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https://mathoverflow.net/questions/14486 | 21 | Speculation and background
--------------------------
Let $\mathcal{C}:=\mathrm{CRing}^\text{op}\_\text{Zariski}$, the affine Zariski site. Consider the category of sheaves, $\operatorname{Sh}(\mathcal{C})$.
According to [nLab](https://ncatlab.org/nlab/show/scheme#translation_between_the_two_approaches), schemes are those sheaves that "have a cover by Zariski-open immersions of affine schemes in the category of presheaves over Aff."
In SGA 4.1.ii.5 Grothendieck defines a further topology on $\operatorname{Sh}(\mathcal{C})$ using a "familles couvrantes", which are families of morphisms $\{U\_i \to X\}$ such that the induced map $\coprod U\_i \to X$ is an epimorphism. Further, he gives another definition. A family of morphisms $\{U\_i \to X\}$ is called "bicouvrante" if it is a "famille couvrante" and the map $\coprod U\_i \to \coprod U\_i \times\_X \coprod U\_i$ is an epimorphism. [Note: This is given for a *general* category of sheaves on a site, not sheaves on our affine Zariski site.]
Speculation: I assume that the nLab definition means that we have a (bi)covering family of open immersions of representables, but as it stands, we do not have a sufficiently good definition of an open immersion, or equivalently, open subfunctor.
It seems like the notion of a bicovering family is very important, because this is precisely the condition we require on algebraic spaces (if we replace our covering morphisms with etale surjective morphisms in a smart way and require that our cover be comprised of representables).
Questions
---------
What does "open immersion" mean precisely in categorical langauge? How do we define a scheme *precisely* in our language of sheaves and grothendieck topologies? Preferably, this answer should not depend on our base site. The notion of an open immersion should be a notion that we have in any category of sheaves on any site.
Eisenbud and Harris fail to answer this question for the following reason: they rely on classical scheme theory for their definition of an open subfunctor (same thing as an open immersion). If we wish to construct our theory of schemes with no logical prerequisites, this is circular.
Once we have this definition, do we require our covering family of open immersions to be a "covering family" or a "bicovering family"?
Further, how can we exhibit, in precise functor of points language, the definition of an algebraic space?
This last question should be a natural consequence of the previous questions provided they are answered in sufficient generality.
| https://mathoverflow.net/users/1353 | Precise definition of a scheme (Key question: How to define an open subfunctor without resorting to classical scheme theory) | Check out the paper of Kontsevich-Rosenberg [noncommutative space.](http://www.mpim-bonn.mpg.de/preprints/send?bid=2331), they defined formally open immersion and open immersion completely functorial way. This definition is nothing to do with "noncommutative"
Definition:
Formally open immersion is formally smooth monomorphism.
But one thing need to point out, they are working on Q-category which is a generalization of Grothendieck topologies(destination is dealing with topology without base change property).But just disregard this notion and work in usual grothendieck topology as you want
| 7 | https://mathoverflow.net/users/1851 | 14566 | 9,778 |
https://mathoverflow.net/questions/14499 | 11 | Forgive the elementary nature of the question. I understand that the second order Peano Axioms are categorical in the sense that all their models are isomorphic. This equivalence class of models is taken to be the definition of natural numbers.
My question is that: Is the theory defined by those second order axioms complete? i.e. is every second-order statement in this theory provable or disprovable from the axioms?
For the first order Peano Arithmetic axioms any statement that is true in all models is provable or disprovable by the completeness of first-order logic, but of course this is in applicable to the second-order axioms.
| https://mathoverflow.net/users/3873 | Is any true sentence in the second-order Peano Axioms provable | Note that there are two different types of models of second-order logic: standard models, where second-order quantified variables range over all subsets of the domain; and Henkin models, where second-order quantified variables are allowed to range over a proper subset of the full power-set.
Henkin proved the completeness theorem for second-order logic for Henkin models, so if a sentence is true in all Henkin models, it is derivable. For completeness to hold, it is not enough to consider only standard models, so even though second-order PA is categorical for standard models, it is not complete (as we would expect, this being a corollary Gödel's theorem).
| 8 | https://mathoverflow.net/users/2004 | 14570 | 9,781 |
https://mathoverflow.net/questions/14568 | 42 | The Poisson summation says, roughly, that summing a smooth $L^1$-function of a real variable at integral points is the same as summing its Fourier transform at integral points(after suitable normalization). [Here](http://en.wikipedia.org/wiki/Poisson_summation_formula) is the wikipedia link.
For many years I have wondered why this formula is true. I have seen more than one proof, I saw the overall outline, and I am sure I could understand each step if I go through them carefully. But still it wouldn't tell me anything about why on earth such a thing should be true.
But this formula is exceedingly important in analytic number theory. For instance, in the book of Iwaniec and Kowalski, it is praised to high heavens. So I wonder what is the rationale of why such a result should be true.
| https://mathoverflow.net/users/3885 | Truth of the Poisson summation formula | It is a special case of the trace formula. Both sides are the trace of the same operator.
| 25 | https://mathoverflow.net/users/927 | 14571 | 9,782 |
https://mathoverflow.net/questions/14586 | 4 | What properties does a distribution (in the generalized function sense) has to have in order to be a function. That is, when is $T(\varphi) = \int f \varphi$ for some $f$?
| https://mathoverflow.net/users/3887 | When can a function be recovered from a distribution? | First of all, $T$ must have order zero, i.e., $|T(\varphi)|\le C(K)\sup|\varphi|$ for any test function $\varphi$ supported on a compact set $K$. By Riesz representation theorem, $T$ is a measure. To be a locally integrable function, it must be absolutely continuous with respect to the Lebesgue measure. One way to express this condition: $C(K)\to 0$ as the Lebesgue measure of $K$ tends to zero, which $K$ staying within a fixed compact set.
| 12 | https://mathoverflow.net/users/2912 | 14591 | 9,797 |
https://mathoverflow.net/questions/14587 | 40 | For all those who are unlikely to have answers to my questions, I provide some
Background:
===========
In some sense, pure motives are generalisations of smooth projective varieties. Every [Weil cohomology](http://en.wikipedia.org/wiki/Weil_cohomology_theory) theory factors through the embedding of smooth projective varieties into the category of [pure Chow motives](http://en.wikipedia.org/wiki/Motive_%28algebraic_geometry%29).
Pure effective motives
----------------------
In the definition of pure motives (say over a field k), the last step is to take the category of pure effective motives and formally invert the Lefschetz motive L.
The category of pure effective motives is the [pseudo-abelian envelope](http://ncatlab.org/nlab/show/Karoubian+category) of a category of correspondence classes, which has as objects smooth projective varieties over k and as morphisms X → Y [cycle](http://en.wikipedia.org/wiki/Algebraic_cycle) classes in X×Y of dimension dim X (think of it as a generalisation of morphisms, where morphisms are included as their graphs), where an [adequate equivalence relation](http://ncatlab.org/nlab/show/adequate+equivalence+relation) is imposed, to have a well-defined composition (therefore the word "classes"). When the adequate relation is rational equivalence, the resulting category is called the category of pure effective Chow motives.
In each step of the construction, the monoidal structure of one step defines a monoidal structure on the next step.
For more background see Ilya's question about the [yoga of motives](https://mathoverflow.net/questions/2146/whats-the-yoga-of-motives).
Definition of the Lefschetz motive
----------------------------------
The Lefschetz motive L is defined as follows:
For each point p in P¹ (1-dimensional projective space over k), there is the embedding morphism Spec k → P¹, which can be composed with the structural morphism P¹ → Spec k to yield an endomorphism of P¹. This is an idempotent, since the other composition Spec k → Spec k is the identity.
The category of effective pure motives is pseudo-abelian, so every idempotent has a kernel and thus [P¹] = [Spec k] + [something] =: 1+L, where the summand [something] is now named Lefschetz motive L.
Properties
----------
The definition of L doesn't depend on the choice of the point p.
[From nLab](http://ncatlab.org/nlab/show/pure+motive) and [Kahn's leçons](http://people.math.jussieu.fr/~kahn/preprints/lecons.pdf) I learned that the inversion of the Lefschetz motive is what makes the resulting monoidal category a [rigid monoidal category](http://ncatlab.org/nlab/show/rigid+monoidal+category) - while the category of pure effective motives is not necessarily rigid.
In the category of pure motives, the inverse $L^{-1}$ is called T, the Tate motive.
Questions:
==========
These questions are somehow related to each other:
>
> 1. Why is this motive L called "Lefschetz"?
> 2. Why is its inverse $L^{-1}$ called "Tate"?
> 3. Why is it exactly this construction that "rigidifies" the category?
> 4. Would another construction work, too - or is this somehow universal?
> 5. How should I think of L geometrically?
>
>
>
I have almost no background in number theory, so even if you have good answers, it may remain totally unclear to me, why the name "Tate" intervenes. I expect however, that the name "Lefschetz" has something to do with the Lefschetz trace formula. I guess that the procedure of inverting L is the only one which makes the category rigid, in some universal way, but I have no idea, why. In addition, I guess there is no "geometric" picture of L.
If I made any mistakes in the background section, feel free to edit. As I'm currently taking a first course on motives, I may now have asked something completely stupid. If this is the case, please point me politely to some document which will then enlighten me or, at least, let me ascend to a higher level of confusion.
**UPDATE:** Thanks so far for the answers, questions 1-4 are now clear to me. It remains, if the "rigidification" could be accomplished by another construction - maybe some universal way to turn a monoidal category into a rigid one? Then one could later identify the Lefschetz motive as some kind of a generator of the kernel of the rigidification functor.
The geometric intuition, to think of L as a curve and of $L^{\otimes d}$ as a d-dimensional manifold, remains fuzzy, but I have the hope that this becomes clear when I have worked a little bit on the classical Lefschetz/Poincare theorems and the proof of Weil conjectures for Betti cohomology (is this hope justified?).
| https://mathoverflow.net/users/956 | Understanding the definition of the Lefschetz (pure effective) motive | The motive $L$ is called Lefschetz because it is the cycle class of the point in ${\mathbb P}^1$, and so underlies (in a certain sense) the Lefschetz theorems about the cohomology of
projective varieties. To understand this better, you may want to read about how the hard Lefschetz theorem for varieties over finite fields follows from the Riemann hypothesis, as well as a discussion of Grothendieck's standard conjectures and how they relate to the Weil
conjectures.
The motvie $L^{-1}$, when converted into an $\ell$-adic Galois representation, is precisely the $\ell$-adic Tate module of the roots of unity. Tensoring by this Galois representation is traditionally called Tate twisting, and so the motive underlying this Galois representation is called the Tate motive.
One needs to have $L^{-1}$ at hand in order for the category to admit duals.
If one were working with just usual singular cohomology, this wouldn't be necessary; Poincare duality would pair $H^i$ with $H^{\text{top}-i}$ into $H^{\text{top}}$, which would be
isomorphic with ${\mathbb Q}$ via the fundamental class.
But motivically, if $X$ (smooth, connected, projective) has dimension $d$, so that the top dimension is $2d$, then $H^{\text{top}} = L^{\otimes d}$, and so $H^i$ and $H^{2d-i}$ pair into
$L^{\otimes d}$. To get a pairing into $\mathbb Q$ (the trivial 1-dim'l motive) we need
to be able to tensor by powers of $L^{-1}$. Traditionally tensoring by the $n$th tensor power of $L^{-1}$ is denoted $(n)$; so we find e.g. that
$H^i$ pairs with $H^{2d -i}(d)$ into ${\mathbb Q}$, and we have our duality.
You can see from the fact that cup product pairs cohomology into powers of $L$ that inverting $L$ is precisely what is needed in order to obtain duals.
Finally, one should think of $L$ as the fundamental class of a curve,
think of $L^{\otimes d}$ as the fundamental class of a smooth projective $d$-dimensional
manifold, and also become comfortable with Poincare duality and the Lefschetz theorems;
these are the basic ideas which will help give solid geometric sense to motivic constructions.
| 39 | https://mathoverflow.net/users/2874 | 14592 | 9,798 |
https://mathoverflow.net/questions/14476 | 2 | This theorem is: let f:X--->Y be a proper morphism of noetherian schemes,and the induced morphism of sheaves f^#:O\_Y---->f\_\*O\_X is isomorphic.Then for any point y belongs to Y,f^-1(y) is nonempty and connected.
I have seen a proof by use of formal schemes.My question is：Is there any proof without this trick？
| https://mathoverflow.net/users/3866 | the proof of "theorem of connectedness" | A typical context in which one has the condition $f\_{\\*}\mathcal O\_X = \mathcal O\_Y$
in that in which $f$ is a birational morphism and $Y$ is normal. In this context, the proof
that the fibres are connected is due to Zariski (I believe that it's the original version
of his ``main theorem'') and certainly predates EGA methods. One can find the paper in his
collected works. (It's been a long time since I looked at it, but I would guess that his main
technical tool is valuation theory; I might check this when I get a chance.)
However, it is worth bearing in mind the evolution of Zariski's work on this kind of question:
his investiations of this sort of connectedness theorem culminated in his proof of his connectedness theorem, to the effect that a specialization of connected varieties is again
connected. His was the first purely algebraic proof of this result, I think. To give the proof, he invented his theory of formally holomorphic functions, which I believe was regarded
at the time as being the most difficult part of the algebraic theory of algebraic geometry
developed by Weil, Chevalley, and Zariski. This theory served as one of the inspirations for the theory of formal schemes, and is the precursor to Grothendieck's theorem on formal functions (the proof via formal schemes mentioned in the question).
If so great a geometer as Zariski was led to introduce these kinds of formal methods to study connectedness problems, it is probably reasonable to regard them as somewhat intrinsic to the problem.
| 4 | https://mathoverflow.net/users/2874 | 14600 | 9,806 |
https://mathoverflow.net/questions/14622 | 6 | My understanding (please correct me if I'm wrong) is that if you have some transitive set M which is an $\epsilon$-model of ZFC, and you take an ultrapower of it using an approprate ultrafilter, you wind up with a new model whose membership relation is not the $\epsilon$ relation of the ambient set theory, but still satisfies ZFC. Furthermore, if the membership relation of the ultrapower is well-founded, one can always use the Mostowski collapse theorem to produce an isomorphic $\epsilon$-model.
My question is this: how could one possibly end up with a model of ZFC which satisfies the axiom of regularity ("every set is disjoint from one of its members"), yet whose membership relation *isn't* well-founded?
I'm struggling to imagine this; the most I can come up with is that for no set in the infinite chain is its transitive closure also a set (of the model). But I'm skeptical about whether or not that can be the case, because it seems like you ought to be able to construct the transitive closure using definition by transfinite recursion, letting $f(0)$ be any set in the chain and $f(n+1)=\bigcup f(n)$ (axiom of union). Then $f(\omega)$ (axiom of infinity) ought to contain all the sets needed to build a contradiction to regularity.
Sorry if this question sounds like I'm arguing with myself. This has been bothering me for a few days now.
| https://mathoverflow.net/users/2361 | How can an ultrapower of a model of ZFC be "ill-founded" yet still satisfy ZFC? | For an ultrapower of $V$ by an ultrafilter $\mathcal{U}$ there is an exact characterization of when the ultrapower will be well-founded: precisely when $\mathcal{U}$ is closed under countable intersections.
As for how a model $M$ could possibly satisfy regularity but not be wellfounded, the problem is that there may be infinite descending chains that $M$ cannot 'see': each individual object may belong to $M$ but the chain itself may not. I can be a little more precise. Let's say $R$ is what $M$ understands to be the $\in$-relation. There may well be $x\_0,\ldots x\_n,\ldots $ belonging to $M$ so that $x\_{n+1}Rx\_n$ for all $n\in\omega$; as long as the sequence $\langle x\_n:n\in\omega\rangle$ does not belong to $M$ the axiom of regularity from $M$'s point of view need not be violated.
You don't need ultrapowers to construct such models. Assuming CON(ZFC) you can build one using the compactness theorem for first order logic.
| 18 | https://mathoverflow.net/users/2436 | 14626 | 9,823 |
https://mathoverflow.net/questions/14613 | 31 | Let's say I start with the polynomial ring in $n$ variables $R = \mathbb{Z}[x\_1,...,x\_n]$ (in the case at hand I had $\mathbb{C}$ in place of $\mathbb{Z}$).
Now the symmetric group $\mathfrak{S}\_n$ acts by permutation on the indeterminates.
The subring of invariant polynomials $R^{\mathfrak{S}\_n}$ has a nice description (by generators and relations) in terms of symmetric functions.
What if I only consider the action of the cyclic group $Z\_n$?
Does anyone know if the ring $R^{Z\_n}$ admits a nice presentation?
(in the case at hand I had $\mathbb{C}[x\_1,x\_2,x\_3]$ and the action of the cyclic group $Z\_3$. Maybe in this case we can use some formula (assuming there is one) for groups splitting as semidirect products?)
| https://mathoverflow.net/users/3701 | Invariant polynomials under a group action (hidden GIT) | The actions of $S\_n$ and $\mathbb Z\_n$ differ in the sense that in the first case the quotient is smooth (it is again $\mathbb C^n$) while in the second case it is singular. This is why in the fist case we have a nice presentation, but in the second not really. For example, the number of generators of the quotient can not be less than the dimension of Zariski tangent space to the singularity at zero of $\mathbb C^n/\mathbb Z\_n$.
Still in principle the presentation can be provided by toric geometry (<https://dacox.people.amherst.edu/toric.html>) because the quotient is the toric singularity. For example, in your case of $\mathbb C^3/\mathbb Z\_3$ let us change the coordinates so that $\mathbb Z\_3$ is acting as $w\_0\to w\_0$, $w\_1\to \mu w\_1$, $w\_2\to \mu^2 w\_2$ (here $\mu^3=1$). Then you can write the minimal set of four generators:
$w\_0, w\_1^3, w\_2^3, w\_1w\_2$, and one obvious relation $(w\_1^3w\_2^3)=(w\_1w\_2)^3$
The case $\mathbb C^n/\mathbb Z\_n$ for $n>3$ will be more involved, but the idea is the same roughly. First you chose the coordinates on $\mathbb C^n$ $w for which the action is diagonal. Then pick the minimal set of monomials (in these new coordinates) that are invariant under the action, and generate the whole set of invariant monomials (of positive degree).
Consider one more case $n=4$, and chose the coordinates $w\_i$, so that $Z\_4$ is acting as $w\_i\to \mu^iw\_i$, $\mu^4=1$. The number of generators is $7$ this time:
$w\_0, w\_1^4, w\_3^4, w\_2^2, w\_1w\_3, w\_1^2w\_2, w\_3^2w\_2$
| 27 | https://mathoverflow.net/users/943 | 14628 | 9,824 |
https://mathoverflow.net/questions/14629 | 14 | I am away from Torsion theory in abelian category for some while. So it might be a stupid question.
The definition of a torsion pair in the category of modules is as follows:
Definition:
A pair $(\mathcal T,\mathcal F)$ of full subcategories of $A-\mathrm{mod}$ is called a torsion pair if following conditions hold:
1. $\mathrm{Hom}\_{A}(M,N)=0$ for all $M \in \mathcal T,
N \in \mathcal F$.
2. $\mathrm{Hom}\_{A}(M,-)|\_{\mathcal F}=0 \Rightarrow M \in \mathcal T$.
3. $\mathrm{Hom}\_{A}(-,N)|\_{\mathcal T}=0 \Rightarrow N \in \mathcal F$.
Condition 2) and 3) means that the pair $(\mathcal T,\mathcal F)$ is maximal for $\mathrm{Hom}\_{A}(M,N)=0$.
This definition is from the book [elements of representation theory of associative algebras](http://www.cambridge.org/us/academic/subjects/mathematics/algebra/elements-representation-theory-associative-algebras-techniques-representation-theory-volume-1)
I found this definition is similar to the definition of t-structures in derived category. I just quote the definition from Dimca, *Sheaves in topology* as follows:
A t-structure on a triangulated category $\mathcal D$ consists in two strictly full subcategories: $\mathcal D^{\leq 0}, \mathcal D^{\geq 0}$ such that the following conditions hold:
1. $\mathrm{Hom}(X,Y)=0$ if $X \in \mathcal D^{\leq 0}, Y \in \mathcal D^{\geq 1}$.
2. $\mathcal D^{\leq 0} \subseteq \mathcal D^{\leq 1}, \mathcal D^{\geq 1} \subseteq \mathcal D^{\geq 0}$.
3. For any $X \in \mathcal D$, there is a distinguished triangle
$$A\rightarrow X\rightarrow B\rightarrow A[+1], \qquad A \in \mathcal D^{\leq 0}, B \in \mathcal D^{\geq 1}.$$
Although the axiom 3) for t-structures looks different to the axioms of torsion pairs. However, there is a proposition of torsion pairs establishing the similar formula:
Let a pair $(\mathcal T,\mathcal F)$ be a torsion pair in $A-\mathrm{mod}$, and let $M$ be an $A$-module. Then there exists a short exact sequence
$$0 \rightarrow tM \rightarrow M \rightarrow M/tM \rightarrow 0, \qquad tM \in \mathcal T, M/tM \in \mathcal F,$$
where $t$ is the idempotent radical (it behaves like radical of module).
My questions
============
1. Is there any relationship between these two constructions?
2. Is there a definition of torsion theory in triangulated categories? If there exists, does it coincide with t-structures in triangulated categories?
3. t-structures played important roles in reconstruction schemes (or go back to abelian category) from derived category. So, is torsion theory in abelian category playing similar roles? (I suspected very much, so it might be stupid.)
Thank you in advance!
| https://mathoverflow.net/users/1851 | What is the relationship between t-structure and Torsion pair? | The two notions are related in the sense that they share a common generalization, namely the notion of torsion pair on a pre-triangulated category (this term has at least two meanings, here we mean a category which has compatible left and right triangulations - it covers several cases including triangulated categories and quasi-abelian categories). The reference for this material is
A. Beligiannis and I. Reiten: ''Homological and Homotopical Aspects of Torsion Theories''
which is available from [Beligiannis' homepage](http://www.math.uoi.gr/~abeligia/). In fact one can take the analogy further and consider the analogy between TTF-triples on an abelian category and recollement of triangulated categories.
There is also another connection given by tilting theory. Suppose that $(\mathcal{T},\mathcal{F})$ is a torsion pair on an abelian category $\mathbf{A}$. Then we can obtain a t-structure on $D= D^b(\mathbf{A})$ by setting
$D^{\leq 0} = \{ X\in D \; \vert \; H^i(X)=0 \; \text{for} \; i>0, H^0(X)\in \mathcal{T} \}$
and
$D^{\geq 0} = \{ X\in D \; \vert \; H^i(X)=0 \; \text{for} \; i<-1, H^{-1}(X)\in \mathcal{F} \}$
For more information on this (in particular for some characterizations of when taking the derived category of the heart obtain from this t-structure is equivalent to $D$) one can see "Tilting in Abelian categories and quasitilted algebras" By Dieter Happel, Idun Reiten, Sverre O. Smalø.
I hope that at least goes some of the way toward answering (1) and (2).
As far as (3) is concerned I am not completely sure what to say. Certainly one can reconstruct a quasi-compact quasi-separated scheme from its derived category using the tensor structure, and if the scheme is particularly nice one can use the Serre functor. I am not aware of (or have forgotten if I knew) a way of reconstructing a scheme via t-structures (I guess one can use strictly localizaing subcategories which are particularly nice t-structures or take the heart of the standard one). One certainly can't just look at all t-structures - even for $D(\mathbb{Z})$ there is a proper class of t-structures.
In the abelian case the closest thing I can think of is taking the spectrum of indecomposable injectives. This is not directly torsion theoretic but it is true that injectives control hereditary torsion theories in the sense that every hereditary torsion theory in a Grothendieck abelian category has as its torsion class the left orthogonal to some injective object.
A particularly nice special case when one can really make the connection precise is the following (due to Krause). Suppose that $\mathbf{A}$ is a locally coherent Grothendieck abelian category i.e., it is a Grothendieck abelian category with a generating set of finitely presented objects and the finitely presented objects form an abelian subcategory. Then one can topologize the spectrum of indecomposable injectives in such a way that there is a bijection between hereditary torsion theories of finite type (those for which the right adjoint to the inclusion also commutes with filtered colimits) and closed subsets of the spectrum.
One last thought for the moment - although one can think of t-structures and torsion theories on abelian categories as common specializations of one more general definition the analogy can be misleading. However, there is a reasonably good analogy between hereditary torsion theories of finite type and smashing subcategories which can be made precise (again this is due to Krause). The heart of this is that every smashing subcategory of a compactly generated triangulated category is generated by an ideal of maps between compact objects. Corresponding to such an ideal there is a hereditary torsion theory of finite type in the category of additive presheaves of abelian groups on the compact objects. Something you may find particularly interesting about this (I certainly do) is that it links the theory of smashing subcategories (and the telescope conjecture) to the spectrum of indecomposable injectives in a nice abelian category.
| 16 | https://mathoverflow.net/users/310 | 14633 | 9,827 |
https://mathoverflow.net/questions/14619 | 8 | In the late 1970's and in the 1980's, Michael Freedman showed a relationship between the topological surgery problem in 4-dimensions, the slice problem for links, and the classification of non-simply-connected 4-manifolds. He also showed the failure of 4-dimensional homology surgery and the homology splitting theorem via a construction I don't really follow (because I haven't read and don't know the original reference, for one thing). I know of these results vaguely but do not understand them (certainly not the proofs). In particular, I don't know the "canonical" references, and MathSciNet doesn't seem to be helping me. They look like basic results in 4-manifold topology, which should surely be standard. Is there a survey paper on this stuff, or is it in some book? What is the best reference?
I should know this (or look it up myself), but perhaps it is more useful to post it here, because perhaps I am not alone in my confusion.
| https://mathoverflow.net/users/2051 | Freedman's work on non-simply-connected 4-manifolds | I would recommend you to look at the reference Freedman-Quinn book *Topology of 4-manifolds* , it might be helpful.
| 6 | https://mathoverflow.net/users/3895 | 14636 | 9,829 |
https://mathoverflow.net/questions/14627 | 34 | I was reading Dieudonne's "On the history of the Weil conjectures" and found two things that surprised me. Dieudonne makes some assertions about the work of Artin and Schmidt which are no doubt correct, but he doesn't give references, and the thought of ploughing through Artin's collected works seems a bit daunting to me, so I thought I'd ask here first.
Background.
-----------
If $V$ is a smooth (affine or projective) curve over a finite field $k$ of size $q$, then $k$ has (up to isomorphism) a unique extension $k\_n$ of degree $n$ over $k$ (so $k\_n$ has size $q^n$) and one can define $N\_n$ to be the size of $V(k\_n)$. Completely concretely, one can perhaps imagine the case where $V$ is defined by one equation in affine or projective 2-space, for example $y^2=x^3+1$ (note that this equation will give a smooth curve in affine 2-space for $p$, the characteristic of $k$, sufficiently large), and simply count the number of solutions to this equation with $x,y\in k\_n\ $to get $N\_n$. Let $F\_V(u)=\sum\_{n\geq1}N\_nu^n$ denote the formal power series associated to this counting function.
Now it turns out from the "formalism of zeta functions" that this isn't the most ideal way to package the information of the $N\_n$, one really wants to be doing a product over closed points of your variety. If $C\_d$ is the number of closed points of $V$ of degree $d$, that is, the number of closed points $v$ of (the topological space underlying the scheme) $V$ such that $k(v)$ is isomorphic to $k\_d$, then one really wants to define
$$Z\_V(u)=\prod\_{d\geq1}(1-u^d)^{-C\_d}.$$
If one sets $u=q^{-s}$ then this is an analogue of the Riemann zeta function, which is a product over closed points of $Spec(\mathbf{Z})$ of an analogous thing.
Now the (easy to check) relation between the $C$s and the $N$s is that $N\_n=\sum\_{d|n}dC\_d$, and this translates into a relation between $F\_V$ and $Z\_V$ of the form
$$uZ\_V'(u)/Z\_V(u)=F\_V(u).$$
This relation also means one can compute $Z$ given $F$: one divides $F$ by $u$, integrates formally, and then exponentiates formally; this works because $f'/f=(\log(f))'$.
The reason I'm saying all of this is just to stress that this part of the theory is completely elementary.
The Weil conjectures in this setting.
-------------------------------------
The Weil conjectures imply that for $V$ as above, the power series $Z\_V(u)$ is actually a rational function of $u$, and make various concrete statements about its explicit form (and in particular the location of zeros and poles). Note that they are usually stated for smooth projective varieties, but in the affine curve case one can take the smooth projective model for $V$ and then just throw away the finitely many extra points showing up to see that $Z\_V(u)$ is a rational function in this case too.
How to prove special cases in 1923?
-----------------------------------
OK so here's the question. It's 1923, we are considering completely explicit affine or projective curves over explicit finite fields, and we want to check that this power series $Z\_V(u)$ is a rational function. Dieudonne states that Artin manages to do this for curves of the form $y^2=P(x)$ for "many polynomials $P$ of low degree". How might we do this? For $P$ of degree 1 or 2, the curve is birational to projective 1-space and the story is easy. For $V$ equals projective 1-space, we have
$$F\_V(u)=(1+q)u+(1+q^2)u^2+(1+q^3)u^3+\ldots=u/(1-u)+qu/(1-qu)$$
from which it follows easily from the above discussion that
$$Z\_V(u)=1/[(1-u)(1-qu)].$$
For polynomials $P$ of degree 3 or 4, the curve has genus 1 and again I can envisage how Artin could have approached the problem. The curve will be birational to an elliptic curve, and it will lift to a characteristic zero curve with complex multiplication. The traces of Frobenius will be controlled by the corresponding Hecke character, a fact which surely will not have escaped Artin, and I can believe that he was now smart enough to put everything together.
For polynomials of degree 5 or more, given that it's 1923, the problem looks formidable.
Q1) When Dieudonne says that Artin verified (some piece of) the Weil conjectures for "many polynomials of low degree", does he mean "of degree at most 4", or did Artin really move into genus 2?
How much further can we get in 1931?
------------------------------------
Now this one really surprised me. Dieudonne claims that in 1931 F. K. Schmidt proved rationality of $Z\_V(u)$, plus the functional equation, plus the fact that $Z\_V(u)=P(u)/(1-u)(1-qu)$, for $V$ an arbitrary smooth projective curve, and that he showed $P(u)$ was a polynomial of degree $2g$, with $g$ the genus of $V$. This is already a huge chunk of the Weil conjectures. We're missing the statement that $P(u)$ has all its rots of size $q^{-1/2}$ (the "Riemann hypothesis") but this is understandable: one needs a fair amount of machinery to prove this. What startled me (in my naivity) was that I had assumed that all this was due to Weil in the 1940s and I am obviously wrong: "all Weil did" was to prove RH. So I have a very basic history question:
Q2) However did Schmidt do this?
---
EDIT: brief summary of answers below (and what I learned from chasing up the references):
A1) Artin didn't do anything like what I suggested. He could explicitly compute the zeta function of an arbitrary given hyperelliptic curve over a given finite field by an elegant application of quadratic reciprocity. See e.g. the first of Roquette's three papers below. The method in theory works for all genera although the computations quickly get tiresome.
A2) Riemann-Roch. Express the product defining $Z$ as an infinite sum and then use your head.
| https://mathoverflow.net/users/1384 | The work of E. Artin and F. K. Schmidt on (what are now called) the Weil conjectures. | Peter Roquette has written four beautiful papers on the history of the zeta-function in characteristic $p$.
[The Riemann hypothesis in characteristic p, its origin and development. Part 1. The formation of the zeta functions of Artin and F.K. Schmidt.](http://www.rzuser.uni-heidelberg.de/~ci3/rv.pdf)
[The Riemann hypothesis in characteristic p, its origin and development. Part 2. The first steps by Davenport and Hasse.](http://www.rzuser.uni-heidelberg.de/~ci3/rv2.pdf)
[The Riemann hypothesis in characteristic p, its origin and development. Part 3: The elliptic case.](http://www.rzuser.uni-heidelberg.de/~ci3/rv3.pdf)
[The Riemann hypothesis in characteristic p, its origin and development. Part 4: Davenport-Hasse fields.](http://www.rzuser.uni-heidelberg.de/~ci3/rv4.pdf)
Relevant to your questions is part 1. From the abstract "This Part 1 is dealing with the development before Hasse's contributions to the Riemann hypothesis. We are trying to explain what he could build upon. The time interval covered will be roughly between 1921 and 1931. We start with Artin's thesis of 1921 where the Riemann hypothesis for function fields was spelled out and discussed for the first time, namely in the case of quadratic function fields. We will describe the activities following Artin's thesis until F.K.Schmidt's classical paper 1931 on the Riemann-Roch theorem and the zeta function of an arbitrary function field. Finally we will review Hasse's paper in 1934 where he gives a summary about all what was known at that time about zeta functions of function fields. "
| 28 | https://mathoverflow.net/users/2275 | 14643 | 9,833 |
https://mathoverflow.net/questions/14639 | 2 | Usually we have axiomatic theory and the we look for model for it - this is book picture. Of course in real math usual one has a "model" that is given structure and looks for proper axiomatizing of it. SO it is interesting question:
Some structure called "model" is given by some (countable) number of axioms.
How many other axiomatic theories it may to be a model?
Are there any different theories ( non-isomorphic) from the first one?
Are this all theories related?
Some intuition: In this picture model is not "example of axiomatic structure" but **"point of intersection of many axiomatic structures"**. How big is space in which this theories may cross?
| https://mathoverflow.net/users/3811 | Given is "model". How many theories may it be a model? | You are using the terms "model" and "theory" in an idiosyncractic way.
In model theory, a *model* is a first-order structure, that is, a set with some functions, relations and perhaps distinguished elements, called constants. A *theory*, in contrast, is a collection of assertions, a set of sentences in this language. A given theory, which can be thought of as a set of axioms in the sense that you mentioned, can give rise to many models. And indeed, the Lowenheim-Skolem theorem says that if a theory has an infinite model, then it has infinite models of arbitrarily large cardinality. (Thus, except in trivial cases one cannot uniquely specify a model by giving "some (countable) number of axioms" as you said, since the same axioms will have models of many different sizes.)
Suppose that M is a model in a language of size κ, meaning that the language has κ many possible assertions. In this case, since any given assertion is either true in M or its negation is true in M, the complete theory of M, that is, the set Th(M) consisting of all sentences true in M, will also have size κ. Any subset S of Th(M) will also be true in M, of course. Thus, there are 2κ many theories true in M. For example, if the language has countably many symbols in it, then any given model in this language will satisfy continuum 2ω many theories.
But this answer counts theories as different, when they are different merely as sets of sentences, even when these theories have the same models. But for the purposes of counting theories, it may be more sensible to use another common definition of *theory*, which is a set of sentences closed under consequence. This amounts to identifying theories that have the same models.
With this second understanding of theory, the answer is a little more subtle. In the empty language, for example, every model is just a naked set, with no structure. There are exactly countably many countable models in this language: one of each finite size and one countably infinite model. If φn is the assertion that there are exactly n objects, then for any set A of natural numbers, we may form the theory TA, which asserts that ¬φn for each n in A. These theories are all inequivalent, and all true in any infinite model. If M is any model, then there are continuum many theories TA that are true in M.
This shows that in fact every model M, in any language, satisfies at least continuum many deductively closed theories.
If the language is larger, with uncountable size κ, then either there are uncountably many relation symbols, uncountably many function symbols or uncountably many constant symbols. In each case, it is a fun exercise to form 2κ many inequivalent theories T in the language. Given any model M, let σ be any sentence false in M. For any theory T containing σ, we may form the theory T' = { σ implies φ | φ in T }. This theory is true in M, since σ is false in M. Thus, by counting theories in this manner, one can show that there are 2κ many inequivalent theories true in M.
| 3 | https://mathoverflow.net/users/1946 | 14646 | 9,836 |
https://mathoverflow.net/questions/14638 | 10 | From seminar on kdV equation I know that for integrable dynamical system its trajectory in phase space lays on tori. In wikipedia article You may read (<http://en.wikipedia.org/wiki/Integrable_system>):
>
> When a finite dimensional Hamiltonian
> system is completely integrable in the
> Liouville sense, and the energy level
> sets are compact, the flows are
> complete, and the leaves of the
> invariant foliation are tori. There
> then exist, as mentioned above,
> special sets of canonical coordinates
> on the phase space known as
> action-angle variables, such that the
> invariant tori are the joint level
> sets of the action variables. These
> thus provide a complete set of
> invariants of the Hamiltonian flow
> (constants of motion), and the angle
> variables are the natural periodic
> coordinates on the torus. The motion
> on the invariant tori, expressed in
> terms of these canonical coordinates,
> is linear in the angle variables.
>
>
>
As I also know that elliptic curve is in fact some kind of tori, then there natural question arises: **Are tori for quasi-periodic motion in action-angle variables of some dynamical systems related in any way to algebraic structure like elliptic curve?** Maybe some small dynamical systems and some elliptic curves are related in some way?
The most interesting in this matter is for me the size of space of elliptic functions: its quite small, every elliptic curve is rational function of Weiestrass function, and its derivative. **Has this property any analogy in integrable dynamical systems theory?**
As isomorphic elliptic curves shares some invariants, it is also interesting it they have any "dynamical meaning".
| https://mathoverflow.net/users/3811 | Integrable dynamical system - relation to elliptic curves | If your system is algebraic, then you bet! More generally, you can get abelian varieties as the fibers for many interesting integrable systems. Google the following for more: algebraic complete integrable Hamiltonian system, Calogero-Moser System, Hitchin System.
As for elliptic curves, they'll only pop out in low dimensional cases, because otherwise, the fibers have to have larger dimension.
As for the latter, it depends what you might want. I've seen the definition of integrable given by "can be solved by a sequence of quadratures" and in this terminology, you can check that an algebraic system you're always working with the global section of the theta function on the abelian variety, which is the unique (up to scaling) global section of the theta divisor on the abelian variety, which for an elliptic curve, is just the Weierstrass function.
| 12 | https://mathoverflow.net/users/622 | 14657 | 9,845 |
https://mathoverflow.net/questions/14624 | 4 | Take the following definition:
"A *parabolic subgroup* of a linear algebraic group defined over $\mathbb{C}$ is a subgroup, closed in the Zariski topology, for which the quotient space is a projective algebraic variety."
My questions are:
(i) Why include *closed* in the definition?
(ii) What is an example of a projective algebraic variety that is the quotient of a linear algebraic group by a non-Zariski-closed subgroup?
(iii) What is an example of a quotient by a parabolic subgroup that is not a flag manifold?
(iv) Elliptic curves cannot be described as quotients of linear algebraic groups. What are other examples of families of varieties that cannot be expressed in this form?
| https://mathoverflow.net/users/1648 | Questions Suggested by the Parabolic Subgroup Definition | If $G={\mathbb C}$ is the additive group and $L\subset G$ is a lattice, then $L$ is not
Zariski closed, $G/L$ exists (in a sense of analytic geometry) and is projective. But of course
$L$ is not a parabolic subgroup of $G$.
| 7 | https://mathoverflow.net/users/4158 | 14663 | 9,849 |
https://mathoverflow.net/questions/14673 | 0 | (I've edited this question)
I'm searching for a continuously differentiable function $f:\mathbb R^2\to\mathbb R$ such that $f(x)+f(x+u+v)\neq f(x+u)+f(x+v)$ for all $x$ and all linearly independent $u$ and $v$.
My original question was about the special case $x=0, f(x)=0$ for merely continuous functions, which turned out to be trivial.
(I was lead to this question when investigating whether one can always find the vertices of a parallelogram (or more specifically, a square) in the graph of a continuously differentiable function $f:\mathbb R^2\to\mathbb R$. The nonexistence of functions such as the above would imply that one cannot always find a parallelogram in the graph of a continuously differentiable function.)
| https://mathoverflow.net/users/814 | Existence of an "anti-additive" (or "never linear") map? | For your new question: functions that satisfy your inequality don't exist
Proof.
Suppose $f(x)+f(x+u+v)> f(x+u)+f(x+v)$ for all $x$ and all linearly independent $u$ and $v$.
Let us get a contradiction from it. Take any square insribed in a circle, and rotate it leaving insrcibed. Rotating continuously for the angle 90 degrees you can exchange to pairs of opposite vertices.
Here is the previous counterexample:
$(x^2+y^2)^{\frac{1}{4}}$
| 9 | https://mathoverflow.net/users/943 | 14675 | 9,857 |
https://mathoverflow.net/questions/14690 | 18 | The fact that the Riemann zeta function $\zeta(s)$ and its brethren have a pole at $s=1$ is responsible for the infinitude of large classes of primes (all primes, primes in arithmetic progression; primes represented by a quadratic form). We cannot hope proving the infinitude of primes $p = a^2+1$ in this way because the series $\sum 1/p$, summed over these primes, converges. This implies that the corresponding Euler product
$$ \zeta\_G(s)= \prod\_{p = a^2+1} \frac1{1 - p^{-s}} $$
converges for $s = 1$. But if we could show that $\zeta\_G(s)$ has a pole at, say,
$s = \frac12$, then the desired result would follow. Now I know that there are heuristics on the number of primes of the form $p = a^2+1$ below $x$ (by Hardy and Littlewood?)
*Can these heuristics be explained by hypothetical properties of $\zeta\_G(s)$ (or a related Dirichlet series), or can the domain of convergence of $\zeta\_G(s)$ be derived from such asymptotics?*
BTW, here's a little known conjecture by Goldbach on these primes: let $A$ be the set of all numbers $a$ for which $a^2+1$ is prime ($A = ${1, 2, 4, 6, 10, $\ldots$}). Then every
$a \in A$ ($a > 1$) can be written in the form $a = b+c$ for $b, c \in A$. I haven't seen this discussed anywhere.
| https://mathoverflow.net/users/3503 | Primes of the form a^2+1 | Hi Franz,
Unfortunately I doubt this Euler product has very good behavior. If you believe the Hardy-Littlewood conjectures, then $\sum\_{n\leq X}\Lambda(n^2+1) \sim cX$ where $c=\prod\_{p>2}(1-\chi\_{4}(p)(p-1)^{-1})$ is some positive constant which is almost certainly transcendental. If $\zeta\_{G}(s)$ reflected this asymptotic behavior, then $\frac{d}{ds}\log{\zeta\_{G}(s)}$ would have a pole at $s=1/2$ of residue equal to $-c$. However, that would imply $\log{\zeta\_G(s)}\sim -c\log{(s-1/2)}$ in a neighborhood of $s=1/2$, so $\zeta\_G(s)$ would behave like $(s-1/2)^{-c}$ near this point. In particular, it would have some kind of branch cut...
People have conjectured that $\sum\_{n\leq X}\Lambda(n^2+1) = cX + O(X^{\frac{1}{2}+\varepsilon})$ is true, which would give continuation of $\zeta\_G(s)$ into the halfplane $\mathrm{Re}(s)>\frac{1}{4}$ after choosing a branch, but I doubt you could get any further.
| 12 | https://mathoverflow.net/users/1464 | 14694 | 9,868 |
https://mathoverflow.net/questions/14696 | 9 | I suspect that the answer to my question is well-known to be no. To be more precise, let
$G$ and $H$ be nonisomorphic finite groups of the same order. Let $S \subseteq G$ and $T \subseteq H$ be subsets satisfying the three properties: (1) the subsets are symmetric, that is $S = S^{-1}$ and $T = T^{-1}$; (2) they are minimal symmetric generating sets; (3) the size of $S$ is equal to the size of $T$. Is it possible for the Cayley graph of the pair $(G,S)$ and the Cayley graph of the pair $(H,T)$ to be isomorphic?
If the answer is yes, what is the smallest such example?
| https://mathoverflow.net/users/2677 | Does a Cayley graph on a minimal symmetric set of generators determine a finite group up to isomorphism? | The truncated cube (polyhedron with eight triangular faces and six octagonal faces) is a Cayley graph of both the symmetric group on four items (generators: transpose first two of the four items, rotate the last three) and of a different group that acts on 3-bit binary strings (generators: rotate the string, flip its first bit). You can tell they're different Cayley graphs because the graph isomorphism does not preserve the Cayley labeling: in one of the two Cayley graphs, the generators labeling the triangles are inverted on half of the triangles compared to the labeling of the other graph. See [this blog post](http://11011110.livejournal.com/116782.html).
| 16 | https://mathoverflow.net/users/440 | 14699 | 9,871 |
https://mathoverflow.net/questions/14679 | -1 | How to multiply this series:
$$(\sum\_{t=-\infty}^{\infty}a\_{t})(\sum\_{k=-\infty}^{\infty}b\_{k})$$
| https://mathoverflow.net/users/3900 | cauchy product for general case | It's not a problem to multiply the series: the product is $\sum\_{(t,k)\in\mathbb Z^2} a\_tb\_k$. The question is how to sum the double series that we have.
For series with nonnegative terms summation is not a problem either: we take the supremum of all finite sums. And since any finite sum is contained in a sufficiently large square, it follows that $\sum\_{(t,k)\in\mathbb Z^2} |a\_tb\_k|$ is finite whenever $\sum\_{t\in\mathbb Z} |a\_t|$ and $\sum\_{k\in\mathbb Z} |b\_k|$ are.
In general, $\sum\_{(t,k)\in\mathbb Z^2} a\_tb\_k=S$ if for any $\epsilon>0$ there is a finite subset $A\subset \mathbb Z^2$ such that $|\sum\_{(t,k)\in B}a\_tb\_k - S|<\epsilon$ whenever $B$ is finite and $B\supset A$. Now if both given series converge absolutely, then the contribution from outside of a large square is small, and it follows that $S$ is the product of two sums.
| 3 | https://mathoverflow.net/users/2912 | 14700 | 9,872 |
https://mathoverflow.net/questions/14714 | 107 | I know the following facts. (Don't assume I know much more than the following facts.)
* The Atiyah-Singer index theorem generalizes both the Riemann-Roch theorem and the Gauss-Bonnet theorem.
* The Atiyah-Singer index theorem can be proven using heat kernels.
This implies that both Riemann-Roch and Gauss-Bonnet can be proven using heat kernels. Now, I don't think I have the background necessary to understand the details of the proofs, but I would really appreciate it if someone briefly outlined for me an extremely high-level summary of how the above two proofs might go. Mostly what I'm looking for is physical intuition: when does one know that heat kernel methods are relevant to a mathematical problem? Is the mathematical problem recast as a physical problem to do so, and how?
(Also, does one get Riemann-Roch for Riemann surfaces only or can we also prove the version for more general algebraic curves?)
**Edit:** Sorry, the original question was a little unclear. While I appreciate the answers so far concerning how one gets from heat kernels to the index theorem to the two theorems I mentioned, I'm wondering what one can say about going from heat kernels **directly** to the two theorems I mentioned. As Deane mentions in this comments, my hope is that this reduces the amount of formalism necessary to the point where the physical ideas are clear to someone without a lot of background.
| https://mathoverflow.net/users/290 | What do heat kernels have to do with the Riemann-Roch theorem and the Gauss-Bonnet theorem? | Here is how the heat kernel proof of Atiyah-Singer goes at a high level. Let $(\partial\_t - \Delta)u = 0$ and define the heat kernel (HK) or Green function via $\exp(-t\Delta):u(0,\cdot) \rightarrow u(t,\cdot)$. The HK derives from the solution of the heat equation on the circle:
$u(t,\theta) = \sum\_n a\_n(t) \exp(in\theta) \implies a\_n(t) = a\_n(0)\cdot \exp(-tn^2)$
For a sufficiently nice case the solution of the heat equation is $u(t,\cdot) = \exp(-t\Delta) \* u(0,\cdot)$.
The hard part is building the HK: we have to compute the eigenstuff of $\Delta$ (this is the Hodge theorem). But once we do that, a miracle occurs and we get the
>
> Atiyah-Singer Theorem: The supertrace
> of the HK on forms is constant: viz.
>
>
> $\mathrm{Tr}\_s \exp(-t\Delta) = \sum\_k (-1)^k \,\mathrm{Tr} \exp(-t\Delta^k) = \mathrm{const}.$
>
>
>
For $t$ large, this can be evaluated topologically; for small $t$, it can be evaluated analytically as an integral of a characteristic class.
**Edit per Qiaochu's clarification**
[This article of Kotake](http://books.google.com/books?id=bw4-P-Pm3FMC&pg=PA137) (really in [here](http://books.google.com/books?id=dqkfAQAAIAAJ) as the books seem to be mixed up) proves Riemann-Roch directly using the heat kernel.
| 47 | https://mathoverflow.net/users/1847 | 14715 | 9,881 |
https://mathoverflow.net/questions/14717 | 19 | Why the Mittag-Leffler condition on a short exact sequence of, say, abelian groups, that ensures that the first derived functor of the inverse limit vanishes, is so named?
| https://mathoverflow.net/users/3903 | Mittag-Leffler condition: what's the origin of its name? | The wording of your question suggests that you're familiar with the "classical" [Mittag-Leffler theorem](http://en.wikipedia.org/wiki/Mittag-Leffler_theorem) from complex analysis, which assures us that meromorphic functions can be constructed with prescribed poles (as long as the specified points don't accumulate in the region).
It turns out - or so I'm told, I must admit to never working through the details - that parts of the proof can be abstracted, and from this point of view a key ingredient (implicit or explicit in the proof, according to taste) is the vanishing of a certain $\lim\_1$ group -- as assured by the "abstract" ML-theorem that you mention.
I'm not sure where this was first recorded - I hesitate to say "folklore" since that's just another way of saying "don't really know am and not a historian". One place this is discussed is in Runde's book *A taste of topology*: see [Google Books](http://books.google.ca/books?id=NIkTtwvZfAYC&pg=PA48&lpg=PA48&dq=mittag-leffler+runde&source=bl&ots=bimMcQjLEM&sig=eXkYcBT5JrHADIVni4PrD715-Z4&hl=fr&ei=g61wS8_jAcaVtgeh6aSBCg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAoQ6AEwAA#v=onepage&q=mittag-leffler%20runde&f=false) for the relevant part.
IIRC, Runde says that the use of the "abstract" Mittag-Leffler theorem to prove the "classical" one, and to prove things like the Baire category theorem, can be found in Bourbaki. Perhaps someone better versed in the mathematical literature (or at least better versed in the works of Bourbaki) can confirm or refute this?
| 14 | https://mathoverflow.net/users/763 | 14721 | 9,885 |
https://mathoverflow.net/questions/14667 | 35 | In complex projective geometry, we have a specified Kähler class $\omega$ and we have a Lefschetz operator $L:H^i(X,\mathbb{C})\to H^{i+2}(X,\mathbb{C})$ given by $L(\eta)=\omega\wedge \eta$. We then define primitive cohomology $P^{n-k}(X,\mathbb{C})=\ker(L^{k+1}:H^{n-k}(X)\to H^{n+k+2}(X))$, and we even have a nice theorem, the Lefschetz decomposition, that says $H^m(X,\mathbb{C})=\oplus\_k L^kP^{n-2k}$. Often, in papers, people just prove their result for primitive classes, as they seem to be easier to work with.
So, what exactly ARE primitive classes? Sure, they're things that some power of $\omega$ kills, but what's the intuition? Why are they an interesting distinguished class? Is there a good reason to expect this decomposition?
| https://mathoverflow.net/users/622 | Intuition for Primitive Cohomology | The primitive classes are the highest weight vectors.
Hard Lefschetz says that the operator $L$ (which algebraic geometers know as intersecting with a hyperplane) is the "lowering operator" $\rho(F)$ in a representation $\rho
\colon \mathfrak{sl}\_2(\mathbb{C})\to End (H^\ast(X;\mathbb{C}))$. The raising operator $\rho(E)$ is $\Lambda$, the restriction to the harmonic forms of the the formal adjoint of $\omega \wedge \cdot$ acting on forms. The weight operator $\rho(H)$ has $H^{n-k}(X;\mathbb{C})$ as an eigenspace (= weight space), with eigenvalue (=weight) $k$.
The usual picture of an irreducible representation of $\mathfrak{sl}\_2(\mathbb{C})$ is of a string of beads (weight spaces) with $\rho(F)$ moving you down the string and decreasing the weight by 2, and $\rho(E)$ going in the opposite direction. The highest weight is an integer $k$, the lowest weight $-k$.
From this picture, it's clear that the space of highest weight vectors in a (reducible) representation is $\ker \rho(E)$. It's also clear that, of the vectors of weight $k$, those which are highest weights are the ones in $\ker \rho(F)^{k+1}$. So the highest weight vectors in $H^{n-k}(X; \mathbb{C})$ are those in $\ker L^{k+1}$.
Of course, all this ignores the rather subtle question of how to explain in an invariant way what this $\mathfrak{sl}\_2(\mathbb{C})$, or its corresponding Lie group, really is.
**Added**, slipping Mariano an envelope. But here's what that group is. Algebraic geometers, brace yourselves. Fix $x\in X$, and let $O\_x = O(T\_x X\otimes \mathbb{C})\cong O(4n,\mathbb{C})$. Then $O\_x$ acts projectively on $\Lambda^\bullet (T\_x X\otimes \mathbb{C})$ via the spinor representation (which lives inside the Clifford action). The holonomy group $Hol\_x\cong U(n)$ also acts on complex forms at $x$, and the "Lefschetz group" $\mathcal{L}$ is the centralizer of $Hol\_x$ in $O\_x$. One proves that $\mathcal{L}\cong GL(\mathbb{C}\oplus \mathbb{C})$. Not only is this the right group, but its Lie algebra comes with a standard basis, coming from the splitting $T\_x X \otimes\mathbb{C} = T^{1,0} \oplus T^{0,1}$. Now, $\mathcal{L}$ acts on complex forms on $X$, by parallel transporting them from $y$ to $x$, acting, and transporting back to $y$. Check next that the action commutes with $d$ and $\*$, hence with the Laplacian, and so descends to harmonic forms = cohomology. Finally, check that the action of $\mathcal{L}$ exponentiates the standard action of $\mathfrak{gl}\_2$ where the centre acts by scaling. (This explanation is Graeme Segal's, via Ivan Smith.)
| 37 | https://mathoverflow.net/users/2356 | 14727 | 9,889 |
https://mathoverflow.net/questions/14730 | 3 | I feel it very unintuitive to understand what an equivariant sheaf is. In the simplest example, L/K is a finite Galois extension, G=Gal(L/K), G acts on Spec L, what are the equivariant sheaves on L?
| https://mathoverflow.net/users/2008 | Examples of Equivariant Sheaves under Group action | I don't think that the example you chose is the simplest one. It might be better to start
with say a vector bundle $\mathcal V \to X$ over some base space $X$. Suppose now that $G$
acts on $X$. A $G$-equivariant structure on $\mathcal V$ is a choice of $G$-action on
$\mathcal V$ (assuming it exists) preserving the vector bundle structure and compatible with
the given $G$-action on $X$.
E.g. if $G$ acts on a space $X$, and $\mathcal V$ is functorially assigned to $X$, e.g.
the tangent bundle or cotangent bundle, then $\mathcal V$ will have a natural $G$-equivariant
structure.
Another example: if $X$ is $n$-dimensional projective space, with its usual action
of $GL\_{n+1}$ (which factors through $PGL\_{n+1}$) and $\mathcal V$ is the tautological line bundle,
then $X$ is naturally $GL\_{n+1}$-equivariant. To see this, note that (if we remove
the zero section) then the tautological line bundle over $\mathbb P^n$ is just
${\mathbb C}^{n+1}\setminus \{0\} \to ({\mathbb C}^{n+1}\setminus \{0\} )/
{\mathbb C}^{\times}\cong {\mathbb P}^n$, and $GL\_{n+1}$-acts compatibly on the members
of this diagram.
---
Another way to phrase the same structure on $\mathcal V$ is to say that for each $g\in G$,
there is a given isomorphism $\alpha\_g: \mathcal V \cong g^\\*\mathcal V$
such that for $g,h \in G$, one has $h^\*(\alpha\_g) \circ \alpha\_h = \alpha\_{g h}$.
This latter condition can then be translated to a corresponding condition on
the sheaf of sections of $\mathcal V$, which finally can be translated to a condition
on any sheaf.
What this amounts to is that one has an action of $G$ on the sections of the sheaf
(let's call it $\mathcal F$) compatible with the $G$-action on $X$. The only thing
one has to be careful about is that since the sheaf may not be determined by its global sections, one has to think about sections over arbitrary open subsets $U$, and then
one has to take into account the fact that $U$ may not be $G$-invariant.
So putting it all together, one gets for any $U$ open in $S$ an isomorphism
$\alpha\_{g,U}: \Gamma(U,\mathcal F) \cong \Gamma(gU,\mathcal F)$ (which is the action
of the element $g \in G$), compatible with restriction to open subsets, and such that $\alpha\_{g,h U} \circ \alpha\_{h,U} = \alpha\_{g h, U}.$
Caveat: I hope I have my composition formulas correct, but if I've gotten things
tangled up, I'm sure they'll be corrected.
---
As for your original question, you should specify in what topology you want to consider your sheaves (Zariski or etale are the two possibilites that come to mind), and perhaps also what kind of sheaves you are thinking about (e.g. locally constant etale, coherent, ... ).
E.g. the structure sheaf of Spec $L$ has a natural equivariant $Gal(L/K)$-structure,
just given by the $Gal(L/K)$-action on $L$. (But note that the terminology here is much
more complicated than the actual facts being discused: Spec $L$ is just a point, and
so the $Gal(L/K)$-action is necessarily trivial on this point, and giving a sheaf is just the same as giving
an abelian group (or set, or ..., depending on what sort of sheaves we are discussing).
So giving a $Gal(L/K)$-equivariant sheaf is just giving an abelian group (or set, or ... ) with
a $Gal(L/K)$-action.)
Added: I see that in the time it took me to write this, it has been clarified that you
mean etale sheaves. Note that if you are thinking of quasi-coherent $\mathcal O$-modules,
then by (Grothendieck's interpretation of) Hilbert's Thm. 90, any sheaf is actually pulled back from the Zariski site, and so one can just think of Zariski sheaves.
One natural requirement in the $\mathcal O$-module context is to ask that
the $Gal(L/K)$-equivariant structure on the $\mathcal O$-module to be
compatible with the given $Gal(L/K)$-equivariant structure on $\mathcal O$ (the structure
sheaf of Spec $L$). It turns out that in this case any such sheaf is just pulled back
from Spec $K$, as I will explain.
By what I've already discussed above, to give a quasi-coherent sheaf on Spec $L$
with a $Gal(L/K)$-equivariant structure, compatible with the $\mathcal O$-module
structure and the given equivariant structure on $\mathcal O$, we are just talking about
$L$-vector spaces with a semi-linear action of $Gal(L/K)$.
By Hilbert's thm. 90 again, any such vector space is in fact spanned by its $Gal(L/K)$-invariants
(which are naturally a $K$-linear subspace), and so
the $L$-vector space is of the form $L\otimes\_K V$, where $V$ is a $K$-vector space, and the $Gal(L/K)$-action is just given by its action on the first factor.
In terms of sheaves, this says that any such sheaf is pulled back from
a quasi-coherent sheaf on Spec $K$.
| 7 | https://mathoverflow.net/users/2874 | 14736 | 9,893 |
https://mathoverflow.net/questions/14735 | 29 | There is, of course, a complete classification for simple complex Lie algebras. Is there a good reference which lists the group of outer automorphisms for each?
| https://mathoverflow.net/users/3513 | Outer automorphisms of simple Lie Algebras | Proposition D.40 of Fulton and Harris' *Representation Theory* states Emerton's comment: the group of outer automorphisms of a simple Lie algebra are precisely the group of graph automorphisms of the associated Dynkin diagram. There is also some discussion of this in Section 16.5 of Humphrey's *Introduction to Lie Algebras and Representation Theory*.
So for $A\_n$ and $n>1$, there is an order 2 automorphism which for $sl\_{n+1}$ amounts to negative transpose.
Type B and C have no outer automorphisms.
For $D\_n$, there is an order two automorphism swapping the two endpoints, and this corresponds to interchanging the two spin representations. On $so\_{2n}$ this is obtained (modulo the inner automorphisms) by conjugating by an orthogonal matrix in $O(2n)$ which has determinant $-1$. For $n=4$, there is also an order 3 automorphism. This is triality, and is discussed in Section 20.3 of Fulton and Harris.
For $E\_6$, there is an order 2 automorphism, though I don't know enough about the exceptional Lie algebras to say anything useful about it. But you can find a discussion of the automorphism group in Section 7 of Jacobson's *Exceptional Lie Algebras* where it is described using Jordan algebras.
For the other 4 exceptional Lie algebras there are no outer automorphisms.
| 25 | https://mathoverflow.net/users/321 | 14743 | 9,896 |
https://mathoverflow.net/questions/14739 | 36 | Given a commutative ring $R$, there is a category whose objects are epimorphisms surjective ring homomorphisms $R \to S$ and whose morphisms are commutative triangles making two such epimorphisms surjections compatible, and the skeleton of this category is a partial order that can be identified with the lattice of ideals of $R$. Now, I have always been under the impression that anything one can say about ideals one can phrase in this purely arrow-theoretic language: most importantly, the intersection of ideals is the product in this category and the sum of ideals is the coproduct. (Since we're working in a partial order, product and coproduct are fancy ways to say supremum and infimum. The direction of the implied ordering on ideals may differ here from the one you're used to, but that's not important.)
However, Harry's made some comments recently that made me realize I don't know how to define the **product** of two ideals purely in terms of this category, that is, via a universal construction like the above. It would be really surprising to me if this were not possible, so maybe I'm missing something obvious. Does anyone know how to do this?
| https://mathoverflow.net/users/290 | How can I define the product of two ideals categorically? | Nice question! The answer is that it's not possible! Let $R=\mathbb{F}\_3[x,y]/(x^2,y^2)$. The lattice of ideals consists of the eight ideals
$(1)$
$(x,y)$
$(x)$ $(y)$ $(x+y)$ $(x-y)$
$(xy)$
$(0)$,
in which each ideal contains all ideals at lower levels. In the middle level, some of the ideals have square zero, and some don't, but you can't tell which ones just from looking at the (unlabeled) lattice.
| 54 | https://mathoverflow.net/users/2757 | 14745 | 9,897 |
https://mathoverflow.net/questions/14742 | 3 | I am reading a paper concerning the action of monoidal category to another category.
Let $k$ be a commutative ring, $R$ is a k-algebra. $A=R-mod$, $B=R^{e}-mod=R\bigotimes \_{k}R^{o}-mod$.
Consider the action:
$B\times A\rightarrow A,(M,N)\mapsto M\bigotimes \_{R}N$ is an action of monoidal category of $R^{e}-mod=B^{~}=(B,\bigotimes \_{R},R)$ on A.
The paper said this action induces the action
$\Phi : D^{-}(B)\times D^{-}(A)\to D^{-}(A)$ of the monoidal derived category $D^{-}(B)$ on $D^{-}(A)$
I know this action should be $(M,N)\mapsto M\bigotimes\_{R}^{L}N$.
But I do not know how is this action of monoidal derived category on the other derived category **induced** by the action of monoidal abelian category. Is there a canonical way(A natural transformation)to get this action?
Notice that the action of monoidal abelian category is defined as follows
$\Psi:=(\Phi ,\phi ,\phi \_{0})$
$\Phi :B=(B,\bigotimes \_{R},R)\rightarrow End(A)$
$\Phi (V)\cdot \Phi (W)\overset{\phi }{\rightarrow}\Phi (V\bigotimes \_{R}W)$
The back ground of this question is localization of differential operator in derived category, so I added the tag"algebraic geometry"
This paper is "Differential Calculus in Noncommutative algebraic geometry I" which is available in MPIM
| https://mathoverflow.net/users/1851 | How is this action of monoidal derived category induced? | The original action takes the form of an additive functor $A \times B \to B$ with notation as in the question (and appropriate coherence conditions giving compatibility with the monoidal structure on $A$ presumably). By additivity this extends to the level of homotopy categories giving $K(A)\times K(B) \to K(B)$ where there is an obvious triangulation on the product category. Left deriving this functor gives the desired action $D(A)\times D(B) \to D(B)$ as uniquely as one can expect (and that there are coherent natural isomorphisms making this act as nicely as one could expect).
Note that taking categories of complexes bounded above is unneccesary.
| 5 | https://mathoverflow.net/users/310 | 14746 | 9,898 |
https://mathoverflow.net/questions/14740 | 6 | As I understand, the lack of indication on how to obtain first integrals in Arnol'd-Liouville theory is a reason why we are interested in bi-Hamiltonian systems.
Two [Poisson brackets](https://en.wikipedia.org/wiki/Poisson_bracket)
$\{ \cdot,\cdot \} \_{1} , \{ \cdot , \cdot \} \_{2}$ on a manifold $M$ are compatible if their arbitrary linear combination
$\lambda \{ \cdot , \cdot \} \_1+\mu\{\cdot,\cdot\} \_2$ is also a Poisson bracket. A bi-Hamiltonian system is one which allows Hamiltonian formulations with respect to two compatible Poisson brackets. It automatically posseses a number of integrals in involution.
The definition of a complete integrability (à la Liouville-Arnol'd) is:
Hamiltonian flows and Poisson maps on a $2n$-dimensional symplectic manifold $\left(M,\{ \cdot, \cdot \}\_M\right)$ with $n$ (smooth real valued) functions $F \_1,F \_2,\dots,F \_n$ such that: (i) they are functionally independent (i.e. the gradients $\nabla F \_k$ are linearly independent everywhere on $M$) and (ii) these functions are in involution (i.e. $\{F \_k,F \_j\}=0$) are called completely integrable.
Now, I would like to understand the connections between these two notions, and because I haven't studied the theory, any answer would be helpful. I find reading papers on these subjects too technical at the moment. Specific questions I have in mind are:
Does completely integrable system always allow for a bi-Hamiltonian structure? Is every bi-Hamiltonian system completely integrable? If not, what are examples (or places where to find examples) of systems that posses one property but not the other?
I apologize for any stupid mistakes I might have made above. Feel free to edit (tagging included).
| https://mathoverflow.net/users/2384 | Connection between bi-Hamiltonian systems and complete integrability | Your understanding is essentially correct. There are three basic (and closely related) approaches to
constructing the integrals of motion required for complete integrability: through separation of variables, through the Lax representation, and through the bi-Hamiltonian representation. The relationship among them is not yet fully understood. See, however, [this paper](https://doi.org/10.1103/PhysRevE.79.056607 "Bi-Hamiltonian representation of Stäckel systems. Phys. Rev. E 79, 056607 (2009)") by M. Błaszak, which, in essence, states that any Hamiltonian system that admits separation of variables is (or, rather, can be extended to) bi-Hamiltonian, and this [survey paper](https://doi.org/10.1023/A:1024080315471 "Separation of Variables for Bi-Hamiltonian Systems. Mathematical Physics, Analysis and Geometry 6, 139–179 (2003)") by G. Falqui and M. Pedroni on separation of variables for bi-Hamiltonian systems. As for the relationship among the Lax representation and bi-Hamiltonian property, see [this paper](https://doi.org/10.1063/1.531771 "Lax–Nijenhuis operators for integrable systems. J. Math. Phys. 37, 6173 (1996)") by F. Magri and Y. Kosmann-Schwarzbach and references therein. Now to your questions.
First of all, the bi-Hamiltonian property as you state it, without further restrictions, does not necessarily lead to integrability, and the claim that a bi-Hamiltonian system automatically possesses some integrals of motion does not hold in full generality, as far as I know. I can't think of a specific example right now, but, roughly speaking, if both your Poisson structures are too degenerate (their rank is too low), the recursion can break down and you will not get enough integrals of motion. An example of this for the infinite-dimensional case can be found in the paper *[Is a bi-Hamiltonian system necessarily integrable?](https://doi.org/10.1016/0375-9601(87)90655-4 "Physics Letters A 123(2), 55-59 (1987)")* by B.A. Kupershmidt. However, if you put in some additional nondegeneracy assumptions, the answer is yes, and dates back to Magri, Morosi, Gelfand and Dorfman. It is nicely summarized e.g. in Theorem 1.1 of [this paper](https://doi.org/10.1063/1.532221 "Magri–Morosi–Gel’fand–Dorfman’s bi-Hamiltonian constructions in the action-angle variables. J. Math. Phys. 38, 6444 (1997)") by R.G. Smirnov. The idea behind this is that the integrals of motion are provided by the traces of powers of the ratio of your Poisson structures.
As for the second question, not any Liouville integrable system is bi-Hamiltonian, at least if you impose some fairly reasonable technical assumptions, see the paper *[Completely integrable bi-Hamiltonian systems](https://doi.org/10.1007/BF02219188 "J Dyn Diff Equat 6, 53–69 (1994)")* by R.L. Fernandes; cf. also the above Smirnov's paper.
| 11 | https://mathoverflow.net/users/2149 | 14747 | 9,899 |
https://mathoverflow.net/questions/14752 | 16 | It's possible I just haven't thought hard enough about this, but I've been working at it off and on for a day or two and getting nowhere.
You can define a notion of "[covering graph](http://en.wikipedia.org/wiki/Covering_graph)" in graph theory, analogous to covering spaces. (Actually I think there's some sense -- maybe in differential topology -- in which the notions agree exactly, but that's not the question.) Anyway, it behaves like a covering space -- it lifts paths and so on.
There's also a "universal cover," which I think satisfies the same universal property as topological universal covers but I'm not sure. Universal covers are acyclic (simply connected) in graph theory, so they're trees, usually infinite. The universal cover doesn't determine the graph; for instance, any two k-regular graphs (k > 1) have the same universal cover. You can construct plenty of other pairs, too.
I'm interested in necessary and sufficient conditions for two graphs $G, H$ to have the same universal cover. One such condition (I'm pretty sure!) is whether you can give a 1-1 correspondence between trails in $G$ and trails in $H$ that preserves degree sequences. Unfortunately this doesn't help me much, since this is still basically an infinite condition. Is there some less-obvious but more easily checkable condition? In particular is it possible to determine if two (finite) graphs have the same universal cover in polynomial time?
| https://mathoverflow.net/users/382 | Checking if two graphs have the same universal cover | Two finite graphs have the same universal cover iff they have a common finite cover. This surprising fact was first proved by Tom Leighton here:
Frank Thomson Leighton, Finite common coverings of graphs. 231-238 1982 33 J. Comb. Theory, Ser. B
I'm quite sure the paper also presents an algorithm for determining if this is the case for two given graphs; essentially you develop a refined "degree" sequence for the graphs, starting from "# of vertices of degree k" and refining to "# of vertices of degree k with so-and-so neighbors of degree l" etc.
As an aside, the reason this result is so surprising is that it says something highly non-trivial about groups acting on trees (any two subgroups of Aut(T) with a finite quotient are commensurable, up to conjugation), and proving this result directly via group-theoretic methods is surprisingly difficult (and interesting). There's a paper of Bass and Kulkarni which pretty much does just that.
Edit: I just ran a quick search and found this sweet overview: "[On Leighton's Graph Covering Theorem](http://arxiv.org/abs/0906.2496)".
| 31 | https://mathoverflow.net/users/25 | 14754 | 9,903 |
https://mathoverflow.net/questions/14748 | 3 | As can be guessed from some of my previous questions, I'm trying to understand, at the moment, the relationship between principal and their associated vector bundles. To this end I've been looking at $\mathbb{CP}^{n} = SU(n+1)/U(n)$ and trying to find the representation of $U(n)$ that gives $\Omega^{(1,0)}(\mathbb{CP}^n)$, for all $n$. For $n=1$, I worked it out using a transition function argument. But for $n>1$ this is proving very cumbersome. Can anyone point me in the direction of a more effective method.
Sketch of transition function method (as requested): So $\Omega^{(1,0)}(\mathbb{CP}^n)$ is the dual of $T^{(0,1)}(\mathbb{CP}^n)$, which is in turn, by definition, the dual of $T^{(1,0)}(\mathbb{CP}^n)$. Now $T^{(1,0)}(\mathbb{CP}^n)$ is defined to be the bundle whose transition functions are the Jacobian of the change-of-coordinate maps. Dual complex bundles have conjugate trans fns, and so, the Jacobian also provides the functions of $\Omega^{(1,0)}(\mathbb{CP}^n)$. These functions $$\phi\_{ij}:U\_i \cap U\_j \to GL(n,\mathbb{C})$$ are elementary to calculate.
Now let $$ \psi:U\_i \cap U\_j \to U(n)$$ be the transition functions of the principal bundle $U(n) \to SU(n+1) \to \mathbb{CP}^n$. For the right representation $\pi:U(n) \to GL(n,\mathbb{C})$, we will have
$$
\pi \circ \psi\_{ij} = \phi\_{ij}.
$$
In the $n=1$, case both sets of transition functions are uncomplicated and it's easy to spot what $\pi$ must be. For $n>1$, it's proving to be more messy, and that's why I'm wondering if there's a smarter way of doing things.
| https://mathoverflow.net/users/2612 | Principal bundles and associated vector bundles, the case of the complex projective space (1,0)-forms | In $SU(n+1)/U(n)$ there is a natural basepoint, the coset of the identity, which is fixed by
the action of $U(n)$ (thought of as acting on the quotient by virtue of being a subgroup
of $SU(n1)$). Since $U(n)$ fixed this point, it acts on the (complexified) cotangent space to this point,
and the problem is then to understand this representation, and in particular, to decompose it
into two pieces, the $(1,0)$ piece and the $(0,1)$ piece.
Now the tangent space is ${\mathfrak su}(n+1)/{\mathfrak u(n)}$ (here I mean complexified Lie algebras), and so we have to decompose
this quotient under the adjoint action of $U(n)$. It has dimension $(n+1)^2-1-n^2 = 2n,$
and in fact it will decompose as the sum of the standard representation of $U(n)$ direct sum
its dual (or equivalently, its complex conjugate). One of these representations will
give the $(1,0)$-subbundle of the tangent bundle, and the other the $(0,1)$-subbundle.
Dualizing (to pass from tangent to cotanget) will give your answer. (I am not going
to actually stipulate which is which, just because I'm likely to blunder while tracing through the constructions and the duality; but it shouldn't
be hard to work out if you sit down with pen and paper.)
[The following is added in response to the comment below, asking for the movitation behind the above calculation; hopefully it is of some help:]
Since we are looking for an $SU(n+1)$-equivariant splitting of the (co)tangent bundle, it is enough to look for a $U(n)$-equivariant splitting of its fibre at the $U(n)$-fixed point. (This is a manifestation of the very reason that $U(n)$-reps. give rise to equivariant bundles on the quotient.) At that fixed point (the identity coset) the tangent space is ${\mathfrak su(n+1)}/{\mathfrak u(n)}$, just because it is the quotient of the tangent spaces at the identity of the corresponding groups. Since this *has* to split into two complex conjugate halves under the $U(n)$ action (we *know* that is breaks up into
a $(1,0)$ and $(0,1)$ part), each of dimension $n$, it's not hard to guess what they must be. A little computation
in the Lie algebra ${\mathfrak su(n+1)}$ confirms the guess.
Maybe a general lesson to be drawn is: when trying to compute a $G$-equivariant
bundle on $G/H$, it is enough to compute the $H$-representation on the fibre at the
identity coset; indeed, passing from $G$-equivariant bundles to this fibre is the quasi-inverse functor
to the one (implicitly) alluded to in the introduction, which associates a $G$-equivariant bundle to an $H$-representation.
[The following was added in response to the question about equivariance and transition functions in the comments:]
Dear Dyke, Let me hide behind Ben's answer, and leave the expression of equivariance in terms of transition functions as an exercise. (Note that, while it is in some sense routine, as Ben indicates, it may also be painful, because if $G$ acts transitively, as in your example, then you won't be able to choose the affine opens on which the bundle is trivialized to be $G$-invariant, and this will complicate things.) Instead, I'll note the following: when you describe a vector bundle by transition functions, you are giving a bunch of open sets that you glue together to get your space, and then the transition functions tell you
how to glue together the trivial bundle on these various opens into a bundle on the space under consideration. But the $G$-equivariant set-up gives a different way to think of the bundle: on $G/H$, since $G$ acts transitively, we just take the trivial bundle at the base point, and then move it around by $G$ to get a bundle over all of $G/H$. The only thing is that this is overdetermined (unless $H$ is trivial); there are lots of ways to go via an element in $G$ to a given point. This overdeterminacy is all encoded in the fact that $H$ stabilizes the base-point: so what we have to do is say, if we take our vector space
at the base-point, and then move it around by an element $h \in H$ (which doesn't actually
move the base-point at all), how we identify the ``moved space'' with the original space.
In other words (making this heuristic precise) we have to describe an $H$-action on
the vector space at the base-point. Ergo, $H$-representations correspond to $G$-equivariant vector bundles on $G/H$.
So in this context, transition functions are not a very natural way to think about how
the vector bundle is constructed; a more representation-theoretic view-point is the way to go.
| 6 | https://mathoverflow.net/users/2874 | 14774 | 9,914 |
https://mathoverflow.net/questions/14776 | 19 | I hope this is serious enough. It is a well-known fact that $\pi\_1(SO(3)) = \mathbb{Z}/(2)$, so $SO(3)$ admits precisely one non trivial covering, which is 2-sheeted.
Another well known fact is that you can hold a dish on your hand and perform two turns (one over the elbow, one below) in the same direction and come back in the original position.
These facts are known to be related, and I more or less can guess why. Some configuration of the system (hand + dish) must draw a path in $Spin(3)$ whose projection in $SO(3)$ is the closed non trivial loop pointed at the identity.
The problem is that I cannot make this precise, since it is not clear to me which is the variety which parametrizes the position of the elbow and the hand. Is there a clean way to see how $Spin(3)$ comes into play?
| https://mathoverflow.net/users/828 | How does $\pi_1(SO(3))$ relate exactly to the waiters trick? | Spin(3) comes into the play only as the covering space of SO(3), I think. You do everything in SO(3). Draw a curve through your body from a stationary point, like your foot, up the leg and torso and out the arm, ending at the dish. Each point along the curve traces out a curve in SO(3), thus defining a homotopy. After you have completed the trick and ended back in the original position, you now have a homotopy from the double rotation of the dish with a constant curve at the identity of SO(3). You can't stop at the halfway point, lock the dish and hand in place, now at the original position, and untwist your arm: This reflects the fact that the single loop in SO(3) is not null homotopic.
| 11 | https://mathoverflow.net/users/802 | 14780 | 9,919 |
https://mathoverflow.net/questions/14781 | 2 | Considering the function $f:\mathbb{R} \to \mathbb{C}$, with $\left| f(x) \right|=1$ for all $x\in \mathbb{R}$.
Considering $g:\mathbb{R} \to \mathbb{C}$ with $\int\_{-\infty}^{\infty}{\left|g(x)\right|^2dx}=1$
I am interested on properties of the amplitude of the Fourier Transform of the product of $f$ and $g$:
$A(k)=\left|FT(f(x)g(x))\right|$
Is there any constraint on $A$ apart from the fact that A is real positive? Considering a fixed $g$, is it possible to attain any $A$ simply by changing $f$?
Thank you for any help
| https://mathoverflow.net/users/3913 | Constraints on the Fourier transform of a constant modulus function | If $g$ happens to be in $L^1$, then the amplitude of the Fourier transform of $fg$ is bounded by the $L^1$ norm of $g$, for any unimodular $f$. This is the only restriction from above since you can always choose $f$ so that $fg\ge 0$, thus bringing the (essential) supremum of $\widehat{fg}$ up to $\|g\|\_{L^1}$.
Another part of the question is how small we can make $A$. I guess "arbitrarily small", but don't have a proof. (Except for special case: if $g$ is in $L^1$, then we can chop it into pieces with disjoint supports and small $L^1$ norm, and then use $f$ to move the Fourier transforms of pieces far from one another.)
| 5 | https://mathoverflow.net/users/2912 | 14784 | 9,921 |
https://mathoverflow.net/questions/14719 | 7 | (Disclaimer: I'm a beginner in this area, so welcome corrections.)
Let $(X,x)$ be a germ of a complex surface (i.e. locally the zero set of some holomorphic functions) and assume that $x$ an isolated singular point. Mumford proved that if the local fundamental group of $X$ at $x$ is trivial, then in fact $x$ is smooth.
All the critters in the above paragraph have algebraic analogues, and the conversion was carried out (I believe) [by Flenner](https://doi.org/10.1007/BF01430965 "Flenner, H. Reine lokale Ringe der Dimension zwei. Math. Ann. 216, 253–263 (1975)"): Let $A$ be a two-dimensional complete local normal domain containing an algebraically closed field of characteristic zero; if the 'etale fundamental group of [EDIT: the punctured spectrum of] $A$ is trivial, then $A$ is regular.
However, Flenner's proof is essentially by reduction to Mumford's theorem [as far as I, a non-German-speaker, can tell], rather than a new algebraic (or algebro-geometric) proof. So:
>
> Does there exist a purely algebraic or algebro-geometric proof of Mumford's theorem?
>
>
>
---
Motivations include: (1) [Mumford's proof](http://www.numdam.org/item/PMIHES_1961__9__5_0/ "Mumford, David. The topology of normal singularities of an algebraic surface and a criterion for simplicity. Publications Mathématiques de l'IHÉS, Tome 9 (1961), pp. 5-22") is completely opaque to me; (2) No, I mean really really opaque; (3) I'm curious about extensions of the theorem to non-isolated singularities [which should probably be another question].
| https://mathoverflow.net/users/460 | An algebraic proof of Mumford's smoothness criterion for surfaces? | I found what I think is the answer, in [a paper by Cutkosky and Srinivasan](https://gdz.sub.uni-goettingen.de/id/PPN358147735_0068?tify=%7B%22pages%22:%5B323%5D%7D "Steven Dale Cutkosky, Hema Srinivasan. Comment. Math. Helvetici 68 (1993), 319–332") called "Local fundamental groups of surface singularities in characteristic $p$". They prove, as Corollary 5: Suppose that $(A, m)$ is a complete normal local domain of dimension two, with algebraically closed residue field $k$ of characteristic zero. (Slightly surprising, given the title of the paper.) Then $\pi\_1(\operatorname{Spec} A -m)=0$ if and only if $A$ is smooth over $k$. They say that this gives "an arithmetic proof of the theorem of Mumford and Flenner."
The proof apparently uses Flenner's paper, but I don't think it uses Mumford's result. They get an expression for the local fundamental group in terms of a tree, and appeal to Flenner's Theorem 2.7 to know that the group is trivial iff $A$ is smooth. I haven't tried to read that section of Flenner's paper yet, but it seems to be independent of Mumford.
| 3 | https://mathoverflow.net/users/460 | 14785 | 9,922 |
https://mathoverflow.net/questions/14799 | 6 | Let $X$, $Y$ and $Z$ be Noetherian schemes.
If $f: Y \to X$ and $g: Z \to X$ are morphisms of finite type, such that at each point of $X$, at least one of the two morphisms is smooth/étale/unramified (at all points of its inverse image), can we conclude that the induced morphism $Y \times\_X Z \to X$ is smooth/étale/unramified everywhere?
If not, which results can we obtain?
(In his textbook on Algebraic Geometry, Liu asks to prove that the answer is always "yes"...)
EDIT. So, indeed, the problem statement in the book is wrong...
| https://mathoverflow.net/users/1107 | products and smooth/étale/unramified morphisms | No. As an extreme example, suppose that $g$ is the identity (which is etale everywhere), and that $f$ is not etale at some point. Then the fibre product is just $f$ again.
But in fact, this is essentially the general case. If $g$ is etale (or smooth) at a point, then it is etale (resp. smooth) in a n.h. of that point, so we may replace $Z$ by the n.h. and so assume that
$g$ is etale everywhere. Then if $f$ is not etale (or smooth) at a point $y \in Y$, the product will not be etale in a n.h. of $y \times Z.$
(Imagine that $Y$ was e.g. a nodal curve with a node at $y$, and that $Z$ is a smooth
curve. (Here $X$ is Spec of the ground field.) Then $Y\times Z$ is the product of a nodal
curve and a smooth curve, which just looks like a cylinder over the nodal curve; it is
singular all along the "cylinder" over the node.)
| 7 | https://mathoverflow.net/users/2874 | 14802 | 9,934 |
https://mathoverflow.net/questions/14737 | 10 | In [Hartshorne, III.3] he proves that injective modules over $R$ give flasque sheaves over $Spec\ R$. I presume that's because they don't give injective sheaves, and flasque is the consolation prize. Is there an easy counterexample?
EDIT: in III.3 he's assuming Noetherian. And he's already proved in II.5.5 the equivalence
of categories of $R$-modules and quasicoherent ${\mathcal O}\_{Spec\ R}$-modules. (And that injective sheaves are flasque, in III.2.)
EDIT: his proof that injectives are flasque uses some non-quasicoherent sheaves. So the ingredients "injective R-modules give injective objects in the category of quasicoherent sheaves [II.5.5]" plus "injective objects in the category of sheaves are flasque [III.2]" isn't enough for the result he gets in III.3, that injective R-modules give flasque sheaves.
| https://mathoverflow.net/users/391 | Can injective modules over R give non-injective sheaves over Spec R? | Let me put this here for the sake of clarity. As was noted by Emerton in a comment above, [this answer to a related Math Overflow question](https://mathoverflow.net/questions/6762/why-is-an-injective-quasi-coherent-sheafs-restriction-to-an-open-subset-still-an/6773#6773) shows that the answer is **no**, for an injective $R$-module $I$, the sheaf $\widetilde{I}$ is not necessarily an injective sheaf.
So if you like that answer, I suggest you click on the link above and upvote that answer.
The reference provided in that answer is to the following:
MR0617087 (82i:13013)
Dade, Everett C.
Localization of injective modules.
J. Algebra 69 (1981), no. 2, 416--425.
Localization of modules over a commutative ring $R$ with respect to a multiplicatively closed subset $S$ of $R$ is an exact functor with a large number of properties, some of which are listed in Theorem 3.76 of J. J. Rotman's book [An introduction to homological algebra, Academic Press, New York, 1979; MR0538169 (80k:18001)]. The fifth property, namely: (LI) the localization $S^{-1}E$ of any injective $R$-module $E$ is an injective $S^{-1}R$-module, is false. Two examples are given here showing that arbitrary $R$ and $S$ need not have the property (LI). Also a positive result is given, showing that (LI) holds for certain non-Noetherian $R$ and certain $S$. In particular, if $R$ is the polynomial ring $k[x\_1,x\_2,\cdots]$ in a countable number of $x\_n$ over a nonzero Noetherian ring $k$, then (LI) holds for all choices of $S$.
| 7 | https://mathoverflow.net/users/1784 | 14806 | 9,936 |
https://mathoverflow.net/questions/14801 | 7 | Suppose $G$ is a simple (linear) algebraic group over an algebraically closed field of characteristic zero, that $n$ is a positive natural number, and that $g\in G$. Can we always find an $h\in G$ such that $h^n=g$?
(It appears to be possible to check this for the classical algebraic groups by direct computations in each case, but covering the exceptional Lie Algebras this way seems like it might be tricky, and anyhow, I'm inclined to think that a case analysis is probably not the optimal way of approaching this problem!)
Note added: Kovalev made a comment showing that the answer is `no' in general. The counterexamples appear to revolve around non-semisimple elements. I wonder whether the answer becomes positive if one restricts oneself to $g$ of finite order?
| https://mathoverflow.net/users/3513 | Taking roots in simple linear algebraic groups | A semisimple element lies in a maximal torus, so you can extract any root from it inside this torus.
| 7 | https://mathoverflow.net/users/3696 | 14816 | 9,941 |
https://mathoverflow.net/questions/14758 | 1 | Hi all,
I am wondering whether someone has considered the definition of the flux homomorphism for manifolds with boundary. More specifically, I am looking at the annulus and I want the diffeomorphism $\phi(r, \theta) = (r, \theta + \alpha)$ for $\alpha$ constant to have non-zero flux.
I was thinking of extending the flux homomorphism using the relative homology like this:
$flux: Symp\_0(A) \times H\_1(A, \partial A) \to R$
where the value of $flux$ on the pair $(\phi, C)$ is given by choosing a 2-cycle $S$ such that $\partial S = C - \phi(C)$ and integrating the symplectic form over this cycle. Now, I think $H\_1(A, \partial A)$ is generated by a circle homologous to the boundary and an arc connecting the two, and if you compute the value of $flux$ on the rotiational diffeomorphism and the latter generator, you get a non-zero result.
My questions are
1. Is this anywhere near correct, and if so, has this been done before?
2. Is there a way to define the flux homomorphism in exactly the same way as for the case of manifolds without boundary, so that the standard results are still valid?
| https://mathoverflow.net/users/3909 | Flux homomorphism for manifolds with boundary | Here are some remarks about your definition.
1) $H\_1(A,\partial A)$ is just one-dimensional, it is generated by a path that joins two sides of $A$.
2) The definition that you gave works for the annulus, and for surfaces with a boundary as well. This will also work for manifolds with boundary $M^n$, in the case
when you consider fluxes of volume-preserving maps (i.e. you work with $\Omega^n$ and $H\_{n-1} (M^n, \partial M^n$). I have not seen this definition before, but it is so natural, that it would be strange if no one considered it.
3) It does not look that this definition will work for higher-dimensional symplectic manfiolds, if you want to study fluxes of symplectomorphisms (and you work with $\omega$ and $H\_1(M^{2n},\partial M^{2n})$), because the restriction of $\omega$ to $\partial$ will be non-zero.
| 2 | https://mathoverflow.net/users/943 | 14817 | 9,942 |
https://mathoverflow.net/questions/14808 | 12 | Given a symmetric monoidal category and a monoid object A in it, one can form the category of modules over this monoid object, i.e. objects are $A \otimes M \rightarrow M$ satisfying the natural properties analogous to modules over a ring and morphisms respecting this. The following seems to be true and I would like to know why:
If the category of modules has a closed symmetric monoidal structure with A as unit object, then A is a *commutative* monoid.
This is how I read the statement right after Proposition 2.3.4 in Hovey/Shipley/Smith's paper "Symmetric Spectra" and it would give an excellent motivation for introducing symmetric spectra...
| https://mathoverflow.net/users/733 | Why is a monoid with closed symmetric monoidal module category commutative? | (I have rewritten parts to respond to potential objections, since someone disliked this answer. I didn't change the thrust of my discussion, but I did alter some places where I might have been slightly too glib. If I missed something, please feel free to correct me in the comments!)
The situation is even better than that! Suppose we are given an $E\_1$-algebra $A$ of a presentable symmetric monoidal $\infty$-category $\mathcal{C}$.
Call an $E\_n$-monoidal structures on the $\infty$-category $\mathbf{Mod}(A)$ of left $A$-modules *allowable* if $A$ is the unit and the right action of $\mathcal{C}$ on $\mathbf{Mod}(A)$ is compatible with the $E\_n$ monoidal structure, so that $\mathbf{Mod}(A)$ is an $E\_n$-$\mathcal{C}$-algebra. Then the space of allowable $E\_n$-monoidal structures is equivalent to the space of $E\_{n+1}$-algebra structures on $A$ itself, compatible with the extant $E\_1$ structure on $A$. (This is even true when $n=0$, if one takes an $E\_0$-monoidal category to mean a category with a distinguished object.) The object $A$, regarded as the unit $A$-module, admits an $E\_n$-algebra structure that is suitably compatible with the $E\_1$ structure an $A$. [Reference: Jacob Lurie, DAG VI, Corollary 2.3.15.]
Let's sketch a proof of this claim in the case Peter mentions. Suppose $A$ is a monoid in a presentable symmetric monoidal category $(\mathbf{C},\otimes)$. Suppose $\mathbf{Mod}(A)$ admits a monoidal structure (not even a priori symmetric!) in which $A$, regarded as a left $A$-module, is the unit. I claim that $A$ is a commutative monoid. Consider the monoid object $\mathrm{End}(A)$ of endomorphisms of $A$ as a left $A$-module; the Eckmann-Hilton argument described below applies to the operations of tensoring and composing to give $\mathrm{End}(A)$ the structure of a commutative monoid object. The multiplication on $A$ yields an isomorphism of monoids $A\simeq\mathrm{End}(A)$.
In the case you mention, the result amounts to the original Eckmann-Hilton, as follows. If $X$ admits magma structures $\circ$ and $\star$ with the same unit (Below, Tom Leinster points out that I only have to assume that each has *a* unit, and it will *follow* that the units are the same. He's right, of course.) with the property that
$$(a\circ b)\star(c\circ d)=(a\star c)\circ(b\star d)$$
for any $a,b,c,d\in X$, then (1) the magma structures $\circ$ and $\star$ coincide; (2) the product $\circ$ is associative; and (3) the product $\circ$ is commutative. That is, a unital magma in unital magmas is a commutative monoid.
| 11 | https://mathoverflow.net/users/3049 | 14840 | 9,960 |
https://mathoverflow.net/questions/12499 | 9 | This is related to [another one of my questions on DM stacks](https://mathoverflow.net/questions/12472/is-the-inertia-stack-of-a-deligne-mumford-stack-always-finite). In Brian Conrad's article 'The Keel-Mori Theorem via Stacks', a sufficient condition on for an Artin stack to have coarse moduli space is that it has finite inertia stack. This does not include DM stacks without finite inertia. My question is that, does every DM stack of finite type over a field have a coarse moduli space? And what's the reference? Thanks.
| https://mathoverflow.net/users/370 | coarse moduli space of DM stacks | No, not every DM-stack has a coarse moduli space. The following is a counter-example (see my paper on geometric quotients):
Let X be two copies of the affine plane glued outside the y-axis (a non-separated scheme). Let G=Z2 act on X by y → –y and by switching the two copies. Then G acts non-freely on the locally closed subset {y=0, x ≠ 0}. The quotient [X/G] is a DM-stack with non-finite inertia and it can be shown that there is no coarse moduli space (neither categorical nor topological) in the category of algebraic spaces.
| 14 | https://mathoverflow.net/users/40 | 14855 | 9,970 |
https://mathoverflow.net/questions/14856 | 6 | Let $S^1$ act on $S^{2n+1}$ via Hopf action and $S^1$ also acts on $\mathbb{R}^2$ via rotation about the origin.
Then $S^1$ acts on $S^{2n+1}\times \mathbb{R}^2$ diagonally.
Let $M$ be the quotient of this diagonal action.
My question is why $ M$ can be viewed as the normal bundle of $\mathbb {CP}^n$ in $\mathbb {CP}^{n+1}$.
I have a feeling that it must be related to the fact that: after removing a $(2n+2)$ disk in $\mathbb {CP}^{n+1}$, the boundary $S^{2n+1}$ is fibered over the $\mathbb {CP}^n$.
But where can I find the proof of the statement.
| https://mathoverflow.net/users/3922 | Normal bundle of $CP^n$ in $CP^{n+1}$ | $S^{2n+1}$ sits in $S^{2n+3}$ with a trivial normal bundle. So the quotient map $S^{2n+3} \to \mathbb CP^{n+1}$ carries the normal bundle of $S^{2n+1}$ in $S^{2n+3}$ to the normal bundle of $\mathbb CP^n$ in $\mathbb CP^{n+1}$.
| 8 | https://mathoverflow.net/users/1465 | 14859 | 9,972 |
https://mathoverflow.net/questions/14632 | 4 | Let $B$ be a ring which is the colimit of rings $B\_\lambda$. Let $X\_\lambda$ be a stack (not necessarily algebraic) over $B\_\lambda$ such that $X\_\lambda \times\_{B\_\lambda} B\_\mu = X\_\mu$ and let $X = X\_\lambda \times\_{B\_\lambda} B$.
If $X$ is an algebraic stack, then does some $X\_\lambda$ have to be algebraic? Are there assumptions we can add to make this true? What if the $X\_\lambda$ are sheaves (so that the question becomes: if $X$ is an algebraic space, then is some $X\_\lambda$ an algebraic space)?
| https://mathoverflow.net/users/28 | Approximation of stacks / algebraic spaces | The answer is no, even for the sheaf-case.
First of all, you would have to make some assumptions such as assume that Xλ→ Spec(Bλ) is locally of finite presentation. For simplicity also assume that the algebraic space X is of finite presentation over Spec(B). Then if U→X is an étale presentation, it descends to a *morphism* Uλ→Xλ. Now the main problem is that even if this morphism is étale after base change to X, we cannot guarantee that it eventually becomes étale. Indeed, this is related to the openness of versality in Artin's algebraization theorem.
For example, let B0 be a ring and let B=colim Bλ be a direct limit of essentially étale algebras such that B is henselian.
If openness of versality holds for Xλ, then Uλ→Xλ is smooth in a neighborhood of the image of U→Uλ and (at least if B0 is noetherian) it follows that Uλ→Xλ is smooth for sufficiently large λ.
But ... openness of versality does not hold for general Xλ. Artin has given a nice bunch of examples where everything but this holds (see "The implicit function theorem in algebraic geometry", S5). For example, let B0=k[x] be the affine line and let B be the henselization (or localization) at the origin. Let X0 be a "bad" sheaf, e.g., the following (Ex. 5.10):
Let X0=colimk X0,k where X0,i=Spec(k[x,y]/y(x-1)(x-2)...(x-k)) — the union of the x-axis and k vertical lines. This is an example of an ind-space, so by definition:
X0(T)=colimk X0,k(T)
for any scheme T. Then clearly, X=X0×B0 Spec(B) is isomorphic to Spec(B) but Xλ is not an algebraic space for any λ.
If you assume that openness of versality holds for Xλ and that the Spec(B)→Spec(Bλ) are smooth, then the answer to your question is likely "yes". However, openness of versality is the most subtle of Artin's criteria so it is probably difficult to assert.
| 7 | https://mathoverflow.net/users/40 | 14862 | 9,973 |
https://mathoverflow.net/questions/14842 | 18 | Given a formal power series $f \in k[[X]]$, where $k$ is a commutative field, is there any good way to tell whether or not $f\in k(X)$?
Edit: To clarify, "good way to tell" means "computable algorithm to tell".
Edit 2: I really screwed up this question, so I am recusing myself from accepting an answer. I will accept an answer after a sufficient number of votes have been cast for the best answer.
| https://mathoverflow.net/users/1353 | When a formal power series is a rational function in disguise | Continued fractions!
To motivate this answer, first recall the continued fraction algorithm for testing whether a real number is rational. Namely, given a real number $r$, subtract its floor $\lfloor r \rfloor$, take the reciprocal, and repeat. The number $r$ is rational if and only if at some point subtracting the floor gives $0$.
Of course, an infinite precision real number is not something that a Turing machine can examine fully in finite time. In practice, the input would be only an approximation to a real number, say specified by giving the first 100 digits after the decimal point. There is no longer enough information given to determine whether the number is rational, but it still makes sense to ask whether up to the given precision it is a rational number *of small height*, i.e., with numerator and denominator small relative to the amount of precision given. If the number is rational of small height, one will notice this when computing its continued fraction numerically, because subtracting the floor during one of the first few steps (before errors compound to the point that they dominate the results) will give a number that is extremely small relative to the precision; replacing this remainder by $0$ in the continued fraction built up so far gives the small height rational number.
What is the power series analogue? Instead of the field of real numbers, work with the field of formal Laurent series $k((x))$, whose elements are series with at most finitely many terms with negative powers of $x$: think of $x$ as being small. For $f = \sum a\_n x^n \in k((x))$, define $\lfloor f \rfloor = \sum\_{n \le 0} a\_n x^n$; this is a sum with only finitely many nonzero terms. Starting with $f$, compute $f - \lfloor f \rfloor$, take the reciprocal, and repeat. The series $f$ is a rational function (in $k(x)$) if and only if at some point subtracting the floor gives $0$.
The same caveats as before apply. In practice, the model is that one has exact arithmetic for elements of $k$ (the coefficients), but a series will be specified only partially: maybe one is given only the first 100 terms of $f$, say. The only question you can hope to answer is whether $f$ is, up to the given precision, equal to a rational function of low height (i.e., with numerator and denominator of low degree). The answer will become apparent when the continued fraction algorithm is applied: check whether subtracting the floor during one of the first few steps gives a series that starts with a high positive power of $x$.
**Bonus:** Just as periodic continued fractions in the classical case correspond to quadratic irrational real numbers, periodic continued fractions in the Laurent series case correspond to series belonging to a quadratic extension of $k(x)$, i.e., to the function field of a hyperelliptic curve over $k$. Abel in 1826 exploited this idea as an ingredient in a method for determining which hyperelliptic integrals could be computed in elementary terms!
| 34 | https://mathoverflow.net/users/2757 | 14874 | 9,982 |
https://mathoverflow.net/questions/3190 | 12 | Suppose G is an algebraic group with an action G×X→X on a scheme. Does the fixed locus (the set of points x∈X fixed by all of G) have a scheme structure? You can obviously define the functor Fix(T)={t∈X(T)|t is fixed by every element of G(T)}. Is this functor always representable?
(This question was "broken off" of [a compound question of mine](https://mathoverflow.net/questions/3124/do-orbits-and-stable-loci-of-group-actions-have-natural-scheme-structures) after Scott Carnahan answered the other part so wonderfully that I had to accept his answer.)
| https://mathoverflow.net/users/1 | Is the fixed locus of a group action always a scheme? | The question gives the "wrong" definition of Fix(T), hence the resulting confusion.
A more natural definition of the subfunctor X^G of "G-fixed points in X" is
(X^G)(T) = {x in X(T) | G\_T-action on X\_T fixes x}
= {x in X(T) | G(T')-action on X(T') fixes x for *all* T-schemes T'}.
(Of course, can just as well restriction to affine T and T' for "practical" purposes.)
By way of analogy with more classical situations, if the base is a field k then a moment's reflection with the case of finite k shows that
{x in X(k) | G(k) fixes x}
is the "wrong" notion of (X^G)(k), whereas
{x in X(k) | G-action on X fixes x}
is a "better" notion, and is what the above definition of (X^G)(k) says.
From this point of view, if (for simplicity of notation) the base scheme is an affine Spec(k) for a commutative ring k then the "scheme of G-fixed points" exists whenever G is affine and X is separated provided that k[G] is k-free (or becomes so after faithfully flat extension on k). So this works when k is a field, or any k if G is a k-torus (or "of multiplicative type"). See Proposition A.8.10(1) in the book [Pseudo-reductive groups](http://math.stanford.edu/~conrad/papers/predfinal.pdf).
| 28 | https://mathoverflow.net/users/3927 | 14876 | 9,984 |
https://mathoverflow.net/questions/14877 | 63 | We may define a topological manifold to be a second-countable Hausdorff space such that every point has an open neighborhood homeomorphic to an open subset of $\mathbb{R}^n$. We can further define a smooth manifold to be a topological manifold equipped with a structure sheaf of rings of smooth functions by transport of structure from $\mathbb{R}^n$, since $\mathbb{R}^n$ has a canonical sheaf of differentiable functions $\mathbb{R}^n\to \mathbb{R}$, with a canonical restriction sheaf to any open subset. This gives a manifold as a locally ringed space. (Of course this definition generalizes to all sorts of other kinds of manifolds with minor adjustments).
Then the questions:
If we totally ignore the definition using atlases, will we at some point hit a wall? Can we fully develop differential geometry without ever resorting to atlases?
Regardless of the above answer, are there any books that develop differential geometry *primarily* from a "locally ringed space" viewpoint, dropping into the language of atlases only when necessary? I looked at Kashiwara & Schapira's "Sheaves on Manifolds", but that's much more focused on sheaves of abelian groups and (co)homology.
Edit:
To clarify (Since Pete and Kevin misunderstood): It's easy to show that the approaches are equivalent, but proofs using charts don't *always* translate easily to proofs using sheaves.
| https://mathoverflow.net/users/1353 | How much of differential geometry can be developed entirely without atlases? | There is the book by Ramanan "Global Calculus" which develops differential geometry relying heavily on sheaf theory (you should see his definition of connection algebra...).
He avoids the magic words "locally ringed space" by requiring the structure sheaf to be a subsheaf of the sheaf of continuous functions (hence maximal ideal of stalks = vanishing functions).
| 32 | https://mathoverflow.net/users/3701 | 14886 | 9,992 |
https://mathoverflow.net/questions/14858 | 4 | Let $V$ be a complex vector space of dimension 6 and let $G\subset {\mathbb P}^{14}\simeq {\mathbb P}(\Lambda^2V)$ be the image of the Plucker embedding of the Grassmannian $Gr(2, V)$.
1. Why the degree of $G$ is 14? or in general, how to calculate the degree of a Plucker embedding?
Let ${\mathbb P}^8\simeq L\subset {\mathbb P}(\Lambda^2V)$ be a generic 8-plane and $S$ be the intersection of $L$ with $G$.
1. How to prove that $S$ is a K3 surface?
Another question: the paper said that this construction depends on 19 parameters. I know that this is the dimension of the deformation family of the polarized K3 we get here. But I think that in this statement 19 is coming from varying the generic 8-plane. How can we obtain this number?
| https://mathoverflow.net/users/2555 | K3 surface of genus 8 | To be able calculate the degree it is worth to read a bit of Griffiths-Harris about Grassmanians (chapter 1 section 5). To prove that $S$ is $K3$ one needs to caluclate the canonical bundle of $G$, use simple facts about Plucker embedding, use adjunction formula and finally the fact that a simply connected surface with $K\cong O$ is $K3$.
I will make the second bit of calculation, that proves that $S$ is a $K3$ (so I don't calculate the degree $14$).
First we want to calculate the canonical bundle of $G$. Denote by $E$ the trivial $6$-dimensional bundle over $G$, and by $S$ the universal (tautological) rank $2$ sub-bundle. Then the tangent bundle to $G$ is $TG=S^\* \otimes (E/S)$ . It follows from the properties of $c\_1$ that $c\_1(TG)=6c\_1(S^\*) $. Similarly for the canonical bundle $K\_G$ we have the expression $K\_G\cong (detS^\*)^{\otimes -6}$.
Now we will use the (simple) statement from Griffiths-Harris that under the Plucker embedding we have the isomorphism of the line bundles $det S^\*=O(1)$. Using the previous calculation we see $K\_G\cong O(-6)$.
Finally, the surface $S$ is an iterated (6 times) hyperplane section of $G$. So by Lefshetz theorem it has same fundamental group as $G$, i.e., it is simply-connected. It suffices now to see that its canonical bundle is trivial. This is done using the adjunction formula $K\_D=K\_X+D|\_D$. Every time we cut $G$ by a hyper-plane we tensor the canonical by $O(1)$, but $O(-6)\otimes O(6)\cong O$.
Added. The calculation of the degree is done by Andrea
Added. The number 19 is obtaied in the following way. The dimension of the grassmanian of $8$-planes in $CP^{14}$ is $6\cdot 9=54$. At the same time the Grassmanian of $2$-planes in $\mathbb C^6$ has symmetires, given by $SL(6,\mathbb C)$, whose dimension is 35. We should quotient by these symmetries and get $54-35=19$
| 15 | https://mathoverflow.net/users/943 | 14899 | 10,001 |
https://mathoverflow.net/questions/14892 | 14 | I am now taking a course focusing on triangulated geometry. The professor has formulated the Beck's theorem for Karoubian triangulated category. The proof is very simple. Just using the universal homological functor(equivalent to Verdier abelianization)back to abelian settings(in particular, Frobenius abelian category), then using the Beck's theorem for abelian category to get the proof.
When he finished the proof, he made a remark that the cohomological descent theory can be taken as a consequence of triangulated version of Beck's theorem.
As we know, the Beck's theorem for abelian category is equivalent to Grothendieck flat descent theory(Beck's theorem may be more general). I have two questions:
1. Is there any reference(other than SGA 4)in English explaining the relationship of usual descent theory and cohomolgical descent theory? What I am looking for is not a very thick book but a lecture notes with some examples.
2. I know Jacob Lurie developed the derived version of Beck's theorem for his infinity category(correct me if I make mistake). But I have never read his paper very carefully. I wonder whether he explained the relationship of Beck's theorem and "cohomogical descent in his settings"(if there exists such terminology).
All the other comments are welcome.
Thank you
| https://mathoverflow.net/users/1851 | Looking for reference talking about relationship between descent theory and cohomological descent | The relationship between cohomological descent and Lurie's Barr-Beck is exactly the same as the relationship between ordinary descent and ordinary Barr-Back. To put things somewhat blithely, let's say you have some category of geometric objects $\mathsf{C}$ (e.g. varieties) and some contravariant functor $\mathsf{Sh}$ from $\mathsf{C}$ to some category of categories (e.g. to $X$ gets associated its derived quasi-coherent sheaves, or as in SGA its bounded constructible complexes of $\ell$-adic sheaves). Now let's say you have a map $p:Y \rightarrow X$ in $\mathsf{C}$, and you want to know if it's good for descent or not. All you do is apply Barr-Beck to the pullback map $p^\ast:\mathsf{Sh}(X)\rightarrow \mathsf{Sh}(Y)$. For this there are two steps: check the conditions, then interpret the conclusion. The first step is very simple -- you need something like $p^\ast$ conservative, which usually happens when $p$ is suitably surjective, and some more technical condition which I think is usually good if $p$ isn't like infinite-dimensional or something, maybe. For the second step, you need to relate the endofunctor $p^\ast p\_\ast$ (here $p\_\*$ is right adjoint to $p^\ast$... you should assume this exists) to something more geometric; this is possible whenever you have a base-change result for the fiber square gotten from the two maps $p:Y \rightarrow X$ and $p:Y \rightarrow X$ (which are the same map). For instance in the $\ell$-adic setting you're OK if $p$ is either proper or smooth (or flat, actually, I think). Anyway, when you have this base-change result (maybe for p as well as for its iterated fiber products), you can (presumably) successfully identify the algebras over the monad $p^\ast p\_\ast$ (should I say co- everywhere?) with the limit of $\mathsf{Sh}$ over the usual simplicial object associated to $p$, and so Barr-Beck tells you that $\mathsf{Sh}(Y)$ identifies with this too, and that's descent. The big difference between this homotopical version and the classical one is that you need the whole simplicial object and not just its first few terms, to have the space to patch your higher gluing hopotopies together.
| 18 | https://mathoverflow.net/users/3931 | 14901 | 10,003 |
https://mathoverflow.net/questions/14897 | 7 | A generic example is ${}\_2 F\_1(\frac{1}{3},\frac{2}{3},\frac{5}{6};\frac{27}{32})=\frac{8}{5}$. So my question: Is there any description of the set of rational points at which the hypergeometric function ${}\_m F\_n$ takes rational values?
The references to the literature where a lot of such examples listed are apreciated very much too.
| https://mathoverflow.net/users/2052 | at which rational points does the Hypergeometric function take rational values | There are theorems that give conditions for the set to be finite even when "rational" is replaced by "algebraic". See the work of Paula Cohen Tretkoff. E.g the following paper is a survey and Theorem 2 is about this:
<http://www.math.tamu.edu/~ptretkoff/martinpub_final.pdf>
| 7 | https://mathoverflow.net/users/2290 | 14905 | 10,006 |
https://mathoverflow.net/questions/14903 | 20 | I am wondering if it is true that **every compact, connected, oriented manifold is cobordant to a simply connected manifold.**
I believe that some sort of surgery will do the trick. Roughly speaking, I want to add handles so that I can kill representative loops. However, I don't know if my surgery process builds a cobordism and it is hard to for me to see what the new boundary is. Another possibility is to build a Morse function that constructs the cobordism for free.
It is hard for me to get an intuition for what is going on, because all the compact, oriented 2-manifolds are boundaries and consequently cobordant to the empty set. CP^2 seems like the "easiest" test case, but it is already simply-connected for cellular reasons. Lens Spaces might be a nice candidate, but the 3 dimensional ones can be realized as boundaries of some disc bundle on S^2.
If possible, I'd prefer a constructive procedure, but any answer that helps elucidate the material is welcome.
| https://mathoverflow.net/users/1622 | Every Manifold Cobordant to a Simply Connected Manifold | Assume that $M^n$ has $\pi\_1$ finitely generated (**Edit:** and n>3). Choose a generator. We will construct (using surgery) a cobordism to $M'$ which kills that generator, and by induction we can kill all of $\pi\_1$. Choose an embedded loop which represents the generator, and choose a tubular neighborhood of the loop. We can view this as a (n-1)-dimensional vector bundle over $S^1$, the normal bundle. Since $M$ is oriented, this is a trivial vector bundle so we can identify this tubular neighborhood with $S^1 \times D^{n-1}$.
Now we build the cobordism. We take $M \times I$, which is a cobordism from $M$ to itself. To one end we glue $D^2 \times D^{n-1}$ along the boundary piece $S^1 \times D^{n-1}$ via its embedding into $M$. This is just attaching a handle to $M \times I$. This new manifold is a cobordism from $M$ to $M'$, where $M'$ is just $M$ where we've done surgery along the given loop.
A van Kampen theorem argument shows that we have exactly killed the given generator of $\pi\_1$. Repeating this gives us a cobordism to a simply connected manifold.
Note that it is essential that our manifold was oriented. $\mathbb{RP}^2$ is a counter example in the non-oriented setting, as all simply connected 2-manifolds are null-cobordant, but $\mathbb{RP}^2$ is not.
---
[I was implicitly thinking high dimensions. Thanks to Tim Perutz for suggesting something was amiss when n=3]
If n=3 then this is "surgery in the middle dimension" and it is more subtle. First of all the normal bundle is an oriented 2-plane bundle over the sphere, so there are in fact $\mathbb{Z} = \pi\_1(SO(2))$ many ways to trivialize the bundle (these are normal framings). Ignoring this, if you carry out the above construction, you will see that (up to homotopy) M' is the union of $M - (S^1 \times D^2)$ and $D^2 \times S^1$ along $S^1 \times S^1$. This can (and does) enlarge the fundamental group.
However a different argument works in dimensions n=1,2,3. The oriented bordism groups in those dimensions are all zero (see the [Wikipedia entry on cobordism](http://en.wikipedia.org/wiki/Cobordism)), so in fact every oriented 3-manifold is cobordant to the empty set (a simply connected manifold). The fastest way to see this is probably a direct calculation of the first few homotopy groups of the Thom spectrum MSO.
| 20 | https://mathoverflow.net/users/184 | 14907 | 10,008 |
https://mathoverflow.net/questions/14861 | 16 | The HKR theorem for cohomology in characteristic zero says that if $R$ is a regular, commutative $k$ algebra ($char(k) = 0$) then a certain map $\bigwedge^\* Der(R) \to CH^\*(R,R)$ (where $\wedge^\* Der(R)$ has zero differential) is a quasi-isomorphism of dg vector spaces, that is, it induces an isomorphism of graded vector spaces on cohomology.
Can the HKR morphism be extended to an $A\_\infty$ morphism? Is there a refinement in this spirit to make up for the fact that it is not, on the nose, a morphism of dg-algebras?
| https://mathoverflow.net/users/132 | Is there a refinement of the Hochschild-Kostant-Rosenberg theorem for cohomology? | Yes there is. It was noted by Kontsevich long time ago that the HKR quasi-isomorphism on cochains can be corrected to give a quasi-isomorphism of dg-algebras and thus induce an $A\_\infty$ quasi-isomorphism of minimal models. The correction is very natural - one needs to compose the HKR map it the contraction by the square root of the Todd class, where the latter is understood as a polynomial of the Atiyah class. This story has been studied in great detail in the past few years and has been generalized further to give Tsygan formality which is a quasi-isomorphism of $\infty$-calculi. This was proven by Dolgushev-Tamarkin-Tsygan and also by Calaque-Rossi-van den Bergh.
The literature on the subject is huge but you should get a good sense of the results if you look at
this [survey](http://arxiv.org/abs/0901.0069) by Dolgushev-Tamarkin-Tsygan and at this [paper](http://arxiv.org/abs/0904.4890) of Calaque-Rossi-van den Bergh. There are also many interesting references listed in these papers, for instance the works of Caldararu on the Mukai pairing.
| 20 | https://mathoverflow.net/users/439 | 14911 | 10,010 |
https://mathoverflow.net/questions/14910 | 2 | Can there exist two non-equivalent equivariant actions of a group $G$ on vector bundle over a $G$ space?
| https://mathoverflow.net/users/2612 | Can there exist two non-equivalent equivariant actions of a group on vector bundle? | Maybe I am miss understanding the question, but it seems the answer is yes.
Take your favorite G-space, mine is $S^1$ with the $\mathbb{Z}/2$-action "flip". Then consider the trivial vector bundles $S^1 \times V$, where $V$ is a $G$-representation. In my favorite example $V = \mathbb{R}$ can be either the trivial representation or the sign representation. Taking the diagonal $G$-action gives an equivariant action of the group on the vector bundle. They are distinct for distinct representations (at least in the $S^1$-example) yet the underlying vector bundles are the same if the representations have the same dimension (they are trivial bundles after all).
| 5 | https://mathoverflow.net/users/184 | 14913 | 10,011 |
https://mathoverflow.net/questions/11633 | 14 | Suppose the continuum is larger than $\aleph\_2$. Does there exist a countably closed notion of forcing that collapses $\aleph\_2$ to $\aleph\_1$, but does not collapse the continuum to $\aleph\_1$? Moreover, does there exist such a forcing notion that is separative and has size continuum? It is known (see below) that the canonical collapse Coll$(\aleph\_1, \aleph\_2)$ collapses the continuum. Trying something like the canonical collapse relativized to some inner model will fail to answer the question, because this forcing will not be countably closed in V.
Background information:
This question came up as a result of my studies of the following theorem.
Let $\kappa < \theta$ be cardinals, with $\kappa$ regular and $\theta^{<\kappa} = \theta$. Then any forcing of size $\theta^{<\kappa}$ which is separative and $<\kappa$ closed and which collapses $\theta$ to $\kappa$ is forcing equivalent to the canonical collapse forcing Coll$(\kappa, \theta)$.
I want to know whether this theorem still holds in the case where $\theta^{<\kappa} = \theta$ fails. The question above is the simplest possible such case.
The reason why Coll$(\aleph\_1, \aleph\_2)$ collapses the continuum (when CH fails) is that we can think of $\aleph\_1$ as $\aleph\_1$ many $\aleph\_0$-blocks. Consider only the elements of Coll$(\aleph\_1, \aleph\_2)$ such that on each $\aleph\_0$ block, they are either fully defined or fully undefined. This is a dense set in Coll$(\aleph\_1, \aleph\_2)$, and it's isomorphic to Coll$(\aleph\_1, \aleph\_2^{\aleph\_0})$ = Coll$(\aleph\_1, \bf{c})$.
| https://mathoverflow.net/users/3183 | Is it possible for countably closed forcing to collapse $\aleph_2$ to $\aleph_1$ without collapsing the continuum? | I got the answer from Stevo Todorcevic last weekend at the MAMLS conference in honor of Richard Laver in Boulder, CO. He told me that it is an unpublished result of his that any semi-proper forcing which collapses $\aleph\_2$ collapses the continuum. As countably closed forcing is semi-proper, the answer to my question is no. Stevo sketched a proof for me, but I do not remember it well enough to reproduce it here.
| 7 | https://mathoverflow.net/users/3183 | 14946 | 10,034 |
https://mathoverflow.net/questions/11505 | 8 | Does there exist a partial order, nontrivial for forcing, that is countably closed, but whose separative quotient is not countably closed? Supposing the answer is yes, then is there a partial order, nontrivial for forcing, that is countably closed, but is not forcing equivalent to any countably closed separative partial order?
For those of you unfamiliar with the separative quotient of a partial order, it is defined as follows. Two elements of a partial order are compatible iff there is some element below both of them. We form the separative quotient of a partial order by taking equivalence classes: x is equivalent to y when x and y are compatible with the exact same things. We then define a new partial order for the separative quotient -- $x \leq y$ iff everything compatible with x is compatible with y.
A partial order is said to be separative if whenever $x \nleq y$, there is $z \leq x$ such that z is incompatible with y. The separative quotient of any partial order is separative.
Some of the ways, order-theoretically speaking, that two partial orders can be forcing equivalent are
(1) They are isomorphic, or more generally,
(2) A dense subset of one of them is isomorphic to a dense subset of the other.
| https://mathoverflow.net/users/3183 | closure of separative quotients | Stevo Todorcevic answered this question for me at the MAMLS conference in honor of Richard Laver last weekend in Boulder, CO. Apparently, the answer is that examples of forcings that are closed, whose separative quotients are not closed, come up frequently, with one particular example being forcings involving semi-selective coideals studied by Ilija and Farah.
| 4 | https://mathoverflow.net/users/3183 | 14948 | 10,036 |
https://mathoverflow.net/questions/9822 | 5 | Supppose there are integers $a\_1,a\_2,\dots$ and a polynomial $p$ so that the integers $p(a\_1),p(a\_2)...$ satisfy some linear recurrence, i.e. $\sum p(a\_i)x^i$ is a rational function of $x$. Must integers $b\_i\in p^{-1}(p(a\_i))$ so that $\sum b\_ix^i$ is a rational function, necessarily exist?
(The answer is no if we ask for the function $\sum a\_i x^i$ to be rational, as can be seen when $p(t)=t^2$ and $a\_i$ being a random sequence of $\pm1$)
| https://mathoverflow.net/users/2384 | Does an inverse polynomial map on the taylor coefficients of a rational function preserve rationality? | If $p(x)=x^d$ then this question coincides with Pisot's d'th root conjecture. A proof is given in [this](http://www.emis.ams.org/journals/Annals/151_1/zannier.pdf) paper of Zannier, I'm not sure if it's the one that Qiaochu was referring to in the comments.
Edit: A more general question was answered in the subsequent article [Equations in the Hadamard ring of rational functions](http://arxiv.org/abs/math/0701772) by Andrea Ferretti and Umberto Zannier.
| 3 | https://mathoverflow.net/users/2384 | 14952 | 10,039 |
https://mathoverflow.net/questions/14950 | 4 | Let $z \in \mathbb{C}$. Consider the following statements:
1. The point $z$ can be constructed with straightedge and compass starting from the points $\{ 0,1\}$.
- There is a field extension $K / \mathbb{Q}$ which has a tower of subextensions, each one of degree 2 over the next, and such that $z \in K$
- The field extension $\mathbb{Q}(z) / \mathbb{Q}$ has a tower of subextensions, each one of degree 2 over the next.
The usual way to prove that a geometric construction is impossible is to use that 1 and 2 are equivalent. My question is: are 2 and 3 equivalent? At first sight this looked like it was going to be true and elementary, but I could not prove it or find a counterexample.
| https://mathoverflow.net/users/3065 | On using field extensions to prove the impossiblity of a straightedge and compass construction | I think they're equivalent. Clearly 3 ==> 2. To see 2 ==> 3, say an extension E/L of fields is 2-filtered if it has a tower of subextensions each of degree 2 over the next. Then note firstly that if E/L and E'/L are 2-filtered, then so is the compositum of E and E' in any extension of L containing both (induct on the length of the tower of, say, E); and secondly for any map f:E-->E' of extensions of L, if E is 2-filtered then so is f(E). Together these two facts imply that in 2 we may assume that K is Galois over Q, say with group G. Then Q(z)/Q corresponds to a subgroup H, and by Galois theory we just need to prove the following group-theoretic lemma:
Lemma: Let G be a 2-group and H a subgroup. Then there is a chain of subgroups connecting H to G where each is index 2 in the next.
Proof: Recall that we can find a central order two subgroup Z of G. If H\cap Z=1, then HZ gets us one up in the filtration. Otherwise H\cap Z=Z, i.e. Z is in H, and we can use induction on |G|, replacing G by G/Z.
| 11 | https://mathoverflow.net/users/3931 | 14954 | 10,040 |
https://mathoverflow.net/questions/14960 | 6 | This seems like a really simple question, but I'm struggling with it. Let $X$ be a separable Banach space, $H$ be a separable Hilbert space, and suppose $i : H \hookrightarrow X$ is a dense, continuous embedding of $H$ into $X$. (This is the abstract Wiener space construction due to Gross, hence the [pr.probability] tag) If we associate $H$ with its dual $H^{\star}$, we have the inclusions $$X^{\star} \hookrightarrow H^{\star} \cong H \hookrightarrow X.$$
**My question:** Is $i^{\star} : X^{\star} \hookrightarrow H^{\star}$ a dense injection?
| https://mathoverflow.net/users/238 | Dense inclusions of Banach spaces and their duals | Yes, if you mean that $i$ is one to one, for an operator $T:X\to Y$ is one to one if and only if $T$\* has weak\* dense range, which means $T$\* has dense range when $X$ is reflexive.
| 7 | https://mathoverflow.net/users/2554 | 14963 | 10,045 |
https://mathoverflow.net/questions/1053 | 10 | Let X be a random variable. Then E(|m-X|^1) is minimized when (as a function of m) when m is the median of X, and E(|m-X|^2) is minimized when m is the mean of x.
A couple weeks ago in a technical stretch of a proof involving the [Lyapunov condition for the central limit theorem](http://en.wikipedia.org/wiki/Lyapunov_condition) I ended up with the expression E(|m-X|^3). Does this statistic have a name, or any nice properties?
**Edit:** Earlier versions of this question had |m-EX| where |m-X| was; this isn't what I meant.
| https://mathoverflow.net/users/143 | What m minimizes E(|m-X|^3) for a random variable X? | The minimizer $m$ is the nearest point projection of $X$ onto the subspace of $L^p$ formed by the constant functions ($p=3$ in your case). This $m$ is sometimes called the $p$-prediction or $p$-predictor of $X$. Apparently, this terminology began with [Andô and Amemiya](http://www.ams.org/mathscinet-getitem?mr=189077). Some of later papers are [Landers and Rogge](http://www.ams.org/mathscinet-getitem?mr=537095) (who wrote a few other papers, e.g. [this one](http://www.ams.org/mathscinet-getitem?mr=567995)), and [Cuesta and Matrán](http://www.ams.org/mathscinet-getitem?mr=894443). The term "generalized (conditional) expectation" also appeared.
| 5 | https://mathoverflow.net/users/2912 | 14965 | 10,046 |
https://mathoverflow.net/questions/14971 | 9 | A standard example for demonstrating the need for genuinely weak n-categories is that a weak 3-category with unique 0- and 1-cells amounts to the same thing as a braided monoidal category (by an Eckmann-Hilton argument), but were one to use a strict 3-category instead, this would automatically become a fortiori symmetric. In trying to get a better intuition for "the right notion" of weak categories (this is still unsettled, right?), I was wondering if anyone could give me a good intuition for what step in the argument for symmetry in the strict case we want to fail in the weak case and why. [I suppose this amounts to more generally giving a good intuition for what higher-dimensional coherence isomorphisms we should refrain from demanding to exist in the definition of weak categories, and why, but this example seems in particular like a usefully illustrative introductory context]
| https://mathoverflow.net/users/3902 | Effects of "weak" vs. "strict" categories in Eckmann-Hilton arguments | [Here](http://cheng.staff.shef.ac.uk/degeneracy/eggclock.pdf) is a nice pictorial proof of the Eckmann-Hilton argument in a higher category (made by Eugenia Cheng). One way to read this is as a proof that in a weak 2-category with one 0-cell and one 1-cell, the composition of 2-cells is commutative: in this case each diagram is equal to the two diagrams next to it, either by a unit law or by an interchange law.
Alternately, we can read it as a proof that in a weak 3-category with unique 0- and 1-cells, the composition on 2-cells is braided. In this case each diagram is *isomorphic* to those next to it, via either a unit isomorphism or an interchange isomorphism. The braiding is the composite isomorphism along the top half (say), and the fact that it isn't a symmetry arises because the composite *all* the way around need not be the identity.
Of course, if both interchange and unit isomorphisms were to be identities, the braiding *would* be a symmetry. It is known in dimension 3 that it's actually good enough (i.e. things don't collapse and become too strict) if *either* the unit *or* interchange law is weak, and the other is strict. If interchange is weak but the unit is strict, we get Gray-categories as in the original Gordon-Power-Street paper on tricategories, while if the interchange is weak but the unit is strict, this is the subject of Simpson's conjecture as mentioned by David (for the n=3 case, see [here](http://arxiv.org/abs/math/0507116)). Note that associativity never enters the picture at all, and at least in dimension 3 it can also be made strict.
There isn't really a "one step" that we want to fail. In a "fully weak" higher category, all of the associativity, unit, and interchange laws would hold only weakly. It so happens that such fully weak categories can be "semi-strictified" to make some, but not (in general) all, of these laws hold strictly, while still being equivalent to the original category in a suitable sense. Such semistrictification can often be technically useful, but I don't regard it as really of foundational importance.
Regarding "intuition for what higher-dimensional coherence isomorphisms we should refrain from demanding to exist," I think it's misleading to view this example in that way. There isn't really a coherence isomorphism that we're refraining from demanding to exist---in a weak 3-category, it's still the case that *all* coherence isomorphisms exist and *all* "formally describable" diagrams commute. Rather, what's happening is that accidentally, if we happen to be considering cells whose source and target are both identities, then there are some structural isomorphisms that we happen to be able to compose in a way "unforeseen" by the general theory of weak n-categories. Therefore, in this particular case, the assertion of that general theory that "all diagrams commute" doesn't apply, since the diagram we're looking at is only well-defined by accident. This is kind of vague, but it can be made precise by the theory of contractible globular operads, where it is true that "all diagrams commute" in the *formal*, operadic, sense, but in some particular algebra over such an operad, there may be accidental "composites" which do not commute. See also [this question](https://mathoverflow.net/questions/4897/where-does-the-easy-definition-of-a-weak-n-category-fail/4906).
| 6 | https://mathoverflow.net/users/49 | 14974 | 10,052 |
https://mathoverflow.net/questions/14961 | 10 | **Background/Motivation**: The facts about the Brauer groups I will be using are mainly in Chapter IV of Milne's book on Etale cohomology (unfortunately it was not in his online note).
Let $R$ be a Noetherian normal domain and $K$ its quotient field. Then there is a natural map $f: Br(R) \to Br(K)$. In case $R$ is regular, a well known result by Auslander-Goldman says that $f$ is injective. The natural question is when can we drop the regularity condition? But anything in the $Cl(R)/Pic(R)$ will be in the kernel of $f$, so we need that quotient to be $0$ to make it interesting.
Also, it is known that even assuming said quotient to be $0$, one has example like $R=\mathbb R[x,y,z]\_{(x,y,z)}/(x^2+y^2+z^2)$ which is UFD, but the kernel contains (I think) the quaternion algebra over $R$. The trouble in this case is that $\mathbb R$ is not algebraically closed. In fact, if $R$ is local, then $ker(f)$ injects into $Cl(R^{sh})/Cl(R)$, here $R^{sh}$ is the strict henselization. Most of the examples with non-trivial kernel I know seem a little ad-hoc, so:
**Question**: Are there more general methods to generate examples of $R$ such that $f$ is not injective?
I would love to see answers with more geometric/arithmetic flavors. I am also very interested in the positive and mixed charateristics case. Thanks in advance.
| https://mathoverflow.net/users/2083 | How to find examples of non-trival kernel of maps between Brauer groups Br(R) -> Br(K) | As R is normal all localizations at height one primes are DVR whence regular and so the question is really one of describing the kernel of the morphism
Br(R) ---> beta(R) = \cap\_P Br(R\_P) (intersection over all ht 1 primes)
beta(R) is called the 'reflexive Brauer group' (see for example ancient papers by Orzech and Hoobler, in a Brauer-group proceedings of a conference in Antwerp, in the Springer LNM 917 begin 80ties)
anyway, I've once given a cohomological description of beta(R) in [this paper](http://win.ua.ac.be/~lebruyn/LeBruyn1985c.pdf) and also one for the exact sequence related to your question
0-->Pic(RT)--->Cl(R)-->BCl(R)--->Br(R)--->beta(R)
here BCl(R) is the 'Brauer-class group'. It consists of reflexive R-modules M such that End(M) is a projectie R-module, made into a group under the modified tensor product (see paper mentioned before).
as you will see in that paper, these cohomological descriptions have everything to do with the smooth locus of R. For example, beta(R) is the Brauer group of the regular open piece and BCl(R) can 'in principle' be calculated from cohomology with support on the singular locus.
| 6 | https://mathoverflow.net/users/2275 | 14976 | 10,053 |
https://mathoverflow.net/questions/14977 | 8 | Yes, this is yet another "foundational" question in valuation theory.
Here's the background: it is a well known classical fact that the dimension (in the purely algebraic sense) of a real Banach space cannot be countably infinite. The proof is a simple application of the Baire Category Theorem: see e.g. [PlanetMath](https://planetmath.org/BanachSpacesOfInfiniteDimensionDoNotHaveACountableHamelBasis).
Suppose now that $(K,| \ |)$ is a complete non-Archimedean (**edit**: nontrivial) normed field. One has the notion of a $K$-Banach space, and the Baire Category Theorem argument works verbatim to show that such a thing cannot have countably infinite $K$-dimension.
Now let $\overline{K}$ be an algebraic closure of $K$. Then $\overline{K}$, by virtue of being a direct limit of finite-dimensional normed spaces over the complete field $K$, has a canonical topology, and indeed a unique multiplicative norm which extends $|\ |$ on $K$.
My question is: does there exist a complete normed field $(K, | \ |)$ such that:
1. $[\overline{K}:K] = \infty$ and
2. $\overline{K}$ is complete with respect to its norm?
As with [a previous question](https://mathoverflow.net/questions/13346/algebraicity-of-the-completion-of-a-field-finiteness), it is not too hard to see that this does not happen in the most familiar cases. Indeed, by the above considerations this can only happen if $[\overline{K}:K]$ is uncountable. But $[\overline{K}:K]$ will be countable if $K$ has a countable dense subfield $F$ [to be absolutely safe, let me also require that $F$ is perfect]. Indeed, the algebraic closure of any infinite field has the same cardinality of the field, so $\overline{F}$ can be obtained by adjoining roots of a countable collection of separable polynomials $P\_i(t) \in F[t]$. It follows from [Krasner's Lemma](https://en.wikipedia.org/wiki/Krasner%27s_lemma) that by adjoining to $K$ the roots of these polynomials one gets $\overline{K}$.
What about the general case?
| https://mathoverflow.net/users/1149 | Can the algebraic closure of a complete field be complete and of infinite degree? | No, there exists no such field (with a non-trivial norm). A proof can be found in Bosch, Güntzer, Remmert: Non-Archimedean Analysis, Lemma 1, Section 3.4.3.
| 6 | https://mathoverflow.net/users/1961 | 14981 | 10,054 |
https://mathoverflow.net/questions/14964 | 8 | I take the bus to work every day. Every bus has a serial number, but unlike in [the German Tank Problem](http://en.wikipedia.org/wiki/German_tank_problem), I don't know if they are numbered uniformly $1...n$.
Suppose the first $k$ buses are all different, but on day $k+1$ I take one I've been on before. What is the best estimate for the total number of buses?
| https://mathoverflow.net/users/40544 | Estimate population size based on repeated observation | Maximum likelihood estimate is the smallest $n$ for which
$$\left( 1+\frac{1}{n} \right)^k \leq \frac{n}{n-k+1},$$
that gives a value of $n$ asymptotically equal to $\frac{k^2}{2}$, consistently with the Birthday Paradox. Not sure whether an unbiased estimate would be better for any practical purpose; maybe you do have an a priori distribution for which a Bayesian estimate makes sense?
| 5 | https://mathoverflow.net/users/2368 | 14988 | 10,058 |
https://mathoverflow.net/questions/14979 | 2 | In a vector space $V$ over a field $F$, a nonempty subset $W$ of $V$ is a subspace if it is closed under addition and scalar multiplication. For a module $M$ over a ring $R$ with identity the similar result is true. Is it true in modules over a nonunital ring?
I should mention that the analogue of the following is not true.
In a vector space $V$ over a field $F$, a nonempty subset $W$ of $V$ is a subspace if for all scalars $\alpha\_1, \alpha\_2$ and for all $w\_1, w\_2 \in W$, $\alpha\_1 w\_1 + \alpha\_2 w\_2 \in W$.
This can be seen by the module $\mathbb{Z}$ over $2\mathbb{Z}$ and the subset $2\mathbb{Z} \cup \{-3,3\}$.
| https://mathoverflow.net/users/3031 | characterization of a submodule | In Hungerford everything is defined without assuming that rings have
identity (or that they are commutative). At least according to Definition IV.1.3 on p. 171 of
Hungerford, the answer to your first question is affirmative.
Quoting:
Definition 1.3. Let $R$ be a ring, $A$ an $R$-module and $B$ a
nonempty subset of $A$. $B$ is a submodule of $A$ provided that $B$
is an additive subgroup of $A$ and $rb\in B$ for all $r\in R$, $b\in
B$.
[I hope that my interpretation of the question was correct. I
understood 'characterization of submodule' as 'definition of submodule.']
EDIT: Google books links to the definition of a module, a submodule, and a ring in Hungerford:
<http://books.google.com/books?id=t6N_tOQhafoC&lpg=PP1&dq=hungerford&hl=iw&pg=PA169#v=onepage&q=&f=false>
<http://books.google.com/books?id=t6N_tOQhafoC&lpg=PP1&dq=hungerford&hl=iw&pg=PA171#v=onepage&q=&f=false>
<http://books.google.com/books?id=t6N_tOQhafoC&lpg=PP1&dq=hungerford&hl=iw&pg=PA115#v=onepage&q=&f=false>
| 2 | https://mathoverflow.net/users/2734 | 14993 | 10,060 |
https://mathoverflow.net/questions/14987 | 4 | The following is a result by C.R. Johnson appearing every now and then in the literature.
Let $A$ be an $n \times n$ inverse $M$-matrix. Then
1. All principal minors of $A$ are positive.
2. Each Principal submatrix of $A$ is an inverse $M$-matrix.
I could verify for $3 \times 3$ matrix. But it does not give any clue for the general case. Could anyone help please!
Love,
Dinesh Karia
| https://mathoverflow.net/users/3031 | inverse m-matrix | Let me begin by admitting that I have no knowledge of the subject area in question: the following is my answer as a googlist, not a mathematician.
Having disclosed that, it seems that the result that you want can be found in Section 2.5 of *Topics in Matrix Analysis* by Horn and [C.R.] Johnson.
The extent of my grasp of the material at the moment is the following: an inverse $M$-matrix is an invertible matrix whose inverse is an $M$-matrix.
**Addendum**: Caveat lector: I looked more closely at the section in question and found the following passage (starting at the bottom of p. 119):
"Exercise. Show that: a) The principal minors of an inverse $M$-matrix are positive; b) every principal submatrix of an inverse $M$-matrix is an inverse $M$-matrix; and c)...Verification of these facts requires some effort."
So one sees the limits of the approach of googling and then quoting from texts: sometimes one has to put some thought into the matter! Probably someone else will enjoy reading this section of the book and working out the exercise, so I'll leave this response up, although it is certainly not an answer in and of itself.
| 6 | https://mathoverflow.net/users/1149 | 14994 | 10,061 |
https://mathoverflow.net/questions/14996 | 5 | Let $S=k[x\_0,...,x\_n]$ be the polynomial ring over a field $k$ and $f\in S$ non-zero and homogeneous. Is it true that $Ext^m(S/(f),S)$ is zero?
This would help me to show that $Ext^m(S/fI,S)\cong Ext^m(S/I,S)(\deg f)$ for $m\geq 2$ and a homogeneous ideal $I$ of codim $\geq 2$. I tried the following approach:
Applying the long exact sequence of $Ext$ to the exact sequence of graded $S$-modules
$$0\to S/I\xrightarrow{\cdot f} S/fI(\deg f)\xrightarrow{\tau}S/(f)(\deg f)\to 0,$$
where $\tau$ is the canonical morphism $s+fI\mapsto s+(f)$,
brings
$$\ldots\to Ext^m(S/(f)(\deg f),S)\to Ext^m(S/fI(\deg f),S)\to Ext^m(S/I,S)\to$$
$$Ext^{m+1}(S/(f)(\deg f),S)\to Ext^{m+1}(S/fI(\deg f),S)\to Ext^{m+1}(S/I,S)\to\ldots$$
and two zeros on the left would suffice.
| https://mathoverflow.net/users/3950 | In what degrees does Ext(S/(f),S) vanish? | It is not true, $Ext^1(S/(f), S)\neq 0$ as Ben pointed out. However, to prove what you want $Ext^m(S/I,S)\cong Ext^m(S/fI,S)(deg(f))$ for $m\geq 2$, just note that
$$Ext^m(S/I,S) = Ext^{m-1}(I,S) $$ for any $I$, any $m\geq 2$ (using $0 \to I \to S\to S/I \to 0$) and $I(-deg(f)) \cong fI$ as $S$-modules.
| 6 | https://mathoverflow.net/users/2083 | 15000 | 10,064 |
https://mathoverflow.net/questions/15005 | 5 | Is there an easy way of remembering the direction of arrows between morphisms in Categories?
The direction of arrows so confuses me: products and co-products, (**EDIT-** Also, pull-backs, pushouts, contra/co-variant functors) and their universal mapping properties. I have to look back into Lang's Algebra and revise it every time.
Please tell if the question is not appropriate.
| https://mathoverflow.net/users/2720 | Remembering arrows' directions in basic Category Theory | I think that the best way to remember these things is to have a good example of each of these constructions in your head. If you remember the direction of the arrows for them, and if the example is "natural" enough the direction will be obvious, you will have the direction in the general case. Just watch out, sometimes two constructions will yield the same things in specific examples so you want to have examples that distinguish between the various notions. An example of this is that finite products and finite co-products have a habit of being the same in categories in which I search for examples (additive categories like categories of representations).
| 11 | https://mathoverflow.net/users/135 | 15006 | 10,066 |
https://mathoverflow.net/questions/15001 | 5 | This question is arose from the question
[Difference between equivalence relations on algebraic cycles](https://mathoverflow.net/questions/11774/difference-between-equivalence-relations-on-algebraic-cycles)
and [the example 3 in lecture 1](http://books.google.com/books?id=-5weWX_YD6sC&lpg=PP1&ots=1r3Rf2JuLx&dq=Lectures%20on%20curves%20on%20an%20algebraic%20surface&pg=PA2#v=onepage&q=&f=false) in Mumford's book Lectures on curves on an algebraic surface.
Here is the example.
Let $E$ be an elliptic curve. A curve $C$ on $\mathbb{P}^1\times E$ is said has projection degree $(d, e)$, if the projections $C\to\mathbb{P}^1$ and $C\to E$ are of degree $d$ and $e$ respectively. In that example Mumford shows that the curves with projection degree (d,e) and $d>0$ forms a $d(e+1)$ irreducible family of curves. Clearly, these curves are algebraic equivalence. But this family is not a linear system.
I can understand why the dimension is $d(e+1)$ with the help that numerical and algebraic equivalence coincide for divisors. But why the equivalences coincide for divisors? Where is the place discussing these equivalence of algebraic cycles? Also what is the irreducible family in this example?
| https://mathoverflow.net/users/2348 | Algebraic equivalence VS Numerical Equivalence - An Example. | Numerical and algebraic equivalence coincide up to torsion for codimension $1$ cycles in non-singular projective varieties over $\mathbb C$. Also, the group $Alg\_{\tau}^1(X)/Alg^1(X)$ can be identified with $H^2(X,\mathbb Z)\_{tor}$ (here $Alg\_{\tau}^1(X)$ is the group of cycles who multiple is in $Alg^1(X)$). This is discussed in 19.3.1 of Fulton's book "Intersection Theory". So you just need to show that $H^2(X,\mathbb Z)\_{tor}=0$, i.e it is free abelian. But presumably this follows from the Kunneth formula.
| 7 | https://mathoverflow.net/users/2083 | 15009 | 10,068 |
https://mathoverflow.net/questions/15016 | 4 | So, in $R-Mod$, we have the rather short sequence
* $\mathrm{Ext}^0(A,B)\cong Hom\_R(A,B) $
* $\mathrm{Ext}^1(A,B)\cong \mathrm{ShortExact}(A,B)\mod \equiv $, equivalence classes of "good" factorizations of $0\in Hom\_R(A,B)\cong\mathrm{Ext}^0(A,B)$, with the Baer sum.
Question:
* $\mathrm{Ext}^{2+n}(A,B) \cong\ ??? $
While I suppose one could pose a conjugate question in algebraic topology/geometry, where the answer might look "simpler", I'm asking for a more directly algebraic/diagramatic understanding of the higher $\mathrm{Ext}$ functors. For instance, I'd expect $\mathrm{Ext}^2(A,B)$ to involve diagrams extending the split exact sequence $A\rightarrow A\oplus B\rightarrow B$, but precisely *what sort* of extension? Or is that already completely wrong?
| https://mathoverflow.net/users/1631 | About higher Ext in R-Mod | They correspond to longer exact sequences under an equivalence relation due to Yoneda. See chapter III.3 (p. 82ff) of MacLane's Homology (or briefly on the wikipedia page for the Ext functor). There are also many online sources for "higher extension modules and yoneda", but MacLane's presentation is clear and describes the Baer addition very nicely. Yoneda also describes a product from Ext^n x Ext^m to Ext^(n+m) that can turn certain Ext's into rings. This is popular to do with Ext^\*(k,k) where k is the trivial module for a k-algebra.
| 11 | https://mathoverflow.net/users/3710 | 15017 | 10,074 |
https://mathoverflow.net/questions/14980 | 3 | I've been reading about how hamming codes are used to 'solve' the [Hat Problem](http://en.wikipedia.org/wiki/Hat_puzzle), and I understand how it 'assigns' one person to be the speaker, and how that speaker knows the answer. Everything I read says that it worked $1-\frac1{2^n}$ times, but what I don't understand is why it fails where it does. Apparently there's one case where every person guesses wrong, but I can't figure out which case that is, or (more importantly) why. Anyone care to explain?
| https://mathoverflow.net/users/3944 | Hat Problem/Hamming Codes | So I looked at Loepp and Wootters and it does seem to answer your question. Here's roughly how the argument goes.
**Strategy** During the strategy session prior to the game the players agree on some binary [$n,n-r$] Hamming code *C* and a check matrix *H* for *C*. The players also number themselves 1 through *n*. The agree on the convention that 0 represents blue and 1 represents red and the convention that the vector $\bar{v}\in \mathbb{Z}^{n}\_{2}$ represents the correct description of the way the hats are distributed. So, for example, suppose $n=3$ and the distribution is that the first two people have red and the third has blue. In this case $\bar{v}=(110)$. After the game begins the $i$th person will knows all of the entries of $\bar{v}$ $except$ for the $i$th entry. This person then forms two possible vectors for $\bar{v}$, e.g. $\bar{v}\_{0}$ if that person has on a blue hat and $\bar{v}\_{1}$ if that person has on a red hat. The $i$th person then does the following:
If $H\bar{v}^{T}\_{0} \ne \bar{0}$ and $H\bar{v}^{T}\_{1} \ne \bar{0}$ he/she passes.
If $H\bar{v}^{T}\_{0} = \bar{0}$ and $H\bar{v}^{T}\_{1} \ne \bar{0}$ he/she guesses red.
If $H\bar{v}^{T}\_{0} \ne \bar{0}$ and $H\bar{v}^{T}\_{1} = \bar{0}$ he/she guesses blue.
The case where $H\bar{v}^{T}\_{0} = \bar{0}$ and $H\bar{v}^{T}\_{1} = \bar{0}$ never occurs.
[Note: $\bar{v}^{T}$ is the transpose of $\bar{v}$.]
**Winning and losing** If they use this strategy the team wins whenever $H\bar{v}^{T}\ne 0$ and they lose whenever $H\bar{v}^{T} = 0$. The probability that the team will win turns out to be $\frac{n}{n+1}$.
| 2 | https://mathoverflow.net/users/3639 | 15026 | 10,080 |
https://mathoverflow.net/questions/14992 | 17 | I am now learning localization theory for triangulated catgeory(actually, more general (co)suspend category) in a lecture course. I found Verdier abelianization which is equivalent to universal cohomological functor) is really powerful and useful formalism. The professor assigned many problems concerning the property of localization functor in triangulated category.He strongly suggested us using abelianization functor to do these problems
If we do these problems in triangulated category, we have to work with various axioms TRI to TRIV which are not very easy to deal with. But if we use Verdier abelianization functor, we can turn the whole story to the abelian settings. Triangulated category can be embedded to Frobenius abelian category(projectives and injectives coincide). Triangulated functors become exact functor between abelian categories. Then we can work in abelian category. Then we can easily go back(because objects in triangulated category are just projectives in Frobenius abelian category, we can use restriction functor). In this way, it is much easier to prove something than Verider did in his book.
My question is:
1. What makes me surprised is that Verdier himself even did not use Abelianization in his book to prove something. I do not know why?(Maybe I miss something)
2. I wonder whether there are any non-trivial application of Verdier abelianization functor in algebraic geometry or other fields?
Thank you
| https://mathoverflow.net/users/1851 | Why do people "forget" Verdier abelianization functor?(Looking for application) | The problem with respect to applications of the abelianization is that the abelian categories one produces are almost uniformly horrible. More precisely they are just too big to deal with. So using them to produce results which aren't formal statements about some class of triangulated categories seems like it would be very challenging.
As a concrete(ish) example suppose that $R$ is a discrete valuation ring and let $D(R)$ be the unbounded derived category of $R$-modules. Then the abelianization $A(D(R))$ is not well-copowered - the image of the stalk complex $R$ in degree zero in $A(D(R))$ has a proper class of quotient objects (the reference for this is Appendix C of Neeman's book Triangulated Categories). Since in this case $R$-Mod is hereditary the derived category is really pretty good - we understand the compacts $D^b(R-mod)$ very well and we know all of the localizing and smashing subcategories of $D(R)$. It even comes with a natural tensor product making it rigidly compactly generated and it has a DG-enhancement. On the other hand the abelianization is kind of crazy.
One can take approximations of the abelianization by abelian categories which are more managable. However, I still don't know of any specific applications that aren't just shadows of facts which work for all sufficiently nice triangulated categories.
| 22 | https://mathoverflow.net/users/310 | 15034 | 10,085 |
https://mathoverflow.net/questions/15003 | 33 | Why is it that the vanishing of the integral first Chern class of a compact Kahler manifold is equivalent to the canonical bundle being trivial? I can see that it implies that the canonical bundle must be topologically trivial, but not necessarily holomorphically trivial. Does proving the equivalence require Yau's theorem, in order to produce a flat connection on the canonical bundle, or is there a more elementary proof?
| https://mathoverflow.net/users/3952 | Two definitions of Calabi-Yau manifolds | I have looked for a while for a proof
which does not use the Calabi-Yau theorem
and nobody seems to know it.
Also, there are plenty of non-Kaehler
manifolds with canonical bundle trivial
topologically and non-trivial as a holomorphic
bundle (the Hopf surface is an easiest
example).
The argument actually uses the Calabi-Yau
theorem, Bochner's vanishing, Berger's classification
of holonomy and Bogomolov's decomposition theorem.
From Calabi-Yau theorem you infer that
there exists a Ricci-flat Kaehler metric.
Since the Ricci curvature is a curvature
of the canonical bundle, this implies
that the canonical bundle admits a flat
connection.
Of course, this does not mean that
it is trivial holomorphically; in fact,
the canonical bundle is flat on Hopf surface
and on the Enriques surface, which are
*not* Calabi-Yau.
For Calabi-Yau manifolds, however,
it is known that the Albanese map
is a locally trivial fibration and
and has Calabi-Yau fibers with trivial
first Betti number. This is shown using the
Bochner's vanishing theorem which implies
that all holomorphic 1-forms are parallel.
Now, by adjunction formula, you prove that
the canonical bundle of the total space is
trivial, if it is trivial for the base and the fiber.
The base is a torus, and the fiber is a Calabi-Yau
with $H^1(M)=0$. For the later, triviality
of canonical bundle follows from Bogomolov's
decomposition theorem, because such a
Calabi-Yau manifold is a finite quotient
of a product of simple Calabi-Yau manifolds
and hyperkaehler manifolds having holonomy
$SU(n)$ and $Sp(n)$. Bogomolov's decomposition
is itself a non-trivial result, and (in this generality)
I think it can be only deduced from the Berger's
classification. The original proof of Bogomolov was
elementary, but he assumed holomorphic triviality
of a canonical bundle, which we are trying to prove.
This argument is extremely complicated; also,
it is manifestly useless in non-Kaehler situation
(and in many other interesting situations).
I would be very interested in any attempt
to simplify it.
**Update:** Just as I was writing the reply,
Dmitri has posted a link to Bogomolov's article, where
he proves that some power of a canonical bundle is
always trivial, without using the Calabi-Yau theorem.
| 25 | https://mathoverflow.net/users/3377 | 15048 | 10,093 |
https://mathoverflow.net/questions/15047 | 1 | I have Z^3/M = Z^3/N = Z\_k where M,N are submodules of Z^3 and Z\_k is cyclic order k.
I would like to say some SL\_3(Z) transformation takes M to N. Is this true? How to show?
| https://mathoverflow.net/users/2031 | equivalence of submodules | It is enough to show that
>
> if $M\subseteq \mathbb Z^3$ be a subgroup such that $\mathbb Z^3/M$ is a cyclic group of order $k$, then there exists $g\in\mathrm{SL}(3,\mathbb Z)$ such that $g(M)=\langle (1,0,0),(0,1,0),(0,0,k)\rangle$.
>
>
>
Let $M\subseteq \mathbb Z^3$ be a subgroup such that $\mathbb Z^3/M$ is a cyclic group of order $k$. Then $M$ is free of rank $3$, and there exists $A\in M(3,\mathbb Z)$ such that $M=A\cdot\mathbb Z^3$. Using the Smith normal form, we know that there exists $3\times 3$ matrices $P$ and $Q$, invertible over $\mathbb Z$, such that $PAQ=D$ with $D=\left(\begin{smallmatrix}a\\\&b\\\&&c\end{smallmatrix}\right)$ and $a\mid b\mid c$.
Then $PM=PAQ\mathbb Z^3=D\mathbb Z^3$.
It follows that $P\in\mathrm{SL}(3,\mathbb Z)$ is such that $PM$ is generated by $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$ with $a\mid b\mid c$. Since $\mathbb Z^3/g(M)$ is cyclic of order $k$, we must have $a=b=1$ and $c=k$. This tells us that the claim above is true.
(I've done everything at the level of generality which your problem needs, and I'll leave the fun of finding the correct general statement for you...)
| 1 | https://mathoverflow.net/users/1409 | 15055 | 10,099 |
https://mathoverflow.net/questions/15041 | 15 | I am looking for a matrix $C$ so that the sequence $tr(C^n)$ is dense in the set of real numbers. Equivalently (in the $2 \times 2$ case), find a complex number $z$ so that the sequence $z^n+w^n$ is dense in $\mathbb{R}$ where $w$ is the conjugate of $z$.
| https://mathoverflow.net/users/3960 | Is there a matrix C so that the trace of C^n is dense in R? | The answer is yes, even in the $2 \times 2$ case. Let $q\_1,q\_2,\ldots$ be an enumeration of the rational numbers. Let $Q\_j$ be the closed interval $[q\_j-1/j,q\_j+1/j]$. Let $I\_0=[0,2\pi]$. Let $z=2e^{i \theta}$ for a $\theta \in I\_0$ to be determined.
By induction, we construct positive integers $n\_1 < n\_2 < \ldots$ and closed intervals $I\_0 \supseteq I\_1 \supseteq \cdots$ such that for each $j$, the trace $z^{n\_j} + \bar{z}^{n\_j}$ is in $Q\_j$ whenever $\theta \in I\_j$. Namely, if $n\_1,\ldots,n\_{j-1},I\_1,\ldots,I\_{j-1}$ have been determined already, then for any sufficiently large $n\_j$, the set of $\theta$ such that $z^{n\_j} + \bar{z}^{n\_j}$ is in $Q\_j$ is a union of closed intervals such that every real number is within $2\pi/n\_j$ of a point inside this union and within $2\pi/n\_j$ of a point outside this union, so if $n\_j$ is chosen large enough, one such interval in this union will be completely contained in $I\_{j-1}$ and we name it $I\_j$.
The intersection of a descending chain of closed intervals is nonempty, so we can choose $\theta$ such that $\theta \in I\_j$ for all $j$. Then $\lbrace z^n+\bar{z}^n : n \ge 1 \rbrace$ contains an element of $Q\_j$ for each $j$, so it is dense in $\mathbb{R}$.
| 21 | https://mathoverflow.net/users/2757 | 15056 | 10,100 |
https://mathoverflow.net/questions/15058 | 8 | How can one construct families of cocompact discrete subgroups of the topological group $\text{SL}\_2(\mathbb{C})$?
Here quaternion algebra's might help, I believe, but I have some difficulties with the construction.
| https://mathoverflow.net/users/1107 | cocompact discrete subgroups of SL_2 | Let $k$ be a number field with one complex place and let $B$ be a quaternion algebra defined over $k$ which ramifies at every real place of $k$ (and perhaps some finite places as well - Hilbert reciprocity implies that the total number of ramified places must be even). Let $\mathcal{O}$ be an order of $B$ and $\mathcal{O}^1$ its elements of reduced norm $1$. Choosing an embedding $k\hookrightarrow \mathbb{C}$ induces an embedding $B\hookrightarrow M\_2(\mathbb{C})$, which in turn restricts to an embedding $\Psi: \mathcal{O}^1\hookrightarrow SL\_2(\mathbb{C})$. A group which is commensurable with $\Psi(\mathcal{O}^1)$ is called *arithmetic*. An arithmetic group is cocompact if and only if the quaternion algebra $B$ is a division algebra (by Wedderburn's Theorem this is equivalent to saying that the set of places of $k$ which ramify in $B$ is nonempty).
| 10 | https://mathoverflow.net/users/nan | 15062 | 10,104 |
https://mathoverflow.net/questions/15063 | 5 | Let $F:R \to S$ be an étale morphism of rings. It follows with some work that $f$ is flat.
However, faithful flatness is another story. It's not hard to show that faithful + flat is weaker than being faithfully flat. An equivalent condition to being faithfully flat is being surjective on spectra.
The question:
Is there any further condition we can require on an étale morphism that implies faithful flatness?
"Faithfully flat implies faithfully flat" or "surjective on spectra is equivalent to faithfully flat" do not count. The answer should in some way use the fact that the morphism is étale (or at least flat).
As you can see by the tag, all rings commutative, unital, etc.
Edit: [Why](http://mathworld.wolfram.com/FaithfullyFlatModule.html) faithfully flat is weaker than faithful + flat.
Edit 2: I resent the voting down of this question without accompanying comments as well as the voting up of the glib and unhelpful answer below. It's clear that some of you are in the habit of voting on posts based on the poster rather than the content, and I think that is shameful. There is nothing I can do because none of you has the basic decency to at least leave a comment. I am completely at your mercy. You've won. I hope it's made you very happy.
Edit 3: To answer Emerton's comment, I asked here after:
a.) Reading this [post](http://sbseminar.wordpress.com/2009/08/06/algebraic-geometry-without-prime-ideals/#comment-6354) by Jim Borger
b.) Asking my commutative algebra professor in an e-mail
Which led me to believe (perhaps due to a flawed reading of said sources) that this was a harder question than it turned out to be.
| https://mathoverflow.net/users/1353 | Weakened conditions for étale + X implies faithfully flat. | The definition of $f:R\to S$ being faithfully flat that I first saw is that $S\otimes\_R-$ is exact and faithful (meaning that $S\otimes\_R M=0$ implies $M=0$). I'm not sure exactly what your definition of "faithfully flat" is, but it looks like you're happy with "flat and surjective on spectra." You get flatness for free from étaleness, so I'll show that the extra faithfulness condition implies surjectivity on spectra.
Upon tensoring with $S$, $f$ becomes $f\otimes\_R id\_S:S\cong S\otimes\_R R\to S\otimes\_R S$, given by $s\mapsto s\otimes 1$. This is injective since it is a section of multiplication $S\otimes\_R S\to S$. By flatness of $S$, this shows that $S\otimes\_R \ker f=0$, so $\ker f=0$. So I'll identify $R$ with a subring of $S$.
Let $\mathfrak p\subseteq R$ be a prime ideal. We wish to show that there is a prime $\mathfrak q\subseteq S$ such that $\mathfrak q \cap R=\mathfrak p$. Let $K$ be the kernel of the morphism $R/\mathfrak p\to S/\mathfrak p S$ of $R$-modules. Upon tensoring with $S$, this morphism becomes injective (as before, it's a section of the multiplication map $S/\mathfrak p S\otimes\_R S/\mathfrak p S\to S/\mathfrak p S$), so by flatness of $S$, we have $S\otimes\_R K=0$, so $K=0$. This shows that $\mathfrak p S \cap R=\mathfrak p$ (if the intersection were any larger, $K$ would be non-zero). So $\mathfrak p$ generates a proper ideal in the localization $(R\setminus \mathfrak p)^{-1}S$. Let $\mathfrak q\subseteq (R\setminus \mathfrak p)^{-1}S$ be a maximal ideal containing $\mathfrak p$. This corresponds to some prime ideal $\mathfrak q$ (slight abuse of notation to use the same letter) of $S$ which contains $\mathfrak p$ but does not intersect $R\setminus \mathfrak p$, so $\mathfrak q\cap R=\mathfrak p$.
See also exercise 16 of Chapter 3 of Atiyah-Macdonald.
| 7 | https://mathoverflow.net/users/1 | 15069 | 10,107 |
https://mathoverflow.net/questions/15068 | 31 | I've just finished reading Ash and Gross's *Fearless Symmetry*, a wonderful little pop mathematics book on, among other things, Galois representations. The book made clear a very interesting perspective that I wasn't aware of before: that a large chunk of number theory can be thought of as a quest to understand $G = \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. For example, part of the reason to study elliptic curves is to describe two-dimensional representations of $G$, and reciprocity laws are secretly about ways to describe the traces of Frobenius elements in various representations. (That's awesome! Why didn't anybody tell me that before?)
Are there number theory textbooks (presumably not introductory, but hopefully not too sophisticated either) which explicitly take this as a guiding principle? I think this is a great idea to organize things like quadratic reciprocity around and I'm wondering if anybody has decided to actually do that at the undergraduate (or introductory graduate, maybe) level.
**Edit:** In response to some comments and at least one downvote, most of the other questions on MO about the absolute Galois group that I can find are about the state of the art, and the references in them seem fairly sophisticated. But it seems to me there are still interesting things to say along the lines of *Fearless Symmetry*, but directed to an undergraduate or introductory graduate-level audience as a kind of "second course in number theory." I'm imagining a textbook like Serre's *Course in Arithmetic.*
| https://mathoverflow.net/users/290 | Number theory textbook based on the absolute Galois group? | If you want to go further in understanding this point of view, I would advise you to begin learning class field theory. It is a deep subject, it can be understood in a vast variety of ways, from the very concrete and elementary to the very abstract, and although superficially it appears to be limited to describing abelian reciprocity laws, it in fact plays a crucial role in the study of non-abelian reciprocity laws as well.
The texts:
Ireland and Rosen for basic algebraic number theory, a Galois-theoretic proof
of quadratic reciprocity, and other assorted attractions.
Cox's book on primes of the form x^2 + n y^2 for an indication of what some of the content of class field theory is in elementary terms, via many wonderful examples.
Serre's Local Fields for learning the Galois theory of local fields
Cassels and Frolich for learning global class field theory
The standard book at the graduate level to learn the arithmetic of non-abelian (at least 2-dimensional) reciprocity laws is Modular forms and Fermat's Last Theorem, a textbook on the proof by Taylor and Wiles of FLT. But it is at a higher level again.
I don't think that you will find a single text on this topic at a basic level (if basic
means Course in arithmetic or Ireland and Rosen), because there is not much to say beyond what you stated in your question without getting into the theory of elliptic curves and/or the theory of modular forms and/or a serious discussion of class field theory.
Also, as basic suggests, you could talk to the grad students in your town, if not at your institution, then at the other one down the Charles river, which as you probably know is currently the world centre for research on non-abelian reciprocity laws (maybe shared with Paris). Certainly there are grad courses offered on this topic there on a regular basis.
| 19 | https://mathoverflow.net/users/2874 | 15071 | 10,108 |
https://mathoverflow.net/questions/15075 | 5 | Is there any 'general' topological invariant to tell the difference between $M$ and $N$, where $M$ has homotopy type of a closed manifold and $N$ has homotopy type of a manifold with boundary.
I meant something like homotopy group/ homology group can 'detect' the obstruction of being homotopic to a closed manifold.
| https://mathoverflow.net/users/3922 | Topological description of Manifold with boundary | If $M$ is a closed manifold, then you can always cross it with an interval to get a
manifold with non-empty boundary that is homotopic to $M$. So the distinction between the two possibilities (closed or non-closed) is not so well-defined in the homotopy category.
On the other hand, suppose that we fix the dimension $n$ of the manifolds we want to consider.
Then,
the most basic difference between closed manifolds and those with boundary is the top dimensional homology, namely: a closed orientable connected manifold of dimension $n$ has $H\_n(M) = {\mathbb Z}$,
while on the other hand, if $N$ has non-empty boundary, then $H\_n(N) = 0$.
| 24 | https://mathoverflow.net/users/2874 | 15080 | 10,112 |
https://mathoverflow.net/questions/15083 | 3 | One example is the Hurewicz theorem which tells us that (e.g) a CW-cx with only one 0-cell has a nontrivial fundamental group if H\_1 is nontrivial. What other examples are there? (The CW-complexes I have in mind always have exactly one 0-cell, but for the sake of a more wide discussion one could assume this is not generally true. However, I am mostly interested in the higher homotopy groups as I already know everything about the fundamental group.)
| https://mathoverflow.net/users/1161 | What can be said about the homotopy groups of a CW-complex in terms of its (co)homology? | Try looking up some references on rational homotopy theory. Rational homotopy theory studies the homotopy groups tensor Q, so basically you kill all torsion information. If we focus only on homotopy groups tensor Q, the question you ask becomes easier. As Steven Sam mentions in the comments, the homotopy groups of spheres are really crazy. But the rational homotopy groups of spheres are quite tractable (in fact completely known, by a theorem of Serre) and can be more or less obtained from cohomology, if I recall correctly.
One particularly impressive theorem, of Deligne-Griffiths-Morgan-Sullivan, says that if your space is a compact Kähler manifold (e.g. a smooth complex projective variety), and if you know its rational cohomology ring, then you can compute for instance the ranks of all of its homotopy groups (maybe you need an extra assumption that the space is simply connected or has nilpotent fundamental group).
| 4 | https://mathoverflow.net/users/83 | 15086 | 10,116 |
https://mathoverflow.net/questions/15070 | 1 | This question has been bugging me for quite some time now.
Say we have some $\beta$ smaller than some $\gamma$ and a sequence
$\beta$$\epsilon$ : $\epsilon$ smaller than cf($\beta$) cofinal in $\beta$ and say
we have some sets $A$n$\epsilon$ and each of these $A$n$\epsilon$ has order type less than $\gamma$$n$.
Now $\forall n$ in $\omega$ let $B$n= $\cup$ $A$n$\epsilon$ for all $\epsilon$ < $\gamma$ and suppose in the end I can write $\beta$ as the union of all the $B$n (but that is not really my problem here)
Why can I deduce that $B$n has order type less than $\gamma$$n+1$ ***only if*** all my sets $A$n$\epsilon$ are disjoint and do not overlap?
(since we have a union of less then $\gamma$ sets each of which is of order type less than $\gamma$$n$)
Why can't I still guarantee that the $B$n will still have order type $\gamma$$n+1$ if all the $A$n$\epsilon$ are not disjoint?
I know that I need to take the $A$n$\epsilon$ to be [$\epsilon$,$\epsilon+1$) so that they are disjoint.
But why does everything in the union have to be in order?
I hope I conveyed my question clearly. Thanks in advance for any help.
| https://mathoverflow.net/users/3859 | Why is it important to have disjoint sets in a union for the union to make sense w.r.t the order types? | You didn't say so, but you are speaking of ordinals here and ordinal exponentiation.
The problem is that if you take the union of sets not in order, then you can't control the order type. Let me give an easy example to illustrate the point. Suppose that An for each n < ω consists of a single ordinal. If the ordinal of An is below the ordinal of Am whenever n < m, then the union set Un An will clearly have order type ω. But if I drop that requirement, then I can get any countable ordinal at all! That is, every set is the uion of its singletons. In particular, if α is a countable ordinal, then α = U { {β} | β < α} is the union of countably many one point orders. Indeed, if you go to the non-ordinal context, then every countable order type, such as Q, is the union of countably many singletons.
A similar problem arises in your specific question. You can't guarrantee that the union of the Anε has small order type, if you allow them to get mixed up all together.
| 2 | https://mathoverflow.net/users/1946 | 15091 | 10,120 |
https://mathoverflow.net/questions/14995 | 16 | Let $G$ be a group and $CG$ the complex group algebra over the field $C$ of complex number. The group von Neumann algebra $NG$ is the completion of $CG$ wrt weak operator norm in $B(l^2(G))$, the set of all bounded linear operators on Hilbert space $l^2(G)$. Let $f:G \to H$ be any homomorphism of groups. My question is: is there a homomorphism of the group von Neumann algebra $NG \to NH$ induced from $f$?.
If $NG$ is replaced with $CG$, it's obvious true. If $NG$ is replaced with $C^\ast\_r(G)$, the reduced group $C^\ast$ algebra, it's not necessary true.
| https://mathoverflow.net/users/1546 | Is the group von Neumann algebra construction functorial? | By the way, here's the "correct" functorial property. If G and H are abelian, and $f:G\rightarrow H$ is a continuous group homomorphism, then we get a continuous group homomorphism $\hat f:\hat H\rightarrow \hat G$ between the dual groups. By the pull-back, we get a \*-homomorphism $\hat f\_\*:C\_0(\hat G) \rightarrow C^b(\hat H)$. We should think of $C^b(\hat H)$ as the multiplier algebra of $C\_0(\hat G)$. Then $C\_0(\hat G) \cong C^\*\_r(G)$, and so we *do* get a \*-homomorphism $C^\*\_r(G) \rightarrow M(C^\*\_r(H))$; the strict-continuity extension of this is a \*-homomorphism $M(C^\*\_r(G)) \rightarrow M(C^\*\_r(H))$ which does indeed send $\lambda(s)$ to $\lambda(f(s))$.
For non-abelian group (in fact, non-amenable groups) it's necessary to work with $C^\*(G)$ instead.
We cannot ensure a map to $C^\*\_r(H)$ itself, as we cannot ensure a map from $C\_0(\hat G)$ to $C\_0(\hat H)$; indeed, this would only happen when $\hat f$ were a [proper map](http://en.wikipedia.org/wiki/Proper_map).
Similarly, we don't get maps at the von Neumann algebra level, as we don't get a map $L^\infty(\hat G) \rightarrow L^\infty(\hat H)$: we would need that $\hat f$ pulled-back null sets in $\hat G$ to null sets in $\hat H$.
| 8 | https://mathoverflow.net/users/406 | 15093 | 10,122 |
https://mathoverflow.net/questions/15094 | 12 | Do You know any **kind of database** of **presentations of groups**?
It may be on-line or off-line in form of tables, ideally case would be integrated in some Computer Algebra System. I am interested the most in infinite group presentations, but feel free to put here also information about tables of presentations of finite groups.
Maybe this thread should be wiki-type because probably there is many good answers to this question, and it is hard to compare for example software system with some kind of book or publication about this matter? I add biglist tag in a hope that it would appear;-)
| https://mathoverflow.net/users/3811 | Database of finite presentations of used groups | [GAP](http://www.gap-system.org) has the following:
* [Finitely Presented Groups](http://www.gap-system.org/Manuals/doc/ref/chap47.html)
* [Presentations and Tietze Transformations](http://www.gap-system.org/Manuals/doc/ref/chap48.html)
An overview of all data libraries in GAP can be found [here](http://www.gap-system.org/Datalib/datalib.html).
The following was a comment, but should be part of the answer:
* There was a lovely book that listed lots of this sort of material
presentations, Group Tables by A. D. Thomas, G. V. Wood. It may not
answer your question but if you can get your hands on a copy is
an interesting perspective.
It was published in the Shiva Mathematics series.
| 7 | https://mathoverflow.net/users/3502 | 15095 | 10,123 |
https://mathoverflow.net/questions/14565 | 8 | In various contexts, I have come across results referred to as "big monodromy." A standard arithmetic example is the open image theorem for the image of Galois action on non-CM elliptic curves. A general setup for such a result in algebraic geometry is:
Given a proper, generically smooth map $\pi:X \rightarrow S$ of relative dimension d, say S is connected. This gives rise to an $l$-adic representations of the etale fundamental group $\pi\_1(U)$ where $U$ is smooth locus of $\pi$ corresponding to higher pushforward $R^d \pi\_\* Q\_l$. One might say it has "big monodromy" if the Zariski closure of the image is as big as it can be given that it has to respect cup-product, etc.
My specific question is what are the geometric consequences of big monodromy? If we know such a result for $\pi$, what does that say about the geometry of the fibration or at the very least is there geometric intuition for what it should mean?
I welcome intuition from number theory, algebraic geometry, or complex geometry.
I have also heard that "one should expect big monodromy unless there is a reason not to" (for example, complex multiplication). What are other examples of things which inhibit big monodromy?
| https://mathoverflow.net/users/81 | Geometric Intuition for Big Monodromy | I found this question a little vague, but let me at least remark on "other examples of things which inhibit big monodromy." Mumford gives an example in section 4 of
D. Mumford, “A note of Shimura’s paper “Discontinuous groups and abelian varieties”,” Math. Ann. 181 (1969), 345–351.
of an abelian variety A whose Galois representation has image strictly smaller than Sp\_{2g}(Z\_p), despite the fact that End(A) = Z. The keyword to look up is "Mumford-Tate group", which is in some sense the answer to the question
How big COULD the Galois representation on an abelian variety be, subject to all geometric 'things which inhibit big monodromy'?
Reference comes from [a paper of Chris Hall](http://w3.uwyo.edu/~chall14/papers/open_image.pdf) which shows how to prove big monodromy results in many cases.
| 3 | https://mathoverflow.net/users/431 | 15102 | 10,128 |
https://mathoverflow.net/questions/15107 | 2 | In Bourbaki an *algebra* over a commutative ring $k$ is defined to be a $k$-module $A$ together with a $k$-bilinear map $A \times A \rightarrow A$. We then have the obvious notion of morphisms of $k$-algebras. This terminology is nice, because e.g. Lie algebras are then a special kind of algebras and so on. But in 95% of my work I am using unital associative $k$-algebras with unital morphisms. Now, is there any better (in particular shorter) terminology available to distinguish these two cases? I don't want to add this "unital associative" and "unital morphisms" all the time. Is perhaps something like *prealgebra* for the first case or another short word used in the literature?
| https://mathoverflow.net/users/717 | Algebra / unital associative algebra: better terminology? | If your work requires you to work mostly with unital assosiative $k$-algebras with unital morphisms, define somewhere prominent in your work 'algebra' and 'morphism of algebras' to mean precisely that, and say 'Lie algebras', 'not necessarily associative algebra', 'possibly non unital morphism of algebra', and so on in the meagre 5% of the remaining cases.
| 16 | https://mathoverflow.net/users/1409 | 15108 | 10,131 |
https://mathoverflow.net/questions/15106 | 11 | I am in this, rather friendly, situation:
I have a complex projective space $\mathbb{P}^n$, and there i have a (possibly non-smooth) hypersurface $S$ defined by one irreducible polynomial $P$ of degree $d$.
What i want is to get information about the existence or not of linear subvarieties of $S$, and their maximal dimension $m$.
I seem to remember the existence of some ways to get bounds on $m$, given $n$ and $d$, but i don't remember anymore and i don't know where to look..
| https://mathoverflow.net/users/3465 | Can projective hypersurfaces contain linear spaces? How big? | I think the key word you are looking for is Fano schemes of $S$. See this
[note](http://www.math.sunysb.edu/~jstarr/papers/appendix3.pdf) by Jason Starr for a reference. For example, if $S$ is smooth and non-degenerate, it can not contain a linear subspace of dimension bigger than half $dim \ S$.
A very interesting related question is when can you find subvarieties of $S$ which was spanned in the Chow group by images of linear subspaces in $\mathbb P^n$. It was discussed in [this paper](http://www.math.neu.edu/%7Elevine/publ/SmallDegree.pdf) by Levine, Esnault and Viehweg.
| 14 | https://mathoverflow.net/users/2083 | 15111 | 10,134 |
https://mathoverflow.net/questions/15121 | 6 | Let $G$ be a graph group and let $S$ be a finitely generated subgroup of $S$.
What torsion can $H\_1(S)$ have?
Let me put this in context: Let $Y$ be a graph, then the corresponding graph group has a generator for each vertex and for each edge we add the relation that the corresponding generators commute. Such groups are also known as right angled Artin groups.
If $Y$ has no edges the graph group is the free group, if $Y$ is a complete graph, then the graph group is a free abelian group.
It is known by work of Crisp and Wiest that non-orientable surface groups embed in graph groups. Such surface groups can have 2-torsion in their abelianization. Is that the only type of torsion which can appear in the abelianization of the subgroup of a graph group?
My interest in this question comes from the recent announcement by Dani Wise that `most' hyperbolic 3-manifold groups embed in a graph group.
| https://mathoverflow.net/users/2985 | subgroups of graph groups | This isn't a proper answer, but here are some remarks.
* Wise's work actually purports to tell you that most (all?) hyperbolic 3-manifold groups *virtually* embed in graph groups - ie, a finite-index subgroup embeds. So you might lose a lot of torsion when you pass to this finite-index subgroup.
* [Haglund and Wise](https://doi.org/10.1007/s00039-007-0629-4) have a very nice criterion for injecting fundamental groups of cube complexes into graph groups. It would be completely reasonable to try to use this to construct interesting examples of torsion in the homology of subgroups of graph groups.
* Because the embeddings in Haglund--Wise are *quasiconvex*, it follows from work of Haglund that every such subgroup H is a *virtual retract* - that is, there is a finite-index subgroup K that contains H and the inclusion map has a left inverse K->H. In particular, any torsion you see in the homology of H will also show up in the homology of K. It would be easy to use a computer algebra package like GAP to look for torsion in finite-index subgroups of graph groups.
UPDATE:
In a comment below, Stefan points out a very nice way of constructing torsion in finite-index subgroups of graph groups, due to Bridson and Miller. As I'm familiar with the argument, I'll explain it. The graph group in question will always be a direct product of two free groups.
Let Q be any group you like and let q:F->Q be a surjection from a free group. Now take the fibre product K of q with itself; in other words, consider the subgroup
$K= \{(g,h)\in F\times F\mid q(g)=q(h)\}~.$
There is a standard five-term exact sequence in homology that derives from the map $q:F\to Q$, which reduces to
$0\to H\_2(Q)\to H\_0(Q,H\_1(\ker f))\to H\_1(F)\to H\_1(Q)$
where, by definition, $H\_0(Q,H\_1(\ker f))$ is the quotient of the abelianisation of $\ker f$ by the natural action of $Q$ by conjugation. On the other hand, projecting onto a factor decomposes $K$ as $\ker f\rtimes F$ (where the action is, again, by conjugation), from which it follows that
$H\_1(K) = H\_0(Q,H\_1(\ker f))\oplus H\_1(F)~.$
Putting these together, we see that $H\_2(Q)$ embeds into $H\_1(K)$.
The fibre product $K$ can also be characterised as the preimage of the diagonal subgroup of $Q\times Q$ in $F\times F$, so if $Q$ is finite then $K$ is of finite index in $F\times F$. This proves the following.
**Proposition.** For any finite group $Q$ there is a finite-index subgroup $K$ of $F\times F$ with $H\_2(Q)\subseteq H\_1(K)$.
This shows that all sorts of torsion can arise in $H\_1$ of subgroups of graph groups.
**Remark.** We took $Q$ to be finite because otherwise $K$ isn't quasiconvex. But the rest of the argument goes through.
| 7 | https://mathoverflow.net/users/1463 | 15126 | 10,141 |
https://mathoverflow.net/questions/15097 | 6 | I was wondering, what a $N$-dimensional simplicial space $X$ should be. Of course the degeneracy maps force the spaces to be nonempty in high dimensions. Currently I have two different versions and i am wondering whether they are equivalent:
1) There are no nondegenerate simplexes above dimension $N$.
2) Let $\Delta|\_N$ be the full subcategory of $\Delta$ consisting of the objects $[0],\ldots ,[N]$. The inclusion induces a restriction functor $R$ from $\Delta-$spaces to $\Delta|\_N$-spaces, which has a left adjoint $L$. There is a canonical map $L(R(X))\rightarrow X$. $X$ should be called $N$-dimensional, iff this map is an isomorphism,i.e. a homeomorphism on each object of $\Delta$.
To make the question more precise: $2)\Rightarrow 1)$ can be easily seen, as the map $L(R(X))\rightarrow X$ cannot hit any nondegenerate simplex above dimension $N$ by construction. The other way round:
Given $1)$, then all the maps occuring in the natural transformation are surjective and continuous. Injectivity should follow from the relations in $\Delta$. So why is the inverse map of sets continuous?
(In the category of simplicial sets (and not spaces) both notions should be equivalent using the same argumentation.)
| https://mathoverflow.net/users/3969 | Notion of finite dimensional simplicial space | Interesting question!
I'm cautiously optimistic that 1) $\Rightarrow$ 2). As you say, if $X$ satisfies 1), then $L(R(X))\to X$ is a continuous bijection. (Because it's a statement about point-sets, and the thing is true for simplicial sets.)
If $N=0$, it should be easy: $L(R(X))$ is the constant simplicial space, whose value at each [k] is $X\_0=X([0])$. The canonical map $L(R(X))\to X$ is the one which at degree $k$ is given by the map $s:X\_0\to X\_k$ defined by the composite of degeneracy operators. But we know that the composite $X\_0\to X\_k\to X\_0$ is the identity, where $d:X\_k\to X\_0$ is a composite of face operators. So if $s$ is a bijection, $d$ is its continuous inverse.
For general $N$, let $Y=L(R(X))$. The functor $R$ also has a right adjoint, which I'll call $M$. Let $Z=M(R(X))$. Just as the space $Y\_k$ looks like a colimit of a certain diagram of the spaces $X\_0,\dots,X\_N$, the space $Z\_k$ looks like a limit of a certain diagram of these spaces.
There are canonical maps $Y=L(R(X)) \to X \to M(R(X))=Z$. I would like to claim that the composite $f:Y\to Z$ gives a homeomorphism of $Y$ onto its image. If you can prove this, that will give your result, since a continuous inverse to $Y\to X$ will be given by $X\to f(Y)\approx Y$.
| 3 | https://mathoverflow.net/users/437 | 15135 | 10,147 |
https://mathoverflow.net/questions/14492 | 20 | Consider the series
$$ S\_f = \sum\_{x=1}^\infty \frac{f}{x^2+fx}. $$
Goldbach showed that, for integers $f \ge 1$,
$$ S\_f = 1 + \frac12 + \frac13 + \ldots + \frac1f $$
(this follows easily by writing $S\_f$ as a telescoping series).
Thus $S\_f$ is rational for all natural numbers $f \ge 1$.
Goldbach claimed that, for all nonintegral (rational) numbers $f$,
the sum $S\_f$ would be irrational.
Euler showed, by using the substitution
$$ \frac1k = \int\_0^1 x^{k-1} dx, $$
that
$$ S\_f = \int\_0^1 \frac{1-x^f}{1-x} dx. $$
He evaluated this integral for $f = \frac12$ and found
that $S\_{1/2} = 2(1 - \ln 2)$ (this also follows easily from
Goldbach's series for $S\_f$). Thus Goldbach's claim holds for all
$f \equiv \frac12 \bmod 1$ since $S\_{f+1} = S\_f + \frac1{f+1}$.
Here are my questions:
1. The irrationality of $\ln 2$ was established by Lambert, who
proved that $e^r$ is irrational for all rational numbers
$r \ne 0$. Are there any (simple) direct proofs?
2. Has Goldbach's claim about the irrationality of $S\_f$ for
nonintegral rational values of $f$ been settled in other cases?
| https://mathoverflow.net/users/3503 | Irrational logs and the harmonic series | Please allow me to put my question on top of the list again by turning my comment into an answer. FC's remarks led me to the article "Transcendental values of the digamma function", J. Number Theory 125, No. 2, 298-318 (2007) by Ram Murty and N. Saradha, where Thm. 9 states that the values of S\_f are transcendental for rational numbers 0 < f < 1. I apologize for not having asked this question in 2006, which is why I have only a bounty to offer (and a reference to FC from MO in Euler's OO).
| 5 | https://mathoverflow.net/users/3503 | 15140 | 10,149 |
https://mathoverflow.net/questions/15139 | 4 | Suppose
* C and D are two ∞-categories (quasi-categories),
* $F : C \to D$ and $G : C \to D$ are two functors (i.e. 0-simplices in the ∞-category of functors Fun(C,D), which is just the simplicial mapping complex),
* $a : F \to G$ and $b : F \to G$ are two natural transformations (i.e. 1-simplices in Fun(C,D)),
* at each object (0-simplex) x of C, $a\_x : F(x) \to G(x)$ and $b\_x : F(x) \to G(x)$ are equivalent.
Does it follow that a and b are equivalent in Fun(C,D)?
| https://mathoverflow.net/users/1100 | When are two natural transformations of infinity-categories equivalent? | No, not necessarily, since a transformation involves "higher structure" in addition to its 1-cell components. For example, let C be the "walking arrow" category with two objects 0 and 1 and one nonidentity morphism from 0 to 1, and let D be an abelian group regarded as a (2,1)-category with one object and one morphism, and thereby as an (∞,1)-category. There is exactly one functor from C to D. An endo-natural-transformation of this functor consists of giving, for each object of C, a morphism in D, and for each morphism in C, a 2-cell in D, subject to certain axioms.
There is only one morphism in D, so any two such transformations will have equal 1-cell components, but their 2-cells might be different. The axioms say that the 2-cell corresponding to an identity morphism is the identity, and that the 2-cell corresponding to a composite is the composite of the 2-cells corresponding to the individual factors; thus for the C and D above, to give a transformation is exactly to give an arbitrary element of the abelian group D (the 2-cell component corresponding to the single nonidentity arrow of C).
Now two transformations are equivalent when there is a modification between them, which consists of for each object of C, a 2-cell in D between the corresponding 1-cell components, and for each morphism of C, a 3-cell relating the 2-cell components etc. Since 3-cells in D are all identities, this latter means that there is a diagram that must commute. Since C has one object, a modification thus consists of giving a single element of the abelian group D, and the diagram which must commute means that it *conjugates* the element corresponding to the first transformation to the element corresponding to the second one. Thus, the equivalence classes of such transformations (all of which have the same 1-cell component) are the conjugacy classes of D.
| 7 | https://mathoverflow.net/users/49 | 15146 | 10,153 |
https://mathoverflow.net/questions/13292 | 15 | Let $\Omega$ be a Banach space; for the sake of this post, we will take $\Omega = {\mathbb R}^2$, but I am more interested in the infinite dimensional setting. Take $\mathcal F$ to be the Borel $\sigma$-algebra, and let $\mathbb P$ be a probability measure on $(\Omega, \mathcal F)$.
Denote by $\vec x = (x,y)$ a point in $\Omega$, and let $\mathcal F\_1$ be the $\sigma$-algebra generated by the first coordinate. Fix $y\_0 \in \mathbb R$ and
$\eta > 0$, and consider $$f(\vec x) = \mathbb P( ~|y - y\_0| \le \eta~ |\mathcal F\_1).$$ (More generally, one can consider $f(\vec x) = \mathbb E( \varphi(\vec x) | \mathcal F\_1)$ for some suitable $\varphi : {\mathbb R}^2 \to \mathbb R$.)
The function $f$ is measurable; that comes from the definition of conditional expectations. I would like to find some reasonable sufficient conditions such that $f$ is continuous and positive.
I feel like this should be relatively elementary material, but unfortunately I'm having trouble finding any references. How should I approach this?
I've included the [fa.functional-analysis] tag because in general I want to consider $\Omega$ to be a space of smooth functions. I'm guessing that to give some additional structure, I'll need to assume that $\mathbb P$ is absolutely continuous with respect to a Gaussian measure, because I don't know any other reasonable measures on function space.
| https://mathoverflow.net/users/238 | Conditional probabilities are measurable functions - when are they continuous? | Since a troll bumped this question to the front page, I might as well answer it. The technology which provides the solution is called [regular conditional probability](http://en.wikipedia.org/wiki/Regular_conditional_probability) or [disintegration](http://en.wikipedia.org/wiki/Disintegration_theorem).
| 3 | https://mathoverflow.net/users/238 | 15149 | 10,154 |
https://mathoverflow.net/questions/15141 | 3 |
>
> **Possible Duplicate:**
>
> [Is there a high-concept explanation for why characteristic 2 is special?](https://mathoverflow.net/questions/915/is-there-a-high-concept-explanation-for-why-characteristic-2-is-special)
>
>
>
There are so many results on primes that either fail for $p=2$ or are not known to be true for $p=2.$ Can anyone give some kind of intuition as to why this is the case?
| https://mathoverflow.net/users/3186 | Why is 2 so odd? | My take on this issue is that p=2 isn't really strange---all small primes are strange, it's just that the smaller you are, the earlier you become troublesome. Look at recent R=T results in the theory of automorphic representations. Nowadays people can prove these sorts of things for $n$-dimensional representations, but they need to assume $p>n+1$ or some such thing. The thing about $p=2$ is that it's so small that it's already causing problems when one is considering $GL(1)$, which is an abelian situation. Now abelian situations are so much easier to understand than the general situation that they are more prevalent in the literature. For example things like quadratic reciprocity can be viewed of as some consequence of class field theory, which is really 1-dimensional representations of Galois groups, and already $p=2$ is causing a problem. Similarly Fontaine's results on commutative group schemes of $p$-power order runs into some trouble when $p=2$ (his basic linear algebra data doesn't give you an equivalence between finite flat group schemes over ${\mathbf Z}\_p$ and "easy semilinear algebra" when $p=2$) and again it's because 2 is just too small. But as people formulate higher-dimensional analogues of these things, they will no doubt have to rule out more primes. So it's not that 2 is behaving badly, it's just that from 2's point of view the theory is more advanced, so you have to deal with more special cases.
| 17 | https://mathoverflow.net/users/1384 | 15150 | 10,155 |
https://mathoverflow.net/questions/15148 | 2 | Suppose that $F$ is a nonarchimedean local field, $G\_1$ and $G\_2$ are connected (linear) algebraic groups over $F$, and $\phi:G\_1\to G\_2$ is a surjective homomorphism of algebraic groups. Suppose $H$ is a hyperspecial maximal compact subgroup of $G\_1$. Is the image $\phi(H)$ necessarily a hyperspecial maximal compact subgroup of $G\_2$?
| https://mathoverflow.net/users/3513 | Image of a hyperspecial subgroup hyperspecial? | If by "surjective" you mean surjective in the usual sense (for example on $\overline{F}$-points) then maybe you have a problem, because $G\_1(F)$ may not surject onto $G\_2(F)$. So for example $SL(2)$ surjects onto $PGL(2)$ but if $R$ is the integers of $F$ then $SL(2,R)$ is hyperspecial max compact but its image in $PGL(2,F)$ isn't (it's not even maximal, as $PGL(2,R)$ strictly contains the image of $SL(2,R)$).
However if $G\_1\to G\_2$ is, say, a $z$-extension, then (by definition) the kernel is central in $G\_1$ and has no $H^1$, so the long exact sequence shows $G\_1(F)\to G\_2(F)$ is surjective. Moreover, if I've got things right, then I think that $G\_1$ unramified forces the kernel to be unramified, and if you take a smooth integral model of $G\_1$ with $G\_1(R)$ equal to the hyperspecial you thought of, then the quotient of this model of $G\_1$ by the Zariski closure of the kernel will also be unramified, and the same cohomology argument shows that $G\_1(R)$ surjects onto $G\_2(R)$, so in this case you win.
| 4 | https://mathoverflow.net/users/1384 | 15152 | 10,156 |
https://mathoverflow.net/questions/15115 | 9 | Hi,
I'm looking for examples of groups that are both Hopfian and Co-Hopfian. I have a non trivial (and beautiful, at least to me) example: $\mathrm{SL}(n,\mathbb{Z})$ (with $n>2$).
Do you know others (non trivial)?
Thank you.
| https://mathoverflow.net/users/3958 | Hopfian and Co-Hopfian groups (examples) | Mapping class groups of closed surfaces are both Hopfian and co-Hopfian (the former follows from residual finiteness, and the latter is due to Ivanov-McCarthy).
Out(F\_n) also has both properties (residual finiteness and a theorem of Farb-Handel).
| 22 | https://mathoverflow.net/users/3977 | 15154 | 10,157 |
https://mathoverflow.net/questions/15151 | 25 | Let $R$ be the ring $$R = \prod\_{p\ \text{prime}} \mathbb{F}\_p$$ where $\mathbb{F}\_p$ is the field having $p$ elements.
Is it true that $R$ has a quotient by a maximal ideal which is a field of characteristic zero and contains $\overline{\mathbb{Q}}$?
Motivation: I like the problem and I can't solve it...
It should have something to do with the Chebotarev density theorem.
| https://mathoverflow.net/users/1107 | product of all F_p, p prime | The answer is Yes, and this is the [ultraproduct](http://en.wikipedia.org/wiki/Ultraproduct) construction. Let U be any nonprincipal ultrafilter on the set of primes. This is simply the dual filter to a maximal ideal on the set of primes, containing all finite sets of primes. (In other words, U contains the Frechet filter.)
The quotient R/U is a field of characteristic 0.
The ultraproduct construction is completely general, and has nothing to do with rings or fields. If Mi for i in I is any collection of first order structures, and U is an ultrafilter on the subsets of I, then we may form the ultraproduct Π Mi/U, which is the set of equivalence classes by the relation f equiv g iff { i in I | f(i) = g(i) } in U. Similarly, the structure is imposed on the ultraproduct coordinate-wise, and this is well-defined. The most important theorem is Los's theorem, which says that Π Mi/U satisfies a first order statement phi([f]) if and only if { i in I | Mi satisfies phi(f(i)) } in U.
In your case, since every Fp is a field, the ultraproduct is also a field. And since the set of p bigger than any fixed n is in U, then the ultraproduct will have 1+...+1 (n times) not equal to 0, for any fixed n > 0. That is, the ultraproduct will have characteristic 0.
---
Edit: I confess I missed the part initially about containing the algebraic numbers, and so there is more to be done, as Kevin points out. What Los's theorem gives you is that something will be true in R/U just in case the set of p for which F\_p has the property is in the ultrafilter U.
What you need to know is that for any finite list of equations, that there is an infinite set of primes p for which the equations have a solution in Fp. This property is equivalent to asking whether every finite list of equations over Z is true in at least one Fp, since one can always add one more equation so as to exclude any particular Fp. Is this true? (I wasn't sure.)
But according to what Kevin says in the comments below, it is true, and this is precisely what you need for the construction to go through. You can form a filter containing those sets, which would form a descending sequence of subsets of primes, and then extend this to an ultrafilter. In this case, any particular equation would have a solution in Fp for a set of p in U, and so the ultrapower R/U would have a solution. In this case, the ultrapower will contain the algebraic numbers.
| 16 | https://mathoverflow.net/users/1946 | 15155 | 10,158 |
https://mathoverflow.net/questions/15153 | 11 | A famous theorem of Euler is that Zeta(2n) is a rational number times pi^(2n). Work of Kummer, Herbrand, Ribet and others shows that the rational multiplier has number theoretic significance.
For more general L-functions attached to motives, the philosophy has emerged (Deligne, Beilinson, Bloch, Kato, etc.) that (in vague terms) their values at certain integers are algebraic multiples of transcendental numbers and that the particular algebraic number that's a multiple of the transcendental number contains information about the motive that the L-function is attached to.
But a (nonzero) algebraic multiple of a transcendental number is again transcendental, so an arbitrary real number does not have a well defined decomposition as a product of an algebraic number and a transcendental number.
Still, because of the theorems of Kummer, et. al. one suspects that powers of pi are (at least close to) the "right" "transcendental parts" of the L-function values to be looking at. Maybe one should really be looking a powers of 2pi? But it seems clear that one should not be looking at powers of 691\*pi because otherwise the statement of Kummer's criterion for the regularity of a prime would have an exceptional clause involving the prime 691.
>
> Is there a conceptually motivated means of picking out the
> "right" "transcendental part" of a special value of an L-function?
>
>
>
Presumably the reason that Euler expressed his theorem in terms of pi is because pi was a commonly used symbol. (I've heard people argue that 2 pi is more conceptually primitive and that a label should have been made for the quantity 2 pi rather than for pi and am not sure what I think about this). In any case, there should be an *a priori* means to pin down the relevant transcendental number down "on the nose" (not up to a rational/algebraic multiple).
I have heard that Beilinson's conjecture give the transcendental number only up to a rational multiple and that the Bloch-Kato conjecture pins down the number. But I don't know enough to understand the statement of either conjecture and so am at present ill equipped to derive insight from reading the paper of Bloch and Kato. Are there more elementary considerations that give insight into how to pick out a particular transcendental number out of the set of all of its algebraic multiples?
| https://mathoverflow.net/users/683 | Periods and L-values | The ingredient in the Beilinson and Bloch--Kato conjectures is a motive (over ${\mathbb Q}$,
say). If we take the integral cohomology of this motive (mod torsion, say) we get an integral lattice. If we take some kind of Neron model, and take the algebraic de Rham cohomology of this, we get a second integral lattice. Now computing the determinant of the pairing of one of these on the other, we get a transcendental number, well defined up to a unit in ${\mathbb Z}^{\times}$, i.e. a sign.
This should give you an idea of how one can attach a canonical period to a motive, and is the
basic idea underlying the construction of periods for motives. (Since one doesn't have
Neron models in general, this idea is just heuristic as it stands, but I think it gives the right idea. If you apply it to ${\mathbb G}\_m$, you should recover the period $2\pi i$.)
EDIT: I should point out that the above really is just a heuristic, explaining how
there are two ways of getting integral structures in cohomology: in singular cohomology,
one just takes integral cycles (i.e. "true" cycles on the motive, with no funny
coefficients), and in de Rham cohomology, one takes algebraic differential forms that
are defined over the integers, like the Neron differential $dx/2y$ on an elliptic curve
with minimal Weierstrass equation $y^2 = f(x)$.
To actually get the correct periods for a given $L$-function, one has to do a little more
manipulation than I indicated; e.g. for an elliptic curve over ${\mathbb Q}$,
one will integrate the Neron differential over the a basis for the real integral cycles
(i.e. the cycles that are fixed by the action of complex conjugation on $E({\mathbb C})$;
these are rank one subgroup of the cohomoloy of $E({\mathbb C})$). But hopefully what
I wrote above gives some intuition for what is going on.
| 14 | https://mathoverflow.net/users/2874 | 15158 | 10,161 |
https://mathoverflow.net/questions/15159 | 6 | Specifically, is it possible for a non-Noetherian ring $R$ to have $R[x]$ Noetherian? Every reference I've seen for the Hilbert basis theorem only states the direction "$R$ Noetherian $\Rightarrow$ $R[x]$ Noetherian", which would certainly seem to imply that the converse is false. Unfortunately, it's tough to think about non-Noetherian rings, and what I'm sure is most people's favorite example of one, $K[x\_1,x\_2,\ldots]$ for a field $K$, is obviously not going to help us here.
| https://mathoverflow.net/users/1916 | Converse to Hilbert basis theorem? | If $A$ is an ideal of $R$, then $A[X]$ is an ideal of $R[X]$, right? So an ascending chain of ideals in $R$ which does not stabilize gives you an ascending chain of ideals in $R[X]$ which doesn't stabilize either?
| 14 | https://mathoverflow.net/users/1107 | 15161 | 10,162 |
https://mathoverflow.net/questions/15162 | 2 | Please consider a random walk on a finite N-dimensional lattice with vectors $(x\_1, ..., x\_N)$. We define the origin to be $(0, ..., 0)$ and the target to be at the point in the lattice furthest away from the origin - i.e. $(||x\_1||, ..., ||x\_N||)$ where $||x\_k||$ is the integer length of the lattice in the $x\_{k}$ dimension. Here, each step of the random walk is a uniformly distributed, strictly positive random integer in each of the N-dimensions with an upper-bound value defined by the requirement that one cannot exceed the dimensions of the lattice.
Is there a nice method, aside from explicit path-counting, to derive the probability density for hitting times provided an arbitrary lattice as defined above?
Some computational results: For the $N=1$ case I expected the target hitting time (defined as the number of steps to reach the target) to fit well with a logarithmic growth function of the form $A\*ln(S)$ where A is a positive real number and $"S"$ is the number of integer steps one takes to reach the target from the origin. Running simulations (averaging 10,000 times) this yielded a decent fit with the value of $A$ ~ 1.146 for $||x|| = 100$, but $A$ decreases to ~1.095 for $||x|| = 1,000$ and decreased further ~1.069 for $||x||=10,000$.
| https://mathoverflow.net/users/3248 | Hitting times for an N-dimensional random walk on a lattice with (strictly positive) random integer steps | Assuming you mean Leonid Kovalev's interpretation, the distribution of the hitting time in the $N = 1$ case is the same as the distribution of the number of cycles of a random permutation of $[n]$.
To be more specific, I'll change coordinates. Let $X\_0 = (x\_0^1, \ldots, x\_0^N) = (S, S, \ldots, S)$. Let $X\_1 = (x\_1^1, \ldots, x\_1^N)$, where $x\_1^j$ is chosen uniformly at random from $0, 1, \ldots, x\_0^j-1$. Define $X\_2$ similarly in terms of $X\_1$, and so on.
Then the sequence $(x\_0^1, x\_1^1, x\_2^1, x\_3^1, \ldots)$ are as follows:
* $x\_0^1 = L$, of course.
* $x\_1^1$ is uniformly distributed on $\{ 0, \ldots, S-1 \}$.
* $x\_2^1$ is uniformly distributed on $\{ 0, \ldots, x\_1^1-1 \}$.
and so on...
In particular the distribution of $x\_1^1$ is the distribution of number of elements in a random permutation on $S$ elements which are {\it not} in the cycle containing 1; In particular the distribution of $x\_1^1$ is the distribution of number of elements in a random permutation on $S$ elements which are {\it not} in any of the $k$ cycles with the smallest minima.
So the distribution of the smallest $k$ for which $x\_k^1 = 0$ is exactly the distribution of the number of cycles of a random permutation of $\{ 1, \ldots, S \}$; this is $1 + 1/2 + \cdots + 1/S \sim \log S + \gamma$, where $\gamma = 0.577\ldots$ is the Euler-Mascheroni constant.
In the higher-dimensional cases, the time to reach $(0, \ldots, 0)$ is just the {\it maximum} of the number of cycles of $N$ permutations chosen uniformly at random; this should still have expectation $\log S$ plus a constant depending on $N$.
| 3 | https://mathoverflow.net/users/143 | 15164 | 10,163 |
https://mathoverflow.net/questions/15181 | 11 | Can you divide one square paper into five equal squares?
You have a scissor and glue. You can measure and cut and then attach as well. Only condition is You can't waste any paper.
| https://mathoverflow.net/users/3056 | Dividing a square into 5 equal squares | The Wallace-Bolyai-Gerwien Theorem theorem says:
Any two simple polygons of equal area are equidecomposable
(where simple means no self intersections and equidecomposable means finitely cut and glued).
For your problem you can take the first polygon to be a unit square and the second to be a sqrt(5) by 1/sqrt(5) rectangle and apply this theorem. Then perform the remaining four cuts.
Also, the generalisation of your question is the 2d analogue of Hilbert's 3rd Problem which asks whether given any two polyhedra with equal volume can one be finitely cut and glued into the other. The answer here, unlike in the 2d case, is "no" which was proved by Dehn using Dehn invariants in 1900.
| 34 | https://mathoverflow.net/users/3623 | 15184 | 10,175 |
https://mathoverflow.net/questions/15147 | 4 | I am trying to optimize a function of the following form:
$L = \int\_{t=0}^{T}(AR-x)dt$, where A is a system parameter
i.e. I am trying to find the optimum x(t) that minimizes L over all admissible x(t)s. R is related to x using a relation:
$\frac{dR}{dt} = axRY - bR$, where a and b are system parameters and $R(0) = R\_{0}$
$\frac{dY}{dt} = -xRY$
I was looking at this problem from mainly a simulation perspective. There is a whole amount of work that went into showing that x can take only two specific values that minimize the function over any given interval. Now, what I was thinking was to convert the integral into its discrete formulation and do the following:
For $t=1$,
Let $x = x\_{min}$ and calculate $L\_{10}$
Let $x = x\_{max}$ and calculate $L\_{11}$
Finally, choose the one that has the min L.
And then continue with t=2 and so on, in the same way. If I visualize this problem, it is nothing but finding a minimum cost path in a binary tree i.e. something of the following form:
---------------------- $L\_{00}$
-----------------------/--\---------
------------------$L\_{10}$----$L\_{11}$----
-----------------/-----\----/----\-------
-------------$L\_{20}$-----$L\_{21}$-$L\_{22}$-$L\_{23}$--
and so on until the last T. I am not sure if my thought process of simulating this is in the right direction. Can someone give me some suggestions?
| https://mathoverflow.net/users/3560 | Minimizing a function containing an integral | I don't think the problem as posed has an optimal solution. This is a problem in optimal control, typically dealt with by solving the [Hamilton–Jacobi–Bellman equation](http://en.wikipedia.org/wiki/Hamilton%E2%80%93Jacobi%E2%80%93Bellman%5Fequation). Your problem here is quite general, namely, a *linear* and *unconstrained* control. Thus the minimum with respect to the control variable ($x$ in your notation, $u$ in the notation of the referenced Wikipedia page) in the HJB equation does not exist, unless the value function $V$ does not depend on the control variable. In other words, an optimum exists only in the trivial case where the control variable does not actually influence the cost function.
(A small caveat: Though I have sat through a number of lectures on optimal control theory, that is the extent of my expertise, so take this with a grain of salt.)
| 1 | https://mathoverflow.net/users/802 | 15187 | 10,177 |
https://mathoverflow.net/questions/15180 | 12 | Suppose I have the group presentation $G=\langle x,y\ |\ x^3=y^5=(yx)^2\rangle$. Now, $G$ is isomorphic to $SL(2,5)$ (see my proof [here](http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328702)). This means the relation $x^6=1$ should hold in $G$. I was wondering if anyone knows how to derive that simply from the group presentation (not using central extensions, etc.). Even nicer would be an example of how software (GAP, Magma, Magnus, etc.) could automate that.
| https://mathoverflow.net/users/1446 | Deriving a relation in a group based on a presentation | The theory (and practice) of automatic groups is the most generally useful systematic way to deal with these things. There is a nice package written by Derek Holt and his associates called kbmag (available for download here: <http://www.warwick.ac.uk/~mareg/download/kbmag2/> ). There is a book "Word Processing in Groups" by Epstein, Cannon, Levy, Holt, Paterson and Thurston that describes the ideas behind this approach. It's not guaranteed to work (not all groups have an "automatic" presentation) but it is surprisingly effective.
I made up a short input file for kbmag, and it immediately came back with a "confluent" system of relations (a particular system which has a technical property that when you just do a series of string substitutions you always get the same answer no matter what order you do them in). For your edification, here they are (xi and yi are x^-1 and yi^-1 respectively, idWord is 1 [edited to show the derivations from kbmag):
```
#Initial equation number 1:
#x*xi -> IdWord
#Initial equation number 2:
#xi*x -> IdWord
#Initial equation number 3:
#y*yi -> IdWord
#Initial equation number 4:
#yi*y -> IdWord
#Initial equation number 5:
#y^4 -> x^3*yi
#Initial equation number 6:
#y*x*y -> x^2
#New equation number 7, from overlap 5, 3:
#x^3*yi^2->y^3
#New equation number 8, from overlap 4, 5:
#yi*x^3->y^4
#New equation number 9, from overlap 6, 3:
#x^2*yi->y*x
#New equation number 10, from overlap 4, 6:
#yi*x^2->x*y
#New equation number 11, from overlap 2, 7:
#xi*y^3->y*x*yi
#New equation number 12, from overlap 2, 9:
#xi*y*x->x*yi
#New equation number 13, from overlap 10, 1:
#x*y*xi->yi*x
#New equation number 14, from overlap 1, 11:
#x*y*x*yi->y^3
#New equation number 15, from overlap 11, 3:
#y*x*yi^2->xi*y^2
#New equation number 16, from overlap 12, 1:
#x*yi*xi->xi*y
#New equation number 17, from overlap 2, 13:
#xi*yi*x->y*xi
#New equation number 18, from overlap 4, 15:
#yi*xi*y^2->x*yi^2
#New equation number 19, from overlap 2, 16:
#xi^2*y->yi*xi
#New equation number 20, from overlap 17, 1:
#y*xi^2->xi*yi
#New equation number 21, from overlap 18, 3:
#x*yi^3->yi*xi*y
#New equation number 22, from overlap 19, 3:
#yi*xi*yi->xi^2
#New equation number 23, from overlap 2, 21:
#xi*yi*xi*y->yi^3
#New equation number 24, from overlap 23, 3:
#yi^4->xi*yi*xi
#New equation number 25, from overlap 3, 24:
#y*xi*yi*xi->yi^3
#New equation number 26, from overlap 25, 2:
#yi^3*x->y*xi*yi
#New equation number 27, from overlap 3, 26:
#y^2*xi*yi->yi^2*x
#New equation number 28, from overlap 27, 4:
#yi^2*x*y->y^2*xi
#New equation number 29, from overlap 3, 28:
#y^3*xi->yi*x*y
#New equation number 30, from overlap 29, 2:
#yi*x*y*x->y^3
#New equation number 31, from overlap 5, 5:
#y*x^3->x^3*y
#New equation number 32, from overlap 5, 6:
#y^3*x^2->x*y*x^2*y
#New equation number 33, from overlap 8, 7:
#yi*x*y^3->x*y^2*x*yi
#New equation number 34, from overlap 7, 8:
#y*x^2*y*x->x^2*y^3
#New equation number 35, from overlap 11, 6:
#y*x*yi*x*y->xi*y^2*x^2
#New equation number 36, from overlap 12, 9:
#xi*y^2*x->x*yi*x*yi
#New equation number 37, from overlap 10, 13:
#yi*x*yi*x->x*y^2*xi
#New equation number 38, from overlap 11, 15:
#y*x*yi*x*yi^2->xi*y^2*xi*y^2
#New equation number 39, from overlap 12, 16:
#xi*y*xi*y->x*yi^2*xi
#New equation number 40, from overlap 17, 13:
#y*xi*y*xi->xi*yi^2*x
#New equation number 41, from overlap 18, 15:
#yi*x*yi^2*xi*y->x*yi^2*x*yi^2
#New equation number 42, from overlap 18, 20:
#yi*xi*y*xi*yi->x*yi^2*xi^2
#New equation number 43, from overlap 19, 20:
#yi*xi^3->xi^3*yi
#New equation number 44, from overlap 17, 21:
#y*xi*yi^3->xi*yi^2*xi*y
#New equation number 45, from overlap 23, 20:
#yi^3*xi^2->xi*yi*xi^2*yi
#New equation number 46, from overlap 22, 24:
#yi*xi^2*yi*xi->xi^2*yi^3
#New equation number 47, from overlap 25, 19:
#yi^3*xi*y->y*xi*yi^2*xi
#New equation number 48, from overlap 22, 26:
#xi^2*yi^2*x->x*yi^2*xi^2
#New equation number 49, from overlap 27, 26:
#yi^2*x*yi^2*x->y*xi*yi^2*x*yi
#New equation number 50, from overlap 49, 1:
#y*xi*yi^2*xi*y->yi^2*x*yi^2
#New equation number 51, from overlap 7, 28:
#x^3*y^2*xi->y^2*x^2
#New equation number 52, from overlap 27, 28:
#yi*x*y^2*xi*y->y^2*xi*y^2*xi
#New equation number 53, from overlap 29, 12:
#yi*x*y^2*x->y^3*x*yi
#New equation number 54, from overlap 31, 7:
#x^3*y^2*x*yi->y*x^2*y^3
#New equation number 55, from overlap 32, 13:
#x*y*x^2*y^2*xi->y^3*x*yi*x
#New equation number 56, from overlap 32, 14:
#x*y*x^2*y^2*x*yi->y^2*x^2*y^2
#New equation number 57, from overlap 2, 56:
#y*x^2*y^2*x*yi->x^2*y^2*xi*y^2
#New equation number 58, from overlap 56, 4:
#y^2*x^2*y^3->x*y*x^2*y^2*x
#New equation number 59, from overlap 57, 4:
#x^2*y^3*x*yi->y*x^2*y^2*x
#New equation number 60, from overlap 33, 34:
#y^2*x^2*y^2*xi*y^2->x*y^2*x^2*y^2*x
#New equation number 61, from overlap 11, 35:
#x^2*y^2*xi*y^2*xi->y*x^2*y^2*xi*y
#New equation number 62, from overlap 35, 25:
#x*yi*x*yi^2*xi->xi*y^2*xi*y
#New equation number 63, from overlap 35, 32:
#x^2*y^2*x^2*y^2*xi->y*x^2*y^2*x^2*y
#New equation number 64, from overlap 33, 35:
#y^2*x^2*y^2*xi*y->yi*x*yi^2*x*yi^2
#New equation number 65, from overlap 64, 3:
#y^2*x^2*y^2->yi^2*x*yi^2
#New equation number 66, from overlap 4, 64:
#y*x^2*y^2*xi*y->xi*yi^2*x*yi^2*xi
#New equation number 67, from overlap 65, 3:
#y*x^2*y->xi*yi^2*xi
#New equation number 68, from overlap 4, 66:
#xi*y*xi*yi^2*xi->x^2*y^2*xi*y
#New equation number 69, from overlap 67, 3:
#xi*yi*xi^2->y*x^2
#New equation number 70, from overlap 4, 67:
#xi^2*yi*xi->x^2*y
#New equation number 71, from overlap 68, 2:
#x^2*y^2*x->xi*y*xi*yi
#New equation number 72, from overlap 1, 69:
#x*y*x^2->yi*xi^2
#New equation number 73, from overlap 69, 2:
#x^3*y->xi*yi*xi
#New equation number 74, from overlap 70, 2:
#x^2*y*x->xi^2*yi
#New equation number 75, from overlap 71, 1:
#xi*yi^3->x^2*y^2
#New equation number 76, from overlap 2, 71:
#yi*xi^2*yi->x*y^2*x
#New equation number 77, from overlap 72, 1:
#xi^3*yi->x*y*x
#New equation number 78, from overlap 73, 3:
#xi^3->x^3
#New equation number 79, from overlap 75, 4:
#x^2*y^3->xi*yi^2
#New equation number 80, from overlap 1, 78:
#x^4->xi^2
#New equation number 81, from overlap 2, 79:
#xi^2*yi^2->x*y^3
#New equation number 82, from overlap 29, 36:
#y^3*x*yi*x*yi->x*y^2*x*yi*x
#New equation number 83, from overlap 7, 37:
#y^3*x*yi*x->yi^2*xi*y*xi
#New equation number 84, from overlap 4, 83:
#xi*yi^2*xi^2->y*x*yi*x
#New equation number 85, from overlap 1, 84:
#yi^2*xi^2->y^3*x
#New equation number 86, from overlap 9, 37:
#yi^3*xi->y^2*x^2
#New equation number 87, from overlap 38, 8:
#x^2*y^2*xi*y^2->xi*yi^2*x*yi^2
#New equation number 88, from overlap 11, 38:
#xi*y^2*xi*y^2*xi*y->y^2*x*yi*x*yi^2
#New equation number 89, from overlap 88, 3:
#xi*y^2*xi*y^2*xi->y*xi*y^2*xi*y
#New equation number 90, from overlap 38, 37:
#y*xi*yi^2*x*yi^2*xi->yi*x*yi^2*x*yi^2
#New equation number 91, from overlap 39, 39:
#x*y^2*x^2->yi*xi*y*xi
#New equation number 92, from overlap 41, 39:
#yi*x*yi^2*x*yi^2*xi->x*yi*x*yi^2*x*yi^2
#New equation number 93, from overlap 42, 47:
#xi*y*xi*yi^2*x*yi^2->x*y^2*xi*y^2*xi*y
#New equation number 94, from overlap 93, 4:
#x*y^2*xi*y^2*xi*y^2->xi*y*xi*yi^2*x*yi
#New equation number 95, from overlap 2, 94:
#y^2*xi*y^2*xi*y^2->x*y^2*x*yi*x*yi
```
#68 eqns; total len: lhs, rhs = 299, 246; 77 states; 0 secs.
max len: lhs, rhs = 8, 8.
#System is confluent.
#Halting with 68 equations.
#Exit status is 0
| 13 | https://mathoverflow.net/users/2784 | 15191 | 10,179 |
https://mathoverflow.net/questions/15186 | 3 | Suppose that $F$ is a nonarchimedean local field, and that $G\_1$, $G\_2$ are connected, simply connected algebraic groups over $F$. Suppose moreover $G\_1$ and $G\_2$ are semisimple. Suppose $H$ is a hyperspecial maximal compact subgroup of $G\_1\times G\_2$. Is $H$ necessarily a product $H\_1\times H\_2$ where the $H\_i$ are hyperspecial maximal compact subgroups of the $G\_i$?
Edited: I had originally written that the Lie algebras of the $G\_i$ were semisimple, which is not such a sensible thing to write. I have changed this to say the $G\_i$ themselves semisimple.
Note: Brian Conrad's answer, below, shows we need assume nothing more than connected reductive hypotheses on $G\_1$ and $G\_2$.
| https://mathoverflow.net/users/3513 | Hyperspecial subgroup of a product of semisimple algebraic groups | To allow all characteristics, the semisimplicity requirement on the Lie algebras should be replaced with the requirement that the $G\_i$ are semisimple as $F$-groups. (In positive characteristic the Lie algebras of connected semisimple groups can often fail to be semisimple.) In fact, the simply connected and semisimplicity hypotheses are stronger than necessary; connected reductive is sufficient. Taking into account the definition of "hyperspecial maximal compact subgroup" (of the group of $F$-rational points), the question in this modified form immediately reduces to a more "algebraic" assertion having nothing to do with local fields, as follows.
Consider a reductive group scheme $\mathcal{G}$ (with connected fibers, following the convention of SGA3) over a normal noetherian scheme $S$, and let $G$ be its fiber over the scheme $\eta$ of generic points of $S$. Assume $G$ decomposes as a direct product $G = G\_1 \times G\_2$ with $G\_i$ necessarily connected reductive over $\eta$. Then does there exist a pair of reductive closed $S$-subgroup schemes (with connected fibers) $\mathcal{G}\_i$ in $G\_i$ such that $\mathcal{G}\_ 1 \times \mathcal{G}\_2 = \mathcal{G}$ compatibly with the given identification on generic fibers? Below is a proof that the answer is ``yes''.
First observe that such $\mathcal{G}\_i$ are necessarily unique if they exist, being the Zariski closures of the $G\_i$ in $\mathcal{G}$ (since $S$ is reduced). In view of this uniqueness, by descent theory it follows that to prove the existence we may work \'etale-locally on the base. (This step tends to ruin connectedness hypotheses, since $S' \times\_S S'$ is generally cannot be arranged to be connected, even if $S$ is connected and $S' \rightarrow S$ is a connected \'etale cover.) Hence, we now assume that $\mathcal{G}$ is $S$-split in the sense that it admits a (fiberwise) maximal $S$-torus $\mathcal{T}$ that is $S$-split. We may also work separately over each connected component of $S$, so we can now assume $S$ is connected (as we will make no further changes to $S$ in the argument).
The maximal $\eta$-torus $T := \mathcal{T}\_ {\eta}$ in $G$ uniquely decomposes as $T = T\_1 \times T\_2$ for necessarily $\eta$-split maximal $\eta$-tori $T\_i$ in $G\_i$. By the classification of split pairs (i.e., fiberwise connected reductive group equipped with a split maximal torus) over any base scheme in terms of root data, to give a direct product decomposition of $(\mathcal{G}, \mathcal{T})$ is the same as to decompose its root datum into a direct product (i.e., direct product of the $X$ and $X^{\vee}$ parts, and corresponding disjoint union for the $R$ and $R^{\vee}$ parts). By connectedness, the normal $S$ is irreducible (i.e., $\eta$ is a single point), so the root data of $(\mathcal{G}, \mathcal{T})$ and its generic fiber $(G,T)$ are canonically identified. QED
| 9 | https://mathoverflow.net/users/3927 | 15192 | 10,180 |
https://mathoverflow.net/questions/15123 | 1 | This question may be trivial, I did not think hard about it.
A friend of mine is looking for an irreducible (reduced) analytic subspace $C \subset \mathbb{C}^2$ with the following property. Let $f \colon C \rightarrow \mathbb{C}$ be the projection on the first factor. He wants that
1) All singular points of $C$ and all ramification points for $f$ lie in a limited set, so removing that set we obtain a topological covering from some open set of $C$ to $\mathbb{C}$ with a ball removed.
2) That covering should be trivial (even better if it is finitely-sheeted).
So the curve $C$ is connected, but only if one passes near the origin. Sufficiently far from that ther should be no way to jump between sheets. Is it possible to find such a $C$?
| https://mathoverflow.net/users/828 | Riemann surface disconnected at infinity | This is an extended version of my comment. Suppose we stay on the surface $z^2+w^2=1$ but away from the origin. The identity
$|z^2+w^2|^2=|z|^4+|w|^4+2Re((z \bar w)^2)$
tells us that the square of $z \bar w$ has negative real part. The set of complex numbers $\zeta$ such that $Re(\zeta^2)<0$ has two connected components: it's the disjoint union of two open sectors. Finally, note that switch from (z,w) to (-z, w) involves going from one component to the other.
| 6 | https://mathoverflow.net/users/2912 | 15194 | 10,182 |
https://mathoverflow.net/questions/15116 | 21 | This is a followup to an earlier question I asked: [Does formally etale imply flat](https://mathoverflow.net/questions/11868/does-formally-etale-imply-flat/)? After some remarks I received on MO I noticed that this was answered to the negative by an answer to an earlier question [Is there an example of a formally smooth morphism which is not smooth](https://mathoverflow.net/questions/195/is-there-an-example-of-a-formally-smooth-morphism-which-is-not-smooth). However, the simple example involves a non-noetherian ring (in fact, a perfect ring; these are seldom noetherian unless they are a field).
So my challenge is to provide an example of a formally etale map of **noetherian** schemes which is not flat, or otherwise proof that for maps of noetherian schemes formally etale implies flat.
| https://mathoverflow.net/users/917 | Does formally etale imply flat for noetherian schemes? | Every formally smooth morphism between locally noetherian schemes is flat; this is a deep result of Grothendieck. Indeed, the formal smoothness is preserved by localization on target and then likewise on the source, so we can assume we're deal with a local map between local noetherian rings. By EGA 0$\_{\rm{IV}}$, 19.7.1 (also proved near the end of Matsumura's book on commutative ring theory) a formally smooth local map between local noetherian rings is flat.
| 41 | https://mathoverflow.net/users/3927 | 15200 | 10,186 |
https://mathoverflow.net/questions/15199 | 2 | While studying for a topological groups course, I wondered if we could define the product of uncountably many topological groups such that the product is still a topological group. That is: let $G\_i$ be a topological group with product law $p\_i$ for each $i \in I$ (with $I$ uncountable). We can give $G = \prod\_{i \in I} G\_i$ the (Tychonoff) product topology and define the product law of $G$ by:
$\pi\_i \circ p = p\_i$ for all $i \in I$.
However, when trying to prove that this mapping is continuous end up needing $I$ to be at most countable or that the topologies of $G\_i$ be discrete.
Is there any way to get around this?
Thanks.
| https://mathoverflow.net/users/3887 | Infinite products of topological groups | You can define the product of an arbitrary family $(G\_i)\_{i \in I}$ of topological groups $G\_i$ by equipping the group-theoretic product $G = \prod\_{i \in I} G\_i$ with the product topology; the product topology is indeed compatible with the group structure (confer Bourbaki, General topology, III.2.9, but it's pretty obvious actually).
Perhaps your problem is the product topology? Note that a basis for the product topology are the sets $(U\_i)\_{i \in I}$ where $U\_i \subseteq G\_i$ is open and $U\_i = G\_i$ for all but finitely many $i \in I$. (confer [wiki](http://en.wikipedia.org/wiki/Product_topology) for the product topology).
| 8 | https://mathoverflow.net/users/717 | 15201 | 10,187 |
https://mathoverflow.net/questions/15202 | 24 | I'm taking my first steps in the language of stacks, and would like something cleared up. The intuitive idea of moduli spaces is that each point corresponds to an object of what we're trying to classify (smooth curves of genus g over ℂ, for example). Fine moduli spaces are defined to be the objects that represent the functor that takes an object and gives you the [set, for schemes; groupoid, as I understand it, for stacks] of ways that that object parametrizes families of the object we want to classify. Now, for schemes - this makes sense in the following way:
Let that functor be F, and let it be represented by M. Then F(Spec ℂ) are the families of (desired) objects parametrized by Spec ℂ (a point), so it corresponds to all desired objects (the ones we want to classify). But F(Spec ℂ) is also Hom(Spec ℂ, M), and so corresponds to the closed points of M. Thus M really does, intuitively, have as points the objects it wishes to classify.
Does this idea go through to moduli stacks? Of course, it probably does, and this is all probably trivial - but I feel like I need someone to assure me that I'm not crazy. So let me put the question like this: Can you formulate how to think of a fine moduli stack (as an object that represents an F as above; also: how would you define this F in the category of stacks?) in a way that makes it clear that it parametrizes the desired objects?
| https://mathoverflow.net/users/3238 | Different interpretations of moduli stacks | I'll assure you that you're not crazy. Not only does the idea go through for stacks, but it's impossible (or at least very hard) to make sense of stacks without that idea.
If you're trying to parameterize wigits, you can build a functor F(T)={flat families of wigits over T}. If there is a space M that deserves to be called the moduli space of wigits, it should represent F. It's not just that the points of M must correspond to isomorphism classes of wigits, so we must have F(Spec ℂ)=Hom(Spec ℂ,M). The points are also connected up in the right way. For example, a family of widits over a curve should correspond to a choice of wigit for every point in the curve in a continuous way, so it should correspond to a morphism from the curve to M.
It happens that if wigits have automorphisms, there's no hope of finding a geometric object M so that maps to M are the same thing as flat (read "continuous") families of wigits. The reason is that any geometric object should have the property that maps to it can be determined locally. That is, if U and V cover T, specifying a map from T to M is the same as specifying maps from U and V to M which agree on U∩V. The jargon for this is "representable functors are sheaves." If a wigit X has an automorphism, then you can imagine a family of wigits over a circle so that all the fibers are X, but as you move around the circle, it gets "twisted" by the automorphism (if you want to think purely algebro-geometrically, use a circular chain of ℙ1s instead of a circle). Locally, you have a trivial family of wigits, so the map should correspond to a constant map to the moduli space M, but that would correspond to the trivial family globally, which this isn't. Oh dear!
Instead of giving up hope entirely, the trick is to replace the functor F by a "groupoid-valued functor" (fibered category), so the automorphisms of objects are recorded. Now of course there won't be a space representing F, since any space represents a set-valued functor, but it turns out that this sometimes revives the hope that F is represented by some mythical "geometric object" M in the sense that objects in F(T) (which should correspond to maps to M) can be determined locally. If this is true, we say that "F is a *stack*" or that "M is a *stack*." Part of what makes your question tricky is that as things get stacky, the line between M and F becomes more blurred. M isn't really anything other than F. We just call it M and treat it as a geometric object because it satisfies this gluing condition. We usually want M to be more geometric than that; we want it to have a cover (in some precise sense) by an honest space. If it does, then we say "M (or F) is an *algebraic* stack" and it turns out you can do real geometry on it.
| 28 | https://mathoverflow.net/users/1 | 15207 | 10,190 |
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