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https://mathoverflow.net/questions/26302 | 13 | Given a smooth manifold *M*, the following procedures yield the differential graded algebra (Ω\*(M),*d*dR) of differential forms:
* Procedure 1 **(synthetic geometry)**.
For each *n*, consider the object of infinitesimal *n*-simplices in *M*. This is a ringed space whose underlying space is the diagonal *M* ⊂ *M**n*+1, and whose structure sheaf is $C^\infty (M^{n+1}) / I^2$, where $I$ is the ideal defining the diagonal.
The objects of infinitesimal *n*-simplices assemble into a simplicial ringed space, and by taking global function one gets a cosimplicial algebra. The normalized Moore complex of that cosimplicial algebra is then canonically isomorphic to the deRham complex of *M*.
* Procedure 2 **(supergeometry)**.
Letting ℝ0|1 denote the odd line, we may consider the internal hom
$Map(\mathbb R^{0|1} , M)$ in the category of supermanifolds. The global functions $C^\infty( Map(\mathbb R^{0|1} , M))$ then form a ℤ/2-graded algebra.
Noting that the supergroup of automorphisms of ℝ0|1 acts on $Map(\mathbb R^{0|1} , M)$, one can then upgrade that ℤ/2-graded algebra to a ℤ-graded algebra with differential *d*. Once again, that procedure recovers the deRham complex of *M*.
>
> *Why do these two recipes produce the same outcome?*
>
>
>
More generally, one can imagine applying Procedures 1 and 2 to objects *M* that are more general than manifolds (supermanifolds, singular spaces, schemes, derived manifolds, differential stacks, infinity stacks, ...). Are those two procedures always going to agree?
If not, when do they agree, when do they disagree, and why?
| https://mathoverflow.net/users/5690 | Two fancy ways of defining differential forms: How does one show that they are equivalent? | I'm afraid I'm not familiar with your first construction, but a very close one is the Alexander complex/Alexander-Spanier cohomology, where we replace your first order jets along the diagonals by all jets along the diagonal (functions on the formal neighborhood). This complex and its identification with de Rham cohomology is explained beautifully in Appendix A of Constantin Teleman's "Borel-Weil-Bott Theory on the moduli of G-bundles over a curve", available eg. [here](http://math.berkeley.edu/~teleman/paperlist.html). Presumably there's a simple relation between this and your construction 1, ie everything reduces to 1-jets, but I haven't thought about this.
One succinct way to describe this Alexander construction is as derived functions on the de Rham stack of M - the quotient of M by the formal neighborhood of the diagonal.
Writing this stack as a simplicial space we get a cosimplicial algebra of functions on it, which is quasiisomorphic to the de Rham complex.
On the other hand, your second construction is a description of the odd tangent bundle, which is a version of the derived loop space of a manifold. In other words, your second construction is a rephrasing of the description of de Rham cohomology as periodic cyclic homology - or more precisely, your $R^{0,1}$ action is the Connes differential on the Hochschild homology of functions on $M$. So I would consider the equivalence of the two approaches the fact that functions on the de Rham space and cyclic homology are two well-known approaches to describing de Rham cohomology. (Edit: A better answer, elaborated on a little below, is that they are Koszul dual in some sense - they're presenting the self-ext complex of the trivial flat connection in two dual setups.)
This approach to de Rham cohomology via loop spaces is the subject eg of my paper with Nadler [Loop Spaces and Connections](http://arxiv.org/abs/1002.3636).
In fact we introduce there a space we call the unipotent loop space of a (derived) stack, which is maps $Map(A^1[1],M)$ from the shifted version of the affine line just as you define (keeping track of Z-gradings here). For a manifold or general scheme this agrees with the shifted tangent bundle (or rather tangent complex if you're singular) $T\_M[-1]$, but this is no longer true for a stack -- for example for $BG$ you get unipotent elements in $G$, up to conjugation. Rather its formal completion is the formal completion of the odd tangent bundle (or if you prefer its relative tangent space is the odd tangent bundle). In any case you can describe de Rham cohomology (or more generally the whole category of D-modules) very succinctly in terms of this mapping space and its translation action by $A^1[1]$ as you mention (this is a linearized form of the circle action on the loop space of $M$).
The relation between the two constructions can be described as a case of Koszul duality -- specifically the groupoid algebra of the de Rham stack is the algebra of differential operators on $M$, which is Koszul dual to the de Rham complex as a dga. (Again this is described in some detail in the above paper - the idea goes back at least to a paper of Kapranov.) This Koszul duality gives an equivalence between sheaves equivariant for the de Rham/Alexander groupoid and dg modules for the de Rham complex. This relation between the two constructions is extremely general - certainly true for arbitrary smooth schemes and stacks, but in fact doesn't require smoothness if you interpret everything correctly. The claim is that the Koszul duality between dg Lie algebras and dg commutative algebras (in an algebroid setting) exchanges the tangent complex (or the algebra of differential operators) and the de Rham complex (or if you prefer, the de Rham complex is the Chevalley-Eilenberg complex of the tangent complex).
In any case, functions on the de Rham stack (construction 1) always calculate the de Rham cohomology of a space (or derived stack or whatever), while functions on maps from the odd line with the translation action as differential gives this answer as long as you're not a stack (i.e. periodic cyclic homology of a stack remembers some things you'd like to forget maybe, namely the cohomomlogy of inertia, once you throw that away the two answers always agree).
| 7 | https://mathoverflow.net/users/582 | 26305 | 17,235 |
https://mathoverflow.net/questions/26306 | 6 | Let $X$ be a complex manifold (for example $\mathbb CP^n$), let $v$ be a holomorphic vector field on $X$, and let $F$ be a coherent sheaf (for example a vector bundle or a structure sheaf of a point). Then $v$ defines an element in $Ext^1(F,F)$. Indeed $v$ generates an action of $\mathbb C$ on $X$ and taking pull-backs of $F$ under this action we get a deformation of $F$, hence an element of $Ext^1(F,F)$.
**Question**. Is there any fancy (or not fancy) way to express the corresponding element of $Ext^1(F,F)$ in terms of $v$ and $F$? Maybe there is some construction with jets? (I understand, this is a bit vague)
**Added.** Two equally nice and far leading answers were given to this question. I would like summarise here what I understood from David's answer in down to earth terms. So, we want to associate an extension $F\to E\to F$ to a vector field $v$. Suppose (just for the sake of been very much down to earth), that $F$ is a vector bundle, and $v$ has no zeros. Then we can consider $1$-jets of sections of $F$ in the direction of $v$ (or if you want, along trajectories of $v$). It is not hard to see that this is a bundle of rank $2rank(F)$, and this is exactly $E$, that we are looking for.
| https://mathoverflow.net/users/943 | Deformations of sheaves via automorphisms. How to express $Ext^1$? | Holomorphic vector fields give elements of the first Hochschild cohomology group of the structure sheaf $$HH^1(O\_X,O\_X)\simeq H^1(X,O\_X)\oplus H^0(X,T\_X)$$ (via an easy part of the Hochschild-Kostant-Rosenberg theorem).
On the other hand the Hochschild cohomology may be identified (or indeed defined as) self-Ext of the identity functor on the category of coherent sheaves. Hence, by definition
of self-ext of the identity, there's a functorial map $$HH^1(O\_X,O\_X)=Ext^1(Id\_{O\_X-mod})\longrightarrow Ext^1(F,F)$$
for any coherent sheaf $F$. This is one fancy way to express your construction. Likewise the second Hochschild cohomology gives deformations of the (derived) category of sheaves, and the corresponding element in $Ext^2(F,F)$ is the obstruction for $F$ to deform.
EDIT: Here's a little more to explain the compatibility with Torsten's answer.
Any sheaf $F$ has a canonical extension by $F\otimes \Omega^1$, given by 1-jets of sections of $F$
$$0\to F\otimes \Omega^1 \to J^1(F) \to F\to 0.$$
We can contract this extension against a vector field to get an $Ext^1$ of $F$ by itself as desired. The functor taking $F$ to its extension is represented (as an integral transform) by the first-order neighborhood of the diagonal, i.e. by the 1-jet of the identity functor (which again can be contracted against any vector field). This is explicitly the map from vector fields to $Ext^1$ of the identity functor, i.e. Hochschild $HH^1$.
This universal extension can be rewritten as an element $$Hom(T\_X[-1]\otimes F,F),$$ which in fact defines a homotopy Lie algebra structure on the shifted tangent sheaf $T\_X[-1]$ (the Atiyah bracket).
The sheaf Hochschild cohomology can then be identified as the universal enveloping algebra of this homotopy Lie algebra. So in some precise sense this universal action of vector fields (as Torsten describes) IS Hochschild cohomology, acting as the derived center of the category.
| 7 | https://mathoverflow.net/users/582 | 26310 | 17,239 |
https://mathoverflow.net/questions/26317 | 8 | A monoid in the Category Cat is a strict monoidal category according to Wikipedia. Is it possible to weaken the monoid so that its realisation in Cat is a weak monoidal category? Do we shift up a dimension, and throw in a 2-morphism associator for the monoid satisfying the analogue of Mac Lane's Pentagon, and similarly for the identities?
Or is it still possible to get a weak monoidal category using a monoid by changing the category we realise it in from Cat to some modification of it?
| https://mathoverflow.net/users/6408 | Is a weak monoidal category a monoid object in some category? | Yes, you can define a *pseudomonoid* in any monoidal 2-category, such that a pseudomonoid in the 2-category Cat is precisely a non-strict monoidal category. The definition of monoidal category, interpreted in terms of functors and natural isomorphisms, i.e. in terms of the 2-category Cat, tells you exactly how to define a pseudomonoid.
| 8 | https://mathoverflow.net/users/49 | 26322 | 17,245 |
https://mathoverflow.net/questions/16226 | 19 | **Background:** The Shannon sampling theorem states that a bandlimited function (an $L^2$ function whose Fourier transform has compact support) can be determined uniquely from sampling it an integer lattice. More precisely, if the Fourier transform\* $\hat{f}$ is supported in $[-\Omega/2, \Omega/2]$ (which we express by writing $f \in B\_{\Omega}$) and $\tau>0$ satisfies $\Omega \tau \leq 1$, then $f$ can be reconstructed from the family $\{f(k\tau)\}\_{k\in \mathbb Z}$. This is proved more or less by periodizing and reducing to the theory of Fourier series. A higher-dimensional analog using lattices in higher dimensions can be proved similarly.
**Question:** What sets can these uniformly spaced lattices be replaced by? Is there a general criterion, or some characterization of the collection of such sets (e.g. measure-theoretic)?
>
> If $\Omega$ is fixed, then for what sets $E$ (say, countable), is the map $B\_{\Omega} \to \mathbb{R}^E$ injective?
>
>
>
**Motivation:** I overheard a conversation between a graduate student and a faculty member; the student wished to determine if one had a ``deformed lattice'' $\{ k\tau + X\_k \}$ where $\tau$ was a very small mesh relative to $\Omega$ and the $X\_k$ was a sequence of reals, for what sequences $\{X\_k\}$ would the samples $\{f(k\tau + X\_k)\}$ determine $f \in B\_{\Omega}$? We could think of the $X\_k$ as independent bounded random variables with a small variance, for instance. He suspected (and wanted to know) whether the set of singular parameters (where this information was insufficient) would be a set of measure zero. Intuitively, the small mesh size suggests something like this should be the case.
It was suggested that one might use some sort of finite-dimensional approximations to the full space $B\_{\Omega}$ and taking finite subsets of the whole deformed lattice, and that the set of admissible $X\_k$ would be some algebraic subvariety of proper codimension, therefore of measure zero, but it wasn't clear how to do this. This led to my more general question boxed above.
Another reason this question may be of interest is that the sampling used in practical applications (e.g. CAT-scans) does not sample based upon a lattice pattern as in the Shannon theorem.
\*I am using the normalization with the factor $2\pi$ for the Fourier transform.
| https://mathoverflow.net/users/344 | What kinds of sets can replace lattices in the Shannon sampling theorem? | One sufficient condition is that, if $|X\_k| <= L$ for some $L < 1/4$ then the sampling map would be an injection, as well as a bounded map from $B$ to $l^2$. This is known as Kadec's theorem.
See
Nonuniform Sampling and Reconstruction in Shift-Invariant Spaces,
Akram Aldroubi and Karlheinz Gröchenig,
SIAM Review, Vol. 43, No. 4 (Dec., 2001), pp. 585-620
for a survey and further references.
| 8 | https://mathoverflow.net/users/1229 | 26333 | 17,248 |
https://mathoverflow.net/questions/26342 | 10 | This is motivated by a recent [question](https://mathoverflow.net/questions/26336/integer-valued-factorial-ratios) by Wadim.
The negative answer should be known, since t is very natural, in this case I would be happy to see any reference.
May Pafnuty Lvovich Chebyshev's approach to distribution of primes lead to PNT itself, if we replace $\frac{(30 n)! n!}{(15 n)! (10 n)! (6 n)!}$ to other integer ratios of factorials? If not, what are the best constants in asymptotic relation
$$
c\_1 \frac{n}{\log n}< \pi(n)< c\_2 \frac{n}{\log n}
$$
which may be obtained on this way?
| https://mathoverflow.net/users/4312 | Chebyshev's approach to the distribution of primes | Erdős and Diamond proved in [**1**] that Chebyshev could have achieved sharper bounds for the asymptotic behavior of the prime counting function. Nevertheless, their proof does not shed any light on the first question that you posed because they took the PNT for granted throughout their note.
References:
[**1**] **H. G. Diamond; P. Erdős.** *On sharp elementary prime number estimates*, Enseign. Math. (2) 26 (1980) 313-321.
| 8 | https://mathoverflow.net/users/1593 | 26344 | 17,254 |
https://mathoverflow.net/questions/26350 | 0 | I'm wondering (hoping) if an inequality is true. Please can anyone help me?
Let $V$ be a complex vector space $dim\_{\mathbb{C}}(V)=n$
with a hermitian scalar product $h$.
Let $v,a, b \in V$.
Is it true that
$(h(v,v)h(a,a)-{|h(v,a)|}^{2})(h(v,v)h(b,b)-{|h(v,b)|}^{2})\geq |(h(v,v)h(a,b)-h(a,v)\overline{h(b,v)}|^{2}$?
With the overline meaning complex conjugate.
Thank you in advance.
| https://mathoverflow.net/users/4971 | Linear algebra inequality | Yes. The case where $v=0$ is trivial so suppose $v\ne0$. Consider
the projection map from $V$ to the hyperplane orthogonal to $v$
and let $a'$ and $b'$ be the images of $a$ and $b$ under this projection.
Then your inequality reduces to
$$h(a',a')h(b',b')\ge\vert h(a',b') \vert^2,$$the Cauchy-Schwarz inequality.
| 4 | https://mathoverflow.net/users/4213 | 26353 | 17,260 |
https://mathoverflow.net/questions/26267 | 154 | While not a research-level math question, I'm sure this is a question of interest to many research-level mathematicians, whose expertise I seek.
At RIMS (in Kyoto) in 2005, they had the best white chalk I've seen anywhere. It's slightly larger than standard American chalk, harder, heavier, and most importantly covered with some enamel-like coating that one must rub through (on the end) to be able to write with. One's hands don't rub through the coating, and thus don't get chalky.
Are there any U.S. manufacturers of such?
EDIT: even though someone has given a link whereby to order this stuff from Japan, I would still be delighted to hear about American products that beat Binney & Smith.
| https://mathoverflow.net/users/391 | Where to buy premium white chalk in the U.S., like they have at RIMS? | You can buy it online [rakuten link deleted]
**Edit (2020-7-10):** Hagoromo Bunku company has been shut down in March 2015 ([Berkeley Alumni Magazine article](https://alumni.berkeley.edu/california-magazine/fall-2019/chalk-market-where-mathematicians-go-get-good-stuff)). There were two successors to their product, because the three custom machines they had for making chalk were sold to two parties.
1. Two machines, together with the rights to the "Hagoromo Fulltouch" name were sold to [Sejongmall](http://en.sejongmall.co.kr), a South Korean company. They have US dollar prices and ship internationally. Incidentally, the splash screen on their site has pictures of Brian Conrad and Andrei Okounkov pulled from a Japanese TV show about the chalk.
2. One machine was sold to Uma-jirushi, a Japanese blackboard maker with plans to expand to chalk. In April 2015 Uma-jirushi announced a new product called DC Chalk Deluxe (DCチョークDX) as "a collaboration with Hagoromo" in its catalog. However, Uma-jirushi [announced on 2020-1-26](https://www.uma-jirushi.co.jp/info/7895) that they would discontinue sales of DC chalk on 2020-5-31.
| 65 | https://mathoverflow.net/users/36665 | 26356 | 17,263 |
https://mathoverflow.net/questions/26368 | 0 | Let G and H be finite groups. Let G' = [G,G] and H' = [H,H] be the corresponding derived groups (commutator subgroups) of G and H. I am looking for an example where G' is isomorphic to H' and G/G' is isomorphic to H/H' but G and H are not isomorphic.
| https://mathoverflow.net/users/4048 | Must finite groups with isomorphic commutators and quotients be isomorphic? | take $G=S\_n$ and $H=A\_n\times Z/2Z$ (here $S\_n$ is the symmetric group and $A\_n$ is the alternating group).
| 13 | https://mathoverflow.net/users/4158 | 26369 | 17,267 |
https://mathoverflow.net/questions/26364 | 4 | Let X be a Calabi-Yau 3-fold, and $D\subset X$, smooth,Fano divisor.
Since $K\_X=0$ we have $N\_D^X=K\_D$. I have seen the following fact in many papers:
By deforming X within Kahler moduli, we can shrink D to a point to get a singular 3-fold Y.
Here are my questions?
1-Why is that possible?(Why it is not possible for non-Fano surfaces)
2-How can we find the type of singularity?(Explicitly for a given surface D)
I have to mention that problem is local, you can consider the total space of $K\_D$ as local model for neighborhood of D, instead of whole X.
| https://mathoverflow.net/users/5259 | Shrinking Fano surfaces to a point in Calabi-Yau 3-folds | **Corrected.** In the first version I stated that CY 3-fold containing Fano surfaces are unknown, this is not at all true as Mohammad pointed out. For example, one can take a product $E\times E\times E=X$ with $E$ an elliptic curve admitting a $\mathbb Z\_3$ action,
and then take crepant resolution of the quotient of diagonal action $X/\mathbb Z\_3$.
On the resolved manifold there will be $27=3^3$ copies of $\mathbb CP^2$.
On the contrary, as Zhiyu pointed out, on smooth quintic in $\mathbb CP^4$ one can not find a plane $\mathbb CP^2$, since the second integral co-homology of the quintic is generated by a hyperplane section (in the previous "*correction*" of this answer I was claiming the opposite).
The rest of the answer was correct, and here it is.
It is also important that $D$ is Fano. In this case the normal budnle to it is negative (i.e. its inverse is ample), and so we can contract holomorphically the divisor by Grauert theorem. Notice that you can find a Kahler form that shrinks $D$ to zero only if it can be contracted holomorphically.
In order to see what singularity you get you just need to study case by case Fano surfaces. For example in the case of $\mathbb CP^2$ you get obrifold singularity $\mathbb C^3/\mathbb Z\_3$ (with diagonal action of
$\mathbb{Z}\_3$ on $\mathbb C^3$). In the case of $\mathbb CP^1\times \mathbb CP^1$ you get the sinuglariy $x^2+y^2+z^2+t^2=0 / \pm 1$. In the case when $D$ is a cubic surface you get just $P\_3(x,y,z,t)=0$, where $P\_3$ is a homogenious polynomial of degree $3$. All other cases of cours were studied and some descripiton of singularities exists. In praticular you can find the model of the singularity if you construct an anti-canonical embedding of your Fano surface in some $\mathbb CP^n$.
No, why can you deform the Kahler form? This is because on the total space of the canonical bundle of a Fano surface the set of Kahler forms is connected, and in particular some of these forms are in the cohomology class $-D+\pi^\*\omega$, where D is the zero section, and $w$ is the pullback of a Kahler form from the total space of the $K$ with zero section contracted. Replacing $-D$ with $-tD$ $(0< t\le 1)$ you get a one parameter family cohomology classes that can be represented by Kahler forms. Finally, for $t=0$ you get a cohomology class vanishing on $D$ (by definition).
| 5 | https://mathoverflow.net/users/943 | 26370 | 17,268 |
https://mathoverflow.net/questions/26338 | 8 | Are there explicit examples of triangulations of exotic 4-spheres?
| https://mathoverflow.net/users/5196 | Triangulations of exotic 4-spheres | Here is my comment expanded to answer form: The question of existence of exotic 4-spheres (i.e., the smooth Poincaré conjecture) is still open, and (according to [Wikipedia](http://en.wikipedia.org/wiki/Generalized_Poincare_conjecture)) the existence of exotic PL structures is equivalent to it. Therefore, the answer is that no such explicit triangulations are known.
In general, explicit triangulations of higher dimensional manifolds seem to be difficult to write down. I've heard from computer algebra specialists that no one has even written an explicit triangulation of $\mathbb{CP}^3$. The chaos surrounding [this earlier question](https://mathoverflow.net/questions/20664/why-is-complex-projective-space-triangulable) might suggest that the problem is subtle.
| 6 | https://mathoverflow.net/users/121 | 26375 | 17,271 |
https://mathoverflow.net/questions/25723 | 17 | This should be straightforward; I'm sorry if it's too much so. Can someone point me to a reference which computes the Dolbeault cohomology of the Hopf manifolds?
Motivation: I'd like to work through a concrete example of the Hodge decomposition theorem failing for non-Kähler manifolds. The textbook I have handy (Griffiths & Harris) doesn't treat this, and the obvious Google search was unhelpful.
| https://mathoverflow.net/users/2819 | Dolbeault cohomology of Hopf manifolds | Even though this question has an accepted answer, the answers so far are not complete or explicit. I kept working on this question, because I have been curious for a long time about the structure of Dolbeault complexes. First of all, the Frölicher spectral sequence does not directly reveal all of the non-Hodge information in the Dolbeault complex of a non-Kähler complex manifold. I learned from Mikhail Khovanov that in the category of bounded double complexes over a field, every object is isomorphic to a unique direct sum of indecomposable objects. Moreover, the indecomposable double complexes can be classified as squares, dots, and zigzags. Here is an example of each type of indecomposable, with the convention that omitted cells and arrows are 0:
$$\begin{matrix} \mathbb{C} & \rightarrow & \mathbb{C} \\\\
\uparrow & & \uparrow \\\\
\mathbb{C} & \rightarrow & \mathbb{C} \end{matrix}\qquad\qquad
\mathbb{C}\qquad\qquad
\begin{matrix} \mathbb{C} & \rightarrow & \mathbb{C} \\\\
& & \uparrow \\\\
& & \mathbb{C} & \rightarrow & \mathbb{C} \end{matrix} $$
This is actually a standard result about an $A\_\infty$ quiver algebra with alternating arrows. To be precise about the Hodge theorem and the Frölicher spectral sequence, the $\partial\bar{\partial}$ lemma says exactly that there are no zigzags (other than length 1, which are then dots), which is then equivalent to the statement that horizontal cohomology is isomorphic to vertical cohomology. The squares are projective objects and do not contribute to any cohomology theory. The odd-length zigzags contribute to the total de Rham cohomology, while the even-length zigzags do not. Meanwhile there are two Frölicher spectral sequences, each of which detects half of the even zigzags. The Frölicher spectral sequences are insensitive to the odd zigzags, but you can still say that a Dolbeault complex is non-Hodge if it has odd zigzags, even though each Frölicher spectral sequence degenerates at $E\_1$ if there are no even zigzags.
The information of all of the zigzags, extracted by discarding only the squares, has been defined in the literature as "Aeppli cohomology". (Serre duality implies, sort-of indirectly, that the Aeppli cohomology of a compact manifold is self-dual; I would be interested to see a direct derivation of this fact.)
Now, the Hopf manifolds. The most standard round Hopf manifold of complex dimension $n$ has an important group action of $U(n) \times S^1$. (For those who aren't familiar with the terminology, the standard [Hopf manifold](http://en.wikipedia.org/wiki/Hopf_manifold) is $(\mathbb{C}^n\setminus 0)/\Gamma\_r$, where $\Gamma\_r$ is generated by rescaling by a real constant $r > 1$.) There is one aspect of my calculation that for me is a conjecture: That a connected, compact Lie group acts trivially on all of the zigzags of a compact manifold. This is true for odd zigzags since the de Rham cohomology has an invariant integer lattice, but I do not have an argument for even zigzags. But let's suppose that it is so.
The invariant part of the Dolbeault complex is algebraically generated by these differential forms of degree $(1,0)$, $(0,1)$, and $(1,1)$:
$$\alpha = \frac{\bar{z} \cdot dz}{z \cdot \bar{z}} \qquad
\bar{\alpha} = \frac{z \cdot d\bar{z}}{z \cdot \bar{z}}
\qquad \omega = \frac{dz \cdot d\bar{z}}{z \cdot \bar{z}}.$$
(I use $z$ and $dz$ as a vector of functions and a vector of 1-forms, so that I can take dot products. I'm leaving out the wedge product symbol.) Then I calculated the following:
$$\partial \alpha = 0 \qquad \bar{\partial} \alpha = \alpha \bar{\alpha} - \omega \qquad \partial \omega = - \alpha \omega \qquad \omega^n = n\alpha \bar{\alpha} \omega^{n-1}.$$
A basis for the invariant part of the Dolbeault complex is given by $\omega^k$, $\alpha \omega^k$, $\bar{\alpha} \omega^k$, and $\alpha \bar{\alpha} \omega^k$ for $0 \le k \le n-1$. The Poincaré series of the invariant complex is a matrix like this one:
$$\begin{matrix} 0 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 2 & 1 \\\\ 0 & 1 & 2 & 1 & 0 \\\\ 1 & 2 & 1 & 0 & 0 \\\\ 1 & 1 & 0 & 0 & 0 \end{matrix}.$$
After calculating the differential, my answer is that this decomposes as a dot at each corner, a zigzag of length 3 next to each each corner, and a progression of squares. Modulo the conjecture that all zigzags are invariant, this is a complete description of the Dolbeault complex. The $n=1$ case is an exception in which the Hopf manifold obviously is Kähler (it's a torus, and all complex curves are Kähler). In this case the invariant complex decomposes as four dots.
| 29 | https://mathoverflow.net/users/1450 | 26380 | 17,275 |
https://mathoverflow.net/questions/26363 | 5 | Suppose that $A$ is a semisimple Hopf algebra with a commutative character ring. Does it follow that $A$ is quasitriangular, i.e $\mathrm{Rep}(A)$ is a braided tensor category?
I think I 've seen this statement in a paper without a proof long time ago. It might be obvious although I don't see how to construct a braiding just knowing non-functorial commutativity of the tensor products.
| https://mathoverflow.net/users/2805 | Semisimple Hopf algebras with commutative character ring | Sebastian,
No, it does not follow.
In [this paper (Example 6.14)](http://lanl.arxiv.org/abs/0704.0195) we proved that if a Tambara-Yamagami fusion category
admits a braiding then its dimension is a power of 2. Note that a Tambara-Yamagami
category
has a commutative Grothendieck ring. Hopf algebras whose representation
category is of Tambara-Yamagami type are classified by Tambara (Representations of tensor categories with fusion rules of self-duality for abelian groups, Isr. J. Math. 118 (2000), 29-60). For example, there is a Hopf algebra $A = k^9 \oplus M\_3(k)$ (so-called Kac-Paljutkin algebra) with commutative chracater ring and $Rep(A)$ admitting no braiding.
| 7 | https://mathoverflow.net/users/3011 | 26381 | 17,276 |
https://mathoverflow.net/questions/26300 | 8 | Recall the following theorem due to Burnside:Let $G$ be a finite group and let $V$ be its irreducible complex representation of dimension greater than 1, then the character
of this representation is $0$ on some element of $G$. Is this statement still correct
if $G$ is any compact Lie group? Thanks.
| https://mathoverflow.net/users/6277 | Is there a generalization of Burnside's theorem for compact Lie groups? | The answer is yes--use the Weyl character formula, for example.
See: [Patrick X. Gallagher, Zeros of group characters. Math. Z. Volume 87 (1965), Number 3.](http://www.springerlink.com/content/l2m14425071477x3/)
| 7 | https://mathoverflow.net/users/6355 | 26384 | 17,277 |
https://mathoverflow.net/questions/26382 | 5 | If we have an extension of bundles $0 \to E \to F \to G \to 0$ on $X$, then to show that this is the zero element in $Ext^1\_X(G,E)$, we need to show that this sequence splits. To produce a splitting is a concrete question in homological algebra. I understand that in general if we have an extension $ 0 \to E \to E\_1 \to \cdots \to E\_k \to G \to 0$, short of showing $Ext^k\_X(G,E)=0$, there is no simple recipe for showing that this particular extension is zero. But are there any known examples where such things have been shown? I am basically interested only in the case $k=2$. Is there a simple recipe in this case?
| https://mathoverflow.net/users/6425 | Showing an Ext^2 element is zero | There is a simple recipe to show that a product of two $Ext^1$ is zero (and it is clear that any $Ext^2$ can be represented as such a product). Namely, let $0 \to E\_0 \to E\_{01} \to E\_1 \to 0$ and $0 \to E\_1 \to E\_{12} \to E\_2 \to 0$ are two exact triples.
The product of the corresponding elements $e\_{01} \in Ext^1(E\_1,E\_0)$ and $e\_{12} \in Ext^1(E\_2,E\_1)$, that is $e\_{01}\circ e\_{12} \in Ext^2(E\_2,E\_0)$, is zero, if and only if there is an object $E$ with a filtration of length 3 on it $F\_0E \subset F\_1E \subset F\_2E = E$ such that the first triple is isomorphic to $0 \to F\_0E \to F\_1E \to F\_1E/F\_0E \to 0$ and the second triple is isomorphic to $0 \to F\_1E/F\_0E \to F\_2E/F\_0E \to F\_2E/F\_1E \to 0$.
Here is a sketch of a proof. Consider the long exact sequence obtained by applying $Hom(E\_2,-)$ to $0 \to E\_0 \to E\_{01} \to E\_1 \to 0$:
$$
\dots \to Ext^1(E\_2,E\_{01}) \to Ext^1(E\_2,E\_1) \to Ext^2(E\_2,E\_0) \to \dots
$$
It is clear that the class $e\_{12}$ is mapped by the second arrow to the class $e\_{01}\circ e\_{12} = 0$, hence there exists a class $e \in Ext^1(E\_2,E\_{01})$ which is mapped to $e\_{12}$ by the first arrow. Let $E$ be the corresponding extension. Then we have an exact sequence $0 \to E\_{01} \to E \to E\_2 \to 0$ and a morphism from this
to the sequence $0 \to E\_1 \to E\_{12} \to E\_2 \to 0$ such that the map $E\_{01} \to E\_1$ is the given one and the map $E\_2 \to E\_2$ is the identity (you can write down a commutative diagram here). It follows that the induced map $E \to E\_{12}$ is a surjection and its kernel is $E\_0$. This is precisely what was claimed ($F\_1E$ is the image of $E\_{01}$ in $E$ and $F\_0E$ is the image of $E\_0 \subset E\_{01}$).
| 10 | https://mathoverflow.net/users/4428 | 26387 | 17,279 |
https://mathoverflow.net/questions/26374 | 10 | I would like to know something more than what is written on wikipedia <http://en.wikipedia.org/wiki/Euler_characteristic>
What would be some large (largest?) class of topological spaces for which $\chi$ is defined, so that all standard properties hold, for example that $\chi(X)=\chi(Y)+\chi(Z)$ if $X=Y \cup Z$, ($Y\cap Z=0$).
ADDED. The answer of Algori indicates that a reasonably large class of spaces for which Euler characteristics can be defined are locally compact spaces $X$, whose one point compactification $\bar X$ is a CW complex. Then we can define $\chi(X)=\chi(\bar X)-1$. For example, the Euler characteristics of an open interval according to this definition is $-1$.
This definition rases a second (maybe obvious) question.
**Question 2.** Suppose $X$ is a locally compact space whose 1 point compactification is a $CW$ complex, and $Y$ is a subspace of $X$ such that both $Y$ and $X\setminus Y$ have this property. Is it ture that $\chi(X)=\chi(Y)+\chi(X\setminus Y)$?
Also, I was thinking, that Euler characteristics is more *fundamental* then homology.But can it be defined for spaces, where homology is not defined?
Finally, Quiaochu pointed out below that a very similar question was already discussed previously on mathoverflow.
| https://mathoverflow.net/users/943 | For which classes of topological spaces Euler characteristics is defined? | The answer to the question as it is stated is that there is probably no "largest" class of spaces for which the Euler characteristic makes sense.
The answer also depends on where you would like the Euler characteristic to take values. Here is the tautological answer (admittedly not a very exciting one): if you have a category $C$ of spaces closed under taking cones and cylinders, then there is the universal Euler characteristic for that category: just take the free abelian group $K(C)$ that has a generator $[X]$ for each $X\in C$ and quotient it by the span of $[X]+[Cone(f)]-[Y]$ for all $X,Y\in C$ and any morphism $f:X\to Y$ in $C$. The Euler characteristic of any $X$ in $C$ is set to be $[X]$. (There may be variations and/or generalizations of this approach.)
The group $K(C)$ is complicated in general but for some choices of $C$ it has interesting quotients. This can happen e.g. when $C$ admits a good "cohomology-like" functor. For example if $C$ is the category of spaces with finitely generated integral homology groups then $K(C)$ maps to $\mathbf{Z}$ and this gives the usual Euler characteristic. If one takes $C$ to be formed by spaces that admit a finite cover with finitely generated integral homology groups (typical examples are the classifying spaces of $SL\_2(\mathbf{Z})$ and more generally of mapping class groups), then $K(C)$ does not map to $\mathbf{Z}$ any more, but it maps to $\mathbf{Q}$. This gives the rational Euler characteristic.
Finally, let me address the last remark by Dmitri. For some categories the group $K(C)$ maps to $\mathbf{Z}$ in several different ways. Let us take e.g. $C$ to be the category formed by spaces whose one-point compactification is a finite CW-complex (with proper maps as morphisms). Then there are (at least) two characteristics; one is obtained using the ordinary cohomology and another one comes from the Borel-Moore homology. On complex algebraic varieties both agree. But the Borel-Moore Euler characteristic of an open $n$-ball is $(-1)^n$.
Here is the answer to the second question: suppose $Y$ is a locally closed subspace of a locally compact space $X$ such that $X,Y,\bar Y,\bar Y\setminus Y, X\setminus\bar Y$ and $X\setminus Y$ are of the form "a finite CW-complex minus a point". Then $\chi(Y)+\chi(X\setminus Y)=\chi(X)$ where $\chi$ is the Euler characteristic computed using the Borel-Moore homology.
The case when $Y$ is closed follows from the Borel-Moore homology long exact sequence. In general we can write $\chi(X)=\chi(X\setminus\bar Y)+\chi(\bar Y)=\chi(X\setminus\bar Y)+\chi(\bar Y\setminus Y)+\chi(Y)$. In the last sum the sum of the first two terms gives $\chi(X\setminus Y)$ since $X\setminus\bar Y$ is open in $X\setminus Y$.
| 15 | https://mathoverflow.net/users/2349 | 26393 | 17,281 |
https://mathoverflow.net/questions/26390 | 6 | The base field will be the field of complex numbers. I have a slightly technical problem concerning the resolution of singularities of a certain variety. Basically, I want to to know if it is projective.
Let $Y$ be a normal projective variety with only quotient singularities. Let $C$ be a smooth projective curve.
Let $p:Y\longrightarrow C$ be a flat projective morphism.
Let $f:X\longrightarrow Y$ be a resolution of singularities. Is $X$ necessarily projective over the base field? I know this is true in dimension 2. Are there any counterexamples in dimension 3?
| https://mathoverflow.net/users/4333 | Is the desingularization of a normal variety with only quotient singularities projective | The following paper might be useful.
*Nagata*, "Existence theorems for nonprojective complete algebraic varieties", Illinois J. Math. 2 1958 490--498.
At the end of the paper Nagata says the following:
"Hironaka recently proved the following: If $V$ is a nonsingular projective variety of dimension not less than 3, then there exists a complete nonsingular nonprojective variety $V'$ such that (1) $V$ is birationally equivalent to $V$, and (2) $V'$ dominates $V$."
So for 3-folds, then this seems to give a negative answer to your question (assuming one can track down the reference, I didn't seem to find it).
In your setting, simply take $Z \to Y$ to be any resolution of singularities. Then take $V'$ as in Hironaka's result. It should also be a resolution of singularities (assuming I am reading it right).
| 6 | https://mathoverflow.net/users/3521 | 26398 | 17,284 |
https://mathoverflow.net/questions/26392 | 1 | Let T be a true theory of arithmetic to which the incompleteness theorems apply. Consider two sentences in the language of T to be equivalent if they are provably equivalent over T. How many equivalence classes are there, under this relation, that contain a true-but-unprovable sentence?
| https://mathoverflow.net/users/1320 | What is cardinality of the set of true undecidable minimal sentences in a formal theory of aritmetic | **Note:** *The top part answers an old version of the question, which is now irrelevant.*
Given a axiomatizable theory T of arithmetic, the set of all statements independent of T is the complement of a computably enumerable set. When nonempty (e.g. when T extends PA) this set is countably infinite. (Trivially, if φ is such then so is φ∧∃x(x=x), etc.) However, there is no general algorithm to produce the *shortest* element of such a set, never mind counting them. (The requirement that the sentence be true is negligible since negation only adds a constant number of symbols depending on syntactic conventions.)
That said, some variants of your question have been actively studied. [Hilbert's Tenth Problem](http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem) says that there are Diophantine equations that have no integer solutions, but this fact is not provable from T. The question of the minimum number variables and minimum degree such diophantine equations have been studied. Over **Z**, Skolem showed that degree 4 is sufficient and Zhi-Wei Sun showed that 11 variables is sufficient. It is unknown whether these results are optimal.
---
Now that I reread your question, I think you wanted to have infinitely many logically inequivalent statements each of which is independent of T. This is true when T has no axiomatizable complete extension, which is guaranteed Gödel's Theorem when T is a consistent axiomatizable theory that extends PA.
Indeed, if there were only finitely many statements φ1,...,φk independent of T, up to T-provable equivalence. Then we could get an axiomatizable extension of T by adding to T each such φi or its negation ¬φi while maintaining consistency. (For example, when the standard model satisfies T, we can pick whichever is true in the standard model.) Since we're only adding finitely many new axioms, the result would be an axiomatizable complete theory even if our finitely many decisions were very complex; this would contradict Gödel's Theorem.
| 6 | https://mathoverflow.net/users/2000 | 26399 | 17,285 |
https://mathoverflow.net/questions/26391 | 8 | Let $X$ be a negatively curved symmetric space. In other words, $X$ is one of the four examples: a hyperbolic space, a complex hyperbolic space, a quaternionic hyperbolic space or the hyperbolic Cayley plane.
Let $B\subset X$ be a compact set (wlog, a large ball). Does there always exist a subgroup $\Gamma$ of the isometry group of $X$ such that $X/\Gamma$ is a compact manifold and the images of $B$ under $\Gamma$ are disjoint (i.e. $B\cap\gamma B=\emptyset$ for every nontrivial $\gamma\in\Gamma$)?
**Remark:** It would suffice to find just one (for each $X$) residually finite discrete co-compact group of isometries. I am sure there is one but I don't know where to look. (Or maybe they all are residually finite?)
**Meta-proof:** If the answer is negative, then, for some of the model geometries (in some dimension), all compact manifolds with this local geometry are "uniformly thin" - all injectivity radii are bounded above by some universal constant. Such a wonderful fact would be widely known.
| https://mathoverflow.net/users/4354 | Are there arbitrarily sparse "lattices" in negatively curved symmetric spaces? | The examples are arithmetic groups, constructed in general by [Borel and Harish-Chandra](http://www.ams.org/mathscinet-getitem?mr=147566).
See also [Dave Witte Morris' preliminary book](http://people.uleth.ca/~dave.morris/books/IntroArithGroups.html).
However, examples in hyperbolic and complex hyperbolic spaces probably go back further to the study of quadratic forms.
For hyperbolic lattices, one can take a quadratic form over a quadratic number field (such as $\mathbb{Q}(\sqrt{2})$), which
is Lorentzian at one place, and definite at the other places (such as $x\_1^2+\cdots +x\_n^2-\sqrt{2} x\_{n+1}^2$), and take the group of matrices $\Gamma$ in $GL(n+1,\mathbb{Z}[\sqrt{2}])$ which preserve this quadratic form. Then [Mahler's compactness theorem](http://en.wikipedia.org/wiki/Mahler%27s_compactness_theorem) (cf. Witte Morris) implies
that the quotient $\mathbb{H}^n/\Gamma$ is compact. Then by Selberg's Lemma and residual finiteness (as Greg points out, Malcev's Theorem), you may find a torsion-free subgroup
of finite-index with as large injectivity radius as you like.
For hyperbolic and complex-hyperbolic spaces, there are other examples which don't come from the arithmetic construction (in fact, most hyperbolic surfaces and 3-manifolds are not arithmetic). These are attributable to [Gromov and Piatetski-Shapiro](http://www.ams.org/mathscinet-getitem?mr=932135) in the hyperbolic case in all dimensions, and there are finitely many examples in the complex hyperbolic case (at least for $\mathbb{C}$-dim >1) going back to Deligne and Mostow (see also [Thurston's paper](http://www.msp.warwick.ac.uk/gtm/1998/01/p025.xhtml)). However, Gromov and Schoen have shown that quaternionic and Cayley-hyperbolic lattices are all arithmetic, so the Borel Harish-Chandra construction is complete in these cases.
Addendum: Related to Protsak's comment, there is a simple example of a negatively curved homogeneous space which has no lattice action. This is Thurston's "9th geometry", which is excluded as a geometry precisely for this reason. One can take the double warped product metric $$ dr^2 + e^{2a r} dx^2 + e^{2b r} dy^2, $$ for $a,b >0$. When $a=b$, this gives hyperbolic space. But when $a\neq b$, the sectional curvatures are $-a^2, -b^2, -ab$. This has a solvable transitive group of isometries, so is homogeneous. But using the solvability, one may see that there is no cocompact action. (Remark: when $a=-b$, this gives sol geometry).
| 11 | https://mathoverflow.net/users/1345 | 26400 | 17,286 |
https://mathoverflow.net/questions/26385 | 37 | It is easy to see that if $A\times B$ is homeomorphic to $A\times C$ for topological spaces $A$, $B$, $C$, then one may not conclude that $B$ and $C$ are homeomorphic (for example, take $C=B^2$, $A=B^{\infty}$). The question is: for which $A$ such conclusion is true? I saw long ago a problem that for $A=[0,1]$ it is not true, but could not solve it, and do not know, where to ask. Hence am asking here. The same question in other categories (say, metric spaces instead topological) also seems to have some sense.
| https://mathoverflow.net/users/4312 | When factors may be cancelled in homeomorphic products? | For $A=[0,1]$, let $B$ be the 2-torus with one hole and $C$ be the 2-disc with two holes.
The products $B\times[0,1]$ and $C\times[0,1]$ can be realized in $\mathbb R^3$: the former as a thickening of the torus, the latter in a trivial way. Each of these products is a handlebody bounded by the pretzel surface (the sphere with two handles). It is easy to deform one to the other "by hand".
| 40 | https://mathoverflow.net/users/4354 | 26404 | 17,288 |
https://mathoverflow.net/questions/26402 | 12 | I know [Chaitin's constant](http://en.wikipedia.org/wiki/Chaitin%2527s_constant) Ω is not computable (and therefore transcendental). Are there other specific, known noncomputable numbers? I am trying to understand what distinguishes a computable transcendental number, such as π, from a noncomputable transcendental number, such as Ω. Is there anything revealing that can be said about the set difference {transcendental numbers} \ {computable transcendental numbers}?
I ask this as a novice. I am re-visiting a [wonderful book](http://books.google.com/books?id=tRdoIhHh3moC&printsec=frontcover&dq=klee+and+wagon&source=bl&ots=BwPb3iorQw&sig=oqbPFWeRYfPcbKnAox_Lvx_wdfY&hl=en&ei=RK8BTO7gB4O0lQe7_OSiCA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBYQ6AEwAA#v=onepage&q&f=false) that sadly can no longer be updated by Victor Klee, in which he and Wagon pose this as an open problem: *If an irrational number is real-time computable, is it then necessarily transcendent?* [Problem 23]
**Update** (*19Jun12*). There is an illuminating discussion under the title
"[Why The Hartmanis-Stearns Conjecture Is Still Open](http://rjlipton.wordpress.com/2012/06/15/why-the-hartmanis-stearns-conjecture-is-still-open/)" at the Lipton-Regan blog.
The Hartmanis-Stearns Conjecture is the open problem mentioned above:
If a number is real-time computable, it is either rational or transcendental.
If true, this has what strikes me as a counterintuitive consequence: that algebraic irrationals like $\sqrt{3}$
are in some sense "more complicated" than transcendentals.
| https://mathoverflow.net/users/6094 | {transcendental numbers} \ {computable transcendental numbers} | Note: Answer is pending update per attached comments.
The difference, stated informally, is that that the non-computable transcendentals in their k-base digit representation are entirely random and non compressible. A computable transcendental, such as e, can be represented by a finite algorithmic description, such as a series expansion, which is a form of compression. For the non-computable numbers no such shorter representation exist. Their shortest computational description is their own infinite digit sequence.
You can read more about computational complexity here:
<http://en.wikipedia.org/wiki/Kolmogorov_complexity>.
There is a wealth of similar numbers to the Ω class of numbers. In general it is "easy" to come up with new definitions for such numbers. These all belong to the countably infinite set of non-computable definable numbers.
To make matters worse, what is left are the non-definable (and therefore also non-computable) numbers. They are the numbers that cannot be described in any way what-so-ever, other than by just iterating through their infinite non-compressible digit sequence. The set of all non-definable numbers is uncountable.
| 2 | https://mathoverflow.net/users/1320 | 26408 | 17,291 |
https://mathoverflow.net/questions/26409 | 8 | On page 164 of his book **Models of Peano Arithmetic**, Kaye states Friedman's Theorem:
Let $M{\vDash}PA$ be nonstandard and countable, let $a\in M$ and let $n\in {\mathbb N}$. Then there is a proper initial segment $I\subseteq\_c M$ ($\subseteq\_c$ means "cofinal in") containing $a$ such that $I\cong M$ and $I<\_{\Sigma\_n} M$.
However, on the next page he writes "Nor can we expect in general to get initial segments $I$ with $M\cong I < M$ and $M\neq I$, i.e., elementary for all formulas. For example if $M=K\_T$ (where $T\neq Th({\mathbb N})$ is a complete extension of PA) then $M$ has no proper elementary substructures, and so certainly has no proper elementary initial segments!"
I am confused. If two models in the same language are isomorphic, are they not elementarily equivalent? Of course the converse need not be true.
An isomorphism of models is a bijective homomorphism of the language's algebraic portion (constants and functions) which preserves **and reflects** all relations of the language (Hodges, **Model Theory**, p5). Therefore by induction on the structure of any ${\mathcal L}\_{\omega,\omega}$ formula, the isomorphism will both preserve and reflect it. So an isomorphism preserves **all** formulas (Hodges, Theorem 2.4.3(c)).
If the proper initial segment is isomorphic as a model to the entire model, how could any first-order sentence possibly be true in one and not in the other?
Thanks,
* a
| https://mathoverflow.net/users/2361 | models of PA which are isomorphic but not elementarily equivalent? | There is a major difference between *elementary equivalence* and *elementary embedding*. Moreover, in this case, the actual embedding is somewhat ambiguous. First, let me recap some often confused terminology.
Two models are *elementary equivalent* if they satisfy the same first-order sentences. Any two isomorphic models are always elementary equivalent. An *elementary embedding* is a map j:A→B such that, for all first-order formulas φ(v1,...,vk) and all a1,...,ak ∈ A, A ⊧ φ(a1,...,ak) iff B ⊧ φ(j(a1),...,j(ak)). An isomorphism is always an elementary embedding.
The notation A ≺ B means that A is an *elementary submodel* of B, i.e. the *inclusion map* A ⊆ B is an elementary embedding from A into B. In your context, the isomorphism (or its inverse) is *not* the proposed elementary embedding, it is the inclusion map which is in question: it is elementary for Σn formulas, but not elementary for all first-order formulas.
| 10 | https://mathoverflow.net/users/2000 | 26412 | 17,293 |
https://mathoverflow.net/questions/26411 | 11 | The Second Incompleteness Theorem says that if $T$ is a consistent (computably) axiomatizable theory which extends IΣ1, then $\mathrm{Con}(T)$ is not provable from $T$. By analogy with computability theory, the stronger theory $T + \mathrm{Con}(T)$ can be thought of as the "jump" of $T$. To abuse this analogy, I will use $T'$ to denote the theory $T + \mathrm{Con}(T)$. I will write $T \leq S$ when $S$ proves every axiom of $T$; I will also write $S \equiv T$ (resp. $T < S$) when $T \leq S$ and $S \leq T$ (resp. $S \nleq T$).
It is well-known that if $T$ is consistent there are plenty of axiomatizable theories $S$ such that $T < S < T'$. In the following questions $H$ will denote an operator (like $\mathrm{Con}$) that uses the computable axiomatization of $T$ to produce a sentence $H(T)$. I will write $T^H$ for the theory $T + H(T)$.
1. Is there a computable operator $H(T)$ such that $T < T^H < T'$ for every consistent axiomatizable theory $T$ extending IΣ1? Is there such an operator which moreover satisfies that $T \equiv S$ implies $T^H \equiv S^H$?
2. Is there a computable operator $H(T)$ such that $(T^H)^H \equiv T'$ for every consistent axiomatizable $T$ extending IΣ1? Is there such an operator which moreover satisfies that $T \equiv S$ implies $T^H \equiv S^H$?
Question 1 asks for a uniform solution to the analogue of Post's Problem for axiomatizable theories. Question 2 asks for a uniform "half-jump" operator.
| https://mathoverflow.net/users/2000 | Uniform solutions to Post's problem for axiomatizable theories | (Note: this has been rewritten to reflect the comments below).
The answer to #1 is basically yes, because the proof that the Lindenbaum algebra above T is atomless is completely constructive.
Start with a (consistent) theory T to which the second incompleteness theorem applies, which means that T + ~Con(T) is also consistent. Then there is a sentence S such that T + ~Con(T) neither proves nor disproves S (using the first incompleteness theorem via Rosser's trick). So T + ~Con(T)$\land$~S is stronger than T + ~Con(T), but is still consistent. This means that T + ~(Con(T)$\lor$S) is consistent, so T + Con(T)$\lor$S is stonger than T.
If T $\vdash$ (Con(T)$\lor$S) $\to$ Con(T) then T $\vdash$ S $\to$ Con(T). But this means T $\vdash$ ~Con(T) $\to$ ~S which is impossible. This shows that T + (Con(T)$\lor$S) < T+ Con(T) .
So we can let TH be T + (Con(T)$\lor$S).
| 8 | https://mathoverflow.net/users/5442 | 26413 | 17,294 |
https://mathoverflow.net/questions/25968 | 11 | Let $R$ be a reduced curve singularity over an algebraically closed field $k$ and $\tilde{R}$ its integral closure in its total ring of fractions.
The $k$-dimension of $\tilde{R}/R$ is finite. If we assume $R$ is non-planar and Gorenstein, then **how small can this number be?**
The ring $R = k[[x,y,z]]/(xy = z^2, z x = y^2)$ is a complete intersection, hence Gorenstein, and the dimension of $\tilde{R}/R$ is $4$. The question is thus "is $2$ or $3$ possible?"
For the sake of concreteness, let's say that a curve singularity is a $1$-dimensional quotient of $k[[x\_1, \dots, x\_n]]$ for some $n$.
**Edit:** I had thought that the $k$-dimension of $\tilde{R}/R$ was widely known as the $\delta$-invariant; I think this the notation Serre uses in *Algebraic Groups and Class Fields*. From the comments, it seems this is non-standard, and I have edited accordingly.
As Graham points out, the number $\operatorname{dim}(\tilde{R}/R)$ is also the colength of the conductor ideal. The number also comes up in computing the (arithmetic) genus of a singular curve.
| https://mathoverflow.net/users/5337 | What is the most simple non-planar Gorenstein curve singularity? | I think Graham's answer already gave most of what you need to prove that $4$ is the smallest possible. Let $V$ be the integral closure of $R$, $n$ be the embedding dimension of $R$, and $e=e(R)$ be the multiplicity.
Claim: If $R=k[[x\_1,\cdots,x\_n]]/I$ is Gorenstein and $n$ is at least $3$, then $\dim\_k(V/R)\geq e$.
Proof: Let $m$ be the maximal ideal of $R$. As Graham pointed out, we have $e = \dim\_k(V/mV)$. So:
$$\dim\_k(V/R) =\dim\_k(V/mR)-\dim\_k(R/mR) \geq \dim\_k(V/mV)-1=e-1$$
We need to rule out the equality. If equality happens, then one must have $mV=mR$. This shows that $m$ is the conductor of $R$. As you already knew, since $R$ is Gorenstein, one must then have $\dim\_k(V/R)=\dim\_k(R/m)=1$. The inequality now gives $e\leq 2$. Abhyankar's inequality (part 2 of Graham's answer) gives $n\leq 2$, so $R$ is planar, contradiction.
Now, one needs to show that for $R$ non-planar, $e\geq 4$. You could use part $3$ of Graham's answer, or arguing as follows: if $n\geq 4$ we are done by Abhyankar inequality. If $n=3$, a Gorenstein quotient of $k[[x,y,z]]$ must be a complete intersetion, and so $I=(f,g)$, each of minimal degree at least $2$ since $R$ is not planar, thus $e$ must be at least $4$.
By the way, one could construct a *domain* $R$ such that $\dim\_k(V/R)=4$ as follows:
Take $R=k[[t^4,t^5,t^6]]$. The semigroup generated by $(4,5,6)$ is symmetric, so $R$ is Gorenstein. The Frobenius number is $7$, and $V/R$ is generated by $t,t^2,t^3,t^7$.
EDIT (references, per OP's request): Abhyankar inequality is standard, for example see Exercise 4.6.14 of Bruns-Herzog "Cohen-Macaulay rings", second edition ([Link to the exact page](http://books.google.com/books?id=ouCysVw20GAC&lpg=PP1&dq=cohen%20macaulay%20rings&pg=PA192#v=onepage&q&f=false)). Or see exercise 11.10 of [Huneke-Swanson book](http://books.google.com/books?id=APPtnn84FMIC&lpg=PP1&ots=2LcMcV9DU0&dq=Huneke%20Swanson%20book&pg=PA233#v=onepage&q&f=false), also available for free [here](http://people.reed.edu/~iswanson/book/index.html). Or Google "rings with minimal multiplicity".
(The original references are now available thanks to Graham, see his comment below)
As for $e=\dim\_k(V/mV)$, I could not find a convenient reference, but here is a sketch of proof using the above reference: First, using the additivity and reduction formula (Theorem 11.2.4 of Huneke-Swanson) to reduce to the domain case. Assume that $R$ is now a complete domain, then $V=k[[t]]$, and $R$ is a subring of $V$. Let $x\in m$ be an element with smallest minimal degree. Then $mV=xV$ ($V$ is a DVR), and it is not hard to see that $e=$ the minimal degree of $x$ $=\lambda(V/xV)$ (see Exercise 4.6.18 of Bruns-Herzog, same page as the link above).
Alternatively, one can use the fact that:
$$e(m,V) = \text{rank}\_RV.e(m,R) = e $$
The second inequality is because $V$ is birational to $R$ so $\text{rank}\_RV=1$. The left hand side can be easily computed by definition to be length of $V/xV$, which equals $\dim\_k(V/mV)$. (use $m^nV=x^nV$ since $V$ is a DVR)
Fun exercise!
| 6 | https://mathoverflow.net/users/2083 | 26418 | 17,297 |
https://mathoverflow.net/questions/26405 | 7 | Let $M\_{n\times n}$ denote the set of $n\times n$ real matrices and let $GL\_n$ be the subgroup of invertible matrices. $GL\_n$ acts on $M\_{n\times n}$ smoothly by conjugation, which means that each conjugacy class (which is an orbit of this action) is an immersed submanifold of $M\_{n\times n}$. However, the action is not proper (e.g. the isotropy groups are not compact) so the orbits may not be embedded submanifolds.
My question is if there are nice conditions on a matrix that guarantee that its conjugacy class is or is not an embedded submanifold. My interest in this question actually comes from trying to understand the space of all complex structures on a real vector space: it can be shown that the set of all complex structures is the conjugacy class of the block matrix $\begin{pmatrix} 0 & -I \\\ I & 0\end{pmatrix}$ and I was wondering if this is an embedded submanifold. So an answer to this question (if the above doesn't have a nice answer) would also be appreciated.
| https://mathoverflow.net/users/4622 | When is a conjugacy class of matrices an embedded submanifold? | If $G$ is an algebraic group (say over $\mathbb C$), acting on an algebraic variety $X$, the orbits are always locally closed smooth algebraic subvarieties of $X$. This is standard, and easy to prove. If $x$ is a point of $X$, and $G \to X$ is the morphism sending $g$ to $gx$, consider the orbit $Gx$ as a subset of its Zariski closure $\overline{Gx}$. By Chevalley's theorem, $Gx$ contains a dense open subset of $\overline{Gx}$. Hence, by homogeneity it is open in $\overline{Gx}$, and smooth.
From this it is easy to deduce that all the orbits of $\mathrm{GL}\_n(\mathbb R)$ in $\mathrm{M}\_n(\mathbb R)$ are embedded submanifolds. The point is that if two real matrices are conjugate as complex matrices, then they are conjugate as real matrices. These means that the orbits of the action of $\mathrm{GL}\_n(\mathbb R)$ in $\mathrm{M}\_n(\mathbb R)$ are of the form $\Omega(\mathbb R)$, where $\Omega$ is an orbit of $\mathrm{GL}\_n(\mathbb C)$ in $\mathrm{M}\_n(\mathbb C)$; since $\Omega$ is a smooth algebraic variety, it follows from the implicit function theorem that $\Omega(\mathbb R)$ is an embedded submanifold of $\mathrm{M}\_n(\mathbb R)$.
| 5 | https://mathoverflow.net/users/4790 | 26428 | 17,303 |
https://mathoverflow.net/questions/26434 | 4 | Hi math people.
I'm in the process of analyzing some data that I collected through an experiment. The data are (somewhat) normally distributed and I represent the different data-sets using [boxplot](http://en.wikipedia.org/wiki/Box_plot), to provide an easy way of visually comparing the mean between the data-sets and the change in the variance.
In Matlab as default, the whiskers are used to represent all samples lying within 1.5 times the [IQR](http://en.wikipedia.org/wiki/Interquartile_range). According to Wikipedia on the boxplot, this is one of several way of using the whiskers. My question is simple why? What special significance does 1.5 times IQR have? Why not e.g. three times sigma?
(NB: I wanted to add the tags "matlab" and "boxplot", but I'm unable to create new tags.)
| https://mathoverflow.net/users/5357 | Why 1.5IQR whiskers in boxplot? | Three sigma has less relevance for asymmetric distributions. Using quartiles keeps it nonparametric. Regarding why not other quantiles of the distribution rather than 1.5\*IQR, you can follow some comments on this [thread](http://tolstoy.newcastle.edu.au/R/help/05/07/8210.html), which basically argues that you want to avoid specifying a fixed fraction of your data to be flagged as 'outliers'.
| 6 | https://mathoverflow.net/users/5282 | 26435 | 17,309 |
https://mathoverflow.net/questions/26438 | 7 | Are there any computer algebra systems (e.g. Macaulay2 og singular) that allows one to compute the Picard number (i.e. the rank of the Neron-Severi group) of a given variety?
| https://mathoverflow.net/users/3996 | Are there any sofware packages for computing Picard numbers? | A more basic question is whether there even exists an algorithm to compute this number.
I've wondered this for a long time, and I honestly don't know what to expect.
Any algorithm would have to be quite subtle. In the early 1980's Shioda had to work quite hard to construct explicit examples of surfaces in $\mathbb{P}^3$, defined over $\mathbb{Q}$, with Picard number 1.
Added: Of course, I should have said Shioda's example have degree >1. As further
evidence of subtlety of the problem: one can decide whether an elliptic curve $E$
has CM by computing the Picard number of $E\times E$.
| 11 | https://mathoverflow.net/users/4144 | 26441 | 17,312 |
https://mathoverflow.net/questions/26443 | 0 | The following combinatorial problem has bothered me quite a bit. I guess people smarter than me have given the problem some taught as the problem has obvious applications (e.g. to the Ising model), but I have not found any solution on the web (this might be because I don't know the proper terminology).
Anyways, here is the problem:
Consider a string of $N$ binary variables, $\uparrow$ and $\downarrow$. The string will have $2^N$ different configurations. Now impose a symmetry to the system; two configurations are equal if you can get from one to the other by cyclic permutation or by reversal of the string (or a combination of these two symmetries). How many unique configurations will the string have?
For 1 $\uparrow$ and $N-1$ $\downarrow$ there will only be 1 unique configuration. For 2 $\uparrow$ and $N-2$ $\downarrow$ there will be $N/2$ configurations if $N$ is even and $(N-1)/2$ configurations if $N$ is odd. But if you take 3 $\uparrow$ and $N-3$ $\downarrow$, it is no longer clear (at least not to me), how one efficiently should count the number of possible configurations.
I would really appreciated some help, or references on relevant literature.
| https://mathoverflow.net/users/4626 | Number of config. of a binary string invariant under cyclic permutation. | <http://en.wikipedia.org/wiki/Necklace_(combinatorics)> will get you started.
| 1 | https://mathoverflow.net/users/6153 | 26445 | 17,314 |
https://mathoverflow.net/questions/26421 | 4 | Is Young's inequality true for an arbitrary measure on $\mathbb R^d$? If so, where can I find a proof of it? In particular, where can I find the proof of the discrete version (i.e the version for $\ell^p$ spaces) of this inequality?
Here is the statement of the inequality (from Wikipedia):
Suppose $f$ is in $L^p(\mathbb R^d)$ and $g$ is in $L^q(\mathbb R^d)$ and
$$ \frac{1}{p} + \frac{1}{q} = \frac{1}{r} + 1$$
with $1\leq p, q, r\leq \infty$
then
$$ || f \* g || \_r \leq ||f||\_p ||g||\_q.$$
| https://mathoverflow.net/users/1229 | Proof of Young's convolutions inequality for a general measure on $\mathbb R^d$ | Note that if your measure is not translation-invariant, the convolution product is not commutative.
This allows for a simple counterexample with $r=q=\infty$, $p=1$,
$d\mu(x)={\bf 1}\_{[0,1]}(x) dx$.
Define $f\*g(x)= \int f(x-s)g(s)d\mu(s) \ (\neq g\*f(x))$.
Take $g\equiv 1, \ f={\bf 1}\_{[1,2]}$.
This gives $||f||\_1=0$, but the spreading in the convolution makes the left term non zero.
$\max\_x \ \ \int\_0^1 {\bf 1}\_{[1,2]}(x-s)ds =1$.
Just add $\varepsilon e^{-|x|}dx$ to $d\mu$ to get a counterexample with a measure of full support. I think that the reasonable setting for a Young inequality is the case of a translation invariant measure on a locally compact group. Jonas Mayer gave a reference (Hewitt and Ross) for that case in the comments.
| 8 | https://mathoverflow.net/users/6129 | 26448 | 17,316 |
https://mathoverflow.net/questions/26461 | 12 | Let $G=\mathbb{Z}/2\mathbb{Z}$. Let $f\colon L \to N$ be a smooth map of connected smooth **closed** $n$-dimensional manifolds such that the induced map
$$f^\* \colon H^\*(N,G) \to H^\*(L,G)$$
is an isomorphism.
**Question**: Are the pull back of the Stiefel-Whitney classes of the tangent bundle of $N$ the Stiefel-Whitney classes of the tangent bundle of $L$?.
This is in fact true for the first Stiefel-Whitney class by considering coverings and degrees, but what about the higher degree classes?
**Motivation**: This came up because (relative) spin is important in defining Floer homology with $\mathbb{Z}$ coefficients. So I am in fact mostly interested in the following sub-question.
**Question**: In particular what about the second Stiefel-Whitney class in the case where both $N$ and $L$ are also assumed to be oriented? and if the answer is negative: what extra conditions do I need to make it positive?
The idea is that I apriori have to use $G$ coefficients, but can prove that it is a $G$-cohomology equivalence, and want to use that to start the argument over again with other coefficients, but for that I need this property of the second Stiefel-Whitney class.
This sub-question and the relation to Floer homology is related to orientations in real $K$-theory and delooping in the following sense: take a map $h\colon X \to U/O$ by delooping we get a map $\Omega h \colon \Omega X \to \Omega U/O \simeq \mathbb{Z}\times BO$ which classifies a virtual bundle over the loop space of $X$. This bundle is oriented iff the original map composed with the canonical map $U/O \to BO$ classified a virtual $0$-dimensional bundle with vanishing second Stiefel-Whitney class. This is the main point of why orientations in Floer homology is initimitely linked with spin! In the case of a Lagrangian sub-manifold $L\subset T^\*N$ the difference of the tangent bundles precisely defines such a map $L \to U/O$ ($U(n)/O(n)$ classifies Lagrangians in $\mathbb{R}^{2n}$) such that the composition to $BO$ classifies the virtual bundle $TN-TL$. So in fact you may add this lifting property as an extra condition to the sub-question if you like, and then I would lose no generality. I believe that this condition implies that all the relative Prontryagin classes vanishes, which may be helpfull.
ADDED: in light of the answer, all this motivation and these extra possible assumptions are not important nor relevant for the actual question.
| https://mathoverflow.net/users/4500 | Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology? | The answer to the question is positive, due to Wu's formula. See e.g. Milnor-Stasheff, Characteristic classes, lemma 11.13 and theorem 11.14. In fact, all one needs to compute the Stiefel-Whitney classes of a smooth compact manifold (orientable or not) is the cohomology mod 2 (as an algebra) and the action of the Steenrod algebra on it. Both structures are preserved under cohomology isomorphisms induced by continuous maps.
| 20 | https://mathoverflow.net/users/2349 | 26469 | 17,328 |
https://mathoverflow.net/questions/26446 | 21 | I'm looking for references on the "algebraic geometry" side of complex analytis, i.e. on complex spaces, morphisms of those spaces, coherent sheaves, flat morphisms, direct image sheaves etc. A textbook would be nice, but every little helps.
Grauert and Remmert's "Coherent analytic sheaves" seems to contain what I want, but it is very dense reading. You could say I'm looking for sources to read on the side as I work through G&R, to get different points of views and examples. For example, B. and L. Kaup's "Holomorphic functions of several variables" talks about the basics of complex analytic geometry, but doesn't go into much detail.
My motivation is twofold. First, I'm studying deformation theory, which necessarily makes use of complex spaces, both as moduli spaces and objects of deformations, so while I can avoid using complex spaces at the moment they're certain to come in handy later. Second, I want to be able to talk to the algebraic geometers in my lab, so I should know what their schemes and morphisms translate to in the analytic case. I like reading as much as I can about what I'm trying to learn, so:
### Do you know of other sources (*anything*: textbooks, lecture notes, survey articles, historical overviews, comparisons with algebraic geometry ...) that talk about complex spaces and their geometry?
| https://mathoverflow.net/users/4054 | References for complex analytic geometry? | Two books that I like a lot:
1) Joseph Taylor's [Several complex variables with connections to algebraic geometry and Lie groups](http://books.google.com/books?id=i8lUNpZ379MC&printsec=frontcover&dq=Taylor,+Several+complex+variables&source=bl&ots=7MVn4Xlu9E&sig=Bwt3jHAJvq5LyuuW8TODoykm9eE&hl=en&ei=0KcCTIL6LsaqlAeEwtGiCA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBYQ6AEwAA#v=onepage&q&f=false) .
2) Constantin Banica and Octavian Stanasila's "Algebraic methods in the global theory of complex spaces" , Wiley (1976)
| 10 | https://mathoverflow.net/users/439 | 26476 | 17,332 |
https://mathoverflow.net/questions/26467 | 2 | Has it been proven that maximal planar graphs are reconstructible?
It seems like an easy result, but I am unable to find it in the literature. Classes of planar graphs that I know are reconstructible are: maximal outerplanar (Manvel 1970), maximal minimally non-outerplanar (has a single interior vertex) (Kunni, Annigeri, 1979), and classes where planar isn't the key property: like trees or cacti (Geller, Manvel, 1969).
Perhaps this has not been explicitly answered because it was only recently (Bilinski, Kwon, Yu, 2007) when it was proven that planar graphs are recognizable, which is the "hard" part of this result.
Bonus question: Are there other classes of planar graphs that are known to be reconstructible?
| https://mathoverflow.net/users/4167 | Reconstructing Maximal Planar Graphs | You may want to look at papers by Fiorini and Lauri. I found the following reference, and I believe there is part II that does reconstruction. But I can't locate it in my bib file, and don't have access to MathSciNet at the moment. Fiorini and Lauri have many papers on other classes of planar graphs. Mostly on edge reconstruction, though. Also look at Bondy's survey Graph Reconstructor's Manual (1991).
@article {MR615313,
AUTHOR = {Fiorini, S. and Lauri, J.},
TITLE = {The reconstruction of maximal planar graphs. {I}.
{R}ecognition},
JOURNAL = {J. Combin. Theory Ser. B},
FJOURNAL = {Journal of Combinatorial Theory. Series B},
VOLUME = {30},
YEAR = {1981},
NUMBER = {2},
PAGES = {188--195},
ISSN = {0095-8956},
CODEN = {JCBTB8},
MRCLASS = {05C60 (05C10 05C35)},
MRNUMBER = {MR615313 (82i:05055a)},
MRREVIEWER = {Thomas Andreae},
}
| 1 | https://mathoverflow.net/users/6438 | 26477 | 17,333 |
https://mathoverflow.net/questions/26482 | 1 | Dear all,
here is a question that has been bothering me. It goes without saying that I would appreciate any help in answering it.
Let {I\_k} be a countable sequence of two sided closed ideals in a C\*-algebra (ring) and J be a two sided closed ideal in the same C\*-algebra (ring).
Then "intersection of {I\_k + J} = (intersection of {I\_k}) + J"
If needed we can relax the above hypothesis by assuming that {I\_k}'s and J are prime ideals.
Thanks in advance,
Audrey Kirilova.
| https://mathoverflow.net/users/6439 | Intersection of ideals in C*-algebra or even rings in general | In the most general form, for arbitrary ideals over rings, this
is false. In the ring $\mathbb{Z}$ let $I\_k$ be generated by $2^k$
and let $J$ be generated by $3$. Then $I\_k+J=\mathbb{Z}$
for all $k$ and so $\cap\_{k=1}^\infty(I\_k+J)=\mathbb{Z}$.
But $\bigcap\_{k=1}^\infty I\_k=\lbrace0\rbrace$ and so
$J+\bigcap\_{k=1}^\infty I\_k=J\ne\mathbb{Z}$.
For an example in a $C^\*$-algebra let $R=C[0,1]$ the continuous
functions on $[0,1]$. Let $(a\_k)$ be an enumeration of
the rationals in $[0,1]$, and $I\_k$ be the ideal of functions
vanishing at $a\_k$. Let $J$ be the ideal of functions vanishing at
$1/\sqrt2$ say. Then $\bigcap\_{k=1}^\infty I\_k=\lbrace0\rbrace$
and $I\_k+J=R$ for all $k$, and the rest proceeds as above.
| 3 | https://mathoverflow.net/users/4213 | 26483 | 17,335 |
https://mathoverflow.net/questions/26473 | 1 | How many positive integer solutions does the equation x^2+y^2+z^2-xz-yz = 1 have? My guess is (1,0,1), (0,1,1) and (1,1,1). What is proof of that fact that there are none other?
| https://mathoverflow.net/users/6437 | Diophantine equation problem | To elaborate on Robin's suggestion, set $x=v+w,y=u+w,z=u+v$, and the equation becomes $$u^2+v^2+2w^2=1,$$ with $u=(y+z-x)/2,v=(x-y+z)/2,w=(x+y-z)/2$ being half-integers. Now a brute force run through the possibilities is feasible, since $|u|\leq 1$ and such.
For the quick answer, you can use Mathematica:
Reduce[x x + y y + z z - x z - y z == 1, Integers]
or even wolfram|alpha:
[solve x x + y y + z z - x z - y z == 1 over the integers](http://www.wolframalpha.com/input/?i=solve+x+x+%2B+y+y+%2B+z+z+-+x+z+-+y+z+%3D%3D+1+over+the+integers)
| 3 | https://mathoverflow.net/users/935 | 26484 | 17,336 |
https://mathoverflow.net/questions/26361 | 2 | Is the following claim true?
**Claim** Let $A, B\in C^{n\times n}$ with $rank(A)=rank(B)=r$. Then there exist nonsingular matrices $P\_1, P\_2, Q\_1, Q\_2$ such that
$$ Q\_1AP\_1=Q\_2BP\_2=\left(\begin{array}{cc}I\_{r}&0\\\
0&0
\end{array}\right)$$ and
$$Q\_1Q\_2^{-1}=\left(\begin{array}{cc}X\_1&0\\\
X\_2&X\_3
\end{array}\right), \qquad
P\_1^{-1}P\_2=\left(\begin{array}{cc}Y\_1&Y\_2\\\
0&Y\_3
\end{array}\right),$$ where $X\_1, Y\_1 \in C^{r\times r}$.
| https://mathoverflow.net/users/3818 | A question on matrix decomposition. | The Gauss-algorithm tells you that you can find matrices $Q\_i$ and $P\_i$ satisfying the first line. For the second line you ask, how can one modify, say, the $Q\_i$ to make the second line come true? One thing you certainly can do, is changing $Q\_1$ by
$$\left(\begin{array}{cc} 1&x\\\ 0&1\end{array}\right)Q\_1$$
and replace $Q\_2$ by $\left(\begin{array}{cc} 1&-y\\\ 0&1\end{array}\right)Q\_2$. The matrix $A=\left(\begin{matrix} a&b\\\ c&d\end{matrix}\right)=Q\_1Q\_2^{-1}$ then changes to
$$\left(\begin{array}{cc} \ast&ay+b+x(cy+d)\\\ \ast&\ast\end{array}\right).$$
Now $A$ being invertible, the matrix $(c\ d)$ has full rank. From this it is easy to see that there exists $y$ such that $(cy+d)$ is invertible.
Hence one can find $x$ such that $ay+bx(cy+d)=0$.
The $P$'s are treated similary.
| 3 | https://mathoverflow.net/users/nan | 26486 | 17,338 |
https://mathoverflow.net/questions/26462 | 7 | Suppose you have the autonomous ordinary differential equation $dx(t)/dt = f(x(t))$ with $x: \mathbb{R} \to \mathbb{R}$ and the initial condition $x(0)=x\_0$. Then, assuming some regularity conditions, you get as solution the
flow $\Phi(x\_0,t):=x(t)$. To give a trivial example: If $f(x)=x$, then $\Phi(x\_0,t)=x\_0 \exp(t)$.
Now, I'm not interested in the trajectories for a given initial condition, that is in $\Phi(x\_0,t)$ with $x\_0$ fixed and $t$ variable; but in the map $x\_0 -> \Phi(x\_0, t)$ for a fixed $t$ (say $t=1$).
Given the function $f$, you can easily (at least in principle, by solving the ODE) get the function $\Phi(\cdot, 1)$. There are a lot of theorems about existence and uniqueness of this problem and analytical and numerical algorithms.
But how can one get $f$ out of $\Phi(\cdot, 1)$? Is this a well posed problem? Are there any theorems?
This problem is closely related to "interpolating" the $n$-fold functional iterates of $g$ (with $g^{[0]} = \mathrm{Id}, g^{[1]} = g, g^{[2]} = g \circ g, g^{[n+m]} = g^{[n]} \circ g^{[m]}$ for $n,m \in \mathbb{N}$) from $n \in \mathbb{N}$ to real values. If such an interpolation succeeds, on can get the ODE out of the flow $\Phi(\cdot, 1)$ by determining $\Phi(\cdot, 1)^{[\alpha]}$ for small $\alpha >0$. I have done some calculation, that give results, but lack
in rigor.
For noninteger iterates of functions, a classical reference is <http://www.math-inst.hu/~p_erdos/1960-07.pdf>.
| https://mathoverflow.net/users/6415 | Noninteger iterates of functions: How to get ODE from flow at a given time? | If g is a real-analytic function defined near x0 with g(x0) = x0 and 0 < λ ≠ 1 where λ := g'(x0), then Koenigs proved that there exists a real-analytic homeomorphism h defined near x0 such that hgh-1(x) = L(x) where
L(x) := x0 + λ(x-x0)
and h is unique up to a constant factor\*.
This allows g to be embedded in a local flow: Φ(x,t) := h-1 Lt h(x), where
Lt(x) := x0 + λt (x-x0),
such that Φ(x,1) = g(x) where defined.
Then Φ satisfies the differential equation dx(t)/dt = V(x(t)) where the velocity V (denoted by f in the Question) is given by
V(x) := ∂Φ(x,t)/∂t |t=0.
Note, however, that the flow Φ(x,t) into which the original function g embeds as the time-1 map need not be unique when there exist more than one fixed point of g. As an example, for x > 0 consider the function
gc(x) := cx.
Then for 1 < c < e1/e the function gc has two distinct fixed points each satisfying the hypotheses of the Koenigs theorem, and these give two distinct flows into which gc embeds as the time-1 map.
Concretely, set c = √2, so that gc(x) = x for both x = 2 and x = 4, with derivatives
gc'(2) = ln(2) and
gc'(4) = ln(4).
Calculating the respective real-analytic flows Φ2(x,t) and Φ4(x,t), both are defined for (x,t) = (3, 1/2).
But Φ2(3, 1/2) and Φ4(3, 1/2) first differ in the 25th decimal place. Hence they are solutions of distinct differential equations. I.e., they have different velocity functions.
---
\* This is actually true in greater generality; see J. Milnor's book *Dynamics in One Complex Variable*, 3rd ed., Princeton University Press, 2006.
| 4 | https://mathoverflow.net/users/5484 | 26495 | 17,343 |
https://mathoverflow.net/questions/26474 | 3 | Suppose we have a complex vector field on $\mathbb{C}^n$ which is analytic and has $|DV| < L$ on ball $B\_r$ with radius r.
I would like to understand:
1) if there exists an analytic flow $\phi\_t(x)$ with complex time $t$ such that $\partial\_t \phi\_t(x)=v(\phi\_t(x))$
2) if such flow is analytic in both $t$ and $x$
3) if the domain of the variable $t$ where $\phi\_t(x)$ is analytic is bounded by $r (\sup\_{B\_r} |v|)^{-1}$ (or something like that).
Are there references about this kind of problems?
Thank you for your attention.
| https://mathoverflow.net/users/6357 | Analytic ODE with complex time | You'll find relevant information in the book
Ordinary differential equations in the complex domain
By Einar Hille
<http://books.google.com/books?id=I1OR4t8UZCgC&printsec=frontcover&dq=Ordinary+Differential+Equations+in+the+Complex+Domain&ei=t8wCTPaXFZLKygT_9e24DA&cd=1#v=onepage&q&f=false>
Fixed point (iteration) results are used to prove local existence, and also to give explicit
lower bounds on the domain of existence, like you want.
| 3 | https://mathoverflow.net/users/5365 | 26502 | 17,348 |
https://mathoverflow.net/questions/26497 | 20 | Hi. From one of the forms of Hilbert's Nullstellensatz we know that all the maximal ideals in a polynomial ring $k[x\_1, \dots, x\_n]$ where $k$ is an algebraically closed field, are of the form $(x\_1 - a\_1, \dots , x\_n - a\_n)$. So that any maximal ideal in this case is generated by polynomials $g\_j \in k[x\_1, \dots, x\_n]$ for $j = 1, \dots , n$, where $g\_j$ only depends on the variable $x\_j$ (Obviously by taking $g\_j = x\_j - a\_j$). Now, I'm interested in the case of a polynomial ring $k[x\_1, \dots, x\_n]$ where $k$ is an arbitrary field (i.e., I can't make use of the Nullstellensatz). I suppose this may no longer be the case, i.e., I don't expect any maximal ideal to be generated by n polynomials, each of them only dependent on one of the variables $x\_j$, but my question is if maybe the maximal ideals can be generated by polynomials $g\_j$ that only depend on the first $j$ variables $x\_1, \dots , x\_j$? If so, does anybody know how to prove this or can anyone suggest me some references that may help me?. Thanks.
| https://mathoverflow.net/users/4170 | Maximal Ideals in the ring k[x1,...,xn ] | The stronger version of the Nullstellensatz states that a maximal
ideal $I$ of $R=k[x\_1,\ldots,x\_n]$ is the kernel of a $k$-homomorphism
from $R$ to $L$ where $L/k$ is a finite extension. Let $a\_1,\ldots,a\_n$
be the images of $x\_1,\ldots,x\_n$ under such a homomorphism.
Then $a\_1$ has a minimal polynomial $m\_1$ over $k$. Let
$f\_1(x\_1,\ldots,x\_n)=m\_1(x\_1)$. Then $f\_1\in I$
Now $a\_2$ has a minimal polynomial $m\_2$
over $k(a\_1)$. We can write the coefficients of $m\_2$ as polynomials
in $a\_1$ over $k$. Doing this, and replacing $a\_1$ by $x\_1$ and the free
variable by $x\_2$ gives a polynomial $f\_2$ in $x\_1$ and $x\_2$. Also $f\_2\in I$.
Keep going. We get a sequence of polynomials $f\_i$ in $x\_1,\ldots,x\_i$
and it's not hard to prove these generate $I$.
| 17 | https://mathoverflow.net/users/4213 | 26503 | 17,349 |
https://mathoverflow.net/questions/26491 | 34 | This is probably common knowledge, alas I have to confess my ignorance.
In simpler more abstract language, does $\mathcal{O}\_K$ being simply connected (having trivial etale $\pi\_1$) imply $\mathcal{O}\_K=\mathbb{Z}$?
| https://mathoverflow.net/users/5756 | Is there a ring of integers except for Z, such that every extension of it is ramified? | Yes, there are examples and Minkowski's proof for ${\mathbf Q}$ can be adapted to find a few of them. Some examples of this kind among quadratic fields $F$, listed in increasing size of discriminant (in absolute value), are
$$
{\mathbf Q}(\sqrt{-3}), \ \ {\mathbf Q}(i), \ \ {\mathbf Q}(\sqrt{5}), \ \ {\mathbf Q}(\sqrt{-7}), \ \ {\mathbf Q}(\sqrt{2}), \ \ {\mathbf Q}(\sqrt{-2}).
$$
A cubic and quartic field $F$ that will come out of the method I describe below are ${\mathbf Q}(\alpha)$ where $\alpha^3 - \alpha - 1 = 0$ and ${\mathbf Q}(\zeta\_5)$.
Now for the details. I suggest when reading this through for the first time that you keep a concrete example in mind, like $F = {\mathbf Q}(i)$. (That's what I did the first time I worked this out.)
Over the rationals, Minkowski showed a number field with degree larger than 1 must have a discriminant whose absolute value is larger than 1. Over other number fields $F$ besides the rationals, the goal is to find sufficient conditions on $F$ so that *any* finite extension $E/F$ with $[E:F] > 1$ has its discriminant ideal ${\mathfrak d}\_{E/F}$ not equal to the unit ideal, and then a prime ideal factor will ramify in $E$.
Rather than show ${\mathfrak d}\_{E/F} \not= (1)$, we will look for a sufficient condition on $F$ which assures us that the norm of this ideal is not 1. That means absolutely the same thing, but it's easier to work with ideal norms since they are positive integers rather than ideals, and moreover it lets us express the problem in terms of discriminants of number fields: the discriminants of $E$ and $F$ are related by $$|d\_E| = {\rm N}({\mathfrak d}\_{E/F})|d\_F|^{[E:F]}.$$
So aiming to show ${\mathfrak d}\_{E/F} \not= (1)$ is the same as *avoiding*
$|d\_E| = |d\_F|^{[E:F]}$, which is the same as *avoiding* $$|d\_E|^{1/[E:{\mathbf Q}]} = |d\_F|^{1/[F:{\mathbf Q}]}.$$ We want sufficient conditions on $F$ to guarantee this equation for any proper finite extension $E/F$ can't take place.
For any number field $K$, the quantity $|d\_K|^{1/[K:{\mathbf Q}]}$ is called the root discriminant of $K$. When $n = [K:{\mathbf Q}]$, Minkowski's lower bound on $|d\_K|$ is
$$
|d\_K| \geq \left(\frac{\pi}{4}\right)^n\frac{n^{2n}}{n!^2},
$$
so we get the root discriminant lower bound
$$
|d\_K|^{1/n} \geq \left(\frac{\pi}{4}\right)\frac{n^{2}}{n!^{2/n}}.
$$
Call the right side $f(n)$, so the Minkowski bound says $|d\_K|^{1/[K:{\mathbf Q}]} \geq f([K:{\mathbf Q}])$.
For $n = 1, 2, 3, 4$ the values of $f(n)$ are $.785, 1.570, 2.140, 2.565$, so we guess $f(n)$ is increasing and it's left as an exercise to prove that. (Hint: use the one-sided Stirling estimate $n! > (n/e)^n\sqrt{2\pi{n}}$.)
Returning to the extension $E/F$, let $m = [F:{\mathbf Q}]$, so $[E:{\mathbf Q}] = [E:F][F:{\mathbf Q}] \geq 2m$ since $E$ is a larger field than $F$. Now *if*
$E/F$ had trivial discriminant ideal, we would have
$$
|d\_F|^{1/[F:{\mathbf Q}]} = |d\_E|^{1/[E:{\mathbf Q}]} \geq f([E:{\mathbf Q}]) \geq f(2m).
$$
This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$. Suppose $F$ is a number field of degree $m$
whose root discriminant is less than $f(2m)$. If $E/F$ is unramified at all primes in $F$
then $E$ and $F$ have equal root discriminants, so the root discriminant of $E$ is less than $f(2m)$. However, we saw above that the root discriminant of $E$ is $\geq f(2m)$ when $[E:F] \geq 2$, so the only choice is $E = F$, i.e., no proper finite extension of $F$ can be unramified at all primes in $F$.
Our goal now is to find examples of number fields $F$ with degree $m$ whose
root discriminant is smaller than $f(2m)$: $|d\_F|^{1/m} < f(2m)$. This is the same as
$$
|d\_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}.
$$
Any $F$ which fits this condition will be an example.
As a reality check, let $F$ be the rationals, so $m = 1$. We have
$f(2) = \pi/2$ and $|d\_{\mathbf Q}| = 1 < \pi/2$, so $\mathbf Q$ has no unramified extensions. We're on the right track.
Taking $m = 2$ we want $|d\_F| < f(4)^2 = 6.57$ and the fields ${\mathbf Q}(i)$, ${\mathbf Q}(\sqrt{-3})$, and ${\mathbf Q}(\sqrt{5})$ all work.
If the root discriminant of $F$ is not below $f(2m)$, where $m = [F:{\mathbf Q}]$,
we can still squeeze out some information, namely an upper bound on the degree of an everywhere (= at finite places) unramified extension of $F$. For instance, ${\mathbf Q}(\sqrt{2})$ has discriminant 8, which is not below 6.57, so this argument doesn't show on its own that every proper extension of ${\mathbf Q}(\sqrt{2})$ is ramified at some prime in ${\mathbf Q}(\sqrt{2})$. However, we can bound the degree of such an extension. The root disc. of ${\mathbf Q}(\sqrt{2})$ is $\sqrt{8} \approx 2.828$, which lies between $f(4)$ and $f(5)$, so any proper unramified extension of ${\mathbf Q}(\sqrt{2})$ must be a quadratic extension of ${\mathbf Q}(\sqrt{2})$. Quadratic extensions are automatically abelian, so we would have an abelian extension of ${\mathbf Q}(\sqrt{2})$ unramified at no finite places, and there's no such thing by class field theory since ${\mathbf Q}(\sqrt{2})$ has wide class number 1 (the class number not involving ramification at infinity). Therefore you can add ${\mathbf Q}(\sqrt{2})$ to the list of number fields with no proper extension unramified at all finite places. The same arguments apply to ${\mathbf Q}(\sqrt{-2})$ and ${\mathbf Q}(\sqrt{-7})$, whose root discriminants are also between $f(4)$ and $f(5)$.
Unfortunately, this method is really limited because although $f(n)$ is increasing,
it's actually bounded. By Stirling's formula, $f(n)$ has limit $\pi e^2/4 \approx 5.803$. So when a number field $F$ has root discriminant exceeding this value, this method won't give us any examples at all (you can't find an $m$ for which the root discriminant is below $f(2m)$). [Edit: Using Odlyzko-type lower bounds on discriminants, one can produce some more examples as Torsten points out.]
This kind of argument using Minkowski's bound does work for a few cubic fields and quartic fields: for cubic fields it works as long as the discriminant of the field (in absolute value) is less than 31.39, and there are two such fields: ${\mathbf Q}(\alpha)$ and ${\mathbf Q}(\beta)$ where $\alpha^3 - \alpha - 1 = 0$ (discriminant -23) and $\beta^3 + \beta + 1 = 0$ (discriminant -31). The next smallest absolute value of a discriminant of a cubic field is 44, which is above the bound. For quartic fields we need the discriminant to be less than 158.32, and I know of three fields which work: ${\mathbf Q}(\gamma)$ where
$\gamma^4 + 2\gamma^3 + 3\gamma + 1 = 0$ (discriminant 117), ${\mathbf Q}(\zeta\_5)$ has discriminant 125, and ${\mathbf Q}(\zeta\_{12})$ has discriminant 144.
| 50 | https://mathoverflow.net/users/3272 | 26504 | 17,350 |
https://mathoverflow.net/questions/26505 | 9 | It's easy enough to build Turing Machines that don't halt. But how complex can we make these? For example, suppose a machine has access to its state transition table and can write to it like a C program could point to its own code page in RAM and poke around. The motivation for the question should clear up the particulars:
Imagine that we've build an intelligent (but deterministic) autonomous robot that can completely self-repair from the environment. Imagine that it's a space probe. We don't want it to shut itself off. Because it can change itself physically, it can also change its own programming. We have no control over that once we launch the thing. It's within the realm of possibility it will go through a series of changes that result in it halting and becoming space junk.
Is there any way to understand the topology of a self-modification "trajectory" so that we could minimize the risk of halting? For example, maybe there's some kind of "attractor" where halting is rare.
Or do we just have to assume that Chaitin's Omega constant applies, and there's an unknown constant probability that the thing will halt?
---
Update: Thanks for the comments--they sent me in new directions. Here is some additional background.
* Microsoft has an [active research project](http://research.microsoft.com/en-us/um/cambridge/projects/terminator/) along these lines.
>
> Turing proved that, in general, proving program termination is undecidable.
> However, this result does not preclude the existence of
> future program-termination proof tools that work 99.9 percent of
> the time on programs written by humans. This is the sort of tool
> that were aiming to make. --Byron Cook, the project leader
>
>
>
* Usually we want programs to halt and give us some output. But for the example I gave, we want it to run forever. Can we build an AI that won't spontaneously turn itself off with high probability, like Shannon's "[Ultimate Machine](http://www.youtube.com/watch?v=cZ34RDn34Ws)"? Supposing that a civilization is effectively computable (a big if, but somewhere to start), is there any way to guard against self-halting? Peter Suber studied this idea, limited to legislative systems, and created the game [Nomic](http://www.nomic.net/~nomicwiki/index.php/GameOfNomic). Paul Krugman gives an [example](http://www.nytimes.com/2010/02/08/opinion/08krugman.html?em) of a government that actually did self-halt. My own thoughts about this are in [this paper](http://arxiv.org/abs/0812.0644), where I assumed Chaitin's Omega would "tax" survival probability of any computable system. This is not very satisfying, however. It implies that we can't do any better than randomly selecting an algorithm.
| https://mathoverflow.net/users/6419 | What are the limits of non-halting? | Your question is about many things, but let me give an answer focused on just one interesting issue, the question of determining how long a program will run.
The [busy beaver](http://en.wikipedia.org/wiki/Busy_beaver)
function exactly measures how long programs of a given size
run before halting (among the ones that do halt). There are
versions of the busy beaver function for any notion of
computation, but let us consider the case of C programs,
since you mentioned them. Note that for any natural number
$n$, there are an enormous number of C programs of size
$n$, measured in kilobytes, say. Nevertheless, this
enormous number is finite. Among all programs of size at
most $n$, some halt and some do not. Define $b(n)$ to be
the running time in clock cycles of the longest-running but
halting C program of size at most $n$.
The interesting thing is that the busy beaver function is
not computable! If we had a way of computing $b$, then we
would be able to solve the halting problem, since given any
C program, we look at its size $n$, compute $b(n)$ and run
the program for that many steps; it it hasn't halted by
then, we know it will never halt. Another way to say this
is that if we have an oracle black-box that allows us
somehow to compute the function $b$, then we would be able
to answer any halting problem query. Since it is impossible
to solve the halting problem, it follows that we cannot
compute the busy beaver function.
**Edit.** In your update, you mention the problem of solving the halting problem 99.99% of the time. The general problem of solving almost all instances of a problem, as opposed to all instances of a problem, gives rise to the subject known as [generic case complexity](http://en.wikipedia.org/wiki/Generic-case_complexity). In particular, the black-hole phenomenon occurs when the difficulty of an unsolvable or infeasible problem is concentrated in a very tiny region, outside of which it is easy. It is not good, for example, to base an encryption scheme on a problem whose difficulty has high worst-case complexity, but whose average-case complexitty is low, for if the robbers can rob the bank 10% of the time, it is good enough for them.
In fact, Alexei Miasnikov and I proved that the halting problem itself admits a black hole---for some of the standard computation models, there is a method to solve the halting problem with probability $1$, using the natural asymptotic density measure on the space of programs. I explain further details in [this MO answer](https://mathoverflow.net/questions/10358/solving-np-problems-in-usually-polynomial-time/10379#10379).
| 8 | https://mathoverflow.net/users/1946 | 26509 | 17,353 |
https://mathoverflow.net/questions/26498 | 5 | While reading a paper, I came across the following peculiar condition:
Let $1 \rightarrow H \rightarrow G \rightarrow G/H \rightarrow 1$ be a short exact sequence, and let $H$ be abelian. We require that any automorphism, $\sigma$, of $G$ that preserves $H$ pointwise and such that $\sigma(g)H=gH$ (preserves cosets pointwise) is trivial.
This doesn't give me a good clue as to what we're talking about. For example: is it true for any semi-direct product of two groups of coprime order (the first one being abelian, as required)? How about for a semi-direct product of cyclic groups of coprime order? Prime coprime orders? What is this condition, and what are some canonical examples of it?
I apologize if this is simple, but my finite group theory is pretty shoddy.
| https://mathoverflow.net/users/5309 | A condition on finite groups |
>
> Short answer: a typical example is G=SL(2,5), H = Z(G) = Z/2Z. If G/H and H are coprime and satisfy the condition, the G = G/H × H is quite dull.
>
>
>
I'll assume you find this interesting, and want to read about it:
Let G be a group (finite is good), H be an abelian normal subgroup of G, and Q be the quotient group.
If σ is an automorphism of a group G such that σ(H) = H, then σ induces automorphisms on H and G/H=Q.
If σ(h)=h for all h in H, then σ(H) = H, so we are interested in those σ that are "invisible" as automorphisms both of H and of Q.
The paper's condition is that no automorphism is invisible, which should seem a reasonable crutch if you want to talk about automorphisms of G in terms of those of H and Q (here it helps if H is also characteristic).
What do invisible automorphisms look like?
Well every element of G can be written as a product q\*h for some q in Q and h in H. (q1\*h1)\*(q2\*h2) = (q1\*q2)\*(h1^q2 \* h2 \* ζ(q1,q2)) where ζ:Q×Q→H is a (set-theoretic) function called a 2-cocycle. If you are only interested in semidirect products, then ζ(q1,q2) = 1H can be ignored.
What does σ do to q\*h? Well it takes products to products, and h to h, but it only takes q to another element of the same coset, q\*δ(q) where δ:Q→H is another set-theoretic function. Hence σ(q\*h) = q\*h\*δ(q).
A good example to keep in mind here are the extra-special groups, like the dihedral group of order 8. They have lots of invisible automorphisms. For instance (x,y)→(x,yz) where x,y are the main generators and z=[x,y] generates the center.
Now clearly not all δ can work, since surely δ(q1\*q2) is related somehow to δ(q1) and δ(q2). Indeed σ(q1\*q2) = (q1\*q2)\*δ(q1\*q2), but it is also equal to σ(q1)\*σ(q2) = (q1\*δ(q1))\*(q2\*δ(q2)) = (q1\*q2)\*( δ(q1)^q2 \* δ(q1) \* ζ(q1,q2) ).
Ignoring ζ for a moment, one gets the equation δ(q1\*q2) = δ(q1)^q2 \* δ(q2), which expresses the fact that δ:Q→H is a **derivation** of the Q-module H. It is not too hard to see that the implications are reversible, and derivations help to define automorphisms fixing H and Q.
Not ignoring ζ does not change things very much, as instead of the subgroup of derivations inside the abelian group of all functions from Q to H, you just take a coset of this subgroup determined by ζ.
At any rate, so in your semi-direct products the condition is that there are no non-identity derivations from Q to H; the only derivation should have δ(q)=1H for all q in Q.
Now of course derivations can exist even when Q and H are coprime: Take G to be the non-abelian group of order 6, H to be its subgroup of order 3. Then δ:Q→H takes the non-identity element of Q to any one of the three elements of H, giving three derivations δ. Checking the automorphism group of G, it is easy to see that every automorphism must take H to H, and must act as the identity on Q, since Aut(Q) = 1. Of the six automorphisms, three act as inversion on H, so are not inivisible, but three must be invisible, one for each δ.
In particular, being cyclic of prime and coprime order is not sufficient. Your coprime condition does, however, severely limit the variety of invisible homomorphisms that are available: they must all be conjugations by elements of H.
An automorphism of G induced by conjugation by an element of H must act trivially on H, since H is abelian. It must act trivially on Q, since hH = 1Q in Q. Hence every automorphism induced by an element of H is invisible.
If G is to have no invisible automorphisms, then H must be central, since q^h = h^-1 \* q \* h = q\* (h^q)^-1 \* h is only the identity on G if h^q = h for all q.
In other words, the paper's condition implies H is central, so you are looking for a group Q with a trivial module H whose first cohomology vanishes, but whose second does not. I think this is reasonably rare. Probably a very standard example is a perfect group Q that is not superperfect, and H to be its Schur multiplier.
For instance, take G=SL(2,5), Q = Alt(5), H = Z/2Z.
| 10 | https://mathoverflow.net/users/3710 | 26510 | 17,354 |
https://mathoverflow.net/questions/26518 | 5 | There are more general definitions, but for my purposes a **Lie algebroid** on a smooth manifold $X$ is a vector bundle $A \to X$, a map $\rho: A \to {\rm T}X$ of vector bundles over $X$, and a bracket $[,]$ making $\Gamma(A)$ into an $\mathbb R$-Lie algebra, such that $\rho$ induces is a map $\Gamma(A) \to \Gamma({\rm T}X)$ of Lie algebras and such that the Leibniz rule $[a,fb] = f[a,b] + (\rho(a)f)b$ is satisfied for $f\in \mathcal C^\infty$ and $a,b\in \Gamma(A)$.
So, suppose I happen to have a Lie algebroid $A\to X$ lying around, and also a smooth map $\phi: Y\to X$. (In my case, $Y \to X$ happens to be a vector bundle, so you can assume some fairly strong properties of the map.) Then I can certainly pull back the vector bundle $A\to X$ to $\phi^\*A \to Y$. Is there a natural Lie algebroid structure I can put on this pullback? The answer is probably "how natural do you want it?": if $Y = \{{\rm pt}\}$ and $\phi({\rm pt}) = y\in X$, then the anchor map for $\phi^\*A \to \{{\rm pt}\}$ must be trivial, but the only part of the fiber $A\_y$ with a Lie algebra structure is $\ker \{A\_y \overset\rho\to {\rm T}\_y X\}$.
So the real question is:
>
> Along what type of maps do Lie algebroids pull back? In particular, can I always pull back a Lie algebroid along a submersion?
>
>
>
| https://mathoverflow.net/users/78 | Do Lie algebroids pull back (along submersions)? | As your example suggests, the vector bundle pullback is perhaps not the right thing to consider when wanting defining the notion of a pullback Lie algebroid. You have to go one step further and take the pullback of $\phi^\*\rho: \phi^\* A \to \phi^\*TX$ along $d\phi: TY \to \phi^\*TX$. The resulting bundle $\phi^{\*\*}A$ (which exists when $d\phi - \phi^\*\rho: TY \oplus \phi^\*A \to \phi^\*TX$ has constant rank, so in particular when $\phi$ is a submersion) then has a Lie algebroid structure with gives the right universal property you would want, if that's what you want. In the case where $Y$ is a point, this gives exactly your example.
This is all detailed in the paper "Algebraic Constructions in the Category of Lie Algebroids" by Higgins and Mackenzie. This may even have something to say about just trying to do something with $\phi^\*A$ itself, but I haven't checked.
| 6 | https://mathoverflow.net/users/2552 | 26521 | 17,361 |
https://mathoverflow.net/questions/26515 | 4 | An orthogonal projection is an Hermitian matrix $P$ such that $P^2=P$.
Denote $U^\*$ the conjugate transpose of a matrix $U$.
It can be easily shown that for two projections $P\_1$ and $P\_2$, there exists a unitary
$U$ such that both $UP\_1 U^\*$ and $UP\_2U^\*$ are block diagonal with blocks of size one
or two (And both resulting matrices have the same block structure).
My question is whether this block decomposition of projections can be generalised,
for more than two projections: Given orthogonal projections $P\_1, P\_2, ..., P\_k$,
Is there a unitary $U$ such that for each $i$, $UP\_i U^\*$ is block diagonal with
blocks of size at most $k$? (The resulting matrices must have the same block structure)
Two weaker questions are:
Is there a bound on the size of the blocks in function of $k$ only, i.e.
in function of the number of projectors independently of their dimensions?
If a block decomposition is not possible, then what about decomposing the projectors
into $k$-diagonal matrices? (All entries of the matrix are zero except (possible) for the diagonal and
the $k$-upper and $k$-lower diagonals)
I would deeply appreciate any help or reference on how to handle these problems.
Best regards,
Mateus
| https://mathoverflow.net/users/6442 | Simultaneous Block decomposition of a set of orthogonal projections | If $P$ is a projection then $I-2P$ is a reflection. Two reflections generate the dihedral
group and all irreducible representations of the dihedral group have dimension at most two.
This explains your observation about two projections.
But the alternating group $Alt(n)$ can be generated by three involutions when $n\ge9$ (a result of Nuzhin). Take the usual permutation representation of $Alt(n)$ with degree $n$. If $R$ is the permutation matrix representing an involution, then $R$ is symmetric and $(I-R)^2 =2(I-R)$. So $\frac12(I-R)$ is a projection.
This permutation representation has one invariant subspace of degree one (the span of the constant vectors) and the orthogonal complement to this is irreducible with dimension $n-1$.
This shows that the size of your blocks is not bounded by the number of projections.
| 8 | https://mathoverflow.net/users/1266 | 26523 | 17,363 |
https://mathoverflow.net/questions/26528 | 5 | In many papers about dynamical system, I found the word " ideal boundary". T don't know what is the definition of ideal boundary.
| https://mathoverflow.net/users/5093 | What is the definition of ideal boundary? | For a [Hadamard space](http://en.wikipedia.org/wiki/Hadamard_space) $X$ there are two kinds of ideal boundaries, the set $Bd(X)$ of horofunctions up to additive constants, and the set $X(\infty)$ of equivalence classes of rays. These two are homeomorphic by the correspondence:
>
> $$\gamma \text{ (a ray)}\to \text{the Busemann function of } \gamma$$
> $$h \text{ (a horofunction)}\to \text{the gradient of } h$$
>
>
>
[Here](http://www.math.psu.edu/petrunin/papers/akp-papers/ballmann-lect.pdf) you can find lecture notes by Ballman, with chapter two treating the boundary at infinity via Buseman functions and via rays. You will also find definitions in [this](http://www.mathnet.or.kr/mathnet/kms_tex/24289.pdf) paper and [this](http://arxiv.org/abs/math/0510105) paper.
| 6 | https://mathoverflow.net/users/2384 | 26532 | 17,369 |
https://mathoverflow.net/questions/26540 | 7 | I read in "Letters to a young mathematician" that 4900 is the only square integer that is the sum of consecutive squares (I believe he meant by that "starting from 1", but maybe that's not even necessary). I did a quick run through with a python script and of course this seems totally devoid of a computational pattern. Why is 4900 (and 1 of course) the only number such that this works?
I did find out that the sum of squares is the following...
$\sum^{n}\_{i=1} i^2 = \frac{n(n+1)(2n+1)}{6}$
| https://mathoverflow.net/users/429 | 4900, a particularly square number | This is a classical Diophantine equation (Mordell, *Diophantine Equations*, p. 258). Apart from n = 0, 1, -1, there is only the solution n = 24. Proofs by G. N. Watson (1919), W. Ljunggren (1952).
There is some history in <http://www.math.ubc.ca/~bennett/paper21.pdf> .
| 14 | https://mathoverflow.net/users/6153 | 26542 | 17,374 |
https://mathoverflow.net/questions/26558 | 4 | **Background**
I have searched a bit for the definition/constructions on how to "semi-localize" a scheme, but have been unsuccessful in finding a good reference; I apologize in advance if this topic has been covered in detail elsewhere (e.g. in a book or article) and would be happy for a reference!
This question arose from a problem I had been working on in finding an étale morphism into affine space.
Much of the terminology here will be from EGA I.
(Aside: The constructions below take place in the Zariski site but I think some of them go through in the étale/Nisnevich site)
**Definitions**
A *local scheme* is the spectrum of a local ring and a *semi-local scheme* the spectrum of a semilocal ring. In the constructions below a ring $O\_{X,C}$ is given, so then the candidate semi-local scheme is $Spec \; O\_{X,C}$.
NB: for some reason I was having trouble with "\varinjlim" here, so I'm using "lim" below to mean direct limit i.e. colimit.
**Question**
Given a scheme $X$ we can *localize* $X$ at a point $x\in X$ by taking $$O\_{X,x} := \lim\_{U\ni x} O\_X(U). $$
Suppose now that we are given a finite set of closed points $x\_1,\ldots, x\_n \in X$.
Let $C :=$ {$x\_1,\ldots, x\_n $}. How can we 'localize' $X$ around $C$?
There are at least three ways I know how to do this procedure and would be happy to hear about other methods as well as comments (especially geometric ones) regarding the following constructions:
$1.$ Define $$O\_{X,C} : = \lim\_{U\supset C} O\_X(U).$$
This construction is similar to the localization construction above in that we take opens $U$ of $X$ containing $C$ and then take the direct limit; the case $n=1, C = \{x\_1\}$ is then a special case. NB: we can 'see' this direct limit in the sense that for each $x\_i$ we find an open $U\_i\ni x\_i$, then taking the (finite!) union of the $U\_i$ we obtain an open $U$ containing $C$. Just as in the local case above, this direct limit is filtered by inclusion.
$2.$ Further assume now that $X$ is locally noetherian and regular. Let $A\_i: = O\_{X,x\_i}$ and then define $$O\_{X,C}:= \prod\_i A\_i .$$
Using the hypothesis that $X$ is regular, we can argue that the maximal ideals here correpsond to the $x\_i$: the maximal ideals in $\prod\_i A\_i$ are generated by elements of the form $(1,1,\ldots, b\_{ij},1,\ldots, 1)$ where the $b\_{ij}$ generate $x\_i$ (here is where we are using the two added hypothesis), i.e. that $(b\_{ij})\_{1\leq j\leq n\_i} = m\_i$ where $m\_i$ is the max ideal corresponding to $x\_i$ and $n\_i = dim O\_{X,x\_i}$.
This construction is more ad-hoc (I think) vs. 1. Moreover, the geometry here is slightly more explicit in that this $Spec \; O\_{X,C}$ is a finite disjoint union of local schemes, whereas in case 1, the topology is less disjoint when looking at neighborhoods of the $x\_i$.
$3.$ With $X$ any scheme (no additional hypothesis as in 2), let $F\_i : = O\_{X,x\_i}/m\_i$ where $m\_i$ is the maximal ideal corresponding to the closed point $x\_i$. Define: $$O\_{X,C}: = \prod\_i F\_i .$$ This construction is the most disjoint of the three in that the spectrum is now we have a finite coproduct of ''points''.
**Closing remarks**
Presently, for me the most useful of the three is 1 and I would appreciate feedback on where the process of semi-localization has been defined. A professor that I admire very much once said (during a lecture) "from now on and for the rest of your life, every time you see something in commutative algebra, try to relate it to geometry, and vice versa" (I'm paraphrasing).
| https://mathoverflow.net/users/4235 | On semi-local schemes | Your construction number (1) seems completely natural and correct to me, for the following reason:
If $A$ is a (Noetherian, commutative, unital) ring and $I$ is an ideal of $A$, then the localaization of $A$ away from $I$ is $A\_I=S^{-1}A$, where
$$S=\{a\in A: a\mod{I}\mbox{ is not a zero divisor is }A/I\}$$
If $I$ is prime, then this is what you expect.
Next, suppose that $I$ is radical and that $\mathfrak{p}\_i$ are the finitely many minimal prime ideals containing it (so $I$ is the intersection of the $\mathfrak{p}\_i$). Then the zero divisors in the reduced ring $A/I$ are exactly the union of the minimal prime ideals of the ring, implying that $S=\bigcup\_i\mathfrak{p}\_i$. From this it follows that the localization $A\_I$ is a semi-local ring whose maximal ideals correspond to the $\mathfrak{p}\_j$. Moreover, $A\_I$ will exactly by the direct limit you describe in (1), if you take $X=\mbox{Spec}A$ and $C=\{\mathfrak{p}\_i\}$.
| 2 | https://mathoverflow.net/users/5830 | 26562 | 17,385 |
https://mathoverflow.net/questions/26557 | 25 | I read that the primitive element theorem for fields was fundamental in expositions of Galois theory before Emil Artin reformulated the subject. What are the differences between pre and post-Artin Galois theory?
| https://mathoverflow.net/users/4692 | What was Galois theory like before Emil Artin? | [The development of Galois theory from Lagrange to Artin](https://doi.org/10.1007/BF00327219) by B. Melvin Kiernan, is a history of pre-Artin Galois theory.
| 20 | https://mathoverflow.net/users/2384 | 26563 | 17,386 |
https://mathoverflow.net/questions/26551 | 18 | Felix Klein, when discussing how the popularity of areas in mathematics rises and falls, mentions that in his youth Abelian functions were at the summit of mathematics, and that later on their popularity plummeted. I could hardly find anything on the net on Abelian functions and Wikipedia thinks that they are barely worth mentioning. What did the 19th century theory of these functions consist of, and why was it so important?
| https://mathoverflow.net/users/4692 | Why were Abelian functions so important in the 19th century? | For a really detailed answer to your question, see *The Legacy of Niels Henrik
Abel*, edited by O.A. Laudal and R. Piene (Springer 2004). In particular, there is a long introductory article by Christian Houzel, most of which can be viewed [here](http://books.google.com.au/books?id=HiXwhBm42hcC&lpg=PA21&ots=ibJuErm5sc&dq=%22christian%20houzel%22%20abel&pg=PA21#v=onepage&q=%22christian%20houzel%22%20abel&f=false).
**Addendum.** The complete article "The Work of Niels Henrik Abel" by
Christian Houzel may be found [here](http://www.abelprisen.no/nedlastning/litteratur/houzel_the_work.pdf).
| 13 | https://mathoverflow.net/users/1587 | 26565 | 17,387 |
https://mathoverflow.net/questions/26538 | 2 | I have some questions about the following exercise in Hartshorne (III.4.7):
Let $f \in k[x\_0,x\_1,x\_2]$ be a homogeneous polynomial of degree $d \geq 1$ and $f \neq 0$ and let $X$ be the closed subscheme of $\mathbb{P}^2\_k$ defined by $f$. Then $\dim H^0(X,\mathcal{O}\_X) = 1, \dim H^1(X,\mathcal{O}\_X) = (d-1)(d-2)/2$. This is done using Cech cohomology.
1 - Hartshorne makes the assumption $f(1,0,0) \neq 0$. Is this necessary?
This implies that $f$ is monic in $x\_0$ and yields a very nice description of the Cech complex (if necessary, I'll add this), which makes the computation possible. But what about the general case?
It's not hard to see that $f$ is mapped by a graded isomorphism of $k[x\_0,x\_1,x\_2]$ to a polynomial, which does not vanish in $(1,0,0)$, if and only if $f$ does not vanish on $k^3$. Thus if $k$ is infinite, you're done. But what happens when $k$ is finite? For example
$f = xy \prod\_{\alpha \in k} (x - \alpha y)$
is a nontrivial homogeneous polynomial of degree $|k|+2$ and vanishes on $k^2$ (and thus on $k^3$).
2 - Is the finite case important for some applications (for example in arithmetic geometry)?
3 - Is it surprising that the cohomology only depends on $d$?
| https://mathoverflow.net/users/2841 | Hartshorne Exercise III.4.7 (cohomology of closed subschemes in $\mathbb{P}^2$) | You can compute the cohomology via the Koszul resolution. If $i:X \to {\mathbb P}^2\_k$ is the embedding then the triple $0 \to O\_{{\mathbb P}^{2}}(-d) \stackrel{f}\to O\_{{\mathbb P}^2} \to i\_\*O\_X \to 0$ is exact. So, you can compute $H^t(X,O\_X) = H^t({\mathbb P}^2\_k,i\_\*O\_X)$ using the long exact sequence associated with this triple.
| 4 | https://mathoverflow.net/users/4428 | 26567 | 17,388 |
https://mathoverflow.net/questions/26492 | 5 | Is there any software that will compute cohomology of vector bundles (or just line bundles) on flag manifolds?
The only one I know of is Macaulay2, via the Schubert2 package, but it works with what it calls "abstract varieties", which are really just the intersection rings over $\mathbb{Q}$, so it's explicitly limited to characteristic zero. I'm interested in (among other things) bad behavior at small prime characteristics.
| https://mathoverflow.net/users/460 | software for computations on flag varieties in arbitrary characteristic | To answer the original question explicitly, there seems to be no relevant software in prime characteristic. Nor is there any on the horizon, unless the theory developed so far becomes much more definitive. In the setting of flag varieties, general principles show that Euler characters in characteristic $p$ are the same as in characteristic 0 for line bundles (etc.) because the objects involved have compatible $\mathbb{Z}$-forms. Kempf even showed that for a dominant line bundle, sheaf cohomology vanishes except for degree 0; so the formal character and dimension of the global section module are given by Weyl's formula. Similarly for the Serre dual, but there are some systematic patterns (based on alcoves for an affine Weyl group relative to $p$) in which some other line bundles have nonvanishing cohomology in multiple degrees. This appears only for weights "close to" Weyl chamber walls and is conjecturally due to failure of "cancellation" in Jantzen-Andersen filtrations near walls.
A moral of the existing work on line bundles and some other vector bundles is that module structure seems needed to understand the vanishing behavior of cohomology. For small $p$ one lacks analogues of Lusztig's conjectures on characters of the simple modules, which may be an added obstacle. Even small calculations are very difficult, for example for small primes in rank 2. Some have occurred in the literature, but there is no mechanical method to generate them.
Most work was done in the 1980s, following a thesis by Mumford's student W.L. Griffith Jr. showing a few counterexamples to the Borel-Weil-Bott picture for some flag varieties and small $p$, as well as Henning Andersen's MIT thesis around the same time. Full references before 1990 appear in my short conference survey:
MR1131312 (92k:20084) 20G10 (14M17 20G05)
Humphreys, J. E. (1-MA),
Cohomology of line bundles on flag varieties in prime characteristic.
Proceedings of the Hyderabad Conference on Algebraic Groups (Hyderabad, 1989), 193–204,
Manoj Prakashan, Madras, 1991.
A couple of recent papers by Steve Donkin focus mostly on SL(3):
MR1958906 (2004f:20083) 20G05 (14M15 20G10)
Donkin, Stephen (4-LNDQM)
Anote on the characters of the cohomology of induced vector bundles on G/B in
characteristic p.
J. Algebra 258 (2002), no. 1, 255–274.
MR2275364 (2008a:20077) 20G10 (14L30 14M15)
Donkin, Stephen (4-YORK),
The cohomology of line bundles on the three-dimensional flag variety.
J. Algebra 307 (2007), no. 2, 570–613.
| 5 | https://mathoverflow.net/users/4231 | 26572 | 17,391 |
https://mathoverflow.net/questions/26555 | 12 | Hello,
Let $X$ and $Y$ be two smooth (probably projective) algebraic varieties defined over $\mathbf{C}$.
What is known in general about the (topological) space of holomorphic maps $\mathrm{Hol}(X(\mathbf{C}),Y(\mathbf{C}))$ from $X(\mathbf{C})$ to $Y(\mathbf{C})$?
In particular, I would be interested to know under what hypothesis on $X$ and $Y$ the space
$\mathrm{Hol}(X(\mathbf{C}),Y(\mathbf{C}))$ is a smooth manifold (and to have a formula to compute the dimension of its connected components).
One example I have in mind is : if $X=Y=\mathbf{P}^1$ (the projective line) then $\mathrm{Hol}(\mathbf{P}^1(\mathbf{C}),\mathbf{P}^1(\mathbf{C}))$, in which case one gets the space of complex rational functions (which has connected components indexed by the positive integers (the degree), but each is a smooth algebraic variety).
I think this might be an (easy?) application of the theory of Grothendieck $\mathrm{Hom}$-schemes, but I don't feel very at ease with this.
I would also be interested to know when $\mathrm{Hom}(X,Y)$ is a smooth algebraic variety.
Many thanks,
K.
| https://mathoverflow.net/users/5239 | Are spaces of holomorphic maps manifolds? | Let $X,Y$ be analytic spaces with $X$ compact reduced and $Y$ arbitrary (i.e. maybe with nilpotents in its structure sheaf). Then Douady showed in his [thesis](http://archive.numdam.org/item/AIF_1966__16_1_1_0/)
that the set of holomorphic maps $Hol(X,Y)$ can be endowed with the structure of an analytic space whose underlying topology is the compact-open topology. If $X,Y$ are compact manifolds, the Zariski tangent space at $f:X\to Y$ is a subspace of the finite-dimensional vector space of sections of the tangent bundle to $Y$ pulled-back to $X$ viz. $T\_f(Hol(X,Y))\subset \Gamma(X,f^\star TY)$.
I don't know any good general criterion for $Hol(X,Y)$ to be smooth at $f$.
**Edit** Here is a class of examples which might interest you, where smoothness occurs. Let $X$ be a Riemann surface of genus $g$. The space of ramified covers $f:X\to \mathbb P^1$ of degree $d$ is non-empty and smooth of dimension $2d+1-g$ as soon as $d\geq g+1$. But there are explicit cases for smaller $d$ where the corresponding space is singular. You can read about these results in [this article](https://projecteuclid.org/journals/proceedings-of-the-japan-academy-series-a-mathematical-sciences/volume-54/issue-7/Examples-of-obstructed-holomorphic-maps/10.3792/pjaa.54.189.full) by Akaohori and Namba.
| 11 | https://mathoverflow.net/users/450 | 26580 | 17,398 |
https://mathoverflow.net/questions/26537 | 7 | It is fundamental to topology that $\mathbb{R}$ is a connected topological space. However, all the topology books that I have ever looked in give the same proof. (the proof I am thinking of can be seen in Munkres's topology or Lee's Introduction to topological manifolds)
This seems strange to me, because for other fundamental results such as the Compactness of $[0,1]$, I can think of several proofs.
Does anyone know any different proofs of the connectedness of $\mathbb{R}$?
| https://mathoverflow.net/users/4002 | Connectedness and the real line | If you've already developed basic facts about compactness you can prove it this way:
Let $[0,1] = A \cup B$ with $A$ and $B$ closed and disjoint. Then since $A \times B$ is compact and the distance function is continuous, there is a pair $(a, b) \in A \times B$ at minimum distance. If that distance is zero, $A$ and $B$ intersect. If not, you get a contradiction by taking any point in the interval from $a$ to $b$: it can't be in either $A$ or $B$ because its distance from $b$ or $a$ is smaller than the minimum.
That shows a compact interval in $\mathbb{R}$ is connected. If $\mathbb{R} = A \cup B$ with $A$ and $B$ closed and disjoint, then for any closed interval $I$ with one endpoint in $A$ and one in $B$, $I = (A \cap I) \cup (B \cap I)$ is disconnection of $I$. Alternatively, you could write $\mathbb{R}$; as a union of closed intervals with a common point.
| 42 | https://mathoverflow.net/users/644 | 26594 | 17,406 |
https://mathoverflow.net/questions/26586 | 10 | This question is on bounding the degree of the Todd class on a complex threefold.
Let $X$ be a smooth compact connected complex surface. Let $c\_i=c\_i(TX)$ be its $i$-th Chern class. Recall the following two facts. These allow one to bound the degree of the Todd class on a surface in terms of $c\_2$.
**1**. If $X$ is not of general type, we have that $c\_1^2$ is bounded absolutely from above by 9. See Table 10 of Chapter VI of *Compact complex surfaces* by Barth, Hulek, Peters and van de Ven.
**2**. If $X$ is of general type, then the Bogomolov-Miyaoka-Yau inequality states that $$c\_1^2 \leq 3c\_2.$$
Now, I am interested in similar results for 3-dimensional smooth projective connected varieties over $\mathbb{C}$.
In this case, the degree of the Todd class of $X$ is the degree of $$\frac{c\_1 c\_2}{24}.$$
>
> **Question**. For 3-dimensional smooth projective connected varieties over $\mathbb{C}$, do there exist any absolute upper bounds on $c\_1c\_2$ (or any bounds for that matter) which are polynomial in $c\_3$?
>
>
>
| https://mathoverflow.net/users/4333 | Can one bound the Todd class of a 3-dimensional variety polynomially in c_3 | The case of complex manifolds of higher dimension is very different from the case of complex surfaces. So the answer to the question about complex $3$ folds is no, there exists a real 6-dimensional simply connected manfiold with integrable complex structures $J\_m$ for all $m\in \mathbb Z^+$ such that $c\_1c\_2=48m$. This is a theorem **A** from the acticle of LeBrun. Though, the manifolds that he constructs are not algebraic
*Topology versus Chern Numbers for Complex 3-Folds*
<http://arxiv.org/PS_cache/math/pdf/9801/9801133v1.pdf>
The question for algebraic manifolds was studied by Kotschick, you may be interested this the following two articles:
*TOPOLOGICALLY INVARIANT CHERN NUMBERS OF PROJECTIVE VARIETIES*
<http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.1587v1.pdf>
*CHERN NUMBERS AND DIFFEOMORPHISM TYPES OF PROJECTIVE VARIETIES*
<http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.2857v2.pdf>
Finally, it should be added that a systematic attempt to construct various complex 3-fold is given in a very nice article of Okonek and Van de Ven "CUBIC FORMS AND COMPLEX 3-FOLDS" of Okonek, Ch. / Van de Ven, A, L'Enseignement Mathématique Volume / Année: 41 / 1995. The link is given in the comments
| 18 | https://mathoverflow.net/users/943 | 26598 | 17,410 |
https://mathoverflow.net/questions/23601 | 18 | My question is based on the following vague belief, shared by many people: It should be possible to use von Neumann algebras in order to define the cohomology theory TMF (topological modular forms) in the same way one uses Hilbert spaces in order to define topological K-theory. More precisely, one expects hyperfinite type $\mathit{III}$ factors to be the analogs of (separable) infinite dimensional Hilbert spaces.
Now, here is a fundamental difference between Hilbert spaces and type $\mathit{III}$ factors: The category of Hilbert spaces has two monoidal structures: direct sum $\oplus$, and tensor product $\otimes$, and both of them preserve the property of being an infinite dimensional Hilbert space.
In von Neumann algebras, the tensor product of two hyperfinite type $\mathit{III}$ factors is again a hyperfinite type $\mathit{III}$ factor, but their direct sum isn't (it's not a factor).
Hence my question: are there other monoidal structures on the category of von Neumann algebras that I might not be aware of? More broadly phrased, how many ways are there of building a new von Neumann algebra from two given ones, other than tensoring them together?
Ideally, I would need something that distributes over tensor product, and that preserves the property of being a hyperfinite type $\mathit{III}$ factor... but that might be too much to ask for.
| https://mathoverflow.net/users/5690 | Monoidal structures on von Neumann algebras | The category of von Neumann algebras W\* admits a variety of monoidal structures of three distinct flavors.
(1) W\* is complete and therefore you have a monoidal structure given by the categorical product.
(2a) W\* is cocomplete and therefore you also have a monoidal structure given by the categorical coproduct.
(2b) I suspect that there is also a “spatial coproduct”, just as there is a categorical
tensor product and a spatial tensor product (see below).
The spatial coproduct should correspond to a certain central projection in the categorical coproduct.
Perhaps the spatial coproduct is some sort of coordinate-free version
of the free product mentioned in Dmitri Nikshych's answer.
(3a) For any two von Neumann algebras M and N
consider the functor F from W\* to Set that sends a von Neumann algebra L
to the set of all pairs of morphisms M→L and N→L with commuting images.
The functor F preserves limits and satisfies the solution set condition, therefore it is representable.
The representing object is the categorical tensor product of M and N.
(3b) There is also the classical spatial tensor product.
I don't know any good universal property that characterizes it except
that there is a canonical map from (3a) to (3b) and its kernel corresponds
to some central projection in (3a). Perhaps there is a nice description of this central projection.
Since your monoidal structure is of the third flavor and you don't want a monoidal
structure of the first flavor, I suggest that you try a monoidal structures of the second flavor.
I suspect that the spatial coproduct of two factors is actually a factor.
You are lucky to work with factors, because in the commutative case 2=3, in particular 2a=3a and 2b=3b.
| 6 | https://mathoverflow.net/users/402 | 26600 | 17,411 |
https://mathoverflow.net/questions/26602 | 2 | I have to say whether or not the following two separation logic statements are valid:
1. $ x \mapsto 3 \* y \mapsto 7 \Longrightarrow x \mapsto 3 \* true $
2. $ true \* x \mapsto 3 \Longrightarrow x \mapsto 3 $
Where $ x \mapsto 3 $ means x (in the stack) points to an abitrary memory location in the heap containing the value 3.
Now as far as I understand the $ true $ in the first statement means that... in fact I'm not sure what it means. I have notes saying:
$ x \mapsto 1 $ == $ h = h1 $
$ y \mapsto 2 $ == $ h = h2 $
$ x \mapsto 1 \* y \mapsto 2 $ == $ h = h1 \* h2 $
$ x \mapsto 1 \* true $ == h1 contained in h
So back to the first statement: is it simply saying that x is pointing to a value 3 *somewhere in the heap* and therefore it is a valid satement because the value 3 is somewhere in the heap and it makes no assertions about y... Does that make any sense?
And the second statement is not valid because... I've never seen the true on the left hand side ??
I've started trying to read this [separation logic overview](http://www.cs.cmu.edu/afs/cs.cmu.edu/project/fox-19/member/jcr/www15818As2007/cs818A3-07.html) as it seems better than our lectures but any other links/pointers would be greatly appreciated.
I'm well aware that I'm rather out of my depth here and could be completely wrong with my assumption/answers above so any feedback would be great.
Thanks in advance
| https://mathoverflow.net/users/5718 | Are these separation logic statements valid? | The trick with separation logic is that the formulas describe resources (heaps) and if a logical implication holds, then both sides of the implication are descriptions of the same heap.
True can correspond to any heap.
$h \* x\mapsto 3$ is almost always a larger heap than $x\mapsto 3$, which consists of just one element. So it is not the case that every heap satisfying $\mathit{True} \* x\mapsto 3$ is also a heap satisfying $x\mapsto 3$.
Thus the first is valid, the second isn't.
It's worth unravelling the semantics of separation logic formula to see this.
| 1 | https://mathoverflow.net/users/2620 | 26603 | 17,412 |
https://mathoverflow.net/questions/26549 | 13 | Edwards, in his book "Divisor theory" says that Kronecker's methods are quite different to Dedekind's and those of today. Is there really much of a difference apart from Kronecker's methods being more constructive?
| https://mathoverflow.net/users/4692 | Is there much difference between Kronecker's and Dedekind's methods in algebraic number theory and commutative algebra? | You can find a nice description of Kronecker's approach in an article by Harley Flanders,
"The Meaning of the Form Calculus in Classical Ideal Theory" (Trans. AMS 95 (1960), 92--100). It is at JSTOR [here](https://www.jstor.org/stable/1993332?seq=1#metadata_info_tab_contents). I found that more to my tastes than Edwards' book.
There is a difference between the two approaches. Kronecker was thinking in very general terms, beyond the "one-dimensional" setting that Dedekind worked in. (Kronecker had a dream -- a second one I suppose -- of unifying number theory and algebraic geometry but the tools to achieve this would take a couple more generations to appear). That accounts in part for Kronecker's multivariable polynomials. He had bigger goals than just unique factorization in rings of integers.
Here is one example of the difference between Kronecker and Dedekind. Suppose ${\mathfrak a}$ is an ideal in the ring of integers of a number field $K$ and I ask you to compute its norm, i.e., the size of ${\cal O}\_K/\mathfrak a$. How would you do it? From Dedekind's point of view, you find a ${\mathbf Z}$-basis of ${\cal O}\_K$ and of ${\mathfrak a}$, write the basis of the ideal in terms of the basis of the ring of integers, and then compute (the absolute value of) the determinant of the matrix expressing the ideal basis in terms of the ring basis. But as you may know, ideals usually are *not* given to us in terms of a ${\mathbf Z}$-basis. More often they are given in terms of just two generators, say ${\mathfrak a} = (\alpha,\beta)$. How can you compute the norm of the ideal in terms of the two generators? In principle it should be possible, since the two generators determine the ideal they generate, so all the data you need is encoded in the numbers $\alpha$ and $\beta$.
There is a Dedekind-style way to write the norm of ${\mathfrak a}$ in terms of the two generators: the norm of an ideal is the gcd of the norms of *all* elements of the ideal.
Watch out: you can't get by using only the gcd of the norms of the two generators.
For example, in the Gaussian integers the ideal $(1+2i,1-2i)$ is the unit ideal $(1)$, so it has norm 1, but the two generators $1+2i$ and $1-2i$ have norm 5, whose gcd is not 1. (Of course the ideal also contains $1+2i - (1-2i) = 4i$, whose norm is 4, and the gcd of that with 5 is one and you're done.) In principle you only need to form the gcd of the norms of a finite number of elements in the ideal, but it's not clear which "finitely many" elements are practically enough. So I think it's fair to say Dedekind's point of view does not easily allow you to find the norm of an ideal in terms of two generators of the ideal, which is how one usually thinks about them concretely.
Now here is how Kronecker would find the norm of the ideal (essentially). Form the polynomial $\alpha + \beta{T}$ in ${\cal O}\_K[T]$. The field extension $K(T)/{\mathbf Q}(T)$ is finite.
Take the field norm of $\alpha + \beta{T}$ down to ${\mathbf Q}(T)$. The result is in ${\mathbf Z}[T]$.
That integral polynomial has finitely many coefficients (which are *not* all norms of elements in $K$, so this isn't some disguised version of the previous paragraph). The gcd of the integral coefficients of ${\rm N}\_{K(T)/{\mathbf Q}(T)}(\alpha + \beta{T})$ is the norm of the ideal. And if the ideal is given to you with more than two generators, just let $f(T)$ be the polynomial with higher degree having those generators as its coefficients, one for each power of $T$ (it doesn't matter what order you use the generators as coefficients) and do the same thing as in the case of two generators: field norm down to ${\mathbf Q}(T)$ and then gcd of the integral coefficients that pop out. I personally was blown away when I saw this method work, since practically no books on algebraic number theory discuss Kronecker's point of view, so this particular result isn't there. (To be honest, you do *not* need Kronecker's multivariable polynomial method to prove this norm formula. Once you know the formula, it can be derived by more orthodox techniques, but of course it leaves out the question of how anyone would have ever discovered this formula in the first place by orthodox methods. Any suggestions?)
In a sense this example is only a "constructive" dichotomy between Kronecker and Dedekind, but I think it still addresses the question that is asked, because each method of solving this problem (Dedekind's ${\mathbf Z}$-bases and Kronecker's polynomials) is constructive but they feel so different from each other.
| 23 | https://mathoverflow.net/users/3272 | 26628 | 17,426 |
https://mathoverflow.net/questions/946 | 5 | One of the basic tools in subfactors is the conditional expectation. If $N\subset M$ is a $II\_1$-subfactor (or an inclusion of finite factors), then there is a unique trace-preserving conditional expectation of $M$ onto $N$. This should be thought of as a (Banach space) projection of norm 1. In fact, it is the restriction of the Jones projection $e\_N$ on $L^2(M)$ to $M$. In the finite index case, we get another conditional expectation (Jones projection...) from the basic construction $M\_1=\langle M, e\_N\rangle$ onto $M$.
In his thesis, Michael Burns showed that if we iterate the basic construction in the infinite index case, we only get half the conditional expectations (we only get the odd Jones projections). The other half of the time, we get a generalization of the conditional expectation called an operator valued weight, originally defined by Haagerup.
Given an inclusion of semifinite von Neumann algebras $(N, tr\_N)\subset (M, tr\_M)$, there is a unique normal, faithful, semi-finite trace-preserving operator valued weight $T\colon M\_+\to \widehat{N\_+}$, where we must take the "extended part" of the positive cone $N\_+$ of $N$.
Edit as per @Dmitri's answer:
Let
$$
n\_T=\{x\in M| T(x^\ast x)\in N\_+\}
$$
and set
$$
m\_T=n\_T^\ast n\_T=span\{x^\ast y| x,y\in n\_T\}.
$$
There is a natural extension of $T$ to $m\_T$. Is there an example of a normal, faithful, semifinite operator valued weight such that
* $N$ is not contained in $T(m\_T)$, and/or
* $1\notin T(M\_+)$?
What about when $M$ and $N$ are factors ($M$ is $II\_\infty$)?
| https://mathoverflow.net/users/351 | Operator Valued Weights | Since $T(m\_T)$ is a non-trivial two sided ideal in $N$, it follows from spectral calculus that $T(m\_T)$ contains a non-zero projection. $T(m\_T)$ then contains all subprojections, hence there exists $x \in m\_T$ such that $p = T(x)$ is a projection with trace Tr$(p) = 1/n$ for some natural number $n$.
We may assume that $pxp = x$, and by considering $(x + x^\*)/2$ we may assume that $x$ is self-adjoint. Since $m\_T$ is spanned by its positive part it follows that we can write $x$ as $x\_1 - x\_2$ where $x\_j \in m\_T$ are positive. Hence $T(x\_1) \geq p$ and so by considering the element $(pT(x\_1)^{-1/2}p)x\_1(pT(x\_1)^{-1/2}p) \in m\_T$ we may assume that $x \geq 0$.
If $N$ is a factor, then since Tr$(p) = 1/n$ we may find a (finite, if $N$ is finite) sequence of partial isometries $v\_k$ such that $v\_k^\*v\_k = p$ for each $k$ and $\Sigma\_k v\_kv\_k^\* = 1$. Then $x\_m = \Sigma\_{k = 1}^m v\_k x v\_k^\*$ is a bounded increasing sequence in $m\_T$ and we have that $T(x\_m)$ increases to $1$. Since the weight is normal we have that $x\_\infty = \Sigma\_k v\_k x v\_k^\* \in m\_T \cap M\_+$ and $T(x\_\infty) = 1$. In particular, since $T(m\_T)$ is an ideal we have $N \subset T(m\_T)$.
If $N$ has infinite dimensional center then as Dmitri explained there are many examples (even bounded ones) for which $1 \not\in T(M\_+)$. I imagine that for finite dimensional center the argument above should still work.
| 6 | https://mathoverflow.net/users/6460 | 26632 | 17,429 |
https://mathoverflow.net/questions/26626 | 4 | Is there any easy proof or/and reference for the following
**Proposition.** If a polynomial $f\in \mathbb{R}[x\_1,\dots,x\_n]$ with zero constant term has isolated local minimum in 0, then $|f|> C (x\_1^2+\dots +x\_n^2)^N$ in some neighborhood of 0, where
$\bullet$ $N$ depends only on degree of $f$ and $n$,
$\bullet$ $C$ depends on uniform bound on $n$, degree of $f$, and bounds for coefficients of $f$.
?
| https://mathoverflow.net/users/4312 | minimum of polynomial | That there is some $C$ and $N$ for which this is satisfied is a case of Lojasiewicz's theorem, which says that if $f(x\_0) = 0$, then there is an open set containing $x\_0$ on which $|f(x)| \geq C|d(x,Z)|^N$ for some $C$ and $N$. Here $d(x,Z)$ denotes the distance from $x$ to the zero set $Z$ of $f(x)$. So in the isolated singularity case it reduces to what you're asking for.
I don't any immediate way to bound the constants in the way you're asking for.
| 2 | https://mathoverflow.net/users/2944 | 26637 | 17,433 |
https://mathoverflow.net/questions/22748 | 8 | Hi,
I am looking at inclusion of discrete groups $H\subset G$ such that $H$ is abelian and $(hgh^{-1},h\in H)$ is infinite if $g\in G-H$. If you have this, $LH\subset LG$ is a maximal abelian subalgebra of a finite von Neumann algebra.
Suppose that $LH\subset LG$ is a Cartan subalgebra, i.e. the group of unitary of $LG$ that normalize the algebra $LH$ generates $LG$. Do we have necessarily that $H$ is a normal subgroup of $G$?
Thanks for your help.
| https://mathoverflow.net/users/2045 | normalizer of algebras and groups | This is true, and in fact more has been shown in the recent preprint <http://arxiv.org/abs/1005.3049> of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:
If $LH \subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] < \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)
Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u\_g$ is orthogonal to $\mathcal N\_{LG}(LH)''$. Since $\mathcal N\_{LG}(LH)''$ is spanned by $\mathcal N\_{LG}(LH)$ it is enought to show that $u\_g$ is orthogonal to this set and so let's fix $v \in \mathcal N\_{LG}(LH)$.
Before we show that $u\_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$.
**Lemma:**
For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A\_0 \subset LH$ such that if $p$ is any minimal projection in $A\_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u\_g^\* p u\_g - \tau(p) \rangle | < \delta \|x \|\_2$ for all $x \in LH$.
**Proof.** This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec\_{LH} L(H \cap gHg^{-1})$ (See Popa's paper <http://www.ams.org/mathscinet-getitem?mr=2231961>).
Let's show this by induction on $n$. For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$. Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec\_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p\_k \in \mathcal P(LH)$ with $\tau(p\_k) = 1/2$ such that $\lim\_{k \to \infty} \| E\_{L(H \cap gHg^{-1})}(1 - 2p\_k ) \|\_2 = 0$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|\_2 < 1$ we have $| \langle x, u\_g^\*p\_ku\_g - \tau(p) \rangle | \leq \| E\_{LH}(u\_g^\* p u\_g - \tau(p) ) \|$ $\_2 = \| E\_{L(H \cap gHg^{-1})} (p\_k - 1/2) \|\_2 < \delta$. The same inequality holds for the other minimal projection $1 - p\_k$.
Once we have produced such an $A\_0$ for $1/2^n$ then given any minimal projection $p \in A\_0$ we again have that $pLH \not\prec\_{pLH} pL(H \cap gHg^\*)$ and so the argument above shows that there exists $p\_1$ and $p\_2$ in $\mathcal P(LH)$ such that $p\_1 + p\_2 = p$, each has half the trace and $| \langle x, u\_g^\* p\_j u\_g - \tau(p\_j) \rangle | < \delta$. This proves the induction step. QED
Now that we have established the above lemma, the fact that $u\_g$ and $v$ are orthogonal follows from a lemma of Popa's in <http://www.ams.org/mathscinet-getitem?mr=703810>. Let's give the proof here.
Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A\_0 \subset LH$ such that if $p$ is any minimal projection in $A\_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u\_g^\*pu\_g - \tau(p) \rangle | < \| x \|\_2 \varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A\_0$ by $p\_k$ where $1 \leq k \leq 2^n$.
Denote by $B\_0$ the commutant of $A\_0$ in $LG$.
Since $v \in \mathcal N\_{LG}(LH)$ we have that $vLHv^\* = LH$, hence $v^\* p\_k v \in LH$ for each $k$. Therefore
$| \langle v, u\_g \rangle |^2 \leq \| E\_{B\_0} ( vu\_g^\*) \|\_2^2$ $=
\| $ $\Sigma\_k$ $ p\_k v u\_g^\* p\_k \|\_2^2 = \Sigma\_k \langle v^\* p\_k v, u\_g^\* p\_k u\_g \rangle < (\Sigma\_k \tau(p\_k)^2 ) + \Sigma\_k \varepsilon/2^{n + 1} < \varepsilon$.
Since $\varepsilon$ was arbitrary we conclude that $u\_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N\_{LG}(LH)'' = L(\mathcal N\_G(H))$.
| 6 | https://mathoverflow.net/users/6460 | 26638 | 17,434 |
https://mathoverflow.net/questions/26601 | 0 | I have been struggling with this problem and hope someone could help. I am trying a variation of non-repetitive combination scenario. I can use the formula n!/r!x(n-r)! to find non-repetitive combinations of size "r" from "n" numbers. However, these combination have repeating elements.
For example:
I have 9 letters - A, B, C, D, E, F, G, H, I
I want to find unique sets of three letters such as:
A B C
D E F
G H I
A D G
B E H
C F I
B E G
etc
If I use the standard non-repetitive combination, I might get sets like A B C, A B D, A B E
. In this case A and B are repeated in all sets.
Following are my questions:
- How to calculate the number of combinations as described above?
- How to calculate the available combinations if we allow k repeating elements.
Example. For a combination of 4 elements, we set the k to 2. This means A B C D, A B E F are allowed but not A B C D and A B C E.
I read a ton of materials on combinations and permutations but none of them seem to be covering this scenario.
I would really appreciate if you can give some pointers and direction.
Thanks
| https://mathoverflow.net/users/6455 | Creating a combinations with unique sets | I have deleted what was here as it was based on a misunderstanding of the problem. My current understanding of the problem is that, given positive integers $k\le n$ one wants the largest number of $k$-element subsets of an $n$-element subset, no two intersecting in more than one element. This has indeed been studied as part of coding theory. In the language of coding theory, we want the biggest binary code of length $n$, all codewords being of weight $k$, the code having minimal distance $2k-2$. The case $k=3$ is discussed at <http://oeis.org/A001839> and the notes there also give references to the general case.
| 1 | https://mathoverflow.net/users/3684 | 26639 | 17,435 |
https://mathoverflow.net/questions/26651 | 16 | Hi, everybody. I'm recently reading W.Bruns and J.Herzog's famous book-Cohen-Macaulay Rings. I personally believe that it would be perfect if the authors provide for readers more concrete examples. After reading the first two sections of this book, I have two questions.
1. Given a non-negative integer n, how can we construct a non-Gorenstein Cohen-Macaulay ring with dimension n?
2. Except by the definitions of Cohen-Macaulay rings, is there a more efficient way to check the Cohen-Macaulayness of local rings?
| https://mathoverflow.net/users/5775 | Is there a simple method to test a local ring to be Cohen Macaulay? | Some interesting examples of Cohen-Macaulay but not Gorenstein rings:
1) Determinantal rings: Let $m\geq n\geq r>1$ be integers. Take $S=k[x\_{ij}]$ with $1\leq i\leq m, 1\leq j\leq n$ and $I$ be the ideal generated by all $r$ by $r$ minors. Then $R=S/I$ is CM, but is Gorenstein if and only if $m=n$. And $R$ has dimension $(m+n-r+1)(r-1)$.
2) Veronese subrings: Let $S=k[x\_1,\cdots,x\_n]$ and $R=S^{(d)}$ be the $k$-subalgebra of $S$ generated by the monomials of degree $d$. Then $R$ is always CM, but is Gorenstein if only if $d$ divides $n$. And $R$ has dimension $n$.
3) Semigroup rings: Let $R=k[t^{a\_1},\cdots, t^{a\_n}]$. $R$ is $1$-dimensional domain, so CM. $R$ is Gorenstein if and only if the semigroup generated by $(a\_1,\cdots,a\_n)$ is symmetric. For higher dimension one can use the trick in BCnrd's comment.
| 13 | https://mathoverflow.net/users/2083 | 26654 | 17,445 |
https://mathoverflow.net/questions/26661 | 3 | Whenever I have seen unique factorization discussed, it is always with respect to the solution of diophantine equations; the equations are solved by splitting the equation into linear functions over a ring and then invoking unique factorization. But the discussions always give the impression that the failure of unique factorization causes all sorts of problems. Apart from the applications to diophantine equations, why is unique factorization such a desirable property?
| https://mathoverflow.net/users/4692 | What goes wrong in a ring that does not have unique factorization? | I think a glance at any elementary number theory textbook will show you numerous topics which rely on uniqueness of factorization. The formulas for the number of divisors, the sum of the divisors, the Euler phi-function, all depend on unique factorization. Going up a level, the Euler product for the Riemann zeta function depends on unique factorization.
| 2 | https://mathoverflow.net/users/3684 | 26662 | 17,450 |
https://mathoverflow.net/questions/26643 | 2 | ZFCfin (ZFC without the axiom of infinity, plus its negation) is biinterpretable with Peano Arithmetic.
Each countable model of PA has what is known as its "standard system": the collection of sets of standard natural numbers which can be coded in that system. Usually the coding is via prime exponentiation: $n$ is in the set coded by $c$ if the $n^{th}$ standard prime divides $c$. The standard system of the standard model has only finite sets of naturals in it; all models with infinite sets in their standard system are nonstandard models.
Nonstandard models of ZFCfin (those not isomorphic to HF) contain "externally infinite" sets; elements of the model for which the number of elements standing in the model's member-of relation to them is greater than any finite number.
Are these "externally infinite" sets of a nonstandard model of ZFCfin somehow connected to the standard system of a nonstandard model of HF?
I ask because I'm starting to work through the literature on nonstandard models of PA and I'm finding it much easier to think of the nonstandard numbers as externally-infinite sets of ZFCfin. I'm wondering how far astray that intuition is likely to lead me.
Edit: I should add that I'm aware the nonstandard sets of ZFCfin are frequently $\epsilon$-illfounded. I still find them easier to think of than infinite numbers (frankly I always thought it strange to rule out such sets in the first place).
| https://mathoverflow.net/users/2361 | "standard system" of a nonstandard model of PA, interpreted in ZFCfin? | Yes, there is a perfect agreement between the standard system of the nonstandard model of ZFCfin, such as a nonstandard version of HF, and the standard system of its corresponding model of PA. The standard systems are identical.
That is, because of the mutual interpretability, from any model of PA we may form a model of ZFCfin and vice versa, and these two models have the same standard system.
The standard system of a model of PA, ZFC or ZFCfin consists simply of the standard parts of the sets of natural numbers that exist in the model. This notion is quite robust and is used in many arguments, particularly in the theory of models of arithmetic. In particular, it admits a variety of equivalent characterizations.
For example, for a nonstandard model M of ZFCfin, the standard system of M is equivalently characterized by any of the following methods. We may regard the standard natural numbers $\mathbb{N}$ as a subclass of $M$, since each one is definable in $M$.
For a set A of natural numbers,
* A is in Ssy(M) if and only if there is an object $a\in M$ such that $A=a\cap\mathbb{N}$, that is, $A$ consists of the elements of $\mathbb{N}$ that are in $a$ in $M$.
* A is in Ssy(M) if and only if there is an object $a\in M$ such that $A=a\cap\mathbb{N}$ and $M$ satisfies that $a$ consists of (nonstandard) natural numbers only.
* A is in Ssy(M) if and only if there is a formula $\varphi$ and parameters $\vec a\in M$ such that $n\in A\iff M\models\varphi(n,\vec a)$. Thus, $A$ is the trace on $\mathbb{N}$ of the set defined by $\varphi(\cdot,\vec a)$.
* A is in Ssy(M) if and only if $A\in\text{Ssy}(\mathbb{N}^M)$, where $\mathbb{N}^M$ is the natural numbers of $M$, as defined in set theory. This is provably a model of PA in ZFCfin.
* A is in Ssy(M) if and only if the characteristic function of $A$ is the standard part of a pseudo-finite binary sequence in $M$.
So it doesn't really matter which coding system you use; you always arrive at the same standard system. Indeed, the first characterization above is a kind of coding-free version. You've already done the coding, in a sense, by working with ZFCfin instead of PA, and so now you just have (nonstandard) sets themselves, rather than codes for such sets. Thus, to get the standard part of a set $a$, you just take the standard elements of it.
These characterizations also apply to ZFC models as well. The Standard System of a ZFC model can also be characterized as the set of standard parts of all reals of $M$. One imagines that the reals of a nonstandard model $M$ of ZFC stick up like overgrown grass, and one gets the standard part by riding a big lawnmower, cutting them down to the standard part.
| 4 | https://mathoverflow.net/users/1946 | 26670 | 17,456 |
https://mathoverflow.net/questions/26644 | 1 | For this question, I am in the smooth finite-dimensional category. All objects are $\mathcal C^\infty$ manifolds and all maps are smooth. Recall the notion of [groupoid](http://www.google.com/search?hl=&q=groupoid), and let's restrict our attention to those for which the source map (and hence also the target map) is a (necessarily surjective) submersion. If $G\rightrightarrows X$ is a groupoid, me define a **subgroupoid over $X$** to be an immersed submanifold $H\hookrightarrow G$ so that $H\rightrightarrows X$ is a groupoid. (If this is not the standard definition, let me know.)
Pick $x\in X$. Does there exist a subgroupoid $H\hookrightarrow G$ so that: (1) $x$ has no automorphisms in $H$, and (2) if $y\in X$ with $x\cong y$ in $G$, then $x\cong y$ in $H$? I.e.: the groupoid $H$ should be as small as possible but still preserve the isomorphism class of $x$. I don't mind if other isomorphism classes shrink in size.
| https://mathoverflow.net/users/78 | Given an object in a Lie groupoid, does there exist a subgroupoid for which the object has no automorphisms but retains its equivalence class? | Consider the case of a group $K$ acting on $X$. Restricting to the orbit of the point $x$, by the orbit-stabilizer theorem $X$ is identified with $K/\text{Stab}(x)$. Your question seems to amount to asking whether the projection $K\times K/\text{Stab}(x)\to K/\text{Stab}(x)\times K/\text{Stab}(x)$ given by $(\gamma,p)\mapsto (\gamma\cdot p,p)$ admits an immersed section. Restricting such a section to a fiber $K/\text{Stab}(x)\times \text{pt}$ would give an section of the map $K\to K/\text{Stab}(x)$, which will not exist in general. (Edit: rewritten. Thanks to David Carchedi for pointing out the mistake in my notation.)
A concrete counterexample: consider the action of $K=\mathbb{R}$ on $X=S^1$ by translation. Then $G$ is diffeomorphic to $\mathbb{R}\times S^1$, and $H$ would have to be diffeomorphic to $S^1\times S^1$, but $S^1\times S^1$ does not immerse in $\mathbb{R}\times S^1$. Moreover the restriction to a fiber $S^1\times \text{pt}$ would be an immersion of $S^1$ into $\mathbb{R}$, which also cannot exist.
But this is not a phenomenon just of fundamental group or of equi-dimensional immersions. For example, consider $K=\text{Isom}(\mathbb{H}^2)=\text{PSL}\_2\mathbb{R}$ acting on a hyperbolic surface $\Sigma$. A section of $\text{PSL}\_2\mathbb{R}\times \Sigma\to \Sigma\times \Sigma$ restricts to an immersed section of $\text{PSL}\_2\mathbb{R}\to \Sigma$. Of course $\Sigma$ *does* immerse, in fact embed, in $\text{PSL}\_2\mathbb{R}$ (it's a 3-manifold). But if we had a continuous section $\varphi:\Sigma\to \text{PSL}\_2\mathbb{R}$, we could translate a nonzero vector $v$ to each point $p$ by the isometry $\varphi(p)$. This would yield a nonzero vector field on $\Sigma$, contradicting the Gauss-Bonnet theorem. The same works for $\text{SO}(3)$ acting on $S^2$ or $\mathbb{R}P^2$, where the fundamental group is not the issue.
| 1 | https://mathoverflow.net/users/250 | 26675 | 17,461 |
https://mathoverflow.net/questions/26665 | 4 | I come across the following problem in my study.
Consider in the real field. Let $ 0\le x\le1 $, $a\_1^2+a\_2^2=b\_1^2+b\_2^2=1$.Is it true
$ (a\_1b\_1+xa\_2b\_2)^2\le\left(\frac{(1-x)+(1+x)(a\_1b\_1+a\_2b\_2)}{(1+x)+(1-x)(a\_1b\_1+a\_2b\_2)}\right)^{2}(a\_1^2+xa\_{2}^{2})(b\_1^2+xb\_{2}^{2})$?
| https://mathoverflow.net/users/3818 | Another plausible inequality. | As Can Hang points out in his response, the inequality does not hold in general. Thanks to his comment to my own post, I stand corrected and claim the inequality is valid at least for the case
$$
a\_1b\_1+xa\_2b\_2\ge0
\qquad(\*)
$$
(and this seems to be a necessary condition as well).
Let me do some standard things. First let
$$
a\_1=\frac{1-u^2}{1+u^2}, \quad
a\_2=\frac{2u}{1+u^2}, \quad
b\_1=\frac{1-v^2}{1+v^2}, \quad
b\_2=\frac{2v}{1+v^2}
$$
where $uv\ge0$.
Substitution reduces the inequality to the following one:
$$
((1-u^2)(1-v^2)+4xuv)^2
\le\biggl(\frac{(uv+1)^2-x(u-v)^2}{(uv+1)^2+x(u-v)^2}\biggr)^2
((1-u^2)^2+4xu^2)((1-v^2)^2+4xv^2).
\qquad{(1)}
$$
Now introduce the notation
$$
A=(1-u^2)(1-v^2)+4xuv, \quad
B=(uv+1)^2, \quad C=x(u-v)^2
$$
and note that $A,B,C$ are nonnegative; the inequality $A\ge0$ is equivalent to the above condition $(\*)$. In addition,
$$
A\le B-C
\qquad{(2)}
$$
because
$$
B-C-A=(1-x)(u+v)^2\ge0.
$$
In the new notation the inequality (1) can be written more compact:
$$
A^2(B+C)^2\le(B-C)^2(A^2+4BC)
$$
which after straightforward reduction becomes
$$
A^2\le(B-C)^2,
$$
while the latter follows from (2).
| 10 | https://mathoverflow.net/users/4953 | 26677 | 17,462 |
https://mathoverflow.net/questions/26680 | 49 | Background
----------
Let $(X,x)$ be a pointed topological space. Then the fundamental group $\pi\_1(X,x)$ becomes a topological space: Endow the set of maps $S^1 \to X$ with the compact-open topology, endow the subset of maps mapping $1 \to x$ with the subspace topology, and finally use the quotient topology on $\pi\_1(X,x)$. This topology is relevant in some situations. A very interesting paper dealing with this topology is:
>
> [1] Daniel K. Biss, A Generalized Approach to the Fundamental Group, The American Mathematical Monthly, Vol. 107
>
>
>
You can find this [online](http://math.mit.edu/~gracelyo/18904/BissCoveringSpaceArticle.pdf). This is somehow an introduction to
>
> [2] Daniel K. Biss, The topological fundamental group and generalized covering spaces , Topology and its Applications, Vol. 124
>
>
>
Question
--------
How can we prove that $\pi\_1(X,x)$ is a topological group? Clearly the inversion map $\pi\_1(X,x) \to \pi\_1(X,x)$ is continuous, since $S^1 \to S^1, z \mapsto \overline{z}$ is continuous and induces this map. But I don't know how to attack the continuity of the multiplication. It's not hard to see that the multiplication on $map((S^1,1),(X,x))$ is continuous, since it is induced by a fold map $S^1 \to S^1 + S^1$. In order to carry this over to $\pi\_1(X,x)$, there are at least two problems which I encounter:
* The quotient map $map((S^1,1),(X,x)) \to \pi\_1(X,x)$ may be not open.
* The product of the quotient maps $map((S^1,1),(X,x))^2 \to \pi\_1(X,x)^2$ may be not a quotient map.
In [1] it is claimed that $\pi\_1(X,x)$ is always a topological group, and this should be proven in [2], but I have no acecss to [2].
An example that products of quotient maps don't have to be quotient maps can be found [here](http://www.uni-bonn.de/~habecker/dokumente/tutorium/prodquotient2.ps). Remark however that this is true in the category of compactly generated spaces.
| https://mathoverflow.net/users/2841 | Fundamental group as topological group | **Update**: A bit of a digital paper chase led me, via [David Robert's thesis](http://ncatlab.org/davidroberts/show/HomePage) (note that in the latest version, it is Chapter 5, section 2 that is most relevant), to [this paper](http://arxiv.org/abs/0910.3685) on the arXiv. The last sentence of the abstract is:
>
> These hoop earring spaces provide a simple class of counterexamples to the claim that $\pi\_{1}^{top}$ is a functor to the category of topological groups.
>
>
>
I recommend reading this article.
(**Added later**: In case it's not clear, the author of that paper is Jeremy Brazas who added an answer afterwards, so if you vote for my answer, you should *definitely* vote for his!)
---
**Original Answer**: *These were my initial thoughts before I found the references above. These were what made me sufficiently intrigued to do the paper chase and find the above-mentioned thesis and article.*
The proof given in the second paper (by Biss) that is mentioned in the question is short enough that I think it reasonable to copy it out here. I shan't copy out the obvious diagram so need to establish some notation first:
1. $m \colon \pi\_1^{Top}(X,x) \times \pi\_1^{Top}(X,x) \to \pi\_1^{Top}(X,x)$ is the multiplication map in question
2. $p \colon \operatorname{Hom}((S^1,1),(X,x)) \to \pi\_1^{Top}(X,x)$ is the quotient map
3. $\overline{m} \colon \operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x)) \to \operatorname{Hom}((S^1,1),(X,x))$ is the "upstairs" multiplication map. (It's a tilde in the original, but that isn't displaying correctly for me so I daren't use it.)
The proof then proceeds:
>
> To show that $m$ is continuous, it suffices to show that $\overline{m}$ is continuous, for then if $U \subset \pi\_1^{Top}(X,x)$ is open, $(p \times p)^{-1} m^{-1}(U) = \overline{m}^{-1}p^{-1}(U)$ is open, but by the definition of a quotient map, $(p \times p)^{-1} m^{-1}(U)$ is open if and only if $m^{-1}(U)$ is.
>
>
>
There then follows a proof that $\overline{m}$ is continuous, a fact that I trust does not need proving.
Comments on your comments:
1. We don't need the quotient map to be open since we are only ever dealing with preimage sets. It is certainly not always true that if $q \colon X \to Y$ is a quotient that $q(U)$ is open in $Y$ for every open $U$ in $X$. But it is true *by definition* that $q^{-1}(U)$ is open in $X$ **if and only if** $U$ is open in $Y$. This is because the topology on $Y$ is precisely that to make this true. So since we are only dealing with sets of the form $(p \times p)^{-1}(A)$ then the assertion is valid *assuming that $p \times p$ is a quotient map*.
2. Here, I find myself worried. A quick back-of-envelope check seems to show that one can't simply assume that the product of quotients is again a quotient in Top (a counterexample eludes me as I don't have *Counterexamples in Topology* to hand and I'm too used to dealing with "nice" spaces). It *may* be the case that for Hom-spaces then there's some magic that can be done (though such is not mentioned in the paper); but again the best that I can do on the back of an envelope is observe that (modulo some basepoint mess) *by construction* $\operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x))$ quotients to $\pi\_1^{Top}((X,x) \times (X,x))$. But to proceed, one would need to know that $\pi\_1^{Top}$ was a product-preserving functor. This is morally the same as saying that it is representable - which looks good since we have an obvious representing object $S^1$! However, this can't be made into a proper argument since although we have a representing object, we *don't* have an enriched Hom-functor $hTop \times hTop \to Top$ which to evaluate at $S^1$.
So I would look for a counterexample to the product of quotients being a quotient, and see where that leads you. Either you'll find a proper counterexample to the proposition in question, or you'll see why *in this special case*, such a counterexample could not occur.
(Of course, I may well be missing something obvious!)
| 30 | https://mathoverflow.net/users/45 | 26684 | 17,468 |
https://mathoverflow.net/questions/26520 | 13 | Skip Garibaldi asks if there is an elementary proof of the following fact that "accidentally" fell out of some high-powered machinery he was working on.
Say that two matrices $A$ and $B$ over the rationals are *rationally congruent* if
there exists a nonsingular matrix $S$ over the rationals such that $S^t A S = B$.
**Theorem** (Garibaldi). Suppose $n \equiv 0 \pmod 4$. Then the diagonal matrices
$$A = diag\left[\binom{n}{0}, \binom{n}{2}, \binom{n}{4}, \ldots, \binom{n}{n/2 - 2}\right]$$
and
$$B = diag\left[\binom{n}{1}, \binom{n}{3}, \binom{n}{5}, \ldots, \binom{n}{n/2 - 1}\right]$$
are rationally congruent. Similarly, suppose $ n \equiv 2 \pmod 4$. Then the matrices
$$A = diag\left[\binom{n}{0}, \binom{n}{2}, \binom{n}{4}, \ldots, \binom{n}{n/2 - 1}\right]$$
and
$$B = diag\left[\binom{n}{1}, \binom{n}{3}, \binom{n}{5}, \ldots, \binom{n}{n/2 - 2}, \frac{1}{2}\binom{n}{n/2}\right]$$
are rationally congruent.
| https://mathoverflow.net/users/3106 | Rational congruence of binomial coefficient matrices | Wadim, isn't that 95% of the proof? First let me correct your first displayed equation (thanks to fherzig for pointing this out): It is not sufficient for the proof, but
$$
\sum\_{i=0}^{n-1}\binom{4n}{2i}P\_i(t)P\_i(s)
=\sum\_{i=0}^{n-1}\binom{4n}{2i+1}\hat P\_i(t)\hat P\_i(s)
$$
is, where $t$ and $s$ are two independent variables.
Let me rename your $P\_i$ as $Q\_{2i}$ and your $\hat{P\_i}$ as $Q\_{2i+1}$, so that your equation
$$
\sum\_{i=0}^{n-1}\binom{4n}{2i}P\_i(t)P\_i(s)
=\sum\_{i=0}^{n-1}\binom{4n}{2i+1}\hat P\_i(t)\hat P\_i(s)
$$
becomes
$$
\sum\_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q\_i(t)Q\_i(s)=0.
$$
Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$. (With some work, the proof below works just as well if $Q$ is any polynomial of degree $n-1$ (not less!), but let me use $t^{n-1}$ for simplicity's sake.) Let $Q\_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$ for every $i\in\mathbb Z$. For any fixed $t$ and $s$, the term $Q\_i\left(t\right)Q\_i\left(s\right)$ is a polynomial in $i$ of degree $2\left(2\left(n-1\right)+1\right)<4n$, and thus satisfies
$$
\sum\_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q\_i(t)Q\_i(s)=0,
$$
since the $4n$-th finite difference of a polynomial of degree $< 4n$ is zero. Due to the symmetry of the function $i\mapsto Q\_i\left(t\right)Q\_i\left(s\right)$ around $i=2n$, and due to $Q\_{2n}\left(t\right)=0$, this becomes
$$
\sum\_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q\_i(t)Q\_i(s)=0.
$$
Now it remains to prove that each of the families $\left(Q\_1,Q\_3,...,Q\_{2n-1}\right)$ and $\left(Q\_0,Q\_2,...,Q\_{2n-2}\right)$ spans the space of all polynomials in $t$ of degree $< n$. This is a particular case of a more general fact: If $x\_1$, $x\_2$, ..., $x\_n$ are $n$ distinct reals, then the polynomials $\left(t-x\_1\right)^{n-1}$, $\left(t-x\_2\right)^{n-1}$, ..., $\left(t-x\_n\right)^{n-1}$ are linearly independent. In order to prove this, assume that they are linearly dependent, take their derivatives of all possible orders, evaluate at $t=0$ (or alternatively, just take their coefficients), and get a contradiction because Vandermonde's determinant is nonzero.
| 10 | https://mathoverflow.net/users/2530 | 26693 | 17,474 |
https://mathoverflow.net/questions/26676 | 9 | The following are a collection of doubts, some of which shall have concrete answers while others may have not. Any kind of help will be welcome.
Reading Peter Smith's "Gödel Without (Too Many) Tears", particularly where he gives a nonstandard model of Q, I began wondering if the reason for the existence of nonstandard models of arithmetic has anything to do with incompleteness theorems.
I do not know if categoricity implies completeness (in the sense of every sentence being decidable by proof), but anyway, it seems reasonable, when one is formalizing a given (informal) theory, to try to "force" somehow the formal theory to talk "almost exclusively" about the intended interpretation. So I started thinking if some axiom (or axiom schema) could be added to PA in order to forbid its most obvious nonstandard models.
The first idea in this line was: ok, we have our class of terms 0, S0, SS0, etc. So, if we found a way to tell that for every x there is some term to which it is equal, we would be done.
But then I realized that our terms are defined inductively and that we are making implicitly the assumption: “and nothing else is a term”, very similar to the desired “and nothing else is a number” we would like to add to PA. This thought sort of worried me: every metatheoretic concept (terms, formulas, and even proofs!) is based on assumptions like these! (I have not still found a way out of these worries).
Leaving that apart. What if we move on to a stronger theory (with different axioms, but with an extension by definitions that proves every axiom of PA), for example ZFC? Natural numbers become then 0 (the empty set) plus the von Neumann ordinals (obtained by Pair and Union) that contain no limit ordinal. The set of natural numbers is obtained from Infinity, just selecting them by Comprehension. Kunen says in page 23 of his “The Foundations of Mathematics” that the circularity in the informal definition of natural number is broken “by formalizing the properties of the order relation on omega”. Could nonstandard models survive this formalization?
Well, I think I've read somewhere that being omega is absolute, so forcing would not be a way to obtain such nonstandard models. Also, I am not sure if (the extension by definitions from) ZFC set theory is a conservative extension of PA, but then it would not be able to prove anything about natural numbers (expressible in the language of arithmetic) that PA alone cannot prove. So somehow it looks like nonstandard models must manage to survive! Maybe due to the notion of being a subset of a given set not being particularly clear (although it looks like it should not be problematic with hereditarily finite sets).
Thank you in advance.
| https://mathoverflow.net/users/6466 | Incompleteness and nonstandard models of arithmetic | Unfortunately, nonstandard models will survive any such attempt. This is guaranteed by the [Löwenheim-Skolem Theorem](http://en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_theorem) which says that if a countable first-order theory T has an infinite model then it has one of every infinite cardinality. Since an uncountable model necessarily has nonstandard elements, this guarantees that there is a nonstandard model of T (and even countable ones).
Actually, in your case you need a "two-cardinal" version of Löwenheim-Skolem. In your ZFC example, you move to a theory which interprets arithmetic inside a definable substructure (the set ω). The definable substructure of such a model which might still be countable even if the model itself is uncountable. Nevertheless, one can still blow up the size of the natural number substructure via the [ultrapower construction](http://en.wikipedia.org/wiki/Ultraproduct), for example.
To evade the Löwenheim-Skolem Theorem, one has to move beyond first-order logic. For example, in [infinitary logic](http://en.wikipedia.org/wiki/Infinitary_logic) one allows infinite disjunctions such as
$$\forall x(x = 0 \lor x = S0 \lor x = SS0 \lor \cdots)$$
which ensures that the model is standard. Also, [second-order](http://en.wikipedia.org/wiki/Second-order_logic) allows quantification over arbitrary sets under the standard interpretation, which again prohibits non-standard models. (See this [related question](https://mathoverflow.net/questions/17483/defining-free-monoid-without-nat).) This is the characterization of **N** most commonly used by working mathematicians.
| 16 | https://mathoverflow.net/users/2000 | 26697 | 17,476 |
https://mathoverflow.net/questions/26689 | 19 | Given an [integral equation](http://en.wikipedia.org/wiki/Integral_equation) is there always a differential equation which has the same (say smooth) solutions?
It seems like not but can one prove this in some example?
**Edit:** Naively I'm hoping for some algorithm which takes an integral equation and applies some operations like taking derivatives, substituting variables for some new ones, adding additional differential equations etc... such that after this procedure you have made all integral signs vanish and obtained a differential equation which has the same solutions as the integral eqution. (maybe similarly to how one can transform any system of PDEs into a system of first order equations)
| https://mathoverflow.net/users/745 | Can an integral equation always be rewritten as a differential equation? | While I second Deane's comment that the author should be a bit more specific about the kind of equations he is interested in, in general the answer is **no** for integral and, more broadly,*integro-differential* equations. However, the latter can be reduced to *functional-differential* equations rather than to purely differential ones. For more details, see Section 6.6 of the book *[Symmetries and Conservation Laws for Differential Equations of Mathematical Physics](http://books.google.com/books?id=lXu1s6sxjYUC)*
(unfortunately the relevant pages appear to be excluded from Google preview).
For instance, I greatly doubt that one could reduce the [Smoluchowski coagulation equation](http://en.wikipedia.org/wiki/Smoluchowski_coagulation_equation) from Example 6.5 of the above book to a differential (as opposed to functional-differential) equation or system thereof.
| 8 | https://mathoverflow.net/users/2149 | 26705 | 17,481 |
https://mathoverflow.net/questions/26718 | 1 | Consider a simple queue model like the one described in <http://en.wikipedia.org/wiki/M/M/1_model>. The article states what the expected waiting time is before a request enters the queue.
Assuming that the actual queue length is unknown, does the expected value of "time left to wait" for a given request change over time while the request is in the queue?
**Clarification**: is it true that the expected time until a request is served stays constant, regardless of how long it has already spent in the queue?
| https://mathoverflow.net/users/6476 | Does waiting in a queue change the expected time left to wait? | I am not entirely sure what you are asking but ...
The model you reference in the wiki article has a memory-less distribution for waiting time and inter-arrival time. Thus, the total time for a request to get processed is not dependent on time.
| 2 | https://mathoverflow.net/users/4660 | 26722 | 17,491 |
https://mathoverflow.net/questions/26318 | 18 | Wonder whether any of you know where it was that the following pearl of topology first appeared:
>
> Prove that at any instant of time you can find three isothermal points on the surface of the Earth that correspond to the vertices of an equilateral triangle.
>
>
>
According to [Léo Sauvé](https://mathoverflow.net/questions/145475/what-is-the-source-of-this-erdos-quote), all problems appeared once in the *Monthly*. Does that *dictum* apply to the above teaser, too?
| https://mathoverflow.net/users/1593 | Points on a sphere | Let me just elaborate a little on the references that Charlie Frohman listed (so this isn't really a separate answer, but it's too long for a comment).
The theorem for equilateral triangles is due to S. Kakutani (1942 Annals). He stated it just for triangles formed by orthonormal bases for $\mathbb R^3$, but the argument applies for all sizes of equilateral triangles. He deduced the interesting corollary that any compact convex set in $\mathbb R^3$ has a circumscribing cube, answering a question posed by Rademacher. Charlie Frohman's proof is similar in spirit to Kakutani's.
The n-dimensional generalization of Kakutani's theorem (and corollary) is due to H. Yamabe and Z. Yujobo (1950 Osaka Math J).
Returning to 3 dimensions, the corresponding result for 4 points at the vertices of a square inscribed in a great circle was shown by F. Dyson (1951 Annals), and G.R. Livesay generalized this to rectangles inscribed in a great circle (1954 Annals). The case of arbitrary triangles was done by E.E. Floyd (1955 PAMS).
It's interesting that this little topic produced three Annals papers, though each one was short -- just 3 pages. (When was the last 3-page Annals paper?)
The techniques used to prove the later theorems varied considerably. It might be interesting to see which ones can be proved by basic algebraic topology arguments as in Kakutani's theorem (which certainly deserves to be included in algebraic topology textbooks!).
| 23 | https://mathoverflow.net/users/23571 | 26731 | 17,495 |
https://mathoverflow.net/questions/26727 | 3 | A. Aubry published a paper entitled "Les Logarithmes avant Neper" in the 1906 edition of L'Enseignement Mathématique signed simply "A. Aubry (Beaugency, Loiret)". Does anyone know what the A. stood for (perhaps Auguste?) or anything more about this A. Aubry?
The only other reference I've found is that it may have been the same A. Aubry that wrote the forward to "Carrés Magiques au degré n" by Général Cazalas but this is pure conjecture.
Thanks in advance for any insight.
Cheers, Scott
| https://mathoverflow.net/users/4111 | Information about A. Aubry | I think it is Auguste Aubry. The *L'Enseignement Mathématique* volume is on archive.org, and there is an earlier paper in it on hyperbolic functions by Aubry. The material and location suggests a school teacher. There is more about him (I presume) in <http://www.univ-lille1.fr/bustl-grisemine/pdf/extheses/50416-1999-Decaillot-Laulagnet.pdf> which is about Edouard Lucas and his associates. On p. 74 of the PDF, Aubry is in a footnote about people who published minor research in the AFAS publications (AFAS is the Association Française pour l'Avancement des Sciences). On p. 96 there it is Auguste Aubry writing something about magic squares in an edition of Fermat. Aubry with A. Gérardin did translation work on some paper of Lucas. He also wrote a paper on factorisation methods (p. 207). More about the Lucas work in <http://www.math.ens.fr/culturemath/histoire%20des%20maths/pdf/Decaillot_textile.pdf> by the same author.
| 4 | https://mathoverflow.net/users/6153 | 26732 | 17,496 |
https://mathoverflow.net/questions/26599 | 11 | This is a question I have thought about and asked a number of people, but have never got an answer beyond "It should be true that..."
Given a finitely generated group $G$ (eg. a link group $G\_L:=\pi\_1(S^3-L)$ for a link $L$) and a finite group $H$ we want to count homomorphisms from $G$ to $H$. For link groups as above, this is an invariant of $L$.
My question: (for which $H$) is there a polynomial-time algorithm (in the number of generators and relations for $G$) for computing $N(G,H):=|Hom(G,H)|$ (particularly for $G\_L$)?
Some things I know:
1) If $L$ is a knot and $H$ is nilpotent then $N(G\_L,H)$ is constant ([M. Eisermann](https://www.ams.org/journals/proc/2000-128-05/S0002-9939-99-05287-9/home.html))
2) [D. Matei; A. I. Suciu,](http://doi:10.1016/j.jalgebra.2005.01.009) have an algorithm for solvable $H$, but the complexity is not clear.
3) The abelianization of $G\_L$ is just $Z^c$, $c$ the number of components, so for $H$ abelian it is easy.
A wild conjecture is that it should always be "FPRASable" i.e. there exists a fully polynomial randomized approximation scheme for the computation.
| https://mathoverflow.net/users/6355 | complexity of counting homomorphisms | For $G$ a knot group, and for $H$ a dihedral group, there should be a simple algorithm for counting the number of homomorphisms. The meridians of $G$ normally generate, and are all conjugate, so they must be sent to conjugate elements in $H$. If they are sent to the cyclic subgroup of index 2, then the image is cyclic, and this is easy to count.
If a meridian is sent to an involution, then an index 2 subgroup of $G$ is sent to a cyclic group. This amounts to computing the homology of the 2-fold branched cover of the knot, together with the action of the involution on this homology. This is certainly polynomial-time computable, and I'm pretty sure one can determine its dihedral quotients easily. In any case, at least this reduces it to the problem of finding dihedral quotients of abelian-by-$\mathbb{Z}/2$ groups.
| 3 | https://mathoverflow.net/users/1345 | 26742 | 17,504 |
https://mathoverflow.net/questions/21657 | 4 | A stupid AG question: could singular (Zarisky) points be dense in a reduced (Noetherian) scheme $S$? If yes, which 'standard' restrictions on $S$ could ensure that this does not happen? For example, should one demand $S$ to be excellent? Will anything change if we want singular points not to be dense in any finite type reduced $S$-scheme?
I would also be gratefull for any ('bad' or 'good') examples.
| https://mathoverflow.net/users/2191 | When singular points of a reduced scheme are not dense in it? | For examples with dense singular locus, see William J. Heinzer and Lawrence S. Levy: Domains of Dimension 1 with Infinitely Many Singular Maximal Ideals, Rocky Mountain J. Math. (2007), 203-214. Their examples are affine and noetherian.
| 3 | https://mathoverflow.net/users/3485 | 26756 | 17,514 |
https://mathoverflow.net/questions/26754 | 0 | I have two variables, $x$ and $y$, and I am using linear regression to predict $y$ from $x$ over a large set of subjects. There are multiple observations per subject.
I have tried several things:1. pool all observations together and fit one model to the entire dataset;
- fit separate models (same specification as (1)) for each subject.
Now, (1) works reasonably well but obviously takes no account of subject-specific effects. On the other hand (2) is prone to overfitting, which manifests itself in poor out-of-sample performance.
If I average the two predictions, the result performs better than both (1) and (2) out of sample. However, this is clearly somewhat *ad hoc*.
My question is: what might be a better way to combine (1) and (2) into a single predictor?
Also, I have reasons to think that "similar" subjects should have similar regression coefficients. Is there any way to make use of this?
**[Edit]** I should have mentioned that there is no natural hierarchy to the subjects. However, I think I can come up with a reasonable similarity metric.
| https://mathoverflow.net/users/6484 | Combining regressions | You should use hierarchical bayesian regression. Google search will provide lots of pointers.
| 0 | https://mathoverflow.net/users/4660 | 26757 | 17,515 |
https://mathoverflow.net/questions/26695 | 3 | We choose our category of spaces to be compactly generated weak Hausdorff spaces for convenience, denoted $CGWH$.
Questions:
1.) Is there any sort of slick argument to verify that CGWH with the Quillen model structure is a right-proper (closed) model category?
2.) If we give the following presentation of the model structure on SSet:
Cofibrations are monomorphisms
Fibrations have the RLP with respect to all horn inclusions $\Lambda^n\_i \subseteq \Delta^n$ for $0\leq i \leq n$.
Or instead of the characterization of cofibrations, we could instead give:
Trivial fibrations have the RLP with respect to all inclusions of the boundary $\partial^n \subseteq \Delta^n$.
(The point of picking a nice presentation is that the (morally) right choice of definition often simplifies a proof.)
Is there any way to verify the model category axioms more easily? The proofs I've seen appeal to all of the hard work done in question 1. It seems like one should be able to verify the axioms for SSet more easily than the case of CGWH spaces.
| https://mathoverflow.net/users/1353 | Slick verification of the model category axioms for Spaces and SSets with the q-model structure? | The Joyal and Tierney notes contain a combinatorial proof as Dan says. They are available [here](http://www.crm.cat/HigherCategories/tierney.pdf). (Wanted to post this in the comments, but it seems impossible to do without sufficient reputation.) I should also mention that it is possible to give a reasonably slick proof of the model category axioms for simplicial sets by using the Cisinski machinery (see his monograph: Ast'erisque vol. 308).
| 6 | https://mathoverflow.net/users/6485 | 26759 | 17,516 |
https://mathoverflow.net/questions/26640 | 24 | Roughly speaking, I want to know whether one-relator groups only have 'obvious' free splittings.
>
> Consider a one-relator group $G=F/\langle\langle r\rangle\rangle$, where $F$ is a free group. Is it true that $F$ splits non-trivially as a free product $A \* B$ if and only if $r$ is contained in a proper free factor of $F$?
>
>
>
**Remarks**
1. One direction is obvious. It is clear that if $r$ is contained in a proper free factor then $G$ splits freely. (We think of $\mathbb{Z}\cong\langle a,b\rangle/\langle\langle b\rangle\rangle$ as an HNN extension of the trivial group, so it's not really a counterexample, even though it might look like one.)
2. A quick search of the literature suggests that the isomorphism problem for one-relator groups is wide open. (I'd be interested in any details that anyone may have.)
3. There is no decision-theoretic obstruction. Magnus famously solved the word problem for one-relator groups. Much more recently, [Nicholas Touikan has shown](http://arxiv.org/abs/0906.3902) that, for any finitely generated group, if you can solve the word problem then you can compute the Grushko decomposition. So one can algorithmically determine whether a given one-relator group splits. If the answer to my question is 'yes' then one can use Whitehead's Algorithm to find this out comparatively quickly.
4. When I first considered this question, it seemed to me that the answer was obviously 'yes' - I don't see how there could possibly be room in a presentation 2-complex for a 'non-obvious' free splitting. But a proof has eluded me, and of course many seemingly obvious facts about one-relator groups are extremely hard to prove.
| https://mathoverflow.net/users/1463 | Free splittings of one-relator groups | $\newcommand{\rank}{\operatorname{rank}}$I think [Grushko](http://en.wikipedia.org/wiki/Grushko_theorem) plus the [Freiheitssatz](http://en.wikipedia.org/wiki/Freiheitssatz) does the trick.
Suppose that $G=A \ast B$ is a one-relator group which splits as a free product non-trivially. By Grushko, $\rank(G)=\rank(A)+\rank(B)=m+n$, where $\rank(A)=m$ and $\rank(B)=n$. If $G$ is not free, then by Grushko there is a one-relator presentation $\langle x\_1,\ldots, x\_m ,y\_1,\ldots, y\_n | R\rangle$, such that $\langle x\_1,\ldots, x\_m \rangle =A \leq G, \langle y\_1, \ldots, y\_n \rangle=B \leq G$ (for this, one has to use the strong version of Grushko that any one-relator presentation is Nielsen equivalent to one of this type). Suppose that $R$ is cyclically reduced, and involves a generator $x\_1 \in F\_m$. Then $\langle x\_2,\ldots , x\_m, y\_1, \ldots, y\_n\rangle$ generates a free subgroup of $G$ by the Freiheitssatz. But this implies that $B = \langle y\_, \ldots, y\_n\rangle$ is free. Moreover, if $R$ involves one of the generators $y\_i$, then one sees that $A=\langle x\_1,\ldots,x\_m\rangle$ is also free, and therefore $G=A\ast B$ is free, a contradiction. So $R \in F\_m$, as required.
| 18 | https://mathoverflow.net/users/1345 | 26767 | 17,524 |
https://mathoverflow.net/questions/21720 | 23 | Hi. Peter Roquette sent me an email asking for an example of a quadratic space in characteristic 2 having certain features. I have no idea on this, but maybe someone reading this does.
He would like an example of a field $K$ of characteristic 2 with the following two properties:
(1) The quaternion algebras over $K$ form a group (within the Brauer group of $K$).
(2) There exists an anisotropic quadratic form of
dimension > 4 over $K$ that is "completely regular in the sense of Arf". That means: there is a quadratic space $(V,Q)$ over $K$ such that
(a) $\dim(V) > 4$
(b) if $Q(v) = 0$ then $v = 0$
(c) for the associated bilinear form $B(v,w) = Q(v+w) - Q(v) - Q(w)$, if
$B(v,w) = 0$ for all $w$ in $V$ then $v = 0$.
Here's the background. Roquette has written a paper with Falko Lorenz on the historical development of the Arf invariant, and they include in the paper a counterexample to the method of proof of a theorem of Arf. (The paper is at Roquette's homepage at <http://www.rzuser.uni-heidelberg.de/~ci3/arf.pdf> and it has also appeared in "Mathematische
Semesterberichte" vol. 57 (2010) pp. 73--102.) Their counterexample to Arf's method of proof is not actually a counterexample to his main theorem. In order to find a counterexample to the main theorem itself they want a quadratic space with the properties above (in characteristic 2). Roquette has asked around but nobody has yet been able to give him an example.
EDIT (Aug. 16): After Roquette learned about the answer posted below, he has posted on his website <http://www.rzuser.uni-heidelberg.de/~ci3/manu.html> a paper (number 46) with Lorenz which discusses the solution to his question in the context of Arf's paper. Anyone interested in this topic is encouraged to look at the paper. He wrote to me "Your idea to put my question on the website Math Overflow has worked wonderfully."
| https://mathoverflow.net/users/3272 | Wanted: Quadratic Space in Characteristic 2 as a Counterexample to a Theorem of Arf | I forwarded this question to Detlev Hoffmann, who says that such examples exist. Specifically, you can produce such an example where there is, say, an anisotropic form of dimension 8 using characteristic 2 analogues of the techniques in Merkurjev's 1992 article "Simple algebras and quadratic forms". He says details can be found in the PhD thesis of his student Frederic Faivre at Université Franche-Comté.
Detlev further explained the source of Arf's confusion, which I will now recap. Over a field $F$ of any characteristic, the tensor product of two quaternion algebras is not a division algebra if and only if the two quaternion algebras contain a common quadratic extension. In characteristic different from 2, this is an essentially complete criterion. But in characteristic 2, one has to wonder if this quadratic extension is separable or inseparable.
Draxl showed that if two quaternion algebras contain a common quadratic extension, you can always find one that is separable. That is, the property of containing a common inseparable extension is much stronger (because it implies that they contain a common separable extension). A nice exposition of this can be found in T.Y. Lam's 2002 article "On the linkage of quaternion algebras".
Detlev asserts: If you demand that every pair of quaternion division algebras over $F$ (of characteristic 2) share an inseparable quadratic extension, then there are no anisotropic regular quadratic forms of dimension > 4. "This is essentially due to Baeza. In fact, in some sense even to Arf, except that he didn't realize there's a difference between [sharing separable and inseparable subfields]." So presumably this is what Arf was claiming to have proved.
In contrast to this, the requirement that the quaternion algebras form a subgroup of the Brauer group is just that every pair of quaternion division algebras share a separable quadratic extension. This is a much weaker hypothesis, and allows for examples of fields like in Faivre's thesis.
Here are some precise references for the theorem "every pair of quaternion algebras over $F$ share an inseparable quadratic extension implies every regular quadratic form of dimension > 4 is isotropic":
* Arf (with confusion mentioned above): Satz 11 in "Untersuchungen über quadratische Formen in Körpern der Charakteristik 2. I." J. reine angew. Math. 183, 148-167 (1941)"
* Baeza: Theorem 3.1 in "Comparing $u$-invariants of fields of characteristic $2$." Bol. Soc. Brasil. Mat. 13 (1982), no. 1, 105--114.
* Faivre's thesis: Proposition 3.3.5 (with complete proof)
| 17 | https://mathoverflow.net/users/6486 | 26772 | 17,529 |
https://mathoverflow.net/questions/26715 | 7 | $\underline{Background}$ :
Suppose $\tau$ is a preadditive category and $R$ a ring. Then one may form a new preadditive category $\tau \otimes R$ in the following way:
$\tau \otimes R$ has the same objects as $\tau$ and for objects $A$ and $B$, set $\tau \otimes R(A,B) := \tau(A,B)\otimes\_{\mathbb{Z}} R$. Composition is given by $\tau(A,B)\otimes\_{\mathbb{Z}} R\otimes\_{\mathbb{Z}}\tau(B,C)\otimes\_{\mathbb{Z}} R \cong \tau(A,B)\otimes\_{\mathbb{Z}}\tau(B,C)\otimes\_{\mathbb{Z}}R\otimes\_{\mathbb{Z}}R \rightarrow \tau(A,C) \otimes\_{\mathbb{Z}}R$, where the last arrow is given by the composition in $\tau$ and multiplication in $R$.
$\tau \otimes R$ is additve if $\tau$ is additive.
$\underline{Question}$ : Suppose $\tau$ is triangulated, is there a (canonical) structure of a triangulated category for $\tau \otimes R$?
Actually, I only need the case $R := \mathbb{Z}[\frac{1}{n}, \theta] \subset \mathbb{C}$, where $\theta$ is a primitive root of unity of order $n$ for some $n \in \mathbb{N}$, but I would also be interested in a general statement/counterexample to the general case.
Thank you for reading.
| https://mathoverflow.net/users/6474 | Is the tensorproduct of a triangulated category with a ring again triangulated? | I would imagine it is false in general that given a triangulated category $T$ the category $T\otimes R$ is also triangulated.
The following is a concrete counterexample. Consider $D^b(\mathbb{Z})$ and let $T = D^b(\mathbb{Z})\otimes \mathbb{Z}[x]$. In order for $T$ to be triangulated the morphism $\mathbb{Z} \stackrel{x}{\to} \mathbb{Z}$ would need to be able to be completed to a distinguished triangle
$$\mathbb{Z}\stackrel{x}{\to}\mathbb{Z}\to C\_x \to \Sigma\mathbb{Z} $$
in $T$. Suppose that such a triangle existed and consider the exact sequence obtained by application of $Hom\_T(\mathbb{Z},-)$
$$0 \to Hom\_T(\mathbb{Z}, \Sigma^{-1}C\_x) \to \mathbb{Z}[x] \stackrel{x}{\to} \mathbb{Z}[x] \to Hom\_T(\mathbb{Z}, C\_x) \to 0 $$
As multiplication by $x$ has no kernel we have
$$Hom\_T(\mathbb{Z}, \Sigma^{-1}C\_x) = 0$$
and the object $C\_x$ would have to satisfy
$$\mathbb{Z} \cong Hom\_T(\mathbb{Z},C\_x) = Hom\_{D^b(\mathbb{Z})}(\mathbb{Z}, C\_x)\otimes\_\mathbb{Z} \mathbb{Z}[x] $$
But decomposing $C\_x$ as a sum of suspensions of free groups and torsion groups this is clearly not possible.
One situation in which something like what you ask for works is the following: let $(T,\square,\mathbf{1})$ be a tensor triangulated category (that is we have a symmetric monoidal structure on $T$ which is exact in each variable). Given a multiplicative subset of even elements $S$ of $End^\*(\mathbf{1})$ one can localize the hom-sets of $T$ at $S$. This can even be realised as a Verdier quotient by a certain subcategory. This is shown in Balmer's "Spectra, spectra, spectra - Tensor triangular spectra versus Zariski spectra of endomorphism rings" section 3.
| 8 | https://mathoverflow.net/users/310 | 26774 | 17,531 |
https://mathoverflow.net/questions/26776 | 32 | The total space of cotangent bundle of any manifold $M$ is a symplectic manifold.
Is it true/false/unknown that for any $M$, $T^\*M$ has Kähler structure?
Please support your claim with reference or counterexample.
| https://mathoverflow.net/users/5259 | Kähler structure on cotangent bundle? | This is true! I assume $M$ compact.
**Method 1.** Real algebraic geometry. Cf. [Fukaya, Seidel, and Smith - Exact Lagrangian submanifolds in simply-connected cotangent bundles](http://arxiv.org/abs/math/0701783). By a version of the Nash–Tognoli embedding theorem, one can realise $M$ as an real affine algebraic variety $V\_\mathbb{R}$, cut out by polynomials $f\_i \in \mathbb{R}[x\_1,\dotsc,x\_N]$. The complex variety $V\_\mathbb{C}$ will then be smooth in a small neighbourhood $U$ of $V\_\mathbb{R}$, hence Kaehler in that region, with $V\_{\mathbb{R}}$ as a Lagrangian submanifold. But $U$ is diffeomorphic to $T^\ast M$. The resulting symplectic structure on $T^\ast M$ may be non-standard; via the Lagrangian neighbourhood theorem, you can take the symplectic form to be the canonical one if you'll settle for a Kaehler structure only near the zero-section.
**Method 2.** Eliashberg's existence theorem for Stein structures. See Cieliebak–Eliashberg's unfinished book, [Symplectic geometry of Stein manifolds](http://math.stanford.edu/~eliash/Public/Site/Eilenberg_Lectures_files/stein7.pdf), Theorem 9.5. We observe that $T^\ast M$ has an almost complex structure $J$ (one compatible with the canonical 2-form, for instance) and a bounded-below, proper Morse function $\phi$ whose critical points have at most the middle index (namely, the norm-squared plus a small multiple of a Morse function pulled back from $M$). In this situation Eliashberg, via an amazing chain of deformations, finds an integrable complex structure $I$ homotopic to $J$ such that $dd^c \phi$ is non-degenerate. This makes $T^\ast M$ Stein! His theorem only applies in dimensions $\geq 6$ (the paper [Constructing Stein manifolds after Eliashberg](http://arxiv.org/abs/0810.4511) of Gompf explains what you have to check in dimension 4), so without doing those checks or appealing to other methods, the case of $M$ a surface is left out.
I think that the more precise version of Eliashberg's theorem, which may not yet be in the book, would tell us that the Stein structure is homotopic to an easy-to-write-down Weinstein structure on $T^\ast M$ involving its canonical symplectic structure $\omega\_\text{can}$, hence that $dd^c\phi$ is symplectomorphic to $\omega\_\text{can}$.
| 41 | https://mathoverflow.net/users/2356 | 26781 | 17,536 |
https://mathoverflow.net/questions/26790 | 3 | Suppose we have a finite $\sigma$ -field $S$, of which $A$ and $B$ are member sets. Since $S$ is closed under union and complementation [by definition], it follows that $(A' \cup B')' = (A \cap B)' \in S$. From closure under complementation, we have that $A \cap B \in S$, implying that
$S$ is closed under intersections.
Does it follow that every finite $\sigma$ -field is a topology?
| https://mathoverflow.net/users/6495 | Finite Topology vs sigma Field | Yes, it is also a topology on its union, the largest member of $S$. Since $S$ is finite, the arbitrary union rquirement amounts to finite union, which you have.
In fact,$S$ is a Boolean algebra, and since it is finite, it is isomorphic to a powerset algebra---the power set of the atoms of $S$ (the minimal non-empty elements of $S$). Every set in $S$ is a union of a finite number of these atoms.
If these atoms are singletons, then $S$ will be the discrete topology on the underlying set. If not, however, then $S$ will clearly not separate those points, and so will not be Hausdorff and so on.
| 6 | https://mathoverflow.net/users/1946 | 26792 | 17,541 |
https://mathoverflow.net/questions/26783 | 6 | Where can I find a comprehensive treatment of this important result at the level of a very advanced undergraduate/beginning graduate student? What works develop the relevant material in a cohesive and readable way? Assume only a familiarity with basic homological algebra and ring theory on the part of the reader in assessing this question, please. Thank you!
| https://mathoverflow.net/users/6131 | Best exposition of the Proof of the Hilbert Syzygy Theorem by Eilenberg-Cartan | The place I learned it from is Chapter 19 of Eisenbud's *Commutative Algebra*. Most of the proofs in that section do not use material from previous chapters if I recall correctly. The route (which I think is what you are looking for) is to construct the Koszul complex of the residue field of a regular (graded) local ring and also prove the symmetry of the Tor functor, and then use these two facts to get finite global dimension which implies Hilbert's syzygy theorem.
| 7 | https://mathoverflow.net/users/321 | 26798 | 17,542 |
https://mathoverflow.net/questions/26797 | 15 | Let $K$ be a finite extension of $\mathbb{Q}\_p$ and $E$ an elliptic curve over $K$ with good ordinary reduction.
The p-adic Tate module $T\_p(E)$ is (after tensoring with $\mathbb{Q}\_p$) a 2-dimensional $\mathbb{Q}\_p$-representation of $\mathop{\mathrm{Gal}}(\bar{K}/K)$.
It is reducible: the kernel of reduction to the residue field is an invariant line.
Does $T\_p(E) \otimes\_{\mathbb{Z}\_p} \mathbb{Q}\_p$ contain another invariant line?
| https://mathoverflow.net/users/1046 | Does the p-adic Tate module of an elliptic curve with ordinary reduction decompose? | Serre has shown that there exists a complementary subspace invariant under the Lie algebra $\mathfrak{g}$ *if and only if* E has complex multiplication. Otherwise the image of Galois is open in the Borel subgroup of $\operatorname{GL}\_2(\mathbb{Q}\_p)$. I learnt this from the paper by Coates and Howson ("Euler characteristics and elliptic curves II", beginning of section 5); they reference Serre's book "Abelian l-adic representations and elliptic curves", but I don't have a copy of that to hand right now to check.
| 17 | https://mathoverflow.net/users/2481 | 26800 | 17,543 |
https://mathoverflow.net/questions/26788 | 6 | Suppose a deck of cards consists of $a\_1+a\_2+\cdots+a\_k$ cards of $k$ types, where there are $a\_i$ indistinguishable cards of each type. How many shuffles does it take, on average, to randomize the deck? For example, $a\_i=4$ and $k=13$ gives a standard deck of playing cards; $a\_i=4$ for $1\le i\le9$, $i\_{10}=16$, $k=10$ gives the cards for blackjack.
In general I would expect that this would be an easier task than shuffling a deck where all cards are indistinguishable. A standard deck has about 226 bits of entropy, while the same deck without suits has only 166 bits of entropy.
---
Consider a standard deck of 52 playing cards. Bayer & Diaconis (famously) showed that, under a certain model, it takes 7 riffle shuffles to sufficiently randomize the deck. Salem applies a different methodology and gets 6 idealized riffle shuffles. Mann uses a much stricter measurement and determines 11.7 as the expected stopping time for the riffle unshuffle (and thus the riffle shuffle) to randomize the deck.
In particular, Mann gives a formula:
$$E(\mathbf{T})=\sum\_{k=0}^\infty\left[1-{2^k\choose n}\frac{n!}{2^{nk}}\right]$$
which lends itself to generalization nicely.
I'm partial to Mann's method, but my question applies broadly.
---
[1] David Aldous and Persi Diaconis, "[Shuffling cards and stopping times](http://www-stat.stanford.edu/~cgates/PERSI/papers/aldous86.pdf)", *The American Mathematical Monthly* **93**:5 (1986), pp. 333-348.
[2] Dave Bayer and Persi Diaconis, "Trailing the dovetail shuffle to its lair", *The Annals of Applied Probability* **2**:2 (1992), pp. 294-313. [JSTOR: 2959752](http://www.jstor.org/stable/2959752)
[3] Brad Mann, [How many times should you shuffle a deck of cards?](http://www.dartmouth.edu/~chance/teaching_aids/Mann.pdf)
[4] Michael P. Salem, [How many shuffles are necessary to randomize a standard deck of cards?](http://cosmos.phy.tufts.edu/~salem/Documents/shuffle.pdf)
| https://mathoverflow.net/users/6043 | Shuffling decks of cards where not all cards are distinguishable | A place to start is the recent preprint:
[Riffle shuffles of a deck with repeated cards](https://arxiv.org/abs/0905.4698)
Sami Assaf, Persi Diaconis, K. Soundararajan.
They also have another preprint about reducing this to a "rule of thumb", and cite some earlier work of Conger and Viswanath, available [here](http://www.math.lsa.umich.edu/%7Edivakar/).
Pulling carelessly from tables in the different papers without checking exactly what I'm saying, it seems that in the total variation distance, (the measure which yields seven shuffles for a standard deck), four shuffles are already sufficient for blackjack, while for the separation distance (the measure yielding eleven shuffles for a standard deck), it still takes nine shuffles.
| 7 | https://mathoverflow.net/users/1102 | 26808 | 17,546 |
https://mathoverflow.net/questions/26810 | 6 | Are there examples of subshifts (that is, closed shift-invariant subsets of the full shift {$1...n$}${}^{\mathbb{Z}}$) on which the shift is topologically mixing, which admit a shift-invariant probability measure of full support, but no invariant ergodic measure of full support ?
I guess the answer is no, but I can't locate a reference.
Also I would be interested in a topologically mixing subshift that has no invariant probability measure of full support.
| https://mathoverflow.net/users/6129 | topologically mixing subshifts without ergodic measures | A construction of a topologically transitive mixing subshift with a fully supported invariant measure, but no fully supported ergodic measure, is given by Benjamin Weiss in the article "Topological transitivity and ergodic measures", *Mathematical Systems Theory* 1971. It gives a direct combinatorial construction, with $n=2$.
It is easy to give an example of a topologically transitive subshift on two symbols with no fully supported measure: just take the orbit closure under the shift of the sequence with all negative entries equal to one, and all non-negative entries equal to two.
| 8 | https://mathoverflow.net/users/1840 | 26813 | 17,547 |
https://mathoverflow.net/questions/26823 | 1 | Trying to solve for the area enclosed by $x^4+y^4=1$. A friend posed this question to me today, but I have no clue what to do to solve this. Keep in mind, we don't even know if there is a straightforward solution. I think he just likes thinking up problems out of thin air.
Anyway, the question becomes more general, since we *think* that
$lim\_{n\to\infty}\int\_0^1{(1-x^n)^{1/n}} = {1\over4}$ (it approaches a square / becomes linear)
can anyone confirm that this is true or not?
| https://mathoverflow.net/users/5562 | Area enclosed by x^4 + y^4 = 1 | I always prefer not to skip $dx$:
$$
I\_n=\int\_0^1(1-x^n)^{1/n}dx.
$$
After the change of variable $t=x^n$, the integral becomes the beta integral,
$$
I\_n=\frac1n\int\_0^1(1-t)^{1/n}t^{1/n-1}dt
=\frac1n\frac{\Gamma(1+1/n)\Gamma(1/n)}{\Gamma(1+2/n)}
=\frac1n\frac{\Gamma(1/n)^2\cdot 1/n}{\Gamma(2/n)\cdot 2/n}
\to1 \quad\text{as $n\to\infty$},
$$
as $1/\Gamma(z)\sim z$ as $z\to 0$.
| 7 | https://mathoverflow.net/users/4953 | 26825 | 17,551 |
https://mathoverflow.net/questions/26815 | 12 | What is the idea behind deformation to the normal cone and what are examples of its applications?
| https://mathoverflow.net/users/nan | deformation to the normal cone | Here's a place to see the normal cone side-by-side with other familiar constructions, that I learned from Fulton's "Intersection Theory". Here $X \subset Y$.
Start with the space $Y \times {\mathbb P}^1$, thought of as a trivial family over ${\mathbb P}^1$. Blow up the subscheme $X \times \infty$. Now we still have a flat family over ${\mathbb P}^1$, in which all the fibers except the one over $\infty$ are still copies of $Y$. The fiber over $\infty$ is reducible: one piece $Z\_1$ is the blowup of $Y$ along $X$, and the other is the projective completion $Z\_2$ of the normal cone to $X$ inside $Y$. These intersect along the projectivization of the normal cone, which appears in $Z\_1$ as the exceptional divisor, and in $Z\_2$ as the stuff added in projective completion.
(An example, for people who like polytopal pictures of toric varieties: let $Y$ be ${\mathbb P}^2$, pictured as a triangle, and $X$ be a point, pictured as a vertex. Then $Y \times {\mathbb P}^1$ is pictured as a triangular prism, and its blowup by cutting a corner off that prism. The fiber over $\infty$ is then pictured as cutting that triangle into a trapezoid $Z\_1$ union a small triangle $Z\_2$, glued along an interval, matching the decomposition above.)
So, if perhaps you don't like degenerating $Y$ (which may be complete) to something noncomplete, you can complete it by including $Z\_1$. The normal cone is $Z\_2 \setminus Z\_1$.
Two other comments. On $Y \times {\mathbb P}^1$ there is a circle action, dilating the ${\mathbb P}^1$. It acts trivially on the $0$ and $\infty$ fibers, moving the rest around. When we blow up $X \times \infty$, the circle action on the new $\infty$ fiber $Z\_1 \cup Z\_2$ is nontrivial on $Z\_2$; it is the dilation of the cone fibers. This is one place to lay blame for the existence of this "cone" structure.
Finally, there's a conformally equivalent way to think about the $Y$ to $Z\_1 \cup Z\_2$ degeneration, at least when $X$ and $Y$ are smooth. Pick a small tubular neighborhood (nonalgebraic!) around $X$, with boundary $S$, a sphere bundle over $X$. Let the metric on $Y$ get very looong nearby $S$, in directions passing through $S$. You might say that $X$ is falling into a black hole, with $S$ the Schwarzschild boundary. In the limit, it gets infinitely long, and $X$ has bubbled off into its own universe.
| 20 | https://mathoverflow.net/users/391 | 26827 | 17,552 |
https://mathoverflow.net/questions/26814 | 1 | I know that the fibre of $A\_{g,n}$ over $\mathbf{F}\_p$ is quasi-projective (of what dimension?). Can one exhibit some smooth projective subvarieties of high dimension in it? What are references for its geometry?
| https://mathoverflow.net/users/nan | projective subvarieties of the moduli space of abelian varieties | The dimension is $g(g+1)/2$. The supersingular locus gives a large projective subvariety but I don't recall whether it is smooth or not. For references, look up the many papers of F. Oort.
| 5 | https://mathoverflow.net/users/2290 | 26835 | 17,556 |
https://mathoverflow.net/questions/10048 | 8 | Does anyone know if there is something that can be said (ideally under at most very mild hypotheses) in group cohomology (let's even restrict to degree 1) that is similar to Serre's twisting, but in the case where the thing you want to twist by is not inner? I'm not expecting the statement to hold word-for-word, just that there is some general and reasonably sharp relationship between the original cohomology set and the twisted one. Thanks!
**Edit** in response to Chris Schommer-Pries:
A complete discussion of Serre's twisting can be found in section 5.3 of his book "Galois Cohomology". It's both a "process" and a proposition I guess. It goes as follows. Start with a group $G$ and a (not necessarily abelian) group $A$ on which $G$ acts, and form the pointed set $H^1(G,A)$. Now pick a $1$-cocycle $c \in H^1(G,A)$. Let $A\_c$ denote the same abstract group $A$, but now it has the "$c$-twisted" action defined by $g\*a = c(g) \cdot (g \cdot a) \cdot c(g)^{-1} $ (the 1st and 3rd $\cdot$ are operations in $A$ and the 2nd $\cdot$ is the original action of $G$). That's the "process". The proposition (cf. Prop 35 bis in the book) is that $d \mapsto d \cdot c $ (product of functions) induces a bijection of pointed sets $ H^1(G,A\_c) \cong H^1(G,A)$ (note that the first set has a different basepoint than the second). It is frequently used to show that some "monomorphism of pointed sets" is actually injective as a function.
**Edit** in response to Mariano Suárez-Alvarez:
By "inner" above I am referring to the fact that, in Serre twisting, the action of $G$ on $A$ was modified by conjugating by an element of $A$. By "outer", I mean the case where you instead replaced the action by $g\*a = f(g)( g \cdot a ) $ where $f : G \rightarrow Aut( A )$ is a prescribed cocycle ($Aut( A )$ has the conjgation action using the original action of $G$ on $A$).
| https://mathoverflow.net/users/2047 | substitute for Serre's twisting when the "twisting" is outer | The short answer is that there is not much to say about the relationship between $H^1(G, B)$ and a twist $H^1(G, B\_c)$ where $c$ is a cocycle taking values in $Aut(B)$. (I am going to write $B\_c$ for the twist instead of Serre's notation $\_cB$ for the sake of easy typesetting.) You can get a good feel for what is possible by soaking in sections I.5.7 and III.1.4 of Serre's *Galois Cohomology*.
Section I.5.7
-------------
One thing you *can* do -- as exhibited in I.5.7 -- is twist all three terms in a short exact sequence of $G$-modules and get a new short exact sequence, assuming the obvious compatibility conditions hold. Serre starts with an exact sequence
>
> $1 \to A \to B \to C \to 1$
>
>
>
where $A$ is assumed central in $B$. Then he fixes a 1-cocycle $c$ with values in $C$ and twists to get an exact sequence
>
> $1 \to A \to B\_c \to C\_c \to 1$.
>
>
>
Note that this twist $B\_c$ is *not* an inner twist of $B$, because $c$ need not be in the image of $H^1(G, B) \to H^1(G, C)$.
This may look like a lame example, in that the twist of $B$ is "pretty close" to being inner. But already here you don't have any results regarding a connection between $H^1(G, B)$ and $H^1(G, B\_c)$. That's a pretty fuzzy statement; Serre says as much as you can say with precision in Remark 1: "it is, in general, false that $H^1(G, B\_c)$ is in bijective correspondence with $H^1(G, B)$."
Section III.1.4
---------------
This section discusses your question for the specific case where $G$ is the absolute Galois group of a field $k$ and $B$ is the group of $n$-by-$n$ matrices of determinant 1 with entries in a separable closure of $k$. Serre explains what you get as $B\_c$ when you twist $B$ by a cocycle with values in $Aut(B)$. You can get, for example, a special unitary group.
You can find explicit descriptions of $H^1(G, B\_c)$ for some $B\_c$'s in *The Book of Involutions*, pages 393 (Cor. 29.4) and 404 (box in middle of page). Note that for $B\_c$ as in Section I.5.7, $H^1(G, B\_c)$ is a group (a nice coincidence) but in the case where you get a true special unitary group, $H^1(G, B\_c)$ does not have a reasonable group structure--it is just a pointed set like you expect.
| 6 | https://mathoverflow.net/users/6486 | 26836 | 17,557 |
https://mathoverflow.net/questions/26838 | 6 | Let $p$ be an odd prime. An old theorem of Jacobi asserts that $p$ has exactly $8(p+1)$ representations as a sum of four squares of integers (solutions counted with order and sign). What is the most effective way to enumerate these solutions computationally? Can it be done in time $p^{1+\varepsilon}$, or even in time $p (\log{p})^A$?
| https://mathoverflow.net/users/1464 | Enumerating representations of an integer as a sum of squares | Form the set $S$ of all squares less than $p$. This has $O(\sqrt{p})$ elements, and writing them down takes $O(\sqrt{p} \log p)$ time. (You don't have to implement fast multiplication to do this; just compute the list of squares by successively adding odd numbers.)
Let $T$ be the set of all integers expressible as the sum of two elements of $S$. This has $O(p)$ elements, and takes $O(p \log p)$ steps to write down.
Sort $T$ and sort $p-T$. This is $O(p \log p)$ steps each. Find all duplicates between the lists $T$ and $p-T$; this takes $O(p)$ steps because they are already sorted.
All in all, $O(p \log p)$ steps, the same size as the output.
| 8 | https://mathoverflow.net/users/297 | 26841 | 17,560 |
https://mathoverflow.net/questions/26832 | 60 | This is an elementary question (coming from an undergraduate student) about algebraic numbers, to which I don't have a complete answer.
Let $a$ and $b$ be algebraic numbers, with respective degrees $m$ and $n$. Suppose $m$ and $n$ are coprime. Does the degree of $a+b$ always equal $mn$?
I know that the answer is "yes" in the following particular cases (I can provide details if needed) :
1) The maximum of $m$ and $n$ is a prime number.
2) $(m,n)=(3,4)$.
3) At least one of the fields $\mathbf{Q}(a)$ and $\mathbf{Q}(b)$ is a Galois extension of $\mathbf{Q}$.
4) There exists a prime $p$ which is inert in both fields $\mathbf{Q}(a)$ and $\mathbf{Q}(b)$ (if $a$ and $b$ are algebraic integers, this amounts to say that the minimal polynomials of $a$ and $b$ are still irreducible when reduced modulo $p$).
I can also give the following reformulation of the problem : let $P$ and $Q$ be the respective minimal polynomials of $a$ and $b$, and consider the resultant polynomial $R(X) = \operatorname{Res}\_Y (P(Y),Q(X-Y))$, which has degree $mn$. Is it true that $R$ has distinct roots? If so, it should be possible to prove this by reducing modulo some prime, but which one?
Despite the partial results, I am at a loss about the general case and would greatly appreciate any help!
[EDIT : The question is now completely answered (see below, thanks to Keith Conrad for providing the reference). Note that in Isaacs' article there are in fact two proofs of the result, one of which is only sketched but uses group representation theory.]
| https://mathoverflow.net/users/6506 | Degree of sum of algebraic numbers | The following answer was communicated to me by Keith Conrad:
See:
>
> M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25
> (1970), 638--641.
>
>
>
>
> [http://alpha.math.uga.edu/~pete/Isaacs70.pdf](http://alpha.math.uga.edu/%7Epete/Isaacs70.pdf)
>
>
>
Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are
relatively prime, then
$K(a,b)$ = $K(a+b)$, which answers the students question in the
affirmative. His proof shows the same conclusion holds under the
weaker assumption that
$[K(a,b):K] = [K(a):K][K(b):K]$.
since Isaacs uses the relative primality assumption on the degrees
only to get that degree formula above, which can occur even in cases
where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.
| 41 | https://mathoverflow.net/users/1149 | 26859 | 17,572 |
https://mathoverflow.net/questions/26857 | 11 | $\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SW{SW}$Let $S\_k$ be the symmetric group. Let $F$ be an algebraically closed field. Let $\Rep(S\_k)$ be the category of representations of $S\_k$ over $F$.
Let $\Rep(\GL\_n(F))$ be the category of algebraic representations of $\GL\_n(F)$.
We can construct a functor $\SW$ from $\Rep(S\_k)$ to $\Rep(\GL\_n(F))$, $\SW(\sigma)=(\otimes^k F^n\otimes \sigma)^{S\_k}$, where $\sigma\in \Rep(S\_k) $.
When $F$ is of characteristic 0, and n>k, it is well-known $\SW$ is a fully faithful and exact functor. Usually it is called Schur-Weyl duality.
My question is:
* When $F$ is of characteristic $p$, is
$\SW$ still a fully faithful and exact
functor?
When $p>k$, the representation theory of $S\_k$ over $F$ behaves exactly the same as characteristic 0 case, but for algebraic representation of $\GL\_n(F)$, it is totally different from characteristic 0 case.
When $p<=k$, representation of $S\_k$ is complicated.
It is well-known problem, to determine the decomposition number in $\Rep(S\_k)$.
* Is it possible to use Schur-Weyl
duality to determine the decomposition number for $S\_k$?
Since for modular representation theory of reductive group, the similar problem is known or almost known by Kazhdan-Lusztig polynomial.
TO sum up, I would like to ask:
* For my purpose, what is the correct formulation for
Schur-Weyl duality in positive
characteristic?
| https://mathoverflow.net/users/5082 | Schur-Weyl duality in positive characteristic | The questions here have certainly been explored (though not definitively) in many recent papers or preprints on arXiv. Look for example at the arXiv paper by Stephen Doty [Link](https://arxiv.org/abs/math/0610591), as well as many others by Steve and/or his collaborators. Most of the arXiv papers have subject listing RT (some also consider quantum analogues under QA). But some predate arXiv; there has been a lot of study of decomposition numbers of symmetric groups in prime characteristic, for example, using what little is known about modular representations of GL$\_n$. Not having gone far with this literature myself, I'd suggest that you start the inquiry with available papers and then maybe raise narrower questions here.
My main point at first has been that a lot of literature exists from the past couple of decades, so the questions should focus on what is in that literature (not just Doty's short conference paper I cited). Concerning the status of modular representations for finite groups of Lie type, what's known is not yet good enough to answer most questions about symmetric groups for small primes. While Lusztig's conjectures promise a good conceptual picture for primes at least the Coxeter number of the Weyl group, even that much can be implemented only in recursive style. For small primes little is known, but it would have immediate applications to $S\_n$. In classical Schur-Weyl duality the dictionary goes the opposite way.
| 5 | https://mathoverflow.net/users/4231 | 26862 | 17,573 |
https://mathoverflow.net/questions/26861 | 17 | Is it possible to construct (without using Axoim of Choice) a totally ordered set S with cardinality larger than $\mathbb{R}$?
**Motivation:** A total ordering is often called a “linear ordering”. I have heard the following explanation: “If you have a total ordering on a set S, you can plot the set on the real line such that elements to the right are greater than elements to the left”. Formally this means that there exist a function $\phi:S\rightarrow \mathbb{R}$ such that for all $ a$ ,$b\in S$, x < y $ \Leftrightarrow \phi(x)$ < $\phi(y)$. This is of course correct if the set is finite or countable (and it gives a good intuition on what a total ordering is), but obviously not if $|S|>|\mathbb{R}|$, and using the axiom of choice it is easy to “construct” a total ordering on, say, the power set of $\mathbb{R}$. But I would prefer to have a more concrete counterexample, and this is why I asked myself this question.
Later I realized that is was possible to construct a total ordering on a set $|S|=|\mathbb{R}|$, such that no such function $\phi$ exist, but I still think that the above question is interesting.
| https://mathoverflow.net/users/2097 | Explicit ordering on set with larger cardinality than R | Yes. By [Hartogs' theorem](http://en.wikipedia.org/wiki/Hartogs_number), there is an ordinal that has no injection into $R$. The minimal such ordinal is the smallest well-ordered cardinal not injecting into $R$. It is naturally well-ordered by the usual order on ordinals. None of this needs AC.
One can think very concretely about the order as follows: Consider all subsets of $R$ that are well-orderable. By the Axiom of Replacement, each well-order is isomorphic to a unique ordinal. Let $\kappa$ be the set of all ordinals that inject into $R$ in this way. One can show that $\kappa$ itself does not inject into $R$, and this is the Hartogs number for the reals.
More generally, of course, there is no end to the ordinals, and they are all canonically well-ordered, without any need for AC.
But in terms of the remarks in your "motivation" paragraph, there are linear orders that do not map order-preservingly into $R$ that are not larger than $R$ in cardinality. For example, the ordinal $\omega\_1$ cannot map order-preservingly into $R$, since if it did so, then there would be an uncountable family of disjoint intervals (the spaces between the successive ordinals below $\omega\_1$), but every such family is countable by considering that the rationals numbers are dense. Another way to see this is to observe that the real line has countable cofinality for every cut, but $\omega\_1$ has uncountable cofinality.
Lastly, there is a subtle issue about your request that the order by "larger than $R$". The examples I give above via Hartog's theorem are not technically "larger than $R$", although they are not less than $R$ in size. The difficulty is that without AC, the cardinals are not linearly ordered, and so these two concepts are not the same. But you can turn the Hartogs argument into a strict example of what you requested by using the lexical order on $R\times\kappa$, where $\kappa$ is the Hartog number of $R$. This order is strictly larger than $R$ in size, and it is canonically linearly ordered by the lexical order.
| 20 | https://mathoverflow.net/users/1946 | 26864 | 17,575 |
https://mathoverflow.net/questions/26860 | 8 | It might be a stupid question.
How to glue perverse sheaves? I am considering the following example. Flag variety of $sl\_2$, which is $P^1$. Consider category of perverse sheaves on $P^1$. Denoted by $Perv(P^1)$. The big cells of flag variety of $sl\_2$ give the affine cover for it. In this case, they should be two $A^1$. We can also consider category of perverse sheaves on $A^1$. My question is:
1. How to glue two pieces of perverse sheaves on 1-dimensional affine spaces to that of projective spaces? More general, is there any gluing machinery which can globalize the perverse sheaves? It seems that Beilinson had a paper talking about this, but what I preferred is some expository notes explaining with some examples.
2. Maybe I need to ask this in another question. How can one define perverse sheaves on noncommutative space. I am aware that there is a paper by Amnon Yekutieli, James J. Zhang talking about [perverse sheaves on noncommutative space](https://arxiv.org/abs/math/0211309 "Amnon Yekutieli, James J. Zhang. Dualizing Complexes and Perverse Sheaves on Noncommutative Ringed Schemes. 2006. https://zbmath.org/?q=an:1137.14300"). However, what they considered was not really a noncommutative space from my understanding. (If I made mistake or bullshit, point out please). They consider quasi coherent sheaves of (not necessarily) commutative algebra on commutative scheme. Which does not fit my need. I am considering the following example: Quantized flag variety of $sl\_2$ i.e. $Proj(O\_q(G/N))$ in the sense of [Lunts–Rosenberg](https://doi.org/10.1007/s000290050044 "Lunts, V., Rosenberg, A. Localization for quantum groups. Sel. math., New ser. 5, 123 (1999). https://zbmath.org/?q=an:0928.16020") (see also [Erik Backelin and Kobi Kremnitzer](https://arxiv.org/abs/math/0401108v2 "Erik Backelin, Kobi Kremnizer. Quantum flag varieties, equivariant quantum D-modules and localization of quantum groups. 2004. https://zbmath.org/?q=an:1165.17304") and [Tanisaki](https://arxiv.org/abs/math/0309349 "Toshiyuki Tanisaki. The Beilinson-Bernstein correspondence for quantized enveloping algebras. 2003. https://zbmath.org/?q=an:1082.17010")). It is a noncommutative scheme. I wonder whether James Zhang has also defined dualizing complexes for this case. **How to define category of perverse sheaves on quantized flag variety**?
The motivation for this question is I think there should be **quantum version of Riemann–Hilbert correspondence**. Which should describe the categorical equivalence:
$Perv(Proj(O\_q(G/N))$ and category of quantum holonomic D-modules on quantized flag variety.
At present, I have more interest to know the answer of question 1. Thank you!
| https://mathoverflow.net/users/1851 | Gluing perverse sheaves? | Beilinson's *How to glue perverse sheaves* explains how one can glue perverse sheaves on a variety from perverse sheaves on a closed subvariety and its open complement (assuming that the closed subvariety is the set of zeros of a regular function on the big variety).
Gluing perverse sheaves from an open covering is a very different problem. There is a paper by Kazhdan and Laumon *Gluing of perverse sheaves and discrete series representations*, devoted to the second problem. Further developments in this direction include the papers [arXiv:math/9811155](https://arxiv.org/abs/math/9811155 "Bezrukavnikov, Polishchuk: Gluing of perverse sheaves on the basic affine space") and [arXiv:math/0104114](https://arxiv.org/abs/math/0104114 "Bezrukavnikov, Braverman, Positselskii: Gluing of abelian categories and differential operators on the basic affine space"). The second one contains some discussion of the general categorical formalism.
| 7 | https://mathoverflow.net/users/2106 | 26872 | 17,581 |
https://mathoverflow.net/questions/26606 | 12 | Let $g \geq 2$, and consider the moduli space $\bar M\_{g,n}$ of stable *n*-pointed curves of genus *g*. There is a natural forgetful map to $\bar M\_g$, which forgets the markings and contracts any resulting unstable component. I am thinking about what the fibers of this map look like. Consider first the *n*-fold fibered product of the universal curve over $\bar M\_g$ -- this would be a moduli space of curves of genus *g* with *n* not necessarily distinct markings. We know that this product is singular at all points where more than one of the markings coincide, or when a marking is placed on a node of a singular base curve. Moreover, we know that $\bar M\_{g,n}$ is a desingularization.
Of course, the difference between the two spaces is that on $\bar M\_{g,n}$, whenever two markings or a marking and a node come together, one ''bubbles off'' a rational curve, and this can be seen as a blow-up of an ambient space in which the curve is embedded. My question is: can one in a similar way explicitly realize the total space $\bar M\_{g,n}$ as a sequence of blow-ups of the *n*-fold fibered product of the universal curve, or a similar construction?
And a related naive question, which may or may not be equivalent to the one above: given a point *[C]* in the interior of $\bar M\_g$, the fiber over it in $\bar M\_{g,n} \to \bar M\_g$ is a compactification of the configuration space of *n* distinct ordered points on *C*. Is this fiber isomorphic to the Fulton-MacPherson compactification of this configuration space?
| https://mathoverflow.net/users/1310 | The fibers of M_{g,n} \to M_g and the Fulton-MacPherson compactification | You don't mean that the forgetful map contracts any rational component, only those touching less than three nodes. Also, the n-fold fibered power is not singular where several markings coincide at a smooth point. After all, a fibered product of smooth morphisms is smooth. It will, however, disagree with $\overline{M}\_{g,n}$ at the locus where three or more markings coincide, so blow-ups will indeed be required.
However, what you wish for in the last paragraph is indeed true. One has to assume that the automorphism group of $C$ is trivial, for otherwise, even the fiber of $\overline{M}\_{g,1}$ over $\overline{M}\_g$ will be not $C$ but $C/ Aut C$. (Working with moduli stacks rather than moduli spaces would cure this problem.) But assuming this, the fiber of $\overline{M}\_{g,n} \to \overline{M}\_g$ is indeed the Fulton-MacPherson configuration space. Indeed, that space is described in the original F-Mac paper as a moduli space of stable configurations of distinct smooth points on a fixed curve $C$ with nodal trees of projective lines attached, modulo projective equivalence on the lines. Here "stable" means that the pointed nodal curve has finite automorphism group, that is, each line carries at least three distinguished points. Those are exactly the configurations that you see in your fiber.
As for your earlier question, it is true in some sense, but it would be quite subtle in practice. Fulton-MacPherson explain how to obtain their space from the product $C^n$ by an explicit sequence of blow-ups. One blows up first the "small diagonal" where all the points come together, and then proper transforms of other diagonals. You could try to blow up all the corresponding loci in the fibered power of $\overline{M}\_{g,n}$ over $\overline{M}\_g$.
But you would have to resolve other loci over the boundary too: if two markings coincide with a node, for example, you want to pull apart the node, glue in an extra line, and draw the two markings on that line. But since there is a 1-parameter family of ways to do this (due to the cross-ratio of the two nodes and the two markings), you want to replace the corresponding points in the fibered power with a $P^1$. Even worse, this would not be a blow-up in the obvious sense. The fibered product of two nodes, in local analytic coordinates, looks like $xy = t = wz$, that is, $xy-wz=0$, the cone on a smooth quadric surface. You can resolve this singularity by blowing up the origin, but then the exceptional divisor is $P^1 \times P^1$, not $P^1$ as you wish. Instead, you should perform one of the "small resolutions" obtained by blowing up one of the Weil divisors $x=w=0$ or $x=z=0$.
Since any birational morphism is the blow-up at some sheaf of ideals, in some abstract sense you are guaranteed that $\overline{M}\_{g,n}$ is obtained from the fibered power by blowing up. But that doesn't mean that you can give an explicit list of smooth centers to be blown up in turn. The example with the two markings at the node suggests that this will be a thorny problem.
| 14 | https://mathoverflow.net/users/6522 | 26873 | 17,582 |
https://mathoverflow.net/questions/26874 | 3 | Given a finite group G, its complexified ring of finite-dimensional complex representations is isomorphic to its algebra of class functions, by the trace map $\mathrm{Tr}\_\rho: g \mapsto \mathrm{Tr}(\rho(g))$. These class functions can in turn can be shown to correspond to the elements in the center of the group algebra, by sending a class function $c:G \to \mathbb{C}$ to the element
$\sum\_{g \in G} c(g) \cdot g$
in the group algebra.
This map from class functions to the group algebra isn't a ring homomorphism. In particular, it doesn't preserve the unit. However, I'm pretty sure that the representation ring and the center of the group algebra are nevertheless isomorphic as algebras. Am I right? Even better, is there a canonical such homomorphism with some group-theoretical importance?
| https://mathoverflow.net/users/799 | Ring homomorphism from the representation ring into group algebra? | Off the top of my head: the representation ring is really indexed by the set of irreps (well, OK, a set of representatives from equivalence classes thereof) and is indeed commutative and semisimple (at least over the complex numbers) so isomorphic to an appropriate direct product of copies of the ground field.
The centre of the group algebra is the (sub)algebra of class functions, so is most naturally indexed by the set of conjugacy classes. Again, it is commutative and semisimple so is isomorphic to a direct product of the appropriate number of copies of the ground field.
Thus, while I agree that the two algebras are isomorphic when you have a finite group, the isomorphisms seems quite heinously uncanonical -- if I am correct in recalling the maxim that there is no canonical bijection between the set of conjugacy classes and a representative set of irreps.
I also have the feeling that if there were some "canonical" homomorphism from the representative ring to the group ring in the case of finite groups, then there should be some analogue for compact groups or abelian groups. But this seems not to be the case: consider, for example, the group of integers with its usual additive structure.
| 6 | https://mathoverflow.net/users/763 | 26882 | 17,586 |
https://mathoverflow.net/questions/26155 | 10 | What is the necessary condition on a ring that guarantees the number of minimal non-zero ideals to be finite? Neither Noetherian or Artinian condition seems sufficient, and the ring being semisimple seems too strong.
| https://mathoverflow.net/users/5292 | Finite number of minimal ideals | (Inspired by Graham's comment) For simplicity I will consider the case $(R,m)$ is a Noetherian, local ring.
A non-zero minimal ideal better be principal. Also, if $(x)$ is such ideal, then for any $y\in m$, the ideal $(xy)$ has to be $0$, so $mx=0$.
Let $I= \{x\in R|mx=0\}$, the *socle* of $R$. Since $Im=0$, $I$ is a vector space over $k=R/m$. You want to know when the set of $1$-dimensional subspaces of $I$ is finite. This happens if and only if $\dim\_kI\leq 1$ or $k$ is a finite field.
If $R$ is also Artinian (i.e. $\dim R=0$), then $\dim\_kI\leq 1$ means precisely that $R$ is [Gorenstein](http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=0331DDC9E848DF83CBCB901AEAE62A4E?doi=10.1.1.53.7033&rep=rep1&type=pdf). So in this case (Noetherian, local of dimension $0$) one has a particularly nice answer: the set of nonzero minimal ideals is finite iff $R$ is Gorenstein or $k$ is finite.
Note that if $\dim R\geq 1$, then $I=0$ iff $\text{depth}\ R\geq 1$.
In general one can localize to get at least necessary conditions. For example, if the height of all maximal ideals is at least $1$, then $R$ satisfying [Serre's condition](https://mathoverflow.net/questions/22228/what-is-serres-condition-s-n-for-sheaves) $(S\_1)$ is certainly sufficient, since the socle when you localize at any maximal ideal will then be $0$, so there is no non-zero minimal ideals.
| 10 | https://mathoverflow.net/users/2083 | 26886 | 17,589 |
https://mathoverflow.net/questions/26892 | 42 | Most of us are at least somewhat curious about what's going on in areas of mathematics outside our own area of research. If a significant breakthrough is made, and can be stated in language that we understand, then we would enjoy hearing about it. Now, for truly fantastic breakthroughs like Fermat's Last Theorem or the Poincaré conjecture, the mathematical grapevine functions quite well and we all hear about it pretty quickly. For anything below that level, however, I for one feel that there is no good way to find out what is going on in fields other than my own. Every four years I seem to have the following conversation: "Did you hear? So-and-so won the Fields Medal!" "No. And, uh, who in the world is so-and-so?" Shameful.
What I'm wondering is, am I doing something wrong? Does anyone else have a better way of keeping abreast, even at a superficial level, of what major advances are happening in other areas of mathematics? At one time, *Mathematical Reviews* would select some articles or books for "Featured Reviews." I really enjoyed Featured Reviews and learned about many interesting results this way. However, MR stopped doing Featured Reviews after a while. No official reason was given, but I have heard that one reason was that some people were treating Featured Reviews as judgments as to which papers/books were "the best," and that MR did not want to accept the responsibility for making such judgments, especially if they were going to be used for tenure and promotion decisions. This is understandable, but unfortunate (from my point of view).
The series of books *What's Happening in the Mathematical Sciences* is also excellent. However, producing an expository article of that quality takes a lot of time, and so *What's Happening* is necessarily limited in scope. I would like to know what's happening in between issues of *What's Happening*. If there are resources out there that others have found useful for keeping abreast of mathematical research news, I would like to hear about them.
| https://mathoverflow.net/users/3106 | How do you find out the latest news in fields other than your own? | It's always hard to follow what's happening, especially in fields other than your own. I don't have any silver bullets, just a few time tested things that require serious work.
1. Talk to colleagues. If you are in a research department, always go to a colloquium. MR can be used both before and after the talk to tie it with what's been going on in the area.
Go to seminar talks.
2. Conferences. Even browsing through the list of abstracts of invited addresses at AMS Sectional Meetings helps to keep you informed. Actually attending these talks and, especially, invited addresses at the Joint Math Meetings is even better: they feature people who have been doing important work recently. I imagine other Math Societies operate in a similar way. For the past few years, the JMM has run a secion on "Current events". Like the Cambridge "Perspectives in Mathematics" conference, they make the texts of all talks available.
3. Browse through new books and journals and read book reviews (the ones in the Bulletin of AMS play a role similar to the MR featured reviews, but are set in a wider context). One of the biggest draws of the JMM is the book sale. By browsing through new publications, one can get the vibes of what is happening; even better, you can probe other people's reaction to anything you find interesting but don't quite understand. Bourbaki Seminar has 18 talks a year on subjects of current research interest.
4. I've heard that some people religiously follow all new submissions on the arXiv, but I can't vouch for the benefits or effectiveness of this method. On the other hand, once you get a wind of some new developments, you can search the arXiv for papers and surveys, many of which never make it to print in spite of being helpful (in some respects, arXiv now fulfills the role formerly played by the Lecture Notes in Mathematics).
5. You can run a self-teach seminar if you have at least one (ideally, 3 or more) interested colleagues. This may be on the subject of a recent paper, survey, or a book. If your colleagues are in a different field, so much the better: they can teach you what's happening there, and you can teach them something in return.
| 18 | https://mathoverflow.net/users/5740 | 26896 | 17,595 |
https://mathoverflow.net/questions/26895 | 3 | Consider the following equation for $X(t)$:
$$X(t)=e^{-bt}X(0)+\sigma\int\_{0}^{b}e^{-b(t-s)}dW(t) \, ,$$
where $0 < b, \sigma\in\mathbb{R} $, $X(0)$ is the initial distribution of $X(t)$, independent of the Brownian motion $W(t)$. I want so show that $dX(t)= -bX(t)dt+\sigma dW(t)$, but I am getting stuck on computing the derivative of $$\sigma\int\_{0}^{b}e^{-b(t-s)}dW(t) \, .$$
Could someone please give me some ideas? Thanks so much for your tim.
PS. the above equation is one of type of the Langevin's equation, more detail could be found here <http://en.wikipedia.org/wiki/Langevin_equation>
| https://mathoverflow.net/users/5136 | how to find derivative of a stochastic process? | If you interpret the stochastic integral in the Ito-sense (often used in finance) you'll have to use Ito's lemma to evaluate it:
See e.g. here: [Ito's lemma](http://en.wikipedia.org/wiki/Ito_lemma)
Alternatively you could interpret it in the Stratonovich-sense (often used in physics):
See e.g. here: [Stratonovich integral](http://en.wikipedia.org/wiki/Stratonovich_integral)
A good introduction to solving these kinds of stochastic differential equations (sde) without the use of measure theory and with lots of intuition is e.g. Wiersema: [Brownian motion calculus](http://books.google.com/books?id=kUFdAQAACAAJ&dq=wiersema&hl=de&ei=_XMHTMGYEdaJ_gabpeEE&sa=X&oi=book_result&ct=result&resnum=20&ved=0CIEBEOgBMBM)
| 9 | https://mathoverflow.net/users/1047 | 26902 | 17,597 |
https://mathoverflow.net/questions/26871 | 17 | This question is inspired by [Relation between Hecke Operator and Hecke Algebra](https://mathoverflow.net/questions/19684/relation-between-hecke-operator-and-hecke-algebra)
I remember having heard of yet another way of looking at Hecke operators acting on the spaces of modular forms for classical congruence subgroups of $SL\_2(\mathbf{Z})$, and I would like to ask if anyone knows a reference for that. Here are some details and a related question.
There is a universal elliptic curve $U$ over $\mathbf{H}$, the upper half-plane. It is an analytic manifold obtained by taking the quotient of $\mathbf{C}\times\mathbf{H}$ by the action of $\mathbf{Z}^2$ given by $(n,m)\cdot (z,\tau)=(z+n+m\tau,\tau)$ where $n,m\in\mathbf{Z},z\in \mathbf{C}$ and $\tau\in\mathbf{H}$.
A torsion free finite index subgroup $\Gamma$ of $SL\_2(\mathbf{Z})$ (congruence or not) acts on $U$. Here is the first question: for which $\Gamma$ is the quotient $U/\Gamma$ algebraic?
In any case $U/\Gamma$ is algebraic if $\Gamma=\Gamma(N)=\ker (SL\_2(\mathbf{Z})\to SL\_2(\mathbf{Z}/N))$ and $N\geq 3$. Denote the corresponding quotient $U/\Gamma(N)$ by $U(N)$. This is the universal elliptic curve with a level $N$ structure. (The notation $U(N)$ may be non-standard, in which case please let me know.)
There is a natural map $U(N)\to Y(N)=\mathbf{H}/\Gamma(N)$. Now take the $n$-th fibered cartesian power $U^n(N)$ of $U(N)$ over $Y(N)$ and let $p:U^n(N)\to Y(N)$ be the projection.
All derived direct images of the constant sheaf under $p$ decompose as direct sums of the Hodge local systems, which correspond to the symmetric powers of the standard representation of $\Gamma(N)$. Using the Eichler-Shimura isomorphism one can construct classes in $H^1(Y(N),V\_k)$ from modular forms of weight $k+2$ for $\Gamma(N)$ where $V\_k$ is the local system on $Y(N)$ that comes the $k$-th symmetric power of the standard representation. (In fact these classes can be interpreted in terms of the Hodge theory, see [Eichler-Shimura isomorphism and mixed Hodge theory](https://mathoverflow.net/questions/9599/eichler-shimura-isomorphism-and-mixed-hodge-theory)) So modular forms give elements in the $E\_2$ sheet of the Leray spectral sequence for $p$.
I would like to ask: can one interpret the Hecke operators acting on modular forms as correspondences acting on $U^n(N)$ over $Y(N)$ (and hence, on the Leray spectral sequence for $p$)?
| https://mathoverflow.net/users/2349 | Hecke operators acting as correspondences? | The answer to your last question is yes, modulo my own misunderstandings. In Scholl's "Motives for modular forms", §4 ([DigiZeitschreiften](http://www.digizeitschriften.de/dms/img/?PID=GDZPPN00210749X)), the Hecke operators are defined in this way, and I think the equivalence of these definitions is implied by the diagram (3.16) and proposition 3.18 of Deligne's "Formes modulaires et representations *l*-adiques" ([NUMDAM](http://www.numdam.org/item?id=SB_1968-1969__11__139_0)).
Let *p* be a prime not dividing *N*, and let *Y(N,p)* be the moduli scheme parametrizing elliptic curves with full level *N* structure and a choice of a cyclic subgroup *C* of order *p*. There are two natural "forgetful" maps *q1* and *q2* from *Y(N,p)* to *Y(N)* -- the former forgets the cyclic subgroup, the latter takes the induced level *N* structure on the quotient by *C*.
From these two maps, one gets two different families of elliptic curves over *Y(N,p)*. Explicitly, the pullback of *Un(N)* along *q1* is isomorphic to the *n*:th fibered power of the universal elliptic curve over *Y(N,p)*; let us denote it *Un(N,p)*. Let *Q(N,p)* be the quotient of *U(N,p)* by the cyclic subgroup *C*, and take its *n*:th fibered power as well. Then similarly *Qn(N,p)* is the pullback of *Un(N)* along *q2*. Finally, the quotient map gives us $\phi : U^n(N,p) \to Q^n(N,p)$.
But this data gives us the right correspondence on *Un(N)* over *Y(N)*, namely, one takes the composite $q\_{1\ast} \phi^\ast q\_2^\ast$ (where I use *q1* and *q2* also for the induced maps on fibered powers of universal curves).
| 13 | https://mathoverflow.net/users/1310 | 26905 | 17,599 |
https://mathoverflow.net/questions/26912 | 3 | I wonder whether it is impossible to write the nth [Motzkin number](http://www.research.att.com/~njas/sequences/A001006) as a sum of a fixed number of, say, hypergeometric terms. To illustrate what I mean: $n!+(2n)!$ is not a hypergeometric term, but it is written as a sum of two hypergeometric terms.
I'd also appreciate other examples, especially if they come from counting weighted Motzkin paths.
| https://mathoverflow.net/users/3032 | "Closed" form for Motzkin and related numbers | Chapter 8 of Petkovsek--Wilf--Zeilberger $A=B$ starts as follows:
"If you want to evaluate a given sum in closed form, so far the tools that have been
described in this book have enabled you to find a recurrence relation with polynomial
coefficients that your sum satisfies. If that recurrence is of order 1 then you are finished;
you have found the desired closed form for your sum, as a single hypergeometric
term. If, on the other hand, the recurrence is of order $\ge2$ then there is more work
to do. How can we recognize when such a recurrence has hypergeometric solutions,
and how can we find all of them?"
"In this chapter we discuss the question of how to recognize when a given recurrence
relation with polynomial coefficients has a closed form solution. We first take the
opportunity to define the term *closed form*."
"A function $f(n)$ is said to be of closed form if it is equal to a linear
combination of a fixed number, $r$, say, of hypergeometric terms. The number $r$ must
be an absolute constant, i.e., it must be independent of all variables and parameters
of the problem."
"Take a definite sum of the form $f(n) = \sum\_k F(n; k)$ where the summand $F(n; k)$
is hypergeometric in both its arguments. Does this sum have a closed form? The
material of this chapter, taken together with the algorithm of Chapter 6, provides a
complete algorithmic solution of this problem."
There are many examples in this chapter which illustrate the algorithm, like
Ap\'ery's numbers
$$
\sum\_k{\binom{n+k}k}^2{\binom{n}k}^2.
$$
Your particular sequence
$$
\sum\_k\frac{n!}{i!(i+1)!(n-2i)!}
$$
falls into the group covered by the algorithm of Chapter 8.
| 5 | https://mathoverflow.net/users/4953 | 26918 | 17,603 |
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