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https://mathoverflow.net/questions/10413 | 26 | Denote Zermelo Fraenkel set theory without choice by ZF. Is the following true: In ZF, every definable non empty class A has a definable member; i.e. for every class $A = \lbrace x : \phi(x)\rbrace$ for which ZF proves "A is non empty", there is a class $a = \lbrace x : \psi(x)\rbrace$ such that ZF proves "a belongs to A"?
| https://mathoverflow.net/users/2689 | Definable collections without definable members (in ZF) | **Update.** (June, 2017) François Dorais and I have completed a paper that ultimately grew out of this questions and several follow-up questions.
>
> F. G. Dorais and J. D. Hamkins, [When does every definable nonempty set have a definable element?](http://jdh.hamkins.org/when-does-every-definable-nonempty-set-have-a-definable-element/) ([arχiv:1706.07285](https://arxiv.org/abs/1706.07285))
>
>
>
> > ***Abstract.** The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi\_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma\_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$.*
>
>
> Read more [at the blog post](http://jdh.hamkins.org/when-does-every-definable-nonempty-set-have-a-definable-element/).
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>
>
Click on the history to see the original answer. Related questions and answers appear at:
* [Can $V\neq\HOD$, if every $\Sigma\_2$-definable set has an ordinal-definable element?](https://mathoverflow.net/q/180850/1946)
* [Is it consistent with ZFC (or ZF) that every definable family of sets has at least one definable member?](https://mathoverflow.net/q/180734/1946)
* We also make use of the $\Sigma\_2$ conception of [Local properties in set theory](https://jdh.hamkins.org/local-properties-in-set-theory).
| 30 | https://mathoverflow.net/users/1946 | 10415 | 7,108 |
https://mathoverflow.net/questions/10404 | 13 | Let C be a stable ∞-category in the sense of Lurie's DAG I. (In particular I do not assume that C has all colimits.) Then C *does* have all finite colimits, the suspension functor on C is an equivalence, and C is enriched in Spectra in a way I don't want to make too precise (basically the Hom functor Cop × C → Spaces factors through Spectra and there are composition maps on the level of spectra).
Now suppose instead that C is an ∞-category which has all finite colimits and comes equipped with an enrichment in Spectra in the above sense. One can show easily that C then has a zero object which allows us to define a suspension on C. Suppose it is an equivalence. Is C then a stable ∞-category? Moreover, is the enrichment on C the one which comes from the fact that it is a stable ∞-category?
| https://mathoverflow.net/users/126667 | Stable ∞-categories as spectral categories | According to Corollary 8.28 in DAG I a pointed $\infty$-category is stable iff it has finite colimits and the suspension functor is an equivalence.
| 8 | https://mathoverflow.net/users/1100 | 10434 | 7,124 |
https://mathoverflow.net/questions/10433 | 6 | Inspired by the two solutions to Harry's question
[Can a topos ever be an abelian category?](https://mathoverflow.net/questions/10290/can-a-topos-ever-be-an-abelian-category)
I was wondering whether all coproducts of 1 in a topos are distinct up to isomorphism? That is $1 + 1 + \dots + 1 \cong 1 + 1 + \dots + 1$ iff there are an equal number of 1s on each side?
**Edit:** In order to make the question (possibly) non-trivial, let's assume that the topos is not equivalent to the terminal category.
| https://mathoverflow.net/users/1106 | Are all coproducts of 1 in a topos distinct ? | At least if you're talking about *finite* coproducts, then the answer is yes. If $n\le m$, then we have a canonical inclusion $\sum\_{i=1}^n 1 \hookrightarrow \sum\_{j=1}^m 1$, which is in fact a complemented subobject with complement $\sum\_{k=1}^{m-n} 1$. If this inclusion is an isomorphism, then its complement is initial, and hence (assuming the topos is nontrivial) $n=m$. Now if we have an arbitrary isomorphism $\sum\_{i=1}^n 1 \cong \sum\_{j=1}^m 1$, then composing with the above inclusion we get a monic $\sum\_{i=1}^m 1 \hookrightarrow \sum\_{j=1}^m 1$. However, one can show by induction that any finite coproduct of copies of $1$ in a topos is Dedekind-finite, i.e. any monic from it to itself is an isomorphism. (See D5.2.9 in "Sketches of an Elephant" vol 2.) Thus, the standard inclusion is also an isomorphism, so again $n=m$.
| 8 | https://mathoverflow.net/users/49 | 10443 | 7,132 |
https://mathoverflow.net/questions/10408 | 28 | This looks like a statement from a calculus textbook, which perhaps it should be.
"Rolle's theorem". Let $F\colon [a,b]\to\mathbb R^n$ be a continuous function such that $F(a)=F(b)$ and $F'(t)$ exists for all $a<t<b$. Then there exist numbers $a < t\_1 < t\_2 < \dots < t\_n < b$ such that the vectors $F'(t\_1),\dots,F'(t\_{n})$ are linearly dependent.
We are all familiar with the case $n=1$. The case $n=2$ is not hard either: pick any $a<t^\ast<b$ and find, using the mean value theorem, numbers $a<t\_1<t^\ast$ and $t^\*<t\_2<b$ such that $F'(t\_j)$ is collinear with $F(t^\ast)-F(a)$. Note that we avoided using the parameter value $t^\ast$, which will be important in a moment. When $n=3$, we pick $a<t^\ast<b$ and project F onto the orthogonal complement of $F'(t^\ast)$, then apply the case $n=2$ to the projection ($t^\ast$ will become the third chosen parameter value). So far so good.
But I get stuck at $n=4$. If the above process is followed, then after $F$ is projected down to two dimensions, we must avoid two particular parameter values. Which is not possible in general: if in two dimensions $F$ parametrizes a triangle and $F(a)$ is a vertex, then one of points $F(t\_j)$ must be one of two other vertices. Presumably this problem can avoided by a generic choice of points of projection, but how?
| https://mathoverflow.net/users/2912 | Rolle's theorem in n dimensions | Here is a solution in the $C^1$ case [but see upd]. Suppose the vectors $F'(t\_1),\ldots, F'(t\_n)$ are linearly independent for all $0\leq t\_1< \cdots < t\_n\leq 1$. Let $L(t\_1,\ldots,t\_{n-1})$ be vector space spanned by the first $n-1$ of these. Let $$t^i=(t\_1^i,\ldots,t\_{n-1}^i),0\leq t^i\_1< \cdots < t^i\_{n-1}\leq 1$$ be a sequence such that $t^i\_{n-1}\to 0$ as $i\to \infty$. Since the space of all vector hyperplanes in $\mathbf{R}^n$ is compact, we can assume the sequence $L(t^i)$ has a limit $L$.
Claim: for any $s,t$ such that $0< s < t\leq 1$ the hyperplane $L$ does not separate $F'(s)$ and $F'(t)$.
Proof of the claim: if $L$ does, then for all sufficiently large $i$ the hyperplane $L(t^i)$ also separates $F'(s)$ and $F'(t)$. Choose $i$ so that moreover $t^i\_{n-1} < s$. Then the determinants of the matrices $(F'(t^i\_1),\ldots, F'(t^i\_{n-1}),F'(s))$ and $(F'(t^i\_1),\ldots, F'(t^i\_{n-1}),F'(t))$ have different signs. This is impossible, since $F'$ is continuous and the determinant is never zero. Claim is proven.
Now one of the two things can happen: either all $F'(t),0 \leq t\leq 1$ are in $L$, in which case $F'(t\_1),\ldots, F'(t\_n)$ are linearly dependent for all $t\_1,\ldots, t\_n$, or there is a $t$ such that for $l(F'(t))\neq 0$ where $l$ is a linear equation defining $L$. Say, $l(F'(t)) > 0$. Then $l(F(0)) < l(F(1))$, so we can't have $F(0)=F(1)$.
upd: here is how one can take care of the case when $F'$ is not assumed continuous. Basically, the only thing that changes is the proof of the claim; the claim itself remains the same except that we assume $t < 1$. Choose $i$ as above and set $g(x),s\leq x\leq t$ to be the determinant of $(F'(t^i\_1),\ldots,F'(t^i\_{n-1}),F(x))$. This function is differentiable and we have $g'(x)=det(F'(t^i\_1),\ldots,F'(t^i\_{n-1}),F'(x))$. So $g'(s)$ and $g'(t)$ have different signs. The claim follows now from the following statement, which is a consequence of the classical Rolle's theorem: if $f:[a,b]\to\mathbf{R}$ is differentiable at each point of $[a,b]$ and $f'(a)$ and $f'(b)$ have different signs, then there is an $x\in (a,b)$ such that $f'(x)=0$.
Then we deduce from the claim that all $F'(t),0 < t < 1$ are in the same half-space with respect to $L$. This suffices.
| 16 | https://mathoverflow.net/users/2349 | 10445 | 7,133 |
https://mathoverflow.net/questions/10409 | 23 | Does anyone know of an introduction and motivation for [W-algebras](https://en.wikipedia.org/wiki/W-algebra)?
Edit: Okay, sorry I try to add some more background. W algebras occur, for example when you study nilpotent orbits: Take a nice algebraic/Lie group. It acts on its Lie-algebra by the adjoint action. Fix a nilpotent element e and make a sl\_2 triple out of it. A W algebra is some modification of the universal enveloping, based on this data.
A precise definition is for example given here: <https://arxiv.org/pdf/0707.3108>.
But this definition looks quite complicated and not very natural to me. I don't see what's going on. Therefor I wonder if there exists an easier introduction.
| https://mathoverflow.net/users/2837 | Introduction to W-Algebras/Why W-algebras? | W-algebras appear in at least three interrelated contexts.
1. **Integrable hierarchies**, as in the article by Leonid Dickey that mathphysicist mentions in his/her answer. Integrable PDEs like the KdV equation are bihamiltonian, meaning that the equations of motion can be written in hamiltonian form with respect to two different Poisson structures. One of the Poisson structures is constant, whereas the other (the so-called *second Gelfand-Dickey bracket*) defines a so-called *classical W-algebra*. For the KdV equation it is the Virasoro Lie algebra, but for Boussinesq and higher-order reductions of the KP hierarchy one gets more complicated Poisson algebras.
2. **Drinfeld-Sokolov reduction**, for which you might wish to take a look at the work of Edward Frenkel in the early 1990s. This gives a homological construction of the classical W-algebras starting from an affine Lie algebra and a nilpotent element. You can also construct so-called *finite W-algebras* in this way, by starting with a finite-dimensional simple Lie algebra and a nilpotent element. The original paper is [this one by de Boer and Tjin](https://arxiv.org/abs/hep-th/9211109). A **lot** of work is going on right on on finite W-algebras. You might wish to check out the work of Premet.
3. **Conformal field theory**. This is perhaps the original context and certainly the one that gave them their name. This stems from [this paper of Zamolodchikov](https://www.ams.org/mathscinet-getitem?mr=829902). In this context, a W-algebra is a kind of vertex operator algebra: the vertex operator algebra generated by the Virasoro vector together with a finite number of primary fields. A review about this aspect of W-algebras can be found in this [report by Bouwknegt and Schoutens](https://www.ams.org/mathscinet-getitem?mr=1208246).
There is a lot of literature on W-algebras, of which I know the mathematical physics literature the best. They had their hey-day in Physics around the late 1980s and early 1990s, when they offered a hope to classify rational conformal field theories with arbitrary values of the central charge. The motivation there came from string theory where you would like to have a good understanding of conformal field theories of $c=15$. The rational conformal field theories without extended symmetry only exist for $c<1$, whence to overcome this bound one had to introduce extra fields (à la Zamolodchikov). Lots of work on W-algebras (in the sense of **3**) happened during this time.
The emergence of matrix models for string theory around 1989-90 (i.e., applications of random matrix theory to string theory) focussed attention on the integrable hierarchies, whose $\tau$-functions are intimately related to the partition functions of the matrix model. This gave rise to lots of work on classical W-algebras (in the sense of **1** above) and also to the realisation that they could be constructed à la [Drinfeld-Sokolov](https://www.ams.org/mathscinet-getitem?mr=760998).
The main questions which remained concerned the *geometry* of W-algebras, by which one means a geometric realisation of W-algebras analogous to the way the Virasoro algebra is (the universal central extension of) the Lie algebra of vector fields on the circle, and the representation theory. I suppose it's this latter question which motivates much of the present-day W-algebraic research in Algebra.
**Added**
In case you are wondering, the etymology is pretty prosaic. Zamolodchikov's first example was an operator vertex algebra generated by the Virasoro vector and a primary weight field $W$ of weight 3. People started referring to this as *Zamolodchikov's $W\_3$ algebra* and the rest, as they say, is history.
**Added later**
Ben's answer motivates the study of finite W-algebras from geometric representation theory and points out that a finite W-algebra can be viewed as the quantisation of a particular Poisson reduction of the dual of the Lie algebra with the standard Kirillov Poisson structure. The construction I mentioned above is in some sense doing this in the opposite order: you first quantise the Kirillov Poisson structure and then you take *BRST cohomology*, which is the quantum analogue of Poisson reduction.
| 22 | https://mathoverflow.net/users/394 | 10448 | 7,136 |
https://mathoverflow.net/questions/10400 | 3 | Although the question is easy to pose, I think some background will help to motivate it, so I'll start with it.
Consider variables $X=(X\_1, \ldots, X\_n)$ over a field $K$ and the elementary symmetric functions $T=(T\_1, \ldots, T\_n)$ in $X$. In other words $X$ are the roots of the polynomial $Y^n + T\_1 Y^{n-1} + \cdots + T\_n$.
A polynomial $f$ in $X$ is symmetric is $f(s X) = f(X)$ for any permutation $s$. Here $s X := (X\_{s(1)}, \ldots, X\_{s(n)})$. Then a basic fact is that if $f(X)$ is symmetric, then $f(X) = g(T)$, for some polynomial $g$.
It is reasonable to define an alternating polynomial to be $f$ that satisfy $f(s X) = sign(s) f(X)$, where $sign(s) = \pm 1$ is the signature. The "elementary" alternating polynomial is the Vandermonde polynomial $V(X) = \prod\_{i<j} (X\_j-X\_i)$, and any other alternating polynomial can be expressed as a polynomial in $T$ and $V$.
Note that $V$ is a square root of the discriminant $\Delta$ of $Y^n + T\_1 Y^{n-1} + \cdots + T\_n$ and the discriminant has an explicit formula in terms of $T$ using the Sylvester matrix.
That definition for alternating polynomials gives nothing interesting in characteristic $2$ (because then $1=-1$). The only definition that makes sense to me in characteristic $2$ is: $f$ is alternating if $f(s X) = f(X) + add.sign(s)$. Here $add.sign(s) = 0,1$ is the additive signature, i.e., equals $1$ if $s$ is odd and $0$ if $s$ is even.
I already figured out what is the "elementary" alternating polynomial $u/V$ and what is the Artin-Schreir equation it satisfies:
$u(X) = \sum\_{s \ {\rm is\ even}} X^{n-1}\_{s(1)} \cdots X^0\_{s(n)}$ and it satisfies the Artin-Schreier equation $X^2 + X = \frac{u(X) u(s\_0 X)}{\Delta}$, where $s\_0$ is any odd permutation (e.g., transposition), and $\Delta$ is again the discriminant. (Note that $u(X) + u(s\_0 X) = V$.)
My question is: Does there exist a nice formula for $\frac{u(X) u(s\_0 X)}{\Delta}$ in terms of $T$?
| https://mathoverflow.net/users/2042 | Explicit expression of an alternating polynomial in characteristic $2$? | (Here is a more detailed version of Felipe's answer.)
In 1976 Elwyn Berlekamp defined characteristic 2 analogues of the discriminant and its square root, related to the expressions you wrote down. Later these were related to what you get if you lift the original polynomial to characteristic 0, compute the usual discriminant, and reduce modulo 8; this should give you a formula, though maybe not a very nice one. Here are the references:
Berlekamp, E. R. An analog to the discriminant over fields of characteristic two. J. Algebra 38 (1976), no. 2, 315--317.
Revoy, Philippe Discriminant d'une extension s�parable d'anneaux. C. R. Acad. Sci. Paris S�r. I Math. 298 (1984), no. 7, 123--126.
Berg�, A.-M.; Martinet, J. Formes quadratiques et extensions en caract�ristique $2$. (French) [Quadratic forms and extensions in characteristic $2$] Ann. Inst. Fourier (Grenoble) 35 (1985), no. 2, 57--77.
Wadsworth, Adrian R. Discriminants in characteristic two. Linear and Multilinear Algebra 17 (1985), no. 3-4, 235--263.
Waterhouse, William C. Discriminants of �tale algebras and related structures. J. Reine Angew. Math. 379 (1987), 209--220.
| 6 | https://mathoverflow.net/users/2757 | 10451 | 7,138 |
https://mathoverflow.net/questions/10453 | 11 | Let $p$ be a prime. Suppose you have an Abelian scheme $A$ over $Spec\ \mathbb{Z}\_p$. How do you prove that if $q$ is another prime, then the $q$-torsion of $A$ injects into the torsion of $A\_p$, under the reduction map?
| https://mathoverflow.net/users/2938 | Torsion of an abelian variety under reduction. | Let me try again at an alternate answer. If $A\_{\mathbb{Z}\_p}$ is an abelian scheme of dimension $g$ and $\ell \neq p$ is a prime, then for any positive integer $n$, the isogeny $[\ell^n]: A \rightarrow A$ is an etale map. [If I am not mistaken, the proof of this does not require formal groups!] Since the special fiber has $\ell^{2gn}$ points over the algebraic closure, by Hensel's Lemma all of the $\ell^n$-torsion on $A$ is defined over the maximal unramified extension, and it follows that the reduction map over the maximal unramified extension is an isomorphism on the $\ell^n$-torsion, hence an injection over $\mathbb{Q}\_p$.
| 11 | https://mathoverflow.net/users/1149 | 10462 | 7,144 |
https://mathoverflow.net/questions/7679 | 7 | Let $K$ be a number field, and let $S\_x$ denote the set of primes of norm at most $x$. Is it possible to find a smaller set of places $T\_x\subset S\_x$ so that a lot of the solutions of the $S\_x$-unit equation $a+b=1$ for $a,b\in S\_x$ are solutions of the $T\_x$-unit equation?
Here's a possible precise statement (although I'd be interested in other formulations as well): Does there exist a constant $0<c<1$, depending on $K$ (but not $x$), so that for each $x$, there is a $T\_x\subset S\_x$ with $|T\_x|\le\sqrt{|S\_x|}$ so that the number of solutions to the $T\_x$-unit equation is at least $c$ times the number of solutions of the $S\_x$-unit equation?
I'm interested in this mostly by analogy: Belabas and Gangl have a bound for the set of places of a number field one must check in order to compute $K\_2$ of the ring of integers. It would be interesting to know if one could at least get a pretty good approximation for $K\_2$ by looking at a much smaller set of places.
| https://mathoverflow.net/users/2046 | S-unit equation and small sets of places | I think that the precise statement you ask for is false, but I have a feeling that current bounds on number of solutions to $S$-unit equations are not good enough to prove this, even for $K=\mathbf{Q}$.
What I can prove is that it is false if you require $T\_x$ to be the subset of the specified size consisting of the *smallest* primes of $S\_x$. (And it seems unlikely that using larger primes would be much better; this is why I think that the precise statement is false.)
**Proof:** Let me assume $K=\mathbf{Q}$ for simplicity. Let $f(x)$ be the number of solutions to the $S\_x$-unit equation. We have $T\_x=S\_y$ where $y=x^{1/2+o(1)}$, so if $c$ exists, we would have $f(x^{1/2+o(1)}) \ge c f(x)$ for all $x$. Iterating this shows that $f(x)$ is bounded by a polynomial in $\log x$ as $x \to \infty$. But $f(x) \ge x-1$ because of the solutions $a+(1-a)=1$ for $a=2,\ldots,x$. This is a contradiction.
| 6 | https://mathoverflow.net/users/2757 | 10464 | 7,145 |
https://mathoverflow.net/questions/10436 | 5 | Is there a theory of Newton-Puiseux type expansions
which works to parameterize singular surface germs $F(x,y,z) =0$?
Ideally, each branch would be the image of map of the form the
$x = u^m, y = v^n, z = \Sigma\_{i + j > n} a\_{i j} u^i v^j$
(after a linear change of the variables $x, y, z$).
| https://mathoverflow.net/users/2906 | Newton Puiseux expansions for singular surfaces? | I don't believe there is anything as general as that, but when your polynomial is over the complex numbers one can do the following. First shift and rotate coordinates so that you're working on a neighborhood of the origin where $F(0,0,0) = 0$ and $\partial\_z^n F(0,0,0) = 0$ for some $n$. Then you can use the Weierstrass preparation theorem and (ignoring a nonvanishing factor) write $F(x,y,z)$ as $z^n + a\_{n-1}(x,y)z^{n-1} + ... + a\_0(x,y)$. If the discriminant of this polynomial, viewed as a function of $x$ and $y$ has normal crossings (i.e. is a monomial times a nonvanishing factor), one can use the Jung-Abhyankar theorem to get a factorization of the form $(z - b\_1(x,y))...(z - b\_n(x,y))$ where the $b\_i$ are analytic in fractional powers of $x$ and $y$ as you want.
Even if the discriminant is not normal crossings, you can first resolve the singularities of the discriminant to make it normal crossings and there are a few ways to do this. One way involves subdividing the $x-y$ space into wedges, and on each wedge there is a way of first doing a coordinate change of the form $(x,y) --> (x,y - g(x))$ then a "blowing up" coordinate change $(x,y) = (x',(x')^m y')$ and get the discriminant as you want. ($m$ may not be an integer, and also you might have to reverse the roles of the $x$ and $y$ variables). Then one proceeds as above. Of course the final result isn't as nice as you wanted, but this is in the nature of things.
One can iterate the above to deal with analogues in higher dimensions, but the formulas get a lot more elaborate as one might guess. There's a paper by Parusinski "On the preparation theorem for subanalytic functions" that discusses a lot of these issues.
| 6 | https://mathoverflow.net/users/2944 | 10472 | 7,151 |
https://mathoverflow.net/questions/10116 | 7 | **Background**:
I am focusing on $G=GL\_{2}(\overline{\mathbb{F\_q}})$ here. If you wonder why I am interested in this, I am trying a problem relating to the Deligne-Lusztig varieties defined over local rings by Stasinksi, and this background theory is relevant there. The definition I am using is this:
1. The first definition I have of Deligne-Lusztig varieties is this: Consider the Lang map $L(g) = g^{-1} F(g)$. Let $T$ be an $F$-stable maximal torus of $G$, and $B$ a Borel subgroup containing $T$ (not necessarily $F$-stable), and $U$ the unipotent radical of this Borel subgroup. Then the Deligne-Lusztig variety is defined as $X = L^{-1}(U)$
**Question**
Roughly: Using these definitions (I am not sure exactly to what extent these two definitions are "compatible"), how do I explicitly compute the Deligne-Lusztig variety for $GL\_{2}(\mathbb{F}\_{q})$ to be the Drinfeldt curve $xy^{q} - yx^{q} = 1$ in the non-split case? More precisely:
1. If I pick a torus that is not maximally split, $T$, and a Borel subgroup $B$ containing $T$, then apply the definition $1$ above - how do we relate this variety to the Drinfeldt curve? Do we explicitly get the Drinfeldt curve, and if not what do we get? If it is not the Drinfeldt curve that we get using this definition, how do we express this variety "nicely" (with a view towards counting $F\_{q}$ points on it).
2. Roughly speaking (not explicitly!) how would I go about doing this for $GL\_{3}(\mathbb{F\_q})$? Is this computationally feasible? Are there any tricks that would help significantly with the computation? Even better, are there any references you know which do this?
3. I understand Deligne-Lusztig varieties, and these tori, correspond in some sense to Weyl group elements. Are there any specific Weyl group elements in $S\_n$ for which the Deligne-Lusztig variety always has a "nice" / "tractable" description, or do they get out of hand very quickly?
**My Attempts at (1)**
(I'm not entirely sure how to write matrices in Latex here, so I did it crudely by writing it as $4$ numbers, Row 1 followed by Row 2).
Pick $\alpha, \beta \in F\_{q^2}$, so that $(x- \alpha)(x- \beta)$ is irreducible over $F\_{q}$. Then a unipotent subgroup of a Borel subgroup for a non-split torus by conjugating the ordinary unipotent subgroup of strictly upper triangular matrices, by the matrix $M$ with entries $(1, 1, \alpha, \beta)$. This is because we can obtain the matrix with entries $(0, 1, -\alpha \beta, \alpha + \beta)$ (lying inside $GL\_{2}(\mathbb{F\_q})$ as $M^{-1} X M$, where $X$ is the matrix with entries $( \alpha, 0, 0, \beta)$. Since when we conjugate the nilpotent matrix with entries $(0, 1, 0, 0)$ by $M$ we get a scalar multiple of the matrix with entries $ ( - \alpha, 1, - \alpha^2, \alpha)$, the end result of this calculation is that our non-split maximal torus consists of matrices of the form $(1-s \alpha, s, - s \alpha^2, 1 + s \alpha)$.
Now if $g = (a,b,c,d)$ (i.e. the matrix with those 4 entries), then $g^{-1} = \frac{1}{ad-bc}( d, -b, -c, a)$, $F(g) = (a^q, b^q, c^q, d^q)$, and $g^{-1} F(g) = \frac{1}{ad-bc} (da^q - bc^q, db^q - bd^q, -ca^q + ac^q, -cb^q + ad^q)$, so equating that $g^{-1} F(g)$ lies in the subgroup calculated in the last paragraph gives the following description of the Deligne-Lusztig variety (let $D = ad-bc$), by comparing entry by entry:
* $ \frac{1}{D} ( da^q - bc^q -cb^q + ad^q ) = 2 $
* $ -ca^q + ac^q = - \alpha^2 (db^q - bd^q) $
This might be easy but I cannot see how to finish off from here: why is this related to the Drinfeldt curve? All I can see from a first glimpse is that the terms $db^q - bd^q$ appears, but I don't know how to get rid of everything else. How can I simplify the defining equations of this variety (hopefully in a suitably simple version, so that counting $F\_{q^{k}}$ points is straightforward, which is what I really need to do).
| https://mathoverflow.net/users/2623 | Explicit computations of small Deligne-Lusztig varieties (e.g. Drinfeld curve) | I found Teruyoshi Yoshida's exposition of the subject very helpful:
<http://www.dpmms.cam.ac.uk/~ty245/Yoshida_2003_introDL.pdf>
As JT commented, the curve you wrote down is really the Deligne-Lusztig variety for SL\_2, not GL\_2. Ben is also right about the curve being $\mathbf{P}^1 - \mathbf{P}^1(\mathbf{F}\_q)$, only he is using a different definition of DL variety from you I would presume. The way Ben has it, the DL variety is a subvariety of G/B, but the curve you want is a subvariety of G/U, where U is the unipotent radical. One formulation is a cover of the other with galois group equal to the rational points on a twist of the torus T. We'll take the G/U point of view here.
So let's start with $G=\text{SL}\_2$ over the field $\mathbf{F}\_q$. We'll let $B$ be the usual Borel and $U$ its unipotent radical. We can then identify $G/B$ with $\mathbf{P}^1$ and $G/U$ with $\mathbf{A}^2$. The latter identification sends $(a,b,c,d)$ to $(a,c)$.
Let $w=(0,1,1,0)$ be the nontrivial Weyl element. We let $X\_w$ be the subvariety of $G/B$ consisting of elements $x$ for which $x$ and $F(x)$ are in relative position $w$, where $F$ is the Frobenius map. This is $\mathbf{P}^1 - \mathbf{P}^1(\mathbf{F}\_q)$ as Ben says.
For cosets $x,y\in G/B$ in relative position $w$, and a coset $gU\in G/U$ for which $gB=x$, we are going to define a new coset $w\_{x,y}(gU)\in G/U$ as follows. First find a $g'\in G$ for which $g'B=x$ and $g'wB = y$. We may further take $g'$ so that $g'U=gU$. (This can be done because of the Bruhat decomposition of $G/B \times G/B$--wait a moment to see how this plays out for $\text{SL}\_2$.) Then define $w\_{x,y}(gU) = g'wU$. (Pardon the abuse of notation of the symbol $w$.) The Deligne-Lusztig variety $Y\_w$ is defined as the set of $gU\in G/U$ for which $F(gU)=w\_{gB,F(gB)}(gU)$.
When does a point $(x,y)\in\mathbf{A}^2=G/U$ lie in $Y\_w$? We need to calculate $w\_{gB,F(gB)}(gU)$, where $g=(x,\*,y,\*)\in G$. We have $gB=g\cdot\infty=x/y$ and $F(gB)=(x/y)^q$. So we must now find $g'\in G$ with $g'U=gU$ and $g'wB=F(g)wB$. The first condition means that $g'=(x,\*,y,\*)$ and the second means that $g'\cdot 0=(x/y)^q$. Thus $g'=(x,ux^q,y,uy^q)$, where $u$ must satisfy $u(xy^q-x^qy)=1$. We find that $w\_{gB,F(gB)}(gU)=g'wU=(ux^q,uy^q)$. The condition that $(x,y)\in Y\_w$ is exactly that $(x^q,y^q)=(ux^q,uy^q)$, which implies that $u^{-1}=x^qy-xy^q=1$. So that's the equation for the Deligne-Lusztig variety.
The equation for the DL variety for the longest cyclic permutation in the Weyl group of $\text{SL}\_n$ is $\det(x\_i^{q^j})=1$, where $0\leq i,j\leq n-1$.
I believe Lusztig calculated the zeta functions of his varieties in a very general setting, but I was never able to trudge through it all. There must be a simple answer for the behavior of the zeta functions for the $\text{GL}\_n$ varieties--if you ever write it up I'd certainly love to read it! I can start you off: for $\text{SL}\_2$ over $\mathbf{F}\_q$, the DL curve has a compactly supported $H^1$ of dimension $q(q-1)$, and the $q^2$-power Frobenius acts as the constant $-q$. (The behavior of the $q$-power Frobenius might be a little subtle--I suspect it has to do with Gauss sums.)
Good luck!
| 6 | https://mathoverflow.net/users/271 | 10477 | 7,154 |
https://mathoverflow.net/questions/10481 | 30 | I hope someone can point me to a quick definition of the following terminology.
I keep coming across *wild* and *tame* in the context of classification problems, often adorned with quotes, leading me to believe that the terms are perhaps not being used in a formal sense. Yet I am sure that there is some formal definition.
For example, the classification problem for nilpotent Lie algebras is said to be wild in dimension $\geq 7$. All that happens is that in dimension $7$ and above there are moduli. In what sense is this wild?
Thanks in advance!
| https://mathoverflow.net/users/394 | When is a classification problem "wild"? | I am not an expert but in the algebra and representation theory the apparently standard definition is [as follows](https://encyclopediaofmath.org/wiki/Representation_of_an_associative_algebra) (see also [Drozd - Tame and wild matrix problems](https://doi.org/10.1007/BFb0088467 "Drozd, J.A. (1980). Tame and wild matrix problems. In: Dlab, V., Gabriel, P. (eds) Representation Theory II. Lecture Notes in Mathematics, vol 832. Springer, Berlin, Heidelberg") and [Belitskii and Sergeichuk - Complexity of matrix problems](https://arxiv.org/abs/0709.2488v1 "Genrich R. Belitskii, Vladimir V. Sergeichuk. Complexity of matrix problems. 2007. doi:10.1016/S0024-3795(02)00391-9. zbMATH review at https://zbmath.org/?q=an:1030.15011")): a problem is wild if it contains a subproblem which is equivalent to the problem of simultaneously reducing to canonical form two linear operators on a finite-dimensional space.
| 26 | https://mathoverflow.net/users/2149 | 10484 | 7,158 |
https://mathoverflow.net/questions/10480 | 8 | In non-commutative geometry, Gelfand duality is the construction of multiplicative linear functionals of a commutative C\*-algebra, which can be viewed as the space of all its irreducible complex representations.
When encountered with a non-commutative C\*-algebra, we can speak of the space of pure states, which is the generalization of multiplicative linear functionals in commutative cases. The GNS construction can then be viewed as a map from the space of states to the space of all representations, and also from the space of pure states to the space of all irreducible representations.
My question is, is the "GNS map" surjective, or injective? What should be viewed as the non-commutative topological space, the space of all irreducible representations, or the space of pure states? And why?
| https://mathoverflow.net/users/2945 | Gelfand duality in NCG | On surjectivity: No, not every representation comes from a state; only the cyclic ones. Every nondegenerate representation of a C\*-algebra is a direct sum of cyclic representations (Zorn), and every cyclic representation comes from a GNS construction. But yes, every irreducible representation (which is also cyclic) comes from a GNS construction.
On injectivity: To even consider this question you probably want to identify representations if they are unitarily equivalent. But no, because for example two unit cyclic vectors in the Hilbert space of a representation will often yield different vector states, but these vector states will yield the same representation.
Which space? That would depend on your applications, and I don't have any absolute answers to that. Instead I'll tell you what comes to mind in the hope that it will help orient you. (Then I'll wait with you to be enlightened by another answer.)
The space of (unitary) equivalence classes of irreducible representations is often called the *spectrum* of a C\*-algebra; a closely related space is the set of kernels of irreducible representations, called the *primitive ideal space*. The primitive ideal space is given the hull kernel topology and is a quotient of the spectrum. I recommend [the book](http://books.google.com/books?id=KV_cC5uYA7EC&dq=morita+raeburn+williams&client=firefox-a&source=gbs_navlinks_s) by Raeburn and Williams; see Appendix A to learn about these spaces, and see the rest of the book for how they're used.
As for the space of pure states, I'm more accustomed to the idea of studying the space of all states, in which the pure states are the extreme points. There is a nice characterization of the state spaces of C\*-algebras in E. Alfsen, H. Hanche-Olsen and F.W. Shultz: State Spaces of C∗-Algebras, Acta Math. 144 (1980) 267–305, which came up at [this other question](https://mathoverflow.net/questions/4848/gelfand-naimark-from-the-category-theoretic-point-of-view). However, there is yet another option which I learned a little about from [Pedersen's book](http://books.google.com/books?id=yBCoAAAAIAAJ&q=gert+pedersen+automorphism+groups&dq=gert+pedersen+automorphism+groups&ei=ACk_S9-nO4m2NfGZ1bEB&client=firefox-a&cd=1), where the quasi-state space Q is used. A quasi-state is a positive functional of norm at most 1. (Other than 0, the extreme points of Q are also the pure states.) A C\*-algebra can be studied as affine functions on Q; see section 3.10 of Pedersen for details.
| 7 | https://mathoverflow.net/users/1119 | 10485 | 7,159 |
https://mathoverflow.net/questions/10488 | 4 | There are $\binom{n}{2}$ distances between $n$ points in $\mathbb{R}^d$. Not all of them can be chosen freely if $n$ exceeds the number $n\_d = d + 1$. If $n = n\_d$ we obviously have $\binom{d+1}{2}$ distances which can be chosen (more or less) independently (restricted only by the triangle inequality).
I see two ways to count $N\_n^d$, the number of independent distances between $n\geq d$ points in $\mathbb{R}^d$, which is given by $nd - \binom{d+1}{2}$
The first one: $nd$ coordinates minus one translation ($d$) minus one rotation ($\binom{d}{2}$):
$N\_n^d = nd - d - \binom{d}{2} = nd - (\binom{d}{2} + d) = nd - \binom{d+1}{2}$
The second one: $\binom{d+1}{2}$ distances between $d + 1$ *base points* plus $d\ (\ n - (d + 1)\ )$ distances between the remaining points and $d$ of the base points (the remaining one serving to say in which half-space with respect to the $d$ base points the point is located):
$N\_n^d = \binom{d+1}{2} + d\ (\ n - (d + 1)\ ) = nd - d (d + 1) + \binom{d+1}{2} = nd - \binom{d+1}{2}$. (Is this sound?)
Observation: The binomials seem to come from two very different directions (with two seemingly different interpretations), also the term $nd$. Does this tell something deep about (Euclidean) geometry? And what?
Are there further "independent" ways to compute $N\_n^d$?
| https://mathoverflow.net/users/2672 | Number of independent distances between n points in d-dimensional Euclidean space? | Here is a little generalisation of your observation.
Suppose we have a manfiold $M$ of dimension d with a metric. The isometry group $I(M)$ of the manifold has dimension at most $\frac{d(d+1)}{2}$. The maximal dimension of $I(M)$ is attained for $R^d$, $H^d$, $S^d$ and $RP^d$ with a constant curvature metric. Now we want to know what is the number of independent distances among $n$ points in $M^d$ with $n\ge d$. This number can be estimated from above by the reasoning identical to yours, namelly it is $nd-dim(I(M))$.
But in reality this is only an estimation from above, because for the universal cover of $M$ the dimension will be the same as for $M$ but its group of symmetries can be larger. This is what happen for $T^n$ and $R^n$ with flat metric.
So I wonder if the following is true. Let $M$ be a simply connected Riemannian manifold of dimension $d$. Let $Dim(M)(n)$ be the dimension of the space of independent distances among $n$ points in $M$. Is it true that $nd-Dim(M)(n)$ is the dimesnion of the group of isometries $dim(I(M))$ of $M$ for $n$ large enough?
| 2 | https://mathoverflow.net/users/943 | 10490 | 7,162 |
https://mathoverflow.net/questions/10487 | 8 | Related to a previous [question](https://mathoverflow.net/questions/10126/reference-for-this-theorem-in-representation-theory) I am asking furthermore a proof
for the following:
Question 1: If $\chi$ is a faithful irreducible character of a
finite group $G$ then the regular character of $G$ is a polynomial
with integer coefficients in $\chi$?
I know this fact is true since there is a generalization of it for
Hopf algebras in Corollary 19 of the paper FSU96-08 from [here](http://www.math.fsu.edu/~aluffi/eprint.archive.html).
The proof from that paper is a little complicated using some
(although elementary) results on norms and inner products.
I was wondering if anyone knows a different proof of this.
Using the Stone - Weierstrass method mentioned in the previous
question, I am asking further if the following is true:
Question 2: If $\chi$ is a faithful irreducible character of a
finite group $G$ does any character of $G$ is a complex polynomial
in $\chi$?
| https://mathoverflow.net/users/2805 | Faithful characters of finite groups | I assume that by "faithful irreducible character" you mean the character of a faithful (i.e., trivial kernel) irreducible representation. In this case, the answer to Question 2 is negative. For instance, the irreducible character $\chi$ of the symmetric group $S\_4$ indexed by the partition (3,1) is faithful and has the same value on two different conjugacy classes of $S\_4$, so the same will be true of any complex polynomial in $\chi$.
| 12 | https://mathoverflow.net/users/2807 | 10498 | 7,167 |
https://mathoverflow.net/questions/10493 | 22 | I am interested in the general form of the Kirchoff Matrix Tree Theorem for weighted graphs, and in particular what interesting weightings one can choose.
Let $G = (V,E, \omega)$ be a weighted graph where $\omega: E \rightarrow K$, for a given field $K$; I assume that the graph is without loops.
For any spanning tree $T \subseteq G$ the weight of the tree is given to be,
$$m(T) = \prod\_{e \in T}\omega(e)$$
and the tree polynomial (or statistical sum) of the graph is given to be the sum over all spanning trees in G,
$$P(G) = \sum\_{T \subseteq G}m(T)$$
The combinatorial laplacian of the graph G is given by $L\_G$, where:
$$L\_G = \begin{pmatrix} \sum\_{k = 1}^n\omega(e\_{1k}) & -\omega(e\_{12}) & \cdots & -\omega(e\_{1n}) \\\ -\omega(e\_{12}) & \sum\_{k = 1}^n\omega(e\_{2k}) & \cdots & -\omega(e\_{2n})\\\
\vdots & \vdots & \ddots & \vdots \\\
-\omega(e\_{1n}) & -\omega(e\_{2n}) & \cdots & \sum\_{k = 1}^n\omega(e\_{nk}) \end{pmatrix} $$
where $e\_{ik}$ is the edge between vertices $i$ and $k$, if no edge exists then the entry is 0 (this is the same as considering the complete graph on n vertices with an extended weighting function that gives weight 0 to any edge not in G). The matrix tree theorem then says that the tree polynomial is equal to the absolute value of any cofactor of the matrix. That is,
$$P(G) = \det(L\_G(s|t))$$
where $A(s|t)$ denotes the matrix obtained by deleting row $s$ and column $t$ from a matrix $A$.
By choosing different weightings one would expect to find interesting properties of a graph G. Two simple applications are to give the weighting of all 1's. Then the theorem allows us to count the number of spanning trees with ease (this yields the standard statement of the Matrix Tree Theorem for graphs). Alternatively, by giving every edge a distinct formal symbol as its label then by computing the relevant determinant, the sum obtained can be read as a list of all the spanning trees.
My question is whether there are other interesting weightings that can be used to derive other interesting properties of graphs, or for applied problems.
| https://mathoverflow.net/users/2947 | The matrix tree theorem for weighted graphs | A very interesting weighting is obtained by just working with directed multigraphs (dimgraphs). 7 or 8 years ago, I applied the matrix-tree theorem applied to dimgraphs in conjunction with the BEST theorem to provide a structure theory for the equilibrium thermodynamics of hybridization for oligomeric (short) DNA strands.
Briefly, the SantaLucia model of DNA hybridization takes a finite word $w$ on four letters (A, C, G, T) and associates to it various thermodynamical characteristics (e.g., free energy $\Delta G$ of hybridization) based on the sequence. Assuming the words are cyclic (which is not fair, but also not a very bad approximation practically), one has $\Delta G (w) = \sum\_k \Delta G (w\_kw\_{k+1})$ where the index $k$ is cyclic and the 16 parameters $\Delta G(AA), \dots, \Delta G(TT)$ are given.
Assuming for convenience that the $\Delta G(\cdot,\cdot)$ are independent over $\mathbb{Q}$, it is not hard to see that $\Delta G$ projects from the space of all words of length $N$ to the space of dimgraphs on 4 vertices (again, A, C, G, T) with $N$ edges, where (e.g.) an edge from A to G corresponds to the subsequence AG. Using matrix-tree and BEST provides a functional expression for the number of words of length $N$ with a given number of AA's, AC's, ... and TT's, and thus for the desired $\Delta G$.
Going a step further, one can use generalized Euler-Maclaurin identities for evaluating sums of functions over lattice polytopes to characterize the space of all (cyclic) words of length $N$ with $\Delta G$ lying in a narrow range. This effectively completes the structure theory and shows how one can construct DNA sequences having desired thermodynamical and combinatorial properties. Among other things, I used this to design a protocol for (as David Harlan Wood put it) "simulating simulated annealing by annealing DNA".
| 11 | https://mathoverflow.net/users/1847 | 10500 | 7,169 |
https://mathoverflow.net/questions/10514 | 25 | What is a good introduction to Teichmuller theory, mapping class groups etc., and relation to moduli space of curves or Riemann surfaces?
| https://mathoverflow.net/users/2938 | Teichmuller Theory introduction | [The primer](http://www.maths.ed.ac.uk/~aar/papers/farbmarg.pdf) on mapping class groups, by Farb and Margalit.
| 20 | https://mathoverflow.net/users/1650 | 10517 | 7,177 |
https://mathoverflow.net/questions/10502 | 8 | In my first algebraic topology class, I remember being told that the simplest reason for homology was to distinguish spaces. For example, if is X=circle and a Y= wedge of a circle and a 2-sphere then X and Y have the same fundamental group, so the fundamental group isn't strong enough to distinguish them. We need to look at the other homotopy groups or homology to tell them apart. I'm looking for a variety of other examples of this nature. The examples I'm wondering about are
1. Same homology groups
2. Same cohomology groups, but different cohomology rings
3. Same cohomology rings (but maybe different Steenrod operations?)
If I put more thought into it, I could come up with others questions like these. Any other examples/thoughts along these lines would be very welcome! (I have examples for the first one, but I'm wondering what others will say.)
| https://mathoverflow.net/users/343 | Examples of the varying strengths of topological invariants | To change up the nature of the responses some, IMO a good theorem to think about is the Kan-Thurston theorem. It states that given any space $X$ you can find a $K(\pi, 1)$ space $Y$ and a map $f : Y \to X$ inducing isomorphisms $f\_\* : H\_i Y \to H\_i X$, $f^\* : H^i X \to H^i Y$ for all coefficients (it can be souped-up to allow local coefficients) and all $i$. The map $\pi\_1 Y \to \pi\_1 X$ is onto.
So from the point of view of cohomology algebras with Steenrod operations, these spaces are the same. One way to "spin" this would be to say the fundamental group is a far stronger invariant than anything (co)homological.
| 6 | https://mathoverflow.net/users/1465 | 10536 | 7,190 |
https://mathoverflow.net/questions/10266 | 5 | Imagine an n-simplex, the solution set for the expression: $a\_1$\*$x\_1$ + $a\_2$\*$x\_2$ + ... + $a\_n$\*$x\_n$ = S, where:
1. $a\_1$ through $a\_n$ are positive bounded integers
2. $x\_1$ through $x\_n$ are positive bounded real numbers
3. 'S' is the sum of the expression
This n-simplex therefore has a single vertex on the origin, as well as a single vertex on each axis at some arbitrary (strictly positive) distance from the origin.
What is the lattice integer-point count?
Can one use Ehrhardt polynomials to compute the integer point count for the n-simplex, perhaps under the restriction that we have vertices strictly at integer coordinates?
* From "Geometry for N-Dimensional Graphics" (by Andrew J. Hanson, CS Dept., Indiana University) we know that the oriented volume for the n-simplex with vertices {$v\_1$, ..., $v\_n$}, or {$a\_1$\*$x\_1$, ..., $a\_n$\*$x\_n$} is:
$V\_n$ = $\dfrac{1}{n!}$ \* det([($v\_1$-$v\_0$), ..., ($v\_n$-$v\_0$)])
(Problems writing LaTeX for matrices here, please see terms as column vectors to obtain square matrix.)
---
Previous formulation of question: Imagine an expression of the form: $a\_1$\*$x\_1$ + $a\_2$\*$x\_2$ + ... + $a\_n$\*$x\_n$ = S, where:
1. $a\_1$ through $a\_n$ are positive bounded integers
2. $x\_1$ through $x\_n$ are positive bounded real numbers
3. 'S' is the sum of the expression
Can we say anything about the maximum value of 'S' (for a given $x\_1$ through $x\_n$) below which there is only one solution for positive integer coefficients $a\_1$ through $a\_n$? For example, given the expression $a\_1$\*98 + $a\_2$\*99 = 'S', where coefficients $a\_1$ and $a\_2$ = [1 through 100], one finds that you can always exactly recover the original $a\_1$ and $a\_2$ if 'S' < 9899. Is there an analytical or more elegant method for obtaining that bound?
[Above such a bound, is there an efficient way to obtain all possible sets of integers $a\_1$ through $a\_n$ that satisfy the relation for a given 'S'? Can the LLL or PSLQ algorithms be used?] <-- This seems to be a restricted/special case of the subset-sum problem, so existing dynamic programming algorithms would work. Can one do better here?
| https://mathoverflow.net/users/2891 | Counting lattice points on an n-simplex | I am informed that you are "counting lattice points inside of a polyhedron."
[Here](http://www.math.ucdavis.edu/~deloera/.../manyaspectsofcountinglatpts.pdf) is a lecture on the subject - the picture on page six looks like the version of the problem you are interested in. To be honest, I found these notes by doing a google search. I am told that this is a huge field!
It might help if you could narrow your problem even further. For example, you say that the $x\_i$ are bounded real numbers. Do you know these to some high precision? Or can you give some information on how the $x\_i$ are given? And can you say the same for $S$?
EDIT:
[Here](http://www.math.ucdavis.edu/~deloera/RECENT_WORK/semesterberichte.pdf) is a survey paper by the same author, Jesús De Loera, covering the same material in greater detail.
| 5 | https://mathoverflow.net/users/1650 | 10551 | 7,203 |
https://mathoverflow.net/questions/10231 | 14 | Let $C$ be the category of $\tau$-algebras for some type $\tau$. Consider the statements:
1. Every monomorphism is regular.
2. Every epimorphism in $C$ is surjective.
It is easy to see that 1. implies 2. what about the converse?
| https://mathoverflow.net/users/2841 | When are epimorphisms of algebraic objects surjective? | Update: the following exchange appeared on the categories mailing list several years ago: <http://article.gmane.org/gmane.science.mathematics.categories/3094>. Walter Tholen's response strongly suggests that the answer to Martin's question is that the converse does not hold, although I don't have access to the four-author article he cites as reference. It's probably worth a look though, and if I learn anything more I'll post another update.
**Second update**: My surmise was correct. Walter Tholen kindly emailed to me the relevant two pages (pp. 88-89) of the four-author paper
* E.W. Kiss, L. Marki, P. Prohle, W. Tholen: Studia Sci. Math. Hungaricum 18
(1983) 79-141.
where the following example is given on page 89: in the category of semigroups with zero such that all 4-fold products are zero, all epimorphisms are surjective but not all monos are regular [a specific nonregular mono is described]. (I can forward this email if you write to me at topological dot musings at gmail dot com.)
Assuming amalgamated products exist (as they do in categories of algebras of a Lawvere theory), a mono $i: A \rightarrowtail B$ is regular if it is the equalizer of the pair of canonical maps from $B$ to the amalgamated product $B \*\_A B$ (i.e., the coprojections of the pushout of $i$ with itself, aka the *cokernel pair* of $i$). The equalizer of the cokernel pair defines a closure operator on the lattice of subalgebras $\mathsf{Sub}(B)$, called the *dominion operator* $\mathsf{Dom}\_B$. So to prove a subalgebra is not regular is to show that it is not $\mathsf{Dom}$-closed. The key technical result needed to prove the claim above is Isbell's Zig-Zag theorem (given in his paper Epimorphisms and Dominions in the 1965 La Jolla conference proceedings on categorical algebra), as recalled in [Peter M. Higgins. "A short proof of Isbell's Zigzag Theorem." Pacific J. Math. 144(1):47–50 (1990)](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-144/issue-1/A-short-proof-of-Isbells-zigzag-theorem/pjm/1102645825.full), which gives a precise and useful criterion for an element to belong to the dominion (i.e., the $\mathsf{Dom}$-closure) of a subalgebra.
Hope this helps. I am voting up your question, Martin, since it's rather nontrivial!
| 13 | https://mathoverflow.net/users/2926 | 10553 | 7,205 |
https://mathoverflow.net/questions/10457 | 3 | Is a good characterization of Spec $\mathbb{Z}[\zeta\_n]$ known? Same question for its unit group.
| https://mathoverflow.net/users/434 | What are the prime ideals in rings of cyclotomic integers? | **Theorem:** Let $\alpha$ be an algebraic integer such that $\mathbb{Z}[\alpha]$ is integrally closed, and let its minimal polynomial be $f(x)$. Let $p$ be a prime, and let
$\displaystyle f(x) \equiv \prod\_{i=1}^{k} f\_i(x)^{e\_i} \bmod p$
in $\mathbb{F}\_p[x]$. Then the prime ideals lying above $p$ in $\mathbb{Z}[\alpha]$ are precisely the maximal ideals $(p, f\_i(\alpha))$, and the product of these ideals (with the multipicities $e\_i$) is $(p)$. ([Theorem 8.1.3](http://modular.math.washington.edu/papers/ant/).)
In this particular case we have $f(x) = \Phi\_n(x)$. When $(p, n) = 1$, its factorization in $\mathbb{F}\_p[x]$ is determined by the action of the Frobenius map on the elements of order $n$ in the multiplicative group of $\overline{ \mathbb{F}\_p }$, which is in turn determined by the minimal $f$ such that $p^f - 1 \equiv 0 \bmod n$ as described in Chandan's answer. (This $f$ is the size of every orbit, hence the degree of every irreducible factor.) When $p | n$ write $n = p^k m$ where $(m, p) = 1$, hence $x^n - 1 \equiv (x^m - 1)^{p^k} \bmod p$. Then I believe that $\Phi\_n(x) \equiv \Phi\_m(x)^{p^k - p^{k-1}} \bmod p$ and you can repeat the above, but you'd have to check with a real number theorist on that. (**Edit:** Indeed, it's true over $\mathbb{Z}$ that $\Phi\_n(x) = \frac{ \Phi\_m(x^{p^k}) }{ \Phi\_m(x^{p^{k-1}}) }$.)
| 12 | https://mathoverflow.net/users/290 | 10555 | 7,207 |
https://mathoverflow.net/questions/10509 | 15 | I'm not a number theorist, so apologies if this is trivial or obvious.
From what I understand of the results of Green-Tao-Ziegler on additive combinatorics in the primes, the main new technical tool is the "dense model theorem," which -- informally speaking -- is as follows:
>
> If a set of integers $S \subset N$ is a dense subset of another "pseudorandom" set of integers, then there's another set of integers $S' \subset N$ such that $S'$ has positive density in the integers and $S, S'$ are "indistinguishable" by a certain class of test functions.
>
>
>
They then use some work of Goldston and Yildirim to show that the primes satisfy the given hypothesis, and note that if the primes failed to contain long arithmetic/polynomial progressions and $S'$ did, they'd be distinguishable by the class of functions. Applying Szemeredi's theorem, the proof is complete.
Obviously I'm skimming over a great deal of technical detail, but I'm led to believe that this is a reasonably accurate high-level view of the basic approach.
My question(s), then: Can one use a similar approach to obtain "Goldbach-type" results, stating that every sufficiently large integer is the sum of at most k primes? Is this already implicit in the Goldston-Yildirim "black box?" If we can't get Goldbach-type theorems by using dense models, what's the central obstacle to doing so?
| https://mathoverflow.net/users/382 | Goldbach-type theorems from dense models? | Yes, this can be done, provided that k is at least 3. A typical example is given in this paper: <http://arxiv.org/abs/math/0701240> . (This uses a slightly older Fourier-based method of Ben that predates his work with me, but is definitely in the same spirit - Ben's paper was very inspirational for our joint work.)
For k=2, unfortunately, the type of indistinguishability offered by the dense model theorem is not strong enough to say anything. The k=2 problem is very similar to the twin prime conjecture (in both cases, one is trying to seek an additive pattern in the primes with only one degree of freedom). If one deletes all the twin primes from the set of primes, one gets a new set which is indistinguishable from the set of primes in the sense of the dense model theorem (because the twin primes have density zero inside the primes, by Brun's theorem), but of course the latter set has no more twins. One can pull off a similar trick with representations of N as the sum of two primes. But one can't do it with sums of three or more primes - there are too many representations to delete them all just by removing a few primes.
| 26 | https://mathoverflow.net/users/766 | 10558 | 7,209 |
https://mathoverflow.net/questions/10370 | 8 | Normal heuristics give that number of k-term arithmetic progressions in [1,N] should be about
$$c\_k\frac{N^2}{\log^kN}$$
for some constant $c\_k$ dependent on k. The paper of Green and Tao gives a similar lower bound for all k (with a much worse constant, but still), and recent work by Green, Tao and Ziegler have established the correct asymptotic for k=3 and k=4.
I am looking for a reference which establishes an upper bound for all k - I'm sure I've heard of one, but I can't find mention of the relevant paper anywhere. Of course, if there is a simple proof, that would appreciated as well.
That is, I am looking for a reference and/or proof which establishes that the number of k-term arithmetic progressions of primes in [1,N] is at most
$c\_k'\frac{N^2}{\log^kN}$.
for some constant $c\_k'$.
| https://mathoverflow.net/users/385 | Upper bound for number of k-term arithmetic progressions in the primes | Well, any standard upper bound sieve (e.g. Selberg sieve, combinatorial sieve, beta sieve, etc.) will give this type of result. I'm not sure where you can find an easily citeable formulation, though. One can get this bound from Theorem D.3 of [this paper of Ben and myself](http://arxiv.org/abs/math.NT/0606088) on page 67 (see in particular the remark at the bottom of that page). But this is certainly not the first place where such a bound appears. (The Goldston-Yildirim papers will give this result too, but this is also not the first place either.)
| 15 | https://mathoverflow.net/users/766 | 10559 | 7,210 |
https://mathoverflow.net/questions/10532 | 10 | Given a moduli problem, it appears that nonexistence of automorphisms is a necessary condition for existence of a fine moduli space(is this strictly true?).
In any case, assuming the above, what additional condition on a moduli problem in algebraic geometry will make sure that a coarse moduli space is in fact a fine moduli space?
In the n-lab page on Deligne-Mumford, the following appears.
Deligne-Mumford stacks correspond to moduli problems in which the objects being parametrized have finite automorphism groups.
Also, a few problematic examples I have heard of, had infinite automorphism groups.
Therefore, is it true that for a moduli problem in which the stack is Deligne-Mumford, and where there are no automorphisms, existence of a coarse moduli space would imply the existence of a fine moduli space?
| https://mathoverflow.net/users/2938 | When is a coarse moduli space also a fine moduli space? | I think this is an instructive question. Here are some partial answers.
A category fibered in groupoids whose fibers are sets (e.g. no automorphisms) is a presheaf. Strictly speaking, I mean equivalent to the fibered category associated to a presheaf. This truly follows from the definitions, and is a good exercise to do when one is learning the basic machinery behind stacks. Similarly, a stack which is equivalent to a presheaf is a sheaf (i.e. the descent condition collapses to the sheaf condition; this is a little harder but still tautological). And a pre-stack with no automorphisms is a separated pre-sheaf.
The other claim is more interesting and not tautological, and reflects the fact that the diagonal of a morphism of stacks is way more interesting than in the case of schemes. For instance, an algebraic stack is DM iff its diagonal is unramified. (I believe this is in Champs Algebriques, but there's a nice discussion in Anton's [notes](http://math.berkeley.edu/~anton/index.php?m1=writings) from Martin Olsson's stacks course.
The point is that the diagonal of a stack carries information about automorphisms of the objects it parameterizes. So for instance the condition (in the definition of a stack) that the diagonal is representable is equivalent to the statement that Isom(X,Y) (and thus Aut(X)) is representable by an algebraic space. Also tautological is the statement that the diagonal being unramified is equivalent to the statement that there are no *infintesmal* automorphisms (e.g. non-trivial automorphisms of an object over $k[\epsilon]/\epsilon^2$ which reduce to the identity map over $k$). So here it is now clear where one uses finiteness: the $k[\epsilon]/\epsilon^2$ points of a finite groups scheme are the same as the $k$ points; on the other hand this fails if the automorphism scheme is, say, $\mathbf{G}\_m$.
Finally, while the tautological answer above answers your question, it is instructive to see how automorphisms cause $M\_g$ (moduli of genus g curves) to not be representable. Let **H** be a hyperelliptic curve given by $y^2 = f(x)$ defined over a field $k$. Then the curve $H\_d$ given by $dy^2 = f(x)$ is not isomorphic to $H$ over $k$ if d is not a square in k. Call this a `twist' of **H**; in general twists of a variety X are given by the Galois cohomology group $H^1(G\_k,Aut X)$ which is non-zero in non-trivial situations (alternatively you can use torsors and etale cohomlogy), and a generic hyperelliptic curve has automorphism group $\{\pm 1\}$. Thus H and $H\_d$ give two different $k$ points of $M\_g$ which become the same point over a finite extension; thus $M\_g$ fails the sheaf condition in the etale topology, (and so in general existence of automorphisms causes, for cohomological reasons, failure of your moduli problem to even be a sheaf).
Last comment (to clarify others' comments): fine moduli space should certainly allow algebraic spaces for a correct answer to your question.
| 12 | https://mathoverflow.net/users/2 | 10562 | 7,212 |
https://mathoverflow.net/questions/10501 | 11 | Let $(R,m)$ be a local complete intersection of dimension $3$. Let $X=Spec(R)$ and $U=Spec(R) -\{m\}$ be the punctured spectrum of $R$. I am trying to understand the following comment by Gabber (see it [here](http://www.mfo.de/programme/schedule/2004/32/OWR_2004_37.pdf) , page 1975-1976):
$Pic(U)$ is torsion-free is equivalent to the flat local cohomology $H\_{\{m\}}^2(X, \mu\_n)=0$ for $n>0$
So let's try to go through a possible argument (this is from my very limited understanding, so feel free to correct me here). The sequence: $ 0 \to \mu\_n \to {\mathbb G\_m} \xrightarrow{t\mapsto t^n} {\mathbb G\_m} \to 0 $
is exact on the flat site over $U$, $U\_{fl}$. Thus the long exact sequence of flat cohomology shows:
$ 0 \to H^1(U\_{fl},\mu\_n) \to H^1(U\_{fl},{\mathbb G\_m}) \xrightarrow{\times n} H^1(U\_{fl},{\mathbb G\_m})$
We know that $ H^1(U\_{fl},{\mathbb G\_m}) \cong Pic(U)$. Now excision gives:
$H^1(X\_{fl},\mu\_n) \to H^1(U\_{fl},\mu\_n) \to H^2\_{\{m\}}(X\_{fl},\mu\_n) \to H^2(X\_{fl},\mu\_n)$
Obviously $H^1(X\_{fl},\mu\_n)=0$ (note that $H^1(X\_{fl},{\mathbb G\_m})\cong Pic(X)=0$, as $R$ is local). EDIT: obviously, it is not true, the argument becomes quite subtle, please see Emerton's answer below.
But why is $H^2(X\_{fl},\mu\_n) =0$? It is true for curves, but Milne warned against using flat cohomology for higher dimensions in his [note](http://www.jmilne.org/math/Books/adt.html) on duality theorems (see the first page of Chapter III). In fact, I could not find any reference about flat cohomology in higher dimensions. So:
Can some one help finish/fix the argument above? Is there a reference from which I can quote the ``facts" I used above?
| https://mathoverflow.net/users/2083 | Flat cohomology and Picard groups | This is not a complete answer by any means, but is intended to get the ball rolling.
First of all, it need not be the case that $H^1(X\_{fl},\mu\_n) = 0.$ Rather, what follows
from the vanishing of $H^1(X\_{fl},{\mathbb G}\_m)$ is that
$H^1(X\_{fl},\mu\_n) = R^{\times}/(R^{\times})^n.$
(This is not always trivial; imagine
e.g. that $R$ is a non-algebraically closed field. You might have been thinking of the case
when $X$ is projective and smooth over an algebraically closed field, when the $H^0$-part of the exact sequence is itself exact, and so can be omitted from consideration. That is not the case here.)
Secondly, this doesn't hurt your arguments, because the same consideration of $H^0$-terms has to be made for the cohomology of $U$. Since $X$ is three dimensional and a complete intersection, restriction induces an isomorphism
$H^0(X,\mathcal O) \cong H^0(U,\mathcal O)$, and so also an isomorphism
$H^0(X,\mathcal O^{\times})\cong H^0(U,\mathcal O^{\times}),$ and so also isomorphisms
$H^0(X\_{fl},\mu\_n) \cong H^0(U\_{fl},\mu\_n)$ and
$H^0(X\_{fl},{\mathbb G}\_m)\cong H^0(U\_{fl},{\mathbb G}\_m).$ Thus in fact one finds that
the $n$-torsion in Pic$(U)$ is equal to the cokernel of the injection
$H^1(X\_{fl},\mu\_n) \hookrightarrow H^1(U\_{fl},\mu\_n).$ And as your analysis shows, this cokernel embeds into $H^2\_{\{m\}}(X\_{fl},\mu\_n)$, with the cokernel of that embedding itself embedding into $H^2(X\_{fl},\mu\_n).$
So what can be said about this latter cohomology group?
Since $H^1(X\_{fl},{\mathbb G}\_m)$ vanishes, as you observed, one finds that $H^2(X\_{fl}, \mu\_n)$ coincides with the $n$-torsion in the cohomological Brauer group $H^2(X\_{fl},{\mathbb G}\_m).$ (Here I am using the fact that since ${\mathbb G}\_m$ is smooth, flat and etale cohomology coincide, so $H^2(X\_{fl},{\mathbb G}\_m) = H^2(X\_{et}, {\mathbb G}\_m).$) So it seems that one wants to kill off the torsion in this Brauer group.
I don't see why this need be true, but what one actually needs is that $H^2(X\_{fl},\mu\_n)
\rightarrow H^2(U\_{fl},\mu\_n)$ is injective. Since $H^2(X\_{fl},\mu\_n)$ embeds into
$H^2(X\_{fl},{\mathbb G}\_m)$, it would be enough to show that the restriction
$H^2(X\_{fl},{\mathbb G}\_m) \to H^2(U\_{fl},{\mathbb G}\_m)$ induces an injection on torsion. Might this be some kind of purity result on Brauer groups of the kind Gabber discusses in his abstract? It would be related to a vanishing of (torsion in) $H^2\_{\{m\}}(X\_{fl}, {\mathbb G}\_m)$. Somewhere (maybe here?) one presumably has to make use of the dimension and lci hypotheses.
P.S. You may well just want to email Gabber to ask him about this. If you do, and you get an answer, please share it!
EDIT: This is an excerpt from the email referred to in Hai Long's comment below:
To learn about these kinds of arguments, my advice is to
do just what you are doing. One works with the exact sequence
linking $\mu\_n$ and ${\mathbb G}\_m$, as you did.
Number theorists (at least of a certain stripe) have some advantages
with this, because the case $X =$ Spec $K$ ($K$ a field) comes up a lot
under the name of Kummer theory, and also Mazur in one of his
famous papers uses a lot of flat cohomology. But in the end, the
formalism is just the one you used in your question.
Then, typically, one has to inject something additional that
is less formal. My suggestion would be to look at de Jong's proof
of Gabber's result showing $Br'(X) = Br(X)$ discussed in his abstract.
(There is a write-up on de Jong's web-page.)
Reading the proof of a result like this might give some insight
into how to work with Brauer groups in a less formal way.
| 11 | https://mathoverflow.net/users/2874 | 10563 | 7,213 |
https://mathoverflow.net/questions/1554 | 21 | Suppose you have an incomplete Riemannian manifold with bounded sectional curvature such that its completion as a metric space is the manifold plus one additional point. Does the Riemannian manifold structure extend across the point singularity?
(Penny Smith and I wrote a paper on this many years ago, but we had to assume that no arbitrarily short closed geodesics existed in a neighborhood of the singularity. I was never able to figure out how to get rid of this assumption and still would like someone better at Riemannian geometry than me to explain how. Or show me a counterexample.)
EDIT: For simplicity, assume that the dimension of the manifold is greater than 2 and that in any neighborhood of the singularity, there exists a smaller punctured neighborhood of the singularity that is simply connected. In dimension 2, you have to replace this assumption by an appropriate holonomy condition.
EDIT 2: Let's make the assumption above simpler and clearer. Assume dimension greater than 2 and that for any r > 0, there exists 0 < r' < r, such that the punctured geodesic ball B(p,r'){p} is simply connected, where p is the singular point. The precludes the possibility of an orbifold singularity.
ADDITIONAL COMMENT: My approach to this was to construct a differentiable family of geodesic rays emanating from the singularity. Once I have this, then it is straightforward using Jacobi fields to show that this family must be naturally isomorphic to the standard unit sphere. Then using what Jost and Karcher call "almost linear coordinates", it is easy to construct a C^1 co-ordinate chart on a neighborhood of the singularity. (Read the paper. Nothing in it is hard.)
But I was unable to build this family of geodesics without the "no small geodesic loop" assumption. To me this is an overly strong assumption that is essentially equivalent to assuming in advance that that differentiable family of geodesics exists. So I find our result to be totally unsatisfying. I don't see why this assumption should be necessary, and I still believe there should be an easy way to show this. Or there should be a counterexample.
I have to say, however, that I am pretty sure that I did consult one or two pretty distinguished Riemannian geometers and they were not able to provide any useful insight into this.
| https://mathoverflow.net/users/613 | Point singularity of a Riemannian manifold with bounded curvature | Once we considered a similar problem but around infinity,
try to look in our paper "Asymptotical flatness and cone structure at infinity".
Let us denote by $r$ the distance to the singular point.
If dimensions $\not= 4$ then the same method shows that at singular point we have Euclidean tangent cone even if curvature is "much less" than $r^{-2}$ (say if $K=O(\tfrac{1}{r^{2-\varepsilon}})$ for some $\varepsilon>0$, but one can make it bit weaker).
In dimension 4 there might be some funny examples: Your singular point has tangent space $\mathbb R^3$,
the $r$-spheres around this point are Berger spheres, so its curvature is very much like curvature of $r\cdot(S^2\times \mathbb R)$, the size of Hopf fibers goes to $0$ very fast.
However if you know that dimension of the tangent space is $4$ then it has to be Euclidean.
All this can happen if curvature grows slowly.
If it is bounded then one can extend Bishop--Gromov type inequality for balls around singular point.
It implies that the dimension of the tangent space is $4$. That will finish the proof.
| 7 | https://mathoverflow.net/users/1441 | 10565 | 7,215 |
https://mathoverflow.net/questions/10567 | 15 | Let $N(x)$ be the number of uniform random variables (distributed in $[0,1]$) that one needs to add for the sum to cross $x$ ($x > 0$). The expected value of $N(x)$ can be calculated and it is a very cool result that $E(N(1)) = e$. The expression for general $x$ is
$E(N(x)) = \sum\_{k=0}^{[x]} (-1)^k \frac{(x-k)^k}{k!} e^{x-k}$
where $[x]$ is the largest integer less than $x$. When $x$ is large, this function $E(N(x))$ grows linearly (as one would expect). Computer simulations suggest that $E(N(x)) \approx 2x + 2/3$. My question is as follows: is it possible to guess this form for $E(N(x))$ without actually computing it? The $2x$ part is intuitive but is there a good intuition for why there is an additive constant and why that value is $2/3$?
My motivation is as follows: When I computed $E(N(x))$ using the formula above for large values of $x$, I came across this asymptotic and found it surprising that a polynomial equation in powers of $e$ gives values that are very close to a rational number. I am very interested in knowing the reason (if any) behind it. Thanks in advance for any help.
| https://mathoverflow.net/users/2878 | Number of uniform rvs needed to cross a threshold | Another way to put it: the expected value of the sum right after it crosses x is x+1/3. If you simply conditioned the sum to be in (x,x+1) then the expectation would be about x+1/2, with the sum almost uniformly distributed. But you also condition on the overshoot being less than the last jump. The expectation of the smaller of two uniform [0,1] iids is 1/3.
| 14 | https://mathoverflow.net/users/2912 | 10568 | 7,217 |
https://mathoverflow.net/questions/10496 | 3 | Let [a,b] = {k integer | a < k <= b}. Further let
* Comp[a,b] = product\_{c in [a,b]} c composite;
* Fact[a,b] = product\_{k in [a,b]} k integer;
* Prim[a,b] = product\_{p in [a,b]} p prime.
*Question*: For n > 2 and n not in {10,15,27,39} is it true that
$$ \text{Comp}[{\left\lfloor n /2 \right\rfloor}, n] < \text{Fact}[1, {\left\lfloor n /2 \right\rfloor}] \ \text{Prim}[{\left\lfloor n /2 \right\rfloor}, n] \ ? $$
*Update*: The state of affairs: Gjergji Zaimi showed that for large enough n the inequality is true. In my answer I affirm that the inequality is true in the range 40 <= n <= 10^5. It remains open whether 10^5 is 'large enough' in the sense of Gjergji's analysis.
| https://mathoverflow.net/users/2797 | An inequality relating the factorial to the primorial. | This answer is just to point out that the result is true for large enough $n$. Let's rewrite it as $$\prod\_{n\le p\le 2n}p > \sqrt{\binom{2n}{n}}$$
Since $\binom{2n}{n}\approx \frac{4^n}{n}$ introducing Chebyshev's functions
$$\theta(x)=\sum\_{p\le x}\text{log}p\quad,\quad \psi(x)=\sum\_{p^{\alpha}\le x}\text{log}p$$
They satisfy $$\psi(x)=\theta(x)+O(\sqrt{x}\text{log}^2x)$$
What we want to prove is
$$\theta(2n)-\theta(n) > n\text{log}2$$
It is a well-known asymptotics that
$$\psi(x)=x+O(x\text{exp}(-c\sqrt{\text{log}x}))$$
for some positive $c$.
In fact under the Riemann Hypothesis it is even true that
$$\psi(x)=x+O(\sqrt{x}\text{log}^2x)$$
but we don't need this refinement. Now
$$\theta(2n)-\theta(n)=n+O(n\text{exp}(-c\sqrt{\text{log}n}))$$ and this proves your assertion for large enough $n$.
A good reference where all these results are proven is for example "Problems in Analytic Number Theory" by M.Ram Murty. I hope this helps, even though I did not mention any thing about possible small counterexamples. To find the smallest $n$ for which this argument works you'd have to look up each of these equations individually and look for specific bounds.
| 9 | https://mathoverflow.net/users/2384 | 10573 | 7,220 |
https://mathoverflow.net/questions/10560 | 26 | Voronin´s Universality Theorem (for the Riemann zeta-Function) according to Wikipedia: Let $U$ be a compact subset of the "critical half-strip" $\{s\in\mathbb{C}:\frac{1}{2}<Re(s)<1\}$ with connected complement. Let $f:U \rightarrow\mathbb{C}$ be continuous and non-vanishing on $U$ and holomorphic on $U^{int}$. Then $\forall\varepsilon >0$ $\exists t=t(\varepsilon)$ $\forall s\in U: |\zeta(s+it)-f(s)|<\varepsilon $.
(Q1) Is this the accurate statement of Voronin´s Universality Theorem? If so, are there any (recent) **generalisations** of this statement with respect to, say, shape of $U$ or conditions on $f$ ? (If I am not mistaken, the theorem dates back to 1975.)
(Q2) Historically, were the Riemann zeta-function and Dirichlet L-functions the first examples for functions on the complex plane with such "universality"? Are there any examples for functions (on the complex plane) with such properties beyond the theory of zeta- and L-functions?
(Q3) Is there any known general argument why such functions (on $\mathbb{C}$) "must" exist, i.e. in the sense of a non-constructive proof of existence? (with Riemann zeta-function being considered as a proof of existence by construction).
(Q4) Is anything known about the structure of the class of functions with such universality property, say, on some given strip in the complex plane?
(Q5) Are there similar examples when dealing with $C^r$-functions from some open subset of $\mathbb{R}^n$ into $\mathbb{R}^m$ ?
Thanks in advance and Happy New Year!
| https://mathoverflow.net/users/1849 | Universality of zeta- and L-functions | Since, I believe, Jonas Meyer provided an answer to Q1, let me just say about the other questions: The concept of universality is much older. It was in fact introduced by Birkhoff, in the case for entire functions, in 1929 (and that is why universal functions are sometimes called Birkhoff functions) "Demonstration d'un theoreme elementaire sur les fonctions entieres." and by Heins, in the case of bounded holomorphic in the unit disk, in 1955.
A possible reference is "Universal functions in several complex variables" by P.S. Chee.
| 7 | https://mathoverflow.net/users/2384 | 10580 | 7,223 |
https://mathoverflow.net/questions/10594 | 9 | As a rule, the various groups and quotients of the divisor group on a variety have coefficients in $\mathbb{Z}$. That is, you take $\mathbb{Z}$-linear combinations of Weil divisors or Cartier divisors, and then to construct other groups you take quotients.
However, in some cases, people tensor with $\mathbb{Q}$ and $\mathbb{R}$. So my question is:
>
> Are these the only rings that people use as coefficients for divisors on a variety?
>
>
>
My vague intuition is that it probably is, because $\mathbb{Z}$ is initial in commutative rings with identity, $\mathbb{Q}$ is a field of characteristic zero, so we can use it to kill torsion, and $\mathbb{R}$ is complete, so we can guarantee that there is an $\mathbb{R}$-divisor, plus with orbifolds, rational coefficients seem to show up naturally. But is this it? More generally, what about for cycles and cocycles? There's an analogy with cohomology and the Chow ring, and we do sometimes take cohomology with coefficients either in an arbitrary ring or in some other rings (finite fields, for instance, when studying things like nonorientable manifolds), which is why I started wondering about this.
| https://mathoverflow.net/users/622 | Is there any value in studying divisors with coefficients in a ring R? | The answer to the question is yes. For example, if $\omega$ is a meromorphic 1-form on a curve (smooth and projective, say) over a field $k$, then one can naturally form a degree zero divisor with coefficients in $k$, namely the residue divisor of $\omega$.
| 12 | https://mathoverflow.net/users/2874 | 10596 | 7,234 |
https://mathoverflow.net/questions/10576 | 6 | The question belongs to elementary category theory, so please forgive me if this is trivial. I think I even read a proof for this some weeks ago, but I can't find it.
In topology, you have the equation $\overline{\overline{A}}=\overline{A}$ for subsets $A$ of a topological space $X$. An analogous theorem in category theory would be: let $X$ be a complete category, and $A$ be a full subcategory of $X$. define $\overline{A}$ as the full subcategory of $X$ consisting of those objects that are limits of small diagrams in $C$, whose objects are in $A$. Do we have $\overline{\overline{A}}=\overline{A}$?
Let's try it: assume $x\_i$ is a diagram in $\overline{A}$ and $(y\_{ij})\_j$ is a diagram in $A$ such that $x\_i = \lim\_j y\_{ij}$. then $\lim\_i x\_i = \lim\_i \lim\_j y\_{ij}$ and we want to interchange limits. But to do this, we have to make $y\_{ij}$ to a diagram in two parameters $i,j$. Perhaps the claim can't be proved in that naive way?
It is easy to see that $\overline{A}$ is closed under products: with the notation above, a morphism $(i,j) \to (i',j')$ corresponds to $i=i'$ and a morphism $j \to j'$. Then $y\_{ij}$ is a diagram in two parameters with limit $lim\_i x\_i$. so it remains to consider equalizers, but how? The problem is here that morphisms between two limits cannot be described in terms of the factors.
Perhaps one should look at examples. Let $X$ be the category of groups, and $A$ the category of finite groups. Then $\overline{A}$ consists of the groups which come from profinite groups, which are exactly the compact, hausdorff, totally disconnected topological groups. Now the category of profinite groups is complete (due to this description) and the forgetful functor to groups preserves limits. Thus in this case, $\overline{\overline{A}} = \overline{A}$. if we put $A=\{\mathbb{Z}/n\}$, $X$ as before, then a similar argument works with the help of $\mathbb{Z}/n$-modules.
EDIT: ok david has given a counterexample. does anyone have an idea how to "fix" this? So what is the "right" definition of $\overline{A}$, so that $\overline{\overline{A}} = \overline{A}$? also, Harry asked for a condition in order this becomes true with my definition.
| https://mathoverflow.net/users/2841 | completion of category is idempotent | The answer is no. Here is my counterexample:
**ARGUMENT SIMPLIFIED, THANKS TO SUGGESTIONS BY Reid, Scott Carnahan AND t3suji**
Let $X$ be the category of abelian groups. Let $A$ be the full subcategory on groups of the form $(\mbox{finite group}) \oplus (\mathbb{Q}-\mbox{vector space})$.
Let $D$ be any diagram in $A$. Every object $G$ in $D$ decomposes as $G\_{\mathrm{fin}} \oplus G\_{\mathbb{Q}}$. Because there are no nonzero homs from a finite group to $\mathbb{Q}$ or vice versa, the diagram $D$ decomposes correspondingly as $D\_{\mathrm{fin}} \oplus D\_{\mathbb{Q}}$. Let $P$ be the limit of $D\_{\mathrm{fin}}$; this is a pro-finite group. Let $V$ be the limit of $D\_{\mathbb{Q}}$; this is a $\mathbb{Q}$ vector space. Then $P \oplus V$ is the limit of $D$.
Fix a prime $p$. Let $R \subset \mathbb{Q}$ be the abelian group of rational numbers whose denominator is relatively prime to $p$. Notice that $R = \mathbb{Q} \cap \mathbb{Z}\_p$, where the intersection takes place in $\mathbb{Q}\_p$. The group $R$ is not an object of $\overline{A}$, because it is not of the form $P \oplus V$ as above.
Let $W$ be the set of all linear maps $\mathbb{Q}\_p \to \mathbb{Q}$ for which the element $1$ of $\mathbb{Q}\_p$ is sent to the element $1$ in $\mathbb{Q}$. Assuming the axiom of choice, the subspace of $\mathbb{Q}\_p$ where all the maps in $W$ coincide is $\mathbb{Q}$.
Consider the diagram in $\overline{A}$ whose objects are $\mathbb{Z}\_p$ and $\mathbb{Q}$, and whose maps are the restrictions to $\mathbb{Z}\_p$ of the maps in $W$. Then the equalizer of this diagram is $\mathbb{Z}\_p \cap \mathbb{Q}$. As observed above, this is $R$, which is not in $\overline{A}$.
| 8 | https://mathoverflow.net/users/297 | 10599 | 7,236 |
https://mathoverflow.net/questions/10578 | 20 | My question, roughly speaking is, what happened to the function fields Langlands conjecture? I understand around 2000 (or slightly earlier perhaps), Lafforgue proved the function fields Langlands correspondence for $GL(n)$ in full generality (proving all aspects of the conjectures). Since then, what's happened to the function fields Langlands conjectures, and what work have people been doing in this direction? (Or has this field died out after Lafforgue's monumental achievement?)
I have been trying but haven't found a good reference for this, but since the function fields Langlands conjectures can be defined for all reductive groups (though I understand in some sense, it is not as "strong" as it is for $GL$ in the general case) - what is the status of these conjectures? Have partial results been obtained?
I understand that geometric Langlands is very intensely researched today, but I consider geometric Langlands as being slightly different (even though it is an analogue of the function fields Langlands correspondence - from reading Frenkel's article on geometric Langlands, my impression was not that geometric Langlands encapsulates all of the information in function fields Langlands conjectures), and I'm asking what work has been done since specifically on function fields Langlands. Or am I misunderstanding things and do the geometric Langlands conjectures actually encapsulate all the information from function fields Langlands as well?
I understand that the Fundamental Lemma has been proven recently, and that Lafforgue is doing some things relating to Langlands functoriality currently. Here's a related thread about Langlands functoriality: [Where stands functoriality in 2009?.](https://mathoverflow.net/questions/1252/where-stands-functoriality-in-2009).
| https://mathoverflow.net/users/2623 | What is the current status of the function fields Langlands conjectures? | (1) Regarding the relationship between geometric Langlands and function field Langlands:
typically research in geometric Langlands takes place in the context of rather restricted ramification (everywhere unramified, or perhaps Iwahori level structure at a finite number of points). There are investigations in some circumstances involving wild ramification (which is roughly the same thing as higher than Iwahori level), but I believe that there is not a definitive program in this direction at this stage.
Also, Lafforgue's result was about constructing Galois reps. attached to automorphic forms. Given this, the other direction (from Galois reps. to automorphic forms), follows immediatly, via
converse theorems, the theory of local constants, and Grothendieck's theory of $L$-functions in the function field setting.
On the other hand, much work in the geometric Langlands setting is about going from local systems (the geometric incarnation of an everywhere unramified Galois rep.) to automorphic sheaves (the geometric incarnation of an automorphic Hecke eigenform) --- e.g. the work of Gaitsgory, Mirkovic, and Vilonen in the $GL\_n$ setting does this. I don't know how much is
known in the geometric setting about going backwards, from automorphic sheaves to local systems.
(2) Regarding the status of function field Langlands in general: it is important, and open, other than in the $GL\_n$ case of Lafforgue, and various other special cases. (As in the number field setting, there are many special cases known, but these are far from the general problem of functoriality. Langlands writes in the notes on his [collected works](http://publications.ias.edu/rpl/series.php?series=51) that "I do not believe that much has yet been done beyond the group $GL(n)$''.) Langlands has initiated a program called [``Beyond endoscopy''](http://publications.ias.edu/rpl/comments.php?paper=263) to approach the general question of functoriality. In the number field case, it seems to rely on unknown (and seemingly out of reach) problems of analytic number theory, but in the function field case there is some chance to approach these questions geometrically instead. This is a subject of ongoing research.
| 22 | https://mathoverflow.net/users/2874 | 10600 | 7,237 |
https://mathoverflow.net/questions/10607 | 7 | Let $E$ be an elliptic curve over $\mathbb Q\_p$. It is possible that $E$ has bad reduction but then when you see $E$ as a curve over a finite extension $K$ of $\mathbb Q\_p$, it obtains good reduction. Let $v$ be the valuation defined on $K$ and $R$ its valuation ring. I was interested in checking $E$ has good reduction over $K$ by hand, using the Weierstrass equation. What that amounts to then is writing down the Weierstrass equation $y^2+a\_1xy + a\_3y = x^3 + a\_2x^2+a\_4x + a\_6$ with the $a\_i \in R$ and considering changes of coordinates $x=u^2x' + r$ and $y=u^3y' + u^2sx' + t$ for $u,r,s,t \in R$ in hopes of finding an equation with $v(\Delta')$ minimized, subject to each $a\_i'$ being in $R$. There are certain congruence conditions that guarantee minimality of the new equation, e.g. $v(\Delta') < 12$, which only depend on the choice of $u$. However, guaranteeing the new equation has coefficients in $R$ requires solving other congruence relations depending on $r,s$ and $t$, e.g. you need $v(a\_1+2s)\geq v(u)$ (because $a\_1' = u^{-1}(a\_1+2s)$). The few times I have done this by hand, I have just had to look at the equations and make some choices until something worked out.
My question is whether or not there exists a general method for obtaining a good change of coordinates $u,r,s,t$ and if not, then how do people go about writing down minimal Weierstrass models. I can't imagine there should be general methods for solving the system of non-linear congruences (higher powers of $u,r,s$ and $t$ appear in the other congruences) in the ring $R$, but if there is then I would also be interested in understanding that as well.
| https://mathoverflow.net/users/389 | Writing down minimal Weierstrass equations | A slightly more theoretical answer: there is an algorithm of Tate, called (unremarkably)
Tate's algorithm, which allows one to compute the minimal model over any local field. I have a vague memory that it's not proved that this algorithm terminates in general, although it is expected to. (Perhaps someone else can say something definitive about this.) I would guess that this is what is implemented in Pari.
| 6 | https://mathoverflow.net/users/2874 | 10614 | 7,246 |
https://mathoverflow.net/questions/10556 | 51 | Given a variety (or scheme, or stack, or presheaf on the category of rings), some geometers, myself included, like to study D-modules. The usual definition of a D-module is as sheaves of modules over a sheaf of differential operators, but for spaces that aren't smooth in some sense, this definition doesn't work that well, and you want to use a different definition. My overall question is how to reinterpret microlocalization in this alternative definition.
deRham spaces
-------------
This definition is that a D-module on $X$ is a quasi-coherent sheaf on a new space $X\_{dR}$, the *deRham space* of $X$. It's easiest to define this is in terms of its functor of points: a map of Spec R to $X\_{dR}$ is by definition a map of Spec $R/J\_R$ to $X$ where $J\_R$ is the nilpotent radical of $R$. So this is not a topological space, but it is a sheaf on the big Zariski site, and I can make sense of a quasi-0-coherent sheaf on one of those. For more details, you can see the [notes of Jacob Lurie](http://www.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/Nov17-19(Crystals).pdf) on these.
More informally $X\_{dR}$ is $X$ "with all infinitesimally close points identified." A sheaf on this space is like a D-module in that a D-module is a sheaf with a connection, i.e. where the fibers of infinitesimally close points are identified. You'll note, I say "space" here, since I want to be vague about what this object is. It's very hard from being a scheme, but I believe it is a (**EDIT:** not actually algebraic!) stack.
microlocalization
-----------------
Now, one of the lovely things about D-modules is that they have a secret life on the cotangent bundle of X. You might think a D-module is a sheaf on X, but this is not the whole picture: there is also a microlocal version of things.
The sheaf of functions on $T^\*X$ has a quantization $\mathcal{O}^h$; this is a non-commutative algebra over $\mathbb{C}[[h]]$ such that $\mathcal{O}^h/h\mathcal{O}^h\cong \mathcal{O}\_{T^\*X}$, defined using [Moyal quantization](http://en.wikipedia.org/wiki/Moyal_product).
There's a ring map $p^{-1}\mathcal{D}\to \mathcal{O}^h[h^{-1}]$, and thus a functor from D-modules to sheaves of $\mathcal{O}^h[h^{-1}]$-modules on $T^\*X$ given by $\mathcal{O}^h[h^{-1}]\otimes\_{p^{-1}\mathcal{D}}\mathcal{M}$, called microlocalization, because it makes D-modules even more local than they were before. This is an equivalence between D-modules and $\mathbb{C}^\*$-equivariant $\mathcal{O}^h[h^{-1}]$-modules.
Given an $\mathbb{C}^\*$ invariant open subset $U$ of $T^\*X$, one can look at $\mathcal{O}^h[h^{-1}]$-modules on $U$, and obtain a *microlocalized category of D-modules*, which has all kinds of interesting geometry one couldn't see before. I'm particularly interested in the semi-stable points for the action of some group $G$ on $X$ (extended to $T^\*X$).
my question:
------------
Now, I'm something of a convert to derived algebraic geometry, so it feels intuitive to me that anything one has to say about D-modules should be sayable using deRham spaces. On the other hand, I have no idea how microlocalization can be phrased in this way. Do any of you out in MathOverflowLand?
| https://mathoverflow.net/users/66 | D-modules, deRham spaces and microlocalization | The first thing to say is that the abelian category of
sheaves on the de Rham space is only a good model
for D-modules if you're in the smooth setting, or very close to it (see for example arXiv:math/0212094 for a setting where all the different notions agree).. so unless you're fully derived you need to be careful with this identification.
In any case, rather than talk about the de Rham space you can talk about crystals,
which give the right notion in general -- see for example chapter 7 (section 10 or 11) of Beilinson-Drinfeld on Quantization of Hitchin for an excellent discussion.
Another picture you might like better is as modules over the enveloping algebroid of the tangent complex -- ie everything is fine if you replace naive differential operators by their correct derived analog. Yet another picture is as dg modules for the de Rham complex. My favorite is as S^1-equivariant sheaves on the derived loop space of X.
(I assume for all of these we're in characteristic zero, otherwise there are many different analogs of D-modules..)
In any case micrlocalization can be said in terms of the deformation to the normal cone of the de Rham groupoid - or the Hodge filtration on nonabelian cohomology, after Simpson. (A good reference for this is Simpson's paper with that title and the awesome preprint on D-modules on stacks by Simpson and Teleman available on the latter's webpage.) Namely there's a canonical deformation (the Hodge stack) from the de Rham space to a stack (not a space this time!) which is the quotient of X by the tangent bundle (acting trivially) -- sheaves on which are the same as quasicoherent sheaves on the cotangent bundle. On the level of sheaves this is just the deformation quantization from sheaves on the cotangent bundle to D-modules with a parameter h - ie we consider modules over the Rees algebra of D rather than over D itself. The category of Rees modules sheafifies over the projectivized cotangent bundle, and so you can define microlocal categories by restricting to your favorite open subsets.
This is not so special to the de Rham stack: we're just saying jets of sections of a formal deformation of the category of sheaves on a variety sheafify over this variety. Once you know in general that D-modules degenerate to sheaves on the cotangent bundle, and this is true in arbitrary generality once you define both sides correctly, you can microlocalize (if you take into account correctly the C^\* equivariance of the deformation).
| 41 | https://mathoverflow.net/users/582 | 10617 | 7,249 |
https://mathoverflow.net/questions/10635 | 48 | I remember one of my professors mentioning this fact during a class I took a while back, but when I searched my notes (and my textbook) I couldn't find any mention of it, let alone the proof.
My best guess is that it has something to do with Galois theory, since it's enough to prove that the characters are rational - maybe we have to find some way to have the symmetric group act on the Galois group of a representation or something. It would be nice if an idea along these lines worked, because then we could probably generalize to draw conclusions about the field generated by the characters of any group. Is this the case?
| https://mathoverflow.net/users/2363 | Why are the characters of the symmetric group integer-valued? | If $g$ is an element of order $m$ in a group $G$, and $V$ a complex representation of $G$, then $\chi\_V(g)$ lies in $F=\mathbb{Q}(\zeta\_m)$. Since the Galois group of $F/\mathbb{Q}$ is $(\mathbb{Z}/m)^\times$, for any $k$ relatively prime to $m$ the elements $\chi\_V(g)$ and $\chi\_V(g^k)$ differ by the action of the appropriate element of the Galois group.
If $G$ is a symmetric group and $g$ an element as above, then $g$ and $g^k$ are conjugate: they have the same cycle decomposition. So $\chi\_V(g)=\chi\_V(g^k)$ whenever $(k,m)=1$, and thus $\chi\_V(g)\in \mathbb{Q}$.
Now, because $\chi\_V(g)$ is an algebraic integer (true for every finite group, every complex character) and a rational number, it is a rational integer, that is, an integer: $\chi\_V(g)\in\mathbb{Z}$.
| 77 | https://mathoverflow.net/users/437 | 10642 | 7,265 |
https://mathoverflow.net/questions/10627 | 18 | Excuse me for the specificity of this question, but this is a silly computation that's been giving me trouble for some time.
I want to explicitly realize the order 21 Frobenius group over ℂ(x), as ℂ(x,y,z) where y3=g(x) and z7=h(x,y). The order 21 Frobenius group is C7⋊C3, where the generator of C3 acts by taking the generator of C7 to its square. Or in other words < a,b|a3=1, b7=1, ba=a(b2) >. Furthermore, I want it to branch at exactly three points, two of which will have 7 preimages, each with ramification 3, and the third will have 3 points over it with ramification 7 each.
This can easily be shown to exist: Take ℙ1 minus three points (say x=6,5,2); look at its algebraic fundamental group (=the profinite completion of < c, d, t| cdt = 1 >), and map it to the Frobenius group surjectively by c goes to a, d goes to b, and t goes to b-1a-1. This gives you a Galois cover of ℙ1, with said ramification (because order(a)=3, order(b)=7, and order(b-1a-1)=3), and group the 21 order Frobenius group.
Of course, this construction is extremely difficult to track because of the topology involved. It would be much easier to deduce the field extension from the ramification behavior.
So: this can be broken down to two cyclic Galois extensions. The first, a ℂ(x,y), of the form y3=(x-2)2(x-6) is pretty easy to deduce (I need it to ramify at x=2 and x=6 and nowhere else -- this must be the equation up to change of variables). The second, a z7=h(x,y) is tricky. I want it to ramify only above x=5. There's some Abhyankar's lemma things going on here, and that makes the guesswork difficult, and my life much harder.
I should note that the distinction of the Frobenius group of order 21, and the reason that I'm at all interested in this example, is that it is the only order 21 group which isn't cyclic. Geometrically, it means that h is a function of both x and y, and not just x.
Thanks in advance.
| https://mathoverflow.net/users/2665 | A Galois Theory Computation | I'd like to change your numbers slightly. (EDIT: Slight adjustment to make the formula nicer and address the correction in the comments, nothing to see here)
One solution is to set $y^3 = x$, triply ramified only over $0$ and $\infty$, and if we want the 7-fold ramification over $x=1$ (which has solutions $y=1,\omega,\omega^2$) we set $z^7 = (1-y)(1 - \omega y)^2(1-\omega^2 y)^4$, which only ramifies over the three preimages. To show this is Galois, it suffices to show that the automorphism $y \mapsto \omega^2 y$ lifts to an automorphism of the whole field. This automorphism maps $(1-y)(1-\omega y)^2(1-\omega^2 y)^4$ to $(1-\omega^2 y)(1- y)^2(1- \omega y)^4$, which has seventh root $z^2/(1-\omega^2 y)$.
| 8 | https://mathoverflow.net/users/360 | 10650 | 7,271 |
https://mathoverflow.net/questions/9874 | 5 | This is a revised version of [a question I already posted](https://mathoverflow.net/questions/9393/can-topologies-induce-a-metric), but which patently was ill posed. Please give me another try.
---
For comparison's sake, the axioms of a metric:
Axiom A1: $(\forall x)\ d(x,x) = 0$
Axiom A2: $(\forall x,y)\ d(x,y) = 0 \rightarrow x = y$
Axiom A3: $(\forall x,y)\ d(x,y) = d(y,x)$
Axiom A4: $(\forall x,y,z)\ d(x,y) + d(y,z) \geq d(x,z)$
---
Let $T$ = {X,T} be a topology, $B$ a base of $T$, $x, y, z$ $\in$ X
Definition D0: $x$ **is nearer to** $y$ **than to** $z$ **with respect to** $B$ ($N\_Bxyz$) iff $(\exists b \in B)\ x, y \in b \ \& \ z \notin b \ \&\ (\nexists b \in B)\ x, z \in b \ \& \ y \notin b$
Definition D1: $B$ is **pre-metric1** iff $(\forall x,y)\ x \neq y \rightarrow N\_Bxxy$
Definition D2: $B$ is **pre-metric2** iff $(\forall x,y,z)\ ((z \neq x\ \&\ z \neq y) \rightarrow N\_Bxyz) \rightarrow x = y$
Definition D3: $B$ is **pre-metric3** iff $(\forall x,y,z)\ N\_Bzyx \rightarrow (N\_Byxz \rightarrow N\_Bxyz)$
---
Definition: $T$ is pre-metrici iff $(\exists B)\ B$ is pre-metrici (i = 1,2,3).
Definition: $B$ is pre-metric iff $B$ is pre-metric1, pre-metric2 and pre-metric3.
Definition: $T$ is pre-metric iff $(\exists B)\ B$ is pre-metric.
Remark: D1 is an analogue of axiom A1, D2 of axiom A2, D3 of axiom A3.
Remark: $T$ is pre-metric1 iff $T$ is T1 *[not quite sure]*.
Remark: If $T$ is induced by a metric, then $T$ is pre-metric.
---
**Question: Can a property *pre-metric4* be defined such that $T$ induces a metric iff $T$ is induced by a metric** with
Definition: $B$ is **metric** iff $B$ is pre-metric and pre-metric4.
Definition: $T$ **induces a metric** iff $(\exists B)\ B$ is metric.
Remark: Property pre-metric4 should be an analogue of A4 (the triangle inequality).
If provably no such property can be defined does this shed a light on the difference (an asymmetry) between topologies and metric spaces? ("It's the triangle inequality, that cannot be captured topologically.")
| https://mathoverflow.net/users/2672 | Can topologies induce a metric? (revised) | I shall prove that there can be no characterization of metrizability along the lines that you seek. (This argument fleshes out and fulfills the expectation of Mariano in the comments.)
Your axioms are all stated in the language with two types of objects: points and basis elements. So let us be generous here, and entertain the possibility of any statement of a similar type, expressed in the same language. You are considering structures consisting of a set X and a collection B of subsets of X, to be used to generate a topology on X. Your language allows you to quantify over points and over basis elements, and to say that a given point is an element of a given basis element, and so on. This is a perfectly fine formal language to work in. One can express that B is indeed a basis for a topology in this language, by the assertion first that every point is in some basis element and second that for every two basis element b and c, and every point x in both b in c, there is a basis element d such that x is in d and for all points z, if z is in d, then it is both b and c. Similarly, one can state the Hausdorff condition and other topological properties in this language.
But I claim that metrizability is simply not expressible in this language. To see why, suppose toward contradiction that there were an axiom Phi expressible in the language of points and basic open sets in the way we have described, which held exactly of the bases that generate metric topologies. Consider now the real line (R,<). This is a linear order, and the natural basis for the order topology, the collection of open intervals, is first-order definable in the language consisting only of the order. Thus, the collection of basic open sets is describable in the language of the linear order. For example, to quantify over basic open sets, one says, "there are two points a<b such that..." and then refers to the interval (a,b). Since the real line is indeed a metric topology, the collection of intervals in the real line (R,<) will satisfy your axiom Phi.
Now comes the problem. By the methods of [nonstandard analysis](http://en.wikipedia.org/wiki/Non-standard_analysis), there is an extension of the real line to the nonstandard real line, denoted (R\*,<), which contains infinitesimal elements and infinite integers, but which has the amazing property that any statement expressible in the language of < which is true in the real line (R,<) is also true in the nonstandard line (R\*,<). In fact, nonstandard analysis is rather powerful, for one can add the field structure and indeed, any structure whatsover to R and get the nonstandard version in R\*. In particular, your axiom Phi will remain true for the nonstandard reals (R\*,<). That is, your axiom Phi, whatever it is, will remain true for the order topology on the nonstandard real line. But this is a serious problem, because there are nonstandard extensions R\* whose order topology is not metrizable. For example, there are instances of R\* where the cofinality of the positive infinitesimals is uncountable, and in these instances, there are no nontrivial convergent sequences in the order topology of R\*, but the topology is not trivial. Thus, the axiom Phi has made a mistake with R\*, contradiction.
Thus, there can be no axiomatization of metrizability along the lines you suggest.
Edit: In case you suggest that the standard basis for the order should not be the witness for Phi in the real line, but instead some other weird basis B is the one satisfying Phi,
then we simply consider the nonstandard (R\*,<) together with B\*. By the transfer principle, B\* is a basis for the order topology on (R\*,<) and continues to satisfy Phi, but R\* is not metrizable, for the same contradiction.
| 10 | https://mathoverflow.net/users/1946 | 10654 | 7,275 |
https://mathoverflow.net/questions/10630 | 58 | Suppose for some reason one would be expecting a formula of the kind
$$\mathop{\text{ch}}(f\_!\mathcal F)\ =\ f\_\*(\mathop{\text{ch}}(\mathcal F)\cdot t\_f)$$
valid in $H^\*(Y)$ where
* $f:X\to Y$ is a proper morphism with $X$ and $Y$ smooth and quasiprojective,
* $\mathcal F\in D^b(X)$ is a bounded complex of coherent sheaves on $X$,
* $f\_!: D^b(X)\to D^b(Y)$ is the derived pushforward,
* $\text{ch}:D^b(-)\to H^\*(-)$ denotes the Chern character,
* and $t\_f$ is some cohomology class that depends only on $f$ but not $\mathcal F$.
According to the [Grothendieck–Hirzebruch–Riemann–Roch theorem](http://en.wikipedia.org/wiki/Grothendieck-Riemann-Roch%20theorem) (did I get it right?) this formula is true with $t\_f$ being the relative Todd class of $f$, defined as the [Todd class](http://en.wikipedia.org/wiki/Todd_class) of relative tangent bundle $T\_f$.
So, let's play at "guessing" the $t\_f$ pretending we didn't know GHRR ($t\_f$ are [not uniquely defined](https://mathoverflow.net/questions/10631/homology-class-orthogonal-to-image-of-chern-characters/10660#10660), so add conditions on $t\_f$ in necessary).
>
> **Question.** Expecting the formula of the above kind, how to find out that $t\_f = \text{td}\, T\_f$?
>
>
>
You don't have to show this choice works (that is, prove GHRR), but you have to show no other choice works.
Also, let's not use [Hirzebruch–Riemann–Roch](http://en.wikipedia.org/wiki/Hirzebruch-Riemann-Roch_formula): I'm curious exactly how and where Todd classes will appear.
| https://mathoverflow.net/users/65 | Why do Todd classes appear in Grothendieck-Riemann-Roch formula? | You look at the case when $X=D$ is a Cartier divisor on $Y$ (so that the relative tangent bundle -- as an element of the K-group -- is the normal bundle $\mathcal N\_{D/X}=\mathcal O\_D(D)$ (conveniently a line bundle, so is its own Chern root), and $\mathcal F=\mathcal O\_D$. And the Todd class pops out right away.
Indeed from the exact sequence $0\to \mathcal O\_Y(-D) \to \mathcal O\_Y\to \mathcal O\_D\to 0$, you get that
$$ch(f\_! \mathcal O\_D)=ch(O\_Y(D))=
ch(\mathcal O\_Y) - ch(\mathcal O\_Y(-D)) = 1- e^{-D}.$$
And you need to compare this with the pushforward of $[D]$ in the Chow group, which is $D$. The ratio
$$ \frac{D}{1-e^{-D}} = Td( \mathcal O(D) )$$
is what you are after. Now you have just discovered the Todd class. I suspect that this is how Grothendieck discovered his formula, too -- after seeing that this case fits with Hirzebruch's formula, that the same Todd class appears in both cases.
| 54 | https://mathoverflow.net/users/1784 | 10655 | 7,276 |
https://mathoverflow.net/questions/10631 | 3 | I had this simple question when formulating the [Todd class question](https://mathoverflow.net/questions/10630/why-todd-classes-appear-in-grothendieck-riemann-roch-formula).
>
> Does there exist an example of proper morphism $f:X\to Y$ together with nontrivial homology class $t\in H^\*(X)$ such that for all coherent sheaves on $X$ the equality $f\_\*(\mathop{\text{ch}}(u)\cdot t) = 0$ holds?
>
>
>
| https://mathoverflow.net/users/65 | Homology class orthogonal to image of Chern characters? | Why not just take a class that is orthogonal to all the algebraic classes on $X$? For instance you can take $Y$ to be a point, and take $X$ to be a generic abelian surface over $\mathbb{C}$, i.e. and abelian surface with $NS(X) = \mathbb{Z}$. The Chern character of any coherent sheaf on $X$ is contained in $H^{0}(X)\oplus NS(X)\oplus H^{4}(X)$. Take now any $t \in NS(X)^{\perp} \subset H^{2}(X)$, i.e. a transcendental cohomology class of degree two.
| 8 | https://mathoverflow.net/users/439 | 10660 | 7,281 |
https://mathoverflow.net/questions/10672 | 4 | Suppose we have a space M with a real-valued, differentiable function F on M. Under what conditions on F will the Euler characteristic of M be expressed as a (signed) sum of Euler characteristics of components of the critical set for F? Can we relax the Morse-Bott requirement? What if the critical set isn't smooth... can we ever say anything?
Thanks! (I've also just asked a longer companion question on similar things).
| https://mathoverflow.net/users/492 | Morse theory and Euler characteristics | You could have a smooth function $f : \Bbb R \to \Bbb R$ whose critical point set is a Cantor set (minima) and the centres of the complementary intervals (local maxima) -- let $f$ be some suitable smoothing of the distance function from the Cantor set (or you could use the smooth Urysohn lemma to construct the function), say. The two sets (local max / local min) don't have the same cardinalities so you've got no hope of a formula.
I suppose the most direct case where it should fail for the critical point set a manifold, is when the Hessian is non-degenerate in the normal directions but having non-constant signature over an individual critical component. Off the top of my head I don't have an example but the proof (Bott's proof) fundamentally breaks down in this situation so this is where I would look first.
| 3 | https://mathoverflow.net/users/1465 | 10676 | 7,288 |
https://mathoverflow.net/questions/10680 | 0 | Among all $n$-vertex graphs with $M$ edges and constant $k$,how to estimate the fraction of graphs of clique less than $k$? Thanks.
| https://mathoverflow.net/users/2976 | How to estimate the fraction of graphs with small clique among the graphs with certain edges | A little further elaboration on what you're looking for would be appreciated; as you've asked it, this could be anything from an elementary probability exercise to what I think might be an open problem!
So if we fix k and take n large, then the answer depends entirely on the size of M compared with n. The term of art is "threshold function"; see [here](http://en.wikipedia.org/wiki/Erd%25C5%2591s%25E2%2580%2593R%25C3%25A9nyi_model) for more details. Specifically, for fixed k, if $M >> n^{2-2/(k-1)}$ then almost every graph with n vertices and M edges has a clique of size k; if $M << n^{2-2/(k-1)}$ then almost no graph with those parameters has a clique of size k. This can be proved by a straightforward application of linearity of expectation and bounding the variance (Chebyshev's inequality), using the fact that every subgraph of $K\_k$ has a smaller ratio of edges to vertices than does $K\_k$ itself. (See Ch. 4 of Alon and Spencer for more details.) When $M$ has the same growth rate as $n^{2-2/(k-1)}$ the analysis is considerably subtler.
If we fix $M/n^2$ to be some constant (say around 1/4), then the size of the largest clique grows like log n, and in fact as n goes to infinity (and taking $M/\binom{n}{2} = 1/2$) the clique number becomes concentrated at two points! This should also not be too difficult, but it's in a chapter of Alon-Spencer (Ch. 10) I haven't actually read yet, so I'm just glancing at the proof -- it doesn't look too bad. If you want more details I can try to read it more closely.
In general, the first approach I'd look at would be to estimate the expected number of k-cliques in a random graph and then bound the variance, but this might work less well depending on what exactly you're trying to do.
| 4 | https://mathoverflow.net/users/382 | 10681 | 7,289 |
https://mathoverflow.net/questions/10678 | 11 | Hi,
what comes to the mind of a physicist, when he hears words like filtered ring and associated graded? What do these guys describe? What are basic/typical/illuminating examples in physics?
Of course also mathematicians may answer from their perspective :)
Edit: Information on my background: My "primeval" motivation to understand filtered rings/graded rings, comes from mathematics: They are the basics for D-Module theory.
Though I would also like to see these rings from a different perspective, therefor I asked the question.
My knowledge in physics is a bit limited, I only attended some undergraduate courses (quantum mechanics, electro dynamics, classical mechanics). Though more sophisticated answers are fine, I don't need to understand every detail, I just want to get the flavor.
| https://mathoverflow.net/users/2837 | What are important examples of filtered/graded rings in physics? | It is debatable that a physicist would use those very words, and if they did one would hope their meaning would be the same as for a mathematician, since it means that they are trying to speak the same language.
Having said, and coming from a Physics background, when I first learnt about filtered objects and associated graded objects, I immediately recognised the following examples from Physics. They all have to do with quantisation/classical limit in one way or another.
1. The Clifford algebra is filtered and its associated graded algebra is the exterior algebra. Under the "classical limit" map which takes the Clifford algebra to the exterior algebra, the first nonzero term in the graded commutator of two elements defines a Poisson structure on the exterior algebra. You can then view the Clifford algebra as the quantisation of this Poisson superalgebra. In Physics the exterior algebra is the "phase space" for free fermions and Clifford modules (=representations of the Clifford algebra) are Hilbert spaces for quantized fermions. Things get more interesting when the underlying vector space is infinite-dimensional, since not all Clifford modules are physically equivalent. (The relevant buzzword is *Bogoliubov transformations*; although you would not guess this from [the wikipedia page](http://en.wikipedia.org/wiki/Bogoliubov_transformation).)
2. The algebra of differential operators on $\mathbb{R}^n$, say, is also filtered and the associated graded algebra is the algebra of functions on $T^\*\mathbb{R}^n \cong \mathbb{R}^{2n}$ which are polynomial in the fiber coordinates (=the "momenta"). Again the first nonzero term in the commutator of two differential operators defines the standard Poisson bracket on $T^\*\mathbb{R}^n$ and one can view the algebra of differential operators as a quantisation of this Poisson algebra. In Physics, this corresponds to quantising $n$ free bosons.
In both cases there is no unique section to the map taking a filtered algebra to the associated graded algebra, but one has to make a choice. There are number of more or less standard ones: Weyl ordering for the bosons, complete skewsymmetrisation for the fermions,...
By the way, this (and a lot more) is explained in the fantastic paper [Symplectic reduction, BRS cohomology, and infinite-dimensional Clifford algebras](http://dx.doi.org/10.1016/0003-4916%2887%2990178-3) by Kostant and Sternberg.
---
**Edit** (inspired by Mariano's answer)
Kontsevich's deformation quantization is not just of interest to physicists, but has a quantum field theoretical [reformulation](http://arxiv.org/abs/math/9902090) due to Cattaneo and Felder. It is basically the perturbative computation of the path integral of the Poisson sigma model. (This is analogous to how the perturbative evaluation of the path integral of Chern--Simons theory gives the Vassiliev invariants of (framed) knots.)
The picture that seems to be emerging is that indeed quantisation (be it deformation or path-integral or what have you) of a classical physical system gives rise to a filtered object, filtered by powers of $\hbar$.
| 19 | https://mathoverflow.net/users/394 | 10683 | 7,291 |
https://mathoverflow.net/questions/10679 | 8 | To head off any confusion: I'm talking about the extremal-combinatorics [Sperner's theorem](http://en.wikipedia.org/wiki/Sperner_family), bounding the sizes of antichains in a Boolean lattice.
So the "canonical proof" of this theorem seems to be essentially Lubell's -- it's formulated in several different ways, but my favorite is this. Let $A$ be an antichain in $2^{[n]}$. Pick an arbitrary saturated chain and consider a random automorphism $\phi$ of the poset. Since $A$ is an antichain, so is the image $\phi(A)$, and the expected number of elements of $\phi(A)$ that lie in the distinguished chain is at most 1. Now if $S \subset [n]$ has k elements, then $\phi$ maps $S$ into our distinguished chain with probability $1/\binom{n}{k}$; linearity of expectation gives us the [LYM inequality](http://en.wikipedia.org/wiki/Lubell%25E2%2580%2593Yamamoto%25E2%2580%2593Meshalkin_inequality) and Sperner's theorem follows.
This is a lovely proof, but reading over the DHJ polymath threads I [heard about another approach](http://gowers.wordpress.com/2009/02/06/dhj-the-triangle-removal-approach/) -- apparently the approach Sperner himself used -- which one commenter described by the wonderful phrase "pushing shadows around." Evocative as this phrase is, I'm not actually sure what it means: presumably the idea is to replace an antichain by another antichain, at least as large and whose elements are "closer to the center?" This idea seems like it should work, but I can't quite get it to go through, which makes me think I'm missing something. Does anyone know the details of this proof?
| https://mathoverflow.net/users/382 | Sperner's theorem and "pushing shadows around" | Let $\mathcal{B}$ be a collection of $k$-sets , subsets of an $n$-set $S$. For $k < n$ define the shade of $\mathcal{B}$ to be $$\nabla \mathcal{B}=\{ D\subset S : |D|=k+1,\exists B\in \mathcal{B},B\subset D\}$$ and the shadow of $\mathcal{B}$ to be
$$\Delta \mathcal{B}=\{ D\subset S : |D|=k-1,\exists B\in \mathcal{B},D\subset B\}$$ Sperner proved the lemma: $|\Delta \mathcal{B}|\geq \frac{k}{n-k+1}|\mathcal{B}|$ for $k > 0$, and $|\nabla \mathcal{B}|\geq \frac{n-k}{k+1}|\mathcal{B}|$ for $k < n$.
This implies that if $k\le \frac{1}{2}(n-1)$, then $|\nabla \mathcal{B}|\geq |\mathcal{B}|$ and if $k\geq\frac{1}{2}(n+1)$ then $|\Delta \mathcal{B}|\geq |\mathcal{B}|$. Now the proof proceeds in the obvious steps: Suppose we have an antichain $\mathcal{A}$, and it has $p\_i$ elements of size $i$. If $i < \frac{1}{2}(n-1)$ then from the above we can substitute them with $p\_i$ elements of size $i+1$. Similarly for the elements of size $i > \frac{1}{2}(n+1)$. You end up with an antichain of elements of size $\lfloor \frac{n}{2}\rfloor$ giving you the proof of Sperner's theorem.
I really enjoyed Combinatorics of finite sets by I. Anderson which treats Sperner's theorem, LYM, Erdos-Ko-Rado, Kruskal-Katona etc. The above is just a sketch but there you can find all the details.
| 10 | https://mathoverflow.net/users/2384 | 10689 | 7,294 |
https://mathoverflow.net/questions/10667 | 60 | Let $Df$ denote the derivative of a function $f(x)$ and $\bigtriangledown f=f(x)-f(x-1)$ be the discrete derivative. Using the Taylor series expansion for $f(x-1)$, we easily get $\bigtriangledown = 1- e^{-D}$ or, by taking the inverses,
$$ \frac{1}{\bigtriangledown} = \frac{1}{1-e^{-D}} =
\frac{1}{D}\cdot \frac{D}{1-e^{-D}}=
\frac{1}{D} + \frac12+
\sum\_{k=1}^{\infty} B\_{2k}\frac{D^{2k-1}}{(2k)!}
,$$
where $B\_{2k}$ are Bernoulli numbers.
(**Edit:** I corrected the signs to adhere to the most common conventions.)
Here, $(1/D)g$ is the opposite to the derivative, i.e. the integral; adding the limits this becomes a definite integral $\int\_0^n g(x)dx$. And $(1/\bigtriangledown)g$ is the opposite to the discrete derivative, i.e. the sum $\sum\_{x=1}^n g(x)$. So the above formula, known as Euler-Maclaurin formula, allows one, sometimes, to compute the discrete sum by using the definite integral and some error terms.
Usually, there is a nontrivial remainder in this formula. For example, for $g(x)=1/x$, the remainder is Euler's constant $\gamma\simeq 0.57$. Estimating the remainder and analyzing the convergence of the power series is a long story, which is explained for example in the nice book "Concrete Mathematics" by Graham-Knuth-Patashnik. But the power series becomes finite with zero remainder if $g(x)$ is a polynomial. OK, so far I am just reminding elementary combinatorics.
Now, for my question. In the (Hirzebruch/Grothendieck)-Riemann-Roch formula one of the main ingredients is the Todd class which is defined as the product, going over Chern roots $\alpha$, of the expression $\frac{\alpha}{1-e^{-\alpha}}$. This looks so similar to the above, and so suggestive (especially because in the Hirzebruch's version
$$\chi(X,F) = h^0(F)-h^1(F)+\dots = \int\_X ch(F) Td(T\_X)$$
there is also an "integral", at least in the notation) that it makes me wonder: is there a connection?
The obvious case to try (which I did) is the case when $X=\mathbb P^n$ and $F=\mathcal O(d)$. But the usual proof in that case is a residue computation which, to my eye, does not look anything like Euler-Maclaurin formula.
But is there really a connection?
---
**An edit after many answers:** Although the connection with Khovanskii-Pukhlikov's paper and the consequent work, pointed out by Dmitri and others, is undeniable, it is still not obvious to me how the usual Riemann-Roch for $X=\mathbb P^n$ and $F=\mathcal O(d)$ follows from them. It appears that one has to prove the following nontrivial
**Identity:** The coefficient of $x^n$ in $Td(x)^{n+1}e^{dx}$ equals
$$\frac{1}{n!}
Td(\partial /\partial h\_0) \dots Td(\partial /\partial h\_n)
(d+h\_0+\dots + h\_n)^n |\_{h\_0=\dots h\_n=0}$$
A complete answer to my question would include a proof of this identity or a reference to where this is shown. (I did not find it in the cited papers.) I removed the acceptance to encourage a more complete explanation.
| https://mathoverflow.net/users/1784 | Euler-Maclaurin formula and Riemann-Roch | As far as I understand this connection was observed (and generalised) by Khovanskii and Puhlikov in the article
A. G. Khovanskii and A. V. Pukhlikov, A Riemann-Roch theorem for integrals and sums of quasipolynomials over
virtual polytopes, Algebra and Analysis 4 (1992), 188–216, translation in St. Petersburg Math. J. (1993), no. 4, 789–812.
This is related to toric geometry, for which some really well written introduction articles are contained on the page of David Cox <http://www3.amherst.edu/~dacox/>
Since 1992 many people wrote on this subject, for example
EXACT EULER MACLAURIN FORMULAS FOR SIMPLE LATTICE POLYTOPES
<http://arxiv.org/PS_cache/math/pdf/0507/0507572v2.pdf>
Or Riemann sums over polytopes
<http://arxiv.org/PS_cache/math/pdf/0608/0608171v1.pdf>
| 27 | https://mathoverflow.net/users/943 | 10691 | 7,296 |
https://mathoverflow.net/questions/10701 | 8 | According to general theory, for a square zero thickening defined by an ideal I: SpecA -> SpecA', there is an obstruction of lifting a smooth scheme X over A to a smooth scheme over A' living in H^2(X,T\_X \otimes I).
Can anyone give an example of being obstructed?
e.g. a smooth scheme over F\_p which does not lift smoothly to Z/(p^2)?
| https://mathoverflow.net/users/1657 | obstruction to smooth lifting of smooth schemes | Ravi Vakil's paper *Murphy's Law in Algebraic Geometry …* gives many references of such things: see Section 2 of
<http://arxiv.org/abs/math/0411469>
The first example is due to Serre:
---
Serre, Jean-Pierre
[Exemples de variétés projectives en caractéristique $p$ non relevables en caractéristique zéro](http://www.jstor.org/stable/70614). (French)
Proc. Nat. Acad. Sci. U.S.A. 47 1961 108--109.
Here is the [MathReview](http://www.ams.org/mathscinet-getitem?mr=132067) by I. Barsotti:
An example of a non-singular projective variety $X\_0$, over an algebraically closed field $k$ of characteristic $p$, which is not the image, $\text{mod}\,p$, of any variety $X$ over a complete local ring of characteristic 0 with $k$ as residue field. The variety $X\_0$ is obtained by selecting, in a 5-dimensional projective space $S$, and for $p>5$, a non-singular variety $Y\_0$ which has no fixed point for an abelian finite subgroup $G$ with at least 5 generators of period $p$, of the group $\Pi(k)$ of projective transformations of $S$, but which is transformed into itself by $G$; then $X\_0=Y\_0/G$. The reason for the impossibility is that $\Pi(K)$, for a $K$ of characteristic 0, does not contain a subgroup isomorphic to $G$. {Misprint: on the last line on p. 108 one should read $s(\sigma)=\exp(h(\sigma)N)$.}
---
**ADDENDUM**:
After looking back at the question, my reference to Serre's paper is inappropriate: this is an example of a variety which does not lift *to characteristic zero*, whereas the poster asked for one which didn't lift (even) to $\mathbb{Z}/p^2 \mathbb{Z}$. It is still true that this sort of thing is discussed in Vakil's paper, but now the canonical primary source seems to be
---
Deligne, Pierre; Illusie, Luc
[Relèvements modulo $p^2$ et décomposition du complexe de de Rham](http://link.springer.com/article/10.1007%2FBF01389078). (French) [Liftings modulo $p^2$ and decomposition of the de Rham complex]
Invent. Math. 89 (1987), no. 2, 247--270.
[MathReview](http://www.ams.org/mathscinet-getitem?mr=894379) by Thomas Zink:
The degeneration of the Hodge spectral sequence $H^q(X,\Omega^p\_{X/k}) \Rightarrow H^n\_{\text{DR}}(X/k)$ for a smooth and projective algebraic variety over a field of characteristic zero is a basic fact in algebraic geometry. Nevertheless, only recently has one found an algebraic proof in connection with the comparison of étale and crystalline cohomology (Faltings, Fontaine, Messing).
In this beautiful paper the authors give a short, elementary, algebraic proof for the degeneration by methods in characteristic $p>0$.
Let $k$ be a perfect field of characteristic $p>0$, and let $X$ be a smooth variety over $k$, which lifts to the Witt ring $W\_2(k)$. Denote by $X'$ the variety obtained by base change via the Frobenius automorphism, and let $F:X\to X'$ be the relative Frobenius morphism …. More precisely, such splittings correspond to liftings of $X'$ to $W\_2(k)$. This theorem implies the degeneration of the Hodge spectral sequence for $X$ in dimension $<p$, and, by Raynaud, a Kodaira vanishing theorem for $X$. The corresponding facts in characteristic zero may be deduced by the usual reduction process.
The authors show that their argument extends to the case where $k$ is replaced by an arbitrary base $S$ of characteristic $p$ and $\Omega^\cdot\_{X/S}$ is replaced by differentials with logarithmic poles.
---
| 12 | https://mathoverflow.net/users/1149 | 10704 | 7,304 |
https://mathoverflow.net/questions/10687 | 2 | Googling for "atomic morphism" gives me only 70 results. Is this concept so fruitless or does it have another standard name?
What I mean is a morphism $f: A \rightarrow B$ such that
$$(\forall g,h)\ f = g \circ h \rightarrow (g = f\ \wedge\ h = id\_A)\ \vee\ (g = id\_B\ \wedge\ h = f).$$
| https://mathoverflow.net/users/2672 | Standard name of "atomic morphisms"? | As mentioned in the comments, I would probably call such a morphism "irreducible" or "prime." A "less [evil](http://ncatlab.org/nlab/show/evil)," and perhaps more useful, version would be to ask that if $f = g \circ h$, then either $g$ or $h$ is an isomorphism. In this form, if you regard the multiplicative monoid of a ring as a category with one object, the (noninvertible) irreducible morphisms are precisely the irreducible elements of the ring.
I agree that in "concrete categories" such morphisms are unlikely to be very common or useful, but one other situation in which they arise is free categories on directed graphs. In such a category, the nonidentity irreducible morphisms are precisely the generators (the images of the edges of the directed graph you started from).
| 8 | https://mathoverflow.net/users/49 | 10711 | 7,310 |
https://mathoverflow.net/questions/10609 | 6 | Let $G$ be a compact lie group and $H$ a closed subgroup and hence think of $G/H$ as a homogeneous space.
Then how are the Killing fields on $G/H$ the projection of the right-invariant vector fields on $G$?
In the same vein I would like to know why the following construction works:
If one looks at the tangent vectors at identity on $G$ which are "transverse" to $H$ and then exponentiate it down and flow along it and project it down to $G/H$ then on $G/H$ you would be flowing along the integral curves of the vielbeins on $G/H$.
This gives a computation approach to writing down the vielbeins on $G/H$.
I am thinking of $G/H$ to have the metric induced on it by the bi-invariant metric on $G$.
| https://mathoverflow.net/users/2678 | Killing fields on homogeneous spaces | I think that if you generalize that statement a little it becomes clearer (also the proof).
Let $G$ be any Lie group (not necessarily compact) with a closed subgroup $H$ and a metric (not necessarily positive definite) on $G$ which is $G$-left-invariant and $H$-right-invariant (not necessarily bi-invariant).
These conditions are equivalent to picking a metric (quadratic form) at $Lie(G)$ (the lie algebra of $G$, thought of as the tangent space at the identity) which is invariant under the Adjoint representation of $G$ restricted to $H$. You extend this metric from the identity to all of $G$ by left translations.
Example: $G=SL(2,R)$, $H=SO(2)$, with the Killing metric on $G$ (bi-invariant but not positive definite). In this case $G/H$ is the hyperbolic plane. Also any semi-simple $G$ with the Cartan-Killing metric and a maximal compact $H$ (then $G/H$ is called a symmetric space).
Another example is $G=SO(3)$, $H=SO(2)$ (standard embedding) with left-invariant metric which is not necessarily right-invariant, but $H$-right-invariant. This is a model for a rigid body motion whose ellipsoid of inertia is axially symmetric.
From these conditions you get that the metric descends to $G/H$ ($G$ modulo right traslations by $H$), and that left translations by $G$, which by definition act by isometries on $G$, descend to isometries on $G/H$ (since left and right translations commute, by associativity).
If you want the metric on $G/H$ to be riemannian (ie positive definite) then you need to ask that $Lie(G)/Lie(H)$ is positive definite. This holds in the examples above.
Next pick any vector $v\in Lie(G)$ and extend it to a **right** invariant vector field $X$ on $G$.
Exercise: the flow of $X$ is given by the action of the 1-parameter subgroup of $G$ generated by $v$, $g\_t=exp(tv)$, acting by **left** translations on $G$.
Since left translations are isometries of $G$ it follows that $X$ is Killing. Since $X$ is right invariant it descends to a vector field $\tilde X$ on $G/H$ and the left translations by $g\_t$ descend to the flow of $\tilde X$, which is by isometries, so $\tilde X$ is Killing.
Note that $v\in Lie(G)$ doesn't have to be transverse to $Lie(H)$. Picking $v\in Lie (H)$ generates Killing fields $\tilde X$ with fixed point $[H]\in G/H$.
Another comment is that this construction doesn't generate in general all the Killing fields on $G/H$.
Take for example $G$ compact with bi-invariant metric and $H$ trivial. The construction misses all the left-invariant vector fields on $G$ (generating right translations).
| 9 | https://mathoverflow.net/users/2991 | 10713 | 7,311 |
https://mathoverflow.net/questions/10724 | 8 | I'm a little confused and in need of some clarification about the
relationship between algebraic and holomorphic differential forms:
(1) What is the exact definition of the module of differential
forms of a complex projective variety?
(2) What is the definition of its differential?
(3) Am I right in assuming that the algebraic forms are a
submodule of the holomorphic forms with the two differentials
coinciding in some obvious sense?
| https://mathoverflow.net/users/1977 | Relationship between algebraic and holomorphic differential forms | Yes, every algebraic differential form is holomorphic and yes, the differential preserves the algebraic differential forms. If you are interested in projective smooth varieties then every holomorphic differential form is automatically algebraic thanks to Serre's GAGA. This answers (3).
Concerning (1) and (2) I suggest that you consult some standard reference as Hartshorne's Algebraic Geometry.
**Edit**: As pointed out by Mariano in the comments below there are subtle points when one compares Kahler differentials and holomorphic differentials. I have to confess that I have not thought about them when I first posted my answer above.
Algebraic differential $1$-forms
over a Zariski open set $U$ are elements of the module generated by $adb$ with $a$ and $b$
regular functions over $U$ (hence algebraic) by the relations $d(ab) = adb + b da$, $d (a + b) = da + db$ and $d \lambda =0$ for any complex number $\lambda$. Since these are the rules of calculus there is a natural map to the module of holomorphic $1$-forms over $U$. **This map is injective**, since the regular functions on $U$ are not very different from quotients of polynomials.
If instead of considering the ring $B$ of regular functions over $U$ one considers the ring $A$ of holomorphic functions over $U$ then one can still consider its $A$-module of Kahler differentials. If $U$ has sufficiently many holomorphic functions, for instance if $U$ is Stein, then one now has a surjective map to the holomorphic $1$-forms on $U$ which is **no longer injective** as pointed out by Georges Elencwajg in this [other MO question](https://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials).
| 9 | https://mathoverflow.net/users/605 | 10725 | 7,320 |
https://mathoverflow.net/questions/10736 | 15 | When I am testing conjectures I have about number fields, I usually want to control the ramification, especially minimize to a single prime with tame ramification. Hence, I usually look for fields of prime discriminant (sometimes positive, sometimes negative).
I get the feeling that I cannot be the only one who does this...
And so, are there families of number fields of prime discriminant for each degree? Or at least degree 3 and 4? (They are the coolest. Except quadratics. Of course.) What about: given a prime - can I find a polynomial of degree d with the prime as its discriminant?
| https://mathoverflow.net/users/2024 | Families of number fields of prime discriminant | [Klueners Malle online](http://www.math.uni-duesseldorf.de/~klueners/groups2.html) might be just the thing you're looking for. Make your own lists! And [here](http://www.math.uni-duesseldorf.de/~klueners/minimum/minimum.html)'s some they made themselves, if you run out of ideas.
| 8 | https://mathoverflow.net/users/1384 | 10740 | 7,326 |
https://mathoverflow.net/questions/10718 | 0 | Are there non-regular strong monics in the category of locales?
| https://mathoverflow.net/users/2884 | Strong monics in the category of locales | No. The category of locales is dual to the category of frames, which is monadic (in fact, equationally presented) over Set. Any such category is Barr-exact, and in particular a regular category, which implies that every strong epic of frames is regular—hence every strong monic of locales is regular.
| 5 | https://mathoverflow.net/users/49 | 10745 | 7,330 |
https://mathoverflow.net/questions/10753 | 3 | I was reading [link text](https://mathoverflow.net/questions/10724/relationship-between-algebraic-and-holomorphic-differential-forms)
and these two much simpler questions occurred to me:
(i) What type of algebraic functions on complex *projective* varieties do the holomorphic functions correspond to? The rational functions?
(ii) What type of algebraic functions on complex *algebraic* varieties do the holomorphic functions correspond to? The affine coordinate ring?
| https://mathoverflow.net/users/2612 | Algebraic and Holomorphic Functions | Well, here's a few theorems that might help:
1: On a complex projective variety, a function that is meromorphic on the whole variety is a rational function. You can get this out of the embedding into projective space.
2: On a compact complex manifold, the only globally holomorphic functions are constant. This follows from the maximum principle.
Additionally, if you're looking locally, then on an affine variety, there are a LOT more holomorphic functions than algebraic functions. On $\mathbb{C}$, you have $\mathbb{C}[z]$ for the algebraic functions, and convergent power series for holomorphic, so $e^z$ is holo but not algebraic. But every algebraic function is holomorphic.
| 8 | https://mathoverflow.net/users/622 | 10756 | 7,334 |
https://mathoverflow.net/questions/10743 | 11 | In the following article : "H. Matsumura, P. Monsky, On the automorphisms of hypersurfaces, J. Math. Kyoto Univ. 3 (1964) 347-361", it is shown that in finite characteristic, automorphism groups of smooth hypersurfaces of $\mathbb{P}^N$ are finite (with known exceptions such as quadrics, elliptic curves, K3 surfaces). However, the question of their reducedness is left open. Does anyone know something about it ?
In fact, we need to show that $H^0(X,T\_X)=0$. In characteristic $0$, you can use Bott's theorem to do that. What can you do in finite characteristic ?
| https://mathoverflow.net/users/2868 | Are automorphism groups of hypersurfaces reduced ? | If $X$ is a smooth hypersurface in $\mathbf{P}^{n+1}$ of degree $d$, where $n \ge 1$, $d \ge 3$, and $(n,d) \ne (1,3)$, then $H^0(X,T\_X)=0$ by Theorem 11.5.2 in Katz and Sarnak, *Random matrices, Frobenius eigenvalues, and monodromy*, AMS Colloquium Publications, vol. 45, 1999. So the connected component of the identity of the automorphism group scheme is trivial in these cases. See also Theorem 11.1 in <http://www-math.mit.edu/~poonen/papers/projaut.pdf> ,
which is Poonen, Varieties without extra automorphisms III: hypersurfaces, *Finite fields and their applications* **11** (2005), 230-268.
| 17 | https://mathoverflow.net/users/2757 | 10759 | 7,336 |
https://mathoverflow.net/questions/10758 | 11 | For any UFD $R$, the concept of a primitive polynomial (gcd of the coefficients is 1) makes sense in $R[x]$. The product of two primitive polynomials is primitive (Gauss's Lemma), and certainly 1 is a primitive polynomial, so the primitive polynomials form a multiplicative subset $S$ of $R[x]$ - hence we can form the ring $S^{-1}R[x]$. What can we say about it? What does this look like geometrically?
| https://mathoverflow.net/users/1916 | Localizing at the primitive polynomials? | A prime ideal of $S^{-1}R[X]$ is the extension of a unique prime ideal of $R$, so that the morphism $Spec(S^{-1}R[X])\to Spec(R)$ is a bijection, and even an homeomorphism. All the extensions of residual fields induced are pure transcendental of transcendence degre $1$.
As an example, if you look at the case $R=\mathbb{Z}$, the morphism of schemes you get "puts in family" the extensions of fields $\mathbb{F}\_p\hookrightarrow\mathbb{F}\_p(X)$.
| 10 | https://mathoverflow.net/users/2868 | 10761 | 7,338 |
https://mathoverflow.net/questions/6253 | 4 | Can anything be said about the measure of the *topological* boundary of a Cacciopoli set in $R^n$? Of course, the reduced boundary has finite (n-1)-dimensional Hausdorff measure, but this does not say anything about the topological boundary, for instance, points with density 0 or 1 can still be part of the boundary.
My precise question: if $E$ is a Caccioppoli set, does there exist a measurable Cacciopoli set $F$, such that $E\triangle F$ and $\partial F$ are both Lebesgue null sets?
| https://mathoverflow.net/users/1969 | Lebesgue measure of boundary of Caccioppoli set | The answer is no. Take countably many disjoint closed balls $B\_i$ contained in the square $Q=[0,1]\times [0,1]$ and such that:
(i) Sum of areas of $B\_i$ is less than 1
(ii) Sum of perimeters of $B\_i$ is finite
(iii) $\bigcup B\_i$ is dense in $Q$
Since the series $\sum \chi\_{B\_i}$ converges in BV norm, the set $E=Q\setminus \bigcup B\_i$ has finite perimeter. It also has positive measure and empty interior. Any representative $F$ of the set $E$ also has empty interior and therefore $\partial F$ is not Lebesgue null.
---
By the way, any Lebesgue measurable set E has a representative F with the property
(\*) $0<|F\cap B(x,r)|<|B(x,r)|$ for all $x\in\partial F$ and all $r>0$.
The proof is straightforward: add the points x for which $|E\cap B(x,r)|=|B(x,r)|$ for some r, and throw out all points x such that $|E\cap B(x,r)|=0$ for some $r$. (See Prop. 3.1 in "Minimal surfaces and functions of bounded variation" by E. Giusti.) By virtue of (\*) the set $F$ has the smallest (w.r.t inclusion) topological boundary among all representatives of $E$, so if this representative doesn't help you, nothing does.
| 6 | https://mathoverflow.net/users/2912 | 10766 | 7,340 |
https://mathoverflow.net/questions/10767 | 2 | Is the sum $$ S= \sum\_{n=2}^\infty \frac{1}{ \log^1n \log^2n \log^3n \cdots\log^{TL(n)}n} $$
convergent?
Here
$\log^i n$ denotes the $i$'th iterate of $\log$ (in base 2) of $n$, so $\log^2n$ means $\log\log n$, etc., and
$T(n)$ is the tower of $n$ (stack of $n$ 2's) defined by $T(1)=2$ $T(n+1)=2^{T(n)}$ for $n\ge1$. We then set
$$TL(n) := \text{the "towerian log"} = \sup \bigl\{ k : T(k) \le n < T(k+1) \bigr\} .$$
**MOTIVATION** : Generalizing the following that are called Bertrand series (I think):The harmonic series $\sum\_{n\ge1} 1/n$ is divergent, as are all of the series
$$
\sum\_{n\ge2} \frac{1}{n\log n},\quad
\sum\_{n\ge2} \frac{1}{n\log n\log^2 n},\quad
\sum\_{n\ge2} \frac{1}{n\log n\log^2n\log^3n},\ldots\,.
$$
Here the product of iterated logs is pushed as far as possible, and its size **depends** on the parameter $n$.
| https://mathoverflow.net/users/3005 | Convergence of a general Bertrand series | The sum diverges. This is [Putnam Problem A4, 2008](http://kskedlaya.org/putnam-archive/).
| 9 | https://mathoverflow.net/users/297 | 10769 | 7,341 |
https://mathoverflow.net/questions/10768 | 3 | Let $F=(F\\_n)\\_n$ be an $\ell$-adic sheaf on $X\\_{et}$, for a variety $X$ over an algebraically closed field $k$ of characteristic not equal to $\ell$. Does the presheaf sending $U$ to $H^i(U,F):=\lim\\_n H^i(U,F\\_n)$ sheafify to zero?
| https://mathoverflow.net/users/370 | sheafifying a projective limit of presheaves | CORRECTED ANSWER: I believe that the answer is no, at least in some contexts.
For example, suppose that
$X = $Spec $k$, with $k$ a field, and $F = {\mathbb Z}\\_{\ell}(1)$. Then
$U = $Spec $l$ for some finite separable extension $l$ of $k$, and $H^1(U,F)
= \ell$-adic completion of $l^{\times}$, which I will denote by $\widehat{l^{\times}}$.
Thus the stalk of the presheaf $U \mapsto H^1(U,F)$ (and hence of the associated sheaf)
at the (unique) geometric point of $X$ is the direct limit over $l$ of $\widehat{l^{\times}}.$
This direct limit need not vanish. For example, if $k$ is finite, then so is $l$,
and $\widehat{l^{\times}}$ is just the $\ell$-Sylow subgroup of $l$. Thus the stalk
in this case is just $\bar{k}^{\times}[\ell^{\infty}],$ the group of $\ell$-power roots of unity in $\bar{k}$.
This fits with a certain intuition, namely that one has to go to smaller and small etale neighbourhoods to trivialize $F\_n$ as $n$ increases, and hence one can't kill of cohomology
classes in $H^i(U,F)$ just by restricting to some $V$.
I think that the answer is yes. Here is a proof (hopefully blunder-free):
It is true for the presheaf $U \mapsto H^i(U,F\\_1).$ In other words, if we fix $U$,
then for each element $h \in H^i(U,F\\_1)$ and each geometric point $x$ of $U$,
there is an etale n.h. $V$ of $x$ such that $h\\_{| V} = 0.$ Since $H^i(U,F\\_1)$
is finite dimensional, there is a $V$ that works for the whole of $H^i(U,F\\_1)$ at once.
I claim that then $H^i(U,F\\_n)$ restricts to $0$ on $V$ as well.
To see this, consider the exact sequence $0 \to F\\_n \to F\\_{n+1} \to F\\_1 \to 0.$
Applying $H^i(U,\text{--})$ to this yields a middle exact sequence
$H^i(U,F\\_n) \to H^i(U,F\\_{n+1}) \to H^i(U,F\\_1).$ Applying $H^i(V,\text{--})$
yields a middle exact sequence
$H^i(V,F\\_n)\to H^i(V,F\\_{n+1}) \to H^i(V,F\\_1).$ Restriction gives a map from the
first of these sequences to the second. It is zero on the two outer terms, by induction
together with the case $n = 1$ proved above, and so is zero on the inner term.
This shows that restricting from $U$ to $V$ kills $H^i(U,F\_n)$ for all $n$, and hence
$H^i(U,F)$, as required.
EDIT: As was noted in the comment below, this proof assumes that $F$ is
${\mathbb Z}\_{\ell}$ -flat. Let me sketch an argument that hopefully handles the general case:
Put $F$ in a short exact sequence $0 \to F\\_{tors} \to F \to F\\_{fl} \to 0.$ The same kind
of argument as above reduces us to checking $F\\_{fl}$ and $F\\_{tors}$ separately. The above proof handles the case of $F\\_{fl}$, while $F\\_{tors} = F\\_{tors,n}$ for some large enough $n$,
and so the projective limit collapses in this case and there is nothing to check.
(Note: I am assuming some basic kind of finiteness assumption on $F$ here, so that the above
makes sense. Constructibility should be enough.)
| 3 | https://mathoverflow.net/users/2874 | 10773 | 7,343 |
https://mathoverflow.net/questions/10726 | 4 | Luca Trevisan [here](http://lucatrevisan.wordpress.com/2008/02/16/approximate-counting/) gives a randomized polynomial-time approximation algorithm for #3-coloring given an NP oracle.
In a similar vein, I was wondering if there were any results on $BPP^{NP}\stackrel{?}{=}$ #P - i.e. outputting a correct count for a #P problem under the presence of an NP oracle with high probability.
The ideal result of course would tell whether they were equal or not but since we don't know whether P=#P or P=BPP, we can't prove the above false. So I'm also interested in any results that provide evidence either way or prove the above is true (which I'm guessing it is unlikely to be).
If there are no such results, then is $BPP^{NP}$ generally believed to be equal to #P?
\**Edit: \** As per Mariano's suggestion, [Here](http://qwiki.stanford.edu/wiki/Complexity_Zoo%3AB#bpp)'s the Complexity Zoo's excellent description of the complexity class BPP. And [here](http://qwiki.stanford.edu/wiki/Complexity_Zoo%3ASymbols#sharpp) is the description of the complexity class #P.
Thanks
| https://mathoverflow.net/users/1612 | BPP being equal to #P under Oracle | First, let's be slightly pedantic and not make statements like P = #P, which cannot possibly be true just because P is a set of decision problems and #P is not. To get a decision version of #P, one can use PP, or something like P#P.
About your question, BPPNP is contained in PPP and P#P by Toda's theorem. On the other hand, if P#P were contained in BPPNP, it would imply that PH is contained in BPPNP, which would collapse the polynomial hierarchy to the third (or second?) level, which is considered unlikely.
In conclusion, P#P is considered to be more powerful than NP, BPP, BPPNP and even NPNPNP.
| 9 | https://mathoverflow.net/users/1042 | 10774 | 7,344 |
https://mathoverflow.net/questions/10778 | 1 | $\mathbb{R}^n$ admits a [tessellation by permutohedra](http://en.wikipedia.org/wiki/Permutohedron#Tessellation_of_the_space). The corresponding identification of facets of a permutohedron therefore gives a well-defined space: call it $X\_n$. For example, $X\_2$, the hexagon with opposite sides identified, can be shown to be a 2-torus (see figure 2 in <http://arxiv.org/pdf/cond-mat/0703326v2>). Is $X\_n \cong (S^1)^n$? Why (or why not)?
| https://mathoverflow.net/users/1847 | The topology of periodic permutohedral boundary conditions | The quotient $X\_n$ is the same as the quotient of $\mathbb R^n$ by a subgroup of $\mathbb Z^n$ acting cocompactly through translations. Such a thing is always a torus.
| 6 | https://mathoverflow.net/users/1409 | 10780 | 7,347 |
https://mathoverflow.net/questions/10776 | 3 | In the category of normed vector spaces in which the morphisms are linear contractions, what do products look like?
| https://mathoverflow.net/users/3007 | What are the products in the category of normed vector spaces with linear contractions? | Two words: sup norm.
I.e., the product of a family is the uniformly bounded subset of the cartesian product of the family, and the norm is the smallest uniform bound.
Explicitly, if $\{X\_i\}\_{i\in I}$ is a family of normed vector spaces with all norms ambiguously denoted $\|\cdot\|$, then the product is $X=\{\{a\_i\}\in\prod X\_i:\sup\|a\_i\|<\infty\}$, and for $\{a\_i\}\in X$, $\|\{a\_i\}\|=\sup\|a\_i\|$.
(My intuition came from products of C\*-algebras, where the $\*$-homomorphisms are automatically contractive and products are defined in this way. So I had a good guess and it is easy to check that it works.)
| 6 | https://mathoverflow.net/users/1119 | 10784 | 7,350 |
https://mathoverflow.net/questions/10793 | 4 | There are examples that show the set of extreme points of a compact convex subset of a locally convex topological vector space need not be closed when the real dimension of the space is at least 3. Is it true that the set of extreme points of a compact convex subset must be closed if the locally convex space in question has dimension 2?
| https://mathoverflow.net/users/792 | Compact Convex sets and Extreme Points | By definition, a non-extreme boundary point lies on an open line segment contained in the set, which happens to be an open subset of the boundary in two dimensions. Hence the set of extreme points is a closed subset of the boundary.
| 9 | https://mathoverflow.net/users/2912 | 10795 | 7,356 |
https://mathoverflow.net/questions/10319 | 4 | The yoneda lemma gives us a characterization of $Psh(\mathcal{C})$ that seems very similar to the theory of distributions. That is, we have a notion of representable presheaves, similar to representable distributions. The ability to talk about presheaves as colimits of representables correlates to the more complicated notion of distributions as derivatives and limits of representables. The whole idea of "test objects" is exactly the same as the notion of "test functions" and so on. Is there a deep connection there or is it just *another* case of stretching the terminology?
| https://mathoverflow.net/users/1353 | Distributions as presheaves? | While maybe not exactly what you were after, here is something that you might enjoy looking into, which relates presheaves and distributions.
There exists a category of sheaves on certain test objects, such that
* this category is a [smooth topos](http://ncatlab.org/nlab/show/smooth+topos) into which the category of smooth manifolds embeds full and faithfully.
* in this topos, there exists not only a notion of [infinitesimals](http://ncatlab.org/nlab/show/infinitesimal+object), as in every smooth topos, but also of invertible infinitesimals, in fact, this topos provides a model for [nonstandard analysis](http://ncatlab.org/nlab/show/nonstandard+analysis).
* Accordingly, in this topos distributions on manifolds are given by actual functions - internally in the topos.
So in a way, this topos makes precise and manifest the intuition that distributions are "generalized functions". They *are* functions in this topos.
The topos that I am talking about is described in great detail in section VI of the textbook [Models for Smooth Infinitesimal Analysis](http://ncatlab.org/nlab/show/Models+for+Smooth+Infinitesimal+Analysis). The test objects in this case, i.e. the objects in the site that the topos is a category of sheaves over, are [smooth loci](http://ncatlab.org/nlab/show/smooth+locus).
Distributions are discussed in section VII,3
| 6 | https://mathoverflow.net/users/381 | 10811 | 7,364 |
https://mathoverflow.net/questions/10789 | 7 | Near the bottom of [the nlab page for Banach space](http://ncatlab.org/nlab/show/Banach+space) I see "To be described: duals (p+q=pq)".
Are $(\mathbb{R}^n)\_p$ and $(\mathbb{R}^n)\_q$ dual objects in the closed symmetric monoidal category of Banach spaces and linear contractions (with the tensor product described on that page)?
**Edit**: take n=2, p=1, q=∞. Then the question becomes whether $V \times V$ (which is $V^2$ with the $l\_\infty$ norm) is isomorphic to $(\mathbb{R}^2)\_\infty \otimes V$. But it seems to me that the functor $V \mapsto V \times V$ does not even commute with coproducts... is that right?
| https://mathoverflow.net/users/126667 | Categorical duals in Banach spaces | My suspicion is "no", because if I recall correctly the map $I \to V \otimes V^\*$ naturally lands in the *injective tensor product*, not the *projective tensor product*, and it is the latter which appears as the ``correct'' tensor product for the SMC category of Banach spaces and linear contractions.
In the toy example given, $V\oplus V$ with the sup norm is the same as continuous maps from a 2-point set to $V$, equipped with sup-norm, and I'm pretty sure that this is indeed isometrically linearly isomorphic to ${\mathbb R}^2 \check{\otimes} V$ i.e. the injective tensor product.
EDIT: as Reid points out my remarks above assume without justification that the inj. t.p. does differ from the proj t.p. in the specific case being considered. I *think* this is indeed the case. Take $V$ to be ${\mathbb R}^2$ with usual Euclidean norm. The projective tensor product of $V$ with $V^\\*$ can be identified with $M\_2({\mathbb R})$ equipped with the trace class norm; the injective tensor product would lead to the `same' underlying vector space, equipped with the operator norm. The 2 x 2 identity matrix has trace class norm 2 and operator norm 1, so the two norms are genuinely different.
My answer is still not as clear as it should be, because due to a sluggish and temperamental internet connection I'm having trouble looking up just what the axioms for categorical duals in a SMC are. But if I recall correctly the natural map from $I \to V \otimes V^\\*$ should be given by multiplying a scalar by the vector $e\_1\otimes e\_1 + e\_2\otimes e\_2$ where $e\_1,e\_2$ is an o.n. basis of ${\mathbb R}^2$ -- and that vector does not have norm 1 in the proj t.p. althought it does have norm 1 in the inj t.p.
| 2 | https://mathoverflow.net/users/763 | 10822 | 7,371 |
https://mathoverflow.net/questions/10819 | 3 | Related to this [question](https://mathoverflow.net/questions/3309/are-there-two-groups-which-are-categorically-morita-equivalent-but-only-one-of-wh) I also had some troubles to understand the classification of module categories over $Rep(G)$. Specifically, on page 12 of Ostrik's [paper](http://arxiv.org/PS_cache/math/pdf/0111/0111139v1.pdf) what is the category $\mathrm{Rep}^1(\tilde{H})$? $k^\\*$ acting as "identity character on V" means $a.v=av$ for all $a \in k^\*$ and $v \in V$? Then what is the structure of module category over $Rep(G)$? Tensor product should be after restricting representations of $G$ to $H$ and then inducing back to $\tilde{H}$?
Concretely, I was thinking about the following example. Let $H$ be a subgroup of $G$. Then $Rep(H)$ is a module category over $Rep(G)$ via tensor product as $H$-modules. What is the decomposition of $Rep(H)$ in indecomposable module categories and what are the corresponding subgroups $H$ and cocyles $\omega \in H^2(H,\;k^\*)$ for each indecomposable subcategory?
| https://mathoverflow.net/users/2805 | Module categories over $Rep(G)$. | Sebastian: your definition of Rep^1(\tilde H) is absolutely correct. If you have
a representation of G you can restrict it to H and consider it as a representation
of \tilde H (this operation is called inflation). Now you can tensor it with any
representation of \tilde H; this tensoring preserves Rep^1(\tilde H); this is
a module category structure (same thing was explained above by t3suji).
The category Rep(H) considered as a module category over Rep(G) is indecomposable.
It corresponds to subgroup H and trivial cocycle \omega (so \tilde H is a direct product
of H and multiplicative group G\_m).
| 7 | https://mathoverflow.net/users/4158 | 10828 | 7,374 |
https://mathoverflow.net/questions/10827 | 9 | Warning: older texts use the word "Hopf algebra" for what's now commonly called "bialgebra", whereas now "Hopf" is an extra condition. So as to avoid any confusion, I'll give my definitions before concluding with my question.
Definitions
-----------
Let $C$ be a category with symmetric monoidal structure $\otimes$ and unit $1$ (and either strictify, or decorate all the following equations with associators and unitators and so on). An (associative, unital) *algebra* in $(C,\otimes)$ is an object $V$ along with maps $e: 1\to V$ and $m: V\otimes V \to V$ satisfying associativity and unit axioms: $m\circ(m\otimes \text{id}) = m\circ (\text{id}\otimes m)$ and $m\circ (\text{id}\otimes e) = \text{id} = m\circ (e\otimes \text{id})$. A (coassociative, counital) *coalgebra* is an object $V$ along with maps $\epsilon: V\to 1$ and $\Delta: V \to V\otimes V$ satisfying coassociativity and counit axioms. A *bialgebra* is any of the following equivalent things:
* A coalgebra in the category of algebras and algebra-homomorphisms ($1$ has its canonical algebra structure coming from the $\otimes$ axioms that $1\otimes 1 = 1$; in the tensor product of algebras, elements in the different multiplicands commute)
* An algebra in the category of coalgebras and coalgebra-homomorphisms
* An object $V$ with maps $e,m,\epsilon,\Delta$ satisfying the axioms above and a compatibility axiom:
$$ \Delta \circ m = (m\otimes m) \circ (\text{id} \otimes \text{flip} \otimes \text{id}) \circ (\Delta \otimes \Delta) $$
A bialgebra can have the property of being *Hopf* (it is a property, not extra data): a bialgebra $V$ is *Hopf* if there exists an *antipode* map $s: V\to V$ satisfying
$$ m \circ (s\otimes \text{id}) \circ \Delta = e\circ \epsilon = m \circ (\text{id} \otimes s) \circ \Delta $$
Naturally, it's better to see these definitions than read them; check e.g. [the Wikipedia article](http://en.wikipedia.org/wiki/Hopf_algebra). If an antipode exists for a bialgebra, it is unique (justifying considering Hopfness a property rather than a structure) and it is an antihomomorphism for both the algebra and coalgebra structures.
Let VECT be the category of vector spaces (over your favorite field), with $\otimes$ the usual tensor product and $1$ the ground field. A ($\mathbb N$-)*filtered vector space* is a sequence $V = \{V\_0 \hookrightarrow V\_1 \hookrightarrow V\_2 \hookrightarrow \dots\}$ in VECT. A morphism of filtered vector spaces $V \to W$ is a sequence of morphisms $V\_n \to W\_n$ so that every square commutes: $\{V\_n \hookrightarrow V\_{n+1} \to W\_{n+1}\} = \{V\_n \to W\_n \hookrightarrow W\_{n+1}\}$. Equivalently, a *filtered vector space* is a space $V \in $VECT along with an increasing sequence of subspaces $V\_0 \subseteq V\_1 \subseteq \dots \subseteq V$ such that $V = \bigcup V\_n$, and a linear map of filtered vector spaces $V \to W$ is *filtered* if the image of $V\_n$ lies in $W\_n$ for each $n$.
Because $\otimes$ is exact in VECT (because every monomorphism splits), to a pair $V,W$ of filtered vector spaces we can define an $\mathbb N^2$-filtered space with $(p,q)$-part $V\_p\otimes W\_q$, and then we can define the $\mathbb N$-filtered space $V\otimes W$ by setting $(V\otimes W)\_n$ to be the colimit of the diagram given by all $V\_p\otimes W\_q$ with $p+q \leq n$. Equivalently, we can take the tensor product in VECT of the unions $V = \bigcup V\_n$ and $W = \bigcup W\_n$, and then filter it by declaring that the $n$th part is the union of the $(p\otimes q)$th parts for $p+q = n$.
A ($\mathbb N$-)*graded* vector space is a sequence $\{V\_0,V\_1,V\_2,\dots\}$ in VECT, or equivalently a space $V$ along with a direct sum decomposition $V = \bigoplus V\_n$. A morphism of graded vector spaces preserves the grading.
Let $V$ be a filtered vector space. Its *associated graded* space $\text{gr}V$ is given by $(\text{gr}V)\_n = V\_n / V\_{n-1}$, where $V\_{-1} = 0$, of course. Then $\text{gr}$ is a symmetric monoidal functor, and so takes filtered bialgebras to graded bialgebras.
Question
--------
Let $V$ be a filtered bialgebra, i.e. a bialgebra in the category of filtered vector spaces. Then $\text{gr}V$ is a graded bialgebra. Suppose that $\text{gr}V$ is Hopf. Does it follow that $V$ is Hopf? I.e. suppose that $\text{gr}V$ has an antipode map. Must $V$ have an antipode map?
(Or perhaps it requires additional hypotheses, e.g. that we be in characteristic 0, or that $V$ is *locally finite* in the sense that each $V\_n$ is finite-dimensional?)
| https://mathoverflow.net/users/78 | If associated-graded of a filtered bialgebra is Hopf, does it follow that the original bialgebra was Hopf? | There is a known theorem which, if I correctly understand your question, answers it (because the Hopf algebra antipode is defined as the $\ast$-inverse of $\mathrm{id}$):
**Theorem 1.** Let $A$ be an algebra and $\left(C,\left(C\_n\right)\_{n\geq 0}\right)$ a filtered coalgebra, i. e. a coalgebra $C$ and a sequence $\left(C\_n\right)\_{n\geq 0}$ such that:
$C\_n$ is a vector subspace of $C$ for every $n\geq 0$;
$C=\bigcup\_{n\geq 0}C\_n$;
$\Delta\left(C\_n\right)\subseteq\sum\_{i=0}^n C\_i\otimes C\_{n-i}$ for every $n\geq 0$.
Let $f:C\to A$ be a linear map such that the restriction $f\mid\_{C\_0}:C\_0\to A$ is $\ast$-invertible. Then, $f$ itself is $\ast$-invertible.
*Proof of Theorem 1.* Since $f\mid\_{C\_0}:C\_0\to A$ is $\ast$-invertible, there exists a map from $C\_0$ to $A$ which is the $\ast$-inverse of $f\mid\_{C\_0}$. Let $g$ be an arbitrary linear extension of this map to the whole $C$. So $g:C\to A$ is a linear map such that $g\mid\_{C\_0}:C\_0\to A$ is a $\ast$-inverse of $f\mid\_{C\_0}:C\_0\to A$. In other words, $\left(f\*g\right)\mid\_{C\_0}=\eta\epsilon\mid\_{C\_0}$ (sorry, I call $\eta$ what you denote by $e$). In yet other words, $\phi\mid\_{C\_0}=0$, where $\phi:C\to A$ is the linear map defined by $\phi = \eta\epsilon - f\*g$. An easy induction (only using $\Delta\left(C\_n\right)\subseteq\sum\_{i=0}^n C\_i\otimes C\_{n-i}$ and $\phi\mid\_{C\_0}=0$) shows that $\phi^i\left(C\_n\right)=0$ for every integers $i$ and $n$ satisfying $i>n\geq 0$, where $\phi^i$ means the $i$-th power of $\phi$ with respect to the convolution $\ast$. Thus, the map $\sum\_{i=0}^{\infty} \phi^i:C\to A$ is well-defined (in fact, the sum $\sum\_{i=0}^{\infty} \phi^i:C\to A$ converges pointwise, as $C=\bigcup\_{n\geq 0}C\_n$). But $\left(\eta\epsilon - \phi\right)\ast\left(\sum\_{i=0}^{\infty} \phi^i\right)=\eta\epsilon$ (by the geometric series formula, since $\eta\epsilon$ is the unity of the ring $\mathrm{Hom}\left(C,A\right)$ with multiplication $\ast$) and $\left(\sum\_{i=0}^{\infty} \phi^i\right)\ast\left(\eta\epsilon - \phi\right)=\eta\epsilon$ (for the same reason). Hence, $\eta\epsilon - \phi$ is $\ast$-invertible. But $\eta\epsilon - \phi=-f\*g$ (by the definition of $\phi$). Thus, $-f\*g$ is $\ast$-invertible. Hence, so is $f\*g$, and thus $f$ has a right-sided inverse. Similarly, $g\*f$ is $\ast$-invertible, and thus $f$ has a left-sided inverse. Therefore, $f$ is invertible. Theorem 1 is proven.
EDIT: Okay, let me explain how to get your assertion from Theorem 1: Since $\mathrm{gr} V$ is a Hopf algebra, the identity $\mathrm{id}:\mathrm{gr} V\to\mathrm{gr} V$ has a $\ast$-inverse. Hence, its restriction $\mathrm{id}\mid\_{V\_0}:V\_0\to\mathrm{gr} V$ to the component $V\_0$ of $\mathrm{gr} V$ also has a $\ast$-inverse. This $\ast$-inverse must have its image in $V\_0$ (because otherwise we can chain it with the projection $\mathrm{gr} V\to V\_0$ and get another $\ast$-inverse of $\mathrm{id}\mid\_{V\_0}:V\_0\to\mathrm{gr} V$, but the $\ast$-inverse is unique when it exists, so it must be the same one). Hence, the map $\mathrm{id}\mid\_{V\_0}:V\_0\to V\_0$ has an $\ast$-inverse. Thus, the map $\mathrm{id}\mid\_{V\_0}:V\_0\to V$ has an $\ast$-inverse. Now, Theorem 1 yields that so does $\mathrm{id}:V\to V$, and we are done.
| 5 | https://mathoverflow.net/users/2530 | 10830 | 7,376 |
https://mathoverflow.net/questions/10831 | 19 | I am looking for an example of a function $f$ that is 1) continuous on the closed unit disk, 2) analytic in the interior and 3) cannot be extended analytically to any larger set. A concrete example would be the best but just a proof that some exist would also be nice. (In fact I am not sure they do.)
I know of examples of analytic functions that cannot be extended from the unit disk. Take a lacuanary power series for example with radius of convergence 1. But I am not sure if any of them define a continuous function on the closed unit disk.
| https://mathoverflow.net/users/2888 | Example of continuous function that is analytic on the interior but cannot be analytically continued? | Let $f(z) = \sum z^n/n^2$, which is continuous and bounded on the closed unit disc but not analytic near $1$. Then consider
$$\sum f(z^n)/n^2.$$
This should have a singularity at every root of unity; and should be analytic in the interior because it is uniformly convergent.
| 19 | https://mathoverflow.net/users/297 | 10837 | 7,379 |
https://mathoverflow.net/questions/10842 | 31 | The [Hawaiian Earring](http://en.wikipedia.org/wiki/Hawaiian_earring) is usually constructed as the union of circles of radius 1/n centered at (0,1/n): $\bigcup\_1^\infty \left[ (0, \frac{1}{n}) + \frac{1}{n}S^1 \right]$. However, nothing stops us from using the sequence of radii $1/n^2$ or any other sequence of numbers $a\_n$.
I will call a Hawaiian Earring for a sequence (of distinct real numbers), $A = \lbrace a\_n \rbrace$, the union of circles of radius $a\_n$ centered at $(0, a\_n )$. Let the union inherit its topological structure from $\mathbb{R}^2$. Are all of these spaces homeomorphic?
If $a\_n$ is a monotone decreasing sequence converging to $0$, is its Hawaiian Earring homeomorphic to that of the sequence $\lbrace 1/n \rbrace$?
| https://mathoverflow.net/users/1358 | Are all Hawaiian Earrings homeomorphic? | The Hawaian earring is the one-point compactification of a countable union of open intervals (with the coproduct or disjoint sum topology). This description is independent of the radii used to construct it.
A beautiful reference about this space is [Cannon, J. W.; Conner, G. R. The combinatorial structure of the Hawaiian earring group. Topology Appl. 106 (2000), no. 3, 225--271 [MR1775709](http://www.ams.org/mathscinet-getitem?mr=MR1775709)) If I recall correctly, they prove my claim there.
| 42 | https://mathoverflow.net/users/1409 | 10843 | 7,384 |
https://mathoverflow.net/questions/10848 | 5 | ### Question
Let X be a smooth, projective curve over the algebraic closure of ℚ. Let f:X->ℙ1 be a meromorphic function. Assume that the zeros and the poles are defined over some number field, K. Then does this imply that cf is defined over K, for some c?
If so, do we have to assume that the zeros and poles are individually defined over K, or would this work if they are collectively defined over K as well?
### Clarification
By being collectively defined I mean that there's some K-model of X, XK, and a closed subscheme YK of XK, such that after base change YK becomes the ramification locus of f.
### Thoughts
Let D:=(f). Obviously, H0(O(D),X) is 1 dimensional. DK:=(fK) will also be degree 0, so all that's left to show is that H0(O(DK),XK) is also nonzero. This smacks of some invariance of cohomology theorem. I couldn't quite find the right one to use.
If this line of argument works, this seems to imply that the ramification locus may be defined merely collectively.
| https://mathoverflow.net/users/2665 | Field of Definition of a Meromorphic Function | It is not sufficient that the subscheme of poles and zeroes is defined together over $K$, as the example of the function $(z+i)/(z-i):\mathbb P^1 \to \mathbb P^1$, defined only over $\mathbb Q(i)$, illustrates.
If poles of $f$, which form together a subscheme $D\_\infty$, are defined over $K$, then the line bundle $\mathcal O(D\_\infty)$ is defined over $K$. $f$ can be defined as the section of this line bundle that has the most poles, so it should be defined over $K$ as well.
More formally, suppose $f$ is not defined over $K$ (where $K$ is a separable extension of base field, which is $\mathbb Q$ here anyway). Then for some element $\sigma$ of Galois group of base field $f$ and $\sigma f$ would be sections of $\mathcal O(D\_\infty)$ that have the same poles and zeroes, so their ratio would be a holomorphic function on $X$, thus constant.
Therefore every $\sigma$ must act as a multiplication by constant. It remains to select any source point $x$ and divide the function by $f(x)$. After this, action of $\sigma$ will have to be multiplication by 1, thus a multiple of $f$ is defined over $K$ indeed.
| 4 | https://mathoverflow.net/users/65 | 10850 | 7,389 |
https://mathoverflow.net/questions/245 | 7 | A pivotal monoidal category is called non-degenerate if the inner product $\left(x,y\right) = Tr\left(xy^{\*}\right)$ (where $y^{\*}$ is the dual map) is non-degenerate. As a rule of thumb non-degenerate is closely related to semisimplicity. For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension $0$). Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.
If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?
| https://mathoverflow.net/users/22 | Are abelian non-degenerate tensor categories semisimple? | I believe this is Proposition 5.7 in [Deligne's](http://www.math.ias.edu/files/deligne/Symetrique.pdf) “La Categorie des Representations du Groupe Symetrique S\_t, lorsque t n’est pas un Entier Naturel”. See also [this question](https://mathoverflow.net/questions/10857/is-tensor-product-exact-in-abelian-tensor-categories-with-duals).
| 2 | https://mathoverflow.net/users/297 | 10851 | 7,390 |
https://mathoverflow.net/questions/10832 | 3 | Given a linear map $T:H\to H$ on an inner-product space $H$ and a subspace $K\subseteq H$, define the map $T\_K = \pi\_K T \pi\_K^\* :K \to K$, where $\pi\_K:H\to K$ is the orthogonal projection.
As an important special case, if $H=\mathbb{R}^n$ and $K$ is a coordinate subspace, then with respect to standard bases, $T\_K$ is represented by a principal submatrix of the matrix of $T$.
Is there a standard, or at least widely recognized, name for $T\_K$ when $K$ is not a coordinate subspace of $\mathbb{R}^n$? The book *Matrix Analysis* by R. Bhatia calls $T\_K$ the *compression* of $T$ to $K$, but I haven't seen that word used in this way elsewhere (and it's a tricky word to google).
| https://mathoverflow.net/users/1044 | Standard name for basis-independent submatrices? | The standard name in operator theory is "compression", and its partner in crime is "dilation". I.e., A is a compression of B if and only if B is a dilation of A (although sometimes "dilation" is reserved for cases where the compression respects powers). The Wikipedia entry is not proof, but [here it is](http://en.wikipedia.org/wiki/Compression_%28functional_analysis%29) anyway. As for searches, you'll get some relevant hits from "compression of an operator" with quotes.
[Here](http://books.google.com/books?id=0RIpoVJtUxMC&lpg=PA165&ots=keUgsBtnyP&dq=compression%2520%2522operator%2520theory%2522&pg=PA165#v=onepage&q=compression%2520%2522operator%2520theory%2522&f=false) [are](http://books.google.com/books?id=C-OnRNnnZXUC&lpg=PA467&dq=compression%2520%2522operator%2520theory%2522&client=firefox-a&pg=PA467#v=onepage&q=compression&f=false) [some](http://books.google.com/books?id=ECZvEPOosk8C&lpg=PA117&dq=compression%2520%2522operator%2520theory%2522&lr=&client=firefox-a&pg=PA5-IA3#v=onepage&q=compression&f=false) [examples](http://arxiv.org/PS_cache/arxiv/pdf/0908/0908.0729v2.pdf).
---
Some further remarks:
Sz.-Nagy and Foiaș in [*Harmonic analysis of operators on Hilbert space*](http://books.google.com/books?id=XyioQwAACAAJ&source=gbs_navlinks_s) (1970) use the notation $\text{pr }T$ for the compression of $T$ onto $K$ (see page 10), but apparently without ever giving it a name. The notation is suggestive of "projection", and that is the terminology used by Sarason in "[Generalized interpolation in $H^\infty$](http://www.jstor.org/stable/1994641)" (1967).
Lebow goes into more detail on terminology in "[A note on normal dilations](http://www.jstor.org/stable/2035600?seq=1)" (1965), saying in particular that Sz.-Nagy used "projection". In fact, this is the terminology used by Sz.-Nagy in the celebrated appendix to Riesz and Sz.-Nagy's *[Functional analysis](http://books.google.com/books?id=jlQnThDV41UC&source=gbs_navlinks_s)* (1955), which in turn refers to Halmos's paper "Normal dilations and extensions of operators" (1950) as the first place where "compression" and "dilation" were used. The terminology "strong compression" may be used when the compression respects powers, and this is the same as saying that $K$ is semi-invariant for $T$ (see Sarason's "On spectral sets having connected complement" (1965)). If $K$ is reducing for $T$, i.e., if both $K$ and $K^\perp$ are invariant subspaces for $T$, then Lebow calls the compression a "reduction".
Dixmier gives some terminology in [*von Neumann algebras*](http://books.google.com/books?id=8xSoAAAAIAAJ&source=gbs_navlinks_s) (translated 1981 printing) for the case when the compression is applied to an entire von Neumann algebra of operators, which clashes somewhat with the terminology of Lebow. A von Neumann algebra compressed to the space of a projection in the algebra is called a "reduced" von Neumann algebra (page 19), even though the space is reducing only if the projection is in the center. The compression of a von Neumann algebra onto the space of a projection in the commutant (in which case the compression is a normal $\*$-homomorphism) is called an "induction". If $P$ denotes the orthogonal projection you called $\pi\_K$, then Dixmier uses the notation $T\_K$ or $T\_P$ for the compression, but without ever giving a name to the construction for single operators. On the other hand, Jones and Sunder use "reduction" for what Dixmier calls "induction", more in tune with Lebow, on page 21 of [*Introduction to subfactors*](http://books.google.com/books?id=z1_e7hlKxhkC&source=gbs_navlinks_s) (1997).
I stand by my answer that by now "compression" is most standard for single operators, and it is satisfying to find out that we have Halmos to thank for this.
| 6 | https://mathoverflow.net/users/1119 | 10852 | 7,391 |
https://mathoverflow.net/questions/10857 | 8 | Suppose we are in an abelian tensor category with duals, where all objects have finite length. Let $0 \to A \to B \to C \to 0$ be a short exact sequence and $Z$ an object of the category. Is
$$0 \to Z \otimes A \to Z \otimes B \to Z \otimes C \to 0$$
exact?
**Motivation**: I am reading the proof of Proposition 5.7 in this paper of [Deligne](http://www.math.ias.edu/files/deligne/Symetrique.pdf) and trying to figure out why the lower sequence at the bottom of page 23 is exact. I believe $\mathcal{H}om(X,Y)$ here is $X^{\vee} \otimes Y$, although I have not actually found the point in the paper where he defines it. What he is trying to prove is that the corresponding sequence of external Hom's is exact, so he can't be using that fact.
There are, of course, tons of abelian tensor categories where $\otimes$ is not exact. For example, modules for any commutative ring $A$ which is not a field, with tensor product $\otimes\_A$. But these don't usually have duals for all of their objects.
| https://mathoverflow.net/users/297 | Is tensor product exact in abelian tensor categories with duals? | Yes, because if Z has a dual, then in particular Z ⊗ – has a left adjoint (Z\* ⊗ –) and hence commutes with limits (and similarly with colimits, but that's automatic if the category is closed monoidal).
| 11 | https://mathoverflow.net/users/126667 | 10859 | 7,396 |
https://mathoverflow.net/questions/10860 | 33 | ### Motivation
I learned about this question from a wonderful article [Rational points on curves](https://www.math.mcgill.ca/darmon/pub/Articles/Expository/12.Clay/paper.pdf) by Henri Darmon. He gives a list of statements (some are theorems, some conjectures) of the form
* the set $\{$ objects $\dots$ over field $K$ with good reduction everywhere except set $S$ $\}$ is finite/empty
One interesting thing he mentions is about [abelian schemes](https://en.wikipedia.org/wiki/Abelian_variety#Abelian_scheme) in the most natural case $K = \mathbb Q$, $S$ empty. I think according to the definition we have a trivial example of relative dimension 0.
### Question
>
> Why is the set of non-trivial abelian schemes over $\mathop{\text{Spec}}\mathbb Z$ empty?
>
>
>
### Reference
This is proven in [Il n'y a pas de variété abélienne sur Z](https://doi.org/10.1007/BF01388584 "Fontaine, JM. Invent Math 81, 515–538 (1985). zbMATH review at https://zbmath.org/?q=an:0612.14043") by Fontaine, but I'm asking because: (1) Springer requires subscription, (2) there could be new ideas after 25 years, (3) the text is French and could be hard to read (4) this knowledge is worth disseminating.
| https://mathoverflow.net/users/65 | Why no abelian varieties over Z? | It's a result related in spirit to Minkowski's theorem that $\mathbb Q$ admits no non-trivial unramified extensions. If $A$ is an abelian variety over $\mathbb Q$ with everywhere good reduction, then for any integer $n$ the $n$-torsion scheme $A[n]$ is a finite flat group scheme over $\mathbb Z$. Although this group scheme will be ramified at primes $p$ dividing $n$, Fontaine's theory shows that the ramification is of a rather mild type: so mild, that
a non-trivial such family of $A[n]$ can't exist.
In the last 25 years, there has been much research on related questions, including by
Brumer--Kramer, Schoof, and F. Calegari, among others. (One particularly interesting recent variation is a joint paper of F. Calegari and Dunfield in which they use related ideas to construct a tower of closed hyperbolic 3-manifolds that are rational homology spheres, but whose injectivity radii grow without bound.)
EDIT: I should add that the case of elliptic curves is older, due to Tate I believe,
and uses a different argument: he considers the equation computing the discriminant of
a cubic polynomial $f(x)$ (corresponding to the elliptic curve $y^2 = f(x)$) and shows
that this solution equation has no integral solutions giving a discriminant of $\pm 1$.
This direction of argument generalizes in different ways, but is related to a result of Shafarevic (I think) proving that there are only finitely many elliptic curves with good reduction outside a finite set of primes. (A result which was generalized by Faltings to abelian varieties as part of his proof of Mordell's conjecture.)
Finally, one could add that in Faltings's argument, he also relied crucially on ramification results for $p$-divisible groups, due also to Tate, I think, results which Fontaine's theory generalizes. So one sees that the study of ramification of finite flat groups schemes and $p$-divisible groups (and more generally Fontaine's $p$-adic Hodge theory) plays a crucial role in these sorts of Diophantine questions. A colleague describes it as the ``black magic'' that makes all Diophantine arguments (including Wiles' proof of FLT as well) work.
P.S. It might be useful to give a toy illustrative example of how finite flat group schemes give rise to mildly ramified extensions: consider all the quadratic extensions of $\mathbb Q$ ramified only at $2$: they are ${\mathbb Q}(\sqrt{-1}),$ ${\mathbb Q}(\sqrt{2})$, and ${\mathbb Q}(\sqrt{-2})$, with discriminants $-4$, $8$, and $-8$ respectively. Thus ${\mathbb Q}(\sqrt{-1})$ is the least ramified, and not coincidentally, it is the splitting field of the finite flat group scheme $\mu\_4$ of 4th roots of unity.
| 39 | https://mathoverflow.net/users/2874 | 10861 | 7,397 |
https://mathoverflow.net/questions/10853 | 4 | Consider a surface $S$ smoothly embedded in $\mathbb{R}^3$. Classically, the (Riemannian) curvature of $S$ is described by the second fundamental form, which is constructed from partial derivatives of a local parameterization.
Alternatively, is there a "nice" variational characterization of surface curvature? (E.g., one that does not depend on local parameterizations but only on the metric $g$.) In other words, is there a scalar functional whose minimizer completely describes the Riemannian curvature?
One idea that comes to mind is that Riemannian curvature is the curvature associated with the Levi-Civita connection -- hence, you might try to construct a functional over the set of metric connections on $(S,g)$ that penalizes torsion.
(This question is motivated by discrete (e.g., piecewise linear or simplicial) differential geometry, where local differential quantities are ill-defined but metric quantities are available nonetheless.)
| https://mathoverflow.net/users/1557 | Variational characterization of curvature? | The curvature *is* a local invariant. There is such a thing as the curvature at a point. The curvature is described as a tensor, after all. It is different in, say, symplectic geometry, where because of the Darboux theorem all symplectic manifolds of the same dimension are locally symplectomorphic; a fact usually paraphrased as "there is no symplectic curvature". This probably means that there is no "global invariant" formulation for the curvature.
As for the variational formulation, one possible line of approach would be to set up an action functional on *algebraic curvature tensors*; that is, sections of $S^2\Lambda^2T^\*M$ which are in the kernel of the Bianchi map
$$S^2\Lambda^2T^\*M \to \Lambda^4T^\*M$$
cooked up in such a way that the Euler-Lagrange equations are the differential Bianchi identities, since then such a tensor would be the Riemann curvature tensor of the metric you use to define the action functional and whose Levi-Civita connection appears in the Euler-Lagrange equations.
Your idea about the action functional on the space of connections is what usually goes by the name of the *Palatini* (or *first-order*) formalism in GR. It is convenient in action functionals to treat the conenction and the soldering forms as independent quantities and let the Euler-Lagrange equations impose the torsion-free condition on the connection.
As a typical example, consider the Palatini action
$$ \int\_M R(e,\omega) \mathrm{dvol} $$
where $R$ is formally the scalar curvature but written in terms of the soldering form $e$ and the connection $\omega$. If you vary the action with respect to $e$ and $\omega$ separately you find that $\omega$ has no torsion and that the $M$ is Ricci-flat. To see what you gain in this formalism you just have to contemplate the calculation of the Euler-Lagrange equations for the Einstein-Hilbert action for the same Ricci-flatness condition, namely,
$$ \int\_M R(e) \mathrm{dvol} $$
where now the connection is written explicitly in terms of $e$.
| 13 | https://mathoverflow.net/users/394 | 10874 | 7,405 |
https://mathoverflow.net/questions/10870 | 46 | My question is about the concept of nonstandard metric space that would arise from a use of the nonstandard reals R\* in place of the usual R-valued metric.
That is, let us define that a topological space X is a *nonstandard metric space*, if there is a distance function, not into the reals R, but into some nonstandard R\* in the sense of [nonstandard analysis](https://en.wikipedia.org/wiki/Nonstandard_analysis). That is, there should be a distance function d from X2 into R\*, such that d(x,y)=0 iff x=y, d(x,y)=d(y,x)
and d(x,z) <= d(x,y)+d(y,z). Such a nonstandard metric would give rise to the nonstandard open balls, which would generate a metric-like
topology on X.
There are numerous examples of such spaces, beginning with R\* itself. Indeed, every metric space Y has a nonstandard analogue Y\*, which is a
nonstandard metric space. In addition, there are nonstandard metric spaces that do not arise as Y\* for any metric space Y. Most of these examples
will not be metrizable, since we may assume that R\* has uncountable cofinality (every countable set is bounded), and this will usually prevent
the existence of a countable local basis. That is, the nested sequence of balls around a given point will include the balls of infinitesimal
radius, and the intersection of any countably many will still be bounded away from 0. For example, R\* itself will not be metrizable. The space R\* is not connected, since it is the union of the infinitesimal neighborhoods of each point. In fact, one can show it is totally disconnected.
Nevertheless, it appears to me that these nonstandard metric spaces are as useful in many ways as their standard counterparts. After all, one can still
reason about open balls, with the triangle inequality and whatnot. It's just that the distances might be nonstandard. What's more, the
nonstandard reals offer some opportunities for new topological constructions: in a nonstandard metric space, one has the standard-part operation,
which would be a kind of open-closure of a set---For any set Y, let Y+ be all points infinitesimally close to a point in Y. This is something
like the closure of Y, except that Y+ is open! But we have Y subset Y+, and Y++=Y+, and + respects unions, etc.
In classical topology, we have various [metrization theorems](https://en.wikipedia.org/wiki/Metrization_theorem), such as Urysohn's theorem that any second-countable regular Hausdorff space is metrizable.
**Question.** Which topological spaces admit a nonstandard metric? Do any of the classical metrization theorems admit an analogue for nonstandard metrizability? How can we tell if a given topological space admits a nonstandard metric?
I would also be keen to learn of other interesting aspects of nonstandard metrizability, or to hear of interesting applications.
I have many related questions, such as when is a nonstandard metric space actually metrizable? Is there any version of R\* itself which is metrizable? (e.g. dropping the uncountable cofinality hypothesis)
| https://mathoverflow.net/users/1946 | Which topological spaces admit a nonstandard metric? | The uniformity defined by a \*R-valued metric is of a special kind.
Let $(n\_i)\_{i<\kappa}$ be a cofinal sequence of positive elements in \*R. We may assume that $i < j$ implies that $n\_i/n\_j$ is infinitesimal.
Given a \*R-valued metric space $(X,d)$ we can define a family $(d\_i)\_{i<\kappa}$ of pseudometrics by $d\_i(x,y) = st(b(n\_id(x,y)))$, where $st$ is the standard part function and $b(z) = z/(1+z)$ to make the metrics bounded by $1$. The topology on $X$ is defined by the family of pseudometrics $(d\_i)\_{i<\kappa}$. Note that these pseudometrics have a the following special property:
(+) If $i < j < \kappa$ and $d\_j(x,y) < 1$ then $d\_i(x,y) = 0$.
Every uniform space can be defined by a family of pseudometrics, but it is rather unusual for the family to have property (+) when $\kappa > \omega$.
On the other hand, given a family of pseudometrics $(d\_i)\_{i<\kappa}$ bounded by $1$ with property (+), then we can recover a \*R-valued metric by defining $d(x,y) = d\_i(x,y)/n\_i$, where $i$ is minimal such that $d\_i(x,y) > 0$.
| 28 | https://mathoverflow.net/users/2000 | 10877 | 7,407 |
https://mathoverflow.net/questions/10881 | 1 | Let $E$ be a holomorphic vector bundle over $\mathbb{P}^n\setminus\begin{Bmatrix}[1,0,0,\cdots,0]\end{Bmatrix}$. Let $D$ be a connection on $E$. Let $\widetilde{E}$ be an extension of $E$. Since $\widetilde{E}$ is reflexive, i.e. double dual of $E$ is isomorphic to itself, then up to isomorphism, $\widetilde{E}$ is unique. My questions are
1)
Is it possible that $\widetilde{E}$ is a vector bundle?
2) If $\widetilde{E}$ is a vector bundle, does it admit a connection $\widetilde{D}$ which is naturally induced by $D$?
Edit:For the first question, I just proved that $\widetilde{E}$ is a vector bundle if and only if $\widetilde{E}$ is splits. I am wondering if this result was already known. If so, does any one know any reference on this result?
| https://mathoverflow.net/users/2348 | Extension of a holomorphic vector bundle | Presumably, you assume $n\ge 2$.
1) Is it possible that $\tilde E$ is a vector bundle? Yes. Is it always a vector bundle for any $E$? No. Unless, of course, you assume that the connection is flat and holomorphic, then
it extends essentially for topological reasons.
2) Does the connection extends to $\tilde E$ if it is a vector bundle? Assuming the connection is holomorphic, the answer is yes: locally, the connection is given by a bunch of holomorphic functions that extend across codimension two.
| 1 | https://mathoverflow.net/users/2653 | 10886 | 7,412 |
https://mathoverflow.net/questions/8497 | 7 | Given a $\sqrt{n}\times\sqrt{n}$ piece of the integer $\mathbb{Z}^2$ grid, define a graph by joining any two of these points at unit distance apart. How many spanning trees does this graph have (asymptotically as $n\to\infty$)?
Can you also say something about the triangular grid generated by $(1,0)$ and $(1/2,\sqrt{3}/2)$?
| https://mathoverflow.net/users/932 | Number of spanning trees in a grid | I think the best way to deal with grids is to find the general eigenfunction of the infinite grid, and then apply appropriate boundary conditions. This is an idea of Kenyon, Propp and Wilson, you can find an outline in the very last section of my Diplomarbeit [link text](http://www.mat.univie.ac.at/~kratt/theses/rubey.ps.gz)
They only do it for the square grid, as far as I remember, but I wouldn't be surprised if the very same Ansatz works with the triangular grid.
I think that Richard Kenyon
also shows how to compute the asymptotics in "Long-range properties of spanning trees in Z^2" (you can find it on his homepage) but I didn't check.
A second trick that might be useful for the triangular grid (due to Knuth), is to observe that the dual of the grid is "almost" regular. You can choose to delete the vertex corresponding to the outer face in the Laplacian when applying the matrix tree theorem, and will get a very nice matrix, I suppose.
**update:**
I just found a reference which proves the asymptotics for the triangular grid:
[On the entropy of spanning trees on a large triangular lattice](http://arxiv.org/abs/cond-mat/0309198). The formulas are gorgeous...
I should have remarked that the given reference contains (exact) expressions for the asymptotics of both lattices: the limit of $1/n \ln \tau(G\_n)$, where $\tau(G\_n)$ is the number of spanning trees of the graph with $n$ vertices, is
$$4/\pi\sum\_{n\geq1} \sin(n\pi/2)/n^2 = 1.166 243 616\dots$$
for the square grid (due to Temperly 1972), and
$$5/\pi\sum\_{n\geq1} \sin(n\pi/3)/n^2 = 1.615 329 736 097\dots$$
for the triangular grid (proved in the reference).
| 9 | https://mathoverflow.net/users/3032 | 10895 | 7,418 |
https://mathoverflow.net/questions/10868 | 12 | Let $G$ be a simple graph (undirected, no loops or parallel edges), with maximum degree $\Delta(G)$. I would like to add edges to the graph to make it regular, *without increasing the maximum degree*.
In general this is not possible. (For example, take the 5-vertex graph formed by taking a triangle ($K\_3$) and adding two pendant edges to different vertices.)
However, what if we are also allowed to add vertices? I think I can see how to do it by creating many copies of the graph - so my question is: what is the least number of vertices we need to add?
| https://mathoverflow.net/users/1028 | Regularizing graphs | It is always enough to add k+2 more vertices where k denotes the maximum degree. This is sharp as shown by the graph which is a cycle of length 5 plus two independent edges.
The proof is the following. Add edges between non adjacent vertices whose degree is smaller than k until we can. After we cannot, we have some vertices with degree k and a clique of size l where l is at most k. This means we have l vertices whose degree is at least l-1 and at most k-1. Let us add k+1 or k+2 new vertices to our graph depending on the parity to be specified later. Connect the new vertices to the ones forming a clique such that the degrees of the new vertices differ by at most one from each other. Now we only have to add some edges among the new vertices, thus we have reduced the problem to [degree sequences](http://en.wikipedia.org/wiki/Degree_%2528graph_theory%2529#Degree_sequence) and it is an easy corollary of the Havel-Hakimi theorem that if the parity is correct, then a sequence consisting of only d-1 and d where d is less than the number of vertices is always a valid degree sequence.
| 6 | https://mathoverflow.net/users/955 | 10898 | 7,420 |
https://mathoverflow.net/questions/10897 | 27 | There's a famous story about an exercise from Lang's Algebra that says something along the lines of *"pick up a homological algebra book and prove all of the theorems yourself"*. I cannot find it in the third revised edition, and I'm wondering if it's still in the third revised edition, if it's only in the older editions, or if it's an urban legend.
| https://mathoverflow.net/users/1353 | "Pick up a homological algebra book and prove all of the theorems yourself" (exercise from Lang's Algebra) | It's real, but only in the first and second editions. (I don't have any electronic proof, but I've seen it in my copy of the second edition and someone else's copy of the first edition.) It's the only exercise in the chapter.
The full quote in the second edition is:
>
> *Take any book on homological algebra,
> and prove all the theorems without
> looking at the proofs given in that
> book.*
>
>
> Homological algebra was invented by
> Eilenberg-MacLane. General category
> theory (i. e. the theory of
> arrow-theoretic results) is generally
> known as abstract nonsense (the
> terminology is due to Steenrod).
>
>
>
If I'm not mistaken, the quote is the same in the first edition. First edition: page 105; Second edition: page 175; so you can look in your library to see if I messed up the quote!
And I do have an electronic copy of the third edition, which I've searched to confirm it is not there. The historical remarks were expanded, written less dismissively and put at the intro to Part Four.
**Update**: I've scanned the evidence. [First](https://drive.google.com/file/d/0B_pEp00B111JbkI4d2E1ZXdFWEU/view?usp=sharing&resourcekey=0-g4OtVvq9uUEM54B5nO8qYA), [Second](https://drive.google.com/file/d/0B_pEp00B111JYWU1NmY4MjktZTNmMS00MDE4LTkxMjQtYzQ4NWIxZjc3ZGY3/view?usp=sharing&resourcekey=0-O2_eiGXtzkhnXxk86ZS_Hg).
| 31 | https://mathoverflow.net/users/1119 | 10899 | 7,421 |
https://mathoverflow.net/questions/10914 | 8 | Is there some existing notation for
>
> `\[f(n)\leq g(n)\]` for sufficiently large n
>
>
>
Apart from just writing that itself?
I'm thinking of something compact like the Landau notation $f\ll g$.
(Apologies if this is too specific for MathOverflow - just close it if so. I was also unsure what tags to add, so just edit it accordingly).
| https://mathoverflow.net/users/385 | Notation for eventually less than | In logic, this relation is called *almost* less than or equal, and is denoted with an asterisks on the relation symbol, like this: $f \leq^\* g$.
For example, the [bounding number](https://mathoverflow.net/questions/8972#9027) is the size of the smallest family of functions from N to N that is not bounded with respect to this relation. Under CH, the bounding number is the continuum, but it is consistent with the failure of CH that the bounding number is another intermediate value.
| 13 | https://mathoverflow.net/users/1946 | 10918 | 7,429 |
https://mathoverflow.net/questions/10731 | 44 | EGA IV 17.1.6(i) states that formal smoothness is a source-local property. In other words, a map $X\to Y$ of schemes is formally smooth if and only if there is an open cover $U\_i$ ($i\in I$) of $X$ such that each restriction $U\_i\to Y$ is formally smooth.
It seems however that there is a gap in the proof. The problem is in the third paragraph on page 59 (Pub IHES v 32). The reference to (16.5.17) does not give the conclusion they need. Corollary (16.5.18) does give this conclusion, but it requires a finite presentation assumption. (So, everything is OK for smoothness instead of formal smoothness.)
Quesiton 1: Can someone give a counterexample or a complete proof of 17.1.6(i)? (My bet would be that there's a counterexample.)
I think the right way of fixing this is to change the definition of formal smoothness. Recall that a map $X\to Y$ is said to be formally smooth if for any closed immersion $X'\to Y'$ of affine schemes defined by a square-zero ideal and for map $X'\to X$ and $Y'\to Y$ making the induced square commute, there is a map $Y'\to X$ commuting with all the other maps in the diagram. I think a better definition would to require only that the map $Y'\to X$ exists *locally* on $Y'$.
If I'm not mistaken, this definition has the following advantages over the old one: a) The definition of smoothness (=formally smooth and locally of finite presentation) would remain unchanged. b) It would make formal smoothness a source-local property. (Or if there is no counterexample to 17.1.16(i), then the argument with the new definition would be much easier than an argument like the one in EGA, in that it would not depend on the facts in scheme theory that the sheaf of lifts $Y'\to X$ is a torsor for a sheaf derivations and that therefore, since $Y'$ is affine, there is always a global section.) In particular, it would probably be better suited for maps of general sheaves of sets on the big Zariski topology, rather than just schemes. d) It's a general rule of thumb that, in sheaf theory, it's easier to work with local existential quantifiers than global ones.
Question 2: Does anyone know of any reason why this new definition would be bad?
| https://mathoverflow.net/users/1114 | Possible formal smoothness mistake in EGA | Let me point out the following remark made by Grothendieck in his book "Catégories Cofibreés Additives et Complexe Cotangent Relatif", 9.5.8
Please excuse my translation:
"Let $f:X \rightarrow Y$ be a morphism of schemes. We say that $f$ is "locally formally smooth" if X can be covered by opens $X\_i$ which are formally smooth over $Y$. Evidently, if $f$ is formally smooth, then it is locally formally smooth; I don't know if the converse is true in general. It was this which was affirmed hastily in EGA IV 17.1.6 but the proof is only valid when we assume that the relative $\Omega^1$ is of finite presentation, for example if $f$ is locally of finite type. The Lemma 9.5.7 [loc. cit., not reproduced] implies the following criterion : the map $f$ is locally formally smooth if and only if one has $N\_{X/Y} = 0$ and $\Omega^1\_{X/Y}$ is "locally projective" in the following sense : one can cover $X$ by open affines $X\_i$ with rings $B\_i$ such that over $X\_i$ the quasi-coherent module $\Omega^1\_{X/Y}$ is given by a projective $B\_i$ module. We will know that this condition implies the formal smoothness of $f$ if we can show that for a commutative ring $B$, every $B$ module $N$ which is locally projective is also projective - or what amounts to the same - it satisfies $H^1(X, \operatorname{Hom}(\tilde{N}, J)) = 0$ for each quasi-coherent module $J$ on $\operatorname{Spec}(B)$."
Apparently, this local nature of projectivity was shown soon thereafter by Raynaud and Gruson ("Critéres de platitude et projectivité"). In fact I think they show that it is an fpqc local condition. I think this implies that formal smoothness is even an étale local property.
| 43 | https://mathoverflow.net/users/397 | 10923 | 7,433 |
https://mathoverflow.net/questions/2497 | 5 | The affine variety $Sym^n(\mathbb{C}^2)$ has a natural quantization as a spherical rational Cherednik algebra. Thus, any primitive ideal of the rational Cherednik algebra has an corresponding ideal in $\mathbb{C}[X\_1,Y\_1..,X\_n,Y\_n]^{S\_n}$. By a theorem of Ginzburg, the zero-set of this ideal is the closure of the elements where the partition given by looking at which points coincide is a particular fixed one.
**My question:** Which paritions can come up this way?
| https://mathoverflow.net/users/66 | What are the "special" strata of Sym^n(C^2)? | Just by coincidence, I was catching up on my arxiv reading and noticed that Losev's [paper](http://arxiv.org/abs/1001.0239) appears to answer your question. See part (3) of Theorem 4.3.1 there. He says it was "known previously".
### Edit
To be kinder to the reader: the Cherednik algebra is really a family of algebras depending on a parameter $c$; it looks like Losev's result is that at $c=k/m$ for relatively prime integers $k$ and $m$ the varieties we want correspond to partitions of $n$ which are of the form $(m,m,\dots,m,1,1,\dots,1)$ for some number of $m$'s and some number of $1$'s.
### 2nd Edit
Having looked more carefully at Losev's paper, it seems the idea of the proof is this: in the spirit of [Bezrukavnikov-Etingof](http://arxiv.org/abs/0803.3639), having fixed a partition $\lambda$ Losev gives (his Theorem 1.3.1) an inductive description of primitive ideals with associated variety corresponding to $\lambda$. The partition $\lambda$ corresponds to a "parabolic" subgroup $W$ of $S\_n$ consisting of those permutations fixing a generic point of the corresponding variety, and the set of primitive ideals corresponding to $\lambda$ is in bijection with the set of primitive ideals of finite codimension in the Cherednik algebra attached to $W$ at the same parameter $c=k/m$. But $W$ is a product of symmetric groups, so the classification due to [Berest-Etingof-Ginzburg](http://arxiv.org/abs/math/0208138) of finite dimensional modules for the type $S\_n$ Cherednik algebra (they exist exactly when $c$ has denominator $n$) implies the result.
It appears that ideas from Losev's work on finite $W$-algebras also work for symplectic reflection algebras. It would therefore be nice to understand the finite dimensional modules outside of the symmetric group case a little better. (And now I'm more motivated to learn about $W$-algebras!)
| 4 | https://mathoverflow.net/users/nan | 10924 | 7,434 |
https://mathoverflow.net/questions/10930 | 2 | is it possible to characterize the elements of a (special) direct limit only using the universal property? in detail:
let's first concentrate on the category of sets. by an element, I mean a morphism defined on the terminal object $\{\*\}$. let $A\_1 \to A\_2 \to ...$ be a sequence of sets, and $A$ their colimit which is defined by the universal property (morphisms on $A$ are compatible morphisms on the $A\_n$). can we prove, **without using an explicit construction** of $A$, that every element of $A$ comes from an element in one of the $A\_n$ and that two such elements are equal iff they are equal in some greater $A\_m$?
for example, it can be shown without using an explicit construction that the coproduct $M+N$ of two sets $M,N$ (or an arbitrary family of sets) consists of the elements of $M$ and of $N$ and they exclude each other: denote $i\_M,i\_N$ the units ("inclusions"), then it is easy to see that $M,N \to im(i\_M) \cup im(i\_N)$ also satisfy the universal property so that $M + N = im(i\_M) \cup im(i\_N)$. now if $x \in im(i\_M) \cap im(i\_N)$, there are $m \in M, n \in N$ such that $x = i\_M(m) = i\_N(n)$. define $M \to \{0,1\}$ by sending $m$ to $0$, and the rest to $1$, and $N \to \{0,1\}$ by sending $n$ to $1$, and the rest to $0$. this yields $M + N \to \{0,1\}$ sending $x$ to $0$ and to $1$, contradiction.
perhaps this is not possible for directed limits because we have to use at least some specific properties of sets. but what about topoi or cosmoi (bicomplete closed symmetric monoidal categories)? I'm mainly interested in sets here, but perhaps more abstractions are needed to make clear what the goal is: a pure categorical approach to the usual constructions.
here is a functorial reformulation: let $... \to A\_2 \to A\_1$ be a sequence of (representable) endofunctors of sets and $A$ their explicitely constructed limit, i.e. $A(M)$ consists of compatible elements of the $A\_n(M)$. how can we prove directly that every natural transformation $A \to id$ factors through one of the projections $A \to A\_n$?
even if it is impossible, let's take this as a sort of axiom and consider algebraic structures (the usual ones, over sets). can we prove that the forgetful functor preserves direct limits of $A\_1 \to A\_2 \to ...$? again, without already knowing it by an explicit construction.
ps: there are lots of related questions. they are in the same spirit as my [earlier one](https://mathoverflow.net/questions/9921/equality-of-elements-in-localization-via-universal-property) which has not been answered (although the rules for the bounty decided to accept an answer).
| https://mathoverflow.net/users/2841 | categorical description of elements in a direct limit | As with that [previous question](https://mathoverflow.net/questions/9921/equality-of-elements-in-localization-via-universal-property), I don't understand precisely what the rules of the game are when you say "without using an explicit construction". But maybe I can say something useful.
---
First I'll answer the first part of the first question. Then I'll use it to explain a little of why I don't think there's going to be a precise way to formulate the "no explicit construction" rule.
We have a sequence $A\_1 \to A\_2 \to \cdots$ in the category of sets, and its colimit $A$, and we wish to show that every element of $A$ comes from an element of some $A\_n$. And we wish to do it without using the explicit formula for colimits. Being a category theorist, I'll write 1 rather than $\*$ for a one-element set.
I'll use the fact that the category of sets is a well-pointed topos, **E** say. (Those hypotheses can probably be weakened.) Take an element $a$ of $A$, that is, a map $a: 1 \to A$. Since colimits are stable under pullback in any topos, pulling the colimit cocone back along $a$ gives a sequential colimit
$$
\mathrm{colim}(X\_1 \to X\_2 \to \cdots) = 1
$$
in the category **E**, together with a map $f\_n: X\_n \to A\_n$ for each $n$, making the evident square commute. (Draw a diagram!) Since any colimit of initial objects is initial, and in a well-pointed topos $1$ is not initial, at least one $X\_n$ is not initial. It's a fact that in a well-pointed topos, every object is either initial or admits an element (map from $1$). So, at least one $X\_n$ admits an element, $x$ say. Then $f\_n x$ is an element of $A\_n$, and a quick diagram chase shows that it maps to $a \in A$.
So, I've answered the first question without apparently using any explicit constructions involving sets --- in the sense that I just assumed certain axioms on the category of sets and did the proof categorically. But the thing is, you can *always* do that. "Explicit constructions" can always be translated into categorical arguments (and though it's not apparent from the proof above, that's a totally mechanical process).
If your point of view on sets is that they are what ZFC says they are, then here's an equivalent categorical formulation: sets and functions form a well-pointed topos with natural numbers object and choice, satisfying a first-order axiom scheme of replacement. ZFC and this entirely categorical axiomatization are in a precise sense equivalent. Anything you can do in one context, you can do in the other.
---
On coproducts: it can be shown that in any topos, coproducts are disjoint and the coprojections are jointly epic. I think that's in Mac Lane and Moerdijk's book *Sheaves in Geometry and Logic*. In a well-pointed topos, epic = surjective and monic = injective (where by sur/injective I'm referring to the elementwise notion implicit in the question). Hence your statements on coproducts hold in any well-pointed topos.
---
On the question about algebraic structures:
Colimits (="direct limits") of sequences are an example of filtered colimits --- see *Categories for the Working Mathematician* Chapter IX, for instance. The forgetful functors **Group**$\to$**Set**, **Ring**$\to$**Set**, etc, all preserve filtered colimits. The jargon for this is: "the free group/ring/... monad is **finitary**". As the terminology hints, preservation of filtered colimits corresponds to the fact that the theory of groups, rings, etc. only involves *finitary* operations.
For example, the theory of groups has an operation with 2 arguments (multiplication), an operation with 1 argument (inverse), and an operation with 0 arguments (identity). The numbers 2, 1, and 0 are all finite, so the theory of groups is finitary, so the forgetful functor **Group**$\to$**Set** preserves filtered colimits.
(A non-example would be the theory of ordered sets in which every countable subset had a supremum, and maps that preserved those suprema. One of the operations in the theory is "take the supremum of a countably infinite subset", so this theory isn't finitary. Correspondingly, the forgetful functor from these ordered sets to **Set** won't preserve filtered colimits.)
| 4 | https://mathoverflow.net/users/586 | 10933 | 7,440 |
https://mathoverflow.net/questions/10934 | 27 | The statement that the class number measures the failure of the ring of integers to be a ufd is very common in books. ufd iff class number is 1. This inspires the following question:
Is there a quantitative statement relating the class number of a number field to the failure of unique factorization in the maximal order - other than $h = 1$ iff $R$ is a ufd?
In what sense does a maximal order of class number 3 "fail more" to be a ufd than a maximal order of class number 2?
Is it true that an integer in a field of greater class number will have more distinct representations as the product of irreducible elements than an integer in a field with smaller class number?
| https://mathoverflow.net/users/2024 | Class number measuring the failure of unique factorization | Theorem (Carlitz, 1960): The ring of integers $\mathbb{Z}\_F$ of an algebraic number field $F$ has class number at most $2$ iff for all nonzero nonunits $x \in \mathbb{Z}\_F$, any two factorizations of $x$ into irreducibles have the same number of factors.
A proof of this (and a 1990 generalization of Valenza) can be found in $\S 22.3$ of [my commutative algebra notes](http://alpha.math.uga.edu/%7Epete/integral.pdf).
This paper has spawned a lot of research by ring theorists on **half-factorial domains**: these are rings in which every nonzero nonunit factors into irreducibles and such that the number of irreducible factors is independent of the factorization.
To be honest though, I think there are plenty of number theorists who think of the class number as measuring the failure of unique factorization who don't know Carlitz's theorem (or who know it but are not thinking of it when they make that kind of statement).
Here is another try [**edit: this is essentially the same as Olivier's response, but said differently; I think it is worthwhile to have both**]: when trying to solve certain Diophantine problems (over the integers), one often gets nice results if the class number of a certain number field is prime to a certain quantity. The most famous example of this is Fermat's Last Theorem, which is easy to prove for an odd prime $p$ for which the class number of $\mathbb{Q}(\zeta\_p)$ is prime to $p$: a so-called "regular" prime.
For an application to Mordell equations $y^2 + k = x^3$, see
[http://alpha.math.uga.edu/~pete/4400MordellEquation.pdf](http://alpha.math.uga.edu/%7Epete/4400MordellEquation.pdf)
Especially see Section 4, where the class of rings "of class number prime to 3" is defined axiomatically and applied to the Mordell equation. (N.B.: These notes are written for an advanced undergraduate / first year grad student audience.)
The Mordell equation is probably a better example than the Fermat equation because:
(i) the argument in the "regular" case is more elementary than FLT in the regular case (the latter is too involved to be done in a first course), and
(ii) when the "regularity" hypothesis is dropped, it is not just harder to prove that there are no nontrivial solutions, it is actually very often false!
| 26 | https://mathoverflow.net/users/1149 | 10939 | 7,443 |
https://mathoverflow.net/questions/10948 | 14 | This is a simple question, but its been bugging me. Define the function $\gamma$ on $\mathbb{R}\backslash \mathbb{Z}$ by
$$\gamma(x):=\sum\_{i\in \mathbb{Z}}\frac{1}{(x+i)^2}$$
The sum converges absolutely because it behaves roughly like $\sum\_{i>0}i^{-2}$.
Some quick facts:
* Pretty much by construction, $\gamma$ is periodic with period $1$.
* As it approaches any integer from the left or right, it goes to positive infinity.
* It is symmetric at every half integer; that is, $\gamma(n+1/2+x)=\gamma(n+1/2-x)$ for all $n\in \mathbb{Z}$ and $x\in \mathbb{R}$.
>
> Can $\gamma$ be expressed in terms of more familiar (presumably trigonometric) functions?
>
>
>
My best guess is $\gamma(x)=\sin^{-2}(\pi x)$, but this is based more on what I *hope* it would be, rather than what it is.
| https://mathoverflow.net/users/750 | What is $\sum (x+\mathbb{Z})^{-2}$? | Use residues! For an entertaining narrative with the correct answer, see [here](https://mathoverflow.net/questions/8741/justifying-a-theory-by-a-seemingly-unrelated-example/8752#8752). For a derivation, see a complex analysis text.
### Added
Here's a link to an outline of the residue method of solving this: [Conway page 122](http://books.google.com/books?id=9LtfZr1snG0C&lpg=PP1&dq=complex%2520analysis%2520conway&pg=PA122#v=onepage&q=&f=true).
| 17 | https://mathoverflow.net/users/1119 | 10949 | 7,449 |
https://mathoverflow.net/questions/10947 | 21 | There's a wonderful analogy I've been trying to understand which asserts that field extensions are analogous to covering spaces, Galois groups are analogous to deck transformation groups, and algebraic closures are analogous to universal covering spaces, hence the absolute Galois group is analogous to the fundamental group. (My vague understanding is that the machinery around etale cohomology makes this analogy precise.)
Does the Hilbert class field (of a number field) fit anywhere into this analogy, and how?
Phrased another way, what does the Hilbert class field of the function field of a nonsinguar curve defined over $\mathbb{C}$ (say) look like geometrically?
| https://mathoverflow.net/users/290 | What's the analogue of the Hilbert class field in the following analogy? | This is a great question. Someone will come along with a better answer I'm sure, but here's a bit off the top of my head:
1) The Hilbert class field of a number field $K$ is the maximal everywhere unramified abelian extension of $K$. (Here when we say "$K$" we really mean "$\mathbb{Z}\_K$", the ring of integers. That's important in the language of etale maps, because any finite separable field extension is etale.)
In the case of a curve over $\mathbb{C}$, the "problem" is that there are infinitely many unramified abelian extensions. Indeed, Galois group of such is the abelianization of the fundamental group, which is free abelian of rank $2g$ ($g$ = genus of the curve). Let me call this group G.
This implies that the covering space of C corresponding to G has infinite degree, so is a non-algebraic Riemann surface. In fact, I have never really thought about what it looks like. It's fundamental group is the commutator subgroup of the fundamental group of C, which I believe is a free group of infinite rank. I don't think the field of meromorphic functions on this guy is what you want.
2) On the other hand, the Hilbert class group $G$ of $K$ can be viewed as the Picard group of $\mathbb{Z}\_K$, which classifies line bundles on $\mathbb{Z}\_K$. This generalizes nicely: the Picard group of $C$ is an exension of $\mathbb{Z}$ by a $g$-dimensional complex torus $J(C)$, which has exactly the same abelian fundamental group as $C$ does: indeed their first homology groups are canonically isomorphic. $J(C)$ is called the **Jacobian** of $C$.
3) It is known that every finite unramified abelian covering of $C$ arises by pulling back an isogeny from $J(C)$.
So there are reasonable claims for calling either $G \cong \mathbb{Z}^{2g}$ and $J(C)$ the Hilbert class group of $C$. These two groups are -- canonically, though I didn't explain why -- Pontrjagin dual to each other, whereas a finite abelian group is (non-canonically) self-Pontrjagin dual. [This suggests I may have done something slightly wrong above.]
As to what the Hilbert class field should be, the analogy doesn't seem so precise. Proceeding most literally you might take the direct limit of the function fields of all of the unramified abelian extensions of $C$, but that doesn't look like such a nice field.
Finally, let me note that things work out much more closely if you replace $\mathbb{C}$ with a finite field $\mathbb{F}\_q$. Then the Hilbert class field of the function field of that curve is a finite abelian extension field whose Galois group is isomorphic to $J(C)(\mathbb{F}\_q)$, the (finite!) group of $\mathbb{F}\_q$-rational points on the Jacobian.
| 14 | https://mathoverflow.net/users/1149 | 10952 | 7,451 |
https://mathoverflow.net/questions/10966 | 38 | Let $M$ be a differentiable manifold of dimension $n$. First I give two definitions of Orientability.
The first definition should coincide with what is given in most differential topology text books, for instance Warner's book.
>
> Orientability using differential forms: There exists a nowhere vanishing differential form $\omega$ of degree $n$ on $M$.
>
>
>
The second one is from Greenberg and Harper, "Algebraic Topology". This is the "fundamental class" approach. Let $x$ be a point on $X$, and let $R$ be a commutative ring and in the following the homologies are with coefficients in $R$.
>
> Local orientability: A local $R$ orientation of $X$ at $x$ is a choice of a generator of the $R$-module $H\_n(X, X-x)$.
>
>
>
By a simple application of Excision, it is seen that the above homology module is indeed isomorphic to $R$. We can also so arrange a neighborhood around every point that this local orientation can be "continued to a neighborhood" and is "coherent". Forgive me for being imprecise here; the detailed lemmas are in the reference given above. With this background in mind, we define:
>
> A Global $R$-orientation of $X$ consists of: 1. A family $U\_i$ of open sets covering of $X$, 2. For each $i$, a local orientation $\alpha\_i \in H\_n(X, X -U\_i)$ of along $X$, such that a "compatibility condition" holds.
>
>
>
Here again I am imprecise about the compatibility condition; please check in the reference given above for details. I mean this basically as a question for those who already know both the definitions, as fully writing down the second definition would take 2-3 pages with all the necessary lemmas.
Also we define "orientation" to be a such a global choice.
Now the question:
>
> How do the two definitions, the first one using differential forms, and the second one using homology, match?
>
>
>
Of course, to match we have to take $\mathbb{Z}$ to be the base ring for homology. A related question is about the meaning of orientability and orientation when we take a base ring other than $\mathbb{Z}$. It is nice when the base ring is $\mathbb{Z}/2\mathbb{Z}$; every manifold is orientable. But what on earth does it mean to have $4$ possible orientations for the circle or real line for instance, when you take the base ring to be $\mathbb{Z}/5\mathbb{Z}$?
Also I ask, are there any additional ways to define orientability/orientation for a differentiable manifold(not just for a vector space)?
| https://mathoverflow.net/users/2938 | Two kinds of orientability/orientation for a differentiable manifold | If $X$ is a differentiable manifold, so that both notions are defined, then they coincide.
The ``patching'' of local orientations that you describe can be expressed more formally as follows: there is a locally constant sheaf $\omega\_R$ of $R$-modules on $X$ whose stalk at a point is $H^n(X,X\setminus\{x\}; R).$ Of course, $\omega\_R = R\otimes\_{\mathbb Z} \omega\_{\mathbb Z}$.
This sheaf is called the orientation sheaf, and appears in the formulation of Poincare duality for not-necessarily orientable manifolds. It is not the case that *any* section of this sheaf gives an orientation. (For example, we always have the zero section.)
I think the usual definition would be something like a section which generates each stalk.
I will now work just with $\mathbb Z$ coefficients, and write $\omega = \omega\_{\mathbb Z}$.
Since the stalks of $\omega$ are free of rank one over $\mathbb Z$, to patch them together you
end up giving a 1-cocyle with values in $GL\_1({\mathbb Z}) = \{\pm 1\}.$ Thus underlying
$\omega$ there is a more elemental sheaf, a locally constant sheaf that is a principal bundle for $\{\pm 1\}$. Equivalently, such a thing is just a degree two (not necessarily connected) covering space
of $X$, and it is precisely the orientation double cover of $X$.
Now giving a section of $\omega$ that generates each stalk, i.e. giving an orientation of $X$, is precisely the same as giving a section of the orientation double cover (and so $X$ is orientable, i.e. admits an orientation, precisely when the orientation double cover is disconnected).
Instead of cutting down from a locally constant rank 1 sheaf over $\mathbb Z$ to just a double cover, we could also build up to get some bigger sheaves.
For example, there is the sheaf $\mathcal{C}\_X^{\infty}$ of smooth functions on $X$.
We can form the tensor product $\mathcal{C}\_X^{\infty} \otimes\_{\mathbb Z} \omega,$
to get a locally free sheaf of rank one over ${\mathcal C}^{\infty}$, or equivalently, the sheaf of sections of a line bundle on $X$. This is precisely the line bundle of top-dimensional forms on $X$.
If we give a section of $\omega$ giving rise to an orientation of $X$, call it $\sigma$, then we certainly get a nowhere-zero section
of $\mathcal{C}\_X^{\infty} \otimes\_{\mathbb Z} \omega$, namely $1\otimes\sigma$.
On the other hand, if we have a nowhere zero section of $\mathcal{C}\_X^{\infty} \otimes\_{\mathbb Z}
\omega$, then locally (say on the the members of some cover $\{U\_i\}$ of $X$ by open balls) it has the form $f\_i\otimes\sigma\_i,$ where $f\_i$ is a nowhere zero real-valued function on $U\_i$ and $\sigma\_i$ is a generator of $\omega\_{| U\_i}.$
Since $f\_i$ is nowhere zero, it is either always positive or always negative; write
$\epsilon\_i$ to denote its sign. It is then easy to see that sections $\epsilon\_i\sigma\_i$
of $\omega$ glue together to give a section $\sigma$ of $X$ that provides an orientation.
One also sees that two different nowhere-zero volume forms will give rise to the same orientation if and only if their ratio is an everywhere positive function.
This reconciles the two notions.
| 35 | https://mathoverflow.net/users/2874 | 10968 | 7,459 |
https://mathoverflow.net/questions/10960 | 9 | Suppose X is dimension two locally Noetherian scheme. Y is a closed subscheme of X, with codimension 1. Denote X' to be the blow up of X along Y. Prove that the structure morphism f:X'-->X is a finite morphism.
It suffices to show it's quasi-finite according to Zariski's main theorem. But I can't exclude the possibility that an irreducible component of $f^{-1}(Y)$ maps to a closed point of Y.
| https://mathoverflow.net/users/2008 | Blow up along codimension one closed subscheme | I think it's not true :
Let $X=Spec(A)$ with $A=k[x,y,z]/(x^2-y^2-z^2)$ be a quadratic cone. Let $Y$ be a line through the origin of the cone : its ideal is $I=(z,x-y)$. We calculate :
$$X'=Proj\_{A}A[t,u]/(zt-(x+y)u,(x-y)t-zu),$$ [**EDIT : THE FORMULA HAS BEEN CORRECTED**]
where, in the graded $A$-algebra $A+I+I^2+....$ we denoted $t$ and $u$ the degree one generators corresponding to $z$ and $x-y$.
Now, quotienting by $x$, $y$, and $z$, we calculate the fiber over the origin of this blow-up It is Proj(k[t,u]), which is a positive-dimensional projective line !
| 14 | https://mathoverflow.net/users/2868 | 10970 | 7,461 |
https://mathoverflow.net/questions/10971 | 17 | I've been trying to get my head around why a likelihood isn't a probability density function. My understanding says that for an event $X$ and a model parameter $m$:
$P(X\mid m)$ is a probability density function
$P(m\mid X)$ is not
It feels like it should be, and I can't find a clear explanation of why it's not. Does it also mean that a Likelihood can take a value greater than 1?
| https://mathoverflow.net/users/3045 | Why isn't likelihood a probability density function? | If $X$ is data and $m$ are the parameters, then the likelihood function $l(m) = p(X | m)$. I.e. it's $p(X | m)$, considered as a function of $m$.
Both $p(X|m)$ and $p(m|X)$ are pdfs: $p(X|m)$ is a density on $X$ and $p(m|X)$ is a density on $m$. But the likelihood is $p(X|m)$, not as a function of $X$ (it would indeed be a density as a function of X), but as a function of m. So it's not a pdf; in particular, it's not necessarily true that $$\sum\_m p(X|m) = 1.$$
Edit: just to clarify, $p(m|X)$ isn't the likelihood. $p(X|m)$ is.
| 9 | https://mathoverflow.net/users/3035 | 10978 | 7,465 |
https://mathoverflow.net/questions/10974 | 23 | Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.
| https://mathoverflow.net/users/3046 | Does homology detect chain homotopy equivalence? | Yes, this is true. Suppose $C\_\*$ is such a chain complex of free abelian groups.
For each $n$, choose a splitting of the boundary map $C\_n \to B\_{n-1}$, so that $C\_n \cong Z\_n \oplus B\_{n-1}$. (You can do this because $B\_{n-1}$, as a subgroup of a free group, is free.) For all $n$, you then have a sub-chain-complex $\cdots \to 0 \to B\_n \to Z\_n \to 0 \to \cdots$ concentrated in degrees $n$ and $n+1$, and $C\_\*$ is the direct sum of these chain complexes.
Given two such chain complexes $C\_\*$ and $D\_\*$, you get a direct sum decomposition of each, and so it suffices to show that any two complexes $\cdots \to 0 \to R\_i \to F\_i \to 0 \to \cdots$, concentrated in degrees $n$ and $n+1$, which are resolutions of the same module $M$ are chain homotopy equivalent; but this is some variant of the fundamental theorem of homological algebra.
This is special to abelian groups and is false for modules over a general ring.
| 28 | https://mathoverflow.net/users/360 | 10983 | 7,469 |
https://mathoverflow.net/questions/10957 | 4 | This is basically an I'm-weak-at-algebraic-geometry question. I asked it as a warm-up question [here](https://mathoverflow.net/questions/10954/what-do-negative-knots-look-like), but Ilya N asked me to break that post up into several questions.
Consider the free commutative monoid $X$ on countably many generators. Let $A$ be the algebra of all functions $X \to \mathbb C$. Then $A$ is a commutative and cocommutative bialgebra, and hence a commutative monoid object in $(\text{CAlg})^{\rm op}$, where $\text{CAlg}$ is the category of commutative algebras. A *global point* of $A \in (\text{CAlg})^{\rm op}$ is an algebra homomorphism $A \to \mathbb C$. The set of global points of $A$ is naturally a commutative monoid that contains $X$ as a submonoid.
What are all the global points of $A$? I'm pretty sure that there are more than just the points of $X$. The form a monoid, because $A$ is a bialgebra: is there a reasonable description of the monoid structure?
| https://mathoverflow.net/users/78 | A hands-on description of a "completion" of the free commutative monoid on countably many generators | Preliminaries
-------------
First part of your question doesn't use the bialgebra structure. That is, you have a space of functions on countable many points which I'll denote $A = \mathbb C\_1\times \mathbb C\_2\times\cdots \times\mathbb C\_n\times\cdots$ equipped with $+$ and $\times$ pointwise. You would like to classify all algebra maps $\varphi: A\to \mathbb C$.
To start, note that idempotents must go to idempotents, of which $\mathbb C$ has only two: $0$ and $1$. Moreover, $1\times1\times\cdots \mapsto 1$ (unless the whole map is trivial).
Classification of points
------------------------
Consider characteristic functions $\lambda\_n$ which take value from point labeled $n$. As $\lambda\_n^2 = \lambda\_n$, we must have $\varphi(\lambda\_n) = 0$ or $1$. There are three cases:
* there are two indices $i, j$ such that $\varphi(\lambda\_i) = \varphi(\lambda\_j) = 1$. This is impossible: $0 = \varphi(\lambda\_i\lambda\_j) = 1$.
* there exists exactly one index $i$ for which $\varphi(\lambda\_i) = 1$. This implies $\varphi(f) = \varphi((1-\lambda\_i)f + \lambda\_if) = f(i)$.
* all functions $\lambda\_i$, and, therefore, all finite sums from $A$, lie in the kernel of $\varphi$.
The maps of the latter type are indeed the "extra" points. Note, however, that they are "wild" and can be easily killed by some extra finitness assumptions, for example the following "limit of zeroes" property: if $\varphi$ restricted to all finite subsets is 0, then $\varphi$ is 0.
Wild maps
---------
You can construct **examples of these wild maps** using the axiom of choice. To do that, make the elements of the form $a\times a \times \cdots$ and all their finite modifications to map to $a$; denote those elements $A\_0$. Next, select an arbitrary $T\_1\in A$ which is transcendental over $A\_0$ and map it to arbitrary $t\_1\in \mathbb C$. This will fix all elements that lie in algebraic closure $A\_1 = A\_0(T\_1)$. Do that again for $T\_2$ and repeat until you have nothing left. At each step you're producing a correct map $A\_n \to \mathbb C$, so you get the final map $A\to \mathbb C$ as the limit of those.
Conversely, any wild map can be constructed by the above process, assuming all necessary set-theoretic things. So, the answer is, the wild points are classified by maps from this terribly non-constructive sequence of $T\_i$ (of cardinality the same as $A$, that is, continuum) to $\mathbb C$. Those are again in cardinality of **continuum**.
Monoid structure
----------------
One now is reminded that $A$ came with a natural basis enumerated by numbers $(n\_1, n\_2, \dots, n\_k, \dots) \in \mathbb Z^+\oplus\mathbb Z^+\oplus\cdots\oplus\mathbb Z^+\oplus\cdots$ (which is isomorphic to $\mathbb Z\_{>0}^\times$). Therefore, it should be possible to add any two points (denoted $\oplus$). The following properties are clear:
* regular points add normally as $n\_k = n\_k' + n\_k''$;
* adding wild point to either regular or wild point results in a wild point.
Now, although you could write explicitly some wild points, nearly all of them are too hard to describe. The best approximation to the resulting picture is probably this: imagine a set of wild points $W$; the whole space is $W \times \mathbb Z^+\times \mathbb Z^+\times\cdots$. The remaining question, therefore, is what monoid $W$ is equivalent to. For that, you need to settle these questions:
* is it true that you cannot have $w\_1 \oplus w\_2 \oplus \cdots\oplus w\_n = r$ where $r$ is a regular point;
* whether you can subtract them;
* whether and how wild points are divisible by naturals.
Operations on wild points
-------------------------
I think now is certainly time to post another question :)
| 4 | https://mathoverflow.net/users/65 | 10986 | 7,472 |
https://mathoverflow.net/questions/10954 | 18 | Background
----------
Recall that a (oriented) *knot* is a smoothly embedded circle $S^1$ in $\mathbb R^3$, up to some natural equivalence relation (which is not quite trivial to write down). The collection of oriented knots has a binary operation called *connected sum*: if $K\_1,K\_2$ are knots, then $K\_1 \# K\_2$ is formed by spatially separating the knots, then connecting them by a very thin rectangle, which is glued on so that all the orientations are correct. Connect sum is commutative and associative, making the space of knots into a commutative monoid. In fact, [by a theorem of Schubert](http://en.wikipedia.org/wiki/Prime_knot), this is the free commutative monoid on countably many generators. A ($\mathbb C$-valued) *knot invariant* is a $\mathbb C$-valued function on this monoid; under "pointwise" multiplication, the space of knot invariants is a commutative algebra $I$, and $\#$ makes $I$ into a cocommutative bialgebra. I.e. $I$ is a commutative monoid object in $(\text{CAlg})^{\rm{op}}$, where $\text{CAlg}$ is the category of commutative algebras.
>
> **Warm-up question:** Any knot $K$ defines an algebra morphism $I \to \mathbb C$, i.e. a *global point* of $I \in (\text{CAlg})^{\rm{op}}$. Are there any other global points?
>
>
>
**Edit:** In response to Ilya N's comment below, I've made this [into its own question](https://mathoverflow.net/questions/10957/a-hands-on-description-of-a-completion-of-the-free-commutative-monoid-on-counta).
Finite type invariants
----------------------
Recall that a *singular knot* is a smooth map $S^1 \to \mathbb R^3$ with finitely many transverse self-intersections (and otherwise it is an embedding), again up to a natural equivalence. Any knot invariant extends to an invariant of singular knots, as follows: in a singular knot $K\_0$, there are two ways to blow up any singularity, and the orientation determines one as the "right-handed" blow-up $K\_+$ and the other as the "left-handed" blow-up $K\_0$. Evaluate your knot invariant $i$ on each blow-up, and then define $i(K\_0) = i(K\_+) - i(K\_-)$. Note that although the connect-sum of singular knots is not well-defined as a singular knot, if $i\in I$ is a knot-invariant, then it cannot distinguish different connect-sums of singular knots. Note also that the product of knot invariants (i.e. the product in the algebra $I$) is not the point-wise product on singular knots.
A *Vassiliev (or [finite type](http://en.wikipedia.org/wiki/Finite_type_invariant)) invariant of type $\leq n$* is any knot invariant that vanishes on singular knots with $> n$ self-intersections. The space of all Vassiliev invariants is a filtered bialgebra $V$ (filtered by type). The corresponding associated-graded bialgebra $W$ (of "weight systems") has been well-studied (some names: Kontsevich, Bar-Natan, Vaintrob, and I'm sure there are others I haven't read yet) and in fact is more-or-less completely understood (e.g. Hinich and Vaintrob, 2002, "Cyclic operads and algebra of chord diagrams", [MR1913297](http://www.ams.org/mathscinet-getitem?mr=1913297), where its graded dual $A$ of "chord diagrams" is described as a sort of universal enveloping algebra). In fact, this algebra $W$ is Hopf. I learned from [this question](https://mathoverflow.net/questions/10827/if-associated-graded-of-a-filtered-bialgebra-is-hopf-does-it-follow-that-the-ori) that this implies that the bialgebra $V$ of Vassiliev invariants is also Hopf. Thus it is a Hopf sub-bialgebra of the algebra $I$ of knot invariants.
I believe that it is an open question whether Vassiliev invariants separate knots (i.e. whether two knots all of whose Vassiliev invariants agree are necessarily the same). But perhaps this has been answered — I feel reasonably caught-up with the state of knowledge in the mid- to late-90s, but I don't know the literature from the 00s.
Geometrically, then, $V \in (\text{CAlg})^{\rm{op}}$ is a commutative group object, and is a quotient (or something) of the monoid-object $I \in (\text{CAlg})^{\rm{op}}$ of knot invariants. The global points of $V$ (i.e. the algebra maps $V \to \mathbb C$ in $\text{CAlg}$) are a group.
Main Questions
--------------
Supposing that Vassiliev invariants separate knots, there must be global points of $V$ that do not correspond to knots, as by [Mazur's swindle](http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle) there are no "negative knots" among the monoid $I$. Thus my question.
>
> **Main question.** What do the global points of $V$ look like?
>
>
>
If Vassiliev invariants do separate knots, are there still more global points of $V$ than just the free abelian group on countably many generators (i.e. the group generated by the free monoid of knots)? Yes: the singular knots. (**Edit:** The rule for being a global point is that you can evaluate any knot invariant at it, and that the value of the invariant given by pointwise multiplication *on knots* is the multiplication of the values at the global point. Let $K\_0$ be a singular knot with one crossing and with non-singular blow-ups $K\_+$ and $K\_-$, and let $f,g$ be two knot invariants. Then $$\begin{aligned} (f\cdot g)(K\_0) & = (f\cdot g)(K\_+) - (f\cdot g)(K\_-) = f(K\_+)\cdot g(K\_+) - f(K\_-)\cdot g(K\_-) \neq \\\\ f(K\_0) \cdot g(K\_0) & = f(K\_+)\cdot g(K\_+) - f(K\_+)\cdot g(K\_-) - f(K\_-)\cdot g(K\_+) + f(K\_-)\cdot g(K\_-)\end{aligned}$$.) What else is there?
What can be said without knowing whether Vassiliev invariants separate knots?
| https://mathoverflow.net/users/78 | What are the points of Spec(Vassiliev Invariants)? | Just some small comments here. Yes, the question of whether or not Vassiliev invariants separate knots is still open.
One broader context for the question is to consider how Vassiliev invariants were first observed -- via the Vassiliev spectral sequence for the space of "long knots". These are knots of the form $\mathbb R \to \mathbb R^3$ which restrict to a standard (linear) inclusion outside of an interval, say $[-1,1]$.
There's a few standard ways in which the space of long knots can be turned into a topological monoid -- a "Moore loop space" construction which is pretty standard, or a "fat knots" construction from my ["little cubes and long knots"](https://arxiv.org/abs/math/0309427) paper (Topology 46 (2007) 1--27.). So this monoid operation is just the connect-sum operation, suitably jazzed-up to be strictly associative.
An observation for which I'm not sure if anyone has written up a complete proof yet is that the Goodwillie embedding calculus (an alternative approach to the Vassiliev spectral sequence, among other things) factors through the group-completion of the space of long knots. By this I mean, if K is the space of long knots, call the "group completion" the map $K \to \Omega BK$. $\Omega BK$ is where the "formal inverse" to a knot lives. The homotopy-type of $\Omega BK$ is computed in ["little cubes and long knots"](https://arxiv.org/abs/math/0309427). In particular, you can think of the Vassiliev spectral sequence as being an invariant of $\Omega BK$, not $K$. From the point of view of Vassiliev invariants this isn't such a major insight as $\pi\_0 \Omega BK$ is the group-completion of the monoid $\pi\_0 K$, so it's just a free abelian group on countably-many generators.
In particular, there are classes in $H^1 K$ for which there are no obvious finite-type approximations. $H\_1 K$ has some torsion classes for which it's pretty unclear how to detect using the Vassiliev spectral sequence. Here is a mass of torsion computations for the homology of the long knot space: Ryan Budney, Frederick Cohen *On the homology of the space of knots*, Geometry & Topology 13 (2009) 99--139, arXiv:[math/0504206](https://arxiv.org/abs/math/0504206).
edit: although "little cubes and long knots" describes the homotopy-type of the group completion of the space of long knots, it would be nice to have a more "concrete" embedding-space type description of the group completion. Mostovoy has an attempt in *[Short Ropes and Long Knots](https://arxiv.org/abs/math/9810040)*. I believe an idea like this should work and likely Mostovoy's idea works as well, but (again) this is something that needs to be written up for which there hasn't been enough time to start the project.
| 11 | https://mathoverflow.net/users/1465 | 10990 | 7,476 |
https://mathoverflow.net/questions/11001 | 3 | Is there a construction that will give a non-abelian group of order $p^mr$ where $p$ is a prime, $r$ and $p$ are relatively prime and $m$ is an arbitrary non-negative integer? I suspect in this generality there is no simple construction so feel free to restrict $m$ and $r$.
I'm reading some notes on group theory and so far I've only seen the group $G=SL(2,p)$ which has order $p(p+1)(p-1)$. This is great because it gives me a class of groups to play with and test out the various theorems. It is a little annoying to have all these theorems and no concrete non-trival examples to test them out on to see all the subtleties since for the abelian groups all these theorems reduce to saying something trivial.
Edit: Mariano makes a good point and I'm not sure how to rule out silly examples like $G\times\mathbb{Z}\_p^{m-3}\times\mathbb{Z}\_r$. These aren't bad per se but what matters for me is an explicit description of $G$, the non-commutative part, so I have some hope of carrying out some calculations. In essence what I would really like is a construction that parametrizes the non-commutative part depending on all the parameters. In Mariano's example the non-commutative part has no dependence on $m$ and this simplifies the structure of the resulting group.
---
Thanks for the examples and references. This gives me a lot more concrete stuff to work with. Now hopefully I can work out some of the reasons for the various assumptions used in the proofs.
| https://mathoverflow.net/users/nan | non-abelian groups of prescribed order | I am also unsure of what "nontriviality" conditions you want to impose. Without any further conditions, the following answers your question:
Call a positive integer $n$ **nilpotent** if every group of order $n$ is nilpotent.
Call a positive integer $n$ **abelian** if every group of order $n$ is abelian.
Suppose that the prime factorization of $n$ is $p\_1^{a\_1} \cdots p\_r^{a\_r}$. Then:
1. $n$ is nilpotent iff for all $i,j,k$ with $1 \leq k \leq a\_i$, $p\_i^k \not \equiv 1 \pmod{p\_j}$.
2. $n$ is abelian iff it is nilpotent and $a\_i \leq 2$ for all $i$.
These results are proved in
---
Pakianathan, Jonathan(1-WI); Shankar, Krishnan(1-MI)
Nilpotent numbers.
Amer. Math. Monthly 107 (2000), no. 7, 631--634.
---
The proofs are constructive: for any $n$ which is not nilpotent (resp. abelian), they give an explicit group of that order which is not nilpotent (resp. abelian).
The paper is available at
[http://alpha.math.uga.edu/~pete/nilpotentnumbers.pdf](http://alpha.math.uga.edu/%7Epete/nilpotentnumbers.pdf)
| 12 | https://mathoverflow.net/users/1149 | 11006 | 7,484 |
https://mathoverflow.net/questions/11020 | 6 | Sorry if I'm missing something here, but what do we call $M$ if the functor $H\_M:N\mapsto Hom(M,N)$ is exact? Is this in fact equivalent to being flat through some adjointness properties?
| https://mathoverflow.net/users/1916 | Tensor product is to flat as Hom is to ? | We call such modules projective. If you take $N\mapsto Hom(N,M)$ then you get injective modules. This is fairly basic, and covered in any homological algebra book, and mentioned on [wikipedia](http://en.wikipedia.org/wiki/Exact_functor).
| 14 | https://mathoverflow.net/users/622 | 11022 | 7,495 |
https://mathoverflow.net/questions/10478 | 4 | Is there a variety $X$ over $\mathbb{Q}$ and a line bundle $L$ over $X$ (other than the trivial line bundle $\mathcal{O}\_X$ ) such that $L\_v$ is the trivial line bundle over $X\_v=X\times\_{\mathbb{Q}}\mathbb{Q}\_v$ for every place $v$ of $\mathbb{Q}$ ?
(Answer known. There is a pun on "locally trivial" in the title.)
| https://mathoverflow.net/users/2821 | An everywhere locally trivial line bundle | The following example was provided to me by Colliot-Thélène some years ago : Let $X$ be the complement in $\mathbb{P}\_{1,\mathbb{Q}}$ of the three closed points defined by $x^2=13$, $x^2=17$, $x^2=221$. Then $\operatorname{Pic}(X)=\mathbb{Z}/2\mathbb{Z}$ but $\operatorname{Pic}(X\_v)=0$ for every place $v$ of $\mathbb{Q}$.
| 6 | https://mathoverflow.net/users/2821 | 11027 | 7,498 |
https://mathoverflow.net/questions/10993 | 14 | In mathematics we often seek to classify objects up to an equivalence relation, where two objects A and B are said to be equivalent if there exists a map $f:A\rightarrow B$ satisfying certain properties. Examples include trying to classify (some class of) n-manifolds up to homeomorphism, or finite groups up to isomorphism, or (some class of) varieties modulo birational equivalence.
> What examples can you give where you can prove equivalence abstractly, but there is no known algorithm to find the map which induces the equivalence?
For example, could you give examples of manifolds you could prove to be homeomorphic (or homotopic, or simple homotopic, or whatever), but where you had no algorithmic way of finding the homeomorphism between them explicitly?
I think such examples are philosophically interesting, because they highlight how much stronger weaker "proving something" might be than "calculating something", with each answer providing an example.
Inspired by [this question](https://mathoverflow.net/questions/840/the-core-question-of-topology/10896#10896).
| https://mathoverflow.net/users/2051 | Can you prove equivalence without being able to calculate it? | If you assume the axiom of choice, then every vector space has a basis, all bases of a given vector space have the same cardinality, and two vector spaces are isomorphic iff they have bases of the same cardinality. Now if the cardinality of the ground field is infinite, but smaller than the cardinality of the basis, then the cardinality of the basis is the same as the cardinality of the vector space! Pithily, we've shown here that "two vector spaces of huge cardinality are isomorphic iff they have the same size".
So now we can just think of a vector space over $\mathbf{Q}$ of cardinality that of the reals, for which we know a basis, for example the vector space of formal finite sums sum\_i q\_i.[r\_i], where r\_i is real, [r\_i] is a symbol, q\_i is a rational (i.e. the formal vector space with basis the real numbers), and we can just think of a vector space over Q of cardinality that of the reals for which we can't find a basis without invoking the axiom of choice, for example the real numbers themselves (one needs AC to find a basis because if we have a basis we can construct a non-measurable set, and yet there are models of ZF where every subset of R is measurable). These two vector spaces are provably isomorphic in ZFC but you'll never "write down an isomorphism" because for any reasonable definition of "write down" this would turn into a proof that they were isomorphic in ZF, and such a proof can't exist.
| 20 | https://mathoverflow.net/users/1384 | 11028 | 7,499 |
https://mathoverflow.net/questions/10913 | 14 | Let $K$ be a finite extension of $\mathbf{Q}\_p$, with integer ring $R$ and residue field $k$. Say $G/R$ is a finite flat (commutative) group scheme of order $p^2$, killed by $p$. Say the special fibre of $G$ is isomorphic to the $p$-torsion in a supersingular elliptic curve over $k$. Is there some finite extension $L/K$ and an elliptic curve $E$ over $R\_L$, the integers of $L$, such that $G$ becomes isomorphic to $E[p]$ over $R\_L$?
The reason I ask is that in a lemma in Vincent Pilloni's thesis (which I unfortunately cannot find online) he proves that if the special fibre of $G$ is as above then the "degree" of such a $G$ is 1 (see Fargues' work on Harder-Narasimhan filtrations, for example, for the definition of "degree") via a brute force calculation. Pilloni's assertion would however also be a consequence of an affirmative answer to my question (and provided the motivation for my question). People are so good at deformation theory of finite flat group schemes nowadays that I thought this might be well-known to some people nowadays.
| https://mathoverflow.net/users/1384 | Lifting the p-torsion of a supersingular elliptic curve. | La réponse est oui. La raison est la suivante. Si $S$ est un schéma sur p est localement nilpotent, par définition (cf. thèse de Messing, chapitre I), un groupe de Barsotti-Tate tronqué d'échelon $1$ sur $S$ est un schéma en groupes fini localement libre sur $S$ annulé par $p$ tel que si $G\_0$ désigne la réduction de $G$ modulo $p$ on ait
$$
Im (F\_{G\_0}) = \ker (V\_{G\_0})
$$
comme faisceaux fppf sur $S\_0$, $S\_0$ désignant la réduction modulo $p$ de $S$. Cependant, il résulte du critère de platitude fibre à fibre de EGA IV que le morphisme de $S\_0$-schémas en groupes de présentation finie
$$
F:G\_0 \rightarrow \ker (V\_{G\_0})
$$
est fidèlement plat si et seulement si il en est de même fibre à fibre sur $S\_0$. De cela on déduit que dans la définition d'un $BT\_1$ on peut remplacer $S\_0$ par $S\_{red}$ !
Dans les considérations précédentes j'ai pris un schéma $S$ sur lequel $p$ est localement nilpotent, mais il en résulte que l'on a le même type de définition-résultat sur une base qui est un schéma formel $p$-adique.
Revenons maintenant à nos moutons, c'est à dire la question de Kevin. On a donc $G$ un schéma en groupes fini et plat sur l'anneau des entiers de $K$
dont la fibre spéciale sur le corps résiduel de $K$ est un $BT\_1$. Le schéme en groupes $G$ est donc un $BT\_1$.
Nous allons maintenant utiliser le théorème suivant (cf. l'article d'Illusie aux journées arithmétiques de Rennes: "Déformations de groupes de Barsotti-Tate (d'après A. Grothendieck)", Astérisque 127). Si $\mathcal{BT}\_1$, resp. $\mathcal{BT}$, désigne le champ des groupes de Barsotti-Tate tronqués d'échelon $1$, resp. des groupes de Barsotti-Tate, sur des bases qui sont des schémas sur lesquels $p$ est localement nilpotent, alors le morphisme "points de $p$-torsion"
$$
\mathcal{BT} \longrightarrow \mathcal{BT}\_1
$$
est formellement lisse. De cela on déduit la chose suivante. Soit $k$ un corps parfait de caractéristique $p$ et $H$ un groupe $p$-divisible sur $k$. Soit $\mathfrak{X}$ l'espace des déformations de $H$, un $spf (W(k))$-schéma formel non-canoniquement isomorphe à
$spf \big (W(k)[[x\_1,\dots,x\_{d(h-d)}]]\big )$ où $h$ désigne la hauteur de $H$ et $d$ sa dimension.
Soit $\mathcal{H}$ la déformation universelle sur $\mathfrak{X}$. Alors, $\mathcal{H}[p]$ est une déformation verselle de $H[p]$.
Pour conclure et répondre à la question de Kevin il suffit maintenant d'invoquer le théorème de Serre-Tate qui montre que si $H=E[p^\infty]$ où $E$ est une courbe elliptique supersingulière sur $k$ alors $\mathfrak{X}$ est également l'espace des déformation de $E$. Appliquant le théorème d'algébrisation de Grothendieck (GAGF) on en déduit que si $\mathfrak{X}= spf (R)$, la déformation universelle de la courbe elliptique $E$ sur $\mathfrak{X}$ provient en fait d'une courbe elliptique sur $spec (R)$. Le résultat s'en déduit par spécialisation sur $\mathcal{O}\_K$.
P.S.: I read the rules for this forum: it is nowhere written the questions and answers should be written in english !!!!!!!!
| 19 | https://mathoverflow.net/users/3052 | 11031 | 7,501 |
https://mathoverflow.net/questions/11025 | 11 | There are two version of the singular simplicial space of a topological space $X$, one discrete and one internal. At least if X is nice, both of them have homotopy equivalent geometric realizations (and are both equivalent to X itself). I want to know why?
Background/Motivation
---------------------
Let $\Delta$ be a skeleton of the category of finite non-empty ordered sets. The objects of $\Delta$ are the ordered sets [n]. A *simplicial object* of a category C is a functor $X: \Delta^{op} \to C$. The category $\Delta$ can be realized as a sub-category of topological spaces (the category of n-simplices, $\Delta^n$) and (via left Kan extension) this gives rise to a *geometric realization functor* from simplicial sets to topological spaces. It lands in the nice category of CW-complexes, and is denoted $|X|$.
The same geometric realization formula works for simplicial spaces and defines a functor from simplicial spaces to topological spaces. It doesn't always land in CW-complexes.
By general non-sense there is a right adjoint to the realization which is the singular functor. It associates to a topological space the simplicial set given by $Sing(X):[n] \mapsto map(\Delta^n, X)$. The realization from simplicial spaces to topological space also has an adjoint which is given by the simplicial space $\underline{Sing}(X): [n] \mapsto \underline{map}(\Delta^n, X)$, where $\underline{map}$ denotes the mapping space with the compactly generated compact open topology (I'm assuming all spaces are compactly generated).
A simplcial set can be viewed as a discrete simplicial space and so we have two different singular functors and a natural map of simplicial spaces between them:
$Sing(X) \to \underline{Sing}(X)$
This gives, on geometric realization a map of spaces: $|Sing(X)| \to |\underline{Sing}(X)|$. When X is sufficiently nice this map is known to be a homotopy equivalence, and both spaces are homotopy equivalent to X.
>
> Why is this the map $|Sing(X)| \to |\underline{Sing}(X)|$ a homotopy equivalence? Can this be deduced from some sort of connectivity estimate between the constituent spaces of these simplicial spaces?
>
>
>
I know there is an indirect way to prove this equivalence by comparing both spaces to X, but I'm interested in generalizing this result to some related constructions and so I would like a direct argument which uses the map between them. I'm willing to replaces "simplicial space" by bisimplicial set or to assume that the simplicial space is sufficiently "good". I'm hoping for something related to [this question](https://mathoverflow.net/questions/7808/connectivity-after-geometric-realization).
| https://mathoverflow.net/users/184 | Why does the internal singular simplicial space realize to the same thing as the discrete singular simplicial set? | There are maps $|Sing(X)| \to |\underline{Sing}(X)| \to X$ which realize to weak homotopy equivalences. The inclusion of the n-skeleton $|Sing(X)|^{(n)}| \to |Sing(X)|$ is n-connected, because this is always true for CW-complexes, and so the map $|Sing(X)|^{(n)} \to X$ is n-connected. You can't really do any better than this estimate because the n-skeleton has zero homology groups in degrees above n.
The simplicial space $\underline{Sing}(X)$ contains the sub-simplicial space of constant simplices $\Delta^n \to X$. This is homeomorphic to $X$ itself and so, if we write $cX$ for the constant simplicial space with value $X$, we get a map $cX \to \underline{Sing}(X)$. This inclusion $X \to \underline{map}(\Delta^n,X)$ is a homotopy equivalence because the simplex is contractible, so this map of simplicial spaces is levelwise a weak equivalence. The geometric realization of $cX$ is $X$ itself, and so is its n-skeleton for all n. An excision argument (which takes some work) will show that the same is true for the simplicial space (at least under good conditions), and so each of the skeleta of $|\underline{Sing}(X)|$ is homotopy equivalent to $X$.
Therefore the map on n-skeleta is n-connected. I realize that this is the "compare with $X$" game that you mentioned, but my point is that because the simplicial space $\underline{Sing}(X)$ is homotopically constant the comparison $|Sing(X)| \to |\underline{Sing}(X)|$ really *is* comparing with $X$. From the point of view of homotopy theory it's not arising from good levelwise structure of the map at all, but comes from the simplicial assemblage.
I don't know whether this helps your generalization.
| 7 | https://mathoverflow.net/users/360 | 11034 | 7,503 |
https://mathoverflow.net/questions/11044 | 8 | Given 4 points ( not all on the same plane ), what is the probability that a hemisphere exists that passes through all four of them ?
| https://mathoverflow.net/users/3056 | What is the probability that 4 points determine a hemisphere ? | See [J. G. Wendell, "A problem in geometric probability", Math. Scand. 11 (1962) 109-111](http://www.mscand.dk/article/view/10655/8676). The probability that $N$ random points lie in some hemisphere of the unit sphere in $n$-space is
$$p\_{n,N} = 2^{-N+1} \sum\_{k=0}^{n-1} {N-1 \choose k}$$
and in particular you want
$$p\_{3,4} = 2^{-3} \sum\_{k=0}^2 {3 \choose k} = {7 \over 8}$$.
**A second solution:** A solution from *The Annals of Mathematics*, 2 (1886) 133-143 (available from [jstor](http://www.jstor.org/stable/1967203)), specific to the (3,4) case, is as follows. First take three points at random, A, B, C; they are all in the same hemisphere and form a spherical triangle. Find the antipodal points to those three, A', B', C'. Now either the fourth point is in the same hemisphere as the first three or it is in the triangle A'B'C'. The average area of this triangle is one-eighth the surface of the sphere.
This gets the right answer, but I'm not sure how I feel about it; why is the average area one-eighth the surface of the sphere? One can guess this from the fact that three great circles divide a sphere into eight spherical triangles, but that's hardly a proof. Generally this solution seems to assume more facility with spherical geometry than is common nowadays.
| 9 | https://mathoverflow.net/users/143 | 11047 | 7,511 |
https://mathoverflow.net/questions/11045 | 19 | A "test category" is a certain kind of small category $A$ which turns out to have the following property: the category $\widehat{A}$ of presheaves of sets on $A$ admits a model category structure, which is Quillen equivalent to the usual model category structure on spaces.
The notion of test category was proposed by Grothendieck, and the above result was proved by Cisinski (Les préfaisceaux comme modèles des types d'homotopie. Astérisque No. 308 (2006)).
Examples of test categories include the category $\Delta$ of non-empty finite ordered sets (i.e., the indexing category for simplicial sets), and $\square$, the indexing category for cubical sets.
It's hard for me to give the precise definition of test category here: it involves the counit of an adjunction $i\_A: \widehat{A} \rightleftarrows \mathrm{Cat} :i\_A^\*$, where the left adjoint $i\_A$ sends a presheaf $X$ to the comma category $A/X$ (where we think of $A\subset \widehat{A}$ by yoneda). An online introduction to test categories, which includes the full definition and an account of Cisinski's results, is given in [Jardine, "Categorical homotopy theory"](http://www.math.uiuc.edu/K-theory/0669/).
I don't really understand how one should try to prove that a particular category is a test category. The example I have in mind is $G$, the (skeleton of) the category of non-empty finite sets, and all maps between them. I believe this should be a test category; is this true?
Note that there is a "forgetful functor" $\Delta\rightarrow G$, which induces some pairs of adjoint functors between $\widehat{\Delta}$ and $\widehat{G}$. If $G$ is really a test category, I would expect one of these adjoint pairs to be a Quillen equivalence.
Another note: $G$ is equivalent to the category of *finite, contractible groupoids*, which is how I am thinking about it.
| https://mathoverflow.net/users/437 | Are non-empty finite sets a Grothendieck test category? | That your G is a test category is stated in the last sentence of 4.1.20 in the paper of Cisinski you mention. This case is also treated in more detail in section 8.3, where it is shown that the left Kan extension/restriction along *both* adjunctions induced by your "forgetful functor" are Quillen equivalences (8.3.8).
(By the way, if I recall correctly, it is a corollary that if we give simplicial sets the Joyal model structure instead, then both adjunctions are still Quillen pairs, and they realize the two adjoints to the inclusion of (∞,0)-categories in (∞,1)-categories.)
| 16 | https://mathoverflow.net/users/126667 | 11051 | 7,514 |
https://mathoverflow.net/questions/10997 | 14 | I am looking at the foundations of homological algebra, e.g. the introduction
of Ext and Tor, and am unsatisfied. The references I look at start with
"this is called a projective module, this is called a projective resolution,
now pick one and use it to define the right derived functors of your
left exact functor". I would like to see a presentation more along the
following lines:
1. The functor Hom(A,\*), applied to a short exact sequence of modules,
doesn't produce another such. An oracle tells us that it does produce
a long exact sequence; what could it be?
2. We already know (from antiquity) that a short exact sequence of
*complexes* induces a long exact sequence on cohomology.
3. But in #1 we put in modules, not complexes. So let's fix that by hoping
that Hom(A,\*) can be extended in a natural way to the category of complexes
(and really, to descend to the derived category).
4. Such an extension might be required to have the following properties: ???
5. Now I'd like it to be easy to see that the extension is unique if it
exists. When is it easy to compute? At this point I'd like the definition
of "projective module" to suggest itself.
6. Finally, the usual boring checks that using projective resolutions
to define it, the extension does indeed exist.
One way to answer this is to say "In part 4, define the derived category,
and its t-structure, then ask that the extension be exact in the
appropriate sense". I'm hoping to avoid going quite *that* far, or at least,
doing it in a way that doesn't involve introducing too many more definitions.
| https://mathoverflow.net/users/391 | Founding of homological without quite involving derived categories | On rereading your question, Jacobson is not what you want after all.
Anton gave a very nice answer along these lines [here](https://mathoverflow.net/questions/1151/sheaf-cohomology-and-injective-resolutions). In comments to that post, Tyler Lawson recommends Cartan and Eilenberg.
| 2 | https://mathoverflow.net/users/297 | 11063 | 7,523 |
https://mathoverflow.net/questions/7775 | 23 | Suppose R → S is a map of commutative differential graded algebras over a field of characteristic zero. Under what conditions can we say that there is a factorization R → R' → S through an "integral closure" that extends the notion of integral closure in degree zero for connective objects, and respects quasi-isomorphism of the map R to S?
I'm willing to accept answers requiring R → S may need to have some special properties. The motivation for this question is that I'd actually like a generalization of this construction to ring spectra; the hope is that if there is a canonical enough construction on the chain complex level, it has a sensible extension to contexts where it's usually very difficult to construct "ring" objects directly.
| https://mathoverflow.net/users/360 | Do DG-algebras have any sensible notion of integral closure? | I like this question a lot. It deserves an answer, and I really wish I had a good one. Instead, I offer the following idea. Maybe it has some merit?
### Background
Let me fix some terminology. Suppose $f:R\to S$ a homomorphism of (classical, commutative) rings.
1. An element $s\in S$ is said to be *integral over* $R$ if there is a monic polynomial $p\in (R/\ker f)[x]$ such that $s$ is a root of $p$; this is equivalent to saying that the subring $(R/\ker f)[s]\subset S$ is finite over $(R/\ker f)$.
2. We say that $f$ is *integrally closed* if it is a monomorphism and if every element of $S$ that is integral over $R$ is in $R$.
3. At the opposite extreme, we say that $f$ is *integrally surjective* if every element of $S$ is integral over $R$. (This turns out to be equivalent to being a colimit of proper homomorphisms of finite presentation.)
4. Among the integrally surjective homomorphisms are the *elementary* integrally surjective homomorphisms, i.e., homomorphisms of the form $R\to (R/\mathfrak{a})[x]/(p)$, where $R$ is of finite presentation, $\mathfrak{a}\subset R$ is a finitely generated ideal, and $p$ is any monic polynomial.
The classical integral closure construction can now be described as a unique factorization of every homomorphism $f:R\to S$ into an integrally surjective homomorphism $R\to\overline{R}$ followed by an integrally closed monomorphism $\overline{R}\to S$.
In §3.6 of Mathieu Anel's (really cool!) [paper](http://arxiv.org/abs/0902.1130), he describes this factorization system and the "proper topology" constructed from it. In particular, he observes that integrally closed monomorphisms are precisely those morphisms satisfying the unique right lifting property with respect to all elementary integrally surjective homomorphisms.
### Making this work for $E\_{\infty}$ ring spectra
Since you expressed interest in getting integral closure off the ground for $E\_{\infty}$ ring spectra, I'll work in that context.
We can use Andre Joyal's theory of factorization systems in ∞-categories (see §5.2.8 of Jacob Lurie's Higher Topos Theory) to try to play this same game in the ∞-category $\mathcal{C}$ of connective $E\_{\infty}$ ring spectra. (For reasons that will become clear, I'm worried about making this fly for nonconnective $E\_{\infty}$ ring spectra.) Once a set $I$ of elementary integrally surjective morphisms is selected, integrally closed monomorphisms are determined as the class of maps that are right orthogonal to $I$, and the integral closure construction is a factorization system on $\mathcal{C}$.
So what should $I$ be? I think there might be some flexibility here, depending on your aims, but here's a proposal:
1. Start with the coherent connective $E\_{\infty}$ ring spectra that are of finite presentation (over the sphere spectrum). (Concretely, these are the connective $E\_{\infty}$ ring spectra $A$ with the following properties: (1) $\pi\_0A$ is of finite presentation, (2) for every integer $n$, $\pi\_nA$ is a finitely presented module over $\pi\_0A$, and (3) the absolute cotangent complex $L\_A$ is a perfect $A$-module.) We'll only need these kinds of $E\_{\infty}$ ring spectra in our construction of $I$.
2. Among these $E\_{\infty}$ ring spectra, consider the set $I'$ of all morphisms $A\to B$ of finite presentation that induce a surjection on $\pi\_0$. Let's call the maps of $I'$ *quotients*.
3. Now we need to enlarge $I'$ to allow ourselves morphisms that act as though they are of the form $A\to A[x]/(p)$ for $p$ monic. For any of our connective $E\_{\infty}$ ring spectra $A$, we can consider any finitely generated and free $E\_{\infty}$-$A$-algebra $A[X]$ (i.e., the symmetric algebra on some free and finitely generated $A$-module), and we can consider quotients (in the sense above) $A[X]\to B$ where $B$ is almost perfect as an $A$-module (equivalently, $\pi\_nB$ is finitely presented as an $\pi\_0A$-module for every integer $n$); let us add the resulting composites $A\to A[X]\to B$ to our set $I'$ to obtain the set $I$.
Now the morphisms that are right orthogonal to $I$ can be called *integrally closed monomorphisms* of $E\_{\infty}$ ring spectra; call the set of them $S\_R$. The morphisms that are left orthogonal to that can be called the *integrally surjective morphisms* of $E\_{\infty}$ ring spectra. The integral closure would be a factorization system $(S\_L, S\_R)$. One shows the existence of a factorization (via a presentation argument), and it follows from general nonsense (more precisely, 5.2.8.17 of HTT) that it is unique.
### Three observations
1. It's not clear that $I$ is big enough for all purposes. One might want to allow shifts of free modules to generate our finitely generated and free $E\_{\infty}$-$A$-algebras in the description above. I haven't thought carefully about this.
2. It's not so obvious how to talk about quotients $A\to B$ when dealing with nonconnective guys. One wants to say that the fiber (in the category of $A$-modules) is "not any more nonconnective than $A$." I'm not quite sure how to formulate this. In any case, that's why I restricted attention to the connective guys above.
3. Predictably, this is definitely not compatible with the usual integral closure: if I take two classical rings $R$ and $S$ and a ring homomorphism $R\to S$, the integral closure of $HR$ in $HS$ is not in general an Eilenberg-Mac Lane spectrum. If $R$ is a $\mathbf{Q}$-algebra, the two notions are compatible, however.
### Making any computation
... seems really hard. But maybe that's not such a big surprise. After all, integral closures are hard to compute classically as well.
| 17 | https://mathoverflow.net/users/3049 | 11064 | 7,524 |
https://mathoverflow.net/questions/11059 | 12 | Let C be the category of topological monoids, that is, the category of [monoids](http://en.wikipedia.org/wiki/Monoid_%2528category_theory%2529) in (Top, $\times$).
1. Can the model category structure on Top (Serre fibrations, cofibrations, weak homotopy equivalence) be transferred to C along the free and forgetful pair of functors ?
2. What are the functorial factorizations in C? Is there a cylinder object in C?
I'm mostly interested in computing a homotopy pushout in the category C, so any ideas how to do that would be helpful too.
| https://mathoverflow.net/users/2536 | Model Structure/Homotopy Pushouts in topological monoids? | The answer to question #1 is yes. You can use Kan's theorem on lifting model structures (11.3.2 in Hirschhorn's book) to obtain a model structure on $C$ such that the weak equivalences (resp., fibrations) are those morphisms of topological monoids that are weak equivalences (resp., Serre fibrations) on the underlying space. The cofibrations here don't seem to admit an elementary description, sadly. (This model category is Quillen equivalent to the category of simplicial monoids. Depending on the particular nature of your homotopy pushout, you may find it easier to compute there.)
| 13 | https://mathoverflow.net/users/3049 | 11068 | 7,526 |
Subsets and Splits