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https://mathoverflow.net/questions/15832 | 7 | This is probably pretty basic, but as I said before I'm just beginning my way in the language of stacks.Say you have an etale cover X->Y of stacks (in the etale site). Is there a standard way to define the degree of this cover? Here's my intuition: if X and Y are schemes, we can look etale locally and then this cover is Yoneda-trivial in the sense of <https://arxiv.org/abs/0902.3464> , meaning that etale locally it is just a disjoint unions of (d many) pancakes. Can we do this generally? Is there some "connectivity" conditions on Y for this to work? Is there a different valid definition for degree of an etale cover of stacks?
### Yoneda-Triviality
I figured since nobody answered so far, maybe I should write down what a possible Yoneda-triviality condition could mean for stacks:
Def: Call f:X->Y (stacks) Yoneda-trivial if there exists a set of sections of f, S, such that the natural map Y(Z)xS->X(Z) is an isomorphism (or maybe a bijection?) for any connected scheme Z.
But I'm still clinging to the hope that there's a completely different definition out there that I'm just not aware of.
| https://mathoverflow.net/users/3238 | Degrees of etale covers of stacks | There is a notion of degree of an integral morphism of DM stacks, and I think it is in Vistoli's "Intersection theory ... " paper [here](https://doi.org/10.1007/BF01388892). Its unrelated to Yoneda triviality.
| 5 | https://mathoverflow.net/users/2 | 15868 | 10,628 |
https://mathoverflow.net/questions/15867 | 0 |
>
> **Motivation**
>
>
> Consider the situation: You know that
> every $x$ that has property $P$ must have property $Q$. $Q$ is a
> rather strong condition but not strong
> enough to fulfill $P$. What is *missing*?
>
>
>
Consider the formulas of a first-order language, a distinguished set of axioms $S$, and the formulas with exactly one free variable, factored out by provable equivalence $P(x) \sim\_S Q(x)$ iff $S \vdash P(x) \leftrightarrow Q(x)$ – for short: *properties*.
There is a partial order on the set of properties by $[P(x)] \Rightarrow\_S [Q(x)]$ iff $S\vdash P(x) \rightarrow Q(x)$ – for short: $P \Rightarrow Q$.
Let $P \Rightarrow Q$, but $Q \not\Rightarrow P$, read: $Q$ *is a proper necessary condition of* $P$.
>
> **Definition**
>
>
> A property $D(x)$ may be
> called a *defect* of $Q(x)$ with
> respect to $P(x)$ if
>
>
> * $D \not\Rightarrow P$
> * $Q \wedge D \Rightarrow P$
>
>
> A defect $D(x)$ may be called *minimal*
> if there is no other
> defect $D'(x)$ with $D(x) \Rightarrow D'(x)$.
>
>
>
---
>
> **Question #1:** Is the search for *(minimal) defects*
> so manifest – in the working mathematician's life – that it is performed every day without needing a proper name? And is the notion ´– thus – of no (meta-)mathematical importance?
>
>
>
---
>
> **Question #2:** ... or is there already a proper –
> and more common – name?
>
>
>
---
>
> **Question #3:** ... or is the definition above and its presuppositions flawed?
>
>
>
---
**Addendum**
The common divisor graph on the natural numbers shares one strong property $Q$ with the [Rado (= random) graph](http://en.wikipedia.org/wiki/Rado_graph): it contains any finite graph as an induced subgraph (which I guess is a quite general necessary condition for randomness). But it's not isomorphic to the Rado graph, i.e. does not fulfill its defining property $P$. I am now interested in the "defect" of the common divisor graph: which property $D$ - as minimal as possible - prevents the common divisor graph from being truly random?
| https://mathoverflow.net/users/2672 | Deficiency of necessary conditions | Normally your "defect" is called an "additional assumption"/"extra condition"/... and the typical phrase is "the inverse implication also holds under the additional assumption that...". Yes, the search for such things is something that mathematicians do on an everyday basis trying to bridge the gap between what is necessary and what is sufficient.
The thing you should understand is that in real life it is not necessarily the first priority to have $D$ as weak as possible from the logical standpoint. What often matters much more is that $D$ is easy to verify, holds in many interesting cases, allows one to give an easy and elegant proof, etc. Actually, Joel has already brought the idea of pure logical minimality to its absurd extreme form, so I hardly need to comment more on this issue.
By the way, the assumption that $P\implies Q$ is completely unnecessary in your definition of the "defect"; it makes just as much sense without it. Indeed, the usual story is that we know something ($Q$), we want to conclude something else ($P$), we suspect that the implication $Q\implies P$ is (may be) not always true, but we want this implication not for its own sake but to figure out something about some object $X$, so we ask what other property $D$ $X$ possesses that together with $Q$ will give us $P$.
| 4 | https://mathoverflow.net/users/1131 | 15870 | 10,630 |
https://mathoverflow.net/questions/15872 | 13 | I thought I had heard or read somewhere that the existence of a non-principal ultrafilter on $\omega$ was equivalent to some common weakening of AC. As I searched around, I read that this is not the case: neither countable choice nor dependent choice are strong enough.
This leads me to two questions:
Where would I find a proof that DC is not strong enough to prove the existence of a non-principal ultrafilter on $\omega$?
Is the assumption that there exists a non-principal ultrafilter on $\omega$ strong enough to show DC or countable choice? I.e. is "there exists a non-principal ultrafilter on $\omega$" stronger than either of countable or dependent choice?
| https://mathoverflow.net/users/2143 | Non-principal ultrafilters on ω | See the [Prime ideal theorem](http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem).
The existence of ultrafilters on every Boolean algebra (which implies non-principal ultrafilters on ω, since these come from ultrafilters on the Boolean algebra P(ω)/Fin) is a set-theoretic principle that follows from AC and is not provable in ZF (if ZF is consistent), but which does not imply full AC. Thus, it is an intermediate weaker choice principle.
Your statement about ultrafilters on ω appears to be even weaker, since it is such a special case of the Prime Ideal Theorem.
Nevertheless, I believe that the method of forcing shows that it is consistent with ZF that there are no non-principal ultrafilters on ω. I believe that some of the standard models of ¬AC, built by using symmetric names for adding Cohen reals, have DC, and hence also ACω, but still have no nonprincipal ultrafilters on ω. In this case, neither DC nor ACω would imply the existence of such ultrafilters.
I'm less sure about finding models that have ultrafilters on ω, but not on all Boolean algebras. But I believe that this is likely the case. These models would show that your principle is strictly weaker even than the Prime Ideal Theorem.
| 9 | https://mathoverflow.net/users/1946 | 15873 | 10,631 |
https://mathoverflow.net/questions/15850 | 2 | Is there a condition for a $G$-equivariant map $X \to Y$ to be a cofibration of $G$-spaces? Here $X$ and $Y$ are CW complexes, the group $G$ is finite, and acts by cellular maps.
I am using the model structure on CW-complexes with G-action where the fibrations and weak equivalences are those maps which are fibrations, weak equivalences respectively when we forget the $G$ action.
| https://mathoverflow.net/users/1676 | characterization of cofibrations in CW-complexes with G-action | In the model structure you describe, the cofibrations should be the retracts of the free relative G-cell maps: i.e., retracts of maps obtained by attaching cells of the form $G \times S^{n} \to G \times D^{n+1}$.
One way to see this is via the following general machine: There is an adjoint pair
$$ G \times -: \mathbf{Top} \leftrightarrow \mathbf{GTop}: forget $$
$\mathbf{Top}$ is a cofibrantly generated model category and one can check that this adjoint pair satisfies the conditions of the standard Lemma for transporting cofibrantly generated model structures across adjoint pairs (see e.g., Hirschorn's "Model categories and their localizations" Theorem 11.3.2). Thus, it gives rise to a model structure on $\mathbf{GTop}$ such that a map in $\mathbf{GTop}$ is an equivalence(resp. fibration) iff its image under the right adjoint (forget) is so. Moreover, the generating (acyclic/)cofibrations are precisely the images under the left adjoint ($G \times -$) of the generating (acyclic/)cofibrations in $\mathbf{Top}$. This yields the description of the cofibrations as retracts of (free G-)"cellular" maps.
Also, some context:
The model structure you describe (which I'd like to call "Spaces over BG") is a localization of a model structure "G-Spaces" (where the weak equivalences are maps inducing weak equivalences on all fixed point sets). An argument along the lines of the above constructs this other model structure and identifies its cofibrations with retracts of (arbitrary) relative G-cell maps: i.e., retracts of maps obtained by attaching cells of the form $G/H \times S^n \to G/H \times D^{n+1}$ for $H$ a closed subgroup of $G$.
| 8 | https://mathoverflow.net/users/1921 | 15880 | 10,638 |
https://mathoverflow.net/questions/13901 | 8 | According to the Baez-Dolan cobordism hypothesis, an extended TQFT is determined by its value on a single point. This value a fully dualizable object of a symmetric monoidal $n$ category (a fully dualizable object is a higher categorical analogue of a finite dimensional vector space). The Alexander polynomial is a quantum invariant, and comes from a TQFT.
>
> How can an "Alexander polynomial" TQFT be put into an extended TQFT, and what is its value at a single point?
The question I just asked is closely related to [this question](https://mathoverflow.net/questions/9847/why-is-the-alexander-polynomial-a-quantum-invariant). I also asked the question on the ldtopology blog [here](http://ldtopology.wordpress.com/2010/02/02/the-alexander-polynomial-as-a-quantum-invariant-part-2/), and Theo Johnson-Freyd suggested that MO might be the place to ask it.
---
Briefly, I will summarize what an extended TQFT is. A TQFT as a symmetric monoidal functor Z:**Cob**(n)-> **Vect**(k) from the tensor category of $n-1$ dimensional manifolds and cobordisms between them to the tensor category of vector spaces over a field k. An extended TQFT is a symmetric monoidal functor Z:**Cob**k(n)->**C** from the n-category of cobordisms to a symmetric monoidal n-category **C**. I vote for the [introduction to Lurie's expository account of his work proving the Baez-Dolan cobordism hypothesis](http://www.math.harvard.edu/~lurie/papers/cobordism.pdf) as the best place to read about why extended TQFT's are natural objects, to understand their motivation, and to understand why people are so excited about them. An extended TQFT assigns a fully dualizable object to a point, and a higher “trace” on this object to a closed n-dimensional manifolds.
| https://mathoverflow.net/users/2051 | What is the Alexander polynomial of a point? | Understanding the Alexander polynomial can get a bit technical, because the quantum dimension is zero. So I don't think I fully understand how you get a TQFT from the Alexander polynomial.
In a more typical situation, like the Jones polynomial (related to SL(2) instead of the more confusing GL(1|1) in the Alexander setting), you have to be a bit careful about your target 3-category. For example, the usual Reshetikhin-Turaev TQFT doesn't come from a 0123 TQFT valued in tensor categories, because it is not the Drinfeld center of a semisimple category. (In this setting obstructions to extending to points are roughly given by the Witt group of Davydov, Mueger, Nikshych, and Ostrik.)
On the other hand, you can extend Reshetikhin-Turaev down to points if you use a different target category, like conformal nets (see <http://arxiv.org/abs/0912.5307>).
| 5 | https://mathoverflow.net/users/22 | 15882 | 10,640 |
https://mathoverflow.net/questions/15884 | 4 | In many places the existence of automorphism is acknowledged as one of the reasons why fine moduli spaces cannot exist. A typical example is the following.
Consider a curve $C$ with a nontrivial automorphism, for instance a hyperelliptic curve with its involution $\phi$. Now let $B$ be any scheme with a free action of (in this case) $\mathbb{Z}/(2)$.
Let $D$ be the quotient of $B \times C$ by the diagonal action of $\mathbb{Z}/(2)$ and let $B'$ be the quotient of $B$ by the action of $\mathbb{Z}/(2)$.
Consider the map $f \colon D \rightarrow B'$. Then the fibers of $f$ are all isomorphic to $C$, but for a suitable choice of $B$ the family $f$ is not isomorphic to the product $C \times B'$.
How can I get an explicit example of the last assertion?
>
> How can I produce $B$ as above such that $f$ is not isomorphic to the trivial family?
>
>
>
| https://mathoverflow.net/users/828 | How to use automorphisms to produce isotrivial non trivial families | Here is an explicit example. Denote by $E$ the elliptic curve $E=\mathbb C/(\mathbb Z+i\mathbb Z)$. Now let $B=C=E$. The involution that we will consider is
$$(z,w)\in (E\times E)\to (z+\frac{1}{2},-w).$$
Notice now that this quotient is not a direct product anymore. You can see this by calculating $H\_1(E\times E)/\mathbb Z\_2$ and verifying that it is not $\mathbb Z^4$.
Alternatively, you can notice that the quotient does not have a non-vanishing holomorphic volume form. Indeed if such form existed it could be lifted to $E\times E$ to a form $cdz\wedge dw$ ($c\ne 0$), which would give a contradiction since $dz\wedge dw$ is anti-invariant.
| 7 | https://mathoverflow.net/users/943 | 15885 | 10,642 |
https://mathoverflow.net/questions/15856 | 1 | A theorem of Erdos states:
"There exists an absolute constant c such that, if n>ck, and if a1/b1, a2/b2, ... are the Farey fractions of order n, then ax/bx and ax+k/bx+k are similarly ordered."
Can someone provide a definition of "similarly ordered" as used here?
Thanks for any insight.
Cheers, Scott
@ARTICLE{Erdos:1943,
author={Erd{\"o}s, Paul},
title={A note on {F}arey series},
journal={Quart. J. Math., Oxford Ser.},
fjournal={The Quarterly Journal of Mathematics. Oxford. Second Series},
volume={14},
year={1943},
pages={82--85},
issn={0033-5606},
mrclass={40.0X},
mrnumber={MR0009999 (5,236b)},
mrreviewer={G. Szeg{\"o}}
| https://mathoverflow.net/users/4111 | Similarly Ordered | Here is a link that gives the definition: <http://qjmath.oxfordjournals.org/cgi/pdf_extract/os-13/1/185>
It's to the paper by Mayer that Erdos refers to. (The link gives you just the first page of the paper, but fortunately the definition is on that page.)
| 3 | https://mathoverflow.net/users/1459 | 15891 | 10,646 |
https://mathoverflow.net/questions/15893 | 12 | This is a strictly technical question on peer-review systems currently employed in the mathematical literature, not a subjective discussion of merits/drawbacks of such systems, so I think/hope it's suitable for MO.
I have noticed that some journals (e.g. PNAS, CRAS, Nonlinearity...) always publish papers with the name of the editor who supervised the refereeing process ("Presented by X", "Recommended by X", "Communicated by X"). Most other journals, while having editors listed explicitly for each area (and hence in theory one could also know in most cases who supervised what), do not make this explicit.
I was wondering was difference it makes, as a junior author, to have an editor's name on a paper:
* is that a strong endorsement of the paper?
* a way to say that the journal is ultra-strict about the refereeing process?
* simply a full-disclosure practise of the journal?
* an incentive to publish there (when some editor is a "big name", or to make sure a specific person read your paper) ?
It's really not clear to me.
| https://mathoverflow.net/users/469 | Editors in peer-review systems | Journals of scientific societies (such as PNAS, CRAS) were, once upon a time, records of meetings and a member of the society would actually present the papers he accepted to the other members, so in this case I believe it's just a tradition. In the case of Nonlinearity it's probably just an affectation. Pay no attention to this stuff.
| 12 | https://mathoverflow.net/users/2290 | 15894 | 10,648 |
https://mathoverflow.net/questions/15888 | 3 | Hi,
I have a superspace spanned by 4 commuting coordinates + 2 anti-commuting ones $\{x^\mu,\theta^\alpha\}$, I have to do the change of coordinates $dx^\mu\to dy^\mu= dx^\mu+d\theta^\alpha \eta\_\alpha^{\;\:\mu}$ where $\eta$ have to be a local function i.e: $\eta\equiv\eta(x)$, and leave the $d\theta$'s unchanged, so how can I translate this change on coordinates themselves? In particular I need to express $\partial\_\mu$ in the new coordinate system.
If the $\eta$ fcts where global it would be simply $x^\mu\to y^\mu= x^\mu+\theta^\alpha \eta\_\alpha^{\;\:\mu}$, so what if $\eta$'s where not global?
| https://mathoverflow.net/users/2597 | Change of coordinates introduced through dx | First of all, I would not call $dx^\mu \mapsto dy^\mu$ a change of coordinates. Not every transformation on 1-forms is going to be the pullback by a smooth map. In fact, your question is whether the transformation $(dx^\mu,d\theta^\alpha) \mapsto (dy^\mu,d\theta^\alpha)$ comes from a change of coordinates.
Pullbacks commute with exterior derivative, so that if $dy^\mu$ is a pullback then applying $d$ to your $dy^\mu$ should give zero. If you do that you find that it is not unless $\eta$ is constant.
| 5 | https://mathoverflow.net/users/394 | 15898 | 10,651 |
https://mathoverflow.net/questions/993 | 70 | I was trying to explain finite groups to a non-mathematician, and was falling back on the "they're like symmetries of polyhedra" line. Which made me realize that I didn't know if this was actually true:
Does there exist, for every finite group G, a positive integer n and a convex subset S of R^n such that G is isomorphic to the group of isometries of R^n preserving S?
If the answer is yes (or for those groups for which the answer is yes), is there a simple construction for S?
I feel like this should have an obvious answer, that my sketchy knowledge of representations is not allowing me to see.
| https://mathoverflow.net/users/625 | Is every finite group a group of "symmetries"? | The permutohedron may have additional symmetries. For example, the order 3 permutohedron $\{(1,2,3),(1,3,2),(2,1,3),(3,1,2),(3,2,1)\}$ is a regular hexagon contained in the plane $x+y+z=6$, which has more than 6 symmetries.
I think we can solve it as follows:
Let $G$ be a group with finite order $n$ thought via Cayley's representation as a subgroup of $S\_n$.
Let $S=\{A\_1,...,A\_n\}$ be the set of vertices of a regular simplex centered at the origin in an $(n-1)$-dimensional real inner product space $V$. Let $r$ be the distance between the origin and $A\_1$. The set of vertices $S$ is an affine basis for $V$.
First unproven claim: If a closed ball that has radius $r$ contains $S$, then it is centered at the origin. Let $B$ be this ball.
The group of isometries that fix $S$ hence contains only isometries that fix the origin and permute the vertices, which can be identified with $S\_n$ in the obvious way. The same is true if we replace $S$ by its convex hull.
Now $G$ can be thought of as a group containing some of the symmetries of $S$.
Let $C=k(A\_1+2A\_2+3A\_3+\cdots+nA\_n)/(1+2+\cdots+n)$, with $k$ a positive real that makes the distance between $C$ and the origin a number $r'$ *slightly smaller* than $r$.
Let $GC=\{g(C) : g \in G\}$. It has $n$ distinct points, as a consequence of $S$ being an affine basis of $V$.
Let $P$ be the convex hull of the points of $S \cup GC$.
Remark: A closed ball of radius $r$ contains $P$ iff it is $B$. The intersection of the border of $B$ and $P$ is $S$.
Second unproven claim: The extremal points of $P$ are the elements of $S \cup GC$.
Claim: $G$ is the group of symmetries of $P$.
If $g$ is in $G$, $g$ is a symmetry of $GC$ and of $S$, and it is therefore a symmetry of $P$.
If $T$ is a symmetry of $P$, then $T(P)=P$, and in particular, $T(P)$ is contained in $B$, and hence $T(0)=0$ (i.e. $T$ is also a symmetry of $B$). $T$ must also fix the intersection of $P$ and the border of $B$, so $T$ permutes the points of $S$, and it can be thought of as an element $s \in S\_n$ sending $A\_i$ to $A\_s(i)$. And since $T$ fixes the set of extremal points of $P$, $T$ also permutes $GC$. Let's see that $s$ is in $G$.
Since $T(C)$ must be an element $g(C)$ of $GC$, we have $T(C)=g(C)$. But since $T$ is linear, $T(C/k)=g(C/k)$. Expanding,
$(A\_{s(1)}+2A\_{s(2)}+\cdots+nA\_{s(n)}/(1+\cdots+n)=(A\_{g(1)}+2A\_{g(2)}+\cdots+nA\_{g(n)})/(1+\cdots+n).$
For each $i \in \{1,...,n\}$ the coefficient that multiplyes $A\_i$ is $s^{-1}(i)/(1+\cdots+n)$ in the left hand side and $g^{-1}(i)/(1+\cdots+n)$ in the right hand side. It follows that $s=g$.
I think that, taking $n$ into account, the ratio $r'/r$ can be set to substantiate the second unproven claim. The first unproven claim may be a consequence of Jung's inequality.
EDIT: With the previous argument, we can represent a finite group of order $n$ as the group of linear isometries of a certain polytope in an $(n-1)$-dimensional real inner product space.
Now, if a finite group $G$ of linear isometries of an $(n-1)$-dimensional inner product space $V$ is given, can we define a polytope that has $G$ as its group of symmetries? Yes. I'll give a somehow informal proof.
Let $G=\{g\_1,...,g\_m\}$. Let $A=\{a\_1,...,a\_n\}$ be the set of vertices of a regular $n$-simplex centered at the origin of $V$. Let $S$ be the sphere centered at the origin that contains $A$, and let $C$ be the closed ball. Notice that $C$ is the only minimum closed ball containing $A$.
(Remark: The set $A$ need not be a regular simplex. It may be any finite subset of $S$ that intersects all the possible hemispheres of $S$. $C$ will then still be only minimum closed ball containing it.)
Remark: An isometry of $V$ is linear iff it fixes the origin.
Before proceeding, we need to be sure that the $m$ copies of $A$ obtained by making $G$ act on it are disjoint. If that is not the case, our set $A$ is useless but we can find a linear isometry $T$ such that $TA$ does the job. We consider the set $M$ of all linear isometries with the usual operator metric, and look into it for an isometry $T$ such that for all $(g,a)$ and $(h,b)$ distinct elements of $GxA$ the equation $g(Ta)=h(Tb)$ does not hold. Because each of the $n\cdot m(n\cdot m-1)$ equations spoils a closed subset of $M$ with empty interior(\*), most of the choices of $T$ will do.
Let $K=\{ga: g \in G, a \in A\}$. We know that it has $n\cdot m$ points, which are contained in the sphere $S$. Now let $e$ be a distance that is smaller than a quarter of any of the distances between different points of $K$. Now, around each vertex $a=a\_i$ of $A$ make a drawing $D\_i$. The drawing consists of a finite set of points of the sphere $S$, located near $a$ (at a distance smaller than $e$). One of the points must be $a$ itself, and the others (if any) should be apart from $a$ and very near each other, so that $a$ can be easily distinguished. Furthermore, for $i=1$ the drawing $D\_i$ must have no symmetries, i.e, there must be no linear isometries fixing $D\_1$ other than the identity. For other values of $i$, we set $D\_i={a\_i}$. The union $F$ of all the drawings contains $A$, but has no symmetries. Notice that each drawing has diameter less than $2\cdot e$.
Now let $G$ act on $F$ and let $Q$ be the union of the $m$ copies obtained. $Q$ is a union of $n\cdot m$ drawings. Points of different drawings are separated by a distance larger than $2\cdot e$. Hence the drawings can be identified as the maximal subsets of $Q$ having diameter less than $2\cdot e$. Also, the ball $C$ can be identified as the only sphere with radius $r$ containing $Q$. $S$ can be identified as the border of $C$.
Let's prove that the set of symmetries of $Q$ is $G$. It is obvious that each element of $G$ is a symmetry. Let $T$ be an isometry that fixes $Q$. It must fix $S$, so it must be linear. Also, it must permute the drawings. It must therefore send $D\_1$ to some $gD\_i$ with $g \in G$ and $1 \leq i \leq n$. But $i$ must be 1, because for other values of $i$, $gD\_i$ is a singleton. So we have $TD\_1=gD\_1$. Since $D\_1$ has no nontrivial symmetries, $T=g$.
We have constructed a finite set $Q$ with group of symmetries $G$. $Q$ is not a polytope, but its convex hull is a polytope, and $Q$ is the set of its extremal points.
(\*) To show that for any $(g,a)$ and $(h,b)$ distinct elements of $G \times A$ the set of isometries $T$ satisfying equation $g(Ta)=h(Tb)$ has empty interior, we notice that if an isometry $T$ satisfies the equation, any isometry $T'$ with $T'a=Ta$ and $T'b\neq Tb$ must do (since $h$ is injective). Such $T'$ may be found very near $T$, provided $\dim V>2$. The proof doesn't work for $n=1$ or $n=2$, but these are just the easy cases.
| 28 | https://mathoverflow.net/users/4118 | 15903 | 10,652 |
https://mathoverflow.net/questions/15897 | 17 | **Background/motivation**
One of the main reason to introduce (algebraic) stacks is build "fine moduli spaces" for functors which, strictly speaking, are not representable. The yoga is more or less as follows.
One notices that a representable functor on the category of schemes is a sheaf in the fpqc topology. In particular it is a sheaf in coarser topologies, like the fppf or étale topologies. Now some naturally defined functors (for instance the functor $\mathcal{M}\_{1,1}$ of elliptic curves) are not sheaves in the fpqc topology
(actually $\mathcal{M}\_{1,1}$ is not even an étale sheaf) so there is no hope to represent them.
Enters the $2$-categorical world and we introduce fibered categories and stacks. Many functors which are not sheaves arise by collapsing fibered categories which ARE stacks, so not all hope is lost. But, as not every fpqc sheaf is representable, we should not expect that every fpqc stack is in some sense "represented by a generalized space", so we make a definition of what we mean by an algebraic stack.
Let me stick with the Deligne-Mumford case. Then a DM stack is a fibered category (in groupoids) over the category of schemes, which
1) is a stack in the étale topology
2) has a "nice" diagonal
3) is in some sense étale locally similar to a scheme.
I don't need to make precise what 2) and 3) mean.
By the preceding philosophy we should expect that DM stacks generalize schemes in the same way that stacks generalize sheaves. In particular I would expect that DM stacks turn out to be stacks in finer topologies, just as schemes are sheaves not only in the Zariski topology (which is trivial) but also in the fpqc topology (which is a theorem of Grothendieck).
**Question**
>
> Is it true that DM stacks are actually stacks in the fpqc topology? And if not, did someone propose a notion of "generalized space" in the context of stacks, so that this result holds?
>
>
>
| https://mathoverflow.net/users/828 | In what topology DM stacks are stacks | The rule of thumb is this: Your DM (or Artin) stack will be a sheaf in the fppf/fpqc topology if the condition imposed on its diagonal is fppf/fpqc local on the target ("satisfies descent").
In other words, in condition 2 you asked that the diagonal be a relative scheme/relative algebraic space perhaps with some extra properties. If there if fppf descent for morphisms of this type (e.g., "relative algebraic space", "relative monomorphism of schemes"), you'll have something satisfying fppf descent. If there is fpqc descent for morphisms of this type (e.g., "relative quasi-affine scheme"), then you'll have something satisfying fpqc descent.
See for instance LMB (=Laumon, Moret-Bailly. Champs algebriques), Corollary 10.7. Alternatively: earlier this year I wrote up some [notes (PDF link)](http://www.tolypreygel.com/notes/note_stacks.pdf) that included an Appendix collecting in one place the equivalences of some standard definitions of stacks, including statements of the type above.
| 23 | https://mathoverflow.net/users/1921 | 15910 | 10,656 |
https://mathoverflow.net/questions/15863 | 5 | Let's say $f$ is a Dirichlet series which converges on the half-plane $\text{Re }s>\sigma$ to a function $f(s)$. Suppose further that $f(s)$ admits an analytic continuation to an entire function, together with the standard sort of functional equation. Let $g\_n$ be a sequence of Dirichlet series, also convergent on $\text{Re }s>\sigma$, which each admit an analytic continuation and functional equation, though their precise FEs may vary. We assume that $g\_n$ converges to $f$ in the following sense: for every $m>0$ there exists an $N$ for which the series $g\_n$ and $f$ match on every term up to the $m$th, for all $n>N$. Note this implies that $g\_n(s)$ converges to $f(s)$ for every $\text{Re }s>\sigma$.
Can it be said that $g\_n(s)$ converges to $f(s)$ for any $s$ outside the domain of convergence?
Perhaps that's too much to hope for, and you can't even expect that $g\_n(s)$ converges to $f(s)$ even for the point $s=\sigma$. I'd certainly be interested in a counterexample which does this!
| https://mathoverflow.net/users/271 | Convergence of a sequence of continuable Dirichlet series | (This answer is a community wiki version of a comment above by FC which answered the question.)
For any integer M, there exists a prime p such that chi\_p(n) = (n/p) = 1 for all n = 1...M. This means that the Dirichlet series L(s,V chi\_p) (for any representation V) "converges" in your sense to L(s,V). but they do not converge at s = 0. If V is trivial, then we are comparing zeta(0) = -1/2 with L(0,chi\_p) which grows without bound by Brauer-Siegel. I think in this class of examples one does get convergence at the critical point.
| 2 | https://mathoverflow.net/users/22 | 15920 | 10,663 |
https://mathoverflow.net/questions/15759 | 11 | I have recently had occasion to investigate the Fourier series of the function $f(x,y)=\log({2+\cos 2\pi x} +\cos{2\pi y})$. Accordingly, define
$I(m,n)=\int\_{0,0}^{1,1}f(x,y)\cos{2\pi mx}\cos{2\pi ny}dxdy$
which is the $(m,n)$th Fourier coefficient. If there is some easy change of variables or flick of the wrist to compute $I$ in closed terms, I haven't been able to find it. One may compute $I(0,0)=\frac{4G}{\pi}-\log 2$, where $G$ is Catalan's constant. As for other values, Mathematica's integration routine returns the values $I(1,1)=\frac{1}{2}+\frac{2}{\pi}$ and $I(2,1)=1-\frac{10}{3\pi}$ (and takes about 10 minutes per computation). I have tried and failed to affirm the following conjecture, which seems reasonable to present here.
**Conjecture**. If $(m,n)\neq(0,0)$, then $I(m,n)=a+\pi^{-1}b$ with $a,b\in \mathbb{Q}$.
Of course, if you could prove this, it would be great to give a closed formula for $a,b$.
Full disclosure: My interest in this stems from a formula of Kasteleyn for the partition function of dimer coverings of squares; in fact the evaluation of $I(0,0)$ given above can be extracted from his original paper.
| https://mathoverflow.net/users/1464 | A two-variable Fourier series and a strange integral | Here is a proof of the conjecture, a proof that also shows how to compute the integrals explicitly.
The proof is somewhat similar to David Speyer's approach, but instead of using multivariable residues, I will just shift a one-variable contour. Without loss of generality, $m>0$. Eliminating the trigonometric functions yields $I(m,n)=(-1/4\pi^2) J(m,n)$, where
$$ J(m,n) := \int\_{|w|=1} \int\_{|z|=1} \log(4+z+z^{-1}+w+w^{-1}) \, z^{m-1} w^{n-1} \, dz \, dw.$$
For fixed $w \ne -1$ on the unit circle, the values of $z$ such that $4+z+z^{-1}+w+w^{-1}$ are a negative real number and its inverse; let $g(w)$ be the value in $(-1,0)$. The function $\log(4+z+z^{-1}+w+w^{-1})$ of $z$ extends to the complement of the closed interval $[g(w),0]$ in a disk $|z|<1+\epsilon$. Shrink the contour $|z|=1$ like a rubber band so that it hugs the closed interval. The upper and lower parts nearly cancel, leaving
$$ J(m,n) = \int\_{|w|=1} \int\_{z=0}^{g(w)} -2\pi i \, z^{m-1} w^{n-1} \,dz \,dw,$$
with the $-2\pi i$ coming from the discrepancy in the values of $\log$ on either side of the closed interval. Thus
$$ J(m,n) = -\frac{2 \pi i}{m} \int\_{|w|=1} g(w)^m w^{n-1} \, dw.$$
Next use the rational parametrization $(g(w),w) = \left( (1-u)/(u+u^2), (u^2-u)/(1+u) \right)$ that David Speyer wrote out. There is a path $\gamma$ from $-i$ to $i$ in the right half of the $u$-plane that maps to $|w|=1$ (counterclockwise from $-1$ to itself) and gives the correct $g(w)$. Integrating the resulting rational function of $u$ gives
$$ J(m,n) = -\frac{2 \pi i}{m} \left.(f(u) + r \log u + s \log(1+u))\right|\_\gamma = -\frac{2 \pi i}{m} (f(i)-f(-i) + r \pi i + s \pi i/2),$$
for some $f \in \mathbf{Q}(u)$ and $r,s \in \mathbf{Q}$. This shows that $I(m,n)=a + b/\pi$ for some $a,b \in \mathbf{Q}$.
**Examples:** *Mathematica* got the answers wrong, presumably because it doesn't understand homotopy very well! Here are some actual values, computed using the approach above:
$$I(1,1) = \frac{1}{2} - \frac{2}{\pi}$$
$$I(2,1) = -1 + \frac{10}{3\pi}$$
$$I(20,10) = -\frac{14826977006}{5} + \frac{56979906453888224582}{6116306625 \pi}.$$
| 21 | https://mathoverflow.net/users/2757 | 15924 | 10,667 |
https://mathoverflow.net/questions/15925 | 3 | In *Wiersema: Brownian Motion Calculus* on p. 205 (in an Annex on Moment Generating Functions (mgf)) the following equation is being presented $${d^k \over d\theta^k} \left ({1\over k!}\theta^k\mathbb{E}[X^k]\right)={1\over k!}\theta^k{d^k \over d\theta^k} \mathbb{E}[X^k]$$ with $X$ being a random variable, $\theta$ the mgf-dummy variable and $\mathbb{E}$ the expectation operator (and $k$ for the $k$'th moment of $X$).
**My question**: Why is it possible to pull the term ${1\over k!}\theta^k$ out and in front of the $k$-times differentiation? Could anyone give me a hint or some intermediate steps? Or is this a typo? (Sorry if this is too elementary but I don't get it anyway - and I want to understand this!)
| https://mathoverflow.net/users/1047 | Moment Generating Function: Pulling a term out of k-times differentiation | That's clearly a typo. Unless I misunderstand the point of this, the right formula should be
$${d^k \over d\theta^k} \left ({1\over k!}\theta^k\mathbb{E}[X^k]\right)={1\over k!}\mathbb{E}\left[{d^k \over d\theta^k} \theta^k X^k\right]$$
In fact, this is what is said in words before the formula in the book you mentioned:
>
> The $k$th moment can be singled out by differentiating $k$ times with respect to $\theta$, and interchanging the differentiation and $\mathbb{E}$ (which is an integration)
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| 3 | https://mathoverflow.net/users/394 | 15929 | 10,671 |
https://mathoverflow.net/questions/15904 | 7 | Let $(X, x\_0)$ be a pointed space. Then we can define the homotopy groups $\pi\_i(X, x\_0)$ for $i \geq 1$. They are abelian groups for $i \geq 2$. It is well-known that the fundamental group $\pi\_1(X, x\_0)$ acts on each of the higher groups $\pi\_i(X, x\_0)$, and that this action generalizes to the [Whitehead Products](http://en.wikipedia.org/wiki/Whitehead_product) which are maps
$$ \pi\_p(X, x\_0) \times \pi\_q(X, x\_0) \to \pi\_{p+q -1}(X, x\_0).$$
The details are given in the wikipedia article I linked to above. Together the Whitehead products turn the graded group $\pi\_\*(X, x\_0)$ (for $\* > 0$) into a graded (quasi-) Lie algebra over $\mathbb{Z}$, where the grading is shifted so that $\pi\_i(X, x\_0)$ is in degree $(i-1)$. Well, it is a little funny since the bottom group is not necessarily abelian.
>
> This is all well and good, but what if we don't want to pick base points? Is there a similar algebraic gadget in that situation?
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If we don't pick base points, then it seems natural to consider the fundamental groupoid $\Pi\_1(X)$. Then the different homotopy groups of $X$ at different base points can be assembled into local systems on $X$. That is for each $i \geq 2$ we have a functor,
$$\pi\_i: \Pi\_1 X \to AB$$
where $AB$ is the category of abelian groups. This already incorporates the action of $\pi\_1$ on the higher homotopy groups but does it in a way which doesn't depend on the choice of base point.
>
> Question: Can we enhance these local systems with a structure which generalizes the Whitehead product, and if so what precisely is this extra structure?
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| https://mathoverflow.net/users/184 | Whitehead Products without Base Points? | As I posted in my comment, I think Paul's suggestion does work. Here's a (sloppy) description of how I think things will work:
The local systems you describe can be obtained, by passing to homotopy groups, from a "local system of loop spaces" $$ \Omega: \Pi\_{\leq \infty} X \to \Omega\mathbf{Spaces}$$
One can imagine that this corresponds under the Grothendieck construction to the free loop-space fibration $\Omega X \to LX \to X$. Alternatively, if we fix a basepoint and identify $X = BG$ for a simplicial group $G$, then this is just encoding the simplicial conjugation action of $G$ on itself.
Rather than think about (strangely-graded) Whitehead products, I prefer to think about (reasonably graded) Samelson products: We think of the structure (Whitehead product) on $\pi\_{\*+1} X $ as really being a structure (Samelson product) on $\pi\_{\*} \Omega X$. I claim that Samelson products give a functor
$$ \pi\_\*: \Omega\mathbf{Spaces} \to \mathbf{grqLie} $$ so that composing with the above gives our desired "local system of graded (quasi-)Lie algebras".
For convenience, I'll replace loop spaces with (strict) simplicial groups. Then, the Samelson product comes from noticing that the commutator map $[,]: G^2 \to G$ is trivial if one of the factors is the identity, and so factors through a pointed map $[,]: G \wedge G \to G$. This pointed map goes on to induce the (quasi-)Lie structure on homotopy. A group homomorphism $H \to G$ preserves commutators and identities, and so induces a map $H \wedge H \to G \wedge G$ compatible with the brackets, so that this construction is indeed functorial.
| 6 | https://mathoverflow.net/users/1921 | 15934 | 10,675 |
https://mathoverflow.net/questions/15902 | 22 | I would like to better understand the simplest case of the correspondence between Galois representations and (phi,Gamma)-modules. Namely, consider 1-dimensional Galois representations of $G\_{Q\_p}$ over $F\_p$ which are in correspondence with 1-dimensional etale (phi,Gamma)-modules over $F\_p((T))$.
There are finitely many such Galois representations. Moreover, their associated (phi,Gamma)-modules are very simple -- the action of phi and Gamma can be described (on some basis element) as scaling by an element of $F\_p$ (as opposed to $F\_p((T))$).
My question: can one see this directly on the (phi,Gamma)-module side? That is, given a 1-dimensional etale (phi,Gamma)-module $D$ over $F\_p((T))$, find a $D'$ isomorphic $D$ such that $D'$ has a basis in which the matrices for phi and elements of Gamma are in $F\_p^\times$.
| https://mathoverflow.net/users/86179 | One dimensional (phi,Gamma)-modules in char p | OK...I think I see how to do this now. In the end, I am seeing $(p-1)^2$ distinct $(\phi,\Gamma)$-modules which matches well with the Galois side.
To do this, let $D$ be any 1-dimensional etale $(\phi,\Gamma)$-module. Let $e$ be a basis, and set $\phi(e)=h(T)e$ with $h(T) \in F\_p((T))^\times$. Write $h(T) = h\_0 T^a f(T)$ with $h\_0 \in F\_p^\times$ and $f(T) \in F\_p[[T]]$ with $f(0)=1$.
Changing basis from $e$ to $u(T)e$ with $u(T) \in F\_p((T))^\times$ gives
$$
\phi(u(T)e) = u(T^p)h(T)e = \frac{u(T^p)}{u(T)} h(T) (u(T)e).
$$
I claim one can find $u(T)$ such that $u(T)/u(T^p)$ equals any element of $1+TF\_p[[T]]$. Indeed, for such an element $g(T)$, the infinite product $\prod\_{j=1}^\infty \phi^j(g(T))$ (which hopefully converges since $g(0)=1$) works.
Thus, we can change basis so that $\phi$ has the form $\phi(e) = h\_0 T^a e$ -- i.e. we can kill off the $f(T)$ term. Further, by making a change of basis of the form $e$ goes to $T^b e$, we may assume that $0 \leq a < p-1$.
Now, we use the fact that the $\phi$ and $\Gamma$ actions commute (which is a strong condition even in dimension 1). Namely, let $\gamma$ be a generator of $\Gamma$, and set $\gamma e = g(T) e$. Then $\gamma \phi e = \phi \gamma e$ implies
$$
((1+T)^{\chi(\gamma)}-1)^a g(T) = g(T^p) T^a.
$$
Comparing leading coefficients, we see this is only possible if $a=0$ and $g(T)$ is a constant.
Thus, $D$ has a basis $e$ so that $\phi(e) = h\_0 e$ and $\gamma(e) = g\_0 e$ with $h\_0,g\_0 \in F\_p^\times$ as desired.
Does this look okay? Any takers for the 2-dimensional case?
| 13 | https://mathoverflow.net/users/86179 | 15940 | 10,680 |
https://mathoverflow.net/questions/15944 | 2 | I call a set X of positive integers *strongly lcm-closed* if a,b ∈ X if and only if lcm(a,b) ∈ X. In the finite case X is the set of divisors of lcmx ∈ Xx. But in the infinite case it is more interesting, for example, $\{a \geq 1: a \not\equiv 0 \pmod p\}$ and $\{p^a:a \geq 0\}$ for any prime p, are strongly lcm-closed sets.
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> Which sets are strongly lcm-closed sets?
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This question arose in my [Ph.D. thesis](https://sites.google.com/site/rknmodn/papers) (p.107) where strongly lcm-closed sets describe where autotopisms of Latin squares give rise to subsquares.
As a side question:
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> Is there a common name for strongly lcm-closed sets?
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| https://mathoverflow.net/users/2264 | Classification of strongly lcm-closed sets | Given a [supernatural number](http://en.wikipedia.org/wiki/Supernatural_numbers) $N$, the set of positive integer divisors of $N$ is a strongly lcm-closed set. And any nonempty strongly lcm-closed set $X$ arises in this way, with $N$ equal to the supernatural lcm of the $x$ in $X$. (See Serre, *Galois cohomology* for the notion of supernatural number: it is a formal product over primes, $\prod\_p p^{n\_p}$, where each $n\_p$ is in $\lbrace 0,1,2, \dots,\infty \rbrace$.)
| 5 | https://mathoverflow.net/users/2757 | 15947 | 10,685 |
https://mathoverflow.net/questions/15951 | 23 | I've been reading up a bit on the fundamentals of formal logic, and have accumulated a few questions along the way. I am pretty much a complete beginner to the field, so I would very much appreciate if anyone could clarify some of these points.
1. A complete (and consitent) propositional logic can be defined in a number of ways, as I understand, which are all equivalent. I have heard it can be defined with one axiom and multiple rules of inferences or multiple axioms and a single rule of inference (e.g. Modus Ponens) - or somewhere inbetween. Are there any advantage/disvantages to either? Which is more conventional?
2. Propositional (zeroth-order) logic is simply capable of making and verifying logical statements. First-order (and higher order) logics can represent proofs (or increasing hierarchial complexity) - true/false, and why?
3. What exactly is the relationship between an nth-order logic and an (n+1)th-order logic, in general. An explanation mathematical notation would be desirable here, as long as it's not *too* advanced.
4. Any formal logic above (or perhaps including?) first-order is sufficiently powerful to be rendered inconsistent or incomplete by Godel's Incompleteness Theorem - true/false? What are the advantages/disadvantages of using lower/higher-order formal logics? Is there a lower bound on the order of logic required to prove all known mathematics today, or would you in theory have to use an arbitrarily high-order logic?
5. What is the role type theory plays in formal logic? Is it simply a way of describing nth-order logic in a consolidated theory (but orthogonal to formal logic itself), or is it some generalisation of formal logic that explains everything by itself?
Hopefully I've phrased these questions in some vaguely meaningful/understandable way, but apologies if not! If anyone could provide me with some details on the various points without assuming too much prior knowledge of the fields, that would be great. (I am an undergraduate Physics student, with a background largely in mathematical methods and the fundamentals of mathematical analysis, if that helps.)
| https://mathoverflow.net/users/602 | Propositional Logic, First-Order Logic, and Higher-Order Logics | This is a long list of questions! These are all related to a certain extent, but you might consider breaking it up into separate questions next time.
1. Proof theorists tend to prefer systems with many rules and few axioms such as natural deduction systems and Gentzen systems. The reason is that these are much easier to manipulate and visualize. (Look at the statement and proof of the Cut Elimination Theorem for an illustrative example.) Model theorists tend to prefer lots of axioms and just modus ponens, like Hilbert systems. The reason is that the semantics of such systems are easy to handle, and semantics is what model theorists really care about.
2. The real difference between propositional and first-order logic is quantification. Quantifiers are naturally more expressive than logical connectives.
3. There are many, many flavors of higher-order logic. The two main classifications are set-based systems and function-based systems. There are variants which don't really fit this division and there are variants which mix both. With strong enough internal combinatorics, these are all equivalent.
The set-based system have a set A of type 0 objects, which are considered atomic. Type 1 objects are elements of the powersets ℘(An). When the base theory has an internal pairing function (like arithmetic and set theory), the exponent n can be dropped. Then, type 2 objects are elements of the second powerset ℘(℘(A)), and similarly for higher types. Set-based higher types are somewhat inconvenient without an internal pairing function.
Function-based systems are similar, with functions An→A as type 1. Again, with an internal pairing function, the higher types streamline to A → A, (A → A) → A, ((A → A) → A) → A, etc. However, it is common to use more complex types in these systems.
4. Be careful how you read Gödel's Incompleteness Theorem. For first-order logic, the hypotheses state that the theory in question must be recusively enumerable and that it must be powerful enough to interpret a reasonable amount of arithmetic. Those are important hypotheses. (And they explain why no such theorem exists for propositional logic.) There are many variants for higher-order systems, but you need to be even more careful when stating them.
You can appreciate the difference between the different levels of higher-order logic by reading about [Gödel's Speed-Up Theorems](http://plato.stanford.edu/entries/goedel/#SpeUpThe).
Most of mathematics today is based on set theory, at least in a theoretical sense. In set theory, higher types are interpreted internally using powersets and sets of functions. They also extend transfinitely and this transfinite hierarchy of higher types was shown necessary to prove theorems very low in the hierarchy, such as [Martin's Borel Determinacy Theorem](http://en.wikipedia.org/wiki/Borel_determinacy_theorem).
5. You may want to read on [categorical logic](http://en.wikipedia.org/wiki/Categorical_logic).
| 35 | https://mathoverflow.net/users/2000 | 15954 | 10,690 |
https://mathoverflow.net/questions/15955 | 8 | Here's a mathematical modeling problem I came across while working on a hobby project.
I have a website that presents each visitor with a list of movie titles. The user has to rank them from most to least favorite. After each visit, I want to create a cumulative ranking that takes into account each visitor's individual ranking. Normally I would just take the mean ordinal rank: e.g., if Person A rated "Avatar" 10th and Person B rated it 20th, its cumulative rank would be 15th. However, new movies will be added to the list as the website grows, so each person will have ranked only a subset of the full movie list.
Any thoughts on how I can define "average rank" when some rankings do not cover the whole set? My best idea so far is to model this as a directed graph, where nodes are movies and weighted edges are preferences (e.g. "10 people ranked 'Avatar' right above 'District 9'"), and then finding sinks and sources. How else could one go about this?
(Sorry if this question is too applied.)
| https://mathoverflow.net/users/4135 | Defining "average rank" when not every ranking covers the whole set | There are a few different ways of approaching the problem. A good reference for this precise problem is '[Rank Aggregation methods For The Web](http://www.eecs.harvard.edu/~michaelm/CS223/rank.pdf)', by Dwork, Kumar, Naor and Sivakumar from the WWW conference in 2001. It's not the most recent work, but it lays out the mathematics nicely.
In general, the way to define an average is to define a metric, and then look at the point that minimizes the sum of distances from the individual lists. If the lists were full (i.e all defined over the same set), you could use [Spearman's footrule distance](http://www-stat.stanford.edu/~cgates/PERSI/papers/77_04_spearmans.pdf), or the [Kendall distance](http://en.wikipedia.org/wiki/Kendall_tau_distance). Since they are not, the general idea is to find some global ranking that's locally optimal (i.e there's no other full ranking that yields a smaller distance to the partial lists and can be obtained by flipping rankings from the current candidate).
As an aside, you can compare a full ranking with a partial ranking by merely projecting the full ranking onto the partial ranking and then computing one of the above mentioned distances.
Most of this is worked out in detail in the paper referenced, so your best bet is to start there.
| 5 | https://mathoverflow.net/users/972 | 15958 | 10,691 |
https://mathoverflow.net/questions/15909 | 6 | I need to calculate (or bound) the probability that one binomial variable is greater than other. Specifically, if $x \leftarrow B(n,p)$ and $y \leftarrow B(n,q)$, what is the probability that $y \geq x$? Informally, I will flip two bent coins with known biases $n$ times each, and I would like a convenient expression/bound on the probability that one beats the other. If it matters, I would specifically like *upper* bounds on $Pr(y \geq x)$ when $q < p$.
I've tried to following strategies:
* Consider $z=y-x$ to be a variable. Bound $Pr(z\geq 0)$ using Hoeffding's inequality. This *works*, but seems to give very weak bounds.
* Approximate $B(n,p)$ and $B(n,q)$ with normal distributions. Call these $P(x)$ and $Q(y)$. Then do the integral $\int\_{x=0}^n \int\_{y=x}^n Q(y) P(x) dy dx$. This gets ugly pretty fast and, even if it succeeded, wouldn't give a tight bound, but only an approximation.
Seems like it would be a well-studied problem, but searching doesn't uncover much. I appreciate any help!
| https://mathoverflow.net/users/4120 | Probability of one binomial variable being greater than another. | Edit: I've filled in a few more details.
The Hoeffding bound from expressing $Y-X$ as the sum of $n$ differences between Bernoulli random variables $B\_q(i)-B\_p(i)$ is
$$Prob(Y-X \ge 0) = Prob(Y-X + n(p-q) \ge n(p-q)) \le \exp\bigg(-\frac{2n^2 (p-q)^2}{4n}\bigg)$$
$$Prob(Y-X \ge 0) \le \exp\bigg(-\frac{(p-q)^2}{2}n\bigg)$$
I see three reasons you might be unhappy with this.
* The Hoeffding bound just isn't sharp. It's based on a Markov bound, and that is generally far from sharp.
* The Hoeffding bound is even worse than usual on this type of random variable.
* The amount by which the Hoeffding bound is not sharp is worse when $p$ and $q$ are close to $0$ or $1$ than when they are close to $\frac12$. The bound depends on $p-q$ but not how extreme $p$ is.
I think you might address some of these by going back to the proof of Hoeffding's estimate, or the Bernstein inequalities, to get another estimate which fits this family of variables better.
For example, if $p=0.6$, $q=0.5$, or $p=0.9$, $q=0.8$, and you want to know when the probability is at most $10^{-6}$, the Hoeffding inequality tells you this is achieved with $n\ge 2764$.
For comparison, the actual minimal values of $n$ required are $1123$ and $569$, respectively, by brute force summation.
One version of the Berry-Esseen theorem is that the Gaussian approximation to a cumulative distribution function is off by at most
$$0.71 \frac {\rho}{\sigma^3 \sqrt n}$$
where $\rho/\sigma^3$ is an easily computed function of the distribution which is not far from 1 for the distributions of interest. This only drops as $n^{-1/2}$ which is unacceptably slow for the purpose of getting a sharp estimate on the tail. At $n=2764$, the error estimate from Berry-Esseen would be about $0.02$. While you get effective estimates for the rate of convergence, those estimates are not sharp near the tails, so the Berry-Esseen theorem gives you far worse estimates than the Hoeffding inequality.
Instead of trying to fix Hoeffding's bound, another alternative would be to express $Y-X$ as a sum of a (binomial) random number of $\pm1$s by looking at the nonzero terms of $\sum (B\_q(i)-B\_p(i))$. You don't need a great lower bound on the number of nonzero terms, and then you can use a sharper estimate on the tail of a binomial distribution.
The probability that $B\_q(i)-B\_p(i) \ne 0$ is $p(1-q) + q(1-p) = t$. For simplicity, let's assume for the moment that there are $nt$ nonzero terms and that this is odd. The conditional probability that
$B\_q(i)-B\_p(i) = +1$ is $w=\frac{q(1-p)}{p(1-q)+q(1-p)}$.
The [Chernoff bound](http://en.wikipedia.org/wiki/Chernoff_bound) on the probability that the sum is positive is $\exp(-2(w-\frac 12)^2tn)$.
$$ \exp(-2\bigg(\frac{q(1-p)}{p(1-q)+q(1-p)} - \frac 12\bigg)^2 \big(p(1-q) + q(1-p)\big) n)$$
is not rigorous, but we need to adjust $n$ by a factor of $1+o(1)$, and we can compute the adjustment with another Chernoff bound.
For $p=0.6, q=0.5$, we get $n \ge 1382$. For $p=0.9, q=0.8$, we get $n \ge 719$.
The Chernoff bound isn't particularly sharp. Comparison with a geometric series with ratio $\frac{w}{1-w}$ gives that the probability that there are more $+1$s than $-1$s is at most
$${nt \choose nt/2} w^{nt/2} (1-w)^{nt/2} \frac {1-w}{1-2w}$$
This gives us nonrigorous bounds of $nt \gt 564.4, n \ge 1129$ for $p=0.6,q=0.5$ and
$nt\gt 145.97, n\ge 562$ for $p=0.9,q=0.8$. Again, these need to be adjusted by a factor of $1+o(1)$ to get a rigorous estimate (determine $n$ so that there are at least $565$ or $146$ nonzero terms with high probability, respectively), so it's not a contradiction that the actual first acceptable $n$ was $569$, greater than the estimate of $562$.
I haven't gone through all of the details, but this shows that the technique I described gets you much closer to the correct values of $n$ than the Hoeffding bound.
| 5 | https://mathoverflow.net/users/2954 | 15961 | 10,693 |
https://mathoverflow.net/questions/11396 | 5 | Let $F$ be a local field of characteristic zero (for simplicity), $\overline{F}$ an algebraic closure of $F$ and $L/F$ a fixed finite Galois extension. If $G$ is a linear algebraic group defined over $F$, then the Galois cohomology group $H^1(F,G)$ can be defined as a direct limit of $H^1(K/F,G)$, where $K$ runs through finite Galois subextensions of $\overline{F}$.
Now the question is: under what conditions is this direct limit just $H^1(L/F,G)$?
I guess there might be restrictions on both $L$ and $G$.
| https://mathoverflow.net/users/1832 | Galois cohomology of linear groups over local fields | As Hunter and Sean noted, since the inflation map ${\rm{H}}^1(L/F,G(L)) \rightarrow {\rm{H}}^1(F,G)$ is
injective and ${\rm{H}}^1(F,G)$ is always finite (Borel-Serre), such an $L$ always exists. Below we give an explicit sufficient condition on $L$ (often satisfied) when $G$ is connected. (One could probably do better with a closer consideration of the Tate local duality aspects of the argument. I am lazy at this step.) This rests on something deeper than the Borel-Serre result: the Kneser-Bruhat-Tits theorem on vanishing of degree-1 Galois cohomology for simply connected semisimple groups over non-archimedean local fields.
First we set up some notation.
Let $U = \mathcal{R}\_u(G)$ denote the unipotent radical of $G$ (this is a good notion over $F$ since $F$ is perfect), and let $G' = G/U$ denote the maximal reductive quotient. The identity component $Z'$ of the center of $G'$ is an $F$-torus, and the derived group $\mathcal{D}(G')$ is a semisimple group, say with simply connected central cover $\mathcal{G} \twoheadrightarrow \mathcal{D}(G')$. The preimage $\mu$ of the central
subgroup $Z' \cap \mathcal{D}(G')$ is a finite $F$-group of multiplicative type.
(It is the kernel of the central covering map if $Z' = 1$.)
Proposition: Assume $G$ is connected and use notation as above.
Let $F'/F$ be a finite Galois splitting field for $G'$ (thus for $Z'$ and dual of
$\mu$). If $L/F$ is a finite Galois extension containing $F'$ with $[L:F']$ divisible by the order of $\mu$ then ${\rm{H}}^1(F,G) = {\rm{H}}^1(L/F,G(L))$.
In particular, if $G$ is a split connected reductive $F$-group then $L/F$ works provided that $[L:F]$ is divisible by the order of the center of the simply connected
central cover of $\mathcal{D}(G)$.
Remark: If $T$ is a maximal $F$-torus in $G$ then it maps isomorphically onto one for $G'$, so could take $F'/F$ to be splitting field for $T$.
Proof: Since $F$ has characteristic 0, the quotient map $G \twoheadrightarrow G'$ admits a section $\sigma$ over $F$, which is to say there exists a connected reductive $F$-subgroup $H \subseteq G$ such that $H \ltimes U \simeq G$ via multiplication (this is a so-called Levi $F$-subgroup of $G$); beware that over any algebraically closed field $k$ with nonzero characteristic Levi subgroups can fail to exist. (A basic natural counterexample is, loosely speaking, ${\rm{SL}} \_2(W \_2(k))$ as a $k$-group, where $W \_2$ denotes length-2 Witt vectors; see Appendix A.6 in the book "Pseudo-reductive groups".)
Using $\sigma$ (or $H$), the natural restriction map ${\rm{H}}^1(F,G) \rightarrow {\rm{H}}^1(F,G')$ is surjective. It is also injective. Indeed, a standard twisting argument (as explained in Serre's book on Galois cohomology) identifies fibers with ${\rm{H}}^1(F,U')$ for various $F$-forms $U'$ of $U$. But since the ground field is perfect, every smooth connected unipotent group is split (i.e., admits a composition series whose successive quotients are $\mathbf{G}\_a$) and hence has trivial ${\rm{H}}^1$. Thus, we get the asserted bijectivity. So far we have not used anything about $F$ other than that it has characteristic 0.
We likewise have ${\rm{H}}^1(E/F,U'(E)) = 1$ for any smooth connected unipotent $F$-group $U'$ and any Galois extension $E/F$, so the same argument gives that ${\rm{H}}^1(E/F,G(E)) \rightarrow {\rm{H}}^1(E/F,G'(E))$ is bijective for any Galois extension $E/F$. Thus, the injective inflation map
$${\rm{H}}^1(E/F,G(E)) \rightarrow {\rm{H}}^1(F,G)$$
is bijective if and only if the same holds for $G'$ in place of $G$.
Consider the central extension structure
$$1 \rightarrow \mu \rightarrow Z' \times \mathcal{G} \rightarrow G' \rightarrow 1$$
over $F$, where the second map uses multiplication. By Kneser-Bruhat-Tits,
we have an exact sequence of pointed sets
$${\rm{H}}^1(F,Z') \rightarrow {\rm{H}}^1(F,G') \rightarrow {\rm{H}}^2(F,\mu)$$
and similarly over $L$, with a commutative diagram using restriction maps.
By local class field theory, the restriction map from ${\rm{H}}^2(F',\mu)$ to
${\rm{H}}^2(L,\mu)$ vanishes, so likewise for ${\rm{H}}^2(F,\mu) \rightarrow {\rm{H}}^2(L,\mu)$. (This is weak; by translating restriction through Tate local duality we can surely give a better sufficient "lower bound" for such an $L$.) For any such $L$, it follows that restriction from ${\rm{H}}^1(F,G')$ to ${\rm{H}}^1(L,G')$ lands in the image of ${\rm{H}}^1(L,Z')$. But this vanishes since the $F$-torus $Z'$ is even $F'$-split, let
alone $L$-split. Thus, ${\rm{H}}^1(F,G') \rightarrow {\rm{H}}^1(L,G')$ vanishes, which is to say that $L$ "works" for $G'$, so the same holds for $G$.
QED
| 14 | https://mathoverflow.net/users/3927 | 15963 | 10,695 |
https://mathoverflow.net/questions/15087 | 35 | Are there any general methods for computing fundamental group or singular cohomology (including the ring structure, hopefully) of a projective variety (over C of course), if given the equations defining the variety?
I seem to recall that, if the variety is smooth, we can compute the H^{p,q}'s by computer -- and thus the H^n's by Hodge decomposition -- is this correct? However this won't work if the variety is not smooth -- are there any techniques that work even for non-smooth things?
Also I seem to recall some argument that, at least if we restrict our attention to smooth things only, all varieties defined by polynomials of the same degrees will be homotopy equivalent. The homotopy should be gotten by slowly changing the coefficients of the polynomials. Is something like this true? Does some kind of argument like this work?
| https://mathoverflow.net/users/83 | Computing fundamental groups and singular cohomology of projective varieties | This is an interesting question. To repeat some of the earlier answers, one should be able to get one's hands on a triangulation algorithmically using real algebro-geometric methods, and thereby compute singular cohomology and (a presentation for) the fundamental group. But this should probably be a last resort in practice. For smooth projective varieties, as people have noted, one can compute the Hodge numbers by writing down a presentation for the sheaf p-forms and then apply standard Groebner basis techniques to compute sheaf cohomology. This does work pretty well on a computer. For specific classes, there are better methods. For smooth complete intersections, there is a generating function for Hodge numbers due to Hirzebruch (SGA 7, exp XI), which is extremely efficient to use.
As for the fundamental group, if I had to compute it for a general smooth projective variety, I would probably use a Lefschetz pencil to write down a presentation.
For singular varieties, one can still define Hodge numbers using the mixed Hodge structure on cohomology. The sum of these numbers are still the Betti numbers. I expect these Hodge numbers are still computable, but it would somewhat unpleasant to write down a general algorithm. The first step is to build a simplicial resolution using resolution of singularities. My colleagues who know about resolutions assure me that this can be done algorithmically now days.
(This is my first reply in this forum. Hopefully it'll go through.)
| 25 | https://mathoverflow.net/users/4144 | 15974 | 10,703 |
https://mathoverflow.net/questions/15979 | 14 | In the search for a Weil cohomology theory $H$ over a field $K$ (with $\text{char}(K)=0$) for varieties in characteristic $p$, a classical argument by Serre shows that the coefficient field cannot be a subfield of $\mathbb{R}$ or of $\mathbb{Q}\_p$; an obvious choice is to take $\mathbb{Q}\_\ell$ for a prime $\ell \neq p$.
Now, we can try to make a Weil cohomology theory by taking the sheaf cohomology with constant sheaves with the Zariski topology, but this does not work as all cohomology vanishes.
Grothendieck's insight was that we can find a different topology, for example the étale topology. Then we can build a Weil cohomology theory with coefficients in $\mathbb{Q}\_\ell$ by taking cohomology with coefficients in the constant sheaves $\mathbb{Z}/ \ell^n\mathbb{Z}$ and then taking the inverse limit with respect to $n$ and tensor with $\mathbb{Q}\_\ell$: this gives $\ell$-adic cohomology.
But it is not so clear to me why the étale topology is best suited at this task. What happens if we repeat the above procedure on other sites? Does the cohomology theory we get fail to be a Weil cohomology theory?
P.S.: Information for fields other than $\mathbb{Q}\_\ell$ would also be nice!
| https://mathoverflow.net/users/362 | Motivation for the étale topology over other possibilities | As observed, the Zariski topology has too few open subsets to compute cohomology with constant coefficients. It had already been observed by Serre that etale covers were enough to trivialize principal bundles for many algebraic groups. That suggested using etale covers. The etale "topology" is the coarsest for which the inverse function theorem holds. The flat topology also gives Weil cohomologies (the same ones as the etale topology), but why use flat covers when etale covers are enough.
| 14 | https://mathoverflow.net/users/930 | 15980 | 10,708 |
https://mathoverflow.net/questions/15981 | 1 | I found the following closed form solution for the abovementioned problem:
$${1\over k^n}\cdot{k!\over (k-m)!}\cdot{\{{n\over m}\}}$$ with ${\{{n\over m}\}}$ being the Stirling Number of the second kind.
Although it seems to have some intuition and seems to work for a sample problem for which I have the solution this closed form is not from a trusted source. Unfortunately I can't find any other source.
**My question**: Could anyone acknowledge this closed form solution and/or give me a hint where to find a citable source.
| https://mathoverflow.net/users/1047 | Probability of n k-sided dice showing exactly m different faces | Applied probability by Kenneth Lange deals with this problem on page 74. It is on Google books, here is the [URL](http://books.google.com/books?id=otp4TDnz6FwC&pg=PA74&lpg=PA74&dq=Stirling+Number+of+the+second+kind+dice&source=bl&ots=YHHbtcknAO&sig=sGd1_d1UXtlNkWGYINVZH6Tadu0&hl=en&ei=boiBS-CpDImKsgOq6en5Aw&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBMQ6AEwAzgK#v=onepage&q=Stirling%2520Number%2520of%2520the%2520second%2520kind%2520dice&f=false).
| 5 | https://mathoverflow.net/users/1098 | 15983 | 10,710 |
https://mathoverflow.net/questions/15987 | 10 | I was feeling a bit rusty on my field theory, and I was reviewing out of McCarthy's excellent book, *Algebraic Extensions of Fields*. Out of Chapter 1, I was able to work out everything "left to the reader" or omitted except for one corollary, stated without proof ([see here](http://books.google.com/books?id=dC5HvUrfy44C&lpg=PP1&pg=PA19#v=onepage&q=&f=false) for the page in the book):
>
> Let $K/k$ be a finite normal
> extension. Then $K$ can be obtained by
> a purely inseparable extension,
> followed by a separable extension.
>
>
>
The text immediately preceding this implies that the intermediate field that's going to make this happen is $F=\{a\in K:\sigma(a)=a$ for all $\sigma\in Gal(K/k)\}$, and I understand his argument as to why $F/k$ is purely inseparable (in fact, that's the theorem, Theorem 21, which this is a corollary to). What I don't understand is why $K/F$ is separable; I don't see how we've ruled out it being non-purely inseparable.
Note that I will be making a distinction between *non-purely inseparable* (inseparable, but not purely inseparable) and *not purely inseparable* (either separable or non-purely inseparable).
Here are some observations / my general approach:
* One big thing that seemed promising was Theorem 11 (at the bottom of [this page](http://books.google.com/books?id=dC5HvUrfy44C&lpg=PP1&pg=PA13#v=onepage&q=&f=false)), which is basically the reverse of the corollary I'm having trouble with:
>
> Let $K$ be an arbitrary algebraic extension of $k$. Then $K$ can be obtained by separable extension followed by a purely inseparable extension.
>
>
>
(the separable extension referred to is of course the separable closure of $k$ in $K$). It seems like we want to use Theorem 11 on $K/F$, and argue that there can't be "any more" pure inseparability, but I couldn't figure out a way of doing this.
* Theorem 21 is actually an "if and only if" (that is, $a\in K$ is purely inseparable over $k$ iff $\sigma(a)=a$ for all $\sigma\in Gal(K/k)$). Because this implies that any $a\in K$ with $a\notin F$ is not purely inseparable over $k$, we have that $F$ is the maximum (not just maximal) purely inseparable extension of $k$ in $K$.
* If any $a\in K$ were purely inseparable over $F$, by Theorem 8 ([see here](http://books.google.com/books?id=dC5HvUrfy44C&lpg=PP1&pg=PA10#v=onepage&q=&f=false)), there is some $e$ for which $a^{p^e}\in F$. But by the same theorem, since $F/k$ is purely inseparable, there is some $b$ for which $(a^{p^e})^{p^b}=a^{p^{e+b}}\in k$. Thus $a$ would be purely inseparable over $k$ by the converse (Corollary 1 to Theorem 9, [see here](http://books.google.com/books?id=dC5HvUrfy44C&lpg=PP1&pg=PA11#v=onepage&q=&f=false)), and hence be in $F$. Thus, $K$ (and any field between $K$ and $F$, besides $F$ itself) is not purely inseparable over $F$.
So, that's why I don't see how we've ruled out $K/F$ being *non-purely inseparable*. Sorry about making lots of references to the book - I'm just not sure what previously established results McCarthy intended to be used, and I wanted to point out what I saw as the important ones for people not familiar with the book. I'm sure I'm missing something obvious here. Does anyone see the last bit of the argument?
| https://mathoverflow.net/users/1916 | If K/k is a finite normal extension of fields, is there always an intermediate field F such that F/k is purely inseparable and K/F is separable? | Edit: My mistake, I misread your post. Here's the correct answer.
<http://books.google.com/books?id=FJmiSW1KRBAC&lpg=PP1&ots=k1ecm3FdbZ&dq=lang%20algebra&pg=PA251#v=onepage&q=&f=false>
Proposition 6.11
| 2 | https://mathoverflow.net/users/1353 | 15988 | 10,713 |
https://mathoverflow.net/questions/15907 | 19 | It is known that there exists a fine moduli space for marked (nonalgebraic) K3 surfaces over $\mathbb{C}$. See for example the book by Barth, Hulek, Peters and Van de Ven, section VIII.12. Of course the marking here is necessary, otherwise the presence of automorphisms can be used to construct isotrivial non trivial families of K3.
Assume that we want to construct an algebraic analogue of this. Of course some structure has to be added; I am thinking of something like torsion points for elliptic curves.
>
> Is there a way to add some structure and actually build a fine moduli space of (K3 + structure) which is defined over $\mathbb{Z}$?
>
>
>
Of course this will only parametrize algebraic K3 surfaces, so it will need to have infinitely many components. A second question, if we do not want to add structure, is
>
> Is there a fine moduli stack of algebraic K3 surfaces which is algebraic (either in the DM or Artin sense)?
>
>
>
It is not difficult to produce K3 surfaces with a denumerable infinity of automorphisms, so I'm not really expecting the answer to be yes, but who knows.
Edit: since there seems to be some confusion in the answers, the point is that I'm asking for a FINE moduli space. I'm aware that one can consider moduli spaces of polarized K3, and this is why I wrote "Of course this will only parametrize algebraic K3 surfaces, so it will need to have infinitely many components."
| https://mathoverflow.net/users/828 | Moduli space of K3 surfaces | I think you are looking for this
<http://arxiv.org/pdf/math/0506120>
which is the same as:
Rizov, Jordan Moduli stacks of polarized $K3$ surfaces in mixed characteristic. Serdica Math. J. 32 (2006), no. 2-3, 131--178.
It builds on an earlier result:
Olsson, Martin C. Semistable degenerations and period spaces for polarized $K3$ surfaces. Duke Math. J. 125 (2004), no. 1, 121--203.
which builds on Friedman's Ph.D. (the algebraic proof of Piatetski-Shapiro and Shafarevich of global Torreli for K3's).
Enough history for one answer I guess.
| 13 | https://mathoverflow.net/users/404 | 15997 | 10,718 |
https://mathoverflow.net/questions/15967 | 4 | I have a camera matrix $P$ which defines a projective transformation $\mathbb{P}^3 \rightarrow \mathbb{P}^2$. In the former space there is a plane $[ x|\pi^Tx=0 ]$. The image of the plane under $P$ does not preserve angles.
How can I find a transformation $H : \mathbb{P}^2 \rightarrow \mathbb{P}^2$ such that a right angle in the plane remains a right angle after applying $H \circ P$?
The application for this problem is extracting texture from a photo of a planar surface where the surface and camera locations are known.
| https://mathoverflow.net/users/818 | Rectifying texture from image | By picking orthogonal coordinates in the given plane you can make an angle preserving projective map $\mathbb{P}^2\to\mathbb{P}^3$ whose image is the given plane. Composing with your camera mapping, you now have a mapping $G\colon\mathbb{P}^2\to\mathbb{P}^2$ that does not preserve angles. Let $H=G^{-1}$. The composition $HG=I$ clearly preserves angles; hence so does $HP$ when restricted to the given plane.
(Reverse the order of composition if you follow the usual computer graphics convention of letting matrices act on the right.)
| 2 | https://mathoverflow.net/users/802 | 16006 | 10,725 |
https://mathoverflow.net/questions/15991 | 3 | In general, the étale topology does not form a topology in the strict sense. However, is there any subcategory of $Sch$ where we can realize the étale topology as an honest topology on some scheme?
| https://mathoverflow.net/users/1353 | Can the étale topology ever be realized as an "honest" topology? | I think Senor Borger speaks truth.
Maybe I am talking non-sense, but maybe this could be a sketch of some way to proceed (maybe not....):
Key feature of a site which we cannot have in a classical topology is that there may be several distinct "inclusion" morphisms from a smaller open into a bigger one.
Hence, we get a "fail" as soon as we encounter a scheme admitting an étale open with several inclusion maps, for example the inclusion of some open into the whole scheme.
Suppose we have a scheme admitting a non-trivial étale covering (i.e. one with non-trivial étale fundamental group), this is also an étale open, but by the action of the étale fundamental group it has several non-equal inclusions into the whole scheme. So we have an effect which is impossible in a classical, "honest" as you say, topology.
In the case our scheme indeed has trivial étale fundamental group, pick a Zariski open which has non-trivial étale fundamental group. Basically, take some finite covering and remove the ramification divisor, the complement is Zariski open. Now compose this Zariski open with the finite covering (which has become étale as the ramification has gone to nowhere land), same problem occurs, so again we get a "fail". Hence, to get honest topology, we need to have trivial étale fund. group for the scheme and all its Zariski opens, well.... that sounds pretty close to being just some closed points again.
| 4 | https://mathoverflow.net/users/3888 | 16008 | 10,727 |
https://mathoverflow.net/questions/16010 | 1 | The terminology would suggest that a separable field extension is so because the resulting field extension has some sort of separable topology, and that a normal extension corresponds to one with a normal topology.
I imagine this is true, or else they wouldn't have named them in such a way.
Also, I'm not sure what subfield this falls under, so if you could suggest additional tags, that would be great as well.
| https://mathoverflow.net/users/96 | Do separable and normal have topological meanings for fields? | This is incorrect. Words like "separable" and "normal" occur in unrelated ways in various parts of mathematics. (Normal subgroup, separable differential equations...)
Other words, too. Like "regular", "perfect" ...
"Separable" in toplogy... Does it mean something can be "separated"? What? I believe it goes back to Fréchet, but what did he mean by it?
| 4 | https://mathoverflow.net/users/454 | 16015 | 10,731 |
https://mathoverflow.net/questions/15990 | 18 | My question points in a direction similar to [Qiaochu's](https://mathoverflow.net/questions/10947/whats-the-analogue-of-the-hilbert-class-field-in-the-following-analogy), but it's not the same (or so I think). Let me provide you with a little bit of background first.
Let E be an elliptic curve defined over some number field K. The Tate-Shafarevich group of E/K consists of certain curves of genus 1, isomorphic to E over some extension, with points everywhere locally. In the simplest case of an element of order 2, such a curve has the form C: y2=f4(x) for some quartic polynomial f4(x) in K[X]; here, C does not have a K-rational point, but has points in every completion of K.
If we look at E in some extension L/K, this curve C still has points everywhere locally, but if it has a global point (in L) we say that the corresponding element in the Tate-Shafarevich group of E/K capitulates. Heegner's Lemma says that elements of order 2 cannot capitulate in extensions of odd degree, which is the analogue of the similarly trivial observation that ideals generating a class of order 2 cannot capitulate (become principal) in an extension of odd degree.
I gave a few talks on the capitulation of Tate-Shafarevich groups more than 10 years ago. A little later the topic became almost fashionable under the name of "visualizing" elements of Sha. I discussed the following question with Farshid Hajir back then, but eventually nothing came out of it. Here it is. For capitulation of ideal classes, there is a "canonical" extension in which this happens: the Hilbert class field. So my question is:
* may we still dream about the existence of a curve with all the right properties, or are there reasons why such a thing should not exist?
We also know that capitulation is not the correct notion for defining the Hilbert class field, which is the largest unramified abelian extension of a number field. These notions do not seem to make any sense for elliptic curves, but we can characterize the Hilbert class field also in the following way: among all finite extensions L/K for which the norm of the class group of L down to K is trivial, the Hilbert class field is the smallest.
Taking the norm of Sha of an elliptic curve defined over L down to K does make sense (just add the equivalence classes of the conjugate homogeneous spaces using the Baer-sum construction or in the appropriate cohomology group). So here's my second question:
* Has this "norm map" been studied in the literature?
(I know that the norm map from E(L) to E(K) was investigated a lot, in particular in connection with Heegner points).
Let me add that I do not assume that such a "Hilbert class curve" can be found among the elliptic curves defined over some extension field; if there is a suitable object, it might be the Jacobian of a curve of higher genus or an abelian variety coming from I don't know where.
| https://mathoverflow.net/users/3503 | What's the Hilbert class field of an elliptic curve? | **EDIT**: This is a completely new answer.
I will prove that your specific suggestion of defining a Hilbert class field of an elliptic curve $E$ over $K$ does not work. I am referring to your proposal to take the smallest field $L$ such that the corestriction (norm) map $\operatorname{Sha}(L) \to \operatorname{Sha}(K)$ is the zero map. (I have to assume the Birch and Swinnerton-Dyer conjecture (BSD), though, for a few particular elliptic curves over $\mathbf{Q}$.)
>
> **Theorem:** Assume BSD. There exists a number field $K$ and an elliptic curve $E$ over $K$ such that there is *no smallest* field extension $L$ of $K$ such that $\operatorname{Cores} \colon \operatorname{Sha}(L,E) \to \operatorname{Sha}(K,E)$ is the zero map.
>
>
>
**Proof:** We will use BSD data (rank and order of Sha) from [Cremona's tables](http://www.warwick.ac.uk/staff/J.E.Cremona//ftp/data/). Let $K=\mathbf{Q}$, and let $E$ be the elliptic curve 571A1, with Weierstrass equation
$$y^2 + y = x^3 - x^2 - 929 x - 10595.$$
Then $\operatorname{rk} E(\mathbf{Q})=0$ and $\#\operatorname{Sha}(\mathbf{Q},E)=4$. Let $L\_1 = \mathbf{Q}(\sqrt{-1})$ and $L\_2 =\mathbf{Q}(\sqrt{-11})$. It will suffice to show that the Tate-Shafarevich groups $\operatorname{Sha}(L\_i,E)$ are trivial.
Let $E\_i$ be the $L\_i/\mathbf{Q}$-twist of $E$. MAGMA confirms that $E\_1$ is curve 9136C1 and $E\_2$ is curve 69091A1. According to Cremona's tables, $\operatorname{rk} E\_i(\mathbf{Q})=2$ and $\operatorname{Sha}(\mathbf{Q},E\_i)=0$, assuming BSD. Thus $\operatorname{rk} E(L\_i) = 0+2=2$ and $\operatorname{Sha}(L\_i,E)$ is a $2$-group. On the other hand, MAGMA shows that the $2$-Selmer group of $E\_{L\_i}$ is $(\mathbf{Z}/2\mathbf{Z})^2$. Thus $\operatorname{Sha}(L\_i,E)[2]=0$, so $\operatorname{Sha}(L\_i,E)=0$.
| 15 | https://mathoverflow.net/users/2757 | 16018 | 10,734 |
https://mathoverflow.net/questions/13856 | 7 | I'm interested in results about functorially-defined subgroups (in a loose sense), especially in the non-abelian case, and would like to know about references I may have missed.
The question, it seems, comes up in its simplest form when noticing a number of common subgroups (the center, commutator subgroup, Frattini subgroup, etc) are characteristic. The characteristicity can be justified by the fact that the object mappings that define of those subgroups give rise to [subfunctors](http://ncatlab.org/nlab/show/subfunctor) of the identity functor, on the [core](http://ncatlab.org/nlab/show/core) category of Grp.
Hence I'm interested in functors F in Grp (or a carefully chosen sub-category) such that
∀ A F(A) ⊆ A and ∀ A,B ∀ f ∈ hom(A, B), f(F(A)) ⊆ F(B)
Does that ring a bell ?
The topic was mentioned a couple of months ago on [Mathoverflow](https://mathoverflow.net/questions/6925/how-can-we-formalize-the-naturality-of-certain-characteristic-subgroups), and can be traced back to (at least) 1945, where Saunders MacLane explains it in some detail in the third chapter of [*A General Theory of Natural Equivalences*](http://www.jstor.org/stable/1990284).
In between, it seems that those functors have been baptised radicals, pre-radicals or subgroup functorials, and studied mostly in the framework of ring theory, notably by A. Kurosh. Among a number of not-so-recent (and therefore quite-hard-to-find) papers mostly dealing with rings, semigroups, or abelian groups, I came across a single reference mentioning the non-abelian case, by B.I. Plotkin : [*Radicals in groups, operations on classes of groups, and radical classes*](http://books.google.com/books?id=DoeD0CKjFlQC&lpg=PA89&ots=akUbzEqFPA&dq=plotkin%20radicals%20%20%22classes%20of%20groups%22&lr=&pg=PA89#v=onepage&q=plotkin%20radicals%20%20%22classes%20of%20groups%22&f=false). Connections seem have been made with closure operators¹, but do not focus much on Grp.
* **Do you have ideas of connections from those functors to other parts of algebra or category theory, other than (pre-)radicals ?**
* **Do you have some pointers to material I may have missed, specially if they mention non-abelian groups ?**
¹: [Categorical structure of closure operators with applications to topology](http://books.google.com/books?id=FeKGFRNTc1sC&lpg=PA1&ots=mSI0HmZkeM&dq=closure%20operators%20dikranjan&lr=&pg=PA51#v=onepage&q=&f=false)
By N. Dikranjan, Walter Tholen, p.51
| https://mathoverflow.net/users/1307 | References on functorially-defined subgroups | After some research, I asked on the [Group-pub mailing list](http://people.bath.ac.uk/masgcs/gpf.html), where [very knowledgeable people roam](http://people.bath.ac.uk/masgcs/folk/folk.html) (the University of Bath, which hosts the mailing-list, also hosted the ['Groups St Andrews' conference in 2009](http://www.groupsstandrews.org/2009/index.html)). [Jan Krempa](http://www.mimuw.edu.pl/badania/publikacje/?LANG=en¶=o&parb=355) pointed me to a recent [seminar on radicals](http://aragorn.pb.bialystok.pl/~piotrgr/BanachCenter/meeting.html), that included a very useful survey paper:
B. J. Gardner, *Kurosh-Amitsur radicals of groups: something for overyone?*
Its lists of references is a goldmine, particularly the included book by the same author (*Radical Theory*). It includes general radical theory that applies to the non-commutative case better than his later book (written with R. Wiegandt :*Radical theory of rings*).
I also got a nice answer from Mike Newman (presumptively [this one](http://wwwmaths.anu.edu.au/~newman/Mike_Newman.html)), who told me:
>
> \*\* An early interest occurs in
>
>
> MR0002876 (2,125i)
> Hall, P.
> Verbal and marginal subgroups.
> J. Reine Angew. Math. 182, (1940). 156--157.
>
>
> This is a brief report of a lecture given at a meeting in Goettingen
> in 1939.
>
>
> \*\* There is recent work which hinges on these sorts of ideas in
>
>
> MR2276769 (2008f:20052)
> Nikolov, Nikolay(4-OXNC); Segal, Dan(4-OXAS)
> On finitely generated profinite groups. I. Strong completeness and uniform bounds. (English summary)
> Ann. of Math. (2) 165 (2007), no. 1, 171--238.
> 20E18 (20E32 20F12)
>
>
>
The link with verbal and marginal subgroups was already hinted at in [the previous mathoverflow question I mentioned](https://mathoverflow.net/questions/6925/how-can-we-formalize-the-naturality-of-certain-characteristic-subgroups).
| 3 | https://mathoverflow.net/users/1307 | 16046 | 10,754 |
https://mathoverflow.net/questions/16048 | 5 | I found myself "naturally" dealing with an object of this form:
X is a complex vector space, with a "product" (a,b) → {aba} which is quadratic in the first variable, linear in the second, and satisfies some associativity conditions. These conditions are actually complicated, but more or less say that {aba} looks like the product (aba) in an alternative algebra Y containing X as a subspace.
For example, the main "associativity condition" I am interested in is:
{a{b{aca}b}a}={{aba}c{aba}}
Examples
1. Symmetric matrices
2. Octonions, or indeed any alternative algebra
3. Let J belong to GL(n,ℂ), with tJ=-J and J²=-Id, and W={w∈M(n×n,ℂ)|JtwJ=-w}
all with the standard product {aba}=aba.
All of these examples are Jordan algebras, with respect to the symmetrized product a∘b=½(ab+ba), but I cannot see any direct link between the Jordan product and my product.
| https://mathoverflow.net/users/1626 | Is there a name for this algebraic structure? | In a Jordan algebra with product $\cdot$, a triple product is defined by $$\{abc\}=(a\cdot b)\cdot c+(b\cdot c)\cdot a-(a\cdot c)\cdot b.$$
In a special Jordan algebra (constructed by symmetrising an associative product) one has $\{aba\}=aba$, and it is easy to show that in such algebras one always has the identity $$\{\{aba\}c\{aba\}\}=\{a\{b\{aca\}b\}a\}.$$ Now, there is an amazing general theorem of Macdonald's that states that any identity in three variables which is of degree at most one in one of them and which is valid in special Jordan algebras actually holds in all Jordan algebras. This is proved in Jacobson's breath-taking *Structure and representations of Jordan algebras*.
So your identity holds in all Jordan algebras. As a consequence, from the information you give it is more or less impossible to distinguish your structure from Jordan algebras, as far as I can see.
By the way, in his book, Jacobson notes that McCrimmon has developed the theory of Jordan algebras based exclusively on the composition $(a,b)\mapsto aba$, and gives [McCrimmon, Kevin. A general theory of Jordan rings. Proc. Nat. Acad. Sci. U.S.A. 56 1966 1072--1079. MR0202783 (34 #2643)] as reference. I do not have access to the paper, though. The paper can be gotten from this [link](http://www.pnas.org/content/56/4/1072.full.pdf) Andrea provided in a comment below.
| 26 | https://mathoverflow.net/users/1409 | 16057 | 10,762 |
https://mathoverflow.net/questions/16047 | 7 | I am looking for examples of non-projective (quasi-projective) varieties $X$ defined over a field of positive characteristic, which have trivial étale fundamental group.
It is well known that the étale fundamental group in positive characteristics is a very difficult object, especially so in the non-projective case due to possibly wild ramification at infinity. I'm not even sure if there are examples of the kind above. Is this known?
| https://mathoverflow.net/users/259 | Simply connected quasi-projective varieties in positive characteristic | This is an answer to Pete's question on simply connected affine varieties (I can not put it in a comment because of space limitation).
I think that in positive characteristic $p$, no affine **irreducible** variety $X$ of positive dimension is simply connected. We can assume $X=\operatorname{Spec}(A)$ integral because $\pi\_1$ is insensible to nipotent elements (SGA IX.4.10). Let $k[t\_1,\ldots, t\_d] \subseteq A$ be a finite extension with minimal degree $[k(A):k(t\_1,\ldots, t\_d)]$. Consider the étale cover $Y\to \mathbb A^d\_k= \operatorname{Spec}(k[t\_1,\ldots, t\_d])$ defined by $s^p-s=t\_1$. Then $X\times\_{\mathbb A^d\_k} Y\to X$ is an étale cover of degree $p$. As $k(Y)$ and $k(X)$ are linearly disjoint over $k(t\_1,\ldots, t\_d)$ ($k(Y)$ is Galois over $k({\bf t}):=k(t\_1,\ldots, t\_d)$ and $k(Y)\cap k(X)=k({\bf t})$), the tensor product $k(Y)\otimes\_{k({\bf t})} k(X)$ is a field. This implies that $X\times\_{\mathbb A^d\_k} Y$ is connected.
| 10 | https://mathoverflow.net/users/3485 | 16060 | 10,765 |
https://mathoverflow.net/questions/16063 | 2 | Given a scheme $X$ with generic point p and a quasi-coherent sheaf $F$ on $X$.
Viewing $X$ as a scheme over $Spec(\mathbb{Z})$, let us assume
$f: X \rightarrow Spec(\mathbb{Z})$ is a proper map.
What conditions have $X$ and $F$ to satisfy, so that one can embed the $\mathbb{Z}$-module $F(X)=H^0(X,F)$ in $F\_p$, respectively when is the restriction map $h: F(X) \rightarrow F\_p$ injective?
Are there some mild conditions, like $X$ integral and $F$ coherent or torsion free?
| https://mathoverflow.net/users/3233 | When is the restriction map on global sections an embedding | If $X$ is integral and $F$ is torsion-free, then for any non-empty affine open subset $U$ of $X$, the canonical map $F(U)\to F\_p$ is injective. So $F(X)\to F\_p$ is injective. You don't need hypothesis on $X \to Spec(\mathbb Z)$. If $X$ is not necessarily reduced, then the flatness of $F$ over $X$ is also enough (same proof).
| 5 | https://mathoverflow.net/users/3485 | 16065 | 10,768 |
https://mathoverflow.net/questions/16066 | 1 | In one paper I saw this equality:
$$\sum\_{\eta=-\infty}^{\infty}\frac{z}{(z+\eta)}=\pi z\cot(\pi z)$$
which is the same as
$$\sum\_{\eta=-\infty}^{\infty}\frac{1}{(z+\eta)}=\pi \cot(\pi z)$$
where summation is understood in the sense of a principal value. What does it mean?
In another paper I found the next expression:
$$\frac{\exp(2\pi iaz)}{\exp(2\pi iz)-1}=\frac{1}{2\pi i}\sum\_{n=-\infty}^{\infty}\frac{\exp(2\pi ina)}{z-n}$$
for $a=0$ it is equivalent to
$$\frac{1}{\exp(2\pi iz)-1}=\frac{1}{2\pi i}\sum\_{n=-\infty}^{\infty}\frac{1}{z+n}$$
which is not exactly the same expression like in the first case.
$$\sum\_{n=-\infty}^{\infty}\frac{1}{z+n}=\pi Cot[\pi z]-i\pi$$
Where is my mistake?
If the second formula is wrong, what is the correct formula for the second case?
$$\sum\_{n=-\infty}^{\infty}\frac{\exp(2\pi ina)}{z+n}=?$$
| https://mathoverflow.net/users/3589 | what is summation in the sense of a principal value? | A principal-value sum (or integral) is usually one in which unconditional summation (or integration) does not converge, so one needs to sum in a particular way to achieve convergence. I suspect that, in this case, the necessary summation is symmetric, so that we consider $\lim\_{N \to \infty} \sum\_{n = -N}^{n = N} f(n)$ instead of $\sum\_{n = 1}^\infty f(-n) + \sum\_{n = 0}^\infty f(n)$.
It's not quite clear to me what your issue is with the two formulæ you mention. Since you are summing different functions ($1/(z + n)$ versus $z/(z + n)$), it is no surprise that the answers are different. What am I missing? (Sorry, I did not notice that you had already factored out the $z$.)
| 2 | https://mathoverflow.net/users/2383 | 16067 | 10,769 |
https://mathoverflow.net/questions/14764 | 16 | I just started to collect the papers of this field and know little things. So if I make stupid mistake, please correct me.
It seems that there are several approaches to localize Kac-Moody algebra(in particular, affine Lie algebra). I just took look at several papers by
**Kashiwara-Tanisaki:(1989)**
They constructed the flag variety of symmetrizable Kac-Moody algebra as ind-scheme.
**Edward Frenkel-B.Feign:**
They constructed the semi-infinite flag manifold and introduce the semi-infinite cohomology.
**Edward Frenkel-Dennis Gaitsgory:**
It seems that they dealt with two kinds of things:
1 Affine Grassmannian
2 Affine flag variety
**Olivier Mathieu**
It seems that he gave the general definition of flag variety of arbitrary Kac-Moody algebra as a stack.(I can read French but not very quickly, so there might be possibilities that I made a mistake to describe his work)
My Question
===========
1. Is there any other definition of flag variety of Kac-Moody algebra(at least for affine case)? What are the relationship between these definitions I mentioned above?
2. What is the relationship between the D-module theory on affine flag variety and D-module theory on affine grassmannian? (Frenkel-Gaitsgory) Why did they consider these two ways to localize affine Lie algebra?
3. It seems all of the construction above are not very easy to deal with(ind-scheme,group ind-scheme which are not locally affine). Is there any existent work to define it as a locally affine space(classical scheme,algebraic space or at least locally affine stack with smooth topology)?
4. From the work of Frenkel-Gaitsgory, they built the derived equivalence between the category of D-modules and full subcategory of modules over enveloping algebra of affine Lie algebra. They claimed in their paper that one can not obtain the equivalence in abelian level. Is there any intuitive explanation for this?
5. I am looking forward to getting some guy who can explain the work of Oliver to me.
Thank you in advance!
| https://mathoverflow.net/users/1851 | What is the recent development of D-module and representation theory of Kac-Moody algebra? | I am not an expert in this but I would of course expect something like ind-scheme approach to be natural. Gerd Faltings used I think ind-schemes to treat Sugawara construction, algebraic loop groups and Verlinde's conjecture in
Gerd Faltings, Algebraic loop groups and moduli spaces of bundles.
J. Eur. Math. Soc. (JEMS) 5 (2003), no. 1, 41--68.
You might also like to work with versions of Kac-Moody GROUPS in analytic approaches.
You could also consult comprehensive and not that old Kumar's book (Kac-Moody groups, their flag varieties and representation theory, Birkhauser) which is written in geometric language.
As far as Frenkel is concerned, not only his work with Feigin but even more I think his paper with Gaitsgory must be relevant (see [arxiv:0712.0788](http://arxiv.org/abs/0712.0788)).
Semi-infinite cohomologies are important but still misterious thing. Some related homological algebra has been recently studied by Positelskii in great generality. Another important thing is relation between the geometry of representations of quantum groups at root of unity and of affine Lie algebras, like in the book of Varchenko and many papers later.
Edit: Frenkel himself I think does not claim (I talked to him at the time) to have intuitive explanation why only derived equivalence. But you should not expect for more: by the correspondence with quantum groups the situation should be like in affine case where one has problems with non-closedness of diagonal in noncommutative geometry what has repercussions on the theory of D-modules. How this reflects in the case of relevant ind-schemes for affine side I do not know but somehow it does.
| 7 | https://mathoverflow.net/users/35833 | 16075 | 10,775 |
https://mathoverflow.net/questions/16095 | 1 | From wikipedia [quantification](http://en.wikipedia.org/wiki/Quantification) has meaning:
>
> In logic, quantification is the
> binding of a variable ranging over a
> domain of discourse
>
>
>
**Is there any formal "definition" of universal quantifier for example using definition of domain of discourse?**
I mean a formula build without universal quantifier, and existential one which has the same meaning if referenced to defined domain of discourse?
For example:
Suppose we use domain of discourse (DoD) given by sentence $ U = \{ x|\phi(x) \}$ for some $\phi(x)$. Then naively we may wrote:
($\forall (x \in U) \Phi(x) ) \equiv ( \{ x|\phi(x) \} => \Phi(x) )$
In words: to say that some property follows for every x in DoD is the same as to say that if x is chosen from DoD then has this property.
We may try also the folowing one:
($\forall (x \in U) \Phi(x) ) \equiv (( \{ x|\phi(x) \} => \Phi(x) ) => (\phi(x) <=> \Phi(x) ))$
In words: to say that some property follows for every x in DoD is the same as to say that $\phi$ and $\Phi$ are evenly spanned.
Do You know any reference for such matter?
---
Gabriel: Yes, I agree that from formal point of view in mathematical practice DoD is a set and to extend it to bigger universe usual is done by pure formal way and may be changed to some additional axioms etc. But this is some kind of mathematical practice: "near every decent theory as far as we know is defined for DoD to be set or smaller but as it works also for proper classes we are trying to write it in a way". But then we omit important statement: **every time DoD has to be defined and additional axioms about it existence, definition,properties has to be added to the theory.** I am only a hobbyist but I do not know any theorem which states: structure to be DoD for formal theory over countable language has to have "this and this" property. Of course for example as in formula $\{ x|\phi(x)\}$ we may require that $\phi(x)$ has some property. For example we may require that it is in first order language. Or in second order. Or in finite order language etc. For me is rather clear that it cannot be whatever I like. As far as I know we do not have any theory for that. But maybe I am wrong?
So my question is: what is that mean "for all" in a context of different definition of DoD ( as well as "there exists"). Do we have clear meaning what it means for very big universes? We use some operator here named "for all" but have we possibility to define its meaning in syntactical way? If not, may we be sure that meaning of sentence "for all" is clearly defined?
I suggest this is example of **Incomplete Inductive reasoning** about possible ways of using general quantifier in mathematics. Moreover I suppose, even after reading something about Hilbert epsilon calculus that quantifiers has usual only intuitive meaning, that is its definition is far from such level of formality as for binary operation $\in$ for ZFC for example, where it may be anything (for example in [von Neumann hierarchy of sets](http://en.wikipedia.org/wiki/Cumulative_hierarchy) "model" of ZFC it is order). When we try to define formal theory we want to abstract from the "meaning" of the symbols and give only pure syntactical rules for them. As far as I know ( but I know not much) I do not know such definition for quantifiers, even in epsilon Hilbert calculus for example, because it **omits the area of possible, acceptable or correct definitions of domain of discourse.**
| https://mathoverflow.net/users/3811 | Is there formal definition of universal quantification? | There is a definition in terms of $\varepsilon$-operator of Hilbert. See [wikipedia](http://en.wikipedia.org/wiki/Epsilon_calculus). If not, either universal quantification or existential quantification is taken as primitive in classical logic, for in classical logic, one is derivable from the other. This is not true in intuitionistic logic, as the proof uses the law of excluded middle.
The [nLab](http://ncatlab.org/nlab/show/choice+operator) also has a page related to Hilbert's operator and its relation to the quantifiers.
| 3 | https://mathoverflow.net/users/1353 | 16096 | 10,788 |
https://mathoverflow.net/questions/16107 | 20 | I have been working on this problem for several months now but have not made much progress. It concerns the set of all [integer partitions](https://en.wikipedia.org/wiki/Integer_partitions) of n.
Let the vertices of the graph G=G(n) denote all the p(n) integer partitions of n. There is an edge between two partitions if and only if one can be transformed into another by only moving one dot between rows in their Ferrers diagram representations.
So, for example, the partitions (3,2,1) and (3,3) of 6 are linked because we can move the dot in the last row to the second row.
```
OOO OOO
OO -------- OOO
O
```
My question: for what values of n does G(n) have a Hamiltonian path from (n) to (1,1,...,1)?
That is, is it possible to go through, without repetition, all the partitions of n by simply moving around the dots in the Ferrers diagrams?
Is there a determinate way to construct such paths?
I have only been able to construct paths for n = 1 to 6.
n=1 (trivial)
n=2:
(2) => (1,1)
n=3:
(3) => (2,1) => (1,1,1)
n=4:
(4) => (3,1) => (2,2) => (2,1,1) => (1,1,1,1)
n=5:
(5) => (4,1) => (3,2) => (2,2,1) => (3,1,1) => (2,1,1,1) => (1,1,1,1,1)
n=6:
(6) => (5,1) => (4,1,1) => (4,2) => (3,3) => (3,2,1) => (2,2,2) => (2,2,1,1) => (3,1,1,1) => (2,1,1,1,1) => (1,1,1,1,1,1)
None of the basic theorems about Hamiltonian paths have not helped me here.
| https://mathoverflow.net/users/4176 | Hamiltonian paths where the vertices are integer partitions | What you want is known as a Gray code for integer partitions.
It exists.
See C. D. Savage, [Gray code sequences of partitions](https://doi.org/10.1016/0196-6774(89)90007-2), *Journal of Algorithms* 10 (1989) 577-595. Disclaimer: I haven't actually read this paper. Other sources such as [this 1997 survey by Savage](https://people.engr.ncsu.edu/savage/AVAILABLE_FOR_MAILING/survey.pdf "A survey of combinatorial Gray codes") and [these class notes](http://homepage.divms.uiowa.edu/%7Esriram/196/fall01/lecture7.pdf) don't say what the construction is but say that it's complicated.
| 18 | https://mathoverflow.net/users/143 | 16112 | 10,798 |
https://mathoverflow.net/questions/16105 | 7 | The answer to this question should be obvious, but I can't seem to figure it out. Suppose we have a surface $F$, and a representation $\rho : \pi\_1(F)\to SU(n)$. We can define the homology with local coefficients $H\_\*(F,\rho)$ straightforwardly as the homology of the twisted complex $$C\_\*(F,\rho):=C\_\*(\widetilde{F};\mathbf{Z})\otimes\_{\mathbf{Z}[\pi\_1(F)]} \mathbf{C}^n$$ where $\widetilde{F}$ is the universal cover, and $\mathbf{Z}[\pi\_1(F)]$ acts on each side in the obvious way.
Now, this complex is actually very easy to compute explicitly: just lift a nice basis of cells in $F$ to $\widetilde{F}$, and write down the boundary maps explicitly. For example, if $F$ is a torus and we take $n=2$, say, we can choose a natural meridian-longitude basis $(x,y)$ for $H\_1(F)$, and the twisted boundary map $\partial\_1:C\_1(F,\rho)=\mathbf{C}^4\to C\_2(F,\rho)=\mathbf{C}^2$ is $$ \left( \begin{array}{ccc}
\rho(x)-Id \newline\rho(y)-Id\end{array} \right)$$
So, here's my question. Since $\rho$ is a unitary representation, we should get a twisted intersection form on $H\_1(F)$, simply by combining the untwisted intersection form with the standard hermitian product on $\mathbf{C}^2$, right? And I would imagine this is also really easy to compute, in a similar basis, say? I can't seem to figure out how it would go. Could anyone help me, even show me how it works for the same torus example?
Or, if I've said anything wrong, tell me where?
| https://mathoverflow.net/users/492 | Intersection form in twisted homology (homology with local coefficients) | For me it is easier to work with cohomology (just for psychological reasons). Also, I will distinguish the representation $\rho$ from the local system $V$ with fibres ${\mathbb C}^2$ that it gives rise to. So where you would write $H^1(F,\rho)$ I will write $H^1(F,V)$.
I will let $\overline{V}$ denote the complex conjugate local system to $V$.
(So it is the same underlying local system of abelian groups, but we give it the
conjugate action of $\mathbb C$.)
The Hermitian pairing on the fibres of $V$ and $\overline{V}$ gives a pairing of local
systems $V \times \overline{V} \to \mathbb R$, where $\mathbb R$ is the constant local system
with fibre the real numbers. If you like we can think of this as an $\mathbb R$-linear map
$V\otimes\_{\mathbb C}\overline{V} \to \mathbb R.$ This pairing will induce a map on
cohomology $H^2(F,V\otimes\_{\mathbb C}\overline{V}) \to H^2(F,\mathbb R)$.
There will also
be a cup product $H^1(F,V) \times H^1(F,\overline{V}) \to H^2(F, V\otimes\_{\mathbb C} \overline{V})$.
Composing this with the previous map on $H^2$ gives your twisted cup product
$H^1(F,V)\times H^1(F,\overline{V}) \to H^2(F,\mathbb R)$.
This gives one perspective on your construction. To compute it, write down the twisted cochains $C^{\bullet}(\tilde{F})\otimes\_{\mathbb Z[\pi\_1(F)]}\mathbb C^2$,
then
write down the cup-product
$$(C^{\bullet}(\widetilde{F})\otimes\_{\mathbb Z[\pi\_1(F)]}\mathbb C^2 ) \times
(C^{\bullet}(\widetilde{F})\otimes\_{\mathbb Z[\pi\_1(F)]}\mathbb C^2)
\to C^{\bullet}(\widetilde{F})\otimes\_{\mathbb Z[\pi\_1(F)]} \mathbb R^2
= C^{\bullet}(F,\mathbb R).$$
The cup product will just be given by the usual formula, and then you will also
pair the $\mathbb C^2$ parts of the cochains using the hermitian pairing.
Hopefully you can follow your nose and do this explicitly for the torus. Then you can
just dualize everything to get to the homology version.
| 5 | https://mathoverflow.net/users/2874 | 16119 | 10,803 |
https://mathoverflow.net/questions/15957 | 22 | We say that a group *G* is in the class *Fq* if there is a CW-complex which is a *BG* (that is, which has fundamental group *G* and contractible universal cover) and which has finite *q*-skeleton. Thus *F0* contains all groups, *F1* contains exactly the finitely generated groups, *F2* the finitely presented groups, and so forth.
My question: For a fixed *q* ≥ 3, is it possible to decide, from a finite presentation of a group *G*, whether *G* is in *Fq* or not? I would assume not, but am not having much luck proving it.
One approach would be to prove that, if *G* is a group in *Fq* and *H* is a finitely presented subgroup, then *H* ∈ *Fq* as well. This would make being in *Fq* a Markov property, or at least close enough to make it undecidable.
Henry Wilton's comment below makes it clear that being *Fq* is not even quasi-Markov, so the above idea won't work. I still suspect that "*G* ∈ *Fq*" is not decidable, but now my intuition is from Rice's theorem:
>
> If $\mathcal{B}$ is a nonempty set of computable functions with nonempty complement, then no algorithm accepts an input *n* and decides whether *φn* is an element of $\mathcal{B}$.
>
>
>
It seems likely to me that something similar is true of finite presentations and the groups they define.
John Stillwell notes below that this can't be true for a number of questions involving the abelianization of G. This wouldn't affect the Rips construction/1-2-3 theorem discussion below if the homology-sphere idea works, since those groups are all perfect.
Any thoughts?
| https://mathoverflow.net/users/4133 | Is any interesting question about a group G decidable from a presentation of G? | It seems to me that the analogue of Rice's theorem fails for finitely presented
groups $G$ because of questions like: is the abelianization of $G$ of rank 3?
The rank of the abelianization of any finitely presented $G$ can be computed
by reducing the abelianization to normal form, so this (slightly) interesting
question can be decided from the presentation of $G$.
| 12 | https://mathoverflow.net/users/1587 | 16122 | 10,805 |
https://mathoverflow.net/questions/16121 | 13 | Can someone give a really concrete example of such a sequence? I am looking at several notes related with such things, but haven't seen any well-calculated example. And I'm really confused at this point.
Besides asking for a good example, I am also wondering about the following two things:
1. There is an exact sequence for elliptic curves defined over a local field $K$, $0 \rightarrow \hat E(m) \rightarrow E(K) \rightarrow \tilde E(k) \rightarrow 0$, where $\hat E(m)$ is the formal group associated to $E$ and $\tilde E(k)$ is the reduction. (See Silverman AEC I, page 118), is this sequence related with connected-etale sequence?
2.Take the p-torsion kernel $E[p]$ of $[p]: E \rightarrow E$ for $E$ defined over $K$ a local field.Is $E[p]$ a finite flat group scheme over $R$ the valuation ring? And if so, what is its connected-etale sequence? (maybe I should change $p$ to an $n$, but I'm also curious what will happen if $p$ is the characteristic of $k$?)
Thank you.
| https://mathoverflow.net/users/1238 | Example of connected-etale sequence for group schemes over a Henselian field? | Regarding 1. : If you pass to the $n$-torsion parts of the members of this exact sequence,
you will get the $R$-valued points
of the connected-etale sequence for $E[n]$.
Regarding 2. : If $E$ has good reduction, then $E[n]$ is a finite flat group scheme.
If the residue char. $p$ of $R$ does not divide $n$ then it is even etale.
In general, by the Chinese remainder theorem, when analyzing $E[n]$ we may assume $n$
is a prime power, so suppose now that $n$ is a power $p^r$, where $p$ is the residue char.
of $R$.
If $E$ has supersingular reduction, then $E[p^r]$ is connected.
If $E$ has ordinary reduction, then the connected and etale parts of $E[p^r]$ each
have order $p^r$. The special fibre of the connected part is the kernel of $Frob^r$.
Note that if $E$ has bad reduction, then $E[n]$ can still have a finite flat extension
over $R$ for some finite number of $n$. (E.g. $X\_0(11)$ has bad reduction at 11, but its
5-torsion is still finite flat at 11.)
To compute an example, I suggest considering elliptic curves in char. 2, and computing the 2-torsion.
Consider the two examples: $y^2 + y = x^3 $ and $y^2 + x y = x^3 + x$. If you compute the 2-division equation it will have degree 4. In the first case, it will be purely inseparable: this case is supersingular. In the second case it will have inseparability degree 2; this case is ordinary. In the first case the 2-torsion group scheme is connected; in the second, it has a connected subgroup scheme of order 2.
| 8 | https://mathoverflow.net/users/2874 | 16130 | 10,810 |
https://mathoverflow.net/questions/16085 | 12 | I recently worked through most of the proof of the rationality of the moduli of genus 3 curves, which seemed to have the following structure:
1. Every nonhyperelliptic genus 3 curve is a smooth plane quartic.
2. The plane quartics form a projective space.
3. Apply GIT to this projective space and the $PGL(3)$ action.
4. Prove that this quotient is rational.
I've seen somewhat similarly structured arguments before. So my question:
>
> When is a GIT quotient rational?
>
>
>
In particular, are quotients of $\mathbb{P}^n$ by $PGL\_k$ rational, under some reasonable hypotheses?
Are there any natural invariants that are preserved by quotients (again, with reasonable conditions, or of the above form)?
| https://mathoverflow.net/users/622 | Rationality of GIT quotients | A useful general result is the 'no-name lemma' stating that when a reductive group $G$ acts linearly on two vector spaces $V$ and $W$ 'almost freely' (that is, the stabilizer subgroup of a general point is trivial), then the GIT-quotients $V/G$ and $W/G$ are stably rational (that is, $V/G \times \mathbb{C}^m$ and $W/G \times \mathbb{C}^n$ are birational for some $m$ and $n$).
Btw. Katsylo used it in the rationality of genus 3 curves you mentioned.
Clearly, the following implications hold:
rational $\implies$ stably rational $\implies$ unirational
and counterexamples to the other implications exist (Artin-Mumford for a unirational non-stably rational variety and Colliot-Thelene, Sansuc and Swinnerton-Dyer for a non-rational stably rational one).
As to $PGL\_n$ : here the 'canonical' example of a vector space having an almost free $PGL\_n$-action is couples of $n\times n$ matrices under simultaneous conjugation. Hence, by the NNL any other almost free GIT-quotient is stably rational to it.
Here the best result known is that when $n$ divides $420=2^2\times3\times5\times7$ then such quotients are stably rational. For couples of matrices under simultaneous conjugation rationality is known for $n\leq 4$ but even for the cases $n=5$ and $n=7$ only stably rationality is known. 'Retract rationality' (a lot weaker than stable rationality) is known for all square-free $n$ by a result of David Saltman.
| 10 | https://mathoverflow.net/users/2275 | 16133 | 10,811 |
https://mathoverflow.net/questions/16104 | 16 | A classical theorem is saying that every smooth, finite-dimensional manifold has a smooth partition of unity. My question is:
1. Which Fréchet manifolds have a smooth partition of unity?
2. How is the existence of smooth partitions of unity on Fréchet manifolds related to paracompactness of the underlying topology?
From some remarks in some literature, I got the impression that not *all* Fréchet manifolds have smooth partitions of unity, but *some* have, e.g. the loop space $LM$ of a finite-dimensional smooth manifold $M$.
For $LM$, the proof seems to be that $LM$ is Lindelöf, hence paracompact.
Is this true for all mapping spaces of the form $C^\infty (K,M)$ for $K$ compact?
| https://mathoverflow.net/users/3473 | Which Fréchet manifolds have a smooth partition of unity? | Use the [source](https://ncatlab.org/nlab/show/The+Convenient+Setting+of+Global+Analysis), Luke.
Specifically, chapters 14 (Smooth Bump Functions) to 16 (Smooth Partitions of Unity and Smooth Normality). You may be particularly interested in:
>
> **Theorem 16.10** If $X$ is Lindelof and $\mathcal{S}$-regular, then $X$ is $\mathcal{S}$-paracompact. In particular, nuclear Frechet spaces are $C^\infty$-paracompact.
>
>
>
For loop spaces (and other mapping spaces with compact source), the simplest argument for Lindelof/paracompactness that I know of goes as follows:
1. Embed $M$ as a submanifold of $\mathbb{R}^n$.
2. So the loop space $LM$ embeds as a submanifold of $L\mathbb{R}^n$.
3. $L\mathbb{R}^n$ is metrisable.
4. So $L M$ is metrisable.
5. Hence $L M$ is paracompact.
(Paracompactness isn't inheritable by **all** subsets. Of course, if you can embed your manifold as a *closed* subspace then you can inherit the paracompactness directly.)
I use this argument in my paper on *[Constructing smooth manifolds of loop spaces](https://arxiv.org/abs/math/0612096)*, Proc. London Math. Soc. **99** (2009) 195–216 (doi:[10.1112/plms/pdn058](https://doi.org/10.1112/plms/pdn058), arXiv:[math/0612096](https://arxiv.org/abs/math/0612096)) to show that most "nice" properties devolve from the model space to the loop space for "nice" model spaces (smooth, continuous, and others). See corollary C in the introduction of the published version.
(I should note that the full statement of Theorem 16.10 (which I did not quote above) is not quite correct (at least in the book version, it may have been corrected online) in that the proof of the claim for strict inductive sequences is not complete. I needed a specific instance of this in my paper *[The Smooth Structure of the Space of Piecewise-Smooth Loops](https://arxiv.org/abs/0803.0611)*, Glasgow Mathematical Journal, **59**(1) (2017) pp27-59. (arXiv:[0803.0611](https://arxiv.org/abs/0803.0611), doi:[10.1017/S0017089516000033](https://doi.org/10.1017/S0017089516000033)) (see section 5.4.2) which wasn't covered by 16.10 but fortunately I could hack together bits of 16.6 with 16.10 to get it to work. This, however, is outside the remit of this question as it deals with spaces more general than Frechet spaces.)
On the opposite side of the equation, we have the following after 16.10:
>
> **open problem** ... Is every paracompact $\mathcal{S}$-regular space $\mathcal{S}$-paracompact?
>
>
>
So the general case is not (at time of publishing) known. But for manifolds, the case is somewhat better:
>
> **Ch 27** If a smooth manifold (which is smoothly Hausdorff) is Lindelof, and if all modelling vector spaces are smoothly regular, then it is smoothly paracompact. If a smooth manifold is metrisable and smoothly normal then it is smoothly paracompact.
>
>
>
Since Banach spaces are Frechet spaces, any Banach space that is not $C^\infty$-paracompact provides a counterexample for Frechet spaces as well. The comment after 14.9 provides the examples of $\ell^1$ and $C([0,1])$.
So, putting it all together: **nuclear** Frechet spaces are good, so Lindelof manifolds modelled on them are smoothly paracompact. Smooth mapping spaces (with compact source) are Lindelof manifolds with nuclear model spaces, hence smoothly paracompact.
(Recall that smooth mapping spaces without compact source aren't even close to being manifolds. I know that Konrad knows this, I merely put this here so that others will know it too.)
| 14 | https://mathoverflow.net/users/45 | 16143 | 10,818 |
https://mathoverflow.net/questions/16146 | 7 | Fix an algebraically closed **†** ground field $k$ of any characteristic. I want to use the classical definition of projective $n$-space $\mathbb{P}^n$ as set quotient of $\mathbb{A}^{n+1}\setminus 0$ by the action of $k^\*$, and for its Zariski topology the definition that closed sets are the "zero sets" of homogeneous polynomials in $n+1$ variables considered as functions from $\mathbb{P}^n$ to {$0,1$}.
I'm not satisfied with my understanding of why this is the same as the quotient topology from the map
$\mathbb{A}^{n+1}\setminus 0\to \mathbb{P}^{n}$. This means proving the map is surjective, continuous, and that a set with closed pre-image is closed, the first two properties being clear. The last part is the least trivial, meaning that:
>
> If a set $S\subseteq \mathbb{P}^n$ has closed preimage $\widehat{S}$ in $\mathbb{A}^{n+1}\setminus 0$, i.e. $\widehat{S}$ is cut out by some polynomials $f\_1,\ldots,f\_r$, then $S$ is closed in $\mathbb{P}^n$, i.e. $\widehat{S}$ is cut out by some *homogeneous* polynomials $g\_1,\ldots,g\_s$.
>
>
>
I'm bad at algebra, so I can't seem to start doing anything here that doesn't look ugly/hard... I really want to "see" what's going on and avoid just quoting results about invariants without understanding how they work in this "toy" example.
Thanks for the help, anyone!
**† Follow up:** the algebraic closed hypothesis is not necessary, since David/Kevin's proof works for any infinite field, and for a finite field our spaces (as defined here) are discrete so the result is trivial.
| https://mathoverflow.net/users/84526 | Elementary proof that projective space is a quotient | Let $f$ be a polynomial which vanishes on $\hat{S}$. Write $f=\sum f\_i$, where $f\_i$ is homogenous of degree $i$. The set $\hat{S}$ is homogenous so, for any $\lambda \in k^\*$, the polynomial $f(\lambda \cdot x) = \sum \lambda^i f\_i$ also vanishes on $f$.
Since $k$ is infinite, we can find more equations of the form $\sum \lambda^i f\_i=0$ than there are terms in $f$. Taking linear combinations, we deduce that each $f\_i$ individually vanishes on $\hat{S}$.
| 7 | https://mathoverflow.net/users/297 | 16147 | 10,820 |
https://mathoverflow.net/questions/16145 | 34 | Let $K$ be a field and $G:=SL\_2(K)$, then $G$ is a $K-$split reductive group (to use some big words). These groups are classified by a based root datum $(X,D,X',D')$. Let $G'$ be group associated to $(X',D',X,D)$, the so called dual group.
Is it correct that $G'=PGL\_2(K)$?
I"m wondering how $PSL\_2(K)$ fits into this picture.
I'm aware of the fact that if C is algebraically closed, then $PSL\_2(C) \cong PGL\_2(C)$ as abstract groups; can this be made into an isomorphism of algebraic groups, i.e. is $PSL$ a $K-$form of $PGL$?
| https://mathoverflow.net/users/3380 | What is the difference between PSL_2 and PGL_2? | Yes, the dual of $SL\_2$ is $PGL\_2$.
But you're not going down the right track with $PSL\_2$. The problem with $PSL\_2$ is that it's not a variety at all! You can quotient out the variety $SL\_2$ by the subgroup $\pm1$ but the quotient is the variety $PGL\_2$ (recall that quotients in the category of sheaves (for these are really fppf sheaves) don't have to be surjective on global sections, so the statement that there's a surjection $SL\_2\to PGL\_2$ does not imply that the induced map $SL\_2(\mathbf{Q})\to PGL\_2(\mathbf{Q})$ is a surjection).
The problem with $PSL\_2$ is that it is a functor from, say, rings to groups, but it's not a representable one, so in particular it's not an algebraic group. If you like, you can imagine $PSL\_2$ as a presheaf quotient, and $PGL\_2$ as the associated (representable) sheaf.
---
Edit: I tried to think of a way to make this observation more enlightening. Let's for example try and build an affine variety over $\mathbf{Q}$ representing the $PSL\_2$ functor. Well we can certainly build an affine variety over $\mathbf{Q}$ representing the $SL\_2$ functor: it's just $A:=\mathbf{Q}[a,b,c,d]/(ad-bc-1)$. Now let's see what happens if we try and quotient out by the group $\pm1$. The quotient is affine, and is represented by the invariants of the action, that is, the subring of $A$ consisting of polynomials in $a$, $b$, $c$, $d$ with the property that every monomial mentioned in the polynomial has total degree even. Now here's the problem: I can see a $\mathbf{Q}$-point of this subring (that is, a map from this subring to $\mathbf{Q}$) that corresponds to the matrix $(s,0;0,1/s)$ for $s=\sqrt{p}$, $p$ a prime number! It's the point that sends $a^2$ to $p$, $ab$ to $0$, and so on and so on, and finally $d^2$ goes to $1/p$ and $ad$ goes to $1$, $bc$ goes to $0$, so $ad-bc=1$. The map from the subring to $\mathbf{Q}$ doesn't extend to a map from the whole ring to $\mathbf{Q}$, so our putative construction has failed because the $\mathbf{Q}$-points of this ring are a group that canonically but strictly contains $PSL\_2(\mathbf{Q})$ (this subgroup being the $\mathbf{Q}$-points that extend to $\mathbf{Q}$-points of $A$).
| 51 | https://mathoverflow.net/users/1384 | 16150 | 10,823 |
https://mathoverflow.net/questions/16141 | 40 | I was looking at Wilson's theorem: If $P$ is a prime then $(P-1)!\equiv -1\pmod P$. I realized this
implies that for primes $P\equiv 3\pmod 4$, that $\left(\frac{P-1}{2}\right)!\equiv \pm1 \pmod P$.
Question: For which primes $P$ is $\left(\frac{P-1}{2}\right)!\equiv 1\pmod P$?
After convincing myself that it's not a congruence condition for $P,$ I found [this sequence in OEIS](http://oeis.org/A058302). I'd appreciate any comments that shed light on the nature of such primes (for example, they appear to be of density 1/2 in all primes that are $3\bmod 4$).
Thanks,
Jacob
| https://mathoverflow.net/users/4181 | Primes P such that ((P-1)/2)!=1 mod P | I am a newcomer here. If p >3 is congruent to 3 mod 4, there is an answer which involves only $p\pmod 8$ and $h\pmod 4$, where $h$ is the class number of $Q(\sqrt{-p})$ .
Namely one has $(\frac{p-1}{2})!\equiv 1 \pmod p$ if an only if either (i) $p\equiv 3 \pmod 8$ and $h\equiv 1 \pmod 4$ or (ii) $p\equiv 7\pmod 8$ and $h\equiv 3\pmod 4$.
The proof may not be original: since $p\equiv 3 \pmod 4$, one has to determine the Legendre symbol
$${{(\frac{p-1}{2})!}\overwithdelims (){p}}
=\prod\_{x=1}^{(p-1)/2}{x\overwithdelims (){p}}=\prod\_{x=1}^{(p-1)/2}(({x\overwithdelims (){p}}-1)+1).$$ It is enough to know this modulo 4 since it is 1 or -1. By developping, one gets $(p+1)/2+S \pmod 4$, where
$$S=\sum\_{x=1}^{(p-1)/2}\Bigl({x\over p}\Bigr).$$ By the class number formula, one has $(2-(2/p))h=S$ (I just looked up Borevich-Shafarevich, Number Theory), hence the result, since $\Bigl({2\over p}\Bigr)$ depends only on $p \pmod 8$.
Edit: For the correct answer see KConrad's post or Mordell's article.
| 37 | https://mathoverflow.net/users/4186 | 16154 | 10,827 |
https://mathoverflow.net/questions/16131 | 12 | My primary reference for this question is the very good book *Quantum Groups and Knot Invariants* by C. Kassel, M. Rosso, and V. Turaev. I'm also drawing from P. Etingof and O. Schiffmann, *Lectures on quantum groups*, another very good book. If you want to see pictures of Lie bialgebras and quasitriangular structures, I will shamelessly self-promote [my notes on Lie bialgebras](https://www.perimeterinstitute.ca/personal/tjohnsonfreyd/GraphicalLanguage.pdf), which I put together while studying for my quals last year.
Pick your favorite field of characteristic $0$, and let $\mathfrak g$ be a finite-dimensional Lie algebra. Abusing the language a bit, let me define a **(quadratic) Casimir** to be any symmetric $\mathfrak g$-invariant element $t\in \mathfrak g^{\otimes 2}$. (Equivalently, $t$ is symmetric and its image in the universal enveloping algebra $\mathcal U\mathfrak g$ is central.) I'd like to compare two constructions of "quantum" categories that start with this Lie theoretic data.
Quantization of infinitesimally-braided categories
--------------------------------------------------
So pick a quadratic Casimir $t$. Let $\mathcal S$ be the category of finite-dimensional $\mathfrak g$-modules. For any two modules $(\pi\_U,U)$ and $(\pi\_V,V) \in \mathcal S$, we have an element $t\\_{U,V} \in {\rm End}\_{\mathfrak g}(U\otimes V)$ given by $t\_{U,V} = (\pi\_U \otimes \pi\_V)(t)$; it is a $\mathfrak g$-morphism because $t$ is $\mathfrak g$-invariant. Moreover, under the canonical "flip" map $\sigma\_{U,V}: U\otimes V \to V\otimes U$, $t\_{U,V}$ maps to $\sigma\_{U,V}t\_{U,V}\sigma\_{V,U} = t\_{V,U}$, because $t$ is a symmetric element of $\mathfrak g^{\otimes 2}$. Along with the rule for how $\mathfrak g$ acts on tensor products (determining the action of $t\_{U,V\otimes W}$ on $U\otimes V\otimes W$), it follows that $t$ defines on $\mathcal S$ the structure of a **infinitesimally braided category**.
Then recall the following very general construction. Let $\Phi$ be a **Drinfeld associator**: i.e., $\Phi(A,B)$ is a formal power series in non-commuting variables of the form $\exp(\text{a Lie series})$, satisfying certain nonlinear conditions (a "pentagon" and two "hexagon"s) that make the following construction work. (Only one Drinfeld associator is explicitly known, given by the solution to the Knizhnik–Zamolodchikov differential equation, and its coefficients are real but transcendental. But the equations defining $\Phi$ at each order are an over-determined linear system in rational coefficients, so if any solution exists, a rational one does. And by a theorem of Le and Murakami, up to equivalence of braided monoidal categories, the output of this construction does not depend on the choice of associator.)
Then define a new category $\mathcal S[[\hbar]]$. The objects are the same as those of $\mathcal S$, and $\hom\_{\mathcal S[[\hbar]]}(V,W) = \hom\_{\mathcal S}(V,W)[[\hbar]]$ (formal power series of morphisms) with composition given simply by the composition in $\mathcal S$ extended $\hbar$-linearly (and adicly), i.e., following the rule for multiplication of formal power series.
Moreover, give $\mathcal S[[\hbar]]$ the tensor product inherited from $\mathcal S$. However, give it nontrivial associativity and braiding constraints. Namely, define the associator by $a\_{123} = \Phi(\hbar t\_{12}, \hbar t\_{23})$ and $c = \sigma \exp(\hbar t / 2) = \exp(\hbar t / 2)\sigma$. The axioms for the Drinfeld associator $\Phi$ imply that $\mathcal S[[\hbar]]$ is a (weak) braided monoidal category.
Quantization of quasitriangular Lie bialgebras
----------------------------------------------
Again let's start with $t$ a quadratic Casimir. But let's suppose a bit more: let's suppose that $t$ is the symmetrization of some element $r\in \mathfrak g^{\otimes 2}$ satisfying the (very over-determined) **classical Yang-Baxter equation** (CYBE). Namely, let $\beta: \mathfrak g^{\otimes 2} \to \mathfrak g$ be the Lie bracket, and use the obvious index notation. Then the CYBE says:
$$ \bigl[(\beta\_{13} \otimes \mathrm{id}\_2 \otimes \mathrm{id}\_4) + (\mathrm{id}\_1 \otimes \beta\_{23} \otimes \mathrm{id}\_4) + (\mathrm{id}\_1 \otimes \mathrm{id}\_3 \otimes \beta\_{24}) \bigr] (r\_{12}\otimes r\_{34}) = 0.$$
A choice of such an $r$ is a **quasitriangular structure** on $\mathfrak g$. Consider the map $\delta: \mathfrak g \to \mathfrak g^{\wedge 2}$ given by antisymmetrizing the output of $x \mapsto \mathrm{ad}\_x(r)$, where $\mathrm{ad}\_x$ is the action of $x\in \mathfrak g$ on $\mathfrak g^{\otimes 2}$. It follows from the CYBE and the fact that $t$ is $\mathfrak g$-invariant that $\delta$ satisfies the co-Jacobi identity.
Then by a general theorem of Etingof and Kazhdan, the data $(\mathfrak g,r)$ as above along with a choice of Drinfeld associator $\Phi$ determines a (noncommutative, noncocommutative) Hopf algebra $\mathcal U\_\hbar \mathfrak g$ (over power series in $\hbar$) and a nontrivial braiding on the (strongly-associative monoidal) category of (finitely-generated topologically free) $\mathcal U\_\hbar\mathfrak g$-modules. As an algebra, $\mathcal U\_\hbar\mathfrak g \cong \mathcal U\mathfrak g[[\hbar]]$, I think, so, as a category (although maybe not as a braided monoidal category?), $\mathcal S[[\hbar]] \cong \text{$(\mathcal U\_\hbar\mathfrak g)$-mod}$.
Questions
---------
The two constructions above are similar: both start with a Lie algebra and a Casimir, and both end up with braided monoidal categories. But they're not the same. The first construction had to deform the associator to make everything work out, whereas in the second the associator is what I called **strongly associative**: the category embeds naturally in the category of vector spaces, and the associator is the same as the essentially-trivial associator there that we all know and love. (So it's almost "strictly associative"; "strong" seems like a good antonym for "weak".) On the other hand, the second construction required more data: it required choosing a classical $r$-matrix.
So: how related are these two constructions? For example, suppose I run the first construction, but happen to know that $t$ comes from an $r$-matrix. Does this tell me anything about more about the structure of $\mathcal S[[\hbar]]$? Conversely, if I take the first construction, and then do some Mac Lane-style strictifying, how close can I get to the second construction?
| https://mathoverflow.net/users/78 | Comparing two similar procedures for quantizing a Casimir Lie algebra | The second construction (Lie bialgebra quantization) in fact also uses a Drinfeld associator. The braided tensor categories obtained in these two ways are equivalent, since the quasitriangular QUE algebra produced by the second construction is obtained by twisting the quasitriangular quasiHopf QUE algebra produced by the first construction. The construction of this twist (which we call $J$) from the Drinfeld associator is in fact the main construction of my paper with Kazhdan "Quantization of Lie bialgebras, I". Categorically, this twist $J$ provides a tensor structure on the forgetful functor on the category produced by the first construction, and the endomorphism algebra of this tensor functor is the Hopf algebra produced by the second construction. Such a tensor functor exists once you find a classical r-matrix r such that $r+r\_{21}=t$ (in fact, such functors bijectively correspond to such classical r-matrices over $k[[h]]$, up to isomorphism).
I'd like to add two comments.
1. If you want the braiding to be $e^{ht/2}$ then the KZ associator $\Phi\_{KZ}$ will not be real, since the KZ equation will have an overall coefficient $1/2\pi i$. So another
associator is the complex conjugate KZ associator $\overline{\Phi\_{KZ}}$.
There is now also a third "explicitly" known associator - the Alexeev-Torossian associator,
see e.g. arXiv:0905.1789, arXiv:0906.0187. This associator is indeed real and depends on $h^2$ (i.e., is even). I wonder if it is the "midpoint" between the KZ associator and its conjugate (the notion of a midpoint on the space of associators makes sense, since the space of associators, according to Drinfeld, has a free transitive action of the Grothendieck-Teichmuller group $GT\_1$, which is prounipotent.) This midpoint is also real and depends on $h^2$.
2. I think the independence of the category in the first construction on the choice of associator is due to Drinfeld, before Le and Murakami; he proves in his paper "Quasi-Hopf algebras" that the associator for any Casimir is unique up to twisting.
| 7 | https://mathoverflow.net/users/3696 | 16155 | 10,828 |
https://mathoverflow.net/questions/16074 | 49 | This is maybe not an entirely mathematical question, but consider it a pedagogical question about representation theory if you want to avoid physics-y questions on MO.
I've been reading Singer's [Linearity, Symmetry, and Prediction in the Hydrogen Atom](http://www.springer.com/mathematics/algebra/book/978-0-387-24637-6) and am trying to come to terms with the main physical (as opposed to mathematical) argument of the text. The argument posits, if I understand it correctly, that a quantum system described by a Hilbert space $H$ on which a group $G$ of symmetries acts by unitary transformations should have the property that its "elementary states" "are" irreducible subrepresentations of the representation of $G$ on $H$. She begins this argument in section 5.1:
> Invariant subspaces are the only physically natural subspaces. Recall from Section 4.5 that in a quantum system with symmetry, there is a natural representation $(G, V, \rho)$. Any physically natural object must appear the same to all observers. In particular, if a subspace has physical significance, all equivalent observers must agree on the question of a particular state's membership in that subspace.
and continues it in section 6.3:
> We know from numerous experiments that every quantum system has \*elementary states\*. An elementary state of a quantum system should be \*\*observer-independent\*\*. In other words, any observer should be able (in theory) to recognize that state experimentally, and the observations should all agree. Secondly, an elementary state should be indivisible. That is, one should not be able to think of the elementary state as a superposition of two or more "more elementary" states. If we accept the model that every recognizable state corresponds to a vector subspace of the state space of the system, then we can conclude that elementary states correspond to irreducible representations. The independence of the choice of observer compels the subspace to be invariant under the representation. The indivisible nature of the subspace requires the subspace to be irreducible. So elementary states correspond to irreducible representations. More specifically, if a vector $w$ represents an elementary state, then $w$ should lie in an \*irreducible\* invariant subspace $W$, that is, a subspace whose only invariant subspaces are itself and $0$. In fact, every vector in $W$ represents a state "indistinguishable" from $w$, as a consequence of Exercise 6.6.
(For people who actually know their quantum, Singer is ignoring the distinction between representations and projective representations until later in the book.)
My first problem with this argument is that Singer never gives a precise definition of "elementary state." My second problem is that I'm not sure what physical principle is at work when she posits that physically natural subspaces and elementary states should be observer-independent (i.e. invariant under the action of $G$). What underlying assumption of quantum mechanics, or whatever, is at work here? Why should a mathematician without significant training in physics find this reasonable? (I have the same question about the identification of elementary particles with irreducible representations of the "symmetry group of the universe," so any comments about this physical argument are also welcome.)
Singer goes on to use this assumption to deduce the number of electrons that fill various electron orbitals, and I won't be able to convince myself that this makes sense until I understand the physical assumption that allows us to use irreducible representations to do this.
| https://mathoverflow.net/users/290 | How is the physical meaning of an irreducible representation justified? | Invariant states are *not* the only meaningful ones. Even in classical mechanics, a baseball traveling 90 mph toward my head is quite meaningful to me, even though it is of no consequence to my fellow mathematician a mile away.
The focus on invariant subspaces comes not from an assumption, but from the way physicists do their work. They want to predict behavior by making calculations. They want to find laws that are universal. They want equations and calculation rules that will be invariant under a change of observers.
Any particular calculation might require a choice of coordinates, but the rules must be invariant under that choice. Once we're talking about one particular baseball trajectory, that trajectory will look different in different coordinate systems; the rules governing baseball flight, however, must look the same in all equivalent coordinate systems.
The natural features of baseballs arise from the equivalence classes of trajectories of baseballs -- equivalence under the group action. Here, if we pretend the earth is flat, gravity is vertical, and air does not resist the baseball, the group is generated by translations and rotations of the plane. Any physically natural, intrinsic property of the baseball itself (such as its mass) or its trajectory (such as the speed of the baseball) must be invariant under the group action. If you don't know a priori what these properties will be, a good way to find them is to pass from individual instances (the baseball heading toward me at 90mph) to the equivalence class generated by individual instances under the group action (the set of all conceivable baseballs traveling at 90mph). Note that the equivalence class is invariant under the group action, and it is exactly this invariance that makes the equivalence class a useful object of the physicists' study.
More generally, if you are studying a physical system with symmetry, it's a good bet that the invariant objects will lead to physically relevant, important quantities. It's more a philosophy than an axiom, but it has worked for centuries.
| 62 | https://mathoverflow.net/users/4188 | 16156 | 10,829 |
https://mathoverflow.net/questions/16134 | 7 | In the paper "Complete topoi representing models of set theory" by Blass and Scedrov, they consider a general notion of Boolean-valued model of set theory, and one of the conditions they impose is that the model contain "no extra ordinals after those of V", i.e. that for all z in the model we have
$$\Vert z \text{ is an ordinal} \Vert = \bigvee\_{\alpha \text{ is an ordinal of } V} \Vert z=\check{\alpha}\Vert $$
where $\Vert-\Vert$ denotes the truth function of the model valued in some complete Boolean algebra.
My question is: do there exist models which *do* contain "extra ordinals" in this sense? I presume so, or they wouldn't have needed to impose this condition. What do such models look like?
(By way of clarification, certainly if the starting model V is a set model in some larger universe, then one can find other set models in that larger universe which contain more ordinals. But I'm interested in just starting with a single model V and building models from it, which can be sets or proper classes.)
| https://mathoverflow.net/users/49 | Can models of set theory contain extra ordinals? | I have two answers.
First, the standard method of building B-valued models of set
theory, where B is any complete Boolean algebra, always
satisfies your condition.
Suppose that B is any complete Boolean algebra, and denote
the original set-theoretic universe by V. One constructs
the B-valued universe VB by building up the
collection of B-names by recursion, so that τ is a
B-name, if it consists of pairs ⟨σ,b⟩ where
σ is a previously constructed name and b ∈ B.
One may impose a B-valued structure on the class
VB of all B-names, by first defining it for
atomic formulas by induction on names and then extending to
all formulas by induction on formulas. This is the usual
way to do forcing with Boolean-valued models, and
VB is the Boolean-valued structure that results.
The remarkable thing, providing the power of forcing, is
that every ZFC axiom gets Boolean value 1 in VB.
In particular, the assertion that any two ordinals are
comparable will have Boolean value 1.
Suppose that z is any B-name. If β is an ordinal above
the Levy rank of z, that is, the place in the
Vβ hierarchy where the name z first exists,
then it is not difficult to see that [[ β ∈ z ]] =
[[ β = z ]] = 0. It follows that [[ z ∈ β ]] =
1. But this latter Boolean value is the same as
Vα<β [[ z = α ]]. Thus, we
have proved your identity
* [[ z is an ordinal ]] = Vα ∈ ORD[[ z =
α ]],
since all of the terms in this join are 0 beyond β and
below β it is the expression we already observed.
There is a subtle point about whether your condition actually expresses "no new ordinals" or not. Suppose that VB is the B-valued model we have constructed and let U be any ultrafilter on B. One may form the quotient model VB/U, and there is a Los theorem, showing that the quotient satisfies φ if and only if [[ φ ]] ∈ U. If U is not V-generic, for example, if U is in V and B is not atomic, then there will be names z such that [[ z is an ordinal ]] = 1, but [[ z = α ]] ∉ U for any ordinal α in V. One way of thinking about this is that VB knows that z is definitely an ordinal, and by your property, Vα ∈ ORD [[ z = α ]] = 1, but VB doesn't know that z is any particular ordinal α. Thus, the ultrafilter U is able to squeeze between these two requirements, and in the quotient, z is a new ordinal. But this doesn't contradict your property.
Now, second, I can give a negative example. It is implicit in your question that the B-valued model somehow includes V, since you refer to the V ordinals α inside the Boolean brackets. Suppose that V is the univese of all sets, and let j:V to M be any elementary embedding that is not an isomorphism. For example, perhaps M is the ultrapower of V by an ultrafilter (M may or may not be well-founded). In particular, not every ordinal of M has the form j(α) for an ordinal α of V. Since M is a model of ZFC, we may regard it as a 2-valued Boolean model, or as a B-valued model for any B, since 2 = {0, 1} is a subalgebra of B. But M has ordinals not of the form j(α) for any V ordinal α. If one identifies V with its image in M, then this would provide a counterexample to the desired property.
| 4 | https://mathoverflow.net/users/1946 | 16157 | 10,830 |
https://mathoverflow.net/questions/16024 | 16 | What kind of role do quantum groups play in modern physics ?
Do quantum groups naturally arise in quantum mechanics or quantum field theories?
What should quantum symmetry refer to ?
Can we say that the "symmetry" of a noncommutative space (quantum phase space) should be a quantum group?
Do quantum groups describe "extended symmetry" ?
| https://mathoverflow.net/users/4155 | What is the relation between quantum symmetry and quantum groups? | Yes, quantum groups naturally arise in many physics problems. E.g. solutions of the quantum Yang-Baxter equation appear as scattering matrices of integrable 2-dimensional quantum field theories (see "Quantum fields and Strings: a course for Mathematicians", p.1179). Also, quantum groups appear in the description of monodromy of the vertex operators in the WZW model of conformal field theory ("the Drinfeld-Kohno theorem"). Thirdly, there are spectacular applications of (infinite dimensional) quantum groups to statistical mechanics, which are described in the book by Jimbo and Miwa "Algebraic analysis of solvable lattice models". Also, quantum groups are useful in construction and studying of certain classes of integrable systems (q-Toda systems, Macdonald-Ruijsenaars systems, etc.)
One of the main mechanisms through which quantum groups appear in physics is the same as for usual Lie groups: if a Hamiltonian of a quantum system has a Lie group symmetry then this helps find its eigenvalues and eigenvectors (which is the main problem in studying a quantum system), because its eigenspaces are representations of this group.
| 16 | https://mathoverflow.net/users/3696 | 16158 | 10,831 |
https://mathoverflow.net/questions/16163 | 11 | Could anyone give me some references where I could find
(a) discrete version(s) of [Ito's lemma](http://en.wikipedia.org/wiki/Ito%27s_lemma)
(b) a proof how it converges to the continuous form in the limit
(c) its usage within stochastic difference equations
(d) a deduction of a discrete version of the Black Scholes model.
Every little bit of information would help.
| https://mathoverflow.net/users/1047 | Discrete version of Ito's lemma | 1. [*Stochastic Calculus for Finance II: Continuous-Time Models*](http://rads.stackoverflow.com/amzn/click/0387401016) by Shreve or
2. Shreve or Øksendal's [*Stochastic Differential Equations*](http://rads.stackoverflow.com/amzn/click/3540047581)
3. Øksendal
4. Williams' [*Probability with Martingales*](http://books.google.com/books?id=e9saZ0YSi-AC) or Shreve
| 11 | https://mathoverflow.net/users/1847 | 16166 | 10,837 |
https://mathoverflow.net/questions/8993 | 5 | This is a question about a name of a very useful lemma,
that permits one in particular to show that smooth birational complex projective
varieties have isomorphic fundamental groups.
If this lemma has no name, I would like at least to have a reference (if it exits).
The lemma can be seen as a
truncated version of the basic fact, that if we have a locally trivial
fibration (say of finite dimensional CW complexes) $F\to E\to B$ then
we get a long exact sequence
$\to \pi\_i(F)\to \pi\_i(E)\to \pi\_i(B)\to \pi\_{i-1}(F)\to$
**Lemma.** Let $E\to B$ be a surjective map of finite dimensional $CW$ complexes,
such that every fiber is connected, simply connected and is a
deformation retract of a small neighbourhood.
Then $\pi\_1(E)=\pi\_1(B)$.
**Question.**
Do you know the name of such a lemma, or of some of its generalizations? Is there a reference for this?
The result about $\pi\_1$ of birationaly equivalent varieties follows
since any birational transformation can be decomposed in blow-ups
and blow downs along smooth submanifolds. And it is not hard
to check that the conditions of lemma are satisfied for such
elementary blow ups.
| https://mathoverflow.net/users/943 | Truncated exact sequence of homotopy groups | Check out the paper "A Vietoris Mapping Theorem for Homotopy," by S. Smale, Proc. Amer. Math. Soc. 8 (1957), 604-610, available at <http://www.jstor.org/stable/2033527> .
Paraphrase of the main theorem: If $f:X\to Y$ is a proper, onto map of 0-connected, locally compact, separable metric spaces, X is $LC^n$, and each point inverse is $LC^{n-1}$ and $(n-1$-connected, then the induced homomorphism $\pi\_r(X)\to \pi\_r(Y)$ is an isomorphism for $r\le n-1$ and surjective for $r=n$.
$LC^n$ is a local connectedness condition surely satisfied by CW complexes, which are locally contractible.
| 2 | https://mathoverflow.net/users/1822 | 16181 | 10,848 |
https://mathoverflow.net/questions/16178 | 13 | We are interested in a solution to the following scheduling problem, or any information about how to find it or its existence. This one comes from real life, so you will not only be helping a mathematician quench his thirst of knowledge!
>
> We have 18 players playing a certain sport (let's say curling) on 3 different alleys (6 players per alley) at the same time. They play 17 games and we want that every combination of 2 players play exactly 5 times together.
>
>
>
(As Douglas Zare points out in a comment below, this is known as a resolvable block design with t=2, v=18, k=6, lambda=5 (and b=51, and r=17)).
We asked around and someone came up with a near solution: almost every pair playing 5 times except for a few 6's and 4's. Brute force seemed too slow so we tried with a genetic algorithm, to no avail (being complete beginners in this, we could not even get close to the near-solution that we had, so we do not draw conclusions from our experiments).
I found the near-solution in my old files, in case anyone wants to tinker a bit.
```
{{1, 2, 3, 4, 5, 6}, {7, 8, 9, 10, 11, 12}, {13, 14, 15, 16, 17, 18}},
{{1, 6, 10, 12, 14, 16}, {2, 3, 8, 11, 15, 17}, {4, 5, 7, 9, 13, 18}},
{{1, 5, 7, 8, 15, 16}, {2, 4, 10, 11, 13, 14}, {3, 6, 9, 12, 17, 18}},
{{1, 4, 8, 9, 14, 17}, {2, 6, 7, 10, 15, 18}, {3, 5, 11, 12, 13, 16}},
{{1, 6, 8, 11, 13, 18}, {2, 4, 9, 12, 15, 16}, {3, 5, 7, 10, 14, 17}},
{{1, 2, 7, 12, 13, 17}, {3, 4, 8, 10, 16, 18}, {5, 6, 9, 11, 14, 15}},
{{1, 3, 9, 10, 13, 15}, {2, 5, 8, 12, 14, 18}, {4, 6, 7, 11, 16, 17}},
{{1, 5, 10, 11, 17, 18}, {2, 6, 8, 9, 13, 16}, {3, 4, 7, 12, 14, 15}},
{{1, 2, 9, 11, 16, 18}, {3, 6, 7, 8, 13, 14}, {4, 5, 10, 12, 15, 17}},
{{1, 4, 8, 12, 15, 18}, {2, 3, 7, 9, 11, 14}, {5, 6, 10, 13, 16, 17}},
{{1, 3, 7, 14, 16, 18}, {2, 5, 8, 9, 10, 17}, {4, 6, 11, 12, 13, 15}},
{{1, 5, 6, 9, 12, 14}, {2, 3, 10, 13, 15, 18}, {4, 7, 8, 11, 16, 17}},
{{1, 3, 10, 11, 12, 16}, {2, 4, 5, 8, 13, 14}, {6, 7, 9, 15, 17, 18}},
{{1, 2, 3, 4, 6, 17}, {5, 7, 11, 12, 13, 18}, {8, 9, 10, 14, 15, 16}},
{{1, 4, 7, 9, 10, 13}, {2, 12, 14, 16, 17, 18}, {3, 5, 6, 8, 11, 15}},
{{1, 2, 5, 7, 15, 16}, {3, 8, 9, 12, 13, 17}, {4, 6, 10, 11, 14, 18}},
{{1, 11, 13, 14, 15, 17}, {2, 6, 7, 8, 10, 12}, {3, 4, 5, 9, 16, 18}}
```
| https://mathoverflow.net/users/4189 | Is there a tournament schedule for 18 players, 17 rounds in groups of 6, which is balanced in pairs? | A collection of $6$-tuples on 18 points with the property that each pair is covered $5$ times is a balanced incomplete block design with $(v,k,\lambda) = (18,6,5)$ and $t=2$. The condition that you can schedule the matches to occur simultaneously in $17$ rounds is that the design is resolvable.
[This article](http://www.iop.org/EJ/abstract/0025-5734/28/3/A05) claims to construct resolvable block designs with parameters including $(18,6,5)$.
| 7 | https://mathoverflow.net/users/2954 | 16182 | 10,849 |
https://mathoverflow.net/questions/16187 | 5 | Let $X \text{~} \text{Binomial}(n, p)$.
What is $\text{P}[X \mod 2 = 0]$? Is it of the form $1/2 + O(1/2^n)$?
| https://mathoverflow.net/users/4197 | Binomial distribution parity | This probability is in fact $1/2 + (1-2p)^n/2$.
Here's a proof. The probability generating function of $X$ is $f(z) = (q+pz)^n$, where $q = 1-p$. That is, the coefficient of $z^k$ in $(q+pz)^n$ is the probability $P(X=k)$.
Now, consider $(f(z)+f(-z))/2$. This polynomial contains just the terms of $f(z)$ which contain $z$ to some even power. If we set $z = 1$, then we get just the sum of those terms. So the probability you want is $(f(1)+f(-1))/2$.
For example, if $n = 3$ we have
$$f(z) = q^3 + 3 p q^2 z + 3 p^2 q z^2 + p^3 z^3$$
$${f(z) + f(-z) \over 2} = q^3 + 3 p^2 q z^2$$
$${f(1) + f(-1) \over 2} = q^3 + 3 p^2 q$$.
Now, we have
$$ {f(1) + f(-1) \over 2} = {(q+p)^n + (q-p)^n \over 2} = {1 + (1-2p)^n \over 2} $$
which is the result.
| 7 | https://mathoverflow.net/users/143 | 16188 | 10,852 |
https://mathoverflow.net/questions/15509 | 10 | Let $\Delta\_k$ be the k-simplex and $\mu$ a non-negative measure over $\Delta\_k$. I want to know if there exists a function $u : \Delta\_k \to \mathbb{R}$ such that $u$ is convex, $u(e\_i) = 0$ for all vertices $e\_i$ of $\Delta\_k$, and $M[u] = \mu$ where
$M[u] = \det\left(\frac{\partial^2 u}{\partial x\_j \partial x\_k}\right)$ is the Monge-Ampère operator. Furthermore, I'd like to know if the solution is unique. Any techniques for how one might solve a specific instance of this problem would be a bonus.
My background is not in PDEs but the closest I've found to an answer seem to be in [1] and [2] where the boundary conditions are more restrictive and the domain is required to be strictly convex for uniqueness.
[1] "On the fundamental solution for the real Monge-Ampère operator", Blocki and Thorbiörnson, Math. Scand. 83, 1998
[2] "The Dirichlet problem for the multidimensional Monge-Ampère equation", Rauch and Taylor, Rocky Mountain Journal of Mathematics, 7(2), 1977.
Any other pointers to solving this type of problem would be greatly appreciated.
| https://mathoverflow.net/users/1915 | Solutions to a Monge-Ampère equation on the simplex | Any set of values of $u$ at the vertices of $\Delta\_k$ can be attained just by adding an affine function to $u$, which does not change $M[u]$. To see that the solution of your problem is not unique, consider $u(x,y)=ax^2+a^{-1}y^2+\mathrm{(affine\ terms)}$ with $a>0$. Clearly $M[u]=4$ for any $a$.
On the other hand, Theorem 1.6.2 in the book *The Monge-Ampère equation* by C. Gutierréz states that there is a unique convex solution of $M[u]=\mu$ (with $M[u]$ properly understood) with prescribed continuous boundary values in a **strictly convex** domain $\Omega$. Without strict convexity we can't allow arbitrary continuous boundary data; it must be at least consistent with some convex function in $\Omega$. (But I don't know if that's enough). The uniqueness part holds without strict convexity; see Corollary 1.4.7 in the same book.
| 6 | https://mathoverflow.net/users/2912 | 16206 | 10,864 |
https://mathoverflow.net/questions/16183 | 17 | I read several times that $(\infty,1)$-categories (weak Kan complexes, special simplicial sets) are a generalization of the concept of model categories. What does this mean? Can one associate an $(\infty,1)$-category to a model category without losing the information on the co/fibrations? How? Why is the $(\infty,1)$-category viewpoint the better one?
$(\infty,1)$-categories are equivalent to simplicial categories (categories enriched over simplicial sets). This is outlined in Lurie's *higher topoi*. A simplicial model category is in particular a simplicial category. Is this the way the association works? Every model category is Quillen equivalent to a simplicial model category and can thus be enriched over simplicial sets.
It would be nice if somebody could help me to clarify this.
**Edit:** Thank you all for the answers. It seems to me that $(\infty,1)$-categories are *not* a generalization of the concept of model categories. A model category is more than a category $C$ together with a class of maps $W$ such that $C[W^{-1}]$ is a category. A model structure data on a category $C$ contains the information on what a cofibration and what a fibration is. This is important for the structure. There exist *different* model structures with the same homotopy category as for example model structures on functor categories.
This means that if there is a kind of functor
$$
F: \{\mbox{model categories}\} \to \{\mbox{($\infty,1)$-categories}\}
$$
it is at least *not* an embedding. In spite of the answers, I still don't see how this functor (if it is really a functor) works. Where is a model category mapped to?
| https://mathoverflow.net/users/4011 | $(\infty,1)$-categories and model categories | Mostly I refer you to my answer [here](https://mathoverflow.net/questions/2185/how-to-think-about-model-categories/2317#2317) and also [this question](https://mathoverflow.net/questions/8663/infinity-1-categories-directly-from-model-categories).
To answer the question about (co)fibrations: No, there is no notion corresponding to (co)fibration in the (∞,1)-category associated to a model category. After all, being a (co)fibration has no homotopical information: every map is equivalent to a (co)fibration. For the sorts of things you need the (co)fibrations to define in model categories, such as homotopy (co)limits, you can give direct definitions in terms of mapping spaces in the (∞,1)-category.
There are two sensible notions of "sameness" of model categories: categorical equivalence, by which I mean an equivalence of categories which preserves each of the three classes of arrows, and Quillen equivalence. This is a lot like the difference between two objects **in** a model category being isomorphic or merely weakly equivalent, though I don't think anyone has a framework in which to make this idea precise. When you consider, say, the projective and injective model structures on a diagram category, these model structures are Quillen equivalent but not categorically equivalent. They have different 1-categorical properties (it's easy to describe left Qullen functors out of the projective model structure and left Quillen functors into the injective model structure) but they model the same homotopy theory. The passage to associated (∞,1)-categories eliminates the distinction between categorical equivalence and Quillen equivalence: two model categories are Quillen equivalent if and only if their associated (∞,1)-categories are equivalent. (Actually, I am not sure whether there are some technical conditions needed for the last assertion, but if so they are satisfied in practice.)
| 9 | https://mathoverflow.net/users/126667 | 16209 | 10,866 |
https://mathoverflow.net/questions/10056 | 20 | Given a topological space **X** and a finite cover **X** = $\cup X\_i$, one can define Cech cohomology of a sheaf of abelian groups **F** with respect to the cover $\{X\_i\}$ in two different ways:
1. (Ordered): The kth term of the Cech
complex is $\bigoplus\_{i\_1 < \ldots
< i\_k} \Gamma(X\_{i\_1} \cap \ldots
\cap X\_{i\_k}, F)$.
2. (Unordered): The kth term of the Cech
complex is $\bigoplus\_{i\_1, \ldots
, i\_k} \Gamma(X\_{i\_1} \cap \ldots
\cap X\_{i\_k}, F)$.
In particular, the second description involves repetition and is non-zero in every degree. These two descriptions give isomorphic cohomology (the first maps you try to write down will likely be homotopy equivalences).
---
**Question**: Is there a canonical reference for this fact?
| https://mathoverflow.net/users/2 | Equivalence of ordered and unordered cech cohomology. | I wrote it up for my algebraic geometry course as a 2-page [handout](http://math.stanford.edu/~conrad/papers/cech.pdf), inspired by EGA $0\_{\rm{III}}$, 11.8.7 (which isn't to say this is a canonical reference; just some written reference...).
| 17 | https://mathoverflow.net/users/3927 | 16213 | 10,870 |
https://mathoverflow.net/questions/16216 | 9 | I am writing an undergraduate thesis on local and global class field theory from a classical (i.e., non-cohomological) approach and am hoping to obtain copies of the early groundbreaking publications in the field. I am primarily interested in finding English translations of articles from Weber, Hasse, Hilbert, Kronecker, and Takagi from 1850 to around 1935 as possible. A fairly comprehensive list of them is found in Hasse's "History of Class Field Theory" in Cassels & Frohlich's ANT, and I can provide a list if needed. Any recommendations for sites that provide translated back issues of *Mathematische Annalen* and/or *Gottinger Nachrichten* from these time periods would be of great utility.
Edit:
Thank you all for responding! You're answers will definitely help me develop the historical portion of my thesis. In response to several comments, I realized when posting that it was probably naive to assume that English translations of all these works were available but presented the question in the above manner for the sake of brevity.
**@KConrad:**
I actually found your cfthistory.pdf file while googling the subject early in my research and have benefited from it greatly! Small world! :)
I would be fine with library copies of the works, of course, but my university would have to order copies from other institutions. As such, I thought I'd first make sure there wasn't some large repository of translated articles from Hilbert et al. online somewhere.
I can't thank you enough for this list of sources, too! I will definitely attempt to get my hands on as many as possible.
Thanks again for helping and posting your history of CFT online!
**@Ben Lenowitz:**
Thank you for pointing me to the Gottingen database. I haven't been in awhile and not while searching for these articles. I will give it a shot.
| https://mathoverflow.net/users/4204 | Where can I find online copies of class field theory publications by Kronecker, Weber, Chevalley, Hasse, Hilbert, Takagi, etc? | First, there's no need to focus on online copies, as asked for in the question. We used to have things called libraries which contain journal articles in them. :) Try looking there.
More seriously, I think your task is to a large extent hopeless. Most of those works were never translated into English. But there are numerous English language sources which describe some aspect of how class field theory was originally developed and you should start there.
Here are some:
G. Frei, Heinrich Weber and the Emergence of Class Field Theory, in ``The History of Modern Mathematics, vol. 1: Ideas and their Reception,'' (J. McCleary and D. E. Rowe, ed.) Academic Press, Boston, 1989, 424--450.
H. Hasse, ``Class Field Theory,'' Lecture Notes # 11, Dept. Math. Univ. Laval, Quebec, 1973. [This is basically adapted from his paper in Cassels and Frohlich, but has some nuggets that were not in C&F.]
K. Iwasawa, On papers of Takagi in Number Theory, in ``Teiji Takagi Collected Papers,'' 2nd ed., Springer-Verlag, Tokyo, 1990, 342--351.
S. Iyanaga, ``The Theory of Numbers,'' North-Holland, Amsterdam, 1975. [The end of the book has a nice exposition of how alg. number theory developed up to class field theory.]
S. Iyanaga, On the life and works of Teiji Takagi, in ``Teiji Takagi Collected Papers,'' 2nd ed., Springer-Verlag, Tokyo, 1990, 354--371.
S. Iyanaga, Travaux de Claude Chevalley sur la th\'eorie du corps de classes:
Introduction, Japan. J. Math. 1 (2006), 25--85. [Are you OK with French?]
M. Katsuya, The Establishment of the Takagi--Artin Class Field Theory, in
``The Intersection of History and Mathematics,'' (C. Sasaki, M. Sugiura, J. W. Dauben ed.),
Birkhauser, Boston, 1995, 109--128.
T. Masahito, Three Aspects of the Theory of Complex Multiplication,
``The Intersection of History and Mathematics,'' (C. Sasaki, M. Sugiura, J. W. Dauben ed.),
Birkhauser, Boston, 1995, 91--108.
K. Miyake, Teiji Takagi, Founder of the Japanese School of Modern Mathematics,
Japan. J. Math. 2 (2007), 151--164.
P. Roquette, Class Field Theory in Characteristic $p$, its Origin and Development,
in ``Class Field Theory -- its Centenary and Prospect,'' Math. Soc. Japan, Tokyo, 2001,
549--631.
H. Weyl, David Hilbert and His Mathematical Work, Bull. Amer. Math. Soc. {\bf 50} (1944), 612--654. [Hilbert's obituary]
I did write up something myself about a year or so ago on the history of class field theory just to put in one place what I was able to cobble together from these kinds of sources. See
<http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/cfthistory.pdf>
which contains the above references as the bulk of the bibliography (I did not just type all those articles references above by hand!) The main thing which had baffled me at first was how they originally defined the local norm residue symbol at ramified primes. I give some examples of how this was determined in the original language of central simple algebras.
| 22 | https://mathoverflow.net/users/3272 | 16222 | 10,876 |
https://mathoverflow.net/questions/15127 | 14 | For several years people have tried to characterize finite groups of maximal order in $\mathrm{GL}(n,\mathbb{Q})$ (or $\mathrm{GL}(n,\mathbb{Z})$) and their orders.
It appear in many articles a reference to an "preprint" article from Walter Feit in 1995 that gave full characterization. And I read a quote that Feit's paper also relies on a unpublished paper from Weisfeiler.
Does anyone know of a (pucblished) paper on this?
| https://mathoverflow.net/users/3958 | The maximum order of finite subgroups in $GL(n,Q)$ | Feit published his paper in the proceedings of the first Jamaican conference, [MR1484185](http://www.ams.org/mathscinet-getitem?mr=1484185). He defines M(n,K) to be the group of monomial matrices whose entries are roots of unity. M(n,Q) is the group of signed permutation matrices.
Theorem A: A finite subgroup of GL(n,Q) of maximum order is conjugate to M(n,Q) and so has order n!2^n except in the following cases... [n=2,4,6,7,8,9,10]. In all cases the finite subgroup of maximum order in GL(n,Q) is unique up to conjugacy.
He notes that the maximum order subgroups of GL(n,Z) need not be unique up to GL(n,Z) conjugacy, since Weyl(Bn) and Weyl(Cn) are GL(n,Q) conjugate but not GL(n,Z) conjugate for n>2.
Theorem B gives a similar result for the cyclotomic fields, Q(l).
Feit published other papers which were very similar, and all of them rely heavily on Weisfeiler's work. However, I believe this is the only published account of his "here is the list" preprint.
| 13 | https://mathoverflow.net/users/3710 | 16223 | 10,877 |
https://mathoverflow.net/questions/16224 | 7 | Can we construct the category of spectra (or maybe just its homotopy category) from the category of pointed topological spaces
using some kind of localization combined with other categorical constructions?
[The first part of the original question was wrong for a trivial reason pointed out by Reid Barton.]
| https://mathoverflow.net/users/402 | Spectra and localizations of the category of topological spaces | [Removed a paragraph relating to an earlier version of the question]
You can construct Spectra categorically by adjoining an inverse to the endofunctor Σ of Top as a presentable (∞,1)-category. Inverting an endofunctor is a very different operation than inverting maps! It's like the difference between forming ℤ[1/p] and ℤ/(p).
Here is one way to verify the claim. To invert the endomorphism Σ of Top we should form the colimit, in the (∞,1)-category Pres of presentable categories and colimit-preserving functors, of the sequence Top → Top → ... where all the functors in the diagram are Σ. A basic fact about Pres is that we can compute such a colimit by forming the diagram (on the opposite index category) formed by the right adjoints of these functors, and taking its *limit* as a diagram of underlying (∞,1)-categories [HTT 5.5.3.18]. The functors in the limit cone will have left adjoints which are the functors to the colimit in Pres. In our case we obtain the sequence Top ← Top ← ... where the functors are Ω, and the limit of this sequence is precisely the classical definition of (Ω-)spectrum: a sequence of spaces Xn with equivalences Xn → ΩXn+1.
| 9 | https://mathoverflow.net/users/126667 | 16231 | 10,881 |
https://mathoverflow.net/questions/16243 | 26 | I am looking for an analogue for the following 2 dimensional fact:
Given 3 angles $\alpha,\beta,\gamma\in (0;\pi)$ there is always a triangle with these prescribed angles. It is spherical/euclidean/hyperbolic, iff the angle sum is smaller than /equal to/bigger than $\pi$. And the length of the sides (resp. their ratio in the Euclidean case) can be computed with the sine and cosine law.
The analogous problem in 3 dimensions would be:
Assign to each edge of a tetrahedron a number in $(0;\pi)$. Does there exists a tetrahedron with these numbers as face angles at those edges. And when is it spherical/euclidean/hyperbolic. Is there a similar Invariant to the angle sum?
And are there formulas to compute the length of the edges?
| https://mathoverflow.net/users/3969 | Tetrahedra with prescribed face angles | The short answer is no - there is no single inequality criterion. Already in $\mathbb{R}^3$ everything is much more complicated. Let me give a sample of inequalities the angles should satisfy. Denote by $\gamma\_{ij}, 1\leq i < j \leq 4$ the six dihedral angles of a Euclidean tetrahedron. Then:
$$
\gamma\_{12}+\gamma\_{23} + \gamma\_{34}+\gamma\_{14} \le 2 \pi
$$
$$
2\pi \le \gamma\_{12} + \gamma\_{13} + \gamma\_{14}+\gamma\_{23} + \gamma\_{24}+\gamma\_{34} \le 3\pi
$$
$$
0 \le \cos \gamma\_{12} + \cos\gamma\_{13} + \cos\gamma\_{14}+ \cos\gamma\_{23} + \cos\gamma\_{24}+ \cos\gamma\_{34} \le 2
$$
(See my [book](https://www.math.ucla.edu/~pak/book.htm) ex. 42.27 for the proofs of these inequalities - they are not terribly difficult, so you might enjoy proving them yourself).
This shows that the set of allowed sixtuples of angles is rather complicated (for spherical/hyperbolic tetrahedra with angles close to $\gamma\_{ij}$, these angles will have to satisfy these inequalities as well). The "invariant" you mention corresponds to the unique equation the angles satisfy in the Euclidean space. The latter is also rather delicate: it is the Gauss-Bonnet equation $\omega\_1+...+\omega\_4=4\pi$, where $\omega\_i$ is the curvature of $i$-th vertex - you need to use spherical cosine theorem to compute it from dihedral angles (see e.g. Prop. 41.3 in my book).
Finally, you might like to take a look at [this](https://arxiv.org/abs/math/0308239) interesting paper by Rivin, to see that a similar generalization of the triangle inequality is just as difficult. To answer your last question (edge lengths from dihedral angles), yes, this is known. I am not an expert on this, but I would start with [this](https://doi.org/10.1007/s11202-006-0079-5) recent paper.
| 19 | https://mathoverflow.net/users/4040 | 16250 | 10,896 |
https://mathoverflow.net/questions/16254 | 1 | This question is not homework, just asked out of curiosity. I wondered how many zeroes could be found at the end of $1990!$ . I computed something that seemed to work and found out 439. So I computed it in Python, and it returned 494 zeroes, so I'm 55 short.
My reasoning went like this : I get 2 zeroes by series of 10 consecutive numbers. The one that can be divided by 10 gives one zero, and I get another one from the multiplication of the one ending with 2 and the one ending with 5. So 199\*2 = 398. But I also have 2 zeroes by hundreds, one comming from the one divisible by 100, and one from XX20 \* XX50. 19\*2 = 38 more. Then same for thousands. I only get 3 zero more, from 1000, 200\*500 and 1200\*1500.
$398 + 38 + 3 = 439$
So here is my question:
How can I compute function $t0(n)$ where $t0(n)$ returns the number of trailing $0$ in $n! = factorial(n)$ ?
(Assume $factorial(n)$ is written in base 10, but even better for any base !)
| https://mathoverflow.net/users/2446 | Counting trailing zeros for factorials | Take any integer $x$, and let $t,f$ represent the highest integers such that $2^t | x$ and $5^f | x$. Then the number of trailing zeros in the base 10 representation of $x$ is $z := \min\{t,f\}$. (One way to see this is to note it must be at least z since you have $(2\*5)^z | x$, so you can write $x = 10^z \* y$ where not both $2,5|y$, and so $y$ can't have any trailing zeros.)
Going back to factorials, $n!$ will always have $t > f$, so to count the zeros, you just have to count the fives. The expression you want is
$$
f = \sum\_{i=1}^\infty \lfloor n / 5^i\rfloor.
$$
But maybe there's a cleaner form.
(for 1990! this gives 495, not 494 as you said. I checked this numerically as well, I guess you have a bug.)
| 4 | https://mathoverflow.net/users/2621 | 16255 | 10,900 |
https://mathoverflow.net/questions/16180 | 7 | Goguen has popularized the initial algebra view of semantics via his "no junk, no confusion" slogan. By "no junk", he means that models of a theory presentation should not have unnecessary elements, and "no confusion" that terms should not be mapped to equal values unless they are provably equal. Sometimes, "no junk" is also interpreted as every element in the model is a denotation of a term, while "no confusion" as two different terms denote different elements in the model. [These are classically equivalent statements, but they are not intuinistically equivalent, so I mention both].
My questions are:
1. What is a 'good' formalization of this slogan? By this I mean an explicit statement of "no junk, no confusion" in the meta-logic (since we're talking about models), where the logical strength of the corresponding statement is well understood.
2. Are there logics in which these requirements can be internalized?
3. What would be the corresponding slogan to "no junk, no confusion" for final coalgebras?
| https://mathoverflow.net/users/3993 | Formalizing "no junk, no confusion" | The way I understand each of the slogans is as follows:
1. "No junk" I just take to mean that an appropriate induction principle is valid -- that is, we should look for initial models in the appropriate category of algebras for the theory. This also implies that every element of the model is in the image of the interpretation of the the algebraic theory.
Presumably the dualization to coalgebras would just be the validity of using bisimilarity to prove equality.
2. "No confusion" is traditionally interpreted to mean that we should look for models in which two elements of the model are semantically equal if and only if the corresponding syntaxes are provably equal. This is the really bizarre requirement, since it amounts to requiring that the model be isomorphic to the term model! And yet Goguen and the algebraic specification community were emphatically *not* happy with decreeing the term model to be the intended model -- they work very hard to get the "right" model.
I personally (ie, I don't know that anyone else believes this) take the way this requirement is phrased to be an artifact of the history of algebraic specification. IIRC, they started out with purely algebraic theories -- that is, theories in which the equational axioms are all pure equalities. (E.g., the axioms for groups.) Now, of course every such algebraic theory has a degenerate model, since the one-element model validates all equalities. So the no-confusion principle is intended to rule out such degenerate models.
These days, of course, the algebraic specification crowd has no problem with theories with inequalities (e.g., the field axioms), and I think this additional freedom lets us state the no-confusion principle in a better way. Namely, we should design algebraic theories whose models are *categorical*. That is, we want theories for which all models are isomorphic. This implies the traditional no-confusion criterion, and also explains why people try to adjust the signature when they can't prove it. (Of course, this is a non-first-order property in general, as you need higher-order logic or set theory to quantify over models.)
| 7 | https://mathoverflow.net/users/1610 | 16263 | 10,905 |
https://mathoverflow.net/questions/16261 | 6 | I need a construction in linear algebraic groups which uses taking quotient by a central finite group subscheme.
My question is, whether it goes through in ``bad'' characteristics, when this group subscheme is not smooth.
First I write this construction in a special case, and then in the general case.
Let $G$ be a connected semisimple $k$-group over a field $k$ of characteristic $p>0$.
We may assume that $k$ is algebraically closed.
Assume that the corresponding adjoint group $G^{ad}$ is $PGL\_n$.
In general, my group $G$ is not special (recall that a $k$-group $G$ is called special if $H^1(K,G)=1$ for any field extension $K/k$).
I want to construct a special $k$-group $H$ related to $G$.
For this end I consider the universal covering $G^{sc}$ of $G$, then $G^{sc}=SL\_n$. Let $Z$ denote the center of $G^{sc}$, then $Z=\mu\_n$.
We have a canonical epimorphism $\varphi\colon SL\_n \to G$. We denote by $C$ the kernel of $\varphi$.
Then $C$ is a group subscheme of $Z$, defined over $k$.
Since $Z=\mu\_n$, there is a canonical embedding $Z\hookrightarrow \mathbb{G}\_m$ into the multiplicative group $\mathbb{G}\_m$.
Thus we obtain an embedding $C\hookrightarrow \mathbb{G}\_m$. Consider the diagonal embedding
$$C\hookrightarrow SL\_n\times \mathbb{G}\_m.$$
I would like to define $H:=(SL\_n\times \mathbb{G}\_m)/C$. Is such a quotient defined, when char($k$) divides $n$ and $C$ is not smooth?
Note that $SL\_n$ embeds into $H$, and we have a short exact sequence
$$1\to SL\_n \to H \to \mathbb{G}\_m \to 1$$
In this exact sequence both $SL\_n$ and $\mathbb{G}\_m$ are special, and from the Galois cohomology exact sequence we see that $H$ is special as well.
In the general case I assume that $G^{ad}$ is a product of groups $PGL\_{n\_i}$, $i=1,\dots s$. Then $G^{sc}$ is the product of $SL\_{n\_i}$.
Let $C$ denote the kernel of the canonical epimorphism $\varphi\colon G^{sc}\to G$, then $C$ is contained in the center $Z$ of $G^{sc}$.
We have $Z=\prod\_{i=1}^s \mu\_{n\_i}$. Again we embed diagonally
$$C\hookrightarrow (\prod\_{i=1}^s SL\_{n\_i}) \times (\mathbb{G}\_m)^s $$
and denote by $H$ the quotient. Again $H$ is special (if it is defined), and again my question is, whether this construction makes sense
when char($k$) divides $n\_i$ for some $i$.
Any help is welcome!
| https://mathoverflow.net/users/4149 | Quotient of a reductive group by a non-smooth central finite subgroup | This is an instance of what I believe is called the $z$-construction, and it is a very useful trick in the arithmetic theory of algebraic groups. (Small correction: your diagonal embedding should really be "anti-diagonal". You are really computing a "central pushout".) However, you may need to restrict your ground field to be local or global (with some caveats in case of real places) to get the cohomological vanishing.
In general, let $G$ be any split connected semisimple group over a field $k$, and $G' \rightarrow G$ the $k$-split simply connected central cover. Let $Z$ be the center of $G'$.This is a $k$-group of multiplicative type, possibly not etale. It is $k$-split since $G'$ is $k$-split, so we can choose a $k$-subgroup inclusion of $Z$ into a $k$-split $k$-torus $T$ (seen using character group). Now just form the central pushout of $G'$ along this inclusion. More precisely, the antidiagonal map
$$Z \rightarrow G' \times T$$
is a central $k$-subgroup scheme, and in SGA3, VI$ \_{\rm{A}}$ quotients are constructed and studied for arbitrary finite type group schemes over a field modulo closed subgroup schemes of finite type (and even more generally). We can therefore form the quotient by this central subgroup scheme, and it has all of the reasonable properties one would want for a quotient. This pushout $H$ is a $k$-group containing $G'$ as a normal $k$-subgroup, and the quotient $H/G'$ is $T/Z$. Being a quotient of $k$-split torus, it is again a $k$-split torus.
Provided the ground field is a local or global function field (with some further restrictions in case of a real place), all simply connected semisimple groups have vanishing degree-1 Galois cohomology. So that does the job for such fields. One can also do a variant for split connected reductive groups, and even a variant without a split hypothesis (but then the conclusion is a little different).
| 7 | https://mathoverflow.net/users/3927 | 16273 | 10,911 |
https://mathoverflow.net/questions/16265 | 14 | In one of the exercises in McDuff and Salamon, they mention homology with compact supports. I know how to define \*co\*homology with compact supports, but I can't picture the homology version. How do I say that a chain has compact support? If I use singular chains, don't they all have compact support anyway?
Google isn't a big help here, so any references would really help me out.
Also, I've added some basic sub-questions that would also help me out tremendously. This must all be pretty simple, but my background in algebraic topology is weak and it completely baffles me!
1. McDuff-Salamon go on to state that for the open annulus $(1/2 < r < 1)$ in the plane, the first compact homology group is generated by the arc $\theta = 0$, $1/2 < r < 1$, which I can understand with hindsight: this is just the generator of the homology rel. the boundary and most likely there will be an isomorphism between the compact and relative homologies, just like there is one between the compact and relative co-homologies.
2. They also implicitly use an isomorphism between the compact homology and compact cohomology in certain dimensions. Should I just use this as the definition for the compact homology? I.e. $H\_{k, c} = (H^k\_c)^\ast$?
| https://mathoverflow.net/users/3909 | homology with compact supports | To elaborate on Ekedahl's comment:
It will be easiest to describe things for a triangulated space, so I can work with simplicial chains and cochains. (But my space could be infinitely triangulated; e.g. think of Escher's famous picture of the infinitely triangulated hyperbolic plane. I will also assume that my triangulation is locally finite.)
As you observed, usual chains have compact support. This is why you can pair them with
cochains (which have arbitrary support). Borel--Moore chains can have unbounded support. These can be paired with not with arbitrary cochains, but only with compactly supported ones.
So Borel--Moore homology is the "homology analogue" of compactly supported cohomology.
(But the support conditions are reversed, since homology is dual to cohomology.)
One can often interpret Borel--Moore homology as relative homology. E.g. if $M$ is a compact manifold
with boundary $\partial M$, then the Borel--Moore homology of $M\setminus \partial M$ (the interior of $M$) is the same as the homology of the pair $(M,\partial M)$.
| 13 | https://mathoverflow.net/users/2874 | 16281 | 10,916 |
https://mathoverflow.net/questions/16312 | 62 | So, I can understand how [non-standard analysis](https://en.wikipedia.org/wiki/Non-standard_analysis) is better than standard analysis in that some proofs become simplified, and infinitesimals are somehow more intuitive to grasp than epsilon-delta arguments (both these points are debatable).
However, although many theorems have been proven by non-standard analysis and transferred via the [transfer principle](https://en.wikipedia.org/wiki/Transfer_principle), as far as I know all of these results were already known to be true. So, my question is:
Is there an example of a result that was *first* proved using non-standard analysis? To wit, is non-standard analysis actually useful for proving *new* theorems?
**Edit**: Due to overwhelming support of François' [comment](https://mathoverflow.net/questions/16312/how-helpful-is-non-standard-analysis#comment30082_16312), I've changed the title of the question accordingly.
| https://mathoverflow.net/users/2233 | How helpful is non-standard analysis? | From the [Wikipedia article](https://en.wikipedia.org/wiki/Non-standard_analysis#Applications):
>
> the list of new applications in
> mathematics is still very small. One
> of these results is the [theorem proven
> by Abraham Robinson and Allen
> Bernstein](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-16/issue-3/Solution-of-an-invariant-subspace-problem-of-K-T-Smith/pjm/1102994835.full) that every polynomially
> compact linear operator on a Hilbert
> space has an invariant subspace. Upon
> reading a preprint of the
> Bernstein-Robinson paper, [Paul Halmos
> reinterpreted their proof using
> standard techniques](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-16/issue-3/Invariant-subspaces-of-polynomially-compact-operators/pjm/1102994836.full). Both papers
> appeared back-to-back in [the same
> issue of the Pacific Journal of
> Mathematics](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-16/issue-3). Some of the ideas used in
> Halmos' proof reappeared many years
> later in Halmos' own work on
> quasi-triangular operators.
>
>
>
| 33 | https://mathoverflow.net/users/1847 | 16314 | 10,937 |
https://mathoverflow.net/questions/16310 | 17 | Morel and Voevodsky construct the motivic stable homotopy category, a category through which all cohomology theories factor and where they are representable, by starting with a category of schemes, Yoneda-embedding it into simplicial presheaves, endowing those with the $\mathbb{A}^1$-local model structure, and then passing to $S^1 \wedge \mathbb{G}\_m$-spectra. The last step ensures that smashing with $S^1$ or with $\mathbb{G}\_m$ induce functors with a quasi-inverse on the homotopy category.
Inverting $S^1$ leads to a triangulated structure on the homotopy category, which is very welcome, but I would like a motivation for inverting $\mathbb{G}\_m$. Since $\mathbb{P}^1$ is $\mathbb{A}^1$-equivalent to $S^1 \wedge \mathbb{G}\_m$ I would also be content with a motivation to invert $\mathbb{P}^1$.
---
I must admit I already know some answers which *certainly are reason enough* to invert $\mathbb{G}\_m$, e.g. (from [slides by Marc Levine](http://seven.ihes.fr/IHES/Scientifique/asie/textes/Motives5.pdf), start at page 64):
1. Inverting $\mathbb{G}\_m$ is necessary to produce a Gysin sequence
2. The algebraic K-theory spectrum appears naturally as a $\mathbb{P}^1$-spectrum
However, I am greedy and would like to hear a motivation like the one for inverting the Lefschetz motive in the construction of pure motives: There one could say that for all envisaged realization functors which should factor through the category of pure motives, the effect of tensoring with the Lefschetz motive can be undone (e.g. is just a change of Galois representation leaving the cohomology *groups* unchanged).
Or, related to this, as Emerton explained in his nice answer [here](https://mathoverflow.net/questions/14587/understanding-the-definition-of-the-lefschetz-pure-effective-motive) one has to invert the Lefschetz motive in order to make the Pure Motives a rigid tensor category. Ideally one would like the triangulated category of motives to arise as derived category of some rigid tensor category - if this was true, would it be reflected in the fact that $\mathbb{P}^1$ or $\mathbb{G}\_m$ are invertible? (in this case of course one should ensure iinvertibility when constructing a candidate for this derived category)
| https://mathoverflow.net/users/733 | Why does one invert $G_m$ in the construction of the motivic stable homotopy category? | I guess there are "internal" and "external" motivations. External for instance -- most natural examples of functors we have from the stable motivic homotopy category to some other category invert G\_m (e.g. any of the usual realizations, or K-theory). Internal for instance -- we want the suspension spectra of varieties to be dualizable under the smash product.
| 12 | https://mathoverflow.net/users/3931 | 16320 | 10,941 |
https://mathoverflow.net/questions/16302 | 4 | Does anyone know of the existence of an archive of the work of J Sutherland Frame?
The Briscoe Center for American History maintains about 100 archives of American mathematics and I have found the folks there to be quite helpful.
Cheers, Scott
| https://mathoverflow.net/users/4111 | Archive of the Work of J Sutherland Frame | The short answer is "no". Frame did interesting work, though usually
outside the conceptual mainstream of representation theory. Some of his
calculations of character tables (such as that of the most exceptional
Weyl group) have permanent value, I think.
| 1 | https://mathoverflow.net/users/4231 | 16322 | 10,943 |
https://mathoverflow.net/questions/16292 | 11 | I was reading Mumford again and I noticed a comment in the beginning of the book:
"Finally, in any study of general preschemes, the varieties are bound, for many reasons which I will not discuss here, to play a unique and central role."
My question is: is there a particular theorem (or series of theorems) Mumford might have been hinting at?
| https://mathoverflow.net/users/3701 | The central role of varieties (a comment from Mumford's Red Book) | Here is a really cool illustration of the principle which Emerton was [outlining](https://mathoverflow.net/a/16294). We know that the Picard group of projective $(n-1)$-space over a field $k$ is $\mathbf{Z}$ ($n \ge 2$), generated by $\mathcal{O}(1)$. This underlies the proof that the automorphism group of such a projective space is ${\operatorname{PGL}}\_n(k)$. But what is the automorphism group of $\mathbf{P}^{n-1}\_A$ for a general ring $A$? Is it ${\operatorname{PGL}}\_n(A)$? That is, there is a natural map
$${\operatorname{PGL}} \_n (A) \rightarrow {\operatorname{Aut}} \_A (\mathbf{P}^{n-1} \_A)$$
(see my [answer](https://mathoverflow.net/a/16175) in the posting called [What is the difference between PSL\_2 and PGL\_2?](https://mathoverflow.net/q/16145)) and we want to know if it is an isomorphism. It's a *really important* fact that the answer is yes. But how to *prove* it? It's a shame that this isn't done in Hartshorne.
By an elementary localization (as in the last step of Emerton's [answer](https://mathoverflow.net/a/16294)), we may assume $A$ is local. In this case we claim that ${\operatorname{Pic}}(\mathbf{P}^{n-1}\_A)$ is infinite cyclic generated by $\mathcal{O}(1)$. Since this line bundle has the known $A$-module of global sections, it would give the desired result if true (since ${\operatorname{PGL}}\_n(A) = {\operatorname{GL}}\_n(A)/A^{\times}$ for local $A$) by the same argument as in the field case. And since we know the Picard group over the residue field, we can twist to get to the case when the line bundle $\mathcal{L}$ is trivial on the special fiber and then we can formulate the problem in two *equivalent* ways: (I) "lift" the generating section there to a generating section over $A$, or (II) prove that $f \_{\ast}(\mathcal{L})$ is invertible in $A$ with the natural map $f^{\ast}(f \_{\ast} \mathcal{L}) \rightarrow \mathcal{L}$ an isomorphism. How to do it?
Step 0: The case when $A$ is a field. Done.
Step 1: The case when $A$ is artin local, via (I): this goes via induction on the length, the case of length 0 being Step 0 and the induction resting on cohomological results for projective space over the residue field.
Step 2: The case when $A$ is complete local noetherian ring. This goes via (I) using Step 1 and the theorem on formal functions (formal schemes in disguise).
Step 3: The case when $A$ is local noetherian. This is faithfully flat descent for (II) from Step 2 applied over $\widehat{A}$.
Step 4: The case when $A$ is local: descent from the noetherian local case in Step 3 via direct limit arguments.
"QED"
| 21 | https://mathoverflow.net/users/3927 | 16324 | 10,945 |
https://mathoverflow.net/questions/16214 | 35 | Recall that a function $f\colon X\times X \to \mathbb{R}\_{\ge 0}$ is a *metric* if it satisfies:
* definiteness: $f(x,y) = 0$ iff $x=y$,
* symmetry: $f(x,y)=f(y,x)$, and
* the triangle inequality: $f(x,y) \le f(x,z) + f(z,y)$.
A function $f\colon X\times X \to X$ is *associative* if it satisfies:
* associativity: $f(x,f(y,z)) = f(f(x,y),z)$.
If $X=\mathbb{R}\_{\ge 0}$, then it might be possible for the same function to be a metric and associative.
**Is there an associative metric on the non-negative reals?**
Note that these demands actually make $X$ into a group. The element $0$ is the identity because $f(f(0,x),x) = f(0,f(x,x)) = f(0,0) = 0$ by associativity and definiteness, so again by definiteness $f(0,x) = x$. Every element is its own inverse because $f(x,x)=0$.
In fact, the following question is equivalent. **Is there an abelian group on the non-negative reals such that the group operation satisfies the triangle inequality?**
Note also that the answer is *yes* if $X=\mathbb{N}$, the non-negative numbers! [Click here for a spoiler](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
"Also known as nim-sum."
The question is originally due to [John H. Conway](http://en.wikipedia.org/wiki/John_Horton_Conway). To my knowledge, the question is unsolved even for $X = \mathbb{Q}\_{\ge 0}$, but he does not seem to care about that case. The spoiler above does extend to the non-negative dyadic rationals $\mathbb{N}[\frac 12]$, but apparently not to $\mathbb{N}[\frac 13]$.
| https://mathoverflow.net/users/1079 | Is there an associative metric on the non-negative reals? | Seems that this is possible. Here is a (non-constructive) proof.
Suggestions are welcome.
The proof is inspired by [Mazurkiewicz's argument](http://www.mathnerds.com/best/mazurkiewicz/index.aspx). This is second version
of the proof: it includes improvements in
the set-theoretic argument suggested by Joel David Hamkins, and also
hopefully clarifies some issues
raised in comments. Thanks for the comments!
Goal: Construct a commutative group structure $\star$ on non-negative
reals
${\mathbb R}\_{\ge 0}$ such that $x\star y\le x+y$ and $x\star x=0$.
Remark: Note that $0$ is automatically a neutral element, and that such a
commutative group is in fact
a vector space over $ {\mathbb F}\_2 $. Also, we automatically have the
triangle inequality:
$$x\star z=x\star y\star y\star z\le x\star y+y\star z.$$
Step 1: Let us order ${\mathbb R}\_{\ge 0}$ in order type $c$ (continuum). Equivalently,
we choose a bijection $\iota:[0,c)\to{\mathbb R}\_{\ge 0}$, where $[0,c)$ is the set
of ordinals smaller than $c$. Note that for any $ \alpha < c $, we have
$$|\iota([0,\alpha))| < c.$$
We may choose $\iota$ so that $\iota(0)=0$, although
it is not strictly necessary.
Plan: For every $\alpha\le c$, we will construct a subset
$S\_\alpha\subset {\mathbb R}\_{\ge 0}$ and a group operation
$\star:S\_\alpha\times S\_\alpha\to S\_\alpha$. The group operation will have the
required properties: $S\_\alpha$ is
a vector space over $F\_2$ with $0$ being the neutral element,
and $x\star y\le x+y$. Besides
it will also have the additional property that $S\_\alpha$ is generated as
a group by $\iota([0,\alpha))$
(in particular, the image is contained in $S\_\alpha$). Moreover, if
$\beta\prec\alpha$, $S\_\beta$ is a subgroup
of $S\_\alpha$.
In particular, we get a group structure with required properties on $S\_c={\mathbb R}\_{\ge 0}$,
as claimed.
Step 2: The construction proceeds by transfinite recursion. The base is
$S\_0=\lbrace 0\rbrace$ (generated by the empty set).
Step 3. Let us now define $S\_\alpha$ assuming that $S\_\beta$ is already defined for
$\beta<\alpha$. If $\alpha$ is a limit ordinal, take
$$S\_\alpha=\bigcup\_{\beta<\alpha}S\_\beta.$$
Therefore, let us assume $\alpha=\beta+1$.
If $\iota(\alpha)\in S\_\beta$, take $S\_\alpha=S\_\beta$.
Step 4. It remains to consider the case when $\alpha=\beta+1$ but $\iota(\alpha)\not\in S\_\beta$.
Since $I=\iota([0,\beta))$ generates $S\_\beta$,
the cardinality of $S\_\beta$ is at most the cardinality of the set
of finite subsets of $I$. Therefore, $|S\_\beta| < c$.
Fix a number $k$ between $0$ and $1$, to be chosen later. Define a
function $f:{\mathbb R}\_{\ge 0}\to{\mathbb R}\_{\ge 0}$ by
$$f(x)=\cases{\iota(\alpha)+k x, \ x \le \iota(\alpha)\cr x+k \iota(\alpha), \ x > \iota(\alpha)}.$$
Now choose $k$ so that $f(S\_\beta)\cap S\_\beta=\emptyset$. This is possible because
for every $x,y\in S\_\beta$, the equation $f(x)=y$ has at most one solution in
$k$, so the set of prohibited values of $k$ has cardinality at most
$|S\_\beta\times S\_\beta|$. (We can use $\iota$ to well-order the interval $(0,1)$;
we can then choose $k$ to be the minimal acceptable value, so as to remove arbitrary choice.)
Step 5. Now define $S\_\alpha=S\_\beta\cup f(S\_\beta)$ and set $\iota(\alpha)\star x=f(x)$ for
$x\in S\_\beta$. The product naturally extends to all of $S\_\alpha$:
$$f(x)\star f(y)=x\star y\qquad f(x)\star y=y\star f(x)=f(x\star y).$$
It is not hard to see that it has the required properties.
First of all, $S\_\alpha$ is an isomorphic image of $S\_\beta\times({\mathbb Z}/2{\mathbb Z})$;
this takes care of group-theoretic requirement. It remains to
check two inequalities:
Step 5a: $$f(x)\star f(y)\le f(x)+f(y)\quad(x,y\in S\_\beta),$$
which is true because $f(x)\ge x$, so
$$f(x)\star f(y)=x\star y\le x+y\le f(x)+f(y).$$
Step 5b: $$f(x)\star y\le f(x)+y\quad(x,y\in S\_\beta),$$
which is true because $f$ is increasing and $f(x+t)\le f(x)+t$, so
$$f(x)\star y=f(x\star y)\le f(x+y)\le f(x)+y.$$
That's it.
| 23 | https://mathoverflow.net/users/2653 | 16328 | 10,948 |
https://mathoverflow.net/questions/16257 | 49 | What I am talking about are reconstruction theorems for commutative scheme and group from category. Let me elaborate a bit. (I am not an expert, if I made mistake, feel free to correct me)
**Reconstruction of commutative schemes**
Given a quasi compact and quasi separated commutative scheme $(X,O\_{X})$ (actually, quasi compact is not necessary), we can reconstruct the scheme from $Qcoh(X)$ as category of quasi coherent sheaves on $(X,O\_{X})$. This is Gabriel-Rosenberg theorem. Let me sketch the statement of this theorem which led to the question I want to ask:
The reconstruction can be taken as geometric realization of $Qcoh(X)$ (abelian category). Let $C\_X$=$Qcoh(X)$. We define spectrum of $C\_X$ and denote it by $Spec(X)$
(For example, if $C\_X=R-mod$.where $R$ is commutative ring, then $Spec(X)$ coincides with prime spectrum $Spec(R)$). We can define Zariski topology on $Spec(X)$, we have open sets respect to Zariski topology. Then we have contravariant pseudo functor from category of Zariski open sets of the spectrum $Spec(X)$ to $Cat$,$U\rightarrow C\_{X}/S\_{U}$,where $U$ is Zariski open set of $Spec(X)$ and $S\_U=\bigcap\_{Q\in U}^{ }\hat{Q}$
(Note:$Spec(X)$ is a set of subcategories of $C\_{X}$ satisfying some conditions, so here,$Q$ is subcategory belongs to open set $U$ and $\hat{Q}$=union of all topologizing subcategories of $C\_{X}$ which do not contain $Q$.)
For each embedding:$V\rightarrow U$, we have correspondence localization functor: $C\_{X}/S\_{U}\rightarrow C\_{X}/S\_{V}$. Then we have **fibered category over the Zariski topology of $Spec(X)$**, so we have given a geometric realization of $Qcoh(X)$ as a **stack** of local category which means the fiber(stalk)at each point $Q$,
$colim\_{Q\in U} C\_{X}/S\_{U}$=$C\_{X}/\hat{Q}$ is a local category.
**Zariski geometric center**
Define a functor $O\_{X}$:$Open(Spec(X))\rightarrow CRings$
$U|\rightarrow End(Id\_{C\_{X}/S\_{U}})$
It is easy to show that $O\_{X}$ is a presheaf of commutative rings on $Spec(X)$, then Zariski geometric center is defined as $(Spec(X),\hat{O\_{X}})$,where $\hat{O\_{X}}$ is associated sheaf of $O\_{X}$.
**Theorem**:
Given **X**=$(\mathfrak{X},O\_{\mathfrak{X}})$ a quasi compact and quasi separated commutative scheme. Then the scheme **X** is isomorphic to Zariski geometric center of $C\_X=Qcoh(X)$. If we denote the fibered category mentioned above by $\mathfrak{F}\_{X}$.
then center of each fiber (stalk) of $\mathfrak{F}\_{X}$ recover the presheaf of commutative ring defined above and hence Zariski geometric center. Moreover, Catersion section of this fibered category is equivalent to $Qcoh(X)$ when $X$ is commutative scheme
Question
--------
It is well known that for a compact group $G$, we can use Tanaka formalism to reconstruct this group from category of its representation $(Rep(G),\bigotimes \_{k},Id)$. Is this reconstruction **Morally** the same as Gabriel-Rosenberg pattern reconstruction theorem?
From my perspective, I think $Qcoh(X)$ and category of group representations are very similar because $Qcoh(X)$ can be taken as "category of representation of scheme". On the otherhand, group scheme is a scheme compatible with group operations. Therefore, I think it should have united formalism to reconstruct both of them. Then I have following questions:
1 Is there a Tanaka formalism for reconstruction of general scheme$(X,O\_{X})$ from $Qcoh(X)$?
2.Is there a Gabriel-Rosenberg pattern reconstruction (geometric realization of category of category of representations of group) to recover a group scheme? I think this formalism is natural (because the geometric realization of a category is just a stack of local categories and the center(defined as endomorphism of identical functor)of the fiber of this stack can recover the original scheme) and has good generality (can be extend to more general settings)
Maybe one can argue that these two reconstruction theorems live in different nature because reconstruction of group schemes require one to reconstruct the group operations which force one to go to monoidal categories (reconstruct co-algebra structure) while reconstruction of
non-group scheme doesn't (just need to reconstruct algebra structure).
However, If we stick to the commutative case. $Qcoh(X)$ has natural symmetric monoidal structure. Then, in this case, I think this argue goes away.
More Concern
------------
I just look at P.Balmer's paper on [reconstruction from derived category](http://www.math.ucla.edu/%7Ebalmer/research/Pubfile/Reconstr.pdf), it seems that he also used Gabriel-Rosenberg pattern in triangulated categories: He defined spectrum of triangulated category as direct imitation of prime ideals of commutative ring. Then, he used triangulated version of geometric realization (geometric realization of triangulated category as a stack of local triangulated category) hence, a fibered category arose, then take the center of the fiber at open sets of spectrum of triangulated categories to recove the presheaf (hence sheaf) of commutative rings. This triangulated version of geometric realization is also mentioned in Rosenberg's lecture notes [Topics in Noncommutative algebraic geometry,homologial algebra and K-theory](http://www.mpim-bonn.mpg.de/preprints/send?bid=3589) page 43-44, it also discuss the relation with Balmer's construction. **But in Balmer's consideration, there is tensor structure in his derived category**
Therefore, the further question is:
**Is there a triangulated version of Tannaka Formalism which can recover P.Balmer's reconstruction theorem.** ? I heard from some experts in derived algebraic geometry that Jacob Lurie developed derived version of Tannaka formalism. I know there are many experts in DAG on this site. I wonder whether one can answer this question.
In fact, I still have some concern about Bondal-Orlov reconstruction theorem but I think it seems that I should not ask too many questions at one time.So, I stop here.All the related and unrelated comments are welcome
Thanks in advance!
*Reconstruction theorem in nLab* [reconstruction theorem](http://ncatlab.org/nlab/show/reconstruction+theorem)
**EDIT**: A nice answer provided by Ben-Zvi for [related questions](https://mathoverflow.net/questions/3446/tannakian-formalism/3467#3467)
| https://mathoverflow.net/users/1851 | How to unify various reconstruction theorems (Gabriel-Rosenberg, Tannaka,Balmers) | The Tannakian reconstruction for schemes and more generally, geometric stacks from QCoh(X) with its tensor structure (due to Jacob Lurie) is explained in my answer to [Tannakian Formalism](https://mathoverflow.net/questions/3446/tannakian-formalism/3467#3467) . One can (and one has) try to extend this to the derived setting, where one ought to get a statement much stronger than Balmer's reconstruction theorem. First of all one needs to replace triangulated categories, which are too coarse to work with effectively, with symmetric monoidal $(\infty,1)$-categories (or symm. monoidal dg categories say in characteristic zero).
There's an obvious functor from such objects C to derived stacks Spec C given by Tannakian reconstruction -- it's defined just as in the abelian case (see my answer above). Namely we build Spec C's functor of points, as a functor from derived rings to spaces or simplicial sets: Spec C(R) = the $\infty$-groupoid of tensor functors from C to R-modules.
In the case that C is the infinity-categorical form of QCoh (X) for X a scheme this recovers X, giving a result
refining Balmer's (this was worked out by Nadler, Francis and myself, and many others I think -- in particular Toen and Lurie understood this long ago). I feel strongly this
is a better point of view than trying to define a locally ringed space from a triangulated category, and has more of a chance to extend to stacks. In particular C tautologically sheafifies (as a sheaf of symmetric monoidal $\infty$-categories) over X.
Things get much more interesting though in the case that C=Qcoh X for X a stack (with affine diagonal, or everything fails immediately) --
eg for C=Reps(G)=QCoh (BG), the setting of usual Tannakian formalism.
There one quickly learns that the above naive picture is false. Namely take G to be just the multiplicative group $G\_m$. Then there are a lot of interesting fiber functors
on the derived Rep G = complexes of graded vector spaces that are not given by
points of BG\_m (ie are not the usual "forgetful" fiber functor) --- namely we can send the defining one-dim rep of $G\_m$ not to the one-dim vector space in degree zero,
but in any degree we want (ie apply shift) - since this is a generator it determines the rest of the fiber functor. This proves that a naive Tannakian theorem fails in the derived setting, without imposing some "connectivity" hypotheses (ie without having t-structures around). I believe a student of Toen's is writing a thesis on this subject, but I don't know the precise statements (and so won't "out" his results).
Of course it's a little disappointing to need BOTH a tensor structure and a t-structure, since morally either one should be enough for reconstruction (having a t-structure means we can recover the abelian category of quasicoherent sheaves, which by Rosenberg recovers X already in the case of schemes without need for tensor structure). But for general stacks I don't know of a better answer.
| 25 | https://mathoverflow.net/users/582 | 16334 | 10,951 |
https://mathoverflow.net/questions/15519 | 19 | Here's a question that has come up in a couple of talks that I have given recently.
The 'classical' way to show that there is a knot $K$ that is locally-flat slice in the 4-ball but not smoothly slice in the 4-ball is to do two things
1. Compute that the Alexander polynomial of $K$ is 1, and so by results of Freedman's you know that $K$ is locally-flat slice.
2. (due to Rudolph) Somehow obtain a special diagram of $K$ (or utilize a more subtle argument) to show that you can present $K$ as a separating curve on a minimal Seifert surface for a torus knot. Since we know (by various proofs, the first due to Kronheimer-Mrowka) that the genus of torus knot is equal to its smooth 4-ball genus (part of Milnor's conjecture), the smooth 4-ball genus of $K$ must be equal to the genus of the piece of the torus knot Seifert surface that it bounds, and this is $\geq 1$.
Boiling the approach of 2. down to braid diagrams, you come up with the slice-Bennequin inequality.
Well, here's the thing. I have this smooth cobordism from the torus knot to $K$, and then I know that $K$ bounds a locally-flat disc. This means that the locally-flat 4-ball genus of the torus knot must be less than its smooth 4-ball genus. So if you were to conjecture that the locally flat 4-ball genus of a torus knot agrees with its smooth 4-ball genus, you would be wrong.
My question is - are there any conjectures out there on the torus knot locally-flat genus? Even asymptotically? Any results? Any way known to try and study this?
Thanks, Andrew.
| https://mathoverflow.net/users/3923 | topological "milnor's conjecture" on torus knots. | Related to the early investigation of the Thom conjecture,the G-signature thm was used circa 1970 to give 4-ball genus bounds for torus knots which asyptocically (in some cases) were a fixed fraction of what we now know to be the smooth category answer. I belive Larry Tayor observed (in the '70s or early 80s) that these G-signature bounds hold in the topologically flat world as well. Thus, I believe, there there are families of torus knots where the the flat-4-ball geunus is known to be at least some known fraction of the smooth 4-ball genus. Sorry I don't have the references at hand.
| 11 | https://mathoverflow.net/users/1643 | 16337 | 10,953 |
https://mathoverflow.net/questions/16330 | 7 | For any finite group, G, we can find a cover of ℙ1ℂ which is G-Galois. The regular inverse Galois problem is equivalent to there existing such a cover that descends with action to ℚ. My question is about the easier problem: given a finite group G, can we find a cover of ℙ1ℂ such that it descends to ℚ as a mere cover (meaning not necessarily with group action)?
From the results that I know, I would be really surprised if this is solved. But what is known? And where is it written?
| https://mathoverflow.net/users/2665 | For a given finite group G, is there a cover of P^1 over Q s.t. over C it's G-Galois? | I don't know about *every* finite group $G$ (I'll guess no), but there are definitely infinitely many finite groups $G$ for which the situation you describe obtains: the extension $K/\mathbb{C}(t)$ has a model over $\mathbb{Q}$ but is not Galois over $\mathbb{Q}$. (And for most of these groups, we do not know how to realize them as Galois groups over $\mathbb{Q}$, regularly or otherwise.)
For instance, this is the situation in a work in progress of John Voight and me:
[http://alpha.math.uga.edu/~pete/triangle-091309.pdf](http://alpha.math.uga.edu/%7Epete/triangle-091309.pdf)
In our slightly different language, there are plenty of situations where the covering itself is defined over $\mathbb{Q}$ but the field of definition of the automorphism group $G$ is strictly larger. (This is equivalent to what you're asking, right? Please let me know.)
[Warning: recently, with the help of Noam Elkies, John and I realized that our arguments as given only work when (in our notation) $a = 2$. This is still a generalization of the setting in which I began this work some years ago: I had $a = 2$, $b = 3$, so I know for sure that there are infinitely many examples of this form.]
| 4 | https://mathoverflow.net/users/1149 | 16344 | 10,956 |
https://mathoverflow.net/questions/16342 | 5 | I was reading HTT 2.1.4, and I just totally lost what was going on. Could someone provide some motivation for this section? Why do we want another model structure?
I'm sorry for not providing motivation, but as you can see, I'm unable to do so.
Here's a link to the ArXiv version: <http://arxiv.org/pdf/math/0608040v4>
| https://mathoverflow.net/users/1353 | Motivation for the covariant model structure on SSet/S | If you have a category $C$, then you can consider the category of functors $Func(C,Sets)$ from $C$ to sets. This comes with the Yoneda functor, $C^{op}\to Func(C,Sets)$, and is a generally useful thing to think about.
In $(\infty,1)$-category land, you want to start with an $(\infty,1)$-category $S$, and build the $(\infty,1)$-category of functors $Func(S,Spaces)$ from $S$ to spaces.
Lurie is describing one way to do this. Given a quasicategory $S$, he produces a closed model category $(sSet/S, covariant)$ on the slice category $sSet/S$ which "models" this functor category.
The analogy is to the "point category" construction. Given a functor $F:C\to Sets$, you can build a category $P\_F$, whose objects are pairs $(c,x)$ where $c$ is an object of $C$, and $x\in F(c)$, and maps $(c,x)\to (c',x')$ are $f:c\to c'$ in $C$ such that $Ffx=x'$. Such a thing comes with a forgetful functor $U:P\_F\to C$, and one can produce an equivalence of categories
$$
Func(C,Sets) \qquad \Leftrightarrow\qquad (\text{certain full subcategory of $Cat/C$}).
$$
The subcategory is that of functors $U:D\to C$ such that given $f:c\to c'$ in $C$ and $d\in D$ such that $U(d)=c$, there is a unique $g: d\to d'$ in $D$ such that $U(g)=f$.
So, fibrant objects in Lurie's covariant model category for $sSet/S$ are supposed to look like left Grothendieck fibrations, and this model category should model "$Func(S,Spaces)$".
**Edit:** originally, I said that the certain full subcategory was that of "left Grothendieck fibrations", but I don't think that's right. You can play the above story a bit more generally, with $Sets$ replaced by $Groupoids$; the functor $P\_F\to C$ corresponding to a (pseudo)functor $F:C\to Groupoids$ is a typical example of what's called a "left Grothendieck fibration" (or "fibered groupoid"), consult nLab for details.
| 8 | https://mathoverflow.net/users/437 | 16345 | 10,957 |
https://mathoverflow.net/questions/16331 | 3 | Does anyone know of generalizations on what Mumford (Red Book) calls "uniformizing parameters"? For example, given a regular quasi-projective scheme over a *finite* field $\mathbb{F}$, is there an etale morphism into affine space over $\mathbb{F}$?.
| https://mathoverflow.net/users/4235 | When does a projective morphism give an etale morphism (into affine space)? (Finite field) (normalization) | [This paper](http://arxiv.org/abs/math/0303382) by Kedlaya might be what you want, since it contains some rearrangement of the words you used, but I can't really tell from the question. If you want a proper F-scheme to have an etale map to affine space, it has to be a disjoint union of finite F-schemes, and the affine space has to be zero dimensional.
| 3 | https://mathoverflow.net/users/121 | 16346 | 10,958 |
https://mathoverflow.net/questions/16360 | 5 | I'm a 2nd year grad student and I'm looking for conferences/summer schools to attend this summer. I checked out the AMS calendar but couldn't find anything I found relevant there. Anyone have any suggestions?
| https://mathoverflow.net/users/4239 | Looking for good conference this summer for homotopy theory | The Georgia Topology Conference will be largely focused on a geometric end of algebraic topology this year. As least, that what I suspect. The homotopy-type of embeddings of smooth manifolds, Goodwillie calculus, configuration spaces, etc, should be a major feature:
<http://www.math.uga.edu/~topology/>
I think there's a big summer school / conference in Pisa, Italy for much of the summer but I haven't found much information on it on-line. I suspect it will be oriented towards configuration spaces but I don't know for sure.
Ah, I found a webpage for the Pisa meeting:
<http://www.crm.sns.it/hpp/events/event.html?id=121;sez=aims#title>
| 5 | https://mathoverflow.net/users/1465 | 16364 | 10,969 |
https://mathoverflow.net/questions/16335 | 11 | The following simple theorem is known as Cauchy's mean value theorem. Let $\gamma$ be an immersion of the segment $[0,1]$ into the plane such that $\gamma(0) \ne \gamma(1)$. Then there exists a point such that the tangent line at that point is parallel to the line passing through $\gamma(0)$ and $\gamma(1)$. So the boundary values of an immersion determine a direction such that for any immersion of a segment with given boundary values there exists a tangent line parallel to the direction.
I would like to propose the following conjecture, generalizing this statement. Instead of immersions of a segment we consider immersions of a compact manifold $M^n$ with non-empty boundary $\partial M$ into ${\mathbf R}^{n+1}$. For a map $f\colon \partial M \to {\mathbf R}^{n+1}$ we consider the space $L(f)$ of all immersions $g\colon M \to {\mathbf R}^{n+1}$ such that $g|\_{\partial M}=f$.
The conjectural claim is the following: If $f$ is sufficiently generic, then for every connected component $L\_0$ of the space $L(f)$ there exists a hyperplane direction $l$ such that for any immersion $g$ from $L\_0$ $l$ is parallel to the tangent plane to $g(M)$ at some point.
If the conjecture is true then it is very interesting how $l$ depends on $g$.
| https://mathoverflow.net/users/2823 | A generalization of Cauchy's mean value theorem. | The answer is no for $n=2$. It is sufficient to construct 3 surfaces with common boundary (say $\Sigma\_i$, $i\in\{1,2,3\}$)
such that there is no choice of points $p\_i\in\Sigma\_i$ with pairwise parallel tangent planes.
Let us take a smooth function $f:S^1\to \mathbb R$, $f(t)\approx\sin(2\cdot t)$ with one little bump near zero
so it has 3 local minima and maxima.
We want to construct three functions $h\_1,h\_2,h\_3$ from unit disc $D$ to $\mathbb R$ such that
each has $f$ as boundary values and
1. if $\nabla h\_1(x)=\nabla h\_2(y)$ then $\nabla h\_1(x)=0$
2. $\nabla h\_3\not=0$ anywhere in the disc.
Then graphs of functions give the needed surfaces.
The graphs of $h\_1$ and $h\_2$ are parts of boundary of
convex hull of graph of $f:\partial D\to\mathbb R$; it is easy to check (1).
The graph of $h\_3$ is a ruled surface which formed by lines passing
through points $(u,f(u))$, $(\phi(u),f(\phi(u))\in\mathbb R^3$, $u\in S^1$ for some involution diffeomorphism $\phi: S^1\to S^1$.
To have the property one has to choose $\phi$ with two fixed points (say at global minima of $f$)
so that if $f(\phi(x))=f(x)$ for some $x$ then $f'(\phi(x)\cdot f'(x)<0$.
The later is easy to arrange, that is the place we need the bump of $f$.
P.S. Hopefully it is correct now :)
| 5 | https://mathoverflow.net/users/1441 | 16366 | 10,970 |
https://mathoverflow.net/questions/15243 | 8 | **Note.** This question had a bounty, so at the end I accepted the best (and only) answer. However, a solution is implied by the answer to [this question](https://mathoverflow.net/questions/15204/space-bounded-communication-complexity-of-identity).
**Question.**
Fix n. We are interested in the biggest t for which there exist two families of functions, $P\_i,Q\_i$, of size t from [n] to [n] such that for any $i,j$ whenever we consider the infinite sequence $P\_i(Q\_j(P\_i(Q\_j\ldots P\_i(3))\ldots)$ (where the number of iterations tends to infinity), it contains no 2's and infinitely many 1's if $i=j$ and it contains no 1's and infinitely many 2's if $i\ne j$.
**A lower bound.**
I know a construction that shows that $t\ge 2^{\frac n2-O(1)}.$ For every subset $S$ of [n] that contains exactly one of $2k$ and $2k+1$ for $2\le k\le \frac n2-2$ we construct a pair of functions, $P\_S,Q\_S$ as follows. For any number m denote by $m^+$ the smallest element of $S$ that is bigger than m or if all elements of $S$ are at most m then define it to be 1.
$P\_S(1)=1, P\_S(2)=2$ and for bigger $m$'s $P\_S(m)=m^+$, while $Q\_S(1)=1, Q\_S(2)=2$ and for bigger $m$'s $Q\_S(m)=m$ if $m\in S$ and $Q\_S(m)=2$ if $m\notin S$. This way we go through all the elements of S and end in 1 if the functions have the same index, but we are pushed to 2 if they differ.
**Upper bound.**
It is of course true that $t\le n^n$. So can you do better than $2^n$?
| https://mathoverflow.net/users/955 | Two [n] to [n] function families | Okay, so I tried to see how this could possibly work. After some thinking I decided that one may as well take $P\_i=Q\_i$, so that the orbit of 3 (under the action of $P\_i$) is a cycle containing 1. If you take the length of this cycle to be roughly $n/2$, send $2\to 3$ and everything else to 2, that's not a bad idea except that it doesn't work for *all* possible $n/2$-subsets; otherwise we would have roughly $\binom{n}{n/2}$ possible $i$, as you wanted to begin with. If you now look at the orbit of 3 under $(P\_iP\_j)$ in this setting you pretty quickly conclude that there is an inherent "even-town theorem" (see Babai-Frankl's [book](http://www.cs.uchicago.edu/research/publications/combinatorics)) and thus $2^{n/2}$ is really the best possible. Of course, in the full generality weird things might be possible - I have no intuition for this, but this doesn't look good and unless the difference is really really important for some applications I wouldn't recommend working on this problem.
| 4 | https://mathoverflow.net/users/4040 | 16371 | 10,973 |
https://mathoverflow.net/questions/16386 | 6 | I read some time ago some papers about proof formalization. Typically, I began whith [this one](http://research.microsoft.com/en-us/um/people/lamport/pubs/lamport-how-to-write.pdf), from Lamport.
Are there more recent works in this field ?
| https://mathoverflow.net/users/2446 | Proof formalization | In general, "formalising mathematical vernacular" is a good source for work relevant to your question. The field generally started in earnest with de Bruijn's Automath project.
Harvey Friedman has done some nice work on leveraging his theory of explicit definitions for set theory (cf. [The Logical Strength of Mathematical Statements I](https://cpb-us-w2.wpmucdn.com/u.osu.edu/dist/1/1952/files/2014/01/hisLogStr1-2azxfn3.pdf), 1976) to give a language for doing mathematics in a way that is fairly natural and that can be easily translated into proofs over ZFC.
I'm afraid I don't have a good reference for this: he gave an invited talk to the ASL 2006 conference in Nijmegen.
**Postscript**
I remembered there were coauthors: the system was written up in a series of papers for a system called [A language for knowledge management](http://www.andrew.cmu.edu/user/avigad/Papers/mkm/), due to Steve Kieffer (for a Master's project), Jeremy Avigad, and Harvey Friedman.
| 4 | https://mathoverflow.net/users/3154 | 16387 | 10,981 |
https://mathoverflow.net/questions/16368 | 7 | Frequently it is useful do deal with countable transitive models M of ZFC, for example in forcing constructions.
The notion of being an ordinal is absolute for any transitive model, so certainly if ($\alpha$ is an ordinal)M then also $\alpha$ is an ordinal. For the same reason, M will contain successors of every ordinal in it.
On the other hand, if M is countable then M cannot contain every countable ordinal; there must be a least (countable) ordinal not in M.
Can anything be said about this ordinal? Does it have any special significance?
| https://mathoverflow.net/users/4241 | Least ordinal not in a countable transitive model of ZFC | The least ordinal not in any transitive model of ZFC can also be described as the supremum of the heights of transitive models of ZFC. It is natural here to consider the class S consisting of all ordinals λ for which there is a transitive model of ZFC of height λ. Thus, the ordinal of your title, when it exists, is simply the supremum of the countable members of S. There are a number of relatively easy observations:
* If M is a transitive model of ZFC, then so is LM, the constructible universe as constructed inside M, and these two models have the same height. Thus, one could equivalently consider only models of ZFC + V = L.
* It is relatively consistent with ZFC that S is empty, that is, that there are no transitive models of ZFC. For example, the least element of S is the least α such that Lα is a model of ZFC. This is sometimes called the *minimal model* of ZFC, though of course it refers to the minimal transitive model. It is contained as a subclass of all other transitive models of ZFC. The minimal model has no transitive models inside it, and so it believes S to be empty.
* The least element of S (and many subsequent elements) is Δ12 definable in V. This is because one can say: a real codes that ordinal iff it codes a well-ordered relation and there is a model of that order type satisfying that no smaller ordinal is in S (a Σ12 property), also iff every well-founded model of ZFC has ordinal height at least the order type of the ordinal coded by z (a Π12 property).
* If S has any uncountable elements, then it is unbounded in ω1. The reason is that if Lβ satisfies ZFC and β is uncountable, then we may form increasingly large countable elementary substructures of Lβ, whose Mostowski collapses will give rise to increasingly large countable ordinals in S.
* In particular, if there are any large cardinals, such as an inaccessible cardinal, then S will have many countable members.
* If 0# exists, then every cardinal is a member of S. This is because when 0# exists, then every cardinal κ is an L-indiscernible, and so Lκ is a model of ZFC. Thus, under 0#, the class S contains a proper class club, and contains a club in every cardinal.
* S is not closed. For example, the supremum of the first ω many elements of S cannot be a member of S. The reason is that if αn is the nth element of S, and λ = supn αn, then there would be a definable cofinal ω sequence in Lλ, contrary to the Replacement axiom.
* S contains members of every infinite cardinality less than its supremum. If β is in S, then we may form elementary substructures of Lβ of any smaller cardinality, and the Mostowski collapses of these structures will give rise to smaller ordinals in S.
* If β is any particular element of S, the we may chop off the universe at β and consider the model Lβ. Below β, the model Lβ calculates S that same as we do. Thus, if β is a limit point of S, then Lβ will believe that S is a proper class. If β is a successor element of S, then Lβ will believe that S is bounded. Indeed, if β is the αth element of S, then in any case, Lβ believes that there are α many elements of S.
* If S is bounded, then we may go to a forcing extension V[G] which collapses cardinals, so that the supremum of S is now a countable ordinal. The forcing does not affect whether any Lα satisfies ZFC, and thus does not affect S.
Reading your question again, I see that perhaps you meant to consider a fixed M, rather than letting M vary over all transitive models. In this case, you will want to look at fine-structural properties of this particular ordinal. Of course, it exhibits many closure properties, since any construction from below that can be carried out in ZFC can be carried out inside M, and therefore will not reach up to ht(M).
| 11 | https://mathoverflow.net/users/1946 | 16394 | 10,985 |
https://mathoverflow.net/questions/16237 | 4 | Part of one of my calculations involves (the innocent looking) expression
$\sum\_{\alpha\in\Sigma} (\alpha,\alpha)$
for simple Lie algebras.
I have two methods of calculating it -- which don't agree. I'm pretty sure that the first one is wrong, but I don't know why. Any help is welcome (which is why I posted here)!
First my starting 'facts' (see e.g. the free book by [Cahn](http://phyweb.lbl.gov/~rncahn/www/liealgebras/book.html) or the comprehensive [Knapp](http://books.google.com.au/books?id=U573NrppkA8C&lpg=PP1&ots=7A2jEnNFSH&dq=Knapp%2520lie%2520groups&pg=PP1#v=onepage&q=&f=false)):
Given a simple Lie algebra $\mathfrak{g}$ and a basis to the Cartan subalgebra $\{h\_i\ ,\ i=1\ldots,r\}$,
the components of the roots are defined by
$$[h\_i,e\_\alpha] = \alpha\_i e\_\alpha$$
The Killing form restricted to the Cartan subalgebra is (Knapp Cor (2.24): but with index ridden notation)
$$ g\_{ij}=\mathrm{tr}h\_i h\_j = \sum\_{\alpha\in\Sigma}\alpha\_i\alpha\_j $$
The inner product on the root space is defined via the Killing form (Knapp eqn (2.28)):
$$(\alpha,\beta) = \alpha^i\beta\_i = \alpha\_i g^{ij} \beta\_j=\mathrm{tr}(h\_\alpha h\_\beta)\ ,\qquad g^{ij}g\_{jk}=\delta^i\_k$$
So we get our first (and probably wrong) way of calculating:
$$\sum\_{\alpha\in\Sigma} (\alpha,\alpha) = g^{ij}\sum\_{\alpha\in\Sigma} \alpha\_i\alpha\_j
= g^{ij}g\_{ij} = \sum\_i \delta^i\_i = \mathrm{rnk} \mathfrak{g}
$$
The second method is to just enumerate and sum over all roots. For a *simply laced* Lie algebra this is easy, because all roots have the same length $(\alpha,\alpha)=l$:
$$\sum\_{\alpha\in\Sigma} (\alpha,\alpha) = l\sum\_{\alpha\in\Sigma} 1
= l\ (\mathrm{dim}\mathfrak{g}-\mathrm{rnk}\mathfrak{g})
$$
These results are not compatible...
| https://mathoverflow.net/users/358 | Sum of all root lengths in simple Lie algebra | It turns out that I was wrong and both methods give correct results. In fact using the two results you can reproduce the scaling parameters for the simple Lie algebras given in the [Broughton paper](http://www.rose-hulman.edu/~brought/Epubs/preprints/killing.pdf) linked to by David.
My mistake was in assuming that the length of the roots is always arbitrary. It's well known that the choice of the invariant bilinear on the algebra is unique up to a scale. The restriction of the bilinear onto the Cartan subalgebra is dualized to give the metric on the root space and thus passes on the choice of scale to the roots.
But I had already chosen the Killing form as my bilinear - a necessary ingredient in the first method I presented. Thus the length of my roots was fixed.
Direct construction of a few low rank cases (something that I should have done before posting) confirmed that everything works out ok.
As a quick check and application, we can reproduce the scaling factor given by Broughton for $A\_n$. We have $\mathrm{rnk}(A\_n)=n$ and $\mathrm{dim}(A\_n)=(n+1)^2-1$, so $l=\frac{\mathrm{rnk}(A\_n)}{\mathrm{dim}(A\_n)-\mathrm{rnk}(A\_n)}=\frac{1}{n+1}$.
| 3 | https://mathoverflow.net/users/358 | 16402 | 10,991 |
https://mathoverflow.net/questions/16403 | 6 | How many ways is there to build an arithmetic expression with fixed number of terms and fixed order? Let’s assume we have only one distinct operation that is neither commutative nor associative. The problem can be reduced to the question of how many different, full, binary trees could be constructed with a fixed number of leaves.
Suppose we have a list of n elements which have to become leaves in a full, binary tree. The root could be chosen between each two sequential numbers. Thus, there is (n-1) different ways to choose the root. If the root is located after the i-th number, we can still construct the left child as a binary tree with i leaves and the right one with (n – i) leaves.
The formula for the number of different ways to construct a full, binary tree is
$ \Phi(n) = \sum^{n-1}\_{i=1}\Phi(i).\Phi(n-i)$
The first value for n=1 ist set to:
$ \Phi(1) = 1 $
Is there a closed formula for this function? Is it - maybe - a popular problem in the Graph Theory with known solutions?
| https://mathoverflow.net/users/4247 | Count of full, binary trees with fixed number of leaves | There is a general algorithm that solves this kind of problem.
1. Calculate the first terms of your sequence by hand.
2. Plug them into Sloane's and see if it is a known sequence.
<http://oeis.org?q=1%2C1%2C2%2C5%2C14&sort=0&fmt=0&language=english&go=Search>
Here you see that your numbers are called Catalan numbers; they are extremely well-studied.
| 20 | https://mathoverflow.net/users/1310 | 16405 | 10,993 |
https://mathoverflow.net/questions/16399 | 6 | What is the relation between characters of a group and its lie algebra?
Roughly,I know that there is a one to one correspondence between representations of a lie algebra and its simple connected lie group by the exp map,and two irreducible representations of a lie group are unequivalent if and only if their characters are different.
Now,then,I think the above statement will still hold for representations of lie algebras, but I am not sure about it .
If someone can give some advises?
Thanks Kevin Buzzard. For represatatons of general lie algebras the character (trace) don't work .
But for compact lie group,characters still work,so I restrict my question on compact lie groups and their lie algebras.
In representation theory of semisimple lie algebras ,we have formal characters/algebra characters which are sums of some formal elements over the weights.
I wonder whether we can realize these algebra characters to be real characters ?
Or can we make them to be functions on cartan subalgebras or maximal torus?
| https://mathoverflow.net/users/4155 | What is the relation between characters of a group and its lie algebra? | The primary reason for studying Lie algebras is the following fundamental fact: the representation theory of a Lie algebra is the same as the representation theory of the corresponding connected, simply connected Lie group.
Of course, the representation theory of a Lie group in general is very complicated. First of all, it might not be connected. Then the Lie algebra cannot tell the difference between the whole group and the connected component of the identity. This connected component is normal, and the quotient is discrete. So to understand the representation theory of the whole group requires, at the very least, knowing the representation theory of that discrete quotient. Even in the compact case, this would require knowing the representation theory of finite groups. For a given finite group, the characters know everything, but of course the classification problem in general is completely intractable.
The other thing that can go wrong is that even a connected group need not be simply connected. Any connected Lie group is a group quotient of a connected, simply connected group with the same Lie algebra, where the kernel is a discrete central subgroup of the connected, simply connected guy. So the representation theory of the quotient is the same as the representations of the simply connected group for which this central discrete acts trivially.
There is a complete classification of connected compact groups. You start with the steps above: classifying the disconnected groups is intractable, but a connected one is a quotient of a connected simply connected one. This simply connected group is compact iff the corresponding Lie algebra is semisimple; otherwise it has some abelian parts. In general, any compact group is a quotient by a central finite subgroup of a direct product: torus times (connected simply connected) semisimple. Torus actions are easy, and the representation theory of semisimples is classified as well. Whether your representation descends to the quotient I'm not entirely sure I know how to read off of the character. When the group is semisimple (no torus part), I do: finite-dimensional representations are determined by their highest weights (which can be read from the character), which all lie in the "weight" lattice; quotients of the simply-connected semisimple correspond exactly to lattices between the weight lattice and the "root" lattice, and you can just check that your character/weights are in the sublattice.
All of this should be explained well in your favorite Lie theory textbook.
| 14 | https://mathoverflow.net/users/78 | 16415 | 11,001 |
https://mathoverflow.net/questions/16401 | 3 | Since a pullback of two functions f and g with common codomain into **Set** category is just a subset of cartesian product like this: {(x,y)/f(x)=g(y)} (with two more functions not important here) could this pullback set be the empty set in some cases (for exemple in the case of constant functions)?
My question is related to find pullbacks for primitive recursive functions, where the functions are all of them total and the domain and codomain are powers of natural numbers set. Which could be the aspect of those supposed pullbacks for constant functions since the empty set is not there available? Do they exist?
| https://mathoverflow.net/users/3338 | Pullbacks for primitive recursive functions. | I'm not sure if this is what you want, but it is not difficult to prove that if f and g are primitive recursive functions, then the set A = { (x,y) | f(x) = g(y) } is a primitive recursive subset of the natural number plane. That is, the characteristic function of this set is primitive recursive. The (trivial) reason is that the characteristic function is simply the composition of the characteristic function of equality with f and g.
One can therefore have primitive recursive access to that set when defining other primitive recursive functions, such as the ones you will need in your commutative diagram.
As you and Reid have observed, when this set is empty, then there can be no pull back in the category of primitive recursive functions. But when it is nonempty, then we can define primitive recursive functions r, s on N2 to make the commutative square, by mapping r(x,y) = f(x) and s(x,y) = g(y), when (x,y) is in A, and otherwise r(x,y) = a and s(x,y) = b for some fixed (a,b) in A. This makes the square commute, and it has the universal property for pullbacks, because if r' and s' from N to N make the square commute (fr' = gs'), then we can define t:N to N2 by t(n) = (r'(n),s'(n)). By commutativity, this is inside A, and the whole diagram commutes.
| 2 | https://mathoverflow.net/users/1946 | 16418 | 11,002 |
https://mathoverflow.net/questions/16423 | 0 | A vertex operator is a linear map associating every state to a operator-valued distributions (quantum field) on a algebra curve, which is also called operator-state correspondence.
Chose a local complex coordinate, we can locally expand quantum fields as operator valued formal Laurent series, this process is called mode expansion (?), the coefficients are called Fourier coefficients.
I confuse the terminology Fourier coefficients, why people give them this name, does mode expansion relate to Fourier transformation?
| https://mathoverflow.net/users/4155 | About vertex algebra, mode expansion | If you expand a meromorphic function in a Laurent series about $z=0$ and now take $z$ on the unit circle in the complex plane, so that $z= e^{i\theta}$, then the Laurent series is a Fourier series.
| 5 | https://mathoverflow.net/users/394 | 16432 | 11,012 |
https://mathoverflow.net/questions/16427 | 14 | How do we define quasi-coherent sheaves on schemes?
Say we start by defining the category of affine schemes Aff as CRing$^{op}$ (the opposite category of unitary commutative rings).
In this context we have an obvious way to define quasi-coherent sheaves:
A quasi-coherent sheaf on an affine scheme X=Spec A is just an A-module.
If we now define schemes as presheaves on Aff (satisfying some condition), how do we define what a quasi-coherent sheaf is?
The same question applies also to the operations of pushforward and pullback, which in Aff have obvious definitions.
| https://mathoverflow.net/users/3701 | Quasi-coherent sheaves in the Functor-of-points approach | The corresponding [nlab page](http://ncatlab.org/nlab/show/quasicoherent+sheaf) has several approaches to the definition of quasicoherent sheaves of O-modules including some in functor of points approach, in various degrees of abstractness. All these definitions while simultaneously applicable define equivalent categories. This works not only for (qcoh modules) over schemes but over more general functors \*(e.g. stacks) on Aff. Look also for some quoted references there, including Orlov's paper mentioned by Zhang in the comment above.
| 10 | https://mathoverflow.net/users/35833 | 16438 | 11,016 |
https://mathoverflow.net/questions/16445 | 5 | Hello everyone, could anyone please tell me where can I find information about the Elton–Odell theorem?
It states:
For any infinite dimensional Banach space $X$ there is a $q > 1$ so that $X$ contains a sequence $(x\_n)$ with $\|x\_n\|=1$ and $\|x\_n-x\_m\|\ge q$, whenever $m\ne n$.
Thanks
| https://mathoverflow.net/users/2737 | Info about Elton–Odell theorem | J. Diestel, Sequences and series in Banach spaces, Springer-Verlag, New York, 1984.
(Chapter XIV, page 241)
| 4 | https://mathoverflow.net/users/3536 | 16447 | 11,022 |
https://mathoverflow.net/questions/16434 | 46 | There is a theorem of Grothendieck stating that a vector bundle of rank $r$ over the projective line $\mathbb{P}^1$ can be decomposed into $r$ line bundles uniquely up to isomorphism. If we let $\mathcal{E}$ be a vector bundle of rank $r$, with $\mathcal{O}\_X$ the usual sheaf of functions on $X = \mathbb{P}^1$, then we can write our line bundles as the invertible sheaves $\mathcal{O}\_{X}(n)$ with $n \in \mathbb{Z}$. Thus, the decomposition can be stated as $$\mathcal{E} \cong \oplus\_{i=1}^r \mathcal{O}(n\_i) \quad n\_1 \leq \dotsb \leq n\_r.$$
If we use the usual open cover of $\mathbb{P}^1$ with two affine lines $U\_0 = \mathbb{P}^1 - \{\infty\}$ and $U\_1 = \mathbb{P}^1 - \{0\}$, note that $\mathcal{O}\_{U\_0 \cap U\_1} = k[x,x^{-1}]$ (with $\mathcal{O}\_{U\_0} = k[x]$ and $\mathcal{O}\_{U\_1} = k[x^{-1}]$).
A vector bundle (up to isomorphism) $\mathcal{E}$ of rank $r$ is then a linear automorphism on $\mathcal{O}\_{U\_0 \cap U\_1}^r$ modulo automorphisms of each $\mathcal{O}\_{U\_i}^r$ for $i = 0,1$. (I am looking at the definition given in Hartshorne II.5.18 where $A = k[x,x^{-1}]$, the linear automorphisms are $\psi\_1^{-1} \circ \psi\_0$ where $\psi\_i: \mathcal{O}\_{U\_i}^r \rightarrow \left.\mathcal{E}\right\rvert\_{U\_i}$ are isomorphisms, and the definition of isomorphism of vector bundles allows us to change bases of $\mathcal{O}\_{U\_i}^r$.
Thinking of this in linear algebra terms, these linear automorphisms on $\mathcal{O}\_{U\_0 \cap U\_1}^r$ are elements of $GL\_r(k[x,x^{-1}])$, and changing coordinates in $\mathcal{O}\_{U\_i}^r$ are elements of $\operatorname{GL}\_r(k[x])$ for $i = 0$ and $\operatorname{GL}\_r(k[x^{-1}])$ for $i = 1$. Thus up to isomorphism, the vector bundles of rank $r$ on $\mathbb{P}^1$ are elements of the double quotient $$ \operatorname{GL}\_r(k[x^{-1}]) \left\backslash \large{\operatorname{GL}\_r(k[x,x^{-1}])} \right/ \operatorname{GL}\_r(k[x]).$$ The decomposition of vector bundles into line bundles SHOULD mean that these double cosets can be represented by matrices of the form $$\begin{pmatrix} x^{n\_1} & & & 0 \\ & x^{n\_2}& & \\ & & \ddots & \\ 0 & & & x^{n\_r}\end{pmatrix} \quad n\_1 \leq n\_2 \leq \dotsb \leq n\_r.$$ I want to know whether there is a way to prove this fact purely via linear algebra (equivalently, if the geometric proof [cf. Lemma 4.4.1 in Le Potier's "Lectures on Vector Bundles"] has a linear algebraic interpretation).
[Note: For the affine case, taking the double quotient $$\operatorname{GL}\_n(k[x]) \left \backslash M\_{n,m}(k[x]) \right/ \operatorname{GL}\_m(k[x])$$ gives the classification of vector bundles over $\mathbb{A}^1\_k$ (and of course, when replacing $k[x]$ with an arbitrary PID, gives the usual structure theorem of finitely generated modules over PID).]
| https://mathoverflow.net/users/1828 | Using linear algebra to classify vector bundles over ℙ¹ | I must admit I have never read this reference, but I remember it from a similar discussion on a German forum, according to which there is a simple proof in
[Michiel Hazewinkel and Clyde Martin, *A short elementary proof of Grothendieck's theorem on algebraic vectorbundles over the projective line*, Journal of pure and applied algebra 25 (1982), pp. 207 - 211](http://www.sciencedirect.com/science/article/pii/0022404982900378).
| 13 | https://mathoverflow.net/users/2530 | 16450 | 11,025 |
https://mathoverflow.net/questions/15913 | 13 | I'm looking for either a few precise mathematical statements about Wiener integrals, or a reference where I can find them.
Background
----------
The Wiener integral is an analytic tool to define certain "integrals" that one would like to evaluate in quantum and statistical mechanics. (Hrm, that's two different mechanics-es....) More precisely, one often wants to define/compute integrals over all paths in your configuration or phase space satisfying certain boundary conditions. For example, you might want to integrate over all paths in a manifold with fixed endpoints. It's conventional to write the integrand as a pure exponential $\exp(f)$. In statistical mechanics, the function $f$ in the exponent is generally real and decays to $-\infty$ (fast enough) in all directions. If the path space were finite-dimensional, you would expect such integrals to converge absolutely in the Riemann sense. In quantum mechanics, $f$ is usually pure-imaginary, so that $\exp(f)$ is a phase, and the integral should not be absolutely convergent, but in finite-dimensional integrals may be conditionally convergent in the Riemann sense. Typically, $f$ is a local function, so that $f(\gamma) = \int L(\gamma(t))dt$, where $\gamma$ is a path and $L(\gamma(t))$ depends only on the $\infty$-jet of $\gamma$ at $t$. In fact, typically $L(\gamma(t))$ depends only on the $1$-jet of $\gamma$, so that $f(\gamma)$ is defined on, for example, continuous piece-wise smooth paths $\gamma$.
For example, one might have a Riemannian manifold $\mathcal N$, and want to define:
$$U(q\_0,q\_1) = \int\limits\_{\substack{\gamma: [0,1] \to \mathcal N \\ {\rm s.t.}\, \gamma(0) = q\_0,\, \gamma(1)=q\_1}} \exp\left( - \hbar^{-1} \int\_0^1 \frac12 \left| \frac{\partial \gamma}{\partial t}\right|^2dt \right)$$
where $\hbar$ is a positive real number (statistical mechanics) or non-zero pure imaginary (quantum mechanics). The "measure" of integration should depend on the canonical measure on $N$ coming from the Riemannian metric, and the Wiener measure makes it precise.
On $\mathcal N = \mathbb R^n$, I believe I know how to define the Wiener integral. The intuition is to approximate paths by piecewise linears. Thus, for each $m$, consider an integral of the form:
$$I\_m(f) = \prod\_{j=1}^{m-1} \left( \int\_{\gamma\_j \in \mathbb R^n} d\gamma\_j \right) \exp(f(\bar\gamma)) $$
where $\bar\gamma$ is the piecewise-linear path that has corners only at $t = j/m$ for $j=0,\dots,m$, where the values are $\bar\gamma(j/m) = \gamma\_j$ (and $\gamma\_0 = q\_0$, $\gamma\_m = q\_1$). Then the limit as $m\to \infty$ of this piecewise integral probably does not exist for a fixed integrand $f$, but there might be some number $a\_m$ depending weakly on $f$ so that $\lim\_{m\to \infty} I\_m(f)/a\_m$ exists and is finite. I think this is how the Wiener integral is defined on $\mathbb R^n$.
On a Riemannian manifold, the definition above does not make sense: there are generally many geodesics connecting a given pair of points. But a theorem of Whitehead says that any Riemannian manifold can be covered by "convex neighborhoods": a neighborhood is *convex* if any two points in it are connected by a unique geodesic that stays in the neighborhood. Then we could make the following definition. Pick a covering of $\mathcal N$ by convex neighborhoods, and try to implement the definition above, but declare that the integral is zero on tuples $\gamma\_{\vec\jmath}$ for which $\gamma\_j$ and $\gamma\_{j+1}$ are not contained within the same convex neighborhood. This would be justified if "long, fast" paths are exponentially suppressed by $\exp(f)$. So hope that this truncated integral makes sense, and then hope that it does not depend on the choice of convex-neighborhood cover.
Of course, path integrals should also exists on manifolds with, say, indefinite "semi-"Riemannian metric. But then it's not totally clear to me that the justification in the previous paragraph is founded. Moreover, really the path integral should depend only on a choice of volume form on a manifold $\mathcal N$, not on a choice of metric. Then one would want to choose a metric compatible with the volume form (this can always be done, [as I learned in this question](https://mathoverflow.net/questions/7817/normal-coordinates-for-a-manifold-with-volume-form)), play the above game, and hope that the final answer is independent of the choice. A typical example: any symplectic manifold, e.g. a cotangent bundle, has a canonical volume form.
One other modification is also worth mentioning: above I was imagining imposing Dirichlet boundary conditions on the paths I wanted to integrate over, but of course you might want to impose other conditions, e.g. Neumann or some mix.
Questions
---------
**Question 0:** Is my rough definition of the Wiener integral essentially correct?
**Question 1:** On $\mathcal N = \mathbb R^n$, for functions $f$ of the form $f(\gamma) = -\hbar^{-1}\int\_0^1 L(\gamma(t))dt$, for "Lagrangians" $L$ that depend only on the $1$-jet $(\dot\gamma(t),\gamma(t))$ of $\gamma$ at $t$, when does the Wiener integral make sense? I.e.: for which Lagrangians $L$ on $\mathbb R^n$, and for which non-zero complex numbers $\hbar$, is the Wiener integral defined?
**Question 2:** In general, what are some large classes of functions $f$ on the path space for which the Wiener integral is defined?
By googling, the best I've found are physics papers from the 70s and 80s that try to answer Question 1 in the affirmative for, e.g., $L$ a polynomial in $\dot\gamma,\gamma$, or $L$ quadratic in $\dot\gamma,\gamma$ plus bounded, or... Of course, most physics papers only treat $L$ of the form $\frac12 |\dot\gamma|^2 + V(\gamma)$.
| https://mathoverflow.net/users/78 | Which functions are Wiener-integrable? | Hi Theo,
0) Your definition is roughly correct, yes. For Wiener measure on paths in vector spaces, see Chapter 3 + Appendix A of the 2nd edition of Glimm & Jaffe. On curved targets, I think Bruce Driver has some good lecture notes. One warning: the rough definition of Wiener measure is misleading in one way: Wiener measure is naturally constructed as a measure on a space of distributions which contains the continuous functions.
1) I don't think there's a general theory. Trying a Lagrangian of the form $F = (\dot{\phi})^{1000}$ will not result in happiness. But: You can define Wiener measure using the standard kinetic term $\int \frac{1}{2}|\dot{\phi}|^2dt$ for any $\hbar$, and you can safely add a potential $V$ which is bounded below and integrable to the kinetic term.
2) Any observable you can write down should give you a Wiener integrable function, in quantum mechanics. This is not true in QFT, however. Most of the work in constructive QFT is proving that the measure on the space of histories actually has moments!
| 6 | https://mathoverflow.net/users/35508 | 16459 | 11,031 |
https://mathoverflow.net/questions/16460 | 33 | Two things today motivated this question.
First, the professor said that in a lecture Thurston mentioned
>
> Any manifold can be seen as the configuration space of some physical system.
>
>
>
Clearly we need to be careful here, so the first question is
>
> 1) What is a precise formulation and an argument to see why the previous statement is true.
>
>
>
Second, the professor went on to say that because of the Poisson Bracket, we see the phase space of a physical system as the Cotangent Bundle of a manifold. I understand that we associate a symplectic form to the cotangent bundle, and that we want to think of Phase Space with a symplectic structure, but my second question is
>
> 2) Could you provide an example of a physical system, give the associated "configuration manifold" show the cotangent space, and explain why this is the phase space of the system.
>
>
>
I pushed the lecturer quite a bit to get this level of detail, so pushing much further would of probably been considered rude. I should also mention he is speaking about these things because we want to quantize the geometry associated to this manifold. So he looks at the cotangent space which apparently has a symplectic structure, and sends the symplectic form to the Lie Bracket. If any of the things I have said are incorrect, please comment with corrections. I am just learning this material and trying to understand how it fits with my current understanding.
One last bonus question(tee hee),
>
> If your manifold is a Lie Group, we get a Lie algebra structure on the Tangent Space. Is there a relationship between this Lie Algebra structure and the one you would get by considering the cotangent space and then quantizing in the fashion of above?
>
>
>
Thanks in advance!
| https://mathoverflow.net/users/348 | How to see the Phase Space of a Physical System as the Cotangent Bundle | Let's start by answering the first question.
Let $M$ be any manifold. Consider a physical system consisting of a point-particle moving on $M$. What are the configurations of this physical system? The points of $M$. Hence $M$ is the configuration space.
Typically one takes $M$ to be riemannian and we may add a potential function on $M$ in order to define the dynamics. (More complicated dynamics are certainly possible -- this is just the simplest example.)
As an example, let's consider a point particle of mass $m$ moving in $\mathbb{R}^3$ under the influence of a central potential
$$V= k/r,$$
where $r$ is the distance from the origin. The configuration space is $M = \mathbb{R}^3\setminus\lbrace 0\rbrace$.
Classical trajectories are curves $x(t)$ in $M$ which satisfy Newton's equation
$$m \frac{d^2 x}{dt^2} = \frac{k}{|x|^2}.$$
To write this equation as a first order equation we introduce the velocity $v(t) = \frac{dx}{dt}$. Geometrically $v$ is a vector field (a section of the tangent bundle $TM$) and hence the classical trajectory $(x(t),v(t))$ defines a curve in $TM$ satisfying a first order ODE:
$$\frac{d}{dt}(x(t),v(t)) = (v(t), \frac{k}{m|x(t)|^2})$$
This equation can be derived from a variational problem associated to a lagrangian function $L: TM \to \mathbb{R}$ given by
$$L(x,v) = \frac12 m v^2 - \frac{k}{|x|}.$$
The fibre derivative of the lagrangian function defines a bundle morphism $TM \to T^\*M$:
$$(x,v) \mapsto (x,p)$$
where
$$p(x,v) = \frac{\partial L}{\partial v}.$$
In this example, $p = mv$. The Legendre transform of the lagrangian function $L$ gives a hamiltonian function $H$ on $T^\*M$, which in this example is the total energy of the system:
$$H(x,p) = \frac{1}{2m}p^2 + \frac{k}{|x|}.$$
The equations of motion can be recovered as the flow along the hamiltonian vector field associated to $H$ via the standard Poisson brackets in $T^\*M$:
$$ \frac{dx}{dt} = \lbrace x,H \rbrace \qquad\mathrm{and}\qquad \frac{dp}{dt} = \lbrace p,H \rbrace.$$
Being integral curves of a vector field, there is a unique classical trajectory through any given point in $T^\*M$, hence $T^\*M$ is a phase space for the system; that is, a space of states of the physical system. Of course $TM$ is also a space of states, but historically one calls $T^\*M$ the phase space of the system with configuration space $M$. (I don't know the history well enough to know why. There are brackets in $TM$ as well and one could equally well work there.)
Not every space of states is a cotangent bundle, of course. One can obtain examples by hamiltonian reduction from cotangent bundles by symmetries which are induced from diffeomorphisms of the configuration space, for instance. Or you could consider systems whose physical trajectories satisfy an ODE of order higher than 2, in which case the cotangent bundle is not the space of states, since you need to know more than just the position and the velocity at a point in order to determine the physical trajectory.
It's late here, so I'll forego answering the bonus question for now.
| 30 | https://mathoverflow.net/users/394 | 16462 | 11,033 |
https://mathoverflow.net/questions/16393 | 24 | Is there any result about the time complexity of finding a cycle of fixed length $k$ in a general graph?
All I know is that [Alon, Yuster and Zwick](http://www.tau.ac.il/~nogaa/PDFS/col5.pdf) use a technique called "color-coding",
which has a running time of $O(M(n))$, where $n$ is the number of vertices of the input graph and $M(n)$ is the time required to multiply two $n \times n$
matrices.
Is there any better result?
| https://mathoverflow.net/users/4248 | Finding a cycle of fixed length | Finding a cycle of any *even* length can be found in $O(n^2)$ time, which is less than any known bound on $O(M(n))$. For example, a cycle of length four can be found in $O(n^2)$ time via the following simple procedure:
>
> Assume the vertex set is $\{1,...,n\}$. Prepare an $n$ x $n$ matrix $A$ which is initially all zeroes. For all vertices $i$ and all pairs of vertices $j, k$ which are neighbors to $i$, check $A[j,k]$ for a $1$. If it has a $1$, output *four-cycle*, otherwise set $A[j,k]$ to be $1$. When this loop finishes, output *no four-cycle*.
>
>
>
The above algorithm runs in at most $O(n^2)$ time, since for every triple $i,j,k$ we either flip a $0$ to $1$ in $A$, or we stop. (We assume the graph is in adjacency list representation, so it is easy to select pairs of neighbors of a vertex.)
The general case is treated by Raphy Yuster and Uri Zwick in the paper:
>
> Raphael Yuster, Uri Zwick: Finding Even Cycles Even Faster. SIAM J. Discrete Math. 10(2): 209-222 (1997)
>
>
>
As for finding cycles of odd length, it's just as David Eppstein says: nothing better is known than $O(M(n))$, including the case where $k=3$.
However, if you wished to detect *paths* of length $k$ instead of cycles, you can indeed get $O(m+n)$ time, where $m$ is the number of edges. I am not sure if the original color-coding paper can provide this time bound, but I do know that the following paper by some random self-citing nerd gets it:
>
> Ryan Williams: Finding paths of length k in O\*(2^k) time. Inf. Process. Lett. 109(6): 315-318 (2009)
>
>
>
| 27 | https://mathoverflow.net/users/2618 | 16464 | 11,035 |
https://mathoverflow.net/questions/16381 | 4 | I'm reading Knutson's book on algebraic spaces, and I stumbled over the quasi-separatedness axiom in his definition of algebraic spaces (Definition 1.1, Chapter II). He defines an algebraic space A as a sheaf on the site of schemes with the étale topology satisfying:
I) Local representability. There exists a representable étale covering $U \rightarrow A$, $U$ a scheme.
II) Quasi-separatedness. The map $U \times\_A U \rightarrow U \times U$ is quasi-compact.
In a technical remark (1.9) at the end of the section, he argues that the quasi-separatedness assumption is needed for the existence of fibre products in the categeory of algebraic spaces. As I see it, this is wrong. For instance, the proof of existence of fibre products for schemes given i Hartshorne carries over to algebraic spaces without problems, even if we just assume local representability.
This makes me wonder, would it be more natural to take only the local representability as requirement for algebraic spaces, or do we run into problems later on?
Sometimes one sees informal definitions of algebraic spaces as the closure of schemes under étale equivalence relations in the category of étale sheaves. In what sence is this true? Here it seems to me that we really do need the quasi-separatedness axiom (or something similar) since we need an étale equivalence relations $R \rightarrow U \times U$ to satisfy effective descent in order to get local representability for its quotient.
| https://mathoverflow.net/users/1084 | Quasi-separatedness for Algebraic Spaces | One can certainly make the basic definitions, and the real issue is to show that the definition "works" using any etale map from a scheme. More precisely, the real work is to show that a weaker definition actually gives a good notion: rather than assume representability of the diagonal, it suffices that $R := U \times\_X U$ is a scheme for *some* scheme $U$ equipped with an etale representable map $U \rightarrow X$. Or put in other terms, we have to show that if $U$ is any scheme and $R \subset U \times U$ is an 'etale subsheaf which is an etale equivalence relation then the quotient sheaf $U/R$ for the big etale site actually has diagonal representable in schemes. Indeed, sometimes we want to *construct* an algebraic space as simply a $U/R$, so we don't want to have to check "by hand" the representability of the diagonal each time.
That being done, then the question is: which objects give rise to a theory with nice theorems? For example, can we always define an associated topological space whose generic points and so forth give good notions of connectedness, open behavior with respect to fppf maps, etc.? (The definition of "point" needs to be modified from what Knutson does, though equivalent to his definition in the q-s case.) The truth is that once the theory is shown to "make sense" without q-s, it turns out that to prove interesting results one has to assume something extra, the simplest version being the following weaker version of q-s: $X$ is "Zariski-locally q-s" in the sense that $X$ is covered by "open subspaces" which are themselves q-s. This is satisfied for all schemes, which is nice. (There are other variants as well.)
In the stacks project of deJong, as well as the appendix to my paper with Lieblich and Olsson on Nagata compactification for algebraic spaces, some of the weird surprises coming out of the general case (allowing objects not Zariski-locally q-s) are presented. (In that appendix we also explain why the weaker definition given above actually implies the stronger definition as in Chris' answer. This was surely known to certain experts for a long time.)
| 11 | https://mathoverflow.net/users/3927 | 16483 | 11,046 |
https://mathoverflow.net/questions/16487 | 6 | I am trying to factorize $\sin(x)\over x$ which by [Taylor series expansion](http://en.wikipedia.org/wiki/Taylor_theorem) and using the roots is $$a \cdot \left(1 - \frac{x}{\pi} \right) \left(1 + \frac{x}{\pi} \right) \left(1 - \frac{x}{2\pi} \right) \left(1 + \frac{x}{2\pi} \right) \left(1 - \frac{x}{3\pi} \right) \left(1 + \frac{x}{3\pi} \right) \cdots$$
Now I was told that this nasty factor $a$ conveniently becomes $1$ due to [Weierstrass’s Factorization Theorem](http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem) which is a transcendental generalization of the [Fundamental Theorem of Algebra](http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra).
**My question**
Could you please show me how $a$ is being neutralized using this theorem? Or don't you even need this theorem to do so?
| https://mathoverflow.net/users/1047 | Using Weierstrass’s Factorization Theorem | The value of this product for small x's is the product of $(1-x^2/(n \pi)^2)$ which, when you take logs (and due to the second power in x), behaves like the sum over n of $-x^2/(n\pi)^2$, which approaches 0 as x approaches 0.
| 5 | https://mathoverflow.net/users/404 | 16493 | 11,053 |
https://mathoverflow.net/questions/16507 | 5 | I have often come across this implicit translation of the classical field of a particle of a given spin into a specific tensor field. But I could not locate any literature from which I could learn this.
In a paper of Avrimidi I found this statement,
"*The tensor fields describe the particles with integer spin while spin-tensor fields describe particles with half-odd spins*"
I would like to know what is the precise mathematical map that he is referring to.
Like if I have a particle of spin s on a space-time manifold M then the classical field of this particle is a section of which bundle on the space-time?
In the same vein I would like to know some expository mathematical reference on what is a "spin connection". I find the usual physics book definition very unconvincing where spin-connection coefficients are just 'defined" analogous to the Christoffel symbols (but with the signs flipped) when the connection is made to act on vierbeins.
| https://mathoverflow.net/users/2678 | Spins as tensor fields | In a nutshell, particles "are" unitary irreducible representations of the Poincaré group, which is the isometry group of Minkowski spacetime on which it acts transitively. Such representations can be constructed using the method of induced representations (cf. Wigner, Bargmann, Mackey,...) as classical fields on Minkowski spacetime subject to certain field equations: wave, Klein-Gordon, Weyl, Dirac, Maxwell, Rarira-Schwinger,... Mathematically these are sections of homogeneous vector bundles associated to certain finite-dimensional representations of the "little group", which is (the maximal compact subgroup of) the stabilizer (in the spin cover of the Lorentz group) of a point on the "mass shell" (=the momenta $p$ with $p^2 = - m^2$, where $m$ is the mass of the particle). The little group for massive representations is isomorphic to $SU(2)$, whereas that of massless particles to a nontrivial double cover of $SO(2)$. Hence massless particles are defined by their helicity (the label of the $SO(2)$ representation from which one induces) and massive particles by their mass and their spin (the label of the irrep of $SU(2)$ from which one induces). The covariant field equations are (the Fourier transform of) the projectors onto irreducible components.
Let's discuss the massive case, since your question mentions spin explicitly.
There are two kinds of irreps of $SU(2)$, those where $-1$ acts trivially and those where it does not. The former case are the integer spin representations whilst the latter are the half-integer representations.
The integer spin representations of $SU(2)$ are contained in the tensorial representations of the Lorentz group, whence this is why the fields for integer-spin massive particles are tensorial.
The irreps of $SU(2)$ with half-integer spin, those where $-1$ does not act trivially, are not contained in tensorial representations of the Lorentz group, since on these representations $SU(2)$ acts by conjugation and $-1$ acts trivially. In order to describe such reps in terms of fields you need to consider spinor fields, which are sections of spinor bundles (possibly twisted by tensor bundles for higher spin fields).
You can read about spinor bundles on any book in Spin Geometry. For example, there's a book by Lawson and Michelsohn of that name. Green-Schwarz-Witten's string theory book (second volume) has a physicsy discussion of this. They will also define the spin connection, which is a connection induced by the Levi-Civita connection on any "spin bundle", which is a principal fibre bundle lifting the bundle of oriented orthonormal frames in such a way that the bundle map between them restricts to the spin covering $\mathrm{Spin} \to \mathrm{SO}$ on the fibres.
It is not hard to show that the connection one-form for the spin connection, when pulled back to the manifold by the lift of a local frame (hence a local 1-form with values in the orthogonal Lie algebra) agrees with the similar expression for the Levi-Civita connection, which is why many books perhaps do not go through the trouble of defining it properly. It also requires introducing quite a bit of formalism, to which many physicists are allergic; although increasingly less so.
| 4 | https://mathoverflow.net/users/394 | 16510 | 11,063 |
https://mathoverflow.net/questions/16466 | 3 | This is may be obvious, but I am not comfortable with ind-schemes.
I have an ind-scheme $X$ over $\mathbb{C}$. Every point has a neighborhood which can be written as an ascending union of regular varieties, which is about as smooth as an ind-scheme can be.
I have an unipotent ind-group $U$. More precisely, $U$ is a group object in the category of ind-schemes, and $U$ has a descending filtration all of whose quotients are $\mathbb{G}\_a$'s. The group $U$ acts freely on $X$.
There is a quasi-projective variety $Y$, and a map $f: X \to Y$, which is a principal $U$-bundle. (Meaning that there is a cover $Y = \bigcup V\_i$ and $f^{-1}(V\_i)$ is isomorphic to $U \times V\_i$.)
>
> Is $X(\mathbb{C}) \to Y(\mathbb{C})$ a weak homotopy equivalence, using the analytic topology on both sides?
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>
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| https://mathoverflow.net/users/297 | Principal bundle for contractible group is weak homotopy equivalence for ind schemes | My recollection is that when you turn these into analytic spaces you get something which is locally contractible topologically. In this case what you are describing is a principal bundle for locally contractible spaces in which the fiber is contractible. If the base is paracompact then this will indeed be a weak equivalence, in fact the space corresponding to X will be a topological product space $X = U \times Y$.
This follows because you can build a global section (trivialization). How do you do this? You start with you local trivializations, and you choose a partition of unity subordinate to this cover. You also choose a contraction of U. You can then patch these together to obtain a global section. The exact method and formula is explained, for example, in the appendix of this paper. (This is probably not the only/first/best reference).
Segal, G. Cohomology of topological groups. 1970 Symposia Mathematica, Vol. IV (INDAM, Rome, 1968/69) pp. 377--387 Academic Press, London
So the real question is whether your space Y is paracompact. I'm pretty sure that your conditions (that Y is quasi-projective) ensure that this is the case.
| 2 | https://mathoverflow.net/users/184 | 16514 | 11,066 |
https://mathoverflow.net/questions/16495 | 14 | The study of the homotopy groups of spheres $\pi\_i(S^n)$ is a major subject in algebraic topology. One knows for example that nearly all of them are finite groups. Some are explicitly known. There is a 'stable range' of indices which one understands better than the unstable part.
I think that there is an analogy (have to be careful with that word) to the distribution of primes: It seems that there exists a general pattern but no one has found it yet. It is a construction producing an infinite list of *numbers* (or groups) but no numbers were put into it. Such a thing always fascinates me.
The largely unknown prime pattern leads to applications in cryptography for example. Are there similar applications of the knowledge (or not-knowledge) of the homotopy groups of spheres? Are there applications to *real* natural sciences or does one study the homotopy groups of spheres only for their inherent beauty?
| https://mathoverflow.net/users/4011 | Applications of homotopy groups of spheres | A few comments on applications that aren't covered by the above Wikipedia article.
I don't know any applications to cryptography. Most cryptosystems require some kind of one-way lossless function and it's not clear how to do that with the complexity of the homotopy groups of spheres. Moreover, the homotopy-groups of spheres have a lot of redundancy, there are many patterns.
There's work by Fred Cohen, Jie Wu and John Berrick's where they relate Brunnian braid groups to the homotopy-groups of the 2-sphere. It's not clear if that has any cryptosystem potential but it's an interesting aspect of how the homotopy-groups of a sphere appear in a natural way in what might otherwise appear to be a completely disjoint subject.
Homotopy groups of spheres and orthogonal groups appear in a natural way in Haefliger's work on the group structure (group operation given by connect sum) on the isotopy-classes of smooth embeddings $S^j \to S^n$. I suppose that shouldn't be seen as a surprise though. Moreover, it's not clear to me that this is always the most efficient way of computing these groups. But I think all techniques that I know of ultimately would require some input in the form of computations of some relatively simple homotopy groups of spheres.
I think one of the most natural applications of homotopy groups of spheres, Stiefel manifolds and orthogonal groups would be obstruction-theoretic constructions. Things like Whitney classes, Stiefel-Whitney classes and general obstructions to sections of bundles. Not so much the construction of the individual classes, more just the understanding of the general method.
| 6 | https://mathoverflow.net/users/1465 | 16515 | 11,067 |
https://mathoverflow.net/questions/16471 | 18 | Consider the set of random variables with zero mean and finite second moment. This is a vector space, and $\langle X, Y \rangle = E[XY]$ is a valid inner product on it. Uncorrelated random variables correspond to orthogonal vectors in this space.
Questions:
(i) Does there exist a similar geometric interpretation for independent random variables
in terms of this vector space?
(ii) A collection of jointly Gaussian random variables are uncorrelated if and only if they
are independent. Is it possible to give a geometric interpretation for this?
| https://mathoverflow.net/users/4267 | A geometric interpretation of independence? | There is a Hilbert space interpretation of independence, which follows from the interpretation of conditional expectation as an orthogonal projection, though it may be more complicated than you had in mind.
Say your underlying probability space is $(\Omega, \mathcal{F}, \mathbb{P})$, and write $L^2(\mathcal{F})$ for the Hilbert space of ($\mathcal{F}$-measurable) random variables with finite variance (with $\Omega$ and $\mathbb{P}$ understood). Denote by $\sigma(X)$ the $\sigma$-algebra generated by the random variable $X$. Now the conditional expectation $\mathbb{E}[X|Y]$ is the orthogonal projection in $L^2(\mathcal{F})$ of $X$ onto the subspace $L^2(\sigma(Y))$ of random variables which are $\sigma(Y)$-measurable. $X$ and $Y$ are independent if and only if $\mathbb{E}[f(X)|Y]=\mathbb{E}f(X)$ for every reasonable function $f$. The functions $f(X)$ span $L^2(\sigma(X))$.
So if I now define $L^2\_0(\sigma(X))$ to be the mean 0, finite variance random variables which are $\sigma(X)$-measurable, I can say: $X$ and $Y$ are independent iff $L^2\_0(\sigma(X))$ is orthogonal to $L^2(\sigma(Y))$ in $L^2(\mathcal{F})$.
| 11 | https://mathoverflow.net/users/1044 | 16517 | 11,069 |
https://mathoverflow.net/questions/16422 | 4 | Let $X$ be a separable Banach space with its Borel $\sigma$-algebra $\mathcal F$. Let $x\_n \to x$ in $X$. Fix a Gaussian covariance operator $K$, and let $\mathbb P\_n$ and $\mathbb P$ be Gaussian measures on $X$ with covariance $K$ and means $x\_n$ and $x$, respectively.
**Question:** How do I show that $\mathbb P\_n \to \mathbb P$ weakly?
Surely this is a theorem or an exercise somewhere; e.g. in Talagrand and Ledoux's *Probability in Banach Spaces* or Vakhania, Tarieladze and Chobanyan's *Probability Distributions in Banach Spaces*.
The characteristic functions of $\mathbb P\_n$ converge to those of $\mathbb P$ (simple exercise). By de Acosta's theorem, this implies that $\mathbb P\_n \to \mathbb P$, provided that the family $\{\mathbb P\_n\}$ is *flatly concentrated*. I'm not so familiar with the concept (hence this question), but I'm guessing this is related to the concentration of measure property of Gaussian measures.
| https://mathoverflow.net/users/238 | Convergence of Gaussian measures | Somehow I didn't register how strong the assumptions Tom was making were, hence the fact that my other answer missed the point.
Unless I'm still missing something, this is very easy. Say $Z$ is a Gaussian random vector in $X$ with covariance $K$ and mean $0$. You want to show that $Z+x\_n \to Z+x$ weakly, i.e. $\mathbb{E} f(Z+x\_n) \to \mathbb{E} f(Z+x)$ for every bounded continuous $f:X\to \mathbb{R}$. Since $f$ is both bounded and continuous, this follows immediately from dominated convergence.
| 5 | https://mathoverflow.net/users/1044 | 16518 | 11,070 |
https://mathoverflow.net/questions/16416 | 30 | I'm looking for a good book in commutative algebra, so I ask here for some advice. My ideal book should be:
-More comprehensive than Atiyah-MacDonald
-More readable than Matsumura (maybe better organized?)
-Less thick than Eisenbud, and more to the point
To put this in context, I'm an algebraic geometer, so I know enough commutative algebra, but I didn't study it systematically (apart from a first course on A-M which I followed as an undergraduate); rather I learned the things I needed from time to time. So I would like to give me an occasion to get a better grasp on the subject.
EDIT: I will be more specific about the level. As I said I already had a course on Atiyah-MacDonald, and I know that material well, so I'm not interested in books of a comparable level. But I'm not completely familiar with Cohen-Macaulay rings and the relationship between regular sequences and the Koszul complex for example. And I know very little of Gorenstein rings and duality. So I'm looking for something a little bit more sophisticated than what has been already proposed. Yes, I know Eisenbud does these things but it's easy to get lost in that book. Something more to the point would be nice.
| https://mathoverflow.net/users/828 | Reference book for commutative algebra | For a reference on Cohen-Macaulay and Gorenstein rings, you can try "Cohen-Macaulay rings" by Bruns-Herzog.
Also, Huneke's [lecture note](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.7033) "Hyman Bass and Ubiquity: Gorenstein Rings" is a great introduction to Gorenstein rings, very easy to read and to the point, I highly recommend it.
EDIT: Since this question is already bumped up, I will take this opportunity to make a longer list.
There are of course some classic references which are still very useful (I find myself having to look in them quite often despite the new sources available): Bourbaki, EGA IV, Serre's "Local Algebras" (very nice read and culminated in the beautiful [Serre intersection formula](http://en.wikipedia.org/wiki/Intersection_theory_%28mathematics%29)).
There has been some work done in commutative algebra since the 60s, so here is a more up-to-date list of reference for some currently active topics (Disclaimer: I am not an expert in any of these, the list was formed by randomly looking at my bookself, and put in alphabetical order (-:). This is community-wiki, so feel free to add or edit or suggest things you found missing.
- Cohen-Macaulay modules, from a representation theory perspective: [Yoshino](http://rads.stackoverflow.com/amzn/click/0521356946) is excellent. Another one is [being written](http://www.leuschke.org/Research/MCMBook).
- Combinatorial commutative algebra: [Miller-Sturmfels](http://rads.stackoverflow.com/amzn/click/0387237070).
- Free resolutions (over non-regular rings): Avramov [lecture note](http://www.math.unl.edu/~lavramov2/papers/resolution.pdf)
- Geometry of syzygies: [Eisenbud](http://rads.stackoverflow.com/amzn/click/0387222324), shorter but free version [here](http://www.msri.org/people/staff/de/ready.pdf).
- Homological conjectures: [Hochster](http://rads.stackoverflow.com/amzn/click/0821816748), [Roberts](http://rads.stackoverflow.com/amzn/click/0521473160) (more connections to intersection theory), Hochster [notes](http://www.math.utah.edu/vigre/minicourses/algebra/hochster.pdf).
- Integral closures: [Huneke-Swanson](http://people.reed.edu/~iswanson/book/index.html), which is available free at the link.
- Intersection theory done in a purely algebraic way: [Flenner-O'Carrol-Vogel](http://rads.stackoverflow.com/amzn/click/3540663193) (for a very interesting story about this, see Eisenbud [beautiful reminiscences](http://www.msri.org/~de/papers/pdfs/1999-002.pdf), especially page 4)
- Local Cohomology: [Brodmann-Sharp](http://rads.stackoverflow.com/amzn/click/0521372860), Huneke's [lecture note](http://www.math.uic.edu/~bshiple/huneke.pdf) (very easy to read), [24 hours of local cohomology](http://rads.stackoverflow.com/amzn/click/0821841262) (I have been told that this one was a pain to write, which is probably a good sign).
- Tight closure and characteristic $p$ method: [Huneke](http://rads.stackoverflow.com/amzn/click/082180412X), Karen Smith's lecture note (more geometric, number 24 [here](http://www.math.lsa.umich.edu/~kesmith/paperlist.html)), and of course many well-written introductions available on Hochster [website](http://www.math.lsa.umich.edu/~hochster/mse.html).
| 31 | https://mathoverflow.net/users/2083 | 16519 | 11,071 |
https://mathoverflow.net/questions/16177 | 3 | Let us assume two samples, A and B, where A are the results obtained with some standard method, and B are the results obtained with a new method, which is not necessarily more accurate, but has additional advantages (eg: lower cost).
So I'm interested in testing if B is "as good as" A, using non-inferiority hypothesis testing, i.e. that
Ho : $A - B >= \delta$
or, equivalently (?)
Ho : $A - \delta >= B$
where $\delta$ is the maximum clinically-acceptable margin of error.
(Please bear with me if my formulation is not perfect. I'm not a statistician; I'm just a researcher trying to statistically get my way out of a paper bag with minimal damage. Suggestions for improvements are welcome.)
I've seen this hypothesis tested with confidence intervals, which I'm weary of using since my sample size is small. Would it be correct to use Student's t-test to test this hypothesis? If no, why not?
Response to sheldon-cooper:
I'm not sure I fully understand what you mean by "the null hypothesis encompasses many values for the mean of one population"
Assuming that testing for Ho: $A >= B$ can validly be tested using Student's two-sample t-test, how does subtracting $\delta$ from $A$ change the validity of the test? In my perhaps naive understanding, you still get two samples from populations with unknown means that we want to test for equivalence, except that one has been artificially penalized.
| https://mathoverflow.net/users/2735 | Can t-test be used for non-inferiority hypothesis testing? | Follow-up on my question, after some more research on the question.
The method of using two one-sided two-sample t-tests for clinical equivalence testing is shown in [Schuirmann 1987](https://doi.org/10.1007/BF01068419 "Schuirmann, D.J. A comparison of the Two One-Sided Tests Procedure and the Power Approach for assessing the equivalence of average bioavailability. Journal of Pharmacokinetics and Biopharmaceutics 15, 657–680 (1987)"), and appears relatively standard in bioclinical statistics (at least it is very widely cited). Some preconditions must be respected for the test to be valid (including same sample size for both groups and normal distribution of the data); I will refer you to the article for the complete list.
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> Under the normality assumption that has been made, the two sets of one-sided hypotheses will be tested with ordinary one-sided t-tests.
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As the non-inferiority test is only one half of the equivalence testing (In my case, I don't care whether B is superior or simply similar to A), I will assume that the method still holds.
| 1 | https://mathoverflow.net/users/2735 | 16520 | 11,072 |
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