant in (0,1) and n going to infinity such that pn is an integer,
consider the distribution on n bits that selects a random subset of pn bits, sets those to 1, and sets the others to 0.
What is the largest k = k(n,p) so that the induced distribution on any k bits is 1/10 close in total variation distance (a.k.a. statistical distance) to the distribution that sets each bit to 1 independently with probability p?
For every p I would like to know k up to a sublinear (i.e. o(n)) additive term.
(For starters, p = 1/8 is good too.)
Does anybody know of a place where this is worked out?
Thanks!
Emanuele
| https://mathoverflow.net/users/3343 | Local view of setting p*n out of n bits to 1 | You want $\frac15 = \sum\_t |P\_1(count=t) - P\_2(count=t)|$.
where $P\_1$ has a binomial distribution and $P\_2$ is hypergeometric.
The difference between these distributions is shown in [this Mathematica demonstration](http://demonstrations.wolfram.com/BinomialApproximationToAHypergeometricRandomVariable/).
I believe both are reasonably well approximated by normal distributions. Both have mean $pk$. The variance for the binomial distribution is $kp(1-p)$, while it is $\frac{n-k}{n-1}\*k(p)(1-p)$ for the hypergeometric distribution.
So, the value of k should be so that the normal distributions $N(0,1)$ and $N(0,\sqrt{\frac{n-k}{n-1}})$ have total variation distance $\frac1{10}$. That should be at about $k=(1-c)n$ where $N(0,1)$ and $N(0,\sqrt{c})$ are $\frac1{10}$ apart. Numerically, it seems that $c$ should be about 0.6605 so $\sqrt{c}$ should be about 0.8127. $k = 0.3395n$.
It appears this is not sensitive to the value of $p$.
| 3 | https://mathoverflow.net/users/2954 | 12160 | 8,235 |
https://mathoverflow.net/questions/11923 | 14 | The question was edited several times. Most recent version, suggested by Fedja:
>
> Does there exist an open set $U\subset \mathbb R^n$ `(n>1)` that contains balls of arbitrarily large radius and such that no polynomial mapping $p\colon \mathbb R^n\to\mathbb R^n$ takes $U$ onto $\mathbb R^n$? (Take $n=2$ if you prefer.)
>
>
>
---
Older version (with bounty) was answered by Fedja in the affirmative:
>
> Does there exist a topological ball $U\subset \mathbb R^n$ of infinite volume such that no polynomial mapping $p\colon \mathbb R^n\to\mathbb R^n$ takes $U$ onto $\mathbb R^n$?
>
>
>
---
Discussion:
Motivated by a [recent question](https://mathoverflow.net/questions/11904/a-polynomial-map-from-n-to-n-mapping-the-positive-orthant-onto-n "question"), I wonder if there is a geometric characterization of open sets $U\subset \mathbb R^n$ that can be mapped onto $\mathbb R^n$ by a polynomial $p$. Let $U$ be a topological ball to simplify matters. The following *tail volume condition* is necessary for the existence of such $p$.
(TVC) $\int\_1^{\infty} r^{m} |U\setminus B(r)|=\infty$ for some $m>0$.
Indeed, the absolute value of the Jacobian of $p$ must have infinite integral over $U$. Since the Jacobian is a polynomial, $\int\_U |x|^N dx=\infty$ for some $N$. The latter can be rephrased in terms of tail volume: $\int\_1^{\infty} r^{N-1} |U\setminus B(r)| =\infty$. The example of $U=\{(x,y)\in \mathbb R^2\colon x>1, 0<y<x^{-M}\}$ shows that (TVC) is somewhat sharp. This particular $U$ can be mapped onto $\mathbb R^2$ by $(x,y)\mapsto (x,x^{M+1}y)$ followed by translation and the power map $(x+iy)\mapsto (x+iy)^8$.
I am interested in other obstructions besides small tail volume, as well as in reasonably general sufficient conditions. Initially I hoped to get a stronger necessary condition from an affirmative answer to the question below, but Bjorn Poonen answered it in the negative.
"If $f\colon U\to\mathbb R^n$ is a polynomial surjection, does there exist $\epsilon>0$ such that $p(U\cap B(r))$ contains $B(r^\epsilon)$ for large $r$?"
| https://mathoverflow.net/users/2912 | Sets that can be mapped onto R^n by a polynomial | It is enough to construct a sequence of pairwise disjoint disks $D\_j$ of infinite total volume so that the image $\bigcup\_j f(D\_j)$ has $0$ density for any polynomial mapping $f:\mathbb R^2\to\mathbb R^2$. Then you can connect them by very thin passageways to get a topological disk and add just a finite area to the image.
The idea is that near each point $x$ with $|x|=r$, the mapping $f$, either $|\mbox{det}\phantom{,}Df| < r^{-5}$, or $f$ is essentially linear in the square $Q$ with side length $r^{-M}$ centered at $x$ where $M$ depends on $f$ only and $f(Q)$ has diameter less than $1$. Now, we can create a locally finite covering of the entire plane with squares $Q$ such that the side length of each $Q$ is less than $e^{-|c(Q)|}$ where $c(Q)$ is the center of $Q$ and put into each square $Q$ a small disk $D$ whose area is o-small of the area of $Q$ as $c(Q)\to\infty$ (which doesn't contradict the infinite volume restriction). If we forget about the image of a large disk centered at the origin depending on $f$, and the image of those squares on which the determinant is small (both have finite area), then the remaining image is the union of small approximate ellipsoids inside much larger but still small approximate parallelograms.
The final observation is that each generic point can be covered by only $N=N(f)$ parallelograms (Bezout's theorem), so the total area of parallelograms intersecting a large disk is just a constant multiple of the area of that disk and the area of the corresponding ellipsoids is much smaller.
This formally answers Leonid's question about TVC but I'm not very happy with this bubble bath construction. So, let us ask whether there is a set containing sequence of disks of radii tending to $\infty$ that cannot be mapped onto $\mathbb R^2$ by a polynomial mapping.
| 11 | https://mathoverflow.net/users/1131 | 12166 | 8,239 |
https://mathoverflow.net/questions/12144 | 28 | Consider a short exact sequence of Abelian groups -- I'm happy to assume they're finite as a toy example:
$$
0 \to H \to G \to G/H \to 0\ .
$$
I want to understand the classifying space of $G$. Since $BH \cong EG/H$, $G/H$ acts on $BH$ and we can write $BG \cong E(G/H) \times\_{G/H} EG/H$. Thus, we have a fiber bundle (which I'll write horizontally)
$$
BH \to BG \to B(G/H)
$$
On the other hand, the central extension is classified by an element of the group cohomology $H^2(G/H,H)$ which is the same as $H^2(B(G/H),H)$. The latter is an element in the homotopy class of maps $[B(G/H),K(H,2)]$ and $K(H,2)\cong BBH$. This map looks like it classifies a principal $BH$ bundle over $B(G/H)$. I find it hard to imagine that this 'principal' $BH$ bundle is not 'the same' as the bundle above, so the question is, how do you see that? From this construction, it's not even obvious to me that the above bundle is a principal bundle.
I would guess (and being a poor physicist, I'm not so up on my homotopy theory), there's a sense that the classifying space of an Abelian group is an 'Abelian group', and taking classifying spaces of an exact sequence gives you back an 'exact sequence'. That gets you a 'principal bundle' (aren't quotation marks fun?), but even then I'm not sure how to see that the classifying map of this bundle is the same as the class in group cohomology.
Any references to the needed background would also be greatly appreciated.
| https://mathoverflow.net/users/947 | Classifying Space of a Group Extension | Yes. The principal bundles are the same and your guess that $BA$ is an abelian group is exactly right. A good reference for this story, and of Segal's result that David Roberts quotes, is Segal's paper:
G. Segal. Cohomology of topological groups, Symposia Mathematica IV (1970), 377- 387.
The functors $E$ and $B$ can be described in two steps. First you form a simplicial topological space, and then you realize this space. It is easy to see directly that $EG$ is always a group and that there is an inclusion $G \rightarrow EG$, which induces the action. The quotient is $BG$. Under suitable conditions, for example if $G$ is locally contractible (which includes the discrete case), the map $EG\rightarrow BG$ will admit local sections and so $EG$ will be a $G$-principal bundle over $BG$. This is proven in the appendix of Segal's paper, above. There are other conditions (well pointedness) which will do a similar thing.
The inclusion of $G$ into $EG$ is a *normal* subgroup precisely when $G$ is abelian, and so in this case $BG$ is again an abelian group.
I believe your question was implicitly in the discrete setting, but the non-discrete setting is relevant and is the subject of Segal's paper. Roughly here is the answer: Given an abelian (topological) group $H$, the $BH$-princical bundles over a space $X$ are classified by the homotopy classes of maps $[X, BBH]$. When $H$ is discrete, $BBH = K(H,2)$. If $X = K(G,1)$ for a discrete group $G$, these correspond to (central) group extensions:
$$H \rightarrow E \rightarrow G$$
If $G$ has topology, then the group extensions can be more interesting. For example there can be non-trivial group extensions which are trivial as principal bundles. Easy example exist when H is a contractible group. However Segal developed a cohomology theory which classifies all these extensions. That is the subject of his paper.
| 10 | https://mathoverflow.net/users/184 | 12167 | 8,240 |
https://mathoverflow.net/questions/12056 | 12 | Background
----------
In [a recent question about Fibonacci numbers](https://mathoverflow.net/questions/11885/nontrivial-question-about-fibonacci-numbers), [it was claimed](https://mathoverflow.net/questions/11885/nontrivial-question-about-fibonacci-numbers/11961#11961) that
>
> every integer can be written in the form $\sum\_{i=1}^6 \epsilon\_i F\_{n\_i}$ with $\epsilon\_i \in \{0,-1,1\}$. The upper limit on the summation isn't a typo: every number is the sum/difference of at most 6 fibonacci's.
>
>
>
I believe this is false, even for larger (but still finite) values of $k$ (not $6$):
First of all, without loss of generality, we may assume that the representations do not repeat any Fibonacci number (i.e., the $n\_i$s are distinct) and moreover, do not contain any two consecutive Fibonacci numbers (i.e., $n\_i \ne n\_j+1$). We may arrive at such a representation by using the following simplifications repeatedly:
* If two consecutive Fibonacci numbers appear with opposite signs, simplify the expansion with the identity $F\_n - F\_{n-1} = F\_{n-2}$.
* If two consecutive Fibonacci numbers appear with the same sign, simplify the expansion with the identity $F\_n + F\_{n-1} = F\_{n+1}$.
* If the same Fibonacci number appears with opposite signs, simply cancel the two terms.
* If the same Fibonacci number appears with the same sign, then use the identity $F\_n + F\_n = F\_{n-2} + F\_{n-1} + F\_n = F\_{n-2} + F\_{n+1}$ to replace them with two non-identical Fibonacci numbers.
The first three operations reduce the number of terms in the expansion and thus strictly simplify the expression (in terms of how many terms there are), but the last may need to be used several times before it "simplifies" the expression (for example, in terms of how many *repeated* terms there are). Nonetheless, this simplification procedure terminates, as it is impossible to get stuck in an infinite loop using the last operation alone. (Proof: we may assume that the $n\_i$ are positive. Then all of the operations either reduce the number of terms, or leaves that unchanged and reduces the sum of the $n\_i$.)
Now, assume we have such a representation (no identical terms, no consecutive terms) and suppose the largest Fibonacci number appearing is $F\_n$. Then the next largest term (in absolute value) that may appear is $F\_{n-2}$, the next largest after that $F\_{n-4}$, and so on. All in all, the sum of the terms excluding $F\_n$ is at most $F\_{n-2} + F\_{n-4} + \cdots \le F\_{n-1}$ (proof by induction: add $F\_n$ to both sides). By the triangle inequality, the sum of *all* the terms must be at *least* $F\_n - F\_{n-1} = F\_{n-2}$. The point of this calculation is that if you want to represent a number that's less than $F\_{n-2}$, you can't use terms that are $F\_n$ or greater.
This leads us to our contradiction. Consider the integers between $0$ and $F\_{n-2}-1$. How many possible representations are there of numbers in this range? Well, we have six terms all of which are 0 or $\pm F\_k$ for $k\lt n$ (from the above discussion), so we have at most $(2n+1)^6$ representations that could possibly fall into the range. (We're over-counting here because it won't matter and this is easier.) However, there are clearly $F\_{n-2}$ different integers in the given range. Assume for contradiction that it were always possible to represent numbers as the sum/difference of at most 6 Fibonacci's. Then we would have
$$ (2n+1)^6 \ge F\_{n-2}. $$
Finally, because the left side grows polynomially while the right side grows exponentially, a large enough value of $n$ will produce a contradiction.
My questions
------------
1. Is the proof above correct? (If not, and the original claim is correct, can you give me a representation of the number 5473?) *Edit*: Please see [Michael Lugo's answer](https://mathoverflow.net/questions/12056/representing-numbers-in-a-non-integer-base-with-few-but-possibly-negative-nonze/12064#12064) for a paper which finds the representation with the fewest nonzero digits in this "signed Fibonacci base". **Please consider the following the actual question here:**
2. Assuming the proof is correct, is the original claim true for *other* non-integer bases? What I mean is the following:
>
> Does there exist a natural number $k$ and a real number $b>1$ such that every integer has a representation as $\sum\_{i=1}^k \epsilon\_i \lfloor b^{n\_i} + \frac12 \rfloor$? That is, does every number have a representation in "base $b$" (because $b$ is probably irrational, we round $b^n$ to the nearest integer) with at most $k$ non-zero "digits", but where the "digits" may be $\pm 1$?
>
>
>
Note that the original claim is an instance of this: $k=6$ and $b=\varphi = \frac{1+\sqrt 5}{2}$.
I don't think my proof works directly because I used special properties of $\varphi$/the Fibonacci numbers. Is it possible to remove this reliance? In particular, the second step of the proof shows that it's not "useful" to have large Fibonacci numbers in the representation of a small number. Is the same true for every base $b$?
*Note*: if the digits were not allowed be negative, then my proof would go through. The main issue is whether or not $\pm b^n$ for large $n$ can cancel and produce small numbers.
Thanks!
| https://mathoverflow.net/users/1079 | Representing numbers in a non-integer base with few (but possibly negative) nonzero digits | This is an answer to your "actual question" (2), building on some of the ideas in Douglas Zare's answer.
**Lemma 1:** Suppose that $0 < r < 1$. Let $S=\lbrace \epsilon r^i : \epsilon = \pm 1 \text{ and } i \in \mathbb{Z}\_{\ge 0} \rbrace$. Fix $k \ge 1$. Let $S\_k$ be the set of sums of the form $s\_1+\cdots+s\_k$ such that $s\_i \in S$ and $|s\_1|=1$ and there is no nonempty subset $I \subset \lbrace 1,\ldots,k \rbrace$ with $\sum\_{i \in I} s\_i = 0$. Then $0$ is not in the closure of $S\_k$.
**Proof:** Use induction on $k$. The base case is trivial: $S\_1=\lbrace -1,1\rbrace$. Now suppose $k \ge 2$. If a sequence $(x\_i)$ in $S\_k$ converges to $0$, then the smallest summand in the sum giving $x\_i$ must tend to $0$, since a lower bound on the absolute values of the summands rules out all but finitely many elements of $S\_k$, which are all nonzero. Discarding the finitely many $x\_i$ for which the smallest summand is $\pm 1$ and removing the smallest summand from each remaining $x\_i$ yields a sequence $(y\_i)$ in $S\_{k-1}$ tending to $0$, contradicting the inductive hypothesis.
Now fix $b>1$ and $k$. Let $T=\lbrace \epsilon \lfloor b^n + 1/2 \rfloor : \epsilon = \pm 1 \text{ and } n \in \mathbb{Z}\_{\ge 0} \rbrace$. Let $T\_k$ be the set of sums of the form $t\_1+\cdots+t\_k$ with $t\_i \in T$.
**Lemma 2:** Each $t=t\_1+\cdots+t\_k \in T\_k$ equals $u\_1+\cdots+u\_\ell+\delta$ for some $\ell \le k$ and some $u\_i \in T$ with $u\_i = O(t)$ and $\delta = O(1)$.
**Proof:** Examine the powers of $b$ used in the $t\_i$. If any nonempty subsum (with signs) of these powers equals $0$, the corresponding $t\_i$ sum to $O(1)$. If $b^n$ is the largest power that remains after removing all such subsums, divide all the remaining $t\_i$ by $b^n$, and apply Lemma 1 with $r=1/b$ to see that $|t|/b^n$ is bounded away from $0$, so all these remaining $t\_i$, which are $O(b^n)$, are $O(t)$.
**Corollary:** The number of elements of $T\_k$ of absolute value less than $B$ is $O((\log B)^k)$ as $B \to \infty$.
**Corollary:** $T\_k \ne \mathbb{Z}$.
| 8 | https://mathoverflow.net/users/2757 | 12177 | 8,248 |
https://mathoverflow.net/questions/12097 | 8 | Is there a notion of a cobordism which is compatible with bundle structure?
That is, if I have bundles $E$ and $F$, is it the case that the manifold $W$ with $E$ and $F$ as boundary components, can be made into a bundle whose bundle structure, when restricted to $E$ or $F$, is the bundle structure of $E$ or $F$.
And, particularly, when can I connect $E$ and $F$ this way (not just when they're cobordant, but when this cobordism is compatable with this structure)? And what can I say about the bundle structure of $W$, knowing what $E$ and $F$ look like? (e.g., if $E$ and $F$ are G-bundles what can I say about the group action on $W$?)
Also, can anyone point me to any particular references which discuss this? I spent a few hours in our (fairly small) math library looking for something like this, but haven't been able to find anything that seems to discuss this. But I may just not know the right catch phrases to search for!
| https://mathoverflow.net/users/3329 | Cobordisms of bundles? | I'll assume you're talking about principal G-bundles. These are classified by maps into $BG$, the base of the universal $G$-bundle, so if we have bundles classified by $f:E \to BG$ and $g:F \to BG$, you are looking for a bordism between $f$ and $g$ - whether there exists a $h : W \to BG$ connecting these classifying maps. So there is a bundle cobordism between the two bundles iff the bordism classes of $f$ and $g$ in $\mathfrak{N}n(BG)$ coincide, and if they do coincide, then the choice of $W$ is parametrized by the bordism group $\mathfrak{N}\_{n+1}(BG)$. I don't know an algorithmic way to obtain the class $[f]$ from $E$, but there is a splitting $\mathfrak{N}\_n(BG) = \oplus H\_j(BG) \otimes \mathfrak{N}\_{n-j}$ which can help identify some bundles' classes.
| 5 | https://mathoverflow.net/users/2368 | 12179 | 8,249 |
https://mathoverflow.net/questions/12180 | 4 | I am glad to see that a general question like [Is there a relationship between model theory and category theory?](https://mathoverflow.net/questions/11974/is-there-a-relationship-between-model-theory-and-category-theory "this") receives quite a lot attention and no down-votes for being too general and unspecific. So I feel encouraged to pose some follow-up questions (firstly restricted to first-order model theory).
### Preliminaries
I hope the following statements are sufficiently sensible, precise and correct.
* Each first-order theory $T$ with signature $\sigma$ unambigously defines a class of (ZF-)models.
* This class of models of $T$ together with the $\sigma$-homomorphisms form a category (the *category of models of $T$*).
* Two first-order theories with two arbitrary signatures may define [equivalent](http://en.wikipedia.org/wiki/Equivalence_of_categories) categories of models.
>
> **Definition**: A first-order theory $T$
> *provides a model of a category $C$* if the
> category of models of $T$ is
> equivalent to $C$.
>
>
>
* Each category $C$ defines a (possibly empty) set of first-order theories: the set of all $T$ which provide a model of $C$.
### Questions
*[Remark: I had to work this question over, since it seemed to be ill-posed.]*
Old version:
Can infinite [concrete categories](http://en.wikipedia.org/wiki/Concrete_category) be specified
other than as categories of (ZF-)models of some (possibly higher-order) theory? Examples?
New version (explicitly restricted to first-order theories):
>
> Given an infinite category of models
> of a first-order theory $T$. Can this
> category - or one equivalent to it - be
> specified/represented/given
> independently of any first-order
> theory $T$ and its (ZF-)models?
>
>
>
Remark: $T$ of course *can* be specified/represented/given independently of its models: as a set of formulas.
---
>
> Why is the notion of *models of a (concrete)
> category* so uncommon? (Maybe because the
> answer to the first question is "No"?)
>
>
>
---
>
> Is there a genuine model-theoretic notion of two
> equivalent theories if these have
> two arbitrary signatures?
>
>
>
---
| https://mathoverflow.net/users/2672 | Can infinite first-order categories be specified other than as categories of models? | Sure, by direct construction. Rings, preorders, the category of paths of a given graph, etc. But that's not what you wanted to know, is it?
| 3 | https://mathoverflow.net/users/3154 | 12181 | 8,250 |
https://mathoverflow.net/questions/12080 | 3 | I was playing around with the Shannon Switching Game for some planar graphs, trying to get some intuition for the strategy, when I noticed a pattern. Since I only played on planar graphs, I'll restrict the problem to those for now, but we can also ask the problem for non-planar graphs.
Call a graph feasible if $E \ge 2V-2$, where $E$ is the number of edges and $V$ the number of vertices on the graph. Call a graph a minimal feasible graph if none of its proper subgraphs containing at least two vertices are feasible.
Is every minimal feasible graph a winning position for Short, regardless of which pair of vertices he has to connect? In other words, for those who don't know the solution to the Shannon Switching Game, is it true that every minimal feasible graph contains two edge-disjoint spanning trees?
If this is true, I think it provides a more "intuitive" way of looking at the Shannon Switching Game in actual play, where you can't spend your time drawing cospanning trees on your notebook - look for the smallest subgraph with more edges than it deserves, mentally collapse it to a point, and repeat.
| https://mathoverflow.net/users/2363 | About the Shannon Switching Game | Yes it is true, it follows easily from Tutte's disjoint tree theorem for k=2:
<http://lemon.cs.elte.hu/egres/open/Tutte%27s_disjoint_tree_theorem>
In every partition class with n\_i vertices you can have at most 2n\_i-3 edges, plus the edges between the partitions, but this together still gives only 2n-2-|P| edges, where |P| is the number of partitions.
(This is joint answer with Filip Moric who does not have a MO account.)
| 4 | https://mathoverflow.net/users/955 | 12193 | 8,258 |
https://mathoverflow.net/questions/1960 | 15 | The number of Dyck paths in a square is well-known to equal the catalan numbers:
<http://mathworld.wolfram.com/DyckPath.html>
But what if, instead of a square, we ask the same question with a rectangle? If one of its sides is a multiple of the other, then again there is a nice formula for the number of paths below the diagonal, but is there a nice formula in general? What is the number of paths from the lower-left corner of a rectangle with side lengths a and b to its upper-right corner staying below the diagonal (except for its endpoint)? I am also interested in asymptotics.
| https://mathoverflow.net/users/955 | Dyck paths on rectangles | Since that Mirko Visontai told me that the answer is ${a+b\choose a}/(a+b)$ if $\gcd(a,b)=1$. The proof is the following (with k=a and l=b):
The number of 0--1 vectors with $k$ 0's and $l$ 1's is ${k+l\choose k}$, so we have to prove that out of these vectors exactly $1/(k+l)$ fraction is an element of $L(k,l)$. The set of all vectors can be partitioned into equivalence classes. Two vectors $p$ and $q$ are equivalent if there is a cyclic shift that maps one into the other, i.e., if for some $j$, $p\_i = q\_{i+j}$ for all $i$. We will prove that exactly one element from each equivalence class will be in $L(k,l)$. This proves the statement as each class consists of $k+l$ elements because $gcd(k,k+l)=1$.
We can view each 0--1 sequence as a walk on $\mathbb R$ where each 0 is a $-l/(k+l)$ step and each 1 is a $+k/(k+l)$ step. Each $(k,l)$ walk starts and ends at zero and each walk reaches its maximum height exactly once, otherwise $ak + bl = 0$ for some $0 < a +b < k+l$ which would imply $\gcd(k,l) \neq 1$.
If we take the cyclic shift that ``starts from the top'', we stay in the negative region throughout the walk, which corresponds to remaining under the diagonal in the lattice path case. Any other cyclic shift goes above zero, which corresponds to going above the diagonal at some point.
| 10 | https://mathoverflow.net/users/955 | 12196 | 8,261 |
https://mathoverflow.net/questions/12190 | 2 | Has anyone ever seen any papers or books including set-theoretic descriptions of formal language theory? Specifically, I'm interested in how one would formalize context-free grammars with sets.
Some of this, I suppose is fairly obvious. For example, strings would use a foundational formalism much like ordered pairs (e.g. Kuratowski's definition or similar) but what about objects like production rules and their semantics?
This isn't really necessary for me to get any actual work done, I just thought it would help me build a better intuition around formal language theory.
Thanks in advance,
* Anthony
Update:
An example of what I'm looking for would be string concatenation. How would one construct this in pure set theroy? If we consider the strings "a" and "b", and consider their set representations to be { { a } } and { { b }} (as per Kuratowski), then the desired result would be { { a }, { a, b } }. Clearly this cannot be accomplished by a defining string concatenation to be set union but there are obviously numerous ways of doing so. The choice at each point in defining how to do something in formal language theory using pure set theory will, in some cases, determine how other notions can be defined. As a consequence, some definitions may be more complex than others (or less elegant than others, one might say). I'm just curious if anyone has done anything like this before.
| https://mathoverflow.net/users/3357 | Set-theoretic foundations for formal language theory? | One can do this using less technology, too...
Let $\Sigma$ be an alphabet, $N$ a set of non-terminals, and $\Sigma^\\*$ and $(\Sigma\cup N)^\\*$ the full languages on $\Sigma$ and $\Sigma\cup N$, respectively. A context-free grammar is a finite subset $G\subset N\times(\Sigma\cup N)^\\*$. Given one such grammar $G$ there is a relation $\mathord\rightarrow\_G\subseteq(\Sigma\cup N)^\\*\times(\Sigma\cup N)^\\*$ which is the least transitive reflexive relation which contains $G$ (notice that $N\times(\Sigma\cup N)^\\*\subseteq (\Sigma\cup N)^\\*\times(\Sigma\cup N)^\\*$, so this makes sense) and such that
$$a\rightarrow\_Gb \wedge a'\rightarrow\_Gb'\implies ab\rightarrow\_Ga'b'.$$ The language generated by $G$ from a non-terminal $n\in N$ is just $L(G, n)=\{w\in \Sigma^\\*:n\rightarrow\_Gw\}$. This is, in fact, the standard way to do this...
| 7 | https://mathoverflow.net/users/1409 | 12201 | 8,265 |
https://mathoverflow.net/questions/12085 | 149 | I would like to ask about examples where experimentation by computers has led to major mathematical advances.
=============================================================================================================
A new look
----------
Now as the question is **five years old** and there are certainly **more examples** of mathematical advances via computer experimentation of various kinds, I propose to consider contributing new answers to the question.
Motivation
----------
I am aware about a few such cases and I think it will be useful to gather such examples together. I am partially motivated by the recent [polymath5](https://gowers.wordpress.com/2010/01/16/the-erds-discrepancy-problem-v/) which, at this stage, have become an interesting experimental mathematics project. So I am especially interested in examples of successful "mathematical data mining"; and cases which are close in spirit to the experimental nature of polymath5. My experience is that it can be, at times, very difficult to draw useful insights from computer data.
Summary of answers according to categories
------------------------------------------
(Added Oct. 12, 2015)
To make the question a useful resource (and to allow additional answers), here is a quick summery of the answers according to categories. (Links are to the answers, and occasionally to an external link from the answer itself.)
**1) Mathematical conjectures or large body of work arrived at by examining experimental data - Classic**
[The Prime Number Theorem](https://mathoverflow.net/a/12206); [Birch and Swinnerton-Dyer conjectures](https://mathoverflow.net/a/12090); [Shimura-Taniyama-Weil conjecture](https://mathoverflow.net/a/123335); [Zagier's conjectures on polylogarithms](https://mathoverflow.net/a/58678); [Mandelbrot set](https://mathoverflow.net/a/14832); [Gosper Glider Gun](https://en.wikipedia.org/wiki/Glider_gun) ([answer](https://mathoverflow.net/a/37473)), [Lorenz attractor](https://mathoverflow.net/a/17190); Chebyshev's bias ([answer](https://mathoverflow.net/a/216002)) ; [the Riemann hypothesis](https://mathoverflow.net/a/23081); the [discovery of the Feigenbaum constant](https://mathoverflow.net/a/13023); (related) [Feigenbaum-Coullet-Tresser universality and Milnor's Hairiness conjecture](https://mathoverflow.net/a/219152/1532); [Solving numerically the so-called Fermi--Pasta--Ulam chain and then of its continuous limit, the Korteweg--de Vries equation](https://mathoverflow.net/a/12087)
**2) Mathematical conjectures or large body of work arrived at by examining experimental data - Current**
"[Maeda conjecture](https://mathoverflow.net/a/123342)"; [the work of Candès and Tao on compressed sensing](https://mathoverflow.net/a/42197); Certain [Hankel determinants](https://mathoverflow.net/a/37515); Weari-Phelan structure; the [connection of multiple zeta values to renormalized Feynman integrals](https://mathoverflow.net/a/17190); [Thistlethwaite's discovery](https://mathoverflow.net/a/13290) of links with trivial Jones polynomial; The [Monstrous Moonshine;](https://en.wikipedia.org/wiki/Monstrous_moonshine) [McKay's account on experimentation leading to mysterious "numerology" regarding the monster.](https://mathoverflow.net/a/220632/1532) ([link to the answer](https://mathoverflow.net/a/12189)); Haiman [conjectures on the quotient ring by diagonal invariants](http://math.berkeley.edu/~mhaiman/ftp/diagonal/diagonal.pdf)
**3) Computer-assisted proofs of mathematical theorems**
Kepler's conjecture ; [a new way to tile the plane](https://mathoverflow.net/a/216002) with a pentagon: [advances regarding bounded gaps between primes](https://mathoverflow.net/a/198721) following Zhang's proof; Cartwright and Steger's work on [fake projective planes](https://mathoverflow.net/a/16732); the Seifert-Weber [dodecahedral space is not Haken](https://mathoverflow.net/a/12157); [the four color theorem, the proof of the nonexistence of a projective plane of order 10;](https://mathoverflow.net/a/12125) [Knuth's work](https://mathoverflow.net/a/17181) on a class of projective planes; The search for Mersenne primes; [Rich Schwartz](http://www.math.brown.edu/~res/)'s [work](https://mathoverflow.net/a/12107); [The computations done by the 'Atlas of Lie groups'](https://mathoverflow.net/a/12216) software of Adams, Vogan, du Cloux and many other; [Cohn-Kumar proof](https://mathoverflow.net/a/13023) for the densest lattice pacing in 24-dim; [Kelvin's conjecture](https://mathoverflow.net/a/27298); (NEW) [$R(5,5) \le 48$ and $R(4,5)=25$](https://mathoverflow.net/a/265783/1532);
**4) Computer programs that interactively or automatically lead to mathematical conjectures.**
[Graffiti](https://mathoverflow.net/a/216222)
**5) Various computer programs which allow proving automatically theorems or generating automatically proofs in a specialized field.**
Wilf-Zeilberger formalism and software; [FLAGTOOLS](https://mathoverflow.net/a/13062)
**6) Computer programs (both general purpose and special purpose) for verification of mathematical proofs.**
[The verification of a proof of Kepler's conjecture.](https://mathoverflow.net/a/216078)
**7) Large databases and other tools**
Sloane's online encyclopedia for integers sequences; the [inverse symbolic calculator](http://mrob.com/pub/ries/).
**8) Resources:**
Journal of experimental mathematics; [Herb Wilf](https://mathoverflow.net/a/12155)'s article: Mathematics, an experimental science in the Princeton Companion to Mathematics, genetic programming applications a fairly comprehensive website [experimentalmath.info](http://www.experimentalmath.info/)
; discovery and experimentation in number theory; Doron Zeilberger's classes called "experimental mathematics":math.rutgers.edu/~zeilberg/teaching.html;
[V.I. Arnol'd's](https://en.wikipedia.org/wiki/Vladimir_Arnold) two books on the subject of experimental mathematics in Russian, [*Experimental mathematics*](http://www.ozon.ru/context/detail/id/3249521/), Fazis, Moscow, 2005, and [*Experimental observation of mathematical facts*](http://biblio.mccme.ru/node/1889), MCCME, Moscow, 2006
**Answers with general look on experimental mathematics:**
[Computer experiments allow new avenues for productive strengthening of a problem](https://mathoverflow.net/a/17181) (A category of experimental mathematics).
---
### Bounty:
There were many excellent answers so let's give the bounty to Gauss...
Related question: [Where have you used computer programming in your career as an (applied/pure) mathematician?](https://mathoverflow.net/questions/23982/where-have-you-used-computer-programming-in-your-career-as-an-applied-pure-math), [What could be some potentially useful mathematical databases?](https://mathoverflow.net/questions/68442/what-could-be-some-potentially-useful-mathematical-databases;) [Results that are easy to prove with a computer, but hard to prove by hand](https://mathoverflow.net/questions/95776/results-that-are-easy-to-prove-with-a-computer-but-hard-to-prove-by-hand) ; [What advantage humans have over computers in mathematics?](https://mathoverflow.net/questions/233483/what-advantage-humans-have-over-computers-in-mathematics;) [Conceptual insights and inspirations from experimental and computational mathematics](https://mathoverflow.net/questions/347490/conceptual-insights-and-inspirations-from-experimental-and-computational-mathema)
| https://mathoverflow.net/users/1532 | Experimental mathematics leading to major advances | The Prime Number Theorem was conjectured by Gauss from looking (*very hard*, one can presume...) at a table of the primes $\leq10^6$. It is not with too much effort that one can read his *Disquisitiones* as a set of tricks to determine primality with as little work as possible, and one can understand the motivation: he was his own computer, in a way :P
(I don't know where Legendre got the statement from, but he must surely have had tables of primes too!)
| 65 | https://mathoverflow.net/users/1409 | 12206 | 8,270 |
https://mathoverflow.net/questions/12200 | 24 | How does one prove that on $S^n$ (with the standard connection) any geodesic between two fixed points is part of a great circle?
For the special case of $S^2$ I tried an naive approach of just writing down the geodesic equations (by writing the Euler-Lagrange equations of the length function) and solving them to gain some insights but even if the equations are solvable I can't see how to show that they are great circles. (the solutions are some pretty complicated functions which don't give me much insight)
I checked the article on Great Circles on Wolfram Mathworld for a coordinate geometry approach to it but that article looked quite cryptic to me!
One knows that on compact semi-simple lie groups any one-parameter subgroup generates a geodesic and $S^n$ is the quotient of 2 compact semi-simple lie groups $SO(n+1)/SO(n)$. Is this line of thought useful for this question?
=================================================================================
After some of the responses came let me put in "a" way of seeing the above for $S^2$ (wonder if it is correct). If $\theta$ and $\phi$ are the standard coordinates on $S^2$ then the equations for the curve are
$$\ddot{\theta} = \dot{\phi}^2 sin(\theta)cos(\theta)$$
$$\dot{\phi}sin^2{\theta} = k$$
where $k$ is some constant set by the initial data of the curve.
Now given the initial point I can choose my coordinate system such that the the initial data looks like $\dot{\phi}=0$, $\theta = \text{some constant}$, $\dot{\theta}=\text{some constant}$, $\phi = \text{some constant}$. Then the differential equations tell me that the $k=0$ and the only way it can happen for times is by having ,
$$\dot{\phi} = 0$$
Which clearly gives me a longitude in this coordinate system.
Hence the geodesic equation gives as a solution a great circle.
Surely not an elegant proof like Bar's reference.
But I hope this is correct.
{As a friend of mine pointed out that this set of coordinates is motivated by the fact that the way the "energy" of the curve is being parametrized the z-component of the angular momentum is conserved which is in fact my second Euler-Lagrange equations}
| https://mathoverflow.net/users/2678 | Geodesics on spheres are great circles | Although Jose has made the essentially the same point, I just want to elaborate (this is really just a comment, but I always run out of room in the comment box).
What nobody else has mentioned explicitly is that you *should* have trouble solving the Euler-Lagrange equation for the length functional. The Euler-Lagrange equation is a second order ODE, but a highly degenerate one. And you know this before you even start. Why? Well, suppose you have a solution. Then if you reparameterize that curve using *any* arbitrary parameterization (i.e., any monotone function of the original parameter), the newly parameterized curve is still a solution to the Euler-Lagrange equation. That means that the ODE has an infinite dimensional space of solutions and it is nothing like any ODE we learned about in our ODE courses or textbooks.
A trick is needed to get around this, namely to use the so-called energy functional $E[\gamma] = \int\_0^1 |\gamma'(t)|^2 dt$ (which is not invariant under reparameterization of the curve) instead of the length functional (which is). The Holder inequality shows that a minimum of the energy functional is necessarily a minimum of the length functional that is parameterized by a constant times arclength, i.e. a constant speed geodesic.
The Euler-Lagrange equation for the energy functional is a nice nondegenerate 2nd order ODE that can be handled by standard ODE techniques and theorems.
As for the standard sphere, there are many different ways to solve for the geodesics. To review the ways already suggested in other answers:
1) I recommend that you first do it without the machinery of Riemannian geometry and using only the Euclidean structure of $R^{n+1}$. Using the discussion above, you should be able to show that a curve on the unit sphere is a constant speed geodesic if and only if its acceleration vector is always normal to the sphere. You should then be able to work out the solutions to this ODE. The suggestion that you assume one point is the north pole and the other lies in a co-ordinate plane is a good one and makes the ODE easy to solve.
2) The other way is to do it all intrinsically. Here, I recommend using stereographic co-ordinates and assuming one point is the origin in those co-ordinates. Again, everything becomes very easy in that situation.
3) And the third way is to view the sphere as a homogeneous space and use formulas for that situation. I don't remember the details myself, but I learned them from the book by Cheeger and Ebin.
I recommend that you work through all 3 different ways, as well as any other way you can find.
As others have noted, the calculations for geodesics on hyperbolic space are identical, except that you are working with a "unit sphere" in Minkowski instead of Euclidean space. There is even a notion of stereographic projection (but onto what?). This is also fun to work out carefully.
Finally, I do want to note that after you work this all out and have it all in your head, it's a really beautiful picture and story. And if you find the right angle, it's all very simple, so you can work out the details yourself and not rely on reading a book line-by-line or having someone else show you all the details. Try to get the essential ideas and necessary tricks (like using the energy functional) from books, lectures, or teachers, but try to work everything else out from scratch (i.e., minimal reliance on theorems you can't prove yourself).
| 29 | https://mathoverflow.net/users/613 | 12217 | 8,277 |
https://mathoverflow.net/questions/12224 | 3 | If $A = (\alpha\_{ij}) \in \mathbb{C}^{nxm}$ we have simple algorithms by which to determine $\mathrm{rank}(A)$. However, is there a polynomial $f \in \mathbb{C}[\alpha\_{ij}]$ where $f \colon \mathbb{C}^{nxm} \to \mathbb{N}$ such that $f(A) = \mathrm{rank}(A)$?
In general, is it possible to determine if a particular algorithm may be expressed as a polynomial or other more general function?
| https://mathoverflow.net/users/3121 | Rank(A) and other algorithms as a polynomial | The preimage of a natural number under a polynomial map is always closed, yet the set of matrices with rank one is not closed. For example, the sequence $(A\_n)\_{n\geq1}$ in $M\\_2(\mathbb C)$ with $A\_n=\left(\begin{smallmatrix}1/n&0\\\0&0\end{smallmatrix}\right)$ has all its items of rank one, yet its limit has rank zero.
| 5 | https://mathoverflow.net/users/1409 | 12225 | 8,283 |
https://mathoverflow.net/questions/12211 | 19 | Finite Ramsey's theorem is a very important combinatorial tool that is often used in mathematics. The infinite version of Ramsey's theorem (Ramsey's theorem for colorings of tuples of natural numbers) also seems to be a very basic and powerful tool but it is apparently not as widely used.
I searched in the literature for applications of infinite Ramsey's theorem and only found
* straight forward generalization of statements that follow from finite Ramsey's theorem (example: Erdos-Szekeres ~> every infinite sequence of reals contains a monotonic subsequence) and some other basic combinatorial applications,
* Ramsey factorization for \omega-words,
* the original applications of Ramsey to Logic.
Where else is infinite Ramsey's theorem used?
Especially are there applications to analysis?
| https://mathoverflow.net/users/3365 | Applications of infinite Ramsey's Theorem (on N)? | The following fact has been called "Ramsey's Theorem for Analysts" by H. P. Rosenthal.
>
> **Theorem.** Let $(a\_{i,j})\_{i,j=0}^\infty$ be an infinite matrix of real numbers such that $a\_i = {\displaystyle\lim\_{j\to\infty} a\_{i,j}}$ exists for each $i$ and $a = {\displaystyle\lim\_{i\to\infty} a\_i}$ exists too. Then there is an infinite sequence $k(0) < k(1) < k(2) < \cdots$ such that $a = {\displaystyle\lim\_{i<j} a\_{k(i),k(j)}}$.
>
>
>
The last limit means that for every $\varepsilon > 0$ there is an $n$ such that $n < i < j$ implies $|a-a\_{k(i),k(j)}| < \varepsilon$. When the matrix is symmetric and ${\displaystyle\lim\_{i\to\infty} a\_{k(i),k(i)}} = a$ too, this is just an ordinary double limit.
The proof is a straightforward applications of the two-dimensional Ramsey's Theorem. The obvious higher dimensional generalizations are also true and they can be established in the same way using the corresponding higher dimensional Ramsey's Theorem. These are used to construct "spreading models" in Banach Space Theory.
| 21 | https://mathoverflow.net/users/2000 | 12228 | 8,285 |
https://mathoverflow.net/questions/12213 | 4 | As is well known, the quantum groups $SU\_q(n)$, amongst others, arise from $R$-matrix solutions of the Yang-Baxter Equation. My question is: For any subalgebra of $GL(n)$, does there exist an $R$-matrix Yang-Baxter solution that q-deforms it?
| https://mathoverflow.net/users/1095 | FTR Quantization for any Subalgebra of $GL(n)$? | There are certainly FRT-type constructions for all the classical simple Lie groups, types A,B,C,D (and a modification of type A to encompass $GL\_n$ instead of $SL\_n$). See e.g. Klymik and Schmudgen, Quantum Groups and Their Representations (I think that's the title). Using the theory of roots, etc. it is also possible to construct subalgebras of $U\_q(\mathfrak{g})$ corresponding to Borel subgroups, parabolic subgroups, and subgroups corrresponding to symmetric pairs (see papers by Dijkhuizen, Joseph, Kolb, Letzter, Noumi, Stokman, etc. with the words "quantum symmetric pairs" in the titles).
More generallly, there seems to be a philosophy that any construction involving simple algebraic groups of classical type has a quantum group generalization, but there are often interesting extra degrees of freedom when you quantize and unexpected subtleties. If you elaborate which subgroups you're interested in, I can try to elaborate where to find sources (it would comprise a whole compendium to list sources for all such constructions =]).
One thing to clarify as regards your question: In the cases I mentioned above, one doesn't seek a new R-matrix to describe the subgroups of $GL\_n$. Remember that the R-matrix is used to define the commutativity relations of the "coordinates" $a^i\_j$. This defines the "quantized coordinate algebra" O\_q(G), which plays the role of O(G) in algebraic groups. finding subgroups of G in terms of the coordinate algebra O(G) means finding quotient Hopf algebras. There is a similar picture for O\_q(G), although depending on the situation, there can be extra difficulties. I believe that the Borel subgroup is quantized in a fairly straightforward way (dually, there is a Hopf subalgebra $U\_q(\mathfrak b) \subset U\_q(\mathfrak g)$). I suspect it's slightly more tricky when you consider $U\_q(\mathfrak p)$ for parabolics, but I believe it is understood by now (perhaps someone more knowledgeable can post another answer with references?). For symmetric pairs, one considers one-sided co-ideal subalgebras of $U\_q(\mathfrak g)$ instead of Hopf subalgebras. The details in each example are technical, but the point I mean to clarify is that one doesn't find a new R-matrix for each sub-algebra, but rather uses the same R-matrix to define commutation relations and then studies quotients (dually, subalgebras) of the original (or dual) algebra.
By the way, I believe that the way one handles real forms, for instance $O\_q(SU\_n)$, is to observe that $O\_q(SL\_n)$ has an involution (discussed in Klymik and Schmudgen, for instance), and one studies $O\_q(SL\_n)$ representations which are compatible with the involution (meaning they have an inner product with a condition relating inner product to convolution). This comes from thinking about $SU\_n$ as a subgroup of $SL\_n$ fixed by $X\mapsto (X^\dagger)^{-1}$.
| 3 | https://mathoverflow.net/users/1040 | 12241 | 8,293 |
https://mathoverflow.net/questions/12249 | 2 | Let's call the following conditions (1): $X$ is a complete metric space with metric $d$, $X = \cup\_{n=1}^\infty A\_n$. Let $\bar{A}$ denote the closure of $A$.
Let's call the following statement (2): at least one of the $\bar{A}\_n$ contains a ball.
Baire category theorem gives:
Fact1:
(1) $\Rightarrow$ (2)
Now take any $x\in X$ and consider the closed ball $B = B(x,\delta)=\{y: d(x,y)\le \delta\}$. This is a complete metric space itself and it is covered by $ A\_n \cap B$. Thus these sets satisfy (1). Fact1 gives us an $n$ such that the closure of $A\_n \cap B$ contains a ball in $B$. Thus, we have strengthened Fact1 to
Fact1': (1) $\Rightarrow$ (2')
where (2') is: For every $x\in X$, every neighborhood of $x$ contains a ball that is contained in one of the $\bar{A}\_n$.
Question: Can (2') be strengthened further? Here are some example statements, both of which are too strong:
* For every $x$, there is a $\delta$ such that $B(x,\delta)$ is contained in one of the $\bar{A}\_n$
* For every $x$, there is an $A\_n$ such that $\bar{A}\_n$ contains an open set $G$ with $d(x,G)=0$.
---
Many thanks for the responses. The motivation for this question was as follows.
1) What does it mean for a set $A$ to have a closure with empty interior? Take an element
$a \in \bar{A}$. This means that $a$ is either in $A$ or there is a sequence in $A$ that converges to $a$. This can be rephrased as $a$ is a point that can be approximated with infinite precision by $A$.' If $\bar{A}$ has no interior, perturbing $a$ by a small amount will give an $a'$, such that $a'$ is a finite distance away from $\bar{A}$. Thus, $a'$ will be a point that can only be approximated by $A$ with finite precision. Then, one can think of $A$ as a multi resolution grid with discontinuous approximation capability: you perturb any point that can be approximated infinitely well by it and you get a point that can only be approximated with finite precision.
2) $X$ itself can be thought of as the perfect multi resolution grid for itself: every point $x\in X$ is well approximated by $X$, and perturbing $x$ will not change this. The way I wanted to think of $X= \cup\_{n=1}^\infty A\_n$ was this: for every $x \in X$, one of the $A\_n$ provides a multiresolution grid around $x$ that has continuous approximation capability. I wanted to think, similar to Leonid's response, that the set of $X$ where this is not possible was to be in some sense to be exceptional.
| https://mathoverflow.net/users/3370 | Baire category theorem | Maybe if you allow an exceptional set, as in "For every $x$ outside of some meagre set, there is a $\delta>0$ such that $B(x,\delta)$ is contained in one of the $\overline{A}\_n$"
(Indeed, the set of all $x$ where the above fails is a closed subset of $X$ with empty interior.)
| 2 | https://mathoverflow.net/users/2912 | 12253 | 8,300 |
https://mathoverflow.net/questions/12248 | 8 | I'm probably missing something obvious, but I've been wondering what the motivation is for requiring the components $A\_\mu$ in a local trivialization of a gauge connection on a smooth principal $G$-bundle to lie in $\mathfrak{g}$, the Lie algebra of $G$. I can see that this gives a couple of nice properties; for example, in a local trivialization it ensures that under a gauge transformation $A'\_\mu=gA\_\mu g^{-1}+g\partial\_\mu g$ lies in $\mathfrak{g}$, and that the curvature form $F=dA+A\wedge A$ lies in $\mathfrak{g}$ (since $\mathfrak{g}$ is closed under the Lie bracket). But is there a more intrinsic or geometric reason that $A\_\mu$ must be in $\mathfrak{g}$? Thanks.
| https://mathoverflow.net/users/3372 | Gauge connections and Lie algebras? | First, I never liked working with principal bundles; vector bundles seem easier and more natural to me. Second, I never like thinking about abstract principal $G$-bundles. I prefer fixing a representation of $G$ and viewing the principal $G$ bundle as a reduced frame bundle associated with a vector bundle.
So let $E$ be a rank $k$ vector bundle and $F$ the bundle of arbitrary frames in $E$ (this is a principal $GL(k)$-bundle). Then $GL(k)$ acts on the right on $F$. Given a subgroup $G$ in $GL(k)$, let $F\_G$ be a subbundle of $F$ such that if $f \in F\_G$, then so is $f\cdot g$ for each $g \in G$.
The primary example is $E = T\_\*M$ and $F\_G$ is the bundle of orthonormal bases of the tangent space with respect to a Riemannian metric.
What is the critical property we want a $G$-connection to satisfy? Well, any connection allows you to parallel translate an arbitrary frame $f \in F$ along a curve. We'd like the $G$-connection to be such that if $f \in F\_G$, then the parallel translation remains in $F\_G$. This leads to the right definition of a $G$-connection.
| 10 | https://mathoverflow.net/users/613 | 12257 | 8,304 |
https://mathoverflow.net/questions/12260 | 8 | I haven't learned that much about primary decomposition, but from I understand about Dedekind domains, we have that all fractional ideals are invertible and all (plain old) ideals factor uniquely into a product of prime ideals, so that Dedekind domains should satisfy this condition. Are these the only such rings, or is there a weaker condition we can give?
| https://mathoverflow.net/users/1916 | When does the group of invertible ideal quotients = the free abelian group on the prime ideals? | First some responses to comments above:
One can (and should) define fractional ideals for any integral domain $R$. A fractional $R$-ideal is a nonzero $R$-submodule $I$ of the fraction field $K$ such that there exists $x \in K \setminus \{0\}$ such that $xI \subset R$. The product of two fractional ideals is again a fractional ideal, and the multiplication is associative and has $R$ itself as an identity: in other words, the set $\operatorname{Frac}(R)$ of fractional ideals of $R$ forms a **commutative monoid**.
We have the notion of a **principal** fractional ideal: this is a submodule of the form $xR$ for $x \in K^{\times}$. The set of principal fractional ideals forms a subgroup $\operatorname{Prin}(R)$ of $R$. One could take the quotient $\operatorname{Frac(R)}/\operatorname{Prin(R)}$, but this is a bit of a false step, as it is not a group in general.
In the case of Dedekind domains there is nothing to worry about:
Theorem: For an integral domain $R$, the following are equivalent:
(i) $R$ is Noetherian, integrally closed of dimension at most one (a Dedekind domain).
(ii) Every nonzero integral ideal of $R$ factors into a product of prime ideals.
(ii)' Every nonzero integral ideal of $R$ factors *uniquely* into a product of prime ideals.
(iii) The fractional ideals of $R$ form a group.
So for a Dedekind domain, certainly $\operatorname{Frac}(R)/\operatorname{Prin(R)}$ is a group, called the **ideal class group** of $R$.
In general, there is an easy way to remedy the problem that $\operatorname{Frac}(R)/\operatorname{Prin(R)}$ need not be a group. Namely, instead of taking the full monoid of fractional ideals, we restrict to the unit group $I(R)$, the **invertible fractional ideals**. For any domain $R$, we may define the Picard group
$\operatorname{Pic}(R) = I(R)/\operatorname{Prin}(R)$.
Because of the theorem above, for a non-Dedekind domain the Picard group isn't capturing any information about the prime ideals in particular. There is however a different construction -- coinciding with $\operatorname{Pic}(R)$ when $R$ is a Dedekind domain -- which does just this. For (a tiny bit of) motivation: even in the case of a Dedekind domain we don't take the free abelian group on *all* the prime ideals: we omit $(0)$.
Now let $R$ be any Noetherian domain. One can define the **divisor class group** $\operatorname{Cl}(R)$ as follows: let $\operatorname{Div}(R)$ be the free abelian group generated by the **height one** prime ideals $\mathfrak{p}$ (these are ideals so that there is no prime ideal $\mathfrak{q}$ properly in between $(0)$ and $\mathfrak{p}$). One can also define, for each $f \in K^{\times}$, a **principal divisor** $\operatorname{div}(f)$. (I don't want to give the exact recipe in the general case: it involves lengths of modules. If $R$ happens to be integrally closed, then the localization $R\_{\mathfrak{p}}$ at a height one prime $\mathfrak{p}$ is a DVR, say with valuation $v\_{\mathfrak{p}}$, and then one takes $\operatorname{div}(f) = \sum\_{\mathfrak{p}} v\_{\mathfrak{p}}(f) [\mathfrak{p}]$.) Again the principal divisors $\operatorname{Prin}(R)$ form a subgroup of $\operatorname{Div}(R)$ and the quotient $\operatorname{Div}(R)/\operatorname{Prin}(R)$ is the divisor class group.
There is a canonical homomorphism $\operatorname{Pic}(R) \rightarrow \operatorname{Cl}(R)$, which is in general neither injective nor surjective. However, the map is an isomorphism in the case that $R$ is a **regular** ring.
These constructions are the affine versions of more familiar constructions in classical algebraic geometry: they are, respectively, Cartier divisors and Weil divisors, which agree on a nonsingular variety but not in general.
Finally, one can also define analogues of these groups for certain non-Noetherian domains (the Noetherianity is used to ensure that $v\_{\mathfrak{p}}(f) = 0$ except for finitely many primes $\mathfrak{p}$), e.g. **Krull domains** and **Prufer domains**. The latter is a domain in which each finitely generated nonzero ideal is invertible. Both are natural and interesting classes of rings.
For more details on this material, see e.g. the (rather rough and incomplete) notes
[http://alpha.math.uga.edu/~pete/classgroup.pdf](http://alpha.math.uga.edu/%7Epete/classgroup.pdf)
[**Addendum**: also see Section 11 of [factorization2010.pdf](http://alpha.math.uga.edu/%7Epete/factorization2010.pdf).]
For *much* more detail see the references cited therein, especially Larsen and McCarthy's *Multiplicative Ideal Theory*.
| 17 | https://mathoverflow.net/users/1149 | 12264 | 8,308 |
https://mathoverflow.net/questions/12266 | 13 | Frobenius Theorem says that a subbundle $E$ of the tangent bundle $TM$ of a manifold $M$ is tangent to a foliation if and only if for any two vector fields $X, Y \subset E$ the bracket $[X,Y]\subset E$. Bracket is a second order operator, hence subbundle $E$ needs to be $C^2$.
Are there any generalizations for subbundle which is $C^1$, $C^{1+smth}$?
Thank you,
Z.
| https://mathoverflow.net/users/3375 | Frobenius Theorem for subbundle of low regularity? | Let me conisder the case when the distribution of planes is of codimension 1 and explain why in this case it is enough to have $C^1$ smoothness in order to ensure the existence of the folitation.
In the case when the distribution is of codimension 1, you can formulate Frobenius Theorem in terms of 1-forms. Namely you can define a non-zero 1-form $A$, whose kernel is the distribution. The smoothness of this 1-form will be the same as the smoothness of the distribution. Now, you can say that the distribution is integrable if $A\wedge dA=0$. This quantity is well defined is A is $C^1$. Let me give a sketch of the proof that $A\wedge dA=0$ garanties existence of the foliation is A is $C^1$.
The proof is by induction
1) Consider the case $n=2$. In this case it is a standard fact of ODE, that for a $C^1$ smooth distribution of directions on the plane the integral lines are uniquelly defined.
2) Conisder the case $n=3$. We will show that the foliation exists locally near any point, say the origin $O$ of $R^3$. The 1-form A, that defines the distribution is non vanishing on one of the coordinate planes, say $(x,y)$ plane in the neighborhood of $O$. Take a $C^1$
smooth vector field in the neigborhood of $O$ that is transversal to planes $z=const$
and satisfies $A(v)=0$. Take the flow correponding to this vector field. The flow is $C^1$ smooth and moreover it preserves the distribution of planes $A=0$. Indeed, dA vanishes on the planes A=0 (by the condition of integrability), and we can apply the formula for Lie derivative $L\_v(A)=d(i\_v(A))+i\_v(dA)=i\_v(dA)$. Finally, we take the integral curve of the restriction of $A=0$ to the plane $(x,y)$ and for evey curve conisder the surface it covers unders the flow of $v$. This gives the foliation.
This reasoning can be repeated by induction.
A good refference is Arnold, Geometric methods of ordinary differential equations. I don't know if this book was transalted to English
| 14 | https://mathoverflow.net/users/943 | 12269 | 8,312 |
https://mathoverflow.net/questions/12236 | 42 | Let $X$ be a regular scheme (all local rings are regular). Let $Y,Z$ be two closed subschemes defined by ideals sheaves $\mathcal I,\mathcal J$. Serre gave a beautiful formula to count the intersection multiplicity of $Y,Z$ at a generic point $x$ of $Y\cap Z$ as:
$$\sum\_{i\geq 0} (-1)^i\text{length}\_{\mathcal O\_{X,x}} \text{Tor}\_i^{\mathcal O\_{X,x}}(\mathcal O\_{X,x}/\mathcal I\_x, \mathcal O\_{X,x}/\mathcal J\_x)$$
It takes quite a bit of work to show that this is the right definition (even that the sum terminates is a non-trivial theorem of homological algebra): it is non-negative, vanishes if the dimensions don't add up correctly, positivity etc. In fact, some cases are still open as far as I know. See [here](http://en.wikipedia.org/wiki/Serre%27s_multiplicity_conjectures) for some reference.
I have heard one of the great things about Lurie's thesis is setting a framework for derived algebraic geometry. In fact, in the introduction he used Serre formula as a motivation (it is pretty clear from the formula that a "derived" setting seems natural). However, I could not find much about it aside from the intro, and Serre formula was an old flame of mine in grad school. So my (somewhat vague):
**Question**: Does any of the desired properties of Serre formula follow naturally from Lurie's work? If so (since things are rarely totally free in math), where did we actually pay the price (in terms of technical work to establish the foundations)? EDIT: Clark's answer below greatly clarifies and gives more historical context to my question, highly recommend!)
| https://mathoverflow.net/users/2083 | Serre intersection formula and derived algebraic geometry? | There are a number of comments to make about Serre's intersection formula and its relation to derived algebraic geometry.
First, we should be a little more cautious about attribution. The idea of using "derived rings" to give an intrinsic version of the Serre intersection formula is not recent. The idea goes back at least to thoughts of Deligne, Kontsevich, Drinfeld, and Beilinson in the 1980s (and possibly earlier). These ideas have been made precise in a number of ways, in particular in work of Kapranov & Ciocan-Fontaine, and Toën & Vezzosi. *EDIT*: As Ben-Zvi reminded me below, one should also mention Behrend and Behrend-Fantechi on DG schemes and virtual fundamental classes. Of course Lurie's work has been the most comprehensive and powerful in its treatment of the foundations of DAG, but it's important to understand that his work arose in the *context* of these fascinating ideas.
Now, just to provide a little context, let me try to recall how Serre's formula arises from DAG considerations. Let's start by using the notation above, but let's assume for simplicity that $X$, $Y$, and $Z$ are all local schemes. (Some of the technicalities of DAG arise in making sheaf theory work with some sort of "derived rings," so our discussion will be easier if we ignore that for now.) So we write $X=\mathrm{Spec}(A)$, $Y=\mathrm{Spec}(B)$, and $Z=\mathrm{Spec}(C)$ for local rings $A$, $B$, and $C$.
Now if our aim is to intersect $Y$ and $Z$ in $X$, we know how to do that algebro-geometrically. We form the fiber product $Y\times\_XZ=\mathrm{Spec}(B\otimes\_AC)$. The tensor product that appears here is really the thing we're going to alter. To do that, we're going to regard $B$ and $C$ as (discrete) *simplicial (commutative) $A$-algebras*, and we're going to form the derived tensor product. This produces a new simplicial commutative ring $B\otimes^{\mathbf{L}}\_AC$ whose homotopy groups are exactly the groups $\mathrm{Tor}^A\_i(B,C)$. The intersection multiplicity is simply the length of $B\otimes^{\mathbf{L}}\_AC$ as a simplicial $A$-module.
As Ben Webster says, the real joy of DAG is in thinking of the geometry of our new derived ring $B\otimes^{\mathbf{L}}\_AC$ as a single unit instead of thinking only of its disembodied homotopy groups. The question you're asking seems to be: does thinking geometrically about this gadget help us to prove Serre's multiplicity conjectures in a more conceptual manner?
The short answer is: I don't know. I do not think a new proof of any of these has been announced using DAG (and it's definitely not in any of Lurie's papers), and in any case I do not think DAG has the potential to make the conjectures "easy." But let me see if I can make a case for the following idea: revisiting Serre's original method of reduction to the diagonal in the context of DAG.
Recall that, if $k$ is a field, if $A$ is a $k$-algebra, and if $M$ and $N$ are $A$-modules, then $$M\otimes\_AN=A\otimes\_{A\otimes\_kA}(M\otimes\_kN).$$
Hence to understand $\mathrm{Tor}^A\_{\ast}(M,N)$, it suffices to understand $\mathrm{Tor}^{A\otimes\_kA}\_{\ast}(A,-)$. This allowed Serre to reduce to the case of the diagonal in $\mathrm{Spec}(A\otimes\_kA)$. The key point here is that everything is flat over $k$, so Serre could only use this to prove the multiplicity conjectures for $A$ essentially of finite type over a field. Observe that the same equality holds if we work in the derived setting: if $M$ and $N$ are simplicial $A$-modules, and $A$ is an $R$-algebra, then the derived tensor product of $M$ and $N$ over $A$ can be computed as
$$A\otimes^{\mathbf{L}}\_{A\otimes^{\mathbf{L}}\_RA}(M\otimes^{\mathbf{L}}\_RN).$$
The gadget on the right (or, strictly speaking, its homotopy) has a name familiar to toplogists; it's the *Hochschild homology* $\mathrm{HH}^R(A,M\otimes^{\mathbf{L}}\_RN)$.
The hope is that we've chosen $R$ cleverly enough that $B\otimes^{\mathbf{L}}\_RC$ is "less complicated" than $B\otimes^{\mathbf{L}}\_AC$. (More precisely, we want the $\mathrm{Tor}$-amplitude of $M$ and $N$ to decrease when we think of them as $R$-modules. There's a particular way of building $R$, but let me skip over this point.)
Has our situation improved? Perhaps only a little: we've turned our problem of looking at the derived intersection $Y\times^h\_XZ$ into the study of the derived intersection of the diagonal inside $X\times^h\_RX$ with some simpler derived subscheme $Y\times^h\_RZ$ thereof. But now we can try to iterate this, working inductively.
I don't know whether this can be made to work, of course.
| 44 | https://mathoverflow.net/users/3049 | 12277 | 8,318 |
https://mathoverflow.net/questions/12169 | 60 | A (discrete) group is **amenable** if it admits a finitely additive probability measure (on *all* its subsets), invariant under left translation. It is a basic fact that every abelian group is amenable. But the proof I know is surprisingly convoluted. I'd like to know if there's a more direct proof.
The proof I know runs as follows.
1. Every finite group is amenable (in a unique way). This is trivial.
2. $\mathbb{Z}$ is amenable. This is not trivial as far as I know; the proof I know involves choosing a non-principal ultrafilter on $\mathbb{N}$. This means that $\mathbb{Z}$ is amenable in many different ways, i.e. there are many measures on it, but apparently you can't write down any measure 'explicitly' (without using the Axiom of Choice).
3. The direct product of two amenable groups is amenable. This isn't exactly trivial, but the measure on the product is at least constructed canonically from the two given measures.
4. Every finitely generated abelian group is amenable. This follows from 1--3 and the classification theorem.
5. The class of amenable groups is closed under direct limits (=colimits over a directed poset). This is like step 2: it seems that there's no *canonical* way of constructing a measure on the direct limit, given measures on each of the groups that you start with; and the proof involves choosing a non-principal ultrafilter on the poset.
6. Every abelian group is amenable. This follows from 4 and 5, since every abelian group is the direct limit of its finitely generated subgroups.
Is there a more direct proof? Is there even a one-step proof?
---
**Update** Yemon Choi suggests an immediate simplification: replace 1 and 4 by
1'. Every quotient of an amenable group is amenable. This is simple: just push the measure forward.
4'. Every f.g. abelian group is amenable, by 1', 2 and 3.
This avoids using the classification theorem for f.g. abelian groups.
Tom Church mentions the possibility of skipping steps 1--3 and going straight to 4. If I understand correctly, this doesn't use the classification theorem either. The argument is similar to the one for $\mathbb{Z}$: one still has to choose an ultrafilter on $\mathbb{N}$. (One also constructs a Følner sequence on the group, a part of the argument which I didn't mention previously but was there all along).
Yemon, Tom and Mariano Suárez-Alvarez all suggest using one or other alternative formulations of amenability. I'm definitely interested in answers like that, but it also reminds me of the old joke:
>
> Tourist: Excuse me, how do I get to Edinburgh Castle from here?
>
>
> Local: I wouldn't start from here if I were you.
>
>
>
In other words, if a proof of the amenability of abelian groups uses a different definition of amenability than the one I gave, then I want to take the proof of equivalence into account when assessing the simplicity of the overall proof.
Jim Borger points out that if, as seems to be the case, even the proof that $\mathbb{Z}$ is amenable makes essential use of the Axiom of Choice, then life is bound to be hard. I take his point. However, one simplification to the 6-step proof that I'd like to see is a merging of steps 2 and 5. These are the two really substantial steps, but they're intriguingly similar. None of the answers so far seem to make this economy. That is, every proof suggested seems to involve two separate Følner-type arguments.
| https://mathoverflow.net/users/586 | Why are abelian groups amenable? | Here is a simpler argument, combining 1--6 into one step.
Let $G$ be a countable abelian group generated by $x\_1,x\_2,\ldots$. Then a Følner sequence is given by taking $S\_n$ to be the pyramid consisting of elements which can be written as
$a\_1x\_2+a\_2x\_2+\cdots+a\_nx\_n$ with $\lvert a\_1\rvert\leq n,\lvert a\_2\rvert\leq n-1,\ldots,\lvert a\_n\rvert\leq 1$.
The invariant probability measure is then defined by $\mu(A)=\underset{\omega}{\lim}\lvert A\cap S\_n\rvert / \lvert S\_n\rvert$ as usual.
A more natural way to phrase this argument is:
1. The countable group $\mathbb{Z}^\infty$ is amenable.
2. All countable abelian groups are amenable, because amenability descends to quotients.
But I would like to emphasize that there is really only one step here, because the proof for $\mathbb{Z}^\infty$ automatically applies to any countable abelian group. This two-step approach is easier to remember, though. (The ideas here are the same as in my other answer, but I think this formulation is much cleaner.)
---
2016 Edit: Here is an argument to see that $S\_n$ is a Følner sequence. It is quite pleasant to think about precisely where commutativity comes into play.
Fix $g\in G$ and any finite subset $S\subset G$. We first analyze the size of the symmetric difference $gS\bigtriangleup S$. Consider the equivalence relation on $S$ generated by the relation $x\sim y$ if $y=x+g$ (which is itself neither symmetric, reflexive, or transitive). We will call an equivalence class under this relation a "$g$-string". Every $g$-string consists of elements $x\_1,\ldots,x\_k\in S$ with $x\_{j+1}=x\_j+g$.
The **first key observation** is that $\lvert gS\bigtriangleup S\rvert$ is **at most twice the number of $g$-strings**. Indeed, if $z\in S$ belongs to $gS\bigtriangleup S$, then $z$ must be the "leftmost endpoint" of a $g$-string; if $z\notin S$ belongs to $gS\bigtriangleup S$, then $z-g$ must be the "rightmost endpoint" of a $g$-string; and each $g$-string has at most 2 such endpoints (it could have 1 if the endpoints coincide, or 0 if $g$ has finite order).
Our goal is to prove for all $g\in G$ that $\frac{\lvert gS\_n \bigtriangleup S\_n\rvert}{\lvert S\_n\rvert}\to 0$ as $n\to \infty$. Since $\lvert abS\bigtriangleup S\rvert\leq\lvert abS\bigtriangleup bS\rvert+\lvert bS\bigtriangleup S\rvert= \lvert aS\bigtriangleup S\rvert+\lvert bS\bigtriangleup S\rvert$, it suffices to prove this for all $g\_i$ in a generating set.
By the observation above, to prove that $\frac{\lvert g\_iS\_n \bigtriangleup S\_n\rvert}{\lvert S\_n\rvert}\to 0$, it suffices to prove that $\frac{\text{# of $g\_i$-strings in $S\_n$}}{\lvert S\_n\rvert}\to 0$. Equivalently, we must prove that the reciprocal $\frac{\lvert S\_n\rvert}{\#\text{ of $g\_i$-strings in $S\_n$}}$ diverges, or in other words that **the average size of a $g\_i$-string in $S\_n$ diverges**.
We now use the specific form of our sets $S\_n=\{a\_1g\_1+\cdots+a\_ng\_n\,|\, \lvert a\_i\rvert\leq n-i\}$. For any $i$ and any $n$, set $k=n-i$ (so that $\lvert a\_i\rvert\leq k$ in $S\_n$). The **second key observation** is that **every $g\_i$-string in $S\_n$ has cardinality at least $2k+1$** unless $g\_i$ has finite order. Indeed given $x\in S\_n$, write it as $x=a\_1g\_1+\cdots+a\_ig\_i+\cdots+a\_ng\_n$; then the elements $a\_1g\_1+\cdots+bg\_i+\cdots+a\_ng\_n\in S\_n$ for $b=-k,\ldots,-1,0,1,\ldots,k$ belong to a single $g\_i$-string containing $x$. If $g\_i$ does not have finite order, these $2k+1$ elements must be *distinct*. This shows that the minimum size of a $g\_i$-string in $S\_n$ is $2n-2i+1$, so for fixed $g\_i$ the average size diverges as $n\to \infty$.
When $g\_i$ has finite order $N$ this argument does not work (a $g\_i$-string has *maximum* size $N$, so the average size cannot diverge). However once $N<2k+1$, the subset containing the $2k+1$ elements above is closed under multiplication by $g\_i$. In other words, once $n\geq i+N/2$ the set $S\_n$ is $g\_i$-invariant, so $\lvert g\_iS\_n\bigtriangleup S\_n\rvert=0$.
I'm grateful to David Ullrich for pointing out that this claim is not obvious, since the quotient of a Følner sequence need not be a Følner sequence (Yves Cornulier gives an example [here](https://math.stackexchange.com/questions/1283824/folner-sets-in-a-quotient-of-a-f-g-amenable-group)).
| 28 | https://mathoverflow.net/users/250 | 12281 | 8,321 |
https://mathoverflow.net/questions/12279 | 20 | In modern valuation theory, one studies not just absolute values on a field, but also **Krull valuations**. The motivation is easy enough:
If $k$ is a field, a **valuation ring** of $k$ is a subring $R$ such that for every $x \in k^{\times}$, at least one of $x, x^{-1}$ is an element of $R$. (It follows of course that $k$ is the fraction field of $R$.) If $| \ |$ is a non-Archimedean norm on a field, then the set $\{x \in k \ | \ |x| \leq 1 \}$ is a valuation ring. However, the converse does not hold, since if $R$ is a valuation ring, then $k^{\times}/R^{\times}$ need not inject into $\mathbb{R}$: rather it is (under a straightforward extension of the divisibility relation on $R$) a totally ordered abelian group. Moreover, a certain formal power series construction shows that for any totally ordered abelian group $\Gamma$, there exists $k$ and $R$ with $k^{\times}/R^{\times} \cong \Gamma$.
My question is this: what are some instances where having the generality of Krull valuations is useful for solving some problem (which is not *a priori* concerned with valuation theory)? How do Krull valuations arise in algebraic geometry?
I can almost remember one example of this. I believe it is possible to give a quick proof of the Lang-Nishimura Theorem -- that having a smooth $k$-rational point is a birational invariant among complete [hmm, valuative criterion!] $k$-varieties. I think I saw this in some of Bjorn Poonen's lecture notes, but I forget where. [Last year at this time, I would have emailed Bjorn. I am trying out this new approach on the theory that Bjorn can reply if he wishes, and if not someone else will surely be eager to tell me the answer.]
Are there other nice examples? Maybe something to do with resolution of singularities?
| https://mathoverflow.net/users/1149 | What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? | Since you asked for it, here is a little bit about the role of valuations in the Lang-Nishimura theorem, one version of which is as follows (my version implies yours):
**Theorem (Lang-Nishimura):** Let $X \to \to Y$ be a rational map between $k$-varieties, where $X$ is integral and $Y$ is proper. If $X$ has a smooth $k$-point $x$, then $Y$ has a $k$-point.
**Sketch of proof:** Let $K$ be the function field of $X$. The rational map gives a $K$-point on $Y$. If $\dim X=1$, then $\mathcal{O}\_{X,x}$ is a valuation ring, so the valuative criterion for properness gives an $\mathcal{O}\_{X,x}$-point of $Y$, which reduces to a $k$-point of $Y$. For $\dim X=n>1$, modify the argument by *embedding* $K$ into a valued field with value group $\mathbb{Z}^n$ and residue field $k$, namely the iterated Laurent series field $F:=k((t\_1))((t\_2))\cdots((t\_n))$, where the $t\_i$ are local parameters at $x$. $\square$
1) If one prefers, one can replace this use of the valuative criterion for a rank $n$ discrete valuation with $n$ uses of the valuative criterion for rank $1$ discrete valuations: prove the lemma that if a proper variety has an $L((t))$-point, then it has an $L$-point, and apply the lemma $n$ times. So for this particular application, you don't really need the fancy valuations.
2) Geometrically, the Lang-Nishimura theorem can be understood as follows: Find a smooth irreducible curve $C$ in $X$ through $x$ such that $C$ is not entirely contained in the locus of indeterminacy of $\phi \colon X \to\to Y$. Then $\phi|\_C$ extends to a morphism, and the image of $x$ is a $k$-point of $Y$. (The existence of $C$ is not completely obvious, though, so the valuation-theoretic proof is cleaner.)
| 18 | https://mathoverflow.net/users/2757 | 12287 | 8,326 |
https://mathoverflow.net/questions/12284 | 28 | In dimensions 1 and 2 there is only one, respectively 2, compact Kaehler manifolds with zero first Chern class, up to diffeomorphism. However, it is an open problem whether or not the number of topological types of such manifolds of dimension 3 (Calabi-Yau threefolds) is bounded. I would like to ask what is known in this direction. In particular, is it known that the Euler characteristic or the total Betti number of Calabi-Yau threefolds can't be arbitrarily large? Are there any mathematical (or physical?) reasons to expect either answer?
As a side question: I remember having heard several times something like "Calabi-Yau 3-folds parametrize (some kind of) vacua in string theory" but was never able to make precise sense of this. So any comments on this point or references accessible to mathematicians would be very welcome.
| https://mathoverflow.net/users/2349 | Topologically distinct Calabi-Yau threefolds | This is a very good question, and I would really love to know the answer since its current state seems to be quite obscure. Below is just a collection of remarks, surely not the full answer by any means. I would like to argue that for the moment there is no any deep mathematical reason to think that the Euler number of CY 3-folds is bounded. I don't believe either that there is any physical intuition on this matter. But there is some empirical information, and I will describe it now, starting by speaking about "how many topological types of CY manifolds we know for the moment".
As far as I understand for today the construction of Calabi-Yau 3-folds, that brought by far the largest amount of examples is the construction of Batyrev. He starts with a reflexive polytope in dimension 4, takes the corresponding toric 4-fold, takes a generic anti-canonical section and obtains this way a Calabi-Yau orbifold. There is always a crepant resolution. So you get a smooth Calabi-Yau. Reflexive polytopes in dimension 4 are classified the number is 473,800,776. I guess, this number let Miles Reid to say in his article "Updates on 3-folds" in 2002 <http://arxiv.org/PS_cache/math/pdf/0206/0206157v3.pdf> , page 519
"This gives some 500,000,000 families of CY 3-folds, so much more impressive than a mere infinity (see the website <http://tph16.tuwien.ac.at/~kreuzer/CY/>). There are certainly many more; I believe there are infinitely many families, but the contrary opinion is widespread"
A problem with the number 500,000,000 in this phrase is that it seems more related to the number of CY orbifolds, rather than to the number of CY manifolds obtained by resolving them. Namely, the singularities that appear in these CY orbifolds can be quite involved and they have a lot of resolutions (I guess at least thousands sometimes), so the meaning of 500,000,000 is not very clear here.
This summer I asked Maximillian Kreuzer (one of the persons who actually got this number 473,800,776 of polytopes), a question similar to what you ask here. And he said that he can guarantee that there exist at least 30108 topological types of CY 3-folds. Why? Because for all these examples you can calculate Hodge numbers $h^{1,1}$ and $h^{2,1}$, and you get 30108 different values. Much less that 473,800,776. As for more refined topological invariants (like multiplication in cohomology) according to him, this was not really studied, so unfortunately 30108 seems to be the maximal number guarantied for today. But I would really love to know that I am making a mistake here, and there is some other information.
Now, it seems to me that the reason, that some people say, that the Euler characteristics of CY 3-folds could be bounded is purely empirical. Namely, the search for CY 3-folds is going for 20 years already. Since then a lot of new families were found. We know that mirror symmetry started with this symmetric table of numbers "($h^{1,1}, h^{2,1})$", and the curious fact is that, according to Maximillian, what happened to this table in 20 years -- it has not got any wider in 20 years, it just got denser. The famous picture can be found on page 9 of the following notes of Dominic Joyce <http://people.maths.ox.ac.uk/~joyce/SympGeom2009/SGlect13+14.pdf> . So, this means that we do find new families of CY manifolds, all the time. But the values of their Hodge numbers for some reason stay in the same region. Of course this could easily mean that we are just lacking a good construction.
Final remark is that in the first version of this question it was proposed to consider complex analytic manifolds with $c\_1=0$. If we don't impose condition of been Kahler, then already in complex dimension 2 there is infinite number of topological types, given by Kodaira surfaces, they are elliptic bundles over elliptic curve. In complex dimension 3 Tian have shown that for every $n>1$ there is a holomorphic structure on the connected sum of n copies of $S^3\times S^3$, with a non-vanishing holomorphic form. Surely these manifolds are non-Kahler. So if you want to speak about any finiteness, you need to discuss say, Kahler 3-folds with non-vanishing holomorphic volume form, but not all complex analytic ones.
| 34 | https://mathoverflow.net/users/943 | 12294 | 8,332 |
https://mathoverflow.net/questions/12303 | 17 | Why is one interested in the mod p reduction of modular curves and Shimura varieties?
From an article I learned that this can be used to prove the Eichler-Shimura relation which in turn proves the Hasse-Weil conjecture for modular curves. Are there similar applications for Shimura varieties?
| https://mathoverflow.net/users/nan | Why is one interested in the mod p reduction of modular curves and Shimura varieties? | The Eichler-Shimura relation doesn't just prove the Hasse-Weil conjecture for modular curves. It e.g. attaches Galois representations to modular forms of weight 2. More delicate arguments (using etale cohomology with non-constant coefficients, machinery that wasn't available to Shimura) attaches Galois representations to higher weight modular forms. These ideas have had many applications (e.g. eventually they proved FLT). In summary: computing the mod p reduction of modular curves isn't *just* for Hasse-Weil.
Doing the same for Shimura varieties is technically much harder because one runs into problems both geometric and automorphic. But the upshot, in some sense, is the same: if one can resolve these issues (which one can for, say, many unitary Shimura varieties nowadays, but by no means all Shimura varieties) then one can hope to attach Galois representations to automorphic forms on other reductive groups, and also to compute the L-function of the Shimura variety in terms of automorphic forms.
Why would one want to do these things? Let me start with attaching Galois reps to auto forms. These sorts of ideas are what have recently been used to prove the Sato-Tate conjecture. Enough was known about the L-functions attached to automorphic forms on unitary groups to resolve the analytic issues, and so the main issue was to check that the symmetric powers of the Galois representations attached to an elliptic curve were all showing up in the cohomology of Shimura varieties. Analysing the reduction mod p of these varieties was just one of the many things that needed doing in order to show this (although it was by no means the hardest step: the main technical issues were I guess in the "proving R=T theorems", similar to the final step in the FLT proof being an R=T theorem; the L-function ideas came earlier).
But to answer your original question, yes: if you're in the situation where you understand the cohomology of the Shimura variety well enough, then analysing the reduction of the variety will tell you non-trivial facts about the L-function of the Shimura variety. Note however that the link isn't completely formal. Mod p reduction of the varieties only gives you an "Eichler-Shimura relation", and hence a polynomial which will annihiliate the Frobenius element acting on the etale cohomology. To understand the L-function you need to know the full characteristic polynomial of this Frobenius element. For GL\_2 one is lucky in that the E-S poly is the char poly, simply because there's not enough room for it to be any other way. This sort of argument breaks down in higher dimensions. As far as I know these questions are still very open for most Shimura varieties.
So in summary, for general Shimura varieties, you can still hope for an Eichler-Shimura relation, but you might not actually be able to compute the L-function in terms of automorphic forms as a consequence.
| 32 | https://mathoverflow.net/users/1384 | 12304 | 8,337 |
https://mathoverflow.net/questions/12226 | 4 | Let $\textbf{HoTop}^\*$ be the homotopy category of pointed topological spaces. In the following, the word "isomorphism" shall always mean isomorphism in $\textbf{HoTop}^\*$, i.e. pointed homotopy equivalence. All constructions like cone or suspensions are pointed/reduced.
A triangle $X\to Y\to Z\to \Sigma X$ is called *distinguished* if it is isomorphic in $\textbf{HoTop}^\*$ to a triangle of the form $X\stackrel{f}{\to} Y\hookrightarrow\text{C}f\to\Sigma X$, where $\text{C}f\to\Sigma X$ is the map collapsing $Y$ to a point.
**Problem:**
Let $\ \ \matrix{X & \to & Y & \to & Z & \to & \Sigma X\cr\downarrow\alpha &&\downarrow\beta&&\downarrow\gamma &&\downarrow&\Sigma\alpha\cr X^{\prime} & \to & Y^{\prime} & \to & Z^{\prime} & \to & \Sigma X^{\prime}}\ \ $ be a morphism of distinguished triangles such that $\alpha$ and $\beta$ are isomorphisms. Is it true that $\gamma$ is an isomorphism, too?
**Suggestions:**
For a morphism of triangles as above (where $\alpha$ and $\beta$ are not necessarily isomorphisms), the morphism $\gamma^\*: [Z^{\prime},-]\to [Z,-]$ is equivariant with respect to $[\Sigma\alpha]^\*: [\Sigma X^{\prime},-]\to [\Sigma X,-]$. (**edit: this is wrong -- see below**) Therefore, I thought one could apply theorem 6.5.3 in Hoveys book on Model Categories. Unfortunately, there seems to be a gap at the end of the proof, as already pointed out [here](https://mathoverflow.net/questions/11977/equivariant-map-preserves-stabilizer).
Therefore, I have the following
**Questions:**
**(1)** Am I misunderstanding something in Hovey's proof of 6.5.3(b), or is there really a gap in it? If it is a gap: Do you have any suggestions on how to fix the proof?
**(2)** If the proof can't be fixed in this generality: Do you have suggestions on how to prove the statement above only for $\textbf{HoTop}^\*$?
**Edit:**
**(1)** The usual proof of this fact for triangulated categories does not work here, because there one uses the fact that $[X,-]$ is abelian-group valued for any $X$ and uses the classical five lemma together with Yoneda to conclude that $\gamma$ is an isomorphism. This doesn't seem to work here.
**(2)** Since partial morphisms of distinguished triangles in $\textbf{HoTop}^\*$ can always be completed to morphisms of triangles, we can reduce to the case where $\alpha$ and $\beta$ both equal the identity. Therefore, we have a commutative diagram (in $\textbf{HoTop}^\*$, i.e. a homotopy commutative diagram in $\textbf{Top}^\*$)
$\matrix{X & \to & Y & \to & Z & \to & \Sigma X\cr\downarrow & \text{id}\_X &\downarrow & \text{id}\_Y&\downarrow&\gamma&\downarrow&\text{id}\_{\Sigma X}\cr X & \to & Y & \to & Z& \to & \Sigma X}$
and we have to prove that $\gamma$ is a homotopy equivalence.
**Hovey's proof**
The way Hovey proceeds in his proof is as follows: We know the following things:
*(1)* $\gamma^\*: [Z,-]\to [Z,-]$ is $[\Sigma X,-]$-equivariant
*(2)* Two maps $c,d\in[Z,W]$ are equal in $[Y,W]$ if and only if they lie in the same $[\Sigma X,W]$-orbit.
From (2) and the commutativity of the middle square it follows that for any $h\in [Z,W]$ there is some $\rho\in[\Sigma X,W]$ such that $\gamma^\*(h)=h.\rho$; in other words $\gamma^\*$ doesn't change the $[\Sigma X,-]$-orbit.
Now, suppose there are $g,h\in [Z,W]$ such that $\gamma^\*(h)=\gamma^\*(g)$. Then, again by the commutativity of the middle square, there is some $\alpha\in [\Sigma X,W]$ such that $g = h.\alpha$. Thus, by (1), $\gamma^\*(g) = \gamma^\*(h).\alpha = \gamma^\*(g).\alpha$, and so $\alpha\in\text{Stab}(\gamma^\*(g))$.
*The point is* that Hovey now wants to show that $\text{Stab}(\gamma^\*(g))=\text{Stab}(g)$; this would imply $\alpha\in\text{Stab}(g)$, and thus $h = g.\alpha^{-1} = g$ as required. The inclusion $\text{Stab}(\gamma^\*(g))\supset\text{Stab}(g)$ is obvious. For the other inclusion, I have no idea how to prove it.
Do you see how one can fix the proof?
**FINAL EDIT**
I made a mistake in proving that for any morphism of triangles $(\alpha,\beta,\gamma)$ the morphism $\gamma^\*$ is equivariant with respect to $(\Sigma\alpha)^\*$. This is wrong.
So *what remains* is the question on how to fix the proof of theorem 6.5.3 in Hovey's book. Any suggestions?
Thank you.
| https://mathoverflow.net/users/3108 | Five lemma in HoTop* and arbitrary pointed model categories | This is false for spaces.
Let $X = S^0, Y = S^1$, and $f:X \to Y$ be the trivial map. Then $Z = Cf$ is $S^1 \vee S^1$. Then $[X,Y]$ is trivial, so then the the truth of this statement would imply: If you have a map $g: Z \to Z$ which is the identity on the first circle, and such that the induced map $S^1 \to S^1$ after collapsing the first circle is homotopic to the identity, then $g$ is a homotopy equivalence.
The map $g$ is a based map from a wedge of two circles to itself, which has fundamental group $F$, the free group on two generators, with generators $x, y$ corresponding to the two circle factors. A self-map is determined up to homotopy by a pair of elements in $x', y' \in F$. The condition that the first circle is mapped by the identity says $x = x'$, and the condition that the induced map on quotients is homotopic to the identity says that the image of $y'$ in $F/\langle x \rangle$ is the same as the image of $y$.
So: Suppose you have a pair of elements $x$ and $y'$ in $F = \langle x,y \rangle$ such that $y \equiv y'$ in $F/\langle x \rangle$. Then do $x$ and $y'$ freely generate $F$?
And the answer is no. For example, if $y' = (y x)^3 y^{-2}$, then $y' \equiv y$ after taking the quotient, but the pair $x, (y x)^3 y^{-2}$ don't generate the free group. This is believeable, but rather than being vague let's give a proof for completeness.
There is a group homomorphism $F \to S\_3$ sending $x$ to the 2-cycle $(1 2)$ and $y$ to $(1 3)$. This homomorphism is surjective because these generate the group, but $y'$ maps to the trivial element because the image of $yx$ has order 3 and the image of $y$ has order 2. Therefore, $x$ and $y'$ don't generate $S\_3$, and so they can't generate $F$.
| 5 | https://mathoverflow.net/users/360 | 12308 | 8,340 |
https://mathoverflow.net/questions/12078 | 5 | I couldn't think of a title for this, but here we go:
Fix $p:S\rightarrow T$, a left fibration of simplicial sets, and an edge $f:\Delta^1 \rightarrow T$. Let $t$ be the first vertex of $f$, and $t'$ be the second vertex.
We name the induced map, $q: S^{\Delta^1}\rightarrow S^{\{1 \}}\times\_{T^{\{1 \}}} T^{\Delta^1}$.
Now let $X$ be the simplicial set of ***sections of the projection*** $S\times\_T \Delta^1\rightarrow \Delta^1$, where the pullback is taken with respect to the map $p:S\rightarrow T$ and the fixed edge $f:\Delta^1 \rightarrow T$.
More notation: We'll denote by $S\_{t'}$ the fiber $S \times\_T \Delta^0$ where $\Delta^0 \rightarrow T$ is given by the inclusion of the vertex $t'$ and $S\rightarrow T$ is given again by p. We give a fiber $q':X \rightarrow S\_{t'}$ of $q$ over the edge $f$ (This is about where I stop understanding what's going on).
What we'd like to show is: $q$ and $q'$ have the same fibers over points of $S^{\{1 \}}\times\_{T^{\{1 \}}} T^{\Delta^1}$ where the second projection is the edge f. Remember that exponentiation denotes the internal Hom.
The problem is this simplicial set of sections. What are its maps out, and why do they naturally go to $S\_{t'}$ and agree with q? I feel like the key to this is understanding how the exponential is mapping into the pullback, but it's not really clear to me how that should work.
This fact is stated in HTT by Lurie in the proof of proposition 2.1.3.1, but I don't really see how it's obvious.
A link to the relevant proof/page: <http://books.google.com/books?id=CTe68E8wK4QC&lpg=PP1&ots=o8qYDiX4mt&dq=lurie%20higher%20topos%20theory&pg=PA67#v=onepage&q=&f=false>
Update:
"Work" I've done thusfar:
$$\ \ \matrix{&S^{\Delta^1}\_f &\to & S^{\Delta^1}&
\cr &\downarrow &Pb &\downarrow
\cr S\_{t'}\cong &L\_f & \to & L & \to & S^{\{1\}} &
\cr &\downarrow &Pb&\downarrow&Pb&\downarrow p
\cr &\Delta^{0} & \to & T^{\Delta^1} & \to & T^{\{1\}}
\cr &&f&&d\_1}\ \ $$
Note that $d\_1$ denotes the face map at the vertex 1. Also, $L:= S^{\{1 \}}\times\_{T^{\{1 \}}} T^{\Delta^1}$, and
The point here is that it should be "morally" the same to give a pullback of $\Delta^1\to T \leftarrow S$ as giving a pullback $\Delta^0\to T^{\Delta^1} \leftarrow S^{\Delta^1}$ with respect to the edge f. So we'd like to show that $X$, the simplicial set of sections of the projection noted before is somehow isomorphic to $S^{\Delta^1}\_f$, since this shows that $q$ and $q'$ agree where they're needed to. So what I'm struggling with at this point is showing this last idea.
| https://mathoverflow.net/users/1353 | A question about fibrations of simplicial sets and their fibers | I've got to agree with you; it's not the best-written proof in HTT. Let me go through it glacially slowly (for my own sake!) to see if I can write something that will help clarify the role of $X$.
Let me write $Map(U,V)$ instead of $V^U$. I find it easier to parse on the internet.
First, what is our $X$? It's the simplicial set of sections of $S\times\_T\Delta^1\to\Delta^1$, that is, it is the fiber product
$Map(\Delta^1,S\times\_T\Delta^1)\times\_{Map(\Delta^1,\Delta^1)}\Delta^0$, where $\Delta^0$ maps in by inculsion of $id$. So (since $Map(\Delta^1,-)$ is a right adjoint) we get:
$$X=Map(\Delta^1,S\times\_T\Delta^1)\times\_{Map(\Delta^1,\Delta^1)}\Delta^0=Map(\Delta^1,S)\times\_{Map(\Delta^1,T)}Map(\Delta^1,\Delta^1)\times\_{Map(\Delta^1,\Delta^1)}\Delta^0$$
$$=Map(\Delta^1,S)\times\_{Map(\Delta^1,T)}\Delta^0$$
where $\Delta^0$ is mapping in by inclusion $f$.
Now we've got our map
$$q:Map(\Delta^1,S)\to Map(\{1\},S)\times\_{Map(\{1\},T)}Map(\Delta^1,T),$$
whose fibers over $f$ we seek. That is, we just include
$$S\_{t'}=Map(\{1\},S)\times\_{Map(\{1\},T)}\Delta^0\to Map(\{1\},S)\times\_{Map(\{1\},T)}Map(\Delta^1,T)$$
(which is itself the pullback of $f:\Delta^0\to Map(\Delta^1,T)$ along the projection, of course) and we pull back to get $q'$.
So the result is a pullback of a pullback. (If I knew how, I'd draw the two pullback squares here.) The composite pullback is the pullback of $f:\Delta^0\to Map(\Delta^1,T)$ along $Map(\Delta^1,S)\to Map(\Delta^1,T)$. But this is what we called $X$. So our result is a map $q':X\to S\_{t'}$, and its fiber over any vertex of $S\_{t'}$ must coincide with the fiber over the corresponding vertex of $Map(\{1\},S)\times\_{Map(\{1\},T)}Map(\Delta^1,T)$, since that will be a pullback of a pullback as well.
Edit (Harry): I typed up the final version of the diagram. If the letters aren't explained, you can deduce what they are just by either looking at Clark's argument or just tracing the pullbacks. Every square is a pullback, so everything is very easy to deal with.
$$\ \ \matrix{
X&\cong&S^{\Delta^1}\_f &\to &Y^{\Delta^1}&\to& S^{\Delta^1}&
\cr &\searrow&\downarrow &Pb &\downarrow&Pb&\downarrow
\cr L\_f&\cong &S\_{t'} & \to &L'&\to& L & \to & S^{\{1\}} &
\cr &&\downarrow &Pb&\downarrow&Pb&\downarrow&Pb&\downarrow p
\cr &&\Delta^{0} & \to &(\Delta^1)^{\Delta^1} &\to& T^{\Delta^1} & \to & T^{\{1\}}
\cr &&&id&&(f)^{\Delta^1}&&d\_1}\ \ $$
| 5 | https://mathoverflow.net/users/3049 | 12311 | 8,342 |
https://mathoverflow.net/questions/12102 | 8 | For local field, the reciprocity map establishes almost an isomorphism from the multiplicative group to the Abelian Absolute Galois group. (In global case the relationship is almost as nice). It is tempted to think that there can be no such nice accident.
Do we know any explanation which suggest that there "should be" a relationship between the multiplicative group and the Galois group?
Actually, my current belief is that the reciprocity map is half accidental. I think that there is a natural extension where we can define a natural action of the multiplicative group. In the local case this is the Lubin-Tate extension (a generalization of cyclotomic extension). The fact that this Lubin-Tate extension is the Abelian Absolute Galois group is an accident.
Do we know something that might support/ reject this view?
Please feel free to edit the question into a form that you think might be better.
| https://mathoverflow.net/users/2701 | How natural is the reciprocity map? | The reciprocity map is completely natural (in the technical sense of category theory). For example, if $K$ and $L$ are two local fields,
and $\sigma:K \rightarrow L$ is an isomorphism, then $\sigma$ induces an isomorphism of
multiplicative groups $K^{\times} \rightarrow L^{\times}$ and also of abelian absolute Galois groups $G\\_K^{ab} \rightarrow G\\_L^{ab}$. The reciprocity laws for $K$ and $L$ are then compatible with these two isomorphisms induced by $\sigma$.
On the other hand the factorization $K^{\times} = {\mathbb Z} \times {\mathcal O}\\_K^{\times}$ is not canonical (it depends on a choice of uniformizer), and the identification of
${\mathcal O}\\_K^{\times}$ with the Galois group of a maximal totally ramified abelian extension of $K$ also depends on a choice of uniformizer (which goes into the construction of the Lubin--Tate formal group, and hence into the construction of the totally ramified extension; different choices of uniformizer will give different formal groups, and different
extensions).
As others pointed out, the local reciprocity map is also a logical consequence of the global Artin map and global Artin reciprocity law (which makes no reference to local multiplicative groups, but simply to the association $\mathfrak p \mapsto Frob\\_{\mathfrak p}$ of Frobenius elements to unramified prime ideals; see the beginning of Tate's article in Cassels--Frohlich for a nice explanation of this). Thus it is natural in a more colloquial sense of the word as well.
Indeed, the idelic formulation of the glocal reciprocity map and the formulation of the local reciprocity map in terms of multiplicative groups are not accidental or ad hoc inventions; they were forced on number theorists as a result of making deep investigations into the nature of global class field theory.
| 11 | https://mathoverflow.net/users/2874 | 12318 | 8,348 |
https://mathoverflow.net/questions/12298 | 3 | Let $P$ be a polynomial in $k$ variables with complex coefficients, and $\deg P=n$. If $k=1$ then there is Bernstein's inequality:$||P'||\le n||P||$, where $||Q||=\max\_{|z|=1} |Q(z)|$.
So, are there similar results for $k\ge 2$? For example, what is the best $f(n,k)$, such that inequality $||\frac{\partial P}{\partial z\_1}||+\cdots + ||\frac{\partial P}{\partial z\_k}||\le f(n,k)||P||$, where $||Q||=\max\_{|z\_1|=\cdots=|z\_k|=1}|Q(z\_1,\cdots,z\_k)|$, holds?
| https://mathoverflow.net/users/1888 | Bernstein inequality for multivariate polynomial | Tung, S. H. Bernstein's theorem for the polydisc. Proc. Amer. Math. Soc. 85 (1982), no. 1, 73--76. MR0647901 (83h:32017)
(from MR review): Let $P(z)$ be a polynomial of degree $N$ in $z=(z\_1,\cdots,z\_m)$; suppose that $|P(z)|\leq 1$ for $z\in U^m$; then $\|DP(z)\|\leq N$ for $z\in U^m$ where $\|DP(z)\|^2=\sum\_{i=1}^m|\partial P/\partial z\_i|^2$.
Here $U^m$ is the polydisc. Same author proved Bernstein-type inequality for the ball,
Tung, S. H. Extension of Bernšteĭn's theorem. Proc. Amer. Math. Soc. 83 (1981), no. 1, 103--106. MR0619992 (82k:32013)
| 3 | https://mathoverflow.net/users/2912 | 12319 | 8,349 |
https://mathoverflow.net/questions/12334 | 12 | This is a sequel to my [earlier question](https://mathoverflow.net/questions/10514/teichmuller-theory-introduction) asking for references for Teichmuller theory and moduli spaces of Riemann surfaces.
In this connection, I have read Chapter 11 of the book [Primer of mapping class groups](https://mathoverflow.net/questions/10514/teichmuller-theory-introduction/10517#10517) by Dan Margalit and Benson Farb.
So I have understood that the moduli space of a Riemann surface is the quotient of the Teichmuller space by the mapping class group, the action is properly discontinuous, the quotient is an orbifold, but it is not in general compact(Mumford's compactness criterion), it has "only one end", etc..
Other than these facts, does Teichmuller theory simplify the study of moduli spaces of Riemann surfaces in any way? Can we do something using Teichmuller theory which we can't do, say, using algebraic geometry? Are we able to prove theorems about moduli spaces, using Teichmuller theory methods? I would be grateful for any examples.
| https://mathoverflow.net/users/2938 | Teichmuller theory and moduli of Riemann surfaces | One of the main "gains" of the Teichmuller theory approach is that you're dealing with a ball. So you're in a situation where you can readily make analytic arguments using fixed-point theory.
Thurston's homotopy-classification of elements in the mapping class group "reducible, (pseudo) anosov, or finite-order" is one example. His argument proceeds roughly along these lines (no real details included): the mapping class group acts on Teichmuller space tautologically. Thurston defined a compactification of Teichmuller space (the "projective measured lamination space") such that the action of the mapping class group extends naturally. In particular, the compactification is a compact ball/disc. So given any element of the mapping class group, you can ask what kind of fixed points it has in this ball. Thurston's theorem is that the fixed point is in the interior if and only if the mapping is finite-order (in the mapping class group). You can think of this part as an elaboration of the theorem that isometry groups of hyperbolic manifolds are finite. There are exactly two fixed points on the boundary (and the automorphism acts as a translation along a line connecting the two points) if and only if the mapping is (isotopic to) a pseudo-anosov. A necessary and sufficient condition to be reducible is that your automorphism of the projective measured lamination space is not of the other two types, i.e. it could have one fixed point on the boundary or any number, so long as it is not precisely two acting as a translation from one to the other.
The proof of geometrization for manifolds that fibre over the circle is of course closely related.
These techniques were used to show mapping class groups satisfy the Tits alternative (which linear groups satisfy) so it was one of the big chunks of "evidence" leading people to ask the question of whether or not mapping class groups are linear.
Another application would be the resolution of the Nielsen Realization problem: <http://en.wikipedia.org/wiki/Nielsen_realization_problem>
The list goes on. But these are really applications of Teichmuller space to other things -- specifically not Moduli space.
| 12 | https://mathoverflow.net/users/1465 | 12337 | 8,361 |
https://mathoverflow.net/questions/12327 | 1 | Hi everyone,
I am looking for some ideas (or references) in order to get an explicit SDE (if it exists) which would have a stylised property extending in some sense the mean-reversion property of SDE of Ornstein-Uhlenbeck type.
More formally, is it possible to have a $n$-means reverting process defined by an SDE ?
I imagine this SDE would have the form like
$dS\_t=f\_1(S\_t,t)dt+...+f\_n(S\_t,t)dt+\sigma dW\_t$
where $f\_i$'s are such that if $S\_t$ is closed to i-th mean $m\_i$ then it stays closed to this point with high probability.
I am sorry to not define the necessary concepts more clearly but as I am only looking for ideas (or refernces) on this, I rather define some intuitive concept than a fully formal framework in order not to close any possibility.
Thank's for the time spend reading those lines
PS : I would like to avoid the n states regime switching technology if possible
| https://mathoverflow.net/users/2642 | Extension of some feature of SDE Ornstein-Uhlenbeck type | I believe your notation is redundant. Let $g = f\_1 + ... + f\_n$. Correct me if I'm wrong, but you just want a function $g$ so that $dS(t) = g(s(t),t)dt + \sigma dW\_t$ has the behavior you specify.
It's not clear exactly what properties you want. One possibility is that you can just let $S$ be a Brownian motion in a potential function $\Phi$ with $n$ local minimums. In that case, you don't need $g$ to depend on $t$: $g = -\Phi'$. You may want the potential function to be approximated by the potential well of an Ornstein-Uhlenbeck process near each minimum.
This would mean S would stay near each minimum for a while, but it would leak out eventually with probability 1. You can increase the potential with time to force the process to stay near the local minimum with positive probability. However, this would no longer resemble a fixed Ornstein-Uhlenbeck process near each minimum.
| 2 | https://mathoverflow.net/users/2954 | 12339 | 8,363 |
https://mathoverflow.net/questions/12366 | 9 | In the following suppose L/K is a finite Galois extension of number fields, (maybe it works for other cases also, I don't know) By the Chebotorev density theorem when Gal(L/K) is cyclic, there are infinitely many primes in K that stay inert during this extension (cf Janus p136, Algerbaic Number Fields.) When L/K is non cyclic, an exercise from Neukirch (somewhere in Chap I) says there are at most finitely many primes that stay inert. I want to say that there are none. The reason is by a cycle description from Janus, p101, Prop 2.8,
In short, that proposition says when $\delta:=Frob(\frac{L/K}{\beta})$, $\beta|p$ is a prime in L, consider $\delta$ act on the cosets of H in G, H=Gal(L/E), $K\subset E\subset L$, then every cycle of length i corresponds to a prime factor in E with residue degree i. In particular, for inert guys we want there is only one cycle in the action. When we take H to be trivial, E=L is Galois over K, and the cosets are just the elements of G themselves.
So we want that there exist an element (the Frobenius element above p) act transitively on G, thus G is cyclic.
I wonder if this is true, then more people should have been aware of it. If it is not, is there a counter example?
| https://mathoverflow.net/users/1877 | How many primes stay inert in a finite (non-cyclic) extension of number fields? | If $L/K$ is a finite, Galois extension of number fields such that $\text{Gal}(L/K)$ is not cyclic, then no prime of K remains inert L. Indeed, one always has an isomorphism $D\_p/I\_p\cong \text{Gal}(L\_p/K\_p)$ of the Decomposition group modulo the Inertia group with the Galois group of the corresponding residue field extension. The latter group is the Galois group of a finite extension of finite fields, hence is cyclic. If the prime p were to remain inert in L, then by definition the Inertia group would be trivial and the Decomposition group would be all of $\text{Gal}(L/K)$. But this would imply that $L/K$ was a cyclic extension - a contradiction.
[**Edit**] I can't help but mention a cute application of this. Let $n$ be any positive integer for which $(\mathbb{Z}/n\mathbb{Z})^\*$ is not cyclic. Then the cycotomic polynomial $\Phi\_n(x)$ is reducible modulo $p$ for every rational prime $p$. Indeed, suppose that $p$ is a rational prime for which $\Phi\_n(x)$ is irreducible modulo $p$. Then by the Dedekind-Kummer theorem, $p$ is inert in the cyclotomic field $\mathbb{Q}(\zeta\_n)$. Then the Galois group of the residue class field extension, which is cyclic, is isomorphic to the Decomposition group, which in this case is $\text{Gal}(\mathbb{Q}(\zeta\_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^\*$. But the latter group is not cyclic - contradiction. Thus $\Phi\_n(x)$ is reducible modulo $p$ for all rational primes $p$.
| 25 | https://mathoverflow.net/users/nan | 12369 | 8,379 |
https://mathoverflow.net/questions/12352 | 21 | In Samuel James Patterson's article titled *Gauss Sums* in *The Shaping of Arithmetic after C. F. Gauss’s Disquisitiones Arithmeticae*, Patterson says
"Hecke [proved] a beautiful theorem on the different of k, namely that the class of the absolute different in the ideal class group is a square. This theorem - an analogue of the fact that the Euler characteristic of a Riemann surface is even - is the crowning moment (coronidis loco) in both Hecke's book and Andre Weil's Basic Number Theory."
About the same matter, J.V. Armitage says (in his review of the 1981 translation of Hecke's book):
"That beautiful theorem deservedly occupies the 'coronidis loco' in Weil's *Basic number theory* and was the starting point for the work on parity problems in algebraic number theory and algebraic geometry, which has borne such rich fruit in the past fifteen years."
>
> What is a reference for learning about the parity
> problems that Armitage alludes to?
>
>
>
It can be impossible to verbalize the reasons for aesthetic preferences, but
>
> Why might Weil, Patterson and Armitage
> have been so favorably impressed by
> the theorem that the ideal class of
> the different of a number field is a
> square in the ideal class group?
>
>
>
Weil makes no comment on why he chose to end *Basic Number Theory* with the above theorem. It should be borne in mind that Weil's book covers the class number formula and all of class field theory, so that the standard against which the above theorem is being measured in the above quotes is high!
| https://mathoverflow.net/users/683 | Context for "Coronidis Loco" from Weil's Basic Number Theory | It is not hard to see that if $L/K$ is an extension of number fields, then the
discriminant of $L/K$, which is an ideal of $K$, is a square in the ideal class group of $K$.
Hecke's theorem lifts this fact to the different. (Recall that the discriminant is the norm of the different.)
If you recall that the inverse different $\mathcal D\_{L/K}^{-1}$ is equal to $Hom\_{\mathcal O\_K}(\mathcal O\_L,\mathcal O\_K),$ you see that the inverse different is the relative dualizing sheaf of $\mathcal O\_L$ over $\mathcal O\_K$; it is analogous to the canonical bundle of a curve (which is the dualizing sheaf of the curve over the ground field). Saying that
$\mathcal D\_{L/K}$, or equivalently $\mathcal D\_{L/K}^{-1}$, is a square is the same as saying that there is a rank 1 projective $\mathcal O\_L$-module $\mathcal E$ such that
$\mathcal E^{\otimes 2} \cong \mathcal D\_{L/K}^{-1}$, i.e. it says that one can take a square
root of the dualizing sheaf. In the case of curves, this is the existence of theta characteristics.
Thus, apart from anything else (and as indicated in the quotation given in the question),
Hecke's theorem significantly strengthens the analogy between rings of integers in number fields and algebraic curves.
If you want to think more arithmetically, it is a kind of reciprocity law. It expresses in some way a condition on the ramification of an arbitrary extension of number fields: however the ramification occurs, overall it must be such that the different ramified primes balance out in some way in order to have $\prod\_{\wp} \wp^{e\_{\wp}}$ be trivial in the class group
mod $2$ (where $\wp^{e\_{\wp}}$ is the local different at a prime $\wp$). (And to go back to the analogy: this is supposed to be in analogy with the fact that if $\omega$ is any meromorphic differential on a curve, then the sum of the orders of all the divisors and poles of $\omega$ is even.) Note that Hecke proved his theorem as an application of quadratic reciprocity in an arbitrary number field.
| 23 | https://mathoverflow.net/users/2874 | 12372 | 8,381 |
https://mathoverflow.net/questions/10671 | 5 | Suppose we have a (closed, oriented) 3-manifold M with a Heegard surface F of genus g. Let F\* denote F with a puncture. Then the space H of representations of pi\_1(F\*) on SU(2) is just SU(2)^2g, and the representation spaces of the two handlbodies sit inside H. Call these spaces Q\_1 and Q\_2 -- we will always think of them as subspaces of H. Finally, the intersection R = Q\_1 \cap Q\_2 is the representation space for M (Note: we haven't quotiented out by conjugation or anything).
Question 1: In the paper <http://www.jstor.org/pss/2001712>, Boyer and Nicas claim that if M is a \Q-homology sphere, the homological intersection [Q\_1 . Q\_2 ] is equal to |H\_1(M)|, and they say it's easy to prove. I can't seem to figure out how to do it though, and I've tried for a bit... it seems like there's some bit of theory I must be missing. Can anyone see how to prove it?
Question 2: Is the Euler characteristic of R (that is, Q\_1 \cap Q\_2) also |H\_1(M)|? If so, how could we prove this? In particular, is there a general relationship between the intersection pairing between two complementary submanifolds, and the Euler characteristic of their intersection (even when the intersection is not a finite number of points)?
The above is reminiscent of Morse-Bott theory, where the differential forms on the critical set of your morse function give a basis for the chain groups of your homology, and therefore the Euler characteristic of the critical set is the Euler characteristic of the manifold (or something like that... do I have this right?) This requires Morse-Bott non degeneracy of the critical set.
Final Question: What's an explicit relationship between the morse theory and the inersection theory? And when we just care about Euler characteristic, can we relax the Morse-Bott non-degeneracy? It seems that Q\_1 and Q\_2 don't always intersect "non-degenerately" (not only non-transversely, but the intersection might not even be smooth, for example) but Boyer and Nicas still claim that the intersection number is something nice (and computable on general grounds). Under what conditions could the same thing happen with a non-Morse-Bott morse function?
Thanks! I hope there aren't too many questions here...
| https://mathoverflow.net/users/492 | Relating Euler characteristic, intersection product, Morse theory (plus SU(2) and 3-manifolds) | It might be helpful to look at the book of Akbulut and McCarthy on Casson's Invariant. I think the answer to Question 1 is fairly clearly explained in Proposition 1.1b of of Chapter III.
| 2 | https://mathoverflow.net/users/3405 | 12379 | 8,386 |
https://mathoverflow.net/questions/11747 | 69 | The question is about characterising the sets $S(K)$ of primes which split completely in a given galoisian extension $K|\mathbb{Q}$. Do recent results such as Serre's modularity conjecture (as proved by Khare-Wintenberger), or certain cases of the Fontaine-Mazur conjecture (as proved by Kisin), have anything to say about such subsets, beyond what Class Field Theory has to say ?
I'll now introduce some terminology and recall some background.
Let $\mathbb{P}$ be the set of prime numbers. For every galoisian extension $K|\mathbb{Q}$, we have the subset $S(K)\subset\mathbb{P}$ consisting of those primes which split (completely) in $K$. The question is about characterising such subsets; we call them **galoisian** subsets.
If $T\subset\mathbb{P}$ is galoisian, there is a unique galoisian extension $K|\mathbb{Q}$ such that $T=S(K)$, cf. Neukirch (13.10). We say that $T$ is **abelian** if $K|\mathbb{Q}$ is abelian.
As discussed [here](https://mathoverflow.net/questions/11688/why-do-congruence-conditions-not-suffice-to-determine-which-primes-split-in-non-a) recently, a subset $T\subset\mathbb{P}$ is abelian if and only if it is defined by congruences. For example, the set of primes $\equiv1\pmod{l}$ [is the same as](https://mathoverflow.net/questions/10457/what-are-the-prime-ideals-in-rings-of-cyclotomic-integers/10473#10473) $S(\mathbb{Q}(\zeta\_l))$. "Being defined by congruences" can be made precise, and counts as a characterisation of abelian subsets of $\mathbb{P}$.
Neukirch says that Langlands' Philosophy provides a characterisation of all galoisian subsets of $\mathbb{P}$. Can this remark now be illustrated by some striking example ?
**Addendum** (28/02/2010) [Berger](http://www.umpa.ens-lyon.fr/~lberger/)'s recent Bourbaki exposé 1017 [arXiv:1002.4111](http://arxiv.org/abs/1002.4111) says that cases of the Fontaine-Mazur conjecture have been proved by Matthew Emerton as well. I didn't know this at the time of asking the question, and the unique answerer did not let on that he'd had something to do with Fontaine-Mazur...
| https://mathoverflow.net/users/2821 | Galoisian sets of prime numbers | I think it is easiest to illustrate the role of the Langlands program (i.e. non-abelian class field theory) in answering this question by giving an example.
E.g. consider the Hilbert class field $K$ of $F := {\mathbb Q}(\sqrt{-23})$; this is a degree 3 abelian extension of $F$, and an $S\_3$ extension of $\mathbb Q$. (It is the splitting field of the polynomial $x^3 - x - 1$.)
The 2-dimensional representation of $S\_3$ thus gives a representation
$\rho:Gal(K/{\mathbb Q}) \hookrightarrow GL\_2({\mathbb Q}).$
A prime $p$ splits in $K$ if and only if $Frob\_p$ is the trivial conjugacy class
in $Gal(K{\mathbb Q})$, if and only if $\rho(Frob\_p)$ is the identity matrix, if and only
if trace $\rho(Frob\_p) = 2$. (EDIT: While $Frob\_p$ is a 2-cycle, resp. 3-cycle, if and only if $\rho(Frob\_p)$ has trace 0, resp. -1.)
Now we have the following reciprocity law for $\rho$: there is a modular form $f(q)$, in fact
a Hecke eigenform, of weight 1 and level 23, whose $p$th Hecke eigenvalue gives
the trace of $\rho(Frob\_p)$. (This is due to Hecke; the reason that Hecke could handle
this case is that $\rho$ embeds $Gal(K/{\mathbb Q})$ as a dihedral
subgroup of $GL\_2$, and so $\rho$ is in fact induced from an abelian character of the
index two subgroup $Gal(K/F)$.)
In this particular case, we have the following explicit formula:
$$f(q) = q \prod\_{n=1}^{\infty}(1-q^n)(1-q^{23 n}).$$
If we expand out this product as $f(q) = \sum\_{n = 1}^{\infty}a\_n q^n,$
then we find that $trace \rho(Frob\_p) = a\_p$ (for $p \neq 23$),
and in particular, $p$ splits completely in $K$ if and only if $a\_p = 2$.
(For example, you can check this way that the smallest split prime is $p = 59$;
this is related to the fact that $59 = 6^2 + 23 \cdot 1^2$.).
(EDIT: While $Frob\_p$ has order $2$, resp. 3, if and only if $a\_p =0$, resp. $-1$.)
So we obtain a description of the set of primes that split in $K$ in terms of
the modular form $f(q)$, or more precisely its Hecke eigenvalues (or what amounts
to the same thing, its $q$-expansion).
The Langlands program asserts that an analogous statement is true for any
Galois extension of number fields $E/F$ when one is given a continuous
representation $Gal(E/F) \hookrightarrow GL\\_n(\mathbb C).$ This is known
when $n = 2$ and either the image of $\rho$ is solvable (Langlands--Tunnell) or $F = \mathbb Q$ and $\rho(\text{complex conjugation})$ is non-scalar (Khare--Wintenberger--Kisin).
In most other contexts it remains open.
| 81 | https://mathoverflow.net/users/2874 | 12382 | 8,388 |
https://mathoverflow.net/questions/12383 | 3 | Motivation: I was reading through Frenkel's article on geometric Langlands program, and the external tensor product of two perverse sheaves occurred in the definition of the geometric Langlands conjectures. There should be a reference somewhere, but the closest I could find is this research note "Exterior Tensor Product Of Perverse Sheaves" by Lyubashenko, which glossed over the definitions far too quickly for me to grasp the content.
**Question**
1. Suppose $X$ and $Y$ are sheaves of vector spaces over two spaces $A$ and $B$ respectively (here "space" means topological space - but I'd also like to know how to do it for varieties or schemes, if that's different in a non-trivial way). My question is, how to define the external tensor product of the sheaves $X$ and $Y$, which should be a sheaf over the direct product $A \times B$?.
I've been thinking about it, and the idea I have is this (similar to how to construct the tensor product of two sheaves over the same space): we need to construct for each open set $U$ in $A \times B$, a vector space $F(U)$, and then sheafify this pre-sheaf. If $U$ is an open set of the form $A\_1 \times B\_1$ where $A\_1, B\_1$ are open in $A, B$, then this is straightforward: simply take the tensor product of the vector spaces corresponding to $A\_1, B\_1$ in the sheaves $X$ and $Y$, and the restriction maps on these are fairly clear. What is not clear to me is how to do these when $U$ is not of that form, but a union of some family of sets of the form $A\_i \times B\_i$. I tried by thinking there should be restriction maps from $F(U)$ to $F(A\_i \times B\_i)$ for each $i$, but I can't see how to explicitly construct the vector space just from that fact.
1. After doing that, how to go from there to an exterior tensor product of two perverse sheaves $X'$ and $Y'$ over $A$ and $B$, but now where $A$ is a variety but $B$ is an algebraic **stack**?
| https://mathoverflow.net/users/2623 | External tensor product of two (perverse) sheaves | You don't need to construct $F(U)$ explicitly for $U$ that are not products.
Any point $(a,b)$ of $A\times B$ has a basis of neighbourhoods of the form $A\_1 \times B\_1$,
so the presheaf defined on just these open sets is enough data to sheafify and obtain
the corresponding sheaf. Since sheafification preserves stalks, you will find that the
stalk of $X\boxtimes Y$ at $(a,b)$ is equal to the stalk of $X$ at $a$ tensored with
the stalk of $Y$ at $b$.
With this construction in hand we can define exterior tensor product for complexes
of sheaves, and then hence for perverse sheaves.
To treat the case where one of $A$ or $B$ is a stack, the main point will be to have
a precise definition of what is meant by a perverse sheaf on a stack. Once this is
understood, the definition of exterior tensor product will proceed in the same way as
for the case of varieties.
| 9 | https://mathoverflow.net/users/2874 | 12388 | 8,391 |
https://mathoverflow.net/questions/12377 | 11 | Let $G$ be a reductive group over a finite field (i.e. finite groups over lie type). The case I am most interested in is $G=GL\_{n}(\mathbb{F}\_{q})$; other classical groups are also interesting I think.
Deligne-Lusztig theory has a lot to say about the irreducible representations and characters of these groups. For $G=GL\_n(\mathbb{F}\_q)$, Green's paper from the 1940's gives the characters explicitly also. The following question I guess, is in part a reference request, since the question has probably been examined in the literature somewhere, but I am unable to find a reference.
**Question:** Let $V$ and $W$ be two irreducible representations of $G$. What can be said about the decomposition of $V \otimes W$ into irreducibles? Specifically:
* Are there any special cases of $V$ for which the decomposition of $V \otimes W$ into irreducibles can always be explicitly determined? (for instance, with the symmetric group $S\_n$, there is some theory which does this for the regular representation of dimension $n-1$, and also I believe work which does this for representations corresponding to two-row partitions).
* Is there anything that can be said for the decomposition of $V \otimes V$ in general?
* What about, if $V$ and $W$ are not actually irreducibles, but instead representations obtained from $l$-adic cohomology; for instance the virtual representation $R\_{T, \theta}$ is defined as alternating sums of various cohomological representations. As an example, consider the representation of $G$ acting on the $i$-th cohomology of the Deligne-Lusztig variety $X\_{T}$ corresponding to a fixed torus $T$; if we tensor together two different cohomological representations corresponding to different tori, and cohomology for different values of $i$, what can we say? Since the $R\_{T, \theta}$ are defined as alternating sums of these, perhaps this question will help with our original problem.
| https://mathoverflow.net/users/2623 | Decomposing tensor products of irreducible representations of reductive groups over a finite field | Theorem 1.4.1 in [arxiv:0810.2076](http://arxiv.org/abs/0810.2076) answers some of your questions for generic semisimple irreducible representations. Emmanuel Letellier has hitherto unpublished results where he does answer your question for all generic irreducible representations in terms of intersection cohomology of certain quiver varieties. We did not know about other results on the representation ring of $GL\_n({\mathbb F}\_q)$. EDIT: but see Victor's answer for related results of Lusztig. EDIT 2 (added 16/03/11) Letellier's paper is now available at: <http://arxiv.org/abs/1103.2759>
| 6 | https://mathoverflow.net/users/1583 | 12391 | 8,393 |
https://mathoverflow.net/questions/12342 | 61 | From time to time, when I write proofs, I'll begin with a claim and then prove the contradiction. However, when I look over the proof afterwards, it appears that my proof was essentially a proof of the contrapositive, and the initial claim was not actually important in the proof.
Can all claims proven by reductio ad absurdum be reworded into proofs of the contrapositive? If not, can you give some examples of proofs that don't reduce? If not all reductio proofs can be reduced, is there any logical reason why not? Is reductio stronger or weaker than the contrapositive?
Edit: Just another minor question (of course this is optional and will not affect me choosing an answer): If they are equivalent, then why would you bother using reductio?
And another bonus question (Like the above, does not influence how I choose the answer to accept.)
Are the two techniques intuitionistically equivalent?
| https://mathoverflow.net/users/1353 | Reductio ad absurdum or the contrapositive? | Although the other answers correctly explain the basic logical equivalence of the two proof methods, I believe an important point has been missed:
* *With good reason*, we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal)
What is the reason? The reason is the *fecundity* of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r1, r2, before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r1 and r2 and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds.
Similarly, when we prove the contrapositive (¬q implies ¬p) directly, we assume ¬q, make intermediary conclusions r1, r2, and then finally conclude ¬p. Thus, we have also established not only that ¬q implies ¬p, but also, that it implies r1 and r2 and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (¬r1 implies q), we learn about many different hypotheses that all imply q.
These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q).
With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue r1, r2, and so on, before arriving at a contradiction. The statements r1 and r2 are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistent, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.
I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof.
---
Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p1, ..., pn are *all* the primes. It follows that since none of them divide the product-plus-one p1...pn+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginners falsely expect after this argument that whenever p1, ..., pn are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p1, ..., pn were *all* the primes. If one organizes the proof, however, as a direct argument showing that whenever p1, ..., pn are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p1...pn+1 has merely a prime divisor not on the original list.
| 135 | https://mathoverflow.net/users/1946 | 12400 | 8,398 |
https://mathoverflow.net/questions/12370 | 8 | I know this sounds dumb, but I can't for the life of me remember how to expand "TC(x)" into a first-order term in the language of set theory (ZFC, not NBG) where epsilon is the only nonlogical symbol.
The obvious definition is an $\omega$-long sentence $x\cup (\bigcup x)\cup (\bigcup\bigcup x)...$, but that isn't in $L\_{\omega\omega}$.
The definition given in Jech, p64 appeals to "the intersection of any class with a set is a set" (p8), which is really expressible only in NBG, right? I'm at a loss to figure out how to turn this into simple ZFC using separation and replacement.
I also don't have much trouble proving that for every set there exists some other set which is its transitive closure; I just can't seem to turn this proof of $(\exists y)\phi$ (for $\phi$ being "y is the transitive closure of x") into an explicit description of the $y$.
I'm starting to suspect that TC(x) isn't definable in ZFC, but that it can be defined as a class-function in NBG (which is a conservative extension of ZFC, so being able to define TC(x) doesn't actually get you any new theorems about sets).
Thanks!
| https://mathoverflow.net/users/2361 | first-order definability transitive closure operator | As Mike Shulman and François G. Dorais correctly point out, the official language of set theory has only the binary relation ε, and so there are no *terms* to speak of in that language beyond the variable symbols.
But no set theorist remains inside that primitive language, and neither is it desirable or virtuous to do so. Rather, as in any mathematical discourse, we introduce new terminology, define notions and introduce terms. What gives? I think the substance of your question is really:
* How can a set theorist (or any mathematician) sensibly and legitimately use terms that are not expressible as terms in the official language of the subject?
The answer is quite general. In any first order theory T, if one can prove that there is a unique object with a certain property, then one may expand the language by adding a term for that object, plus the defining axiom that that term has the desired property. The resulting theory T+ will be a *conservative extension* of T, meaning that the consequences of T+ that are expressible in the old language are exactly the same as the consequences of T. The reason is that any model M of T can be (uniquely) expanded to a model of T+, simply by interpreting the new term in M by its definition. This is why we may freely introduce symbols for emptyset or ω (or Q and R) and so on to set theory. Similarly, if T proves that for every x, there is a unique object y such that φ(x,y), then we may introduce a corresponding symbol fφ(x), with the defining axiom ∀x φ(x,fφ(x)). This new theory, in the expanded language with fφ, is again conservative over T.
This is what is going on with the term TC(x) for the transitive closure of x. Although there is no official term for the transitive closure of x in the basic language of set theory, we may *introduce* such a term, once we prove that every set x does indeed have a transitive clsoure. And once having done so, the term becomes officially part of the expanded language.
To see that every set x has a transitive closure, one needs very little of ZFC, and as Dorais mentions in the comments to your question, you don't need to build the Vα hierarchy. For example, every set has a transitive closure even in models of ZF- (and much less), where the power set axiom fails and so the Vα hierarchy does not exist. Simply define a sequence x0 = x and xn+1 = U xn. By Replacement, the set { xn | n ε ω } exists, and the union of this set is precisely TC(x).
In summary, we should feel free to introduce defined terms, and there is absolutely no reason not to write TC(x) on the chalkboard, as you mentioned. In particular, we should not feel compelled to express our beautiful mathematical ideas in a primitive language with only ε, like some kind of machine code, just because it is possible in principle to do so.
| 15 | https://mathoverflow.net/users/1946 | 12405 | 8,402 |
https://mathoverflow.net/questions/12364 | 5 | I'm not quite sure the best way to ask this, so bear with me: Does anyone know of a subset of integers such that, for any odd prime p, the subset only occupies (p-1)/2 equivalence classes mod p (and does so uniformly)?
For example, take the subset of squares. Elementary number theory shows that they (as quadratic residues) occupy (p+1)/2 equivalence classes mod p. But the answer to the above is not to take the non-residues since being a non-residue is a local property, not a property of an integer.
It is possible to construct such a set of integers one element at a time in an ad hoc manner using some initial members, a whole lot of CRT, and making a somewhat arbitrary choice at each step. But is there a more ``well-known'' set that has this property?
| https://mathoverflow.net/users/3400 | Integer subset that only occupies (p-1)/2 equivalence classes mod p? | See section 4.3 of Helfgott and Venkatesh, ["How small must ill-distributed sets be?"](http://math.stanford.edu/~akshay/research/hs.pdf)
for an example of a subset of [1..N] of size about log N with small projections onto Z/pZ,
and section 4.2 for a "guess" about what such subsets might look like in general. They speculate that such a set might have to be either very small (say, of size N^eps) or highly correlated with a "thin set," say, the values of a polynomial (i.e. x^2, as in the first case you describe.)
| 6 | https://mathoverflow.net/users/431 | 12407 | 8,404 |
https://mathoverflow.net/questions/12394 | 26 | There have been a couple questions recently regarding metric spaces, which got me thinking a bit about representation theorems for finite metric spaces.
Suppose $X$ is a set equipped with a metric $d$. I had initially assumed there must be an $n$ such that $X$ embeds isometrically into $\mathbb{R}^n$, but the following example shows that this doesn't quite work:
Take for $X$ the vertex set of any graph, and let $d(x,y)$ be the length (in steps) of the shortest path connecting $x$ to $y$. Then for a minimal path connecting $x$ to $y$, the path must map to a straight line in $\mathbb{R}^n$. This is because $\mathbb{R}^n$ has the property that equality in the triangle inequality implies colinearity.
So take a graph such that $x$ and $y$ have $d(x,y) \geq 2$ and *two* minimal paths between them; a plain old square will do the trick. The two minimal paths must each get mapped to the same line in $\mathbb{R}^n$, so the map cannot be an isometry (nor even an embedding, for that matter).
This gives us one obstruction to representability: a finite metric space cannot be representable unless it satisfies the property $d(x,y) = d(x,z) + d(z,y) \wedge d(x,y) = d(x,z') + d(z',y) \wedge d(x,z) = d(x,z') \implies z = z'$. In the graph case this means "unique shortest paths"; I'm not clear if there is a snappy characterization like that in the general case.
**Question #1:** Is this the only obstruction to representability?
In a slightly different direction, you could get around the problem above by trying to represent the finite metric spaces on some surface instead of $\mathbb{R}^n$. This at least works in the graph case by replacing the points with little discs and the edges with very fat ribbons of length 1. Then compactifying the whole thing should give a surface into which the graph embeds isometrically. This suggests the answer to
**Question #2a:** Does every finite metric space have a representation on a surface?
is yes, as long as the answer to
**Question #2b:** Does every finite metric space have a global scaling which embeds $\epsilon$-isometrically into a graph?
is also yes. The $\epsilon$ is to take care of finite metric spaces with irrational distances.
Of course, there is also the important
**Question #0:** Is there some standard place I should have looked for all this?
| https://mathoverflow.net/users/2510 | Representability of finite metric spaces | Since the paper referred to by Hagen Knaf is published by Springer, it may not be available to one and all. The (publicly viewable) MathSciNet reference is: [MR355836](http://www.ams.org/mathscinet-getitem?mr=355836).
It's a very short paper (7 pages) and the main theorem is:
**Theorem** A metric space can be embedded in Euclidean n-space if and only if the metric space is flat and of dimension less than or equal to n.
Clearly, the two terms "flat" and "dimension" need expanding. To define these, Morgan considers *simplices* in the metric space; that is, an n-simplex is simply an ordered (n+1)-tuple of elements of the metric space. Given such an n-tuple, say $(x\_0, \dots, x\_n)$, Morgan defines $D(x\_0,\dots,x\_n)$ to be the determinant of the matrix whose $(i,j)$th entry is
$$
\frac{1}{2}\left(d(x\_0,x\_i)^2 + d(x\_0,x\_j)^2 - d(x\_i,x\_j)^2\right)
$$
A metric space is **flat** if this is positive for all simplices. If it is flat, its **dimension** (if this exists) is the largest n for which there is an n-simplex with this quantity positive.
The argument (on a skim read) is quite cunning. Rather than go for a one-shot embedding, Morgan defines a map into $\mathbb{R}^n$ of the appropriate dimension and then uses that map to define an inner product on $\mathbb{R}^n$ with respect to which the map is an embedding. Standard linear algebra then completes the argument.
Now finite dimension, in this sense, is clearly very strong. It basically says that the metric is controlled by n points. The flatness condition says that those n points embed properly (and presumably rules out the examples in the question and the first answer). But then, that's probably to be expected since embeddability into Euclidean space is similarly a strong condition.
| 23 | https://mathoverflow.net/users/45 | 12409 | 8,405 |
https://mathoverflow.net/questions/12410 | 5 | Many people are familiar with the notion of an additive category. This is a category with the following properties:
(1) It contains a zero object (an object which is both initial and terminal).
This implies that the category is enriched in pointed sets. Thus if a product $X \times Y$ and a coproduct $X \sqcup Y$ exist, then we have a canonical map from the coproduct to the product (given by "the identity matrix").
(2) Finite products and coproducts exist.
(3) The canonical map from the coproduct to the product is an equivalence.
A standard exercise shows this gives us a multiplication on each hom space making the category enriched in commutative monoids (with unit).
(4) An additive category further requires that these commutative monoids are abelian groups.
>
> I want to know what standard terminology is for a category which satisfies the first three axioms but not necessarily the last.
>
>
>
I can't seem to find it using Google or Wikipedia. An obvious guess, "Pre-additive", seems to be standard terminology for a category enriched in abelian groups, which might not have products/coproducts.
| https://mathoverflow.net/users/184 | Terminology: Is there a name for a category with biproducts? | One name that I have seen used is [semiadditive category](http://ncatlab.org/nlab/show/biproduct#semiadditive_categories_3).
| 7 | https://mathoverflow.net/users/49 | 12419 | 8,412 |
https://mathoverflow.net/questions/12416 | 22 | First of all, let me apologize in advance for the terseness of this question.
It seems that by now there are well-developed techniques (the "Taylor-Wiles-Kisin" method) for proving modularity lifting theorems over totally real fields. For example, it is now known that many elliptic curves over totally real fields are modular. I am curious: what exactly is the stumbling block to proving such results over non-totally real fields? It seems to be common knowledge among experts that the usual techniques for comparing a universal deformation ring and a Hecke algebra in this situation break down quite badly, but I cannot find a reference for this in print. It would be great if someone could illustrate the problem here.
I understand that there are some obvious difficulties. For example, for $GL\_2$ over an imaginary quadratic field, the locally symmetric spaces on which the relevant modular forms live is $SL\_2(\mathbb{C})/SU(2)$, and arithmetic quotients of this are certainly not Shimura varieties. I am asking about more fundamental obstructions to applying the Taylor-Wiles-Kisin method, i.e strange behavior in the Hecke algebra and the universal deformation ring...
| https://mathoverflow.net/users/1464 | The difficulties in proving modularity lifting theorems over non-totally real fields | Note: This is a fairly precise and detailed question about an important but technical aspect of algebraic number theory. My answer is written at a level that I think is appropriate for the question; it assumes some familiarity with the topic at hand.
---
The most basic difficulty is that there is not a map $R \rightarrow {\mathbb T}$ in general
(i.e. one typically doesn't know how to create Galois representations attached to automorphic forms).
The second difficulty is that in the TWK method, one must argue with auxiliary primes (the primes typically labelled $Q$), and show that as you add these primes, ${\mathbb T}$ grows in a reasonable way (basically, is free over $\mathcal O[\Delta\_Q],$ where $\Delta\_Q$ is something like the $p$-Sylow subgroup of $({\mathbb Z}/Q{\mathbb Z})^{\times}.)$
One shows this (or some variant of it) by considering the analogous queston about cohomology of the arithmetic quotients. Suppose for a moment we are in the Shimura variety context, or perhaps the compact at infinity context. Then it will be the middle dimensional cohomology that is of interest, and if we localize at a non-Eisenstein maximal ideal we might hope to kill all other cohomology. Then we can replace a computation of middle dimensional cohomology by an Euler characteristic computation, and its easy to see that the Euler char. will multiply by $|\Delta\_Q|$ when we add the auxiliary primes $Q$.
But in more general contexts, there won't be a single middle dimension in which the maximal ideal of interest is supported (even if it is non-Eisenstein), and computing Euler characteristics will just give $0$, which is not much use. It's not clear that it's even true that adding the auxiliary primes forces the approriate growth of cohomology, and possible torsion in the cohomology just adds to the complication.
There is much current work, by various groups of researchers, with various different approaches, aimed at breaking this barrier.
---
I should add that one can now handle certain questions about non-totally real field,
say question related to conjugate self-dual Galois reps. over CM fields, because
these are still related to a Shimura variety context. This plays a role in the recent
progress on Sato--Tate for higher weight forms by Barnet-Lamb--Geraghty--Harris--Taylor
and Barnet-Lamb--Gee--Geraghty, and is also the basis for a recent striking theorem
of Calegari showing that if $\rho:G\_{\mathbb Q} \to GL\_2({\mathbb Q}\_p)$ is ordinary
at $p$ and de Rham with distinct Hodge--Tate weights (and probably $\overline{\rho}$ should satisfy some technical conditions), then $\rho$ is necessarily odd!
---
| 22 | https://mathoverflow.net/users/2874 | 12429 | 8,416 |
https://mathoverflow.net/questions/12428 | 8 | In a two-column double complex, one gets from the associated spectral sequence short exact sequences $0\to E\_2^{1,n-1}\to H^n\to E\_2^{0,n}\to 0$, where $H^n$ is the cohomology of the total complex, but I have never seen the construction of this sequence. Any text I've seen merely states it as a fact, or leaves it as an exercise which I have had no luck trying to solve. Can anyone give a construction or good reference?
| https://mathoverflow.net/users/1481 | How does one get the short exact sequence in a two-column spectral sequence? | This follows precisely from the very definition of convergence of the spectral sequence, once one has identified the $\infty$-term. It is done with some details in McLeary's *User Guide*---which is, in my opinion, a very good reference for both the technicalities and the pragmatics of dealing with spectral sequences.
Now, if you are *starting* with a two column double complex (as opposed to starting with an arbitrary double complex whose spectral sequence has two contiguous columns), you can get the short exact sequences very much 'by hand'.
Indeed, suppose your double complex is $T^{\bullet,\bullet}=(T^{p,q})\_{p,q\geq0}$ and that $T^{p,q}\neq0$ only if $p\in\{0,1\}$. If we define complexes $X^\bullet$ and $Y^\bullet$ with $X^q=T^{0,q}$ and $Y^q=T^{1,q}$, with differentials coming from the vertical differential $d$ of $T^{\bullet,\bullet}$, then the horizontal differential of $T^{\bullet,\bullet}$ can be seen as a map of complexes $\delta:X^\bullet\to Y^{\bullet}$.
Now, in the spectral sequence induced by the filtration by columns we clearly have $E\_1^{0,q}=H^q(X^\bullet)$, $E\_1^{1,q}=H^q(Y^\bullet)$ and the differential on the $E\_1$ page is induced by the horizontal differential in $T^{\bullet,\bullet}$. In other words, the $E\_1$ page is more or less the same thing as the map $H(\delta):H(X^\bullet)\to H(Y^\bullet)$. It follows that we have short exact sequences $$0\to E\_2^{0,q}\to H^q(X^\bullet)\xrightarrow{H^q(\delta)} H^q(Y^\bullet)\to E\_2^{1,q}\to 0,$$ and the spectral sequence dies at the second act for degree reasons.
On the other hand, there is a short exact sequence of complexes $$0\to Y[-1]^\bullet\to\mathrm{Tot}\;T^{\bullet,\bullet}\to X^\bullet\to 0,$$
from which we get a long exact sequence $$\cdots\to H^{q-1}(Y^\bullet)\to H^q(\mathrm{Tot}\; T^{\bullet,\bullet})\to H^q(X)\to H^q(Y)\to\cdots,$$ in which you can compute directly that the map $H^q(X)\to H^q(Y)$ is precisely $H^q(\delta)$. Since the first four-term exact sequence identifies for us the kernel and the cokernel of $H^q(\delta)$, exactness of the second exact sequence provides the short exact sequence $$0\to E\_2^{1,q-1}\to H^q(\mathrm{Tot}\; T^{\bullet,\bullet})\to E\_2^{0,q}\to 0$$ that you wanted.
(It is an extremely instructive exercise to try to see what can one do in this spirit for a three-column double complex, and fighting with this is a great prelude to an actual exposition to spectral sequences...)
| 14 | https://mathoverflow.net/users/1409 | 12430 | 8,417 |
https://mathoverflow.net/questions/12426 | 38 | **Background**
Assuming ZFC is consistent, then by downward Löwenheim–Skolem, there is a countable model (M,$\in$) of ZFC. Since the universe M is countable, we may as well think of it as actually being the set of natural numbers, so $\in$ will be some binary relation on the natural numbers.
>
> Can such a relation ever be computable?
>
>
>
**Partial results**
~~One can show that the class of binary relations $R$ on the natural numbers such that $(\mathbb{N},R) \models ZFC$ forms a $\Pi\_0^1$ class, and will be nonempty so long as ZFC is consistent. This already gives us some interesting results. For example, by the low basis theorem, there is a low $R$ such that $(\mathbb{N},R) \models ZFC$. But I have been unable to determine whether such a function can be made computable; the best I can do is show that if such a function is computable, then there is no effective way of finding, given a finite set D of natural numbers, the element n such that D={m : mRn}.~~
| https://mathoverflow.net/users/3410 | Is there a computable model of ZFC? | The Tennenbaum phenomenon is amazing, and that is totally correct, but let me give a direct proof using the idea of [computable inseparability](http://en.wikipedia.org/wiki/Effectively_separable).
**Theorem**. There is no computable model of ZFC.
Proof: Suppose to the contrary that M is a computable model of ZFC. That is, we assume that the underlying set of M is ω and the membership relation E of M is computable.
First, we may overcome the issue you mention at the end of your question, and we can computably get access to what M thinks of as the
nth natural number, for any natural number n. To see this, observe first that there is a particular natural number z, which M believes
is the natural number 0, another natural number N, which M believes to the set of all natural numbers, and another natural number s, which M
believes is the successor function on the natural numbers. By decoding what it means to evaluate a function in set theory using ordered pairs, We
may now successively compute the function i(0)=z and i(n+1) = the unique number that M believes is the successor function s of i(n). Thus,
externally, we now have computable access to what M believes is the nth natural number. Let me denote i(n) simply by **n**. (We
could computably rearrange things, if desired, so that these were, say, the odd numbers).
Let A, B be any [computably inseparable](http://en.wikipedia.org/wiki/Effectively_separable) sets. That is, A and B are disjoint computably
enumerable sets having no computable separation. (For example, A is the set of TM programs halting with output 0 on input 0, and B is the set of
programs halting with output 1 on input 0.) Since A and B are each computably enumerable, there are programs p0 and p1 that
enumerate them (in our universe). These programs are finite, and M agrees that **p0** and **p1** are TM programs that
enumerate a set of what it thinks are natural numbers. There is some particular natural number c that M thinks is the set of natural numbers
enumerated by **p0** before they are enumerated by **p1**. Let A+ = { n | **n** E c }, which is the set
of natural numbers n that M thinks are enumerated into M's version of A before they are enumerated into M's version of B. This is a computable
set, since E is computable. Also, every member of A is in A+, since any number actually enumerated into A will be seen by M to have
been so. Finally, for the same reason, no member of B is in A+, because M can see that they are enumerated into B by a (standard)
stage, when they have not been enumerated into A. Thus, A+ is a computable separation of A and B, a contradiction. QED
Essentially this argument also establishes the version of Tennenbaum's theorem mentioned by Anonymous, that there is no computable nonstandard
model of PA. But actually, Tennenbaum proved a stronger result, showing that *neither* plus nor times individually is computable in a nonstandard model of PA. And this takes a somewhat more subtle argument.
**Edit (4/11/2017).** The question was just bumped by another edit, and so I thought it made sense to update my answer here by mentioning a generalization of the result. Namely, in recent joint work,
* M. T. Godziszewski and J. D. Hamkins, [Computable quotient presentations of models of arithmetic and set theory](http://jdh.hamkins.org/computable-quotient-presentations-of-models-of-arithmetic-and-set-theory/), click through for pdf at the arxiv.
we prove that not only is there no computable model of ZFC, but also there is no computable quotient presentation of a model of ZFC, and indeed there is no c.e. quotient presentation of such a model, by an equivalence relation of any complexity. That is, there is no c.e. relation $\hat\in$ and equivalence relation $E$, of any complexity, which is a congruence with respect to $\hat\in$, such that the quotient $\langle\mathbb{N},\hat\in\rangle/E$ is a model of ZFC.
| 36 | https://mathoverflow.net/users/1946 | 12434 | 8,420 |
https://mathoverflow.net/questions/12436 | 48 | I know there was a question about good algebraic geometry books on here before, but it doesn't seem to address my specific concerns.
\*\*
Question
\*\*
Are there any well-motivated introductions to scheme theory?
My idea of what "well-motivated" means are specific enough that I think it warrants a detailed example.
\*\*
Example of what I mean by well motivated
\*\*
The only algebraic geometry books I have seen which cover schemes seem to leave out essential motivation for definitions. As a test case, look at Hartshorne's definition of a separated morphism:
*Let $f:X \rightarrow Y$ be a morphism of schemes. The diagonal morphism is the unique morphism $\Delta: X \rightarrow X \times\_Y X$ whose composition with both projection maps $\rho\_1,\rho\_2: X \times\_Y X \rightarrow X$ is the identity map of $X$. We say that the morphism $f$ is separated if the diagonal morphism is a closed immersion.*
Hartshorne refers vaguely to the fact that this corresponds to some sort of "Hausdorff" condition for schemes, and then gives one example where this seems to meet up with our intuition. There is (at least for me) little motivation for why anyone would have made this definition in the first place.
In this case, and I would suspect many other cases in algebraic geometry, I think the definition actually came about from taking a topological or geometric idea, translating the statement into one which only depends on morphisms (a more category theoretic statement), and then using this new definition for schemes.
For example translating the definition of a separated morphism into one for topological spaces, it is easy to see why someone would have made the original definition. Use the same definition, but say topological spaces instead of schemes, and say "image is closed" instead of closed immersion, i.e.
*Let $f:X \rightarrow Y$ be a morphism of topological spaces. The diagonal morphism is the unique morphism $\Delta: X \rightarrow X \times\_Y X$ whose composition with both projection maps $\rho\_1,\rho\_2: X \times\_Y X \rightarrow X$ is the identity map of $X$. We say that the morphism $f$ is separated if the image of the diagonal morphism is closed.*
After unpacking this definition a little bit, we see that a morphism $f$ of topological spaces is separated iff any two distinct points which are identified by $f$ can be separated by disjoint open sets in $X$. A space $X$ is Hausdorff iff the unique morphism $X \rightarrow 1$ is separated.
So here, the topological definition of separated morphism seems like the most natural way to give a morphism a "Hausdorff" kind of property, and translating it with only very minor tweaking gives us the "right notion" for schemes.
Is there any book which does this kind of thing for the rest of scheme theory?
Are people just expected to make these kinds of analogies on their own, or glean them from their professors?
I am not entirely sure what kind of posts should be community wiki - is this one of them?
| https://mathoverflow.net/users/1106 | Motivation for concepts in Algebraic Geometry | I would say that the book you're looking for is probably "The Geometry of Schemes" by Eisenbud and Harris. It is very concrete and geometric, and motivates things well (though I don't think it does so in quite the detail of proving that a topological space is Hausdorff iff $X\to 1$ is separated, but I believe it does discuss separatedness and why it is good and why it captures the intuition of Hausdorff space)
| 33 | https://mathoverflow.net/users/622 | 12437 | 8,422 |
https://mathoverflow.net/questions/12432 | 11 | This was asked as part of an [earlier question](https://mathoverflow.net/questions/10966/orientability-and-orientation-for-a-differentiable-manifold). But since this part did not attract many answers, I am asking it separately.
We consider the homology definition of an orientation for a manifold, as you define fundamental class., ie as some generator of some homology modules, satisfying some compatibility conditions. See for instance the book of Greenberg and Harper. What does it mean to say that a manifold is orientable, over rings other than $\mathbb Z$?
It is nice when the base ring is $\mathbb{Z}/2\mathbb{Z}$; every manifold is orientable here, and has a unique orientation. And thus you can do Poincare duality, etc.. But what on earth does it mean to have $4$ possible orientations for the circle or real line for instance, when you take the base ring for homology to be $\mathbb{Z}/5\mathbb{Z}$?
Maybe it is just a formalism; maybe we do not really have to bother about orientations except the ones given by $+1$ and $-1$ in a ring, and the rest are just matters of additional generators giving some extra vacuous information. But I keep wondering. I hope somebody can clarify.
| https://mathoverflow.net/users/2938 | Meaning of orientation/orientability over rings other than the integers | I just wanted to mention that while orientability for cohomology with arbitrary coefficients is governed solely by cohomology with coefficients in ℤ, there are other cohomology theories for which is is not true. For example, if you have an action of $\pi\_1(X)$ on an abelian group M, then you can talk about (co)homology with twisted coefficients in M. For any vector bundle there is a coefficient module such that the bundle is orientable with respect to these twisted coefficients (or, to paraphrase Matthew Ando, "every bundle is orientable if you're twisted enough").
Also, one can ask whether a vector bundle is orientable with respect to topological K-theory, real or complex, or many other generalized cohomology theories, which capture interesting information about the manifold.
So while ℤ/m-coefficients may not be the most interesting coefficient systems to study orientability in, they're part of a larger systematic family of questions (and they don't take much extra work if you're already doing ℤ and ℤ/2-coefficients).
Finally, for something like real coefficients you might think of orientations that differ by a real scalar geometrically, e.g. according to some volume form.
| 10 | https://mathoverflow.net/users/360 | 12441 | 8,424 |
https://mathoverflow.net/questions/12438 | 20 | I have encountered iterated integrals on papers dealing with multizeta values, polylogarithms etc.. Since then I am trying to figure out the motivations and purpose of the theory.
It seems the defintions and methods go back to K.-T. Chen. The integrals seem to converge like an exponential series. He published many papers on this topic. Some of these(as seen in his collected works) seem to relate to path spaces, loops spaces etc., and their homology/cohomology. Many notions in algebraic topology seem to be carried out in this context using the analytic tool of iterated integral. He calls it a "de Rham theoretical approach" to the fundamental group, etc.. Is this a "de Rham homotopy theory"? Are we able to capture topological properties by repeated integration? In particular I have in mind the "iterated path integrals" paper of K.-T. Chen in mind. There are lots of others too, and some of them are in the Annals; so one cannot question the mathematical importance of the topic.
I am sorry for asking a vague question. I am a beginner on a topic struggling to understand the concepts and motivations behind them. I will be grateful for any pointers towards more understanding, so that I can get started.
| https://mathoverflow.net/users/2938 | Understanding iterated integrals | The theory of iterated integral gives a mixed Hodge structure on rational homotopy of a variety. In the case of the fundamental group, as far as I know, one can only detect the
nilpotent completion of the fundamental group. (At least, this is the only part of $\pi\_1$ that people work with in a motivic context --- see e.g. Deligne's paper on the thrice punctured sphere, many papers of Dick Hain, and perhaps Sullivan's original paper from 1977 on rational homotopy theory.)
See recent work of Minhyong Kim for application of these ideas to studying rational points
on curves over number fields.
| 14 | https://mathoverflow.net/users/2874 | 12445 | 8,426 |
https://mathoverflow.net/questions/12448 | 8 | As an amusement at the start of [this talk](http://videolectures.net/sicgt07_rosenfeld_falkp/), Moshe Rosenfeld poses the following question.
>
> Suppose that there are n salmon which
> begin at distinct points on a unit
> circle, each facing either clockwise
> or counterclockwise. On a signal, each salmon moves around the circle in its chosen direction at a constant speed (the same for all salmon). When two salmon meet, they both instantly reverse directions. If any salmon ever returns to its starting point, it dies. (If two salmon meet at one of their starting points, there is a death and no change of direction; as Rosenfeld says, "Death comes first.")
>
>
> 1. Is it true that all the salmon will eventually die?
> 2. (assuming the answer to part 1 is yes) Give an algorithm to find the last survivor.
>
>
>
I spent a certain amount of time on buses and planes tinkering with this. It's quite easy to show that every configuration is preperiodic, as a start. I have some ideas about how one might finish. Eventually I decided just to look for more information on the problem, with no real success.
One of the themes of his talk is how some problems become popular and some gather dust on the shelf. Is the latter what happened to this problem?
His second question is a bit mysterious. The problem setup itself is algorithmic in nature, so what does it mean? Is there anything besides "elegance" that would distinguish the kind of answer we should have in mind from a stupid answer like "just watch the salmon"? ("Running time" could be an answer, but it seems likely that just letting the salmon swim wouldn't take all that long.)
I am really asking three subquestions on this topic.
>
> 1. Did this question ever get solved or taken up seriously? If so, where?
> 2. Is there a natural, nonvacuous interpretation of the second part of
> the question?
> 3. What is the solution? (This is actually the subquestion I am *least*
> interested in, but it felt wrong not
> to ask it.)
>
>
>
(Please feel free to re-tag, still getting used to things here.)
| https://mathoverflow.net/users/2559 | Moshe Rosenfeld's Salmon Problem | Ah, irony. Now that I've publicly asked the question, I practically trip over a reference.
I just found [this paper](http://amc.imfm.si/index.php/amc/article/view/25/43) of Rosenfeld from 2008, which has large overlap with the talk mentioned in my original post. In the paper it is shown that there are initial configurations which do not lead to extinction, though little progress is made on how one might recognize these special configurations in advance.
(I'm still interested in answers to subquestion 2 above.)
| 3 | https://mathoverflow.net/users/2559 | 12450 | 8,429 |
https://mathoverflow.net/questions/12443 | 3 | It is well known that the first Kronecker limit theorem gives the Laurent expansion of the Eisenstein series $E(z,s)$ over $SL(2,Z)$ at $s=1$; see, for example, Serge Lang's book Elliptic Curves, Section 20.4.
My question is, is there an analogous formula for the Eisenstein series over congruence subgroups? This seems a natural question and I believe that it must be hidden somewhere in the literature, but I cannot find a reference.
Any help will be appreciated. Thanks!
Remark: Thanks to Anweshi's help, we can find the following papers.
To be more precise, the above mentioned papers are
1. MR0318065 (47 #6614) Goldstein, Larry Joel, Dedekind sums for a Fuchsian group. I. Nagoya Math. J. 50 (1973), 21--47. 10D10 (10G05). [This paper gives the generalization of Kronecker's first limit formula to Eisenstein series over a Fuchsian group of the first kind at any cusp.]
MR0347739 (50 #241) Goldstein, Larry Joel, Errata for ``Dedekind sums for a Fuchsian group. I'' (Nagoya Math. J. 50 (1973), 21--47). Nagoya Math. J. 53 (1974), 235--237. 10D15 (10G05)
1. MR0347740 (50 #242) Goldstein, Larry Joel, Dedekind sums for a Fuchsian group. II. Nagoya Math. J. 53 (1974), 171--187. 10D15 (10G05). [This paper gives the generalization of Kronecker's second limit formula for "generalized Eisenstein series".]
| https://mathoverflow.net/users/3416 | Eisenstein series and the Kronecker limit theorem | Yes, there is a generalization. It was done by Larry Joel Goldstein, in the paper "Dedekind sums for a Fuchsian group". The paper has two parts and was published in the Nagoya Journal in around 1973-74.
| 3 | https://mathoverflow.net/users/2938 | 12458 | 8,434 |
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