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https://mathoverflow.net/questions/17794 | -1 | I am looking for a book/paper which has the proof of the Rellich-Nicas identity.
**[EDIT by Yemon Choi]** It seems that what was meant is "the Rellich-Necas identity", although the original poster hasn't really clarified or expanded on the request.
| https://mathoverflow.net/users/4540 | Rellich-Necas identity | (Boo! I tried to post this in a comment to Ady, but the HTML Math won't parse right. So here goes. Sorry about the really long equation being broken up not very neatly.)
Googling Rellich-Necas turns up a bunch of recent papers by LUIS ESCAURIAZA in which the identities are used. But as far as I can tell the identity is just a simple differential equality obtained from symbolic manipulation of terms. The following seems to be a straight-forward version of the identity: let $A = (A\_{ij})$ be a symmetric bilinear form (with variable coefficients) on RN, $v$ a vector field, $u$ a function, and $\delta$ denoting the Euclidean divergence, we have
$ \delta( A(\nabla u,\nabla u) v) = 2 \delta( v(u) A(\nabla u)) + \delta(v) A(\nabla u,\nabla u)$
$- 2A(\nabla u) \cdot \nabla v \cdot \nabla u - 2 v(u) \delta(A(\nabla u)) + v(A)(\nabla u,\nabla u)$
Where $v(u)$ is the partial derivative of $u$ in the direction of $v$, and $A(\nabla u)\cdot\nabla v \cdot \nabla u$ is, in coordinates, $\partial\_i u A\_{ij} \partial\_j v\_k \partial\_k u$ with implied summation, and $v(A)$ is the symmetric bilinear form obtained by taking the $v$ partial derivative of the coefficients of $A$.
Verifying that the identity is true should just be a basic application of multivariable calculus.
| 4 | https://mathoverflow.net/users/3948 | 17863 | 11,930 |
https://mathoverflow.net/questions/17822 | 2 | I have been looking around, unsuccessfully, for generalizations of universal algebra based on higher-order logic (rather than first order) and where the relations are not purely equational. Motivation: I need a "theory of syntax" for presentations of higher-order, non-equational theories. Furthermore, I want to be able to specify 'combinators' over these presentations, rigorously.
I am aware of [Lawvere theories](http://en.wikipedia.org/wiki/Lawvere_theory), but these are still equational (and neither particularly higher-order, though the multi-sorted generalization seems straightforward enough). There is a beginning of model theory done in a logical independent way, i.e. [model theory over an institution](http://en.wikipedia.org/wiki/Institutional_model_theory); but that seems to concentrate on the model-theoretic aspects, rather than the universal algebra aspects. Perhaps what I am looking for are [sketches](http://en.wikipedia.org/wiki/Sketch_%28mathematics%29)?
[Edit:] From the various answer below, it seems I should be asking the question "how can I view type theory as a theory of syntax"? Somehow, that seems like an 'implementation' (as it requires a fair bit of 'encoding'); for example, to express the 'theory of categories' [i.e. (Obj, Mor, id, src, trg, $\circ$) and 5-6 axioms, I need a dependent record. Plus what is a sort (and sort constructor for Mor), what is an operation, and what is in Prop? Universal algebra cleanly separates these.
A good question was asked: what theorems do I want? Well, whatever operations I make on theories, well-formedness of the results will require discharging some obligations -- these obligations should all be finitely expressible (and automatically well-formed). Furthermore, the resulting syntactic objects and their morphisms should form a finitely co-complete category. Note that I expect that deciding if a given (presentation of a ) theory has a model to be undecidable.
| https://mathoverflow.net/users/3993 | Higher-order, multi-sorted, non purely equational version of universal algebra ? | Lawvere theories generalize to higher-order logic in a straightforward way. A first-order hyperdoctrine is a functor $\mathcal{P} : C^{\mathrm{op}} \to \mathrm{Poset}$ where $C$ has products and is used to interpret terms in context, plus a small herd of conditions to make substitution and quantifiers work out right.
If you want a hyperdoctrine that can interpret higher-order predicate logic, what you additionally want is to (a) require $C$ to be cartesian closed, and (b) $C$ should have an internal heyting algebra $H$ with the property that for each $X$ in $C$, $\mathit{Obj}(\mathcal{P}(X)) \simeq C(X, H)$. Basically, $H$ models the sort of propositions, and the cartesian closure lets you freely interpret lambda-terms denoting predicates and relations. The bijection $\mathit{Obj}(\mathcal{P}(X)) \simeq C(X, H)$ tells you that the morphisms into $H$ actually do correspond to truth values in context $X$, so you can interpret a higher-order term by interpreting a proposition in $\mathrm{Poset}$ via the functor $\mathcal{P}$, and then transporting it back into $C$.
(I would have made this a comment, but it was too long.)
| 3 | https://mathoverflow.net/users/1610 | 17873 | 11,937 |
https://mathoverflow.net/questions/17870 | 8 | If a four-legged, rectangular table is rickety, it can nearly always be stabilised just by turning it a little. This is very useful in everyday life! Of course it relies on the floor being the source of the ricketiness; if the table's legs are different lengths, it doesn't work (this is why I said 'nearly always').
Here is a quick 'proof': Label the feet A, B, C, D in order, and integrate the function F(A) + F(C) - F(B) - F(D) while the table is turned through 180° (here F is the height of the floor). The result is **∫** F - **∫** F = 0 (these integrals are over 360°), so the function must be zero somewhere; at this point, the table is stable.
This proof can easily be adapted for any cyclic quadrilateral ABCD.
The proof only works if the slope S of the floor is small enough that S2 can be ignored. If not, and if the stable solution is not horizontal, then the feet will not lie on the projection of the circle that we integrated over. So a more complicated argument is required:
Suppose ABCD is a square, and suppose that |F(x)-F(y)| <= S|x-y| everywhere, where S = sin-1(1/√3). Fix a vertical line V. Then ABCD can always be positioned with its centre on V, so that each corner touches the floor.
To see this, consider the two diagonals AC and BD separately. Choose a starting position such that
```
(i) their endpoints lie on the floor,
(ii) their midpoints are on V,
(iii) they are perpendicular to each other.
```
We are not requiring that their centres coincide (so we will have to dismantle the table to achieve this). We may suppose that in the starting position, the centre of AB lies above the centre of BD. Now rotate them so that constraints (i)-(iii) remain satisfied, until AC occupies the starting position of BD and vice versa. (The constraint on the slope ensures that we can do this continuously.) Now the centre of AB lies below the centre of BD, so the two centres must have coincided at some time.
(This proof generalises to rectangles, but the maximum slope depends on the rectangle. It does not generalise to cyclic quadrilaterals.)
**My first question** is this: Is the condition on the slope (or a less stringent slope condition) necessary? Or can we always stabilise the table, whatever the slope of the floor?
**My second question** (if the answer to the first question is that we can always stabilise the table): Given any partition of $\mathbb R$3 into {L,U} with L bounded above and U bounded below, and a vertical line V, can we always find a unit square whose corners lie in the closures of both L and U, and whose centre lies on V?
**Update** The paper referenced below by Q.Q.J. answers my first question: a rectangular table can always be stabilised, if the floor function F is continuous. But it can only be stabilised by a *smooth rotation* if the floor satisfies the slope condition. (I was wrong about different rectangles requiring different slope conditions.)
| https://mathoverflow.net/users/767 | Stable Tables on Fluctuating Floors | ["Mathematical table turning revisited"](http://arxiv.org/abs/math/0511490) by Baritompa, L"owen, Polster, and Ross
I am no expert on what is or isn't possible but there are at least two different groups who have looked at this type of problem and this article contains a number of references that are probably relevant to you.
| 10 | https://mathoverflow.net/users/3623 | 17874 | 11,938 |
https://mathoverflow.net/questions/17875 | 6 | *(This is a follow-up to my previous questions [Natural models of graphs?](https://mathoverflow.net/questions/11647/natural-models-of-graphs).)*
Erdös in [The Representation of a Graph by Set Intersections](http://www.renyi.hu/~p_erdos/1966-21.pdf) (1966) states:
>
> **Theorem**. Let $G$ be an arbitrary
> graph. Then there is a set $S$ and a
> family of subsets $S\_1, S\_2, ...$ of
> $S$ which can be put into one-to-one
> correspondence with the vertices of
> $G$ in such a way that $x\_i$ and $x\_j$ are joined by an
> edge of $G$ iff $i \neq j$
> and $S\_i \cap S\_j \neq \emptyset$.
>
>
>
If we identify $S$ with a set of prime numbers and each $S\_i$ with the product of its members we get the following:
>
> **Corollary**. Let $G$ be an arbitrary finite
> graph. Then there is a sequence of natural numbers $(n\_1, n\_2, ..., n\_k)$
> which can be put into one-to-one
> correspondence with the vertices of
> $G$ in such a way that $x\_i$ and $x\_j$ are joined by an edge iff $i \neq j$ and GCD$(n\_i, n\_j) > 1$.
>
>
>
We can choose the prime numbers (the elements of $S$, from which the $n\_i$ are built) arbitrarily, and so the question arises, whether they can always be choosen in such a way, that the set $(n\_1, n\_2, ..., n\_k)$ is an [arithmetic sequence](http://en.wikipedia.org/wiki/Arithmetic_progression).
Of course every *complete* graph on $k$ nodes can be represented by an arithmetic sequence: just take some consecutive sequence of even numbers. [Green-Tao's Theorem](http://en.wikipedia.org/wiki/Green%25E2%2580%2593Tao_theorem) guarantees that also every *empty* graph on $k$ nodes can be represented by an arithmetic sequence $(p\_1, p\_2, ..., p\_k)$ of primes.
>
> **Question:** Can every graph on $k$ nodes be represented by an arithmetic sequence
> of natural numbers such that $n\_i$ and $n\_j$ are joined by an edge iff $n\_i \neq n\_j$ and GCD$(n\_i, n\_j) > 1$
>
>
>
This would be one kind of *natural model of a graph*, that I was looking for, originally.
Maybe some references?
**Added**: Due to Kevin's concise answer and Thomas' comment, I'd like to add the following question:
>
> **Question:** If not every graph on $k$ nodes can be represented by an arithmetic sequence
> of natural numbers such that $n\_i$ and $n\_j$ are joined by an edge iff $n\_i \neq n\_j$ and GCD$(n\_i, n\_j) > 1$: **Are there interesting classes of graphs with this property?**
>
>
>
| https://mathoverflow.net/users/2672 | Can every finite graph be represented by an arithmetic sequence of natural numbers? | OK so take the unique tree on 3 vertices. Claim: you can't encode this with an arithmetic progression (AP). For if the AP is $a,a+d,a+2d$ then (because we have two edges) either vertices 1 and 2 are joined, or vertices 2 and 3 are joined (or both). Hence there is some $p>1$ such that either $p$ divides both $a$ and $a+d$, or $p$ divides both $a+d$ and $a+2d$. In either case, $p$ then divides $d$, so it divides $a$, so it divides everything, so the graph is complete.
| 11 | https://mathoverflow.net/users/1384 | 17879 | 11,942 |
https://mathoverflow.net/questions/17893 | 0 | Suppose $V$ is a no-where zero vector field on $S^n$ ($n$ odd). Let $p \in S^n$. Let $\gamma\_p$ be the unique curve on $S^n$ through $p$ and tangential to $V$ everywhere along it. Is it true that $\gamma\_p$ is a closed curve $\forall p \in S^n$? If so, is it true that the length of $\gamma\_p$ must be finite?
Alternatively, is it possible to find a curve $\gamma$ on $S^n$ of infinate length such that the tangent vectors of $\gamma$ can be extended to form a continuous, no-where zero tangential vector field of $S^n$?
| https://mathoverflow.net/users/3121 | Following curves on S^n | This is the [Seifert conjecture](http://en.wikipedia.org/wiki/Seifert_conjecture). There are nowhere-zero vector fields on $S^3$ with no closed orbits, and there are both smooth and real-analytic constructions.
| 8 | https://mathoverflow.net/users/121 | 17896 | 11,952 |
https://mathoverflow.net/questions/17892 | 2 | In forcing, we take a collection of forcing conditions and impose a partial order on them. The convention is that if $p$ is stronger than $q$, then we say $p < q$. This is perfectly fine, but it seems intuitively backwards to me. If I were designing the notation for forcing, I would want the stronger condition to be larger. (Something I read says, I think, that Shelah uses the opposite convention that I find more intuitive. Is this so?)
Further, if we are forcing with a collection of partial functions (as we often do), we want the stronger condition to be the partial function with the larger domain. This leads us to a definition of the poset order whereby $f < g$ iff $f \supset g$. This seems notationally awkward.
Nonetheless, Cohen must have had some good reasons choosing the order that he did. What is/was the rational for Cohen's notational convention? Does it have benefits today, or is it just an artifact of a older approach to forcing?
| https://mathoverflow.net/users/4087 | When forcing with a poset, why do we order the poset in the order that we do? | The reason is that in the corresponding Boolean algebra, 0 is less than 1. That is, stronger conditions correspond to lower Boolean values in the Boolean algebra. The trivial condition (which is often the empty function in the cases you mention), corresponds to the element 1 in the Boolean alebra.
We definitely want to regard lower elements of the Boolean algebra as stronger, since they have more implications in the Boolean algebra sense. (After all, 0 = false is surely the strongest assumption you could make, right?)
Meanwhile, you can be comforted by the fact that Shelah and many researchers surrounding him (and a few others) use the alternative forcing-upwards notation. Nevertheless, the forcing-downwards notation is otherwise nearly universal.
This difference in culture sometimes causes some funny problems when authors from opposing camps collaborate. Sometimes a compromise is struck to never officially to use the order explicitly, and to write "stronger than" or "weaker than" in words, rather than take sides. Another alternative is the use the forcing turnstyle symbol itself as the order, but this solution suffers from the fact that it only works when the order is separative.
---
Edit. I looked at [Cohen's PNAS 1963 article](http://www.worldscibooks.com/etextbook/4800/4800_chap1.pdf), and in that article, he does not use the forcing-downwards notation at all. Rather, he uses the containment symbol $\supset$ explicitly. Thus, the assumption in the question that Cohen did indeed use the downward-oriented relation may be unwarranted. (Perhaps this view is a little softened by the observation that he consistently uses $\supset$ rather than $\subset$.)
Here is my theory. In logic and set theory there has been a long-standing tradition of consistently using the relation ≤ in preference to ≥, presumably to avoid the problems associated with mixing up the greater-than less-than order. Perhaps this goes back to Cantor? Now, in the case of forcing, it is usually the case that you have a condition P already, and you want to ask whether there is Q stronger than P with a certain property (one rarely asks for weaker conditions this way). Thus, if you have the downward-oriented relation, you can economically say "there is Q ≤ P such that..." This is just how Cohen's text reads, since he says "there is Q \supset P such that ...". Generalizing Cohen's containment order to an arbitrary partial order, one then wants to interpret containment as ≤. And then the further support for this convention arrives with the fact that it agrees with the Boolean algebra order a few years later, so it became standard (except for the Shelah school and a few others).
| 9 | https://mathoverflow.net/users/1946 | 17897 | 11,953 |
https://mathoverflow.net/questions/17736 | 49 | Consider the Sobolev spaces $W^{k,p}(\Omega)$ with a bounded domain $\Omega$ in n-dimensional Euclidean space. When facing the different embedding theorems for the first time, one can certainly feel lost. Are there certain tricks to memorize the (continuous and compact) embeddings between the different $W^{k,p}(\Omega)$ or into $C^{r,\alpha}(\bar{\Omega})$ ?
| https://mathoverflow.net/users/3509 | Way to memorize relations between the Sobolev spaces? | Sobolev norms are trying to measure a combination of three aspects of a function: height (amplitude), width (measure of the support), and frequency (inverse wavelength). Roughly speaking, if a function has amplitude $A$, is supported on a set of volume $V$, and has frequency $N$, then the $W^{k,p}$ norm is going to be about $A N^k V^{1/p}$.
The uncertainty principle tells us that if a function has frequency $N$, then it must be spread out on at least a ball of radius comparable to the wavelength $1/N$, and so its support must have measure at least $1/N^d$ or so:
$V \gtrsim 1/N^d.$
This relation already encodes most of the content of the Sobolev embedding theorem, except for endpoints. It is also consistent with dimensional analysis, of course, which is another way to derive the conditions of the embedding theorem.
More generally, one can classify the integrability and regularity of a function space norm by testing that norm against a bump function of amplitude $A$ on a ball of volume $V$, modulated by a frequency of magnitude $N$. Typically the norm will be of the form $A N^k V^{1/p}$ for some exponents $p$, $k$ (at least in the high frequency regime $V \gtrsim 1/N^d$). One can then plot these exponents $1/p, k$ on a two-dimensional diagram as mentioned by Jitse to get a crude "map" of various function spaces (e.g. Sobolev, Besov, Triebel-Lizorkin, Hardy, Lipschitz, Holder, Lebesgue, BMO, Morrey, ...). The relationship $V \gtrsim 1/N^d$ lets one trade in regularity for integrability (with an exchange rate determined by the ambient dimension - integrability becomes more expensive in high dimensions), but not vice versa.
These exponents $1/p, k$ only give a first-order approximation to the nature of a function space, as they only inspect the behaviour at a single frequency scale N. To make finer distinctions (e.g. between Sobolev, Besov, and Triebel-Lizorkin spaces, or between strong L^p and weak L^p) it is not sufficient to experiment with single-scale bump functions, but now must play with functions with a non-trivial presence at multiple scales. This is a more delicate task (which is particularly important for critical or scale-invariant situations, such as endpoint Sobolev embedding) and the embeddings are not easily captured in a simple two-dimensional diagram any more.
I discuss some of these issues in my lecture notes
<http://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/>
EDIT: Another useful checksum with regard to remembering Sobolev embedding is to remember the easy cases:
1. $W^{1,1}({\bf R}) \subset L^\infty({\bf R})$ (fundamental theorem of calculus)
2. $W^{d,1}({\bf R}^d) \subset L^\infty({\bf R}^d)$ (iterated fundamental theorem of calculus + Fubini)
3. $W^{0,p}({\bf R}^d) = L^p({\bf R}^d)$ (trivial)
These are the extreme cases of Sobolev embedding; everything else can be viewed as an interpolant between them.
EDIT: I decided to go ahead and draw the map of function spaces I mentioned above, at
<http://terrytao.wordpress.com/2010/03/11/a-type-diagram-for-function-spaces/>
| 90 | https://mathoverflow.net/users/766 | 17906 | 11,959 |
https://mathoverflow.net/questions/17886 | 27 | This question is perhaps somewhat soft, but I'm hoping that someone could provide a useful heuristic. My interest in this question mainly concerns various derived equivalences arising in geometric representation theory.
**Background**
For example, Bezrukavnikov, Mirkovic, and Rumynin have proved the following: Let $G$ be a reductive algebraic group over an algebraically closed field of positive characteristic. Then there is an equivalence between the bounded derived category of modules for the sheaf $\cal D$ of crystalline (divided-power) differential operators on the flag variety, and the bounded derived category of modules with certain central character for the enveloping algebra $\cal U$ of Lie($G$). What is interesting is that it is *not* true that this equivalence holds on the non-derived level: The category of $\cal D$-modules is not equivalent to the category of $\cal U$-modules with the appropriate central character. This is true in characteristic 0 (this is the Beilinson-Bernstein correspondence), but something is broken in positive characteristic: there are certain "bad" sheaves that are $\cal D$-modules which make the correspondence not hold.
There are other results in geometric representation theory of this form. For example, Arkhipov, Bezrukavnikov, and Ginzburg have proved that there is an equivalence (in characteristic 0) between the bounded derived category of a certain block of representations for the quantum group associated to $G$, and the bounded derived category of $G \times \mathbb C^\*$-equivariant sheaves on the cotangent bundle of the flag variety of $G$. Again, this equivalence does not hold on the non-derived level.
In general, there are a number of results in geometric representation theory that hold on the derived level, but not the non-derived level.
**Question**
Here's my question: Why would one be led to expect that a derived equivalence holds, when the non-derived equivalence does not? It seems as though the passage to the derived level in some sense fixes something that was broken on the non-derived level; how does it do that?
| https://mathoverflow.net/users/1528 | Why would one expect a derived equivalence of categories to hold? | One crude answer is that passing to derived functors fixes one obstruction to being an equivalence. Any equivalence of abelian categories certainly is exact (i.e. it preserves short exact sequences), though lots of exact functors are not equivalences (for example, think about representations of a group and forgetting the G-action).
What derived functor does is fix this problem in a canonical way; you have to replace short exact sequences with exact triangles, but you get a functor which is your original "up to zeroth order," exact, and uniquely distinguished by these properties.
So, what BMR do is take a functor which is not even exact (and thus obviously not an equivalence), and show that the lack of exactness is "the only problem" for this being an equivalence.
**EDIT:** Let me just add, from a more philosophical perspective, that derived equivalences are just a lot more common. There are just more of them out in the world. Given an algebra A, Morita equivalences to A are classified essentially by projective generating A-modules, whereas derived Morita equivalences of dg-algebras are in bijection with all objects in the derived category of $A-mod$ which generate (in the sense that nothing has trivial Ext with them): you look at the dg-Ext algebra of the object with itself. If you have an interesting algebra (say, a finite dimensional one of wild representation type), there are a lot more of the latter than the former in a very precise sense. Of course, the vast majority of these are completely uncomputable an tell you nothing, but there are enough of them in the mix to make things interesting.
| 28 | https://mathoverflow.net/users/66 | 17907 | 11,960 |
https://mathoverflow.net/questions/17916 | 1 | It is a theorem that the category of compactly generated weakly Hausdorff (CGWH) spaces is Quillen equivalent to the category of simplicial sets with the Kan model structure. However, I know next to nothing about the model structure on CGWH spaces.
Questions:
What are the analogs of Kan fibrations, trivial fibrations, cofibrations, and weak cofibrations in CGWH?
Are there analogs of left, right, inner fibrations in CGWH (along with their corresponding anodyne maps)?
| https://mathoverflow.net/users/1353 | Analogs of left, right, inner, and Kan fibrations in CGWH | The analogs are:
* Serre fibrations (map with lifting property with respect to $I^n\times 0\to I^{n+1}$)
* trivial Serre fibrations (map with lifting property with respect to $S^{n-1}\to D^n$)
* retracts of maps built by attaching $S^{n-1}\to D^n$
* retracts of maps built by attaching $I^n\times 0\to I^{n+1}$.
There are any number of references to the Quillen model structure on Top, starting with Quillen's book; you can google for the article by Dwyer and Spalinksi on "Homotopy Theories and Model Categories".
Hovey's book "Model Categories" also does this, and gives significant attention to the cases of k-spaces and CGWH; the description of the fibrations/cofibrations is the same in these categories, but some care is needed to make sure the proof that you get a model category goes through. Hirschhorn's book may also do this, though somebody seems to have my copy, so I can't check.
I can't imagine you can define left/right/inner fibrations in Top or CGWH. Topologically, an n-simplex has no "left" or "right"; you always have a homeomorphism $\Delta^n\to \Delta^n$ which permutes the vertices however you please.
| 10 | https://mathoverflow.net/users/437 | 17930 | 11,973 |
https://mathoverflow.net/questions/17821 | 10 | Let G be a finite group of Lie type. Assume G is also of universal type. Is the Steinberg representation of G generic, i.e., does the Steinberg representation admit a Whittaker model?
A Whittaker model for a representation of G is defined in a similar fashion as in the case of GL(2, F) in Bump's "Automorphic Forms and Representations." I am interested in the genericity of the Steinberg representation of a group of matrices over a finite field.
| https://mathoverflow.net/users/4544 | Steinberg Representations of Finite Groups of Lie Type | I think that for finite groups of Lie type, the analogue of "having a Whittaker model" is that the representation occurs in a Gelfand-Graev representation: these are the representations obtained by inducing a "regular" character from the unipotent subgroup of a rational Borel. Such representations are multiplicity free and so constitute a "model" (in the sense I think people say "Whittaker model"). Now when the center of $G$ is connected, all regular characters are conjugate under the action of the maximal torus of the Borel, so the Gelfand-Graev representation is unique (otherwise there is a family of such representations). In their famous paper, Deligne and Lusztig decompose the Gelfand-Graev representation in this case and show that there is exactly one constituent in each "geometric conjugacy class" of irreducible representations (which can be thought of as a semisimple conjugacy class in the dual group). The Steinberg representation is then the representative in the conjugacy class of the identity element -- that is the representative among the "unipotent representations".
To focus more on the actual question (!) the character of the Steinberg representation is explicitly known, and it is easy to check from this that its restriction to $U$ is the regular representation, so it certainly occurs in the Gelfand-Graev representation.
| 11 | https://mathoverflow.net/users/1878 | 17934 | 11,976 |
https://mathoverflow.net/questions/17937 | 21 | Let $X$ be a nice variety over $\mathbb{C}$, where nice probably means smooth and proper.
I want to know: How can we show that the hypercohomology of the algebraic de Rham complex agrees with the hypercohomology of the analytic de Rham complex (equivalently the cohomology of the constant sheaf $\mathbb{C}$ in the analytic topology)? Does this follow immediately from GAGA? If not, how do you prove it?
I think that this does not follow immediately from GAGA because, while the sheaves $\Omega\_X^i$ are coherent, the de Rham $d$ is not a map of coherent sheaves (it is not multiplicative). Am I correct in my thinking?
| https://mathoverflow.net/users/83 | Algebraic de Rham cohomology vs. analytic de Rham cohomology | I don't think you can get this directly from GAGA. The reference that I know for this result is Grothendieck, [On the de Rham cohomology of algebraic varieties](http://www.ams.org/mathscinet-getitem?mr=199194). It is short, beautiful, and in English.
| 13 | https://mathoverflow.net/users/297 | 17939 | 11,979 |
https://mathoverflow.net/questions/17938 | 3 | Does there exist an orthonormal basis of square-integrable functions (either $L^2(\mathbb{R})$ or $L^2(\mathbb{C})$) such that the sequence of functions has bounded variance, and also the sequence consisting of the Fourier transform of each function also has bounded variance?
Some background:
This question came up in a [comment on SciRate](http://scirate.com/who.php?id=1003.2133&what=comments) regarding a recently translated [paper by von Neumann](http://arXiv.org/abs/1003.2133). There the commenter, Matt Hastings, points out some related results.
In particular, the [Balian-Low theorem](http://en.wikipedia.org/wiki/Balian%E2%80%93Low_theorem) states that this can't exist for any Gabor basis, i.e. one which is composed of time and frequency translates of a given fiducial $L^2$ function. If there were a generalization of this theorem to arbitrary bases, it would prove that such a sequence can't exist.
| https://mathoverflow.net/users/1171 | Simultaneous time-frequency concentration of orthonormal sequences? | Such orthonormal bases do exist, as proved in:
Bourgain, J. A remark on the uncertainty principle for Hilbertian basis. J. Funct. Anal. 79 (1988), no. 1, 136--143 ([MathSciNet link](http://www.ams.org/mathscinet-getitem?mr=950087)).
The theorem says that for each $\rho>1/2$ there is an orthonormal basis for $L^2(\mathbb{R})$ such that all of the variances of the basis elements and their Fourier transforms are less than $\rho$. After the statement Bourgain remarks:
>
> Thus Balian’s strong uncertainty principle does not hold for a nonperiodic
> basis.
>
>
>
It is remarkable that this appears to have been discovered (rediscovered?) well after von Neumann's time. Powell [proved](http://www.springerlink.com/content/j365534350378065/) more recently the result that Matt Hastings mentioned, namely that in such a case the sequence of means of the orthonormal basis is unbounded.
---
My old answer, posted before reading Matt Hastings's comment led me to the correct question, was to the question of whether all of the variances can be finite. It was this:
Yes, because you can take an orthonormal basis in the Schwartz space by applying Gram-Schmidt to a countable $L^2$ dense subset of the Schwartz space.
| 5 | https://mathoverflow.net/users/1119 | 17942 | 11,982 |
https://mathoverflow.net/questions/17872 | 4 | Does there exist a constant $A$ giving an upper bound on the absolute value of the regulator for infinitely many totally real number fields?
| https://mathoverflow.net/users/4556 | Totally real number fields with bounded regulators | I believe the question is: "Does there exist a constant $A$ such that there exists infinitely many totally real number fields with regulator less then $A$?" Ignore this response if that's incorrect.
I think the answer is no. The place to go for questions like this are the work of Zimmert and the survey paper of Odlyzko (which, as it turns out, would've pointed you to Zimmert anyway). You'll have to dig into the references to make sure I haven't misinterpreted a convention or notation, but I believe a proof goes as follows:
In Odlyzko's "Bounds for disciminants...", he cites Zimmert as giving a lower bound (equation 5.5) for the regulator of a number field which grows exponentially in the degree of the number field. So if the answer to your question were yes, your infinitely many number fields would have to be of bounded degree. But for number fields of a fixed degree, Theorem C in Friedman's "Analytic Formula for the Regulator of a Number Field" (attributed to Silverman, from an interesting-looking paper that I'd never heard of before) gives a lower bound for the regulator in terms of the discriminant D. This lower bound goes to infinity with D under the condition that all proper subfields of your number field have a strictly smaller unit rank, which is true for totally real number fields. The nail in the coffin is then that there are only finitely many number fields with discriminant under any given bound, and so only finitely many totally real number fields with regulator under any given bound.
Note that you can't cite the Silverman result right off the bat since the lower bound appears to me to go to 0 as $D$ and $n$ (discriminant and degree) simultaneously go to infinity.
Hope that helps.
| 6 | https://mathoverflow.net/users/35575 | 17948 | 11,987 |
https://mathoverflow.net/questions/17951 | 8 | Recall that a **chain complex** is a (finite) diagram of the form
$$ V = \{ \dots \to V\_3 \overset{d\_3}\to V\_2 \overset{d\_2}\to V\_1 \overset{d\_1}\to V\_0 \to 0 \} $$
where the $V\_n$ are (finite-dimensional) vector spaces and for each $n$, $d\_n \circ d\_{n+1} = 0$. If $V$ and $W$ are chain complexes, a **chain map** $f: V \to W$ is a map $f\_n : V\_n \to W\_n$ for each $n$ such that all the obvious squares commute — "$[d,f]=0$" — and the pair (chain complexes, chain maps) defines a category. In fact, it is a 2-category: the 2-morphisms between $f,g : V \rightrightarrows W$ are the **chain homotopies**, i.e. a system of maps $h\_n: V\_n \to W\_{n+1}$ such that "$[d,h] = f-g$". The category of chain complexes has a **biproduct** (both a product and a coproduct) $\oplus$ given by the pointwise direct sum.
I thought I knew what the tensor product of chain complexes was. Namely, if $V$ and $W$ are chains, then the usual thing is to define
$$ (V\otimes W)\\_n = \bigoplus\_{k=0}^n V\_k \otimes W\_{n-k} $$
and the chain maps are the sums of the obvious tensor products of differentials, decorated with signs.
But now I'm not sure why this is the tensor product picked. Namely, if I have a linear category, I think that a **tensor product** $V \otimes W$ should satisfy the following universal property: for any $X$, $\hom(V \otimes W,X)$ should be naturally isomorphic to the space of *bilinear* maps $V \times W \to X$. Now, I've never really known how to write down the word "bilinear" in a general category, without refering to individual points. But I think I do know what the "set" $V \times W$ is when $V$ and $W$ are chains — it's the set underlying $V \oplus W$ — and then I think I do know what bilinear maps should be.
In any case, then it's clear that the usual tensor product is not this. For example, if $V,W$ have no non-zero terms above degree $n$, then the bilinear maps $V \times W \to X$ I think cannot be interesting above degree $n$, whereas the above $\otimes$ has terms in degree $2n$.
In any case, in [HDA6](http://arxiv.org/abs/math/0307263v5), Baez and Crans consider two-term chain complexes $V\_1 \to V\_0$ (they argue that these are the same as "2-vector-spaces"), and then construct a different tensor product, given by:
$$ V\otimes W = \{ (V\_1 \otimes W\_1) \oplus (V\_1 \otimes W\_0) \oplus (V\_0\otimes W\_1) \to (V\_0 \otimes W\_0) \} $$
where the differential is the sum of the obvious tensor products of differentials and identity maps. They then assert that this tensor product satisfies the correct universal property, although they leave the details to the reader.
This leads naturally to:
>
> **Question:** What is the precise universal property that $\otimes$ *ought* to have, and what "product" of chain maps satisfies this universal property?
>
>
>
| https://mathoverflow.net/users/78 | What tensor product of chain complexes satisfies the usual universal property? | I think you need to revise your treatment of degree when working with bilinear maps, since the notion of "bilinear" requires more information from $V \times W$ than just the fact that it is an object in the category. For a simple case, try forgetting the differentials, and just work with graded vector spaces, or comodules over $k[\mathbb{Z}]$, where $k$ is your base field. The correct notion of tensor product forces a bilinear map from $V \times W$ to respect total degree.
| 5 | https://mathoverflow.net/users/121 | 17957 | 11,989 |
https://mathoverflow.net/questions/17946 | 26 | In 1986 G.X. Viennot published "Heaps of pieces, I : Basic definitions and combinatorial lemmas" where he developed the theory of heaps of pieces, from the abstract: a geometric interpretation of Cartier-Foata's commutation monoid. This theory unifies and simplifies many other works in Combinatorics : bijective proofs in matrix algebra (MacMahon Master theorem, inversion matrix formula, Jacobi identity, Cayley-Hamilton theorem), combinatorial theory for general (formal) orthogonal polynomials, reciprocal of Rogers-Ramanujan identities, graph theory (matching and chromatic polynomials).
In the references the subsequent articles "Heaps of pieces, 4 and 5" are listed as "in preparation" where the applications of the theory to solving the directed animal problem and statistical physics are supposed to be developed. I know that these parts of the theory have appeared in literature but I am sort of puzzled as to why the series of papers was not continued (searching for Heaps of pieces II or III or IV doesn't give results). Is there any survey of the full theory somewhere else?
Also, since I didn't feel like asking this in a separate question, is there any paper that proves classical theorems of dimers (Kasteleyn's theorem, Aztec diamond etc.) using Viennot's theory?
| https://mathoverflow.net/users/2384 | What (if anything) happened to Viennot's theory of Heaps of pieces? | Ok, this I know. Viennot basically invents lots of great stuff but rarely publishes his work. About "heaps of pieces" - this is a pretty little theory with very few original consequences. It is really equivalent to Cartier-Foata partially commutative monoid (available [here](http://www.mat.univie.ac.at/~slc/books/cartfoa.html)). See Christian Krattenthaler's [article](http://www.mat.univie.ac.at/~kratt/artikel/heaps.html) for the connection and details. See there also some references to other recent papers.
Now, for many of Viennot's unpublished results, see his "Orthogonal polynomials..." book, which was unavailable for years, but is now on his [web page](http://www.xavierviennot.org/xavier/Bienvenue.html). See there also several of his video lectures (mostly in French), where he outlined some interesting bijections based on the heaps (some related to various lattice animals were new to me, even if he may have come up with them some years ago - take a look). The reciprocal of R-R identities is an elegant single observation which he published separately: [MR0989236](http://www.ams.org/mathscinet-getitem?mr=989236). It really does not reprove the R-R identities, just gives a new combinatorial interpretation for one side using heaps of dimers (various related results were obtained by Andrews-Baxter a bit earlier).
About some recent applications outside of enumerative combinatorics. Philippe Marchal describes [here](http://algo.inria.fr/seminars/sem00-01/marchal.html) that heaps of pieces easily imply [David Wilson's theorem](http://dbwilson.com/ja/tau.ps) on [Loop-erased random walks](http://en.wikipedia.org/wiki/Loop-erased_random_walk) giving random spanning trees. Ellenberg and Tymoczko give a [beautiful application](http://arxiv.org/abs/math.GR/0510506) to the diameter bound of certain Cayley graphs. Finally, in [my paper](http://arxiv.org/abs/math/0607737) with Matjaz Konvalinka, we use a heaps-of-pieces style bijection to give the "book proof" of the (non- and q-commutative) [MacMahon Master theorem](http://en.wikipedia.org/wiki/MacMahon_Master_theorem).
On your followup question regarding Kasteleyn's theorem and the Aztec diamond theorem - no, these are results of different kind, heaps don't really apply, at least as far as I know.
| 23 | https://mathoverflow.net/users/4040 | 17958 | 11,990 |
https://mathoverflow.net/questions/17960 | 96 | Hi all!
Google published recently questions that are asked to candidates on interviews. One of them caused very very hot debates in our company and we're unsure where the truth is. The question is:
>
> In a country in which people only want
> boys every family continues to have
> children until they have a boy. If
> they have a girl, they have another
> child. If they have a boy, they stop.
> What is the proportion of boys to
> girls in the country?
>
>
>
Despite that the official answer is 50/50 I feel that something wrong with it. Starting to solve the problem for myself I got that part of girls can be calculated with following series:
$$\sum\_{n=1}^{\infty}\frac{1}{2^n}\left (1-\frac{1}{n+1}\right )$$
This leads to an answer: there will be ~61% of girls.
The official solution is:
>
> This one caused quite the debate, but
> we figured it out following these
> steps:
>
>
> * Imagine you have 10 couples who have 10 babies. 5 will be girls. 5
> will be boys. (Total babies made: 10,
> with 5 boys and 5 girls)
> * The 5 couples who had girls will have 5 babies. Half (2.5) will be
> girls. Half (2.5) will be boys. Add
> 2.5 boys to the 5 already born and 2.5 girls to the 5 already born. (Total
> babies made: 15, with 7.5 boys and 7.5
> girls.)
> * The 2.5 couples that had girls will have 2.5 babies. Half (1.25) will
> be boys and half (1.25) will be girls.
> Add 1.25 boys to the 7.5 boys already
> born and 1.25 girls to the 7.5 already
> born. (Total babies: 17.5 with 8.75
> boys and 8.75 girls).
> * And so on, maintianing a 50/50 population.
>
>
>
Where the truth is?
| https://mathoverflow.net/users/4568 | Google question: In a country in which people only want boys | The proportion of girls in one family is a [biased estimator](http://en.wikipedia.org/wiki/Bias_of_an_estimator) of the proportion of girls in a population consisting of many families because you are underweighting the families with a large number of children.
If there were just 1 family, then your formula would be wrong, but the average of the percentage of girls you would observe would be
$$\sum\_{n=0}^\infty \frac{1}{2^{n+1}} \bigg(\frac{n}{n+1}\bigg) = 1-\log2 = 30.69\%.$$
Half of the time, you would observe $0\%$ girls.
If you have multiple families, the average of the observed percentage of girls in the population will increase.
For 2 families, the average percentage of girls would be
$$\sum\_{n=0}^\infty \frac{n+1}{2^{n+2}} \bigg(\frac{n}{n+2}\bigg) = \log 4 - 1 = 38.63\%.$$
More generally, the average percentage for $k$ families is
$$\sum\_{n=0}^\infty \frac{n+k-1 \choose k-1}{2^{n+k}} \bigg(\frac{n}{n+k}\bigg) = \frac{k}{2}\bigg(\psi\left(\frac{k+2}2\right)-\psi\left(\frac{k+1}2\right)\bigg)$$
where $\psi$ is the [digamma function](http://en.wikipedia.org/wiki/Digamma_function) which satisfies
$$ \psi(m) = -\gamma + \sum\_{i=1}^{m-1} \frac1i = -\gamma + H\_{m-1}$$
$$ \psi\left(m+\frac12\right) = -\gamma -2\log 2 + \sum\_{i=1}^m \frac{2}{2i-1}.$$
With a little work, one can verify that this goes to $1/2$ as $k\to \infty$. So, for a large population such as a country, the official answer of $1/2$ is approximately correct, although the explanation is misleading. In particular, for $10$ couples, the expected percentage of girls is $10 \log 2 - 1627/252 = 47.51\%$ contrary to what the official answer suggests. With $k$ families, the expected proportion is about $1/2 - 1/(4k)$.
It is not enough to argue that the expected number of boys equals the expected number of girls, since we want $E[G/(G+B)] \ne E[G]/E[G+B].$ Expectation is linear, but not multiplicative for dependent variables, and $G$ and $G+B$ are not independent even though $G$ and $B$ are.
| 165 | https://mathoverflow.net/users/2954 | 17963 | 11,993 |
https://mathoverflow.net/questions/17953 | 27 | Let $\mathcal C$ be a category. Recall that a morphism $f : X \to Y$ is **epi** if $$\circ f: \hom(Y,Z) \to \hom(X,Z)$$ is injective for each object $Z \in \mathcal C$. ($f$ is **mono** if $f\circ : \hom(Z,X) \to \hom(Z,Y)$ is injective.)
Let $\mathcal C,\mathcal D$ be categories. Then $\hom(\mathcal C,\mathcal D)$, the collectional of all functors $\mathcal C \to \mathcal D$, is naturally a category, where the morphisms are **natural transformations**: if $F,G: \mathcal C \to \mathcal D$ are functors, a natural transformation $\alpha: F \Rightarrow G$ assigns a morphism $\alpha(x) : F(x) \to G(x)$ in $\mathcal D$ for each object $x \in \mathcal C$, and if $f: x \to y$ is a morphism in $\mathcal C$, then $\alpha(y) \circ F(f) = G(f) \circ \alpha(x)$ as morphisms in $\mathcal D$.
>
> Given a natural transformation, can I check whether it is epi (or mono) by checking pointwise? I.e.: is a natural transformation $\alpha$ epi (mono) iff $\alpha(x)$ is epi (mono) for each $x$?
>
>
> If not, is there an implication in one direction between whether a natural transformation is epi and whether it is pointwise-epi?
>
>
>
A more general question, one that I never really learned, is what types of properties of a functor are "pointwise" in that they hold for the functor if they hold for the functor evaluated at each object. E.g.: is the (co)product of functors the pointwise (co)product?
| https://mathoverflow.net/users/78 | Can epi/mono for natural transformations be checked pointwise? | Theo, the answer is basically "yes". It's a qualified "yes", but only very lightly qualified.
Precisely: if a natural transformation between functors $\mathcal{C} \to \mathcal{D}$ is pointwise epi then it's epi. The converse doesn't *always* hold, but it does if $\mathcal{D}$ has pushouts. Dually, pointwise mono implies mono, and conversely if $\mathcal{D}$ has pullbacks.
The context for this --- and an answer to your more general question --- is the slogan
>
> (Co)limits are computed pointwise.
>
>
>
You have, let's say, two functors $F, G: \mathcal{C} \to \mathcal{D}$, and you want to compute their product in the functor category $\mathcal{D}^\mathcal{C}$. Assuming that $\mathcal{D}$ has products, the product of $F$ and $G$ is computed in the simplest possible way, the 'pointwise' way: the value of the product $F \times G$ at an object $A \in \mathcal{C}$ is simply the product $F(A) \times G(A)$ in $\mathcal{D}$. The same goes for any other shape of limit or colimit.
For a statement of this, see for instance 5.1.5--5.1.8 of [these notes](http://www.maths.ed.ac.uk/~tl/msci/). (It's probably in *Categories for the Working Mathematician* too.) See also sheet 9, question 1 at the page linked to. For the connection between monos and pullbacks (or epis and pushouts), see 4.1.31.
You do have to impose this condition that $\mathcal{D}$ has all (co)limits of the appropriate shape (pushouts in the case of your original question). Kelly came up with some example of an epi in $\mathcal{D}^\mathcal{C}$ that isn't pointwise epi; necessarily, his $\mathcal{D}$ doesn't have all pushouts.
| 35 | https://mathoverflow.net/users/586 | 17977 | 12,004 |
https://mathoverflow.net/questions/17612 | 2 | Suppose I have a continuous map $f:X\rightarrow Y$. Then one can wonder, whether for every open set $U\subset X$ the set
$U':=\{x\in X|f^{-1}(f(x))\subset U\}$ is open again. This is not true in general, as the example
$U:=\{(x,y)|\quad |y|<\frac{1}{x^2+1}\}\subset \mathbb{R}^2$ and $f:\mathbb{R}^2\rightarrow \mathbb{R} \qquad (x,y)\mapsto y$
shows. Here $U'$ is just $\mathbb{R}\times \{0\}$.
So the question is: Which continuous maps $f:X\rightarrow Y$ have the property, that for every open set $U\subset X$, the set $U'$ is also open?
| https://mathoverflow.net/users/3969 | Finding saturated open sets | The definition of $U'$ doesn't depend on the Topology on $Y$.
Given any map $f:X\rightarrow Y$, we can consider the equivalence relation $x'\sim x$, iff $f(x)=f(x')$ on $X$ and factor the map $f$ as $X\rightarrow X/\sim \quad \rightarrow Y$. Call the first $f'$. Note that,the second map is injective. Hence we get the same set $U'$ for any set $U$, if we replace $f$ by $f'$.
So we might assume without restriction, that $f:X\rightarrow X/\sim $ is a quotient map. Then the upper statement is equivalent to saying, that $f$ is closed.
Suppose $f$ is closed. Then one gets for the complements $U'^c=f^{-1}(f(U^c))$. And hence $U'^c$ must be closed.
There is the [closed map lemma](http://en.wikipedia.org/wiki/Open_and_closed_maps), which shows, that under the nice conditions mentioned in the comments above properness is also sufficient.
Now suppose $f:X\rightarrow X/\sim$ is not closed and a quotient map. Then there is a closed set $A\subset X$, such that its image is not closed. By definition of the quotient topology, a subset of $X/\sim$ is closed, if and only if its preimage is closed. So $f^{-1}(f(A))$ is not closed and hence $U=A^c$ gives the desired counterexample.
So we get: A map $f:X\rightarrow Y$ has the property, if and only if the induced map $X\rightarrow X/\sim$ is closed.
| 1 | https://mathoverflow.net/users/3969 | 17978 | 12,005 |
https://mathoverflow.net/questions/10937 | 3 | Let $G$ be a simply connected semi-simple algebraic group over an algebraically closed field of positive characteristic. The Steinberg tensor product theorem gives a tensor product decomposition of an irreducible rational $G$-module $S(\lambda)$ with heighest weight $\lambda$ according to the $p$-adic expansion of $\lambda$.
I am trying to understand the proof of this theorem as given by Cline, Parshall, Scott in Journal of Algebra 63, 264-267 (1980). I have two major problems with this proof:
1. In theorem 1 where it is proven that any irreducible $\mathcal{U}$-module over the restricted universal enveloping algebra extends uniquely to a rational $G$-module, how do I get this representation $\rho:G \rightarrow PGL(V)$? I was able to see that I have a morphism $G \rightarrow \mathrm{Aut}\_{\_kAlg}(\mathrm{End}\_k(V))$. But what next? It seems to be well-known that the group of algebra-automorphisms of the endomorphism ring is just $PGL(V)$, but why? Moreover, why do I get a morphism of *algebraic* groups?
2. In the proof of theorem 2, why acts the Lie algebra $L(G)$ trivially on $\mathrm{Hom}\_{L(G)}(S \_1, S)$? (Here $S$ is some irreducible $G$-module, $S \_1$ is an irreducible $L(G)$-submodule of $S$ and $S \_1$ also denotes the unique extension to a rational $G$-module). This is not explicitly mentioned, but I think it is used in b). This must have something to do with theorem 1!?
Any hints and ideas are welcome.
| https://mathoverflow.net/users/717 | Proof of Steinberg's tensor product theorem | The 1980 CPS paper is short but not easy to read without enough background.
They gave the first conceptual alternative to Steinberg's somewhat opaque
and computational proof of the tensor product theorem in 1963 (which built
on the 1950s work of Curtis on "restricted" Lie algebra representations
coming from the algebraic group plus the older work of Steinberg's teacher
Richard Brauer on rank 1). Steinberg relied quite a bit on working with
covering groups and projective representations.
CPS already realized the importance of
getting beyond the Lie algebra by using Frobenius kernels. The best modern
source is the large but well-organized 1987 book by J.C. Jantzen,
Representations of Algebraic Groups (expanded AMS edition in 2003). Here
the foundations are worked out thoroughly and the CPS proof is given an
efficient treatment in part II, 3.16-3.17. While CPS had in mind the
analogy with Clifford theory for finite groups, Jantzen gives a self-contained
treatment avoiding use of projective representations or Skolem-Noether.
Apart from sources, the essential goal is to single out the finitely many
"restricted" simple modules for the Lie algebra among the infinitely many
simple (rational) modules for the ambient algebraic group, then realize the
latter modules as twisted tensor products of the former. This requires a notion of
Frobenius morphism for each power of the prime, Frobenius kernels being
infinitesimal group schemes. The Lie algebra just plays the role of
first Frobenius kernel (a normal subgroup scheme), so the Clifford theory
analogue developed by Ballard and CPS makes sense here.
| 5 | https://mathoverflow.net/users/4231 | 17983 | 12,007 |
https://mathoverflow.net/questions/17979 | 5 | Let $Q$ and $R$ be two acyclic quivers which differ only in the directions of their arrows (i. e., the underlying undirected graphs are the same).
**1.** Does there exist an isomorphism of additive categories $\mathrm{Rep}Q\to\mathrm{Rep}R$ ?
At the moment, I am not even 100% sure about the weaker statement that there exists an equivalence of additive categories $\mathrm{Rep}Q\to\mathrm{Rep}R$. (My intuition for this is pretty much zero.)
In before reflection functors - they are not isomorphisms. (But yes, it's a nice exercise to see that we can obtain $R$ from $Q$ by a sequence of admissible reflections, where an "admissible reflection" means choosing some sink in $Q$ and switching all the arrows to $Q$. Unfortunately, the corresponding reflection functors may turn some nonzero representations of $Q$ to zero.)
**2.** Are the representation rings of $Q$ and $R$ isomorphic? The isomorphy of their additive groups follows from Gabriel's theorem, but it is not clear to me what this does to tensor products.
**1+2.** Does there exist an isomorphism of tensor categories $\mathrm{Rep}Q\to\mathrm{Rep}R$ ?
**3.** If some of these answers are No, what if we restrict ourselves to Dynkin quivers?
I am new to quivers, so I'm sorry if this has been already talked over a dozen of times here.
| https://mathoverflow.net/users/2530 | Acyclic quivers differing only in arrow directions: functorial isomorphism of representation categories? | The categories are not equivalent. In fact, an acyclic quiver is determined (up to non-unique isomorphism) by the equivalence class of its category of representations. The construction is simple: Find the isomorphism classes of simple objects. These are in bijection with the vertices. For any two simple objects $S$ and $S'$, the number of arrows from vertex $S$ to vertex $S'$ is the dimension of $\mathrm{Ext}^1 (S, S')$. The nonuniqueness is because we need to choose a basis of $\mathrm{Ext}^1 (S, S')$.
---
There is a lot more to say on this subject, but it doesn't go in the direction your questions are pointing. When the underlying graphs of $Q$ and $R$ are trees, then it is true that the bounded **derived** categories of $\mathrm{Rep} \ Q$ and $\mathrm{Rep} \ R$ are isomorphic. The basic point here is that, although the reflection functors are not equivalences of categories, the derived reflection functors are equivalences of derived categories. Of course, every Dynkin diagram is a tree, so in particular this is true for Dynkin diagrams.
There is a version of this for non-trees, but I don't know a reference for it nor the exact statement and I don't want to hunt one down until I know whether derived categories are something that you love or that you fear. The proof, of course, is completely different.
Here is a result of Kac you might be happier with. Let our ground field be a finite field $\mathbb{F}\_q$ and fix a dimension vector $d$. Then the number of isomorphism classes of representations of $Q$ and $R$, of dimension $d$, over $\mathbb{F}\_q$, is the same. ([Infinite root systems, representations of graphs and invariant theory](http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=171997), Theorem 1.)
Morally, one wants to work over an arbitrary field. The statement then is a statement about the stacks of $Q$ and $R$ representations of dimension $d$. But, again, to formulate this you need to know about a lot of machinery: stacks, perverse sheaves, derived categories again.
---
In short, the categories are not equivalent. There are very close relations between them, but the best formulations of those relations use sophisticated category theory. You can often see shadows of these relations by counting points over $\mathbb{F}\_q$.
| 13 | https://mathoverflow.net/users/297 | 17987 | 12,010 |
https://mathoverflow.net/questions/18002 | 18 | Suppose $G\_i$ are finite groups for $i=1,2$ and G is the direct product of $G\_i$. If V is a finite dimensional irreducible representation of $G$, then it is well known that $V$ is a tensor product of $V\_i$,$i=1,2$ and each $V\_i$ is an irreducible representation of $G\_i$.
The question I have is when $V$ is given, is there a canonical way to construct $V\_i$ from $V$?
| https://mathoverflow.net/users/1832 | decomposition of representations of a product group | I agree with David Speyer's answer, and furthermore there is no canonical way to construct $V\_i$ from $V$. This is a subtle and oft-overlooked point in representation theory, in my opinion. Many texts prove that an irrep of $G\_1 \times G\_2$ is isomorphic to a tensor product of an irrep of $G\_1$ with an irrep of $G\_2$. The typical slick proof relies on character theory -- kind of a cheat, in my view, since it only says something about isomorphism classes.
Here's a categorical explanation of the theorem: Let $G\_1$ and $G\_2$ be finite groups, and let $\pi$ be an irrep of $G\_1 \times G\_2$ on a complex vector space $V$.
Then, for every pair $(\rho\_1, W\_1)$, $(\rho\_2, W\_2)$ of representations of $G\_1, G\_2$, one gets a complex vector space:
$$H\_\pi(\rho\_1, \rho\_2) := Hom\_G(\rho\_1 \boxtimes \rho\_2, \pi).$$
In fact, this extends to a contravariant functor:
$$H\_\pi: Rep\_{G\_1} \times Rep\_{G\_2} \rightarrow Vec.$$
Here we use categories of finite-dimensional complex representations and vector spaces.
This is also functorial in $\pi$, yielding a functor:
$$H: Rep\_G \rightarrow [ Rep\_{G\_1} \times Rep\_{G\_2}, Vec ],$$
where the right side of the arrow denotes the category of functors (for categories enriched in $Vec$). What this demonstrates is that the canonical thing is to take representations of $G$ to objects of an appropriate functor category related to $Rep\_{G\_1}$ and $Rep\_{G\_2}$. By Yoneda's lemma (for categories enriched in $Vec$), there is an embedding of categories:
$$Rep\_{G\_1} \times Rep\_{G\_2} \hookrightarrow [ Rep\_{G\_1} \times Rep\_{G\_2}, Vec ].$$
It turns out -- and this is where some finiteness is important, and a proof necessarily uses some counting, character theory, or the like -- that for any irrep $\pi$ of $Rep\_G$, the functor $H\_\pi \in [Rep\_{G\_1} \times Rep\_{G\_2}, Vec]$ is representable. It is not uniquely representable, but it is uniquely representable up to natural isomorphism.
Practically, what this means is that given an irrep $\pi$ of $G$, there exists an isomorphism $\iota: \pi \rightarrow \pi\_1 \boxtimes \pi\_2$ for some irreps $\pi\_1, \pi\_2$ of $G\_1, G\_2$, respectively. The pair $(\pi\_1, \pi\_2)$ is not unique, but the triple $(\iota, \pi\_1, \pi\_2)$ is unique up to unique isomorphism. This is usually good enough.
| 23 | https://mathoverflow.net/users/3545 | 18007 | 12,021 |
https://mathoverflow.net/questions/18022 | 6 | I was wondering if anyone can suggest a reference which treats the Krull topology. Most of the books I have found don't go into any kind of detail.
It is my understanding that the Krull topology arises primarily as a way of obtaining a correspondence theorem for infinite Galois extensions. Are there other instances where this topology arises naturally?
Any pointers to books or papers will be warmly received.
| https://mathoverflow.net/users/1146 | References and applications involving the Krull Toplogy | I think what you really want to read about, to see things in a bigger context, are profinite groups. A profinite group is a certain class of compact groups which includes Galois groups with their Krull topology. See Chapter 1 of Serre's "Galois Cohomology" and Chapter 6 of Karpilovsky's "Topics in Field Theory" for a discussion of profinite groups aimed at later use in the setting of infinite Galois theory. Also look at Lenstra's treatment of profinite groups at <http://websites.math.leidenuniv.nl/algebra/Lenstra-Profinite.pdf>.
There are whole books on the topic of profinite groups, e.g., by Wilson and by Ribes and Zalesskii. Among profinite groups, the pro-p groups play the basic role that p-groups do among finite groups, and a book-length treatment on an important class of pro-p groups is the book "Analytic pro-p Groups" by Dixon, du Sautoy, Mann, and Segal. Reading a whole book on such topics might be more than you're ready to swallow at this point.
Concerning your question of places other than Galois theory where the Krull topology naturally arises, a better question to ask is where else besides Galois theory the concept of a profinite group is used (because the label "Krull topology" is limited to automorphism groups of field extensions). There are applications of profinite groups in both number theory and group theory. Do a search on the Golod-Shafarevich theorem and its applications.
Returning to Galois groups and their Krull topology, to really understand what's going on with that topology, you ought to study closely an example where the topology can also be experienced in a different way. The simplest important example for that is $p$-power cyclotomic extensions of $\mathbf Q$: ${\rm Gal}({\mathbf Q}(\mu\_{p^\infty})/{\mathbf Q})$ is isomorphic in a really concrete way to
the group ${\mathbf Z}^\times\_p$ of units in the $p$-adic integers, and in particular the Krull topology on the Galois side matches the $p$-adic topology on the $p$-adic side: two automorphisms are close when the $p$-adic units they correspond to are $p$-adically close. (This isomorphism generalizes the natural isomorphism from ${\rm Gal}(\mathbf Q(\mu\_{p^n}))/\mathbf Q)$ to $({\mathbf Z}/p^n{\mathbf Z})^\times$.)
Try to see concretely how the action of ${\mathbf Z}\_p^\times$ on $p$-power roots of unity works and why the subgroup $1 + p^n{\mathbf Z}\_p$ in $\mathbf Z\_p^\times$ corresponds to the intermediate field ${\mathbf Q}(\mu\_{p^n})$. I don't think you can really get a feel for infinite Galois theory without understanding this basic example, so in particular if you don't know what the $p$-adic integers are then stop and learn about them, and then go back to your Krull topology studies.
| 13 | https://mathoverflow.net/users/3272 | 18025 | 12,033 |
https://mathoverflow.net/questions/18034 | 16 | If you have two manifolds $M^m$ and $N^n$, how does one / can one decompose the diffeomorphisms $\text{Diff}(M\times N)$ in terms of $\text{Diff}(M)$ and $\text{Diff}(N)$? Is there anything we can say about the structure of this group? I have looked in some of my textbooks, but I haven't found any actual discussion of the manifolds $\text{Diff}(M)$ of a manifold $M$ other than to say they are "poorly understood."
Can anyone point me to a source that discusses the manifold $\text{Diff}(M)$? My background is in physics, and understanding the structure of these kinds of groups is important for some of the the things we do, but I haven't seen any discussion of this in my differential geometry textbooks.
| https://mathoverflow.net/users/3329 | Can we decompose Diff(MxN)? | The homotopy-type of the group of diffeomorphisms of a manifold are fairly well understood in dimensions $1$, $2$ and $3$. For a sketch of what's known see Hatcher's "Linearization in three-dimensional topology," in: Proc. Int. Congress of. Math., Helsinki, Vol. I (1978), pp. 463-468.
Similarly, the finite subgroups of $Diff(M)$ are well understood in dimensions $3$ and lower. Hatcher's paper is a good reference for that as well, when combined with a few semi-recent theorems.
If you're interested in general subgroups of $Diff(M)$, there's still a fair bit of discussion going on just for subgroups of $Diff(S^1)$, as it contains a pretty rich collection of subgroups.
In high dimensions there's not much known. For example, nobody knows if $Diff(S^4)$ has any more than two path-components. See for example [this](http://garden.irmacs.sfu.ca/?q=op/what_is_the_homotopy_type_of_the_group_of_diffeomorphisms_of_the_4_sphere) little blurb. Some of the rational homotopy groups of $Diff(S^n)$ are known for $n$ large enough.
I wrote a survey on what's known about the spaces $Diff(S^n)$, and spaces of smooth embeddings of one sphere in another $Emb(S^j,S^n)$ a few years ago: [*A family of embedding spaces* (Budney, 2006)](https://arxiv.org/abs/math/0605069).
Getting back to your earlier question, groups of diffeomorphisms of connect-sums can be pretty compicated objects. In dimension $2$ it's already interesting. For example, $Diff(S^1 \times S^1)$ has the homotopy-type of $S^1 \times S^1 \times GL\_2(\mathbb Z)$. Diff of a connect-sum of $g$ copies of $S^1 \times S^1$ has the homotopy-type of a discrete group provided $g>1$, this is called the mapping class group of a surface of genus $g$. It's a pretty complicated and heavily-studied object. In the genus $g=2$ case this group is fairly similar to the braid group on $6$ strands.
In dimension $3$, it's an old theorem of Hatcher's that $Diff(S^1 \times S^2)$ doesn't have the homotopy-type of a finite-dimensional CW-complex, as it has the homotopy-type of $O\_2 \times O\_3 \times \Omega SO\_3$. I've been spending a lot of time recently, studying the homotopy-type of $Diff(M)$ when $M$ is the complement of a knot in $S^3$, and knot complements in general. The paper of mine I linked to goes into some detail on this.
From the perspective of differential geometry, the homotopy-type of $Diff(S^n)$ is rather interesting as it's closely related to the homotopy-type of the space of "round Riemann metrics" on $S^n$. This is a classic construction, is outlined in my paper but it goes like this: $Diff(S^n)$ has the homotopy type of a product $O\_{n+1} \times Diff(D^n)$ where the diffeomorphisms of $D^n$ are required to be the identity on the boundary -- this is a local linearization argument. $Diff(D^n)$ has the homotopy-type of the space of round metrics on $S^n$. The idea is that any two round metrics are related by a diffeomorphism of $S^n$. So $Diff(S^n)$ acts transitively on the space of round metrics (with a fixed volume, say), and the stabilizer of a round metric is $O\_{n+1}$ basically by the definition of a round metrics. Kind of silly but fundamental.
edit: I should add, there are some nice theorems about $\pi\_0 Diff(S^1 \times D^n)$ for $n$ at least 5, and similarly $\pi\_0 Diff( (S^1)^n )$, due to Hatcher and Wagoner. They derive their results in some sense indirectly, by getting a strong understanding of the pseudo-isotopy diffeomorphisms of $S^1 \times D^n$. One way to think about their work, is that the isotopy-classes of diffeomorphisms of $S^1 \times D^n$ (fixing the boundary pointwise) is governed by three groups: (1) $\pi\_0 Diff(D^n)$, (2) $\pi\_0 Diff(D^{n+1})$ and (3) $\pi\_0 Emb(D^n, S^1 \times D^n) / Diff(D^n)$. Think of this last group as the isotopy-classes of embedded n-discs in $S^1 \times D^n$ that agree with a standard linear embedding on the boundary. i.e. these are submanifolds without parametrization. It turns out this is a group with a stacking construction. Hatcher and Wagoner show that $\pi\_0 Diff(S^1 \times D^n)$ is the direct sum of these three groups, with this group of embedded discs being an infinitely-generated $2$-torsion group.
Recently David Gabai and I were able to give a weak analogue to this Hatcher-Wagoner theorem but in dimension $4$, i.e. for $\pi\_0 Diff(S^1 \times D^3)$. While we have not managed any new results about $\pi\_0 Diff(D^4)$, we can show $\pi\_0 Emb(D^3, S^1 \times D^3)$ is infinitely-generated, even rationally. From a certain perspective our embeddings of $D^3$ in $S^1 \times D^3$ are fairly similar to the Hatcher-Wagoner embeddings, but we rely on slightly different geometry than they do. Roughly speaking, our embeddings stem more from the ability for $S^2 \sqcup S^1$ being able to **link** in $S^4$, while Hatcher and Wagoner's construction has more to do with $S^i \sqcup S^j$ for $i+j=n$ being able to **Hopf link** in $S^n$ when $i,j \geq 3$. In a vague sense that is the key difference between our diffeomorphisms being rationally independent, while theirs are $2$-torsion.
| 28 | https://mathoverflow.net/users/1465 | 18035 | 12,038 |
https://mathoverflow.net/questions/1072 | 7 | I've been trying to find a definition of an *infinite permutation* on-line without much success. Does there exist a canonical definition or are there various ways one might go about defining this?
The obvious candidate I guess would be a bijection p : {1,2,...} -> {1,2,...} between the natural numbers. One might also try to use the Robinson-Schensted correspondence between permutations of length n and pairs of standard Young tableaux of size n. Then one would need a definition of infinite Young tableaux.
Another correspondence that might be used is between permutations and permutation matrices.
| https://mathoverflow.net/users/340 | Definition of infinite permutations | There are two closely related definitions which satisfy the properties you want.
First, consider the group $\Sigma\_k$ of all bijections $\pi: \Bbb Z \to \Bbb Z$ such that $\pi(x+k) = \pi(x)+k$ for all $x$. Note that $S\_k$ is a subgroup in $\Sigma\_k$ - simply take any permutation of $\{1,\ldots,k\}$ and extend it periodically to all $x$. This group ([introduced by Lusztig](http://www.ams.org/journals/tran/1983-277-02/S0002-9947-1983-0694380-4/)) is finitely generated and is closely related to affine Lie algebra $\widehat A\_k$. The RSK algorithm does not exactly work here, but Lusztig does study the shape of Young diagrams (of what would be resulting two tableaux). The shape is a partition of $k$, and can be described using decreasing subsequences, extending Curtis Greene's theorem (I forgot if this is in Lusztig's paper or my own easy observation).
Second, a somewhat related definition is the group $\Phi\_k$ of bijections $\pi: \Bbb N \to \Bbb N$ such that $\pi(x+k) = \pi(x)+k$ for all $x$ large enough. I studied this definition in [this paper](http://www.math.ucla.edu/~pak/papers/inf2.pdf). This group $\Phi\_k$ is also finitely generated. It is very suitable for RSK, which is not always, but sometimes invertible. The asymptotic shape I defined is essentially the same as Lusztig's. Neither I nor anyone else studied the infinite matrix extension. The infinite permutation version is already difficult enough.
| 8 | https://mathoverflow.net/users/4040 | 18049 | 12,047 |
https://mathoverflow.net/questions/18060 | 9 | I had one or two little fights with correspondences in the context of algebraic geometry where an elementary correspondence $C:X\to Y$ of connected smooth $k$-Schemes seems to be defined as an irreducible closed (reduced) subscheme $C\hookrightarrow Y\times\_k X$, such that the projection to $X$ is finite an surjective.
Further, I have vaguely heard of correspondences in topology (invented by Lefschetz?) where it seems to me, that such a thing is a cohomology class in $H^\*(Y\times X)$ for compact, oriented manifolds $X,Y$. Using Poincare duality and the cohomology pushforward functor $(-)\_!$ I can associate a cohomology class $(\Delta\_f)\_!(1)$ in $H^\*(Y\times X)$ to a map $X\to Y$.
My questions are:
1. I can not really see the analogy of the concepts, except that in booth cases one can associate a corresponces to a morphism by its graph. So, what (or how deep) is the analogy?
2. I know (only a very few) applications of correspondences in algebraic geometric, but of none in topology. What are they good for? Where can I find applications?
| https://mathoverflow.net/users/2146 | Correspondences in Topology | For simplicity and definiteness, let's assume that $X$ and $Y$ are smooth and compact (and orientable, which will always be the case if they are complex varieties),
and let $n$ be the dimension of $Y$.
First of all, it might help to note that $H^n(X\otimes Y) \cong H^\*(X)\otimes H^{n-\*}(Y) \cong
Hom(H^\*(Y),H^\*(X))$, where for the final assertion I am using that $Y$ is smooth and compact, so that its cohomology satisfies Poincare duality. Thus if $Z$ is a cycle in $X\otimes Y$,
of dimension equal to that of $X$ (and so of comdimension $n$) it induces a cycle class in $H^n(X\times Y)$, which in turn induces a map from cohomology of $X$ to that of $Y$.
If $f:X\to Y$ and $Z = \Gamma\_f$ is the graph of $f$ then this map is just the pull-back of cohomology classes by $f$.
So correspondences in the sense of physical cycles on $X\times Y$ induce correspondences in the sense of cohomology classes on $X\times Y$, which in turn
induce morphisms on cohomology. If you like, you can strengthen the analogy with the $\Gamma\_f$ case by thinking of a correspondence as a multi-valued function. Functions induce morphisms on cohomology;
but since cohomology is linear (you can add cohomology classes), correspondences also induce
morphisms on cohomology (you can simply add up the multiples values!). This gives the same construction as
the more formal one given above.
Ben Webster notes in his answer that geometric representation theory provides a ready supply of correspondences. So does the theory of arithmetic lattices in Lie groups and the associated symmetric space. (I am thinking of Hecke correspondences and the resulting action of Hecke operators on cohomology.)
A very general framework, which I think covers both contexts, is as follows: suppose that
a group $G$ acts on a space $X$, and that $H \subset Aut(X)$ is another subgroup commensurable with $G$, i.e. such that $G \cap H$ has finite index in each of $G$ and $H$.
Then (perhaps under some mild assumptions) $(G\cap H)\backslash X \hookrightarrow (G \backslash X \times H \backslash X)
$ is a correspondence (in the physical, geometric sense) which will give a correspondence
in cohomology via its cycle class. The resulting maps on cohomology are (a very general form of) Hecke operators.
| 8 | https://mathoverflow.net/users/2874 | 18065 | 12,052 |
https://mathoverflow.net/questions/18031 | 23 | A couple weeks ago I attended a talk about the Keel-Mori theorem regarding existence of coarse moduli spaces for Deligne-Mumford stacks with finite inertia. Here are some questions that I have been wondering about since then: What are some applications of this theorem? What does it matter if a DM stack has a coarse space? What are examples of things that we can do with the coarse space that we maybe can't do with the stack? Given (for instance) a moduli problem, what does the existence of a coarse moduli space tell us that the existence of a DM moduli stack doesn't tell us?
Since the coarse space, if it exists, is probably determined by the stack (is it?), I should probably be asking instead: What can we do more easily or more directly with a coarse space than with a stack?
Here is a bad answer: If we are interested in intersection theory (as in e.g. Gromov-Witten theory), then the existence of the coarse space can help us to circumvent having to develop an intersection theory for stacks. But clearly this is a pretty lame answer.
| https://mathoverflow.net/users/83 | What can we do with a coarse moduli space that we can't do with a DM moduli stack? | An example is Deligne's theorem on the existence of good notion of quotient $X/G$ of a separated algebraic space $X$ under the action of a finite group $G$, or relativizations or generalizations (with non-constant $G$) due to D. Rydh. See Theorem 3.1.13 of my paper with Lieblich and Olsson on Nagata compactification for algebraic spaces for the statement and proof of Deligne's result in a relative situation, and Theorem 5.4 of Rydh's paper "Existence of quotients..." on arxiv or his webpage for his generalization.
Note that in the above, there is no mention of DM stacks, but they come up in the proof! The mechanism to construct $X/G$ (say in the Deligne situation or its relative form) is to prove existence of a coarse space for the DM stack $[X/G]$ via Keel-Mori and show it has many good properties to make it a reasonable notion of quotient. Such quotients $X/G$ are very useful when $X$ is a scheme (but $X/G$ is "only" an algebraic space), such as for reducing some problems for normal noetherian algebraic spaces to the scheme case; cf. section 2.3 of the C-L-O paper. I'm sure there are numerous places where coarse spaces are convenient to do some other kinds of reduction steps in proofs of general theorems, such as reducing a problem for certain DM stacks to the case of algebraic spaces.
Also, Mazur used a deep study of the coarse moduli scheme associated to the DM stack $X\_0(p)$ in his pioneering study of torsion in and rational isogenies between elliptic curves over $\mathbf{Q}$ (and these modular curves show up in numerous other places). But those specific coarse spaces are schemes and can be constructed and studied in more concrete terms without needing the fact that they are coarse spaces in the strong sense of the Keel-Mori theorem, so I think the example of Deligne's theorem above is a "better" example.
| 13 | https://mathoverflow.net/users/3927 | 18067 | 12,053 |
https://mathoverflow.net/questions/18058 | 3 | Are there any known statements that are ***provably*** independent of $ZF + V=L$? A similar question was asked [here](https://mathoverflow.net/questions/11480/on-statements-independent-of-zfc-vl) but focusing on "interesting" statements and all examples of statements given in that thread are not provably indepedent of $ZF + V=L$, they all raise the consistency strength bar. For example, the claim that "there exists an inaccessible" is independent of $ZF + V=L$, is really just an assumption. Because of Gödel's second incompleteness theorem, we cannot prove this. It is well possible that $ZFC$ proves "there is no inaccessible". The same holds for $Con(ZF)$ or "there is a set model of ZF". Those are assumed to be independent of $ZF + V=L$, but this cannot be proved without large cardinal assumptions.
So my question is: Is there any known (not necessarily "interesting" or "natural") statement $\phi$ and an elementary proof of $Con(ZF) => Con(ZF + V=L + \phi) \wedge Con(ZF + V=L + \neg\phi)$? Or is there at least a metamathematical argument that such statements should exists? (Contrast this with the situation of $ZFC$ and $CH$!)
And if not: Might $ZF + V=L$ be complete in a weak sense: There is no statement provably independent of it?
What is known about this?
| https://mathoverflow.net/users/4607 | On statements provably independent of ZF + V=L | The Incompleteness theorem provides exactly the requested independence. (But did I sense in your question that perhaps you thought otherwise?)
The Goedel Incompleteness theorem says that if T is any consistent theory interpreting elementary arithmetic with a computable list of axioms, then T is incomplete. Goedel provided a statement σ, such as the "I am not provable" statement, which is provably equivalent to Con(T), or if you prefer, the Rosser sentence, "there is no proof of me without a shorter proof of my negation", such that T does not prove σ and T does not prove ¬σ.
This establishes Con(T) implies Con(T + σ) and Con(T + ¬σ), as you requested. [Edit: one really needs to use the [Rosser sentence](http://en.wikipedia.org/wiki/Rosser%27s_trick) to make the full conclusion here.]
In particular, this applies to the theory T = ZFC+ V=L, since this theory interprets arithmetic and has a computable list of axioms. Thus, this theory, if consistent, is incomplete, using the corresponding sentences σ above. Since it is also known (by another theorem of Goedel) that Con(ZF) is equivalent to Con(ZFC + V=L). This establishes the requrested implication:
* Con(ZF) implies Con(ZFC + V=L + σ) and Con(ZFC + V=L + ¬σ)
The Incompleteness theorem can be proved in a very weak theory, much weaker than ZFC or even PA, and this implication is similarly provable in a very weak theory (PA suffices).
One cannot provably omit the assumption Con(ZF) of the implication, since the conclusion itself implies that assumption. That is, the existence of an independent statement over a theory directly implies the consistency of the theory. So since we cannot prove the consistency of the theory outright, we cannot prove the existence of any independent statements. But in your question, you only asked for relative consistency (as you should to avoid this triviality), and this is precisely the quesstion that the Incompleteness theorem answers.
| 8 | https://mathoverflow.net/users/1946 | 18076 | 12,061 |
https://mathoverflow.net/questions/17711 | 10 | I'm looking at algorithms to construct short paths in a particular Cayley graph defined in terms of quadratic residues. This has led me to consider a variant on Lagrange's four-squares theorem.
The Four Squares Theorem is simply that for any $n \in \mathbb N$, there exist $w,x,y,z \in \mathbb N$ such that
$$
n = w^2 + x^2 + y^2 + z^2 .
$$
Furthermore, using algorithms presented by Rabin and Shallit (which seem to be state-of-the-art), such decompositions of $n$ can be found in $\mathrm{O}(\log^4 n)$ random time, or about $\mathrm{O}(\log^2 n)$ random time if you don't mind depending on the ERH or allowing a finite but unknown number of instances with less-well-bounded running time.
I am considering a Cayley graph $G\_N$ defined on the integers modulo $N$, where two residues are adjacent if their difference is a "quadratic unit" (a multiplicative unit which is also quadratic residue) or the negation of one (so that the graph is undirected). Paths starting at zero in this graph correspond to decompositions of residues as sums of squares.
It can be shown that four squares do not always suffice; for instance, consider $N = 24$, where $G\_N$ is the 24-cycle, corresponding to the fact that 1 is the only quadratic unit mod 24. However, finding decompositions of residues into "squares" can be helpful in finding paths in the graphs $G\_N$. The only caveat is that only squares which are relatively prime to the modulus are useable.
So, the question: let $p$ be prime, and $n \in \mathbb Z\_p ( := \mathbb Z / p \mathbb Z)$. Under what conditions can we efficiently discover multiplicative units $w,x,y,z \in \mathbb Z\_p^\ast$ such that $n = w^2 + x^2 + y^2 + z^2$? Is there a simple modification of Rabin and Shallit's algorithms which is helpful?
Edit: In retrospect, I should emphasize that my question is about *efficiently finding* such a decomposition, and for $p > 3$. Obviously for $p = 3$, only $n = 1$ has a solution. Less obviously, one may show that the equation is always solvable for $n \in \mathbb Z\_p^\ast$, for any $p > 3$ prime.
| https://mathoverflow.net/users/3723 | Lagrange four-squares theorem: efficient algorithm with units modulo a prime? | As indicated by Felipe (primarily in his responses to my comments of his solution above), the problem is actually easy modulo a prime $p > 3$. Here I outline an explicit random poly-time solution, depending on ideas contributed by him.
First, the special case $p = 5$. We can only express 0 as a sum of an even number of quadratic units (which in this case are $\pm 1$), and can only express the quadratic units themselves using exactly one or at least three quadratic units. We set this case aside and assume $p > 5$.
Second, for any $p \equiv 3 \pmod{4}$, the quadratic residues do not include $-1$; therefore $0$ cannot be formed as a quadratic residue. It suffices however to represent any $n \ne 0$ as a sum of two quadratic units, in which case we may easily reduce the problem to expressing $n \in \mathbb Z \setminus$ { 0 } as a sum of two quadratic units.
A classical result is that modulo $p$, and for $p > 5$, the number of ordered pairs $(q,q+1)$ of consecutive quadratic residues is asymptotically a large constant fraction of $p$ (specifically 1/4); and similarly for the number of pairs $(q,q+1)$ for which $q$ is a quadratic unit and $q+1$ a "non-quadratic" unit. We may then express
$$ n = nq(q+1)^{-1} + n(q+1)^{-1} $$
where we choose $q+1$ to have the same "residuacity" as $n$; this is then a sum of two quadratic units. Because the suitable pairs $(q,q+1)$ occur a large constant fraction of the time, we may easily generate such a pair whether $n$ is a quadratic unit or not; and by efficiently finding roots for $nq(q+1)^{-1}$ and $n(q+1)^{-1}$, we may then find unconditionally find solutions in random polynomial time.
(I post this answer in order to provide an explicit record, and to emphasize that it can be done unconditionally; however I'm upvoting his answer as the one which contributed the useful ideas.)
| 2 | https://mathoverflow.net/users/3723 | 18081 | 12,065 |
https://mathoverflow.net/questions/17927 | 8 | Let $D\subset\mathbb R^2\subset\mathbb R^n$ be a unit planar disc in $\mathbb R^n$. Let $S$ be an orientable two-dimensional surface in $\mathbb R^n$ such that $\partial S=\partial D$. Of course, we have $area(S)\ge area(D)$. Assume that $area(S)< area(D)+\delta$ where $\delta>0$ is small.
Then $S$ is close to $D$ in the following sense: there is a 3-dimensional surface $F$ filling the gap between $S$ and $D$ such that $volume(F)<\varepsilon(\delta)$ where $\varepsilon(\delta)\to 0$ as $\delta\to 0$ ($n$ is fixed). "Filling the gap" means that $\partial F=S-D$.
This fact immediately follows from the compactness theorem for flat norms. But this proof is indirect and does not answer the following questions (I am especially interested in the second one):
1) Are there explicit upper bounds for $\varepsilon(\delta)$? How do they depend on $\delta$ and $n$?
2) Can $\varepsilon(\delta)$ be independent of $n$? Or, equivalently, does the above fact hold true in the Hilbert space?
In the unlikely event that 2-dimensional surfaces are somehow special, what about the same questions about $m$-dimensional surfaces, for a fixed $m$?
Remarks: "Surfaces" here are Lipschitz surfaces or rectifiable currents or whatever you prefer to see in this context. Rather than talking about the filling surface $F$, one could equivalently say that the integral flat norm of $S-D$ is less than $\varepsilon(\delta)$.
| https://mathoverflow.net/users/4354 | Estimating flat norm distance from a planar disc | There is [Almgren's isoperimetric inequality](http://www.ams.org/bull/1985-13-02/S0273-0979-1985-15393-5/S0273-0979-1985-15393-5.pdf):
>
> Let $\Sigma$ be a $k$-surface in $\mathbb R^n$. Assume $vol \_k \Sigma \le vol\_k S^k$. Then one can fill $\Sigma$ by a $(k+1)$-surface with volume $\le vol\_{k+1} B^{k+1}$. (Here the "surfaces" might have singularities.)
>
>
>
I will use it to show that there is an estimate $\epsilon(\delta)$ which does not depend on $n$.
Take $r$-nbhd $Z\_r$ of $D$.
Note\* that one can give an explicite estimate of $r$, independent of $n$ so that total area of $S$ outside of $Z\_r$ is very small. Moving a bit $r$, one can make the length of intersection curve $\gamma=\partial Z\_r\cap S$ sufficiently small.
Use Almgren to fill $\gamma$ by a surface;
it breaks $S$ into two pieces $S=S\_1+S\_2$;
* the surface $S\_1$ lies in $Z\_r$ and $\partial S\_1=\partial D$,
* the surface $S\_2$ has small area and $\partial S\_2=0$.
Fill both $S\_1$ and $S\_2$ separately:
* taking all segments from point on $S\_1$ to its projection on $D$ gives a filling of $S\_1-D$
* fill $S\_2$ using Almgren again.
---
(\*)There is a map $\mathbb R^n\to D$ which decrease distances by some factor $k=k(r)<1$ outside of $Z\_r$ and $k(r)$ can be found explicitly.
So if an essential piece of $S$ is outside of $Z\_r$ then the area of $S$ is essentially bigger that $area(D)$.
Say, take $f(x)=$ "sum of maximal and minimal distance to the points in $D$".
This function is convex and it is constant on $D$.
Take Sharafutdinov retruction for the level sets of this function.
| 4 | https://mathoverflow.net/users/1441 | 18083 | 12,067 |
https://mathoverflow.net/questions/18008 | 7 | According to some form of Tannakian reconstruction, given a finite tensor category with a fiber functor to the category of vector spaces, one determines a Hopf algebra by considering tensor endomorphisms of the fiber functor. As far as I know, a similar procedure is used to reconstruct a group from its symmetric tensor category of representations.
I am curious about what happens if one is given a finite tensor category $\mathcal{C}$ and a tensor functor $\mathcal{C} \to Rep(G)$ for $G$ a finite group. It follows that there should exist a Hopf algebra $H$ (by the previous reconstruction business applied to the composition of this tensor functor with the forgetful functor $Rep(G) \to \mathrm{Vect}$) and homomorphism $\mathbb{C}[G] \to H$.
>
> Under what conditions will $H$ be a
> semidirect product of $G$ with some Hopf algebra?
>
>
>
| https://mathoverflow.net/users/344 | Is there a relative version of Tannakian reconstruction? | Akhil,
Let $\mathcal{C}$ be a tensor category, and let $(A,\mu) \in \mathcal{C}-Alg$ be an algebra in $\mathcal{C}$. So $\mu:A\otimes A\to A$ is a morphism in $\mathcal{C}$ and $\otimes$ here means the $\otimes$ in $\mathcal{C}$ (of course there isn't another one around at this point, but I mean to emphasize it's not just the $\otimes$ of Vect).
In this context, it makes sense to talk about $A$-modules in $\mathcal{C}$, whose definition you can guess. These form a k-linear abelian category $D$ with a forgetful functor to $\mathcal{C}$ which forgets the $A$ action.
Now if $\mathcal{C}$ has a fiber functor F, then $\mathcal{C}$ is realized as the Hopf algebra $End(F)$, as you said (well $End(F)^{op}$ I think , but nevermind). The algebra $A$ can be pushed forward by $F$ to an ordinary algebra $F(A)$ in Vect. However, $D$ is not the category of $A$-modules, but a well-known proposition tells that $D$ is the category of $A\rtimes H$-modules, the semi-direct product you asked about.
Notice that no part of the discussion so far asked for any symmetric structure on $\mathcal{C}$, and also $A$ is only an algebra. To define a bialgebra in $\mathcal{C}$, however, one needs $\mathcal{C}$ to be braided, because the compatibility between $\Delta$ and $\mu$ will use the braiding. In your case braiding just means symmetry. I never worked this out in detail, but I imagine that if $A$ is actually a bi-algebra in $\mathcal{C}$, then $D$ gets endowed with a monoidal structure, and that $D$ is the category of $A\rtimes H$-modules, where $A\rtimes H$ is a bi-algebra. Likewise if $A$ is Hopf in $\mathcal{C}$, then $D$ is tensor and $A\rtimes H$ is a Hopf algebra, and $D\cong A \rtimes H$-mod.
---
Thus ends the part where I'm pretty sure I'm not saying anything too incorrect. Below I will try to answer your actual question. I would not trust it though until somebody smarter agrees with it.
---
So, now your question becomes (let's revert to considering algebras at the top and not Hopf algebras, since it should be clear how to extend): Given a functor $F:D\to \mathcal{C}$, when is $F$ the forgetful functor corresponding to some algebra $A \in \mathcal{C}$? I think for this, it will be enough to assume (in addition, of course, to assuming that $F$ is faithful and exact) that $D$ has a projective generator $M$ (although maybe this is guaranteed by a lesser assumption?). This is definitely necessary to be able to realize $D$ as some category of modules of a ring, as you desire, and I imagine that you then let $A=\underline{Hom}(M,M)$ (meaning $\mathcal{C}$-internal homs, which are distinct from $Hom\_\mathcal{C}(M,M)$!), which will be an algebra in $\mathcal{C}$, and you can plug into the above.
You should definitely read Ostrik's <http://arxiv.org/abs/math/0111139> and other papers by Etingof, Nikshych, and Ostrik about fusion and finite tensor categories.
| 7 | https://mathoverflow.net/users/1040 | 18086 | 12,068 |
https://mathoverflow.net/questions/18037 | 9 | Consider the number of fixed points in a permutation chosen uniformly at random from the symmetric group on $n$ elements - this gives a probability distribution. For $k < n$, the $k$-th moments of this distribution are the same as the Poisson distribution with $\lambda = 1$.
Since you can't pick uniformly randomly from an infinite set, it doesn't make sense to ask if you get exactly the Poisson distribution in the limit of the symmetric group on a countably infinite set. But can you get there in another way? Is there a known way to take limits in some category of distributions and groups that leads to the Poisson distribution for the fixed points of some countably infinite group of permutations? Alternately, is there a known structure that can be put on $|S\_\omega|$ that gives rise to a Poisson process that is analogous to the finite case?
Also, are there other infinite families of finite permutation groups that give rise to a similarly elegant characterization of the distribution of fixed points? Say by taking one of the families of finite simple groups and applying the same finite extension to each of them?
(Please excuse me if I accidentally posted this twice - I didn't see the first attempt go through.)
| https://mathoverflow.net/users/4594 | Symmetric groups and Poisson processes | This isn't a problem I've looked at before, but I've been thinking about it since reading your post, and there does seem to be an interesting limit. The following looks like it should all work out, but I haven't gone through all the details yet.
Embed the set $\{1,...,n\}$ into the unit interval $I=[0,1]$ by $\theta(i)=\frac in$. Then look at $\theta(A)$ for the fixed points $A\subset \{1,..,n\}$ of a random permutation. Increasing n to infinity, the distribution of $\theta(A)$ should converge weakly to the [Poisson point process](http://en.wikipedia.org/wiki/Poisson_random_measure) on I with intensity being the standard Lebesgue measure.
That is only the fixed points though. You can look at the limiting distribution of the set of all points contained in orbits of bounded size (≤ m, say), and at the action of the random permutation on this set. This should have a well defined limit, and you can then take the limit m → ∞ to to give a random countable subset of I and a random permutation on this, such that all orbits are finite.
Let In={1/n,2/n,...,n/n} ⊂ I, and consider a random permutation π of this. The probability that any point P ∈ In is an element of an orbit of size r is (1-1/n)(1-2/n)...(1-(r-1)/n)(1/n). The expected number of points lying in such orbits is (1-1/n)...(1-(r-1)/n) and the expected number of orbits of size r is (1-1/n)...(1-(r-1)/n)/r, which tends to 1/r as n → ∞.
Now represent orbits of size r of the permutation π by sequences of distinct elements of the interval I, (x1,...,xr), where π(xi) = xi+1 and π(xr) = x1. We'll identify cyclic permutations of such sequences, as they represent the same orbit and the same action of π on the orbit. So (x1,...,xr) = (xr,x1,...,xr-1). Let I(r) be the space of such sequences up to identification of cyclic permutations.
Then, every permutation π of In gives a set of fixed points A1 ⊂ I(1)=I, orbits of size 2, A2 ⊂ I(2), orbits of size 3, A3 ⊂ I(3), etc.
Then, I propose that A1, A2, A3,... will jointly converge to the following limit; (α1,α2,α3,...) are independent Poisson random measures on I(1),I(2),I(3),... respectively such that the intensity measure of αr is uniform on I(r) with total weight 1/r.
Taking the union of all these orbits gives a random countable subset of I and a random permutation of this.
I used the unit interval as the space in which to embed the finite sets {1,...,n} but, really, any finite measure space (E,ℰ,μ) with no atoms will do. Just embed the finite subsets uniformly over the space (i.e., at random).
| 5 | https://mathoverflow.net/users/1004 | 18087 | 12,069 |
https://mathoverflow.net/questions/18084 | 60 | Wikipedia defines the [Jaccard distance](https://en.wikipedia.org/wiki/Jaccard_index) between sets *A* and *B* as $$J\_\delta(A,B)=1-\frac{|A\cap B|}{|A\cup B|}.$$ There's also a [book](https://books.google.com/books?id=DGjbibiS-S0C&pg=PA38&dq=jaccard+similarity&as_brr=3&ei=iNGbS96kNYfIywTVjZ2cCg&cd=1#v=onepage&q=jaccard%20similarity&f=false) claiming that this is a metric. However, I couldn't find any explanation of why $J\_\delta$ obeys the triangle inequality. The naive approach of writing the inequality with seven variables (e.g., $x\_{001}$ thru $x\_{111}$, where $x\_{101}$ is the number of elements in $(A\cap C) \backslash B$) and trying to reduce it seems hopeless for pen and paper. In fact it also seems hopeless for Mathematica, which is trying to find a counterexample for 20 minutes and is still running. (It's supposed to say if there isn't any.)
Is there a simple argument showing that this is a distance? Somehow, it feels like the problem shouldn't be difficult and I'm missing something.
| https://mathoverflow.net/users/840 | Is the Jaccard distance a distance? | The trick is to use a transform called the Steinhaus Transform. Given a metric $(X, d)$ and a fixed point $a \in X$, you can define a new distance $D'$ as
$$D'(x,y) = \frac{2D(x,y)}{D(x,a) + D(y,a) + D(x,y)}.$$
It's known that this transformation produces a metric from a metric. Now if you take as the base metric $D$ the symmetric difference between two sets and empty set as $a$, what you end up with is the Jaccard distance (which actually is known by many other names as well).
For more information and references, check out Ken Clarkson's survey [Nearest-neighbor searching and metric space dimensions](https://web.archive.org/web/20161027161008/http://kenclarkson.org/nn_survey/p.pdf) (Section 2.3).
| 60 | https://mathoverflow.net/users/972 | 18090 | 12,071 |
https://mathoverflow.net/questions/18085 | 8 | I heard that $Ext(M,N)$ is naturally isomorphic to $Ext(M^\*\otimes N,1)$ where 1 is the trivial representation and $M,N$ some representations of a group $G$.
Can anyone explain why?
Is there an explicit construction of a map from one to the other or does it just follow from some general considerations about derived functors?
Thanks.
| https://mathoverflow.net/users/4477 | Question about Ext | Perhaps the best way to think about this is as follows: pick your favorite injective resolution for N and favorite projective resolution of M. Then $\mathrm{Ext}(M,N)$ is given by taking Hom between these complexes (NOT chain maps, just all maps of representations between the underlying modules), and putting a differential on those in the usual way.
Now, use the usual identification of $\mathrm{Hom}(A,B)\cong \mathrm{Hom}(A\otimes B^\*,1)$ on this complex. So you see, it's the same as if we had tensored the projective resolution of $M$ with the dual of the injective resolution of $N$, which is a projective resolution of $N^\*$, and then taken Hom to 1. Of course, the tensor product of two projective resolutions is a projective resolution of the tensor product, so we see this complex also computes $\mathrm{Hom}(N^\*\otimes M,1)$.`
It also follows by abstract nonsense in one line: isomorphic functors have isomorphic derived functors.
| 11 | https://mathoverflow.net/users/66 | 18092 | 12,073 |
https://mathoverflow.net/questions/18089 | 27 | I am aware that assigning the type of Type to be Type (rather than stratifying to a hierarchy of types) leads to an inconsistency. But does this inconsistency allow the construction of a well-typed term with no normal form, or does it actually allow a proof of False? Are these two questions equivalent?
| https://mathoverflow.net/users/2185 | What is the manner of inconsistency of Girard's paradox in Martin Lof type theory | Girard's paradox constructs a non-normalizing proof of False. You could read Hurken's "A simplification of Girard's paradox", or maybe Kevin Watkin's [formalization in Twelf](http://www.cs.cmu.edu/~kw/research/hurkens95tlca.elf).
In general, these questions are not equivalent, though they often coincide. A "reasonable" type theory will by inspection have no *normal* proofs of False, and so then normalization implies consistency. The inverse (non-normalization => proof of False) is much less obvious, and it is certainly possible to construct reasonable paraconsistent type theories, where non-termination is confined under a monad and does not result in a proof of False.
| 27 | https://mathoverflow.net/users/1015 | 18097 | 12,077 |
https://mathoverflow.net/questions/18074 | 5 | Given $n \in \mathbf{N}$,is always possible to construct a monic polynomial in $\mathbf{Z}[x]$ of degree $2n$, whose roots are in $\mathbf{C} \setminus \mathbf{R}$ and whose Galois group over $\mathbf{Q}$ is $S\_{2n}$?
I have an approximate idea of how to solve the problem for the Galois group (I immagine something related to the Hilbert irreducibility theorem), but I have no idea for the condition on the roots.
Furthermore, is it possible to give an explicit example?
| https://mathoverflow.net/users/1967 | A special integral polynomial | An easy way to ensure that a polynomial $g$ of degree $m$ over $\mathbf{Z}$ has
Galois group $S\_m$ is to take primes $p\_1$, $p\_2$ and $p\_3$
with $g$ irreducible modulo $p\_1$, a linear times an irreducible
modulo $p\_2$ and a bunch of distinct linears times an
irreducible quadratic modulo $p\_3$. Then the Galois group
must be doubly transitive and have a transposition, so it's $S\_m$.
Now take $m=2n$ and a polynomial $f$ over $\mathbf{Q}$ with no real
roots (e.g. $(x^2+1)^n$). Replacing the coefficients of $f$ by
close rationals won't create any real roots. So replace the $x^k$
coefficient of $f$ by a sufficient close rational $a\_k/b\_k$
where $a\_k$ and $b\_k$ are congruent modulo $p\_1 p\_2 p\_3$
respectively to the $x^k$ coefficient of $g$ and to $1$. Then
the new polynomial has rational coefficients, no real roots
and Galois group $S\_{2n}$. You can easily convert it to one
with these properties and integer coefficients should you wish.
| 9 | https://mathoverflow.net/users/4213 | 18106 | 12,083 |
https://mathoverflow.net/questions/18041 | 68 | Peter May said famously that algebraic topology is a subject poorly served by its textbooks. Sadly, I have to agree. Although we have a freightcar full of excellent first-year algebraic topology texts - both geometric ones like Allen Hatcher's and algebraic-focused ones like the one by Rotman and more recently, the beautiful text by tom Dieck (which I'll be reviewing for MAA Online in 2 weeks, watch out for that!) - there are almost no texts which bring the reader even close to the frontiers of the subject.
GEOMETRIC topology has quite a few books that present its modern essentials to graduate student readers - the books by Thurston, Kirby and Vassiliev come to mind - but the vast majority of algebraic topology texts are mired in material that was old when Ronald Reagan was President of The United States. This is partly due to the youth of the subject, but I think it's more due to the sheer vastness of the subject now. Writing a cutting edge algebraic topology textbook - TEXTBOOK, not MONOGRAPH - is a little like trying to write one on algebra or analysis. The fields are so gigantic and growing, the task seems insurmountable.
There are only 2 "standard" advanced textbooks in algebraic topology and both of them are over 30 years old now: Robert Switzer's *Algebraic Topology: Homology And Homotopy* and George Whitehead's *Elements of Homotopy Theory*. Homotopy theory in particular has undergone a complete transformation and explosive expansion since Whitehead wrote his book. (That being said, the fact this classic is out of print is a crime.) There is a recent beautiful textbook that's a very good addition to the literature, Davis and Kirk's *Lectures in Algebraic Topology* - but most of the material in that book is pre-1980 and focuses on the geometric aspects of the subject.
We need a book that surveys the subject as it currently stands and prepares advanced students for the research literature and specialized monographs as well as makes the subject accessible to the nonexpert mathematician who wants to learn the state of the art but not drown in it. The man most qualified to write that text is the man to uttered the words I began this post with. His beautiful concise course is a classic for good reason; we so rarely have an expert give us his "take" on a field. It's too difficult for a first course, even for the best students, but it's "must have" supplementary reading. I wish Dr. May - perhaps when he retires - will find the time to write a truly comprehensive text on the subject he has had such a profound effect on. Anyone have any news on this front of future advanced texts in topology?
I'll close this box and throw it open to the floor by sharing what may be the first such textbook available as a massive set of online notes. I just discovered it tonight; it's by Garth Warner of The University Of Washington and available free for download at his website. I don't know if it's the answer, but it sure looks like a huge step in the right direction. Enjoy. And please comment here.
<http://www.math.washington.edu/~warner/TTHT_Warner.pdf>
| https://mathoverflow.net/users/3546 | Algebraic topology beyond the basics: any texts bridging the gap? | At the moment I'm reading the book [Introduction to homotopy theory](http://books.google.com/books?id=CGld8QJPFrgC&printsec=frontcover&dq=paul+selick+introduction+to+homotopy+theory&source=bl&ots=ujT4ahxSI2&sig=uNhStJZkZR5RaeDHwEvvEjK3iSI&hl=en&ei=bwycS7OkB8bt-QbDn5DjAQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAUQ6AEwAA#v=onepage&q=&f=false) by Paul Selick. It is quite short but covers topics like spectral sequences, Hopf algebras and spectra. This is the first place I've found explanations (that I understand) of things like Mayer-Vietoris sequences of homotopy groups, homotopy pushout and pullback squares etc.. The author writes in the preface that the book is inteded to bridge the gap which the OP talks about.
| 16 | https://mathoverflow.net/users/1123 | 18113 | 12,089 |
https://mathoverflow.net/questions/18108 | 10 | Perhaps this is basic knowledge in Riemannian geometry, but I can't seem to figure out the answer. Here is the precise statement of my question.
Let $M$ be a Riemannian manifold, $p$ a point in $M$. Let $R$ be small enough that $exp\_p$ restricts to a diffeomorphism on the ball $B\_R(0)$ of radius $R$ centered at the origin, and let $U\_R$ be the intersection of $B\_R(0)$ and any two dimensional plane through the origin in $T\_p M$. Question: does there exist $R$ such that $exp\_p(U\_R)$ is geodesically convex, in the sense that for every two points of $exp\_p(U\_R)$ the unique geodesic segment connecting them lies entirely in $exp\_p(U\_R)$?
It would be really convenient for me if the answer is yes. If so, I am curious to know if the statement is still true if $P$ is replaced by a subspace of higher dimension, but I only need the result for planes.
Thanks!
| https://mathoverflow.net/users/4362 | Must a surface obtained by exponentiating a plane in a tangent space of a Riemannian manifold be geodesically convex? | No, a generic Riemannian metric does not have totally geodesic 2-dimensional submanifolds at all. The property that you ask for is very rare. For example, it implies that $R(X,Y)Y$ belongs to the linear span of $X$ and $Y$, for every $p\in M$ and every $X,Y\in T\_pM$. This means point-wise constant sectional curvature if I remember correctly.
| 13 | https://mathoverflow.net/users/4354 | 18118 | 12,093 |
https://mathoverflow.net/questions/18054 | 4 | Hello all, the $\Delta$ operator on functions $\mathcal{N} \to \mathbb{R}$
(where $\mathcal N$ denotes $\lbrace 1,2, \ldots , \rbrace$ )defined by
$\Delta(f)(n)=f(n+1)-f(n)$ is well-known and
it is not very hard to show by induction that
$f$ is a polynomial of degree $\leq k$ iff $\Delta^{k+1}(f)$ is identically zero, where
$\Delta^{k+1}$ denotes $\Delta$ iterated $k+1$ times. Now I say that
a function $f : \mathcal{N} \to \mathbb{R}$ is "almost polynomial" iff
$\Delta^{k}(f)$ is a bounded function for some $k$.
My question is : let $\lambda >0$ be a non-integer, and let
$f(n)=n^{\lambda}$. Is it true that $f$ is almost polynomial ?
| https://mathoverflow.net/users/2389 | Is it true that all the "irrational power" functions are almost polynomial ? | One can also get this from a standard form of the remainder in Taylor's theorem. Namely, if $k>\lambda$, and $T\_{k-1}(x)$ is the degree $k-1$ Taylor polynomial of $f(x)$ at $x=n$, then
$$f(x) = T\_{k-1}(x) + \frac{f^{(k)}(\xi\_x)}{k!} (x-n)^k$$
for some $\xi\_x \in [n,x]$. Applying $\Delta^k$ kills the Taylor polynomial term, and when we evaluate the result at $x=n$, the linear combination of the remainders is small when $n$ is large relative to $k$, because each $f^{(k)}(\xi\_x)$ is $O(n^{\lambda-k})$ as $n \to \infty$.
| 10 | https://mathoverflow.net/users/2757 | 18129 | 12,101 |
https://mathoverflow.net/questions/18134 | 3 | I am reading the proof of Theorem Poincaré duality in "principles in AG" of Griffith.
They constructed "dual cell decomposition" of a polyhedra decomposition of manifold $M$ and the cochain complex of this dual cells.
I don't know why this cohomology group of this cochain complex is the singular cohomology group of $M$?
Someone help me give answer? Thanks in advance.
| https://mathoverflow.net/users/4621 | Poincare duality | For any CW complex $X$ one defines a chain complex $C\_\*(X)$: choose an orientation of each cell; the group $C\_n(X)$ is the free abelian group with a basis whose elements correspond to the $n$-cells of $X$ and the differential $C\_n(X)\to C\_{n-1}(X)$ is defined by $c\mapsto \sum\_{c'\subset\partial c} (c,c')c'$ where $(c,c')$ is the incidence number of $c$ and $c'\subset\partial c$ defined as follows.
By the definition of a CW complex one can extend the homeo of an open $n$-ball to $c$ to a map of the closed ball; compose the restriction of this map to the boundary $S^{n-1}$ of the closed ball with the map $X\_{n-1}\to S^{n-1}$ obtained by collapsing all cells of the $n-1$ skeleton of $X$ but $c'$ to a point; the incidence number of $c$ and $c'$ is the degree of the resulting map $S^{n-1}\to S^{n-1}$ where the first sphere is oriented using the "outgoing normal first" rule and the orientation of the second one is induced from $c'$.
This generalizes the chain complex of a simplicial set. The homlogy of $C\_\*(X)$ is isomorphic to the singular homology of $X$, see e.g. Hatcher, Algebraic topology, p. 137 (freely available online) or Milnor, Stasheff, Characteristic classes, Appendix A.
| 3 | https://mathoverflow.net/users/2349 | 18137 | 12,105 |
https://mathoverflow.net/questions/17964 | 7 | * **Suppose You ask a question beginning from "Why some structure is..." or "Why some object has property..." and several
answers arises. Which criteria do You
use to qualify which answer is correct?**
For example [here](http://www.cs.ru.nl/~freek/mizar/) You may find interesting picture (gzipped postscript file) of [proofs formalized in Mizar system.](http://www.cs.ru.nl/~freek/mizar/mml.ps.gz). Mizar library of formalized theorems is really [huge](http://www.mizar.org/JFM/mmlident.html). On the picture You may see, that theorems arises from other and are used in proofs of another ones, forming big graph of structure of theorems formalized in Mizar so far. If I may read something from this graph, there is no theorem which will have more that 3 or 4 incoming edges what means there is no theorem which is used in more that 3 or 4 proofs. Of course there are some with 5 incoming edges, but in fact there is many theorems which have more or less equal number of incoming edges, which may mean that most theorems are equivalently important. Maybe it should be measured by tree deepness? Maybe there is something like Google page rank algorithm for theorems?
Probably we would like to have such relation: "theorems recognized as important should be influential, or foundational for broad area of theory".
* **I understand that one may believe that this is a real state of matter,
but are there any strict results
based on real data in this matter?**
By real data I mean at best **proof theory** analysis, or even **citation analysis**, but not someone opinion (which of course may be enlightening and inspirational). I would like to learn something **about structure of deductive theories**, and not about "real practice". It is the same as in real life: we try to measure risks, and income rate not based on someone opinion but on facts. Could we know the facts here?
It seems from Mizar graph ( which is the only one accessible for me in this area) I could not find any object which will correspond to our intuition of importance of theorems. Maybe this is effect of present Mizar state of affair, and in bigger/other system, some theorems begin central one? Are there any conditions to state such position?
* **What about other proof assistants, as
Isabelle or COQ. Is there any similar
graph from other systems suitable for
such analysis?**
| https://mathoverflow.net/users/3811 | Is there a known way to formalise notion that certain theorems are essential ones? | Although your question is vague in certain ways, one robust answer to it is provided by the subject known as [Reverse Mathematics](http://en.wikipedia.org/wiki/Reverse_mathematics). The nature of this answer is different from what you had suggested or solicited, in that it is not based on any observed data of mathematical practice, but rather is based on the provable logical relations among the classical theorems of mathematics. Thus, it is a mathematical answer, rather than an engineering answer.
The project of Reverse Mathematics is to reverse the usual process of mathematics, by proving the axioms from the theorems, rather than the theorems from the axioms. Thus, one comes to know exactly which axioms are required for which theorems. These reversals have now been carried out for an enormous number of the classical theorems of mathematics, and a rich subject is developing. (Harvey Friedman and Steve Simpson among others are prominent researchers in this area.)
The main, perhaps surprising conclusion of the project of Reverse Mathematics is that it turns out that almost every theorem of classical mathematics is provably equivalent, over a very weak base theory, to one of five possibilities. That is, most of the theorems of classical mathematics turn out to be equivalent to each other in five large equivalence classes.
For example,
* Provable in and equivalent to the theory RCA0 (and each other) are: basic properties of the natural/rational numbers, the Baire Cateogory theorem, the Intermediate Value theorem, the Banach-Steinhaus theorem, the existence of the algebraic closure of a countable field, etc. etc. etc.
* Equivalent to WKL0 (and each other) are the Heine Borel theorem, the Brouer fixed-point theorem, the Hahn-Banach theorem, the Jordan curve theorem, the uniqueness of algebraic closures, etc. , etc. etc.
* Equivalent to ACA0 (and each other) are the Bolzano-Weierstraus theorem, Ascoli's theorem, sequential completeness of the reals, existence of transcendental basis for countable fields, Konig's lemma, etc., etc.
* Equivalent to ATR0 (and each other) are the comparability of countable well orderings, Ulm's theorem, Lusin's separation theorem, Determinacy for open sets, etc.
* Equivalent to Π11 comprehension (and each other) are the Cantor-Bendixion theorem and the theorem that every Abelian group is the direct sum of a divisible group and a reduced group, etc.
The naturality and canonical nature of these five axiom systems is proved by the fact that they are equivalent to so many different classical theorems of mathematics. At the same time, these results prove that those theorems themselves are natural and essential in the sense of the title of your question.
The overall lesson of Reverse mathematics is the fact that there are not actually so many different theorems, in a strictly logical sense, since these theorems all turn out to be logically equivalent to each other in those five categories. In this sense, there are essentially only five theorems, and these are all essential. But their essential nature is mutable, in the sense that any of them could be replaced by any other within the same class.
I take this as a robust answer to the question that you asked (and perhaps it fulfills your remark that you thought ideally the answer would come from proof theory). The essential nature of those five classes of theorems is not proved by looking at their citation statistics in the google page-rank style, however, but by considering their logical structure and the fact that they are logically equivalent to each other over a very weak base theory.
Finally, let me say that of course, the Reverse Mathematicians have by now discovered various exceptions to the five classes, and it is now no longer fully true to say that ALL of the known reversals fit so neatly into those categories. The exceptional theorems are often very interesting cases which do not fit into the otherwise canonical categories.
| 26 | https://mathoverflow.net/users/1946 | 18141 | 12,108 |
https://mathoverflow.net/questions/18094 | 49 | If $p\_n$ is the $n$'th prime, let $A\_n(x) = x^n + p\_1x^{n-1}+\cdots + p\_{n-1}x+p\_n$. Is $A\_n$ then irreducible in $\mathbb{Z}[x]$ for any natural number $n$?
I checked the first couple of hundred cases using Maple, and unless I made an error in the code those were all irreducible.
I have thought about this for a long time now, and asked many others, with no answer yet.
| https://mathoverflow.net/users/4614 | Polynomial with the primes as coefficients irreducible? | I will prove that $A\_n$ is irreducible for all $n$, but most of the credit goes to Qiaochu.
We have
$$(x-1)A\_n = b\_{n+1} x^{n+1} + b\_n x^n + \cdots + b\_1 x - p\_n$$
for some positive integers $b\_{n+1},\ldots,b\_1$ summing to $p\_n$. If $|x| \le 1$, then
$$|b\_{n+1} x^{n+1} + b\_n x^n + \cdots + b\_1 x| \le b\_{n+1}+\cdots+b\_1 = p\_n$$
with equality if and only $x=1$, so the only zero of $(x-1)A\_n$ inside or on the unit circle is $x=1$. Moreover, $A\_n(1)>0$, so $x=1$ is not a zero of $A\_n$, so every zero of $A\_n$ has absolute value greater than $1$.
If $A\_n$ factors as $B C$, then $B(0) C(0) = A\_n(0) = p\_n$, so either $B(0)$ or $C(0)$ is $\pm 1$. Suppose that it is $B(0)$ that is $\pm 1$. On the other hand, $\pm B(0)$ is the product of the zeros of $B$, which are complex numbers of absolute value greater than $1$, so it must be an empty product, i.e., $\deg B=0$. Thus the factorization is trivial. Hence $A\_n$ is irreducible.
| 75 | https://mathoverflow.net/users/2757 | 18148 | 12,113 |
https://mathoverflow.net/questions/7052 | 27 | What would the slice-ribbon conjecture imply for 4-dimensional topology?
I've heard people speak of the slice-ribbon conjecture as an approach to the 4-dimensional smooth Poincare conjecture, and to the classification of homology 3-spheres which bound homology 4-balls. But I've never understood what they were talking about.
| https://mathoverflow.net/users/2051 | What would the slice-ribbon conjecture imply? | I think of the ribbon-slice conjecture as a **wish** that would simplify certain 4D questions. Let me explain this in 3 examples.
1. Given an embedded "ribbon disk" in 4-space (where the Morse function has no local maxima) one can push it up into 3-space and obtain an immersed disk (whose boundary is still the given knot) where the singularities are mild: these are the so called "ribbon singularities", arcs of double points such that on one of the sheets, the arc lies in the interior. (Picture...) One would actually call this immersed disk in 3-space a "ribbon" (that is allowed to cut through itself). It contains the information about the embedded disk in 4-space by pushing one sheet of each ribbon singularity into the forth dimension.
There is a fairly obvious algorithm how to create all such ribbons in 3-space, starting from the unlink and adding bands.
No such simple 3D-picture exists for arbitrary slice disks and one may **wish** that any slice knot is ribbon.
2. A knot K is slice if and only if there is a ribbon knot R such that the connected sum K # R is ribbon. One may **wish** that one didn't have to stabilize.
3. Consider the monoid M of oriented knots under connected sum. If -K is the reversed mirror image of K then K # (-K) is ribbon. So it's very tempting to try to turn M into a group (where -K would become the inverse of K) by identifying two knots K' and K if K' # (-K) is ribbon. But the **wish** doesn't come true: this is not an equivalence relation and if we force it to be one then by 2 we end up with the knot concordance group (where two knots K' and K are identified if K' # (-K) is slice).
It is amazing that there are no proposed counter examples to this conjecture, not even for links.
| 23 | https://mathoverflow.net/users/4625 | 18154 | 12,118 |
https://mathoverflow.net/questions/18155 | 2 | For $f\in L^2(0,\infty),$ define $(Tf)(x)=x^{-1}\int\_0^x f(s)ds,$ for $x\in(0,\infty),$ then from hardy's inequality, $T\in B(L^2),$ my question is how to show that $T$ is not compact?
| https://mathoverflow.net/users/3801 | Show a linear operator is not compact | For a natural number $j$ let $f\_j$ be the indicator function of the interval $[0,1/j]$ times the square root of $j$.
Then the $L^2$-norm of $f\_j$ is one.
A simple calculation shows that one has
$||Tf\_i-Tf\_j||^2\ge\int\_0^{1/j}(\sqrt i-\sqrt j)^2dx= (1-\sqrt{i/j})^2$ for $i\le j$, which implies that no subsequence can be Cauchy.
It is natural to look for an example around $x=0$ since that is where the kernel of $T$ fails to be $L^2$.
| 10 | https://mathoverflow.net/users/nan | 18159 | 12,120 |
https://mathoverflow.net/questions/18140 | 9 | I am wondering if there is any mathematical (or physical, besides the fact that classical quantum mechanics uses complex numbers) justification for why the complexified (1,3) Clifford algebra is used in Dirac's equation. A (the?) key point of special relativity is that spacetime is a real 4-d vector space with an inner product of signature (1,3). But by complexifying the signature becomes irrelevant-- all complex Clifford algebras in a given dimension are isomorphic where for real Clifford algebras, even signatures (p,q) and (q,p) are not isomorphic in general (as a side question: can there be a physical significance to this fact? or do only the spin group and the even subalgebra of the Clifford algebra, which are the same for (p,q) and (q,p), matter? I only hear about spinor bundles and spin structures, never pinor bundles or pin structures).
Thanks and I hope this isn't too physicsy of a question!
| https://mathoverflow.net/users/4622 | Clifford Algebra in Dirac Equation | This problem was investigated by Cecile DeWitt-Morette et al. This is a [review article](http://arxiv.org/PS_cache/math-ph/pdf/0012/0012006v1.pdf) describing the role of Pin groups in Physics. This article includes also a historical survey and a comprehensive list of references about Pin groups.
Becides the fact that the Clifford algebras Cl(3,1) and Cl(1,3) are nonisomorphic, there is the question if the two types can be experimentally distinguished.
The article gives a positive in the neutrinoless double beta decay experiment where it was observed that the neutrino has to be a Pin(1,3) particle.
The article also describes solutions of the Dirac equation in topologically
nontrivial spaces where the vacuum expectations of the Fermi currents are different in the two types of groups. (Thus one might conclude that this difference can be used to get information about the space-time topology).
The article also refers to Choquet-Bruhat, DeWitt-Morette and Dillard-Bleick's book about obstructions to the construction of Pin bundles.
| 18 | https://mathoverflow.net/users/1059 | 18160 | 12,121 |
https://mathoverflow.net/questions/17732 | 21 | On the one hand, Wikipedia suggests that every distribution defines a Radon measure:
* <http://en.wikipedia.org/wiki/Distribution_(mathematics)#Functions_as_distributions>
On the other hand, Terry Tao and LK suggest not:
* <http://www.math.ucla.edu/~tao/preprints/distribution.pdf>
* [When can a function be recovered from a distribution?](https://mathoverflow.net/questions/14586/when-can-a-function-be-recovered-from-a-distribution)
Can someone please clarify this for me?
| https://mathoverflow.net/users/3676 | Difference between measures and distributions | This is a summary of what I've learned about this question based on the answers of the other commenters.
[\*] Any positive distribution defines a positive Radon measure.
I had naively assumed a result for distributions like The Hahn Decomposition Theorem[1] for measures, i.e. I assumed that a distribution could be expressed as the difference of two positive distributions. If it *could* be, then applying Theorem [\*] would yield the result that any distribution is a signed measure.
However, this is not the case. The derivative of the delta function, i.e. δ', satisfies
δ'(f) = -f'(0). This is not a measure. I can't find any way of proving it's not the difference of two positive distributions, other than by contradiction using the above result.
[1] <http://en.wikipedia.org/wiki/Hahn_decomposition_theorem>
| 3 | https://mathoverflow.net/users/3676 | 18166 | 12,126 |
https://mathoverflow.net/questions/18163 | 9 | Geometric group theory is mainly concerned with topological and geometric properties of groups, spaces on which they act etc., so the ideas employed in GGT are mainly algebraic/geometric/topological. Is there any subfield of GGT where methods from analysis find applications? I once heard that analytical tools, e. g. geodesic flows, are used in studying ends of groups, but that's all I know.
| https://mathoverflow.net/users/2192 | Geometric group theory and analysis | Analytic ideas enter into several parts of geometric group theory. Tom already mentioned amenability, so I'll skip that.
1) Complex analysis and the theory of quasiconformal mappings plays an important role in understanding the mapping class group, which is one of the most important groups studied by geometric group theorists. For information on this, see Farb and Margalit's forthcoming book "A primer on mapping class groups", available on either of their webpages.
2) The study of groups with property (T) ends up using quite a bit of analysis. See the book "Kazhdan's Property (T)" by de la Harpe, Bekka, and Valette for information on this.
3) Analysis (together with ideas from ergodic theory, which is of course quite analytic) plays an important role in proving various rigidity theorems. The most famous is the Mostow Rigidity Theorem, whose original proof uses lots of analysis : quasiconformal mapping in high dimensions, the fact that Lipschitz functions are differentiable almost everywhere, ergodic theory, etc.
| 13 | https://mathoverflow.net/users/317 | 18168 | 12,127 |
https://mathoverflow.net/questions/18174 | 3 | Everyone knows the result by Kronecker: if $r$ is a real number not rational and $\epsilon>0$ then there exist a natural number $N$ such that $\{Nr\}<\epsilon$.
There must be such a result for pairs (and even for any other quantity) of real numbers:
let $r\_1$, $r\_2$ ` be real numbers independent over $\mathbb{Q}$ and $\epsilon>0$ then there exist a natural number $N$ such that $\{Nr\_1\}<\epsilon$ and $\{Nr\_2\}<\epsilon$.
I heard this result more than 10 years ago but i still don't know the proof.
I just guess this problem is related to everywhere density of trajectory on torus.
| https://mathoverflow.net/users/4475 | Independence over Q and Kroneckers result | This is on wikipedia. See [Kronecker's theorem](http://en.wikipedia.org/wiki/Kronecker%27s_theorem). It was proved by Kronecker in 1884.
The necessary and sufficient condition for integral multiples of a point $(r\_1,\dots,r\_n)$ in the $n$-torus $(\mathbf R/\mathbf Z)^n$ to be dense is not that the $r\_i$'s are all irrational: that is necessary but far from sufficient. Consider, for example, integral multiples of $(\sqrt{2},1+\sqrt{2})$ in the 2-torus. The correct necessary and sufficient condition is that $1, r\_1, \dots,r\_n$ are linearly independent over $\mathbf Q$. (For $n = 1$, this recovers the irrationality condition as being necessary and sufficient for denseness of integral multiples on a circle.)
A proof of this theorem can be found in Hardy and Wright's Introduction to the Theory of Numbers (first in one dimension and then in general; see Chapter 23). It can also be proved by ideas from ergodic theory: the hypothesis that $1,r\_1,\dots,r\_n$ are linearly independent over $\mathbf Q$ implies translation on the $n$-torus by $(r\_1,\dots,r\_n)$ is ergodic and the orbit of *any* point in a compact topological group under a left or right translation that's ergodic is dense in the group. (Initially one can say only that almost every point in a compact group -- in the sense of its Haar measure -- has a dense orbit under an ergodic transformation, but left and right translation by a fixed element is a pretty special transformation: if such a translation has one dense orbit then all the orbits of that translation are dense.)
Note: The linear independence of $1, r\_1,\dots,r\_n$ over $\mathbf Q$ is actually *equivalent* to the ergodicity of translation by $(r\_1,\dots,r\_n)$ on the $n$-torus.
Weyl's equidistribution theorem strengthens Kronecker's theorem: that sequence of integral multiples isn't just dense in the $n$-torus but in fact is uniformly distributed in the $n$-torus. This quantifies Kronecker's theorem in the same way the ergodic theorem quantifies the Poincare Recurrence Theorem.
| 21 | https://mathoverflow.net/users/3272 | 18175 | 12,132 |
https://mathoverflow.net/questions/17917 | 5 | Let $E$ be your favorite elliptic curve, and let $Tor^m$ be the moduli stack of torsion sheaves of degree $m$ on $E$. This sounds horrible, but it's not so bad; it's a global quotient of a smooth Quot scheme (the space of rank m subbundles of degree -m in $\mathcal{O}\_E^{\oplus m}$) modulo the obvious action of GL(m).
For various nefarious reasons of my own, I would like to have an explicit presentation of the cohomology ring of this stack (i.e. the equivariant cohomology of this Quot scheme for GL(m)).
>
>
> >
> > Does anyone know of such a presentation?
> >
> >
> >
>
>
>
**EDIT:** Since Torsten asked in comments, I'd be perfectly happy just to understand the rational cohomology, though obviously integral would be even better. I would also mention that I really want the ring structure, not just the vector spaces.
| https://mathoverflow.net/users/66 | Is there a presentation of the cohomology of the moduli stack of torsion sheaves on an elliptic curve? | It seems that one can obtain the additive structure of rational cohomology
without too much effort (in no way have I checked this carefully so caveat
lector applies). As Allen noticed, for rational cohomology it is enough to
compute $T$-equivariant cohomology and then take $\Sigma\_m$-invariants (if this is to
work also for integral cohomology a more careful analysis would have to be made I think). Now, the space $Tor^m\_C$ of length $m$ quotients of $\mathcal O^m\_C$ (I use $C$
as everything I say will work for any smooth and proper curve $C$) is smooth and proper so we may use the Bialinski-Birula analysis (choosing a general $\rho\colon \mathrm{G}\_m \to T$) and we first look at the fixed point locus of $T$.
Now, for every sequence $(k\_1,\ldots,k\_m)$ of non-negative integers with
$k\_1+\ldots+k\_m=m$ gives a map $S^{k\_1}C\times\cdots\times S^{k\_m}C \to Tor^m\_C$, where $S^kC$ is the
symmetric product interpreted as a Hilber scheme, and the map takes $(\mathcal
I\_1,\ldots,\mathcal I\_m)$ to $\bigoplus\_i\mathcal I\_i \hookrightarrow \mathcal O^m\_C$. It is
clear that this lands in the $T$-fixed locus and almost equally clear that this
is the whole $T$-fixed locus (any $T$-invariant submodule must be the direct sum of its
weight-spaces).
We can now use $\rho$ to get a stratification parametrised by the sequences
$(k\_1,\ldots,k\_m)$. Concretely, the tangent space of $Tor^m\_C$ to a point of
$S^{k\_1}C\times\cdots\times S^{k\_m}C$ has character $(k\_1\alpha\_1+\cdots+k\_m\alpha\_m)\beta$, where
$\beta=\sum\_i\alpha\_i^{-1}$ (and we think of characters as elements of the group ring
of the character group of $T$ and the $\alpha\_i$ are the natural basis elements of
the character group). This shows that $\rho$ can for instance be chosen to be $t \mapsto
(1,t,t^2,\ldots,t^{m-1})$. In any case the stratum corresponding to $(k\_1,\ldots,k\_m)$ has
as character for its tangent space the characters on the tangent space on which
$T'$ is non-negative and its normal bundle consists of those on which $T'$ is
non-negative (hence with the above choice, the character on the tangent space is
$\sum\_{i\geq j}k\_i\alpha\_i\alpha\_j^{-1}$ and in particular the dimension of the stratum is
$\sum\_iik\_i$). We now have that each stratum is a vector bundle of the
corresponding fixed point locus so in particular the equivariant cohomology of
it is the equivariant cohomology of the fixed point locus and in particular is
free over the cohomology ring of $T$. Furthermore, if we build up the cohomology
using the stratification (and the Gysin isomorphism), at each stage a long exact
sequence splits once we have shown that the top Chern class of the normal bundle
is a non-zero divisor in the cohomology of the equivariant cohomology of the
component of the fixed point locus (using the Atiyah-Bott criterion).
However, the non-zero divisor condition seems more or less automatic (and that
fact should be well-known): The normal bundle $\mathcal N$ of the
$(k\_1,\ldots,k\_m)$-part, $F$ say, of the fixed point locus splits up as direct sum
$\bigoplus\_\alpha \mathcal N\_\alpha$, where $T$ acts by the character $\alpha$ on $\mathcal N\_\alpha$. Then
the equivariant total Chern class of $\mathcal N\_\alpha$ inside of
$H^\ast\_T(F)=H^\ast(F)\bigotimes H^\ast\_T(pt)$ is the Chern polynomial of $\mathcal N\_\alpha$
as ordinary vector bundle evaluated at $c\_1(\alpha)=\alpha\in H^2\_T(pt)$. Hence, if we quasi-order
the characters of $T$ by using $\beta \mapsto -\rho(\beta)$ (``quasi'' as many characters get the
same size), then as $-\rho(\alpha)>0$ (because $\alpha$ appears in the normal bundle) we get
that $1\otimes\alpha^{n\_\alpha}\in H^\ast\_T(F)$, where $n\_\alpha$ is the rank of $\mathcal N\_\alpha$, is the term of
$c\_n(\mathcal N\_\alpha)$ of largest order. Hence, we get that for the top Chern class
of $\mathcal N$ which is the product $\prod\_\alpha c\_{n\_\alpha}(\mathcal N\_\alpha)$ its term of
largest order has $1$ as $H^\ast(F)$-coefficient and hence is a non-zero divisor.
As the cohomology of $S^nC$ is torsion free we get that all the involved
cohomology is also torsion free and everything works over the integers but as
I've said to go integrally from $T$-equivariant cohomology to
$\mathrm{GL}\_m$-equivariant cohomology is probably non-trivial.
If one wants to get a hold on the multiplicative structure one could use that
the fact that the Atiyah-Bott criterion works implies that that $H^\ast\_T(Tor^m\_C)$
injects into the equivariant cohomology of the fixed point locus. The algebra
structure of the cohomology of $S^nC$ is clear (at least rationally) so we get
an embedding into something with known multiplicative structure. The tricky
thing may be to determine the image. We do get a lower bound for the image by
looking at the ring generated by the Chern classes of of the tautological bundle
but I have no idea how close that would get us to the actual image. (It is a
well-known technique anyway used in for instance equivariant Schubert calculus
so there could be known tricks.)
There is another source of elements, namely we have the map $Tor^m\_C \to S^mC$. This map becomes
even better if one passes to the $\Sigma\_m$-invariants.
[Later] Upon further thought I realise that the relation between $T$- and $G=GL\_m$-equivariant cohomology
is simpler than I thought. The point is (and more knowledgeable people certainly know this) that the map $EG\times\_TX \to EG\times\_GX$ is a $G/T$ bundle and $G$ being special $H^\*\_T(X)$ is free as a $H^\*\_G(X)$-module (with $1$ being one of the basis elements). That means that $H^\*\_G(X) \to H^\*\_T(X)$
is injective but more precisely $H^\*\_T(X)/H^\*\_G(X)$ is torsion free with $\Sigma\_m$-action without invariants so that $H^\*\_G(X)$ is the ring of $\Sigma\_m$-invariants of $H^\*\_T(X)$.
| 8 | https://mathoverflow.net/users/4008 | 18181 | 12,134 |
https://mathoverflow.net/questions/17989 | 1 | *(This is a follow-up to my previous question [Can every finite graph be represented by an arithmetic sequence of natural numbers?](https://mathoverflow.net/questions/17875/can-every-finite-graph-be-represented-by-an-arithmetic-sequence-of-natural-number))*
Since it is obviously false that every finite graph can be represented by an *arithmetic* sequence (with an edge between two vertices/numbers $n\_i, n\_j$ iff GCD$(n\_i, n\_j)>1$) I'd like to reformulate my question:
Consider a family $F$ of parametrized computable functions $\lbrace f\_{\alpha}:\mathbb{N}^k\mapsto\mathbb{N}\rbrace\_{\alpha}$ with $\alpha \in \mathbb{N}^{n}$ for some $k,n \in \mathbb{N}$.
A sequence $(n\_1,....,n\_k,n\_{k+1},...,n\_{k+l})$ is an $F$-sequence if $n\_{i+k} = f\_\alpha(n\_i,n\_{i+1},...,n\_{i+k-1})$ for some fixed $f\_\alpha \in F$.
An arithmetic sequence is an $F$-sequence for $F = \lbrace f\_\alpha(n) = n + \alpha\rbrace\_{\alpha \in \mathbb{N}}$.
>
> **Question:** Is there a family $F$ of functions $f\_{\alpha}:\mathbb{N}^k\mapsto\mathbb{N}$ (as above) such that every graph on $n > k$ nodes can be represented by an $F$-sequence
> such that $n\_i$ and $n\_j$ are joined by an edge iff $n\_i \neq n\_j$ and GCD$(n\_i, n\_j) > 1$
>
>
>
My first question was genuinely "first-order" and admitted a straight-forward "first-order" answer (even though negative). *This* question is genuinely "second-order" and furthermore existential and probably doesn't admit a straight-forward definite answer. Sorry for that (and I tagged it as a "soft question"). But maybe someone can give a hint how to try to attack it?
| https://mathoverflow.net/users/2672 | Can every finite graph be represented by one prescribed sequence of natural numbers? | The answer is "yes": there is such a family of $F$ functions. In fact, a single computable function, acting on a single integer argument, suffices. We may do this by storing essentially complete information about the graph, and about the process of "constructing" the graph $G$ (that is, the process of computing suitable integers $n\_j$ representing vertices $v\_j \in V(G)$), in the integers $n\_j$ themselves.
**Edit:** *My original answer contained an error in which the correct adjacency relations were not properly produced. I correct this below, and have taken the opportunity to make other minor revisions.*
We may specify a graph on $n$ vertices using a single integer by a number of different methods, such as using binary representation to use an $n^2$ bit integer to give the adjacency matrix of the graph. (The leading 1 represents an entry in the diagonal, which may serve to define the size of the graph without denoting any edges if we consider only simple graphs.) Denote this binary representation of the graph $G$ by $g \in \mathbb N$. We compute the integers $n\_j$ by carrying $g$ as the exponents of different primes, and using other prime factors to "induce" adjacency between the integers representing different integers.
Let $p\_j$ denote the $j^{\textrm{th}}$ prime, with $p\_1 = 2$, $p\_2 = 3$, etc. We define
$$ N\_j = p\_j^g $$
which will act as a "label" of sorts for our vertices; each vertex $v\_j \in V(G)$ (for an arbitrary ordering of the vertices) will be represented by an integer which is divisible by the prime $p\_j$, but not by any other prime $p\_k$ for $1 \le k \le n$. At the same time, this labelling carries with it an entire description of the graph, as well as the vertex ordering (in this case, given by the ordering of the rows/columns of the adjacency matrix of $G$). The smallest prime factor of the integer $n\_j$ corresponding to the vertex $v\_j$ will indicate which vertex it is in the order, and carry a complete description of the graph $G$.
We may induce adjacency among the vertices by giving them appropriate common prime factors. A simple way to do this is to associate a prime to each edge, and give any two vertices belonging to a common edge the corresponding prime factor. (Vertices which do not share an edge in common will have no common prime factors, and thus be coprime.) We order the possible edges by considering the lexicographic ordering on all ordered pairs $(v\_j, v\_k)$ such that $v\_j < v\_k$: thus the possible edge $v\_j v\_k$ (for $j < k$) will be edge number $\varepsilon(j,k) = \binom{k-1}{2} + j$ in the enumeration. More generally, we may define
$$ \varepsilon(j,k) = \binom{\max \{j,k\} - 1}{2} + \min \{j,k\} .$$
We may represent the adjacency of two vertices $v\_j, v\_k$ by giving their corresponding integers $n\_j, n\_k$ a common prime factor, namely the prime $p\_{n+\varepsilon(j,k)}$. The exponents of the primes $p\_{n+1}$ through $p\_{n+\binom{n}{2}}$ in the integers $n\_j$ then form the incidence matrix of edges to vertices in $G$.
Thus, we may let $F(x)$ be the function which computes the following:
1. Determine the smallest prime $p\_{j-1} \mid x$, and the corresponding index $j$.
- Determine the exponent, $g$, of the largest power of $p\_{j-1}$ which divides $x$.
- Extract from $g$ complete information about the graph $G$, including its size $n$.
- Determine the next largest prime $p\_j > p\_{j-1}$.
- Compute $N\_j = p\_j^g$.
- Compute $M\_j = p\_{n+j}$.
- For each $1 \le k \le n$, $k \ne j$:
* Compute $\varepsilon(j,k)$.
* If $v\_j$ is adjacent to $v\_k$ in $G$, set $M\_j \leftarrow M\_j p\_{n+\varepsilon(j,k)}$.- Return $y = N\_j M\_j$.
To produce an appropriate $F$-sequence which represents $G$, it then suffices to compute $n\_1$, which is
$$ n\_1 = p\_1^g \prod\_{k \ne j} {p\_{n+\varepsilon(j,k)}}^{A\_{j,k}} , $$
where $A$ is the adjacency matrix of $G$; this suffices to iteratively produce the other integers $n\_k$ using $F$.
The above can be readily extended to accommodate hypergraphs (allow more than two vertices per 'edge') and other generalizations; one only needs to choose a suitable representation, and store it in an integer in some way which can be extracted. These are standard tricks in computability theory; we are just adapting Godel numbering for a special purpose in this case.
If you want an "interesting" way of representing graphs by models in integer sequences, which does not just amount to packing and unpacking of data structures in the integers, you're going to have to impose some non-trivial bounds --- on the integer sizes, on the computational efficiency of the procedure, etc. --- and then, you will almost certainly have to be satisfied with obtaining only some special class of graphs. But that special class may still be an interesting one, if your computational constraints are well-chosen.
| 2 | https://mathoverflow.net/users/3723 | 18184 | 12,137 |
https://mathoverflow.net/questions/18203 | 21 | First, pick a commutative ring $k$ as the "ground field". Everything I say will be $k$-linear, e.g. "algebra" means "unital associative algebra over $k$". Then recall the following construction due to Grothendieck:
Let $A$ be a commutative algebra, and $X$ an $A$-module. Then the **differential operators on $X$** is the filtered algebra $D = D(A,X)$ given inductively by:
$$ D\_{\leq 0} = \text{image of $A$ in }\hom\_k(X,X) $$
$$ D\_{\leq n} = \{ \phi \in \hom\_k(X,X) \text{ s.t. } [\phi,a] \in D\_{\leq n-1}\\, \forall a\in A\} $$
$$ D = \bigcup\_{n=0}^\infty D\_{\leq n} $$
(**Edit:** In the comments, Michael suggests that $D\_{\leq 0} = \hom\_A(X,X)$ is the more standard definition, and the rest is the same.)
Then the following facts are more or less standard:
* If $A$ acts freely on $X$, then $D\_{\leq 1}$ acts on $A$ by derivations. (It's always true that $D\_{\leq 1}$ acts on $D\_{\leq 0}$ by derivations; the question is whether $D\_{\leq 0} = A$ or a quotient.) If $X = A$ by multiplication, then $D\_{\leq 1}$ splits as a direct sum $D\_{\leq 1} = \text{Der}(A) \oplus A$.
* If $k=\mathbb R$, $M$ is a finite-dimensional smooth manifold, and $A = C^\infty(M) = X$, then $D$ is the usual algebra of differential operators generated by $A$ and $\text{Vect}(M) = \Gamma(TM \to M)$.
* If $A,X$ are actually sheaves, so is $D$.
Thus, at least in the situation where $A = X = C^{\infty}(M)$, the algebra $D$ is acting very much like the universal enveloping algebra of $U (\text{Vect}(M))$; in particular, the map $U(\text{Vect}(M)) \to D$ is filtered and is (almost) a surjection: it misses only the non-constant elements of $A$. So when $A = X = C^\infty(-)$ are sheaves on $M$, it's very tempting to think of $D$ as a sheafy version of $U(\text{Vect}(-))$. Note that $U(\text{Vect}(-))$ is not a sheaf: its degree $\leq 0$ part consists of constant functions, not locally constant, for example, and there are non-zero elements in $U\_{\leq 2}$ that restrict to $0$ on an open cover. I think that it cannot be true that the sheafification of $U(\text{Vect}(-))$ is $D$, as the sheafification of $U\_{\leq 0}$ is the locally-constant sheaf, not $C^\infty$.
So: is there a description of $D$ that makes it more obviously like a universal enveloping algebra? E.g. is there some adjunction or other categorical description? Is it really true that $D$ is a "sheafy" version of $U$ in a precise sense, or is this just a chimera?
| https://mathoverflow.net/users/78 | Is there a "categorical" description of Grothendieck's algebra of differential operators? | The construction you're looking for is the universal enveloping algebra of a [Lie](http://en.wikipedia.org/wiki/Lie_algebroid) [algebroid](http://ncatlab.org/nlab/show/Lie+algebroid).
A Lie algebra is a Lie algebroid on a point. It's universal enveloping algebra as an algebroid is the same as the usual.
Every smooth manifold has a Lie algebroid structure on its tangent bundle, which differential operators are the universal enveloping algebra of.
Another cool examples is that TDOs (sheaves of twisted differential operators) can be constructed from central extensions of the tangent Lie algebroid by the structure sheaf (you take UEA, and then identify the two copies of the structure sheaf).
(I'll just note: Mariano's answer and mine are essentially equivalent. See the nLab page linked above for brief explanation).
| 16 | https://mathoverflow.net/users/66 | 18204 | 12,150 |
https://mathoverflow.net/questions/18223 | 1 | For an algebraic group that cannot be embedded into $GL\_n$, is there a nice definition for congruence subgroup? Do we just define it as the compact open subgroup of $G(A\_f)$, where $A\_f$ is the finite adele?
| https://mathoverflow.net/users/2008 | Definition of congruence subgroup for non-matrix groups | Even though every linear algebraic group (understood to mean affine of finite type) can be embedded into ${\rm{GL}}\_ n$, if we change the embedding then the notion of "congruence subgroup" may change (in the sense of a "group of integral points defined by congruence conditions"). So a more flexible notion is that of arithmetic subgroup, or $S$-arithmetic for a non-empty finite set of places $S$ (containing the archimedean places), in which case we can make two equivalent definitions: subgroup commensurable with intersection of $G(k)$ with a compact open subgroup of the finite-adelic points, or subgroup commensurable with intersection of $G(k)$ with $S$-integral points of some ${\rm{GL}}\_ n$ relative to a fixed closed subgroup inclusion of $G$ into ${\rm{GL}}\_ n$ (i.e., if $\mathcal{G}$ denotes the schematic closure in ${\rm{GL}}\_ n$ over $\mathcal{O}\_ {k,S}$ of $G$ relative to such an embedding over $k$, we impose commensurability with
$\mathcal{G}(\mathcal{O}\_ {k,S})$).
Personally I like to view the ${\rm{GL}}\_ n$-embedding as just a quick-and-dirty way to make examples of flat affine integral models (of finite type) over rings of integers (via schematic closure from the generic fiber). However, in some proofs it is really convenient to reduce to the case of ${\rm{GL}}\_ n$ and doing a calculation there. It is not evident when base isn't a field or PID whether flat affine groups of finite type admit closed subgroup inclusions into a ${\rm{GL}}\_ n$ (if anyone can prove this even over the dual numbers, let me know!), so in the preceding definition it isn't evident (if $\mathcal{O}\_ {k,S}$ is not a PID) whether the $\mathcal{G}$'s obtained from such closure account for *all* flat affine models of finite type over the ring of $S$-integers. In other words, we really should be appreciative of ${\rm{GL}}\_ n$.
| 3 | https://mathoverflow.net/users/3927 | 18231 | 12,167 |
https://mathoverflow.net/questions/18245 | 3 | On Pages 1-3 of Cours 2 of Toën's [Master Course on Stacks](http://www.math.univ-toulouse.fr/~toen/cours2.pdf), he defines the notion of a Geometric context with a rather extensive list of axioms (they take up about two pages over and above the definition of a Grothendieck topology, which isn't even included). Maybe it's just my prejudices talking, but it seems like there should be a way to simplify this definition using sieves or some other kind of functorial machinery (think about the definition of a sheaf in terms of sieves compared to the definition using the gluing axiom). The reason I ask this is that the important properties of class the geometric structure morphisms (what Toën calls $P$) allow us to define closed and open subsheaves, representable covers, atlases, etc, which all seem like things that sieves were meant to do.
Question: Can we simplify the statements of those axioms using sieves or some other kind of functorial machinery? If not, why not?
Edit: I just remembered that "geometric morphism" already means something else, so I've replaced it. "The name 'geometric structure morphism' is a word that I coined myself, spending a week thinking of nothing else."
| https://mathoverflow.net/users/1353 | Simplifying the definition of a geometric context using sieves? | First of all, the "extensive list of axioms" is largely a list of *definitions* (of rather natural terminology). Anyway, he seems to be just getting at the issue of the use of fiber squares in arguments involving properties of morphisms, such as come up in many of the basic constructions in algebraic geometry (Hilbert schemes, quotients by group actions, descent theory, etc.). In contrast, one of the points of the sieve language is to carry out Grothendieck topological stuff without assuming the existence of fiber products. So asking for a sieve formulation may be contrary to the spirit of what he is trying to do. The real test is to see how he *uses* this general nonsense before deciding if it is unsuitable for the task to which he intends to apply it. Have you checked out any such uses?
When I think of all of the *examples* which I care about, his axioms are completely trivial to verify and so the framework seems to just be setting up a way to prove all of the "general nonsense" in one go to later apply it to the usual interesting examples. It only seems long in the same way that the definition of an algebraic group would seem long if one wrote out the definition of every ingredient that goes into it and didn't already know what an algebraic variety is. Is there any interesting *example* for which his actual axioms seem non-trivial to verify? If so, it should be noted in the question. If not, why is the question being asked? (I assume you have at least checked his axioms for etale and fppf morphisms of schemes to see how easily verified and natural they are for "real" examples.)
| 9 | https://mathoverflow.net/users/3927 | 18254 | 12,183 |
https://mathoverflow.net/questions/18246 | 12 | Hello all, this question is a variant (and probably a more difficult one)
of a (promptly answered ) question that I asked here, at [Is it true that all the "irrational power" functions are almost polynomial ?](https://mathoverflow.net/questions/18054/is-it-true-that-all-the-irrational-power-functions-are-almost-polynomial).
For $n\geq 1$, let $f(n)$ denote the "integer part"
(largest integer below) $n^{\frac{3}{2}}$, and let
$g(n)=f(n+2)-2f(n+1)+f(n)$.
Question : Is it true that $g(n)$ is always in $\lbrace -1,0,1\rbrace$
(excepting the initial value $g(1)=2$) ? I checked this up to
$n=100000$.
It is not too hard to check that if $t$ is large enough compared to $r$ (say
$t\geq \frac{3r^2+1}{4}$),
$f(t^2+r)$ is exactly $t^3+\frac{3rt}{2}$ (or $t^3+\frac{3rt-1}{2}$ if $t$
and $r$ are both odd ) and similarly
$f(t^2-r)$ is exactly $t^3-\frac{3rt}{2}$ (or $t^3-\frac{3rt+1}{2}$ if $t$
and $r$ are both odd ). So we already know that the answer is "yes" for
most of the numbers.
| https://mathoverflow.net/users/2389 | Growth of the "cube of square root" function | Here is a proof that $|g(n)|\le 1$ for all but finitely many $n$. You can extract an explicit bound for $n$ from the argument and check the smaller values by hand.
If $f(n)=n^{3/2}$ without the floor, then $g(n)\sim \frac{3}{4\sqrt n}$, so it is positive and tends to 0. When you replace $n^{3/2}$ by its floor, $g(n)$ changes by at most 2, hence the only chance for failure is to have $g(n)=2$ when the fractional parts of $n^{3/2}$ and $(n+2)^{3/2}$ are very small and the fractional part of $(n+1)^{3/2}$ is very close to 1 (the difference is less than $\frac{const}{\sqrt{n}}$).
Let $a,b,c$ denote the nearest integers to $n^{3/2}$, $(n+1)^{3/2}$ and $(n+2)^{3/2}$. Then $c-2b+a=0$ because it is an integer very close to $(n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}$. Denote $m=c-b=b-a$. Then $(n+1)^{3/2}-n^{3/2}<m$ and $(n+2)^{3/2}-(n+1)^{3/2}>m$. Observe that
$$
\frac{3}{2}\sqrt{n}<(n+1)^{3/2} - n^{3/2} < \frac{3}{2}\sqrt{n+1}
$$
(the bounds are just the bounds for the derivative of $x^{3/2}$ on $[n,n+1]$. Therefore
$$
\frac{3}{2}\sqrt{n} < m < \frac{3}{2}\sqrt{n+2}
$$
or, equivalently,
$$
n < \frac49 m^2 < n+2.
$$
If $m$ is a multiple of 3, this inequality implies that $n+1=\frac49 m^2=(\frac23m)^2$, then $(n+1)^{3/2}=(\frac23m)^3$ is an integer and not slightly smaller than an integer as it should be. If $m$ is not divisible by 3, then
$$
n+1 = \frac49 m^2 + r
$$
where $r$ is a fraction with denominator 9 and $|r|<1$. From Taylor expansion
$$
f(x+r) = f(x) +r f'(x) +\frac12 r^2 f''(x+r\_1), \ \ 0<r\_1<r,
$$
for $f(x)=x^{3/2}$, we have
$$
(n+1)^{3/2} = (\frac49 m^2 + r)^{3/2} = \frac8{27}m^3 + mr + \delta
$$
where $0<\delta<\frac1{4m}$.
This cannot be close to an integer because it is close (from above) to a fraction with denominator 27.
| 9 | https://mathoverflow.net/users/4354 | 18260 | 12,188 |
https://mathoverflow.net/questions/18042 | 9 | Can one prove the Cantor-Bernstein (or Schröder-Bernstein) theorem without using the Axiom of Infinity?
| https://mathoverflow.net/users/4600 | Axiom of Infinity needed in Cantor-Bernstein? | Actually, the usual proof of the Cantor-Schröder-Bernstein Theorem does not use the Axiom of Infinity (nor the Axiom of Powersets).
By the usual proof, I mean the one found on [Wikipedia](http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem#Proof), for example. Using the notation from that proof, the main point of contention is whether we can form the sets $C\_n$ and $C = \bigcup\_{n=0}^\infty C\_n$. These sets exist by comprehension:
$C\_0 = \{x \in A: \forall y \in B\,(x \neq g(y))\}$
$C\_n = \{x \in A: \exists s\,(s:\{0,\dots,n\}\to A \land s(0) \in C\_0 \land (\forall i < n)(s(i+1) = g(f(s(i)))) \land s(n) = x\}$,
where abbreviations such as $s:\{0,\dots,n\}\to A$ should be replaced by the equivalent (bounded) formulas in the language of set theory. The definition of $C\_n$ is uniform in $n$, and so
$C = \{x \in A: \exists n\,(\mathrm{FinOrd}(n) \land x \in C\_n)\}$,
where $x \in C\_n$ should be replaced by the above definition and $\mathrm{FinOrd}(n)$ is an abbreviation for "$n$ is zero or a successor ordinal and every element of $n$ is zero or a successor ordinal." An alternate definition of $C$ is
$C = \{x \in A: \forall D\,(C\_0 \subseteq D \subseteq A \land g[f[D]] \subseteq D \to x \in D)\}$,
which shows that $C$ is $\Delta\_1$-definable. The rest of the proof uses induction on finite ordinals, but since the definition of the sets $C\_n$ is uniform these are special cases of transfinite induction.
In conclusion, it looks like the proof could work in [Kripke-Platek Set Theory](http://en.wikipedia.org/wiki/Kripke%E2%80%93Platek_set_theory) — which has neither the Axiom of Infinity nor the Axiom of Powersets — provided that the two definitions of $C$ given above are provably equivalent. I haven't tried to check whether the two definitions are provably equivalent but, in any case, the proof can be carried out in Kripke-Platek Set Theory with $\Sigma\_1$-Comprehension.
| 10 | https://mathoverflow.net/users/2000 | 18267 | 12,193 |
https://mathoverflow.net/questions/18275 | 1 | Consider sequence $s(n) = \sin{nx}$. Are there values of $x$ for which the following holds: For every $y \in \[-1,1\]$ there is a subsequence of $s(n)$ converging to $y$? (Or perhaps just for the open interval...) Someone hypothesised that the answer is yes, and further that every $x$ that is relatively irrational with $\pi $ has this property.
The question I am more interested in is the generalised version of this to arbitrary sequences. What are necessary and sufficient conditions for a sequence having subsequences converging to any point in the set of values the sequence visits? Does it have anything to do with properties like the function $f(n)$ being ergodic or mixing?
(suggestions for tags welcome in comments)
| https://mathoverflow.net/users/4076 | Existence of convergent subsequences for all values in range? | More conventional language: Are there values of $x$ such that the sequence $\sin(nx)$ is dense in the interval $[-1,1]$. The answer is yes, almost all $x$ have this property, in particular all $x$ such that $x/\pi$ is irrational.
See Weyl's Criterion
<http://en.wikipedia.org/wiki/Weyl%27s_criterion>
for something (equidistributed) that implies much more than merely dense. And $nx$ mod 1 is equidistributed in $[0,1]$ if $x$ is irrational.
| 5 | https://mathoverflow.net/users/454 | 18278 | 12,198 |
https://mathoverflow.net/questions/18268 | 7 | There is [a question that was asked on stackoverflow](https://stackoverflow.com/questions/2441506/how-to-generate-correlated-binary-variables/2447678#2447678) that at first sounds simple but I think it's a lot harder than it sounds.
Suppose we have a stationary random process that generates a sequence of random variables x[i] where each *individual* random variable has a [Bernoulli distribution](http://en.wikipedia.org/wiki/Bernoulli_distribution) with probability p, but the correlation between any two of the random variables x[m] and x[n] is α|m-n|.
How is it possible to generate such a process? The textbook examples of a Bernoulli process (right distribution, but independent variables) and a discrete-time IID Gaussian process passed through a low-pass filter (right correlation, but wrong distribution) are very simple by themselves, but cannot be combined in this way... can they? Or am I missing something obvious? If you take a Bernoulli process and pass it through a low-pass filter, you no longer have a discrete-valued process.
(I can't create tags, so please retag as appropriate... stochastic-process?)
| https://mathoverflow.net/users/1305 | discrete stochastic process: exponentially correlated Bernoulli? | Here is a construction.
* Let $\{Y\_i\}$ be independent Bernouilli random variables with probability $p$.
* Let $N(t)$ be a [Poisson process](http://en.wikipedia.org/wiki/Poisson_process) chosen so that $P(N(1)=0)=\alpha$.
* Let $X\_i = Y\_{N(i)}$.
In words, we have some radioactive decay which tells us when to flip a new (biased) coin. $X\_n$ is the last coin flipped at time $n$. The correlation between $X\_m$ and $X\_n$ comes from the possibility that there are no decays between time $m$ and time $n$, which happens with probability $\alpha^{|m-n|}$.
The conditional correlation between $X\_m$ and $x\_n$ is $1$ if $N(m) = N(n)$, and $0$ if $N(m)\ne N(n)$, so $\text{Cor}(X\_n,X\_m) = P(N(m)=N(n)) = \alpha^{|m-n|}.$
You can simplify this by saying that $N(i) = \sum\_{t=1}^i B\_i$ where $\{B\_i\}$ are independent Bernoulli random variables which are $0$ with probability $\alpha$.
| 11 | https://mathoverflow.net/users/2954 | 18287 | 12,204 |
https://mathoverflow.net/questions/18280 | 4 | More specifically, is it true that a representation of $\dim < p+1$ of the algebraic group $SL\_2(\mathbb{F}\_p)$ is always completely reducible? (of course above this dimension there are non completely reducible examples)
More general results that might help in this direction are also welcome.
Thanks
| https://mathoverflow.net/users/4477 | Are low dimensional modular representations of SL2(Fp) completely reducible? | The essential work in this direction was published from 1994 on by J.-P. Serre
and J.C. Jantzen, concerning both algebraic groups and related finite groups
of Lie type. Related papers by R. Guralnick and G.J. McNinch followed. There are uniform dimension bounds for complete reducibility, stricter in rank 1. For a finite group of simple type over a field of $q$ elements in characteristic $p$, Jantzen's upper bound is $p$ for rank at least 2 but $p-2$ in your case. The best I can do is list a few references:
MR1635685 (99g:20079) 20G05 (20G40)
Jantzen, Jens Carsten (1-OR)
Low-dimensional representations of reductive groups are semisimple.
Algebraic groups and Lie groups, 255–266, Austral. Math. Soc. Lect. Ser., 9, Cambridge Univ.
Press, Cambridge, 1997.
MR1753813 (2001k:20096)
McNinch, George J.(1-NDM)
Semisimple modules for finite groups of Lie type.
J. London Math. Soc. (2) 60 (1999), no. 3, 771--792.
MR1717357 (2000m:20018) 20C20
Guralnick, Robert M. (1-SCA)
Small representations are completely reducible.
J. Algebra 220 (1999), no. 2, 531–541.
| 6 | https://mathoverflow.net/users/4231 | 18291 | 12,207 |
https://mathoverflow.net/questions/18282 | 3 | Suppose you start at position 0. You then roll 2 6-sided dice. You move to the integer, call it z, that is the sum of the two dice. You then roll again. If the result of the roll is z', you move to z+z'. You then continue in this fashion. I am looking for formulas (recursive or non-recursive) for the probability of eventually landing on spot n (where n is a positive integer). For small n, n < 10 say, these probabilities are relatively easy to compute by just checking all cases. Note, as n grows, the functions will converge relatively quickly. So, for say n > 40, this may not be a very interesting question.
| https://mathoverflow.net/users/4250 | Probabilities and rolling 2 dice | The probabilities do converge to 1/7. One way to see this is to start from Tony Huynh's comment: the probability that $n$ is hit is the coefficient of $t^n$ in $$f(t) = {1 \over (1-(t+t^2+t^3+t^4+t^5+t^6)^2/36)}$$. The denominator is a polynomial of degree 12; its roots are $t = 1$ and eleven points $r\_1, \ldots, r\_{11}$ which are outside the unit circle. Thus we can write
$$ f(t) = {A \over 1-t} + \sum\_{k=1}^{11} {C\_k \over 1 - t/r\_k} $$
where $A$ and $C\_1, C\_2, \ldots, C\_{11}$ are (complex) constants. This is just the usual partial fraction expansion of a rational function. Taking the $z^n$ coefficient of both sides of the above equation gives
$$p(n) = A + \sum\_{k=1}^{11} C\_k r\_k^{-n}.$$
If we want to show that $\lim\_{n \to \infty} p(n) = 1/7$, we just need to show that $A = 1/7$. This is easy: $A = \lim\_{t \to 1} f(t) (1-t)$. The denominator in $f(t)$ is divisible by $1-t$, so do the polynomial division and substitute $t = 1$.
I wouldn't call the closed form above a "nice" closed form for $p(n)$, though.
| 6 | https://mathoverflow.net/users/143 | 18293 | 12,208 |
https://mathoverflow.net/questions/18298 | 17 | This is a pure curiosity question and may turn out completely devoid of substance.
Let $G$ be a finite group and $H$ a subgroup, and let $V$ and $W$ be two representations of $H$ (representations are finite-dimensional per definitionem, at least per my definitions). With $\otimes$ denoting inner tensor product, how are the two representations $\mathrm{Ind}^G\_H\left(V\otimes W\right)$ and $\mathrm{Ind}^G\_HV\otimes \mathrm{Ind}^G\_HW$ are related to each other? There is a fairly obvious map of representations from the latter to the former, but it is neither injective nor surjective in general. I am wondering whether we can say anything about the decompositions of the two representations into irreducibles.
| https://mathoverflow.net/users/2530 | Induction of tensor product vs. tensor product of inductions | Surely you mean "former to latter"?
I think the natural map is injective. Let $V$ and $W$ have
bases $v\_1,\ldots,v\_r$ and $w\_1,\ldots,w\_s$ respectively.
Let $g\_1,\ldots,g\_t$ be coset representatives for $H$ in $G$.
Then a basis for $\mathrm{Ind}\_H^G V\otimes \mathrm{Ind}\_H^G W$
consists of the $(v\_i g\_k)\otimes(w\_j g\_l)$. The image of
the natural injection from $\mathrm{Ind}\_H^G(V\otimes W)$
is spanned by the $(v\_i g\_k)\otimes(w\_j g\_l)$ with $k=l$.
There are exactly the right number of these.
| 8 | https://mathoverflow.net/users/4213 | 18310 | 12,221 |
https://mathoverflow.net/questions/18313 | 22 | I am currently doing a self study on algebraic geometry but my ultimate goal is to study more on elliptic curves. Which are the most recommended textbooks I can use to study? I need something not so technical for a junior graduate student but at the same time I would wish to get a book with authority on elliptic curves. Thanks
| https://mathoverflow.net/users/4171 | What are the recommended books for an introductory study of elliptic curves? | [Silverman and Tate](http://books.google.com/books?id=mAJei2-JcE4C) to start, then [Silverman](http://books.google.com/books?id=6y_SmPc9fh4C), and finally [Silverman again](http://books.google.com/books?id=dnZ0Vdo-7BsC). These are basically canonical references for the subject.
| 42 | https://mathoverflow.net/users/1847 | 18314 | 12,223 |
https://mathoverflow.net/questions/18307 | 11 | What generalizations of Seiberg-Witten theory to 4-manifolds *with boundary* do exist?
I would be especially interested in theories which "behave good" under gluing along the boundary (comparable to TQFTs).
| https://mathoverflow.net/users/3816 | Seiberg-Witten theory on 4-manifolds with boundary | Hi Fabian! Kronheimer and Mrowka's book [Monopoles and three-manifolds](http://books.google.com/books?id=k6AcxmuTNTIC&printsec=frontcover&dq=kronheimer+mrowka+monopoles&source=bl&ots=Epxl6Ooz7g&sig=pqvqz_jr1zJaMw4aa0_GwnMlB24&hl=en&ei=vKueS-yBFYb6NffRiI0K&sa=X&oi=book_result&ct=result&resnum=2&ved=0CAsQ6AEwAQ#v=onepage&q=&f=false) lays out comprehensively the construction of a Seiberg-Witten TQFT, called monopole Floer homology.
It is conjectured to be isomorphic to the [Heegaard Floer homology](http://arxiv.org/abs/math/0101206) TQFT of Ozsváth-Szabó.
There are also beautiful constructions due to [Froyshov](http://arxiv.org/abs/0809.4842) and to [Manolescu](http://arxiv.org/abs/math/0311342) which do not apply in quite so much generality.
The structure of monopole Floer homology is as follows. The TQFT is a functor on the cobordism category $COB\_{3+1}$ whose objects are *connected*, smooth, oriented 3-manifolds. In fact, the TQFT consists of a trio of functors, denoted $\widehat{HM}\_{\bullet}$, $\overline{HM\_\bullet}$ and $\check{HM}\_{\bullet}$ (the last of these is such a sophisticated invariant that you have to download a special LaTeX package just to typeset it properly). These are $\mathbb{Z}[U]$-modules; there's a story about gradings that's too long to be worth summarising here. There are natural transformations which, for any connected 3-manifold $Y$, define the maps in a long exact sequence
$$ \cdots\to \widehat{HM} \_{\bullet}(Y) \to \overline{HM\_\bullet}(Y)\to \check{HM}\_{\bullet}(Y) \to \widehat{HM}\_{\bullet}(Y) \to \cdots
$$
Why this structure? Well, the theory is based on the Chern-Simons-Dirac functional $CSD$ on a global Coulomb gauge slice through a space of (connection, spinor) pairs. $CSD$ is a $U(1)$-equivariant functional, and $\check{HM}\_{\bullet}$ is, philosophically, its $U(1)$-equivariant semi-infinite Morse homology. $\overline{HM}\_\bullet$ is the part coming from the restriction of $CSD$ to the $U(1)$-fixed-points, and $\widehat{HM}\_\bullet$ is the equivariant homology relative to the fixed point set.
Now here's a subtlety for the TQFT enthusiasts out there to get your teeth into (axiomatize, explain...)! The invariant of a closed 4-manifold $X$ in any of the three theories is... zero. The famous SW invariant of a 4-manifold with $b\_+>0$ comes about via a secondary operation, not part of the TQFT itself. Delete two balls from $X$ to get a cobordism from $S^3$ to itself. When $b\_+(X)>0$, there are generically no reducible SW monopoles on this cobordism, and this implies that the TQFT-map $\widehat{HM}\_\bullet(S^3) \to \widehat{HM}\_\bullet(S^3)$ lifts canonically to a map $\widehat{HM}\_\bullet(S^3) \to \check{HM}\_\bullet(S^3)$; it is this lift that carries the SW invariant.
| 18 | https://mathoverflow.net/users/2356 | 18315 | 12,224 |
https://mathoverflow.net/questions/18301 | 5 | Yet another question of the form 'How to apply the decomposition theorem?' The example that I am considering ought to have a simple answer, but I'm getting confused and I would appreciate if someone could point out where I'm going astray. The confusing point can be stated briefly, at the end of observation 3. But I'll give an explanation of what I do understand, hoping that this will be helpful to other people and make clear what I am missing.
Let $Y$ be a quasi-projective 3-fold with a unique singular point $0 \in Y$ and suppose that the blow-up at $0$ is a resolution $p: X \rightarrow Y$ and the exceptional divisor $p^{-1}(0)=S$ is a smooth projective surface.The goal is to understand the summands of $Rp\_{\ast}IC\_{X} \simeq \bigoplus\_{i}
{}^{\mathfrak{p}}\mathcal{H}^{i}(Rp\_{\ast}IC\_{X})[-i]$, where the decomposition is into perverse cohomology sheaves given by the decomposition theorem.
Observations:
1. By base change, the fact that $p$ is an isomorphism over the open set $U=Y\setminus 0$
implies that $Rp\_{\ast} IC\_{X}$ restricted to $U$ is just $IC\_{U}$, so ${}^{\mathfrak{p}}\mathcal{H}^{0}(Rp\_{\ast}IC\_{X})\simeq IC\_{Y} \oplus E$ for some sky-scraper
$E$ at $0$. Further, the other perverse cohomology sheaves ${}^{\mathfrak{p}}\mathcal{H}^{i}(Rp\_{\ast}IC\_{X})[-i]$ must be supported at $0$ and so consist of shifted sky-scrapers. Thus we just have to understand the stalk of $Rp\_{\ast} IC\_{X}$ at $0$.
2. By base change and the fact that $p^{-1}(0)=S$ a smooth projective surface, we have that the stalk of $Rp\_{\ast} IC\_{X}$ at $0$ is $H^{0}(S,\mathbb{Q})[3]\oplus H^{1}(S,\mathbb{Q})[2]\oplus H^{2}(S,\mathbb{Q})[1]\oplus H^{3}(S,\mathbb{Q})\oplus H^{4}(S,\mathbb{Q})[-1]$. Since $IC\_{Y}$
is concentrated in degrees $-3,-2,-1$ with respect to the standard t-structure,
$E \simeq H^{3}(S,\mathbb{Q})\otimes \mathbb{Q}\_{0}$ and
${}^{\mathfrak{p}}\mathcal{H}^{1}(Rp\_{\ast}IC\_{X})[-1] \simeq H^{4}(S,\mathbb{Q})\otimes \mathbb{Q}\_{0}[-1]$.
Further, there are no higher perverse cohomology sheaves, for degree reasons.
1. By Verdier duality and self-duality of $Rp\_{\ast}IC\_{X}$, the only other perverse cohomology
sheaf in the decomposition is ${}^{\mathfrak{p}}\mathcal{H}^{-1}(Rp\_{\ast}IC\_{X})[1]$, which
must be dual to
${}^{\mathfrak{p}}\mathcal{H}^{1}(Rp\_{\ast}IC\_{X})[-1] \simeq H^{4}(S,\mathbb{Q})\otimes \mathbb{Q}\_{0}[-1]$.
I would think that it should look like
$H^{0}(S,\mathbb{Q})\otimes \mathbb{Q}\_{0}[1]$, but then the degree in which $H^{0}(S)$ sits is off by $-2$ from what appears in observation 2. The only sky-scraper sitting in degree $-1$
is $H^{2}(S)$, but that isn't dual to $H^{4}(S)$ in general.
Presumably I am having a problem with applying Verdier duality, but I don't see where the problem lies. Any comments are very welcome.
| https://mathoverflow.net/users/4659 | Decomposition theorem and blow-ups | In this example we have $p : X \to Y$ and we may assume, wlog, that $X$ is isomorphic to the total space of the normal bundle to the surface, and $p$ is the contraction of the zero section.
Then, by the Deligne construction, $IC(Y) = \tau\_{\le -1} j\_\* \mathbb{Q}[3]$, where $j : Y^0 \hookrightarrow Y$ is the inclusion of the smooth locus (which is isomorphic to $X^0$ the complement of the zero section in $X$).
In order to work this out, we can use the Leray-Hirsch spectral sequence
$E\_2^{p,q} = H^p(S) \otimes H^q(\mathbb{C}^\*) \Rightarrow H^{p+q}(X^0)$
this converges at $E\_3$ and we get that the degree 0, 1 and 2 parts of the cohomology of $X^0$ is given by the *primitive* classes in $H^i(S)$ for $i = 0, 1, 2$. Note that this is everything in degrees 0 and 1, but in degree two the primitive classes form a codimension one subspace $P\_2 \subset H^2(S)$.
The Deligne construction above, gives us that $IC(Y)\_0 = H^0(S)[3]
\oplus H^1(S)[2] \oplus P\_2[1]$.
(This is a general fact: whenever you take a cone over a smooth projective variety, the stalk of the intersection cohomology complex at 0 is given by the primitive classes with respect to the ample bundle used to embed the variety. This follows by exactly the same arguments given above.)
Then the decomposition theorem gives
$p\_\* \mathbb{Q} = \mathbb{Q}\_0[1] \oplus ( IC(Y) \oplus H^3(S) ) \oplus H^4(S)[-1]$.
EDIT: fixed typos pointed out by Chris.
| 4 | https://mathoverflow.net/users/919 | 18317 | 12,225 |
https://mathoverflow.net/questions/18258 | 3 | Let $S$ be a string over some alphabet $\Sigma$. It is well known that a substring of $S$ is commonly defined as a sequence of contiguous elements from $S$, while a subsequence of $S$ is a sequence made of non necessarily contiguous elements from $S$ (e.g. if $S="123465835"$, then $"4658"$ is a substring of $S$ while $"1236"$ is a subsequence of $S$). But is there a word to refer to subsequences that can be obtained from $S$ by removing an arbitrary substring (e.g. $"12835"$)?
(this concept seems complementary to that of a subsequence, hence the conjectured "co-subsequence" in the title -- although "co-substring" might be a good choice too)
| https://mathoverflow.net/users/3356 | Strings and "co-subsequences" | Since you are taking the complement of a substring, and it appears that there may be no firmly established terminology, I propose:
* a *substring complement* is what remains after deleting a substring,
* and more generally, a *subsequence complement* is what remains after deleting a subsequence.
Thus, one may refer to the substring complement of s in t, and use the notation t - s, or $t \setminus s$, with the same notation for the subsequence.
I would prefer this natural language terminology over the alternative co-substring and co-subsequence, which sound unnecessarily technical to my ear, but this difference may be slight.
It does seem worthwhile, however, to distinguish between the two cases, and so I would argue against using the term co-subsequence, as you suggested in your question, to refer to the substring complement.
| 4 | https://mathoverflow.net/users/1946 | 18322 | 12,228 |
https://mathoverflow.net/questions/18227 | 46 | First time poster, long time lurker here. I have a really basic question that has been bugging me for sometime. Specifically, I'm not exactly sure what the 'correct' category theoretic definition of a matroid should be. The only definition I know involves heavy use of set-theory, and is kind of clumsy:
Given a set $E$, a *matroid* $\mathcal{I} \subseteq 2^E$ is a non-empty collection of subsets which satisfy the following axioms:
1. (Heredity) If $X \in \mathcal{I}$ and $X' \subset X$, then $X' \in \mathcal{I}$.
2. (Exchange) If $X, Y \in \mathcal{I}$ and $|X| > |Y|$, then there exists some $b \in X \backslash Y$ such that $Y \cup \{ b \} \in \mathcal{I}$.
Given that both categories and matroids were introduced around the same time and both were studied by MacLane, it stands to reason that someone ought to have thought about this before. Also it is obvious from the Heredity axiom that each matroid is a category, since the containment relation is reflexive and transitive. The second property is a bit more difficult to model, as I am not sure how to get rid of the ugly element / cardinality operators.
In the optimal solution, it would be nice to get rid of the set $E$ entirely, and instead view the specific interpretation of the abstract matroid as a functor from $\mathcal{I} \to 2^E$, the power-set lattice. This would also suggest a functorial interpretation of the graph theoretic and linear algebra applications of matroids. I strongly suspect that someone has already done this, but am having great difficulty locating any references. (Of course I may also be totally wrong headed here too...)
| https://mathoverflow.net/users/4642 | Category theoretic interpretation of matroids? | If I understand your question correctly, I believe that the problem is still open. That is, if we let $\mathcal{M}$ be the category of (simple) matroids, where the morphism are given by strong maps, then it is still open how to describe $\mathcal{M}$ by a nice set of axioms. However, partial progress has been made in this [paper](http://www.emis.de/journals/HOA/IJMMS/Volume26_12/770.pdf).
| 18 | https://mathoverflow.net/users/2233 | 18333 | 12,236 |
https://mathoverflow.net/questions/11153 | 9 | Is there a nice characterization of topological spaces with the property that the product with any paracompact space is paracompact?
All compact spaces have this property (this can be shown from the tube lemma). But somebody once gave me an example (that I cannot locate) of a non-compact space with the property. I didn't check the example carefully, so I cannot vouch for its accuracy.
If a characterization is too hard, an example of a non-compact space would also be great.
[NOTE: I don't assume Hausdorffness in my definitions of compact and paracompact, though it would be nice if the example were a Hausdorff space.]
ADDED LATER: I forgot to mention this, but a product of paracompact spaces need not be paracompact. The standard example is the Sorgenfrey line (the real line with the lower limit topology), which is paracompact, whose product with itself, the Sorgenfrey plane, is not paracompact.
| https://mathoverflow.net/users/3040 | Space whose product with paracompact space is paracompact | One interesting conjectured characterization is due to Rastislav Telgársky. In [*Spaces defined by topological games*](http://matwbn.icm.edu.pl/ksiazki/fm/fm88/fm88120.pdf) (Fund. Math. 88, 1975), Telgársky coined several games and provided partial results regarding this class of paracompact spaces (among other things). In *Some remarks on a Telgarsky’s conjecture concerning products of paracompact spaces* (Topology Appl. 156, 2009, [MR2512606](http://www.ams.org/mathscinet-getitem?mr=2512606)), Kazimierz Alster made some significant progress in showing that one of Telgársky's games completely characterizes the class of spaces whose product with every paracompact space is paracompact.
Looking at MathSciNet, it looks like Alster is getting close to answering this question ([MR2243730](http://www.ams.org/mathscinet-getitem?mr=2243730) and [MR2502008](http://www.ams.org/mathscinet-getitem?mr=2502008)).
| 6 | https://mathoverflow.net/users/2000 | 18334 | 12,237 |
https://mathoverflow.net/questions/18336 | 56 | Background
----------
Model categories are an axiomization of the machinery underlying the study of topological spaces up to homotopy equivalence. They consist of a category $C$, together with three distinguished classes of morphism: Weak Equivalences, Fibrations, and Cofibrations. There are then a series of axioms this structure must satisfy, to guarantee that the classes behave analogously to the topological maps they are named after. The axioms can be found [here](http://en.wikipedia.org/wiki/Model_category).
(As far as I know...) The main practical advantage of this machinery is that it gives a rather concrete realization of the localization category $C/\sim$ where the Weak Equivalences have been inverted, which generalizes the homotopy category of topological spaces. The main conceptual advantage is that it is a first step towards formalizing the concept of "a category enriched over topological spaces".
A discussion of examples and intuition can be found at [this question](https://mathoverflow.net/questions/2185/how-to-think-about-model-categories).
The Question
------------
The examples found in the answers to [Ilya's question](https://mathoverflow.net/questions/2185/how-to-think-about-model-categories), as well as in the introductory papers I have read, all have a model category structure that could be expected. They are all examples along the lines of topological spaces, derived categories, or simplicial objects, which are all conceptually rooted in homotopy theory and so their model structures aren't really surprising.
I am hoping for an example or two which would elicit disbelief from someone who just learned the axioms for a model category. Along the lines of someone who just learned what a category being briefly skeptical that any poset defines a category, or that '$n$-cobordisms' defines a category.
| https://mathoverflow.net/users/750 | What are surprising examples of Model Categories? | Here is an [example](http://arxiv.org/abs/0906.4087) that surprised me at some time in the past.
Bisson and Tsemo introduce a nontrivial model structure on the topos of directed graphs.
Here a directed graph is simply a $4$-tuple $(V,E,s,t)$ where an arc $e \in E$ starts at $s(e) \in V$ and ends at $t(e) \in V$. Fibrations are maps that induce a surjection on the set of outgoing arcs of each vertex, cofibrations are embeddings obtained by attaching a bunch of trees, and weak equivalences are maps that induce a bijection on the sets of cycles.
In this model structure fibrant objects are graphs without sinks and
cofibrant objects are graphs with exactly one incoming arc for every vertex.
Cofibrant replacement replaces a graph by the disjoint union of its cycles
with the obvious morphism into the original graph.
We have a chain of inclusions of categories $A\to B \to C\to D$, where $D$ is the topos of directed graphs, $C$ is the full subcategory of $D$ consisting of all graphs with exactly
one incoming arc for each vertex, $B$ is the full subcategory of $C$ consisting of all graphs with exactly one outgoing arc for each vertex, and $A$ is the full subcategory of $B$ consisting of all graphs such that $s=t$.
Each functor is a part of a Quillen adjunction and total left and right derived
functors compute nontrivial information about graphs under consideration.
Two finite graphs are homotopy equivalent iff they are isospectral iff their zeta-functions coincide.
| 49 | https://mathoverflow.net/users/402 | 18337 | 12,239 |
https://mathoverflow.net/questions/18341 | 2 | Let $\mathbb S$ be $\mathbb C^{\times}$'s restriction of scalar to $\mathbb R$. To give a real Hodge structure on an $\mathbb Q$ vector space $V$ is to give a real representation of $\mathbb S$ on $V\_{\mathbb R}$. Let $\mathbb G\_m\rightarrow \mathbb S\rightarrow GL(V\_{\mathbb R})$ be the weight homomorphism. If it is actually defined over $\mathbb Q$ we say the hodge structure is rational. But can we say that the weight homomorphism is always algebraic, that is, defined over $\overline{\mathbb Q}$? Every resource claims this, but I can't see why...
| https://mathoverflow.net/users/2008 | The algebraicity of Hodge structure map | It looks false to me. Let $V=\mathbb{Q}^{2}$, and let $V(\mathbb{R})=V^{0}\oplus V^{2}$ where $V^{0}$ is the line defined by $y=ex$ and $V^{2}$
is the line defined by $y=\pi x$. Give $V^{0}$ the unique Hodge structure of
type $(0,0)$ and $V^{2}$ the unique Hodge structure of type $(1,1)$. To say
that $w$ is defined over the subfield $\mathbb{Q}^{\mathrm{al}}$ of
$\mathbb{C}$ means that the gradation $V(\mathbb{R})=V^{0}\oplus V^{2}$
arises from a gradation of $V(\mathbb{Q}{}^{\mathrm{al}})$ by tensoring up,
but this isn't true. Perhaps the all the "resources" have additional
conditions, or perhaps they are all ...
Added: When you are defining a Shimura variety, the weight homomorphism w factors through a Q-subtorus of GL(V), and then it is true that w is defined over the algebraic closure of Q (because, for tori T,T', the group Hom(T,T') doesn't change when you pass from one algebraically closed field to a larger field).
| 6 | https://mathoverflow.net/users/930 | 18343 | 12,241 |
https://mathoverflow.net/questions/18319 | 7 | I'm a bit embarrassed to admit that:
a) This is a rather unmotivated question.
b) I can't remember whether or not I've asked this before, but searching doesn't seem to turn anything up so ...
Consider some "shape" function $\phi: \mathbf{R} \to \mathbf{R}$. Then given some function $f: \mathbf{R} \to \mathbf{R}$, one can ask whether the "difference quotient",
$\lim\_{y\to x} \frac{f(y)-f(x)}{\phi(y-x)}$,
exists at various points $x$. Letting $\phi(x) = x$ corresponds to taking normal derivatives, and intuitively when the limit exists this means that near $x$, the function $f$ "looks like" $\phi$ does near 0.
However, if the ratio $\phi(x)/x$ is not bounded above or away from 0 as $x\to 0$ (I'm mostly thinking of the case when it is neither, so that $\phi$ is "wildly oscillating" in some sense), then anywhere the above limit exists and is nonzero, the function $f$ is necessarily non-differentiable.
My question: If $\phi$ is some wildly oscillating function as described above (pick your favorite), can there be an $f$ for which this limit exists everywhere?
(Edit: I suppose I really want $\phi$ and $f$ to be continuous functions.)
| https://mathoverflow.net/users/4402 | Can a continuous, nowhere differentiable function have specified "shape" at every point? | Assume WLOG that $\phi(x)>0$ when $x>0$. Since the limit described exists for all $x$ in the source of $f$. We get for any $x$ the bound:
$f(x+\delta)-f(x) \leq C\phi(\delta)$
for $0 < \delta < \delta\_0$ for some $C,\delta\_0>0$ which may depend on $x$.
diving by $\delta$ we get by the assumptions on $\phi$ that
$\underline{\lim}\_{\delta \to 0} ( \frac{f(x+\delta) - f(x)}{\delta}) \leq 0$
This is one the four derivatives of $f$, and proposition 2 chapter 5 in Real Analysis by H.L. Royden states that if $f$ is continuous then it is (non-strictly) decreasing. Similar for increasing. So $f$ is constant.
| 6 | https://mathoverflow.net/users/4500 | 18345 | 12,242 |
https://mathoverflow.net/questions/18352 | 64 | As a non-native English speaker (and writer) I always had the problem of understanding the distinction between a 'Theorem' and a 'Proposition'. When writing papers, I tend to name only the main result(s) 'Theorem', any auxiliary result leading to this Theorem a 'Lemma' (and, sometimes, small observations that are necessary to prove a Lemma are labeled as 'Claim'). I avoid using the term 'Proposition'.
However, sometimes a paper consists of a number important results (which by all means earn to be called 'Theorem') that are combined to obtain a certain main result. Hence, another term such as 'Proposition' might come in handy, yet I don't know whether it suits either the main or the intermediate results.
So, my question is: When to use 'Theorem' and when to use 'Proposition' in a paper?
| https://mathoverflow.net/users/3919 | Theorem versus Proposition | The way I do it is this: main results are theorems, smaller results are called propositions.
A Lemma is a technical intermediate step which has no standing as an independent result.
Lemmas are only used to chop big proofs into handy pieces.
| 70 | https://mathoverflow.net/users/nan | 18356 | 12,245 |
https://mathoverflow.net/questions/18357 | 10 | Hello. I am currently searching for some nice examples of proofs by induction in the finite case, that can be generalized to the infinite case using transfinite induction (and dont become trivial there).
Apparently, the only proof that comes in my mind is that every Vector-Space has a base (mostly this is proven by the Zorn-Lemma, but in the finite case this can be proven by induction, and the same proof can be used in the infinite case with transfinite induction).
Do you have any other examples?
| https://mathoverflow.net/users/3118 | Examples of inductive proofs that can be generalized by transfinite induction | A nice example arises in structural proof theory. You can prove cut-elimination of the sequent calculus for first-order logic by an induction on the size of the cut formula, and the sizes of the proofs you are cutting into and cutting from. By adding rules for induction over the natural numbers, you need transfinite induction up to $\epsilon\_0$ -- but because each rule of the sequent calculus only mentions a principal formula, all the other cases are left untouched in the new proof.
| 5 | https://mathoverflow.net/users/1610 | 18362 | 12,248 |
https://mathoverflow.net/questions/18351 | 7 | What can be said about the relation between the complex and the real K-theory of a CW complex? An $n$-dimensional complex vector bundle is an $2n$-dimensional real vector bundle but not vice versa. You get a map
$$
K(X)\to KO(X).
$$
What can be said about that?
| https://mathoverflow.net/users/2625 | Relation between $KO$ and $K$ | There is a long exact sequence of (reduced) K-groups
$$
K^{n-1}(X) \to KO^{n+1}(X) \to^\eta KO^n(X) \to^c K^n(X) \to^f KO^{n+2}(X) \to \cdots
$$
The map $c$ is induced by complexification, sending a real vector bundle $\xi$ to the associated complex vector bundle. The map $\eta$ is multiplication by the Hopf element (the Mobius band bundle) in
$$
KO^{-1}(S^0) = KO^0(S^1) = \pi\_1(BO).
$$
The map $f$ is induced by the forgetful map from complex vector bundles to real vector bundles, together with the Bott periodicity isomorphism:
$$
K^n(X) \cong K^{n+2} \to KO^{n+2}(X)
$$
The maps $c$, $f$, and Bott periodicity can be used to produce the splitting of $KO(X)$ off $K(X)$ after inverting the prime 2 that Andrew mentioned.
| 7 | https://mathoverflow.net/users/360 | 18369 | 12,253 |
https://mathoverflow.net/questions/18375 | 8 | I know it's likely that, given a finite sequence of digits, one can find that sequence in the digits of pi, but is there a proof that this is possible for all finite sequences? Or is it just very probable?
| https://mathoverflow.net/users/2576 | Is there any finitely-long sequence of digits which is not found in the digits of pi? | [This article](http://www.lbl.gov/Science-Articles/Archive/pi-random.html) contains the following statements.
>
>
> >
> > Describing the normality property, Bailey explains that "in the familiar base 10 decimal number system, any single digit of a normal number occurs one tenth of the time, any two-digit combination occurs one one-hundredth of the time, and so on. It's like throwing a fair, ten-sided die forever and counting how often each side or combination of sides appears."
> >
> >
> > Pi certainly seems to behave this way. In the first six billion decimal places of pi, each of the digits from 0 through 9 shows up about six hundred million times. Yet such results, conceivably accidental, do not prove normality even in base 10, much less normality in other number bases.
> >
> >
> > In fact, not a single naturally occurring math constant has been proved normal in even one number base, to the chagrin of mathematicians. While many constants are believed to be normal -- including pi, the square root of 2, and the natural logarithm of 2, often written "log(2)" -- there are no proofs.
> >
> >
> >
>
>
>
Edit: We seem to have lost Gerald Edgar's comment on another answer, which pointed out that the normality property implies but is not equivalent to the property of containing every finite sequence as a substring. For example, consider the sequence enumerating all possible finite strings, separated by increasingly huge oceans of zeros.
| 12 | https://mathoverflow.net/users/1946 | 18378 | 12,258 |
https://mathoverflow.net/questions/18374 | 10 | Let $K$ be a field and $L$ an extension of $K$. I wonder how much larger the multiplicative group $L^\times$ of $L$ is than the multiplicative group $K^\times$ of $K$.
I know that if $L=K(t)$ and $t$ is transcendental over $K$, then $L^\times/K^\times$ is isomorphic to the direct product of an infinite number of copies of the integers: Indeed, any (monic) rational function can be uniquely written as the product of a finite number of powers of irreducible polynomials.
In particular, $L^\times/K^\times$ is not finitely generated.
What happens when $L=K(t)$ and $t$ is algebraic over $K$? I can handle the case when $K$ is finite, but what happens in the infinite case?
Even for $L=Q(\sqrt{2})$ it's not immediate to me if $L^\times/K^\times$ is finitely generated or not.
| https://mathoverflow.net/users/3380 | If L is a field extension of K, how big is L*/K*? | You're looking for
* A. Brandis,
*Über die multiplikative Struktur von Körpererweiterungen*,
Math. Z. 87 (1965), 71-73
Brandis proved that $L^\times/K^\times$ is not finitely generated whenever $K$ is infinite and $L \ne K$ (thanks Pete).
The claim is reduced to finite algebraic extensions of global fields, for which there are infinitely many prime ideals in $K$ that do not remain inert in $L$, and this does it.
| 14 | https://mathoverflow.net/users/3503 | 18380 | 12,259 |
https://mathoverflow.net/questions/18393 | 3 | An arbitrary union, or a finite intersection, of open sets in a topological space is again open. What name is given to the hypothetical property that an *arbitrary* intersection of open sets is open?
As an example, consider a partially ordered set $X$. Call a subset $U\subseteq X$ open if $y\le x\in U$ implies $y\in U$. (Bonus question: Are there other interesting examples?)
| https://mathoverflow.net/users/802 | What do you call a topology that is closed under arbitrary intersections? | [Alexandrov spaces](http://en.wikipedia.org/wiki/Alexandrov_topology).
| 16 | https://mathoverflow.net/users/1409 | 18394 | 12,269 |
https://mathoverflow.net/questions/18348 | 2 | When we say, that, say, a surface contains ${\infty}^{k}$ lines, do we mean that it contains a k-parameter family of lines? Do we assume that this family is parametrized by a $P^{k}$, say, or we use this term more informally?
This is certainly a standard notation, but I didn't see its explanation in standard modern textbooks.
| https://mathoverflow.net/users/4668 | The meaning of ${\infty}^{k}$. | The terminology is fairly common in classical works on projective algebraic/differential geometry. I am not aware of its origins. Anyway it is used rather informally and only means
that you have a $k$-dimensional "continuous" family of objects. The parameter space usually is only local and should be thought as a small ball in $\mathbb C^k$.
| 3 | https://mathoverflow.net/users/605 | 18396 | 12,270 |
https://mathoverflow.net/questions/18358 | 2 | I'm a programmer and I came a across an interesting problem. I'm sure there is a mathematical method or an algorithm to solve it, but I don't know where to start with the search nor which literature to read. If anyone can please tell me what method to use to solve this kind of problem, I would appreciate it much.
Here's a simplified example:
```
a + b + c >= 100
d + e + f >= 50
g + h >= 30
x = 1
y = 1
z = 1
5xa + 8yb + 5zc + 3xd + 2ye + 2zf + 6xg + 7yh = w.
```
I need to find the values for a, b, c, d, e, f, g and h that minimize w.
Of course, those numbers are just thought-off; the real numbers are input into my computer program by the user, and number of variables (a,b,c...) might be different for each run.
This was the simple case. In step 2, I also need to add rules like this:
```
if (8yb + 2ye + 7yh > 620) then y = 0.8 else y = 1.0
if (d > 35) then x = 0.9 else x = 1.0
etc.
```
I understand I could write a set of formulas for each rules (by replacing the value for x, y or z and adding another (in)equation), however those rules are such that they might be met, but not necessary - it is only important to minimize W, and only those rules that can "help" should be "activated". I wonder if there is some way to write "if" in mathematics?
I believe I've seen this kind of problem when I was in grad school, but that was some 10 years ago and I can't remember, so I don't even know what to look for, and I also don't know how to classify this problem. Please re-tag.
| https://mathoverflow.net/users/4669 | Which method to apply to this problem? | If you really have equations like
>
> $ 5xa + 8yb + 5zc + 3xd + 2ye + 2zf + 6xg + 7yh \leq w $
>
>
>
and all of the letters above are variables, then in the most general case you have an instance of Multivariate Quadratic Equations, which is $NP$-complete. (Even without the "if...then" rules.) The hardness is really independent of the domain of the variables. As long as each variable domain takes on at least two distinct possible values, you are in the land of $NP$-hard. Note you can always force a variable $x$ to take on exactly two of the possible values $v\_1$, $v\_2$ by imposing that
>
> $(x-v\_1)(x-v\_2) = 0$.
>
>
>
As mentioned in other comments, a tractable special case would be if $x$, $y$, $z$ are fixed and $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ vary over the rationals. In this case your problem is an instance of Linear Programming, known to be solvable in polynomial time (and efficiently in practice; note these two properties do not always coincide!).
Another potentially tractable case (in practice, not in theory) is when $x$, $y$, $z$ are fixed and some of the $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ can only be $0$ or $1$. (Typically this condition can be translated to: some variables in the range $a$ through $g$ can only take on one of two possible values such as $a=0.9$ or $a=1$, through a linear transformation.) This case is called Integer Linear Programming. Although it is also $NP$-complete, there is software available that can sometimes solve instances of these fairly efficiently. Moreover, depending on the "if...then" rules you have, you may be able to cleverly translate them into integer linear programming constraints. (Note an "if...then" rule has one of two possible outcomes for a variable, and similarly an integer-valued variable will have one of two possible outcomes.)
Here's a simple example of what I mean, though I don't think this example will help you directly. Suppose $x$ is a variable that's either 0 or 1 and $d$ is a coefficient. You want to translate: "if ($d \geq 35$) then $x=0$ else $x=1$". Look at the inequalities $d \geq (d-35)x + 35$, $d \leq (34.999-d)x+d$. When $x=0$, we have $d \geq 35$, $d \leq d$. When $x=1$, we have $d \geq d$, $d \leq 34.999$.
To get any more specific about things, I would need to know more properties of the problems you are trying to solve. Hopefully by searching for the names of the above problems you can find more relevant references. Good luck!
| 2 | https://mathoverflow.net/users/2618 | 18398 | 12,271 |
https://mathoverflow.net/questions/18404 | 12 | Given two infinite sets $A$, $B$ of natural numbers, write $A\preceq B$ if $B\setminus A$ is a finite set. Define the equivalence relation $A\sim B$ if $A\preceq B$ and $B\preceq A$, and let $\partial\mathbb{N}$ be the set of equivalence classes of infinite sets under this equivalence relation. Write $[A]$ for the equivalence class of $A$.
Now define a topology on the disjoint union $\overline{\mathbb{N}}=\mathbb{N}\cup\partial\mathbb{N}$ as follows: A set $U\subseteq\overline{\mathbb{N}}$ is open if and only if, for every $[A]\in U\cap\partial\mathbb{N}$, $[B]\in U$ whenever $B\prec A$, and moreover $A'\subset U$ for *some* $A'\sim A$.
>
> Is this a known topology? Does it have a name?
>
>
>
It is not hard to see that $\overline{\mathbb{N}}$ is compact (hence the question title): For any neighbourhood of $[\mathbb{N}]$ is the entire space minus a finite subset of $\mathbb{N}$. In particular, it is very much a non-Hausdorff space. Also, $\overline{\mathbb{N}}$ contains $\mathbb{N}$ as an open subset with the discrete topology on the latter. The subspace $\partial\mathbb{N}$ is indeed the boundary of $\mathbb{N}$ in this topology (hence my chosen notation), and it is an [Alexandrov space](https://mathoverflow.net/questions/18393/what-do-you-call-a-topology-that-is-closed-under-arbitrary-intersections).
I came up with this topology while thinking about sequences, subsequences, and their limits. It seems rather natural, so I don't think I am the first one to ever think of it.
| https://mathoverflow.net/users/802 | Is this a known compactification of the natural numbers? | Your set $\partial\mathbb{N}$ is also intensely studied in set theory and known as P(ω)/Fin. What you have done is mod out by the ideal of finite sets. People study more general properties P(X)/I, taking the quotient by many other ideals (or by an arbitrary ideal). P(X)/I is a Boolean algebra, and many forcing arguments can be viewed as forcing with this Boolean algebra. The topological properties are very much used in that forcing context, since the generic filters are exactly those containing elements from every ground model dense set. The finite sets become equivalent to the point [emptyset] in this algebra.
Perhaps Lusin was the first to study P(ω)/Fin seriously, and found the phenomenon of Lusin gaps. A gap in P(ω)/Fin is a cut in the order, where the left side increases and the right side decreases, and everything on the left is below everything on the right, with respect to almost-inclusion. Lusin found gaps of various types, including ones with uncountable cofinality.
A particularly interesting case is P(ω1)/I, where I is the ideal of non-stationary sets, and many set theoretic hypotheses, some engaging with large cardinals, interact with the topological properties of that situation.
A few quick examples:
* [Farah, How many Boolean algebras P(ω)/I are there?](http://www.math.yorku.ca/~ifarah/Ftp/ch.pdf)
* [Farah, Luzin gaps](http://www.jstor.org/stable/3844945)
* [Just and Krawczyk, Isomorphisms of P(ω)/I](http://www.jstor.org/stable/1999489)
* [Jech, Precipitous Ideals](http://www.jstor.org/stable/2273349)
* [Matsubara, Saturated Ideals and the SCH](http://www.jstor.org/stable/2275442)
* There are many others, including probably some good introductory surveys.
| 14 | https://mathoverflow.net/users/1946 | 18405 | 12,276 |
https://mathoverflow.net/questions/18411 | 31 | The
[Baumslag-Solitar groups](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-68/issue-3/Some-two-generator-one-relator-non-Hopfian-groups/bams/1183524561.full)
$BS(m,n) = \langle b,s\mid s^{-1}b^ms = b^n\rangle$, with $mn\neq 0$,
are important examples (more often, counter-examples) in group theory.
They are residually finite if, and only if, either $m$ and $n$
are equal in absolute value, or one of $m$ and $n$ has absolute value equal to
$1$. In that case, they are Hopfian, and also when $m$ and $n$ have the same
prime divisors. Otherwise, they are non-Hopfian.
For $m=n$, we get examples of one-relator groups with non-trivial center.
The group $BS(2,2)$ is an example in which the Howson property fails.
I can only recall having seen these groups defined by means of a presentation. I would like to know whether these groups (apart from the obvious special cases such as the metabelian ones) can be realized by some other fairly elementary and concrete mechanism. My question is:
>
> Does $BS(m,n)$ occur "in nature"?
>
>
>
(For example, the Sanov matrices
$\left(\begin{matrix} 1&2 \\ 0&1 \end{matrix}\right)$
and
$\left(\begin{matrix} 1&0 \\ 2&1\end{matrix}\right)$
generate a free subgroup of $SL(2,\mathbb{Z})$, so I would say that free groups occur "in nature".)
Obviously, since the Baumslag-Solitar groups are often non-Hopfian, they
cannot be constructed as groups of matrices. But, perhaps there is some other concrete realization of these groups.
If there isn't a general construction, it would still be useful to get a concrete realization for the non-Hopfian group $BS(2,3)$.
| https://mathoverflow.net/users/1392 | Do the Baumslag-Solitar groups occur in nature? | For $m,n \gt 0, m\ne n$, $BS(m,n)$ acts on $\mathbb H^2$ by isometries with a common ideal fixed point. In particular, you can represent it by the action on the upper half plane of
$S = \bigg(\begin{matrix}\sqrt{m/n} & 0 \\\ 0 &\sqrt{n/m}\end{matrix}\bigg),$
$B = \bigg(\begin{matrix}1 & \alpha \\\ 0 &1\end{matrix}\bigg)$
where $\alpha$ is arbitrary.
There are elements of $BS(m,n)$ which act trivially, but you can lift this to a free action on a topological $T\times \mathbb R$ where $T$ is a $n+m$-regular directed tree with outdegree $m$ and indegree $n$ so that for any path $P \subset T$, $P\times \mathbb R$ is a copy of $\mathbb H^2$ so that the sets $\{p\}\times \mathbb R$ are concentric horocycles (like horizontal lines in the upper half plane).
This is related to the tilings of the hyperbolic plane by horobricks, e.g., there are tiles with arbitrarily small diameter which tile $\mathbb H^2$ which are really fundamental domains of $BS(m,n)$, and analogously there are polyhedra of arbitrary Dehn invariant which tile $\mathbb H^3$.
| 29 | https://mathoverflow.net/users/2954 | 18417 | 12,283 |
https://mathoverflow.net/questions/18423 | 31 | I've got a group $G$ that I'm trying to prove is free. I already know that $G$ is torsion-free. Moreover, I can "almost" prove what I want : I can find a finite index subgroup $G'$ of $G$ that is definitely free.
This leads me to the following question. Can anyone give me an example of a torsion-free group $G$ that is not free but contains a free subgroup of finite index? I've tried pretty hard to find groups like this, but i can't seem to avoid introducing torsion. Thanks!
| https://mathoverflow.net/users/4682 | Proving that a group is free | It's a theorem of Stallings and Swan that a group of cohomological dimension one is free.
By a theorem of Serre, torsion-free groups and their finite index subgroups have the same cohomological dimension.
So, a torsion-free group is free if and only if one of its finite index subgroups are free.
(Here are the references. For Stallings-Swan, see
John R. Stallings, "On torsion-free groups with infinitely many ends", Annals of Mathematics 88 (1968), 312–334.
and
Richard G. Swan, "Groups of cohomological dimension one", Journal of Algebra 12 (1969), 585–610.
Serre's theorem is in Brown's book "Cohomology of Groups.")
| 52 | https://mathoverflow.net/users/1335 | 18425 | 12,288 |
https://mathoverflow.net/questions/16888 | 31 | If we have a nonconstant map of nonsingular curves $\varphi:X\rightarrow Y$, then Hartshorne defines a map $\varphi^\* Div(Y)\rightarrow Div(X)$ using the fact that codimension one irreducibles are just points, and looking at $\mathcal{O}\_{Y,f(p)} \rightarrow \mathcal{O}\_{X,p}$. My question is if we don't have a nice map of curves, what conditions can we put on the morphism so that we may pull divisors back? Clearly it's not true in general, since we can take a constant map and then topologically the inverse image doesn't even have the right codim.
Thinking about this in terms of Cartier divisors (and assuming the schemes are integral), it seems like we just need a way to transport functions in $K(Y)$ to functions in $K(X)$. If $\varphi$ is dominant, then we'll get such a map. Is this sufficient? Also is there something we can say when $\varphi$ is not dominant? Something like we have a way to map divisors with support on $\overline{\varphi(X)}$ to divisors on $X$?
| https://mathoverflow.net/users/3261 | When do divisors pull back? | If you want to pull back a Cartier divisor $D$, you can do that provided the image of $f$ is not contained in the support of $D$: just pull back the local equations for $D$.
If this does not happen, on an integral scheme, you can just pass to the associated line bundle $\mathcal{O}\_X(D)$ and pull back that, obtaining $f^{\*} \mathcal{O}\_X(D)$; of course you lose some information because a line bundle determines a Cartier divisor only up to linear equivalence.
Fulton invented a nice way to avoid this distinction. Define a pseudodivisor on $X$ to be a triple $(Z, L, s)$ where $Z$ is a closed subset of $X$, $L$ a line bundle and $s$ a nowhere vanishing section on $X \setminus Z$, hence a trivialization on that open set. Then you can simply define the pullback of this triple as $(f^{-1}(Z), f^{\*} L, f^{\*} s)$, so you can always pull back pseudo divisors, whatever $f$ is.
The relation with Cartier divisors is the following: to a Cartier divisor $D$ you can associate a pseudodivisor $(|D|, \mathcal{O}\_X(D), s)$, where $s$ is the section of $\mathcal{O}\_X(D)$ which gives a local equation for $D$.
This correspondence is not bijective. First, a pseudodivisor $(Z, L, s)$ determines a Cartier divisor if $Z \subsetneq X$; note that in this case enlarging $Z$ will not change the associated Cartier divisor, so to obtain a bijective correspondence with Cartier divisors you have to factor out pseudodivisors by an equivalence relation, which I leave to you to formulate. But if $Z = X$, you only obtain a line bundle on $X$, and you have no way to get back a Cartier divisor.
If you want to know more about this, read the second chapter of Fulton's intersection theory.
| 21 | https://mathoverflow.net/users/828 | 18426 | 12,289 |
https://mathoverflow.net/questions/11249 | 3 | By Fermat's little theorem we know that
$$b^{p-1}=1 \mod p$$
if p is prime and $\gcd(b,p)=1$. On the other hand, I was wondering whether
$$b^{n-1}=-1 \mod n$$
can occur at all?
Update: sorry, I meant n odd. Please excuse.
| https://mathoverflow.net/users/3032 | b^(n-1)=-1 mod n | There are no solutions to $b^{n-1}\equiv-1\pmod n$ with $n$ odd.
Let $n>1$ be odd. Every prime dividing $n$ can be written as $2^km+1$ for some positive $k$ and some odd integer $m$. Among those primes, let $p$ have the minimal value of $k$. Then $n-1=2^kr$ for some integer $r$. If
$b^{n-1}\equiv-1\pmod n$ then $b^{n-1}\equiv-1\pmod p$ so $b^{(n-1)m}\equiv(-1)^m\equiv-1\pmod p$ and
$\gcd(b,p)=1$. But $b^{(n-1)m}=b^{2^kmr}=b^{(p-1)r}\equiv1\pmod p$ by little Fermat. Contradiction, QED.
| 10 | https://mathoverflow.net/users/3684 | 18428 | 12,291 |
https://mathoverflow.net/questions/18410 | 13 | I am wondering about the following problem: for which (say smooth, complex, connected) algebraic varieties $X$ does the statement *any regular map $X\to X$ has a fixed point* hold?
MathSciNet search does not reveal anything in this topic.
This is true for $\mathbb{P}^n$ (*because its cohomology is $\mathbb{Z}$ in even dimensions
and $0$ otherwise, and the pullback of an effective cycle is effective, so all summands
in the Lefschetz fixed point formula are nonnegative, and the 0-th is positive* -- is this a correct argument?). Is it true for varieties with cohomology generated by algebraic cycles (i.e. $h^{p,q}(X)=0$ unless $p=q$ and satisfying Hodge conjecture), for example for **Grassmannians**, **toric varieties**, etc.? This is not at all clear that the traces of $f$ on cohomology will be nonnegative.
Probably you have lots of counterexamples. What about positive results?
| https://mathoverflow.net/users/3847 | Fixed Point Property in Algebraic Geometry | By demand I expand a little on my answer. The holomorphic Lefschetz fixed point formula (aka the Woods-Hole formula) considers an endomorphism $f\colon M \to M$ of a smooth and compact complex manifold $M$ (or proper
smooth algebraic variety) with only isolated fixed points which are also assumed to be non-degenerate (i.e., the
tangent map of $f$ at a fixed point does not have eigenvalue $1$. Then the alternating trace of the action of $f$ on $H^\*(M,\mathcal O\_M)$ is equal to a sum over the fixed points $p$ of $1/det(1-T\_p(f)$. In particular if there
are no fixed points the alternating trace is equal to $0$. However, in the case when also $H^i(M,\mathcal O\_M)=0$ for $i>0$ then the alternating trace is equal to $1$ so the assumption that there are no fixed points gives a contradiction. Note, that the dimension of $H^i(M,\mathcal O\_M)$ is just $h^{0,i}$ so that an assumption that the Hodge numbers vanish off the diagonal gives the required vanishing by a good margin. Furthermore, the vanishing of just $h^{0,i}$ for $i>0$ is much much weaker, it is for instance a birational condition whereas blowing up a smooth curve of genus $>0$ in a variety of dimension at least 3 always give
off diagonal Hodge numbers.
As for the question of whether a birational involution of $\mathbb C^n$ always has a fixed point this seems trickier. It is true that the involution can be made to act regularly on a smooth and proper model and hence by the above has a fixed point. It is not clear however that the fixed point will map to a point of $\mathbb C^n$ as $\mathbb C^n$ is not proper.
| 12 | https://mathoverflow.net/users/4008 | 18430 | 12,292 |
https://mathoverflow.net/questions/18433 | 9 | More specificaly, is there a haussdorf non-discrete topology on $\mathbb{Z}$ that makes it a topological group with the usual addition operation?
| https://mathoverflow.net/users/4619 | Is there a non-trivial topological group structure of $\mathbb{Z}$? | Yes. Take, for example, the subgroups $p^k\mathbb{Z}$, for $k>0$ and a fixed prime $p$, as a basis of neighborhoods of the identity.
| 14 | https://mathoverflow.net/users/1392 | 18435 | 12,295 |
https://mathoverflow.net/questions/18421 | 58 | In an unrelated thread Sam Nead intrigued me by mentioning a formalized proof of the Jordan curve theorem. I then found that there are at least two, made on two different systems. This is quite an achievement, but is it of any use for a mathematician like me? (No this is not what I am asking, the actual question is at the end.)
I'd like to trust the theorems I use. To this end, I can read a proof myself (the preferred way, but sometimes hard to do) or believe experts (a slippery road). If I knew little about topology but occasionally needed the Jordan theorem, a machine-verified proof could give me a better option (and even if I am willing to trust experts, I could ensure that there are no hidden assumptions obvious to experts but unknown to me).
But how to make sure that a machine verified the proof correctly? The verifying program is too complex to be trusted. A solution is of course that this smart program generates a long, unreadable proof that can be verified by a dumb program (so dumb that an amateur programmer could write or check it). I mean a program that performs only primitive syntax operations like "plug assertions 15 and 28 into scheme 9". This "dumb" part should be independent of the "smart" part.
Given such a system, I could check axioms, definitions and the statement of the theorem, feed the dumb program (whose operation I can comprehend) with these formulations and the long proof, and see if it succeeds. That would convince me that the proof is indeed verified.
However I found no traces of this "dumb" part of the system. And I understand that designing one may be hard. Because the language used by the system should be both human-friendly (so that a human can verify that the definitions are correct) and computer-friendly (so that a dumb program can parse it). And the definitions should be chosen carefully - I don't want to dig through a particular construction of the reals from rationals to make sure that this is indeed the reals that I know.
Sorry for this philosophy, here is the question at last. Is there such a "dumb" system around? If yes, do formalization projects use it? If not, do they recognize the need and put the effort into developing it? Or do they have other means to make their systems trustable?
UPDATE: Thank you all for interesting answers. Let me clarify that the main focus is interoperability with a human mathematician (who is not necessarily an expert in logic). It seems that this is close to interoperability between systems - if formal languages accepted by core checkers are indeed simple, then it should be easy to automatically translate between them.
For example, suppose that one wants to stay within symbolic logic based on simple substitutions and axioms from some logic book. It seems easy to write down these logical axioms plus ZF axioms, basic properties (axioms) of the reals and the plane, some definitions from topology, and finally the statement of the Jordan curve theorem. If the syntax is reasonable, it should be easy to write a program verifying that another stream of bytes represents a deduction of the stated theorem from the listed axioms. Can systems like Mizar, Coq, etc, generate input for such a program? Can they produce proofs verifiable by cores of other systems?
| https://mathoverflow.net/users/4354 | How do they verify a verifier of formalized proofs? |
>
> Is there such a "dumb" system around? If yes, do formalization projects use it? If not, do they recognize the need and put the effort into developing it? Or do they have other means to make their systems trustable?
>
>
>
This is called the "de Bruijn criterion" for a proof assistant -- just as you say, we want a simple proof checker, which should be independent of the other machinery. The theorem provers which most directly embody this methodology are those in the LCF tradition, such as Isabelle and HOL/Light. They actually work by generating proof objects via whatever program you care to write, and sending that to a small core to check. Systems based on dependent type theory (such as Coq) tend to have more complex logical cores (due to the much greater flexibility of the underlying logic), but even here a core typechecker can fit in a couple of thousand lines of code, which can easily be ([and have been](http://matita.cs.unibo.it/)) understood and reimplemented.
| 31 | https://mathoverflow.net/users/1610 | 18436 | 12,296 |
https://mathoverflow.net/questions/18444 | 3 | Consider the structure $\langle HF,\epsilon\rangle$ (the hereditarily finite sets with the epsilon-relation). An ultrapower of this structure will have externally-infinite elements -- elements not generated by a finite number of applications of the (definable) singleton+binary-union operations.
Can anybody give me a starting point for literature on the properties of these externally-infinite sets?
Thanks!
| https://mathoverflow.net/users/2361 | Behavior of externally-infinite elements in ultrapowers of $\langle HF,\epsilon\rangle$ | As I pointed out in the comments, $(HF,{\in})$ is biinterpretable with $\mathbb{N}$, which means that the corresponding ultrapowers are biinterpretable too. So you will find all you need in the vast literature on nonstandard arithmetic.
A nice interpretation of $(HF,{\in})$ in $\mathbb{N}$ is given by defining $m \in n$ if the $m$-th binary digit of $n$ is $1$. For example, here are the first few coded sets:
* $0$ codes $\varnothing$
* $1 = 2^0$ codes $\{\varnothing\}$
* $2 = 2^1$ codes $\{\{\varnothing\}\}$
* $3 = 2^0 + 2^1$ codes $\{\varnothing,\{\varnothing\}\}$
You can similarly interpret the ultrapower $HF^\omega/\mathcal{U}$ in the ultrapower $\mathbb{N}^{\omega}/\mathcal{U}$. If $\bar{m},\bar{n} \in \mathbb{N}^\omega$, the $\bar{m}$-th binary digit of $\bar{n}$ is first interpreted term-by-term giving a sequence $\bar{b} \in \{0,1\}^\omega$ where each $b\_i$ is the $m\_i$-th binary digit of $n\_i$. In $\mathbb{N}^\omega/\mathcal{U}$, $\bar{b}$ is evaluated as either $0$ or $1$, depending on which value occurs $\mathcal{U}$-often. This value tells you whether $\bar{m} \in \bar{n}$ according to the above interpretation.
Of course, you can simply compute things directly. Given sequences $\bar{x},\bar{y} \in HF^\omega$, we have $\bar{x} \in \bar{y}$ in the ultrapower $HF^\omega/\mathcal{U}$ if and only if $\{i : x\_i \in y\_i \} \in \mathcal{U}$. This gives exactly the same structure as interpreting sets in $\mathbb{N}^\omega/\mathcal{U}$ as described above.
---
It just occurred to me that you may be looking for a more set-theoretic description of the nonstandard elements of $HF^\omega/\mathcal{U}$.
The wellfounded part of $HF^\omega/\mathcal{U}$ is precisely $HF$ and no more. A sequence $\bar{x} \in HF^\omega$ will represent a wellfounded set in $HF^\omega/\mathcal{U}$ if and only if it has bounded rank mod $\mathcal{U}$, i.e. $\{i : \mathrm{rk}(x\_i) < n\} \in \mathcal{U}$ for some $n < \omega$, in which case it will be constant mod $\mathcal{U}$ since there are only finitely many sets of rank less than $n$. If this is not the case, then $\langle\mathrm{rk}(x\_i)\rangle$ evaluates to a nonstandard ordinal $N$ in $HF^\omega/\mathcal{U}$. This means that in $HF^\omega/\mathcal{U}$, we can (externally) find an infinite descending ${\in}$-chain starting with the evaluation of $\bar{x}$. So it is impossible to describe the evaluation of $\bar{x}$ as a real set.
Without peering into the depths of the nonstandard ${\in}$ relation, there is not much to elements of $HF^{\omega}/\mathcal{U}$. Every element of $HF^\omega/\mathcal{U}$ has a bijection with a (possibly nonstandard) ordinal, so it looks exactly like an internal initial segment of the nonstandard ordinals.
| 2 | https://mathoverflow.net/users/2000 | 18452 | 12,307 |
https://mathoverflow.net/questions/18440 | 50 | This summer my wife and one of my friends (who are both programmers and undergraduate math majors, but have not learned any algebraic geometry) want to learn some algebraic geometry from me, and I want to learn how to program from them, so we were planning on working on some computational algebraic geometry together. While there are several books which we could work through, I thought it might be more fun and productive if we had the goal of developing a usable new algorithm, or at least implementing an algorithm which no one has implemented before. I do not have any ideas, but I thought that some mathoverflowers might have had an idea for an algorithm they would like to see implemented but have never had the time to work through the details. Keep in mind that my wife and friend will have to learn any mathematics past a first course in topology and abstract algebra as we go.
So does anyone have any ideas for an algorithm they would like to use which is within the reach of my "team" to implement within a summer? We are planning on working on this stuff between 2 and 3 hours a day for about 3 months.
| https://mathoverflow.net/users/1106 | What algorithm in algebraic geometry should I work on implementing? | Just a thought, but maybe you should have a look at [sage](http://sagemath.org). It's a big open source project that is currently under very active development. If you're interested in contributing, I would suggest that you post to the sage-devel Google group with this same question. Some thoughts for things to do would be to improve the support for relative extensions of number fields and for function fields.
| 23 | https://mathoverflow.net/users/434 | 18453 | 12,308 |
https://mathoverflow.net/questions/18221 | 1 | Let S = {1, .. ,n}.
Let H = (S, E) be the m-uniform hypergraph with r edges.
Let F(H, k) = #{B | |B| = k, $\exists R \in E, B \cap R = \emptyset$ } -
a number of k-subsets, that doesn't intersect with edges of H.
I am interested in following two problems:
>
> Let r, m, k be fixed parameters.
>
>
> 1. For which hypergraph H, F(H, k) be minimal?
> 2. For which H, F(H, k) be maximal?
>
>
>
I have a conjecture for the first problem, but can't prove it.
Introduce a total order $\preceq$ on m-element subsets of S: $a \preceq b$ iff $\exists s$, such that $s \in a, s \notin b$ and $\forall i < s, i \in a \cap b \mbox{ or } i \notin a \cup b$.
Example: 3-subsets from 5 element set in ascending order according to $\preceq$.
11100
11010
11001
10110
10101
10011
01110
01101
01011
01111
>
> **Conjecture** For all k minimal value of F(H, k) achieved on hypergraph, whose edges are r first sets from ${S \choose m}$ according to $\preceq$ order.
>
>
>
| https://mathoverflow.net/users/4641 | Some extremal problem for uniform hypergraph with fixed number of edges. | The answer to your first question should follow from a more or less straightforward application of Kruskal-Katona.
Namely, if you let $H^c=(S,E')$ be the family of complements, i.e. $X \in E'$ iff $S \setminus X \in E$, then $F(H,k)$ is simply the number of $k$-element subsets contained in edges of $H^c$, which is simply the $l$-th shadow of $H^c$, where $l=(n-m) - k$. You should get that the minimal value of $F(H,k)$ corresponds to the initial segment under the colex ordering for $H^c$; taking the complements of that family should resolve your conjecture.
| 2 | https://mathoverflow.net/users/2739 | 18455 | 12,309 |
https://mathoverflow.net/questions/18447 | 5 | I have two "vague questions" which are the following: if you have two $n$-dimensional $\ell$-adic Galois representations of a number field $K$ (with the standard ramification conditions) that have the same char. polynomial in a set of primes of density $1$, then they are isomorphic. The same is true if I replace Galois representations by automorphic representations and I ask the local representations to be the same in a set of density $1$. So the question is: "what is the best bound for such result"? (assume $n=2$ and $K= \mathbb Q$ if you want).
So for example, is it true that given $\epsilon >0$ there exist two representations that are not isomorphic but whose Frobenius/local components are the same in a set of density $1-\epsilon$? (at least in the case $n=1$?)
The second somehow related question (but has nothing to do, just in spirit) is if I have a restricted tensor product of local automorphic representations (say $K=\mathbb Q$, $n=2$) then there is no reason for it to be modular, but is there any result saying "there exists $\delta <1$ such that if you change the local component appropiately in a set of primes of density $\delta$ then it is an automorphic form"?
Of course, if one starts with a Galois representation and construct a restricted tensor product via the local langlands correspondence, one would like to say that such representation $\Pi$ is modular (under the appropiate assumptions on the Galois representation), so my question can be viewed as "how far (in terms of density) are we from proving modularity in general"
| https://mathoverflow.net/users/4685 | Density results for equality of Galois/automorphic representations | Firstly, at the beginning of the question you are missing irreducibility/cuspidality assumptions. If $\rho\_1$ and $\rho\_2$ are $\ell$-adic Galois reps with the same char poly in a set of primes of density 1, then you can only deduce their semisimplifications are isomorphic. A counterexample to your statement would be given by $\rho\_1=1+\omega$ ($\omega$ the cyclotomic character) and $\rho\_2$ some non-split extension coming from Kummer theory (taking $\ell$-poower roots of some prime number $p$ for example). Of course you knew that already.
On the automorphic side you make "the same slip", well, a related slip. If $\chi$ and $\psi$ are two Grossencharacters and their ratio is the norm character at some place $v$ (or possibly even infinitely many, or even all $v$), then the induction of $\chi+\psi$ from the Borel to $GL\_2(\mathbf{A})$ is reducible, and any Jordan-Hoelder factor (which by definition means a tensor product of J-H factors of the local inductions) ramified at all but finitely many places is an automorphic representation. So now you can build two non-cuspidal automorphic representations which are isomorphic at all but one place (and such that the local components don't even have the same dimension at the bad place) easily. Of course for cuspidal representations you have strong multiplicity 1 theorems.
Passing remark: it is a source of some confusion to me as to why these errors are "similar" but on the other hand they don't "biject". The problem is that on the Kummer theory side, if you allow ramification at two primes, you get a whole projective line of non-isomorphic Galois representations (all with the same semisimplificiation). On the automorphic side the amount of control you have is much more combinatorial (you can change a finite set of places from trivial to Steinberg and that's it). So $\pi$ and $\rho$s don't match up (so Toby Gee and I only conjecture that given any (EDIT: algebraic) $\pi$ one expects a semi-simple $\rho$, and nothing more, and for $GL\_n$ this observation is no doubt much older).
OK so after these pedantic remarks, that you no doubt knew anyway, but could have avoided if you'd put "irreducible" and "cuspidal" in the appropriate places, we move on to the far more interesting question of whether one can do better than multiplicity 1. And your hunch is correct. First you should do the following exercise on the Galois side: if $\rho\_1$ and $\rho\_2$ are two irreducible 2-dimensional representations of a finite group $G$, and their traces agree on a set $S$ in $G$ of density greater than 7/8, then $\rho\_1$ and $\rho\_2$ are isomorphic. Big hint: orthogonality relations. Next you should convince yourself that 7/8 is optimal in this result Big hint: D\_8 x D\_8. (EDIT: slightly simpler is D\_8 x C\_2---here D\_8 has 8 elements). So now for silly Artin representation reasons you can't expect to beat 7/8 (like you can't expect to beat 1/2 in the GL\_1 case as in Jared's comment).
So now the great news is that Dinakar Ramakrishnan proved an analogue on the automorphic side! At least in some cases. See the appendix to "$l$-adic representations associated to modular forms over imaginary quadratic fields. II." by Richard Taylor (Inventiones 116).
As for your second question though, it's completely hopeless. Even for $GL\_1$ your hope is (provably) way out, so for $GL\_2$ one can perhaps construct "Eisenstein" counterexamples (induced from non-automorphic characters of $GL\_1$). The first objection is that $\pi\_p$ can be ramified for infinitely many $p$, making you dead in the water. But even if $\pi\_p$ is unramified for all $p$ you've still got no chance. Let me stick to $GL\_1/\mathbf{Q}$. The point is that every Grossencharacter for $GL\_1/\mathbf{Q}$ is the product of a finite order character $\chi$ (a Dirichlet character) and $||.||^s$, so you can recursively construct characters of $\mathbf{Q}\_p^\times$ each of which is totally at odds with everything that came before. For example lets send $Frob\_2$ to 1. Now let's write down all the solutions to $2^s.\zeta=1$, with $\zeta$ a root of unity and $s$ a complex number. There are only countably many values of $s$ in this list. So choose $s$ not in the list and send $Frob\_3$ to $3^s$. Now knock off countably many more $s$ and send $Frob\_5$ to $5^s$ for $s$ in neither list. We are making a collection of representations here that are pairwise completely incompatible! This is only one of the many obstructions. For example, for cuspidal automorphic representations there are Weil bound obstructions (Ramanujan conjecture) and arithmeticity obstructions in the holomorphic case---all deep theorems or conjectures about automorphic forms in some cases that can easily be violated if we're allowed to build $\pi$ locally. These arguments trivially show that the set of automorphic $\pi$s have density zero (and "a very small zero" at that) amongst all the $\pi$s. In fact here's a much cleaner objection for $GL\_2$: if you stick to cuspidal automorphic representations with a fixed central character then there are only countably many! But you have uncountably many choices at each local place! So it's completely hopeless I think.
| 9 | https://mathoverflow.net/users/1384 | 18469 | 12,319 |
https://mathoverflow.net/questions/18475 | 5 | The Rado graph contains every finite graph as an induced subgraph. It surely contains *some* finite graphs infinitely often as an induced subgraph, e.g. $K\_2$. Does it contain *all* finite graphs infinitely often as an induced subgraph? Or can an example of a graph be given that is *not* contained infinitely often?
| https://mathoverflow.net/users/2672 | Rado graph containing infinitely many isomorphic subgraphs | It must contain every finite subgraph infinitely often
as an induced subgraph. For a finite graph $G$ and the positive integer
$n$ consider the graph $H$ consisting of $n$ vertex-disjoint copies of $G$.
As $H$ is an induced subgraph of Rado then there are $n$ vertex-disjoint
induced subgraphs of Rado isomorphic to $G$.
According to Wikipedia, Rado also has every countable graph
as an induced subgraph (I wasn't aware of this until now). Then
the above argument will work for countable graphs too.
| 11 | https://mathoverflow.net/users/4213 | 18476 | 12,322 |
https://mathoverflow.net/questions/18368 | 11 | Background
----------
Suppose we have a finite-dimensional Hilbert space $H = \mathbb{C}^s$ (for a natural number s) and we construct the symmetric (or *bosonic*) Fock space built from it: $$F(H):= \mathbb{C} \oplus H \oplus S(H \otimes H) \oplus S(H \otimes H \otimes H) \oplus \ldots$$
where S is the symmetrising operator.
Vectors in F are sequences of vectors $\psi = (\psi\_0, \psi\_1,\psi\_2,\ldots)$ such that
$\psi\_0 \in \mathbb{C}$, $\psi\_1 \in H$, $\psi\_2 \in S(H \otimes H)$ etc such that
$\sum\_{n=0}^\infty ||\psi\_n||\_n^2 < \infty$ where || ||n denotes the appropriate norm.
For any vector f $\in$ H we can define a pair of unbounded densely defined operators $a^\dagger(f)$ and $a(f)$ acting on F. These are called the "creation and annihilation operators". They are mutually adjoint and satisfy a commutation relation of the form: $$a(f) a^\dagger(g) - a^\dagger(g) a(f) = \langle f, g\rangle $$
where $\langle f, g\rangle $ is the inner-product of f, g $\in$ H.
The best reference for all this is M. Reed, B. Simon, "Methods of Mathematical Physics, Vol 2", section X.7 p207-212. This is partially available on Google books here: <http://books.google.co.uk/books?id=Kz7s7bgVe8gC&lpg=PA141&dq=reed%20and%20simon%20x.7&client=firefox-a&pg=PA210#v=onepage&q=&f=false>
The sum $\phi(f) = a(f) + a^\dagger(f)$ is self-adjoint (more properly the closure of their sum is self-adjoint) and is called the Segal quantisation of f (up to a factor of $\sqrt{2}$).
>
> Since $\phi(f)$ is self-adjoint we can apply the spectrum theorem to it. The question is, what is its spectral decomposition? Or more loosely, what are its eigenvalues and eigenvectors? or what can we tell from about its spectral decomposition?
>
>
>
| https://mathoverflow.net/users/4673 | Spectral theory for self-adjoint field operators on a symmetric Fock space | One convenient way to do analysis on the symmetric Fock space is to use its isomorphism to
the Bargmann (reproducuing Kernel Hilbert) space (sometimes called the Bargmann-Fock pace)
of analytic functions on $\mathbb C^s$ (with respect to the Gaussian measure) defined in the classical paper:
*Bargmann, V.*, [**On a Hilbert space of analytic functions and an associated integral transform**](http://dx.doi.org/10.1002/cpa.3160140303), Commun. Pure Appl. Math. 14, 187-214 (1961). [ZBL0107.09102](https://zbmath.org/?q=an:0107.09102).
An introduction to the Bargmann space may be found in chapter 4 of the
[book](https://www.mat.univie.ac.at/%7Eneretin/lectures/lectures.htm) by Uri Neretin
*Neretin, Yurii A.*, [**Lectures on Gaussian integral operators and classical groups**](http://dx.doi.org/10.4171/080), EMS Series of Lectures in Mathematics. Zürich: European Mathematical Society (EMS) (ISBN 978-3-03719-080-7/pbk). xii, 559 p. (2011). [ZBL1211.22001](https://zbmath.org/?q=an:1211.22001).
On the Bargmann space the creation and anihilation operators
are just the multiplication $a\_j = z\_j$ and the derivation $a^\*\_j = d/dZ\_j$
and consequently, the theory of several complex variables can be used for the analysis on this space,
for example the trace of (a trace class) operator can be represented as an integral on its symbol.
Remark: The isomorphism between the symmetric Fock and Bargmann spaces is not proved in the Book. It can be found for example in the references of the following:
*Stochel, Jerzy B.*, Representation of generalised creation and annihilation operators in Fock space, Zesz. Nauk. Uniw. Jagiell. 1208, Univ. Iagell. Acta Math. 34, 135-148 (1997). [ZBL0949.47027](https://zbmath.org/?q=an:0949.47027).
Regarding the question about $a(f)+a^\*(f)$, it is proportional to the position operator of quantum mechanics.
This is an unbounded operator, its spectrum is the whole real line, but it does not have
eigenfunvectors within the Fock space (Loosly speaking, they are Dirac delta functions), however one can find a series of vectors which approximate arbitrarily closely its eigenvectors. Using the corresponding projectors, one can approximate the spectral decomposition of this operator.
The case of the momentum operator $i(a(f)-a^\*(f))$ is used more frequently, a possible choice of the approximate eigenvectors is by means of wave packets.
| 4 | https://mathoverflow.net/users/1059 | 18478 | 12,323 |
https://mathoverflow.net/questions/18482 | 1 | I have some code where the "hot part" relies on an inefficient solution to this problem.
Problem: I have 3 inputs:
a. A collection of N points on the surface of a sphere.
b. A line segment on the sphere.
c. A distance X (distance can be on the surface or in 3D as it's trivial to map between them)
Output:
Find the subset of the line segment which is more than distance X from the collection of points.
(My problem actually involves a curve on the sphere, but I can reduce it to a line segment by chopping it up into smaller pieces.)
At present, I parametrize the line and created a distance function, subtract X then slam it into a method that finds the times when a function is positive. *VERY* slow.
Also, precomputations count in this algorithm. That is, the set does change over time, but it changes less frequently than I need this result for different line segments. Maybe 5 queries to every set change? That's worst case.
X is fixed over time.
| https://mathoverflow.net/users/4694 | Find the subset of a line on a sphere "far" from a set of points on the sphere. | By a line on the sphere, I assume you mean a part of a great circle spanning less than 180°.
For each of the *N* points, find what part of the line is closer than *X* to that point. This is at most a single segment. Find their union *C* as a disjoint union of segments. This is probably easier if you sort them by their starting point, and the obvious algorithm gives you the union as a sorted list as well – which makes the last part trivial: Find the complement of *C*.
Oh, and the first part can be done with some simple linear algebra.
| 5 | https://mathoverflow.net/users/802 | 18490 | 12,328 |
https://mathoverflow.net/questions/17629 | 2 | Given a Riemann surface $S$, e.g. $\mathbb{P}^1(\mathbb{C})$, with complex conjugation on the coordinates and a holomorphic vector bundle $E$ over $S$.
The complex conjugation $f$ is not holomorphic, but $C^{\infty}$. So we can look at $f^\*E$ as a smooth real vector bundle. Since $f$ is an involution $E$ and $f^\*E$ are isomorphic as smooth real bundles (this is most likely wrong). But somehow there should be more structure on $f^\*E$. I mean $f$ is antiholomorphic, that should be useful somehow.
Is there a way to define a (canonical?) complex structure on $f^{\*}E$, such that for any point $p$ we have an antilinear isomorphism $E\_{p} \rightarrow f^{\*}E\_{p}=E\_{\overline{p}}$?
I think we don't get a linear iso, since the complex conjugation itself is only antilinear.
| https://mathoverflow.net/users/3233 | Definition of a complex structure on a vector bundle | For any antiholomorphic Diffeomorphism $f\colon S\to S$ we get a canonical identification $f^\star\bar K=K,$ $ K $ and $\bar K$ being the canonical and anticanonical bundle of the Riemann surface. A holomorphic structure on a complex vectorbundle $E$ is the same as an complex operator
$D\colon\Gamma(E)\to\Gamma(\bar KE)$ satisfying the (Cauchy Riemann) Leibnitz rule. (holomorphic sections are exactly the one in the kernel of D). (For higher dimensions this is not true.)
Now, $f^\* E$ has a natural complex structure (it's just i).
Therefore one gets an anti-holomorphic structure $\bar D\colon\Gamma(f^\star E)\to\Gamma(Kf^\* E)$ satisfying the antiholomorphic Cauchy Riemann equation.
But the complex conjugate bundle $\bar E$ also has a anti-holomorphic structure, since $\overline{\bar K E}=K\bar E.$ Therefore, $f^\* \bar E$ has a natural holomorphic structure.
These two holomorphic structures are not isomorphic in general:
In the case of a line bundle $L=E$ of degree $0$ one might see this as follows: every holomorphic structure $D$ gives rise to an unique unitary flat connection $\nabla$ such that
$D=1/2(\nabla+i\*\nabla).$ Then the anti-holomorphic structure on $\bar L$ is given by $1/2(\nabla-i\*\nabla)$ and, the unitary flat connection corresponding to the holomorphic structure on $f^\* L$ is the connection $f^\* \nabla.$ But this connection is not gauge equivalent to $\nabla$ in general: For example, on a square torus with f given by $z\mapsto \bar z$ the connection $d+c idx$ is not gauge equivalent to $d+ci dy$ for $c\in R\setminus 2\pi Z.$
| 4 | https://mathoverflow.net/users/4572 | 18495 | 12,330 |
https://mathoverflow.net/questions/18496 | 16 | Let $R$ be a Noetherian domain, and let $\mathfrak{p}$ be a prime ideal; consider the completion $\hat R\_{\mathfrak{p}}$ of $R$ at $\mathfrak{p}$ (the inverse limit of the system of quotients $R/\mathfrak{p}^n$). If $R$ is a PID, it is easy to see that $\hat R\_{\mathfrak{p}}$ is a domain.
Someone asked in sci.math if $\hat R\_{\mathfrak{p}}$ would always be a domain. I thought it would, but looking at Eisenbud's "Commutative Algebra", I found a reference to a theorem of Larfeldt and Lech that says that if $A$ is any finite-dimensional algebra over a field $k$, then there is a Noetherian local integral domain $R$ with maximal ideal $\mathfrak{m}$ such that $\hat{R\_{\mathfrak{M}}}\cong A[[x\_1,\ldots,x\_n]]$ for some $n$; and so this completion will not be a domain if $A$ is not a domain. I would like to know an example directly, if possible.
Does someone know an easy example of a noetherian domain $R$ and a prime ideal $\mathfrak{p}$ such that $\hat R\_{\mathfrak{p}}$ is not a domain? Thanks in advance.
| https://mathoverflow.net/users/3959 | Example of the completion of a noetherian domain at a prime that is not a domain | Let $R=\mathbb{C}[x,y]/(y^2-x^2(x-1))$. This is the nodal cubic in the plane. Look at the prime $\mathfrak{p}=(x,y)$, corresponding to the nodal point. The completion here is isomorphic to $\mathbb{C}[[x,y]]/(xy)$.
| 35 | https://mathoverflow.net/users/622 | 18498 | 12,332 |
https://mathoverflow.net/questions/18465 | 3 | Suppose that $K$ is an $\textit{alternating}$ knot in $S^3$, and let $R\_0$ be the space of homomorphisms from $\pi\_1(S^3 - K)\to SU(2)$ which send meridians to trace free matrices. Denote the subset of $R\_0$ consisting of metabelian representations as $R\_m \subset R\_0$.
Question: when $K$ is prime, is there any reason to think that $R\_0$ retracts to $R\_m$? Could at least $H\_\*(R\_0;\mathbb{Z})=H\_\*(R\_m;\mathbb{Z})$? Does anyone know of anything about such questions in the literature?
Might there be similar analyses for more general $SU(2)$ representations of knot groups?
| https://mathoverflow.net/users/492 | SU(2) representations of alternating knot groups | I doubt there is such a retraction. Such representations are representations of the $\pi$-orbifold, obtained by killing the square of the meridian (at least if one quotients by $\pm Id$). If one takes a Montesinos knot, these orbifolds are Seifert fibered, and such representations should factor through the quotient orbifold. Such oribfolds can have several non-metabelian isolated representations into $SO(3)$, so I expect that the answer is no even on $H\_0$.
| 4 | https://mathoverflow.net/users/1345 | 18503 | 12,336 |
https://mathoverflow.net/questions/18491 | 3 | I've come across this infinite series: $\sum\_{n=0}^\infty x^{n^\alpha}$, with $0<x<1$ and $\alpha > 0$.
Does this series have a name and/or is there a method for computing it (besides brute force, obviously)? Thanks!
| https://mathoverflow.net/users/4507 | Is there any numerical technique to sum x^(n^alpha), n=0,1,...? | For $\alpha > 1$, this sum will converge extremely rapidly, even if $x$ is fairly close to $1$ (you can't express in floating point numbers those numbers close enough to $1$ where this would converge slowly). The only case that is difficult, convergence wise, is when $\alpha$ is very close to 0.
In that case, your best bet are Levin's U-transform and related algorithms. Even better, just link in to [GSL](http://www.gnu.org/software/gsl/), which [already implements that](http://www.gnu.org/software/gsl/manual/html_node/Series-Acceleration-References.html).
| 4 | https://mathoverflow.net/users/3993 | 18506 | 12,338 |
https://mathoverflow.net/questions/18505 | 9 | This question was something I considered when looking into CW-structures on Grassmannians, but I found no general treatment of this in the literature:
Question: Assume that $X$ is an $n-1$ dimensional finite CW complex. and assume that $X'$ is given as a a set by the disjoint union of $X$ and a single open cell $e$ of dimension $n$. I.e. $e$ is an open subspace of $X'$ homeomorphic to the open $n$ disc (and of course $X$ is homeomorphic to the complement). Assume also that $X'$ is compact Hausdorff. Is $X'$ homeomorphic to a CW complex given by attaching a single $n$ cell to $X$?
Remark: Maybe I am missing some obvious counter example!
| https://mathoverflow.net/users/4500 | What is enough to conclude that something is a CW complex? | Let $X$ be the closed subspace of $R^2$ which is the union of $0\times [-1,1]$ and the point $(1, \sin(1))$. Let $e$ be the graph in the plane of $f(x)=\sin(1/x)$ for $x\in (0,1)$ (the "topologists sine curve"), and let $X'=X\cup e$, viewed as a subspace of $R^2$.
I believe that $X'$ is closed, and so is compact Hausdorff, $e$ is an open subset of $X'$ homeomorphic to the open interval, and $X$ is a CW-complex. But $X'$ is *not* obtained by attaching a $1$-cell to $X$: you can't produce a map $\Phi\colon[0,1]\to X'$ which restricts to a homeomorphism $(0,1)\to e$.
This is not quite a counterexample to your problem, because you wanted the new cell $e$ to have greater dimension than the cells of $X$. But it seems like you could use this kind of idea to make a counterexample.
**Added.** Let $S^2$ be the unit sphere in $R^3$, pick a point $p$ on $S^2$, and consider the function $f\colon (S^2-\{p\})\to R$ given by $f(x)=\sin(1/|x-p|)$. Let $X'$ be the closure of the graph of $f$ inside $S^2\times [-1,1]$; this should consist of $X=\{p\}\times [-1,1]$ (a CW complex) and $e=$ graph of $f$ (homeomorphic to an open $2$-disk). This would seem to be the counterexample you want.
| 6 | https://mathoverflow.net/users/437 | 18507 | 12,339 |
https://mathoverflow.net/questions/18508 | 15 | Given a group $G$ we may consider its group ring $\mathbb C[G]$ consisting of all finitely supported functions $f\colon G\to\mathbb C$ with pointwise addition and convolution. Take $f,g\in\mathbb C[G]$ such that $f\*g=1$. Does this imply that $g\*f=1$?
If $G$ is abelian, its group ring is commutative, so the assertion holds. In the non-abelian case we have $f\*g(x)=\sum\_y f(xy^{-1})g(y)$, while $g\*f(x)=\sum\_y f(y^{-1}x)g(y)$, and this doesn't seem very helpful.
If $G$ is finite, $\dim\_{\mathbb C} \mathbb C[G]= |G|<\infty$, and we may consider a linear operator $T\colon \mathbb C[G]\to\mathbb C[G]$ defined by $T(h) = f\*h$. It is obviously surjective, and hence also injective. Now, the assertion follows from $T(g\*f)=f=T(1)$.
What about infinite non-abelian groups? Is a general proof or a counterexample known?
| https://mathoverflow.net/users/4698 | Is a left invertible element of a group ring also right invertible? | A ring is called Dedekind-finite if that property holds. Semisimple rings are Dedekind finite, so this covers $\mathbb CG$ for a finite group $G$; this is easy to do by hand. It is a theorem of Kaplansky that this also holds $KG$ for arbitrary groups $G$ and arbitrary fields $K$ of characteristic zero. See [Kaplansky, Irving. Fields and rings. The University of Chicago Press, Chicago, Ill.-London 1969 ix+198 pp. [MR0269449](http://www.ams.org/mathscinet-getitem?mr=MR0269449)] It is open, I think, for general fields.
| 21 | https://mathoverflow.net/users/1409 | 18509 | 12,340 |
https://mathoverflow.net/questions/18513 | 11 | Which of the statements is wrong:
1. a generalized cohomology theory (on well behaved topological spaces) is determined by its values on a point
2. reduced complex $K$-theory $\tilde K$ and reduced real $K$-theory $\widetilde{KO}$ are generalized cohomology theories (on well behaved topological spaces)
3. $\tilde K(\*)= \widetilde{KO} (\*)=0$
But certainly $\tilde K\neq \widetilde{KO}$.
| https://mathoverflow.net/users/2625 | K-theory as a generalized cohomology theory | 1 is doubly wrong. First, you need to distinguished generalized cohomology theories and *reduced* generalized cohomology theories. If you want to work with the latter, you should replace "a point" in 1 by "$S^0$", and then the corrected version of 3 no longer holds. But even this new version 1' is false; a generalized cohomology theory is not determined by its coefficients, unless they are concentrated in a single degree (example: complex K-theory vs. integer cohomology made even periodic).
| 17 | https://mathoverflow.net/users/126667 | 18515 | 12,342 |
https://mathoverflow.net/questions/18512 | 2 | Hello everyone. I'm in a little trouble trying to find the proof of a theorem stated by W. T. Gowers. It is the Lemma 1.6 in his article 'An infinite Ramsey theorem and some Banach space dichotomies' (Annals of Mathematics, 156 (2002)). He refers Lindenstrauss but I couldn't find somehting clear and detailed about the proof either.
If anyone knows about another reference about this proof I would appreciate it.
Thanks.
| https://mathoverflow.net/users/2737 | A proof about an unconditional basis theorem | Dan, I doubt that you will find a proof of this in the literature. To get (ii) from (i), note that (i) gives you a block basic sequence $x\_1,....,x\_m$ in the block subspace $Y$ s.t. $\|\sum\_{i=1}^m x\_i\|= 1$ and for some choice $a\_i$ of signs,
$\|\sum\_{i=1}^m a\_i x\_i\| > C$. WLOG $a\_1=-1$. Now group together maximal blocks all of whose signs are
the same to rewrite
$\sum\_{i=1}^m a\_i x\_i$ as $\sum\_{j=1}^n (-1)^j y\_j$, where $y\_j$ is the sum of the $x\_i$'s in the $j$-th maximal block.
I assumed real coefficients in the above, but the complex case is only a bit more involved.
If this explanation isn't sufficient, I suggest that you study further the section in [LT] on bases. I addressed the only point that I thought might be troubling to someone who had studied [LT].
| 3 | https://mathoverflow.net/users/2554 | 18517 | 12,344 |
https://mathoverflow.net/questions/18483 | 13 | For a positive integer $k$, let $d(k)$ be the number of divisors of $k$. So $d(1)=1$, $d(p)=2$ if $p$ is a prime, $d(6)=4$, and $d(12)=6$.
What are the precise asymptotics of $\sum\_{k=1}^n 1/(k d(k))$?
Background:
-----------
1) This came up on the side in the [polymath5](http://gowers.wordpress.com/2010/03/13/edp12-representing-diagonal-maps/#comment-6697) project.
2) There, Tim Gowers wrote: If nobody knows the answer, maybe that’s one for MathOverflow, where I imagine a few minutes would be enough.
3) Asked: 14:17 Jerusalem time. (The first accurate answer: 17:44 Jerusalem time.)
4) Looking only at primes or only at integers with a typical number of divisors suggested a $\log\log n$ behavior, but looking at semiprimes indicates the sum is larger. I don't know how much larger.
5) I couldn't find an answer on the web. If there is an easy way searching for an answer that I missed this will be interesting too.
Follow up:
==========
Great answers! Thanks. What about the sum $\sum\_{k=1}^n 1/(kd^2(k))$ ?
| https://mathoverflow.net/users/1532 | An elementary number theoretic infinite series | The idea (from the Selberg-Delange) method to doing this problem is the following steps:
1) Let $F(s) = \sum\_{n\ge 1} \frac{1}{n^s d(n)} = \prod\_{p} \left(1 + \sum\_{k=1}^{\infty} \frac{1}{(k+1) p^{ks}} \right)$. The latter is by multiplicativity of $d(n)$.
2) If we look, instead at $G(s) = \prod\_p \left( 1 + \frac{1}{2 p^s} \right)$ we can see that $F(s)/G(s)$ has a non-zero limit as $s \rightarrow 1$ from above. $G(s)$ corresponds in our original sum to restricting $n$ to be square-free.
3) $G(s)^2$ almost looks like $\zeta(s)$. Show that $G(s)^2/\zeta(s)$ also has a non-zero limit at $s \rightarrow 1$.
4) You then use some Tauberian theorems to show that since $H\_n \sim \log n$ (which is the sum associated with $\zeta(s)$ then the corresponding sum for $G(s)$ (i.e. over the square-free $n$) is $\sim to \sqrt{\log n}$.
| 14 | https://mathoverflow.net/users/2784 | 18518 | 12,345 |
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