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https://mathoverflow.net/questions/13072 | 12 | This question is prompted by [another one](https://mathoverflow.net/questions/12920/).
I want to motivate the definition of a scheme for people who know about manifolds(smooth, or complex analytic). So I define a manifold in the following way.
Defn: A smooth $n$-manifold is a pair $(X, \mathcal{O}\_X)$, where $X$ is a topological space and $\mathcal{O}\_X$ is a sheaf of rings on it, such that, every point $x \in X$ has a neighborhood $U\_x$ which is homeomorphic to an open set $V\_x$ in $\mathbb{R}^n$ and $\mathcal{O}\_X$ restricted to $U\_x$ is isomorphic to the sheaf of ring of smooth functions on $V\_x$ and its open subsets.
This agrees with the usual definition using charts and atlases, for all except the requirement that a manifold is a (separable) Hausdorff space. But indeed it seems that many things in differential topology can be proven without using the Hausdorff property. In a fleeting conversation in a brief encounter, a personal I shall refer to as O., informed me that even Stokes' theorem can be done this way. But I am unable to ask O. again about this, as he is not physically around.
If the above is true, then this is a really good point to mention when introducing schemes to a person who knows about differentiable manifolds.
So my main question:
>
> Is it true that the proof of Stokes' theorem for smooth manifolds can be proven without the Hausdorff condition on the manifold? If so, is it done in any well-known reference?
>
>
>
Aside question:
What are some crucial propositions/theorems in differential topology that use Hausdorff condition, except those involving imbedding in some $\mathbb{R}^m$ for high enough $m$, for which of course Hausdorffness is a necessary condition(together with separable property)?
Tom Church answers below that partitions of unity does not work, for instance on the example of the line with the double point. However I believe that one can still make sense of integration of differential forms even with such pathologies, because by introducing a measure for instance, we can ignore sets of measure $0$.
| https://mathoverflow.net/users/2938 | Stokes' theorem etc., for non-Hausdorff manifolds | The existence of flows in the direction of a vector field seems to require Hausdorff; indeed, consider the vector field $\frac{\partial}{\partial x}$ on the line-with-two-origins. We have no global existence of a flow for any positive t, even if we make our space compact (that is, considering the circle-with-one-point-doubled). If the nonexistence of the flow is not visibly clear, consider instead the real line with the interval [0,1] doubled.
Also, partitions of unity do not exist; for example, in the line with two origins, take the open cover by "the line plus the first origin" and "the line plus the second origin". There is no partition of unity subordinate to this cover (the values at each origin would have to be 1).
For me, a basic example of the beauty of this function-theoretic approach is the definition of a vector field as a derivation $D\colon C^\infty(M)\to C^\infty(M)$. The proof that such a derivation defines a vector field hinges upon the fact that $Df$ near a point p only depends on $f$ near the point p. To prove this fact you use the fineness of your sheaf $\mathcal{O}\_X$, i.e. the existence of partitions of unity. (It is true though that the failure of fineness in the non-Hausdorff case is of a different sort and might not break this particular theorem.) I feel that the existence of partitions of unity, and the implications thereof, is one of the basic fundamentals of approaching smooth manifolds through their functions; more importantly, a good handle on how partitions of unity are used is important to understand the differences that arise when the same approach is extended to more rigid functions (holomorphic, algebraic, etc.).
---
Now that the question has been edited to ask specifically about Stokes' theorem, let me say a bit more. Stokes' theorem will be false for non-Hausdorff manifolds, because you can (loosely speaking) quotient out by part of your manifold, and thus part of its homology, without killing all of it.
For the simplest example, consider dimension 1, where Stokes' theorem is the fundamental theorem of calculus. Let $X$ be the forked line, the 1-dimensional (non-Hausdorff) manifold which is the real line with the half-ray $[0,\infty)$ doubled. For nonnegative $x$, denote the two copies of $x$ by $x^\bullet$ and $x\_\bullet$, and consider the submanifold $M$ consisting of $[-1,0) \cup [0^\bullet,1^\bullet] \cup [0\_\bullet,1\_\bullet]$. The boundary of $M$ consists of the three points $[-1]$ (with negative orientation), $[1^\bullet]$ (with positive orientation), and $[1\_\bullet]$ (with positive orientation); to see this, just note that every other point is a manifold point.
Consider the real-valued function on $X$ given by "$f(x)=x$" (by which I mean $f(x^\bullet)=f(x\_\bullet)=x$). Its differential is the 1-form which we would naturally call $dx$. Now consider $\int\_M dx$; it seems clear that this integral is 3, but I don't actually need this. Stokes' theorem would say that
$\int\_M dx=\int\_M df = \int\_{\partial M}f=f(1^\bullet)+f(1\_\bullet)-f(-1)=1+1-(-1)=3$.
This is all fine so far, but now consider the function given by $g(x)=x+10$. Since $dg=dx$, we should have
$\int\_M dx=\int\_M dg=\int\_{\partial M}g=g(1^\bullet)+g(1\_\bullet)-g(-1)=11+11-9=13$. Contradiction.
It's possible to explain this by the nonexistence of flows (instead of $df$, consider the flux of the flow by $\nabla f$). But also note that Stokes' theorem, i.e. homology theory, is founded on a well-defined boundary operation. However, without the Hausdorff condition, open submanifolds do not have unique boundaries, as for example $[-1,0)$ inside $X$, and so we can't break up our manifolds into smaller pieces. We can pass to the Hausdorff-ization as Andrew suggests by identifying $0^\bullet$ with $0\_\bullet$, but now we lose additivity. Recall that $M$ was the disjoint union of $A=[-1,0)$ and $B=[0^\bullet,1^\bullet] \cup [0\_\bullet,1\_\bullet]$. So in the quotient $\partial [A] = [0]-[-1]$ and $\partial [B] = [1^\bullet]-[0]+[1\_\bullet]-[0]=[1^\bullet]+[1\_\bullet]-2[0]$, which shows that $\partial [M]\neq \partial [A]+\partial [B]$. This is inconsistent with any sort of Stokes formalism.
Finally, I'd like to point out that Stokes' theorem aside, even rather nice non-Hausdorff manifolds can be significantly more complicated than we might want to deal with. One nice example is the leaf-space of the foliation of the punctured plane by the level sets of the function $f(x,y)=xy$. The leaf-space looks like the union of the lines $y=x$ and $y=-x$, except that the intersection has been blown up to four points, each of which is dense in this subset. In general, any finite graph can be modeled as a non-Hausdorff 1-manifold by blowing up the vertices, and in higher dimensions the situation is even more confusing. So for any introductory explanation, I would strongly recommend requiring Hausdorff until the students have a lot more intuition about manifolds.
| 27 | https://mathoverflow.net/users/250 | 13084 | 8,844 |
https://mathoverflow.net/questions/13044 | 3 | Hi,
I'm studying an ODE with a small parameter $\epsilon$ and I'm trying to find the solution in terms of a zeroth-order term and a boundary layer. The zeroth-order term has a logarithmic behavior near $x=0$ while the boundary layer term has an exponential (special function Bi) behavior at $+\infty$. To get all the constants I need to do some sort of asymptotic matching between the two solutions but I'm a bit at a loss as to whether this is possible.
More specifically,
the solutions that span the outer solution are the Bessel functions $J\_0$ and $y\_0$ and the inner solutions are the integral of a sum of the Airy functions $Ai$ and $Bi$
The problem of-course pertains to the $y\_0$ and the $Bi$. The only way out that I see now is to set the coefficients in front of both functions to zero due to this problem and only use the other two. But then there's a problem with the boundary conditions.
Any advice or a reference would be greatly appreciated.
Cheers,
Yossi.
| https://mathoverflow.net/users/3578 | Asymptotic Matching of an logarithmic Outer solution to an exponential growing inner solution | It's been a long time since I did any singular perturbations, but when I did, the text we used was [Kevorkian and Cole](http://rads.stackoverflow.com/amzn/click/0387942025); it definitely covers this type of problem. I think that Bender and Orszag's Advanced Mathematical Methods for Scientists and Engineers (sorry, MO won't let me add another link) has a section on it also, but not in as much depth.
FWIW, I remember Bender and Orszag being a better book in general.
| 2 | https://mathoverflow.net/users/2293 | 13085 | 8,845 |
https://mathoverflow.net/questions/13107 | 5 | As I was studying the Möbius $\mu$ function and Gram series,
I got myself some pretty nice books:
**Ribenboim - The New Book of Prime Number Records**
**Apostol - Introduction to Analytic Number Theory**
**Niven, Zuckerman, Montgomery - An Introduction to the Theory of Numbers**
**Iwaniec and Kowalski - Analytic Number theory**
All of them deal with the Möbius $\mu$ function.
But none of them dealt with the subject in details other than giving a few theorems and problems...
So I would like to know, If you guys know of some good books that deal exclusively with arithmetic functions?
| https://mathoverflow.net/users/2865 | Good books on arithmetic functions? | A MathSciNet search set to Books and with "arithmetic functions" entered into the "Anywhere" field yields 148 matches. Some of the more promising ones:
>
> The theory of arithmetic functions.
> Proceedings of the Conference at Western Michigan University, Kalamazoo, Mich., April 29--May 1, 1971. Edited by Anthony A. Gioia and Donald L. Goldsmith. Lecture Notes in Mathematics, Vol. 251. Springer-Verlag, Berlin-New York, 1972. v+287 pp.
>
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$ $
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> Narkiewicz, Wƚadysƚaw Elementary and analytic theory of algebraic numbers. Monografie Matematyczne, Tom 57. PWN---Polish Scientific Publishers, Warsaw, 1974. 630 pp. (errata insert).
>
>
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$ $
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> Babu, Gutti Jogesh
> Probabilistic methods in the theory of arithmetic functions.
> Macmillan Lectures in Mathematics, 2. Macmillan Co. of India, Ltd., New Delhi, 1978.
>
>
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$ $
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> Elliott, P. D. T. A. Arithmetic functions and integer products. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 272. Springer-Verlag, New York, 1985.
>
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$ $
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> Sivaramakrishnan, R. Classical theory of arithmetic functions. Monographs and Textbooks in Pure and Applied Mathematics, 126. Marcel Dekker, Inc., New York, 1989.
>
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$ $
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> Schwarz, Wolfgang; Spilker, Jürgen Arithmetical functions. An introduction to elementary and analytic properties of arithmetic functions and to some of their almost-periodic properties. London Mathematical Society Lecture Note Series, 184. Cambridge University Press, Cambridge, 1994.
>
>
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$ $
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> Tenenbaum, Gérald Introduction to analytic and probabilistic number theory. Translated from the second French edition (1995) by C. B. Thomas. Cambridge Studies in Advanced Mathematics, 46. Cambridge University Press, Cambridge, 1995.
>
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>
| 4 | https://mathoverflow.net/users/1149 | 13111 | 8,862 |
https://mathoverflow.net/questions/13122 | 5 | In most physics situations one gets the metric as a positive definite symmetric matrix in some chosen local coordinate system.
Now if on the same space one has two such metrics given as matrices then how does one check whether they are genuinely different metrics or just Riemannian Isometries of each other (and hence some coordinate change can take one to the other).
If in the same coordinate system the two matrices are different then is it proof enough that they are not isometries of each other? (doing this test over say a set of local coordinate patches which cover the manifold)
Asked otherwise, given two ``different" Riemannian Manifolds how does one prove the non-existence of a Riemannian isometry between them?
There have been two similar discussions on mathoverflow at [this](https://mathoverflow.net/questions/12461/changing-coordinates-so-that-one-riemannian-metric-matches-another-up-to-second) and [this one](https://mathoverflow.net/questions/8023/when-is-a-riemannian-metric-equivalent-to-the-flat-metric-on-mathbb-rn).
And [this article](http://deltaepsilons.wordpress.com/2009/11/12/the-test-case-flat-riemannian-manifolds/) was linked from the later.
In one of the above discussions Kuperberg had alluded to a test for local isometry by checking if the Riemann-Christoffel curvatures are the same locally. If the base manifold of the two riemannian manifolds is the same then one can choose a common coordinate system in which to express both the given metrics and then given softwares like mathtensor by Mathew Headrick this is probably not a very hard test to do.
So can one simply patch up such a test through out a manifold to check if two given riemannian manifolds are globally isometric?
How does this compare to checking if the metrics are the same or not in a set of common coordinate patches covering the manifold?
I somehow couldn't figure out whether my query above is already getting answered by the above two discussions.
| https://mathoverflow.net/users/2678 | Testing for Riemannian isometry | This goes by the name of the "equivalence problem" in riemannian geometry and it is an important problem in the classification effort of solutions to gravity field equations.
Malcolm MacCallum has many results in this area. For example [this paper from 1985](https://doi.org/10.1007/3-540-15984-3_242) might be a place to start reading about it.
| 4 | https://mathoverflow.net/users/394 | 13141 | 8,880 |
https://mathoverflow.net/questions/13133 | 3 | Is it possible that $\mathbb{P}^n$ is an algebraic vector bundle over some algebraic variety? This is an interesting question that my friend asked in a student seminar. I believe that the answer is NOT. Because the only global sections of $\mathcal{O}\_{\mathbb{P}^n}$ are constants. However as a total space of vector bundles $E$ over $X$, the global sections are all in $\Gamma(X, \oplus Sym^n E)=\oplus \Gamma(X,Sym^n E)$ which may not be $\mathbb{C}$ in general. The trouble case is when $\Gamma(X, E)=0$? Am I correct? Is there any other explanation?
Edit: Thanks to everyone for your comments and answer. It seems that compactness is the correct way to follow.
However, I still wondering that why the argument on sections doesn't work. In other words, is there an example of a non-proper algebraic variety whose structure sheaf only has constant global sections.
| https://mathoverflow.net/users/2348 | Can $\mathbb{P}^n$ be regarded as an algebraic vector bundle over some algebraic variety? | Stephen Griffeth's argument works over any field. The total space of a vector bundle is never proper (follows by, e.g., valuative criterion for properness). On the other hand, $P^n$ is always proper.
Here is an argument that the total space of a vector bundle is not proper: A fiber of a vector bundle is isomorphic to $A^n$ . Moreover, the fiber considered as a subscheme is a closed subscheme. You can map $(A^1 \setminus 0)$ into the fiber by a map like $f(x)=1/x$. This map can't extend over zero, because if it did, then zero would be sent to something in the closure of the fiber. But the fiber is already closed, so zero would be sent to the fiber, contradiction.
---
Edit: Of course, amaanush's answer is a much better answer than mine!
| 6 | https://mathoverflow.net/users/83 | 13144 | 8,883 |
https://mathoverflow.net/questions/13106 | 34 | I've attempted going past basic number theory several times, and always got lost in its vastness. Do any of you, perhaps, know a good review that pieces together the many concepts involved (Hecke algebras, SL2(ℤ), Fuchsian groups, L-functions, Tate's thesis, Ray class groups, Langlands program, Fourier analysis on number fields, cohomological versions of CFT, Iwasawa theory, modular forms, ...)?
Thanks.
| https://mathoverflow.net/users/3238 | Map of Number Theory | The book you are looking for exists!! And indeed it contains ALL the buzzwords in your question!
It is Manin/Panchishkin's "Introduction to Modern Number Theory". This is a survey book that starts with no prerequisites, contains very few proofs, but nicely explains the statements of central theorems and the notions occurring therein and gives motivations for the questions that are being pursued. You should take a look, at least it can help you decide what you want to study in more detail.
| 44 | https://mathoverflow.net/users/733 | 13145 | 8,884 |
https://mathoverflow.net/questions/13139 | 21 | Given a semisimple Lie group G, let T be a maximal torus, W be the Weyl group defined as the quotient N(T)/C(T), where N(T) denotes the normalizer of T and C(T) denotes the centralizer.
Two questions are:
1. How many ways are there we can realize W as a subset of G?
2. Can we always realize W as a subgroup of G?
| https://mathoverflow.net/users/1832 | Can we realize Weyl group as a subgroup? | In general it is not possible to embed the Weyl group $W$ in the group $G$: already you can see this for $SL\_2(\mathbb C)$, where the Weyl group has order $2$: if the torus fixes the lines spanned by $e\_1$ and $e\_2$ respectively, you want to pick the linear map taking $e\_1$ to $e\_2$ and $e\_2$ to $e\_1$, but this has determinant $-1$. A lift of $W$ to $N(T)$ must be an element of order $4$ not $2$, say $e\_1 \mapsto -e\_2$ and $e\_2 \mapsto e\_1$.
In fact, Tits has shown ([Normalisateurs de Tores I. Groupes de Coxeter Étendus (Journ. Alg. 4, 1966, pp. 96-116](https://doi.org/10.1016/0021-8693(66)90053-6)) that this is essentially the only obstruction: the Weyl group can always be lifted to a group $\tilde{W}$ inside $G$ which is an extension of $W$ by an elementary abelian $2$-group of order $2^l$ where $l$ is the number of simple roots. If I recall correctly, this lift is then unique up to conjugation.
| 30 | https://mathoverflow.net/users/1878 | 13146 | 8,885 |
https://mathoverflow.net/questions/13120 | 3 | What is a lower bound for the Jacobian of the exponential map from the skew-symmetric matrices to the orthogonal matrices near the origin?
| https://mathoverflow.net/users/1170 | Lower bound for Jacobian of matrix exponential map near origin | For simplicity, I work with $2n \times 2n$ matrices, the odd by odd case is similar.
Summary: If $B$ is a skew symmetric matrix with eigenvalues $\pm i \theta\_1$, $\pm i \theta\_2$, ..., $\pm i \theta\_n$ then the Jacobian matrix of the exponential near $B$ has eigenvalues
$$\frac{1-e^{i(\mp \theta\_j \mp \theta\_k)}}{i(\pm \theta\_j \pm \theta\_k)}, \quad 1 \leq j < \leq n$$
as well has having the eigenvalue $1$ with multiplicity $n$. (This is $4 \binom{n}{2}+n=\binom{2n}{2}$ eigenvalues in total.)
The determinant of the Jacobian matrix is thus
$$\prod\_{1 \leq j < k \leq n} \frac{1-e^{i(\mp \theta\_j \mp \theta\_k)}}{i(\pm \theta\_j \pm \theta\_k)} = \prod\_{1 \leq j < k \leq n} \frac{16 \sin^2 \frac{\theta\_j-\theta\_k}{2} \sin^2 \frac{\theta\_j+\theta\_k}{2}}{(\theta\_j^2-\theta\_k^2)^2}.$$
Take whatever sort of bound you have on $B$ and turn it into a lower bound on the above quantity. Notice that, if $\theta\_j + \theta\_k$ gets as large as $2 \pi$, the above quantity is zero, so there is no nontrivial lower bound in that case.
---
A practical note on minimizing the above quantity: the log of the above is the sum of many terms of the form $f(\phi) := \log \sin(\phi) - \log \phi$, where $\phi$ is a linear function. By the double derivative test, $f$ is concave. So the sum of many terms of the form $f(\mbox{linear function})$ will be concave. This means that, on any convex region, the minimum will occur somewhere on the boundary.
---
Notation: We write $\mathfrak{so}$ for the vector space of Skew symmetric matrices. We fix $B$ and $\theta\_j$ as above.
Explanation: Let me first point out why the question makes sense. The orthogonal matrices are a manifold, not a vector space, so one might be tempted to wonder whether it even makes sense to speak of a Jacobian; let alone to speak of the eigenvalues of the Jacobian matrix.
There are two ways to fix this, a naive way, and a sophisticated way, and they both give the same answer. The naive way is to point out that the orthogonal matrices are contained in the $n \times n$ matrices. So we certainly have a $\binom{2n}{2} \times (2n)^2$ matrix, giving the Jacobian matrix of the exponential map as a map from skew-symmetric matrices to all matrices. This matrix is not square; its image is the tangent plane at $e^B$ to the space of orthogonal matrices. Explicitly, that tangent plane is $e^B \mathfrak{so}$. We can rotate that tangent plane by the orthogonal matrix $e^{-B}$, giving us a map from $\mathfrak{so}$ to itself; it is now sensible to discuss the eigenvalues of that map.
The sophisticated way is say that the Jacobian matrix is a map from $\mathfrak{so}$ to the tangent space of $SO$ at $e^B$. But that tangent space is canonically identified with the tangent space of $SO$ at the identity, and the latter tangent space is $\mathfrak{so}$.
Either way, we are being asked to consider the following map from $\mathfrak{so}$ to itself:
$$E \mapsto e^{-B} \lim\_{t \to 0} (e^{B+tE} - e^B)/t. \quad (\*)$$
Let $A$ be the map $X \mapsto [B,X]$ from $\mathfrak{so}$ to itself. By the Baker-Campbell-Hausdorff formula, $(\*)$ is
$$\frac{1-e^{-A}}{A} E$$
where $(1-e^{-A})/A$ must be understood as the power series $1-A/2+A^2/6-\cdots$. If written out as matrices, $A$ would be a $\binom{2n}{2} \times \binom{2n}{2}$ matrix and $E$ would be a vector of length $\binom{2n}{2}$.
Now, if the eigenvalues of $B$ are $\pm i \theta\_j$, as above, then the eigenvalues of $A$ are $0$ with multiplicity $n$ and $i(\pm \theta\_j \pm \theta\_k)$. (Because the root system of type $D\_n$ is $\{ \pm e\_j \pm e\_k \}$, or because it is an easy computation.) If $\alpha\_1$, $\alpha\_2$, ..., $\alpha\_N$ are the eigenvalues of $A$, then the eigenvalues of $(1-e^{-A})/A$ are $(1-e^{-\alpha\_j})/\alpha\_j$, where $(1-e^{-0})/0$ is interpreted as $1$.
Putting this all together, we get the above formulas.
| 3 | https://mathoverflow.net/users/297 | 13147 | 8,886 |
https://mathoverflow.net/questions/11777 | 9 | Given two non-isogenous elliptic curves $E\_1$ and $E\_2$ over $\mathbb{C}$.
Set $A:=E\_1 \times E\_2$. Given a nontrivial sheaf of quaternion algebras $D$ over $A$, what is the dimension of the vector space $H^1(A,D)$?
If one thinks of $D$ as an element in the Brauer group $Br(A)$, then it is $2$-torsion, hence belongs to $Br(A)[2]$. Since the curves are non-isogenous there is an isomorphism $Pic(E\_1)[2] \otimes Pic(E\_2)[2] \to Br(A)[2]$. So there should be a connection between such quaternions and $2$-torsion line bundles on the curves, but i cannot find an explicit description for this isomorphism. If there is one, i thought one could use the Künneth formula to compute $H^1(A,D)$ in terms of the cohomology of the line bundles on the curves.
For now i could only work out the bound $d=dim(H^1(A,D)) \geq 2$: using Hirzebruch-Riemann-Roch and simplifying terms one gets $d=c\_2(D)+2$. After a result of M.Lieblich one has $c\_2(D)\geq 0$.
Does anyone see/have an explicit description of the isomorphism mentioned above? Is the idea using Künneth a promising approach to this problem at all? Or does anyone have another approach? Are there some calculations regarding this in the literature (i couldn't find one)?
Another question in this context is: what is the image of such an algebra under the map $Br(A) \rightarrow Br(\mathbb{C}(A))$. This should be nontrivial $\mathbb{C}(A)$-quaternions, since the map "looking at the genric point $\eta$" is injective, i.e. $D\_{\eta}$ is generated by elements $i,j$ with $i^2=a, j^2=b and ij=-ji$. But what are a resp. b? I think they should have something to do with functions h such that 2\*Y=div(h), where Y defines one of the line bundles. Is this true?
| https://mathoverflow.net/users/3233 | Cohomology of quaternions on an abelian variety | For the description of the quaternion algebra associated to a pair of torsion line bundles, try the following. Take line bundles ${\cal L}\_i$ on $E\_i$ equipped with isomorphisms ${\cal L}\_i^{\otimes 2} \to {\cal O}$, and pull these back to $A$. Define
`$$D = {\cal O} \oplus {\cal L}_1 \oplus {\cal L}_2 \oplus {\cal L}_1 \otimes {\cal L}_2$$`
with multiplication induced by the maps ${\cal L}\_i^{\otimes 2} \to {\cal O}$, ${\cal O}$ being the unit, and the elements of ${\cal L}\_1$ and ${\cal L}\_2$ anticommuting.
ADDED: I wish I had a more conceptual explanation for why this represents the cup-product in the Brauer group, but here is a cocycle description along the lines of what Oren suggested.
Suppose $X$ is given with 2-torsion line bundles $\cal L$ and $\cal M$. Choose cocycles representing these, in the form of a cover (either open in the analytic case, or etale in the algebraic case) $U\_\alpha$ of $X$ together with sections $s\_\alpha$ of $\cal L$ and $t\_\alpha$ of $\cal M$ on $U\_\alpha$ such that $s\_\beta / s\_\alpha = u\_{\alpha \beta} \in \{\pm 1\}$ and similarly $t\_\beta / t\_\alpha = v\_{\alpha \beta}$; these latter two are the representing cocycles.
Then D has basis $\{1,s\_\alpha, t\_\alpha, s\_\alpha t\_\alpha\}$ on $U\_\alpha$, where $s\_\alpha^2 = t\_\alpha^2 = 1$, and you can explicitly make this isomorphic to a matrix algebra. The change-of-basis sends $s\_\alpha$ to $s\_\beta = u\_{\alpha \beta} s\_\alpha$ and similarly for $t$. This can be achieved by conjugation by the element $t\_\alpha^{(1 - u\_{\alpha \beta})/2} s\_\alpha^{(1 - v\_{\alpha \beta})/2} = g\_{\alpha\beta} \in D \cong M\_2(\mathbb{C})$. These change-of-basis matrices reduce to a cocycle in $PGL\_2(\mathbb{C})$ representing the algebra, and so the image in the Brauer group is represented by the coboundary $(\delta g)\_{\alpha \beta \gamma} \in \{\pm 1\}$.
EDIT: fixed up following description of the coboundary so that it correctly described where the cup product lands.
Explicit computation finds $(\delta g)\_{\alpha \beta \gamma}$ is equivalent to the cocycle $v\_{\alpha \beta} \otimes u\_{\beta \gamma}$ with coefficients in $\{\pm 1\} \otimes \{\pm 1\} \cong \{\pm 1\}$ (I may have mixed the indices, if I did please let me know and I'll correct it), which is precisely the formula for the cup product of the cocycles $u$ and $v$.
So based on your description of the cup product inducing an isomorphism between 2-torsion in the Brauer group and cup products of 2-torsion elements in the Picard group, this genuinely should provide you with the bundles you're looking for.
| 2 | https://mathoverflow.net/users/360 | 13160 | 8,895 |
https://mathoverflow.net/questions/12951 | 9 | Most calculations of étale cohomology in Milne's book deal with constructible or torsion sheaves. Are there references where the cohomology of varieties with $\mathbf{G}\_m$ coefficients are calculated? I'm especially interested in the top dimension $2\mathrm{dim}(X)$ ($+ 1$). I found some calculations in Le groupe de Brauer (In: Dix Exposés sur la Cohomologie des Schémas), but they don't help me.
Edit: Assume $X$ is a variety over a finite field.
| https://mathoverflow.net/users/nan | étale cohomology with G_m coefficients | I found calculations in S. Lichtenbaum, Zeta functions of varieties over finite fields at s = 1, Arithmetic and geometry, Vol. I, 173–194 Progr. Math., 35, Birkhauser Boston, Boston, MA, 1983, especially Proposition 2.1 and Theorem 2.2--2.4.
| 5 | https://mathoverflow.net/users/nan | 13163 | 8,897 |
https://mathoverflow.net/questions/13124 | 3 | Dear Colleagues,
This is a math question for people who know the rules of (American) football.
Every year my barber runs a “football squares” game. He finds 100 customers, each put in 20 dollars, and each person is assigned a square on a 10 by 10 grid. After all squares are sold he picks numbers out of a hat to label each row and column 0 thru 9. So each contestant winds up being assigned with an ordered pair of numbers – in my case (3,7) this year. Now prizes are awarded by inspecting the last digit of the score of each team in the Super Bowl at the end of the first, second, third quarters and the final game score. Each winner gets $500.
For example, suppose that the scores are AFC 14, NFC 10 at the end of the first quarter. Then the person with (4,0) wins $500.
Now this is a fair bet, in the sense that each square is assigned randomly (and I believe that my barber doesn’t cheat.) However, it would seem that some numbers are “better” than others. Football scores are not random. For instance, at the end of the first quarter it is extremely unlikely that (5,5) will win. On the other hand, ((0,7) would seem like a good number. What is needed is some probabilistic analysis based upon the actual scoring patterns in football together with looking at actual scores of many pro football games. I am unable to find any analysis of this on the internet. I have tagged this as a probability question but since I work in operator algebras for a living this may be mis-tagged, and I ask you to pardon my error.
Let’s make it precise. Let f: {0,1,2,…, 9} x {0,1,2,…, 9} x {1,2,3,final} \to [0,1] be the function that to a point (x,y,z) assigns the probability that the score at the end of the z’th quarter of the Super Bowl will be equal to (x,y) mod 10. Find the function f. Where does f achieve its max?
How about it, colleagues? Inquiring minds want to know!
CS
| https://mathoverflow.net/users/3577 | Football Squares | Alternative (and rough) idea for a model: Consider a random walk on score vectors $[x,y]$ where x and y represent the scores of the two teams. Set $P\_0=[0,0]$. Let $P\_1=P\_0+v\_1$ where $v\_1$ is with probability 1 chosen (uniformly distributed or otherwise) from $[0,7]$, $[7,0]$, $[3,0]$, or $[0,3]$. Let $P\_2=P\_1+v\_2$ where $v\_2$ is randomly chosen from the same set of vectors but with probability $1/1.1$ (adding the zero vector with probability $.1/1.1$). Repeat: $P\_i=P\_{i-1}+v\_i$ where $v\_i$ is chosen from those four vectors with probability $1/(1.1)^i$ or zero otherwise. Letting $i\rightarrow\infty$ and repeating gives you a distribution of possible football scores, avoiding the problem of finding an "expected value of a game", which would be nearly irrelevant to the trailing digit.
Quick notes:
* The constant $1.1$ is chosen so that the expected total number of scores is 10, which I found online (and thus must be true) as the average number of scores in an NFL game.
* Truncating after fewer steps would gives you distributions for quarter scores instead of final scores.
* Probabilities should probably not be uniform, either between the two teams (so this model could build in one team being a favorite) or between field goals and touchdowns.
* A fun tweak would be to build in 8's for two-point conversions context-dependent on the current score. Safeties are probably negligible.
* Easily modifiable and poorly written SAGE code below:
```
def rungame():
v=vector(ZZ,[0,0])
p = random()
enter code here
def iterate(v,i):
p = random()
if (0<p<.25/(1.1^i)):
return v+vector(ZZ,[7,0])
if (.25/(2^i)<=p<.50/(1.1^i)):
return v+vector(ZZ,[3,0])
if (.5/(2^i)<=p<.75/(1.1^i)):
return v+vector(ZZ,[0,3])
if (.75/(2^i)<=p<=1/(1.1^i)):
return v+vector(ZZ,[0,7])
return v;
for i in range(0,20):
v=iterate(v,i);
print v
```
EDIT: Here's the result of 10,000 trials. The 117 in the top left is the number of times the game ended with (0,0) digits, etc.
[117 91 116 121 117 51 89 118 116 86]
[121 54 91 47 91 110 43 115 117 50]
[ 42 90 111 42 47 52 48 83 118 85]
[119 85 39 99 90 54 122 92 88 120]
[ 89 48 91 44 120 89 92 119 97 42]
[112 109 87 114 86 0 117 121 73 113]
[ 89 99 114 119 47 33 118 121 41 114]
[ 99 88 120 88 88 89 40 120 98 84]
[ 99 120 100 49 91 44 40 43 32 115]
[ 90 100 82 89 98 91 50 115 111 100]
| 2 | https://mathoverflow.net/users/35575 | 13165 | 8,899 |
https://mathoverflow.net/questions/13174 | 11 | It is well known that if a category has all coequalizers and all (small) coproducts then in fact it has all (small) colimits. More important is the proof which shows that every colimit can be built by using coproducts and coequalizers. This implies that if a functor commutes with coproducts and coequalizers, then it must commute with all (small) colimits as well.
>
> Is there homotopical analog of this? If I have a functor which commutes with all small (homotopy) coproducts and all homotopy coequalizers, does it necessarily commute with all homotopy colimits in general?
>
>
>
This question makes sense for general model categories, but I am particularly interested in the usual model structure on spaces.
| https://mathoverflow.net/users/184 | Do h-coequalizers and coproducts give all h-colimits? | There is an analogue, but one should replace coequalizers by geometric realizations (homotopy colimits over Δop). If $F : I \to M$ is a diagram in a model category, one has
$$\operatorname{hocolim}\_I F = \operatorname{hocolim}\_{k \in \Delta^{\operatorname{op}}} \coprod\_{i\_0 \to \cdots \to i\_k \in I} F(i\_0).$$
For instance, see section 2 of [<http://www.math.uchicago.edu/~eriehl/hocolimits.pdf>](http://www.math.uchicago.edu/~eriehl/hocolimits.pdf) for the simplicial model category case.
| 13 | https://mathoverflow.net/users/126667 | 13177 | 8,904 |
https://mathoverflow.net/questions/13181 | 2 | What is the variance of $1/(X+1)$ where $X$ is Poisson-distributed with parameter $\lambda$! The series for the second moment is horrible!
$E({1\over (X+1)^2})=\sum\_{k=1}^{\infty}\frac{1}{k^{2}}\frac{\lambda^{k}e^{-\lambda}}{k!}$
Is there an easy way to do it?
| https://mathoverflow.net/users/3589 | variance of $1/(X+1)$ where $X$ is Poisson-distributed with parameter $\lambda$ | Sorry, I gave a moronic answer before. Let me try to give a better one.
There should be no expression for $f(\lambda) := \sum\_{k \geq 1} \lambda^k/(k^2 k!)$ in elementary functions. If there were, then $g(\lambda) = \lambda f'(\lambda) = \sum\_{k \geq 1} \lambda^{k}/(k \cdot k!)$ would also be elementary. But $g(\lambda)=\int\_0^{\lambda} \frac{e^t-1}{t} dt$ and $e^t/t$ is a standard example of a function without an elementary antiderivative.
| 2 | https://mathoverflow.net/users/297 | 13189 | 8,911 |
https://mathoverflow.net/questions/13162 | 15 | Let $S$ be a scheme of positive characteristic $p$ and $X$ a smooth $S$-scheme. Let $F:X\rightarrow X^{(p)}$ denote the relative Frobenius. A result by Cartier (often called Cartier descent or Frobenius descent) then states that the category of quasi-coherent $\mathcal{O}\_{X^{(p)}}$-modules is equivalent to the category of quasi-coherent $\mathcal{O}\_X$-modules $(E,\nabla)$ with integrable connection of $p$-curvature $0$ (which means $\nabla(D)^p-\nabla(D^p)=0$ for all $S$-derivations $D:\mathcal{O}\_X\rightarrow \mathcal{O}\_X$).
The equivalence is given by
$$ (E,\nabla)\longmapsto E^\nabla$$
and
$$ E\mapsto (F^\*E,\nabla^{can})$$
where $\nabla^{can}$ is the canonical connection locally given by $f\otimes s\mapsto (1\otimes s)\otimes df$, for
$$f\otimes s\in \mathcal{O}\_X(U)\otimes E(U).$$
(tensor over the sections of the structure sheaf of $X^{(p)}$, somehow jtex can't handle that)
The proof of this theorem can be found in 5.1. in Katz' paper ["Nilpotent connections and the monodromy theorem"](http://www.numdam.org/numdam-bin/item?id=PMIHES_1970__39__175_0)
> My question is: As $X/S$ is smooth, the relative Frobenius is faithfully flat (at least it is if $S$ is the spectrum of a perfect field), can the above theorem be interpreted as an instance of faithfully flat descent along $F$? In other words, does the connection $\nabla$ give rise to a descent datum for $E$ with respect to $F$?
I know that connections are "first-order descent data", i.e. modules with connection descend along first order thickenings, but I don't see how this applies here.
| https://mathoverflow.net/users/259 | Frobenius Descent | I believe that the answer is yes, and that this may have been one of Grothendieck's motivations for developing the general theory of flat descent. (If I remember correctly,
in the first (?) expose of FGA, in which he explains flat descent, Grothendieck has a reference to work of Cartier involving descent in the context of inseparable extensions,
and I would guess that it is a reference to this Cartier, or Frobenius, descent. Can
anyone cofirm this?)
To put a connection on $E$ is to extend the action of $\mathcal O\_X$ to an action
of $\mathcal D\_X$, the ring of differential operators generated (in local coordinates) by
$\partial /\partial x\_1,\ldots,\partial/\partial x\_n.$ (This is not the same as the full ring
of differential operators in charateristic $p$.) The $p$-curvatures generate an ideal
in this ring; I think it is just the ideal generated by $(\partial/\partial x\_i)^p$.
So if $E$ has vanishing $p$-curvature, the action of $\mathcal D\_X$ factors through
the quotient by this ideal. One can now interpret this information in terms
of descent data.
A precise description is given in Prop. 2.6.2 of Berthelot's book
[D-modules arithmetiques II; Descente par Frobenius](http://perso.univ-rennes1.fr/pierre.berthelot/).
| 13 | https://mathoverflow.net/users/2874 | 13193 | 8,915 |
https://mathoverflow.net/questions/13169 | 11 | Suppose $E$ is a topos, and consider the operations $0,1,+,\times$ (denoting the initial and terminal objects, the coproduct, and the product), and recall that $E$ satisfies the usual arithmetic laws, such as the distributive law.
For the unfamiliar, one should think of objects in $E$ as sets, but of course they don't have to be. An "element" of a "set" $A$ means a map $B\rightarrow A$ for some object $B$. If $B=1$, we say the element $1\rightarrow A$ is a unit element of $A$, and you can think of these instead if you prefer, but note that the "generalized" elements $B\rightarrow A$ of an object $A$ completely define $A$, whereas the unit ones don't.
Suppose $A$ is an object of the topos $E$. A "linked list of type A" is a new object $L$ of the topos in which an element is either the word "null" (meaning empty list) or a pair $(a,l)$ where $a$ is an element of $A$ and $l$ is an element of $L$. Equationally:
(\*) L=1+AxL.
By this point, you should understand these things. Ok, now I'm going to tell you the weird puzzle. Suppose we want to "solve" for $L$, so that we can see what it really is. To do this, I'm going to cheat. In a topos, there is no such thing as subtraction nor division.
$$L=1+A\times L$$
$$L-(A\times L)=1$$
$$L\times(1-A)=1$$
$$L=1/(1-A)$$
$$L=1 + A + A^2 + A^3 + \cdots$$
So a linked list of type $A$ is "either the empty list, or an element of $A$, or an ordered pair of elements of $A$, or a triple in $A$, etc."
This faulty computation has led to the correct answer. Once this answer is found, one can check topos-theoretically that it is correct (although note that toposes aren't guaranteed to have infinite coproducts). My question is:
Q: where is this computation actually taking place?
Clearly, as the topos of finite sets is the background for the usual arithmetic of natural numbers, it stands to reason that one would generalize toposes as natural numbers were generalized to rational ones. Has or can this be done? Can this calculation make sense in some appropriate context?
| https://mathoverflow.net/users/2811 | "Linked List" puzzle | The theory of species is the full answer to your question -- but in this specific case all that is needed are some very basic properties of polynomial functors.
Let's focus on you fixed-point equation $L=1+A \times L$, and observe that by unrolling it you can generate your infinite sequence $L = 1 + A + (A \times A) + \ldots$, without using subtraction or division.
Now, you mentioned you were working with objects, and so what you want to do is find the object which is the fixed point of the endo-*functor* $F(X) = 1 + A \times X$. IIRC, Arbib and Manes proved back in the 1970s that in any topos, any polynomial functor (of one variable) will have a least fixed point (that is, the category of $F$-algebras will have an initial algebra). (Morally this is because trees can be encoded with natural numbers.)
Species only start coming into their own when you start considering non-free combinatorial constructions. E.g., think about what the "generating functor" for *unordered* sequences should be -- I won't say because it will spoil the surprise.
| 10 | https://mathoverflow.net/users/1610 | 13197 | 8,917 |
https://mathoverflow.net/questions/13194 | 3 | Suppose we have a square $n\times n$ real matrix $A$ of full rank such that the squares of the elements in each row sum to 1, an $n\times 1$ vector of variables $x$, and an $n\times 1$ real vector $a$, such that $A\cdot x = a$. We can of course take the inverse of $A$ to solve uniquely for $x$.
My question is as follows: suppose we do not know $a$ exactly, but only up to additive error epsilon: that is, we know $a'$ such that $a' = a + error$, where $error$ is a real $n\times 1$ vector with each component in the range $[-\epsilon,\epsilon]$. Vector $x$ is no longer uniquely determined. However, we can solve for some $x'$ such that $x' = x + error'$. My question is, what can we say about the magnitude of the components of $error'$, and how they relate to $\epsilon$?
| https://mathoverflow.net/users/3590 | Solving a noisy set of linear equations. | To be a little more precise, the assumption here is that $\| error \|\_\infty \le \epsilon$ and it seems you want to bound $\| error' \|\_\infty$. So from $error' = A^{-1} error$ it follows that $\| error'\|\_\infty \le \| A^{-1} \|\_\infty \epsilon$. Here $\| A^{-1} \|\_\infty$ is the operator norm of $A^{-1}$ induced by the $\infty$-norm on $\mathbb{R}^n$, which is easily expressed in terms of the entries of $A^{-1}=[b\_{ij}]$:
$$\| A^{-1} \|\_\infty = \max\_{1\le i\le n} \sum\_{j=1}^n |b\_{ij}|.$$
| 2 | https://mathoverflow.net/users/1044 | 13213 | 8,927 |
https://mathoverflow.net/questions/13205 | 48 | One often hears in popular explanations of the failure to find a "Grand Unified Theory" that "Gravity goes off to infinity, but cutting off the edges gives us wrong answers", and other similar mathematically vague statements. Clearly, this issue has some kind of mathematical explanation, but I'm not really qualified to read the corresponding physics work, so I'm wondering what actually fails mathematically.
| https://mathoverflow.net/users/1353 | Mathematical explanation of the failure to quantize gravity naively | Other people have said that the problem is that GR isn't renormalizable. I want to explain what that means in measure-theoretic terms. What I say won't be 100% rigorous, but it should get the general story across.
Quantum field theories are generally defined using a Feynman path integral measure. This measn that you compute correlation functions of observables by summing over all histories of your system, weighting each history by e^{-S}, where S is an functional on the space of fields, called the action. In a field theory, these histories are functions on some spacetime manifold.
Just as you define the ordinary integral as a limit of Riemann sums, you define the Feynman path integral as a limit of "regularized" path integral measures. There are a lot of ways to do this; one of the most popular is the lattice regularization. We choose a finite set of points in spacetime, living on a lattice whose nearest neighbors are a distance a apart. Then we choose a "microscopic" action on our space of fields and discretize it, replacing derivatives with finite difference quotients, and approximate the path integral's weighted sum over histories with a sum over functions defined on the lattice.
Let's grant for a moment that, for any fixed lattice spacing a, this procedure defines a quantum theory, meaning that you can use the moments of the measure to compute correlation functions for your observables. Let's also grant that the expectation values of observables satisfy classical equations of motion (so that when we approach the classical limits, and the probability distributions become concentrated at their expected value, we get deterministic evolution of these values).
We usually want to take a continuum limit, making the lattice spacing smaller and smaller. If we do this, while keeping fixed the coupling constants in the microscopic action we used to define our measure, in most cases, we run into a problem: the coupling constants in the classical equations of motion depend on the lattice spacing a, and become infinite as a goes to zero. So that's probably not the limit we want.
And indeed, in most circumstances, we know the classical physics, and are trying to find a quantum theory that reproduces it. So we'll try something else: make the coupling constants in the microscopic action depend on the lattice spacing, and hope that we can tune them in a way that keeps the classical physics fixed.
Sometimes this works; sometimes it doesn't. It can happen that there isn't any way of tuning the coupling constants so as to reproduce a nice classical limit; when this happens, the theory is said to be non-renormalizable. This appears to be what happens in General Relativity, if you use the discretized Einstein-Hilbert action as your microscopic action.
| 49 | https://mathoverflow.net/users/35508 | 13214 | 8,928 |
https://mathoverflow.net/questions/13209 | 8 | As objects which are minimal (in some respect), this seems entirely plausible, but I'm not sure what category we should be working in, and what restrictions we would need, to actually have a situation where minimal surfaces would be characterized by a universal property, if they ever can be. An uneducated guess on one possible setup where minimal surfaces would be universal: the objects are surfaces whose boundary is a given simple closed curve, and the morphisms are the area-decreasing isometries - it seems like a minimal surface should be a final object, though we would probably need to introduce an equivalence relation on the morphisms to get the maps to be unique?
I'm also curious about the same question, but for geodesics. Perhaps for them, we would use the collection of paths from point $x$ to point $y$ on a given surface, and use the length-decreasing homotopies?
Being a final object isn't the only option - maybe, for any surface, some kind of map will factor through a minimal surface associated to it?
EDIT: I'm worried this is perhaps too soft a question for MathOverflow - I'm not sure there's really a "right" answer.
| https://mathoverflow.net/users/1916 | Can minimal surfaces be characterized by some universal property? | I'm not sure if this answer provides you with the universal property that you desire, but there is such a category that unifies these concepts that you are after.
Cohen, Jones and Segal introduced a concept known as the "Flow Category" in the paper *Morse Theory and Classifying Spaces*, which associates to any manifold with a Morse Function a category whose objects are the critical points of the Morse function and whose morphisms are the gradient trajectories of some gradient-like vector field. Here is the reference:
<http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.5003>
You can get the paper on Ralph Cohen's page if you don't have university access:
<http://math.stanford.edu/~ralph/papers.html>
Recall that Morse Theory was invented by Marston Morse to study geodesics on manifolds. Geodesics correspond precisely to critical points of the Energy functional. I imagine that any variational problem fits into this framework.
As a word of caution, understanding the space of gradient trajectories lies at the heart of Floer Theory, so if you want to understand Morse Theory on infinite dimensional spaces, prepared to get your hands dirty with some serious analysis. Comment if you want more references. Also, most of the above article is concerned with proving a very elegant result about the classifying space of this category for certain situations. It is very slick!
| 10 | https://mathoverflow.net/users/1622 | 13218 | 8,930 |
https://mathoverflow.net/questions/13130 | 14 | The analytic continuation and functional equation for the Riemann zeta function were proved in Riemann's 1859 memoir "On the number of primes less than a given magnitude." What is the earliest reference for the analytic continuation and functional equation of Dirichlet L-functions? Who first proposed that they might satisfy a Riemann hypothesis? Dirichlet did none of these things; his paper dates from 1837, and as far as I know he only considered his L-functions as functions of a real variable.
| https://mathoverflow.net/users/1464 | Historical question in analytic number theory | Riemann was the first person who brought complex analysis into the game, but if you ask just about functional equations then he was not the first. In the 1840s, there were proofs of the functional equation for the $L$-function of the nontrivial character mod 4, relating values at $s$ and $1-s$ for real $s$ between 0 and 1, where the $L$-function is defined by its Dirichlet series. In particular, this happened before Riemann's work on the zeta-function. The proofs were due independently to Malmsten and Schlomilch. Eisenstein had a proof as well (unpublished) which was found in his copy of Gauss' Disquisitiones. It involves Poisson summation. Eisenstein's proof is dated 1849 and Weil suggested that this might have motivated Riemann in his work on the zeta-function.
For more on Eisenstein's proof, see Weil's "On Eisenstein's Copy of the Disquisitiones" pp. 463--469 of "Alg. Number Theory in honor of K. Iwasawa" Academic Press, Boston, 1989.
| 25 | https://mathoverflow.net/users/3272 | 13219 | 8,931 |
https://mathoverflow.net/questions/13195 | 3 | I'm afraight this might be obviously true or false, but anyway: Let $({\mathcal A},{\mathcal E})$ be a Frobenius category and $X,Y\in{\mathcal A}$. If there exist projective-injective $P,Q\in{\mathcal A}$ such that $X\oplus P\cong Y\oplus Q$ in ${\mathcal A}$, then $X\simeq Y$ in $\underline{\mathcal A}$. Is there converse true?
Considering the case $Y=0$ the condition $X\simeq 0$ in $\underline{\mathcal A}$ is equivalent to the existence of a projective-injective object $P$ and maps $p: P\to X$, $\sigma: X\to P$ such that $p\circ\sigma=\text{id}\_X$. Thus if ${\mathcal A}$ is idempotent complete, then this implies that $X$ is a summand of $P$ and thus itself projective-injective. How can one proceed in the general case ?
| https://mathoverflow.net/users/3108 | When do two objects become isomorphic in the stable category? | Suppose $X$ and $Y$ are stably isomorphic, so that there exist a morphism $f:X\to Y$ whose image $\underline f:X\to Y$ in the stable category is an isomorphism. Then $\underline f$ has an inverse: there exists $g:Y\to X$ such that $\underline g\circ\underline f=1\_X$ and $\underline f\circ\underline g=1\_Y$, and this means in particular that there is a projective-injective $P$ and maps $r:X\to P$ and $s:P\to X$ such that $g\circ f-1\_X=s\circ r$.
This gives us maps $F=\left(\begin{smallmatrix}f\\\a\end{smallmatrix}\right):X\to Y\oplus P$ and $G=\left(\begin{smallmatrix}g&-b\end{smallmatrix}\right):Y\oplus P\to X$ such that $G\circ F=1\\_X$. *If now we assume that $\mathcal A$ has all its idempotents split*, then we can conclude that there is a $Q$ such that there is an isomorphism $H:X\oplus Q\xrightarrow{\cong} Y\oplus P$ such that $F=H\circ\iota:X\to Y\oplus P$, with $\iota:X\to X\oplus Q$ the canonical map.
Notice that $\underline F$ and $\underline H$ are isomorphisms in $\underline{\mathcal A}$, so that also $\underline\iota$ is an isomorphism there. If $p:X\oplus Q\to X$ is the projection, then $\underline p\circ\underline\iota=1\\_X$ in $\underline{\mathcal A}$, so in fact $(\underline\iota)^{-1}=\underline p$, and in consequence the composition $\iota\circ p:X\oplus Q\to X\oplus Q$ is the identity of $X\oplus Q$ in $\underline{\mathcal A}$. In other words, there exists a projective-injective $R$ and morphisms $u:X\oplus Q\to R$ and $v:R\to X\oplus Q$ such that $p\circ\iota-1\_{X\oplus Q}=v\circ u$.
If now $j:Q\to X\oplus Q$ and $q:X\oplus Q\to Q$ are the canonical maps, we have $q\circ v\circ u\circ j=-1\\_Q$, so that the morphism $\underline{1\\_Q}:Q\to Q$ is zero. This implies that in fact $Q\cong 0$ in $\underline{\mathcal A}$. By what you showed in your question, this implies that $Q$ is a projective-injective in $\mathcal A$.
All in all, we have shown that there exists projective-injectives $P$ and $Q$ such that $X\oplus Q\cong Y\oplus P$ in $\mathcal A$, as you wanted.
(I do not think your question will have a positive answer when $\mathcal A$ does not have all its idempotents split... I do not have a counterexample, though)
| 4 | https://mathoverflow.net/users/1409 | 13222 | 8,932 |
https://mathoverflow.net/questions/13224 | 5 | Is there an online catalog available of Ramsey numbers, preferably one that for unknown values documents the known upper/lower bounds?
| https://mathoverflow.net/users/3499 | Where can I find a catalog of known Ramsey numbers? | MathWorld has a pretty decent list (scroll down in the link) and cites numerous papers with good bounds
<http://mathworld.wolfram.com/RamseyNumber.html>
| 4 | https://mathoverflow.net/users/934 | 13225 | 8,933 |
https://mathoverflow.net/questions/13203 | 7 | Following, e.g., [Wikipedia's definitions](http://en.wikipedia.org/wiki/Quantum_cohomology), the (small) quantum cohomology ring of $X$ is defined over a "Novikov ring" consisting of formal power series of the form $$ \sum\_{\beta \in H\_2(X;\mathbb{Z})} a\_\beta e^\beta,$$
where the $a\_\beta$ are in some fixed ring (which is probably usually $\mathbb{Q}$ or $\mathbb{C}$).
On the other hand, in the papers of Fukaya and company for instance, there seems to be a different "Novikov ring", consisting of power series of the form $$\sum\_{i=1}^\infty a\_i T^{\lambda\_i},$$
where $a\_i \in \mathbb{Q}$, $\lambda\_i \in \mathbb{R}$, and $\lim\_{i \to \infty} \lambda\_i = \infty$.
What's up with this? Why are there apparently two different Novikov rings? How do I reconcile this? I'm guessing that the $\lambda\_i$ in the second definition should correspond to something like $\int\_\beta \omega$ where $\beta$ is as in the first definition (and where $\omega$ is the symplectic form), but beyond this I'm not sure...
| https://mathoverflow.net/users/83 | Different definitions of Novikov ring? | If you read around, you'll find plenty of other variants of the Novikov ring... The underlying point is that to apply the Gromov compactness theorem (or its algebraic counterpart) you need a bound on the energy (= symplectic area) of ridid holomorphic spheres. Since there's no a priori bound in general, you instead count curves according to their area, or according to some refinement thereof, such as homology class.
Your idea is correct: if we fix a base field $k$, the homomorphism $H\_2(X)/tors. \to \mathbb{R}$ given by pairing with $[\omega]$ induces a homomorphism from the completed group $k$-algebra on $H\_2(X)/tors.$ to the "universal Novikov field" (that is, the field of series $\sum\_{\lambda \in \mathbb{R}}{a(\lambda)T^\lambda}$ where the support of $a \colon \mathbb{R}\to k$ contains only finitely many reals less than any given $C$). After making a small perturbation to $\omega$, changing its cohomology class, we can make this homomorphism injective. If memory serves, the smaller Novikov ring is a PID, and the universal Novikov field is flat over it because it's a torsion-free module over a PID. So you can safely tensor with it after passing to cohomology.
The advantages of the universal Novikov field are that it's universal (meaning biggest, essentially), that it's a field, and that it comes with a natural valuation. The disadvantage is that it's bigger than is strictly necessary.
| 7 | https://mathoverflow.net/users/2356 | 13226 | 8,934 |
https://mathoverflow.net/questions/13220 | 5 | In "Higher algebraic K-theory I" Quillen defines a morphism inverting functor to be a functor from a category C to the category Sets which maps "arrows" in C to isomorphisms in Sets.
Proposition 1:
The category of covering spaces of BC is canonically isomorphic to the category of morphism-inverting functors $F: C\rightarrow Sets$.
[For $C$ a small category, its classifying space $BC$ is the geometric realization of its nerve, $NC$]
This proposition plays an essential role in Quillen's Theorem 1 showing that his Q-construction agrees with Grothendieck's construction for $K\_0$.
Theorem 1:
$\pi\_1(B(QC))$ is canonically isomorphic to the Grothendieck group $K\_0(M)$
Questions: Have morphism-inverting functors played an important role in other contexts? Is there a more modern incarnation of morphism-inverting functors related to the fundamental groupoid of an infinity-category?
| https://mathoverflow.net/users/874 | Quillen's Morphism Inverting Functors | Proposition 1 is extremely straightforward to prove (provided you have some facts like the quillen adjunction between SSet and CGWH). Sing(|S|) gives you a simplicial set where all of the edges "forget" their direction, and when you apply the inverse of the nerve functor, you get back a copy of C with all of its arrows as isomorphisms. Covering spaces are equivalent to (etale) bundles (of sets) on a topological space, which by a theorem in Mac Lane (Sheaves in Geometry and Logic) is equivalent to taking sheaves on the space, so by unraveling these equivalences, you get your result. The last equivalence is probably one you're familiar with as the espace \'etal\'e. (While in general, the nerve functor does not have an inverse, the nerve of a category has some nice properties that make the total singular complex (the $Sing$ functor) pull back intact, modulo directedness of edges. If you think about the actual graph of the nerve of an ordinary category, it's not hard to see why this is true. This is precisely because the geometric realization "forgets" some information.)
The construction you're describing is generalized by a functor in HTT called the unstraightening functor, which you can read about in HTT Ch 2.2. With a number of more sophisticated results, we can generalize the adjunction between $Sing$ and $| \cdot |$ to a Quillen equivalence between SSet-Cat and CGWH-Cat.
HTT is Higher topos theory by J. Lurie.
| 2 | https://mathoverflow.net/users/1353 | 13227 | 8,935 |
https://mathoverflow.net/questions/13105 | 2 | First, the (simple!) setup:
I have a Markov chain X t on some finite state space Ω with stationary distribution π, and a function f from Ω to R. I'd like to estimate the integral of f with respect to π, which I'll write E π (f). There are theorems which say that
$\frac{1}{n} \Sigma\_{t=1}^{n} f(X\_{t})$ converges to E π (f) as n goes to infinity.
Now, if the $X\_{t}$ were iid, then the Berry-Esseen theorem would give error rates in terms of n and (say) the maximum value of f.
Are there similar theorems which give error rates in terms of n, the maximum value of f, and one (or several) of the frequently computed statistics of finite state Markov chains, like relaxation time, mixing time, covering time, etc?
I'm vaguely aware of Sanov-type theorems for Markov chains, which give large-deviation results, but not in terms of these sorts of quantities, and I don't see how to convert the bounds immediately. Alternatively, I'd be very happy if anyone can give a reference for places that people have actually computed the sorts of error terms that do show up in statements of Sanov's theorem for some simple random walks.
EDIT: Added Mark's comments, so that the question might actually make some sense now. In particular, fixed a missing f, and the rather more important mistake that in fact the CLT doesn't give any sort of quantitative bounds by itself.
FURTHER EDIT: I accepted D. Zare's answer, since it certainly works. If anybody is interested in this question, I have since seen a bunch of articles, the latest of which is 'Optimal Hoeffding Bounds for Discrete Reversible Markov Chains' by C. Leon, which are a bit more specialized to the Markov chain case. I have also been told that Brad Mann's thesis is worth reading on the subject, but haven't yet picked up a copy myself.
| https://mathoverflow.net/users/2282 | An Easy Sanov-Type Theorem for Markov Chains? | The sequence of states of a Markov chain with a finite state space is a good example of a sequence of weakly dependent random variables. A convolution like a moving average is another. There are plenty of versions of [central limit theorems for weakly dependent sequences](http://en.wikipedia.org/wiki/Central_limit_theorem#Under_weak_dependence), and even versions of the [Berry-Esseen theorem with weak dependence](http://www.jstor.org/pss/3318596).
The variance of the sum picks up some covariance terms (the cross terms of the second moment don't vanish, although they can be estimated in terms of the mixing time and $f\_{max} - f\_{min}$), and the Berry-Esseen-like bounds get a little worse.
| 1 | https://mathoverflow.net/users/2954 | 13229 | 8,937 |
https://mathoverflow.net/questions/13230 | 26 | Introduction:
Let A be a subset of the naturals such that $\sum\_{n\in A}\frac{1}{n}=\infty$. The [Erdos Conjecture](https://en.wikipedia.org/wiki/Erd%C5%91s_conjecture_on_arithmetic_progressions) states that A must have arithmetic progressions of arbitrary length.
Question:
I was wondering how one might go about *categorizing* or *generating* the divergent series of the form in the introduction above. I'm interested in some particular techniques and I list some examples below:
If we let $S$ be the set of such divergent series: $S=\left[ A: \sum\_{n\in A}\frac{1}{n}=\infty, \ A\in\mathbb{N} \right]$, what kind of operations are there that would make S a group, or at the very least a semigroup? I'm rather vague on what the operatons should be for a reason, because although I presume trivial operations exist, their usefulness in understanding the members of $S$ would be questionable.
Alternately, can one look at these divergent sums through the technique of Ramanujan summation (think: $1+2+3+\ldots =^R -\frac{1}{12}$, $R$ emphasizing Ramanujan summation)? The generalizations of Ramanujan summation (a good reference [here](http://algo.inria.fr/seminars/sem01-02/delabaere2.pdf) ) allow one to assign values to some of these series and give some measure of what kind of divergence is occurring. Moreover, basic series manipulations that hold for convergent series tend to carry over to Ramanujan summation, so can one perhaps look at the set $S$ above as a set of equivalence classes in the sense of two elements being equivalent if they share the same Ramanujan summation constant.
Thanks in advance for any input!
| https://mathoverflow.net/users/934 | Erdos Conjecture on arithmetic progressions | Most of the point of this answer is to promote a piece of terminology:
Three years ago I first taught a number theory course at UGA in which I made the following definition: a subset $A$ of the positive integers is **substantial** if
$\sum\_{n \in A} \frac{1}{n} = \infty$.
A little bit of discussion of this concept occurs in Section 4 of
[http://alpha.math.uga.edu/~pete/4400primes.pdf](http://alpha.math.uga.edu/%7Epete/4400primes.pdf)
I do mention the Erdos(-Turan?) conjecture about arithmetic progressions in substantial sets. For the purpose of constructing examples, possibly the remark I make about any set with positive upper density being substantial will be of most use to you.
Those notes don't contain a proof of that, but a proof of this and more can be found in a very (very!) nice final project done in this class by (then) undergraduate Alex Rice. Sadly he never gave me an electronic copy in a form that I was able to upload to my webpage. If you want to see his writeup, let me know and I'll bug him about this again: the winds of fate have blown him around a bit, but he is now again a UGA (graduate) student taking a number theory course from me.
Finally, to answer one of your questions in a cheap way: yes, the set of substantial subsets of $\mathbb{Z}^+$ certainly forms a semigroup under union. This seems like a completely unhelpful observation, but who knows...
| 14 | https://mathoverflow.net/users/1149 | 13235 | 8,941 |
https://mathoverflow.net/questions/13233 | 7 | In the introduction of his class field theory notes Milne mentions that some famous mathematicians failed to ask if the Artin isomorphism is canonical (between $Gal(L/K)$ and $C\_m/H$ where $H$ is generated by the split primes in $L$). Does this mean:
1)in category theory terms: there is a natural transformation between the functors from abelian extensions over K to abelian groups given by $Gal(?/K)$ and $C\_m/H?$ (where H? is generated by the primes split over $?/K$).
2)or some kind of vaguer statement about whether we need to make choices along the definition of the map.
or maybe 2) is precisely encoded in the definition of 1).
| https://mathoverflow.net/users/1645 | remark in milne's class field theory notes | The point is that it is one thing to show that two mathematical objects are isomorphic; it is another (stronger) thing to give a particular isomorphism between them. A rather concrete instance of this is in combinatorics, where if $(A\_n)$ and $(B\_n)$ are two families of finite sets, one could show that $\# A\_n = \# B\_n$ by finding formulas for both sides and showing they are equal, but it is preferred to find an actual family of bijections $f\_n: A\_n \rightarrow B\_n$.
This is not just a matter of fastidiousness or a general belief that constructive proofs are better. When considering functorialities between various isomorphic objects, the choice of isomorphism matters. For instance, often one wants to put various isomorphic objects into a diagram and know that the diagram commutes: this of course depends on the choice of isomorphism.
In the case of class field theory, these functorialities take the form of maps between the abelianized Galois groups / norm cokernel groups / idele class groups of different fields. The isomorphisms of class field theory can be shown to be the unique ones which satisfy various functoriality properties (and some "normalizations" involving Frobenius elements), and this uniqueness is often just as useful in the applications of CFT as the existence statements.
All of this, by the way, is explained quite explicitly in Milne's (excellent) notes: you just have to read a bit further. See for instance Theorem 1.1 on page 20: "There exists a unique homomorphism...with the following properties [involving Frobenius automorphisms and functoriality]..."
As a final remark: it is important to note that the word "canonical" in mathematics does not have a canonical meaning. To say that two objects are canonically isomorphic requires further explanation (as e.g. in the Theorem I mentioned above). Even the "unique isomorphisms" that one gets from universal mapping properties are not unique full-stop [generally!]; they are the unique isomorphisms satisfying some particular property.
| 9 | https://mathoverflow.net/users/1149 | 13239 | 8,944 |
https://mathoverflow.net/questions/13240 | 11 | Something that seems to be pretty standard in every introductory treatment is that the infinite places correspond to embeddings into $\mathbb{C}$. Do the finite places correspond to embeddings as well? I can envision two possibilities. My first guess is that the primes sitting above $p \in \mathbb{Q}$ correspond to embeddings into $\overline{\mathbb{Q}\_p}$, and thus also to embeddings into $\mathbb{C}$ by some messy non-canonical field isomorphism. My second guess, which I think would imply the first, is that the places of $\mathbb{Q}[\alpha]$ above $p \in \mathbb{Q}$ correspond to embeddings into $\mathbb{Q}\_p[\alpha]$. I've never been able to find a precise statement about this in any of the texts I've been studying (mostly Milne's notes and Frohlich & Taylor) and would appreciate if anyone could let me know where to learn more about this -- or if I'm just plain wrong.
One other thing is that the embeddings into $\mathbb{C}$ play a central role in analyzing the basic structure of a number field by way of Minkowski theory. Is there some analog for the finite places, or does that even make any sense?
| https://mathoverflow.net/users/434 | Do finite places of a number field also correspond to embeddings? | The Archimedean places of a number field K do not quite correspond to the embeddings of K into $\mathbb{C}$: there are exactly $d = [K:\mathbb{Q}]$ of the latter, whereas there are
$r\_1 + r\_2$ Archimedean places, where:
if $K = \mathbb{Q}[t]/(P(t))$, then $r\_1$ is the number of real roots of $P$ and $r\_2$ is the number of complex-conjugate pairs of complex roots of $P$. In other words, $r\_1$ is the number of degree $1$ irreducible factors and $r\_2$ is the number of degree $2$ irreducible factors of $P(t) \in \mathbb{R}[t]$.
There is a perfect analogue of this description for the non-Archimedean places. Namely, the places of $K$ lying over the $p$-adic place on $\mathbb{Q}$ correspond to the irreducible factors of $\mathbb{Q}\_p[t]/(P(t))$; or equivalently, to the prime ideals in the finite-dimensional $\mathbb{Q}\_p$-algebra $K \otimes\_{\mathbb{Q}} \mathbb{Q}\_p$.
More generally: if $L = K[t]/P(t)/K$ is a finite degree field extension and $v$ is a place of $K$ (possibly Archimedean), then the places of $L$ extending $v$ correspond to the prime ideals in $L \otimes\_K K\_v$ or, if you like, to the distinct irreducible factors of $P(t)$ in $K\_v$, where $K\_v$ is the completion of $K$ with respect to $v$.
See e.g. Section 9.9 of Jacobson's *Basic Algebra II*.
By coincidence, this is exactly the result I'm currently working towards in a course I'm teaching at UGA. I'll post my lecture notes when they are finished. (But I predict they will bear a strong resemblance to the treatment in Jacobson's book.)
| 9 | https://mathoverflow.net/users/1149 | 13243 | 8,946 |
https://mathoverflow.net/questions/13074 | 9 | Just a d\*mb question on Lie algebras:
Given a Dynkin diagram of a root system (or a Cartan Matrix), how do I know which combination of simple roots are roots?
Eg. Consider the root system of G\_2, let a be the short root and b be the long one, it is clear that a, b, b+a, b+2a, b+3a are positive roots. But it is not clear to me that 2b+3a is a root just from the Dynkin diagram.
| https://mathoverflow.net/users/1657 | Figure out the roots from the Dynkin diagram | Here's an answer in the simply-laced case. Its proof, and generalization to non-simply-laced, are left to the reader.
1) Start with a simple root, and think of it as a labeling of the Dynkin diagram with a 1 there and 0s elsewhere.
2) Look for a vertex whose label is < 1/2 the sum of the surrounding labels. Increment that label. You've found a root!
3) Go back to (2), unless there is no such vertex anymore. You've found the highest root!
Take the union over all such games, and you get all the positive roots. Include their negatives, and you have all roots.
If you start with a non-Dynkin-diagram, the game doesn't terminate. This is part of a way to classify the Dynkin diagrams.
[EDIT: yes it can terminate. There's a variant, where you replace the vertex label by the sum of the surrounding, minus the original label. *That* game terminates only for Dynkin diagrams.]
BTW at the highest root, most of the vertices have labels = 1/2 the surrounding sum. If you put in a new vertex, connected to those vertices with > 1/2 the sum, you get the affine Dynkin diagram.
| 33 | https://mathoverflow.net/users/391 | 13259 | 8,956 |
https://mathoverflow.net/questions/13278 | 6 | I apologize that this is perhaps not adequate for mathoverflow but I have struggled with this for days now and become desperate...
The reduced K-group $\tilde{K}(S^0)$ of the zero sphere is the ring $\mathbb{Z}$ as being the kernel of the ring morphism $K(S^0)\to K(x\_0)$. The ring structure on $K(S^0)$ and $K(x\_0)$ comes from the tensor product $\otimes$ of vector bundles.
If $H$ is the canonical line bundle over $S^2$ then $(H-1)^2=0$ where the product comes from $\otimes$. The Bott periodicity theorem states that the induced map $\mathbb{Z}\left[H\right]/(H-1)^2\to K(S^2)$ is an isomorphism of rings. So $\tilde{K}(S^2)\cong \mathbb{Z}\left[H-1\right]/(H-1)^2$, I think, and every square in $\tilde{K}(S^2)$ is zero.
The reduced external product gives rise to a map $\tilde{K}(S^0)\to \tilde{K}(S^2)$ which is a ring (?) isomorphism (see e.g. Hatcher Vector Bundles and K-Theory, Theorem 2.11.) but not every square in $\tilde{K}(S^2)$ is zero then. How can this be?
Aside from that I do not understand the relation of $\otimes:K(X)\otimes K(X)\to K(X)$ and the composition of the external product with map induced from the diagonal map
$K(X)\otimes K(X)\to K(X\times X)\to K(X)$.
| https://mathoverflow.net/users/2625 | Understanding the product in topological K-theory | Reduced $K$-groups are ideals of the standard $K$-groups. $\tilde K(X) \subset K(X)$ is the ideal of virtual-dimension-zero elements.
In particular, the reduced K-theory $\tilde K(S^2)$ is not $\mathbb{Z}[H]/(H-1)^2$, but rather the ideal of this generated by $(H-1)$. In particular, any element in this group does square to zero.
Additionally, the "exterior product" isomorphism $f: \tilde K(X) \to \tilde K(X \wedge S^2)$, which is an isomorphism, is not a ring map: it takes an element $x$ to the exterior product $x \wedge (H-1)$. Instead, it satisfies $f(x) f(y) = (H-1) f(xy) = 0$. This is because the suspension is covered by two contractible open subsets, and so all products must vanish.
| 16 | https://mathoverflow.net/users/360 | 13280 | 8,965 |
https://mathoverflow.net/questions/13274 | 0 | Define a formal tautology as a statement where by the nature of its atomic components there exists no truth-value assignment where it is not true. A contingent statement is a statement that is true by facts of the world. A statement is necessarily true if the statement is true in all possible worlds. A necessarily true statement is not contingent and a contingent statement is not necessarily true. A formal tautology is necessarily true. But a necessarily true statement is not always a formal tautology.
Are there mathematical theorems which are contingent? Are all mathematical theorems necessarily true?
| https://mathoverflow.net/users/3611 | Are all mathematical theorems necessarily true? | Apart from the storm of comments, let me just try to answer the question.
There are several ways in a which a mathematical theorem
can be contingent.
* First, the independence phenomenon in set theory shows the striking ubiquity of
contingency in mathematics. For example, the Continuum Hypothesis is true
is some set-theoretic
worlds and false in others (and we can control it
exactly). There are hundreds of other examples of
statements with the same independence status---they are
true in some worlds and false in others. The method of
forcing has been used to spectacular effect in proving
many of these independence results.
* The Incompleteness phenomenon of Goedel can be used to show that
whether a statement is provable or not from a given
axiom system (in classical logic) can be contingent.
Specifically, the Incompleteness theorem says that no
theory T, if consistent, can prove its own consistency.
Thus, if ZFC is consistent, then there are models of ZFC
in which ZFC is thought to be inconsistent. In such a
model, ZFC is thought to prove any statement at all! But
in our world, not all these statements will be theorems.
Thus, in this sense, even the question of whether a given statement is a theorem or not
can be contingent.
* The large cardinal hierarchy in set theory provides
numerous examples of statements transcending the
consistency strength of weaker statements. If large cardinals are consistent, then there are some set-theoretical universes in which ZFC proves that there are no inaccessible cardinals and other universes where it does not.
There are also several ways in which contingency is ruled
out.
* First, one of the most important properties of a proof
system is *soundness*,
which means that any statement provable in the system from
true hypotheses will remain true. Of course, this is an
expected feature of any proof system worthy of the name. A
*theorem* is a statement having a proof in such a system. Once we have adopted a
given proof system that is sound, and the axioms are all
necessarily true, then the theorems will also all be necessarily
true. In this sense, there can be no contingent theorems.
* Second, one of the profound achievements of Goedel was
his *Completeness theorem*, which states that any
statement that holds in all models of a given first order
theory T, actually has a proof from the theory. For
example, every statement in the language of group theory
that happens to be true in all groups, actually has a
finite proof from the group axioms (using any of several
proof systems). This is far from obvious, and I find it profound.
But it answers a dual version of a question you might
have asked, which I find interesting, namely:
Is every necessary truth a theorem?
The answer is Yes, and this is just what the Completeness
theorem expresses.
These last two points together explain that if one takes the possible worlds to be all models of a given theory, then the necessary truths are precisely the theorems of that theory.
| 18 | https://mathoverflow.net/users/1946 | 13288 | 8,970 |
https://mathoverflow.net/questions/13257 | 40 | A morphism $f: V \rightarrow X$ of schemes is a locally closed immersion if it can be factored into a closed immersion followed by an open immersion. It is not hard to show that if $f$ is an open immersion followed by a closed immersion, then it is a locally closed immersion, but the converse is at the very least not clear (to me). For a number of reasons, this choice as the definition of locally closed immersion (rather than the opposite) is the right one (e.g. it is then not hard to see that compositions of locally closed immersions are locally closed immersions).
>
> Is there some $f: V \rightarrow X$ that can be factored into a closed immersion followed by an open immersion, that cannot be factored into an open immersion followed by a closed immersion?
>
>
>
*Warning:* it isn't too hard to show that there is no example with $V$ reduced or with $f$ quasicompact, so any example has to be a little strange-looking. *Back-story:* I've been confronted with this question when learning algebraic geometry with a class (conventionally known as "teaching"); it seems a natural question. And any counterexample would likely be a handy example to have for other reasons as well: a very limited stock of counterexamples tends to refine my intuition, and to warn me what can go wrong.
| https://mathoverflow.net/users/299 | A closed subscheme of an open subscheme that is not an open subscheme of a closed subscheme? | Hi Ravi,
There is an example in [Tag 01QW](http://math.columbia.edu/algebraic_geometry/stacks-git/locate.php?tag=01QW) in Johan's stacks project.
Jarod
| 23 | https://mathoverflow.net/users/42 | 13295 | 8,974 |
https://mathoverflow.net/questions/13293 | 0 | For a real algebraic variety, is the integral of the product of the Chern classes of two line bundles equal to the intersection number of the two corresponding divisors?
| https://mathoverflow.net/users/1648 | Divisor Intersections and Chern Class Products | I'm not sure what you mean by a line bundle on a real algebraic variety and its Chern classes, but for smooth complex analytic manifolds the Chern class of a line bundle corresponding to a divisor is Poincare dual to the homological class of the divisor, as explained e.g. in Griffiths-Harris, Chapter 1, Chern classes of line bundles. So the $c\_2([D\_1]\oplus [D\_2])=c\_1[D\_1]c\_1[D\_2]$ is indeed the indeed Poincare dual to the intersection class of $D\_1$ and $D\_2$. If the manifold is a surface, then we get the a number.
Maybe this not what you meant, but in that case you should really add some more details to your question.
| 3 | https://mathoverflow.net/users/2349 | 13300 | 8,978 |
https://mathoverflow.net/questions/13303 | 7 | The Riemann-Roch theorem is a result about Riemann Surfaces that was extended to the Hirzebruch–Riemann–Roch theorem, a result about compact complex manifolds. The Hodge Index theorem is a result about Riemann surfaces (I'm just worried about the complex case) that is proved using Riemann-Roch. Has the Hirzebruch–Riemann–Roch theorem been used to extend the Hodge Index theorem to a result about compact complex manifolds.
| https://mathoverflow.net/users/1977 | Hodge Index theorem for Complex Manifolds | The Hodge index theorem IS a result on compact Kahler manifolds of complex dimension 2n.
It states that the signature of the intersection form on $H^{2n}(X, \mathbb{R})$ equals $\sum (-1)^a h^{a, b}(X)$, where $h^{a, b}$ are the Hodge numbers.
See Voisin, Hodge theory and complex algebraic geometry I, theorem 6.33
By the way, the result you probably have in mind is on complex surfaces, which are NOT Riemann surfaces (these are complex curves).
| 15 | https://mathoverflow.net/users/828 | 13312 | 8,985 |
https://mathoverflow.net/questions/13305 | 8 | I don't know about schemes and every definition of a Hilbert scheme (quite naturally!) involves schemes. But, the Hilbert scheme of points on a complex surface is known to be smooth (Fogarty). So is there a concrete description of it as a complex manifold? (For instance in the case of n=2 it is a blowup of XxX along the diagonal)
| https://mathoverflow.net/users/3709 | Hilbert scheme of points on a complex surface | Given a codimension $d$ ideal $I$ in $R = {\mathbb C}[x,y]$, the quotient ring $R/I$ can be thought of as a $d$-dimensional vector space with actions of two commuting operators $x,y$ and a "cyclic" vector $1$ that generates it as an $R$-module.
Consequently, if your surface is the plane you can think of the Hilbert scheme as the space of pairs of commuting matrices on $d$-space, but then take the (open) set in there of pairs $(x,y)$ that admit a cyclic vector, and then divide that variety by the conjugation action of $GL(d)$. To see smoothness, you might first show this open set is smooth, then that the $GL(d)$ action is free and proper.
All this is spelled out in Nakajima's book (which I'm guessing is the one Andrea Ferretti meant to reference), except I think he shows smoothness by analyzing the tangent spaces.
| 6 | https://mathoverflow.net/users/391 | 13331 | 8,999 |
https://mathoverflow.net/questions/13176 | 6 | When I first learned about the etale fundamental group, there was a mythical theorem going around that in the algebraic case all we need to look at is the finite covers, because the infinite degree algebraic covers are inverse limits of the finite ones (obviously unlike the topological case). But I've never seen a convincing source for this theorem.
It seems reasonable that the statement would be: "every flat unramified map of a connected scheme onto a quasi-projective curve is an inverse limit of finite etale covers". Is this true? Do you have a reference for this?
| https://mathoverflow.net/users/2665 | Is every flat unramified cover of quasi-projective curves profinite? | Any modification of the theorem where the definition of "cover" you give is local on the base and contains inverse limits of finite etale covers (e.g. flat plus unramified as in the original question) will also be false because the property of being an inverse limit of finite etale covers is not local on the base.
To see this, proceed similarly to Scott Carnahan's example, but instead of gluing a chain of $\mathbb{P}^1$'s together, glue together $\mathbb{P}^1$'s "indexed by $\operatorname{Spec} \mathbb{C}[...,x\_{-1},x\_0,x\_1,...]/\langle x\_i^2-1\rangle$." Explicitly, let the base curve $B$ be two $\mathbb{P}^1$'s glued together at two distinct points. Over each $\mathbb{P}^1$ consider the affine morphism corresponding to the sheaf of algebras $\mathcal{O}\\_{\mathbb{P}^1}[...,x\_{-1},x\_0,x\_1,...]/\langle x\_i^2-1\rangle$. At one of the points, glue together the two possible $x\_i$'s. At the other, glue $x\_i$ to $x\_{i-1}$. Over each $\mathbb{P}^1$, the resulting morphism is an inverse limit of finite covers, but over all of $B$, it is not. This is written down fully in Warning 2.5b of <http://math.harvard.edu/~kwickelg/papers/VW.pdf> -- Kirsten Wickelgren
| 9 | https://mathoverflow.net/users/3619 | 13332 | 9,000 |
https://mathoverflow.net/questions/13318 | 6 | ... aka locally linear compact vector spaces. The one reference I know is <http://www.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/Nov3-10(CentExt).pdf>. Does anyone know another good reference?
| https://mathoverflow.net/users/788 | Reference for Tate vector spaces | 1. Beilinson, Drinfeld. *Chiral Algebras* section 2.7 (I think)
2. Beilinson, Feigin, Mazur. *Notes on Conformal Field Theory (Incomplete)* available on Mazur's web page.
Also: Tate, *Residues of differentials on curves*
| 6 | https://mathoverflow.net/users/121 | 13345 | 9,010 |
https://mathoverflow.net/questions/13350 | 2 | Suppose A is a matrix with coefficient in $Q\_{\ell}$, and all the coefficients of its char. polynomial are in $Z$ (thus an integral polynomial). Prove that the char. polynomial of $A^n$ is also integral. (This question probably has nothing to do with the base field $Q\_{\ell}$)
This question actually comes from a remark in Serre's book "abelian l-adic representations", so allow me to tag it with "number theory"...
| https://mathoverflow.net/users/1238 | an exercise on integrality of characteristic polynomials | The coefficients of the characteristic polynomial of $A^n$ are symmetric functions of the roots of the characteristic polynomial of $A$, so the result follows from the Fundamental Theorem of Symmetric Functions.
| 5 | https://mathoverflow.net/users/1409 | 13352 | 9,014 |
https://mathoverflow.net/questions/13356 | 16 | Is it true that for any $n$, there exists a $n \times n$ real orthogonal matrix with all coefficients bounded (in absolute value) by $C/\sqrt{n}$, $C$ being an absolute constant ?
Some remarks :
* If we want $C=1$, the matrix must be a Hadamard matrix.
* The complex analogue has an easy answer: the Fourier matrix $(\exp(2\pi \imath jk/n)/\sqrt{n})\_{(j,k)}$. Forgetting the complex structure gives a positive answer to the question in the real case when $n$ is even.
* A random matrix doesn't work (the largest entry is typically of order $\sqrt{\log(n)}/\sqrt{n}$).
| https://mathoverflow.net/users/908 | Orthogonal matrices with small entries | Here's an idea which I think might be expandable to a solution once some details are filled in. (I am rather tired at the moment, though, so apologies if there is a cretinous error in what follows.)
We'll do the case $n=4m-1$ where $m$ is an integer; the case $n=4m-3$ is similar.
Let $C$ be a $2m\times 2m$ matrix which has the required form. Let $A$ be the $n\times n$ matrix with $C$ in the top left corner, $1$ on the remaning $2m-1$ diagonal entries, and zero elsewhere. Let $B$ be the $n\times n$ matrix with $C$ in the bottom right corner, $1$ on the remaining $2m-1$ diagonal entries, and zero elsewhere.
$A=\left[\begin{matrix} C & 0 \\\\ 0 & I\_{2m-1} \end{matrix} \right]\quad,\quad B= \left[\begin{matrix} I\_{2m-1} & 0 \\\\ 0 & C \end{matrix} \right]$
Both $A$ and $B$ will be real orthogonal since $C$ is.
Consider the matrix $AB$, which being the product of real orthogonal matrices will also be orthogonal. I claim that the entries will all be $O(\sqrt{n})$ as required.
In more detail:
-- If both $i$ and $j$ are $\leq 2m-1$, then $(AB)\\_{ij}=A\\_{ij}=C\\_{ij}$ which is small by our choice of $C$; by symmetry, we can dispose of the case where both $i$ and $j$ are $\geq 2m+1$ in a similar way.
-- If $i\leq 2m-1$ and $j\geq 2m+1$, then on considering $\sum\\_r A\\_{ir}B\\_{rj}$ we see that the only nonzero contribution comes when $r\leq 2m$ and $r\geq 2m$, i.e. when $r=2m$ and so $(AB)\\_{ij}=A\\_{i,2m}B\\_{2m,j}$ is small.
-- If $i=2m$ or $j=2m$ then a similar analysis shows that $(AB)\\_{ij}$ can't be bigger than the entries of $C$ (at least up to some constant independent of $m$).
-- If $i\geq 2m+1$ and $j\leq 2m-1$ then $(AB)\\_{ij}=0$.
That should handle the case $n=4m-1$. The case $n=4m-3$ can be done in a similar fashion, but this time we will have extra factors of $3$ floating around since we have $3\times n$ and $n\times 3$ regions to consider, rather than just $1\times n$ and $n\times 1$ regions.
| 7 | https://mathoverflow.net/users/763 | 13365 | 9,019 |
https://mathoverflow.net/questions/13269 | 1 | I want to learn more about numerical algorithms that use mixed-precision computational models (where instead of everything being 32/64 bit floating points, we can do lower precision calculations at lower costs).
Does anyone know of good articles/books on this? All I can find are various haphazard fpga-implementation articles on google scholar.
Thanks!
| https://mathoverflow.net/users/3609 | Numerical algorithms on mixed-precision computational models. | Check also *arithmetic filters*, eg in [http://dx.doi.org/10.1016/S0166-218X(00)00231-6](http://dx.doi.org/10.1016/S0166-218X%2800%2900231-6).
| 2 | https://mathoverflow.net/users/532 | 13368 | 9,022 |
https://mathoverflow.net/questions/12486 | 32 | EDIT, 9 March 2014: when I asked this in 2010, I did not have the courage of my convictions, and so did not ask for an if and only if proof, as Kevin Buzzard quite properly pointed out. Such problems are now somewhat known open problems, as I told them to some experts in 2011. Probably the easiest of the bunch: It is easy to describe a set of integers that are not represented by $4 x^2 + 2 x y + 7 y^2 - z^3;$ I even sent that in as a Monthly problem (December 2010, problem 11539), only one solver, Robin Chapman(December 2012). **Open problem**: can we prove that the polynomial integrally represents every other number? There is a similar open problem for each discriminant of positive binary quadratic forms with class number three, including the other direction for Kevin Buzzard's answer below.
ORIGINAL: The following problem is my variant of something Irving Kaplansky noticed when we worked together. I do not think it is by nature a difficult problem, it is simply too hard for me to finish.
Suppose we have an integer $C > 0$ that is not divisible by 2 or 3, while there is another integer $F > 0$ such that
$ 27 C^2 - 23 F^2 = 4.$
For any integers $x,y,z,$
is it true that $ 2 x^2 + x y + 3 y^2 + z^3 - z \neq C $ and
$ 2 x^2 + x y + 3 y^2 + z^3 - z \neq -C ,$ or together
$ 2 x^2 + x y + 3 y^2 + z^3 - z \neq \pm C $ ?
I have proved it for the four smallest values of $C,$ those being
1, 599, 14951, 9314449. The case $C=1$ comes directly from the
Hudson and Williams paper below. Note that the even values of $C$ fail miserably, they seem to all be values of $ z^3 - z.$ The polynomial $ g(x,y,z) = 2 x^2 + x y + 3 y^2 + z^3 - z $
represents every other number $n$ with
$ -10,000,000 \leq n \leq 10,000,000,$ according to my computer.
The Spearman and Williams article (see below) is explicit class field theory, not a topic I know.
I should point out that, in retrospect, what I proved for the four smallest (odd) $C$ seems to amount to the statement that $z^3 - z + C$ is irreducible $\pmod q$ for any prime $ q = 2 u^2 + u v + 3 v^2.$
(27 January 2010) I finally got smart and decided to do a part check of the "irreducible" version. For the next three values of $C,$ those being 232488049, 144839681351, 3615189146999, I factored $z^3 - z + C \pmod q$ and found it to be irreducible for primes $q < 1000$ and $ q = 2 u^2 + u v + 3 v^2.$
The two main references are:
Blair K. Spearman and Kenneth S. Williams,
"The Cubic Conguence $x^3 + {A} x^2 + {B} x + {C} \equiv 0 \bmod p $
and Binary Quadratic Forms",
Journal of the London Mathematical Society,
volume 46,
1992,
pages 397-410
Richard H. Hudson and Kenneth S. Williams",
"Representation of primes by the principal form of
discriminant $-{D}$ when the classnumber $h(-{D})$ is $3$",
Acta Arithmetica,
volume 57,
1991,
pages 131-153.
One needs this Lemma: if an integer $n$ has an integer representation as
$ n = 2 x^2 + x y + 3 y^2,$ then $n$ is divisible by some prime
$ q = 2 u^2 + u v + 3 v^2.$
Everything I know about this problem is in pdf's at (Feb. 2018):
<http://zakuski.utsa.edu/~jagy/inhom.cgi>
including a proof of the preceding Lemma in jagy\_division.pdf
and the proof for the four smallest $C$ in jagy\_conjecture\_23.pdf
and a list of intimately related problems in jagy\_list.pdf
.
I welcome individual responses to this along with posted answers or comments. One of my email addresses can be found using the search feature at
<http://www.ams.org/cml>
| https://mathoverflow.net/users/3324 | Integers not represented by $ 2 x^2 + x y + 3 y^2 + z^3 - z $ | EDIT: Hendrik Lenstra emailed me a proof of Conjecture 2. I'll append it below. So Jagy's question is now solved.
---
OK so I think that Jagy wants to make the following conjecture:
CONJECTURE 1: an integer $C$ is not representable by the form F(x,y,z)=2x^2+xy+3y^2+z^3-z if, and only if, $C$ is odd and $27C^2-4=23D^2$ with $D$ an integer.
[EDIT/clarification: Jagy only asks one direction of the iff in his question, and this answer below gives a complete answer to the question Jagy asks. I came back to this question recently though [I am writing this para a year after I wrote the original answer] and tried to fill in the details of the argument in the other direction (proving that if C was not an odd integer solution to $27C^2-4=23D^2$ then $C$ was represented by the form) and I failed. So the "hole" I flag in the answer below still really is a hole, and this post still remains an answer to Jagy's question, but not a complete proof of Conjecture 1, which should still be regarded as open.]
I have a proof strategy for this. I am too lazy to fill in some of the details though, so maybe a bit of it doesn't work, but it should be OK. However, I am also reliant on a much easier-looking conjecture (which I've tested numerically so should be fine, but I can't see why it's true):
CONJECTURE 2: if $C$ is odd and $27C^2-4=23D^2$, then there's no prime p
dividing D of the form $2x^2+xy+3y^2$.
So I am claiming Conj 2 implies the "only if" version of Conj 1. I don't know how to prove Conj 2
but it looks very accessible [edit: I do now; see below]. Note that the Pell equation is related to units
in $\mathbf{Q}(\sqrt{69})$ and the $2x^2+xy+3y^2$ is related to factorization
in $\mathbf{Q}(\sqrt{-23})$. I've seen other results relating the arithmetic
of $\mathbf{Q}(\sqrt{D})$ and $\mathbf{Q}(\sqrt{-3D})$.
---
Ok, so assuming Conjecture 2, let me sketch a proof of the "only if" part of Conjecture 1.
The Pell equation is intimately related to the recurrence relation
$$t\_{n+2}=25t\_{n+1}-t\_n$$
with various initial conditions. For example the positive $C$s which
are solutions to $27C^2-4=23D^2$ are all generated by this recurrence
starting at $C\_1=C\_2=1$, and the $D$s are all generated by the same
recurrence with $D\_1=-1$ and $D\_2=1$. Note that $C\_n$ is even iff $n$
is a multiple of 3, and (by solving the recurrence explicitly) one
checks easily that $C\_{3n}=(3C\_{n+1})^3-(3C\_{n+1})$, so we've represented
the even solutions to the Pell equation as values of $F$ (with $x=y=0$).
Let's then consider the odd solutions to the Pell equation. Say $C$
is one of these. We want to prove that there is no solution in
integers $x,y,z$ to
$$2x^2+xy+3y^2=z^3-z+C.$$
Let's do it by contradiction. Consider the polynomial $Z^3-Z+C$. First
I claim it's irreducible. This is because it is monic, of degree 3,
and has no integer root, because $C$ is odd. Next I claim that
the splitting field contains $\mathbf{Q}(\sqrt{-23})$. This is
because of our Pell assumption and the fact that the discriminant
of $Z^3-Z+C$ is $4-27C^2$. Next I claim that the splitting
field of $Z^3-Z+C$ is in fact the Hilbert class field of
$\mathbf{Q}(\sqrt{-23})$. I only know an ugly way of seeing this:
if $\theta$ is a root of $Z^3-Z+1=0$ then I know recurrence relations
$e\_n$, $f\_n$ and $g\_n$ (all defined using the relation above but with
different initial conditions) with $e\_n\theta^2+f\_n\theta+g\_n$ a root of
$Z^3-Z+C\_{3n+1}$, and other relations giving roots of $Z^3-Z+C\_{3n+2}$
and $Z^3-Z-C\_{3n+1}$ and $Z^3-Z-C\_{3n+2}$. Most unenlightening but it
does the job because it embeds $\mathbf{Q}(\theta)$ into the splitting
field, and the Galois closure of $\mathbf{Q}(\theta)$ is the Hilbert
class field of $\mathbf{Q}(\sqrt{-23})$.
Right, now for the contradiction, assuming Conjecture 2. Let's assume
that $C$ is a solution to the Pell, and $z^3-z+C$ can be written $2x^2+xy+3y^2$.
Now $C$ is odd so $z^3-z+C$ isn't zero, and hence it's positive,
so it's the norm of a non-principal ideal~$I$ in the integers $R$ of
$\mathbf{Q}(\sqrt{-23})$. This ideal $I$ is a product of prime ideals,
and $I$ isn't principal, so one of the prime ideals had better also not
be principal. Say this prime ideal has norm $p$. We conclude that $p$
divides $z^3-z+C$ and $p$ is of the form $2x^2+xy+3y^2$. Note in
particular that this implies $p\not=23$. Also $p\not=3$, because $C$
is odd and (because of general Pell stuff) hence prime to 3.
CASE 1: $p$ is coprime to $D^2$ (with $27C^2-4=23D^2$). In this
case the polynomial $Z^3-Z+C$ has non-zero discriminant mod $p$
(because $p\not=23$) and furthermore has a root $Z=z$ mod $p$.
Hence mod $p$ the polynomial either splits as the product of a linear
and a quadratic, or the product of three linears. This tells us
something about the factorization of $p$ in the splitting field
of $Z^3-Z+C$: either $p$ remains inert in $\mathbf{Q}(\sqrt{-23})$,
or it splits into 6 primes in the splitting field and hence splits
into two principal primes in $\mathbf{Q}(\sqrt{-23})$ (because the
principal primes are the ones that split completely in the Hilbert
class field). In either case $p$ can't be of the form $2x^2+xy+3y^2$,
so this case is done.
CASE 2: This is simply Conjecture 2.
In both cases we have our contradiction, and
so we have proved, so far, assuming Conjecture 2, that a solution $C$ to $27C^2-4=23D^2$
is representable as $2x^2+xy+3y^2+z^3-z$ iff it's even.
Note that Conjecture 2 can be verified by computer for explicit values
of $C$, giving unconditional results---for example I checked in just
a few seconds that any odd $C$ with $|C|<10^{72}$ and satisfying the
Pell equation was not representable by the form, and that result
does not rely on anything. At least that's something concrete for Jagy.
---
OK so what about the other way: say $27C^2-4$ is not 23 times a square.
How to go about representing $C$ by our form? Well, here I am going to
be much vaguer because there are issues I am simply too tired to deal
with (and note that this is not the question that Jagy asked anyway).
Here's the idea. Look at the proof of Theorem 2 in Jagy's pdf Mordell.pdf.
Here Mordell gives a general algorithm to represent certain integers
by (quadratic in two variables) + (cubic in one variable). If you
apply it not to the form we're interested in, but to the following
equation:
$$x^2+xy+6y^2=z^3-z+C$$
then, I didn't check all the details, but I convinced myself that they
could easily be checked if I had another hour or two, but I think that
the techniques show that whatever the value of $C$ is, this equation
has a solution. The idea is to fix $C$, let $\theta$ be a root of
the cubic on the right (which we can assume is irreducible, as if it
were reducible then we get a solution with $x=y=0$), to rewrite the right
hand side as $N\_{F/\mathbf{Q}}(z-\theta)$, with $F=\mathbf{Q}(\theta)$
and now to try and write $z-\theta$ as $G^2+GH+2H^2$ with
$G,H\in\mathbf{Z}[\theta]$. Mordell does this explicitly (in a slightly
different case) in the pdf. The arguments come out the same though,
and we end up having to check that a certain cubic in four variables
has a solution modulo~23 with a certain property. I'll skip the painful
details. The cubic depends on $C$ mod 23, and so a computer calculation
can deal with all 23 cases.
Once this is done properly we have a solution to $x^2+xy+6y^2=z^3-z+C$,
so we have written $z^3-z+C$ as the norm of a principal ideal in
the integers of $\mathbf{Q}(\sqrt{-23})$. What we need to do now is
to write it as the norm of a non-principal ideal, and of course we'll
be able to do this if we can find some prime $p$ dividing $z^3-z+C$
which splits in $\mathbf{Q}(\sqrt{-23})$ into two non-principal
primes, because then we replace one of the prime divisors above $p$
in our ideal by the other one. What we need then is to show that
if the discriminant of $z^3-z+C$ is not $-23$ times a square,
then there *is* some prime $p$ of the form $2x^2+xy+3y^2$ dividing
some number of the form $z^3-z+C$ which is the norm of a principal
ideal. This should follow from the Cebotarev density theorem, because
Mordell's methods construct a huge number of solutions to $x^2+xy+6y^2=z^3-z+C$
which are "only constrained modulo 23", and so one should presumably
be able to find a prime which splits in $\mathbf{Q}(\sqrt{-23})$,
splits completely in the splitting field of $z^3-z+C$ and doesn't
split completely in the splitting field of $z^3-z+1$. I have run out
of energy to deal with this point however, so again there is a hole here.
This issue seems analytic to me, and I am not much of an analytic guy.
[edit: I came back to this question a year later and couldn't do it,
so this should not be regarded as a proof of the "if" part of Conj 1]
---
EDIT: OK so here, verbatim, is an email from Lenstra in which he establishes
Conjecture 2.
(EDIT: dollar signs added - GM)
Fact. Let $\theta$ be a zero of $X^3-X+1$, let $\eta$ in ${\bf Z}[\theta]$ be
a zero of $X^3-X+C$ with $C$ in $\bf Z$ odd, and let $p$ be a prime
number that is inert in ${\bf Z}[\theta]$. Then $p$ does not divide
index$({\bf Z}[\theta]:{\bf Z}[\eta])$.
Proof. By hypothesis, ${\bf Z}[\theta]/p{\bf Z}[\theta]$ is a field of size $p^3$.
Let $e$ be the image of $\eta$ in that field. Since $X^3-X+C$ is
irreducible in ${\bf Z}[X]$ (even mod 2), it is the characteristic
polynomial of $\eta$ over $\bf Z$. Hence its reduction mod $p$ is the
characteristic polynomial of $e$ over ${\bf Z}/p{\bf Z}$. If now $e$ is in
${\bf Z}/p{\bf Z}$, then that characteristic polynomial also equals $(X-e)^3$,
so that in ${\bf Z}/p{\bf Z}$ we have $3e = 0$ and $3e^2 = -1$, a contradiction.
Hence $e$ is not in ${\bf Z}/p{\bf Z}$, so $({\bf Z}/p{\bf Z})[e] = {\bf Z}[\theta]/p{\bf Z}[\theta]$,
which is the same as saying ${\bf Z}[\theta] = {\bf Z}[\eta] + p{\bf Z}[\theta]$. Then
$p$ acts surjectively on the finite abelian group ${\bf Z}[\theta]/{\bf Z}[\eta]$,
so the order of that group is not divisible by $p$. End of proof.
| 34 | https://mathoverflow.net/users/1384 | 13369 | 9,023 |
https://mathoverflow.net/questions/13322 | 123 | A very important theorem in linear algebra that is rarely taught is:
>
> A vector space has the same dimension as its dual if and only if it is finite dimensional.
>
>
>
I have seen a total of one proof of this claim, in Jacobson's "Lectures in Abstract Algebra II: Linear Algebra". The proof is fairly difficult and requires some really messy arguments about cardinality using, if I remember correctly, infinite sequences to represent $\mathbb{N}\times\mathbb{N}$ matrices. Has anyone come up with a better argument in the 57 years since Jacobson's book was published, or is the noted proof still the only way to prove this fact?
Edit: For reference, the proof is on pages 244-248 of Jacobson's
*Lectures in Abstract Algebra: II. Linear Algebra*.
| https://mathoverflow.net/users/1353 | Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional | Here is a simple proof I thought, tell me if anything is wrong.
First claim. Let $k$ be a field, $V$ a vector space of dimension at least the cardinality of $k$ and infinite. Then $\operatorname{dim}V^{\*} >\operatorname{dim}V$.
Indeed let $E$ be a basis for $V$. Elements of V\* correspond bijectively to functions from $E$ to $k$, while elements of $V$ correspond to such functions with finite support. So the cardinality of $V^{\*}$ is $k^E$, while that of $V$ is, if I'm not wrong, equal to that of $E$ (in this first step I am assuming $\operatorname{card} k \le \operatorname{card} E$).
Indeed $V$ is a union parametrized by $\mathbb{N}$ of sets of cardinality equal to $E$. In particular $\operatorname{card} V < \operatorname{card} V^{\*}$, so the same inequality holds for the dimensions.
Second claim. Let $h \subset k$ two fields. If the thesis holds for vector spaces on $h$, then it holds for vector spaces on $k$.
Indeed let $V$ be a vector space over $k$, $E$ a basis. Functions with finite support from $E$ to $h$ form a vector space $W$ over $h$ such that $V$ is isomorphic to the extension of $W$, i.e. to $W\otimes\_h k$. Every functional from $W$ to $h$ extends to a functional from $V$ to $k$, hence
$$\operatorname{dim}\_k V = \operatorname{dim}\_h W < \operatorname{dim}\_h W^\* \leq \operatorname{dim}\_k V^\*.$$
Putting the two claims together and using the fact that every field contains a field at most denumerable yields the thesis.
| 139 | https://mathoverflow.net/users/828 | 13372 | 9,025 |
https://mathoverflow.net/questions/13371 | 8 | A scheme is *separated* if the diagonal inclusion $X \to X \times X$ is a closed immersion. I what to know if there is a good generalization of `separated' for algebraic stacks?
My usual stack reference, Anton Gerashchenko's [stack notes](http://stacky.net/files/written/Stacks/Stacks.pdf), doesn't seem to provide an answer.
In a [previous MO question](https://mathoverflow.net/questions/9043/why-is-this-not-an-algebraic-space) several related notions came up. The most similar is quasi-separated where you require the diagonal to be quasi-compact. You can check wikipedia for some relevant [algebraic geometry terminology](http://en.wikipedia.org/wiki/Glossary_of_scheme_theory). How does this compare to separatedness?
The main obstacle that I can see in defining separated for stacks is that the property of a map of schemes $X \to Y$ being separated does not appear to be local in the target. Since maps between affines are separated, it seems that every map of schemes is *locally* separated. This means that we shouldn't expect the usual trick of replacing an algebraic stack by a scheme which covers it to work very well.
| https://mathoverflow.net/users/184 | Is there a good notion of `Separated Stack'? | Look at Def. 4.7 of Deligne--Mumford for the definition when $X$ is DM: they define $X$
to be separated if $X \to X \times X$ is proper (or equivalently, finite).
| 9 | https://mathoverflow.net/users/2874 | 13373 | 9,026 |
https://mathoverflow.net/questions/13346 | 11 | At the end of my 8410 class today (see [http://alpha.math.uga.edu/~pete/MATH8410.html](http://alpha.math.uga.edu/%7Epete/MATH8410.html) if you care), one of my students asked me the following very interesting question:
Let $(K,|\ |)$ be a [normed field](http://eom.springer.de/n/n067360.htm), with completion $(\hat{K},| \ |)$. Suppose $\hat{K}$ is algebraic over $K$. Must we then have $\hat{K} = K$?
As I have mentioned here before, I feel very lucky to be getting such penetrating questions. This one I was not able to answer on the spot, although I remarked that it is true in all of the most familiar examples and that the (possible) lack of algebraicity of the completion is a key motivation for considering the **Henselization** instead.
**Edit**: the answer is **no**, as I have just heard from one of my students. I have encouraged him to come to this site and register the answer.
To make the question more interesting, suppose we ask whether $\hat{K}/K$ can be finite and nontrivial?
| https://mathoverflow.net/users/1149 | Algebraicity of the completion of a field? Finiteness? | (I'll delete this if your student came up with the same answer.)
Choose a ring-theoretic automorphism of the complex numbers that doesn't fix the reals (I'm pretty sure any nontrivial automorphism other than complex conjugation will work), and consider the image of the reals in it. A similar trick should work for any real closed field with transcendence degree at least 1 over Q. I'm not sure what I was thinking with the last sentence, but it's clearly false.
However, a similar trick should work for any finite Galois extension of complete normed fields such that the overfield has a discontinuous automorphism. For example, if we hit $\mathbb{C}((t))$ with some discontinuous non-$\mathbb{C}$-linear automorphism, I think the subfield $\mathbb{C}((t^3))$ is sent to a dense subfield.
| 10 | https://mathoverflow.net/users/121 | 13378 | 9,030 |
https://mathoverflow.net/questions/13270 | 5 | Hello,
I've been working deriving the orthogonality relation for quadratic Dirichlet characters $\chi\_d(n)$ (or real primitive characters). The statement I'm trying to prove is
$$\lim\_{X \rightarrow \infty} \frac{1}{D} \sum\_{0 < |d| \leq X} \chi\_d(n)= \begin{cases}
\prod\_{p|n} \left(1 + \frac{1}{p}\right)^{-1} \quad &\text{if $m$ is a perfect square,} \newline
0 \quad &\text{otherwise,}
\end{cases}$$
where the sum ranges over fundamental discriminants $d$ and $D$ is the number of terms appearing in the sum.
You could recast this as
$$\lim\_{X \rightarrow \infty} \frac{1}{D} \sum\_{0 < |d| \leq X} \chi\_d(n)\chi\_d(m)= \begin{cases}
\prod\_{p|nm} \left(1 + \frac{1}{p}\right)^{-1} \quad &\text{if $mn$ is a perfect square,} \newline
0 \quad &\text{otherwise,}
\end{cases}$$
which has the standard form of an orthogonality relation for characters. In the first case, $\chi\_d(n) = 1$ unless $\gcd(d,n) > 1$, so essentially i'm trying to count fundamental discriminants. But this still seems a bit tricky to me.
My approach is to try and count fundamental discriminants by using the generating function for squarefree numbers, i.e. $$\frac{\zeta(s)}{\zeta(2s)} = \sum\_{n=1}^\infty \frac{|\mu(n)|}{n^s}.$$ I know that Jutila proves this in his paper On the Mean Values of $L(1/2,\chi\_d)$ for Real Characters, but I would like to prove this lemma equation using analytic methods. Can anyone help me.
| https://mathoverflow.net/users/3610 | orthogonality relation for quadratic Dirichlet characters | So I think I solved half of the problem. Suppose that $n$ is a perfect square. Then $\chi\_d(n) = 1$ unless $\gcd(d,n) > 1$, in which case its $0$. So, for $\gcd{d,n} = 1$, we are simply pulling out the subset of fundamental discriminants having no common divisor with $n$. To quantify the size of this subset, we must first count fundamental discriminants.
The set of fundamental discriminants consists of all square-free integers congruent to $1$ modulo $4$ (i.e. odd fundamental discriminants) and all such numbers multiplied by $-4$ and $\pm 8$ (i.e. even fundamental discriminants). The odd fundamental discriminants may be counted by considering the series $$\sum\_{\text{$d$ odd}} \frac{1}{|d|^s}.$$ In fact, this is a Dirichlet series. For observe that $$\sum\_{\text{$d$ odd}} \frac{1}{|d|^s} = 1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \cdots = \prod\_{p>2} \left(1 + \frac{1}{p^s}\right) = \frac{\zeta(s)}{\zeta(2s)} \left(1 + \frac{1}{2^s}\right)^{-1},$$
where $$\frac{\zeta(s)}{\zeta(2s)} = \sum\_{n=1}^\infty \frac{|\mu(n)|}{n^s},$$ is the Dirichlet series which generates the square-free numbers. So, by following the definition of fundamental discriminants given above, we can count fundamental discriminants by using the Dirichlet series $$\left(1 + \frac{1}{4^s} + \frac{2}{8^s}\right) \frac{\zeta(s)}{\zeta(2s)} \left(1 +\frac{1}{2^s}\right)^{-1} = \left(1 + \frac{1}{4^s} + \frac{2}{8^s}\right) \underbrace{\prod\_{p>2} \left(1 + \frac{1}{p^s}\right)}\_{l\_p(s)}.$$
Now, to omit those discriminants with $\gcd(d,m) > 1$, we just omit the corresponding factors in $l\_p(s)$. What's missing is $$\prod\_{\substack{p>2 \\ p|m}} \left(1 + \frac{1}{p^s}\right),$$ so the relative density (compared to all fundamental discriminants $d$) is can be quantified by the expression $$\frac{1}{D} \cdot \prod\_{\substack{p>2 \\ p\nmid m}} \left(1 + \frac{1}{p^s}\right) = \prod\_{\substack{p>2 \\ p|m}} \left(1 + \frac{1}{p^s}\right)^{-1}.$$ As in the proof of the prime number theorem, the main contribution here comes from the simple pole at $s=1$ (of $\zeta(s)$). In fact, $p=2$ also fits at $s=1$ since $1 + \frac{1}{4} + \frac{2}{8} = 1 + \frac{1}{2}$. Thus, in the end we obtain $$\lim\_{X\rightarrow \infty} \frac{1}{D} \sum\_{0 < |d| \leq X} \chi\_d(m) = \prod\_{p|m} \left(1 + \frac{1}{p}\right)^{-1},$$ as desired.
Now, does anyone know how to explain the other case. I assume it just follows from the symmetry of the roots of unity. But can anyone validate and perhaps clarify this. Thank you.
| 4 | https://mathoverflow.net/users/3610 | 13382 | 9,033 |
https://mathoverflow.net/questions/13337 | 6 | Let $k$ be a field and $A$ be a finitely generated (commutative) algebra over $k$. If $A\_1$ and $A\_2$ are finitely generated $k$-subalgebras of $A$, is it true that $A\_1 \cap A\_2$ is also finitely generated (as an algebra) over $k$? What if $A$ is a polynomial ring?
**Update** (for the sake of completeness, April 1, 2017): [This paper](https://arxiv.org/abs/1301.2730) (disclaimer: it's mine) describes the smallest dimensional counterexample to the second question in zero characteristic. In positive characteristic the answer seems to be unknown.
| https://mathoverflow.net/users/1508 | Intersection of finitely generated subalgebras also finitely generated? | [Thomas Bayer](http://www.emis.de/journals/BAG/vol.43/no.2/19.html) has found a counter-example using rings of invariants inside polynomial rings.
| 6 | https://mathoverflow.net/users/297 | 13399 | 9,044 |
https://mathoverflow.net/questions/13400 | 11 | I'm turning here (a variation of) a question asked by a friend of mine. For the purposes of this question I will say that a compact complex manifold is projective if it is isomorphic to a subvariety of $\mathbb{P}^n$ and algebraic if it is isomorphic to the complex analytic space associated to a scheme.
One often finds many examples of compact complex manifolds which are not projective. For instance Hopf and Inoue surfaces, some K3, some tori... Usually the proof shows that either this manifold is not Kahler, or the Kahler cone does not intersect the lattice of integral cohomology. But of course this does not tell us anything about their being algebraic in the sense outlined above.
So there are two questions. First, what is an example of a compact complex manifold which is not algebraic? Probably this is standard, but I don't have any reference in mind. I think that the complex analytic space associated to a smooth algebraic space which is not a scheme will do, but I'm not expert of algebraic spaces, so I don't have even such an example in mind.
The second, subtler, question is: are the examples above of nonprojective complex manifold also non algebraic? How one can prove such a statement?
| https://mathoverflow.net/users/828 | Nonalgebraic complex manifolds | I wrote a [blog post](http://sbseminar.wordpress.com/2008/02/14/complex-manifolds-which-are-not-algebraic/) about some of the standard examples of nonalgebraic compact complex manifolds.
| 10 | https://mathoverflow.net/users/297 | 13402 | 9,046 |
https://mathoverflow.net/questions/13342 | 8 | I believe that there is a statement along the following lines (I would, of course, love to be corrected): a formal power series is the Taylor expansion of a rational function if and only if the coefficients eventually satisfy a linear relationship.
Let's suppose that I understand what "satisfy a linear relationship" means, because it's not the part I actually want to ask about (although clarifications are very welcome!). What I would like to know is what conditions on a rational function are equivalent to all the Taylor coefficients being nonnegative integers. For example, I happen to know that $1/(1-kx) = \sum (kx)^n$, and so any sum or product of such functions works. In particular, I can try playing around with partial-fraction decompositions to see if I can write a given rational function in this way. But I have no idea if this is all of them.
Put another way, there is a map $\mathbb R(x) \to \mathbb R[x^{-1},x]]$ (rational functions to Laurent series). I would like to understand the inverse image of $\mathbb N[x^{-1},x]]$.
(Oh, also, I have no idea how to tag this, and I think "general mathematics" is probably an inappropriate tag for MO. So please re-tag as you see fit.)
| https://mathoverflow.net/users/78 | What characterizes rational functions with nonnegative integer Taylor coefficients? | [This paper (?) of Gessel](http://people.brandeis.edu/~gessel/homepage/papers/nonneg.pdf) might help you out, although it is mostly about combinatorics. There are two natural ways to write down rational functions with non-negative integer coefficients in combinatorics, one coming from transfer matrices / finite automata and one coming from regular languages. The two give the same class of rational functions, but there exist rational functions with non-negative integer coefficients which provably don't arise in this way, so the situation seems complicated.
Your question seems to indicate you're not familiar with this class of rational functions, so here are two equivalent definitions: it is, on the one hand, the class of all non-negative linear combinations of entries of matrices of the form $(\mathbf{I} - \mathbf{A})^{-1}$ where $\mathbf{A}$ is a square matrix with entries in $x \mathbb{N}[x]$, and on the other hand the minimal class of rational functions containing $1, x$, and closed under addition, multiplication, and the operation $f \mapsto \frac{1}{1 - xf}$.
**Edit:** One reason there isn't likely to be a particularly nice classification is that one can start with any rational function with integer coefficients and add a polynomial and a term of the form $\frac{1}{1 - kx}$ for $k$ such that $\frac{1}{k}$ is smaller than the smallest pole.
| 3 | https://mathoverflow.net/users/290 | 13406 | 9,049 |
https://mathoverflow.net/questions/13414 | 6 | Can someone provide examples of Kähler manifolds which are not algebraic?
This question came to my mind seeing [the post of Andrea Ferretti](https://mathoverflow.net/questions/13400/).
| https://mathoverflow.net/users/2938 | Kähler manifold which is not algebraic | generic complex tori in complex dimension 2 or higher.
MR
| 13 | https://mathoverflow.net/users/4696 | 13416 | 9,053 |
https://mathoverflow.net/questions/13396 | 2 | Landau and Streater proved that a set of Kraus operators, Ai, is extremal if and only if the set
$\{A\_{k}^{\dagger}A\_{l}\}\_{k,l \ldots N}$
are linearly independent. I have seen very convincing arguments both for and against. You can even see two PDFs of Mathematica notebooks "proving" *both* answers here: <http://quantummoxie.wordpress.com/2010/01/28/a-quirky-mathematical-problem-in-need-of-explanation/>
What is missing from these proofs?
| https://mathoverflow.net/users/2576 | Are the Gell-Mann matrices extremal when used as Kraus operators for a quantum channel? | Hi Jon. Actually, the requirement that the set $A\_{i}^{\dagger}A\_{j} \oplus A\_{j}A\_{i}^{\dagger}$ be linearly independent is specifically for extremal *unital* channels. If the requirement on unitality is relaxed, Landau and Streater showed that only the set $A\_{i}^{\dagger}A\_{j}$ need be linearly independent.
jc: I see your point, but then can you tell me what is wrong with my Mathematica code? In other words, there must be something wrong with it if it returns a "True" for linear independence. Maybe someone has a better way to check these in Mathematica.
| 1 | https://mathoverflow.net/users/3639 | 13418 | 9,055 |
https://mathoverflow.net/questions/13413 | 21 | Given a variety V and a locally free (coherent) sheaf $\mathcal{F}$ of rank 1 (equivalently a line bundle $L$), I can do a Cech cohomology on it. Then $H^0(V; \mathcal{F})$ are just global sections. Is there a similarly understandable meaning to elements of $H^1(V; \mathcal{F})$?
Thanks!
| https://mathoverflow.net/users/3637 | Interpretation of elements of H^1 in sheaf cohomology. | $H^1(V;\mathcal{F})$ is the space of bundles of affine spaces modeled on $\mathcal{F}$. An affine bundle $F$ modeled on $\mathcal{F}$ is a sheaf of sets that $\mathcal{F}$ acts freely on as a sheaf of abelian groups (i.e., there is a map of sheaves $F\times \mathcal{F}\to F$ which satisfies the usual associativity), and on a small enough neighborhood of any point, this action is regular (i.e., the action map on some point gives a bijection). You should think of this as a sheaf where you can take differences of sections and get a section of $\mathcal{F}$.
This matches up with what Anweshi said as follows: given such a thing, you can try to construct an isomorphism to $\mathcal{F}$. This means picking an open cover, and picking a section over each open subset and declaring that to be 0. The Cech 1-cochain you get is the difference between these two sections on any overlap, and if an isomorphism exists, the difference between the actual zero section and the candidate ones you picked is the Cech 0-chain whose boundary your 1-cochain is.
Another way of saying this is that a Cech 1-cycle is exactly the same sort of data as transition functions valued in your sheaf, so if you have anything that your sheaf acts on (again, as an abelian group), then you can use these transition functions to build a new sheaf; a homology between to 1-cycles (i.e. a 0-cycle whose boundary is their difference) is exactly the same thing as an isomorphism between two of these.
I'll note that there's nothing special about line bundles; this works for any sheaf of groups (even nonabelian ones). For example, if you take the sheaf of locally constant functions in a group, you will classify local systems for that group. If you take continuous functions into a group, you will get principal bundles for that group. If you take the sheaf $\mathrm{Aut}(\mathcal{O}\_V^{\oplus n})$, you'll get rank $n$ locally free sheaves. A particularly famous instance of this is that line bundles are classified by $H^1(V;\mathcal{O}\_V^\*)$.
| 16 | https://mathoverflow.net/users/66 | 13426 | 9,060 |
https://mathoverflow.net/questions/13380 | 6 | Suppose I have a collection of $n$ vectors $C \subset \mathbb{F}\_2^n$. They are of course spanned by the canonical set of $n$ basis vectors.
What I would like to find is a much smaller (~ $\log n$) collection of basis vectors that span a collection of vectors which well approximate $C$. That is, I would like basis vectors $b\_1,\ldots,b\_k$ such that for every $v \in C$, there exists a $u \in span(b\_1,\ldots,b\_k)$ such that $||u-v||\_1 \leq \epsilon$.
When is this possible? Is there a property that $C$ might posses to allow such a sparse approximation?
| https://mathoverflow.net/users/3282 | Sparse approximate representation of a collection of vectors | By triangle inequality, preserving the property you wish for means that you can find "representatives" for each $v$ so that the $\ell\_1$ distances between any $v, v'$ are preserved to within 2$\epsilon$ additive error.
There is a general result by Brinkman and Charikar that says that in general, for a collection of $n$ vectors in an $\ell\_1$ space, there's no way to construct a set of $n$ vectors in a smaller (e.g $\log n$) dimensional space) that preserves distances approximately even multiplicatively (let alone additively). This distinction is important if you have vectors in the original space that are $O(\epsilon)$ apart, but otherwise doesn't matter greatly.
Brinkman, B. and Charikar, M. 2005. On the impossibility of dimension reduction in l1. J. ACM 52, 5 (Sep. 2005), 766-788. DOI= <http://doi.acm.org/10.1145/1089023.1089026>
So I'm guessing that the answer to your first question should be no.
| 4 | https://mathoverflow.net/users/972 | 13439 | 9,069 |
https://mathoverflow.net/questions/13393 | 0 | Lemma: Let $A\_1,\ldots,A\_n$ are events $n\in\mathbb{N}$ then
$$
\sum\_{i=1}^n \mathbb{P}(A\_i) = \mathbb{P}(\cup\_{i=1}^n A\_i)
$$
if and only if $A\_1,\ldots,A\_n$ are mutually exclusive.
Both ways are shown by an easy induction.
However, I think that we are assuming that the probability spaces are finite. Does this lemma still hold if we have a probability space $(\Omega,\mathcal{F},\mathbb{P})$ where $\mathcal{F}$ is countable or uncountably infinite.
Thanks in advance.
| https://mathoverflow.net/users/2011 | Equality in the union bound. | Your "lemma" is false for finite probability spaces, e.g.,
$\Omega = \{a,b\}, \mathbb P(\{a\})=0,\mathbb P(\{a,b\})=1, \mathbb P(\{a\} \cup \{a,b\})=1.$
After you fix it, a cannon to swat the fly is [inclusion-exclusion](http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle), or more specifically, the [Bonferroni inequalities](http://en.wikipedia.org/wiki/Boole's_inequality).
I think people are confusing your question as stated with the natural and very elementary question of whether countably additive probabilities must be uncountably additive, and the example of Lebesgue measure on $[0,1]$ shows this this is not the case.
You should very rarely do anything with the sample space itself in any intrinsic probability question. See the answer gowers gave to [this question](https://mathoverflow.net/questions/12516/a-random-variable-is-it-a-function-or-an-equivalence-class-of-functions-closed) and this Tao [blog entry](http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/). It's ok to have a sample space when you apply probability to something like an analysis question (e.g., proving the Weierstrauss approximation theorem using probability) or use the probabilistic method in combinatorics.
| 1 | https://mathoverflow.net/users/2954 | 13441 | 9,070 |
https://mathoverflow.net/questions/13454 | 7 | I realized that I am very confused about a certain sign in the definition of a Poisson group. I will give some definitions, and then point out my confusion.
Definitions
-----------
### Group objects
Let $\mathcal C$ be a category with Cartesian products. Recall that a *group object* in $\mathcal C$ is an object $G \in \mathcal C$ along with chosen maps $e: 1\to G$ and $m: G\times G \to G$ (choose an initial object $1$ and a particular instance of the categorical product, and they imply all the others), such that (i) the two maps $G^{\times 3} \to G$ agree, (ii) the three natural maps $G\to G$ agree, and (iii) the map $p\_1 \times m: G^{\times 2} \to G^{\times 2}$ is an isomorphism, where $p\_1$ is the "project on the first factor" map $G^{\times 2}\to G$.
You may be used to seeing axiom (iii) presented slightly differently. Namely, if $p\_1 \times m: G^{\times 2} \to G^{\times 2}$ is an isomorphism, then consider the map $i = p\_2 \circ (p\_1\times m)^{-1} \circ (e\times \text{id}) : G = 1\times G \to G$. It satisfies the usual axioms of the inverse map. Conversely, if $i: G\times G$ satisfies the usual axioms, then $p\_1 \times (m \circ (i\times \text{id}))$ is an inverse to $p\_1 \times m$. I learned this alternate presentation from [Chris Schommer-Pries](https://mathoverflow.net/users/184/).
### Poisson manifolds
A *Poisson manifold* is a smooth manifold $M$ along with a smooth bivector field, i.e. a section $\pi \in \Gamma(\wedge^2 TM)$, satisfying an axiom. Recall that if $v\in \Gamma(TM)$, then $v$ defines a (linear) map $C^\infty(M) \to C^\infty(M)$ by differentiating in the direction of $v$. Well, if $\pi \in \Gamma(\wedge^2 TM)$, then it similarly defines a map $C^\infty(M)^{\wedge 2} \to C^\infty(M)$. The axiom states that this map is a Lie brackets, i.e. it satisfies the Jacobi identity.
A *morphism of Poisson manifolds* is a smooth map of manifolds such that the induced map on $C^\infty$ is a Lie algebra homomorphism.
[The category of Poisson manifolds has products](http://en.wikipedia.org/wiki/Poisson_manifold#Product_manifold) (Wikipedia).
### Poisson groups
A *Poisson Group* is a manifold $G$ with a Lie group structure $m : G\times G \to G$ and a Poisson structure $\pi \in \Gamma(\wedge^2 TG)$, such that $m$ is a morphism of Poisson manifolds. Recall that a *Lie group* $G$ is a group object in the category of smooth manifolds.
Recall also that a Lie group is almost entirely controlled by its Lie algebra $\mathfrak g = T\_eG$. Then it is no surprise that the Poisson structure can be described infinitesimally. Indeed, by left-translating, identify $TG = \mathfrak g \times G$. Consider the adjoint action of $G$ on the abelian Lie group $\mathfrak g$. Then we can define a Lie group structure $\mathfrak g^{\wedge 2} \rtimes G$ on $\wedge^2 TG$. Recall that a section $\pi \in \Gamma(\wedge^2 TG)$ is just a manifold map $G \to \wedge^2 TG$ that splits the projection $\wedge^2 TG \to G$. Then a Poisson manifold $(G,\pi)$ is a Poisson group if and only if $\pi : G \to \wedge^2 TG$ is a map of Lie groups.
Thus, a Poisson group structure is precisely the same as a Lie algebra $d\pi : \mathfrak g \to \wedge^2 \mathfrak g \rtimes \mathfrak g$ splitting the obvious projection (here $\wedge^2 \mathfrak g$ is an abelian Lie algebra, and $\mathfrak g$ acts on it via the adjoint action), and such that $(d\pi)^\* : \wedge^2\mathfrak g^\* \to \mathfrak g^\*$ satisfies the Jacobi identity. (Any failure of $G$ to be simply-connected, which might prevent such a map from lifting, also fails in $\wedge^2 TG$, so this really is a one-to-one identification of Poisson group structures on $G$ and "Lie bialgebra" structures on $\mathfrak g$.)
From this perspective, then, it is more or less clear that [the inverse map $i:G\to G$ is not a morphism of Poisson manifolds](http://en.wikipedia.org/wiki/Poisson%E2%80%93Lie_group) (Wikipedia). Indeed, infinitesimally, $di = -1: \mathfrak g \to \mathfrak g$, which takes $d\pi$ to $-d\pi$ (as $d\pi$ has one $\mathfrak g$ on the left and two on the right). Instead, $i$ is an "anti-Poisson map". The monoid $(\mathbb R,\times)$ acts on the category of Poisson manifolds by doing nothing to the underlying smooth manifolds and rescaling the Poisson structures; a smooth map is *anti-Poisson* if it becomes Poisson after twisting by the action of $-1$.
The unit map $e: 1 \to G$, on the other hand, is Poisson; it follows from the axioms of a Poisson group that $\pi(e) = 0$, and the terminal object in the category of Poisson manifolds is $1 = \{\text{pt}\}$ with the trivial Poisson structure. ($C^\infty(\{\text{pt}\}) = \mathbb R$ can only support this Poisson structure.)
My question
-----------
Suppose that $G$ is a Poisson group. Then $p\_1 \times m: G^{\times 2} \to G^{\times 2}$ is a Poisson map, and an isomorphism of smooth manifolds. Thus, I would expect that it is an isomorphism of Poisson manifolds. On the other hand, in the first section above I constructed the inverse map $i : G\to G$ out of this isomorphism and the other structure maps, all of which are Poisson when $G$ is a Poisson group. And yet $i$ is not a Poisson map. So where am I going wrong?
| https://mathoverflow.net/users/78 | Is a Poisson Group a group object in the category of Poisson Manifolds? | I think the problem is that the product of Poisson manifolds is not actually a categorical product. This is due to the fact (if I remember correctly) that two Poisson maps $f: X \to Y$ and $g: X \to Z$ give a Poisson map $f \times g: X \to Y \times Z$ only when the images of $f^\*$ and $g^\*$ Poisson commute in $C^\infty(X)$. In particular, $p\_1 \times m$ doesn't seem to be Poisson.
| 6 | https://mathoverflow.net/users/2552 | 13458 | 9,081 |
https://mathoverflow.net/questions/13436 | 11 | Moduli spaces of pseudoholomorphic curves do not carry the structure of a (compact) differentiable manifold in general (due to transversality issues). Nevertheless one would like to at least associate a fundamental class to the moduli space in question.
It looks like two approaches dominate: Kuranishi structures and polyfolds.
Both seem to be mammoth projects. Before diving seriously into one of them I may ask:
What are the advantages/drawbacks of these approaches? What is their motivation? Are they rigourosly proved? Are there reasonable alternatives?
| https://mathoverflow.net/users/3509 | Kuranishi structures vs polyfolds | Kuranishi models are a traditional - and beautiful - technique for describing the local structure of moduli spaces cut out by non-linear equations whose linearization is Fredholm. A more elaborate version, "Kuranishi structures", are used by Fukaya-Oh-Ohta-Ono (FOOO) and Akaho-Joyce to handle transversality for moduli spaces of pseudo-holomorphic polygons with Lagrangian boundary conditions, and the compactifications of these spaces. FOOO's [book](http://www.mrlonline.org/bookstore-getitem/item=AMSIP-46) is the result of a decade of dedicated thought by a superb team, but few have assimilated it (I certainly haven't).
[Polyfolds](http://arxiv.org/abs/0809.3753), Hofer's "infrastructure project", are designed with the severe demands of symplectic field theory (SFT) foremost in mind. This is a more radical rethink of how to handle transversality, whose aims include absorbing the difficulties of the lack of canonical coordinates when gluing Morse-Floer trajectories. (What difficulties? Try to prove that the moduli space of unparametrized broken gradient flow-lines of an ordinary Morse function is a smooth manifold with corners, and you'll find out.) Several papers into the project, it's still not completely clear how efficiently it will work in applications, especially those outside SFT. I'd hope that polyfolds will help us set up Cohen-Jones-Segal Floer homotopy-types, for instance - but there may still be severe difficulties. I've also never heard a compelling argument that Kuranishi structures are insufficient for SFT.
[This paper](http://arxiv.org/abs/math/0702887) of Cieliebak-Mohnke suggests an intriguing alternative approach.
My view would be that these mammoths are worth chasing only if you have *very* clear aims in mind. There are many excellent problems in symplectic topology that don't need such gigantic foundations. If you're interested in Fukaya categories, there's lots to be proved using the definition from Seidel's book, which deals with exact symplectic manifolds. If you want to prove things about contact manifolds, try using symplectic cohomology, a close cousin of SFT requiring less formidable analysis.
| 14 | https://mathoverflow.net/users/2356 | 13460 | 9,083 |
https://mathoverflow.net/questions/13463 | 6 | If $n>2$, does the impossibility of solving $x^n +y^n=z^n$ with $x, y, z$ rational integers imply the same with $x, y, z$ algebraic integers?
Rather, If insolvability in algebraic integers does follow, then does it follow from simple considerations, or is it still an interesting question?
| https://mathoverflow.net/users/493 | Fermat over Number Fields | This is mostly an amplification of Kevin Buzzard's comment.
You ask about points on the Fermat curve $F\_n: X^n + Y^n = Z^n$ with values in a number field $K$.
First note that since the equation is homogeneous, any nonzero solution with $(x,y,z) \in K^3$ can be rescaled to give a nonzero solution $(Nx,Ny,Nz) \in \mathbb{Z}\_K$, the ring of algebraic integers of $K$ -- here $N$ can taken to be an ordinary positive integer.
Thus you have two "parameters": the degree $n$ and the number field $K$.
If you fix $n$ and ask (as you have seemed to) whether there are solutions in *some* number field $K$, the answer is trivially yes as Kevin says: take $x$ and $y$ to be whatever algebraic integers you want; every algebraic integer has an $n$th root which is another algebraic integer, so you can certainly find a $z$ in some number field which gives a solution. Moreover, if you take $x$ and $y$ in a given number field $K$ (e.g. $\mathbb{Q}$), then you can find infinitely many solutions in varying number fields $L/K$ of degrees at most $n$. But it is interesting to ask over *which* number fields (or which number fields of a given degree) there is a nontrivial solution.
On the other hand, if you fix the number field $K$ and ask for which $n$ the Fermat curve
$F\_n$ has a solution $(x,y,z) \in K$ with $xyz \neq 0$, then you're back in business: this is a deep and difficult problem. (You can ask such questions for any algebraic curve, and many people, myself included, have devoted a large portion of their mathematical lives to this kind of problem.) So far as I know / remember at the moment, for a general $K$ there isn't that much which we know about this problem for the family of Fermat curves specifically, and there are other families (modular curves, Shimura curves) that we understand significantly better. But there are some beautiful general results of Faltings and Frey relating the plenitude of solutions (in fact not just over a fixed number field but over all number fields of bounded degree) to geometric properties of the curves, like the least degree of a finite map to the projective line (the "gonality").
| 12 | https://mathoverflow.net/users/1149 | 13466 | 9,087 |
https://mathoverflow.net/questions/13428 | 73 | Kidding, kidding. But I *do* have a question about an $n$-line outline of a proof of the first case of FLT, with $n$ relatively small.
Here's a result of Eichler (remark after Theorem 6.23 in Washington's Cyclotomic Fields): If $p$ is prime and the $p$-rank of the class group of $\mathbb{Q}(\zeta\_p)$ satisfies $d\_p<\sqrt{p}-2$, then the first case of FLT has no non-trivial solutions. Once you know Herbrand-Ribet and related stuff, the proof of this result is even rather elementary.
The condition that $d\_p<\sqrt{p}-2$ seems reminiscent of rank bounds used with Golod-Shafarevich to prove class field towers infinite. More specifically, a possibly slightly off (and definitely improvable) napkin calculation gives me that for $d\_p>2+2\sqrt{(p-1)/2}$, the $p$-th cyclotomic field $\mathbb{Q}(\zeta\_p)$ has an infinite $p$-class field tower. It's probably worth emphasizing at this point that by the recent calculation of Buhler and Harvey, the largest index of irregularity for primes less than 163 million is a paltry 7.
So it seems natural to me to conjecture, or at least wonder about, a relationship between the unsolvability of the first case of FLT and the finiteness of the $p$-class field tower over $\mathbb{Q}(\zeta\_p)$. Particularly compelling for me is the observation that regular primes (i.e., primes for which $d\_p=0$) are precisely the primes for which this tower has length 0, and have obvious historical significance in the solution of this problem. In fact, the mechanics of the proof would probably/hopefully be to lift the arithmetic to the top of the (assumed finite) p-Hilbert class field tower, and then use that its class number is prime to $p$ to make arguments completely analogously to the regular prime case.
I haven't seen this approach anywhere. Does anyone know if it's been tried and what the major obstacles are, or demonstrated why it's likely to fail? Or maybe it works, and I just don't know about it?
**Edit**: Franz's answer indicates that even for a relatively simple Diophantine equation (and relatively simple class field tower), moving to the top of the tower introduces as many problems as it rectifies. This seems pretty compelling. But if anyone has any more information, I'd still like to know if anyone knows or can come up with an example of a Diophantine equation which *does* benefit from this approach.
| https://mathoverflow.net/users/35575 | Please check my 6-line proof of Fermat's Last Theorem. | Suppose you have a diophantine problem whose solution is connected with the structure of the p-class group of a number field K. Then you have the following options:
1. Use ideal arithmetic in the maximal order OK
2. Replace OK by a suitable ring of S-integers with trivial p-class group
3. Replace K by the Hilbert class field, which (perhaps) has trivial p-class group.
Experience with descent on elliptic curves has shown me that ultimately, the equations you have to solve in methods 1 and 2 are the same; moreover, the approach using ideals is a lot less technical than using factorial domains in S-integers (the class group relations come back in through the larger rank of the group of S-units). I am certain that the route via the Hilbert class field is even more technical: again, the unit group in the class field will produce more difficulties than a trivial class group will eliminate.
**Edit.** As an example illustrating my point in a *very* simple example, let me solve the diophantine equation $x^2 + 5y^2 = z^2$ in several different ways. I will always assume that $\gcd(x,y) = 1$.
**1. Elementary Number Theory**
The basic idea is factoring: from $5y^2 = (z+x)(z-x)$. Since $d = \gcd(z-x,z+x) = \gcd(2z,z+x) \mid 2$ we have $d = 1$ or $d = 2$; moreover we clearly have $z-x > 0$.
This gives $z+x = da^2$, $z-x = 5db^2$ or $z+x = 5da^2$, $z-x = db^2$. Solving for $x$ and $z$ yields
$$ x = \pm \frac d2 (a^2 - 5b^2), \quad y = dab. $$
**2. Parametrization**
Set $X = \frac xz$ and $Y = \frac yz$; then $X^2 + 5Y^2 = 1$. Take the line $Y = t(X+1)$ through the obvious point $(-1,0)$; the second point of intersection is given by
$$ X = \frac{1-5t^2}{1+5t^2}, \quad Y = \frac{2t}{1+5t^2}. $$
Dehomogenizing using $t = \frac ba$ and $X = \frac xz$ etc. gives
the projective parametrization
$$ (x:y:z) = (a^2-5b^2:2ab:a^2+5b^2). $$
If $ab$ is odd, all coordinates are even, and we find
$$ x = \frac12(a^2 - 5b^2), \quad y = ab; $$
if $a$ or $b$ is even we get
$$ x = a^2 - 5b^2, \quad y = 2ab $$
as above.
**3. Algebraic Number Theory**
Consider the factorization
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
in the ring of integers of the number field $K = {\mathbb Q}(\sqrt{-5}\,)$.
The class number of $K$ is $2$, and the ideal class is generated by
the prime ideal ${\mathfrak p} = (2,1+\sqrt{-5}\,)$.
The ideal $(x + y\sqrt{-5}, x - y\sqrt{-5}\,)$ is either $(1)$ or
${\mathfrak p}$; thus
$$ (x + y\sqrt{-5}\,) = {\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) =
{\mathfrak b}^2 $$
in the first and
$$ (x + y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak a}^2, \quad
(x - y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak b}^2 $$
in the second case.
The second case is impossible since the left hand side as well as
${\mathfrak a}^2$ are principal, but ${\mathfrak p}$ is not. We
could have seen this immediately since $x$ and $y$ cannot both be odd.
In the first case, assume first that ${\mathfrak a} = (a + b\sqrt{-5}\,)$
is principal. Since the only units in ${\mathcal O}\_K$ are $\pm 1$,
this gives $x + y \sqrt{-5} = \pm(a+b\sqrt{-5}\,)^2$ and hence
$$ x = \pm (a^2 - 5b^2), \quad y = \pm 2ab. $$
If ${\mathfrak a}$ is not principal, then
${\mathfrak p}{\mathfrak a} = (a+b\sqrt{-5}\,)$ is,
and from $({\mathfrak p}{\mathfrak a})^2 = 2(x+y\sqrt{-5}\,)$ we
similarly get
$$ x = \pm \frac12(a^2 - 5b^2), \quad y = \pm ab. $$
**4. S-Integers**
The ring $R = {\mathbb Z}[\sqrt{-5}\,]$ is not a UFD, but $S = R[\frac12]$ is;
in fact, $S$ is even norm-Euclidean for the usual norm in $S$
(the norm is the same as in $R$ except that powers of $2$ are dropped).
It is also easily seen that $S^\times = \langle -1, 2 \rangle$. From
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
and the fact that the factors on the left hand side are
coprime we deduce that $x + y\sqrt{-5} = \varepsilon \alpha^2$ for some unit
$\varepsilon \in S^\times$ and some $\alpha \in S$. Subsuming squares into
$\alpha$ we may assume that $\varepsilon \in \{\pm 1, \pm 2\}$. Setting
$\alpha = \frac{a + b\sqrt{-5}}{2^t}$, where we may assume that $a$
and $b$ are not both even, we get
$$ x + y \sqrt{-5} = \varepsilon \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{2^{2t}}. $$
It is easily seen that we must have $t = 0$ and $\varepsilon = \pm 1$ or
$t = 1$ and $\varepsilon = \pm 2$; a simple calculation then yields the
same formulas as above.
**5. Hilbert Class Fields**
The Hilbert class field of $K$ is given by $L = K(i)$. It is not
too difficult to show that $L$ has class number $1$ (actually it is
norm-Euclidean), and that its unit group is generated by $i = \sqrt{-1}$
and $\omega = \frac{1+\sqrt{5}}2$ (we only need to know that these
units and their product are not squares). From
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
and the fact
that the factors on the left hand side are coprime in ${\mathcal O}\_K$
we deduce that $x + y \sqrt{-5} = \varepsilon \alpha^2$. Subsuming
squares into $\alpha^2$ we may assume that
$\varepsilon \in \{1, i, \omega, i\omega \}$. Applying the nontrivial
automorphism of $L/K$ to $x + y \sqrt{-5} = \varepsilon \alpha^2$ we find
$\varepsilon \alpha^2 = \varepsilon' {\alpha'}^2$. Since the ideal
${\mathfrak a} = (\alpha)$ is fixed and since $L/K$ is unramified,
the ideal ${\mathfrak a}$ must be an ideal in ${\mathcal O}\_K$.
Thus either ${\mathfrak a} = (a+b\sqrt{-5}\,)$ is principal in $K$,
or ${\mathfrak p} {\mathfrak a} = (a+b\sqrt{-5}\,)$ is; in the second
case we observe
that ${\mathfrak p} = (1+i)$ becomes principal in ${\mathcal O}\_L$.
Thus either
$$ x + y \sqrt{-5} = (a+b\sqrt{-5}\,)^2 \quad \text{or} \quad
x + y \sqrt{-5} = i \Big(\frac{a+b\sqrt{-5}}{1+i}\,\Big)^2, $$
giving us the same formulas as above.
Avoiding ideal arithmetic in $K$ and only using the fact that
${\mathcal O}\_L$ is a UFD seems to complicate the proof even more.
**Edit 2** For good measure . . .
**6. Hilbert 90**
Consider, as above, the equation $X^2 + 5Y^2 = 1$.
It shows that the element $X + Y \sqrt{-5}$ has norm $1$;
by Hilbert 90, we must have
$$ X + Y \sqrt{-5} = \frac{a+b\sqrt{-5}}{a-b\sqrt{-5}}
= \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{a^2 + 5b^2}. $$
Dehomogenizing via $X = \frac xz$ and $Y = \frac yz$ yields the same
projective parametrization as above, and we end up with the
familiar formulas.
**7. Binary Quadratic Forms**
The equation $x^2 + 5y^2 = z^2$ tells us that the form $Q\_0(X,Y) = X^2 + 5Y^2$
with fundamental discriminant $\Delta = -20$ represents a square;
this implies that $Q\_0$ lies in the principal genus (which is trivial
since $Q\_0$ is the principal form), and that the representations of
$z^2$ by $Q\_0$ come from composing representations of $z$ by forms
$Q\_1$ with $Q\_1^2 \sim Q\_0$ with themselves.
There are only two forms with discriminant $\Delta$ whose square is
equivalent to $Q\_0$: the principal form $Q\_0$ itself and the form
$Q\_1(X,Y) = 2X^2 + 2XY + 3Y^2$. Thus either
$$ z = Q\_0(a,b) = a^2 + 5b^2 \quad \text{or} \quad
z = Q\_1(a,b) = 2a^2 + 2ab + 3b^2. $$
The formulas for Gauss composition of forms immediately provide us with
expressions for $x$ and $y$ in terms of $a$ and $b$, but they can also
be checked easily by hand. In the first case, we get
$$ x^2 + 5y^2 = (a^2 + 5b^2)^2 = (a^2 - 5b^2)^2 + 5(2ab)^2, $$
and in the second case we can reduce the equations to this one
by observing that $2Q\_1(a,b) = A^2 + 5b^2$ with $A = 2a+b$, which gives
$$ x^2 + 5y^2 = \frac14\Big(A^2 + 5b^2\Big)^2
= \Big(\frac{A^2 - 5b^2}2\Big)^2 + 5(Ab)^2. $$
| 106 | https://mathoverflow.net/users/3503 | 13469 | 9,088 |
https://mathoverflow.net/questions/13478 | 7 | This may be a silly question - but are there interesting results about the invariant: the minimal size of an open affine cover? For example, can it be expressed in a nice way? Maybe under some additional hypotheses?
| https://mathoverflow.net/users/3238 | Minimal size of an open affine cover | Consider for simplicity smooth projective varieties defined over $\mathbb C$. In this case, the minimal size equals $n+1$ where $n$ is the dimension of the variety.
Proof. Let $M^n$ be such a variety. Take $n+1$ generic very ample divisors $D\_0,...,D\_n$. Then such divisors don't have a common intersection. At the same time for any $i$ $M^n\setminus D\_i$ is affine.
In order to show that you do need $n+1$ open affine sub-varieties, notice that the complement to an open affine sub-variety is a closed variety of dimension $n-1$. Now proceed by induction.
For projective irreducible varieties exactly the same reasoning should hold, and it can be generalised further I guess
| 5 | https://mathoverflow.net/users/943 | 13482 | 9,095 |
https://mathoverflow.net/questions/13480 | 18 | All texts I have read on set-theoretic independence proofs consider several different sorts of constructions separately, such as Boolean-valued models (equivalently, forcing over posets), permutation models, and symmetric models. However, the topos-theoretic analogues of these notions—namely topoi of sheaves on locales, continuous actions of groups, and combinations of the two—are all special cases of one notion, namely the topos of sheaves on a site. Is there anywhere to be found a direct construction, in the classical world of membership-based set theory, of a "forcing model" relative to an arbitrary site?
| https://mathoverflow.net/users/49 | Set-theoretic forcing over sites? | To the best of my knowledge, this has never been "officially" described in the set theoretic literature. This has been described by Blass and Scedrov in *Freyd's models for the independence of the axiom of choice* (Mem. Amer. Math. Soc. 79, 1989). (It is of course implicit and sometimes explicit in the topoi literature, for example Mac Lane and Moerdijk do a fair bit of the translation in *Sheaves in Geometry and Logic*.) There are certainly a handful of set theorists that are well aware of the generalization and its potential, but I've only seen a few instances of crossover. In my humble opinion, the lack of such crossovers is a serious problem (for both parties). To be fair, there are some important obstructions beyond the obvious linguistic differences. Foremost is the fact that classical set theory is very much a classical theory, which means that the double-negation topology on a site is, to a certain extent, the only one that makes sense for use classical set theory. On the other hand, although very important, the double-negation topology is not often a focal point in topos theory.
---
Thanks to the comments by Joel Hamkins, it appears that there is an even more serious obstruction. In view of the main results of Grigorieff in
*Intermediate submodels and generic extensions in set theory*, Ann. Math. (2) 101 (1975), it looks like the forcing posets are, up to equivalence, precisely the small sites (with the double-negation topology) that preserve the axiom of choice in the generic extension.
| 9 | https://mathoverflow.net/users/2000 | 13488 | 9,100 |
https://mathoverflow.net/questions/13410 | 4 | Suppose I have a divisor $D$ on a curve $X$ (Hartshorne curve - smooth, projective, dimension one over an algebraically closed $k$). If the complete linear system $|D|$ is basepoint free then I get a map $\varphi:X\rightarrow\mathbb{P}^n\_k$. My question is, say for simplicity our map ends up being to $\mathbb{P}^1\_k$, what if anything is the relationship between the degree of the divisor $D$ and the degree of the morphism $\varphi$?
It seems for many cases that we have $deg(\varphi)=deg(K)$, however I can't find anywhere that proves that this is always the case.
Thanks
| https://mathoverflow.net/users/3261 | Degree of divisors and degrees of the corresponding maps to projective space | Here's how I think about it.
Let's assume we are in the case that $\dim\varphi(X)=\dim X$. Then $\varphi : X\to \varphi(X)$ is an generic finite map. Let $d$ be the degree of this map which is defined as the degree of field extension $[k(X):k(\varphi(X))]$. The degree of $\varphi(X)$ is given by $\varphi(X)\cdot H^{\dim X}$ where $H$ is a general hyperplane of $\mathbb{P}^n$. Pulling $H$ back to $X$, we get $D$. Then, by projection formula, $D^{\dim X} = X\cdot D^{\dim X}=d\cdot(\varphi(X)\cdot H^{\dim X})$. In the case that $X$ is a curve, $D^{\dim X}$ is noting but the degree of $D$. So, the degree of $D$ equals that the degree of image times the degree of the map.
However, in higher dimension, $D^{\dim X}$ may not be the degree of $D$. For example, $D$ is a irreducible degree 2 curve in $\mathbb{P}^2$. The degree of $D$ is 2 which is not equal to $D\cdot D=4$ by Bézout's theorem.
Edit: I think in higher dimension, to define the degree of a divisor $D$, we need to choose a very ample divisor $A$ at first and then define the degree as the intersection number $D\cdot A^{\dim D}$.
| 3 | https://mathoverflow.net/users/2348 | 13489 | 9,101 |
https://mathoverflow.net/questions/13423 | -3 | I find when I read a paper, Costello" The Gromov-Witten potential associated to TCFT"
| https://mathoverflow.net/users/2391 | Why is a partition function of a Topological Conformal Field Theory related to Deligne-Mumford space | As I said in the comments, you should read AJ's answer to [this question](https://mathoverflow.net/questions/1312/gromov-witten-theory-and-compactifications-of-the-moduli-of-curves).
If you haven't read Costello's paper "Topological conformal field theories and Calabi-Yau categories", then you should definitely take a look at it, as the paper you reference is the sequel to this paper. You should also take a look at Lurie's TFTs paper.
In Costello's work and in Lurie's work, you will notice that TCFTs, *by definition*, involve moduli spaces of nonsingular Riemann surfaces (or nonsingular algebraic curves).
On the other hand, in order to do Gromov-Witten theory, we also need to consider moduli spaces of certain *singular* Riemann surfaces ("stable" Riemann surfaces). This is where Deligne-Mumford spaces come into play. So Gromov-Witten theory and TCFT are not exactly the same thing; they involve different moduli spaces. The idea of Costello (and Kontsevich) is that sometimes we can take a TCFT and extend the theory from the uncompactified moduli space to the compactified moduli space, thus getting something which is "a Gromov-Witten theory" associated to the TCFT.
One of Costello's and Kontsevich's motivations comes from mirror symmetry. The idea is that the Fukaya category of a compact symplectic manifold $X$ should give a TCFT. This is why I asked [this question](https://mathoverflow.net/questions/13114/are-fukaya-categories-calabi-yau-categories). Then, we should be able to extend this TCFT to the DM boundary and obtain the Gromov-Witten theory of the manifold. On the mirror side, for example the derived category of coherent sheaves on a Calabi-Yau variety $Y$ should also give a TCFT. Again, if we extend this TCFT to the DM boundary, we should get "a Gromov-Witten theory", which will not be the Gromov-Witten theory of $Y$, but it should at least be related to the Gromov-Witten theory of whatever $Y$ is mirror to.
I might be wrong about this, but I think that in some sense we *have* to consider compactifications of the moduli spaces, such as the Deligne-Mumford compactification (but there are [other possible compactifications](https://mathoverflow.net/questions/1312/gromov-witten-theory-and-compactifications-of-the-moduli-of-curves)), because in order to obtain things like partition functions or the Gromov-Witten potential function, we must do integrals over the moduli spaces in question. But if the moduli spaces are non-compact, which they are, there may be issues in defining these integrals. So one way to get around this is to compactify.
In any case that is at least vaguely the broad picture. If you want to know more details you will have to clean up your question and make it more specific.
| 6 | https://mathoverflow.net/users/83 | 13512 | 9,114 |
https://mathoverflow.net/questions/13506 | 2 | This is a question posed to me in private communication by [this user](https://mathoverflow.net/users/3582/amaanush).
Given a scheme $T$, let $\Gamma (T) = Mor (T, \mathbb{A}^1)$ be the ring
of global sections. Note that there is a canonical map
$\phi : T \rightarrow Spec (\Gamma (T))$.
Is $\phi$ a closed mapping onto the image, ie. is $im\ (Z)$ a
closed subset of $im(T)$ for all closed subsets $Z$ of $T$ ?.
| https://mathoverflow.net/users/2938 | Is the mapping from a scheme to its global sections a closed map? | Here is a ''natural'' example as expected by Martin. Let $T$ be the projective line over ${\mathbb Z}$, minus a rational point $x\_0$ of the closed fiber at some prime $p$. Then $O(T)=\mathbb Z$ (direct computation or use Zariski's extension theorem for normal schemes), $\phi$ is just the structural morphism and is onto. Let $Z'$ be a section of the projective line passing through $x\_0$. Then $Z=Z'\cap T$ is closed in $T$, but $\phi(Z)$ is not closed in $\phi(T)$ because it is the image of
the composition $Z\to {\rm Spec} O(Z)\to {\rm Spec} O(T)$, and as $Z$ is affine, the image is just the complementary of the closed point $p$.
| 9 | https://mathoverflow.net/users/3485 | 13515 | 9,117 |
https://mathoverflow.net/questions/13511 | 5 | "Three circles packed inside a triangle such that each is tangent to the other two and to two sides of the triangle are known as Malfatti circles" (for a brief historical account on this topic, see [here](http://mathworld.wolfram.com/MalfattiCircles.html) and [here](http://mathworld.wolfram.com/MalfattisProblem.html) on MathWorld).
Consider the triangle formed by the centers of these circles, one can draw a new set of smaller Malfatti circles in this triangle. What is the limiting point of this process?
One thing sort of discouraging is that I tried on an isosceles triangle, unfortunately did not find the limiting point matching any of the known relevant points (e.g., incenter or [the first Ajima-Malfatti point](http://mathworld.wolfram.com/Ajima-MalfattiPoints.html)).
| https://mathoverflow.net/users/3350 | Malfatti Circles - Limiting point | I don't know the answer to your question, but it should be easy enough to compute this limit point numerically for an arbitrary triangle and use the result to [search the Encyclopedia of Triangle Centers](http://faculty.evansville.edu/ck6/encyclopedia/search.html).
| 5 | https://mathoverflow.net/users/440 | 13517 | 9,118 |
https://mathoverflow.net/questions/13510 | 8 | Let $\varphi(n)$ denote Euler's phi-function. If we let
$$ \sum\_{n\leq x} \varphi(n) = \frac{3}{\pi^2}x^2+R(x),$$
then it is not hard to show that $R(x)=O(x\log x)$. What is the best known bound for $R(x)$ assuming the Riemann Hypothesis?
| https://mathoverflow.net/users/3659 | Question concerning the arithmetic average of the Euler phi function: | There is information on page 68 of Montgomery and Vaughan's book, and also on page 51 of "Introduction to analytic and probabilistic number theory" by Gérald Tenenbaum. Briefly, Montgomery has established that
$$
\limsup\_{x \rightarrow +\infty}\frac{R(x)}{x\sqrt{\log\log(x)}} > 0
$$
and similarly with the limit inferior. So there is only modest room for improvement. Unfortunately I cannot find any reference to an upper bound conditional on RH. On page 40 Tenenbaum has a reference to page 144 of Walfisz' book on exponential sums. Walfisz uses Vinogradov's method to show that
$$
R(x) = O\left(x\log^{2/3}(x)(\log\log(x))^{4/3}\right).
$$
I don't own a copy of Walfisz' book, so I have no further details.
| 8 | https://mathoverflow.net/users/3304 | 13519 | 9,119 |
https://mathoverflow.net/questions/13516 | 5 | This is a rather nice question I got from [this user](https://mathoverflow.net/users/3582/) via private communication.
Let $\mathcal{C} = Top$ the category of topological spaces. Let $\mathcal{C}^\prime$ be the category $Funct(\mathcal{C}^{op}, Sets)$. Say $F \in \mathcal{C}^\prime$ is a
sheaf, with the usual patch up condition. Does this imply that F is
representable?
An elementary way of putting this without using the category theory terminology: Say a set $X$ is endowed with the following data:
For every topological space $Y$, a subset $M(Y,X) \subset Maps(Y,X)$, such that,
1. whenever $f: Y \rightarrow X$ in $M(Y,X)$, and $g : Z \rightarrow Y$ continuous,
$f \circ g : Z \rightarrow X$ belongs to $M(Z,X)$.
2. If ${U\_i}$ is an open cover of $Y$, and $f: Y \rightarrow X$ is a set map,
then $f \in M(Y,X)$ if and only if $f|\_{U\_i} \in M (U\_i, X)$ for all $i$.
Now, endow $X$ with the finest topology such that for all $f: Y \rightarrow X \in
M (Y,X)$ [for all topological space $Y$], $f$ becomes continuous. In
other words, declare $U \subset X$ to be open if for all $f \in M(Y,X)$
for all $Y \in ob (Top)$, $f^{-1} (U)$ is open. [it is clear that that
is a topology].
It is clear that under this topology, $f \in M(Y,X)$ is continuous. Does
the converse hold? ie. if $f: Y \rightarrow X$ is continuous, then is
$f \in M(Y,X)$?
| https://mathoverflow.net/users/2938 | Sheaf condition and representability in the category Top | There are lots of counterexamples. Take a property of functions defined between topological spaces, such as bounded, and define the sheaf of functions which have this property locally. This works in most cases, I think.
Define $F(Y) := \{f : Y \to \mathbb{R} : f \text{ is locally bounded}\}$. Clearly this is a subsheaf of $Maps(-,\mathbb{R})$. Assume $F$ is representable, then $X:=F(pt)=\mathbb{R}$ is the representing object and carries the finest topology, making all the $f \in F(Y)$ continuous. Assume $U$ is a nonempty open subset of $X$, and take $Y$ to be an arbitrary bounded interval around which intersects $U$, endowed with the indiscrete topology, and $f : Y \to X$ the inclusion. Then $f \in F(Y)$, thus $f$ is continuous, meaning $Y \subseteq U$. As we can vary $Y$, we get $U = X$. Thus $X$ carries the indiscrete topology. But then $X$ represents the functor $Maps(Y,\mathbb{R})$, which is larger than $F$.
There is another reason that $F$ is not representable, namely $F$ does not preserves colimits. A function $f : Y/\sim \to \mathbb{R}$, whose composition with $Y \to Y/\sim$ is locally bounded, does not have to be locally bounded. Take $Y = \mathbb{R}$ and collapse $\mathbb{Z}$ to one point, and let the supremum of $\mathbb{R} \to \mathbb{R}$ around $z \in \mathbb{Z}$ be $|z|$, but taking the same values at $z$.
A reasonable question is now: Let $F : Top^{op} \to Sets$ a continuous sheaf. Is then $F$ representable? If the canonical map $F \to Maps(-,F(pt))$ is injective, the solution set condition is fulfilled and we may apply Freyds Representability Theorem.
EDIT: Ok it follows from SAFT.
| 9 | https://mathoverflow.net/users/2841 | 13521 | 9,121 |
https://mathoverflow.net/questions/13533 | 5 | Suppose $C \subset \lbrace 0,1\rbrace^{n}$ is a binary code with distance $\delta \* n$, for $1/2 < \delta < 1$. Can $|C|$ be arbitrarily large (if I allow n to be arbitrarily large)? Note that the Hadamard code gives you $\delta = 1/2$.
| https://mathoverflow.net/users/630 | Binary codes with large distance | No. If we take $\{-1/\sqrt n,1/\sqrt n\}^n$ instead of $\{0,1\}^n$, the problem reduces to asking if we can have many unit vectors $v\_j$ with pairwise scalar products $-\gamma$ or less where $\gamma>0$ is a fixed number. But if we have $N$ such vectors, then the square of the norm of their sum is at most $N-\gamma N(N-1)$. Since this square must be non-negative, we get $N-1\le\gamma^{-1}$ regardless of the dimension.
| 14 | https://mathoverflow.net/users/1131 | 13546 | 9,139 |
https://mathoverflow.net/questions/13486 | 8 | A lambda-ring $R$ is called "special" if it satisfies the $\lambda^i\left(xy\right)=...$ and $\lambda^i\left(\lambda^j\left(x\right)\right)=...$ relations, or, equivalently, if the map $\lambda\_T:R\to\Lambda\left(R\right)$ given by $\lambda\_T\left(x\right)=\sum\limits\_{i=0}^{\infty}\lambda^i\left(x\right)T^i$ (where the $\sum$ sign means addition in $R\left[\left[T\right]\right]$, not addition in $\Lambda\left(R\right)$) is a morphism of lambda-rings. If you are wondering what the hell I am talking about, most likely you belong to the school of algebraists that denote only special lambda-rings as lambda-rings at all.
Anyway, let $A$ and $B$ be two special lambda-rings, and for every $i>0$, let $\Psi\_A^i$ and $\Psi\_B^i$ be the $i$-th Adams operations on $A$ and $B$, respectively. Let $f:A\to B$ be a ring homomorphism such that $f\circ\Psi\_A^i=\Psi\_B^i\circ f$ for every $i>0$. Does this yield that $f$ is a lambda-ring homomorphism, i. e. that $f\circ\lambda\_A^i=\lambda\_B^i\circ f$ for every $i>0$ ?
Note that this is clear if both $A$ and $B$ are torsion-free as additive groups (i. e., none of the elements $1$, $2$, $3$, ... is a zero-divisor in any of the rings $A$ and $B$), but Hazewinkel, in his text [Witt vectors, part 1](http://arxiv.org/abs/0804.3888) (Lemma 16.35), claims the same result for the general case. I am writing a list of errata for his text, and I would like to know whether this should be included - well, and I'd like to know the answer anyway, as I am writing some notes on lambda-rings as well.
For the sake of completeness, here is a definition of Adams operations: These are the maps $\Psi^i:R\to R$ for every integer $i>0$ (where $R$ is a special lambda-ring) defined by the equation
$\sum\limits\_{i=1}^{\infty} \Psi^i\left(x\right)T^i = -T\frac{d}{dT}\log\left(\lambda\_{-T}\left(x\right)\right)$ in the ring $R\left[\left[T\right]\right]$ for every $x\in R$.
Here, even if the term $\log\left(\lambda\_{-T}\left(x\right)\right)$ may not make sense (since some of the fractions $\frac{1}{1}$, $\frac{1}{2}$, $\frac{1}{3}$, ... may not exist in $R$), the logarithmic derivative $\frac{d}{dT}\log\left(\lambda\_{-T}\left(x\right)\right)$ is defined formally by
$\displaystyle \frac{d}{dT}\log\left(\lambda\_{-T}\left(x\right)\right)=\frac{\frac{d}{dT}\lambda\_{-T}\left(x\right)}{\lambda\_{-T}\left(x\right)}$.
| https://mathoverflow.net/users/2530 | Is every Adams ring morphism a lambda-ring morphism? | Here is a more general point of view on Charles's example, which someone might find helpful.
Let $M$ be an abelian group, and let $\mathrm{Z}[M]$ denote $\mathrm{Z}\oplus M$ with the usual split ring structure $(a,m)(a',m')=(aa',am'+a'm)$. There is a simple description of the (special) $\lambda$-ring structures on such rings. For any prime number $p$, let $\theta\_p$ denote the symmetric function $(\psi\_p-e^p)/p$, where $e=x\_1+x\_2+\cdots$ and $\psi\_p=x\_1^p+x\_2^p+\cdots$. Observe that $\theta\_p$ has integral coefficients and therefore defines a natural operation on any $\lambda$-ring, and for any element $x$ in any $\lambda$-ring, we have $\psi\_p(x)=x^p+p\theta\_p(x)$. In particular, for any $\lambda$-structure on $\mathrm{Z}[M]$, the Adams operation $\psi\_p$ satisfies $\psi\_p(m)=p\theta\_p(m)$ for all $m\in M$.
Exercise: Given a commuting family of additive maps $\theta\_p:M\to M$, there is a unique $\lambda$-ring structure on $\mathrm{Z}[M]$ whose $\theta\_p$ operators on $M$ are the given ones. Conversely, suppose $\mathrm{Z}[M]$ has a $\lambda$-ring structure. Then each $\theta\_p$ preserves the ideal $M$, and resulting map $\theta\_p:M\to M$ is additive.
Now it is a general fact that a ring map between two $\lambda$-rings is a $\lambda$-ring map if and only if it commutes with the $\theta\_p$ operators. (This is because such symmetric functions and all their compositions under plethysm generate the whole ring of symmetric functions.) Thus, in our case, a $\lambda$-ring map $\mathrm{Z}[M]\to\mathrm{Z}[N]$ is the same as a linear map $M\to N$ that commutes with each $\theta\_p$. But that is not the same as commuting with each $p\theta\_p$, i.e. the Adams operations, and it is easy to make counterexamples. For instance Charles's counterexample is with $M=N=\mathrm{Z}/2\mathrm{Z}$, where $\theta\_p$ on $M$ is the identity for all $p$, but $\theta\_p$ on $N$ is the identity for all odd $p$ but is zero for $p=2$. The identity map $M\to N$ therefore commutes with each $p\theta\_p$ and so is an Adams map. It commutes with each $\theta\_p$ with $p$ odd, but it doesn't commute with $\theta\_2$. Therefore it is not a $\lambda$-map.
| 12 | https://mathoverflow.net/users/1114 | 13551 | 9,143 |
https://mathoverflow.net/questions/13553 | 0 | An equalizer in a category $\mathcal{C}$ is a couple $(E,eq)$ consisting in an object $E$ and a morphism $eq:E\longrightarrow X$ so that $f\circ eq=g\circ eq$ for every pair of parallel morphisms $f,g:X\longrightarrow Y$ and if for every other object $O$ and morphism $m:O→X$ there exists a unique morphism $u:O→E$ so that $eq\circ u=m$.
In the category $Set$, by taking sets and functions between them, an equalizer is the set of elements of the common domain where the functions are equal, that is: $Eq(f,g)=\{x\in X/f(x)=g(x)\}$ with $X$ a set and $(f,g)$ a couple of parallel morphism in Set.
My question is: can we say that the equalizer set is minimal among all the *equalizer sets* (like $O$ in the definition)?
Thanks for participate.
| https://mathoverflow.net/users/3338 | Equalizer objects in Set. | Given two parallel morphisms $f,g:X\to Y$ in some category $\mathcal{C}$, let us consider the category $\mathcal{E}\_{f,g}$ :
**Objects :** all pairs $(E,e)$, where $E$ is an object of $\mathcal{C}$ and $e:E\to X$ is a morphism in $\mathcal{C}$ such that $f\circ e=g\circ e$,
**Morphisms :** from $(E',e')$ to $(E,e)$ are just morphisms $\varphi:E'\to E$ in $\mathcal{C}$ such that $e'=e\circ\varphi$.
Now an equaliser of $f,g$ is just a final object in the category $\mathcal{E}\_{f,g}$. Final objects in any category are unique (provided they exist), up to a unique morphism; we may then talk of "the" equaliser of $f,g$.
When $\mathcal{C}$ is the category of sets, the equaliser always exists (and is therefore uniquely unique); as you say, it is the largest subset of the common source where the two maps coincide.
It is fine to think of it as a "maximal" object in $\mathcal{E}=\mathcal{E}\_{f,g}$, but one must realise that it is also a "minimal" object in the opposite category $\mathcal{E}^\circ$ in the sense of being an initial object therein.
| 5 | https://mathoverflow.net/users/2821 | 13556 | 9,144 |
https://mathoverflow.net/questions/13555 | 31 | I want to know if there's any book that categorizes problems by subjects of Functional Analysis.
I'm studying Functional Analysis now a days and I really need to solve some problems in order to assure myself that I've really understood the concepts and definitions.
For example: problems related to the Hahn-Banach theorem or Banach Spaces or Hilbert Spaces or related subjects.
| https://mathoverflow.net/users/3124 | A book for problems in Functional Analysis | Another classical book is [Theorems and problems in functional analysis](http://books.google.com/books?id=XAApQAAACAAJ) by Kirillov and Gvishiani.
| 30 | https://mathoverflow.net/users/2149 | 13563 | 9,148 |
https://mathoverflow.net/questions/13518 | 10 | Let C be the model category of simplicial commutative monoids (with underlying weak equivalences and fibrations), or equivalently the (∞,1)-category PΣ(Top), where T is the Lawvere theory for commutative monoids. In C, as in any pointed (∞,1)-category with finite limits and colimits, we can define adjoint functors ΣC and ΩC as ΣCX = hocolim [• ← X → •] and ΩCX = holim [• → X ← •].
The category CommMon of commutative monoids sits inside C as a full subcategory (as the constant objects, or the objectwise-discrete presheaves). Consider the two functors CommMon → C given by sending M to ΩCΣCM and to the group completion of M, respectively. Is there a natural equivalence between these functors?
(This question is closely related to [Chris's question here](https://mathoverflow.net/questions/430/homological-algebra-for-commutative-monoids). A thorough answer to that question would probably yield this immediately.)
| https://mathoverflow.net/users/126667 | Is ΩΣ in {simplicial commutative monoids} group completion? | I think an answer is given by the arguments that Segal gives in Section 4 of his paper on "Categories and Cohomology Theories" (aka, the $\Gamma$-space paper), in Topology, v.13. I'll try to sketch the main idea, translated into the context of simplicial commutative monoids. I'll show that if $M$ is a discrete simplicial commutative monoid, then it's group completion is homotopically discrete; according to the comments, this should answer the question.
Given a commutative monoid $M$, we can define a simplicial commuative monoid $M'$ as the nerve of the category whose objects are $(m\_1,m\_2)\in M\times M$, and where morphisms $(m\_1,m\_2)\to (m\_1',m\_2')$ are $m\in M$ such that $m\_im=m\_i'$. We can prolong this to a functor on simplicial commutative monoids.
Let $H=H\_\*|M|=H\_\*(|M|,F)$ (the homology of the geometric realization of $M$, with coefficients in some field $F$), viewed as a commutative ring under the pontryagin product. Then Segal shows that $H\_\*|M'|\approx H[\pi^{-1}]$, where $\pi$ denotes the image of $\pi\_0|M|$ in $H\_0|M]$. His proof amounts to computing the homology spectral sequence for a simplicial space whose realization is $M'$, and whose $E\_2$-term is $\mathrm{Tor}\_i^H(H\otimes H,F)$, and observing that the higher tor-groups vanish.
This means that if $M$ is discrete, then $H\_\*|M'|$ is concentrated in degree $0$. Since $|M'|$ is a grouplike commutative monoid, the Hurewicz theorem should tell us that $|M'|$ is weakly equivalent to a discrete space, namely the group completion of the monoid $M$.
Segal goes on to show that $BM\to BM'$ is a weak equivalence, using the above homology calculation and another spectral sequence. Since $M'$ is weakly equivalent to a group, $\Omega BM\approx \Omega BM'\approx M'$.
| 10 | https://mathoverflow.net/users/437 | 13565 | 9,150 |
https://mathoverflow.net/questions/13582 | 1 | How to compute this integral in general case?
$$t(x)=\int\_{-\infty}^{\infty}\frac{\exp(ixy)}{1+y^{2q}}dy$$
Mathematica can compute it when q is known. For example,for q=1 this integral is
$$\exp(-{\left|x\right|})\pi$$
But even in this case, I don't really know how to get this result.
| https://mathoverflow.net/users/3589 | How can I calculate the characteristic function of these distributions? [previously: difficult integral] | If $q$ is a positive integer, then I think one can find this in any one of several *undergraduate* textbooks on complex analysis, where it's usually one of the standard examples to show the power of contour integration. I dimly remember something like this in Priestley's little OUP book, for instance. For arbitrary positive real values of $q$, I can't remember how this works I'm afraid.
(This is probably the sort of question which you could try out on fellow colleagues/students first, in my view.)
| 2 | https://mathoverflow.net/users/763 | 13583 | 9,159 |
https://mathoverflow.net/questions/13571 | 8 | A while ago, I was reading Majid's book *Foundations of quantum group theory*, and Section 9.4 has a rather fascinating description of a Tannaka-Krein reconstruction result for quantum groups. In particular, there seems to be the claim that if $H$ is a quasi-triangulated quasi-Hopf algebra, then the braided endomorphisms of the identity functor on (a suitably large category of) $H$-comodules form a Hopf algebra object $H'$ in $H$-comodules, in a way that identifies $H'$-comodules with $H$-comodules. Furthermore, $H'$ is commutative and cocommutative with respect to the braided structure on the category of $H$-comodules, and under some nondegeneracy assumptions, it is self-dual. This is called "transmutation" because $H'$ appears to have nicer properties than $H$ (although it may live in a strange category). Some examples are given, e.g., $U\_q(g)$ and quantum doubles of finite groups. Unfortunately, the arguments in the proof are given in a diagrammatic language that I was unable to fathom.
>
> Why does this result seem problematic?
>
>
>
The first problem comes from reasoning by analogy. If I want to describe a Hopf algebra object in a monoidal category, I need some kind of commutor transformation $V \otimes W \to W \otimes V$ to even express the compatibility between multiplication and comultiplication, e.g., that comultiplication is an algebra map. In operad language, I need (something resembling) an E[2]-structure on the category to describe compatible E[1]-algebra and E[1]-coalgebra structures. If you think of the spaces in the E[k] operad as configurations of points in $\mathbb{R}^k$, this is roughly saying that you need two dimensions to describe compatible one-dimensional operations. In the above case, the category of $H$-comodules has an E[2]-structure, but I'm supposed to get compatible E[2]-algebra and E[2]-coalgebra structures. Naively, I would expect an E[4]-category to be necessary to make sense of this, but I was unable to wrestle with this successfully.
The second problem comes from a construction I've heard people call Koszul duality, or maybe just Bar and coBar. If we are working in an E[n]-category for n sufficiently large (like infinity, for the symmetric case), then there is a "Bar" operation that takes Hopf algebras with compatible E[m+1]-algebra and E[k]-coalgebra structures, and produces Hopf algebras with compatible E[m]-algebra and E[k+1]-coalgebra structures. There is a "coBar" operation that does the reverse, and under some conditions that I don't understand, composing coBar with Bar (or vice versa) is weakly equivalent to the identity functor. In the above case, I could try to apply Bar to $H'$, but the result cannot have an E[3]-coalgebra structure, since E[3] doesn't act on the category. Applying Bar then coBar implies the coalgebra structure on $H'$ is a priori only E[1], and applying coBar then Bar implies the algebra structure on $H'$ is a priori only E[1]. It is conceivable (in my brain) that the E[2]-structures could somehow appear spontaneously, but that seems a little bizarre.
>
> Question
>
>
>
Am I talking nonsense, or is there a real problem here? (or both?)
| https://mathoverflow.net/users/121 | Transmutation versus operads | Scott, I believe the source of your confusion is that Majid doesn't claim that the braided Hopf algebra he constructs is both braided commutative and braided co-commutative in C. Just as in the usual case, the Hopf algebra one constructs is braided co-commutative (like U(g)) or braided commutative (like O(G)) if you work dually, but not both. I think your arithmetic about E[n]-operads is correct as to why that would be unusual.
Can you point where in Chapter 9.4 is it claimed that the resulting Hopf algebra is both commutative and co-commutative? For instance in my copy on page 481, he explains U(C) is braided co-commutative but doesn't mention commutative anywhere. On page 477 of my copy, he says "One can use the term 'braided group' more strictly to apply to braided-Hopf algebras which are 'braided-commutative' or 'braided-cocommutative' in some sense." (keyword "or").
By the way, in the Section 3 of <http://arxiv.org/abs/0908.3013> is an exposition (not original) reconstructing the algebra A (transmutation of O\_q(G)), which uses language probably more familiar to you. Also there is a Remark 3.3 (explained to us by P. Etingof) which gives a more concise description of A (and can be used to derive its key properties like braided-commutativity) by considering module categories and internal homs.
(apologies in advance if i'm incorrect; I read that book awhile ago and opened it up briefly to address your question)
| 6 | https://mathoverflow.net/users/1040 | 13586 | 9,162 |
https://mathoverflow.net/questions/11716 | 21 | In his answer to [this question](https://mathoverflow.net/questions/11610/examples-of-poisson-schemes), Scott Carnahan mentions "mirror symmetry mod p". What is that?
(Some kind of) Gromov-Witten invariants can be defined for varieties over fields other than $\mathbb{C}$. Moreover other things that come up in mirror symmetry, like variation of Hodge structure, and derived categories of coherent sheaves, also make sense. (Though I can't imagine that it's possible to talk about Fukaya categories...) Can we formulate any sort of sensible mirror symmetry statement, similar to say that of Candelas-de la Ossa-Green-Parkes relating Gromov-Witten invariants of a quintic threefold to variation of Hodge structure of the mirror variety, when the varieties are over some field other than $\mathbb{C}$? In particular, can we do anything like this for fields of positive characteristic?
I googled "arithmetic mirror symmetry" and "mirror symmetry mod p", and I found some stuff about the relationship between the arithmetic of mirror varieties, but nothing about Gromov-Witten invariants. I did find [notes](http://math.arizona.edu/~swc/aws/04/04Notes.html) from the Candelas lectures that Scott referred to, but I wasn't able to figure out what was going on in them.
More generally, there are many examples of mathematical statements about complex algebraic varieties which come from physics/quantum field theory/string theory. Some of these statements (maybe with some modification) can still make sense if we replace "variety over $\mathbb{C}$ with "variety over $k$", where $k$ is some arbitrary field, or a field of positive characteristic, or whatever. Are there any such statements which have been proven?
Edit: I'm getting some answers, and they are all sound very interesting, but I'm still especially curious about whether anybody has done anything regarding Gromov-Witten invariants over fields other than $\mathbb{C}$.
| https://mathoverflow.net/users/83 | Mirror symmetry mod p?! ... Physics mod p?! | For fixed integers $g,n$, any projective scheme $X$ over a field $k$, and a linear map $\beta:\operatorname{Pic}(X)\to\mathbb Z$, the space $\overline{M}\_{g,n}(X,\beta)$ of stable maps is well defined as an Artin stack with finite stabilizer, no matter the characteristic of $k$. You can even replace $k$ by $\mathbb Z$ if you like.
Now if $X$ is a smooth projective scheme over $R=\mathbb Z[1/N]$ for some integer $N$, then $\overline{M}\_{g,n}(X,\beta) \times\_R \mathbb Z/p\mathbb Z$ is a Deligne-Mumford stack for almost all primes $p$. For such $p$, $\overline{M}\_{g,n}(X,\beta) \times\_R \mathbb Z/p\mathbb Z$ has a virtual fundamental cycle, and so you have well-defined Gromov-Witten invariants. This holds for all but finitely many $p$. Nothing about $\mathbb C$ here, that is my point, the construction is purely algebraic and very general.
It is when you say "Hodge structures" then you better work over $\mathbb C$, unless you mean $p$-Hodge structures.
As far as mirror symmetry in characteristic $p$, much of it is again characteristic-free. For example Batyrev's combinatorial mirror symmetry for Calabi-Yau hypersurfaces in toric varieties is simply the duality between reflexive polytopes. You can do that in any characteristic, indeed over $\mathbb Z$ if you like.
| 9 | https://mathoverflow.net/users/1784 | 13591 | 9,165 |
https://mathoverflow.net/questions/13587 | 5 | I apologize for asking something that might well be found in a mathematical dictionary, but the similarity of the French word to an English one is frustrating my attempts to Google the answer (and the library is shut at time of typing). I suspect the answer should be obvious to those who, unlike me, know some basic Lie group/Lie algebra terminology.
Some context: I am reading an old paper of Dixmier from 1969, which has the following construction/definition. Let $\mathfrak g$ be a Lie algebra (characteristic zero, finite-dimensional), let $\mathfrak n$ be its largest nilpotent ideal -- the *nilradical* -- and put ${\mathfrak h}=[{\mathfrak g},{\mathfrak g}]+{\mathfrak n}$. Dixmier calls ${\mathfrak h}$ "le nilradicalisé de ${\mathfrak g}$".
Literal translation would surely be "the nilradicalised", but that sounds more like a mopey university indie band than a mathematical object. So what is the usual name for this object in English?
| https://mathoverflow.net/users/763 | Translation of "le nilradicalisé de g" | I suspect pretty strongly that this is idiosyncratic terminology; I've never seen that subalgebra used, and the term has no google hits other than this post.
| 4 | https://mathoverflow.net/users/66 | 13604 | 9,174 |
https://mathoverflow.net/questions/13601 | 15 | Define a sequence $(a\_n)\_{n \geq 1}$ by $$na\_n = 2 + \sum\_{i = 1}^{n - 1} a\_i^2.$$
(In particular, $a\_1 = 2$.)
How can you show - preferably **without** using a pc! - that not all terms of the sequence are integral?
And which will be the first such term?
Motivation: nothing interesting to say, it's a random problem which I got from someone - I have no reference - and which interested me. Usually one has to prove that all terms are integral :)
Thoughts: nothing interesting. The terms are quickly getting enormous...
| https://mathoverflow.net/users/1107 | a weird sequence with a non-integral term | Sequences like this are sometimes called Somos sequences (and sometimes Gobel sequences) and you can find information about them at Problem E15 in Guy, Unsolved Problems in Number Theory and in the references Guy gives; also I'm sure typing Somos or Gobel into your favorite search engine will turn up something.
| 13 | https://mathoverflow.net/users/3684 | 13606 | 9,176 |
https://mathoverflow.net/questions/13581 | 6 | A quantum channel is a mapping between Hilbert spaces, $\Phi : L(\mathcal{H}\_{A}) \to L(\mathcal{H}\_{B})$, where $L(\mathcal{H}\_{i})$ is the family of operators on $\mathcal{H}\_{i}$. In general, we are interested in CPTP maps. The operator spaces can be interpreted as $C^{\*}$-algebras and thus we can also view the channel as a mapping between $C^{\*}$-algebras, $\Phi : \mathcal{A} \to \mathcal{B}$. Since quantum channels can carry classical information as well, we could write such a combination as $\Phi : L(\mathcal{H}\_{A}) \otimes C(X) \to L(\mathcal{H}\_{B})$ where $C(X)$ is the space of continuous functions on some set $X$ and is also a $C^{\*}$-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between $C^{\*}$-algebras. Note, however, that these are not necessarily the same $C^{\*}$-algebras. Since the channels are represented by square matrices, the input and output $C^{\*}$-algebras must have the same dimension, $d$. Thus we can consider them both subsets of some $d$-dimensional $C^{\*}$-algebra, $\mathcal{C}$, i.e. $\mathcal{A} \subset \mathcal{C}$ and $\mathcal{B} \subset \mathcal{C}$. Thus a quantum channel is a mapping from $\mathcal{C}$ to itself.
**Proposition** A quantum channel given by $t: L(\mathcal{H}) \to L(\mathcal{H})$, together with the $d$-dimensional $C^{\*}$-algebra, $\mathcal{C}$, on which it acts, forms a category we call $\mathrm{\mathbf{Chan}}(d)$ where $\mathcal{C}$ is the sole object and $t$ is the sole arrow.
**Proof**: Consider the quantum channels
$\begin{eqnarray\*}
r: L(\mathcal{H}\_{\rho}) \to L(\mathcal{H}\_{\sigma}) &
\qquad \textrm{where} \qquad &
\sigma=\sum\_{i}A\_{i}\rho A\_{i}^{\dagger} \\
t: L(\mathcal{H}\_{\sigma}) \to L(\mathcal{H}\_{\tau}) &
\qquad \textrm{where} \qquad &
\tau=\sum\_{j}B\_{j}\sigma B\_{j}^{\dagger}
\end{eqnarray\*}$
where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite $t \circ r: L(\mathcal{H}\_{\rho}) \to L(\mathcal{H}\_{\tau})$ where
$\begin{align}
\tau & = \sum\_{j}B\_{j}\left(\sum\_{i}A\_{i}\rho A\_{i}^{\dagger}\right)B\_{j}^{\dagger} \notag \\
& = \sum\_{i,j}B\_{j}A\_{i}\rho A\_{i}^{\dagger}B\_{j}^{\dagger} \\
& = \sum\_{k}C\_{k}\rho C\_{k}^{\dagger} \notag
\end{align}$
and the $A\_{i}$, $B\_{i}$, and $C\_{i}$ are Kraus operators.
Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. $$\sum\_{k} C\_{k}^{\dagger}C\_{k}=\mathbf{1}.$$ For a similar methodology see Nayak and Sen (<http://arxiv.org/abs/0605041>).
We take the identity arrow, $1\_{\rho}: L(\mathcal{H}\_{\rho}) \to L(\mathcal{H}\_{\rho})$, to be the time evolution of the state $\rho$ in the absence of any channel. Since this definition is suitably general we have that $t \circ 1\_{A}=t=1\_{B} \circ t \quad \forall \,\, t: A \to B$.
Consider the three unital quantum channels $r: L(\mathcal{H}\_{\rho}) \to L(\mathcal{H}\_{\sigma})$, $t: L(\mathcal{H}\_{\sigma}) \to L(\mathcal{H}\_{\tau})$, and $v: L(\mathcal{H}\_{\tau}) \to L(\mathcal{H}\_{\upsilon})$ where $\sigma=\sum\_{i}A\_{i}\rho A\_{i}^{\dagger}$, $\tau=\sum\_{j}B\_{j}\sigma B\_{j}^{\dagger}$, and $\eta=\sum\_{k}C\_{k}\tau C\_{k}^{\dagger}$. We have
$\begin{align}
v \circ (t \circ r) & = v \circ \left(\sum\_{i,j}B\_{j}A\_{i}\rho A\_{i}^{\dagger}B\_{j}^{\dagger}\right) = \sum\_{k}C\_{k} \left(\sum\_{i,j}B\_{j}A\_{i}\rho A\_{i}^{\dagger}B\_{j}^{\dagger}\right) C\_{k}^{\dagger} \notag \\
& = \sum\_{i,j,k}C\_{k}B\_{j}A\_{i}\rho A\_{i}^{\dagger}B\_{j}^{\dagger}C\_{k}^{\dagger} = \sum\_{i,j,k}C\_{k}B\_{j}\left(A\_{i}\rho A\_{i}^{\dagger}\right)B\_{j}^{\dagger}C\_{k}^{\dagger} \notag \\
& = \left(\sum\_{i,j,k}C\_{k}B\_{j}\tau B\_{j}^{\dagger}C\_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag
\end{align}$
and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). $\square$
**Question 1**: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct.
| https://mathoverflow.net/users/3639 | Quantum channels as categories: question 1. | Phrasing this in terms of categories is kind of misleading: A category with a single object is just a monoid (associative binary operation with identity). So, per Yemon Choi's correction, you are just trying to demonstrate that the set of quantum channels $L(\mathcal{H}) \to L(\mathcal{H})$ forms a monoid. [Here I'm assuming that "channel" implies CPTP.]
This requires three things:
1. Closure: The product of two channels is a channel.
2. Identity: The identity operator is a channel.
3. Associativity: Multiplication of channels is associative.
**1:**
There are two different ways to prove closure. You used the characterization of channels as maps given by that Kraus operator form ($\sum\_{i} A\_{i}\rho A\_{i}^{\dagger}$). This works, although I don't feel you made the construction of the $C\_k$ from the $A\_i$ and $B\_j$ clear enough. It also requires that you've already proven this characterization (CPTP linear map $\Leftrightarrow$ Kraus operator form).
You could instead directly use the characterization of a channel as a CPTP linear map. This way is probably easier: It is immediately clear that if *r* is (a linear map which takes positive matrices to positive matrices and preserves trace) and *t* is (a linear map which takes positive matrices to positive matrices and preserves trace) then $t\circ r$ will be (a linear map which takes positive matrices to positive matrices and preserves trace). That does it.
**2:**
I really wish you hadn't said "time evolution". :-) But you basically have the right idea: the identity in this situation is, well, the identity map, which is obviously linear and CPTP.
**3:**
Practically speaking, you almost never have to prove associativity. This is because as long as your maps are functions deep down on the inside, associativity is an immediate consequence of associativity of function composition. This is one of those cases.
So in response to your question "Is there a clearer or more concise proof?" I would say "Absolutely."
---
But again, I think it's unnecessary to put this in the context of categories. UNLESS you plan on generalizing to channels between distinct spaces. *Then* the concept of a category gains its power.
Good luck!
| 7 | https://mathoverflow.net/users/3685 | 13611 | 9,179 |
https://mathoverflow.net/questions/13616 | 28 | Are there enough interesting results that hold for general locally ringed spaces for a book to have been written? If there are, do you know of a book? If you do, pelase post it, one per answer and a short description.
I think that the tags are relevant, but feel free to change them.
Also, have there been any attempts to classify locally ringed spaces? Certainly, two large classes of locally ringed spaces are schemes and manifolds, but this still doesn't cover all locally ringed spaces.
| https://mathoverflow.net/users/1353 | A book on locally ringed spaces? | In addition to the examples mentioned in the question, of manifolds and schemes, other commonly occuring types of locally ringed spaces are formal schemes and complex analytic spaces.
I don't know how extensive the taxonomy of locally ringed spaces is. For example,
if $A$ is a local ring, we can form the locally ringed space consisting of a single point,
with $A$ sitting on top of it. These are the topologically simplest locally ringed spaces
(after the empty space).
If $A$ is a field, one obtains a scheme. If $A$ is a complete local ring, one obtains a formal scheme. In general, this doesn't fit into any particular taxonomic grouping that I know of.
Incidentally, it might be worth mentioning that the various taxonomic classes can interact:
for example, analytification of schemes over ${\mathbb C}$ is conveniently described in terms of
maps (in the category of locally ringed spaces) to complex analytic spaces.
| 18 | https://mathoverflow.net/users/2874 | 13618 | 9,181 |
https://mathoverflow.net/questions/13590 | 4 | Recall that a *braided monoidal category* is a category $\mathcal C$, a functor $\otimes: \mathcal C \times \mathcal C \to \mathcal C$ satisfying some properties, and a natural isomorphism $b\_{V,W}: V\otimes W \to W\otimes V$ satisfying some properties. Recall also that a *monoidal category* (just $\mathcal C,\otimes$ and their properties) is the same as a one-element 2-category: the objects of $\mathcal C$ become the morphisms, and the monoidal structure becomes composition.
Thus, is there a natural definition of "1-braided 2-category"? I'm calling it "1-braided" because the braiding acts on 1-morphisms (as opposed to "braided monoidal 2-category", where the braiding acts on the 0-morphisms).
I realize, of course, that if $V,W$ are morphisms of a 2-category so that $V\circ W$ is defined, then generally $W\circ V$ is not defined, so a priori asking for any relationship $V\circ W \cong W\circ V$ doesn't make sense. On the other hand, consider [Aaron Lauda's categorification of $U\\_q(\mathfrak{sl}\\_2)$](http://arxiv.org/abs/0803.3652). It is a 2-category, but different hom-sets can be more-or-less identified.
| https://mathoverflow.net/users/78 | Candidate definitions for "1-braided 2-category"? | In the Baez-Dolan periodic table a braided monoidal category is just a 2-monoidal category (that is a 3-category with one object and one 1-morphism). If you just think that a braiding means that the structure is inherently 3-dimensional, then you might just want to think about a 1-monoidal 2-category (that is a 3-category with one object).
| 6 | https://mathoverflow.net/users/22 | 13629 | 9,191 |
https://mathoverflow.net/questions/4894 | 9 | Hi,
the following question was posed to me, it apparently has applications for linear codes. Let *n>1*, and $K = \rm{GF}(2^n)$. Let $k$ be coprime to $2^n-1$. Does there always exist $a \neq 0$ in $K$ such that the curve
$y^2+y = x^k+ax$
has exactly $2^n$ affine solutions? (I ran some computer checks [although only for quite small *n* and *k*] without finding a counterexample.)
| https://mathoverflow.net/users/1310 | Existence of hyperelliptic curve with specific number of points in a family | I think that this is equivalent to a known open question. Here are
the details. For $K:=\mathbb{F}\\_{2^n}$, the function $f:y\mapsto
y+y^2:K\to K$ is $\mathbb{F}\_2$-linear, and its kernel $\{0,1\}$ has
dimension 1. The image is therefore of dimension $n-1$, and for $z$ in
the image, the fiber $f^{-1}(z)$ has exactly 2 elements.
Hence, to prove that $y^2+y=x^k+ax$ has exactly $2^n$ solutions
for some fixed $a\in K$, we have to show that $|\{x\in K|x^k+ax\in
\mathrm{Im}(f)\}|=2^{n-1}$.
Since $\sigma:y\mapsto y^2$ is a generator of the Galois group of
$K/\mathbb{F}\\_2$, Hilbert's Theorem 90 (in additive form) says that
$z\in \mathrm{Im}(f)$ if and only if $\mathrm{Tr}(z)=0$, where
$\mathrm{Tr}$ stands for the trace map from $K$ to $\mathbb{F}\_2$.
So the problem is equivalent to showing that there exists an $a\neq 0$ in $K$
such that $|\{x\in K|\mathrm{Tr}(x^k+ax)=0\}|=2^{n-1}$. In other
words, we would like to show that there exists a nonzero $a\in K$ such that
$$
S\_k(a):=\sum\_{x\in K}(-1)^{\mathrm{Tr}(x^k+ax)}=0.
$$
Apparently, this question was addressed in the coding community. In
detail, in [1, p. 258], the following conjecture (of Helleseth) is
mentioned:
``Conjecture 3. For any $m$ and $k$ such that
$\mathrm{gcd}(2^m-1,k)=1$, the sum
$\sum\_{x\in\mathbb{F}\_{2^m}}(-1)^{\mathrm{Tr}(x^k+ax)}$ is null for at
least one nonzero $a$.'' (Note that $n$ in the current question is
$m$ in [1](http://www-rocq.inria.fr/secret/Pascale.Charpin/Charpin-jct04.pdf)).
It seems that in [1, Corollary 1, p. 253], Conjecture 3 is proved for
even $m$ and for certain values of $k$ (the ``Niho exponents,''
defined on p. 252 of [1](http://www-rocq.inria.fr/secret/Pascale.Charpin/Charpin-jct04.pdf)).
Interestingly, at least at a first glance it seems that [1](http://www-rocq.inria.fr/secret/Pascale.Charpin/Charpin-jct04.pdf) has
nothing to say on $k\in\{1,\ldots,2^{n-1}\}$, but to me it seems that
this case is trivial (am I missing something?): Consider a normal
basis for $K/\mathbb{F}\\_2$, that is, a basis $B$ consisting of an orbit of
an element $\gamma\in K$ under the Galois group of $K/\mathbb{F}\_2$
(the $i$th element of $B$ is $b\_i:=\gamma^{2^i}$ for
$i\in\{0,\ldots,n-1\}$).
From the linearity of the trace and the fact that the trace is onto,
we must have $\mathrm{Tr}(b)=1$ for at least one element $b\in B$, and
from $\mathrm{Tr}(b^2)=\mathrm{Tr}(b)$ we then have
$\mathrm{Tr}(b)=1$ for all $b\in B$. So the trace of an element in $K$
is just the modulo-2 sum of the coefficients in its decomposition
according to the basis $B$.
Let $a$ be any element in the
trace-dual basis of $B$, say $\mathrm{Tr}(ab\_i)=\delta\_{i,0}$. Then
for $k=2^j$, if we write $x=\sum\_i \alpha\_i b\_i$, we
get: $\mathrm{Tr}(ax)=\alpha\_0$,
$\mathrm{Tr}(x^k)=\mathrm{Tr}(x)=\sum \alpha\_i$ (sum in
$\mathbb{F}\_2$). These agree for half of the $x\in K$, as required.
That's about it. I hope at least some of this makes sense :) I also
hope that the original person asking this question didn't actually want
to solve the above open question by converting it to a question about
curves, for then this answer is useless.
[1](http://www-rocq.inria.fr/secret/Pascale.Charpin/Charpin-jct04.pdf) P. Charpin, ``[Cyclic codes with few weights and Niho exponents](http://www-rocq.inria.fr/secret/Pascale.Charpin/Charpin-jct04.pdf),''
Journal of Combinatorial Theory, Series A 108 (2004)
247--259.
| 2 | https://mathoverflow.net/users/2734 | 13630 | 9,192 |
https://mathoverflow.net/questions/13614 | 23 | Recall that a *(smooth) manifold with corners* is a Hausdroff space that can be covered by open sets homeomorphic to $\mathbb R^{n-m} \times \mathbb R\_{\geq 0}^m$ for some (fixed) $n$ (but $m$ can vary), and such that all transition maps extend to smooth maps on open neighborhoods of $\mathbb R^n$.
I feel like I know what a "differential form" on a manifold with corners should be. Namely, near a corner $\mathbb R^{n-m} \times \mathbb R\_{\geq 0}^m$, a differential form should extend to some open neighborhood $\mathbb R^{n-m} \times \mathbb R\_{> -\epsilon}^m$. So we can set up the usual words like "closed" and "exact", but then Stokes' theorem is a little weird: for example, the integral of an exact $n$-form over the whole manifold need not vanish.
In any case, I read in [D. Thurston, "Integral Expressions for the Vassiliev Knot Invariants", 1999](http://arxiv.org/abs/math/9901110v2), that "there is not yet any sensible homology theory with general manifolds with corners". So, what are all the ways naive attempts go wrong, and have they been fixed in the last decade?
As always, please retag as you see fit.
**Edit:** It's been pointed out in the comments that (1) I'm not really asking about general (co)homology, as much as about the theory of De Rham differential forms on manifolds with corners, and (2) [there is already a question about that](https://mathoverflow.net/questions/12920/stokes-theorem-for-manifolds-with-corners). Really I was just reading the D. Thurston paper, and was surprised by his comment, and thought I'd ask about it. But, anyway, since there is another question, I'm closing this one as exact duplicate. I'll re-open if you feel like you have a good answer, though. -Theo **Edit 2:** Or rather, apparently OP can't just unilaterally close their own question?
| https://mathoverflow.net/users/78 | Is there a good (co)homology theory for manifolds with corners? | I suppose you are talking about deRham cohomology. Then it would be wise to take a look at the work of Richard Melrose, e.g. his book *[The Atiyah-Patodi-Singer Index Theorem](http://www-math.mit.edu/~rbm/book.html)*.
On page 65 he discusses deRham cohomology for manifolds with boundary (which can be easily generalized to the corner case, as was also done by him!). On manifolds with corners something interesting happens: there are different versions of reasonable vector fields (and—by duality—differential forms), e.g.
1. extendible vector fields (like you mentioned)
2. tangent vector fields (tangent to any boundary hypersurface)
3. "zero" vector fields (vanishing on all boundary hypersurfaces)
(It can be shown that $d$ preserves the classes 1.-3., giving a deRham complex whose cohomology can be computed)
Melrose points out (compact with boundary case) that in cases 1. and 3. the deRham cohomology is canonically isomorphic to the singular cohomology of the underlying topological space. For the 2nd case also the cohomology of the boundary enters (via a degree shift).
I should also point out that there is also a working Morse theory on manifolds with corners, see for example
>
> M. Shida, *Fundamental Theorems of Morse Theory for Optimization on Manifolds with Corners*, Journal of Optimization Theory and Applications **106** (2000) pp 683-688, doi:[10.1023/A:1004669815654](https://doi.org/10.1023/A:1004669815654)
>
>
>
and [[this broken link](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V1K-45TY6KH-1&_user=10&_coverDate=11%2F30%2F2002&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1188262351&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=8b2c4ec9a50cdb034dbdd9f078f0335b) **EDIT** perhaps someone else can extract a result -DR]
Furthermore it is easy to construct "invariants" of the manifold with corners by also taking into account its corners (but be careful with respect to which transformations this is an invariant)!
| 16 | https://mathoverflow.net/users/3509 | 13637 | 9,195 |
https://mathoverflow.net/questions/13634 | 0 | **Background:** There are 7 "bricks" used in the game of Tetris. These are the 7 *unique* combinations of 4 unit squares in which every square shares at least one edge with another square. ("unique" in this case refers to the idea that no brick can be rotated in 2-D space to become another brick.)
**Question:** Using 5 unit cubes, how many *unique* "bricks" could be formed in which each cube shares at least one face with another cube? (Please provide a proof to this in your answer if you can find one.)
| https://mathoverflow.net/users/3690 | Tetris in 3D with 5 units | There are 29 distinct 5-cube bricks (counting mirror images as distinct). Together with one 1-cube brick, one 2-cube brick, two 3-cube bricks, and eight 4-cube bricks, these constitute the brick set for the highly addictive 3-D Tetris game Blockout II, available at <http://www.blockout.net/blockout2/>. Source code is also available.
I think a proof just involves writing out the cases.
| 6 | https://mathoverflow.net/users/767 | 13659 | 9,210 |
https://mathoverflow.net/questions/13648 | 8 | In spite of the fact that the matrix ring $\mathbb{C}^{n \times n}$ is not a field, is it still possible to talk about it being 'algebraically closed' in the sense that $\forall f \in \mathbb{C}^{n \times n}[x]$ does $\exists A \in \mathbb{C}^{n \times n}$ such that $f(A) = 0$? If so, then is it 'algebraically closed'?
Are there any other non-field sets that this idea can be extended to?
| https://mathoverflow.net/users/3121 | Is the matrix ring $\mathrm{Mat}_n(\mathbb{C})$ "algebraically closed"? | The matrix
$\left( \begin{array}{cc}
0 & 1 \\\\
0 & 0
\end{array} \right)$ has no square root.
Polynomials make sense for continuous complex functions on a space. If that space is $\mathbb R$, then polynomial equations with complex coefficients are solvable. If that space is $\mathbb C$ or $S^1$ then $g^2 = f$ may not be solvable.
| 20 | https://mathoverflow.net/users/2954 | 13661 | 9,211 |
https://mathoverflow.net/questions/13665 | 0 | The integral I need:
$$t(x)=\int\_{-K}^{K}\frac{\exp(ixy)}{1+y^{2q}}dy$$
$K<\infty$, q natural number
For q=1 this integral is
$$\pi/2-\int\_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy $$
Where Arc has radius $K$
Upper bound is $$K\pi/(K^2-1)^2$$
Can I obtain a better expression for the integral?
One more question about this integral. For K<1 this integral is just
$$-\int\_{Arc}\frac{\exp(ixy)}{1+y^{2}}dy?$$
| https://mathoverflow.net/users/3589 | An integral arising in statistics | Are you asking if this integral can be expressed in terms of elementary functions?
Most likely no. The reason is that there's a fairly straight forward way of expressing it using exponential integrals, which are not elementary functions. To do that, expand the rational part $1/(1+y^{2q})$ in partial fractions. Each term should have a simple pole. Shift the pole to zero and use the definition of the [exponential integral](http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/07/01/01/).
Or are you interested in some asymptotic expression for the integral in the limit of large/small K or x? The answer would then depend on the limits you are interested in.
| 1 | https://mathoverflow.net/users/2622 | 13673 | 9,218 |
https://mathoverflow.net/questions/13672 | 5 | I have a polynomial of degree 4 $f(t) \in \mathbb{C}[t]$, and I'd like to know when it has two repeated roots, in terms of its coefficients.
Phrased otherwise I'd like to find the equations of the image of the squaring map
$sq \colon \mathbb{P}(\mathbb{C}[t]^{\leq 2}) \rightarrow \mathbb{P}(\mathbb{C}[t]\_{\leq 4})$.
(for some reason the first lower index wouldn't parse, so I put it on top).
Of course I can write the map explicitly and then find enough equations by hand, but this looks cumbersome. I'm not an expert in elimination theory, so I wondered if there is some simple device to find explicit equations for this image. For instance one can detect polynomials with one repeated root using the discriminant, but I don't know how to proceed from this.
| https://mathoverflow.net/users/828 | Polynomial with two repeated roots | It seems to me that this example is easy to do by hand. By the standard tricks, we can assume your polynomial is of the form
$$x^4+ c x^2 + dx +e.$$
A polynomial of this form is a square if and only if $d=0$ and $4e=c^2$.
| 7 | https://mathoverflow.net/users/297 | 13675 | 9,220 |
https://mathoverflow.net/questions/13632 | 1 | For a complex manifold $M$, the transition functions of the tangent bundle $T(M)$ come from the Jacobian of the change-of-coordinate maps. Does there exist a related description of the transition functions of $T^{(0,1)}$ and $T^{(1,0)}$? An example would also be nice, maybe $\mathbb{CP}^1$.
| https://mathoverflow.net/users/1648 | Transition Functions and Complex Structure | For a real manifold $M$ the transition functions of the tangent bundle $T(M)$ come from the Jacobian of the change-of-coordinate maps.
When $M$ is complex, it has a complex tangent bundle $T\_{\mathbb{C}}M$, which can be identified with the holomorphic vector bundle $T^{(1,0)} \subset TM \otimes \mathbb{C}$. The transition functions on $T\_{\mathbb{C}}M$ are given by the (complex) Jacobian of the change-of-coordinate maps, so the same is true for $T^{(1,0)}$.
Since $T^{(0,1)}$ is the complex conjugate bundle, its transition functions are the complex conjugate of the same Jacobian.
| 4 | https://mathoverflow.net/users/828 | 13677 | 9,222 |
https://mathoverflow.net/questions/13361 | 0 | What is the name of the Lie algebra decomposition where the positive root vectors are upper triangular and the negative root vectors are lower triangular?
| https://mathoverflow.net/users/3623 | Name of upper triangular/lower triangular Lie Algebra decomposition | I think people use the term "triangular decomposition" or sometimes "polarization"
| 4 | https://mathoverflow.net/users/3696 | 13704 | 9,241 |
https://mathoverflow.net/questions/13705 | 0 | A beginner's question:
We know: "Since order-equivalence is an equivalence relation, it partitions the class of all sets into equivalence classes." (from [Wikipedia](http://en.wikipedia.org/wiki/Order_type))
This holds since every set can be (well-)ordered by the Axiom of Choice.
But there can be many (well-)orderings of a given set. Especially, the Axiom of Choice doesn't tell us, *what* the choice function is and thus, what the well-ordering is: there can be many.
Thus, a set can belong to many order types and order-equivalence isn't an equivalence relation anymore.
>
> What's wrong with this (presumably dummy) line of thoughts?
>
>
>
| https://mathoverflow.net/users/2672 | Axiom of Choice and Order Types | Order equivalence is an equivalence relations on ordered sets, not on sets. It is just the isomorphism relation on ordered structures. An ordered structure is a set, together with an order.
The Axiom of Choice says that every set has a well-order. Since the order-types of well-orders are well-ordered (given any two, one of them is uniquely isomorphic to a unique initial segment of the other), it follows under AC that for every set, we can associate to it the smallest order-type of a well-order on that set. This is called the *cardinality* of the set.
There is another more general concept of cardinality, which does not rely on AC or on orderings at all, and it is just the equinumerosity class of the set.
| 3 | https://mathoverflow.net/users/1946 | 13706 | 9,242 |
https://mathoverflow.net/questions/13684 | 13 | A colleague in algebra asked me this, and I couldn't answer it. On the [Wikipedia page for "epimorphism"](http://en.wikipedia.org/wiki/Epimorphism) it is claimed that in the category of Hausdorff spaces and continuous maps, a function is epi if and only if it has dense range. The "if" case is easy, but I couldn't justify the "only if" case.
This boils down to: let Y be a Hausdorff space, and let X in Y be a closed subset not equal to Y, and not empty. Can you find a Hausdorff space Z and functions f,g:Y->Z such that f and g agree on X, but are not equal. I think, by using a quotient argument, you can assume that X is just a point.
| https://mathoverflow.net/users/406 | Functions separting points in Hausdorff spaces | Let $Y$ be a Hausdorff space, and let $X\subset Y$ be a closed subspace. Consider disjoint union of two copies of $Y$, and let $Z$ be the coequalizer of two embeddings of $X$ into it
(that is, we glue two copies of $Y$ along $X$). Clearly, the two natural maps $\iota\_{1,2}:Y\to Z$ coincide only on $X$. It is easy to see that $Z$ is Hausdorff.
Indeed, take $z\_1,z\_2\in Z$, $z\_1\ne z\_2$. The map $p:Z\to Y$ is continuous, so if $p(z\_1)\ne p(z\_2)$, we can take preimages of open neighborhoods of $p(z\_1)$ and $p(z\_2)$ to separate $z\_1$ and $z\_2$. It remains to deal with the case $z\_{1,2}=\iota\_{1,2}(y)$ for $y\in Y-X$.
But the neighborhoods $\iota\_{1,2}(Y-X)$ work.
| 13 | https://mathoverflow.net/users/2653 | 13711 | 9,245 |
https://mathoverflow.net/questions/13679 | 0 | I have a problem.
I'm trying to recover a bounding volume (actually line segments that form the bounding volume) from a kDop definition (in a 3D space). (its to draw the kDop on screen)
In my kDOP structure i have the Min/Max values calculated for each axis. (And well, i know the axis used)
I tried before coming here 3 ways to resolve it :
1) Intersecting 3 neighboring planes to find a point and then connecting them (Sometimes the intersection is out of the k-DOP, i didn't manage to find a way to figure the correct planes to use)
2) Starting from the lines of the AABB (DOP6) and then intersecting the lines with each of the k-DOP planes, but when a line is completely behind a plane (i.e. not visible), we have to add two or more new lines and i don't know which is to be drawed and which not. :P (and i get too many lines drawed)
3) Brute force method. Intersect each plane with each other plane to get a set of lines and then intersect these lines with each plane to find the correct line segment (or discard the line) (too slow :()
What's the best way to go? Can you give some explanation on it?
Thanks
EDIT : About the slowness. Well yes, i need to do it for 100-120 objects under 16,667 ms (n^3 method takes 0,96 ms (and i've optimized pretty much everything))
EDIT 2: Answering the questions from TonyK :
1) (1,0,0)(0,1,0)(0,0,1) // AABB (Axis Aligned Bounding Box)
(1,1,1)(1,-1,1)(1,1,-1)(1,-1,-1) // Corners
(1,1,0)(1,0,1),(0,1,1),(1,-1,0),(1,0,-1),(0,1,-1) // Edges
These are the axes used. Are fixed and I can happen to use a subset of those. (only AABB (DOP6) or AABB + Corners (DOP14) or AABB+Corners+Edges (DOP26)). Yes, the only thing changing is the Min/Max values.
(The normals for each plane are prestored (26 normals in the worst case (DOP26)))
2) My point of view is the camera of the simulation and "can" change. (the line segment obtained will be multiplied by World/View/Projection matrix to get the final position on the Viewport, but that's a GPU detail)
3) Mhn...well, there are Axes that do not intersect each other.
4) What do you mean by "treating the bounding plane as opaque"?
EDIT 3:
Regarding the other question, No, I treat them as wireframe so the lines under the kDOP surface will appear.
| https://mathoverflow.net/users/3681 | Finding a bounding volume (line segments) from a kDop definition. | OK, it seems like your $N=52$ then and the normals to the planes are fixed. Then you are right that $N^3$ method takes 1ms time. Indeed, to find the parametric coefficients of the intersection of any 2 planes once you prestore the line directions and the inverse matrices takes just 6 multiplications and 3 additions, to find the cut takes 52 cycles of 6 multiplications, 5 additions, one division, and two comparisons each (keeping the max and the min), and doing all 1326 pairs that requires 1326\*(9+52\*14)=977262 elementary operations, which puts your processor speed at 10^9 multiplications per second, which is about right. There is an $O(N^2)$ algorithm if you know a point inside the polytope, which, once you are talking about kDop, you, probably do, but we need to count operations very carefully to see if it beats the $N^3$ one.
The key problem is that in the worst case when every plane matters, you may have about 3\*52=156 edges which is just one tenth of the maximal possible amount and if you spend the same 9+52\*14 operations on each edge, you will be just on the edge of what you need. Still, it is worth trying. It works under the assumption that no 4 planes intersect at one point. To make sure that it is the case, just perturb your Mins and Maxes by some random numbers that are much bigger than the precision with which you can compute the points but much less than what is visible to the eye (say, $10^{-4}$ times the object size, also make sure that you decrease Min's and increase Max's) but do all comparisons with much higher precision (like $10^{-12}$ times the object size) where the object size is just the maximal span in your directions, which you can find in linear time.
Start with finding one edge. To this end, just take any point inside the polytope, compute the vertices of the dual polytope with respect to that point, choose any direction, and take the furthest vertex of the dual polytope from the origin. Now, you know that this plane certainly matters. It will take you just linear time, so I'll not even count that. Once you figured out one plane, you can find an edge on that plane in the time 50\*(9+52\*14)=36850 elementary operations just as in the $N^3$ method. Keep track on which planes give you the boundary points.
Now, at each moment, keep track of edges you already have, and the vertices you have with their planes. For each vertex also keep the track of whether you know each of the adjacent edges (according to the pair of planes) completely. In the beginning, you have the list of 2 vertices with one known and two unknown edges.
Take the first vertex in the list with at least one unknown adjacent edge. You know the pair of planes for this edge, so you can find the new edge in time (9+26\*15)=737, where you have 26 instead of 52 because you can find whether you'll need Max or Min in each direction in one comparison. Once you have found the edge, see if its other end is one of the already found vertices and mark the corresponding edge at that vertex as known (about 156\*2=312 operations). If not, add its other vertex to the end of the list. This gives you 1049 operations per edge.
Thus, your total is 36850+156\*1049=200494 operations, which is 5 times better than my count for the $N^3$ algorithm but still a bit slower than you need.
There is a room for improvement. The most obvious one is that you may want to keep the list of already found vertices for each pair of planes (the update costs 3 operations and reduces 312 to just 6. Thus, you are down to 36850+156\*153=154318, which is within the limits, but just barely. Another thing is that you can try to reduce the time you need to find the first edge. but I do not see how to do it at the moment.
---
All right, first thing first. The main part of the cure is to construct the admissibility table that, for each pair of directions, lists all directions that may form a vertex with the given two. Surprisingly, it is very short. I'll keep the directions in the 0,$\pm 1$ format like you (not normalized). I assume that you keep an array of these directions p of length 26. The only thing I want you to make sure about the order is that p[2k] and p[2k+1] are opposite for each k=0,1,...,12. This will spare a few operations in the edge tracing algorithm.
First, let us exclude all degenerate pairs. That is pretty easy. The pair (A,B) is degenerate iff A+B is a multiple of some other direction in the list. In practical terms, it means that all non-zero entries of $A+B$ are equal in absolute value. The first thing
is to create the table of pairs of directions and precompute some information for them.
To make it working for both algorithms, I suggest that for each pair (i,j) of indices, you create the structure Q[i,j] with the following fields
Boolean nondegeneracy; true if the pair is non-degenerate, false if it is degenerate.
You should get 288 non-degenerate pairs if the order of i,j matters (144 with i
vector normal; normal=$p[i]\times p[j]$ is the direction orthogonal to both p[i] and p[j]
iprojector, jprojector; those are just found as $v=p[j]\times\operatorname{normal}$; iprojector$=v/(v\cdot p[i])$ and similarly for jprojector. This will allow you to find a point X with given scalar products $a=p[i]\cdot X$ and $b=p[j]\cdot X$ as a times iprojector plus b times jprojector (6 multiplications and 3 additions)
an array goodindices of structures. Each structure element goodindices[m] will have 3 fields: integer k such that p[i],p[j],p[k] may make a vertex, and 3 scalar products iscal=iprojector$\cdot$p[k],jscal=jprojector$\cdot$p[k],and nscal=normal$\cdot$p[k]. To construct it, you take your pair a=p[i],b=p[j] of directions and include the structure with the first field k (once you know k, you can compute the other three fields) into goodindices if the following two conditions hold:
1)the determinant det=det(a,b,c) is not 0
2) for every index m different from i,j,k at least one of the determinants det(a,b,d), det(a,d,c), and det(d,b,c) where d=p[m] has sign opposite to the sine of det, i.e., if, say, det > 0, one of them is strictly less than 0 (since all coordinates are integer, compare to -0.5). This checks that there is no direction in the cone formed by a,b,c.
Funnily enough, the total number of indices in goodindices is never greater than 6.
Also, let Max[i] be the maximal scalar product of your object points with p[i]
(I prefer to keep all 26 directions and Max array to keeping 13 and Min and Max arrays just to avoid any casework anywhere)
Now, when the table is ready, it is very easy to run the N^3 algorithm. For each pair i,j with i < j do the following.
If Q[i,j].nondegeneracy=false, just skip the pair altogether
Determine tmin as the maximum over all structures S in Q[i,j].goodindices such that S.nscal<0 of the ratios (M[S.k]-M[i]S.iscal-M[j]S.jscal)/S.nscal .
Determine tmax as the minimum over all structures S in Q[i,j].goodindices such that S.nscal>0 of the ratios (M[S.k]-M[i]S.iscal-M[j]S.jscal)/S.nscal.
If tmin < tmax, put X=M[i]Q[i,j].iprojector+M[j]Q[i,j].jprojector.; compute the points X+tmin Q[i,j].normal and X+tmax Q[i,j].normal and draw the edge between them (or store it for later drawing).
You should optimize here updating tmin and tmax when looking at each k and stopping immediately if tmax gets less than or equal to tmin. This cycle over k is the most time-consuming part when defining the edge, so how you write it determines the real speed of this algorithm. Optimize as much as possible here to make sure that you do not do any extraction from array twice, etc. Each little extra operation counts and memory is not a problem. That's why I suggested to prestore the scalar products as well, though it might seem a bit silly when we are talking of cycle of length 6 within a cycle of length 144
That's all. It should run fast enough to be acceptable already and it is robust in the sense that you do not need to perturb your maxes, check for paralellness anywhere, etc., etc.
The other algorithm is still about 4 times faster but I should check that it has no stability issues before posting it. Try this first and tell me the running time.
| 4 | https://mathoverflow.net/users/1131 | 13731 | 9,260 |
https://mathoverflow.net/questions/9799 | 96 | Hi Everyone,
Famous anecdotes of G.H. Hardy relay that his work habits consisted of working no more than four hours a day in the morning and then reserving the rest of the day for cricket and tennis. Apparently his best ideas came to him when he wasn't "doing work." Poincare also said that he solved problems after working on them intensely, getting stuck and then letting his subconscious digest the problem. This is communicated in another anecdote where right as he stepped on a bus he had a profound insight in hyperbolic geometry.
I am less interested in hearing more of these anecdotes, but rather I am interested in what people consider an appropriate amount of time to spend on doing mathematics in a given day if one has career ambitions of eventually being a tenured mathematician at a university.
I imagine everyone has different work habits, but I'd like to hear them and in particular I'd like to hear how the number of hours per day spent doing mathematics changes during different times in a person's career: undergrad, grad school, post doc and finally while climbing the faculty ladder. "Work" is meant to include working on problems, reading papers, math books, etcetera (I'll leave the question of whether or not answering questions on MO counts as work to you). Also, since teaching is considered an integral part of most mathematicians' careers, it might be good to track, but I am interested in primarily hours spent on learning the preliminaries for and directly doing research.
I ask this question in part because I have many colleagues and friends in computer science and physics, where pulling late nights or all-nighters is commonplace among grad students and even faculty. I wonder if the nature of mathematics is such that putting in such long hours is neither necessary nor sufficient for being "successful" or getting a post-doc/faculty job at a good university. In particular, does Malcom Gladwell's 10,000 hour rule apply to mathematicians?
Happy Holidays!
| https://mathoverflow.net/users/1622 | How Much Work Does it Take to be a Successful Mathematician? | I agree that hard work and stubbornness are very important (I think we should all take after Wiles and Perelman as much as we can). But it is also important how you spend the many hours you dedicate to mathematics. For instance, choice of problems is quite important: it is important to make sure that when you work on something, you spend your time usefully, i.e. you not only make progress on this particular problem, but also learn something new about mathematics in general. It is also important not to get hyperfocused on a fruitless attempt to solve a problem; after some time and effort spent on it, it becomes addictive. In such a situation, it is sometimes better to stop and ask for help/read something or switch to another problem for a while. Often, you'll wake up one morning a month or a year later and see that the insurmountable obstacle has magically disappeared! Or maybe this "Aha!" moment will come during a discussion with another mathematician, or while listening to a talk. For many people it is also helpful to have many simultaneous projects, so that when you get stuck on one, you can work on another. To summarize, I think that not only the number of hours matters, but also how efficiently you spend them, not only in terms of publishable results,
but also in terms of your personal growth as a mathematician.
| 112 | https://mathoverflow.net/users/3696 | 13734 | 9,261 |
https://mathoverflow.net/questions/13660 | 13 | This seems unrealistic, because the topology on a manifold doesn't have anything to do with the properties its structure sheaf, but I figured I might as well ask. This wouldn't be the first time I was pleasantly surprised about something like this. If not, is there any sort of way to attack differential geometry with abstract nonsense?
Even though schemes have singularities, "it's better to work with a nice category of bad objects than a bad category of nice objects". Manifolds seem to be perfect illustration of this fact.
Edit: Apparently my question wasn't clear enough. The actual question here is if we can define manifolds entirely as "functors of points" like we can with schemes (sheaves on the affine zariski site). There is no fully categorical and algebraic description of the category of smooth manifolds. When I say a "comprehensive functor of points approach", I mean a fully categorical description of the category of smooth manifolds.
| https://mathoverflow.net/users/1353 | A comprehensive functor of points approach for manifolds | Here are two things that I think are relevant to the question.
First, I want to support Andrew's suggestion #5: synthetic differential geometry. This definitely constitutes a "yes" to your question
>
> is there any sort of way to attack differential geometry with abstract nonsense?
>
>
>
--- assuming the usual interpretation of "abstract nonsense". It's also a "yes" to your question
>
> Can we describe it as some subcategory of some nice grothendieck topos?
>
>
>
--- assuming that "it" is the category of manifolds and smooth maps. Indeed, you can make it a *full* subcategory.
Anders Kock has two nice books on synthetic differential geometry. There's also "A Primer of Infinitesimal Analysis" by John Bell, written for a much less sophisticated audience. And there's a brief chapter about it in Colin McLarty's book "Elementary Categories, Elementary Toposes", section 23.3 of which contains an outline of how to embed the category of manifolds into a Grothendieck topos.
Second, it's almost a categorical triviality that there is a full embedding of **Mfd** into the category **Set**${}^{U^{op}}$, where $U$ is the category of open subsets of Euclidean space and smooth embeddings between them.
The point is this: $U$ can be regarded as a subcategory of **Mfd**, and then every object of **Mfd** is a colimit of objects of $U$. This says, in casual language, that $U$ is a *dense* subcategory of **Mfd**. But by a standard result about density, this is equivalent to the statement that the canonical functor **Mfd**$\to$**Set**${}^{U^{op}}$ is full and faithful. So, **Mfd** is equivalent to a full subcategory of **Set**${}^{U^{op}}$.
There's a more relaxed explanation of that in section 10.2 of my book [Higher Operads, Higher Categories](http://www.maths.gla.ac.uk/~tl/book.html), though I'm sure the observation isn't original to me.
| 10 | https://mathoverflow.net/users/586 | 13740 | 9,267 |
https://mathoverflow.net/questions/13733 | 23 | The question is, loosely put, what is known about wild ramification?
Is there a semi-well-established theory of wild ramification that can be furthered in various specific situations? Or maybe there are some situations where we know a lot, but what we know doesn't apply to other situations? Or are all results about wild ramification ad hoc for whatever the author is interested in?
In short: what is there to know, and where is it written?
| https://mathoverflow.net/users/2665 | Wild Ramification | The structure of wildly ramified *abelian* extensions of local fields is given by local class field theory (and conversely is where most of the content of LCFT resides): see Milne's notes or Serre's *Corps Locaux*.
Wildly ramified nonabelian extensions of local fields are "understood" in at least the following two senses [neither one of which connotes perfect understanding to me]:
1) The absolute Galois group of a local field is a topologically finitely presented profinite group, with known generators and relations.
2) Local Langlands for GLn (as proved by Harris-Taylor and Henniart) has something deep to say about the structure of wildly ramified extensions.
It would not be fruitful for me to elaborate, since there are other active MOers who are much more knowledgeable in these matters. This was really just a long comment.
**Addendum**: upon request, here is some bibliographical material for 1) above:
>
> Jannsen, Uwe
> Über Galoisgruppen lokaler Körper. (German) [On Galois groups of local fields]
> Invent. Math. 70 (1982/83), no. 1, 53--69.
>
>
>
>
> Jannsen, Uwe; Wingberg, Kay
> Die Struktur der absoluten Galoisgruppe p-adischer Zahlkörper. (German) [The structure of the absolute Galois group of p-adic number fields]
> Invent. Math. 70 (1982/83), no. 1, 71--98.
>
>
>
>
> Wingberg, Kay
> Der Eindeutigkeitssatz für Demuskinformationen. (German) [The uniqueness theorem for Demushkin formations]
> Invent. Math. 70 (1982/83), no. 1, 99--113.
>
>
>
Let $k$ be a finite extension of ${\bf Q}\_p$ with $p\neq 2$ and $\overline k$ be the algebraic closure of $k$. The study of the Galois group $G\_k = G(\overline k/k)$ was initiated by K. Iwasawa [Trans. Amer. Math. Soc. 80 (1955), 448--469; MR0075239 (17,714g)] and continued mainly in several papers of A. V. Yakovlev and the reviewer. Yakovlev [Izv. Akad. Nauk SSSR Ser. Mat. 32 (1968), 1283--1322; MR0236155 (38 #4453)] succeeded in the description of $G\_k$ as a profinite group with generators and relations. But his result was unsatisfactory, since the structure of one relation which comes from Demushkin's relation for maximal $p$-extensions was rather complicated, especially in the case $n = [k: Q\_p]\equiv 1\ (\text{mod}\,2)$. Therefore the reviewer considered the question once more [Dokl. Akad. Nauk SSSR 238 (1978), 19--22; MR0472776 (57 #12466)] and gave a cohomological characterisation of $G\_k$ as a filtered group with the filtration by the inertia group $T\_k$ and the $p$-Sylow group $V\_k$ of the inertia group. The three papers under review can be considered as the final answer to the question of the structure of $G\_k$. The progress with respect to the above-mentioned earlier papers consists in the following: (1) The authors give a description of $G\_k$ with two relations, instead of three relations as in the earlier work on $G\_k$, which simplifies the situation. (2) For the first time they give a satisfactory description of the case $n\equiv 1\ (\text{mod}\,2)$. (3) The study of $G\_k$ is based on the notion of Demushkin formations, which was introduced by the reviewer [op. cit.]. The proof of the uniqueness of a Demushkin formation with given invariants which characterizes $G\_k$ as a filtered group is given in full detail, based on structure theorems in module theory. Finally the authors give an explicit description of a Demushkin formation with given invariants by means of $n+3$ generators and two relations. The case $p=2$ remains open.
Reviewed by Helmut Koch
| 16 | https://mathoverflow.net/users/1149 | 13743 | 9,268 |
https://mathoverflow.net/questions/13742 | 6 | As far as I understand it the [closing lemma](https://en.wikipedia.org/wiki/Pugh%27s_closing_lemma) implies that closed geodesics on surfaces of negative curvature are dense. So: how can they be constructed in general?
A concrete answer that dovetails with the construction of such surfaces with constant negative curvature and genus $g$ from regular hyperbolic $(8g-4)$-gons along lines indicated by [Adler and Flatto](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-25/issue-2/Geodesic-flows-interval-maps-and-symbolic-dynamics/bams/1183657182.full) and gives the endpoints of the geodesics in the Poincaré disk model would be ideal. More useful still would be a way to construct all the closed geodesics that cross the boundaries of translates of the fundamental $(8g-4)$-gon some specified number of times (I am pretty sure this ought to be a finite set, but I couldn't say why off the top of my head).
| https://mathoverflow.net/users/1847 | How can generic closed geodesics on surfaces of negative curvature be constructed? | If you think of your surface as the upper half plane modulo a group of Moebius transformations $G$, start by representing each of your Moebius transformations $ z \longmapsto \frac{az+b}{cz+d}$ by a Matrix.
$$A = \pmatrix{ a & b \\\ c & d}$$
And since only the representative in $PGL\_2(\mathbb R)$ matters, people usually normalize to have $Det(A) = \pm 1$.
The standard classification of Moebius transformations as elliptic / parabolic / hyperbolic (loxodromic) is in terms of the determinant and trace squared. You're hyperbolic if and only if the trace squared is larger than $4$. Hyperbolic transformations are the ones with no fixed points in the interior of the Poincare disc, and two fixed points on the boundary, and they are rather explicitly "translation along a geodesic".
Elliptic transformations fix a point in the interior of the disc so they can't be covering transformations. Parabolics you only get as covering transformations if the surface is non-compact, because parabolics have one fixed point and its on the boundary -- if you had such a covering transformation it would tell you your surface has non-trivial closed curves such that the length functional has no lower bound in its homotopy class.
So your covering tranformations are only hyperbolic. That happens only when $tr(A)^2 > 4$. So how do you find your axis? It's the geodesic between the two fixed points on the boundary, so you're looking for solutions to the equation:
$$ t = \frac{at+b}{ct+d}$$
for $t$ real, this is a quadratic equation in the real variable $t$. If I remember the quadratic equation those two points are:
$$ \frac{tr(A) \pm \sqrt{tr(A)^2 - 4Det(A)}}{2c}$$
Is this what you're after?
| 7 | https://mathoverflow.net/users/1465 | 13757 | 9,278 |
https://mathoverflow.net/questions/13662 | 10 | Let $p>2$ be prime. Then for abstract reasons the special linear group $\text{SL}\_2({\mathbb F}\_p)$ possesses a free action on some sphere (one has to check that any abelian subgroup of $\text{SL}\_2({\mathbb F}\_p)$ is cyclic and that there's at most one element of order $2$).
Does somebody know a concrete example for such a free action for general $p$? (For $p=5$, for example, $\text{SL}\_2({\mathbb F}\_p)$is the binary icosahedral group which is a subgroup of ${\mathbb S}^3$ thus acting freely on it by multiplication; I'd like to know if there's one single action that can be written down for all $p$ simultaneously).
| https://mathoverflow.net/users/3108 | Free action of SL_2(F_p) on a sphere | Apparently, a linear free action exists only for $p=5$ (if $p\ge 5$), see paper by C. Thomas "Almost linear actions by $SL\_2(p)$ on $S^{2n-1}$". There is a weaker notion of an "almost linear" action, and it seems that constructing such actions is a fairly complicated business, using state-of-the-art differential geometry and topology; see arXiv:math/9911250. It seems that simple explicit actions for higher $p$ are not expected (also, of course, the sphere must be odd dimensional, since an orinentation preserving self-map of an even dimensional sphere has a fixed point by the Lefschetz theorem).
| 13 | https://mathoverflow.net/users/3696 | 13767 | 9,288 |
https://mathoverflow.net/questions/13769 | 4 | Let $Aut(\bar{Q})$ be the automorphism group on the field of algebraic complex numbers. The order of an element $f \in Aut(\bar{Q})$ is the least natural number $n$ (if there exists one) such that $(f)^{n}$ is identity. What are the possible orders of elements of $Aut(\bar{Q})$?
| https://mathoverflow.net/users/2689 | Orders of field automorphisms of algebraic complex numbers | This amounts to the Artin-Schreier theorem, which has come up several times already on MO (c.f. [Examples of algebraic closures of finite index](https://mathoverflow.net/questions/8756/examples-of-algebraic-closures-of-finite-index)):
if $K/F$ is a field extension with $K$ algebraically closed and $[K:F] < \infty$, then
$[K:F] = 1$ or $2$, and in the latter case, $F$ is real-closed.
Thus the answer here is that $n$ can be $1$, $2$ or $\infty$, and all possibilities occur: the field of real algebraic numbers gives an index $2$ subfield of $\overline{\mathbb{Q}}$.
(Also, just to be sure, there are elements of infinite order! E.g., if not then every element would have order $1$ or $2$, so the absolute Galois group would be abelian, and thus every finite Galois group over $\mathbb{Q}$ would be abelian, and this is certainly not the case.)
| 13 | https://mathoverflow.net/users/1149 | 13771 | 9,289 |
https://mathoverflow.net/questions/13774 | 6 | It is well known that a smooth cubic surface $X\subset \mathbb{P}^3$ has exactly 27 lines in it. Furthermore, it is easy to check that Picard group $$Pic(X)\cong \mathbb{Z}^7$$ Here the generators are lines which are $\mathbb{P}^1$ topologically. Furthermore, it is easy to check that $$\chi\_{top}(X)=2+H\_2(X)=9$$Topologically speaking, notice that as smooth manifold $X$ has no 1-skeleton. This makes the 2-dimensional cells glue to points along their bounday, getting spheres $\mathbb{P}^1$ as the result of this gluing process.
*My first question is **why the other 20 lines do not contribute to the Euler characteristic of $X$.***
Going further, if $X\subset \mathbb{P}^3$ has degree 4, it is known that $X$ sometimes has lines, sometimes it does not. However, $$\chi\_{top}(X)=2+H\_2(X)=24$$ meaning that, despite the fact that $X$ can perfectly have no lines, we still have homology $H\_2$ which are spheres topologically! meaning, there are in fact spheres (due to the argument above which says that $X$ has no 1-skeleton). Besides $\chi\_{top}$ is constant even though $X$ may have $64$, $32$ or even $0$ number of lines in it. There are spheres whose existence is not being noticed by $\chi\_{top}$ at all. Here let me be vague please. What is going on!?
Any type of editing to make this clearer will be welcome.
References highly appreciated.
| https://mathoverflow.net/users/1547 | Interaction of topology and the Picard group of Algebraic surfaces | Even if there is no lines or rational curves, there can still be spheres representing classes in $H^2$. There is no contradiction here since these spheres are maps from $S^2$ to the surface which are not necessarily holomorphic. Some of the spheres may be representable by holomorphic maps. Counting how many in each homology class (sometimes in some generalized sense) is a very interesting problem in enumerative geometry, which is connected to the theory of Mirror Symmetry.
| 5 | https://mathoverflow.net/users/3696 | 13779 | 9,295 |
https://mathoverflow.net/questions/13770 | 9 | It is a theorem of Solovay that any stationary subset of a regular cardinal, $\kappa$ can be decomposed into a disjoint union of $\kappa$ many disjoint stationary sets. As far as I know, the proof requires the axiom of choice. But is there some way to get a model, for instance a canonical inner model, in which ZF + $\neg $C holds and Solovay's Theorem fails?
I am interested in this problem because Solovay's theorem can be used to prove the Kunen inconsistency, that is, that there is no elementary embedding j:V -->V, where j is allowed to be any class, under GBC. The Kunen inconsistency may be viewed as an upper bound on the hierarchy of large cardinals. Without choice, no one has yet proven the Kunen inconsistency (although it can be proven without choice if we restrict ourselves to definable j). So if there is hope of proving Solovay's Theorem without choice, we could use this to prove the Kunen inconsistency without choice.
| https://mathoverflow.net/users/3183 | Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? | The Axiom of Determinacy (AD) implies that the club filter on $\omega\_1$ (the subsets of $\omega\_1$ containing a club) is an ultrafilter. Certainly if that is the case then we can't even decompose $\omega\_1$ into two disjoint stationary sets, because one of them would have to contain a club. Assuming sufficient large cardinal hypotheses (infinitely many Woodin cardinals and a measurable cardinal above them) one has that $L(\mathbb{R})$ satisfies AD, and hence that is a canonical inner model of the form I think you are looking for.
What about above $\omega\_1$? I believe it is a theorem of John Steel's that (again under large cardinal assumptions) in $L(\mathbb{R})$ for any regular $\kappa$ below $\Theta$, the $\omega$-club filter on $\kappa$ is an ultrafilter. (An $\omega$-club is an unbounded set closed under countable limits). So for such $\kappa$ the stationary set of ordinals of countable cofinality cannot be partitioned into two disjoint stationary sets. I don't know about getting Solovay's theorem to fail at cardinals higher than that.
Also, that all assumes some large cardinals. I do not know if large cardinal assumptions are necessary to get the failure of Solovay's Theorem.
| 9 | https://mathoverflow.net/users/2436 | 13783 | 9,297 |
https://mathoverflow.net/questions/13768 | 33 | This is a rather technical question with no particular importance in any case of actual interest to me, but I've been writing up some notes on commutative algebra and flailing on this point for some time now, so I might as well ask here and get it cleared up.
I would like to define the Picard group of an arbitrary (i.e., not necessarily Noetherian) commutative ring $R$. Here are two possible definitions:
(1) It is the group of isomorphism classes of rank one projective $R$-modules under the
tensor product.
(2) It is the group of isomorphism classes of invertible $R$-modules under the tensor product, where invertible means any of the following equivalent things [Eisenbud, Thm. 11.6]:
a) The canonical map $T: M \otimes\_R \operatorname{Hom}\_R(M,R) \rightarrow R$ is an isomorphism.
b) $M$ is locally free of rank $1$ [**edit**: in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M\_{\mathfrak{p}} \cong R\_{\mathfrak{p}}$.]
c) $M$ is isomorphic as a module to an invertible fractional ideal.
What's the difference between (1) and (2)? In general, (1) is stronger than (2), because projective modules are locally free, whereas a finitely generated locally free module is projective iff it is finitely presented. (When $R$ is Noetherian, finitely generated and finitely presented are equivalent, so there is no problem in this case. This makes the entire discussion somewhat academic.)
So, *a priori*, if over a non-Noetherian ring one used (1), one would get a Picard group that was "too small". Does anyone know an actual example where the groups formed in this way are not isomorphic? (That's stronger than one being a proper subgroup of the other, I know.)
Why is definition (2) preferred over definition (1)?
| https://mathoverflow.net/users/1149 | What is the right definition of the Picard group of a commutative ring? | For what it's worth, I think in Bourbaki's Algèbre Commutative, this is chapter II, section 5.4 (or so), but I don't have a copy in front of me. (Pete confirms that it's II.5.4, Theorem 3.)
| 18 | https://mathoverflow.net/users/3049 | 13787 | 9,300 |
https://mathoverflow.net/questions/7853 | 14 | This is a clarification of [another post of mine](https://mathoverflow.net/questions/7817/normal-coordinates-for-a-manifold-with-volume-form).
Fix $n$ a positive integer. Let $SL(n)$ have its usual matrix representation, so that it really is the codimension-one subset of $M(n) = \mathbb R^{n^2}$ cut out by the degree-$n$ condition that the determinant is $1$. So we have $n^2$ coordinate functions $A^i\_j$ on $SL(n)$, $i,j = 1,\dots,n$.
Let $U$ be a domain in $\mathbb R^n$, with coordinates $x\_1,\dots,x\_n$. Consider the set $\mathcal S$ of smooth functions $f: U \to SL(n)$ satisfying the differential equation $\frac{\partial f^i\_j}{\partial x^k} = \frac{\partial f^i\_k}{\partial x^j}$ for each $i,j,k = 1,\dots,n$ (of course, $f^i\_j = A^i\_j \circ f$ is the $(i,j)$th coordinate of $f$).
(Why would you care about $\mathcal S$? Because a smooth map $g: U \to \mathbb R^n$ is volume-preserving if and only if $\frac{\partial g^i}{\partial x^j} \in \mathcal S$, and every element of $\mathcal S$ arises this way; indeed, $\mathcal S$ is the space of volume-preserving maps up to translations.)
Let's agree that a *smooth path* in $\mathcal S$ is a smooth function $F: [0,1] \times U \to SL(n)$ such that for each $t\in [0,1]$, $F(t,-) \in \mathcal S$.
**Question:** Is $\mathcal S$ smooth-path-connected? I.e. given $f\_0, f\_1 \in \mathcal S$, does there exist a smooth path $F$ so that $F(0,-) = f\_0$ and $F(1,-) = f\_1$?
If the answer is "no" in general, is it "yes" for sufficiently nice domains $U$ (contractible, say, or with compact closure and require that each $f\in \mathcal S$ extend smoothly to a neighborhood of the closure, or...)?
| https://mathoverflow.net/users/78 | Is the space of volume-preserving maps path-connected? | The answer is yes, on a contractible domain $U$. Suppose that $f^i\_j$ satisfies $\partial\_k f^i\_j = \partial\_j f^i\_k = g^i\_{jk}$. Then $g$ satisfies $g^i\_{jk} = g^i\_{kj}$ and $g^i\_{ik} = 0$. Conversely, if $g^i\_{jk} = g^i\_{kj}$ and $g^i\_{ik} = 0$, then since $U$ is contractible, there exists $f^i\_j$ with $\partial\_kf^i\_j = g^i\_{jk}$. Indeed, pick a basepoint $u \in U$; then $f$ is determined by $g$ and $f(u)$.
However, the conditions on $g$ are linear. Let $f\_1,f\_2$ be two matrix-valued functions with $\partial\_k{f\_1}^i\_j = \partial\_j{f\_1}^i\_k$ and $\partial\_k{f\_2}^i\_j = \partial\_j{f\_2}^i\_k$. Then set ${g\_a}^i\_{jk} = \partial\_k{f\_a}^i\_j$ for $a=1,2$. Introduce a time variable $t$, and let $G^i\_{jk}(t) = (1-t){g\_1}^i\_{jk} + t{g\_2}^i\_{jk}$. Then for each $t$, $G(t)$ satisfies the integrability conditions that $g$ did in the previous paragraph. For the basepoint $u\in U$, let $F(u,t)^i\_j$ be any smooth path connecting the values $F(u,0) = f\_1(u)$, $F(u,1) = f\_2(u)$. Then we can find extend $F(u,-)$ to $F(-,-)$ satisfying $\partial\_k F(t)^i\_j = G(t)^i\_{jk}$. And this does the job.
| 0 | https://mathoverflow.net/users/78 | 13791 | 9,303 |
https://mathoverflow.net/questions/11082 | 10 | Let $\mathbb K$ be a field (of characteristic 0, say), $\mathfrak g$ a [Lie bialgebra](http://en.wikipedia.org/wiki/Lie_bialgebra) over $\mathbb K$, and $\mathcal U \mathfrak g$ its usual universal enveloping algebra. Then the coalgebra structure on $\mathfrak g$ is equivalent to a co-Poisson structure on $\mathcal U \mathfrak g$, i.e. a map $\hat\delta : \mathcal U \mathfrak g \to (\mathcal U \mathfrak g)^{\otimes 2}$ satisfying some axioms. A *formal quantization* of $g$ is a Hopf algebra $\mathcal U\_\hbar \mathfrak g$ over $\mathbb K[[\hbar]]$ (topologically free as a $\mathbb K[[\hbar]]$-module) that deforms $\mathcal U \mathfrak g$, in the sense that it comes with an isomorphism $\mathcal U\_\hbar \mathfrak g / \hbar \mathcal U\_\hbar \mathfrak g \cong \mathcal U \mathfrak g$, and moreover that deforms the comultiplication in the direction of $\hat\delta$: $$\Delta = \Delta\_0 + \hbar \hat\delta + O(\hbar^2),$$ where $\Delta$ is the comultiplication on $\mathcal U\_\hbar \mathfrak g$ and $\Delta\_0$ is the (trivial, i.e. which $\mathfrak g$ is primitive) comultiplication on $\mathcal U\mathfrak g$. This makes precise the "classical limit" criterion: "$\lim\_{\hbar \to 0} \mathcal U\_\hbar \mathfrak g = \mathcal U \mathfrak g$"
I am wondering about "the other" classical limit of $\mathcal U\_\hbar \mathfrak g$. Recall that $\mathcal U\mathfrak g$ is [filtered](http://en.wikipedia.org/wiki/Filtered_algebra) by declaring that $\mathbb K \hookrightarrow \mathcal U\mathfrak g$ has degree $0$ and that $\mathfrak g \hookrightarrow \mathcal U\mathfrak g$ has degree $\leq 1$ (this generates $\mathcal U\mathfrak g$, and so defines the filtration on everything). Then the associated graded algebra of $\mathcal U\mathfrak g$ is the symmetric (i.e. polynomial) algebra $\mathcal S\mathfrak g$. On the other hand, the Lie structure on $\mathfrak g$ induces a Poisson structure on $\mathcal S\mathfrak g$, one should understand $\mathcal U \mathfrak g$ as a "quantization" of $\mathcal S\mathfrak g$ in the direction of the Poisson structure. Alternately, let $k$ range over non-zero elements of $\mathbb K$, and consider the endomorphism of $\mathfrak g$ given by multiplication by $k$. Then for $x,y \in \mathfrak g$, we have $[kx,ky] = k(k[x,y])$. Let $\mathfrak g\_k$ be $\mathfrak g$ with $[,]\\_k = k[,]$. Then $\lim\\_{k\to 0} \mathcal U\mathfrak g\_k = \mathcal S\mathfrak g$ with the desired Poisson structure.
I know that there are functorial quantizations of Lie bialgebras, and these quantizations give rise to the Drinfeld-Jimbo quantum groups. So presumably I can just stick $\mathfrak g\_k$ into one of these, and watch what happens, but these functors are hard to compute with, in the sense that I don't know any of them explicitly. So:
>
> How should I understand the "other" classical limit of $\mathcal U\_\hbar \mathfrak g$, the one that gives a commutative (but not cocommutative) algebra?
>
>
>
If there is any order to the world, in the finite-dimensional case it should give the dual to $\mathcal U(\mathfrak g^\\*)$, where $\mathfrak g^\\*$ is the Lie algebra with bracket given by the Lie cobracket on $\mathfrak g$. Indeed, B. Enriquez has a series of papers (which I'm in the process of reading) with abstracts like "functorial quantization that respects duals and doubles".
On answer that does not work: there is no non-trivial *filtered* $\hbar$-formal deformation of $\mathcal U\mathfrak g$. If you demand that the comultiplication $\Delta$ respect the filtration on $\mathcal U\mathfrak g \otimes \mathbb K[[\hbar]]$ and that $\Delta = \Delta\_0 + O(\hbar)$, then the coassociativity constraints imply that $\Delta = \Delta\_0$.
This makes it hard to do the $\mathfrak g \mapsto \mathfrak g\_k$ trick, as well. The most naive thing gives terms of degree $k^{-1}$ in the description of the comultiplication.
| https://mathoverflow.net/users/78 | The other classical limit of a quantum enveloping algebra? | The answer is essentially given in Kassel and Turaev, "Biquantization of Lie bialgebras", Pacific Journal of Mathematics, 2000 vol. 195 (2) pp. 297-369, [MR1782170](http://www.ams.org/mathscinet-getitem?mr=1782170). They do the following: To a finite-dimensional Lie bialgebra $\mathfrak g$ over $\mathbb C$, they define a biassociative bialgebra $A\_{u,v}(\mathfrak g)$, (topologically) free over $\mathbb C[u][[v]]$, such that:
1. $A\_{u,v}(\mathfrak g)$ is commutative module $u$ and cocommutative modulo $v$.
2. $A\_{u,v}(\mathfrak g) / (u,v) = \mathcal S\mathfrak g$, the symmetric algebra, with its induced Poisson and co-Poisson structures.
3. $A\_{u,v}(\mathfrak g) / (u)$ is a commutative Poisson bialgebra and its cobracket quantizes $\mathcal S(\mathfrak g)$ in the co-Poisson direction.
4. $A\_{u,v}(\mathfrak g) / (v)$ is a cocommutative co-Poisson bialgebra and its bracket quantizes $\mathcal S(\mathfrak g)$ in the Poisson direction. Indeed, $A\_{u,v}(\mathfrak g) / (v,u-1) = \mathcal U\mathfrak g$.
5. $A\_{u,v}(\mathfrak g)$ is essentially dual to $A\_{v,u}(\mathfrak g^\*)$.
Thus the Etingof-Kazhdan quantization is $A\_{u,v}(\mathfrak g) / (v-\hbar,u-1)$. More generally, we have $\hbar = uv$.
I still don't know if there's some way to look at this construction without variables $u,v$, and rather with (co-)filtrations. But, then again, I've only read the introduction to the paper. I also don't know if it works over other fields. Given that Kassel and Turaev rely on the Etingof-Kazhdan methods, which in turn rely on a Drinfeld associator, I assume that the method requires working over a $\mathbb Q$-algebra.
| 3 | https://mathoverflow.net/users/78 | 13794 | 9,305 |
https://mathoverflow.net/questions/13773 | 7 | Suppose $\kappa\_0$ is a measurable cardinal and $\mu\_0$ is a normal measure on $\kappa\_0$. $M\_1$ is the transitive collapse of $Ult(V,\mu\_0)$, $j\_{0,1}:V\rightarrow{M\_1}$ is the elementary embedding induced by the ultrapower. In $M\_1$, $\kappa\_1=j\_{0,1}(\kappa\_0)$ is a measurable cardinal and $\mu\_1$ is a normal measure on $\kappa\_1$ in $M\_1$ such that $\mu\_1$ is not in the range of $j\_{0,1}$. $M\_2$ is the transitive collapse of $Ult(M\_1,\mu\_1)$, $j\_{1,2}:M\_1\rightarrow{M\_2}$ is the elementary embedding induced by the ultrapower. $j\_{0,2}=j\_{1,2}\circ{j\_{0,1}}$.
Is it true that: ``Suppose $N$ is an inner model, $i:V\rightarrow{N}$ and $k:N\rightarrow{M\_2}$ are elementary embeddings such that $k\circ{i}=j\_{0,2}$. Then $k''N=j\_{0,2}''V$ or $k''N=j\_{1,2}''M\_1$ or $k''N=M\_2$''?
| https://mathoverflow.net/users/3692 | A question on ultrapower | This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ on a measurable cardinal κ is uniquely factored by the steps of the iteration itself.
The answer is **Yes**.
Let me denote κ0 just by κ and j02 by j. Since
μ1 is a measure in M1, it has the
form j01(m)(κ), where m =
(να | α < κ). Since you
have said that μ1 is not in
ran(j01), we may choose the
να to be all different, and different
from μ0. In this case, there is a partition
of κ as the disjoint union of Xα,
with Xα in να and none
in μ0. Let x = (Xα | α
< κ). Note that κ is not in
j01(Xα) for any α <
κ, and similarly κ1 is not in
j(Xα). But κ is in
j01(x)(β) for some β <
κ1, since this is a partition of
κ1. Apply j12 to conclude that
κ1 is in j(x)(β) for this β.
Thus, there is some β in the interval [κ,
κ1) having the form β =
j(f)(κ1) for the function f that picks the
index. From this, it follows from normality of
μ0 that we can write κ =
j(g)(κ1) for some function g, since any
β < κ1 generates κ via
j01. In my favored terminology, the seed
κ1 generates κ via j and in fact
generates all β in [κ,κ1) via
j.
Similarly, suppose that δ is in the interval
[κ1,j(κ)). We know δ =
j12(f)(κ1) for some function f
on κ1 in M1. We also know f =
j01(F)(κ) for some F in V. Thus, δ =
j(F)(κ, κ1). In Y, let
(α,β) be the smallest pair with δ =
j(F)(α,β). It cannot be that both are below
κ1, since this would be inside
ran(j12) and so the least pair must have β
= κ1. Thus, δ generates
κ1, which we already observed generates
κ.
To summarize, every ordinal in the interval
[κ1,j(κ)) generates
κ1, which generates all the ordinals
β in [κ,κ1), any of which
generate κ and all the other such β.
This is enough to answer your question. The k " N in your
question is just an arbitrary elementary substructure of
M2 containing ran(j), so suppose we have Y
elementary in M2 and ran(j) subset Y. The case Y
= ran(j) is one of your cases. Otherwise, Y has something
not in ran(j). Every object in M2 has form
j(h)(κ,κ1) for some function h, so
by looking at the smallest pair of ordinals to generate a
given object with j(h), we see that there must be ordinals
below j(κ) in Y. If Y contains any ordinal δ in
the interval [κ1,j(κ)), then it will
contain both κ and κ1, since we
observed that any such δ generates these ordinals. In
this case, Y = M2, since those two ordinals
generate everything. So we assume that Y contains no such
δ. In this last case, Y must contain some ordinal
β in the interval [κ,κ1). Since
any such β generates κ, Y contains all such
ordinals. It follows that ran(j12) subset Y and
in fact = Y, since if Y contained anything more it would
have to have an additional ordinal δ in
[κ1,j(κ)).
So we've seen that your three cases are the only
possibilities. And like your previous question, there is no
need to assume that Y or N is somehow internally definable.
By the way, this was a problem that I had solved many years
ago for my dissertation, although perhaps other people had
also thought about it. I was interested in understanding
which pairs of ordinals (α,β) generate product
measures via an embedding j, and this question is very much
related to that.
(Click the edit history to see my previous answer, which was just about the case when μ1 is in the range of j01, a case for which the answer is no.)
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