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https://mathoverflow.net/questions/6165 | 1 | I have a general question in Riemannian geometry:
Let M be a compact manifold and $\partial M \neq \emptyset$. Then shoot a geodesic from any boundary point perpendicularly into the interior of M. How can one prove it will end at boundary? If so, it induces a transformation of $\partial M$, does anyone know any result about this transformation? For example, one can ask rigidity property, i.e. if the transformation is an isometry, can one say anything about M?
| https://mathoverflow.net/users/1947 | Transformations induced by geodesics of boundary | It doesn't seem true to me that the geodesic will always return to the boundary. It may end up accumulating around a closed geodesic in the interior.
For example, consider the hyperboloid of one sheet
$$x^2+y^2-z^2=1$$
in $\mathbb{R}^3$. This is a surface of revolution and so one can talk of the angular momentum of a path with respect to the axis. For a geodesic this must be constant (Clairault's theorem). If a geodesic has momentum exactly one (in some well-chosen units!) it will accumulate on the "waist" of the surface, where $z=0$.
Of course this is a non-compact example, but if a geodesic accumulates on the waist, it stays inside some compact region. So it is easy to imagine a compact surface with boundary which contains the relevant compact region and for which this accumulating geodesic meets the boundary at a right angle. This gives an example where the geodesic doesn't return to the boundary. (I think it's even possible to find an example of a metric on the disc of this sort.)
(In case you're curious, a geodesic on the hyperboloid starting with $z>0$, pointing "down" and with momentum less than 1 will pass through the waist and end up asymptotic to a meridian at z = -infinity. If the momentum is bigger than 1 it will not reach the waist and swing back upwards, ending up asymptotic to a meridian at z = +infinity.)
| 9 | https://mathoverflow.net/users/380 | 6167 | 4,199 |
https://mathoverflow.net/questions/6180 | 39 | In the same vein as [Kate](https://mathoverflow.net/questions/544/why-are-subfactors-interesting) and [Scott](https://mathoverflow.net/questions/2046/how-do-i-describe-a-fusion-category-given-a-subfactor)'s questions, why are fusion categories interesting? I know that given a "suitably nice" fusion category (which probably means adding adjectives such as "unitary," "spherical," and "pivotal"), we get a subfactor planar algebra which, in turn, gives us a subfactor. Also, I vaguely understand that these categories give us Turaev Viro TQFTs.
What else do fusion categories do?
What's a good reference for the Turaev Viro stuff?
| https://mathoverflow.net/users/351 | Why are fusion categories interesting? | Fusion categories (over $\mathbb{C}$) are a natural generalization of finite groups and their behavior over $\mathbb{C}$. The complex representation theory of a finite group is a fusion category, but there are many others. In fact, you can think of a fusion category as a non-commutative, non-cocommutative generalization of a finite group. A finite-dimensional Hopf algebra is that too, but they don't have to be semisimple, while the semisimple ones give you many fusion categories, but again not by any means all of them.
Many of the basic results about the structure and representation theory of finite groups generalize, or seem like they could generalize, to fusion categories. This principle has been worked out to a very incomplete but interesting extent by Etingof and others. For instance there is an analogue of the theorem that the dimension of complex irrep of a finite group $G$ divides $|G|$. (Addendum: A qualified analogue, as Scott and Noah point out. If the category is braided, it is a strict analogue; otherwise it is an analogue of dividing $|G|^2$.) There are also semisimple Hopf algebras and other fusion categories that look a lot like $p$-groups.
You can think of the whole theory as a rebooted theory of finite groups. However, we are miles and miles away from any fusion category equivalent of the classification of finite simple groups. It is a struggle to make fusion categories that are not derived very closely from finite groups, or do not come from quantum groups at roots of unity. Only a few types of examples are known, and who knows what else is out there.
One enticing thing that does change is that dimensions of irreducible objects in a fusion category don't have to be integers. For instance, one of the simplest fusion categories is the Fibonacci category. It has two irreducible objects, the trivial one $I$ and the other object $F$. The dimension of $F$ is the golden ratio, as you can infer from the branching equation $F \otimes F \cong F \oplus I$. (But the dimensions are algebraic integers, and even cyclotomic algebraic integers. Hence divisibility is still a sensible question.)
You could also ask, why the semisimple case. As you learn in undergraduate or basic graduate representation theory, the semisimple representation theory of a finite group is much cleaner than the modular representation theory in positive characteristic.
And yes, you also get 3-manifold invariants and subfactors.
---
For references: Really Turaev and Viro's original paper, [state sum invariants of 3-manifolds and 6j symbols](http://www.math.uu.se/~oleg/6j.pdf), is pretty good. The generalization to spherical categories is due to Barrett and Westbury, [Invariants of Piecewise-Linear 3-manifolds](http://arxiv.org/abs/hep-th/9311155). And there is a discussion in Turaev's book.
A sketch: Recall that a basis-independent expression in tensor calculus has the structure of a graph with vertices labelled by tensors and edges labelled by vector spaces. A monoidal category allows the evaluation of similar expressions, except that the graph must be planar and acyclic. In a rigid pivotal category, there are good duals and the graph just needs to be planar. In a spherical category, left trace equals right trace, so a closed graph can be drawn on a sphere. If it is spherical, rigid, and semisimple, then you can use the graph of a tetrahedron to make a local interaction on the tetrahedra of a triangulated 3-manifold, and the result up to normalization is the Turaev-Viro 3-manifold invariant. (In this setting you should dualize the tetrahedra, so that a tensor morphism in the category is associated to a face of the tetrahedron.)
| 48 | https://mathoverflow.net/users/1450 | 6183 | 4,208 |
https://mathoverflow.net/questions/6186 | 4 | I need to compute the number of solutions to the equation $x^{p+1} = y^4$ in the field with $p^2$ elements (for p sufficiently large). The form of the equation suggests to me that the solution would depend on the congruence class of p mod 4, but I have reason to believe that the answer is a single polynomial in p.
I feel as if this should be easy, and I'm missing an obvious approach. Can anyone help me out?
| https://mathoverflow.net/users/10273 | Counting solutions to x^{p+1}=y^4 in a finite field | Let g be a generator of the multiplicative group of the field; assuming x and y are nonzero, we can write x=ga and y=gb with 0 <= a,b < p2-1, and then xp+1=y4 becomes ga(p+1)=g4b, or equivalently a(p+1) = 4b (mod p2-1).
From this we see that p+1 | 4b is necessary, and if 4b=k(p+1) then (a,b) gives a solution iff a=k (mod p-1). Since a can range from 0 to p2-2, then, there are either 0 solutions or p+1 solutions for any fixed b. The total number of nonzero solutions is therefore (p+1)\* #{b | p+1 divides 4b}, and then (x,y)=(0,0) is the remaining solution.
Now if p is 1 (mod 4) we have p+1 | 4b iff b is a multiple of (p+1)/2, and there are 2(p-1) such b up to p2-1, so there are 2(p-1)(p+1)+1 = 2p2-1 solutions.
On the other hand, if p is 3 (mod 4) then p+1 | 4b iff b is a multiple of (p+1)/4, so we have 4(p-1) such b and there are 4p2-3 solutions.
| 17 | https://mathoverflow.net/users/428 | 6196 | 4,217 |
https://mathoverflow.net/questions/6194 | 15 | Okay, that's a misleading title. This is a somewhat subtler problem than undergraduate linear algebra, although I suspect there's still an easy answer. But I couldn't resist :D.
Here's the actual problem: We're given a *black-box* linear transformation from $V \rightarrow W$, where $V, W$ are vector spaces of dimensions m, n respectively (say m < n), and we want to know if it has full rank. (Numerical considerations aren't an issue; if you want, say it's over a finite field.) This is easy to do in time $O(m^2n)$ and with m calls to the black-box function, just by computing the image of a basis in m and using Gaussian elimination. It's also immediately obvious that we can't do better than m calls to the function in a deterministic algorithm, and I'm pretty sure but haven't quite managed to prove that you can't beat Gaussian elimination asymptotically either.
But can we do better if we just want a probabilistic algorithm? If we're allowed to make as many function calls as we want? What's the best lower bound we can get, probabilistically? These are probably pretty trivial questions (since everything's linear-algebraic and nice), but I just don't know how to approach them.
| https://mathoverflow.net/users/382 | How to compute the rank of a matrix? | I think there would be a problem if the transformation was almost independent. If one vector were a combination of the others but otherwise there was independence. I think you would have to compute the image of basis to test for this.
If you want to have a high probability for any every black box function it will have to deal with a distribution with either full rank or rank n-1 and that specific case of rank n-1 with no dependent set of rows smaller than n-2 which looks hard.
I have found a paper on randomized algorithms for computing the rank of a matrix here:
[www.emis.de/journals/ELA/ela-articles/articles/vol11\_pp16-23.ps](http://www.emis.de/journals/ELA/ela-articles/articles/vol11_pp16-23.ps)
| 8 | https://mathoverflow.net/users/1098 | 6197 | 4,218 |
https://mathoverflow.net/questions/6168 | 3 | Is there a closed form for $E(Y^n)$, where $Y$ is a random variable with a gamma distribution with parameters $\alpha$, $\beta$?
| https://mathoverflow.net/users/1646 | Expected value of a gamma-distributed random variable to the n-th power? | If the shape parameter is $\alpha$ and the scale parameter $\beta$, then $E(Y^r) = \beta^r \Gamma(\alpha + r)/\Gamma(\alpha)$ for real $r > 0$.
| 1 | https://mathoverflow.net/users/136 | 6199 | 4,220 |
https://mathoverflow.net/questions/6139 | 12 | There is a [wikipedia article](https://en.wikipedia.org/wiki/Trace_diagram). There is [a paper by Elisha Peterson](https://arxiv.org/abs/0910.1362). I tried reading these but they don't seem to click for me.
Are there books or other resources for learning how to do linear algebra with trace diagrams?
| https://mathoverflow.net/users/812 | How can I learn about doing linear algebra with trace diagrams? | The best resource I can point a beginner to is the first few chapters of Stedman's book "Group Theory". He focuses on the specific example of 3-vector diagrams, and does a good job of including lots of sample calculations. Unfortunately, it's not available online. I have found Cvitanovic' book fascinating but tough to internalize. You might also try looking at some of the work of Jim Blinn (a compilation of his work is at <http://research.microsoft.com/pubs/79791/UsingTensorDiagrams.pdf>), which has a lot of examples worked out. Another text that I commonly referred to is "The classical and quantum 6j-symbols" (<https://books.google.com/books?id=mg8ISMd5mO0C>), although this is limited to a special case of the diagrams.
As for learning about the diagrams themselves, I think the only way to really get comfortable with it is to work out lots of examples. I filled endless chalkboards at the University of Maryland with the doodles... it's one of the fun parts of the subject. :)
The paper you mentioned is focused on the applications of diagrams to ideas in traditional linear algebra. I have not found any other source that focuses exclusively on this use of diagrams, although Cvitanovic' book (for example) mentions without proof that one of his equations corresponds to the Cayley-Hamilton Theorem. This is probably because many mathematicians do not see much use in reproving old results (particularly if one must learn new notation to do so). I personally feel that there is sufficient beauty and elegance (once the notation is understood) in diagrammatic proofs of these "old proofs" to make them interesting. I also think that a deeper understanding of diagrammatic techniques is a worthy goal in itself. Others have mentioned some of the existing applications.
The term "trace diagrams" originated in my thesis, so you won't find it in many published papers. I use it to mean the particular class of diagrams that are labeled by matrices. There are many other names. I first learned about them in the special case of "spin networks" (a special case), and Penrose has the strongest claim to historical priority, hence "Penrose tensor diagrams".
| 13 | https://mathoverflow.net/users/498 | 6203 | 4,223 |
https://mathoverflow.net/questions/6175 | 16 | Let $A$ be an associative algebra over a commutative ring $k$. I've read statements saying that Hochschild (co)homology is the "right" notion of (co)homology for associative algebras. When $A$ is projective over $k$, the Hochschild cohomology, say, can be written as $Ext^\*\_{A \otimes A^{op}}(A,A)$, where $A^{op}$ is the opposite algebra, i.e., $A$ with $xy$ redefined to be $yx$.
On the other hand, when $A$ is augmented, the ext-group $Ext^\*\_A(k,k)$ is also referred to as the cohomology of $A$. What is the difference between these two notions of cohomology, and why would I choose one over the other?
| https://mathoverflow.net/users/1800 | Cohomology of associative algebras | For good algebras, Hochschild cohomology computes ‘all’ other interesting cohomologies. For example, it is already in Cartan-Eilenberg that if $M$ and $N$ are left $A$-modules, then $\mathrm{Ext}\_A^\bullet(M,N)=H^\bullet(A,\hom(M,N))$, where on the right $H^\bullet(A,\mathord-)$ is Hochschild cohomology with coefficients, and $\hom(M,N)$ is the space of homomorphisms over the base field turned into an $A$-bimodule using the left $A$-module structures of $M$ and $N$.
One general-nonsense explanation of the fact that Hochschild cohomology is somehow preferred is that an associative algebra is an algebra over the $\mathcal{A}ss$ operad, which is a Koszul operad, and a Koszul operad determines a canonical cohomology theory for its algebras: in the case of $\mathcal{A}ss$ you obtain in this way Hochschild cohomology. (Likewise, the operad $\mathcal{L}ie$ whose algebras are Lie algebras picks the usual Lie algebra cohomology, and $\mathcal{C}omm$, the operad of commutative algebras, picks the Harrison cohomology. This explanation breaks for, say, Hopf algebras—which are not the algebras of an operad—and then you do not have a clear winner among the cohomologies: then you have $\mathrm{Ext}\_H^\bullet(k,k)$ and the Gerstenhaber-Schack cohomology as alternatives, both quite ‘preferred’...)
| 14 | https://mathoverflow.net/users/1409 | 6212 | 4,231 |
https://mathoverflow.net/questions/6200 | 83 | I would like to know what quantization is, I mean I would like to have some elementary examples, some soft nontechnical definition, some explanation about what do mathematicians quantize?, can we quantize a function?, a set?, a theorem?, a definition?, a theory?
| https://mathoverflow.net/users/1651 | What is Quantization ? | As I'm sure you'll see from the many answers you'll get, there are lots of notions of "quantization". Here's another perspective.
Recall the primary motivation of, say, algebraic geometry: a geometric space is determined by its algebra of functions. Well, actually, this isn't quite true --- a complex manifold, for example, tends to have very few entire functions (any bounded entire function on C is constant, and so there are no nonconstant entire functions on a torus, say), so in algebraic geometry, they use "sheaves", which are a way of talking about local functions. In real geometry, though (e.g. topology, or differential geometry), there are partitions of unity, and it is more-or-less true that a space is determined by its algebra of total functions. Some examples: two smooth manifolds are diffeomorphic if and only if the algebras of smooth real-valued functions on them are isomorphic. Two locally compact Hausdorff spaces are homeomorphic if and only if their algebras of continuous real-valued functions that vanish at infinity (i.e. for any epsilon there is a compact set so that the function is less than epsilon outside the compact set) are isomorphic.
(From a physics point of view, it should be taken as a definition of "space" that it depends only on its algebra of functions. Said functions are the possible "observables" or "measurements" --- if you can't measure the difference between two systems, you have no right to treat them as different.)
So anyway, it can be useful to recast geometric ideas into algebraic language. Algebra is somehow more "finite" or "computable" than geometry.
But not every algebra arises as the algebra of functions on a geometric space. In particular, by definition the multiplication in the algebra is "pointwise multiplication", which is necessarily commutative (the functions are valued in R or C, usually).
So from this point of view, "quantum mathematics" is when you try to take geometric facts, written algebraically, and interpret them in a noncommutative algebra. For example, a space is locally compact Hausdorff iff its algebra of continuous functions is commutative c-star algebra, and any commutative c-star algebra is the algebra of continuous functions on some space (in fact, on its spectrum). So a "quantum locally compact Hausdorff space" *is* a *non*-commutative c-star algebra. Similarly, "quantum algebraic space" is a *non*-commutative polynomial algebra.
Anyway, I've explained "quantum", but not "quantization". That's because so far there's just geometry ("kinetics"), and no physics ("dynamics").
Well, a noncommutative algebra has, along with addition and multiplication, an important operation called the "commutator", defined by $[a,b]=ab-ba$. Noncommutativity says precisely that this operation is nontrivial. Let's pick a distinguished function H, and consider the operation $[H,-]$. This is necessarily a differential operator on the algebra, in the sense that it is linear and satisfies the Leibniz product rule. If the algebra were commutative, then differential operators would be the same as vector fields on the corresponding geometric space, and thus are the same as differential equations on the space. In fact, that's still true for noncommutative algebras: we define the "time evolution" by saying that for any function (=algebra element) f, it changes in time with differential [H,f]. (Using this rule on coordinate functions defines the geometric differential equation; in noncommutative land, there does not exist a complete set of coordinate functions, as any set of coordinate functions would define a commutative algebra.)
Ok, so it might happen that for the functions you care about, $[a,b]$ is very small. To make this mathematically precise, let's say that (for the subalgebra of functions that do not have very large values) there is some central algebra element $\hbar$, such that $[a,b]$ is always divisible by $\hbar$. Let $A$ be the algebra, and consider the $A/\hbar A$. If $\hbar$ is supposed to be a "very small number", then taking this quotient should only throw away fine-grained information, but some sort of "classical" geometry should still survive (notice that since $[a,b]$ is divisible by $\hbar$, it goes to $0$ in the quotient, so the quotient is commutative and corresponds to a classical geometric space). We can make this precise by demanding that there is a vector-space lift $(A/\hbar A) \to A$, and that $A$ is generated by the image of this lift along with the element $\hbar$.
Anyway, so with this whole set up, the quotient $A/\hbar A$ actually has a little more structure than just being a commutative algebra. In particular, since $[a,b]$ is divisible by $\hbar$, let's consider the element $\{a,b\} = \hbar^{-1} [a,b]$. (Let's suppose that $\hbar$ is not a zero-divisor, so that this element is well-defined.) Probably, $\{a,b\}$ is not small, because we have divided a small thing by a small thing, so that it does have a nonzero image in the quotient.
This defines on the quotient the structure of a *Poisson algebra*. In particular, you can check that $\{H,-\}$ is a differential operator for any (distinguished) element $H$, and so still defines a "mechanics", now on a classical space.
Then **quantization** is the process of reversing the above quotient. In particular, lots of spaces that we care about come with canonical Poisson structures. For example, for any manifold, the algebra of functions on its cotangent bundle has a Possion bracket. "Quantizing a manifold" normally means finding a noncommutative algebra so that some quotient (like the one above) gives the original algebra of functions on the cotangent bundle. The standard way to do this is to use Hilbert spaces and bounded operators, as I think another answerer described.
| 133 | https://mathoverflow.net/users/78 | 6216 | 4,235 |
https://mathoverflow.net/questions/6217 | 14 | To motivate my question, I will describe a related problem and then give a solution to it. My question will then be a variant of this problem.
N individuals sit around a table and want to compute the average of their salaries. They wish to do this in a manner such that no private information is leaked. This is to say no one obtains any information (regarding the other's salaries) that he couldn't deduce from the public information.
More formally we assume: (1) all of the salaries are non-negative integers bounded by B (2) everyone behaves honestly and doesn't attempt to halt the process (3) no subset of individuals will collude (4) there are secure private lines of communications between all participants (5) all of this information is well known (6) there is no outside trusted party.
Question 1: Is it possible for the N individuals to collectively compute the average without leaking any information? We say information is leaked if any individual has any information at the end of the process regarding anyone else's salary that he couldn't have deduced from knowing his own salary and the average.
The answer is Yes. It suffices to compute the sum of the salaries. Set S = 10\*N\*B. Now the first individual (Alice) chooses a uniformly random number between 0 and S-1 and adds this to her salary mod S. She then passes the sum to her neighbor, Bob, who adds his salary. This continues around the table until Zoey (the last participant) passes the number back at Alice. Alice subtracts off the random number and announces the sum to the group.
Here are two related questions:
Question 2: Is it possible for the group to compute the maximum salary (subject to the constraints above) without leaking any information?
Question 3: Can we remove the assumption that a bound on the size of the salaries are known in advance from the algorithm given above.
Additional Note: In Question 2 we want to compute only the maximum without providing any other information. One can note that, say, the entire distribution of salaries could be computed and communicated to the group by computing moments of the sequence via the method above. This would give the maximum (however a lot of other information as well).
| https://mathoverflow.net/users/630 | Computing the maximum salary | These questions (and many others) are studied in the literature under the heading of secret-sharing or common-knowledge protocols. A nice but short review appears in chapter 4 of David Gale's "Tracking the Automatic Ant".
The "sum protocol" you presented can be modified to determine how many people have salary x (without revealing their identity). Just have each participant communicate 0 or 1 according to whether or not she has salary x, and by scanning all x's in the (presumably known) range of salaries you can learn the distribution, as well as the maximum (by scanning down from some upper bound). However, this protocol reveals not only the max salary, but also how many people earn that max.
Such protocols are called $t$-private if they do not reveal any additional information unless $t$ people 'collude' and discuss their knowledge with each other. The protocol you mentioned is, in fact, $n$-private - unless everyone cooperates, they are all in the dark (EDIT: this is false, of course, as pointed out in the comments. The correct $n$-private protocol is described below). The sum is, essentially, the only function that can be computed $n$-privately. The maximum (without the extra knowledge of how many people earn it), the product etc. can all be computed $t$-privately for $t < n/2$, but not for $t \geq n/2$. The existence is proved by Ben-Or, Goldwasser and Wigderson in STOC 1988; the non-existence by Chor and Kushilevitz (STOC 1989) for Boolean functions and by Beaver for general integer-valued functions. This is all extracted from Gale's book.
How to compute the sum $n$-privately: Each person breaks up their salary into a sum of $n$ numbers, chosen at random except for the constraint that their sum is $n$. Each person now communicates the $j$th part to the $j$ person (including "communicating" one of the parts to herself). They then all announce the sums of all the pieces that were communicated to them. It's a fun exercise to show that no $k$ people can figure anything out other than whatever can be derived from their own salaries, for any $k < n$.
| 12 | https://mathoverflow.net/users/25 | 6219 | 4,236 |
https://mathoverflow.net/questions/6222 | 15 | What can one say about the set of continued fractions $[0;a\_1,a\_2,\ldots]$, where $a\_1,a\_2,\ldots$ are a *permutation* of the set of natural numbers?
| https://mathoverflow.net/users/824 | Continued fractions using all natural integers | Yeah, it has measure 0 by what Qiaochu said, although I strongly suspect there's probably a more elementary way to prove this (or at least give a strong heuristic argument.)
It's also uncountable, which means that in particular: 1. There exist numbers of this type that are transcendental, 2. There exist numbers of this type that are uncomputable, 3. It's very very very very unlikely that you'll be able to come up with a nontrivially different characterization of them.
They don't form any algebraic structure of any significance as far as I can tell (not a field, not a ring, I'm 100% sure but not completely convinced not a group), they're obviously not dense in the reals...
In essence, these are pretty much a completely typical uncountable set of measure 0, as far as I can tell.
| 6 | https://mathoverflow.net/users/382 | 6228 | 4,242 |
https://mathoverflow.net/questions/6248 | 3 | I've been attending a series of lectures on Cryptography from an engineering perspective, which means that most of the assertions made are supplied without proof... here's one that the lecturer couldn't recall the reason for, nor original source of.
Given an unfactored $n=pq$, computing $\phi(n)$ is as hard as finding $p,q$; this is the key idea of various "RSA-like" cryptosystems. One presented had a step in which for a secret $k$ and a random $t$, $k-t\phi(n)$ is transmitted. The claim was then that this process should only be applied once, as if an attacker sees $k-t\phi(n)$ and $k-u\phi(n)$ then they can recover $(t-u)\phi(n)$, and it's alleged that this makes it easier to compute $\phi(n)$.
So my question is, why is it easier to compute $\phi(n)$ given a random multiple of it, assuming we're at "cryptographic size"? (that is, $p,q$ sufficiently large that it's not feasible to try and factor $n$ and $\phi(n)$)
| https://mathoverflow.net/users/987 | Recovering $\Phi(n)$ from a multiple? | While I cannot immediately see an easy way to find Φ(n) from (t-u)Φ(n), assuming t-u is also of "cryptographic size" of course, an attacker will probably not have to.
Depending on how RSA-like the cryptosystem is, knowing a multiple of Φ(n) might well be enough to decrypt. After all, given the public exponent e, the private key in RSA is only an integer d with the property that ed = 1 mod Φ(n). Knowing (t-u)Φ(n) will allow the attacker to find an integer d satisfying ed = 1 mod (t-u)Φ(n) as long as gcd(t-u,e)=1. This d will also satisfy ed = 1 mod Φ(n), so it will work as a private key.
| 13 | https://mathoverflow.net/users/1972 | 6254 | 4,260 |
https://mathoverflow.net/questions/6227 | 15 | I hope I'm using the terminology correctly. What I mean is this: fix $K = \mathbb{R}$ or $\mathbb{C}$ (I'm interested in both cases). Which topological spaces $X$ have the property that for every open set $U$, every continuous function $f : U \to K$ is a quotient of continuous functions $\frac{g}{h}$ where $g, h : X \to K$ and $h \neq 0$ on $U$?
| https://mathoverflow.net/users/290 | Which topological spaces have the property that their sheaves of continuous functions are determined by their global sections? | This isn't a complete answer, but I think that whatever the family is, it contains compact metric (metrisable) spaces. With a paracompactness argument, I suspect that it would extend to locally compact, and I would not be surprised if one could replace "metrisable" by something weaker (though I think that it would need that separation property one-above-normal which I can never remember the name of: namely that every closed set is the zero set of a continuous function).
Here's a proof (I hope): Let $M$ be a compact metric space, $U \subseteq M$ an open subset, $f : U \to \mathbb{R}$ a continuous function. Let's write $K$ for the complement of $U$ in $M$. For each $n \in \mathbb{N}$, let $C\_n \subseteq U$ be the subset consisting of points at least distance $1/n$ away from $K$. Then $C\_n$ is closed in $M$, hence compact, and $\bigcup C\_n = U$. Let $h\_0 : M \to \mathbb{R}$ be the "distance from $K$" function (so that $C\_n = h\_0^{-1}([1/n,\infty))$). Let $V\_n$ be the complement of $C\_n$.
As $C\_n$ is compact, $f$ is bounded on $C\_n$. Let $a\_n = \max\{|f(x)| : x \in C\_n\}$, then $(a\_n)$ is an increasing sequence. Let $(b\_n)$ be a decreasing sequence that goes to $0$ faster than $(a\_n)$ increases, specifically that $(a\_nb\_n) \to 0$. Let $r : [0,\infty) \to [0,\infty)$ be a continuous decreasing function such that $r(1/n) = b\_{n+1}$ (as $(b\_n) \to 0$ (this always exists) and let $h = r \circ h\_0$. Then for $x \in V\_{n-1}$, $h\_0(x) \lt 1/(n-1)$ so $h(x) \lt b\_n$.
Then $h : M \to \mathbb{R}$ is a continuous function. Moreover, $h f$ (the product, with $h$ restricted to $U$) has the property that for $x \in C\_n \setminus C\_{n-1} = V\_{n-1} \setminus V\_n$,
$$
|(f h)(x)| = |f(x)| |h(x)| \le a\_n b\_n
$$
Thus as $x \to K$, $(f h)(x) \to 0$ and so we can extend $f h$ to a continuous function $g : M \to \mathbb{R}$ by defining it to be $0$ on $K$.
Then on $U$, $f = g/h$.
(I made this up, so obviously, there may be something I've overlooked in this so please tell me if I'm not correct.)
**Edit:** This one's been bugging me all weekend. I've even gone so far as to look up [perfectly normal](http://en.wikipedia.org/wiki/Perfectly_normal_space).
This property holds for perfectly normal spaces. In a perfectly normal space, every closed set is the zero set of a function (to $\mathbb{R}$, and this characterises perfectly normal spaces according to Wikipedia).
Here's the proof. Let $X$ be a perfectly normal space. Let $U \subseteq X$ be an open set, and $f : U \to \mathbb{R}$ a continuous function. Let $r : X \to \mathbb{R}$ be such that the zero set of $r$ is the complement of $U$. Let $s : \mathbb{R} \to \mathbb{R}$ be the function $s(t) = \min\lbrace 1, |t|^{-1}\rbrace$.
The crucial fact is that if $p : U \to \mathbb{R}$ is a **bounded** function then the *pointwise* product $r \cdot p : U \to \mathbb{R}$ (technically, $p$ should be restricted to $U$ here) extends to a continuous function on $X$ by defining it to be zero on $X \setminus U$.
From this, the rest follows easily.
1. The composition $s \circ f$ is bounded on $U$, hence $r \cdot (s \circ f)$ extends to a continuous function on $X$, say $h$.
2. The product $(s \circ f) \cdot f$ is also bounded on $U$, since $(s \circ f)(x) = \min\lbrace 1, |f(x)|\rbrace)$. Hence $r \cdot (s \circ f) \cdot f$ extends to a continuous function on $X$, say $g$.
3. As $s(t) \ne 0$ for all $t \in \mathbb{R}$, $(s \circ f)(x) \ne 0$ for all $x \in X$. Hence $h(x) \ne 0$ for all $x \in U$.
4. Finally, on $U$, $g(x) = h(x) \cdot f(x)$, whence, as $h$ is never zero on $U$, $f = g/h$ as required.
This isn't a complete characterisation of these spaces. Essentially, this result holds if there are **enough** continuous functions (as above) on $X$ and if there are too few.
As an example of the latter, consider a topological space $X$ where every pair of non-trivial open sets has non-empty intersection. Then there can be no non-constant functions to $\mathbb{R}$, either on $X$ or on any open subset thereof. Hence every continuous function on an open subset of $X$ trivially extends to the whole of $X$.
However, there's probably some argument that says that once you have sufficient continuous functions (say, if the space is functionally Hausdorff - i.e. continuous functions to $\mathbb{R}$ separate points) then it would have to be perfectly normal. The difficulty I have with making this into a proof is that there's no requirement that the function $g$ be zero on the complement.
Finally, note that metric spaces are perfectly normal so this supersedes my earlier proof. I leave it up, though, in case it's of use to anyone to see the workings as well as the current state. (Actually, for the record I ought to declare that initially I thought that this was *false* for almost all spaces. However, once I'd examined my counterexample closely, I realised my error and now I'm having difficulty thinking of a *reasonable* space where it does not hold.)
| 6 | https://mathoverflow.net/users/45 | 6257 | 4,263 |
https://mathoverflow.net/questions/6250 | 26 | Are there any good examples of theorems in reasonably expressive theories (like Peano arithmetic) for which it is substantially easier to prove (in a metatheory) that a proof exists than it is actually to find the proof? When I say "substantially easier," the tediousness of formalizing an informal proof should not be considered. In other words, a rigorous but informal proof doesn't count as a nonconstructive demonstration that a formal proof exists, since there is no essential difficulty in producing a formal proof besides workload. If there is some essential barrier, then there's something wrong either with the proof or with the definition of "formal proof." (Although the concept of "informal proof" is by nature vague, it seems reasonable to say that the procedure transforming a typical informal proof to a formal proof is a primitive recursive computation, even though it's probably impossible to make that precise enough to prove.)
I don't know what such a nonconstructive meta-proof could look like, but I also don't see why one couldn't exist. The only near-example I can think of right now is in the propositional calculus: you can prove that a propositional formula is a theorem by checking its truth table, which does not explicitly provide a deduction from the axioms; however, in most interesting cases, that probably isn't much easier (if at all) than exhibiting a proof, just maybe more mechanical. (Of course, estimating the difficulty of proving or disproving a candidate for a propositional theorem is a huge open problem.) Also, I think there is actually a simple way to convert a truth-table proof into an actual deduction from the axioms of propositional calculus, although I don't remember and didn't retrace all of the details. (Anyway, I'm not even so interested in the propositional calculus for this question, since it's not very expressive.)
Another thing that seems vaguely relevant is Godel's completeness theorem, which states that a formula is a theorem of a first-order theory if and only if it is true in every model of that theory. I don't know in what cases it would be easier, though, to show that some formula were true in every model than it would be just to prove the theorem in the theory.
| https://mathoverflow.net/users/302 | Are there any good nonconstructive "existential metatheorems"? | Set theory provides a good example. It is often convenient in set theory to work with the concept of "classes" and treat them as mathematical objects of their own kind. The standard axiomatization of set theory with classes is called Goedel-Bernays set theory, denoted GBC, whereas the usual ZFC axioms have only set objects. But there is a general theorem that any statement purely about sets that is proved by using classes in GBC can be proved without them purely in ZFC. This is what it means to say that GBC is a [*conservative* extension](https://en.wikipedia.org/wiki/Conservative_extension) of ZFC. To prove this general theorem, it suffices to observe that any ZFC model can be expanded to a model of GBC, by adding only the classes definable from parameters.
There are many other examples of conservative extensions in mathematics, and all of them would seem to be examples of the type that you seek. For example, PA has a conservative extension to the analogous theory true in the collection HF of hereditarily finite sets. Thus, to prove a theorem about numbers in PA, one can freely use heredtiarily finite sets (e.g. sequences of numbers, sequences of sets of numbers, etc.), not just as coded by numbers via Goedel coding, but as actual mathematical objects. And surely this makes proofs much easier.
Perhaps another type of example arises in the absoluteness phenomenon in set theory. For example, the [Shoenfield Absoluteness theorem states](https://en.wikipedia.org/wiki/Absoluteness#Shoenfield.27s_absoluteness_theorem) that any $\Sigma^1\_2$ statement has the same truth value in any two models of set theory with the same ordinals. In particular, to prove that a particular $\Sigma^1\_2$ statement is true in ZFC, it suffices to prove it under the assumption also that V=L, where one also has all kinds of additional structure available. The Absoluteness Theorems (and there are many) can therefore be viewed under the rubric you mention.
But of course, in all these cases, we have an actual proof in the weaker theory. To prove that there is a proof, is a proof, so I believe ultimately there will be no way to avoid the quibbling over whether it is easy or hard to translate the high-level proof into a low level proof, since in principle it will always be possible to do this.
| 26 | https://mathoverflow.net/users/1946 | 6268 | 4,272 |
https://mathoverflow.net/questions/6278 | 12 | Suppose $\cal{X}$ is a DM-stack, and *X* its coarse moduli space. Let *F* be a sheaf on $\cal{X}$, and $\pi : \mathcal{X} \to X$ the projection. In all examples I have seen, it has been true that
$H^i(\mathcal{X},F) = H^i(X,\pi\_\ast F)$.
Is there a simple example where this fails? Are there easy conditions where this is true?
| https://mathoverflow.net/users/1310 | When can cohomology be calculated on the coarse moduli space? | Let k be a field and your DM-smack be [Spec(k)//G] for a trivial action of a group G. A sheaf on this stack is roughly a sheaf with a group action, and cohomology is group cohomology. If you consider a sheaf where multiplication by |G| is invertible, then group cohomology vanishes in high degrees, but otherwise there is a lot of information not coming from the cohomology of Spec(k).
EDIT: Under amenable circumstances when the left adjoint $\pi^\\*$ is exact, you can get a sufficient condition for your isomorphism to hold. (I'm not sure what kind of topology you want to work with, or whether you're working with simply sheaves of abelian groups or sheaves of $\mathcal{O}$-modules or quasicoherent modules or...) In this case, $\pi\\_\\*$ preserves injectives and so there's a Grothendieck spectral sequence
$$
H^p(X, {\mathbb R}^q \pi\\_\* F) \Rightarrow H^{p+q}(\mathcal{X},F)
$$
and so a sufficient condition is for the higher direct image functors $\mathbb{R}^q \pi\\_\* F$ to vanish for q > 0. These are the sheaves associated to the presheaves $U \mapsto H^q(\pi^{-1} U, F)$, and often their stalks are related to the cohomology of the fibers of $\pi$ - which are, in this case of a DM-stack, pretty much the group cohomology of the stabilizers.
| 11 | https://mathoverflow.net/users/360 | 6286 | 4,284 |
https://mathoverflow.net/questions/6287 | 1 | Let $A$ be a ring and $E$ a module. If $\mathrm{spec} A$ is connected, then so is $\mathrm{spec} S^\bullet E$. If this is not true in general, then what are some minimal conditions that make it true?
| https://mathoverflow.net/users/21 | Is the total space of a module connected? | If I'm not mistaken the geometric fibers are nonempty affine spaces (hence connected, even if E is the zero module), and the augmentation gives you a zero section everywhere. I think that means the answer is yes.
| 2 | https://mathoverflow.net/users/121 | 6291 | 4,288 |
https://mathoverflow.net/questions/6279 | 10 | The graduate students here at MIT have been thinking about questions like the following: Over $\mathbb{F}\\_q$, how many symmetric matrices are there with nonzero determinant and $0$'s on the diagonal? They are doing brute force computer searches; checking every point in $\mathbb{A}^N(\mathbb{F}\\_q)$ and seeing whether it is in the variety.
Is there a better way? Their examples are, like the above, low codimension subvarieties of high dimensional affine spaces. I think they are usually looking at smooth varieties, but I wouldn't swear to it in every case. The ideal answer, of course, would be preexisting software.
| https://mathoverflow.net/users/297 | Counting points on varieties of low codimension | For a general variety and for a fixed small value of $q$, there isn't going to be a very good algorithm. That is because you can encode a Boolean formula in a single polynomial equation with minimal overhead. You are therefore counting solutions to a general logical expression, which is not only #P-hard, but also morally there is often nothing much better than exhaustive search.
On the other hand, if $q$ grows for a fixed formula, then you can use zeta function facts that algebraic geometers and number theorists know or conjecture to extrapolate from small values of $q$. For instance if $q = p^k$, you can use the Weil conjectures.
Since your example variety is far from general, you can try to exploit special structure to chip away at an exponential search (or count) and make it a better exponential search or count. I'll stick to your example problem, on the assumption that the others that they are looking at are similar.
First idea: There is a torus action on the set of symmetric matrices with vanishing diagonal. Thus you can assume that the first row and column is entirely 0s and 1s, and multiply by the size of the orbit. (Refinement: You can assume that the leading term in each row and column is 1.)
Second idea: Quadratic hypersurfaces in $\mathbb{A}^n$ are classified and you know how many points they have. So you do not have to complete the matrix; you can instead complete all but one row and column of the matrix. This idea combines with the first idea.
Third idea, much better than the other two: For each vector subspace $V \subseteq \mathbb{A}^n$, you can count the symmetric matrices which vanish on the diagonal and which annihilate $V$ (and maybe other vectors), because the equations for that are linear. You can then apply Möbius inversion on the lattice of subspaces to count the matrices that do not annihilate any vectors. You can loop over the vector spaces $V$ by assuming the unique basis in RREF form. Moreover, many subspaces with the same RREF pivot positions have to give you the same answer, for one reason because of the torus action.
| 9 | https://mathoverflow.net/users/1450 | 6295 | 4,291 |
https://mathoverflow.net/questions/6292 | 20 | Seriously. As an undergrad my thesis was on elliptic curves and modular forms, and I've done applied industrial research that invoked toric varieties, so it's not like I'm a partisan here. But this can't be a representative cross-section of mathematical questions. How can this be fixed? (I mean the word "fixed".)
| https://mathoverflow.net/users/1847 | Why is algebraic geometry so over-represented on this site? | I agree that the founder effect is a significant factor, but I also have another (more crackpot-ish) theory. I think some disciplines, like algebraic geometry, are harder to pick up in a traditional classroom setting, or out of a book. In practice, they are more often learned like a language, through repeated exposure and watching other people do it. Of course, to some extent this is true for most disciplines, but I think its more true for algebraic geometry than for analogously hard disciplines. If you grant my point, then there would be an over-representation of these disciplines on the internet, because they are looking for explanations and intuition unavailable in technical contexts. I think a similar statement is true for category theory, for instance.
| 17 | https://mathoverflow.net/users/750 | 6297 | 4,293 |
https://mathoverflow.net/questions/6281 | 50 | When I think of a "simplicial complex", I think of the geometric realization of a simplicial set (a simplicial object in the category of sets). I'll refer to this as "the first definition".
However, there is another definition of "simplicial complex", e.g. [the one on wikipedia](http://en.wikipedia.org/wiki/Simplicial_complex): it's a collection $K$ of simplices such that any face of any simplex in $K$ is also in $K$, and the intersection of two simplices of $K$ is a face of both of the two simplices. There is also the notion of "[abstract simplicial complex](http://en.wikipedia.org/wiki/Abstract_simplicial_complex)", which is a collection of subsets of $\{ 1, \dots, n \}$ which is closed under the operation of taking subsets. These kinds of simplicial complexes also have corresponding geometric realizations as topological spaces. I'll refer to both of these definitions as "the second definition".
The second definition looks reasonable at first sight, but then you quickly run into some horrible things, like the fact that triangulating even something simple like a torus requires some ridiculous number of simplices (more than 20?). On the other hand, you can triangulate the torus much more reasonably using the first definition (or alternatively using the definition of "Delta complex" from Hatcher's algebraic topology book, but this is not too far from the first definition anyway).
I believe you can move back and forth between the two definitions without much trouble. (I think you can go from the first to the second by doing some barycentric subdivisions, and going from the second to the first is trivial.)
Due to the fact that the second definition is the one that's listed on wikipedia, I get the impression that people still use this definition. My questions are:
1. Are people still using the second definition? If so, in which contexts, and why?
2. What are the advantages of the second definition?
| https://mathoverflow.net/users/83 | Definition of "simplicial complex" | Simplicial sets and simplicial complexes lie at two ends of a spectrum, with Delta complexes, which were invented by Eilenberg and Zilber under the name "semi-simplicial complexes", lying somewhere in between. Simplicial sets are much more general than simplicial complexes and have the great advantage of allowing quotients and products to be formed without the necessity of subdivision, as is required for simplicial complexes. In this way simplicial sets are like CW complexes, only more combinatorial or categorical. The price to pay for this is that simplicial sets are perhaps less geometric, or at least not as nicely geometric as simplicial complexes. So the choice of which to use may depend in part on how geometric the context is. In some areas simplicial sets are far more natural and useful than simplicial complexes, in others the reverse is true. If one drew a Venn diagram of the people using one or the other structure, the intersection might be very small.
Delta complexes, being something of a compromise, have some of the advantages and disadvantages of each of the other two types of structure. When I wrote my algebraic topology book I had the feeling that Delta complexes had been largely forgotten over the years, so I wanted to re-publicize them, both as a pedagogical tool in introductory algebraic topology courses and as a sort of structure that arises very naturally in many contexts. For example the classifying space of a category is a Delta complex.
Incidentally, I've added 5 pages at the end of the Appendix in the online version of my book going into a little more detail about these various types of simplicial structures. (I owe a debt of thanks to Greg Kuperberg for explaining some of this stuff to me a couple years ago.)
| 75 | https://mathoverflow.net/users/23571 | 6302 | 4,298 |
https://mathoverflow.net/questions/6303 | 8 | I am currently following a course on differential equations and difference equations (recurrence relations).
The teacher tries to make parallels between the two concepts, because the methods for solving both of these kinds of equations are essentially the same(sub-question : is there a deeper reason of why this is so?).
For example, to find solutions for linear differential equations or linear difference equations when the coefficients are constants, you find the roots of the caracteristic polynomial.
The teacher introduced the Wronskian
$$W(f,g)=
\begin{vmatrix}
f& g\\\\ f'&g'\\\\
\end{vmatrix}$$
That is used to know if two solutions of a differential equation are linearly independent (it is zero if they are dependent, and everywhere non-zero if not)
He told us that the equivalent of the Wronskian for difference equations was the "Carosatian" (my course is in french, the exact term was "Carosatien") which is the determinant:
$$C\_n(x,y)=
\begin{vmatrix}
x\_n & y\_n \\\\ x\_{n+1} & y\_{n+1}\\\\
\end{vmatrix} $$
for two sequences x, y. It works in the same way, in that this Carosatian is 0 only if x and y are linearly dependent.
When I search google for carosatian or carosatien, I get exactly 0 results, which is very surprising usually for things that actually exist. I was wondering if there was a more popular name for this concept?
(edit : I didn't get the matrices to work, but they're supposed to be 2x2)
(edit 2 : I got a very fast answer, but I would still be very happy to get an answer as to why are the two kinds of equations solvable with the same methods)
| https://mathoverflow.net/users/1619 | What is the term analogous to "Wronskian" for difference equations? | You mean "Casoratian"?
| 14 | https://mathoverflow.net/users/441 | 6304 | 4,299 |
https://mathoverflow.net/questions/4661 | 16 | [The Sylvester-Gallai theorem](http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem) asserts that for every collection of points in the plane, not all on a line, there is a line containing **exactly** two of the points.
One high dimensional extension asserts that for every collection of points not all on a hyperplane in a d-dimensional space there is a [d/2]-space L whose intersection with the collection is a spanning set of cardinality [d/2]+1
**My question is:** Given a k-dimensional real algebraic variety V [perhaps of a certain kind] embedded in n-dimensional affine space (whose image spans this space) can one find an affine r-dimensional space L so that $V \cap L$ spans L and is topologically "simple".
**Remarks:** 1) For smooth complex varieties the Lefschetz hyperplane theorem describes sort of the opposite phenomenon: the homology of the hyperplane section is (more or less) as complicated as the homology of the original manifold all the way to half the dimension.
2) There is an analog of the Sylvester-Gallai Theorem over the complex numbers (there, if the points are not colinear you can always find a line containing 2 or 3 points, and if the points are not coplanar you can find a line containing precisely two points. The later statement was a conjecture by Serre first proved by Kelly. See [this post](http://konradswanepoel.wordpress.com/2007/12/12/complex-and-hypercomplex-sylvester-gallai-theorems/) in Konard Swanepoel's Blog).
So the difference between real AG and complex AG is not the only issue at hand (another issue seems to be how reducible the manifold is), and we can ask for sections with simple topology for complex varieties as well. Are there any results known in this direction?
3) Note that the whole point in the Sylvester-Gallai theorem and the proposed generalization is about non-generic embeddings and about intersection with non-generic flats.
However, the behavior of "generic embedding" (regarding intersection with non-generic flats) is sort of a role for what we expect for non-generic embedding. One difficulty is that *I am not aware of a definition of "generic embedding" of a real algebraic semi variety in a large Euclidean space.* Is there such a definition?
---
Let me formulate an open problem (and a few variations) which takes into account the discussion so far. We say that an embedding of a real semi algebraic variety V into an Euclidean space is genuine if its image span the space.
**Problem** (special case): Let V be a 2-dimensional variety genuinly embedded into n-space. (n cannot be too small, but maybe n=4 will suffice.) Then there is a plane L whose intersection with V is 1-dimensional, $V \cap L$ affinely span V, and $V \cap L$ is of very restricted topological type. (Maybe belong to a finite list of homotopy types.)
**Problem** (general case): The same conclusion (V \cap L belongs to a small set of homotopy types) when V is k dimensional, L is r dimensional, the intersection between V and L is j-dimensional and the dimension of the ambient space n is sufficiently large (but perhaps being moderately large suffices.) Greg's example shows that n cannot be too small. (For k=1,r=3,j=1 we cannot take n=4.) We know also that for k=j=0 we need n>=2r.
**Problem** (special case) Consider the case where V is an arrangement of subspaces.
**Problem** (complex analog) Consider the case that V is a complex variety; (and the special case of an arrangement of subspaces).
| https://mathoverflow.net/users/1532 | The Sylvester Gallai Theorem and Sections of Varieties with "Simple Topology". | There are many cases of the question as stated that follow quickly from the standard Sylvester-Gallai theorem. If $V$ is an $r$-dimensional variety, then its intersection with a generic $(n-r)$-plane is a finite set of points. You can then apply the standard Sylvester-Gallai theorem, or the high-dimensional generalization stated here.
There are cases where nothing can be said for singular varieties. As a warm-up, let's consider a set which is not an algebraic variety but a union of line segments. Then it could be the union of all of the interior diagonals of a convex polytope $P$ with complicated facets. For instance you could take all of the interior diagonals of the Cartesian product of two $n$-gons. Any 3-plane that intersects $P$ 3-dimensionally has to intersect many of the edges.
A line segment is not a real algebraic variety. However, it can be replaced by a thin needle with cusps at the ends that is a real algebraic variety. You can replace all of the diagonals with these needles, as long as you skip the edges of $P$ itself, and the result will still lie in the convex hull of $P$.
A needle of this type can have a cross-section of any dimension and very complicated topology. If you asked for a hyperplane that specifically intersects in more than a finite set, then the diagonal-needle construction can force a lot of topology.
You could specifically look at non-singular varieties. I don't have a rigorous result here, but the smooth restriction would make it difficult to avoid hyperplanes that do something at the boundary of the convex hull of $V$. The Sylvester-Gallai theorem is more about things that have to happen in the interior if they do not happen at the boundary of the convex hull.
You could bound the degree of the variety $V$. Then a simple compactness argument bounds the complexity of intersects, and there are a lot of interesting bounds on the topology of $V$ itself. But that also goes against the spirit of Sylvester-Gallai, because the number of points in that result is not bounded.
Maybe a more interesting variation is to keep a finite intersection, but replace the hyperplane with a $V$ with bounded degree. However, that is no longer the question posted.
The question is a bit open-ended. I can think of several constructions that seem to ask for a less open-ended question, or a question which is open-ended in a different way.
| 7 | https://mathoverflow.net/users/1450 | 6310 | 4,304 |
https://mathoverflow.net/questions/6332 | 6 | As is very well known, the algebraic variety $S^2$ is isomorphic to projective variety $\mathbb{CP}^1$ as a complex manifold. As is also well known, the coordinate ring of $S^2$ is given by $< x,y,z > / < x^2 + y^2 +z^2 - 1 >$ and the function field of $\mathbb{CP}^1$ is $\mathbb{C}(\mathbb{CP}^1)$ (its coordinate ring being $\mathbb{C}$). My question is how the coordinate ring of $S^2$ and the function field of $\mathbb{CP}^1$ are related? Presumably this relation is a special case of a general variety-complex manifold relationship.
| https://mathoverflow.net/users/1977 | The 2-sphere and $\mathbb{CP}^1$ | Any projective variety is also a real affine variety, by using the real and imaginary parts of the coordinates $x\_{jk} = z\_j\overline{z\_k}$. You should first normalize the projective coordinates to have Hermitian-Euclidean length 1. Ordinarily the projective coordinates are sections of a line bundle that is only defined up to a scalar. But with the length normalization, the only ambiguity is the phase. The phase cancels out in the definition of $x\_{jk}$, so it is a well-defined function rather than just a section.
This realification of $\mathbb{CP}^n$ is a map to $(n+1) \times (n+1)$ Hermitian matrices. It is important in quantum probability: The image of the map is the set of pure states of a quantum system; its convex hull is the set of all states. This convex region is the quantum analogue of the simplex of states (= measures = distributions) for classical probability on a finite set. The matrices have unit trace, so the image lies in an $(n^2+2n)$-dimensional real subspace. You can check that it is a sphere when $n=1$.
Another viewpoint that is important is that of toric varieties. The phase part of the toric action on $\mathbb{CP}^1$ is rotation about the $z$ axis, and the moment map is the projection onto the $z$ coordinate. This too generalizes to any projective toric variety.
| 10 | https://mathoverflow.net/users/1450 | 6335 | 4,319 |
https://mathoverflow.net/questions/6316 | 24 | Some background on GIT
----------------------
Suppose G is a reductive group acting on a scheme X. We often want to understand the quotient X/G. For example, X might be some parameter space (like the space of possible coefficients of some polynomials which cut out things you're interested in), and the action of G on X might identify "isomorphic things", in which case X/G would be the "moduli space of isomorphism classes of those things."
It happens that the quotient X/G often doesn't exist, or is in some sense bad. For example, if the closures of two G-orbits intersect, then those orbits must get mapped to the same point in the quotient, but we'd really like to be able to tell those G-orbits apart. To remedy the situation, the idea is to somehow remove the "bad locus" where closures of orbits could intersect. For reasons I won't get into, this is done by means of choosing a *G-linearized line bundle* L on X (i.e. a line bundle L which has an action of G which is compatible with the action on X). Then we define the *semi-stable* and *stable* loci
>
> Xss(L) = {x∈X|there is an invariant section s of some tensor power of L such that x∈Xs (the non-vanishing locus of s) and Xs is affine}
>
> Xs(L) = {x∈Xss(L)| the induced action of G on Xs is closed (all the orbits are closed)}
>
>
>
Note that Xss(L) and Xs(L) are G-invariant. Then the basic result is Theorem 1.10 of [Geometric Invariant Theory](http://books.google.com/books?id=dFlv3zn_2-gC&lpg=PP1&pg=PA38#v=onepage&q=&f=false):
>
> **Theorem.** There is a good quotient of Xss(L) by G (usually denoted X//LG, I believe), and there is a geometric (even better than good) quotient of Xs(L) by G. Moreover, L descends to an ample line bundle on these quotients, so the quotients are quasi-projective.
>
>
>
---
My Question
-----------
Are there some conditions you can put on G, X, L, and/or the action/linearization to ensure that the quotient X//LG or Xs(L)/G is projective?
Part of the appeal of the GIT machinery is that the quotient is automatically quasi-projective, so there is a natural choice of compactification (the projective closure). The problem is that you then have to find a modular interpretation of the compactification so that you can actually compute something. Is there some general setting where you know that the quotient will already be compact? If not, do you have to come up with a clever trick for showing that a moduli space is compact every time, or do people always use the same basic trick?
| https://mathoverflow.net/users/1 | When are GIT quotients projective? | I'm not sure if this is the sort of thing you are after but one can say the following.
Suppose we work over a base field $k$. If $X$ is proper over $k$ and the $G$-linearized invertible sheaf $L$ is ample on $X$ then the uniform categorical quotient of $X^{ss}(L)$ by $G$ is projective and so it gives a natural compactification of $X^s(L)/G$.
I believe this is mentioned in Mumford's book but no proof is given. It goes via considering $\operatorname{Proj}R\_0$ where $R\_0$ is the subring of invariant sections of
$$\bigoplus\_i H^0(X,L^{\otimes i})$$
and showing that this is the quotient.
This is stated in Theorem 1.12 [in these notes](http://www.cimat.mx/Eventos/c_vectorbundles/newstead_notes.pdf) of Newstead. I haven't found a proof written down - but either showing it directly or doing it by reducing to the case of projective space shouldn't be too hard I don't think.
| 6 | https://mathoverflow.net/users/310 | 6336 | 4,320 |
https://mathoverflow.net/questions/6343 | 16 | Let $\{A\_i\}$ be a collection of $m$ hyperplanes in $\mathbb{C}^n$ which all pass through the origin (a **central hyperplane arrangement**). Such an arrangement is called **Coxeter** if reflecting across any hyperplane in $\{A\_i\}$ sends the arrangement to itself (and so the reflections automatically will generate a Coxeter group).
Now, I will define a rather random-seeming condition on an arbitrary central arrangement. Choose a normal vector $n\_i$ to each $A\_i$. Consider the function on $\gamma$ on $\mathbb{C}^n$ given by
$$ \gamma(v) = \sum\_{1\leq i< j\leq m}(n\_i,n\_j)\left(\prod\_{k\neq i,j} (n\_k,v) \right) $$
The function $\gamma$ depends on the specific choice of normal vectors; however, whether $\gamma$ vanishes does not depend on the choice of normal vectors (since scaling a normal vector will scale the output). Call a central hyperplane arrangement **puzzling** if $\gamma(v)=0$ for all $v$.
The 'puzzling' condition came up in studying a very specific research problem. However, both the context and a day's worth of experimentation have lead to the following conjecture.
**Conjecture:** The puzzling arrangements are exactly the Coxeter arrangements.
It's worth noting that I can't show either direction of the conjecture. Brute force computation says that for $n=2$, the conjecture is true.
Just from the form, it kind of reminds me of a Weyl character formula-type identity, but I don't really know much about those. My hope is that this kind of identity is pretty well known to people who work with such things.
---
**Edit:** There's another way of stating this identity, that's closer to the context in which I encountered it (differential operators). Let $n\_i^\*$ denote the function $(n\_i,-)$, and let $d\_i$ denote differentiation along the vector $n\_i$. Then
$$ \gamma = \left( \sum\_i (n\_i^\*)^{-2}(n\_i^\*d\_i-(n\_i,n\_i))\right) \prod\_j n\_j $$
Therefore, the 'puzzling' condition is then a statement about the function $\prod\_jn\_j$ being killed by a particular rational differential operator.
---
**Edit 2:** Fixed the definition of Coxeter arrangements... I want all the reflections, not just a generating set.
| https://mathoverflow.net/users/750 | Coxeter Arrangements and an Identity | **Proof that Coxeter arrangements obey this identity:** Group together summands according to the two-plane spanned by $n\_i$ and $n\_j$. For any two-plane $H$, every summand coming from that two-plane is divisible by $\prod\_{n\_k \not \in H} \langle n\_k, v \rangle$. Factoring out this common summand, the contribution from $H$ is
$$\sum\_{n\_i \neq n\_j,\ n\_i, n\_j \in H} \langle n\_i, n\_j \rangle \prod\_{n\_k \in H,\ n\_k \neq n\_i, n\_j} \langle n\_k, v \rangle.$$
This is the two dimensional example you've already done.
**Proof that only Coxeter arrangements obey this identity:**
Consider $H$, a two plane spanned by some $(n\_i, n\_j)$. Our first goal is to show that $H \cap \{ n\_k \}$ is a dihedral root system.
Let $r$ be the number of hyperplanes in your arrangement. Let $S$ be the ring of polynomial functions and let $I$ be the ideal generated by the functions $\langle n, \ \rangle$, for $n \in H$. Note that every term of your sum which does not come from $(n\_i, n\_j)$ with $n\_i$, $n\_j \in H$ lies in $I^{r-1}$. So, the sum of the terms with $n\_i$, $n\_j \in H$ must be zero modulo $I^{r-1}$.
As before, all of those terms are divisible by $\prod\_{n\_k \not \in H} \langle n\_k, v \rangle$. This is not a zero divisor in $S/I^{r-1}$. So we can factor it out and deduce that
$$\sum\_{n\_i \neq n\_j,\ n\_i, n\_j \in H} \langle n\_i, n\_j \rangle \prod\_{n\_k \in H,\ n\_k \neq n\_i, n\_j} \langle n\_k, v \rangle \equiv 0 \ \mathrm{mod} \ I^{r-1}$$
But the left hand side is degree $r-2$, so it must be identically zero. By the two dimensional example which you have already done, this shows that $\{ n\_k: n\_k \in H \}$ is a root system.
So, for any $n\_i$ and $n\_j$, the set of $n\_k$ in the span of $(n\_i, n\_j)$ is a root system. In particular, the reflection of $n\_i$ by $n\_j$ is some $n\_k$. So your whole set of vectors is a root system.
**Warning:** I have not, myself, checked the two dimensional case which I am relying on.
| 15 | https://mathoverflow.net/users/297 | 6344 | 4,326 |
https://mathoverflow.net/questions/6179 | 15 | More precisely, does there exist an unbounded sequence $a\_0, a\_1, ... \in \mathbb{N}$ of primes such that the function
$\displaystyle O(z) = \sum\_{n \ge 0} a\_n z^n$
is meromorphic on $\mathbb{C}$?
[A previous version of the question also asked about the exponential generating function of $(a\_n)$. However, such a function can trivially be entire. - GJK]
| https://mathoverflow.net/users/290 | Does there exist a meromorphic function all of whose Taylor coefficients are prime? | Borel proved the following much stronger result: if a power series with integer coefficients
represents a function f(z) that is meromorphic in a disk of radius >1, then f(z) extends to a rational function on all of C. I found this result without a reference on page 3 of www.mathematik.uni-bielefeld.de/~anugadre/Adeles.pdf.
| 19 | https://mathoverflow.net/users/2807 | 6347 | 4,327 |
https://mathoverflow.net/questions/6346 | 1 | Notation question:
What does $(\Delta^1)^{ \{1, \ldots,n-1 \}}$ denote? **UPDATE**: I (David Speyer) tried to fix the LaTeX. Please see if I got it right.
Vocabulary question:
Suppose $z:\Delta^{n+1} \rightarrow S$ is a morphism of simplicial sets. What does the following translate to in algebraic terms: $z|\Delta^{ \{0,\ldots,n \} }$ is a constant simplex at a vertex $x$.
So mainly, I just don't know what that is supposed to mean, "is a constant simplex at the vertex x". Everything else makes fine sense.
I've searched through a number of books on homotopy theory, algebraic topology, etc. and I've been unable to find these precise usages.
I ask these questions only because I'm reading HTT by Lurie, and these usages come up and they're quite confusing.
| https://mathoverflow.net/users/1353 | Simplicial set notation and vocabulary question. | Okay, I've found the relevant notation in Higher Topos Theory. The first is at the end of 1.1.5.10, and is the simplicial set of maps from an n-1 element set to the 1-simplex (i.e., an n-1-cube). The second is at the end of warning 1.2.2.2, and describes a constant map of simplicial sets whose image is x. The curly braces give a reference to the specific ordered set that defines the the simplex.
A simplex at a vertex is a degenerate simplex. You take the simplicial subset defined by x, and demand that the map from your simplex to the target factor through the inclusion of x.
**Edit in response to comment:** You can think of vertices in (at least) two ways. One way is as an element of S0, i.e., a zero-simplex of the simplicial set. Another way is as a simplicial subset X of S, such that X0 is the chosen element of S0, and all Xi have a single element, namely the image of X0 under the unique degeneracy map. The statement is that the map Z takes a particular nondegenerate n-dimensional face of $\Delta^{n+1}$ to the unique element of Xn.
| 5 | https://mathoverflow.net/users/121 | 6349 | 4,329 |
https://mathoverflow.net/questions/4961 | 3 | I am looking for some simple and concrete -- but still non-trivial and illustrative -- applications of Atiyah-Bott localization in the context of equivariant cohomology.
Do you know any good ones?
| https://mathoverflow.net/users/83 | Simple applications of Atiyah-Bott localization | You want to read [A Lefschetz Fixed Point Formula for Elliptic Complexes: II. Applications](http://www.jstor.org/pss/1970721), Atiyah and Bott, Ann. of Math., Vol. 88, No. 3 (Nov., 1968), pp. 451-491
| 2 | https://mathoverflow.net/users/297 | 6372 | 4,346 |
https://mathoverflow.net/questions/6373 | 12 | First a little background for those unaware. The Stasheff polytopes (or associahedra) are certain convex polytopes that arise in the theory of $A\_\infty$-algebras. There is one polytope for each $n\geq 2$ and is denoted by $K\_n$. The $K\_n$'s essentially encode the homotopies, higher homotopies and so on of the associativity relation. One way to describe $K\_n$, which is has dimension $(n-2)$, is to take all rooted binary trees with $n$ leaves and take a *suitable* convex hull. For example, $K\_2=\{\ast\}$, $K\_3$ is an interval while $K\_4$ is a pentagon.
It is known that the number of vertices $v$ of $K\_n$ is the $(n-1)^{th}$ Catalan number, i.e., $v=\frac{1}{n}{2n-2 \choose n-1}$. What can one say about the number of edges of $K\_n$ and generally about counting faces of all codimension?
| https://mathoverflow.net/users/1993 | Combinatorics of the Stasheff polytopes | After a change of variables, the answer is [sequence A033282](http://www.research.att.com/~njas/sequences/A033282) in the Encyclopedia of Integer Sequences: $T(n,k)$ is the number of diagonal dissections of a convex $n$-gon into $k+1$ regions. The page gives the wonderful formula,
$$T(n,k) = \frac{1}{k+1}\binom{n-3}{k}\binom{n+k-1}{k},$$
for relevant values of $k$.
I find it easier to also keep track of the dual Stasheff polytope, which can be realized as a simplicial complex based on dissections of a convex polygon. The polytope $K\_n^\*$ has a vertex for each diagonal of an $(n+1)$-gon, and it has a face for every collection of disjoint diagonals. So, in terms of your original parameters, $K\_n$ has $T(n+1,n-2-k)$ faces of dimension $k$.
Also: One reason that I like the dual Stasheff polytope is that it has an amazing infinite generalization called the Hatcher-Thurston arc complex. You again take collections of disjoint arcs that connect marked points, but in the generalization you can take any surface with or without boundary, as long as it has at least one marked point total and at least one on each boundary component. (And I suppose in the disk case it needs at least three marked points.) Each isotopy class of arcs is a vertex, and each disjoint collection is a face. It is a combinatorial model of Teichmüller spaces or moduli spaces of curves (with the requisite marked points).
---
Gil Kalai in the comments asks for a few more details of the Hatcher-Thurston arc complex, and he gives a reference to one of the original papers, "On triangulations of surfaces, Topology Appl. 40 (1991), 189–194," by Allen Hatcher. Briefly: Suppose that $\Sigma$ is a fixed surface with some marked points. There should be enough marked points so that there exists at least one generalized triangulation of $\Sigma$ whose vertex set is the marked points. A question that could be taken as motivation is the following: Can you find a complete set of moves on triangulations, moves on moves, moves on moves on moves, etc.? Whether the set of moves is complete is not entirely a rigorous question, but there is an interesting answer. The moves and higher moves just come from erasing edges of the triangulation. The main theorem is that the resulting simplicial complex is contractible. The mapping class group acts on the complex, and it acts freely on the high-dimensional simplices.
More precisely, the arc complex has a vertex $v$ for every isotopy class of an arc between two of the marked points, among properly embedded arcs that miss all of the marked points in the interior. If a collection of arcs can be made disjoint after isotopy, then the corresponding vertices subtend a simplex. The disk case is an exception in which the arc complex is not quite contractible, but rather a sphere.
| 18 | https://mathoverflow.net/users/1450 | 6374 | 4,347 |
https://mathoverflow.net/questions/6376 | 50 | for forgetful functors, we can usually find their left adjoint as some "free objects", e.g. the forgetful functor: AbGp -> Set, its left adjoint sends a set to the "free ab. gp gen. by it". This happens even in some non-trivial cases. So my question is, why these happen? i.e. why that a functor forgets some structure (in certain cases) implies that they have a left adjoint? Thanks.
| https://mathoverflow.net/users/1657 | Why forgetful functors usually have LEFT adjoint? | Forgetful functors usually have a left adjoint because they usually preserve limits. For example, the underlying set of the direct product of two groups is the direct product of the underlying sets, and similarly for equalizers (that gives you all finite limits).
However, functors that preserve limits don't have to have left adjoints, because once in a while what you want to do to construct a free object results in a proper class. An example is [complete lattices](http://en.wikipedia.org/wiki/Complete_lattice).
Freyd's [Adjoint Functor Theorem](http://en.wikipedia.org/wiki/Adjoint_functors) gives a necessary and sufficient condition for a limit-preserving functor to have a left adjoint. The proof and related results is discussed in section 1.9 of [Toposes, Triples and Theories](http://www.tac.mta.ca/tac/reprints/articles/12/tr12.pdf).
| 44 | https://mathoverflow.net/users/342 | 6378 | 4,349 |
https://mathoverflow.net/questions/6379 | 154 | What is an **integrable system**, and what is the significance of such systems? (Maybe it is easier to explain what a non-integrable system is.) In particular, is there a dichotomy between "integrable" and "chaotic"? (There is an interesting [Wikipedia article](https://en.wikipedia.org/wiki/Integrable_system), but I don't find it completely satisfying.)
**Update** (Dec 2010): Thanks for the many excellent answers. I came across another quote from Nigel Hitchin:
"Integrability of a system of differential equations should manifest
itself through some generally recognizable features:
* the existence of many conserved quantities
* the presence of algebraic geometry
* the ability to give explicit solutions.
These guidelines would be interpreted in a very broad sense."
(If there are some aspects mentioned by Hitchin not addressed by the current answers, additions are welcome...)
Closely related questions:
* *[What does it mean for a differential equation "to be integrable"?](https://mathoverflow.net/questions/224327/what-does-it-mean-for-a-differential-equation-to-be-integrable)*
* *[basic questions on quantum integrable systems](https://mathoverflow.net/questions/114264/basic-questions-on-quantum-integrable-systems)*
| https://mathoverflow.net/users/1532 | What is an integrable system? | This is, of course, a very good question. I should preface with the disclaimer that despite having worked on some aspects of integrability, I do not consider myself an expert. However I have thought about this question on and (mostly) off.
I will restrict myself to integrability in classical (i.e., hamiltonian) mechanics, since quantum integrability has to my mind a very different flavour.
The standard definition, which you can find in the wikipedia article you linked to, is that of Liouville. Given a Poisson manifold $P$ parametrising the states of a mechanical system, a *hamiltonian* function $H \in C^\infty(P)$ defines a vector field $\lbrace H,-\rbrace$, whose flows are the classical trajectories of the system. A function $f \in C^\infty(P)$ which Poisson-commutes with $H$ is constant along the classical trajectories and hence is called a *conserved quantity*. The Jacobi identity for the Poisson bracket says that if $f,g \in C^\infty(P)$ are conserved quantities so is their Poisson bracket $\lbrace f,g\rbrace$. Two conserved quantities are said to be in involution if they Poisson-commute. The system is said to be *classically integrable* if it admits "as many as possible" independent conserved quantities $f\_1,f\_2,\dots$ in involution. Independence means that the set of points of $P$ where their derivatives $df\_1,df\_2,\dots$ are linearly independent is dense.
I'm being purposefully vague above. If $P$ is a finite-dimensional and symplectic, hence of even dimension $2n$, then "as many as possible" means $n$. (One can include $H$ among the conserved quantities.) However there are interesting infinite-dimensional examples (e.g., KdV hierarchy and its cousins) where $P$ is only Poisson and "as many as possible" means in practice an infinite number of conserved quantities. Also it is not strictly necessary for the conserved quantities to be in involution, but one can allow the Lie subalgebra of $C^\infty(P)$ they span to be solvable but nonabelian.
Now the reason that integrability seems to be such a slippery notion is that one can argue that "locally" any reasonable hamiltonian system is integrable in this sense. The hallmark of integrability, according to the practitioners anyway, seems to be coordinate-dependent. I mean this in the sense that $P$ is not usually given abstractly as a manifold, but comes with a given coordinate chart. Integrability then requires the conserved quantities to be written as local expressions (e.g., differential polynomials,...) of the given coordinates.
| 70 | https://mathoverflow.net/users/394 | 6382 | 4,352 |
https://mathoverflow.net/questions/6377 | 42 | I'm studying Steenrod operations from Hatcher's book. Like homology, one can use them only knowing the axioms, without caring for the actual construction. But while there are plenty of intuitive reasons to introduce homology, I cannot find any for the Steenrod operations. I can follow the steps in the proofs given by Hatcher, but I don't understand why one introduces all these spaces like $\Lambda X$, $\Gamma X$ and so on (in Hatcher's notation, I don't know if it's universal). Does anyone know how to get an intuitive grasp of what's going on?
| https://mathoverflow.net/users/828 | Why does one think to Steenrod squares and powers? | Steenrod operations are an example of what's known as a *power operation*. Power operations result from the fact that cup product is "commutative, but not too commutative". The operations come from a "refinement" of the operation of taking $p$th powers (squares if $p=2$), whose construction rests on this funny version of commutativity.
A cohomology class on $X$ amounts to a map $a: X\to R$, where $R = \prod\_{n\geq0} K(F\_2,n)$. So the cup product of $a$ and $b$ is given by
$$X\times X \to R\times R \xrightarrow{\mu} R.$$
In other words, the space $R$ carries a product, which encodes cup product. (There is another product on $R$ which encodes *addition* of cohomology classes.)
You might expect, since cup product is associative and commutative, that if you take the $n$th power of a cohomology class, you get a cohomology class on the quotient $X^n/\Sigma\_n$, where $\Sigma\_n$ is the symmetric group, i.e.,
$$X^n \xrightarrow{a^n} R^n \rightarrow R$$
should factor through the quotient $X^n/\Sigma\_n$. This isn't quite right, because cup product is really only commutative up to infinitely many homotopies (i.e., it is an "E-infinity structure" on $R$). This means there is a contractible space $E(n)$ with a free action of $\Sigma\_n$, and a product map:
$$\mu\_n' : E(n)\times R^n\to R$$
which is $\Sigma\_n$ invariant, so it factors through $(E(n)\times R^n)/\Sigma\_n$. Thus, given $a: X\to R$, you get
$$P'(a): (E(n)\times X^n)/\Sigma\_n \to (E(n)\times R^n)/\Sigma\_n \to R.$$
If you restrict to the diagonal copy of $X$ in $X^n$, you get a map
$$P(a):E(n)/\Sigma\_n \times X\to R.$$
If $n=2$, then $E(2)/\Sigma\_2$ is what Hatcher seems to call $L^\infty$; it is the infinite real proj. space $RP^\infty$. So $P(a)$ represents an element
in $H^\* RP^\infty \times X \approx H^\*X[x]$; the coefficients of this polynomial in $x$ are the Steenrod operations on $a$.
Other cohomology theories have power operations (for K-theory, these are the Adams operations).
You can also describe the steenrod squares directly on the chain level: the account in the book by Steenrod and Epstein is the best place to find the chain level description.
| 58 | https://mathoverflow.net/users/437 | 6384 | 4,354 |
https://mathoverflow.net/questions/5108 | 3 | Is this possible to do constructively? The only sources that I have for the possibility of this construction is an exercise in Lang's Algebra (on p. 598, I believe) which states that one can be constructed, and then a construction given in Milnor and Hussemoller's book which only applies in the case that the elementary divisors are all 1.
In my case, I have a rank 4 lattice of with a symplectic form of type (1, n), and I have some elements which span an index n sublattice. I would like to somehow relate them to a symplectic basis on the whole lattice, but it seems to me that I would need to have a constructive method for creating such a basis for this to be of any use whatsoever.
| https://mathoverflow.net/users/1703 | How do you construct a symplectic basis on a lattice? | Here is an algorithm, it may not be a good one. I will only explain how to find a basis $e\_i$, $f\_i$ such that $\langle e\_i, e\_j \rangle = \langle f\_i, f\_j \rangle =0$ and $\langle e\_i, f\_j \rangle = c\_i \delta\_{ij}$ for some constants $c\_i$. I will punt on explaining how to make sure that $c\_1$ divides $c\_2$ which divides $c\_3$ and so forth. This presentation is closely based on the algorithm in Wikipedia for computing [Smith normal form](http://en.wikipedia.org/wiki/Smith_normal_form#Algorithm).
Find $e$ and $f$ so that $\langle e,f \rangle \neq 0$ **(EDIT)** and such that the lattice spanned by $e$ and $f$ is the intersection of the whole lattice with the vector space spanned by $e$ and $f$.
Set $d := \langle e,f \rangle$. Complete $(e,f)$ to a basis $(e,f,g\_1, g\_2, \ldots, g\_{2n})$ of our lattice.
**Case 1:** We have $\langle e,g\_i \rangle = \langle f, g\_i \rangle =0$ for all $i$. Take $e\_1=e$, $f\_1=f$ and apply our algorithm recursively to the sublattice spanned by the $g\_i$.
**Case 2:** For all $i$, the integer $d$ divides $\langle e,g\_i \rangle$ and $\langle f, g\_i \rangle =0$. Replace $g\_i$ by
$$g\_i - \frac{1}{d}\langle g\_i, f \rangle e- \frac{1}{d}\langle e, g\_i \rangle f.$$
We have now reduced to Case 1.
**Case 3:** There is some $g\_i$ so that $k:=\langle e, g\_i \rangle$ is not divisible by $d$. Then, for some $q$, we have $0 < k-qd < d$. Set $f'=g\_i-kf$ and $g'\_i=f$. Then $(e,f',g\_1,g\_2,\ldots, g'\_i, \ldots, g\_n)$ is a basis, and $\langle e, f' \rangle$ is less than $d$. Return to the beginning with this new basis.
**Case 4:** There is some $g\_i$ so that $k:=\langle f, g\_i \rangle$ is not divisible by $d$. Just like Case 3, with the roles of $e$ and $f$ switched.
Since $d$ decreases at every step, eventually we will hit Case 1 and be able to reduce the dimension.
| 9 | https://mathoverflow.net/users/297 | 6391 | 4,359 |
https://mathoverflow.net/questions/6388 | 8 | If we ask which natural numbers n are not expressible as $n = ab + bc + ca$ ($0 < a < b < c$) then this is a well known open problem. Numbers not expressible in such form are called Euler's "numerus idoneus" and it is conjectured that they are finite.
If we omit the condition $a < b < c$ and assume $0 < a \leq b \leq c$ then it was proved (assuming Generalized Riemann Hypothesis) that there is only a finite number of such numbers $n$.
I am interested in the problem of expressing a prime number $p$ as $p = ab + ac + bc$ for $a \geq 1$ and $b,c \geq 2$.
Anybody knows if there is some known result related to expressing prime numbers in such form? This would yield (as a corollary) a very beautiful theorem related to spanning trees in graphs.
| https://mathoverflow.net/users/1737 | Prime numbers $p$ not of the form $ab + bc + ac$ $(0 < a < b < c )$ (and related questions) | Looking at the encyclopedia of integer sequnces, I've found that the set of numbers not expressible as ab + bc + ac 0 < a < b < c is finite.
<http://oeis.org/A000926>
Chowla showed that the list is finite and Weinberger showed that there is at most one further term.
| 10 | https://mathoverflow.net/users/1737 | 6408 | 4,373 |
https://mathoverflow.net/questions/6394 | 31 | Next term I am supposed to teach a course on representation of finite groups. This is a third year course for undegrads. I was thinking to use the book of Grodon James and Martin Liebeck "Representations and characters of groups", but also looking for other references.
The question is: could you advise some other books (or lecture notes)? Maybe you had a nice experience of teaching or listening to a course with a similar title? It would be really nice if this book (notes) has also exercises.
ADDED. I would like to thank everybody who answered the question, very helpful answers!!! The answer of John Mangual below contains a "universal" reference.
For the moment my favourites are Serre (very clear and short introduction of main ideas), some bits from notes of Teleman and Martin, and Etingof for beautiful exposition. My last problem is to have enough of exercises, in particular to write down a good exam. So I would like to ask if there are some additional references for exercises (with or without solutions)?
| https://mathoverflow.net/users/943 | Lecture notes on representations of finite groups | Some material from the undergrad rep theory course in Cambridge: [Example sheets](https://www.dpmms.cam.ac.uk/study/II/RepresentationTheory/), A [recent set of notes](http://tartarus.org/gareth/maths/notes/ii/Repn_Theory.pdf) (by Martin), and a [less recent (but very nice) set of notes](http://math.berkeley.edu/~teleman/math/RepThry.pdf) (by Teleman).
| 15 | https://mathoverflow.net/users/349 | 6409 | 4,374 |
https://mathoverflow.net/questions/6071 | 13 | In Connes work on the Riemann Hypothesis he talks about constructing $\overline{\text{Spec}\mathbb{Z}}$ as a curve over the field with one element. I just want to know what Spec means. Is the same as spectrum from algebraic geometry?
| https://mathoverflow.net/users/1867 | What is $\overline{\text{Spec}\mathbb{Z}}$? | If you remove the overline, you have the affine scheme Spec Z. It is the spectrum of a noetherian domain of Krull dimension one. This description also holds for any affine algebraic curve over a field, so we have the basis for an analogy. Z has many structural features in common with the ring of polynomials with coefficients in a field (e.g., basic number-theoretic machinery like a Euclidean algorithm), so there are good reasons to think of Spec Z as a curve.
However, there are several differences. First, Z is the initial object in the category of commutative rings, so Spec Z is the final object in the category of locally ringed spaces. In particular, Spec Z is not a curve over any base field. Second, there are maps from the spectra of fields of many different characteristics into Spec Z. This makes objects like zeta functions and L-functions much more transcendental (in appearance), since the logarithm doesn't behave as nicely as it does over finite fields. You can also find a definition of genus of a number field in Neukirch's *Algebraic Number Theory*, and while it is zero for Q, it tends to be transcendental in general (essentially due to the presence of logs).
Despite the visible flaws in the analogy, there are good reasons for thinking that Spec Z should have a compactification, in a manner similar to the compactification of Spec F[t] into the projective line. The most basic is that Z has a valuation that is not captured by the points of the topological space, namely the usual Archimedean absolute value. For the ring of polynomials in a field, this translates to the degree. By adding this extra valuation you find that if you take all of the absolute values of a rational number (normalized appropriately) and take their product, you get one. Logarithmically, the sum of valuations is zero just like the residue theorem in the function field case. A more sophisticated reason comes from arithmetic intersection theory. If you take a curve defined over Z, you can compute a notion of intersection between two integral points, in an manner analogous to computing the intersection class of two curves in a surface. In the topological setting, you need some conditions for this to be well-behaved, e.g., the surface should be compact, so you can't push the intersections off the end of the surface. This is achieved over Z using Arakelov theory - the intersection over infinity is computed by base changing the curve to the complex numbers, making some distribution on the resulting Riemann surface using the integral points, and computing the integral of that distribution.
I don't think there is general agreement on how to make a geometric object with all of the properties we want from a compactification of Spec Z. In particular, there doesn't seem to be a satisfactory theory of the "base field" yet. There is a way to approximate the compactification using Berkovich spaces, which are an analytic refinement of schemes. The Berkovich spectrum of Z already comes with an Archimedean branch, and if you weaken the triangle inequality, you can have points corresponding to arbitrary non-negative powers of the Archimedean absolute value. Compactification means adding the point $|-|^\infty\_\infty$. The local ring at this point is the closed interval [-1,1] in R, with the multiplication operation (i.e., you have some kind of logarithmic structure, instead of an actual ring), and the global functions on this object are given by the multiplicative monoid {-1,0,1}, which is a rather boring looking candidate for the field with one element.
| 18 | https://mathoverflow.net/users/121 | 6417 | 4,380 |
https://mathoverflow.net/questions/6418 | 12 | What is the fastest algorithm that exists to solve a particular NP-Complete problem? For example, a naive implementation of [travelling salesman](http://en.wikipedia.org/wiki/Travelling_salesman_problem) is $O(n!)$, but with dynamic programming it can be done in $O(n^2 2^n)$. Is there any "easier" NP-Complete problem that has a better running time?
Note that I'm curious about exact solutions, not approximations.
| https://mathoverflow.net/users/1646 | Best-case Running-time to solve an NP-Complete problem | If P is an NP-complete problem, then define Pk = instances of P in which the instances have been blown up from size n to size nk by padding them with blanks. Then Pk is also NP-complete, but if P takes time exp(p(n)) to solve where p is some polynomial then Pk can be solved in time essentially exp(p(n1/k)) (there's a little more time required to check that the input really does have the right amount of padding but unless the running time is polynomial this is a negligable fraction of the total time). So there is no "easiest" problem: for every problem you name this construction gives another easier but still NP-complete problem.
As for non-artificial problems: most hard graph problems like Hamiltonian circuit, that are hard when restricted to planar graphs, can be solved in time exponential in √n or in (√n)(log n) by dynamic programming using a recursive partition by graph separators.
| 25 | https://mathoverflow.net/users/440 | 6420 | 4,381 |
https://mathoverflow.net/questions/6438 | 7 | Prove or counterexample: If A is a commutative ring and $A\_p$ is a finitely generated algebra over A for all prime ideal p of A, then A is a product of local rings.
| https://mathoverflow.net/users/2008 | When is a localization of a commutative ring finitely generated as an algebra? | Here is a counterexample. Let p and q be distinct prime numbers, let S denote the complement of {p,q} in the set of all primes, and let A denote Z[1/S]. Then the prime ideals of A are pA, qA, and {0}, the first two of which are maximal. Also, since A is domain (being a subring of Q) with two maximal ideals, it's not a product of local rings. On the other hand, the local rings at the various prime ideals are A[1/q], A[1/p], and A[1/p,1/q], which are clearly finitely generated as A-algebras.
| 11 | https://mathoverflow.net/users/1114 | 6451 | 4,393 |
https://mathoverflow.net/questions/6450 | 1 | If $\sum\_{n=1}^\infty \frac{a\_n}{n^s} $ converges, does
$\sum\_{n=1}^\infty \frac{a\_n}{(n+1)^s} $ also converge?
| https://mathoverflow.net/users/2003 | Shifted Dirichlet series | Yes, because $(n+1)^{-s} = n^{-s} + sn^{-s-1} + O(|s|^2n^{-\sigma - 2})$. The first series necessarily converges in the open half-plane strictly to the right of s, and converges absolutely in the half-plane strictly to the right of s + 1. I hope I am not doing homework from a course in analytic number theory here.
| 4 | https://mathoverflow.net/users/3304 | 6454 | 4,395 |
https://mathoverflow.net/questions/6457 | 15 | I just wonder about Kapranov's ["Analogies between Langlands Correspondence and topological QFT"](http://books.google.de/books?id=TOPa9irmsGsC&pg=PA119&lpg=PA119&dq=Kapranov+%22%22Analogies+between+Langlands+Correspondence+and+topological+quantum+field+theory%22%22&source=bl&ots=kGZ-Wnk_Fu&sig=G-E6CK1Rza5dfp2niaBOmIdoN0w&hl=de&ei=bSQJS7WAMZaInQPV45jACg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBAQ6AEwAA#v=onepage&q=Kapranov%20%22%22Analogies%20between%20Langlands%20Correspondence%20and%20topological%20quantum%20field%20theory%22%22&f=false). I would like to read a more detailed exposition and how one turns that analogy into concrete conjectures. Do there exist texts from courses or seminars on that? Has [this book](http://www.worldscibooks.com/mathematics/7164.html) by İkeda already been published and reviewed?
| https://mathoverflow.net/users/451 | Kapranov's analogies | There are notes from more recent (2000) lectures of Kapranov on the subject on [my webpage](http://www.math.utexas.edu/users/benzvi/GRASP/lectures/Langlands.html).
Despite this, as far as I know, there isn't any convincing argument at this point that there should be a general higher dimensional Langlands theory. What we learn from physicists is that one can expect deep dualities for algebraic surfaces and threefolds, but probably not in general. And we also learn that the shape of these dualities won't be one
you are likely to guess from analogies with curves and with higher dimensional class field theory, but for which string theory is a great guide. One way to put this is that we now (post 2005) understand that geometric Langlands is an aspect of four-dimensional topological field theory (more specifically, maximally supersymmetric gauge theory) — so one ought to look for higher dimensional analogs of this, and they are very very few and very special.
Perhaps the main issue in making direct generalizations (that Kapranov addresses) is the great complexity in describing Hecke algebras in higher dimensions — in fact this is one of the most exciting areas of current research, under the name Hall Algebras (cf work of Joyce, Kontsevich-Soibelman and others).
It turns out that Hall algebras of 3-dimensional (Calabi-Yau) varieties have a remarkably rich representation theoretic structure, which (upon dimensional reduction to surfaces or curves) encapsulates much of classical representation theory and geometric Langlands etc.
In any case the most exciting thing around as far as I'm concerned is a 6-dimensional supersymmetric field theory (which lacks a good name — it's called the (0,2) theory or the M5 brane theory) which seems to enfold everything we know in all of the above examples. It is a conformal field theory, and "explains" Langlands duality simply as the SL2(Z) conformal symmetry of tori (when one considers it on R^4 times a torus).
One can expect this theory to lead to an interesting Langlands-like program for algebraic surfaces, which includes in particular the interesting recent work in this direction of Braverman-Finkelberg ([arXiv:0711.2083](http://arxiv.org/abs/0711.2083) and [0908.3390](http://arxiv.org/abs/0908.3390)).
I should mention also that since
Kapranov's paper there has been a lot of work, by Kazhdan, Braverman, Gaitsgory and Kapranov in particular, on representation theory for higher dimensional local fields — it's just that there isn't yet a clear Langlands type picture for this.
Again one should expect something special to happen for surfaces, of which we have many glimpses — in particular the beautiful theory of Cherednik's double affine Hecke algebra and its Fourier transform, which is I think accepted as a clear hint at a Langlands picture for surfaces. But in any case I strongly believe that the current evidence indicates we should be learning from the physicists, reading papers by Gaiotto etc, to figure out what should happen for surfaces rather than just believing that we should try to extrapolate from what we know for curves.
| 26 | https://mathoverflow.net/users/582 | 6466 | 4,404 |
https://mathoverflow.net/questions/6427 | 4 | $$F\\,(s) = 1/1^s + 0/2^s + 1/3^s + 1/4^s + 0/5^s + 1/6^s + 0/7^s + 1/8^s + ... $$
The coefficients of this Dirichlet series are the base-2 digits of $\sqrt2$.
Does it have an analytic continuation into the critical strip?
Assuming so, can anything be said about the locations of its poles in this region? Are there any? Are they all on the critical line?
What about $\pi$ and $e$ and $\log 2$?
Are all of these sequences "random" in the same sense as the Liouville and Moebius functions?
| https://mathoverflow.net/users/2003 | Dirichlet series whose coefficients are the bits of sqrt(2) | David Speyer is right. A Dirichlet series with coefficients that are random choices from
$0,1$ does not count as a random Dirichlet series. Random choices from $-1,1$ would count as a random Dirichlet series. Taking his insight into account, I correct my prediction: With probability $1$ the Dirichlet series $\sum b\_n(x)n^{-s}$ where $b\_n(x)$ are the binary digits
of $x \in [0,1)$ will have a meromorphic continuation to $\sigma > 1/2$ with a single simple pole at $s = 1$ with residue $1/2$. Also I now remember a reference: Armand Denjoy *L'Hypothese de Riemann sur la distribution...* C. R. Acad. Sci. Paris 192 (1931), 656-658. This paper deals with the situation where the Dirichlet series has coefficients chosen at random from $-1,1$.
| 5 | https://mathoverflow.net/users/3304 | 6467 | 4,405 |
https://mathoverflow.net/questions/6325 | 12 | So, let's say we have a symplectic variety over $\mathbb{C}$, $M$, of dimension $2n$, and $f\_1,\ldots,f\_n$ Poisson commuting functions with $df\_1\wedge\ldots\wedge df\_n$ generically nonzero. Further assume that the fibers of the map $f:M\to\mathbb{C}^n$ determined by the $f\_i$ is an open subset of an abelian variety and the vector fields $X\_{f\_i}$ are linear. We call such a thing an algebraically completely integrable Hamiltonian system.
Now, I'm told that there's a definition of integrable system in PDEs that acts as some sort of stability condition, though I don't understand it. I've also been told that there's a way to, from an integrable system of PDEs, construct an integrable Hamiltonian system (just drop the condition that fibers be in abelian varieties), and that these two types of objects should be equivalent.
My questions:
1) What's the correct formulation for PDEs to make something like this work out? (I know virtually nothing about PDEs, and would be quite grateful just to be pointed at a good reference, if there's too much I need to read to get a quick, understandable answer)
2) Is there a general method of going from an algebraically completely integrable Hamiltonian system, which is algebro-geometric in nature to working out the PDEs explicitly? Does it help if the symplectic variety is known to be the cotangent bundle of something? How about if the base is unirational? rational?
| https://mathoverflow.net/users/622 | Equations for Integrable Systems | I'm no expert on this, but since nobody answers:
one book which should definitely help is [this big introductory one by Babelon, Bernard and Talon](http://books.google.fr/books?id=c6JN6Gp4RBQC&pg=PP1&dq=INTRODUCTION+TO+CLASSICAL+INTEGRABLE+SYSTEMS#v=onepage&q=&f=false). There's also a very technical older [paper by Ben-Zvi and Frenkel](http://arxiv.org/abs/math.AG/9902068) where apparently some sort of general construction is made, and there's a more readable [paper by Inoue, Vanhaecke and Yamazaki](http://www-math.univ-poitiers.fr/~vanhaeck/art/art31/www31.pdf) on algebraic complete integrability and integrable hierarchies of PDEs intended for your second question (especially section 6 for the relationship).
Just glancing at all this I'm not so sure a general explicit method to obtain the PDEs exists (I could be wrong) but some cases seem to be understood. Hope this helps...
| 5 | https://mathoverflow.net/users/469 | 6469 | 4,407 |
https://mathoverflow.net/questions/6455 | 17 | Is there a game-theoretic interpretation of nimber multiplication? There is such for addition (a single move in a+b is either a move in a or a move in b).
| https://mathoverflow.net/users/2014 | Nimber multiplication | As Alex says, there's no good one. One way to see the problem is to compare nimbers with (surreal) numbers. Addition in both is just game addition, so you can consider nimbers and numbers together and add them consistently. But there is no consistent multiplication. (What would the unit be?) A good game theoretic multiplication would explain how to multiply arbitrary games.
Instead, we have nimber multiplication, which is constructed ad hoc and based on the algebraic structure of the nimbers: nimbers include all additive groups of characteristic 2, and the multiplication makes them into the universal field of characteristic 2 (in the sense that any field of characteristic 2 embeds in the nimbers); any totally ordered field embeds in the surreal numbers.
| 12 | https://mathoverflow.net/users/78 | 6470 | 4,408 |
https://mathoverflow.net/questions/6447 | 10 | If we have the complete countably infinite bipartite graph $K\_{\omega,\omega}$ and we colour the edges with just two colours. Should we expect to get a monochromatic copy of $K\_{\omega,\omega}$.
Infinite Ramsey theorem gives us infinitely many edges of one colour but this is no enough.
| https://mathoverflow.net/users/2011 | Ramsey Theory, monochromatic subgraphs | The example given by Konstantin Slutsky is, however, essentially unique, in the following sense. Let $G$ be the complete bipartite graph with both vertex sets equal to (copies of) $\mathbb{N}$ and colour its edges red or blue. For each $m< n$, colour the set $\{m,n\}$ according to whether the edge $(m,n)$ is red or blue and whether the edge $(n,m)$ is red or blue. (So this is a 4-colouring of the complete graph on $\mathbb{N}$.) By Ramsey's theorem we can find a subset $X$ of $\mathbb{N}$ such that all pairs $\{m,n\}$ with $m,n\in X$ have the same colour. If the colour is RED-RED or BLUE-BLUE then we have a monochromatic bipartite subgraph by choosing two disjoint subsets $Y$ and $Z$ of $X$ (to avoid the problem when $m=n$). If the colour is RED-BLUE then we can again pass to two disjoint subsets, which we could even make alternate, and the edge $(m,n)$ will then be RED if $m< n$ and BLUE if $m> n$, and similarly if the colour is BLUE-RED. So we either find a monochromatic subgraph or we find Konstantin Slutsky's example. This is an example of a "canonical Ramsey theorem" -- you can't find a monochromatic structure but you can find a simple "canonical" structure.
| 15 | https://mathoverflow.net/users/1459 | 6472 | 4,410 |
https://mathoverflow.net/questions/6276 | 23 | I'd be grateful for a good reference on this, it feels like a classic subject yet I couldn't find much about it.
Polynomials in one variable of the form $x^n+a\_{n-1}x^{n-1}+\dots +a\_1 x+a\_0$ can be transformed into simpler expressions. For instance [it is apparently well-known](http://homepage.mac.com/ehgoins/ma598/lecture_27.pdf) that the Tschirnhaus transformation allows to bring *any* quintic into so-called Bring-Jerrard form $x^5+ax+b$, while for degree 6 one needs at least three coefficents $x^6+ax^2+bx+c$.
Is there a name for such "generalized Bring-Jerrard form", and what is known about it? In particular there is a cryptic footnote of Arnold (page 3 of [this lecture](http://www.pdmi.ras.ru/~arnsem/Arnold/arnlect1.ps.gz)) where he says roughly that the degrees for which more coefficients are needed occur along "a rather strange infinite sequence": could someone please describe what those degrees are (I had a look at the OEIS but I believe that sequence is different from [Hamilton numbers](https://oeis.org/A000905), and couldn't find a relevant one).
| https://mathoverflow.net/users/469 | Least number of non-zero coefficients to describe a degree n polynomial | You might have a look at [Polynomial Transformations of Tschirnhaus, Bring and Jerrard](http://www.apmaths.uwo.ca/~djeffrey/Offprints/Adamchik.pdf) ([Internet Archive](http://web.archive.org/web/20180816061604/http://www.apmaths.uwo.ca/~djeffrey/Offprints/Adamchik.pdf)). It gives more explicit detail on why you can remove the first three terms after the leading term (covering the cases of degree 5 and 6 you mention above), but it does concentrate on degree 5.
Hamilton's [1836 paper](http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/Jerrard/Jerrard.pdf) ([Internet Archive](http://web.archive.org/web/20171126173729/https://www.maths.tcd.ie/pub/HistMath/People/Hamilton/Jerrard/Jerrard.pdf)) on Jerrard's original work has an elementary explanation of the technique (much of the paper concentrates on showing that certain other reductions Jerrard proposed, including a general degree 6 polynomial to a degree 5, were "illusory"). It also explains Jerrard's trick for eliminating the 2nd, 3rd and 5th terms. Finally, Jerrard has a method for eliminating the second and fourth terms, while bringing the third and fifth coefficients into any specified ratio: this only works in degree 7 or above (Jerrard had mistakenly thought this worked generally, and thus solved the general quintic by reducing it to de Moivre's solvable form -- this all predates Abel's work!)
If by "Bring-Jerrard" form you just mean a certain number of the initial terms (after the first) have been eliminated, then the Hamilton numbers you linked to are indeed exactly what you want.
| 16 | https://mathoverflow.net/users/3 | 6474 | 4,412 |
https://mathoverflow.net/questions/6144 | 53 | The Chern character is often seen as just being a convenient way to get a ring homomorphism from K-theory to (ordinary) cohomology.
The most usual definition in that case seems to just be to define the Chern character on a line bundle as $\mathrm{ch}(L) = \exp(c\_1(L))$ and then extend this; then for example $\mathrm{ch}(L\_1 \otimes L\_2) = \exp(c\_1(L\_1 \otimes L\_2)) = \exp(c\_1(L\_1) + c\_2(L\_2)) = \mathrm{ch}(L\_1) \mathrm{ch}(L\_2)$; then we can use this to define a Chern character on general vector bundles.
This all seems a bit ad-hoc, and it doesn't give much insight as to why such a thing exists anyway.
An explanation I like a lot better comes from even complex oriented cohomology theories. Given any complex oriented periodic cohomology theory, such as K-theory or periodic (ordinary) cohomology, we have $H(\mathbb{CP}^\infty) \cong H(P)[[t]]$ for $P$ a point. Seeing as $\mathbb{CP}^\infty$ is the classifying space for line bundles, this gives us a way of having "generalised Chern classes" for any line bundle corresponding to any cohomology theory, and even for any vector bundle.
We have a link between complex oriented periodic cohomology theories and formal group laws, corresponding to what corresponds to $c\_1(L\_1 \otimes L\_2)$ in $\mathbb{CP}^\infty$: for ordinary cohomology, as above, we get that $c\_1(L\_1 \otimes L\_2) = c\_1(L\_1) + c\_1(L\_2)$ which gives the additive formal group law, and for K-theory we get $c\_1(L\_1 \otimes L\_2) = c\_1(L\_1) + c\_1(L\_2) + c\_1(L\_1) c\_2(L\_2)$ which is the multiplicative formal group law. The fact that over $\mathbb{Q}$ (but not over $\mathbb{Z}$) there is an isomorphism between the formal group laws given by the exponential map, and this reflects in the cohomology, giving the Chern character $K(X) \otimes\_\mathbb{Z} \mathbb{Q} \to \prod\_n H^{2n}(X,\mathbb{Q})$.
I'm not too sure what the exact formulation in that second case is, but more importantly I was wondering if there are any other, cleaner interpretations of the Chern character (I've been hearing about generalised Chern characters, and I have no idea where they would come from in this case). It seems like there should be a way to link the Chern character to things like the genus of a multiplicative sequence, and tie it in with other similar ideas for example the Todd genus or the L-genus given by similar formal power series. I guess the trouble is that I don't see how these related ideas all fit in together.
| https://mathoverflow.net/users/362 | Explanation for the Chern character | There is a nice discussion about multiplicative sequences, &c., in Lawson and Michelsohn's book "Spin Geometry". It discusses things like the Todd genus, the A-hat genus, and so on, but also the Chern character and the ring homomorphism from K-theory to ordinary cohomology. It is a readable exposition and perhaps "connects the dots" in a way that would be helpful to you.
| 12 | https://mathoverflow.net/users/1938 | 6479 | 4,414 |
https://mathoverflow.net/questions/6481 | 13 | This is a follow-up to the following answer:
[Solvable class field theory](https://mathoverflow.net/questions/4379/solvable-class-field-theory/4386#4386)
in which it is stated as a "folklore" conjecture that the maximal solvable extension of Q is pseudo algebraically closed (this means, in particular, any geometrically connected variety over Q has a point over a solvable extension).
I am curious what evidence there is to support such a conjecture.
In addition, what can be said for the analogous statement for global function fields?
| https://mathoverflow.net/users/81 | Evidence for $Q^{\operatorname{solv}}$ being pseudo-algebraically-closed | So far as I know, there is no compelling evidence to support this conjecture. (And some leading arithmetic geometers think it is false.) Rather, there are some very interesting consequences of this conjecture, e.g. a solution of the Inverse Galois Problem over Q^{solv}: in other words, for any finite group G, there exists a tower of radical extensions
K\_0 = Q, K\_1 = K\_0(a\_0^{1/n\_0}) < K\_2 = K\_1(a\_1^{1/n\_1}) < ... <= K\_n
and a Galois extension L/K\_n with Galois group isomorphic to G. It also shows that geometrically irreducible algebraic varieties "acquire rational points" in a very different way from irreducible zero-dimensional varieties (which have a unique minimal
splitting field which need not be solvable).
Of course, interesting things which follow from a conjecture are, if anything, evidence *against* the truth of the conjecture, although they support the claim that the *question* is interesting.
There is one impressive result towards this conjecture, namely the Ciperiani-Wiles theorem: let C\_{/Q} be a genus one curve with points everywhere locally and semistable Jacobian elliptic curve E. Then C(Q^{solv}) is nonempty.
On the negative side, there is a paper of Ambrus Pal which constructs, for each sufficiently large integer g, a curve C of genus g over a field K which does not admit any points over the maximal solvable extension of K. (Here K is not a number field.)
On the other hand, as far as I know, it is still open to find an absolutely irreducible variety V/Q which fails to have rational points over the maximal *metabelian extension* of Q, i.e., over (Q^{ab})^{ab}. For some thoughts about this, see
<http://alpha.math.uga.edu/~pete/abeliantalk.pdf>
ADDENDUM:
I forgot to address the last part of the question: what about global function fields?
As I alluded to above, there are counterexamples over the function field of a sufficiently complicated ground field, like Q. Of course you mean a finite extension of F\_q(T), in which case I think absolutely nothing is known. In particular, I believe the analogue of Ciperiani-Wiles is open here, and might not be a straightforward adaptation, since C-W uses results on the modularity of elliptic curves. This could make a nice thesis problem...but I would talk to Mirela Ciperiani before doing any serious work on it.
| 8 | https://mathoverflow.net/users/1149 | 6486 | 4,417 |
https://mathoverflow.net/questions/6476 | 4 | What is the relationship between the Hausdorff dimension and cardinality of a set?
Specifically, assuming the Continuum Hypothesis, if a set has Hausdorff dimension greater than zero does, that imply that its cardinality is equal too or greater than that of $2^{\aleph\_0}$?
Or, does the negation of CH, imply the existence of a set with positive Hausdorff dimension and cardinality strictly between $\aleph\_0$ and $2^{\aleph\_0}$?
| https://mathoverflow.net/users/1320 | Hausdorff dimension vs. cardinality | As stated, countable sets have Hausdorff dimension 0.
So any set $S$ with $\mathrm{HD}(S)>0$ has power $\ge \aleph\_1$.
No need for continuum hypothesis.
Without CH, though, we cannot say whether power $ \ge c = 2^{\aleph\_0}$ is required. But this is not about Hausdorff dimension, it is the same question for positive Lebesgue measure in the line. It is consistent with ZFC (follows from Martin's Axiom) that any set with power $< c$ has Lebesgue measure zero, or (for the same reason, or with the same proof, or even consequently) any set with power $< c$ has Hausdorff dimension zero. However, without CH (and without Martin's Axiom) there could be sets of reals of power $< c$ but with positive outer Lebesgue measure, and thus Hausdorff dimension 1.
| 6 | https://mathoverflow.net/users/454 | 6501 | 4,428 |
https://mathoverflow.net/questions/6493 | 4 | Hi,
this is kind of continuation of [this thread](https://mathoverflow.net/questions/6429/asymptotically-multiplicative-functions-and-matrices) to concentrate on a specific problem from linear algebra and analysis that, I think, is rather interesting for itself.
Here we go:
1) Main problem: Let $(c\_n)\_{n\in\mathbb{N}}$ be some fixed sequence of complex numbers. Is there a sequence of matrices $A\_n=(a(n)\_{ij})\in M\_n(\mathbb{C})$, $n\in\mathbb{N}$, such that $a(n)\_{ij}\widetilde{a(n)\_{ij}}=c\_i c\_j$, where $\widetilde{a(n)\_{ij}}$ denotes the cofactor of $a(n)\_{ij}$? (For a fixed $n\in\mathbb{N}$ this actually establishes a non-linear system of $n^2$ equations with $n^2$ unknowns.)
2) "Inverse" problem: Let $A\_n=(a(n)\_{ij})\in M\_n(\mathbb{C})$, $n\in\mathbb{N}$, be some given sequence of complex quadratic matrices of increasing order. What are necessary and sufficient conditions for such a matrix sequence to have a sequence of complex numbers $(c\_n)\_{n\in\mathbb{N}}$ such that $\forall n\in\mathbb{N}$ $\forall 1\leq i \leq n, 1\leq j \leq n: a(n)\_{ij}\widetilde{a(n)\_{ij}}=c\_i c\_j$ (notation as above)?
Any input is welcome and I highly appreciate any references to similar or related problems.
Thanks,
efq
| https://mathoverflow.net/users/1849 | factorization of the product of a matrix element and its cofactor | For the main problem, we can use expansion by minors along the ith row to compute
$\det(A\_n) = (-1)^{i+1}(a(n)\_{i1}\widetilde{a(n)\_{i1}} - a(n)\_{i2}\widetilde{a(n)\_{i2}} + a(n)\_{i3}\widetilde{a(n)\_{i3}} - \dots)$
or det(An) = (-1)i+1ci(c1 - c2 + c3 - ...).
This is true for any i <= n, so whenever c1 - c2 + ... - (-1)ncn != 0 we must have ci = (-1)i+1c1 for all i <= n. Therefore when cn != (-1)n+1c1 for the first time, not only must we have c1 - c2 + ... - (-1)ncn = 0 but this alternating sum must vanish for all greater n, meaning that cj = 0 for all j > n.
In conclusion: this is only possible if ck = (-1)k+1c1 for all k < K (K constant but possibly infinite) and then ck = 0 for all k > K.
| 3 | https://mathoverflow.net/users/428 | 6502 | 4,429 |
https://mathoverflow.net/questions/6440 | 7 | Suppose we fix a universe $U$ and a $U$-small category $C$. The regular Yoneda lemma gives us some locally small (not necessarily locally U-small?) functor category $C'=[C^{op},Sets]$ with a fully faithful embedding $C\rightarrow C'$ and the canonical bijection between $Nat(F,Hom(-,x))$ and $F(X)$. Suppose we consider now, the U-small Yoneda lemma, that is, we look at $[C^{op},U-Sets]$. This is well-behaved since even though it is not U-small, $Ob([C^{op},U-Sets])$ is still a set.
So the main question I have is: Are there any useful properties of the standard Yoneda lemma that we cannot reproduce with the $U$-small Yoneda lemma for some $U$?
| https://mathoverflow.net/users/1353 | Sets, Universes, and the small Yoneda Lemma | Well, for any universe U, the U-small sets satisfy the axioms of ZFC (or whatever your preferred set theory is). Therefore, anything that we can prove in ZFC about the Yoneda embedding into "all" presheaves will also be true about the Yoneda embedding into U-small presheaves. So on that score, the answer would seem to be "no."
There might be other interpretations that would make the answer "yes," since in vanilla ZFC every class is definable, whereas not every U-large set is definable in terms of U-small ones. Thus an informal statement like "for every large thingumbob X, ..." might be true in ZFC, but no longer provable relative to a universe, since the meaning of "large" has changed (unless we change the quantifier to "for every U-small-definable large thingumbob X"). This doesn't contradict the first observation, since such a statement cannot be a single theorem of ZFC, only a meta-theorem, and each instance of the meta-theorem is about a particular definable class and therefore still true about the corresponding U-small-definable U-large set. However, right now I can't think of any interesting or useful properties of the Yoneda embedding that would fall under this heading.
So I think that probably the answer is still "no."
| 4 | https://mathoverflow.net/users/49 | 6505 | 4,432 |
https://mathoverflow.net/questions/5678 | 7 | Let $\ell$ be a positive integer greater than 1. The problem is to find a set of $n$ real positive numbers $x\_i$ and $n+1$ numbers $y\_i$ such that
$$\sum\_{i=1}^n x\_i^k= \sum\_{i=1}^{n+1} y\_i^k$$
for $k=\ell,\cdots,2\ell-1$. These $2n+1$ numbers need to be upper/lower bounded by a constant independent of $\ell$ [thus $x\_i,y\_i=\Theta(1)$] and also I suspect that it is possible to do so with just $n=\ell$ or $n=O(\ell)$. [$\ell$ equations with $2\ell$ unknowns, why not!] An existential proof suffices but a constructive proof or a recipe would be really nice.
For me it is useful to find a bounded from below solution that scales in polynomially in the following sense: There exist *positive* $c$, and $s$ such that $c\le x\_i,y\_i$ and
$$\sum\_i x\_i+\sum\_i y\_i=O(\ell^s)$$.
The problem is related to a follow up on this paper of mine: arxiv:0908.1526 .
| https://mathoverflow.net/users/1837 | Simultaneous Equations Involving Power Sums | Actually Darsh gave an almost full solution. Let me fill in the minor technical details.
1) We need the following quantitative form of the inverse function theorem. Suppose that $F:\mathbb R^n\to \mathbb R^n$. Assume also that $\|DF(X)^{-1}\|\le C\_1$, that $\max\_{Y\in B(X, \delta)}\|D^2F(Y)\|\le C\_2$, and that $C\_1C\_2\delta\le\frac 12$. Then $F(B(X,\delta))\supset B(F(X),\frac{\delta}{2C\_1})$.
2) Take $n=2\ell-1$ and consider the mapping $F:\mathbb R^n\to \mathbb R^n$ given by $F(y\_1,\dots,y\_n)\_k=\sum\_{j=1}^n y\_j^k$ where $k=1,2,\dots,n$. Take $X=(x\_1,\dots,x\_n)$ where $x\_j=\frac{n+j}{n}$ for $j=1,\dots,n$.
3) Note that in $B(X,1)$, we have $\|D^2F\|\le A^n$ for some absolute $A>1$.
4) Note also that $DF(X)^\*$ is the linear operator that maps the vector $(c\_1,\dots,c\_n)$ to the vector $p(x\_1),\ldots,p(x\_n)$ consisting of the values of the polynomial $p(x)=\sum\_{k=1}^n c\_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.
5) Thus, taking $\delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,\delta)$ contains a ball of radius $\frac\delta{2A^n}\ge C^{-n}$ with some absolute $C>2$.
6) In particular, it contains two points with the difference $(0,0,\dots,0,D^{-\ell},D^{-(\ell+1)},\dots,D^{-(2\ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.
| 11 | https://mathoverflow.net/users/1131 | 6509 | 4,435 |
https://mathoverflow.net/questions/6498 | 3 | One way to define the free abelian group on a set $S$ is as $F(S) := \mbox{Hom}\_{\text{Set}}(S, \mathbb{Z})$ with the group operation coming by composition with the map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. The problem with this definition is that it's contravariant, whereas it seems to me that $F$ should really be a covariant functor since it should take functions $S \to T$ to homomorphisms $F(S) \to F(T)$. (Or should it?)
One way to get a covariant functor is to actually define $F(S) := \mbox{Hom}\_{\mbox{Ab}}( \mbox{Hom}\_{\text{Set}}(S, \mathbb{Z}), \mathbb{Z})$. It's interesting to think of the free abelian group as a "double dual" construction, but now I'm not sure that one gets the group operation in a natural way. Which of these definitions is "correct" (if either), and why?
**Edit:** Let me rephrase my question, then. One way to explicitly describe the free abelian group on a set $S$ is as the direct sum of $|S|$ copies of $\mathbb{Z}$. Is there a categorical description of this construction, i.e. one which doesn't refer to elements, only morphisms?
(If it helps, I'm trying to digest more thoroughly the relationship between chain and cochain groups.)
| https://mathoverflow.net/users/290 | What is the "right" definition of the free abelian group on a set? | As I understand things, the desire to find a more categorical *construction* of the free abelian group on a set S is not itself fully in the categorical spirit. Rather, for an object which is defined by a universal mapping property, you only need to convince yourself that it exists; you don't need to be bothered by looking at any particular construction. The point is that anything that you want to know about this object will follow most transparently from the universal mapping property.
In this case, if we have a universal map S -> FreeAb(S) and a map of sets S -> T, we can define a map FreeAb(S) -> FreeAb(T) without "looking under the hood" at how FreeAb's are
constructed. Namely, by composing S -> T and T -> FreeAb(T) we get a set map S -> FreeAb(T). By the universal mapping property, this factors through a homomorphism FreeAb(S) -> FreeAb(T), which is what we wanted. Done!
| 11 | https://mathoverflow.net/users/1149 | 6510 | 4,436 |
https://mathoverflow.net/questions/6480 | 8 | Are there any good books out there that can serve as an introduction to thermodynamical formalism in dynamical systems?
I know only Zinsmeister's short "Thermodynamical formalism and holomorphic dynamical systems", which is concerned mostly with holomorphic dynamics, and Ruelle's "Thermodynamical formalism", which is, alas, too dense and has very little intuition. Any suggestions?
| https://mathoverflow.net/users/1121 | Suggested reading for thermodynamic formalism | If you are interested in learning mathematical statistical physics (e.g. you would like to read Ruelle) then Minlos "Introduction to mathematical statistical physics" is a good choice for introduction to the area.
If you would like to learn the part of ergodic theory which is called "thermodanamic formalism" and you don't care much about statistical physics then Omri Sarig's lecture notes is a great source
<http://www.math.psu.edu/sarig/TDFnotes.pdf>
| 5 | https://mathoverflow.net/users/2029 | 6528 | 4,449 |
https://mathoverflow.net/questions/6535 | 2 | I recently was wondering if there was a name for sheaves which were locally constant on the open simplexes in a simplicial complex. After some googling I stumbled across simplicial sheaves. I am alright with the definition as the presheaves to simplicial sets, but now I wonder, is this the answer to my question?
Are simplicial sheaves related to the locally-constant-on-simplex etale sheaves of a simplicial complex? If not, does this concept have a different name? Are there interesting places this sort of thing appears?
What are the other interesting places where simplicial sheaves appear?
I know that this question is fairly general, but I hope you understand what I am asking.
| https://mathoverflow.net/users/348 | Simplicial Sheaves? | If I understand correctly, these are constructible sheaves with respect to the stratification of your simplicial complex by its skeleta. I think by a theorem of MacPherson the category of such sheaves is equivalent to the category of functors from the poset of faces and face inclusions to whatever category your sheaves take values in (maybe there are some conditions on the target category).
[Simplicial sheaves](http://ncatlab.org/nlab/show/simplicial+presheaf) are something else entirely—they're *(pre)sheaves of simplicial sets* on, say, a category equipped with a Grothendieck topology.
| 6 | https://mathoverflow.net/users/126667 | 6537 | 4,455 |
https://mathoverflow.net/questions/6475 | 25 | The following things are all called holonomic or holonomy:
1. A holonomic constraint on a physical system is one where the constraint gives a relationship between coordinates that doesn't involve the velocities at all. (ie, $r=\ell$ for the simple pendulum of length $\ell$)
2. A D-module is holonomic (I might not have this one quite right) if the solutions are locally a finite dimensional vector space, rather than something more complicated depending on where on the manifold you look (this is apparently some form of being very over-determined)
3. A smooth function is holonomic if it satisfies a homogeneous linear ODE with polynomial coefficients.
4. On a smooth manifold $M$, with vector bundle $E$ with connection $\nabla$, the holonomy of the connection at a point $x$ is the subgroup of $GL(E\_x)$ given by the transformations you get by parallel transporting vectors along loops via the connection.
Now, all of these clearly have something to do with differential equations, and I can see why 2 and 3 are related, but what's the common thread? Is 4 really the same type of phenomenon, or am I just looking for connections by terminological coincidence?
| https://mathoverflow.net/users/622 | What is the relationship between various things called holonomic? | Not a complete answer, but you might wish to look at Historical Remarks at the beginning of Bryant's [Recent advances in the theory of holonomy](http://arxiv.org/pdf/math/9910059v2), which seems to offer a relationship between (1) and (4).
| 5 | https://mathoverflow.net/users/394 | 6550 | 4,462 |
https://mathoverflow.net/questions/6554 | 28 | A lot of the terminology of category theory has obvious antecedents in analysis: limits, completeness, adjunctions, continuous (functors), to name but a few. However, analysis and category theory *seem* to be at opposite poles of the spectrum.
Is there anything deep here, or is it a case of "it has wings, so let's call it a duck"?
This was partly inspired by the top-rated answer to the question [What is a metric space?](https://mathoverflow.net/questions/5957/what-is-a-metric-space) and by the (slightly unsatisfactory answers to the) question [Can adjoint linear transformations be naturally realized as adjoint functors?](https://mathoverflow.net/questions/476/can-adjoint-linear-transformations-be-naturally-realized-as-adjoint-functors). In particular, in the first case - the categorical view of metric spaces - there does seem to be an obvious route between the two worlds, do the terminologies correspond there?
| https://mathoverflow.net/users/45 | Terminology in category theory | Names in category theory are often born when someone realizes that a concept in one particular topic can be generalized in a categorical way. The generally-defined concept is then named after the original narrowly-defined one.
The case of metric spaces provides a slightly notorious example. As discussed in that [other question](https://mathoverflow.net/questions/5957/what-is-a-metric-space), metric spaces can be viewed as an example of enriched categories. So, given any concept in metric space theory, you can try to generalize it to the context of enriched categories. This happened with the property of *completeness* of metric spaces, which one might call *Cauchy-completeness* since it's about Cauchy sequences. This concept turns out to generalize very smoothly to enriched categories, and to be a useful and important property there.
Many people call the property "Cauchy-completeness" in the general context of enriched categories too. But a significant minority disagree with this choice, feeling that it's stretching the terminology too far. For example, when applied to ordinary (**Set**-enriched) categories, the property merely says that every idempotent morphism in the category splits. This doesn't "feel" like the completeness condition on metric spaces. So there are other names in currency too, such as "Karoubi complete" (especially popular in the French school).
It's true that many pieces of categorical terminology do come from analysis, but maybe all that says is that analysis is an old and venerable subject. *Exact* is another example. It's used to mean several slightly different things in category theory, confusingly, but the most common usage is that a functor is "left exact" if it preserves finite limits. Now that comes from homological algebra, where one talks about exact sequences; a functor between abelian categories preserves left exact sequences iff it preserves finite limits. And that in turn, I believe, comes from the terminology of differential equations.
| 23 | https://mathoverflow.net/users/586 | 6564 | 4,470 |
https://mathoverflow.net/questions/6552 | 40 | After realising that I don't have an intuitive understanding of adjoint **functors**, I then realised that I don't have an intuitive understanding of adjoint **linear transformations**!
Again, I can use 'em, compute 'em, and convolute 'em, but I have no real intuition as to what is going on. My best attempts (when teaching them) were:
1. The adjoint allows us to shift stuff from one side of the inner product to the other, thus, in a fashion, moving it out of the way while we do something and then moving it back again.
2. Nice behaviour with respect to the adjoint (say, normal or unitary) translates into nice behaviour with respect to the inner product.
But neither of those is really about what the adjoint **is**, just what it does.
| https://mathoverflow.net/users/45 | What is an intuitive view of adjoints? (version 2: functional analysis) | Just to add to Yemon's answer, in the case of inner product spaces I think it may be helpful to understand the case of operators of finite rank. So let's start with the shockingly simple case where $\eta$ is a single vector in the inner product space $H$. I like to identify this with the linear map $t\mapsto t\eta$ from $\mathbb{C}$ to $H$, whose adjoint is the map $\xi\mapsto\langle\xi,\eta\rangle$ – in other words, just the functional associated with $\xi$. Going the other way (assuming $H$ is complete), the adjoint of a linear functional “is”, of course, just the vector provided to us by the Riesz representation theorem. To physicists, who write the inner product backwards (with good reason, btw) we are mere looking at the bra and ket version of the same vector, so $\langle\eta|$ and $|\eta\rangle$ will be each others' adjoints.
Next, you can tackle rank one operators between Hilbert spaces, which will be of the form $\xi\mapsto\langle\eta,\xi\rangle\zeta$ for fixed vectors $\eta$ and $\zeta$. And once you have a fairly good idea how that works, it's a short way to adding a finite number of these to get any finite rank operator.
Admittedly, from here to infinite rank is something of a stretch, but I think it might help anyhow.
| 22 | https://mathoverflow.net/users/802 | 6573 | 4,478 |
https://mathoverflow.net/questions/6517 | 60 | This is something of a followup to the question ["Kapranov's analogies"](https://mathoverflow.net/questions/6457/kapranovs-analogies), where a connection between Cherednik's double affine Hecke algebras (DAHA's) and Geometric Langlands program was mentioned.
I am interested to hear about known connections of the double affine Hecke algebras, or their various degenerations called rational and trigonometric degenerate DAHA's (or simply Cherednik algebras) to other areas of mathematics. Affine Hecke algebras answers are also interesting; I once asked a friend if he planned to attend a talk about DAHA, and was told "nah, that sounds like a little too much affine Hecke algebra for my tastes." As prompts for answers, I'll share my limited understanding in the hopes a kind reader will elaborate:
1. I understand that they were introduced to solve the so-called [Macdonald conjectures](http://en.wikipedia.org/wiki/Macdonald_polynomial), though I don't know much about this and would be delighted to hear anyone's thoughts.
2. I understand that they can be defined in terms of certain "Dunkl-Opdam" differential operators, and can be thought of as the algebra of differential operators on a non-commutative resolution of $\mathfrak{h}/W$. I'd be interested to hear more about this.
3. I understand that they have an action by automorphisms of $\widetilde{SL\_2(Z)}$ coming alternately from their connection to configurations of points on a torus, or from their description via differential operators. If anyone has another take on this action, that would be interesting to hear about.
4. Apparently they also relate to the geometric Langlands program, which is what really prompts my post; can anyone elaborate on this?
| https://mathoverflow.net/users/1040 | Double affine Hecke algebras and mainstream mathematics | Well the first thing to say is to look at the very enthusiastic and world-encompassing papers of Cherednik himself on DAHA as the center of the mathematical world (say his 1998 ICM).
I'll mention a couple of more geometric aspects, but this is a huuuge area..
There are at least three distinct geometric appearances of DAHA, which you could classify by the number of loops (as in loop groups) that appear - two, one or zero.
(BTW for those in the know I will mostly intentionally ignore the difference between DAHA and its spherical subalgebra.)
Double loop picture: See e.g. Kapranov's paper arXiv:math/9812021 (notes for lectures of his on it available on my webpage) and the related arXiv:math/0012155. The intuitive idea, very hard to make precise, is that DAHA is the double loop (or 2d local field, such as F\_q((s,t)) ) analog of the finite (F\_q) and affine (F\_q((s)) ) Hecke algebras. In other words it appears as functions on double cosets for the double loop group and its "Borel" subalgebra. (Of course you need to decide what "functions" or rather "measures" means and what "Borel" means..) This means in particular it controls principal series type reps of double loop groups, or the geometry of moduli of G-bundles on a surface, looked at near a "flag" (meaning a point inside a curve inside the surface). The rep theory over 2d local fields that you would need to have for this to make sense is studied in a series of papers of Kazhdan with Gaitsgory (arXiv:math/0302174, 0406282, 0409543), with Braverman (0510538) and most recently with Hrushovski (0510133 and 0609115). The latter is totally awesome IMHO, using ideas from logic to define definitively what measure theory on such local fields means.
Single loop picture: Affine Hecke algebras have two presentations, the "standard" one (having to do with abstract Kac-Moody groups) and the Bernstein one (having to do specifically with loop groups). These two appear on the two sides of Langlands duality (cf eg the intro to the book of Chriss and Ginzburg). Likewise there's a picture of DAHA that's dual to the above "standard" one. This is developed first in Garland-Grojnowski (arXiv:q-alg/9508019) and more thoroughly by Vasserot arXiv:math/0207127 and several papers of Varagnolo-Vasserot. The idea here is that DAHA appears as the K-group of coherent sheaves on G(O)\G(K)/G(O) - the loop group version of the Bruhat cells in the finite flag manifold (again ignoring Borels vs parabolics). Again this is hard to make very precise. This gives in particular a geometric picture for the reps of DAHA, analogous to that for AHA due to Kazhdan-Lusztig (see again Chriss-Ginzburg).
[EDIT: A
[new survey](http://arxiv.org/abs/0911.5328) on this topic by Varagnolo-Vasserot has just appeared.]
Here is where geometric Langlands comes in: the above interp means that DAHA is the Hecke algebra that acts on (K-groups of) coherent sheaves on T^\* Bun\_G X for any Riemann surface X -- it's the coherent analog of the usual Hecke operators in geometric Langlands.
Thus if you categorify DAHA (look at CATEGORIES of coherent sheaves) you get the Hecke functors for the so-called "classical limit of Langlands" (cotangent to Bun\_G is the classical limit of diffops on Bun\_G).
The Cherednik Fourier transform gives an identification between DAHA for G and the dual group G'. In this picture it is an isom between K-groups of coherent sheaves on Grassmannians for Langlands dual groups (the categorified version of this is conjectured in Bezrukavnikov-Finkelberg-Mirkovic arXiv:math/0306413). This is a natural part of the classical limit of Langlands: you're supposed to have an equivalence between coherent sheaves on cotangents of Langlands dual Bun\_G's, and this is its local form, identifying the Hecke operators on the two sides!
In this picture DAHA appears recently in physics (since geometric Langlands in all its variants does), in the work of Kapustin (arXiv:hep-th/0612119 and with Saulina 0710.2097) as "Wilson-'t Hooft operators" --- the idea is that in SUSY gauge theory there's a full DAHA of operators (with the above names). Passing to the TFT which gives Langlands kills half of them - a different half on the two sides of Langlands duality, hence the asymmetry.. but in the classical version all the operators survive, and the SL2Z of electric-magnetic/Montonen-Olive S-duality is exactly the Cherednik SL2Z you mention..
Finally (since this is getting awfully long), the no-loop picture: this is the one you referred to in 2. via Dunkl type operators. Namely DAHA appears as difference operators on H/W (and its various degenerations, the Cherednik algebras, appear by replacing H by h and difference by differential). In this guise (and I'm not giving a million refs to papers of Etingof and many others since you know them better) DAHA is the symmetries of quantum many-body systems (Calogero-Moser and Ruijsenaars-Schneiders systems to be exact), and this is where Macdonald polynomials naturally appear as the quantum integrals of motion.
The only thing I'll say here is point to some awesome recent work of Schiffmann and Vasserot arXiv:0905.2555, where this picture too is tied to geometric Langlands..
very very roughly the idea is that H/W is itself (a degenerate version of an open piece of) a moduli of G-bundles, in the case of an elliptic curve. Thus studying DAHA is essentially studying D-modules or difference modules on Bun\_G in genus one (see Nevins' paper arXiv:0804.4170 where such ideas are developed further). Schiffman-Vasserot show how to interpret Macdonald polynomials in terms of geometric Eisenstein series in genus one..
enough for now.
| 53 | https://mathoverflow.net/users/582 | 6575 | 4,480 |
https://mathoverflow.net/questions/6543 | 11 | Algebraic geometry allows one to think of an $A$-module $M$ geometrically as a module of functions on the $A$-scheme $\mathrm{Spec}(\mathrm{Sym}(M))$.
I'm wondering if anything is lost in just replacing $M$ by this geometric object. Since nothing is lost in taking $\mathrm{Spec}$, this amounts to asking:
>
> (1) Can two non-isomorphic $A$-modules $M$ , $N$ have isomorphic symmetric $A$-algebras $\mathrm{Sym}(M)$ ,
> $\mathrm{Sym}(N)$?
>
>
>
(Clearly they are not isomorphic as *graded* $A$-algebras.)
If the answer is "No", great! If "Yes", I would like to see a specific example.
It may be interesting to have a second interpretation, even if it doesn't help solve the problem. Since we have the adjunction (of set-valued functors)
$hom\_{A-alg}(\mathrm{Sym}(M),B) \simeq hom\_{A\mathrm{-mod}}(M,B),$
by Yoneda's lemma, an equivalent question would be:
>
> (2) If the (set-valued) functors $hom\_{A-\mathrm{mod}}(M,-)$ and $hom\_{A-\mathrm{mod}}(N,-)$ agree on $A$-algbras, do they agree on $A$-modules?
>
>
>
**Edit:** I emphasized "set-valued" above, thanks to a comment from Buzzard. Also, partially in response to Mark Hovey's comment, I removed "Is it safe to think of modules geometrically" from the quesiton statement, since I don't want to assert that this is "the correct" geometric interpretation of a module.
| https://mathoverflow.net/users/84526 | Can different modules have the same symmetric algebra? (answered: no) | I now believe a-fortiori's argument: translations are a problem, but, as a-fortiori observed, they are the only problem. Let me spell it out.
Say $f:Sym(M)\to Sym(N)$ is an isomorphism. For $m\in M$ write $f(m)=f\_0(m)+f\_1(m)+f\_{\geq2}(m)$ with obvious notation: $f\_0(m)$ is in $A$, $f\_1(m)$ is in $N$ and $f\_{\geq2}(m)$ is all of the rest. Now here's another $A$-algebra map $g:Sym(M)\to Sym(N)$. To define $g$ all I have to do is to say where $m\in M$ goes so let's say $g(m)=f\_1(m)+f\_{\geq2}(m)$.
Claim: $g$ is an $A$-algebra isomorphism.
The proof is that $g$ is just $f$ composed with the isomorphism $Sym(M)\to Sym(M)$ sending $m$ to $m-f\_0(m)$ (one needs to check that this is an isomorphism but it is because there's an obvious inverse).
Claim: the isomorphism of rings inverse to $g$ also has "no constant terms", i.e. it's of the form $h:Sym(N)\to Sym(M)$ where $h(n)=h\_1(n)+h\_{\geq2}(n)$ with no constant term.
The proof is that $g$ sends terms of degree $d$ to terms of degree $d$ or higher, so applying $g$ to $h(n)=h\_0(n)+h\_1(n)+h\_{\geq2}(n)$ we see $n=h\_0(n)+f\_1(h\_1(n))+$(higher order terms).
Claim: $f\_1$ and $h\_1$ are mutual inverses. This is easy now.
So in fact all the ideas are in a-fortiori's comments.
| 6 | https://mathoverflow.net/users/1384 | 6582 | 4,483 |
https://mathoverflow.net/questions/6576 | 15 | Given a univariate polynomial with real coefficients, p(x), with degree n, suppose we know all the zeros xj, and they are all real. Now suppose I perturb each of the coefficients pj (for j ≤ n) by a small real perturbation εj. What are the conditions on the perturbations (**edit:** for example, how large can they be, by some measure) so that the solutions remain real?
Some thoughts: surely people have thought of this problem in terms of a differential equation valid for small ε that lets you take the known solutions to the new solutions. Before I try to rederive that, does it have a name? This seems like a pretty general solution technique, but perhaps it is so general as to be intractable practically, which might explain why I don't know about it.
If you ignore the smallness of the perturbation, then there is a general question here which seems like it might be related to Horn's problem: given two real polynomials p(x) and q(x) of degree n and their strictly real roots, what can you infer about the roots of p(x)+q(x)? This question is very interesting and I would love to hear what people know about it. But I'm also happy with the perturbed subproblem above, assuming it is indeed simpler.
| https://mathoverflow.net/users/1171 | Finding the new zeros of a "perturbed" polynomial | Assume that the roots of $p(x)$ are real and distinct. Then we may let $\mu > 0$ denote the minimum distance between any two roots. Let $q(x)$ be any polynomial of degree $< n$ such that $|q(\alpha)| < (\mu/2)^n$ for any root $\alpha$ of $p(x)$. Then I claim that $g(x) = p(x) + q(x)$ has real roots.
Note that for a root $\alpha$ of $p(x)$, we have $|g(\alpha)| = |q(\alpha)| < (\mu/2)^n$. Yet
$$|g(\alpha)| = \prod |\alpha - \beta\_i|$$
where the $\beta\_i$ are the roots of $g$. It follows that $g(x)$ has a root $\beta$ such that
$|\alpha - \beta| < \mu/2$. By the triangle inequality, $\alpha$ is uniquely determined by $\beta$ and this inequality.
In particular, $g(x)$ has exactly one root within $\mu/2$ of
each root of $p(x)$. Since $g(x)$ and $p(x)$ have the same degree, this exhausts all the roots of $g(x)$. Yet if $\beta$ was complex, then $|\alpha - \beta| = |\alpha -
\overline{\beta}| < \mu/2$, a contradiction.
If the roots of $p(x)$ are not distinct, then one is in trouble, as the example
$p(x) = x^2$ shows.
This can be thought of as an application of the $\mathbf{R}$-version of Krasner's Lemma, and is, in particular, an (exact) analog of the argument that the splitting field of a separable polynomial $f(x)$ over the $p$-adics is locally constant.
---
EDIT:
I am confused about your second question. Let $f(x)$ be any polynomial of degree $n$ with real coefficients. Let $A$ and $B$ denote the maximum and minimum values of $f(x)$ on the interval $[0,n]$. Choose any $M > \max(|A|,|B|)$.
A polynomial of degree $n$ may be defined by specifying $n+1$ of its values.
Let $p(x)$ be the polynomial of degree $n$ such that
the values $p(k)$ for $k = 0,\ldots,n$ are alternatively $-M$ and $+M$.
By the intermediate value theorem, $p(x)$ has $n$ real roots.
Let $q(x) = f(x) - p(x)$. The signs of $q(k)$ also alternate for $k = 0,
\ldots,n$ by construction. It follows that $q(x)$ also has $n$ real roots.
Yet $p(x) + q(x) = f(x)$, and thus the sum of two polynomials with real roots does not satisfy any restrictions.
| 15 | https://mathoverflow.net/users/nan | 6585 | 4,485 |
https://mathoverflow.net/questions/6560 | 10 | The Brauer-Nesbitt theorem (well, one of them) says that if $k$ is a field and I have two semisimple representations (on finite-dimensional $k$-vector spaces) $r\_1$ and $r\_2$ of a $k$-algebra $A$ with the property that the char polys of $r\_1(a)$ and $r\_2(a)$ coincide for all $a\in A$, then the representations are isomorphic.
Is it the case that if the char poly of $r\_1(a)$ divides the char poly of $r\_2(a)$ for all $a\in A$, then the smaller representation is a direct summand of the bigger?
[I came up against this recently, but fortunately in the case I was considering $A$ was commutative and $k$ had characteristic zero, and I convinced myself it was surely fine in this case (base change up to an alg closure of $k$ and convince yourself that the semisimple representations kill all the nilpotent elements, so WLOG $A$ is etale and now do it by hand). If $k$ is finite then I'm still not sure which way to bet. If $A$ were a group ring and we knew only that one char poly divided the other for all elements of the group, then my gut feeling is that this isn't enough in characteristic $p$ but maybe I'm wrong. If $k$ has characteristic zero then I am betting on yes but then again I'm no algebraist.]
| https://mathoverflow.net/users/1384 | Version of Brauer-Nesbitt for summands | The result is true, regardless of characteristic.
Lemma: Let A be a k algebra and M a semi-simple A-module which is finite dimensional as a k-algebra. Then the image of A in $\mathrm{End}\_k(M)$ is a semi-simple ring.
So, by the Artin-Wedderburn theorem, this image is a direct sum of matrix algebras over division rings.
Proof: Call this image S. Since S is finite dimensional, it is artinian. Let $M= \bigoplus V\_i$ and let $t=(t\_i)$ lie in the Jacobson radical of S. For each $V\_i$, the condition that $V\_i$ is a simple A-module implies that it is a simple S-module. So t must act trivially on $V\_i$, and thus $t\_i=0$. But we have proved this for all i, so $t=0$ and the Jacobson radical of S is trivial. QED.
We can now reduce to the case that A=S, and is a direct sum of division rings. Say $A = \bigoplus \mathrm{Mat}\_{n\_i}(\Delta\_i)$. So every A-module is of the form
$$\bigoplus (\Delta\_i^{n\_i})^{k\_i}$$
for some $k\_i$ and the corresponding characteristic polynomial is
$$\chi(\lambda, g) = \prod \chi\_i(\lambda, g\_i)^{k\_i}$$
where, for $h \in \mathrm{Mat}\_{n\_i}(\Delta\_i)$, the polynomial $\chi\_i(\lambda, h)$ is the characteristic polynomial of $h$ acting on $\Delta\_i^{n\_i}$.
A much better argument, suggested by buzzard's comment below. (I am not sure whether or not the original can be fixed.) Let $M = \bigoplus (\Delta\_i^{n\_i})^{k\_i}$ and $N = (\Delta\_i^{n\_i})^{\ell\_i}$. Suppose M is not a summand of N, so $k\_i > \ell\_i$ for some $i$. Let g be 1 on the i component and 0 everywhere else. Then the characteristic polynomials of M and N are of the form $(\lambda-1)^{n\_i k\_i} \lambda^{\bullet}$ and $(\lambda-1)^{n\_i \ell\_i} \lambda^{\bullet}$. So the former does not divide the latter.
We just need to show that the polynomials $\chi\_i$, as polynomial functions on $\overline{k} \times A$, are relatively prime to each other. This is easy enough. Let $t\_a$ be a basis for the $k$-linear functions on $\mathrm{Mat}\_{n\_i}(\Delta\_i)$ and $u\_b$ a basis for the $k$-linear functions on $\mathrm{Mat}\_{n\_j}(\Delta\_j)$. Then $\chi\_i(\lambda, g\_i)$ is a homogenous polynomial in $\lambda$ and $t\_a$; while $\chi\_j$ in homogenous in $\lambda$ and $u\_b$. Their GCD must be homogenous in both ways, hence, it is of the form $\lambda^m$.
But, if $\lambda | \chi\_i(\lambda)$, this means that there is no $g$ which acts invertibly on $\Delta\_i^{n\_i}$; contradicting that the identity does so act. So the GCD is 1, and the result is true.
| 6 | https://mathoverflow.net/users/297 | 6587 | 4,487 |
https://mathoverflow.net/questions/6593 | 14 | A fairly simple question: I've read in multiple sources that Godel proved that if we accept the axiom of constructibility in ZFC, then we can create an explicit formula that well-orders the real numbers. I tried searching for a paper or some other source that explains what this formula is, but I came up empty-handed. Can someone explain what this formula is, or perhaps point me to a resource that does?
| https://mathoverflow.net/users/1455 | V=L and a Well-Ordering of the Reals | The order is very easy. Under $V=L$, the set-theoretic universe is built according the hierarchy $(L\_\alpha \mid \alpha \in \mathrm{Ord})$, where $L\_0$ is empty, $L\_{\alpha+1}$ consists of all definable subsets of $L\_\alpha$, and $L\_\lambda$ is the union of all earlier $L\_\alpha$ when $\lambda$ is a limit ordinal.
Since we can order the definitions used to go from $L\_\alpha$ to $L\_{\alpha+1}$, we obtain a definable well-ordering of the entire universe. Namely, $x$ is less than $y$ iff
1. $x$ appears before $y$ in the hierarchy or
2. they appear at the same stage, but $x$ appears with an earlier definition than $y$.
If one analyzes the complexity of the resulting definition for real numbers, it has complexity $\Delta^1\_2$ in the descriptive set theoretic hierarchy.
| 29 | https://mathoverflow.net/users/1946 | 6600 | 4,494 |
https://mathoverflow.net/questions/6592 | 8 | I came along the following question while trying to understand and apply some ideas of Dugger's article *Universal Homotopy Theories*.
Suppose, we are given a nice model category $\mathcal{C}$, say left proper and cellular or combinatorial, so we have a good theory of localization. I am primarly thinking here of the category of presheaves of simplicial sets on some site with the projective model structure, where weak equivalences and fibrations are defined "pointwise".
Suppose furthermore, $S$ is a class of morphisms in $\mathcal{C}$ we can (left Bousfield) localize at [e.g. the class required for descent for hypercovers] and $T$ is an arbitrary set of morphisms in $\mathcal{C}$.
Now, let's consider a fibrant object $X$ in $\mathcal{C}[S^{-1}]$, i.e. some object which is $S$-local (and $\mathcal{C}$-fibrant), and take it's fibrant replacement $X^f$ in $\mathcal{C}[S^{-1}][T^{-1}]$.
Is it now reasonably to expect under some circumstances or even generally true that the map $X \to X^f$ is a weak equivalence in $\mathcal{C}[T^{-1}]$?
| https://mathoverflow.net/users/2039 | Localizing Model Structures | (Your question is basically about presentable (∞,1)-categories, so I will take the liberty of writing my answer in that language. Hopefully the translations to model category language will be straightforward.)
Inside $\mathcal{C}$ we have the full subcategories of $S$-local objects, $T$-local objects, and $(S \cup T)$-local objects, which are all reflexive subcategories; denote the associated localizations on $\mathcal{C}$ by $L\_S$, $L\_T$, $L\_{S \cup T}$ respectively. Your question is whether for $X$ an $S$-local object, the map $X \to L\_{S \cup T}X$ is a $T$-local equivalence, i.e., whether $L\_T X \to L\_T L\_{S \cup T} X = L\_{S \cup T} X$ is an equivalence in $\mathcal{C}$. In other words, the question is whether, if I start with an $S$-local object $X$, the localization $L\_T X$ is still $S$-local.
In general this will not hold. For example, take $I$ to be the category $\ast \to \ast$ and let $\mathcal{C}$ be the diagram category $\mathrm{Fun}(I, \mathrm{Spaces})$. Write a typical object $X$ of $\mathcal{C}$ as $[X\_1 \to X\_2]$. There are sets $S$ and $T$ of morphisms such that the $S$-local objects are the ones for which the map $X\_1 \to X\_2$ is an equivalence and the $T$-local objects are the ones for which $X\_2$ is a point (I think they're $S = \{[\emptyset \to \ast] \to [\ast \to \ast]\}$ and $T = \{[\emptyset \to \emptyset] \to [\emptyset \to \ast]\}$). Then $L\_S[X\_1 \to X\_2] = [X\_2 \to X\_2]$ and $L\_T[X\_1 \to X\_2] = [X\_1 \to \ast]$. Clearly the $T$-localization of an $S$-local object need not be $S$-local.
One situation where I think the statement would hold arises from looking at models of a (finitary) essentially algebraic theory inside an ∞-topos. The idea is that the topos is a left exact localization of a category of presheaves of spaces, so that localization preserves the finite limits used to define the theory. However, that localization would need to be $T$, not $S$ as in your example, so I'm not sure whether this is the kind of example you had in mind.
| 9 | https://mathoverflow.net/users/126667 | 6610 | 4,503 |
https://mathoverflow.net/questions/6608 | 10 | Let $B$ be the graded ring $\bigoplus\_i B^i$ (with $B^k B^l \subset B^{k+l}$), and $B\_f$ the multiplicative group of all formal sums $1 + b\_1 + b\_2 + \cdots$ where $b\_i \in B^i$ for all $i$.
The idea when talking about genera (such as the Todd genus or the $L$ genus) is that we can take $B$ to be $H^{2 \bullet}(X,\mathbb{Q})$ for example, and then a typical element of $B\_f$ is something like the total Chern class.
Now a genus corresponds to a multiplicative sequence $(K\_n)$, where each $K\_n$ is a polynomial in n variables over $B$ that is homogeneous (with respect to the grading), so that $K\_n(t x\_1, t^2 x\_2, \ldots, t^n x\_n) = t^n K\_n(x\_1, \ldots, x\_n)$. This corresponds to the idea that for an $n$ dimensional complex manifold, the evaluation of $K\_n$ on the fundamental class should be some number, so that we want this homogeneity to ensure that $K\_n$ corresponds to some element of the top cohomology group.
Given this, we can form an element $K(b)$ in $B\_f$ for each element $b = 1 + b\_1 + b\_2 + \cdots$ of $B\_f$ by $K(b) = 1 + K\_1(b\_1) + K\_2(b\_1,b\_2) + \cdots$.
Then multiplicativity of the sequence $(K\_n)$ means that $K(bc) = K(b)K(c)$, which is what you would want in algebraic topology to get multiplicativity of the numbers you get, from multiplicativity of the total Chern classes.
Now, all the books I have been reading about this say that there is an essentially unique way to associate such a multiplicative sequence to a formal power series, and that all multiplicative sequences come from this. But I don't understand precisely how; I don't even know over what ring the formal power series should be defined and even less how one can find a multiplicative sequence out of that.
It's easy enough when $B = H^{2 \bullet}(X,\mathbb{Q})$ and you're working with Chern classes, as the Chern classes of a direct sum of lines bundles are given by the symmetric polynomials in the first Chern classes. Then, given a formal power series $Q(x) \in \mathbb{Q}[[x]]$, one can form the product $\prod\_i Q(x\_i)$ in $\mathbb{Q}[[x\_1, \ldots, x\_n]]$ and get a multiplicative sequence as YBL says in his answer, by taking $1 + \sum\_j K\_j = \prod\_i Q(x\_i)$, where the $K\_j$ are taken as polynomials in the elementary symmetric polynomials (ie the Chern classes).
But I don't see how you can do this without appealing to this decomposition (or indeed in a more general setting without any idea of Chern classes).
For example, in *Characteristic Classes* by Milnor and Stasheff, they do the same explanation as I did above, but when they get to the formal power series part it seems that they just assume $B = \Lambda[t]$ for some commutative ring $\Lambda$ and from there I lose track of what is happening; this gets me really quite confused as to what's going on.
| https://mathoverflow.net/users/362 | How are multiplicative sequences related to formal power series and genera of manifolds? | I think the idea is that using the splitting principle everything reduces to the first Chern class of line bundles: Chern classes of a general bundle $E$ are symetric functions of $c\_1(L\_i)$ where $\bigoplus L\_i = E$.
If $Q(z)$ is a power series with constant term 1, you can define $K\_n$ by the formula:
$$
\sum K\_n(x\_1,\ldots,x\_n) = \prod Q(z\_j)
$$
where $x\_i$ is the $i$-th elementary symetric function of the variables $z\_j$. Hogomeneity corresponds to the fact that the $z\_j$ have degree 1.
I think the statement that every multiplicative sequence comes from such a power series is only true in caracteristic 0. A multiplicative sequence with coefficients in $A$ corresponds to a ring homomorphism from the Lazard ring $\mathbb{L} = \Omega^\*(pt)$ to $A$ that is to a formal group law $F(t\_1,t\_2) \in A[[t\_1,t\_2]]$. There is a natural action of power series $f(z)$ satistfying $f(0) = 0$ and $f'(0) = 1$ on the Lazard ring; just change of coordinate of formal group laws $(F^f)(t\_1,t\_2) := f^{-1}(F(f(t\_1),f(t\_2))$. Now in caracteristic zero, this action is simply transitive: every law is equivalent to the additive one $(t\_1,t\_2) \mapsto t\_1+t\_2$ because we can define the logarithm of a law by formally integrating an invariant differential. This should correspond to the fact that every multiplicative sequence is defined by a power series $Q(z) = z/f(z)$.
| 6 | https://mathoverflow.net/users/1985 | 6614 | 4,507 |
https://mathoverflow.net/questions/6618 | 56 | I was wondering whether there is some notion of "vector bundle de Rham cohomology".
To be more precise: the k-th de Rham cohomology group of a manifold $H\_{dR}^{k}(M)$ is defined as the set of closed forms in $\Omega^k(M)$ modulo the set of exact forms. The coboundary operator is given by the exterior derivative.
Let now $E \rightarrow M$ be a vector bundle with connection $\nabla^E$ over $M$, and consider the $E$-valued $k$-forms on $M$: $\Omega^k(M,E)=\Gamma(\Lambda^k TM^\ast \otimes E)$.
If $E$ is a **flat** vector bundle, we get a coboundary operator $d^{\nabla^E}$ (since $d^{\nabla^E} \circ d^{\nabla^E} = R^{\nabla^E}=0$, with $R^{\nabla^E}$ being the curvature) and we can define
$$H\_{dR}^{k}(M,E) := \frac{ker \quad d^{\nabla^E}|\_{\Omega^k(M,E)}}{im \quad d^{\nabla^E}|\_{\Omega^{k-1}(M,E)}}$$
So my question: Is this somehow useful? I mean can one use this definition to make some statements about $M$ or $E$ or whatever? Or is the restriction of $E$ to be a flat vector bundle somehow disturbing? Or is this completely useless?
| https://mathoverflow.net/users/675 | de Rham cohomology and flat vector bundles | Warning: The first paragraph of the following is outside my expertise.
I am told this construction is very useful in PDE's. If you have a PDE on some manifold $M$, you can often formulate the vector space of solutions as the kernel of some flat connection on a vector bundle. In particular, I believe that the analytic side of the [Atiyah-Singer index theorem](http://en.wikipedia.org/wiki/Atiyah-Singer_index_theorem#The_analytical_index) is the Euler characteristic of the deRham theory you have described.
I can tell you that the analogous construction is very important in complex algebraic geometry. Given a holomorphic vector bundle on a complex manifold, there is a natural way to define a $d$-bar connection on it. (This mean $\nabla\_X$ is only defined when $X$ is a $(0,1)$ vector field.) The cohomology of the resulting deRham-like complex, which is called the Dolbeault complex in this setting, is the same as the cohomology of the sheaf of holomorphic sections of the vector bundle. See Wells' *Differential Analysis on Complex Manifolds* or the early parts of Voisin's *Hodge Theory and Algebraic Geometry*.
| 21 | https://mathoverflow.net/users/297 | 6625 | 4,514 |
https://mathoverflow.net/questions/6541 | 2 | Does anyone know of good references for nonstandard set theories and their applications to various branches of mathematics like category theory, algebra, geometry, etc.?
Edit: What I mean by "nonstandard set theory" is a formalization of the naive notion of sets that allows direct arguments about certain intuitive notions like infinitesimals and infinite integers without recourse to model theoretic constructions like ultrafilters and ultraproducts. Infinitesimal number is the usual application I keep seeing but I'm sure there must be other applications and that's the intent of my question. I'm not sure if this is precise enough.
| https://mathoverflow.net/users/nan | nonstandard set theories | It seems that Lars Brünjes and Christian Serpé have a whole program for introducing non standard mathematics in algebraic geometry; they wish to play with non standard contructions, seen internally and externally (the interest of this game consists precisely to look at the non-classical logic (or internal) point of view and at the classical (or external) one at the same time). For instance, for an infinite prime number $P$, the ring $\mathbf{Z}/P\mathbf{Z}$ behaves internally like a finite field, while externally, it is a field of characteristic zero which contains an algebraic closure of $\mathbf{Q}$. Non-standard constructions can often be interpreted in a precise way as standard ones using ultraproducts and ultrafilters. Their purpose is to develop all the tools of classical algebraic geometry (homotopical and homological algebra, stacks, étale cohomology, algebraic K-theory, higher Chow groups...) in a non-standard way, in order to prove facts in the classical setting. Most of their papers can be found [here](http://wwwmath.uni-muenster.de/u/serpe/arbeiten-en.html) (their papers contains more precise ideas on the possible interpretations and explanations). Brünjes and Serpé see non-standard mathematics as an enlargement of standard mathematics, and their work deals a lot in making this precise. However, they seem to have quite few concrete problems in mind. For instance, they have found sufficient conditions on cohomology classes to be algebraic in a very classical sense (see arXiv:0901.4853).
| 7 | https://mathoverflow.net/users/1017 | 6628 | 4,516 |
https://mathoverflow.net/questions/6636 | 0 | Let $S\_n$ denote the symmetric group on $n$ letters, and let $S\_n(p)$ denote a Sylow $p$-subgroup. Why is the image of $H\_i(S\_n(p))$ in $H\_i(S\_n)$ the $p$-primary part of $H\_i(S\_n)$?
| https://mathoverflow.net/users/2046 | Homology of symmetric groups | That is the content of the first part of theorem 10.1 in Cartan-Eilenberg (up to the fact that their $\hat H^i$ is the same as $H\_{-i-1}$ for $i<-1$). In particular, this is true for all finite groups, not just the symmetric ones.
| 3 | https://mathoverflow.net/users/1409 | 6638 | 4,522 |
https://mathoverflow.net/questions/6637 | 11 | I'm reading Madsen and Tornehave's "From Calculus to Cohomology" and tried to solve this interesting problem regarding knots.
Let $\Sigma\subset \mathbb{R}^n$ be homeomorphic to $\mathbb{S}^k$, show that $H^p(\mathbb{R}^n - \Sigma)$ equals $\mathbb{R}$ for $p=0,n-k-1, n-1$ and $0$ for all other $p$. Here $1\leq k \leq n-2$.
Now the case $p=0$ is obvious from connectedness and the two other cases are easily solved by applying the fact that
$H^{p+1}(\mathbb{R}^{n+1} - A) \simeq H^p(\mathbb{R}^n - A),~~~~p\geq 1$
and
$H^1(\mathbb{R}^n - A) \simeq H^0(\mathbb{R}^n - A)/\mathbb{R}\cdot 1$
So what is my problem, really?
Now instead let's look at this directly from Mayer-Vietoris. If $\hat{D}^k$ is the open unit disk and $\bar{D}^k$ the closed. Then $\mathbb{R}^n - \mathbb{S}^k = (\mathbb{R}^n - \bar{D}^k)\cup (\hat{D}^k)$ and $(\mathbb{R}^n - \bar{D}^k)\cap (\hat{D}^k) = \emptyset$
Now
$H^p(\mathbb{R}^n - \bar{D}^k) \simeq H^p(\mathbb{R}^n - \{ 0 \})$ since $\bar{D}^k$ is contractible. And $H^p(\mathbb{R}^n - \{ 0 \})$ is $\mathbb{R}$ if $p=0,n-1$ and $0$ else. Since $\hat{D}^k$ is open star shaped we find it's cohomology to be $\mathbb{R}$ for $p=0$ and $0$ for all other $p$.
This yields and exact sequence
$\cdots\rightarrow 0\overset{I^{\ast}}\rightarrow H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \overset{J^{\ast}}\rightarrow \mathbb{R} \rightarrow 0\cdots$
So due to exactness I find that $\ker(J^\*) = \text{Im}(I^\*) = 0$ and that $J^\*$ is surjective, hence
$H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \simeq \mathbb{R}$.
But ... If I apply the exact same approach to $p = n-k-1$ my answer would be $0$ for $H^{n-k-1}(\mathbb{R}^n - \mathbb{S}^k)$.
Where does this last approach fail?
| https://mathoverflow.net/users/2044 | The De Rham Cohomology of $\mathbb{R}^n - \mathbb{S}^k$ | To apply the Mayer-Vietoris sequence, you need subspaces whose *interiors* cover your space (see e.g. [Wikipedia](http://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence), or [Hatcher](http://www.math.cornell.edu/~hatcher/AT/ATpage.html), p. 149). This is not true in your example, because a *k*-disk in Rn has empty interior for k<n.
You might also enjoy deriving this result from [Alexander duality](http://en.wikipedia.org/wiki/Alexander_duality).
---
Edit: Carsten makes an excellent point, which is that the hard part is to show that the homology of the complement is independent of the embedding. You did say that this is proved in your book, but I wanted to point out that this is quite difficult and arguably surprising.
1) One embedding that satisfies the conditions of the theorem is the [Alexander horned sphere](http://mathworld.wolfram.com/AlexandersHornedSphere.html), a "wild" embedding of S2 into S3 (this [animation](http://www.youtube.com/watch?v=d1Vjsm9pQlc) is quite nice too). While it's true that the outer component of the complement has the homology of a point, it is very far from being simply connected -- in fact its fundamental group is not finitely generated. (You can find an explicit description of its fundamental group in Hatcher, p. 170-172.)
2) Every knot is an embedding of S1 into S3. The fundamental group of the complement is a strong knot invariant, and is usually much more complicated than just Z. Since H1 of the knot complement is the abelianization of the knot group, the result you are using implies that all knot groups have infinite cyclic abelianization. This is true (it can be seen nicely from the [Wirtinger presentation](http://books.google.com/books?id=s4eGEecSgHYC&lpg=PA56&ots=GGa4d52ITK&pg=PA56)), but it's not obvious.
3) It is important that the ambient space is a sphere (or equivalently Rn). For a simple example where the theorem breaks down, consider embeddings of S1 into a surface Σg of genus g≥2. Taking g=2 for simplicity, we see that there are three topologically inequivalent ways of embedding a circle into Σ2: A) a tiny loop enclosing a disk; B) a loop encircling the waist of the surface and separating it into two components, each of genus 1; and C) a loop going through one of the handles, which does not separate the surface at all.
The homology groups of Σ2 are H0=Z, H1=Z4, and H2=Z. For both A) and B), the complement of S1 has homology groups H0=Z2 because the curve separates, and H1=Z4. However, for C) we have H0=Z because the complement is connected, and H1=Z3 because we have "interrupted" one of the elements [you can see where it went by looking at the Mayer-Vietoris sequence]. Thus we see that the homology of the complement depends essentially on the embedding into the surface Σg, in contrast with the classical case of embedding a circle into the sphere S2.
| 15 | https://mathoverflow.net/users/250 | 6639 | 4,523 |
https://mathoverflow.net/questions/6647 | 13 | Let $H\leq G$ be an inclusion of finite groups. Define a map $E\colon \mathbb{C}[G]\to \mathbb{C}[H]$ to be the $\mathbb{C}$-linear extension of
$$
E(g)=\begin{cases}
g &\text{if } g\in H\\\
0 &\text{else,}
\end{cases}
$$
i.e., $E$ is the projection onto $\mathbb{C}[H]$. A finite subset $B\subset \mathbb{C}[G]$ will be called a left basis for $G$ over $H$ if
$$
x=\sum\limits\_{b\in B} b E(b^\ast x)
$$
for all $x\in \mathbb{C}[G]$, where $\ast$ is the anti-linear extension of the map $g\mapsto g^{-1}$. For an example, take $B$ to be a set of left-coset representatives. Similarly, we can define a right basis to be a finite subset $B\subset \mathbb{C}[G]$ such that
$$
x=\sum\limits\_{b\in B} E(x b^\ast)b
$$
for all $x\in\mathbb{C}[G]$.
Note that there exist groups for which there is a basis which is both a left and right basis, but $H$ is not a normal subgroup of $G$. One can take the subgroup of the symmetric group $S\_n$ ($n\geq 3$) which fixes $1$. Then a set of left and right coset representatives is given by
$$
\{ (1 j)|j=1,\dots,n\}.
$$
Does there always exist a basis which is both a left and right basis, or are there inclusions of groups for which there is no simultaneous left and right basis?
The motivation for this question is another question from subfactor theory: if $N\subset M$ is a finite index, extremal $II\_1$-subfactor, does there always exist a Pimsner-Popa basis which is both a left and right basis? The subgroup subfactor is an example of such a subfactor, and the question posed above is a watered-down version of the subfactor question, where perhaps an answer is already known or more easily obtainable.
| https://mathoverflow.net/users/351 | Do subgroups have "two sided bases"? | If $H$ is a subgroup of finite index in a group $G$, there is a subset $\mathcal B$ of $G$ which serves both as a set of representatives for the left cosets of $H$ in $G$ and as a set of representatives for the right cosets of $H$ in $G$. (See, for example, Theorem 3, §4, Chap. I, in the book *The Theory of groups* by H. Zassenhaus) That should do it, no?
| 12 | https://mathoverflow.net/users/1409 | 6652 | 4,529 |
https://mathoverflow.net/questions/314 | 6 | Is there a good monoidal structure on a category of integrable representations of a quantum affine algebra? In the ordinary affine Kac-Moody case, there is the usual tensor product (symmetric, adds charges) and a fusion structure (braided, comes from G-bundles on curves, preserves central charge). In the quantum case, there is the usual tensor product (braided meromorphic-braided$^\ast$), but all I see in the literature about fusion is vague comments that it can't exist. I guess my question should be "what is the major malfunction?"
$^\ast$ **Edit:** The meromorphic property (in the sense of Soibelman's *Meromorphic tensor categories*) seems to be a first hint at problems, and I should have paid better attention to it.
| https://mathoverflow.net/users/121 | Are there interesting monoidal structures on representations of quantum affine algebras? | The fusion product for affine Lie algebras is closely related to the existence of "evaluation homomorphisms" from the loop algebra to the finite-dimensional semisimple Lie algebra g, which split the natural inclusion of g as the subalgebra of constant loops. In the quantum case there is no evaluation map from the quantum affine algebra to the finite-type quantum algebra outside of type A (this is proved - at least for Yangians - in Drinfeld's original paper I'm pretty sure).
You see consequences of this in lots of places: e.g. for representations of g, the evaluation homomorphisms mean any irreducible representation for g can be lifted to an irreducible representation of the affine Lie algebra Lg. On the other hand, irreducible representations of the associated quantum groups do not (necessarily) lift to representations of quantum affine algebras, and so one asks about "minimal affinizations" -- irreducible finite dimensional representations of the quantum affine algebra which have the given irreducible as a constituent when restricted to the finite-type quantum group.
That said, the "ordinary" tensor product for finite dimensional representations of quantum affine algebras is pretty interesting -- it's not braided any more for example.
| 5 | https://mathoverflow.net/users/1878 | 6655 | 4,532 |
https://mathoverflow.net/questions/6651 | 13 | Gauss's Lemma on irred. polynomial says,
Let R be a UFD and F its field of fractions. If a polynomial f(x) in R[x] is reducible in F[x], then it is reducible in R[x].
In particular, an integral coefficient polynomial is irreducible in Z iff it is irreducible in Q. For me this tells me something on how the horizontal divisors in the fibration from the arithmetic plane SpecZ[x] to SpecZ intersects the generic fiber: a prime divisor (the divisor defined by the prime ideal (f(x)) in Z[x]) intersect the generic fiber exactly at one point (i.e. the prime ideal (f(x)) in Q[x]) with multiplicity one.
Now here is my question:
**Give a ring R, with Frac(R)=F, and a polynomial f(x) in R[x] such that f(x) is reducible in F[x], but is irreducible in R[x].**
Of course, R should not be a UFD.
I'd like to see an example for number fields as well as a geometric example (where R is the affine coordinate ring of an open curve or higher dimensional stuff). Thanks
| https://mathoverflow.net/users/1657 | "Counter"-example for Gauss's Lemma on irreducible polynomials | Gauss' Lemma over a domain R is usually taken to be a stronger statement, as follows:
If R is a domain with fraction field F, a polynomial f in R[T] is said to be *primitive* if the ideal generated by its coefficients is not contained in any proper principal ideal. One says that Gauss' Lemma holds in R if the product of two primitive polynomials is primitive. (This implies that a polynomial which is irreducible over R[T] remains irreducible over F[T].) Say that a domain is a *GL-domain* if Gauss' Lemma holds.
It is known that this property is intermediate between being a GCD-domain and having irreducible elements be prime (which I call a EL-domain; this is not standard). Here is a relevant MathSciNet review:
---
MR0371887 (51 #8104)
Arnold, Jimmy T.; Sheldon, Philip B.
Integral domains that satisfy Gauss's lemma.
Michigan Math. J. 22 (1975), 39--51.
Let $D$ be an integral domain with identity. For a polynomial $f(x)\in D[X]$, the content of $f(X)$, denoted by $A\_f$, is the ideal of $D$ generated by the coefficients of $f(X)$. The polynomial $f(x)$ is primitive if no nonunit of $D$ divides each coefficient of $f(X)$ (or equivalently, if $D$ is the $v$-ideal associated with $A\_f$). On the other hand, $f(X)$ is superprimitive if $A\_f{}^{-1}=D$. The authors study, among other things, the relation between the following four properties on an integral domain: (1) each pair of elements has a greatest common divisor; (2) each primitive polynomial is superprimitive; (3) the product of two primitive polynomials is primitive; (4) each irreducible element is prime. In an integral domain $D$, the implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) hold, while no reverse implication holds in general. On the other hand, the properties (2), (3) and (4) are equivalent in $D[X]$.
---
On the other hand, when R is Noetherian, all of these conditions are equivalent, and equivalent to being a UFD: see, e.g., Theorem 17 of
[http://alpha.math.uga.edu/~pete/factorization.pdf](http://alpha.math.uga.edu/%7Epete/factorization.pdf)
Thus a Noetherian domain satisfies Gauss' Lemma iff it is a UFD. In particular, such rings must be integrally closed, but this condition is not sufficient: e.g. take the ring of integers of any number field which is not of class number one (for instance Z[\sqrt{-6}]).
| 22 | https://mathoverflow.net/users/1149 | 6659 | 4,536 |
https://mathoverflow.net/questions/6635 | 9 | I am looking for a certain masa in a $II\_1$ factor which is singular and has nontrivial Takesaki invariant.
For this I am looking for an example of an inclusion of groups $H\subset G$ such that:
* $G$ is a countable icc (infinite conjugacy class) group
* $H$ is abelian
* $\forall g\in G-H,\{ hgh^{-1} |h\in H \}$ is infinite
* $ |H\backslash G/H| \geq 3$
* there exists $g\in G-H$ and $h\_1\neq h\_2\in H$ with $h\_1 g=gh\_2$.
Does such an example exist?
| https://mathoverflow.net/users/2045 | Does such a subgroup exist? | [generalization of Agol's answer]
Take $H$ a group and let $K$ act on $H$ by isomorphisms (write the action as $\sigma$) and consider $G=H\rtimes\_\sigma K$. Then
* condition 2 is satisfied if $H$ is abelian
* condition 4 is satisfied if $K$ contains at least 3 elements
* condition 5 is satisfied if $K$ acts non-trivially
* condition 3 is satisfied if $\{h^{-1}\sigma\_k(h) : h\in H\}$ is infinite for all $k \in K$
* condition 1 is satisfied if $K$ acts with infinite orbits on $H$ and condition 3 is satisfied.
| 7 | https://mathoverflow.net/users/2055 | 6682 | 4,552 |
https://mathoverflow.net/questions/6685 | 5 | To establish the Weil conjectures for $n$-dimensional projective space over a finite field is elementary. Does there exist a simple direct proof of the conjectures for finite field Grassmannians?
| https://mathoverflow.net/users/1095 | Weil Conjectures for Grassmannians | The first result on the google search "zeta function of grassmannian" seems to contain quite a direct and not too long derivation of the zeta function for a grassmannian over a finite field:
<http://www.math.mcgill.ca/goren/SeminarOnCohomology/GrassmannVarieties%20.pdf>
From the zeta you see that it is rational, of course get the zeros (which are none), but you don't immediately get confirmation of the functional equation. Though, from the very simple combinatoric representation of the zeta function, it might be easy to prove directly, I will try with pen and paper later.
I'm glad I searched this, I didn't know the zeta was so simple in this case as well
| 10 | https://mathoverflow.net/users/2024 | 6688 | 4,557 |
https://mathoverflow.net/questions/6674 | 9 | Given a number field K of dimension d over Q, and galois closure of dimension d! over Q (i.e galois group Sd), can we relate the discriminant of the galois closure to that of the discriminant of K? Assume no special ramification happens in the closure or the other subfields, for example if the discriminant of K is a prime.
Tests in sage indicate that the discriminant of the galois closure is $\Delta\_K^{\frac{d!}{2}}$, and that the discriminants of the other subfields are also powers of $\Delta\_K$, but the power has something to do with the corresponding subgroup of Sd. (Not just the size of the subgroup)
Is there a way to prove the first indication, and thoughts about the second?
| https://mathoverflow.net/users/2024 | How do I calculate the discriminant of a galois closure and its other subfields? | It's true for $d = 2$. Even for $d = 3$, it fails miserably. If $K = \mathbf{Q}(p^{1/3})$ and $p \ne 3$ then $p^2$ exactly divides $\Delta\_{K}$, whereas $p^4$ exactly divides the discriminant of the Galois closure. (As David points out, things are even worse for $p = 3$.) Not to mention the fact that $p$ doesn't divide to any power the discriminant of $\mathbf{Q}(\sqrt{-3})$ which is contained inside the Galois closure of $K$.
About the only time the power of $p$ dividing the discriminant of the Galois closure is $n!/2$ times the power of $p$ dividing the discriminant is when $p$ *exactly* divides $\Delta\_K$. In this case, the power of $p$ dividing the fixed field of $H$ in $S\_n$ is equal to
$$(n-2)! \cdot \frac{\text{the number of $2$-cycles which do not lie in $H$}}{|H|}$$
This is always an integer, and it answers your question when $\Delta\_K$ is squarefree. (The general case having already seen to be false.)
| 11 | https://mathoverflow.net/users/nan | 6714 | 4,574 |
https://mathoverflow.net/questions/6711 | 50 | Is there a (preferably simple) example of a function $f:(a,b)\to \mathbb{R}$ which is everywhere differentiable, such that $f'$ is not Riemann integrable?
I ask for pedagogical reasons. Results in basic real analysis relating a function and its derivative can generally be proved via the mean value theorem or the fundamental theorem of calculus. Proofs via FTC are often simpler to come up with and explain: you just integrate the hypothesis to get the conclusion. But doing this requires $f'$ (or something) to be integrable; textbooks taking such an approach typically stipulate that $f'$ is continuous. Proofs via MVT can avoid such unnecessary assumptions but may require more creativity. So I'd like an example to show that the extra work does actually pay off.
Note that derivatives of everywhere differentiable functions cannot be arbitrarily badly behaved. For example, they satisfy the conclusion of the intermediate value theorem.
| https://mathoverflow.net/users/1044 | Integrability of derivatives | I believe this answers the question:
---
MR0425042 (54 #13000)
Goffman, Casper
A bounded derivative which is not Riemann integrable.
Amer. Math. Monthly 84 (1977), no. 3, 205--206.
In 1881 Volterra constructed a bounded derivative on $[0,1]$ which is not Riemann integrable. Since that time, a number of authors have constructed other such examples. These examples are generally relatively complicated and/or involve nonelementary techniques. The present author provides a simple example of such a derivative $f$ and uses only elementary techniques to show that $f$ has the desired properties.
---
The paper is available [here](http://alpha.math.uga.edu/%7Epete/Goffman77.pdf):
| 48 | https://mathoverflow.net/users/1149 | 6716 | 4,576 |
https://mathoverflow.net/questions/6701 | 31 | Is there a canonical definition of the concept of inner products for vector spaces over arbitrary fields, i.e. other fields than $\mathbb R$ or $\mathbb C$?
| https://mathoverflow.net/users/2058 | Definition of inner product for vector spaces over arbitrary fields | As Qiaochu wrote, the answer to the question is "not really, but..." Let me amplify on the "but" part.
Positive-definiteness inherently requires an ordering on your field. Conversely, if you have an ordered field, then the theory of inner products goes through verbatim. (The ordering does not have to be Archimedean, so this indeed gives lots more examples.)
Let's assume that you have a field K of characteristic different from 2 which cannot be ordered: by a theorem of Artin-Schreier, this is equivalent to -1 being a sum of squares in the field. Then you don't have "positive definiteness". What is more, the "standard inner product"
$$q(x\_1,...,x\_n) = x\_1^2 + ... + x\_n^2$$
will be isotropic for sufficiently large $n$, i.e., there will exist nonzero vectors
$v = (x\_1,...,x\_n)$ such that $q(v) = 0$. For instance, if K is finite, this occurs as soon as $n \ge 3$.
Let me remark that "isotropic inner products" are not inherently worthless. I have a preliminary version of a wonderful book, "Linear Algebra Methods in Combinatorics" by Laszlo Babai, which indeed makes nice use of the above inner product over finite fields, even in characteristic 2.
(See <http://www.cs.uchicago.edu/research/publications/combinatorics>. Unfortunately it seems that the book never came to fruition. I got my copy more than 10 years ago when I took an undergraduate course in combinatorics from Babai.)
On the other hand, to any quadratic form over a field K of characteristic not 2, you can associate a symmetric bilinear form. See (among infinitely other references) p. 2 of
[http://alpha.math.uga.edu/~pete/quadraticforms.pdf](http://alpha.math.uga.edu/%7Epete/quadraticforms.pdf)
As above, it is plausible that an algebraic substitute for "inner product space" is "vector space endowed with an anisotropic quadratic form", i.e., a regular quadratic form without nonzero vectors v for which $q(v) = 0$. Witt discovered that you can do a lot of "geometry" in this case: especially, he defined reflection through the hyperplane determined by any anisotropic vector: see (e.g....) pp. 17-18 of the above reference. More is true here than is included in my introductory notes: for instance the orthogonal group of an anistropic quadratic form has the "compactness properties" of the standard real orthogonal group O(n) (that is, it contains no nontrivial split subtorus).
| 28 | https://mathoverflow.net/users/1149 | 6721 | 4,579 |
https://mathoverflow.net/questions/6704 | 45 | There are a few questions about CM rings and depth.
1. Why would one consider the concept of depth? Is there any geometric meaning associated to that? The consideration of regular sequence is okay to me. (currently I'm regarding it as a generalization of not-a-zero-divisor that's needed to carry out induction argument, e.g. as in $\operatorname{dim} \frac{M}{(a\_1,\dotsc,a\_n)M} = \operatorname{dim} M - n$ for $M$-regular sequence $a\_1,\dotsc,a\_n$; correct me if I'm wrong!) But I don't understand why the length of a maximal regular sequence is of interest. Is it merely due to some technical consideration in cohomology that we want many $\operatorname{Ext}$ groups to vanish?
2. What does CM rings mean geometrically? As I read from Eisenbud's book, there doesn't seem to be an exact geometric concept that corresponds to it. Nonetheless I would still like to know about any geometric intuition of CM rings. I know that it should be locally equidimensional. Some examples of CM rings come from complete intersection (I read this from wiki). But what else?
3. Why do we care about CM rings? If I understand it correctly, CM rings ⇔ unmixedness theorem holds for every ideal for a noetherian ring, which should mean every closed subschemes have equidimensional irreducible components (and there's no embedded components). This looks quite restrictive.
| https://mathoverflow.net/users/nan | How to think about CM rings? | "Life is really worth living in a Noetherian ring $R$ when all the local rings have the property that every s.o.p. is an R-sequence. Such a ring is called Cohen–Macaulay (C–M for short).": [Hochster, "Some applications of the Frobenius in characteristic 0", 1978](https://doi.org/10.1090/S0002-9904-1978-14531-5).
Section 3 of that paper is devoted to explaining what it "really means" to be Cohen–Macaulay. It begins with a long subsection on invariant theory, but then gets to some algebraic geometry that will interest you.
In particular, he points out that if $R$ is a standard graded algebra over a field, then it is a module-finite algebra over a polynomial subring $S$, and that $R$ is Cohen–Macaulay if and only if it is free as an $S$-module. Equivalently, the scheme-theoretic fibers of the finite morphism $\operatorname{Spec} R \to \operatorname{Spec} S$ all have the same length.
At the end of section 3, Hochster explains that the CM condition is exactly what is required to make intersection multiplicity "work correctly": If $X$ and $Y$ are CM, then you can compute the intersection multiplicity of $X$ and $Y$ without all those higher $\operatorname{Tor}$s that Serre had to add to the definition.
He gives lots of examples and explains "where Cohen–Macaulayness comes from" (or doesn't) in each one. The whole thing is eminently readable and highly recommended.
| 32 | https://mathoverflow.net/users/460 | 6724 | 4,580 |
https://mathoverflow.net/questions/6719 | 21 | I know that for a finitely presented $A$-module $M$ ($A$ a commutative ring), TFAE:
1. $M$ is *projective*;
2. $M$ is *max-locally free*, meaning that $M\_{\mathfrak m}$ is free for every maximal ideal $\mathfrak m$;
3. $M$ is *prime-locally free*, meaning that $M\_{\mathfrak{p}}$ is free for every prime ideal $\mathfrak p$;
4. $M$ is *Zariski-locally free*, meaning that there are some $f\_1,\ldots,f\_n$ generating the unit ideal in $A$ such that each $M\_{f\_i}$ is free.
(Reference: Eisenbud commutative algebra, p. 136 / end of chapter 4).
I know that (1) implies (2) without finite presentation: see [Kaplansky (1958): *Projective Modules*](https://www.jstor.org/stable/1970252), p. 374. (He doesn't even assume $A$ is commutative, and uses an awesome lemma that any projective module is a direct sum of countably-generated submodules.) Finite presentation is used to prove (3) implies (4), as is often the case when passing from stalks of a sheaf to actual open sets.
So now I'm wondering in particular if you need finite presentation to prove (4) implies (1), and more generally,
>
> If $M$ is *Zariski-locally projective* (meaning there are some $f\_1,\ldots,f\_n$ generating the unit ideal in $A$ such that each $M\_{f\_i}$ is projective), is it *projective*?
>
>
>
If so, how can I see this directly / commutative-algebraically?
---
**Follow up:** I checked out Bhargav's reference, [Raynaud and Gruson: *Critères de platitude et de projectivité*](https://doi.org/10.1007/BF01390094). It turns out (on p. 81) they actually use the same technique as Kaplansky in the paper I linked above, of writing a module as a transfinite union with countably generated successive quotients, which they call a "Kaplansky division" when these quotients are direct summands. The conclusion that projectiveness is Zariski-local is stated as Example 3.1.4(3) on the bottom of page 82.
Tricky stuff!
| https://mathoverflow.net/users/84526 | Is projectiveness a Zariski-local property of modules? (Answered: Yes!) | Being projective is indeed a local property for the Zariski topology. In fact, it is even local for the fpqc topology --- this is a famous theorem of Raynaud and Gruson (see MR0308104).
| 13 | https://mathoverflow.net/users/986 | 6738 | 4,592 |
https://mathoverflow.net/questions/6508 | 28 | If doing geometry over $\mathbb F\_p$ means also using its algebraic closure, it must be interesting to talk about the algebraic closure of $\mathbb F\_1$ - the field with one element.
I saw that the finite extensions of $\mathbb F\_1$ are considered as $\mu\_n$, but an article by Connes et al says that it is unjustified to think of the direct limit of these.
In their paper, the group ring $\mathbb Q[\mathbb Q/\mathbb Z]$ appears a lot. Maybe it's one of $\mathbb Q/\mathbb Z$, $\mathbb Q[\mathbb Q/\mathbb Z]$, $\mathbb Z[\mathbb Q/\mathbb Z]$ ?
What is the algebraic closure of the field with one element?
And then, what is $\overline{\mathbb F\_1} \otimes\_{\mathbb F\_1}\mathbb Z$?
This seems like a very interesting question...
| https://mathoverflow.net/users/2024 | What is the algebraic closure of the field with one element? | There have been several questions on mathoverflow about the field with one element. Of course, such a field doesn't really exist and the discussion must fray sooner or later. So here is a different kind of answer.
Besides finite fields, which are 0-manifolds, there are only two fields which are manifolds, $\mathbb{C}$ and $\mathbb{R}$. There is a generalization of cardinality for manifolds and similar spaces, namely the geometric Euler characteristic. (This is as opposed homotopy-theoretic Euler characteristic; they are equal for compact spaces.) The geometric Euler characteristic of $\mathbb{C}$ is 1, while the geometric Euler characteristic of $\mathbb{R}$ is -1. In this sense, $\mathbb{C} = \mathbb{F}\_1$ while $\mathbb{R} = \mathbb{F}\_{-1}$.
It works well for some of the motivating examples of the fictitious field with one element. For instance, the Euler characteristic of the Grassmannian $\text{Gr}(k,n)$ over $\mathbb{F}\_q$ is then uniformly the Gaussian binomial coefficient $\binom{n}{k}\_q$.
In this interpretation, $\mathbb{F}\_1$ is algebraically closed. It is also a quadratic extension of $\mathbb{F}\_{-1}$; the generalized cardinality squares, as it should.
| 12 | https://mathoverflow.net/users/1450 | 6745 | 4,597 |
https://mathoverflow.net/questions/6668 | 2 | Preface: I think this is interesting (and hopefully at least one other mathematician will agree!), but it's entirely possible that y'all will consider this too low-brow for MO. There isn't a completely definite answer, but I think there can probably be a near-consensus. If you're able, feel free to re-title if you think you have something more appropriate. Also, I have no idea what tag(s) to apply. Community-wiki?
==========================================================
I'm trying to set up a winner-takes-all bet with four other friends about the order in which we end up getting married. So that nobody has to simultaneously shell out $100 and face the prospect of living alone for the rest of their life, it seems prudent to end the bet and determine the winner after 3 of us have gotten married. Here was my original scoring scheme:
123 > 132 > 213 > 231 > 312 > 321 > 12\* > 13\* > 1\*2 > 1\*3 > 21\* > 23\* > 2\*1 > 2\*3 > 31\* > 32\* > 3\*1 > 3\*2 > \*12 > \*13 > \*21 > \*23 > \*31 > \*32 > \* 1 \* > \* 2 \* > \* 3 \* > \*\*1 > \*\*2 > \*\*3
[For example, the first ordering 123 > 132 means the following. Suppose I list 1: Aaron, 2: Ben, 3: Carlos. Then, I am better off if they get married in the order Aaron, Ben, Carlos than if they come out in order Aaron, Carlos, Ben.]
[Note that if you want to put asterisks around a number you need to include spaces, otherwise they'll disappear and the number will be italicized. Although I'm sure there's a way to avoid this.]
I made this based off of the naive initial assumption that getting the top three picks should beat anything else. But one friend pointed out that probably it should be that 12\*>321 (for example), and so the question becomes:
* **What are reasonable parameters for the ordering?**
* **How should they be prioritized?**
* **What ordering (if any) does this yield?**
With regards to the third question, I wouldn't be at all surprised if some variant of Arrow Impossibility comes into play, but we'll cross that bridge when we get there...
| https://mathoverflow.net/users/303 | determining a fair betting scheme | I think it is easier if each person chooses an order for all 5 people (12345) but you still call it after only three people get married. To determine the winner, you play out the best possibility for 4th and 5th, and then you are rewarded 1 point for every relationship you get correctly.
In other words, you choose one person to be 1, and when you choose that person, you are saying that you believe he will beat 2-5. Every time you are right (1 before 2, 1 before 3, 1 before 4, 1 before 5) you get a point. Same with all other numbers.
So I label 12345. The actual occurrence is: 132. So,
13245
1 beats 3, 2, 4, 5 = four points
2 beats 4, 5 = two points
3 beats 4, 5 = two points
4 beats 5 = one point
Total = nine points.
That would beat 125, for example, which would get eight points, but it would tie with 124, which would get nine points. I think this relationship idea gets at the heart of the bet (who gets married before who).
The biggest issue with this is that you can easily get ties. There are 60 possible 3 number outcomes, 11 possible point scores. You could choose something like a lexical tie breaker, so 125>132. But you probably want to choose something more fun like, in the event of a tie, the winners take shots until only one man is left standing. Literally.
| 1 | https://mathoverflow.net/users/2065 | 6752 | 4,601 |
https://mathoverflow.net/questions/6751 | 2 | ### Background
Consider an electron with mass $1$ moving in $\mathbb R^n$ in under the influence of a static electromagnetic field. Up to identifying vector fields with differential forms, Maxwell's equations state that the electric force is given by a closed one-form and the magnetic field is given by a closed two-form. Since $\mathbb R^n$ is contractible, we can pick antiderivatives of each of these: let $C$ be the electric potential (a function on $\mathbb R^n$) and $B$ the magnetic potential (a one-form on $\mathbb R^n$). Again identifying vectors with covectors, the equations of motion of the electron are:
$$ \ddot q = dB\cdot \dot q + dC \quad\quad \text{(EOM)} $$
where $q(t)$ is the position at time $t$, I have identified one-forms with vector fields, and $\cdot$ is the pairing that takes the two-form $dB$ and the vector $\dot q$ to a covector. (Pick your favorite sign, perhaps swapping $B$ for $-B$ below.)
Pick an open set $U \subseteq \mathbb R^n$ and let $Q: U \times [0,1] \to \mathbb R^n$ satisfy: $Q(x,0) = 0$, $Q(x,1) = x$, and $Q(x,-)$ is a solution to (EOM) for each $x \in \mathbb R^n$; i.e. $Q$ is a family of solutions to (EOM) starting at $0$ and parameterized by the value at $t=1$. Define the *Hamilton principal function* $S$ on $U$ by
$$ S(x) = \int\_0^1 \left( \frac12 \left( \frac{\partial Q}{\partial t}(x,t)\right)^2 + B\bigl(Q(x,t)\bigr) \cdot \frac{\partial Q}{\partial t}(x,t) + C\bigl(Q(x,t)\bigr) \right) dt $$
$S$ depends on the choice of antiderivatives $B,C$ from the first paragraph. It satisfies:
$$ \frac{\partial S}{\partial x}(x) = \frac{\partial Q}{\partial t}(x,1) + B(x) $$
up to identifying vectors and covectors. So the differential $dS = \frac{\partial S}{\partial x}$ does not depend on the choice $C$ of antiderivative of the electric field. By differentiating again, the Hessian $\frac{\partial^2 S}{\partial x^2}$ does not depend on the choice $B$ of antiderivative of the magnetic field.
### My question
I know how to prove that the Hessian of $S$ is nondegenerate on the open set $U$. I would like to know whether the Hessian is necessarily positive-definite?
Suppose that $B = 0$. (See my answer below when $B \neq 0$; sorry about bringing it into play before, I was confused.) Does it follow that the Hessian of $S$ is positive-definite?
### Some examples
I can work very few examples. In particular, I know the answer when $B = 0$ and $C(q)$ is homogeneous quadratic. Pick a basis in which $C(q)$ is diagonalized, and let the eigenvalues be $C\_1,\dots,C\_n$. Then the Hessian is diagonalized in the same basis. If $C\_i = 0$, then the $i$th eigenvalue of the Hessian is $1$; if $C\_i > 0$, then the $i$th eigenvalue of the Hessian is $C\_i / \sinh^2\!\! \sqrt{C\_i}$. If $C\_i < 0$, then the $i$th eigenvalue is $|C\_i| / \sin^2\!\! \sqrt{|C\_i|}$. (Actually, this is true even if $C(q)$ is quadratic but not homogeneously so; let $C\_i$ be the eigenvalues of the homogeneous part.)
But the other examples I've thought of seem to require either coupled ODEs or elliptic integrals, so I haven't solved them directly.
### Bonus question
I used the metric on $\mathbb R^n$ twice: once in (EOM) to identify the vector $\ddot q$ with the covector on the RHS, and once in the definition of $S$ to square the vector $\frac{\partial Q}{\partial t}$. But consistently changing the metric in both places allows everything to be defined still. Up to changing bases, the only way to change a metric is to change its signature. Anyway, in the quadratic case, changing the signature of the metric changes the signature of the Hessian in exactly the same way. So I expect that the correct statement is that the ratio of the Hessian to the metric is positive-definite. But I'm not sure.
### Bonus bonus question
Experts know that one can equations of motion and Hamilton principal functions for much more general Lagrangian functions $L: \mathbb R^{2n} \to \mathbb R$ — suppose that the matrix $\frac{\partial^2 L}{\partial v^2}(v,q)$ is invertible for every $(v,q)$, so that the equations of motion define a nondegenerate second-order ODE. Then is the Hessian of the action necessarily positive-definite?
| https://mathoverflow.net/users/78 | Is the Hessian of Hamilton's function positive-definite? | The answer to your bonus bonus question is negative, I'm afraid. The Euler-Lagrange equations only extremise the action in general, not minimise it. Nondegeneracy of the lagrangian and positive-definiteness of the Hessian are two separate notions: the former indicating only the absence of constraints. Just take a massive particle moving on the line with a potential which is not bounded below. The lagrangian
$$L = \frac12 m v^2 - V(q)$$
is nondegenerate, since $\frac{\partial^2 L}{\partial v^2} = m$ is nonzero, but the Hessian depends on the second derivative of the potential.
| 2 | https://mathoverflow.net/users/394 | 6760 | 4,608 |
https://mathoverflow.net/questions/6759 | 10 | I'm inspired by Yuhao's question. The functor that takes a scheme S to the set of k-dimensional vector subbundles of C^n x S (understanding "subbundle" to mean that the quotient by it is another vector bundle) is represented by the Grassmannian G(k,n). What functor is represented by the Schubert subvarieties of G(k,n)?
The Schubert variety associated to a sequence of numbers d = (d\_0,d\_1,...,d\_n) is the collection of k-planes that meet the standard i-plane in a subspace of dimension $\geq$ d\_i. (The d for which the associated Schubert variety is nonempty are naturally parameterized by k-element subsets of {1,2,...,n}, or more usefully by partitions that fit in a k x (n-k) box.)
It's tempting to say that the functor represented by a Schubert variety should take a scheme S to the set of subbundles of C^n x S that, fiber-by-fiber, meet the standard i-plane in a subspace of dimension $\geq$ d\_i. But I don't know what this should mean on a non-reduced scheme.
One way to look for a moduli interpretation for the functor represented by a variety X is to find one or several "universal" or "tautological" families of things over X. Then one hopes that families of things of the same type over another scheme S will be pulled back along maps from S to X. The natural-looking tautological objects over a Schubert variety are the families of intersections $V \cap \mathbf{C}^i$, where V runs through k-planes belonging to the Schubert variety. These families are not equidimensional, and in particular not flat; this might be an obstacle.
...
It's really clear what's going on from Steven's answer, but I decided to follow it to the end and write down a functor. Beyond saying that these things are zero loci of interesting sections of vector bundles, this point of view probably doesn't much illuminate Schubert varieties. But here it is.
Steven's answer is cleanest if we regard the Grassmannian and its subvarieties as parameterizing quotients rather than subspaces of C^n. So a point in the Grassmannian G(k,n) is an isomorphism class of surjective maps C^n --> V, or equivalently of tuples (V,v\_1,...,v\_n) where V is a k-dimensional vector space and v\_1,...,v\_n span V.
The equivalence class of (V,v\_1,...,v\_n) belongs to the Schubert variety corresponding to r = (r\_1,...,r\_n) if for each i the first i terms on the list of vs span a subspace of dimension $\leq$ r\_i. As in the answer we can use exterior powers to make sense of these rank conditions over a scheme S. So an S-valued point of the Schubert variety corresponding to r is an equivalence class of tuples (E,s\_1,...,s\_n) where
* E is a vector bundle
* s\_1,...,s\_n is a list of sections of E
* these sections generate E in each fiber, and given any i and any (r\_i + 1)-tuple of sections chosen from (s\_1,...,s\_i), we have
$$s\_{j\_1} \wedge s\_{j\_2} \wedge \cdots \wedge s\_{j\_{r\_i + 1}} = 0$$
For example, if S = Spec(C[x]/x^2), and E is the rank 2 free sheaf on S spanned by e\_1 and e\_2, then (x.(ae\_1 + be\_2), x.(ce\_1 + de\_2), e\_1, e\_2) is an S-valued point of the Schubert variety corresponding to r = (1,1,2,2). The complex numbers a,b,c, and d parameterize the 4-dimensional Zariski tangent space to the singular point on this 3-dimensional variety.
There's a loose end. I think it's clear that the functor
$$S \mapsto \{(E,s\_1,\ldots,s\_n)\}/\sim$$
represents a subscheme of the Grassmannian with the same set-theoretic support as a Schubert variety. Why does it represent the actual Schubert variety? It would be enough to know that the representing object is reduced. But I don't even know how to express "X is reduced" in terms of the functor of points Hom(-,X).
| https://mathoverflow.net/users/1048 | What functor does a Schubert variety represent? | Everything you said should be fine. As for the case of not necessarily reduced schemes, we have to be careful, but I think the following will work.
Say E is our given vector bundle of rank n which is a subbundle of V which is some trivial bundle, and fix a trivial subbundle W of V. The condition $\dim(E(x) \cap W(x)) \ge k$ (here x is a closed point and E(x) denotes fiber) is equivalent to saying that rank$(E(x) \to V/W(x)) \le n - k$, or equivalently, that the map on exterior products
$\bigwedge^{n-k+1} E(x) \to \bigwedge^{n-k+1} V/W(x)$
is zero. Well, this map can be thought of as a section s of
$(\bigwedge^{n-k+1} E)^\vee \otimes \bigwedge^{n-k+1} V/W$.
So to define the locus of points where $\dim(E(x) \cap W(x)) \ge k$ scheme-theoretically, we can just ask for the zero locus of s (is this what it's called?).
So if you're asking for several intersection conditions, we can take the scheme-theoretic intersection of all these zero loci and ask if it's equal to our scheme.
This seems to be the reasonable thing to think about since those sections can be pulled back if you have a map to the Grassmannian whose image is contained in a certain Schubert variety, since the Schubert variety can be defined by the intersection of certain zero loci in the Grassmannian.
| 5 | https://mathoverflow.net/users/321 | 6767 | 4,611 |
https://mathoverflow.net/questions/6765 | 24 | I'm trying to understand Milnor's proof of the existence of exotic 7-spheres.
Milnor finds his examples among $S^{3}$ bundles over $S^{4}$ (with structure group $SO(4)$ ). Such a bundle can be described as follows:
Given $M$, an $S^{3}$ bundle over $S^{4}$, if we restrict $M$ to the northern (or southern) hemisphere of $S^{4}$, it must trivialize since each hemisphere is contractible. Hence, we can build $M$ by specifying, for each point $p$ in $S^{3}$ = equator of $S^{4}$ = intersection of northern and southern hemispheres, an element of $SO(4)$ which glues $p\times S^{3}$ in the northern hemisphere to $p\times S^{3}$ in the southern hemisphere.
This defines a function $f:S^{3}\rightarrow SO(4)$, which is known as the clutching function for $M$. By usual fiber bundle theory, the isomorphism type of $M$ only depends on the homotopy class of $f$.
$SO(4)$ is double covered by $S^3\times S^3$, and hence $\pi\_3(SO(4)) = \mathbb{Z}\oplus\mathbb{Z}$. Thus, $f$ is really determined (at least, up to homotopy) by an ordered pair of integers (i,j).
Now, as the bundles have structure group $SO(4)$, it makes sense to talk about the Pontryagin classes of $M$. In Milnor's proof of the existence of exotic spheres, he needs to argue that $p\_1(M) = \pm 2(i-j)$. His first step in this argument is that "clearly $p\_1(M)$ is a linear function of $i$ and $j$."
It IS clear to me that the Pontragin classes associated to $(ni, nj)$ for $n\in \mathbb{Z}$ will depend linearly on $n$. For, if we let $N\_{i,j}$ denote the principal $SO(4)$ bundle over $S^{4}$ corresponding to $(i,j)$, then $N\_{ni,nj}$ is clearly obtained as the pullback of $N\_{i,j}$ via a degree $n$ map from $S^{4}$ to itself.
However, it's not clear to me why $p\_1(M)$ is additive in $(i,j)$. Am I missing something simple?
And while we're talking about it, is more true? That is, For any sphere bundle over a sphere, say, $S^{k}\rightarrow E\rightarrow S^{n}$, should any characteristic classes (Pontryagin, Stiefel-Whitney, Euler) be linear in terms of the clutching function?
For example, we can think of $p\_1$ as a map from $\pi\_{n-1}(SO(k+1))\rightarrow H^{4}(S^{n})$. Is this map a homomorphism? How about for the other characteristic classes?
| https://mathoverflow.net/users/1708 | Characteristic classes of sphere bundles over spheres in terms of clutching functions | There is a way to explain it that's similar to what you said about multiplication by $n$.
Let $G$ be a Lie group, and let $f\_1$ and $f\_2$ be any two clutching functions that describe $G$-bundles $E\_1$ and $E\_2$ on $S^n$. Suppose that $f\_1$ and $f\_2$ agree at a base point of $S^{n-1}$. Let $c$ be some characteristic class of $G$-bundles of degree $n$; it could even be a cup product of standard classes such as Chern or Pontryagin or whatever. Let $f\_3$ be the combination of $f\_1$ and $f\_2$ on the one-point union $S^{n-1} \vee S^{n-1}$. It is the clutching function of a bundle $E\_3$ on the suspension $\Sigma(S^{n-1} \vee S^{n-1})$, which is the union of two $n$-spheres along an interval and thus homotopy equivalent to $S^n \vee S^n$.
Whether you define $c$ the old-fashioned way by obstructions, or the more modern way by pullbacks from classifying spaces, it is easy to argue that $c(E\_3) = c(E\_1) \oplus c(E\_2)$. I.e., it's just the ordered pair of the characteristic classes of its parts. Now addition in $\pi\_n$ is modeled by a map $S^n \to S^n \vee S^n$, and the induced map on $H^n$ takes $a \oplus b$ to $a+b$. (Your generalized question about $H^k(S^n)$ is of course non-trivial only when $k=n$.)
---
A shorter, more modern summary of the same story is as follows. The $X$ is a space and $LX$ is its loop space, then $\pi\_{n-1}(LX) \cong \pi\_n(X)$. The loop space of the classifying space $B\_G$ is homotopy equivalent to $G$ itself, so $\pi\_{n-1}(G) \cong \pi\_n(B\_G)$. A characteristic class of degree $n$ is any cohomology class in $H^n(B\_G)$. The linearity that Milnor uses is the transposed form of the fact that the Hurewicz homomorphism $\pi\_n(B\_G) \to H\_n(B\_G)$ is linear. (If you expand this it out more explicitly, it isn't really different from what I say above.)
| 24 | https://mathoverflow.net/users/1450 | 6769 | 4,613 |
https://mathoverflow.net/questions/6695 | 12 | If we replace projective variety with algebraic variety in the statement of the Weil conjectures what happens? To me it seems the statement still makes sense. But is it still true?
| https://mathoverflow.net/users/1095 | Weil Conjectures for nonprojective algebraic varieties | Correctly restated, the conjectures hold for any variety $V$ (not necessarily complete or nonsingular) over a finite field $k$.
Dwork proved that the zeta function $Z(V,t)$ of $V$ is a rational function of $t$.
Grothendieck (et al.) expressed $Z(V,t)$ as the alternating product of the characteristic
polynomials of the Frobenius map $F$ acting on the etale cohomology groups of $V$ with compact support.
Deligne showed (Weil II) that for each positive integer $i$ and each eigenvalue $a$ of $F$ acting on the $i$th etale cohomology group of $V$ with compact support, there exists an integer $j\leq i$ such that all the complex conjugates of $a$ have absolute value $q^{j/2}$ where $q=|k|$.
When $V$ is nonsingular and complete, these statements together with Poincare duality, give Weil's original conjectures.
| 22 | https://mathoverflow.net/users/930 | 6777 | 4,617 |
https://mathoverflow.net/questions/6762 | 10 | X is a Noetherian scheme, F is an injective object in the category of quasi-coherent sheaves on X. U is an open subset of X. Why F's restriction on U is still an injective object in the category of quasi-coherent sheaves on U?
| https://mathoverflow.net/users/2008 | Why is an injective quasi-coherent sheaf's restriction to an open subset still an injective object? | The restriction-by-zero type arguments can actually be made to work, with some effort and an extra hypothesis. Suppose $X$ is *locally Noetherian*, $j: U \to X$ the inclusion of an open subscheme.
Let $Mod(X)$ and $QCoh(X)$ be the categories of $O\_X$-modules, and quasi-coherent $O\_X$-modules, respectively.
The "some effort" is the following Lemma
**Lemma** If $X$ is locally Noetherian, then the injective objects in $QCoh(X)$ are precisely the injective objects of $Mod(X)$ which are quasi-coherent as sheaves of modules.
**Pf**: Any injective object of $Mod(X)$ which is quasi-coherent must certainly be injective in the smaller category $QCoh(X)$. For the converse, it suffices to show that any injective object $I$ of $QCoh(X)$ injects into some $I'$ which is a quasi-coherent injective object of $Mod(X)$, for then $I$ will be a retract of $I'$ and so injective in $Mod(X)$. This seems tricky, but is proved in Theorem 7.18 of Hartshorne's "Residues and duality".
---
Now, let's prove the result using the Lemma: If $J$ is an injective object in $QCoh(X)$, then the hard direction of the Lemma implies that it is injective in $Mod(X)$. The restriction-by-zero argument applies in this category, allowing us to conclude that $j^\* J$ is injective in $Mod(U)$. It's clearly quasi-coherent, so applying the easy direction of the Lemma we see that it is injective in $QCoh(U)$ as desired.
[Aside: On a *Noetherian* scheme, any quasi-coherent sheaf is a union of its coherent subsheaves and one can "extend" coherent sheaves on U to coherent sheaves on X (see e.g., Hartshorne Ex. II.5.15). Using these facts, one should be able to give a more direct argument in the Noetherian case.]
| 25 | https://mathoverflow.net/users/1921 | 6778 | 4,618 |
https://mathoverflow.net/questions/6780 | 9 | Theorem (Triangle-free Lemma). For all $\eta>0$ there exists $c > 0$ and $n\_0$ so that every graph $G$ on $n>n\_0$ vertices, which contains at most $cn^3$ triangles can be made
triangle free by removing at most $\eta\binom{n}{2}$ edges.
I am trying to find some information related to this topic, I am unable to access the orignal paper by Ruzsa & Szemeredi.
Does anyone know any useful papers/books on the triangle-free lemma?
| https://mathoverflow.net/users/2011 | Triangle-free Lemma | Possibly an even better place to look is in surveys on graph property testing; this is by far the most common use nowadays of the triangle removal lemma, and any sufficiently good introduction to the subject should have some information on it. (I haven't actually read it, but I believe the most recent edition of Alon and Spencer's *The Probabilistic Method* has a chapter on property testing. If you have access to a copy, assuming the new chapter is anywhere near as good as the rest of the book, that would be my first recommendation.)
If you want a proof and some applications to more traditional combinatorics, Tim Gowers has a wonderful two-paragraph sketch and some discussion [here](http://gowers.wordpress.com/2009/01/30/background-to-a-polymath-project/) (about halfway down the post).
| 7 | https://mathoverflow.net/users/382 | 6788 | 4,625 |
https://mathoverflow.net/questions/6810 | 8 | Does anyone know a nice description of a Seifert surface of a torus knot? I can construct such surfaces in band projection, but what I get is ugly and unwieldy. Is there some elegant description for Seifert surfaces for such knots which I'm missing? (I'm not sure precisely what I mean by elegant...)
| https://mathoverflow.net/users/2051 | Seifert surfaces of torus knots | There's the usual description of the Seifert surface for a general cable obtained by taking copies of a Seifert surface for the knot and a fiber for the cable in the solid torus. See [Ken Baker's discussion](http://sketchesoftopology.wordpress.com/2009/11/18/cabling-a-knots-surface/).
| 10 | https://mathoverflow.net/users/1335 | 6815 | 4,647 |
https://mathoverflow.net/questions/5982 | 3 | This is theorem 14.C on p.84 of Matsumura's commutative algebra.
>
> Let $A$ be a noetherian domain, and let $B$ be a finitely generated overdomain of $A$. Let $P \in Spec(B)$ and $p = P \cap A$. Then we have
> $ht(P) \leq ht(p) + tr.d.\_{A} B - tr.d.\_{K(p)} K(P)$ with equality holds when $A$ is universally catenary or if $B$ is a polynomial ring over $A$.
>
>
>
Question: How should one understand this formula? I'm hazarding a guess that this factor, $tr.d.\_{A} B - tr.d.\_{K(p)}K(P)$, can somehow measure how primes of $B$ will be identified when they are restricted back to $A$. But this sounds woefully wrong and I just want to know how I should view this result or whether there is any (geometric) intuition behind the result.
Thanks!
| https://mathoverflow.net/users/nan | Intuition for Nagata's altitude formula? | Put dim B=n for the dimension of the variety with coordinates ring B. Then
n-ht P ≥ ((n-tr deg A B)- ht p)+ tr deg *k*(p) *k*(P)
The first member of the inequality indicates the dimension of the subvariety definited by P.
The term (n-tr deg A B) in the second member is the dimension of the variety with coordinates ring A: it looses
tr deg A B dimensions with respect the other variety with coordinate ring B.
Then ((n-tr deg A B)- ht p) represent the dimension of the subvariety definited by p.
The term tr deg *k*(p) *k*(P) is a corrector term because blow up can occur.
| 2 | https://mathoverflow.net/users/2040 | 6816 | 4,648 |
https://mathoverflow.net/questions/6802 | 17 | Usually when people talk on the absolute Galois group *G*ℚ of ℚ they have in mind two elements they can describe explicitly, namely the identity and complex conjugation (clearly, everything is up to conjugation), although the cardinality of the group is uncountable.
Can you describe other elements of *G*ℚ?
| https://mathoverflow.net/users/2042 | Element in the absolute Galois group of the rationals | See the last of this extended answer. I'm going to part company with everyone else and say that you **can** describe other elements of $\text{Gal}(\mathbb{Q})$. In other words, I claim that you can identity a specific element of $\text{Gal}(\mathbb{Q})$ in a wide range of ways, together with an algorithm to compute the values of that element as a function on $\mathbb{Q}$. You can use either a synthetic model of $\overline{\mathbb{Q}}$, or its model as a subfield of $\mathbb{C}$. Although this is all doable, what's not so clear is whether these explicit elements are interesting.
The other two parts of the answer raise interesting issues, but they are moot for the original question.
---
This is not exactly the question, but it is related. To begin with, it is difficult to "explicitly" describe $\overline{\mathbb{Q}}$ except as a subfield of $\mathbb{C}$. I found a paper, [Algebraic consequences of the axiom of determinacy](http://www.ams.org/mathscinet-getitem?mr=756898) (in English translation of the title) that establishes that $\mathbb{C}$ does not have any automorphisms other than complex conjugation in ZF plus the axiom of determinacy (AD). So you need some part of the axiom of choice (AC) for this related question.
As for the smaller field $\overline{\mathbb{Q}}$, the Wikipedia page for the fundamental theorem of algebra suggests that you might not even be able to construct it in the first place without the axiom of countable choice. (I say "suggests" because I'm not entirely sure that that is a theorem. Note that AC and AD both imply countable choice even though they are enemy axioms.) Any construction with countable choice isn't truly "explicit". On the other hand, if you allow countable choice, then I suspect that you can build $\overline{\mathbb{Q}}$ synthetically by induction rather than as a subfield of $\mathbb{C}$, and that you can build many automorphisms of it as you go along.
So the questions for logicians is whether there is a universe over ZF in which $\overline{\mathbb{Q}}$ does not exist, or a universe in which it does exist but has no automorphisms.
---
I got email about this from Kevin Buzzard that made me look again at the paper referenced by Wikipedia, [A weak countable choice principle](http://math.fau.edu/richman/docs/wcc.pdf) by Bridges, Richman, and Schuster. According to this paper, life is pretty strange without countable choice. You want to make the real numbers as the metric completion of the rationals. However, there is a difference between general Cauchy sequences and what they called "modulated" sequences, which are sequences of rationals with a promised rate of convergence. They cite a result of Ruitenberg that the modulated complex numbers are algebraically closed in ZF. Hence $\mathbb{Q}$ has an algebraic closure in ZF.
But it still seems possible that without countable choice, algebraic closures of $\mathbb{Q}$ need not be unique up to isomorphism, and that the complex analysis model of $\overline{\mathbb{Q}}$ might not have automorphisms other than complex conjugation.
---
A better and hopefully final technical answer: As mentioned, $\overline{\mathbb{Q}}$ exists explicitly (in just ZF) as a subfield of $\mathbb{C}$. You can also construct it synthetically as follows: Consider the monic Galois polynomials over $\mathbb{Z}$. These are the polynomials such that the Galois group acts freely transitively on the roots; equivalently the splitting field is obtained by adjoining just one root. The Galois polynomials can be written in a finite notation and enumerated. Beginning with $\mathbb{Q}$, formally adjoin a root of $p\_n(x)$, the $n$th monic Galois polynomial, for each $n$ in turn. If $p\_n(x)$ factors over the field constructed so far, the factors can also be expressed in a finite notation; take the first irreducible factor. The result is an explicit, synthetically constructed $\overline{\mathbb{Q}}$.
For comparison, let $\widetilde{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. Each element of it is computable: Its digits can be generated by an algorithm, even with an explicit bound on its running time. As we build $\overline{\mathbb{Q}}$, we can also build an isomorphism between $\widetilde{\mathbb{Q}}$. We can do this by sending the formal root of $p\_n(x)$ to its first root in $\mathbb{C}$, using some convenient ordering on $\mathbb{C}$. Or we could just as well have used its last root, its second root if it has one, etc. Composing these many different isomorphisms between $\overline{\mathbb{Q}}$ and $\widetilde{\mathbb{Q}}$ gives you many field automorphisms.
| 12 | https://mathoverflow.net/users/1450 | 6830 | 4,658 |
https://mathoverflow.net/questions/6827 | 3 | As is well known, the set
$\{a^ib^jc^k | i,j,k \in \mathbb{Z}\\_{\geq 0},k>0\} \cup \{b^lc^md^n | l,m,n \in \mathbb{Z}\\_{\geq 0}\}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
| https://mathoverflow.net/users/1867 | Basis of quantum SU(n) | [edit: Following John's helpful comments below, I made this answer much more complete.]
Yes, this is the statement that $O\_q(G)$ is a flat deformation of $O(G)$ for any semi-simple group G. See the book by Klimyk and Schmuedgen, "Quantum Groups and Their Representations" for a proof of this: on page 311 they state the relevant theorem for $Mat\_q(n)$ (although the proof is just a reference to the original source). In the following section, they prove that det\_q is central, which allows us to identify $O\_q(SL\_N)$ with $Mat\_q(n)/(det\_q-1)$. The OP asked about $SU(N)$, but in the context of algebraic groups one studies SL\_N, which has a compact real form $SU(N)$, and morally the same representation theory.
In general we have to be careful when either inverting or specializing to a scalar any element in a noncommutative algebra, because this can in general drastically change the size of the algebra relative to what you'd expect from the commutative situation (it is bigger in the former case and smaller in the latter than expected). For inverting, you need the element to lie in a "denominator set", which assures that you don't have to add too many more things to invert it (imagine inverting $y$ in the free algebra $k((x,y))$ on two generators x and y: it would be a lot bigger than the vector space $k((x,y))[1/y]$). [edit: I can't get carot's or braces to work, hence the awkward symbol for free algebra; I hope it's clear.] For specializing, your element should honestly lie in the center of the noncommutative algebra, since it's image in the quotient will be a scalar (thus central). For instance, if you take $A^2\_{q}=k((x,y))/(yx=qxy)$, this has the same basis as $A^2=k[x,y]$. However, quotienting A\_q by y-1 forces x=0, which doesn't happen in A.
So far as I remember, the standard proof of the PBW theorem in this example (and many examples) relies on a technical lemma called the diamond lemma, Lemma 4.8 from KS, which gives an ordering on the monomials of O\_q(G) compatible with the defining relations, allowing one to prove the existence of PBW basis.
| 8 | https://mathoverflow.net/users/1040 | 6831 | 4,659 |
https://mathoverflow.net/questions/6820 | 94 | Is there an **infinite** field $k$ together with a polynomial $f \in k[x]$ such that the associated map $f \colon k \to k$ is not surjective but misses only finitely many elements in $k$ (i.e. only finitely many points $y \in k$ do not lie in the image of $f$)?
For finite fields $k$, there are such polynomials $f$. If such a poynomial $f$ exists, then $k$ cannot be algebraically closed; the field $\mathbb{R}$ doesn't work either.
| https://mathoverflow.net/users/296 | Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points? | Since such a polynomial would have to have degree at least 2, its existence implies that
the set of k-rational points of the affine line over k is *thin* in the sense of Serre's *Topics In Galois Theory*. It follows from the results presented in that book that this cannot be the case over any *Hilbertian field*. This includes finite extensions of Q, finite extensions of F(t) for any field F, and many other fields.
What about p-adic fields?
| 33 | https://mathoverflow.net/users/1149 | 6832 | 4,660 |
https://mathoverflow.net/questions/6776 | 2 | I wonder whether following statements holds
If A is an abelian category(or quasi abelian category) having enough projectives, then category of pointed diagram(which means diagram has final object,or for simplicity, one assume the diagram is finite)(A\_D=(D--->A))has enough projectives.
I want to construct a pair of adjoint functor between this two category. Then use left adjoint of exact functor maps projectives to projectives
Other methods to prove this statement is welcomed
| https://mathoverflow.net/users/1851 | How to construct pair of adjoint functors from category A to category A_D(category of diagrams) | If D is small and A has enough projectives and has infinite sums then $A^D$ has enough projectives. For the proof, see Weibel, "An introduction to homological algebra", 2.3.13 on p.43. It contains the adjoint you are apparently looking for.
The proof is a version of Godement's argument that the category of sheaves of abelian groups has enough injectives.
| 2 | https://mathoverflow.net/users/1784 | 6836 | 4,662 |
https://mathoverflow.net/questions/6833 | 11 | What is the difference between
connected
strongly-connected and
complete?
My understanding is:
**connected**: you can get to every vertex from every other vertex.
**strongly connected**: every vertex has an edge connecting it to every other vertex.
**complete**: same as strongly connected.
Is this correct?
| https://mathoverflow.net/users/1983 | Difference between connected vs strongly connected vs complete graphs | * *Connected* is usually associated with undirected graphs (two way edges): there is a **path** between every two nodes.
* *Strongly connected* is usually associated with directed graphs (one way edges): there is a **route** between every two nodes.
* *Complete graphs* are undirected graphs where there is an **edge** between every pair of nodes.
| 22 | https://mathoverflow.net/users/507 | 6837 | 4,663 |
https://mathoverflow.net/questions/6789 | 147 | Of course "flatness" is a word that evokes a very particular geometric picture, and it seems to me like there should be a reason why this word is used, but nothing I can find gives me a reason!
Is there some geometric property corresponding to "flatness" (of morphisms, modules, whatever) that makes the choice of terminology obvious or at least justifiable?
| https://mathoverflow.net/users/382 | Why are flat morphisms "flat?" | A lot of people will tell you that flatness means "continuously varying fibres" in some sense, and that flatness was invented to have correspondingly nice *consequences*, which is true. But there is a way to expect this (vague) interpretation *a priori* from an alternative, equivalent definition:
An $A$-module $M$ is flat $\iff$ $I \otimes\_A M \to IM$ is an isomorphism for every ideal $I$.
I would prefer to present this as the definition of flatness, and present the fact that tensoring with $M$ preserves exact sequences as a theorem. Why?
Thinking "geometrically", $I$ just corresponds (uniquely) to a closed subscheme $Z=Z(I)=$
$=Spec(A/I)\subseteq Spec(A)$. If we think of $M$ in the usual geometric way as a module of generalized functions on $X$ (like sections of a bundle), and $M/IM \simeq M\otimes\_A A/I$ as its restriction to $Z$, then the above definition of flatness can be interepreted directly to mean that **$M$ restricts nicely to closed subschemes $Z$**.
More precisely, it says that what we lose in this restriction, the submodule $IM$ of elements which "vanish on $Z$", is easy to understand: it's just formal linear combinations of elements $i\otimes m$, with no surprise relations among them, i.e. the tensor product $I \otimes\_A M$.
In topology, continuous functions "restrict nicely" to points and closed sets (by taking limits), so you can see, without much experience at all, how this definition corresponds in an intuitive way to continuity.
Having this motivation in place, the best thing to do is to check out examples along the lines of Dan Erman's answer to see the analogy with continuity and limits at work.
| 189 | https://mathoverflow.net/users/84526 | 6844 | 4,668 |
https://mathoverflow.net/questions/6845 | 5 | A random walk matrix has largest eigenvalue 1 with multiplicty 1 - why?
Let $G$ be a non-directed, regular connected graph with degree $d$. Let $A$ be its random walk matrix, i.e. it's adjacency matrix, with each entry divided by $d$.
i) It is easy to observe that $A$ is symmetric, hence normal, that it has real eigenvalues only and can be diagonalized by a pair of orthogonal matrices (at least if don't mix up something from my past course in linear algebra)
ii) Second, one can observe that the all-one vector scaled by $1/n$ is an eigenvector of $A$ for the eigenvalue $1$
iii) Furthermore, by observing that for any natural $k$, $A^k$ is doubly-stochastic, too, and applying Gelfands formula with $l^1$-norm, we can see that the spectral norm is $1$
It remains to show that $1$ has multiplicity $1$. After hours I couldn't manage to figure this out, although it seems rather simple at first sight. So probably I simply don't know the 'trick', which yields this result.
Can somebody help me?
| https://mathoverflow.net/users/2082 | A random walk matrix has eigenvalue 1 with multiplicty 1 - why? | Here is a simple proof.
Suppose $Ax = x$. Consider the entry of $x$ with the largest absolute value; lets use $x\_k$ to denote this entry (e.g. if $x=[1,2,-4,3]^T$, then $k=3, x\_k=-4$). Consider the $k$'th row of the equation $Ax=x$; it's telling you
that $x\_k$ is a convex combination of the $x\_i$'s of its neighbors $i$ in the graph $G$. This immediately implies that $x\_k=x\_i$ for all neighbors $i$ of $k$ in $G$.
Now you iterate this argument and apply it to each neighbor $i$ of $k$. Using connectivity of $G$, eventually you get the conclusion that every entry of $x$ equals $x\_k$. Thus the only solutions to $Ax=x$ are multiples of the all ones vector.
Observe that some of the conditions you imposed were not used: the above proof did not use
the fact that $A$ is symmetric or that the graph is regular.
| 8 | https://mathoverflow.net/users/1407 | 6856 | 4,676 |
https://mathoverflow.net/questions/6271 | 2 | Following on from my last two questions [link text](https://mathoverflow.net/questions/5865/classical-calculi-as-universal-quotients) and [link text](https://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials): Is it correct (and useful) to say that the relationship between Connes' cyclic cohomology approach to de Rham cohomology and Woronowicz's differential calculi approach, is a noncommutative generalisation of the difference between ordinary differentials and Kahler differentials respectively?
| https://mathoverflow.net/users/1867 | Connes v Woronowicz - Cyclic Cohomology v Diff Calculi | I'll answer here instead of in a comment, because of the character limit...
If $A$ is the coordinate algebra of an affine variety which is *smooth* and the base field $k$ contains $\mathbb{Q}$, then $$HC\\_n(A) \cong \Omega^n\\_{A/k} / d\Omega^{n-1}\\_{A/k} \oplus H\\_{\mathrm{dR}}^{n-2}(A) \oplus H\\_{\mathrm{dR}}^{n-4}(A) \oplus \cdots$$ for all $n\geq0$. Here $\Omega^n\\_{A/k}$ is the $n$-th exterior power of the $A$-module $\Omega^1\\_{A/k}$ of Kähler differentials of $A$ over $k$, and $H\\_{\mathrm{dR}}^\bullet(A)$ denotes the cohomology of the complex $$A\to \Omega^1\\_{A/k} \to \Omega^2\\_{A/k} \to \Omega^3\\_{A/k} \to\cdots $$ whose differential is the exterior differential.
The summand $\Omega^n\\_{A/k} / d\Omega^{n-1}\\_{A/k}$ appearing in $HC\\_\bullet(A)$ is slightly ugly. If we consider instead periodic cyclic homology, we get instead $$HC^{\mathrm{per}}\\_n(A) \cong \bigoplus\\_{i\in\mathbb{Z}}H\\_{\mathrm{dR}}^{n+2i}(A),$$ which is manifestly nicer.
(If $k$ is not of characteristic zero you only have a spectral sequences going from de Rham cohomology to the cyclic homology). If, on the other hand, $A$ is not smooth then André-Quillen cohomology intervenes, and everythng is rather more complicated.)
You should really take a look at Loday's book.
| 3 | https://mathoverflow.net/users/1409 | 6859 | 4,678 |
https://mathoverflow.net/questions/6442 | 9 | This is more of a request for pointers to *relevant* literature than a question per se. I am, erm, looking at a paper which uses a kind of iterated pushout construction to obtain a commutative monoid with certain desired properties. The particular "gluing" construction the authors want to do is handled by quite direct means, and while that's probably fine for this particular problem, it would be nice to put it in the proper context.
My copy of Howie's "Introduction to Semigroup Theory" discusses the general notion of "semigroup amalgams" but doesn't seem to say anything about the pushouts in the smaller category of commutative monoids. The closest I can find via MathSciNet is a 1968 paper of Howie, which seems to be working in the category of commutative semigroups, but I can't get hold of a copy at the moment.
Anyway: I'm hoping that someone reading might know of a sensible place to look for a summary of some general results. (The issue is whether the constituent pieces "embed" into the pushout, not the mere existence of the pushout.)
**EDIT:** On further reflection, while pushouts are undoubtedly relevant, for the purposes of the paper I'm looking at, faithfulness of the embedding is more important than the universal property of pushouts. So I'm changing the title of the question to reflect this. I also think that while the suggestions below have been useful in a broad sense they don't quite address the case I need - which could just be an illustration that said case falls between the two extremes most commonly looked at by semigroup theorists.
| https://mathoverflow.net/users/763 | References/literature for pushouts in category of commutative monoids? [ed. - amalgams] | Arthur Ogus wrote a book on logarithmic geometry, apparently soon to be published, and there is a [preprint version](http://math.berkeley.edu/~ogus/preprints/log_book/) on his webpage. The first chapter is about commutative monoids, and in particular, it has a bit about pushouts (starting on page 12, and you can tell your computer to search for other instances of the word).
General pushouts can be quite pathological, but you can say interesting things if the monoids satisfy some properties such as integrality. For example, if all of the monoids are integral and one of the monoids is a group, then the pushout is integral, and its group completion is the pushout of the group completions. This reduces Reid's example to a calculation with groups.
| 5 | https://mathoverflow.net/users/121 | 6863 | 4,681 |
https://mathoverflow.net/questions/6851 | -3 | Now there are n independent projects, each project is composed by several steps, each step is labeled, there are k workers are going to work on these projects.
Assume that each person can do only one step, and each step can only be done by a specific worker(it will be declared in input file). Each project has a order for its steps, for example, workers can operate the second step only if the first step has been done. when one person finished one step, then he can be assigned to another step immediately.
Task:Input the number of projects, and the number of steps of each project, and the worker corresponds to each step, and how much time it will take to finish the step. Try to find an algorithm to calculate the optimized time to finish all projects.
Example inputs:
2 3(2 projects, 3 workers)
3 1 2 2 3 3 5(for project 1, there are 3 steps, the first will be done by worker 1 takes 2 hours, the second will be done by worker 2 by 3 hours, the third will be done by worker 3 by 5 hours)
2 2 3 3 2(for project 2, there are 2 steps, the first will be done by worker 2 by 3 hours, the second will be done by worker 3 by 2 hours)
Output
11 ( the minimized hours to finish the whole project)
| https://mathoverflow.net/users/1956 | A problem seeking for algorithm | I think this is a member of a class of problems called scheduling problems. In particular I think it is the job shop scheduling problem. There are machines instead of people but the basic structure is the same. If I am right about this the prospects of a solution of this type of problem is bleak as problems with three machines have been proven to be strongly NP-hard. I have found a reference for this \_Handbook of scheduling: algorithms, models, and performance analysis\_By Joseph Y-T. Leung I was able to look at parts of it on google books.
| 3 | https://mathoverflow.net/users/1098 | 6869 | 4,687 |
https://mathoverflow.net/questions/6870 | 66 | Consider an [elliptic curve](http://en.wikipedia.org/wiki/Elliptic_curve) $y^2=x^3+ax+b$. It is well known that we can (in the generic case) create an addition on this curve turning it into an abelian group: The group law is characterized by the neutral element being the point at infinity and the fact that $w\_1+w\_2+w\_3=0$ if and only if the three points $w\_j$ are the intersections (with multiplicity) of a line and the elliptic curve.
Groups can be hard to work with, but in most cases proving that the group is in fact a group is easy. The elliptic curve is an obvious exception. Commutativity is easy, but associativity is hard, at least to this non-algebraist: The proof looks like a big calculation, and the associativity seems like an algebraic accident rather than something that *ought* to be true.
So this is my question, then: Why is the group law on an elliptic curve associative? Is there some good reason for it? Is the group perhaps a subgroup or a quotient of some other group that is easier to understand? Or can it be constructed from other groups in some fashion?
I gather that historically, the group law was discovered via the addition law for the [Weierstrass $\wp$-function](http://en.wikipedia.org/wiki/Elliptic_curve). The addition law is itself not totally obviuos, plus this approach seems limited to the case where the base field is $\mathbb{C}$. In any case, I'll elaborate a bit on this shortly, in a (community wiki) answer.
| https://mathoverflow.net/users/802 | Why is an elliptic curve a group? | Everything I am writing below is carried out explicitly in Chapter III of Silverman's book on elliptic curves. In the earlier chapters, he defines the Picard group.
For any curve over any field, algebraic geometers are interested in an associated group called the Picard group. It is a certain quotient of the free abelian group on points of the curve. It consists of formal sums of points on the curve modulo those formal sums that come from looking at the zeroes and poles of rational functions. It is a very important tool in the study of algebraic curves.
The very special thing about elliptic curves, as opposed to other curves, is that they turn out to be in natural set-theoretic bijection with their own Picard groups (or actually, the subgroup $\operatorname{Pic}^0(E)$). The bijection is as follows: let $O$ be the point at infinity. Then send a point $P$ on the elliptic curve to the formal sum of points $[P] - [O].$ (It is not obvious that this is a bijection, but the work to prove it is all "pure geometric reasoning" with no computations.) So there is automatically a group law on the points of $E.$ Then it requires no messy formulas to show that under this group law, the sum of three collinear points is $O.$ So for free, you also get that this group law is the same as the one you defined in the question and that the one you defined is associative!
| 71 | https://mathoverflow.net/users/1018 | 6872 | 4,689 |
https://mathoverflow.net/questions/6874 | 33 | In the early 1990's, Gil Kalai introduced me to a very interesting generalization of homology theory called intersection homology, which existed for like 10 years back then I believe. Defined initially by Goresky and MacPherson, this is a version of homology which agrees with ordinary homology on manifolds, but also retains crucial properties like Poincare Duality and Hodge Theory on singular (non-)manifolds. The original definition was combinatorial, but it was later re-interpreted in sheaf-theoretic terms (perverse sheaves?).
Back then it certainly looked like an exciting new development. So, I'm curious - where does the field stand today? Is it still thriving, or has it been merged with something else, or just faded away?
| https://mathoverflow.net/users/25 | What (if anything) happened to Intersection Homology? | Intersection homology and cohomology are still around, but as a topic they have just substantially been renamed. They are part of the theory of perverse sheaves, which are widely used in the Langlands program, in algebraic geometry approaches to categorification, and elsewhere in algebraic geometry.
To the extent that intersection homology was intended for topology, it has stoked relatively less interest than in algebraic geometry. On the one hand, there has been a trend away from homological algebra in geometric topology. On the other hand, singularities are part of the structure of intersection homology. Singularies are more germane to geometric topology than to algebraic topology in the sense of homotopy theory. Both singularities and homological algebra are major aspects of algebraic geometry.
| 37 | https://mathoverflow.net/users/1450 | 6875 | 4,691 |
https://mathoverflow.net/questions/6895 | 2 | (Edit: The first formulation is wrong. See the second answer) Does every totally ordered set contain an unbounded countable subset. In other words: If S is a totally ordered set, can we find a (edit: at most) countable subset A, such that for every $s \in S$, there is a $a \in A, a\geq s$?
| https://mathoverflow.net/users/2097 | Unbounded countable subset | There is a counterexample in the long line L. It is totally ordered and every sequence has a limit in L. see the following:
[http://en.wikipedia.org/wiki/Long\_line\_(topology)](http://en.wikipedia.org/wiki/Long_line_%28topology%29)
| 4 | https://mathoverflow.net/users/1098 | 6900 | 4,707 |
https://mathoverflow.net/questions/6890 | 33 | The famous Birkhoff-von Neumann theorem asserts that every doubly stochastic matrix can be written as a convex combination of permutation matrices.
The question is to point out different generalizations of this theorem, different "non-generalizations" namely cases where an expected generalization is false, and to briefly describe the context of these generalizations.
A related MO question: [Sampling from the Birkhoff polytope](https://mathoverflow.net/questions/73805/sampling-from-the-birkhoff-polytope)
| https://mathoverflow.net/users/1532 | Generalizations of the Birkhoff-von Neumann Theorem | I am cheating a little to give this answer, because I am fairly sure that it is part of Gil's motivation in asking the question. The most natural generalization of the Birkhoff hypothesis to quantum probability is only true for qubits. (It might also be true for a qubit tensor a classical system; I did not check that case.)
A quantum measurable space is a von Neumann algebra. We are most interested in the finite-dimensional case, where classically "measurable space" is just a fancy name for the random variables on a finite set. A finite-dimensional von Neumann algebra is a direct sum of matrix algebras. In particular, $M\_2$ is called a qubit and $M\_d$ is called a qudit.
To make a long story short, the Birkhoff hypothesis can be stated for a direct sum of $a$ copies of $M\_b$, or $aM\_b$. In this setting, a doubly stochastic map $E$ is a linear map from $aM\_b$ to itself that preserves trace, that preserves the identity element, and that is completely positive. In this setting, $E$ is completely positive if it takes positive semidefinite elements of $aM\_b$ to positive semidefinite elements, and if $E \otimes I$ also has that property on the algebra $aM\_b \otimes N$ for another von Neumann or $C^\*$-algebra $N$. The natural analogue of permutation matrices are the \*-algebra automorphisms of $aM\_b$. These are permutations of the matrix blocks, composed with maps of the form $E(x) = uxu^\*$, where $u$ is a unitary element of $aM\_b$. The question as before is whether the doubly stochastic maps are the convex hull of the automorphisms.
This Birkhoff hypothesis is true for $M\_2$, false for $M\_d$ for $d \ge 3$, and I should check it for $nM\_2$. It is true for $aM\_1 = a\mathbb{C}$, because then it is the usual Birkhoff-von Neumann theorem.
I am left wondering about two infinite classical versions of Birkhoff's theorem, for the algebras $\ell^\infty(\mathbb{N})$ and $L^\infty([0,1])$. In the former case, one would ask whether any stochastic map that preserves counting measure (even though counting measure is not normalized) is an infinite convex sum of permutations of $\mathbb{N}$. In the latter case, whether any stochastic map that preserve Lebesgue measure is a convex integral of measure-preserving permutations of $[0,1]$. **Addendum:** At least the discrete infinite case is addressed, with generally positive results, in [this review](https://doi.org/10.1016/j.jfa.2004.09.001) and in [this older review](http://www.springerlink.com/content/un75627561733g22/). The older paper also raises the continuous question but with no results. However, with some more Googling I found [this counterexample paper](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-99/issue-2/Counterexamples-to-some-conjectures-about-doubly-stochastic-measures/pjm/1102734023.full).
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Since Gil asks for a reference, a recent one is [Unital Quantum Channels - Convex Structure and Revivals of Birkhoff's Theorem](https://arxiv.org/abs/0806.2820), by Mendl and Wolf.
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Here also is a more orthodox combinatorial generalization of the Birkhoff theorem, and also another case that I once encountered that is between a generalization and a non-generalization. Since Gil now offers a bounty, maybe it's better to merge this answer with the other one.
A doubly stochastic matrix can be interpreted as a flow through a directed graph, with unit capacities. (See [Unimodular matrix](https://en.wikipedia.org/wiki/Unimodular_matrix) in Wikipedia; I learned about this long ago from Jesus de Loera.) Any such graph has a polytope of flows, called a network flow polytope. Any network flow polytope has integer vertices, because it is a totally unimodular polytope.
A totally unimodular polytope is a polytope whose facets have integer equations, and with the property that any maximal, linearly independent collection of facets intersects in an integer point because their matrix has determinant $\pm 1$. In particular the vertices are such intersections, so the vertices are all integral. This is a vast generalization of Birkhoff's theorem that comes from generalizing one of the proofs of Birkhoff's theorem.
Example: An alternating-sign matrix is equivalent to a square ice orientation of a square grid. The square ice orientations can be defined by a network flow, so you obtain an alternating-sign-matrix polytope. The generalized Birkhoff theorem in this case says that every vertex of the polytope is an alternating-sign matrix, in fact that every integer point of the $n$-dilated polytope is a sum of $n$ alternating-sign matrices.
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The other case that I encountered was the polytope of fractional perfect matchings of a non-bipartite set with $2n$ elements. By contrast, the Birkhoff polytope is the case of a bipartite set with $n$ elements of each type. By definition, it is the polytope of non-negative weights assigned to the edges of the complete graph on $2n$ vertices, such that the total weight at each vertex is 1. Strictly speaking, the Birkhoff theorem is false; not every vertex is a perfect matching. Instead, all of the vertices are combinations of matched pairs, and odd cycles with weight $\frac12$.
At first glance this looks like bad news for the application of computing a perfect matching or the optimum perfect matching of a graph. Indeed, if instead you take the convex hull of the perfect matchings, the result is a polytope with exponentially many facets. However, a good algorithm exists anyway; there is a version of the simplex algorithm that only ever uses polynomially many of the facets.
| 14 | https://mathoverflow.net/users/1450 | 6904 | 4,709 |
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