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https://mathoverflow.net/questions/46343 | 4 | Hello!
This is a very short question:
**Given a local graded Noetherian ring $R\_{\bullet}$, is it true that any graded projective module over $R\_{\bullet}$ is free?**
In the ungraded case, this is true, but I do not know where the graded case is considered. Are there any references?
Thank you!
Hanno
| https://mathoverflow.net/users/3108 | Kaplansky's theorem for graded local rings | There is a proof of the ungraded result on page 10 of Matsumura's "Commutative Ring Theory", and you can insert gradings everywhere in a straightforward way to prove the graded result.
The main reason why you cannot just appeal to the ungraded result is as follows: a local graded ring has (by definition) precisely one homogeneous ideal that is maximal among proper homogeneous ideals, but typically there will be many maximal inhomogeneous ideals, so the underlying ungraded ring will not be local.
| 9 | https://mathoverflow.net/users/10366 | 46349 | 29,329 |
https://mathoverflow.net/questions/46362 | 6 | The Gaussian algorithm tells us, that for any field $k$ a $n\times n$-matrix over $k$ can written as a product of at most $C$ elementary matrices ($C\sim n^2$).
I am wondering, whether such a constants also exists for other rings - like $\mathbb{Z}$.
Given a matrix $A\in SL\_2(\mathbb{Z})$, one can basically use the Euclidean algorithm to find such a decomposition. However if we take a the following [matrix](http://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form) involving the Fibonacci numbers, the algorithm takes about $n$-steps and hence we get a decomposition in $\sim n$ factors. But there might still be a better decomposition.
So is there for every $n$ a matrix $A \in SL\_2(\mathbb{Z})$, that cannot be written as a product of elementary matrices ?
I guess the construction with the Fibonacci numbers might be a candidate, I don't know how to prove, that it is impossible to decompose it in a better way.
| https://mathoverflow.net/users/3969 | length of decompositions into elementary matrices | Yes. The group $SL\_2(\mathbb Z)$ is virtually free and is not boundedly generated by any finite generating set because of that. You can look at the (very nice) slides of Dave Witte-Morris' talks on bounded generation [here.](http://people.uleth.ca/~dave.morris/talks.shtml)
| 4 | https://mathoverflow.net/users/nan | 46365 | 29,338 |
https://mathoverflow.net/questions/46367 | 0 | Let $X$ be a smooth projective variety and $E\longrightarrow X$ a vector bundle of rank $n$. For any $0\leq k\leq n$ the associated Grassmann bundle $G\_k(E)\longrightarrow X$ yields and we have the so-called "basis theorem" (see Fulton "Intersection theory", Proposition 14.6.5) which asserts that for any $s\geq 0$,
$$CH\_s(G\_k(E))=\bigoplus\_{\lambda}CH\_{s-k(n-k)+|\lambda|}(X)$$
where $\lambda$ runs over all partitions $\lambda=(\lambda\_1,...,\lambda\_k)$ with $n-k\geq \lambda\_1\geq...\geq\lambda\_k\geq 0$.
I would like to know if there is some similar result for fiber product of Grassman bundles, i.e. consider a vector bundle $E\longrightarrow X$ of rank $n$ and $0\leq k\_1\leq k\_2\leq n$ two integer : does the $s$-dimensional Chow group $CH\_s(G\_{k\_1}(E)\times\_{X} G\_{k\_2}(E))$ is isomorphic to a direct sum of $\bigoplus\_{p\in \mathcal{P}}CH\_p(X)$ for some set $\mathcal{P}$ ?
I tried to look a an answer considering the immersion $G\_{k\_1}(E)\times\_X G\_{k\_2}(E)\longrightarrow G\_{k\_1k\_2}(E\times\_X E)$ without success.
This question studying the case where $X=\mathbb{P}^1$ and $E$ the tautological vector bundle.
| https://mathoverflow.net/users/10541 | Chow group of a fiber product of grassmann bundles | You just need to repeat the same procedure twice. Indeed, let $E'$ be the pullback of $E$ from $X$ to $G\_{k\_1}(E)$. Then $G\_{k\_1}(E)\times\_X G\_{k\_2}(E) \cong G\_{k\_2}(E')$.
| 2 | https://mathoverflow.net/users/4428 | 46372 | 29,343 |
https://mathoverflow.net/questions/46388 | 5 | Does anyone know of the link to an online library of of unlabeled, connected graphs on n vertices? I remember looking at such an archive a few years ago while at a Macaulay 2 workshop, but I've been unable to find it (or any other one) since then.
The page I remember seeing only had enumerated unlabeled graphs up to n=11,12, or 13 vertices, and the graph I'm looking for data on is much larger, so links to repositories of larger (special) graphs.
The most specific part of this request: The graph I'm looking to find a list of edges for is the 1-skeleton of the 600-cell, if anyone happens to just have that information on-hand (or readily available.)
| https://mathoverflow.net/users/5002 | Online Library of Unlabeled Connected Graphs on n Vertices | Is this it? It is a searchable database of
"[Small Simple Graphs:
Connected, undirected, and unlabeled](http://gfredericks.com/sandbox/graphs)."
And an explicit list of the vertices, edges, and faces of the 600-cell can be found
at [Paul Bourke's site](http://local.wasp.uwa.edu.au/~pbourke/geometry/platonic4d/).
| 4 | https://mathoverflow.net/users/6094 | 46389 | 29,349 |
https://mathoverflow.net/questions/46404 | 0 | Given a flat and projective morphism of noetherian schemes, $f: X \rightarrow Y$ and $F$, $G$ two coherent $O\_X$-modules, flat over $Y$. Furthermore given a morphism $u: Y' \rightarrow Y$ of noetherian schemes and a quasi coherent $O\_{Y'}$-module $\mathcal{M}$. Then we have $X':=X\times\_YY'$ with $p\_i$ the i-th projection.
1) Take a locally free resolution $P\_{\cdot} \rightarrow F \rightarrow 0$. Now the article says: "Since $f$ and $F$ are flat over $Y$, $p\_1^{\\*}P\_{\cdot}$ is a resolution of $p\_1^{\\*}F$". Don't we need $u$ to be flat for this? Why is this implied by the flatness of $f$ and $F$?
2) Assume $Y$ and $Y'$ are affine, say $Y=Spec(A)$ and $Y'=Spec(A')$. Given an $A'$-module $M$. For a descending induction, the article says there is an exact sequence:
$0\rightarrow p\_1^{\\*}G'(n)\otimes\_{A'} M \rightarrow \bigoplus\limits\_{i=1}^r O\_{X'}(-k+n)\otimes\_{A'} M \rightarrow p\_1^{\\*}G(n)\otimes\_{A'} M\rightarrow 0$.
I see that we have $0\rightarrow G'(n) \rightarrow \bigoplus\limits\_{i=1}^r O\_{X}(-k+n)\rightarrow G(n)\rightarrow 0$ on X, now if $p\_1$ would be flat (see (1)), we'd have $0\rightarrow p\_1^{\\*}G'(n) \rightarrow \bigoplus\limits\_{i=1}^r O\_{X'}(-k+n) \rightarrow p\_1^{\\*}G(n)\rightarrow 0$ on $X'$, but why is $\otimes\_{A'} M$ exact in this case? Or is there any other way to construct such a sequence?
Background: I'm reading the article "Universal families of extensions" by Herbert Lange (J. Algebra 83 (1983), 101–112), where he constructs base change homomorphisms for relative $Ext$-sheaves on schemes. I get the basic idea of the construction, but now i'm trying to understand the details, and these two questions stayed open.
| https://mathoverflow.net/users/3233 | Subtleties in the construction of base change morphisms | It is easy, $p\_1^\*G = G\otimes\_A A'$ is flat over $A'$ since $G$ is flat over $A$. Hence $Tor\_1^{A'}(p\_1^\*G,M) = 0$, hence your sequence is exact after tensoring with $M$. The same argument works for the first question as well.
| 1 | https://mathoverflow.net/users/4428 | 46408 | 29,360 |
https://mathoverflow.net/questions/46413 | 2 | What is the best reference for someone (i.e. me) trying to learn Instanton Floer homology? Assume I already know symplectic Floer homology.
| https://mathoverflow.net/users/9052 | Instanton homology - reference request | Donaldson's book
<http://books.google.com/books?id=CbMq-dh8nEoC&pg=PA106&dq=simon+donaldson+floer+yang&hl=en&ei=hFPkTMWtM4WnnAep1LzXDg&sa=X&oi=book_result&ct=result&resnum=6&ved=0CEoQ6AEwBQ#v=onepage&q&f=false>
| 4 | https://mathoverflow.net/users/3874 | 46416 | 29,364 |
https://mathoverflow.net/questions/46414 | 7 | Suppose you launch $n$ point-particles on
distinct reflecting nonperiodic [billiard trajectories](http://en.wikipedia.org/wiki/Dynamical_billiards)
inside a convex polygon. Assume they all have the same speed.
Define an *$\epsilon$-cluster* as a configuration of the particles
in which they all simultaneously lie within a disk of radius $\epsilon$.
It is my understanding that *[Poincaré's Recurrence Theorem](http://en.wikipedia.org/wiki/Poincar%25C3%25A9%2527s_recurrence_theorem)*
implies that at some time after launch, the particles
will form an $\epsilon$-cluster somewhere.
(Please correct me if I am wrong here, in which case the remainder
is moot.)
Picturesquely, if I sit in my office long enough, all the air molecules
will cluster into a corner of the room. :-)
The reason I specify that the trajectories be *distinct* is to
exclude
the particles being shot in a stream all on the same trajectory.
The reason I specify *nonperiodic* is to exclude sending
the particles on parallel periodic trajectories whose length
ratios are rational, in which case no clustering need occur.
My question is:
>
> How long must one wait for an $\epsilon$-cluster to occur?
>
>
>
Essentially I am seeking a quantitative version of
Poincaré's Recurrence Theorem, quantitative enough
to actually make a calculation.
I would like to put a number of years to the air-molecule example
(air molecules move perhaps 700 mph or 300 m/s).
It could serve as a useful pedagogical anecdote.
I found a beautiful paper that should help
me answer this question:
Benoit Saussol,
"An Introduction to Quantitative Poincaré Recurrence in Dynamical Systems,"
*Reviews in Mathematical Physics*, Volume 21, Issue 08, pp. 949-979 (2009).
But I am having difficulty making the leap from the abstract
theorems to an explicit calculation.
Any help or additional pointers would be appreciated!
**Addendum**. Vaughn's analysis, although leaving a few loose ends (as he notes), largely
answers my question. Thanks to all for the astute comments and responses!
| https://mathoverflow.net/users/6094 | How quickly will billiard trajectories cluster? | I'll offer a few partial answers, which may eventually lead to a complete answer.
As observed in Saussol's paper (Theorem 3, Kac's lemma), if you have an ergodic invariant measure $\mu$ then the mean return time to a set $A$ is equal to $1/\mu(A)$. (Note, however, that this is the mean return time *for trajectories that begin in $A$*, rather than for an arbitrary initial condition, for which you'd need the *hitting time*).
So we need to know what the ergodic invariant measure $\mu$ is. Many topological dynamical systems have lots of invariant measures lying around, in which case it's a non-trivial task to pick the one you're really interested in. However, if we assume that your convex polygon satisfies the Veech dichotomy (see, for example, [this article](http://www.math.bgu.ac.il/~barakw/papers/veech.pdf) by John Smillie and Barak Weiss), then it's true that the flow in every direction is either periodic or uniquely ergodic. Regular polygons satisfy the Veech dichotomy, so in that setting your non-periodicity condition would be enough to guarantee that Lebesgue measure is the only invariant measure.
Of course, Lebesgue measure is always invariant for the billiard flow, so the main thing we need this for is the fact that Lebesgue measure is *ergodic* under these assumptions. (There may be an easier way to get ergodicity, but this is the first thing that came to my mind.)
If we write $X\_\theta$ for the phase space of the billiard flow with angle $\theta$, and let $\theta\_1,\dots,\theta\_n$ be the angles that your $n$ particles are launched at, then the above discussion implies that the billiard flow on each $(X\_{\theta\_n},\text{Leb})$ is ergodic. The phase space for your entire system is $\prod\_n X\_{\theta\_n}$, and the direct product of the $n$ different Lebesgue measures is an invariant measure for your overall system.
What we'd like to do next is say that this measure is actually ergodic. Unfortunately, it's not quite that simple, since the direct product of two ergodic systems need not be ergodic (just consider $R\_\alpha\times R\_\alpha$, the direct product of two circle rotations by irrational multiples of $\pi$). So to say that Lebesgue measure is ergodic for your whole system requires something more, which is where your condition that the trajectories be distinct ought to come in. I'm not sure exactly how this step should go, but you should be able to get something along the lines of "for almost every set of angles $(\theta\_1,\theta\_2,\dots,\theta\_n)$, Lebesgue measure is ergodic for the whole system". Billiard flows have something to do with interval exchange transformations (IETs), and so [this paper](http://arxiv.org/abs/0905.2370) by Jon Chaika may well have the result that's needed at this point.
Once you know that (product) Lebesgue measure is invariant for the whole system, you're in the clear: given a ball of radius $\epsilon$, the (normalised) Lebesgue measure of that ball in each $X\_\theta$ is $\pi \epsilon^2 / C$, where $C$ is the area of the polygon, and so the set in $\prod\_n X\_{\theta\_n}$ corresponding to those configurations for which all particles lie in this ball is just $(\pi \epsilon^2 / C)^n$. The inverse of this is your expected return time, provided we have appropriate hypotheses to justify all the above steps. (Expected hitting time might be a different story, I'd need to think a bit.)
| 5 | https://mathoverflow.net/users/5701 | 46422 | 29,368 |
https://mathoverflow.net/questions/46230 | 2 | What does it exactly mean to say that in a certain category pushouts and pullbacks "commute"? Is it the same to say that they "distribute"?
| https://mathoverflow.net/users/2625 | Distributivity / commutativity of pushouts and pullbacks | The condition Martin and Todd mention is indeed a sort of distributivity condition. It is also often called {\em stability} of pushouts. I think that it should not be called commutativity.
Let D and C be small categories, and A a category with D-shaped limits and C-shaped colimits. Then D-limits commute in A with C-shaped colimits when the functor
$[D,A]\to A$ which calculates the limit preserves C-colimits. This is equivalent to saying
that the functor $[C,A]\to A$ which calculates the colimit preserves D-limits. The most famous example is commutativity of finite limits with filtered colimits.
Note, however, that commutativity of pullbacks and pushouts in this sense is rare, and is probably not what was meant.
| 3 | https://mathoverflow.net/users/10862 | 46426 | 29,371 |
https://mathoverflow.net/questions/46399 | 13 | Let R be the 2-periodic complex K-theory spectrum, or any other naturally occuring 2-periodic E-infty ring spectrum. The suspend-once functor gives an autoequivalence of the category of R-module spectra, and since R is 2-periodic applying it twice is isomorphic to the identity functor. In some coarse sense this defines an action of Z/2 on the category of R-modules.
There is a better, more complicated and more correct notion of Z/2-action on a collection of module spectra. Module spectra form an infty-category, and we can ask if Z/2 acts on this infty-category. One way to put it: the suspend-once functor and the homotopy between suspend-twice and the identity give us a map from RP^2 into the classifying space of automorphisms of this infty-category. Does this extend to a map from BZ/2 = RP^infty?
| https://mathoverflow.net/users/1048 | How often does suspension define an action of Z/2 on a category of module spectra? | At least for complex K-theory, I believe the answer is yes. If you want an explicit
construction, you can use the fact that there's a symmetric monoidal functor from
the 2-groupoid of Clifford algebras and Morita equivalences to the groupoid of invertible K module spectra. So it suffices to construct a monoidal functor from Z/2 into this 2-groupoid, which is a much more concrete problem.
| 15 | https://mathoverflow.net/users/7721 | 46443 | 29,382 |
https://mathoverflow.net/questions/46299 | 9 | it is motivated by [Density of congruence classes covered by a set](https://mathoverflow.net/questions/46249/density-of-congruence-classes-covered-by-a-set/46277#46277)
Let say just "$n$-gon" for the set of vertices of a regular $n$-gon inscribed in a unit circle, "2-gon" for the set of two opposite points.
is it true that for given positive integers $1 < b\_1 < b\_2 < \dots < b\_k$ the union of $b\_i$-gons has the minimal cardinality when they all have a common vertex?
In other words, if $G\_i$ are subgroups of the finite cyclic group $G$, is it true that
$$|\cup\_{i=1}^k x\_iG\_i|\ge |\cup\_{i=1}^k G\_i|$$
for arbitrary cosets $x\_iG\_i$?
| https://mathoverflow.net/users/4312 | union of regular polygons | Let me attempt a proof using the group-theoretic formulation. I will use the additive notation for the group operation.
The proof is by induction on $n=|G|$, with the base being trivial. Let $n=p^rm$ for some prime $p$ with $\gcd(p,m)=1$.
Consider $G' = pG$. Our goal is to reduce the problem for $G$ to its instance for $G'$. Let $R\_j:=G' + jm$. Then $\{R\_0,R\_1,\ldots, R\_{p-1}\}$ is a partition of $G$ into $G'$-cosets.
Let $G\_i=d\_iG$ be a subgroup of $G$ with $d\_i | n$. If $p|d\_i$ then $G\_i \subseteq G'$ and every $G\_i$ coset belongs to some $R\_j$. In this case
we say that $G\_i$ is *of the first kind*. Otherwise, $d\_i | m$, and translating $G\_i$ by $m$ does not change $G\_i$. We say that such $G\_i$ is *of the second kind*.
Let $S = \cup\_{i=1}^k (G\_i+x\_i)$ be the union under consideration, let $S\_1$ be the union of cosets of the first kind among cosets comprising $S$, and $S\_2$ -- of the second. Let $T\_j=S\_1 \cap R\_j$ for $0\leq j \leq p-1$. Then $T\_j$ is actually union of some of our cosets, and the sets $T\_0,T\_1,\ldots,T\_{p-1}$ are disjoint. Let $T\_j'=T\_j-jm$: we shift all the cosets in $T\_j$ from $R\_j$ to $R\_0=G'$. Note that $S\_2-jm=S\_2$ and therefore the intersection of $T\_j$ and $S\_2$ shifts with $T\_j$. In particular, $|T\_j'-S\_2|=|T\_j-S\_2|$.
The set $S'=S\_2 \cup (\cup\_{j=0}^{p-1}T'\_j) $ is still a union of cosets of $G\_i$'s. We have
$|S'| \leq |S\_2|+ \sum\_{j=1}^{p-1}|T'\_{j}-S\_2| = |S\_2|+ \sum\_{j=1}^{p-1}|T\_{j}-S\_2|=|S|$.
Thus it suffices to consider $S'$, which means that we may assume that $S\_1 \subseteq G'$. Let $G\_i'=G\_i \cap G'$ and let
$x\_i'$ be chosen so that $G\_i'+x\_i'= (G\_i+x\_i) \cap G'$. By the induction hypothesis applied to $G'$, we get
$a\_1:=| \cup\_{i=1}^k G\_i' | \leq | \cup\_{i=1}^k (G\_i' + x\_i')|=: b\_1$
and also
$a\_2:=|\cup\_{i: G\_i \not \subseteq G'} G\_i'| \leq |\cup\_{i: G\_i \not \subseteq G'} (G\_i' + x\_i')|=:b\_2$,
where in this second inequality we restrict our attention to $G\_i$'s of the second kind. The set $S\_2$ is the disjoint union of $p$ translates of its intersection with $G'$, which intersection is present on the right side of the inequality directly above. It follows that
$|S|=|S \cap G'|+|S\_2 - G'|=b\_1 + (p-1)b\_2,$
while similarly we have
$|\cup\_{i=1}^k G\_i|=a\_1 + (p-1)a\_2$.
It follows that $|S| \geq |\cup\_{i=1}^k G\_i|$, as desired.
---
Finally, let me note that the inequality does not hold for non-cyclic groups. Already for $G = \mathbf{Z}\_2 \times \mathbf{Z}\_2$ the union of three distinct subgroups of $G$ of size $2$ is $G$, while it is possible to choose their cosets with the union of size $3$.
| 4 | https://mathoverflow.net/users/8733 | 46451 | 29,385 |
https://mathoverflow.net/questions/46448 | 0 | Let x,y,z be points taken exclusively from the positive orthant.
For the scaling transformation
x'=x/(x+y+z)
y'=y/(x+y+z)
z'=z/(x+y+z)
where each function is a linear fractional transform
how can I interpret this?
That is, does this sort of componentwise LFT enjoy all the usual LFT properties?
I am trying to figure out if this transformation of mine preserves the cross ratio
and am hoping to use the properties of an LFT to do so.
Is this going to work out like I want it to?
I am having no luck finding anything relevant in the literature. Any suggestions there?
| https://mathoverflow.net/users/10917 | geometric interpretation of componentwise linear fractional transformation(LFT) | As mentioned, this is a central projection to the $x+y+z=1$ plane, or, in the positive orthant, you can also view it as an $L\_1$-norm normalization.
Preservation of circles:
* Circles in planes parallel to $x+y+z=1$ are preserved.
* Circles (and any other shapes) in planes ~~normal to $x+y+z=1$~~ *passing through the origin* are projected to line segments.
* Other circles are projected to ellipses.
Cross ratios:* Component-wise cross ratios are obviously preserved.
* Geometric cross-ratios (i.e. for 4 collinear points the cross-ratios of distances between them) are also preserved, as they would be with any conic or linear projection (with the exception of the case when the points are on a line ~~normal to the $x+y+z=1$ plane~~ *passing through the origin*.)
If you're interested in some other kind of cross-ratio, you have to define how you intend to "multiply" and "divide" the differences of points.
**Edit:** See corrections.
| 1 | https://mathoverflow.net/users/10876 | 46456 | 29,390 |
https://mathoverflow.net/questions/25411 | 33 | There are two books by Matsumura on commutative algebra. The earlier one is called *Commutative Algebra* and is frequently cited in Hartshorne. The more recent version is called *Commutative Ring Theory* and is still in print. In the preface to the latter, Matsumura comments that he has replaced a section from a previous (Japanese?) edition because it "did not substantially differ from a section in the second edition of my previous book Commutative Algebra." This suggests that Matsumura considered the two books complementary, and certainly did not intend *Commutative Ring Theory* to replace *Commutative Algebra*.
Basically, my question is
>
> Why are there two books, by the same author, apparently on the same subject?
>
>
>
A more practical question is whether both books are equally appropriate as references when reading a book like Hartshorne. However, I am curious to know the answer to the broader question as well.
| https://mathoverflow.net/users/5094 | Matsumura: "Commutative Algebra" versus "Commutative Ring Theory" | By comparing the tables of contents, the two books seem to contain almost the same material, with similar organization, with perhaps the omission of the chapter on excellent rings from the first, but the second book is considerably more user friendly for learners. There are about the same number of pages but almost twice as many words per page. The first book was almost like a set of class lecture notes from Professor Matsumura's 1967 course at Brandeis. Compared to the second book, the first had few exercises, relatively few references, and a short index. Chapters often began with definitions instead of a summary of results. Numerous definitions and basic ring theoretic concepts were taken for granted that are defined and discussed in the second. E.g. the fact that a power series ring over a noetherian ring is also noetherian is stated in the first book and proved in the second. The freeness of any projective modules over a local ring is stated in book one, proved in the finite case, and proved in general in book two. Derived functors such as Ext and Tor are assumed in the first book, while there is an appendix reviewing them in the second. Possibly the second book benefited from the input of the translator Miles Reid, at least Matsumura says so, and the difference in ease of reading between the two books is noticeable. Some arguments in the second are changed and adapted from the well written book by Atiyah and Macdonald. More than one of Matsumura's former students from his course at Brandeis which gave rise to the first book, including me, themselves prefer the second one. Thus, while experts may prefer book one, for many people who are reading Hartshorne, and are also learning commutative algebra, I would suggest the second book may be preferable.
Edit: Note there are also two editions of the earlier book Commutative algebra, and apparently only the second edition (according to its preface) includes the appendix with Matsumura's theory of excellent rings.
| 40 | https://mathoverflow.net/users/9449 | 46457 | 29,391 |
https://mathoverflow.net/questions/46418 | 8 | Hello,
I'm trying to understand the relation between the points of view of
log geometry (monoids) and toric geometry (fans).
Suppose that $k$ is a field and $P$ is a finitely generated monoid.
Then $k[P]$ has a natural log structure and furthermore, any choice
of generators $\mathbf N^r\to P$ induces a closed embedding
$Spec(k[P])\subset\mathbf A^r$.
On the other hand, starting from a cone $\sigma$ satisfying some properties in a lattice
$N\otimes\mathbf R$, where $N = \mathbf Z^r$, we obtain a monoid
$P' = \sigma^\vee\cap M$, where $M = Hom(N,\mathbf Z)$ and $\sigma^\vee$ is the set of
all $x\in M\otimes\mathbf R$ such that $x(\sigma) \geq 0$.
Question: if we start with $P$ (and a choice of generators as above), can one write a
corresponding cone so as to recover $P$ by the construction in the previous paragraph?
Thanks!
| https://mathoverflow.net/users/10580 | relation between toric geometry and log geometry | Not all finitely generated monoids $P$ will come from the cone construction. You need to assume that:
1. $P^{gp}$ is torsion-free: If $x \in P^{gp}$ and $n\cdot x = 0$ then $x = 0$.
2. $P$ is cancellative: If $x + y = x + y'$, then $y = y'$. This is equivalent to saying that the map $P \rightarrow P^{gp}$ is injective.
3. $P$ is saturated: If $x \in P^{gp}$ and $x^n \in P$, then $x \in P$. Assuming the previous two proporties, this is equivalent to $k[P]$ being normal.
Here, $P^{gp}$ refers to the group formed by inverting all the elements of $P$.
If $P$ is finitely generated and satisfies 1, then $P^{gp}$ is a lattice, i.e. isomorphic to $\mathbb Z^r$ for some $r$, and this is the lattice $M$ from the cone construction. The dual lattice $N$ is $\textrm{Hom}(M, \mathbb Z)$, and $\sigma$ can be taken to be those $\lambda$ in $N \otimes\_{\mathbb Z} \mathbb R = \textrm{Hom}(M, \mathbb R)$ such that $\lambda(x) \geq 0$ for all $x \in P$.
| 4 | https://mathoverflow.net/users/8914 | 46460 | 29,394 |
https://mathoverflow.net/questions/46433 | 1 | Let $\pi:E\to M$ be a vector bundle over a closed smooth manifold and supose $\Pi:F\to E$ is a fibre bundle over the total space of $\pi$. I'd like to know if, restricted to $E\_p$, the second bundle $Pi$ is trivializable (moreover, homogeneous) as a fibre bundle over the vector space $E\_p$. I belive this is true if $\Pi$ is a vector bundle over $E$, but I haven't written down a proof yet. I'll appreciate any help. Thanks.
| https://mathoverflow.net/users/10328 | Is a fibre bundle over a vector bundle trivializable on each fibre? | If $E \to M$ is a vector bundle then it is a weak equivalence. (being a vector bundle means it is a fibration and then look at the LES in homotopy and as Somnath points out the fiber is contractible.) So having a bundle on E is the same as having a bundle on M up to homotopy.
But as for your question, up to isomorphism there is only one vector bundle over a contractible space.
It is definitely trivial over each fiber of the vector bundle you started with.
| 2 | https://mathoverflow.net/users/3901 | 46461 | 29,395 |
https://mathoverflow.net/questions/46358 | 2 | Assume that $G$ is a semi-simple linear algebraic group defined over $\mathbb{Q}$, which is $\mathbb{Q}$-simple, and that $G(\mathbb(R)$ is non-compact, without $\mathbb{R}$-factors of rank 1. Then by Margulis's works, the arithmetic subgroups of $G(\mathbb{R})$ are the same as discrete lattice in $G(\mathbb{R})$. Here a lattice is a discrete subgroup $\Gamma$ such that the quotient $\Gamma\backslash G(\mathbb{R})$ is of finite volume with respect to the measure deduced from the left Haar measure. In particular, an arithmetic subgroup in $G(\mathbb{R})$ is Zariski dense in $G\_\mathbb{R}$.
Conversely, a discrete subgroup
$\Gamma$ in $G(\mathbb{R})$
is given, such that $\Gamma$ is also dense in $G\_\mathbb{R}$ for the Zariski topology, what condition should one impose to make it arithmetic? Shall I assume $\Gamma$ to be finitely generated, or stable under certain actions such as $Aut(\mathbb{R/Q})$? I feel that such kind of results are more or less available in the literature, bu I'm far from an expert in this field.
Many thanks!
| https://mathoverflow.net/users/9246 | arithmetic groups VS. Zariski dense discrete subgroups? | Arbitrary Zariski-dense subgroups in a semisimple group can be very small from a real-analytic point of view. It seems that algebra cannot distinguish between "small" and "large" Zariski-dense subgroups, so most criteria to distinguish between the two have a strong non-algebraic flavour. (Of course one can also characterize arithmetic groups algebraically, but this has even less to do with the line of argument you seem to suggest.) From a dynamical point of view, the key difference between lattices and arbitrary Zariski-dense subgroups is that the former act transitively on the product of the Furstenberg boundary of the ambient Lie group with itself ("double ergodicity"). This is a sort of "largeness" property. There are various ways to capture this property, the most systematic way seems to me the concept of a generalized Weyl group due to Bader and Furman.
| 5 | https://mathoverflow.net/users/9927 | 46468 | 29,398 |
https://mathoverflow.net/questions/46485 | 4 | Is there a simple way to parametrize the orthogonal group O(3) of 3 by 3 orthogonal matrices?
| https://mathoverflow.net/users/10621 | Parametrization of O(3) | The general element is $\pm\exp(A)$ where $A$ is skew-symmetric.
(This gives each element infinitely often). This trick essentially
works for all compact Lie groups.
There is also the Cayley parameterization: $(I+A)(I-A)^{-1}$
for skew-symmetric $A$
is the general element of $SO(3)$ which lacks an eigenvalue $-1$
(so isn't a half-turn.) This parameterizes all such matrices
once each.
| 11 | https://mathoverflow.net/users/4213 | 46487 | 29,410 |
https://mathoverflow.net/questions/46508 | 3 | Let $\mathbb{Z}/2\mathbb{Z}$ act on $\mathbb{A}^1$ as $x \mapsto -x$, and let $\mathscr{X}$ be the quotient stack. It has coarse moduli space $\mathbb{A}^1$ and a residual gerbe $B\mathbb{Z}/2\mathbb{Z}$ at 0. There is then a surjective map from cartesian powers $\mathscr{X}^n$ to $\mathbb{A}^n$. What are the closed substacks of $\mathscr{X}^n$? Will there be anything other than pullbacks of closed subschemes of $\mathbb{A}^n$?
| https://mathoverflow.net/users/2234 | closed substacks of cartesian powers of a stack | A substack of $\mathcal{X}$ is given by a $Z/2Z$ invariant subscheme of $\mathbb{A}^1$ and hence given by a $Z/2Z$ invariant ideal $I\subset k[x]$. So for example, the ideals $(x^n)$ give substacks, but they only pullback back from subschemes of the coarse space for $n$ even: the coarse space is given by $Spec(k[x^2])$. The substack given by $(x)$ here is just a copy of $B(Z/2Z)$ embedded at the origin.
| 4 | https://mathoverflow.net/users/9617 | 46511 | 29,422 |
https://mathoverflow.net/questions/46502 | 17 | Does anyone know of any good resources for the proof of the number of Archimedean solids (also known as semiregular polyhedra)?
I have seen a couple of algebraic discussions but no true proof. Also, I am looking more at trying to prove it topologically, but for now, any resource will help.\*
\*I worked on this project a bit as an undergraduate and am just now getting back into it.
| https://mathoverflow.net/users/10918 | On the number of Archimedean solids | A proof of the enumeration theorem for the Archimedean solids (which basically dates back to Kepler) can be found in the beautiful book [*"Polyhedra"*](http://books.google.co.uk/books?id=OJowej1QWpoC&printsec=frontcover&dq=cromwell+polyhedra&source=bl&ots=R2YEvXquUw&sig=eujgkMXZQRwcRiv9DIimWU7tn90&hl=en&ei=B2TlTM-zNsidOreixYcK&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDUQ6AEwAw#v=onepage&q&f=false) by P.R. Cromwell (Cambridge University Press 1997, pp. 162-167).
| 18 | https://mathoverflow.net/users/5371 | 46512 | 29,423 |
https://mathoverflow.net/questions/46480 | 5 | Let $p(x)$ be the chromatic polynomial of a special graph. Performing a certain type of operation on the graph changes $p$ by shifting it and adding a constant, say to: $q(x)=p(x+a) + b$.
I have noticed that this operation always results in $q$ having the same discriminant and splitting field as $p$. This would of course be obvious if we were just shifting $p$ by $a$, but it seems unusual given that we are also adding $b$.
I realise it is difficult to comment without knowing more details, but all I really want to know is whether or not this might be significant. The polynomials in question are irreducible cubics. How common is it that two cubics - which are not just shifted versions of each other - have the same discriminant? Does anybody have any idea what this might signify?
(edit: since posting this I have noticed that the chromatic polynomial in question is of the form $p(u,v,x)$, where $u$ and $v$ are given constants, and the discriminant of $p$ is a symmetric polynomial in $u$ and $v$. The graph operation switches $u$ and $v$...this is why the discriminant remains the same. So my question could be formulated as: what is the significance - if any - of a discriminant which is a symmetric polynomial in 2 variables?)
| https://mathoverflow.net/users/4078 | polynomials with the same discriminant | [Substantial edit: As I mentioned previously, cubic fields can have the same discriminant but not be isomorphic; but I've revised my answer to better address the author's question.]
There are a lot of polynomials that will generate the same cubic field. Here is the basic idea, due to Delone and Faddeev. Write down the cubic form $f(x, y) = ax^3 + bx^2 y + cxy^2 + dy^3$, where setting $y = 1$ gives a generating polynomial for the cubic field. Then there is an action of $GL\_2$ on such cubic fields; if $\gamma$ is a 2x2 matrix with entries $\alpha, \beta, \gamma, \delta$ then
$\gamma f(x, y) = f(\alpha x + \gamma y, \beta x + \delta y)$, optionally with a factor of $(det \ \gamma)^{-1}$.
Anyway, if two cubic forms are $GL\_2(\mathbb{Q})$-equivalent then the corresponding polynomials (if irreducible and nondegenerate) generate the same field. Moreover, if two cubic forms are $GL\_2(\mathbb{Z})$-equivalent then the corresponding polynomials generate the same order over $\mathbb{Z}$.
See [here](http://www.math.ucsd.edu/%7Ewgan/G2.ps), page 10 of Section 4, for a good reference.
[Previous answer:]
I think the underlying question is how often nonisomorphic cubic *fields* have the same discriminant. This happens, but it doesn't seem to be very common.
You might find it interesting to browse through [this database](http://hobbes.la.asu.edu/NFDB/) which lists cubic (and other) fields, their discriminants, and their minimal polynomials.
Theoretically, I believe it is known that this happens infinitely often, although I don't recall the reference. For upper bounds, it is known that the number of cubic fields of discriminant n is $O(n^{1/3 + \epsilon})$, due to Ellenberg and Venkatesh; the "correct" bounds are believed to be much sharper.
| 5 | https://mathoverflow.net/users/1050 | 46513 | 29,424 |
https://mathoverflow.net/questions/46431 | 6 | This is kind of a spin-off of the question asked [here](https://mathoverflow.net/questions/46414/how-quickly-will-billiard-trajectories-cluster). Take the interval $X:=[0,1]$ with $\mu$ being standard Lebesgue measure. Let $f$ be a measure preserving map $f:[0,1]\rightarrow [0,1]$. The Poincaré Recurrence theorem tells us that if I pick a measurable set $E\subset [0,1]$, then under iterations of $f$ almost every point in $E$ returns to $E$ infinitely often, i.e.
$\mu\left({x\in E: \ \exists N, \mbox{ such that } \forall n\geq N, \ f^n(x)\notin E}\right)=0$
Call the set of exceptions above $M$. My question is:
>
> If I specify an $f$, for what class of sets $E$ in $X$ is $M$ not dense? I am interested in two cases:
>
> 1) "dense" with respect to $X$, if $E$ is dense in $X$
>
>
> 2) "dense" with respect to $E$
>
>
>
For example, let $E=\mathbb{Q}\cap[0,1]$ and $f(E):=E+\phi$ where $\phi$ is irrational. Then $M=E$, which of course is not a contradiction since $\mu(M)=\mu(E)=0$. On the other hand, throw in the set $H:=\{n\cdot \phi: n\in \mathbb{N}\}$, so that $E':=H\cup E$. In this case we still have $M=E$. As well, it looks like the class of sets I'm interested in is $H\cup {\mbox{not a dense set in X}}$.
Note: My original motivation for asking this was to try and conceptualize the Poincaré recurrence theorem for a human physicist. If I were looking at the phase plot of balls on a billiard table at a specific time, I would only be able to give imprecise measurements of both position and velocity. In this case, it seems that in order to invoke Poincaré recurrence, I would need small intervals around every point to recur perfectly, in the sense that $M=\emptyset$. Perhaps this IS the case if $f$ arises from some nice ODE, but I'm interested in a more general setting. I also don't really want to require that $\mu(M)>0$, which is why I feel asking about denseness is more appropriate.
| https://mathoverflow.net/users/934 | Poincaré Recurrence and Dense Sets | If $f$ is minimal, i.e. every orbit is dense, then $E$ is either empty or dense. So it remains to decide if your set is empty. Clearly it is non-empty for positive measure sets. If $\mu(E) = 0$. Then also
$$
\mu(\bigcup\_{n} f^{-n} E) =0
$$
and for this set the exceptional set is non-empty (and trivially dense).
| 1 | https://mathoverflow.net/users/3983 | 46525 | 29,431 |
https://mathoverflow.net/questions/46524 | 18 | Are there examples of non-Kahler complex manifolds with holomorphically trivial canonical bundle?
| https://mathoverflow.net/users/10934 | Non-Kahler "Calabi-Yau"? | Yes, you might look at the following paper by J. Fine and D. Panov: <http://arxiv.org/abs/0905.3237>
| 16 | https://mathoverflow.net/users/10675 | 46528 | 29,433 |
https://mathoverflow.net/questions/46503 | 1 | I'm reading a paper on complex semi-simple algebraic group geometry at the moment, but finding the going a bit tough since I'm missing alot of the prerequisites. Firstly, the author refers to a *principal embedding* of one Lie group into another. I guess that this is an embedding of one group into another that is maximal in some sense. The paper states that in the dual Lie algebra setting this means that ${\frak g}$ is obtained from ${\frak g\_0}$ by adding a node to the Dynkin diagram. The family of examples used is that of the embedding of $U\_{N-1}$ into $SU\_{N}$. I've searched the web but can't find a principal embedding defn, could someone point me in the right direction.
Secondly, if $G\_0$ is principal embedded into $G$, is there some result about the quotient $G/G\_0$ having a complex structure? This works for example $CP^{N-1} = SU\_N/U\_{N-1}$.
| https://mathoverflow.net/users/1095 | Lie Group Principal Embedding | This doesn't answer your question completely, but at least it's a start.
If $G$ is a complex Lie group, then its quotient $G/H$ by a complex subgroup $H$ is always a complex manifold. In case $G$ is semisimple (and connected), then the *compact* quotients $G/H$ come from what are called parabolic subgroups $H$. For simplicity, let's assume that $G$ is simple. Then, up to conjugacy, the parabolic subgroups of $G$ lie in one-to-one correspondence with subsets of the nodes of the Dynkin diagram of $\mathfrak{g}$. It's easy to find a description of this bijection in the literature, so I will say no more about it. Let me just mention that a parabolic subgroup corresponding to omitting one node from the Dynkin diagram is called maximal parabolic.
Now suppose that $K$ is a compact real form of the complex semisimple group $G$. Then if the quotient space $G/H$ is compact (i.e., if $H$ is parabolic), a theorem of Montgomery asserts that $K$ acts transitively on $G/H$, and so we obtain an identification $G/H = K/(K\cap H)$. The a priori real manifold $K/(K\cap H)$ is thus endowed with a complex structure.
In summary, we have the following result:
>
> Let $K$ be a compact semisimple Lie group with complexification $G$. Then, for a closed subgroup $S$ of $K$, the homogeneous space $K/S$ is complex if $S = K \cap P$ for some parabolic subgroup $P$ of $G$.
>
>
>
Here's an example. Let $G = \rm{SL}(n,\mathbb{C})$. In terms of the bijection between parabolic subgroups of $G$ and nodes, it's not hard to see that the maximal parabolic corresponding to deleting the $k$th node is the subgroup of $G$ preserving a $k$-dimensional subspace of $\mathbb{C}^n$. In particular, if we let $P$ denote the maximal parabolic corresponding to the deletion of the first node, we get $G/P = \mathbb{CP}^{n-1}$. Now the maximal compact subgroup of $G$ is $K = \rm{SU}(n)$. In an appropriate basis, $P$ is the subgroup of $G$ of the form
$$ \begin{pmatrix} \ast & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \ast \\ \vdots & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \ast \end{pmatrix}. $$
Thus $K \cap P = \rm{U}(n-1)$ (to see this quickly, just recall that unitary matrices leave invariant the perp of an invariant subspace), where I'm thinking of $\rm{U}(n-1)$ as sitting in $K$ as
$$ \begin{pmatrix} (\det A)^{-1} & 0 \\ 0 & A \end{pmatrix}. $$
So by the result stated above we obtain the identification $\rm{SU}(n)/\rm{U}(n-1) = \mathbb{CP}^{n-1}$ given in the OP. Using the other maximal parabolics we can get a similar description for the complex Grassmannian of $k$-planes in $\mathbb{C}^n$.
---
Here are a couple of useful references:
1. Wang, Hsien-Chung, *Closed manifolds with homogeneous complex structure.* Amer. J. Math. **76** (1954). 1–32.
2. Wolf, Joseph A, *The action of a real semisimple group on a complex flag manifold. I. Orbit structure and holomorphic arc components.* Bull. Amer. Math. Soc. **75** (1969), 1121–1237.
Both these papers address the issue of finding complex structures on *compact* homogeneous spaces (so their results apply in particular to homogeneous spaces coming from compact Lie groups). There's a standing assumption of simply connectedness in Wang, but if I remember correctly, there is no such restriction in Wolf's paper.
---
**Edit:** Here's some more information that might be helpful. Let $K$ and $G$ be as above. A parabolic subgroup $P$ of $G$ admits "Levi decompositions" $P=LU$, where $U$ is the unipotent radical of $P$ and $L$ is a reductive group (called a Levi factor). In particular, every Levi factor $L$ has a compact real form $L\_{\mathbb{R}}$. The different Levi factors of $P$ are all conjugate. There is a choice that is compatible with the choice of maximal compact $K$, and in this case we have $P \cap K = L\_{\mathbb{R}}$.
| 2 | https://mathoverflow.net/users/430 | 46532 | 29,435 |
https://mathoverflow.net/questions/46507 | 5 | I am looking for a reference for the following fact.
Let $G$ be simple undirected connected planar graph with $\geq 2$ vertices. Then $G$ contains an edge $\{u,v\}$ such that $|N(u) \cap N(v)| \leq 2$, in other words the number of common neighbors of $\{u,v\}$ is at most $2$.
Here's a sketch of the proof. Any edge incident on a vertex of degree $\leq 3$ satisfies this requirement. If there is a vertex $v$ of degree $4$ for which none of its edges satisfy the requirement, then $N[v]$ must be a $K\_5$ which is not possible; if there is a vertex of degree 5 for which none of its edges satisfy then $N[v]$ must contain a $K\_{3,3}$. Since any planar graph contains a vertex of degree $\leq 5$ this is sufficient.
I would like to be able to skip the proof in my paper, using a reference to known work instead. Can anyone give me a reference for the above fact?
| https://mathoverflow.net/users/5200 | Every connected planar graph contains adjacent vertices with at most 2 common neighbors | it is not what you are asking for (not a reference), but just maybe slightly easier proof, without searching for $K\_{3,3}$. We may take any vertex $x$ of degree $d$, if any of $d$ vertices adjacent to $x$ has at least three neighbours between those $d$ vertices, we get at least $d+3d/2$ edges between $d+1$ vertices, hence $5d/2\leq 3(d+1)-6=3d-3$, $d\geq 6$. So degree of any vertex is not less then $6$, what is impossible.
By the way, we may even fix vertex $u$, and then find $v$. Indeed. if $v\_1$, $v\_2$, $\dots$, $v\_d$ are labeled counterclockwise neighbours of $u$, then without loss of generality $v\_i$ and $v\_{i+1}$ are joined ($v\_{d+1}=v\_1$), and $u$ is outside the cycle $v\_1-v\_2-\dots-v\_d$. But any triangulation of the (topological) polygon contains a vertex of degree 2.
| 3 | https://mathoverflow.net/users/4312 | 46536 | 29,438 |
https://mathoverflow.net/questions/46495 | 6 | In the vector space $V$ of $3\times 3$ symmetric real matrices, we can define a nonassociative algebra structure by the multiplication
$$A \bullet B = \frac12 (AB +BA).$$
This turns $V$ into a Jordan algebra.
### Question
>
> What is the minimum number of generators of this Jordan algebra? And could you give me one set of such generators?
>
>
>
Thanks a lot!
| https://mathoverflow.net/users/3945 | Generators for the Jordan algebra of symmetric 3-by-3 matrices | The minimum number of generators is 2. First, it is easy to check that one generator is not enough: every symmetric matrix is diagonalizable, so the subalgebra it generates has dimension at most 3.
Next, the claim is that $$A:=S\_{11}+2S\_{22}+3S\_{33}$$ and $$B:=S\_{12}+2S\_{13}$$ generate the algebra, where $S\_{ij}$ is the usual notation for the symmetric matrices. In fact the powers of $A$ generate the subalgebra $D$ of diagonal matrices (by Vandermonde) and the powers $B$ satisfy $B^2=S\_{23}$ mod $D$ and $B^3=S\_{12}+\frac12S\_{13}$ mod $D$.
| 9 | https://mathoverflow.net/users/10675 | 46537 | 29,439 |
https://mathoverflow.net/questions/45602 | 18 | One form of Vopenka's principle (a large cardinal axiom) states that no locally presentable category contains a full subcategory which is large (= a proper class) and discrete (= contains no nonidentity morphisms). In terms of this definition, my question is:
>
> Can one define a particular locally presentable category C and write down an explicit formula $\phi(x)$ in the first-order language of set theory such that *if* Vopenka's principle fails, then $\{ x | \phi(x) \}$ is a large discrete full subcategory of C?
>
>
>
But feel free to use any equivalent statement of Vopenka's principle and answer a suitably equivalent version of the question.
| https://mathoverflow.net/users/49 | Can Vopenka's principle be violated definably? | **Update.** My new article grows out of and extends my 2010 answer to this question. The new part is the conservativity result, showing that the Vopěnka principle has the same first-order consequences as the strictly weaker Vopěnka scheme.
* Joel David Hamkins, [The Vopěnka principle is inequivalent to but conservative over the Vopěnka scheme](http://jdh.hamkins.org/vopenka-principle-vopenka-scheme/), manuscript under review. ([arxiv](http://arxiv.org/abs/1606.03778))
**Abstract.** The Vopěnka principle, which asserts that every proper class of first-order structures in a common language admits an elementary embedding between two of its members, is not equivalent over GBC to the first-order Vopěnka scheme, which makes the Vopěnka assertion only for the first-order definable classes of structures. Nevertheless, the two Vopěnka axioms are equiconsistent and they have exactly the same first-order consequences in the language of set theory. Specifically, GBC plus the Vopěnka principle is conservative over ZFC plus the Vopěnka scheme for first-order assertions in the language of set theory.
The *Vopěnka principle* is the assertion that for every proper class $\mathcal{M}$ of first-order $\mathcal{L}$-structures, for a set-sized language $\mathcal{L}$, there are distinct members of the class $M,N\in\mathcal{M}$ with an elementary embedding $j:M\to N$ between them. In quantifying over classes, this principle is a single assertion in the language of second-order set theory, and it makes sense to consider the Vopěnka principle in the context of a second-order set theory, such as Godel-Bernays set theory GBC, whose language allows one to quantify over classes. In this article, GBC includes the global axiom of choice.
In contrast, the first-order *Vopěnka scheme* makes the Vopěnka assertion only for the first-order definable classes $\mathcal{M}$ (allowing parameters). This theory can be expressed as a scheme of first-order statements, one for each possible definition of a class, and it makes sense to consider the Vopěnka scheme in Zermelo-Frankael ZFC set theory with the axiom of choice.
Because the Vopěnka principle is a second-order assertion, it does not make sense to refer to it in the context of ZFC set theory, whose first-order language does not allow quantification over classes; one typically retreats to the Vopěnka scheme in that context. The theme of my article is to investigate the precise meta-mathematical interactions between these two treatments of Vopěnka's idea.
**Main Theorems.**
1. If ZFC and the Vopěnka scheme holds, then there is a class forcing extension, adding classes but no sets, in which GBC and the Vopěnka scheme holds, but the Vopěnka principle fails.
2. If ZFC and the Vopěnka scheme holds, then there is a class forcing extension, adding classes but no sets, in which GBC and the Vopěnka principle holds.
It follows that the Vopěnka principle VP and the Vopěnka scheme VS are not equivalent, but they are equiconsistent and indeed, they have the same first-order consequences.
Statement 1 is proved by class forcing to add a club class $C$ avoiding the regular cardinals. This destroys the assertion "Ord is Mahlo" and therefore destroys the Vopěnka principle, while preserving the Vopěnka scheme because it does not add sets.
Statement 2 is proved by class forcing of the global axiom of choice. The difficult part is to show that the Vopěnka principle holds true with respect the new classes definable from the generic filter. The proof involves the concept of a *stretchable* set $g\subset\kappa$ for an $A$-extendible cardinal, one which has the property that for every cardinal $\lambda<\kappa$ and every extension $h\subset\lambda$ with $h\cap\kappa=g$, there is an elementary embedding $j:\langle V\_\lambda,\in,A\cap V\_\lambda\rangle\to\langle V\_\theta,\in,A\cap V\_\theta\rangle$ such that $j(g)\cap\lambda=h$. Thus, the set $g$ can be stretched by an $A$-extendibility embedding so as to agree with any given $h$. This stretchability property is the $A$-extendibility analogue of the master condition technique in other large cardinal contexts.
**Corollaries.**
1. Over GBC, the Vopěnka principle and the Vopěnka scheme, if consistent, are not equivalent.
2. Nevertheless, the two Vopěnka axioms are equiconsistent over GBC.
3. Indeed, the two Vopěnka axioms have exactly the same first-order consequences in the language of set theory. Specifically, GBC plus the Vopěnka principle is conservative over ZFC plus the Vopěnka scheme for assertions in the first-order language of set theory. $$\text{GBC}+\text{VP}\vdash\phi\qquad\text{if and only if}\qquad\text{ZFC}+\text{VS}\vdash\phi$$
See the edit history for my 2010 answer.
| 21 | https://mathoverflow.net/users/1946 | 46538 | 29,440 |
https://mathoverflow.net/questions/44850 | 12 | A continuous representation $\hat{\mathbb{Z}} \rightarrow GL\_n(\mathbb{Q}\_p)$ is determined by the image of $1$. But the image of $1$ does not always defines such a representation (consider for example the representation which sends $1$ on $p$ from $\mathbb{Z}$ to $GL\_1(\mathbb{Q}\_p)$). So my question is : what are the conditions on the image of $1$ ?
For example if $n=1$, then I know that $1$ must be sent on an element of $\mathbb{Z}\_p^\times$, but I don't know if the converse is true.
EDIT: Correction about the example.
| https://mathoverflow.net/users/10427 | What are the $p$-adic representations of $\hat{\mathbb{Z}}$ ? | I am going to write a community wiki answer here which people can vote up.
(See [this meta thread](http://mathoverflow.tqft.net/discussion/780) concerning the Mathoverflow user,
which bumps questions with no voted-up answer.)
Main result: A homomorphism $f: \mathbb Z \to GL\_n(\mathbb Q\_p)$ extends continuously to
$\hat{\mathbb Z}$ if and only if the image of $f$ can be conjugated into $GL\_n(\mathbb Z\_p)$.
Proof:
If $f:\hat{\mathbb Z} \to GL\_n(\mathbb Q\_p)$ is continuous, the image is compact, hence contained
in a maximal compact subgroup, which can be conjugated into $GL\_n(\mathbb Z\_p)$.
Conversely, if $f:\mathbb Z \to GL\_n(\mathbb Q\_p)$ lands in a compact subgroup,
then the closure of the image is compact, hence profinite (any compact subgroup of $GL\_n(\mathbb Q\_p)$ is profinite), and hence $f$ extends to $\hat{\mathbb Z}$
(since $\hat{\mathbb Z}$ is precisely the profinite completion of $\mathbb Z$).
QED
As noted in the comments, to tell if a matrix (e.g. $f(1)$) can be conjugated into
$GL\_n(\mathbb Z\_p)$, one simply has to look at the characteristic polynomial,
and ask that all the coefficients lie in $\mathbb Z\_p$, with the constant term being
a unit. Thus to apply the theorem in practice, one simply computes the characteristic polynomial of $f(1)$ and see if its satisfies these conditions.
EDIT: Now actually made community wiki; sorry about that --- I thought I had already clicked the CW box,
but obviously not. (The point is that the above argument is just a rephrasing of what is in the comments.)
| 19 | https://mathoverflow.net/users/2874 | 46542 | 29,442 |
https://mathoverflow.net/questions/46534 | 8 | The Lovász $\vartheta$-function of a graph $G$, $\vartheta(G)$, is well-known to be "sandwiched" between the independence number of the graph, $\alpha(G)$, and the chromatic number of its complement, $\chi (\overline{G})$.
When $G$ is perfect then $$\alpha(G)=\vartheta(G)=\chi (\overline{G}).$$
I would like to know if there are graphs $G$, such that $$\alpha(G)< \vartheta(G)= \chi (\overline{G}).$$
| https://mathoverflow.net/users/nan | When the Lovász theta-function saturates its upper bound | Suppose $G$ is a $k$-regular graph on $n$ vertices, with least eigenvalue $\tau$.
Lovasz proved that
$$
\theta(G) \le \frac{n}{1-\frac{k}{\tau}}.
$$
Further if the automorphism group of $G$ acts arc-transitively, then equality holds.
In fact equality holds if G is a single class in a homogeneous coherent configuration,
for example, if $G$ is a strongly regular graph.
Class 1: Latin square graphs. Take the graph whose vertices are the $n^2$ cells of
and $n\times n$ Latin square, where two cells are adjacent if they are in the same row,
same column, or have the same contents. This graph is regular with valency $3(n-1)$
and least eigenvalue $-3$ and so $\theta(G)=n$. But if the Latin square is the multiplication
table of a cyclic group of even order then $\alpha(G) < n$ (because cyclic Latin squares
do not have transversals, but you can find a short proof on page 225 of one of my favorite
books on algebraic graph theory). The vertices in a given column form a clique of size
$n$ and so we see that $\chi(\bar{G})=n$ for any Latin square of order $n$.
Class 2: generalized quadrangles: A GQ with parameters $(s,t)$ is a collection of points
and lines satisfying some axioms, in particular each line contains exactly $s+1$ points.
Deem two points adjacent if they are collinear.
If $s,t>1$ this gives a strongly regular graph on $(s+1)(st+1)$ vertices with valency
$s(t+1)$ and least eigenvalue $-t-1$. Hence $\theta(G) = st+1$ and a coclique of size $st+1$
is known as an *ovoid*. The points on line forms a clique of size $s+1$, and a set
of lines that partition the point set is called a *spread*. Hence you want
GQ's with spreads but no ovoids.
For actual examples, I refer you to Payne and Thas's book on GQ's---the GQ's $Q(5,q)$ work.
| 14 | https://mathoverflow.net/users/1266 | 46547 | 29,446 |
https://mathoverflow.net/questions/45855 | 37 | In their famous 'Primes is in P' paper Agrawal, Kayal and Saxena stated the following conjecture:
>
> If for coprime integers $n$ and $r$ the equality $(X-1)^n = X^n - 1$ holds in $\mathbb{Z}\_n[X]/(X^r-1)$ then either $n$ is prime or $n^2 = 1 \pmod{r}$.
>
>
>
If true this would give a beautiful characterization of primes that could be easily transformed into a fast ($O(\log^{3+\epsilon}{n})$) and deterministic primality test.
Shortly after publishing 'Primes is in P' Hendrik Lenstra noticed that the conjecture may not be valid for $r=5$ and $n$ of the very special form (see [Lenstra's and Pomerance's note](http://www.aimath.org/WWN/primesinp/primesinp.pdf), p.30). It was unknown whether any such $n$ existed but Carl Pomerance gave a heuristic argument convincing that there should be infinitely many $n$'s sharing these, apparently rare, properties. I'm not aware of any strict proof for this.
It may also happen that the conjecture in a modified form (if we restrict $r$ to be greater than $\log{n}$) can be still true.
Martin Mačaj (see [Some remarks and questions about the AKS
algorithm and related conjecture](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.144.7953&rep=rep1&type=pdf)) gave another version of this conjecture together with a proof that relied on yet another unsolved problem.
Does anyone know if there were any advances in this area in the recent years?
| https://mathoverflow.net/users/10792 | What is the current status of Agrawal's conjecture? | I had some students look at this problem during an REU. A group proved that the conjecture is true if $r > n/2$, that's not too hard and can certainly be improved. Another group tried to find a counterexample using the computer for $r=5$, without success. I agree with Lenstra and Pomerance that the conjecture should be false in general.
Link to REU page: <http://www.ma.utexas.edu/users/voloch/reu.html>
| 19 | https://mathoverflow.net/users/2290 | 46564 | 29,457 |
https://mathoverflow.net/questions/46561 | 4 | I am currently a mathematics graduate student at Western Kentucky University in Bowling Green, KY. I am looking for some kind of summer opportunity to participate in during summer 2011.
Does anyone have any suggestions of good opportunities or a good list of opportunities?
I would really appreciate it!
EDIT: I am more interested in pure mathematics than I am in applied mathematics. My favorite areas are set theory, complex variables/analysis, topology, difference equations, difference(discrete) calculus, and time scale calculus.
I am really looking to broaden my horizons though with new and interesting topics.
Naturally, I am going to speak with my advisor and professors, but I would like to get insight from others as well.
| https://mathoverflow.net/users/10918 | Mathematics Graduate Student Summer Opportunities | NSA has a summer program. Should be somewhere at <http://www.nsa.gov>
| 2 | https://mathoverflow.net/users/2290 | 46565 | 29,458 |
https://mathoverflow.net/questions/46566 | 32 | The existence and uniqueness of algebraic closures is generally proven using Zorn's lemma. A quick Google search leads to a [1992 paper of Banaschewski](http://www3.interscience.wiley.com/journal/113463992/articletext?DOI=10.1002%2Fmalq.19920380136), which I don't have access to, asserting that the proof only requires the ultrafilter lemma. Questions:
* Is it known whether the two are equivalent in ZF?
* Would anyone like to give a quick sketch of the construction assuming the ultrafilter lemma? I dislike the usual construction and am looking for others.
| https://mathoverflow.net/users/290 | Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma? | Qiaochu, using the [link](http://consequences.emich.edu/file-source/htdocs/conseq.htm) I provided in my answer to [this question](https://mathoverflow.net/questions/45928/does-arzela-ascoli-require-choice), you find that this question is still open (or was, as of the mid 2000s, and I haven't heard of any recent results in this direction).
(According to the site's notation, the existence of algebraic closures is form 69, the ultrafilter theorem is form 14, uniqueness of the algebraic closure (in case they exist) is form 233; these numbers can be found by entering appropriate phrases in the last entry form in the page linked to above.)
It is known that uniqueness implies neither existence nor the ultrafilter theorem.
It is open whether existence implies uniqueness or the ultrafilter theorem, and also whether (existence and uniqueness) implies the ultrafilter theorem.
(Enter 14, 69, 233 in Table 1 in the link above for these implications/non-implications.)
Jech's book on the axiom of choice should provide the proofs of the known implications and references, and the book by Howard-Rubin (besides updates past the publication date of Jech's book) provides references for the known non-implications.
---
Here are some details on Banaschewski's paper:
`1.` First, lets see that the ultrafilter theorem can be used to prove **uniqueness** of algebraic closures, in case they exist.
Let $K$ be a field, and let $E$ and $F$ be algebraic closures. We need to show that there is an isomorphism from $E$ onto $F$ fixing $K$ (pointwise).
Following Banaschewski, denote by $E\_u$ (resp., $F\_u$) the splitting field of $u\in K[x]$ inside $E$ (resp., $F$); we are not requiring that $u$ be irreducible. We then have that if $u|v$ then $E\_u\subseteq E\_v$ and $F\_u\subseteq F\_v$. Also, since $E$ is an algebraic closure of $K$, we have $E=\bigcup\_u E\_u$, and similarly for $F$.
Denote by $H\_u$ the set of all isomorphisms from $E\_u$ onto $F\_u$ that fix $K$; it is standard that $H\_u$ is finite and non-empty (no choice is needed here). If $u|v$, let $\varphi\_{uv}:H\_v\to H\_u$ denote the restriction map; these maps are onto.
Now set $H=\prod\_{u\in K[x]} H\_u$ and for $v|w$, let
$$ H\_{vw}=\{(h\_u)\in H\mid h\_v=h\_w\upharpoonright E\_v\}. $$
Then the Ultrafilter theorem ensures that $H$ and the sets $H\_{vw}$ are non-empty. This is because, in fact, Tychonoff for compact Hausdorff spaces follows from the Ultrafilter theorem, see for example the exercises in Chapter 2 of Jech's "The axiom of choice." Also, the sets $H\_{vw}$ have the finite intersection property. They are closed in the product topology of $H$, where each $H\_u$ is discrete.
It then follows that the intersection of the $H\_{vw}$ is non-empty. But each $(h\_u)$ in this intersection determines a unique embedding $h:\bigcup\_uE\_u\to\bigcup\_u F\_u$, i.e., $h:E\to F$, which is onto and fixes $K$.
`2.` **Existence** follows from modifying Artin's classical proof.
For each monic $u\in K[x]$ of degree $n\ge 2$, consider $n$ "indeterminates" $z\_{u,1},\dots,z\_{u,n}$ (distinct from each other, and for different values of $u$), let $Z$ be the set of all these indeterminates, and consider the polynomial ring $K[Z]$.
Let $J$ be the ideal generated by all polynomials of the form
$$ a\_{n-k}-(-1)^k\sum\_{i\_1\lt\dots\lt i\_k}z\_{u,i\_1}\dots z\_{u,i\_k} $$
for all $u=a\_0+a\_1x+\dots+a\_{n-1}x^{n-1}+x^n$ and all $k$ with $1\le k\le n$.
The point is that any polynomial has a splitting field over $K$, and so for any finitely many polynomials there is a (finite) extension of $K$ where all admit zeroes. From this it follows by classical (and choice-free) arguments that $J$ is a proper ideal.
We can then invoke the ultrafilter theorem, and let $P$ be any prime ideal extending $J$. Then $K[Z]/P$ is an integral domain. Its field of quotients $\hat K$ is an extension of $K$, and we can verify that in fact, it is an algebraic closure. This requires to note that, obviously, $\hat K/K$ is algebraic, and that, by definition of $J$, every non-constant polynomial in $K[x]$ split into linear factors in $\hat K$. But this suffices to ensure that $\hat K$ is algebraically closed by classical arguments (see for example Theorem 8.1 in Garling's "A course in Galois theory").
`3.` The paper closes with an observation that is worth making: It follows from the ultrafilter theorem, and it is strictly weaker than it, that countable unions of finite sets are countable. This suffices to prove uniqueness of algebraic closures of countable fields, in particular, to prove the uniqueness of $\bar{\mathbb Q}$.
| 26 | https://mathoverflow.net/users/6085 | 46568 | 29,459 |
https://mathoverflow.net/questions/46541 | 70 | In the coming spring semester I will be teaching for the first time an introductory (graduate) course in Commutative Algebra. As many people know, I have been plugging away for a while at this subject and keeping my own lecture notes. So I feel relatively prepared to teach the course in the sense that I know more than enough theorems and proofs to cover. However, there is more to teaching a graduate course than just theorems and proofs. Commutative algebra has a reputation for being somewhat dry and unmotivated. After something like 10 years of hard work (over a period of about 15 years!), I am at a point where I find the subject both interesting in and of itself and useful. But how to communicate this to students?
Looking through my notes, the technicalities begin with the introduction of various fundamental classes of modules, especially free, projective and flat modules (but also including other things like finite generation and finite presentation). I remember well that when I first learned this material, projective and (especially) flat modules were a tough sell: for instance, the first time I picked up Bourbaki's *Commutative Algebra* out of curiosity, I saw that the very first chapter was on flat modules, and I put it down in horror.
How would you introduce these concepts to an early career graduate student? What are the fundamental differences between these classes of modules, and why do we care?
Here are some of my preliminary thoughts:
1) Free modules are of course an easy sell: for such things the usual notions of linear algebra work well, including that of dimension (called "rank" in this case: note that my rings are commutative!). One can easily prove that for a commutative ring R, all R-modules are free iff R is a field, so the need to move beyond free modules is easily linked to the ideal theory of R.
2) Projective modules are important for at least the following reasons.
a) Geometric: A finitely generated module over a ring R is projective iff it is locally free (in the stronger sense of an open cover of $\operatorname{Spec} R$). In other words, projective modules are the way to express **vector bundles** in algebraic language. I plan to drive this point home by discussing Swan's theorem on modules over $C^{\infty}(M)$.
b) Homological: projective modules play a distinguished role in homological algebra, i.e., in the construction of left-derived functors.
c) K-theoretic: It is of interest to know to what extent finitely generated projective modules must be free. This leads to the construction of $K\_0(R)$ -- i.e., stable isomorphism classes of projective modules -- and in the rank one case, to the Picard group $\operatorname{Pic}(R)$.
(After writing the above, I feel that in some sense it is a good enough answer -- certainly these are three important reasons for studying projective modules. But I'm not sure how to explain them to beginning students. Let's be clear that this is part of the question.)
3) Flat modules: this is harder to explain!
a) Geometric: Flatness is the "right" condition for things to vary nicely in families, but this is more of a mantra than an explanation. I think that most people hear this at one point and come to believe and understand it slowly over time.
b) Homological: Flat modules are those which are acyclic for the Tor functors. But this is not a homological algebra course: I would be happiest not to mention Tor at all.
c) Near equivalence with projective modules in the finitely generated case: the difference between finitely generated flat and finitely generated projective modules is very subtle. Recall for instance:
Theorem: For a finitely generated flat module $M$ over a commutative ring $R$, TFAE:
(i) $M$ is projective.
(ii) $M$ is finitely presented.
(iii) The associated rank function is locally constant.
Thus in almost all of the cases of importance to a beginning algebra student -- e.g. if $R$ is Noetherian or is an integral domain -- finitely generated flat modules are projective.
d) However infinitely generated flat modules are a much bigger class, since flatness is preserved under direct limits. In particular, there are large classes of domains $R$ for which a module is flat iff it is torsionfree (namely this holds iff $R$ is a Prufer domain; even the case $R = \mathbb{Z}$ is useful). Perhaps this is significant?
Your insight will be much appreciated.
---
**Update**: the prevailing sentiment of the answers thus far seems to be that a very little bit of homological algebra will go a long way in presenting the basic definitions in a unified and useful way. Duly noted.
| https://mathoverflow.net/users/1149 | How to introduce notions of flat, projective and free modules? | Hi Pete, this sounds like a lot of fun! I wish I could be there (-:
Here is a concrete and useful property of flatness, you can explain it without using Tor. Suppose $R\to S$ is a flat extension.
Then if $I$ is an ideal of $R$, tensoring the exact sequence:
$$ 0 \to I \to R \to R/I \to 0$$
with $S$ gives that $I\otimes\_RS = IS$. The left hand side is somewhat abstract object, but the right hand side is very concrete.
There are very natural extensions which are flat but not projective. For example, if $R$ is Noetherian and $\dim R>0$, then $S=R[[X]]$ is flat but [never projective](http://arxiv.org/abs/math/0509180) over $R$.
| 21 | https://mathoverflow.net/users/2083 | 46570 | 29,461 |
https://mathoverflow.net/questions/46291 | 3 | Let $\bar{\mathcal{P}}$ denote the closed $n$-dimensional convex polytope subtended by the origin and the lattice points {$b\_{i} \textbf{e} \_ {i}$}, where {$\mathbf{e}\_{i}$} is the standard basis of $\mathbb{R}^{n}$. Define the Ehrhart function $L \_{\bar{\mathcal{P}}}(t) = | t \bar{\mathcal{P}} \cap \mathbb{Z} ^{n}|$, where $t \bar{ \mathcal{P} }$ denotes the $t$-dilate of $\bar{ \mathcal{P} }$.
It is known that $L\_{\bar{ \mathcal{P} } }(t)$ is a polynomial of degree $n$ in $t \in \mathbb{N}$ if $\bar{\mathcal{P}}$ is integral, i.e., {$b\_{i}$} are positive integers. My question is about the meaning of $L\_{\bar{ \mathcal{P} } }(t)$ when $t$ is *rational*.
**Question**: Suppose I'd like to calculate the number of non-negative integer solutions of
\begin{eqnarray}
\frac{x\_1}{b\_1} + \cdots + \frac{x\_n}{b\_n} \leq t
\end{eqnarray}
for some positive rational $t$. Is there a way to compute this number from the aforementioned Ehrhart polynomial (via some interpolation method, etc.) or is some other machinery necessary?
Thanks in advance!
| https://mathoverflow.net/users/10280 | Rational dilates of integral convex polytopes | R. Diaz, S. Robins, and I studied your question for the inequality $b\_1 x\_1 + \dots + b\_d x\_d \le t$ for integral $t$ (which gives a rational polytope) in
* *The Frobenius problem, rational polytopes, and Fourier-Dedekind Sums*, Journal of Number Theory 96 (2002), 1–21, doi:[10.1006/jnth.2002.2786](https://doi.org/10.1006/jnth.2002.2786), arXiv:[math.NT/0204035](http://arxiv.org/abs/math.NT/0204035).
The case where $t$ is truly a rational variable is more complicated. A starting point is
* Eva Linke, *Rational Ehrhart quasi-polynomials*, Journal of Combinatorial Theory, Series A **118** Issue 7 (2011) 1966-1978, doi:[10.1016/j.jcta.2011.03.007](https://doi.org/10.1016/j.jcta.2011.03.007), arXiv:[1006.5612](http://arxiv.org/abs/1006.5612)
| 5 | https://mathoverflow.net/users/3193 | 46595 | 29,478 |
https://mathoverflow.net/questions/46387 | 3 | (Sorry I'm outsider in this field.)
I need to count the number of integral points in a convex polytope in $\mathbf{R}^3$. The cones in the dual fan are not necessarily regular (does it create any problem?)
So, I need the $c\_1$ coefficient of the Ehrhart polynomial. There're some formulas (complicated enough for me) but I heard one can express $c\_1$ as the sum (over all the edges of the polytope) of the integral lengths of the edges times some correction factors.
1. Can someone give the formula? (In the simple English, please.) Or a reference to something very down-to-earth?
2. In fact I do not need the precise expression but only a very good lower bound. Does something like this exist?
| https://mathoverflow.net/users/2900 | Counting integral points of a polytope in R^3 (the c_1 coefficient of Ehrhart polynomial) | Jamie Pommersheim gave a general formula for $c\_1$ in his 1993 paper in Math. Ann. for the case of a tetrahedron. So if you can easily triangulate your polytope, this might be useful.
If you're "only" interested in bounds, you might want to look at the Ehrhart series instead, i.e., the generating function of the Ehrhart polynomial. This is a rational function with denominator (1-x)^4, so the information encoded in the numerator is the same information encoded in the Ehrhart polynomial (in fact, it's a linear transformation). A famous theorem of Stanley states that the numerator coefficients of the Ehrhart series are nonnegative, and of course this gives you inequalities among the coefficients of the Ehrhart polynomial. For more inequalities of this type, check out Alan Stapledon's work (e.g., <https://arxiv.org/abs/0904.3035> or <https://arxiv.org/abs/0801.0873>).
(And thanks for your kind words, Andres.)
| 10 | https://mathoverflow.net/users/3193 | 46596 | 29,479 |
https://mathoverflow.net/questions/46597 | 15 | I am under the impression that in the definition of the Grothendieck group $K\_0(R)$ of a (non-commutative) ring it doesn't matter whether we apply the usual $K\_0$ construction to the exact category of all finitely generated projective *left* $R$-modules, or if we apply the construction to the category of all finitely generated projective *right* $R$-modules.
Is this obvious? To be honest my first reaction is disbelief but given that certain other notions for rings (such as semisimplicity of a ring or Morita equivalence for rings) are left/right-independent, I guess I can believe it. But it is not clear to me why this is true in the case of $K\_0$.
I've taken a look in the standard references for classical algebraic $K$-theory but none of them seem to mention this point and either deal just with left $R$-modules throughout, or make no mention at all of the handedness of their modules.
This circle of ideas leads to a more general and vague follow up question: Does anyone have any intuition for *why* certain notions for non-commutative rings (such as the examples mentioned above: semisimplicity, Morita equivalence, $K\_0$) do not depend on whether we look at properties of the collection of left modules over the ring or whether we look at properties of the collection of right modules over the ring.
| https://mathoverflow.net/users/1148 | Why does the Grothendieck group $K_0(R)$ of a ring not depend on our choice of using left modules instead of right modules? | First of all, right $R$-modules are the same as left $R^{op}$-modules. Hence you are asking whether the $K$-theory changes if you pass form $R$ to $R^{op}$. The answer is: It does not change.
The reason for the independence is that one has a very concrete picture.
$$K\_0(R) = Gr(V(R))$$
where $Gr$ denotes the Grothendieck group, and $V(R)$ denotes the monoid of conjugacy classes of idempotents in $\cup\_{n \in \mathbb N} M\_n(R)$. Clearly, neither the notion of *idempotent* nor *conjugacy class* depends on the order in which your multiplication is performed. Hence, the $K$-theories of $R$ and $R^{op}$ are canonically isomorphic.
The argument for $K\_1$ is even easier, since $K\_1$ is defined as the abelianization of $GL\_{\infty}(R)$.
| 17 | https://mathoverflow.net/users/8176 | 46598 | 29,480 |
https://mathoverflow.net/questions/46574 | 5 | Does there exist a field $k$ and a subring $R$ of $S = M\_2(k)$ such that $R$ is not finitely generated over its center, $S=kR$ and $1\_R = 1\_S$? ($S$ is the algebra of $2 \times 2$ matrices over $k$.)
| https://mathoverflow.net/users/10943 | M_2(k) as a central extension | I think the answer is "yes". Let $A$ be a non-Noetherian integral domain (for example a polynomial ring in infinitely many variables over a field), let $I$ denote a non-finitely-generated ideal, and let $k$ be the field of fractions of $A$. Let $R$ denote the ring of $2\times 2$ matrices with coefficients in $A$ and with bottom left hand entry in $I$.
I think this ticks all the boxes. For example $kR=M\_2(k)$ because I can scale any element of $M\_2(k)$ until it's in $M\_2(R)$ and then again so that all entries are in $I$.
However, I don't think $R$ can be finitely-generated over its centre (which is easily checked to be $A$). For if $r\_1,r\_2,\ldots,r\_n$ are finitely many elements of $R$ then the ring they generate over $A$ will be contained in the $2\times 2$ matrices with coefficients in $A$ and bottom left hand entry in $J$, the finitely-generated ideal generated by the bottom left hand entries of the $r\_i$, and this is a proper subset of $I$.
| 7 | https://mathoverflow.net/users/1384 | 46600 | 29,481 |
https://mathoverflow.net/questions/46576 | 8 | Write $M\_n = S^n \cup\_2 D^{n+1}$. I know, as a matter of folklore, that the identity map $\mathrm{id}\_M$, considered as an element of the group $[M\_n, M\_n]$, has order $4$ (for $n > 3$, let's say).
What is a reference for this? And is there a simple and pretty argument?
| https://mathoverflow.net/users/3634 | Order of the identity map of a Moore space. | Alternatively: if the order were $2$ then $M\_n\wedge M\_n$ would be $M\_{2n}\vee M\_{2n+1}$. The mod $2$ cohomology of $M\_n$ has a generator $a$ in degree $n$ and a generator $b=Sq^1(a)$ in degree $n+1$ and nothing else. It follows that the cohomology of $M\_n\wedge M\_n$ has generators $a\otimes a$, $a\otimes b$, $b\otimes a$ and $b\otimes b$ and we find using the Cartan formula that $Sq^2(a\otimes a)=b\otimes b$. However, $Sq^2$ is zero on the cohomology of $M\_{2n}\vee M\_{2n+1}$.
| 5 | https://mathoverflow.net/users/10366 | 46603 | 29,484 |
https://mathoverflow.net/questions/46577 | 13 | If $\Gamma=\Gamma\_1(N)$, or $\Gamma=\Gamma\_0(N)$, the Hecke operator $[\Gamma diag(1,l) \Gamma]$ for $l$ a prime (acting on the space of cusp forms of level $\Gamma$ and some weight $k$) is in general denoted $T\_l$ when $l$ does not divide $N$, but $U\_l$ otherwise. It is well-known that the Hecke operators $T\_l$ are normal (that is commute with their adjoint) for the Petersson's inner product, but not the $U\_l$. Or are they ?
Is there any value of the level $N$, the weight $k$, and a prime $l$ dividing $N$ such that $U\_l$ is normal ?
Of course, this would imply that $U\_l$ is diagonalizable, but this is conjectured (known if $k=2$) to happen if $l^3$ does not divide $N$ (cf. Coleman-Edixhoven, Math. Annalen, 1998).
This question is surely easy, but my problem is that I can't really compute the adjoint of $U\_l$. Of course, it is the Hecke operator $\Gamma diag(l,1) \Gamma$ and it's not hard to write this double class as a union of simple class, but the formulas I get look awful.
For example consider this simplest case. Let $f$ be form of weight $k$ and level $1$, normalized eigenform for all $T\_l$. Choose a prime $p$. As it is well known,
the space of form of level $\Gamma\_0(p)$ with same eigenvalues as $f$ for all the $T\_l$ with $l \neq p$ is two dimensional, generated by $f(z)$ and $f(pz)$, or preferably
generated by $f\_\alpha(z) = f(z) - \beta f(pz)$ and $f\_\beta(z) = f(z)-\alpha f(pz)$ where $\alpha$ and $\beta$ are the two roots of $X^2-a\_pX + p^{k-1}$, with $T\_p f = a\_p f$ (I suppose $\alpha \neq \beta$, which is always conjectured and often known).
The interest of this basis is that $U\_p$ is diagonal in it: $U\_p f\_\alpha = \alpha f\_\alpha$ and $U\_p f\_\beta = \beta f\_\beta$
Can you compute the matrix of the adjoint of $U\_p$ in the basis $f\_\alpha$, $f\_\beta$?
| https://mathoverflow.net/users/9317 | Adjoint of Atkin-Lehner's U_p | The following is essentially taken from Miyake's "On Automorphic Forms on GL2 and Hecke Operators", with some help interpreting notation by looking at Gelbart's "Automorphic forms on adele groups".
From the point of view of representation theory, you can define Hecke operators at any prime (well, up to a normalizing factor) by the integral operator
$$(T\_pf)(g)=\int\_{K\_p(N)}\omega^{-1}(k)f\Big(g\cdot k\cdot \Big(\matrix{p&0\cr 0&1}\Big)\Big)\ dk$$
where $f$ is your cusp form, $K\_p(N)$ is the p-adic version of $\Gamma\_0(N)$ (it is the largest subgroup under which f is invariant under right translation), and $\omega$ is the nebentypus (central character) of f. This is equivalent to convolution in $L^2(Z\_p\backslash G\_p,\omega)$ with the characteristic function of $K\_p(N) \Big(\matrix{p&0\cr 0&1}\Big) K\_p(N)$.
Now the adjoint of "convolution with a function $\phi(g)$" on a Hilbert space is convolution with $\overline {\phi(g^{-1})}$, which in our case turns out to be
$$(T\_p^\*f)(g)=\int\_{K\_p(N)}\omega^{-1}(k)f\Big(g\cdot k\cdot \Big(\matrix{p^{-1}&0\cr 0&1}\Big)\Big)\ dk$$
Since it is possible to write the integral operator version of $T\_p$ in classical terms, you should be able do the same with the adjoint . . .
Miyake proves (he says he is generalizing Hecke and Atkin-Lehner) that the largest subspace of $S\_k(N,\omega)$ on which the algebra generated by all the Hecke operators and their adjoints act commutatively (and normality is the only obstruction to this) is the space of newforms. This implies that whenever there are oldforms, there is at least one prime for which the Hecke operator is not normal. In particular, in your simplest case of prime level $p$, $U\_p$ is not normal. (Though, I guess, since $p^3$ does not divide $p$, it is diagonalizable)
Two papers I came across: Li, "Diagonalizing modular forms" adjusts some $U\_p$ to get a family of Hecke operators at all $p$ which diagonalize the whole space of modular forms. Choie and Kohnen, "Diagonalizing 'bad' Hecke operators, etc" shows that the $U\_p$ are almost always diagonalizable on $S\_k(pN)$, for $N$ square-free, prime to $p$.
To summarize: 1) you can write down the adjoint if you work adelically, 2) normality basically fails, 3) diagonalizability often succeeds.
Disclaimer: I don't spend much time working with classical modular forms: until tonight, I didn't know how the $U\_p$ were defined, and I think this might be the first time I've typed "nebentypus".
UPDATE: Further googling led me to [this page](http://maths.dept.shef.ac.uk/magic/course.php?id=48) of power point slides by Jens Funke. In lecture 19, he claims that the adjoint of $U\_n$ is $n^kV\_n$, where $V\_ng(z)=g(nz)$, though this is left "as an exercise".
| 6 | https://mathoverflow.net/users/6753 | 46606 | 29,486 |
https://mathoverflow.net/questions/46605 | 1 | randomized SVD decomposes a matrix by extracting the first k singular values/vectors using k+p random projections.
my question concerns the singular values that are output from the algorithm. why aren't the values equal to the first k-singular values if you do the full SVD?
Below I have a simple implementation in R.
```
rsvd = function(A, k=10, p=5) {
n = nrow(A)
y = A %*% matrix(rnorm(n * (k+p)), nrow=n)
q = qr.Q(qr(y))
b = t(q) %*% A
svd = svd(b)
list(u=q %*% svd$u, d=svd$d, v=svd$v)
}
> set.seed(10)
> A <- matrix(rnorm(500*500),500,500)
> svd(A)$d[1:15]
[1] 44.94307 44.48235 43.78984 43.44626 43.27146 43.15066 42.79720 42.54440 42.27439 42.21873 41.79763 41.51349 41.48338 41.35024 41.18068
> rsvd.o(A,10,5)$d
[1] 34.83741 33.83411 33.09522 32.65761 32.34326 31.80868 31.38253 30.96395 30.79063 30.34387 30.04538 29.56061 29.24128 29.12612 27.61804
```
| https://mathoverflow.net/users/10950 | randomized SVD singular values | They should be approximation to the true singular values (with suitable hypotheses, with good probability...).
Intuitively, it is like trying to infer the singular values of a $n\times n$ matrix by looking at its leading $(k+p)\times (k+p)$ submatrix only and replacing the rest with zeros: it is cheaper to compute, but yields only an approximate result.
Where does the "randomized" part come into play? Well, depending on the specific matrix, the $(k+p)\times (k+p)$ submatrix may not be representative of the whole matrix: so, we conjugate everything with a random orthogonal matrix $Q$ that mixes up everything and ensures that there is "nothing special" regarding the first $k+p$ entries with respect to the rest.
| 4 | https://mathoverflow.net/users/1898 | 46612 | 29,490 |
https://mathoverflow.net/questions/46505 | 28 | What do I do if I have to solve the usual quadratic equation $X^2+bX+c=0$ where $b,c$ are in a field of characteristic 2? As pointed in the comments, it can be reduced to $X^2+X+c=0$ with $c\neq 0$.
Usual completion of square breaks. For a finite field there is [Chen Formula](http://www.ams.org/mathscinet-getitem?mr=680147) that roughly looks like $X=\sum\_{m} c^{4^m}$. I am more interested in the local field $F((z))$ or actually an arbitrary field of characteristic 2.
| https://mathoverflow.net/users/5301 | How to solve a quadratic equation in characteristic 2 ? | I think this solves $X^2+X+c=0$ over $F((t))$:
I want to assume that $c\in F[[t]]$. If not, say $c=at^{-m}+...$, then the quadratic has no solutions when $m$ is odd or $a$ is not a square, and otherwise the substitution $X\mapsto X+\sqrt{a}t^{-m/2}$ gives a new equation with smaller $m$. So, after finitely many steps $c=c\_0+c\_1t+...$ is integral.
Because $X^2+X+c$ has derivative $1$, by Hensel's lemma the equation has a solution if and only the constant term $c\_0$ is of the form $d^2+d$ for some $d$ in $F$. And if it is, Hensel's approximations are obtained by starting with an approximate solution $x\_0=d$ and recursively computing $x\_{m+1}=x\_m-f(x\_m)/f'(x\_m)=x\_m^2+c$. This gives
$$
x = d + \sum\_{n=0}^\infty (c-c\_0)^{2^n}
$$
as the solution (the partial sums are the $x\_m$). Actually, the approach seems to work over any complete field, reducing the problem to the residue field. Hope this helps.
| 17 | https://mathoverflow.net/users/3132 | 46616 | 29,492 |
https://mathoverflow.net/questions/46553 | 9 | Which would be the most efficient way (in computational time) to compute tr(inv(H)), where H is a (dense) square matrix?
In my particular problem I also have a LU decomposition of H already available, which was used in a previous context to solve a system of linear equations. My current approach is to simply use the LU to compute the inverse and then calculate the trace. Is there any other more efficient way to achieve this, considering I already have this matrix factorized?
I could have first computed the inverse and then made a multiplication by the inverse to solve my previous system of linear equations, but I was trying to avoid multiplication by the inverse to avoid numerical inaccuracies.
Edit: Forgot to mention that under normal conditions H should be symmetric.
| https://mathoverflow.net/users/4430 | Fast trace of inverse of a square matrix | Given that the poster has specified that his matrix is symmetric, I offer a general solution and a special case:
1. Eigendecomposition actually becomes more attractive here: the bulk of the work is in reducing the symmetric matrix to tridiagonal form, and finding the eigenvalues of a tridiagonal matrix is an O(n) process. Assuming that the symmetric matrix is nonsingular, summing the reciprocals of the eigenvalues nets you the trace of the inverse.
2. If the matrix is positive definite as well, first perform a Cholesky decomposition. Then there are [methods](http://books.google.com/books?id=myzIPBwyBbcC&pg=PA119) for generating the diagonal elements of the inverse.
| 7 | https://mathoverflow.net/users/7934 | 46620 | 29,495 |
https://mathoverflow.net/questions/46625 | 4 | Is there an infinite (finite degree) transitive amenable hyperbolic graph ?
| https://mathoverflow.net/users/10957 | hyperbolic amenable graph | I think the answer is that every such graph should be quasi-isometric to the infinite line. Look at [this paper.](http://arxiv.org/PS_cache/math/pdf/9806/9806129v1.pdf) Theorem 9 there states that even a quasi-transivite amenable graph should have trivial Poisson boundary (no non-constant harmonic functions). On the other hand, a hyperbolic transitive graph, if it is not quasi-isometric to the Cayley graph $\mathbb Z$, always has non-trivial Poisson boundary. [Here](http://www.cmat.edu.uy/~lessa/tesis/Kaimanovich%20-%20groups%20with%20hyperbolic.pdf) it is proved for groups, but I think it works for transitive hyperbolic graphs as well.
**Edit.** As R W says, the first reference in my answer is not relevant. It is nice that the statement is still true, and that it can be proved using Poisson boundaries.
| 1 | https://mathoverflow.net/users/nan | 46629 | 29,499 |
https://mathoverflow.net/questions/46628 | 5 | Let $\Omega\subset \mathbb{R}^d$ be an open and bounded domain with Lipschitz smooth boundary. Let $\delta>0$ and
$
\Omega\_\delta = \{ x\in\Omega : \inf\_{y\in\partial\Omega} \left\|x-y\right\|\_{L\_2(\Omega)}\geq \delta \}
$
Is there a $\hat{\delta}>0$ such that $\forall 0<\delta<\hat{\delta}$ the space $\Omega\_{\delta}$ has a Lipschitz smooth boundary?
This statement seems like it should be true. I would really appreciate some help.
Thanks in advance.
| https://mathoverflow.net/users/2011 | Shrinking a Lipschitz smooth domain. | My answer is **Yes**. Of course, I presume that you assume $\Omega$ is on one side only of its boundary.
The boundary $\partial\Omega$ is compact. By assumptions, it has an atlas with finitely many charts, each one corresponding to a piece $\Gamma\_j$ which is the graph of a Lipschitz function: in appropriate coordinates, $\Gamma\_j$ is given by $x\_d=\phi\_j(x\_1,\ldots,x\_{d-1})$.
Let us choose $\delta$ smaller than one tenth of
$$\delta\_0:=\min\_{x\in\bar\Omega}\\,\max\_j\{d(x,\partial\Gamma\_j)\}>0$$
Let $\bar x$ be on the boundary of $\Omega\_\delta$, not on $\partial\Omega$. Let $j$ be such that $B(\bar x;9\delta)\cap\partial\Omega\subset\Gamma\_j$. There exists a point $\bar y\in\partial\Gamma\_j$ such that $d(\bar x,\bar y)=\delta$. Wlog, we have $\bar y=0$ and the graph is locally $x\_d=\phi\_j(x\_1,\ldots,x\_{d-1})$. Let $M$ be the Lipschitz constant of $\phi\_j$. Locally, the boundary of $\Omega\_\delta$ is the upper envellop of the balls $B(y;\delta)$ with $y\in \Gamma\_j$; as a matter of fact, for $x\in B(\bar x;\delta)$, the closest points of $\partial\Omega$ to $x$ belong to $\Gamma\_j$. It is easy to see that it is the graph of a function $\phi\_j^\delta$ whose Lipschitz constant is at most $M$.
| 4 | https://mathoverflow.net/users/8799 | 46631 | 29,501 |
https://mathoverflow.net/questions/46567 | 11 | I'm wondering if a classification of analytic functions, $f\,$ (it may be that $C^1$ is enough, but I'm not taking any chances, if you have a reason why I only need to consider a larger class of functions, I would enjoy that as well) with the property $F(x):=\int f(x) dx = \mathcal{O}(xf(x))$ as $x\to\infty$ AND $xf(x)=\mathcal{O}(F(x))$ where $\mathcal{O}$ is big-O notation. I'm not sure if it's standard or not, but I'll denote this condition by
$$\mathcal{O}(F(x))=\mathcal{O}(xf(x))\qquad (\*)$$
I'm motivated by the naïve notion of integration from the mistake many first semester students in calculus make when trying to take anti-derivatives and keep on thinking the same thing over and over: the power rule.
It is straightforward to check that $f(x)=x^n\quad n\ne -1$ and $f(x)=\log^n x\quad n\ge 0$ satisfy the condition $(\*)$, the first just by checking and the second by induction on $n$ and integration by parts. It's also easy to see that this is not the case for $x^ne^x,\; n\in\mathbb{Z}$ again by integration by parts.
A preliminary investigation yields some interesting first starts:
1) A valid refomulation of the problem is ${F\over f}$ has a slant asymptote, i.e. ${F\over f}=kx+\mathcal{o}(1)$, and if you know that the derivative of this little o function is also little o of 1, and say $k=1$ then you can rephrase this as $1-{d\over dx}(\log f(x))\cdot {F\over f}(x)=1+\mathcal{o}(1)$
2) If the function is increasing we can get half of this inequality since $F(x)\le xf(x)$ since $F(x)=\int\_a^xf(x)$, but the other half fails.
3) A convex $\mathcal{o}(1)$ in (1) might imply that the derivative is also $\mathcal{o}(1)$
| https://mathoverflow.net/users/4701 | Functions whose antiderivative behaves like xf(x) | Note: This is a major rewrite of my earlier answer, to include necessary and sufficient conditions applicable to an even wider class of functions.
Instead of expanding to the class of all analytic functions (where the asymptotics can be hard to get control over, due to oscillatory behavior), my inclination would be to focus on large classes of functions with well-behaved asymptotics, including all the functions that arise in ordinary asymptotic analysis. The usual buzz phrase for this is "Hardy field" (mentioned for example in my answer [here](https://mathoverflow.net/questions/45284/examples-of-sequences-whose-asymptotics-cant-be-described-by-elementary-function/45368#45368)), which by definition is an ordered field of germs at infinity of $C^\infty$ functions.
I will describe several classes of such functions. The first is the class of all functions which are first-order definable in the structure $(\mathbb{R}, +, \cdot, <, \exp)$ together with all real numbers adjoined as constants. This class contains all functions that are constructible from polynomials, $\exp$, $\log$ and closed under the usual arithmetic operations and composition. It thus contains all the functions that usually arise in asymptotic analysis, and many more besides. This class enjoys the following strong model-theoretic property (as developed more fully in the [theory of o-minimal structures](http://en.wikipedia.org/wiki/O-minimal_theory)):
>
> (O) The zero set of any function $F: [a, \infty) \to \mathbb{R}$ in this class is a finite union of points and intervals (finite or infinite in extent).
>
>
>
Condition O ensures that every such function $F$ is either eventually positive ($F(x) > 0$ for all sufficiently large $x$), eventually zero, or eventually negative. As a result, the ring of germs at infinity of the definable functions in this class forms an ordered field, i.e., is a Hardy field.
Also, if $F$ is definable, then $F'$ is also first-order definable (and its domain can be shown to be the domain of $F$ save for finitely many points). Applying condition O to $F'$, every definable $F$ in this class is either eventually increasing, eventually decreasing, or eventually constant.
**Proposition:** A function $F$ in this class satisfies $F(x) = O(xF'(x))$ and $xF'(x) = O(F(x))$ if and only if there exist $n, N$, both positive or both negative, for which $x^n < |F(x)| < x^N$ for all sufficiently large $x$.
**Proof:** WLOG we may assume $F$ is eventually positive, and is not eventually constant. Thus $F$ is either eventually increasing or eventually decreasing, say eventually increasing. If $F$ is eventually bounded above by some $x^N$, $N > 0$, then by increasing $N$ if necessary we may assume $F(x)/x^N$ tends to zero, whence it is eventually decreasing. Taking the derivative, we conclude that eventually
$$x^N F'(x) - Nx^{N-1}F(x) < 0$$
whence $xF'(x) < NF(x)$, i.e., $|xF'(x)| < N|F(x)|$ or $xF'(x) = O(F(x))$. However, if $F(x)$ is eventually bounded above by *every* positive-power function $x^N$ (think $N$ small!), this also shows $xF'(x) = o(F(x))$, so that $F(x)$ is not $O(xF'(x))$.
By a similar argument, if $F(x)$ is eventually bounded below by some positive-power function $x^n$, we get $nF(x) < xF'(x)$ eventually, so that $F(x) = O(xF'(x))$. This also shows that if $F(x)$ is bounded below by *every* positive-power function (think $n$ large!), then $F(x) = o(xF'(x))$, so $xF'(x)$ is not $O(F(x))$.
Thus, if $F(x)$ is positive and eventually increasing, a necessary and sufficient condition that $F'$ satisfy condition ($\ast$) in the question is that there exist two positive-power functions that $F$ is eventually squeezed between. An entirely similar analysis shows that if $F(x)$ is positive and eventually decreasing, a necessary and sufficient condition that $F'$ satisfy condition ($\ast$) is that there exist two negative-power functions that $F$ is eventually squeezed between. Thus the proposition is proved.
---
The almost freshman-level triviality of this proof testifies to the great power of condition O (which is a special case of the o-minimality axiom), from which all flows.
Thus it is of interest to know of classes of functions which satisfy it. I will mention an extraordinary result in this regard, due largely to Patrick Speissegger (The Pfaffian closure of an o-minimal structure, J. Reine Angew. Math. 508 (1999), 189--211):
There is an o-minimal expansion of the ordered exponential field $\mathbb{R}$ (thus, including the class of functions described above) so large that
* The structure includes the restriction of any analytic function to a compact box,
* If $f: [a, \infty) \to \mathbb{R}$ is first-order definable within this structure, then so is any antiderivative $F$ (even though general antiderivatives are not definable by a first-order construction),
This may fit better with Adam Hughes's formulation in terms of antiderivatives. Since condition O is satisfied (according to the more general o-minimality condition), the same analysis as above applies.
| 13 | https://mathoverflow.net/users/2926 | 46635 | 29,504 |
https://mathoverflow.net/questions/46643 | 8 | In the book "profinite groups, arithmetic, and geometry" of Shatz, the index $(G:H)$ of a closed subgroup $H$ of a profinite group $G$ is defined to be the supernatural number $lcm\big((G/U):(H/(H\cap U))\big)$ where $U$ runs over the open normal subgroups of $G$. There is an exercise following this definition saying that "$(G:H)=lcm(G:U)$ where $U$ runs over those open normal subgroups of $G$ containing $H$.
If $G$ is a finite group with discrete topology, then the index given is nothing but the number of elements in the coset space $G/H$. However, if we take $G$ to be a finite simple group having a non-trivial proper subgroup $H$, e.g. $Alt\_n$ for a suitable $n$, the only normal subgroup contating $H$ is $G$ itself and $\big((G/G):(H/(H\cap G))\big)=1$.
I am not sure if the claim in the exercise is true for infinite profinite groups as they are necessarily non-simple, which means they don't admit trivial counter-examples. But at least the exercise seemed me wrong for finite case. Am I missing something, or this is a well-known misprint which I don't know?
| https://mathoverflow.net/users/10469 | index of a closed subgroup of a profinite group | (The first time around I had read your question too quickly and not properly appreciated it. Sorry about that.)
You are right: the exercise on p. 12 of Shatz's book is false, because of the example you suggest. You asked if there were also counterexamples among infinite profinite groups. Certainly: let $n \geq 5$, let $p$ be a prime number greater than $n$, and consider $G = \mathbb{Z}\_p \times A\_n$. Then the problem persists: take $H = \mathbb{Z}\_p \times H'$, where $H'$ is a proper nontrivial subgroup of $A\_n$. (Use [Goursat's Lemma](http://en.wikipedia.org/wiki/Goursat_lemma).)
I checked that this exercise does not appear in Serre's *Galois Cohomology*. Have you found that it is used at any point of Shatz's book?
It seems plausible to me that you could recover a statement like this by working prime-by-prime with the Sylow subgroups of the groups in question -- certainly there are enough normal subgroups of $p$ groups to detect indices -- but I haven't thought carefully about that.
| 8 | https://mathoverflow.net/users/1149 | 46644 | 29,508 |
https://mathoverflow.net/questions/46648 | 3 | I have a few really basic questions about minimal surfaces. Does a smooth or piecewise smooth injection $S^1$ into $\mathbb R^3$ always give a unique minimal surface or are there instances with discrete distinct solutions? Can it not be the case that a 1-parameter family of minimal surfaces exists for a given "frame"? Do linear maps preserve minimal surfaces? I'm guessing no, but I don't have a good example in mind.
Also, if I have a simplicial decomposition of $S^1$ and map it into $\mathbb R^3$ with a simplicial map, is it known that there is a unique minimal surface spanning this piecewise linear frame? Is the formula readily given?
These questions came up defining a surface $f(s,t)=(1-s)(1-t)v\_0+(1-s)tv\_1+s(1-t)v\_3+stv\_4$
to interpolate the simplicial map defined by $f(0,0)=v\_0, f(0,1)=v\_1, f(1,1)=v\_2, f(1,0)=v\_3$ which maps $\partial I^2$ into $\mathbb R^3$. I believe this surface is minimal, and wonder how this works for polygons with more vertices than the square.
Thanks
| https://mathoverflow.net/users/2031 | minimal surfaces | In $\mathbb{R}^3$ there is always some minimal surface spanning a given connected simple curve (assuming the curve is not too horrible, say Lipschitz). Here surface needs to be broadly understood as being possibly immersed and of arbitrary topology. This can be seen using the machinery of geometric measure theory and the direct method in the calculus of variations (i.e. by minimizing area). Alternatively, since there is a close connection to harmonic mappings in this context, one can always find a minimal surface spanning the curve that is topologically a disk (this is sometimes referred to as the Douglas-Rado solution). This is also constructed by direct methods but here one minimizes Dirichlet energy of the map.
If you know more about the bounding curve one can get more information about possible minimal surfaces spanning the curve. For instance if the curve lies on the boundary of a convex set then it spans at least one embedded minimal surface.
In general there is no uniqueness (there are in fact an example due to F. Morgan where a given curve bounds a continuum of minimal surfaces). Though if you know that the bounding curve is analytic and of total geodesic curvature less than $4\pi$ then Nitsche shows that the curve bounds a unique minimal disk (I don't remember if it can also bound surfaces of other topological types).
In general I doubt you could get an explicit parameterization for the surface even if the curve is simplicial. Though you might be able to get something if you look at the Enneper-Weierstrass representation.
| 11 | https://mathoverflow.net/users/26801 | 46651 | 29,513 |
https://mathoverflow.net/questions/29638 | 3 | What I'm really looking for is a good set of examples for a semi-direct product of two (coprime) cyclic groups, with a non-trivial (not $1$ and not everything) center. What resource would you recommend for that purpose?
| https://mathoverflow.net/users/5309 | List of centers of finite groups | If you would like a complete classification following Jack's strategy, say your group is $G=C\_n\rtimes C\_m$ with $n,m$ coprime, $C\_n=\langle c\rangle$, $C\_m=\langle x\rangle$ and $xcx^{-1}=c^j$. The order of $j$ in $({\mathbb Z}/n{\mathbb Z})^\times$ has to divide $m$, and every such triple $(n,m,j)$ gives a valid semi-direct product $G$. Now
>
> $G$ has a non-trivial centre if and only if either
>
>
> * The order of $j$ is less than $m$, or
> * $j$ is congruent to 1 modulo a prime $p|n$.
>
>
>
In the first case, $C\_m$ does not act faithfully on $C\_n$, and the elements in the kernel commute with both $C\_n$ and $C\_m$, so they are in the centre. In the second case, $C\_p\lt C\_n$ is central. If neither condition holds, then the centre must be contained in $C\_n$ (by faithfullness of the action) and cannot contain any $C\_p$ for $p|n$ (by the failure of the second condition), so it has to be trivial.
| 5 | https://mathoverflow.net/users/3132 | 46652 | 29,514 |
https://mathoverflow.net/questions/46633 | 3 | Let $f \colon X \to Y$ be a proper morphism of (Noetherian) schemes, $\mathcal{F} \in \mathop{Coh}(X)$. Let $i\_Z \colon Z \hookrightarrow Y$ be a closed subscheme and take the inverse image $W := X \times\_Y Z \overset{i\_W}{\hookrightarrow} X$.
If we assume that $\dim X\_y \leq k - 1$ for all $y \in Y \setminus Z$, then the sheaf $R^k f\_{\*}(\mathcal{F})$ is concentrated on $Z$ by flat base change for the open inclusion $Y \setminus Z \to Y$.
>
> Is it true in this case that $$R^k f\_{\*}(\mathcal{F}) \cong i\_{Z\*} R^k (f|\_W)\_{\*}(\mathcal{F}|\_{W})$$ where $\mathcal{F}|\_{W} = i\_W^{\*}(\mathcal{F})$?
>
>
>
I was not able to derive this from the standard results about base change, nor to find any counterexamples.
**Special case:** What I actually need is a rather special case of the former. Namely in the case I am insterested in:
* all objects are varieties over an algebraically closed field
* $X$, $Z$ and $W$ are smooth
* $\mathcal{F}$ is a line bundle
* $k = 1$
* the map $X \setminus W \to Y \setminus Z$ is an isomorphism.
It interesting to know something about the general case, though.
| https://mathoverflow.net/users/828 | A form of cohomology and base change | Here are more details for the second example from comments. Set $Y={\mathbb A}^2$, $Z=0$ (the origin), $X=$ blow-up of $Y$ and $Z$, so that $W$ is the exceptional divisor. Take $F\_n=O\_X(nW)$. Since the self-intersection of $W$ is $-1$, we see that $F\_n/F\_{n-1}=F\_n|\_W\simeq O\_W(-n)$
(more properly, the direct image of this sheaf to $X$). Looking at the short exact sequence
$$0\to F\_{n-1}\to F\_n\to F\_n/F\_{n-1}\to 0\qquad(n>0),$$
we see that $R^0f\_\*(F\_n)=R^0f\_\*(F\_{n-1})$, while $R^1f\_\*(F\_n)$ is an extension
$$0\to R^1f\_\*(F\_{n-1})\to R^1f\_\*(F\_n)\to R^1f\_\*(F\_n/F\_{n-1})\to 0.$$
The right term of this exact sequence is $i\_{Z\*}R^1f\_\*(F\_n|\_W)$, so your condition fails as long as $R^1f\_\*(F\_{n-1})\ne0$. But this happens for $n>2$ (again, follows from the same sequence).
| 4 | https://mathoverflow.net/users/2653 | 46654 | 29,516 |
https://mathoverflow.net/questions/35736 | 17 | I have heard that the canonical divisor can be defined on a normal variety X since the smooth locus has codimension 2. Then, I have heard as well that for ANY algebraic variety such that the canonical bundle is defined:
$$\mathcal{K}=\mathcal{O}\_{X,-\sum D\_i}$$
where the $D\_i$ are representatives of all divisors in the Class Group.
I want to prove that formula or I want to find a reference for that formula, or I want someone to rephrase it in a similar way if they heard about it.
Why do I want to prove it? Well, I use the definition that something is Calabi Yau if its canonical bundle is 0. In the case of toric varieties, $\sum D\_i$~0 if all the primitive generators for the divisors lie on a hyperplane. Then the sum is 0 and therefore the toric variety is Calabi-Yau.
Can someone confirm or fix the above formula? I do not ask for a debate on when something is Calabi-Yau, I handle that OK, I just ask whether the above formula is correct. A reference would be enough. I have little access to references at the moment.
| https://mathoverflow.net/users/1887 | The canonical line bundle of a normal variety |
>
> **Edit** (11/12/12): I added an explanation of the phrase "this is essentially equivalent to $X$ being $S\_2$" at the end to answer aglearner's question in the comments.
> [See also [here](https://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/45354#45354) and [here](https://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/45616#45616)]
>
>
>
Dear Jesus,
I think there are several problems with your question/desire to define a canonical divisor on *any* algebraic variety.
First of all, what is *any* algebraic variety? Perhaps you mean a quasi-projective variety (=reduced and of finite type) defined over some (algebraically closed) field.
OK, let's assume that $X$ is such a variety. Then what is a *divisor* on $X$? Of course, you could just say it is a formal linear combination of *prime divisors*, where a prime divisor is just a codimension 1 irreducible subvariety.
OK, but what if $X$ is not equidimensional? Well, let's assume it is, or even that it is irreducible.
Still, if you want to talk about divisors, you would surely want to say when two divisors are *linearly equivalent*. OK, we know what that is, $D\_1$ and $D\_2$ are linearly equivalent iff $D\_1-D\_2$ is a *principal divisor*.
But, what is a principal divisor? Here it starts to become clear why one usually assumes that $X$ is normal even to just talk about divisors, let alone defining the canonical divisor. In order to define principal divisors, one would need to define something like the *order of vanishing* of a regular function along a prime divisor. It's not obvious how to define this unless the local ring of the general point of any prime divisor is a DVR. Well, then this leads to one to want to assume that $X$ is $R\_1$, that is, regular in codimension $1$ which is equivalent to those local rings being DVRs.
OK, now once we have this we might also want another property: If $f$ is a regular function, we would expect, that the zero set of $f$ should be 1-codimensional in $X$. In other words, we would expect that if $Z\subset X$ is a closed subset of codimension at least $2$, then if $f$ is nowhere zero on $X\setminus Z$, then it is nowhere zero on $X$. In (yet) other words, if $1/f$ is a regular function on $X\setminus Z$, then we expect that it is a regular function on $X$. This in the language of sheaves means that we expect that the push-forward of $\mathscr O\_{X\setminus Z}$ to $X$ is isomorphic to $\mathscr O\_X$. Now this is essentially equivalent to $X$ being $S\_2$.
So we get that in order to define *divisors* as we are used to them, we would need that $X$ be $R\_1$ and $S\_2$, that is, *normal*.
Now, actually, one can work with objects that behave very much like divisors even on non-normal varieties/schemes, but one has to be very careful what properties work for them.
As far as I can tell, the best way is to work with *Weil divisorial sheaves* which are really reflexive sheaves of rank $1$. On a normal variety, the sheaf associated to a Weil divisor $D$, usually denoted by $\mathcal O\_X(D)$, is indeed a reflexive sheaf of rank $1$, and conversely every reflexive sheaf of rank $1$ on a normal variety is the sheaf associated to a Weil divisor (in particular a reflexive sheaf of rank $1$ on a regular variety is an invertible sheaf) so this is indeed a direct generalization. One word of caution here: $\mathcal O\_X(D)$ may be defined for Weil divisors that are not Cartier, but then this is (obviously) not an invertible sheaf.
Finally, to answer your original question about canonical divisors. Indeed it is possible to define a canonical divisor (=Weil divisorial sheaf) for all quasi-projective varieties. If $X\subseteq \mathbb P^N$ and $\overline X$ denotes the closure of $X$ in $\mathbb P^N$, then the dualizing complex of $\overline X$ is
$$
\omega\_{\overline X}^\bullet=R{\mathscr H}om\_{\mathbb P^N}(\mathscr O\_{\overline X}, \omega\_{\mathbb P^N}[N])
$$
and the canonical *sheaf* of $X$ is
$$
\omega\_X=h^{-n}(\omega\_{\overline X}^\bullet)|\_X=\mathscr Ext^{N-n}\_{\mathbb P^N}(\mathscr O\_{\overline X},\omega\_{\mathbb P^N})|\_X
$$
where $n=\dim X$. (Notice that you may disregard the derived category stuff and the dualizing complex, and just make the definition using $\mathscr Ext$.) Notice further, that
if $X$ is normal, this is the same as the one you are used to and otherwise it is a reflexive sheaf of rank $1$.
As for your formula, I am not entirely sure what you mean by "where the $D\_i$ are representatives of all divisors in the Class Group". For toric varieties this can be made sense as in Josh's answer, but otherwise I am not sure what you had in mind.
>
> (Added on 11/12/12):
>
>
>
**Lemma** A scheme $X$ is $S\_2$ if and only if for any $\iota:Z\to X$ closed subset of codimension at least $2$, the natural
map $\mathscr O\_X\to \iota\_\*\mathscr O\_{X\setminus Z}$ is an isomorphism.
**Proof**
Since both statements are local we may assume that $X$ is affine.
Let $x\in X$ be a point and $Z\subseteq X$ its closure in $X$. If $x$ is a codimension at most $1$ point, there is nothing to prove, so we may assume that $Z$ is of codimension at least $2$.
Considering the exact sequence (recall that $X$ is affine):
$$
0\to H^0\_Z(X,\mathscr O\_X) \to H^0(X,\mathscr O\_X) \to H^0(X\setminus Z,\mathscr O\_X) \to H^1\_Z(X,\mathscr O\_X) \to 0
$$
shows that $\mathscr O\_X\to \iota\_\*\mathscr O\_{X\setminus Z}$ is an isomorphism
if and only if
$H^0\_Z(X,\mathscr O\_X)=H^1\_Z(X,\mathscr O\_X)=0$ the latter condition is equivalent to
$$
\mathrm{depth}\mathscr O\_{X,x}\geq 2,
$$
which given the assumption on the codimension is exactly the condition that $X$ is $S\_2$ at $x\in X$. $\qquad\square$
| 51 | https://mathoverflow.net/users/10076 | 46663 | 29,519 |
https://mathoverflow.net/questions/46650 | 1 | Let $f: \Omega\rightarrow \mathbb{R}$ where $\Omega\subset\mathbb{R}^d$ is bounded with lipschitz smooth boundary. Further suppose that $f\in\mathcal{H}^{\tau}(\Omega)$, $\tau>0$ and that $f$ decays rapidly to $0$ on the boundary.
Let $$ \Omega\_{\delta} = \{ x\in\Omega : \inf\_{y\in\partial\Omega} \left\|x-y\right\|\_2 > \delta \} .$$ Where $\delta>0$ is small enough to preserve smoothness in the boundary of $\Omega\_{\delta}$. See: [Shrinking a Lipschitz smooth domain.](https://mathoverflow.net/questions/46628/shrinking-a-lipschitz-smooth-domain)
Are there any known bounds on $\left\|f\right\|\_{L\_2(\Omega\setminus\Omega\_{\delta})}$? i.e. bounding $f$ near the boundary of $\Omega$.
Note:
$
\left\|f\right\|\_{L\_2(\Omega)}= \left(\int\_{\Omega} |f|^2\right)^{\frac{1}{2}}
$
I ideally would like some bound of the form: Given $f\in\mathcal{H}^\tau(\Omega)$, $\tau>d/2$ which is zero on the boundary and $\delta$ sufficiently small then:
$\left\|f\right\|\_{L\_2(\Omega\setminus\Omega\_{\delta})} \leq C\delta^\alpha\left\|f\right\|\_{L\_2(\Omega)}$ with $\alpha>0$ as large as possible (hopefully $\alpha=1$) and $C$ is a constant not depending on $\delta$ or $f$.
| https://mathoverflow.net/users/2011 | Bounding a smooth function near the boundary | For sure $\|f\|\_ {L^2(\Omega\setminus\Omega\_\delta)}=o(1)$ as $\delta\to0$ for any $f\in L^2(\Omega)$ (this, even if $\Omega$ was not bounded). For $f\in L^\infty(\Omega)$ you have $\|f\|\_ {L^2(\Omega\setminus\Omega\_\delta)}=O(\delta)$, for the Lebesgue measure of $\Omega\setminus\Omega\_\delta$ is bounded by $\delta\mathcal{H}^{n-1}(\partial\Omega),$ as a consequence of the coarea formula applied to the distance function, or directly, on the lines of Denis Serre's construction in the linked answer. For the same reason, if $f$ is Hölder continuous of exponent $0\leq \alpha\le1$ (for instance, it is in a Sobolev space in the hypothesis of the Morrey-Sobolev embedding) and vanishes on $\partial\Omega$, you have $\|f\|\_ {L^2(\Omega\setminus\Omega\_\delta)}=O(\delta^{1+\alpha}).$ Finally, it is not completely clear what you mean exactly by "rapidly decaying to 0", but certainly any bound on $|f|$ on $\Omega\setminus\Omega\_\delta$ gives a bound on the norm as said, and in few words, everything is like in the case $n=1$.
| 3 | https://mathoverflow.net/users/6101 | 46665 | 29,521 |
https://mathoverflow.net/questions/46662 | 5 | As is well known (see Kassel), when $q$ is not a root of unity, the centre or the quantum enveloping algebra $U\_q({\mathfrak sl}\_2)$ of ${\mathfrak sl}\_2$ is generated by the element
$$
C\_q = EF + \frac{q^{-1}K+qK^{-1}}{(q-q^{-1})^2}.
$$
The element is called the quantum Casimir. My questions are as follows:
(i) Does this situation extend to the general setting of $U\_q({\mathfrak sl}\_N)$?
(ii) If it does, is there a general formula for $C\_q$?
(iii) How would this formula relate to the usual formula for the classical Casimir? (The uasual formula I refer to is $\sum X^iX\_i$, for some basis $X\_i$ and its dual $X^i$, see wikipedia for details.)
| https://mathoverflow.net/users/2612 | Does There Exists a General Quantum Casimir Extending the $U_q({\mathfrak sl}_2$ Case? | The centers of the Drinfeld-Jimbo quantum groups $U\_q(\mathfrak{g})$ are well-understood and quite analogous to the classical case. See the book by Klimyk and Schmüdgen, Section 6.3, where in particular the quantum Casimirs are constructed.
| 8 | https://mathoverflow.net/users/10756 | 46668 | 29,524 |
https://mathoverflow.net/questions/46660 | 7 | Suppose a square, dense, symmetric matrix $A$ has been factorized into $L$ and $U$ components by performing a LU decomposition. Now let $B = A+\lambda I$. Is there any way to efficiently compute the LU decomposition of $B$ by reusing $L$ and $U$, avoiding the cost of a new decomposition?
If the cost of reusing $L$ and $U$ becomes too similar to the cost of recomputing the entire decomposition, would there be any other decomposition more suitable in this case?
**Edit:** I have found my question to be very similar to [solving series of linear systems with diagonal perturbations](https://mathoverflow.net/questions/36743/solving-series-of-linear-systems-with-diagonal-perturbations). I am also trying to solve a series of linear systems where every system is in the form $(A+\lambda I)x=b$ where $\lambda$ is a real number and all $b$s are the same. However, unlike the aforementioned question, my matrices are always dense.
| https://mathoverflow.net/users/4430 | Adding a multiple of the Identity to a LU factorized matrix | Such an efficient computation is unlikely. Suppose you can do it, with factors $U\_\lambda$ and $L\_\lambda$. Then by $\det(A+\lambda I)=\det U\_\lambda\cdot\det L\_\lambda$, you obtain the characteristic polynomial for free. Thus the cost of the calculation you are looking for cannot be smaller than the cost of the calculation of the characteristic polynomial.
For the latter problem, the best algorithm today is in $O(n^{7/2})$, if you employ standard multiplication of matrices (see the new edition of my book "Matrices ; theory and applications, section 3.10). It becomes $O(n^3)$ if you employ fast methods of multiplication (strictly better than Strassen's), but this is only theoritical, since fast matrix multiplaction is recursive, and extremely uncomfortable (if not impossible) to code.
| 2 | https://mathoverflow.net/users/8799 | 46670 | 29,526 |
https://mathoverflow.net/questions/41102 | 10 | The mutual information $I(\mathfrak A\_1;\mathfrak A\_2)$ of two complete $\sigma$-algebras $\mathfrak A\_1$ and $\mathfrak A\_2$ in a Lebesgue probability space $(X,m)$ is the integral of the logarithm of the Radon-Nikodym derivative $dP/d(P\_1\otimes P\_2)$, where $P\_i$ are the quotient measures on the factor-spaces $X\_i$ determined by the $\sigma$-algebras $\mathfrak A\_i$, respectively, and $P$ is the joint distribution on $X\_1\times X\_2$. If $\mathfrak A\_i$ correspond to countable measurable partitions $\alpha\_i$ of the space $X$, then $I(\mathfrak A\_1;\mathfrak A\_2)$ are expressed in terms of the associated entropies as $H(\alpha\_1)+H(\alpha\_2)-H(\alpha\_1\vee\alpha\_2)$. I need an explicit reference for the following fact (I know how to prove it): if $\mathfrak B\_n$ is a decreasing sequence of $\sigma$-algebras converging to a limit $\mathfrak B$, and $\mathfrak A$ is another $\sigma$-algebra such that $I(\mathfrak A;\mathfrak B\_1)<\infty$, then $I(\mathfrak A;\mathfrak B\_n)$ converges to $I(\mathfrak A;\mathfrak B)$.
| https://mathoverflow.net/users/8588 | Continuity of the mutual information | The last proof of this continuity property (and of its analogue for increasing sequences) is given in [this paper](http://www.ams.org/mathscinet-getitem?mr=2472012) by Harremöes and Holst. It also contains a pretty comprehensive list of references to earlier work. Apparently, first this property was established by Pinsker in his 1960 book "Information and Information Stability of Random Variables and Processes".
| 5 | https://mathoverflow.net/users/8588 | 46672 | 29,528 |
https://mathoverflow.net/questions/46686 | 3 | I am told that finite groups have unique factorization under direct product. That is, call a nontrivial group "indivisible" if it is not isomorphic to a direct product of nontrivial groups. Then every finite group can be "factored" (by direct product) into a unique collection of indivisible groups.
In particular, if $G$ and $H$ are finite groups so that $G\times G\cong H\times H$, then $G\cong H$.
Can anyone provide a reference to a proof of these results? What is known in the infinite case? Thanks.
| https://mathoverflow.net/users/1079 | Unique factorization of finite groups under direct sum? | About the first fact see [this](https://groupprops.subwiki.org/wiki/Direct_product_is_cancellative_for_finite_groups) page (the Krull–Remak–Schmidt theorem). For infinite (even finitely generated) groups the situation is different because there exists an infinite f.g. group isomorphic to its [direct square.](https://groupprops.subwiki.org/wiki/Group_isomorphic_to_its_square)
**Update.** [Hirshon](https://doi.org/10.1007/BF01229716 "Hirshon, R. The cancellation of an infinite cyclic group in direct products. Arch. Math 26, 134–138 (1975). zbMATH review at https://zbmath.org/?q=an:0303.20022"), found two non-isomorphic finitely generated nilpotent (infinite) groups $G,H$ such that $G\times G\cong H\times H$.
| 7 | https://mathoverflow.net/users/nan | 46690 | 29,537 |
https://mathoverflow.net/questions/46701 | 16 | I have heard references to "hard" vs. "soft" analysis. What is the difference? It seems to do with generality versus nitty-gritty estimates, but I haven't gotten any responses more clear than that.
| https://mathoverflow.net/users/10946 | What is the difference between hard and soft analysis? | Disclaimer: I'm no expert-this is really a question for the analysts and historians of mathematics.
As far as I know,the terminology came into existence in the early 20th century distinguish the classical "calculus" type analysis (hard analysis) from the new point set topology/functional analysis approach (soft analysis). A hard analytic argument uses a direct calculation or construction of an exact estimate bounds of specific function or function types to prove a statement. A soft analytic arguement uses the general topological or geometric properties of a space in which a function or function class is defined to prove a result indirectly without a precisely calculated "bound".
For example, the fact that the Cantor set has measure zero is a "hard" analytic arguement; it uses an epsilon-delta arguement to show the limit of the sequence of "slices" of the lengths of it's component intervals on the real line converges to 0.
An example of a "soft" analytic arguement: (IVT) Let $f$ be a continuous function defined on a connected subset of the real line i.e. an interval with a well defined least upper bound and greatest lower bound. Then the function is defined at every point inbetween the lub and the glb. A soft proof would be as follows: Since an interval $I$ of $\Bbb R$ is a connected subset of $\Bbb R$ and $f$ is continuous, then $f(I)$ is also connected. Therefore, for every $x \in I$, $f(x)$ is in $f(I)$. Notice this proof does not involve a direct computation of bounds that proves $f(x)$ is in the image set of $f$ (although it certainly COULD be proven that way).
Anyway, that's how Gerald Itzkowitz taught it to me and I learned a long time ago to trust him on these matters........
| 12 | https://mathoverflow.net/users/3546 | 46705 | 29,547 |
https://mathoverflow.net/questions/46700 | 19 | ### Background
One of my favourite elementary results in group theory is [Goursat's Lemma](http://en.wikipedia.org/wiki/Goursat%27s_lemma). This lemma characterises the subgroups of a direct product of groups in terms of fibred products.
Indeed, let $L$ and $R$ be groups and let $G < L \times R$ be a subgroup of their direct product. We have natural projections $\pi\_L : L \times R \to L$ and $\pi\_R: L \times R \to R$. We may assume without loss of generality that $\pi\_L$ and $\pi\_R$ are surjective when restricted to $G$. Let $L\_0 = \pi\_L(G \cap \ker\pi\_R)$ and $R\_0 = \pi\_R(G \cap \ker\pi\_L)$ denote, respectively, normal subgroups of $L$ and $R$. Goursat's Lemma is the observation that $G$ defines an isomorphism $L/L\_0 \cong R/R\_0$.
*Proof*: If $x \in L$, let $y \in R$ be such that $(x,y) \in G$. Such a $y$ exists because of surjectivity of $\pi\_L : G \to L$, but it need not be unique. Nevertheless the coset $y R\_0$ is well defined. This procedure then defines a homomorphism $L \to R/R\_0$ whose kernel is precisely $L\_0$ and which is surjective because $\pi\_R : G \to R$ is.
If we let $\lambda: L/L\_0 \to F$ and $\rho: R/R\_0 \to F$ be isomorphisms to the same abstract group $F$, then we may identify $G$ with the fibred product
$$ G = \lbrace (x,y) \in L \times R ~|~ \lambda(x L\_0) = \rho(y R\_0) \rbrace .$$
There are similar results for Lie subalgebras of a direct sum of Lie algebras, and probably also in other categories. This suggests the following categorical slogan:
>
> "subobjects of a product are pullbacks"
>
>
>
(Well, at least subobjects with the property that the composition with the epis in the product are also epis.)
Of course, this is not going to be true in all categories, which prompts the following question.
### Question
Let $\mathcal{C}$ be a category and $L,R$ be objects whose product $L \times R$ exists. Let $G \to L \times R$ be a monomorphism such that the compositions $G \to L \times R \to L$ and $G \to L \times R \to R$ are epimorphisms.
What must we demand of $\mathcal{C}$ so that there exist epimorphisms $L \to F$ and $R \to F$ such that
$$\begin{matrix} G & \rightarrow & L \cr
\downarrow & & \downarrow \cr
R & \rightarrow & F \end{matrix}
$$
is a pullback?
### Epilogue
One would be tempted to call these categories *Goursat categories*, but alas the name seems to be taken already for what seems like a different concept.
Thanks in advance.
| https://mathoverflow.net/users/394 | For which categories does one have a Goursat Lemma? | To make life simple, suppose that finite limits and finite colimits exist. If we work with regular epimorphisms rather than epimorphisms then your condition is equivalent to saying that for any two (regular) epimorphisms $G\to L$ and
$G\to R$, if you form the pushout $F$ then the canonical comparison from $G$ to the pullback $L\times\_F R$ is a regular epimorphism.
This is true in any exact Mal'cev category: see Theorem 5.7 of
>
> Carboni, Kelly and Pedicchio, Some remarks on Maltsev and Goursat categories, Applied Categorical Structures 1:385-421, 1993.
>
>
>
Here exact means that the category
(i) has finite limits
(ii) has regular epi-mono factorizations
(iii) the pullback of a regular epi is a regular epi
(iv) any equivalence relation is the kernel pair of some map (one can choose the map to be the coequalizer)
and Mal'cev can be characterized in many ways. For example, it says that if R and S are equivalence relations on some object A, then RS=SR.
In fact if the category is regular, in the sense that (i)-(iii) hold, then your condition is equivalent to being exact and Mal'cev.
By the way, the Goursat categories you mention are only slightly weaker: they have
RSR=SRS rather than RS=SR. You can still prove your condition for Goursat categories if you suppose that at least
one of $G\to L$ and $G\to R$ is a split epimorphism (i.e. has a section), and in fact this can be used to characterize Goursat categories. See [link text](http://www.springerlink.com/content/a6x8448227q13062/)
| 16 | https://mathoverflow.net/users/10862 | 46709 | 29,550 |
https://mathoverflow.net/questions/46687 | 4 | It is known for Hilbertian fields that all groups that are abelian, solvable, $A\_n$ or $S\_n$ are realizable over them. $\mathbb{Q}(x)$ is one such field, but it's not obvious that the extensions that are guaranteed by this theorem will be regular extensions.
$A\_n$ and $S\_n$ are regularly realizable through the symmetric polynomials. Is it known whether all solvable (or even abelian) groups are *regularly* realizable over $\mathbb{Q}(x)$?
Remark: for those who aren't familiar with the terminology, by regularly realizable over $\mathbb{Q}(x)$ I mean that there's an extension of it, $L$, such that $L$ over $\mathbb{Q}(x)$ is $G$-Galois, and such that the algebraic closure of $\mathbb{Q}$ in $L$ is $\mathbb{Q}$.
| https://mathoverflow.net/users/5309 | Are all solvable groups *regularly* realizable over Q(x)? | It is known that all finite abelian groups are regularly realisable over $\mathbb{Q}(x)$. See e.g. B.H. Matzat, Konstruktive Galoistheorie, p. 224, M. Fried and M. Jarden, Field Arithmetic, Lemma 24.46, or J.P Serre, Topics in Galois Theory, p. 36. It is also known that, e.g., the set of regularly realisable groups is closed under wreath products, which gives you some non-abelian soluble groups.
But according to Jack Sonn, [Brauer groups, embedding problems, and nilpotent groups as Galois groups](http://www.springerlink.com/index/Y0874238L4X4R417.pdf), Israel Journal of Mathematics 85, which, alas, is 16 years old, it is not even known whether all finite nilpotent groups are regularly realisable over $\mathbb{Q}(x)$. I don't know of any more recent developments (there has been quite a lot of work over "large" fields instead of $\mathbb{Q}$).
| 6 | https://mathoverflow.net/users/35416 | 46710 | 29,551 |
https://mathoverflow.net/questions/46736 | 1 | The question is in subject.
**Update:** See Andreas Thom's answer.
| https://mathoverflow.net/users/4807 | Do separable $C^*$-algebras form a set? | It is not so clear what you mean.
However, every separable $C^\ast$-algebra embeds in $B(\ell^2 \mathbb N)$. Hence, the isomorphism classes of separable $C^\ast$-algebras form a set.
| 7 | https://mathoverflow.net/users/8176 | 46737 | 29,566 |
https://mathoverflow.net/questions/46752 | 13 | Let me be more specific. Let $M$ be a Kahler manifold with Riemannian metric $g$ and complex structure $I$. Then $T^\ast M$ will also be Kahler with metric and complex structure induced from $M$ (I will give them the same name). It is also holomorphic symplectic, with canonical holomorphic symplectic form $\Omega \_\mathbb C$.
If $M$ was an affine space with the standard metric I could define $\omega \_J$ and $\omega \_K$ on $T^\ast M$ by taking the real and imaginary parts of $\Omega \_\mathbb C$ which would define a hyperkahler structure on $T^\ast M$ (everything is covariantly constant with constant coefficients).
**Question 1**: Does this work for a general kahler manifold $M$?
It seems a bit unreasonable to me, as the construction of $\Omega \_\mathbb C$ does not depend on the metric (but does depend on the complex structure, which is compatible with the metric...)
I also know that every hyperkahler manifold is holomorphic symplectic (with $\Omega \_\mathbb C = \omega \_J + I\omega \_K$) and Yau's theorem implies that every *compact* holomorphic symplectic manifold is hyperkahler.
**Question 2**: Does $T^\ast M$ admit a hyperkahler metric, with the associated holomorphic symplectic form the canonical one (coming from the cotangent bundleness)?
**Question 3**: Is $g$ a hyperkahler metric for $T^\ast M$ at all? Or, does $T^\ast M$ admit a hyperkahler metric at all?
I don't know much about this sort of thing, but it seemed like a natural question to me, and I couldn't find an answer anywhere.
| https://mathoverflow.net/users/7762 | Is the cotangent bundle to a Kahler manifold hyperkahler? | Such hyper Kaehler metrics do exist near the zero section, e.g. in a formal or an analytic tubular neighborhood of the zero section. After that one can use some homogeneity to spread them on the whole cotangent bundle but typically the resulting metrics are non-complete. One gets nice global metrics on the cotangent bundles of Hermitian symmetric spaces but this is pretty much it. This question was studied extensively. There are two different proofs of the existence: in [this work](http://www.google.com/url?sa=t&source=web&cd=4&sqi=2&ved=0CCMQFjAD&url=http%3A%2F%2Fpeople.maths.ox.ac.uk%2Fhitchin%2Fhitchinstudents%2Ffeix.ps.gz&rct=j&q=Birte%20Feix&ei=ogXoTKeTFsnMswahpLyyCw&usg=AFQjCNEfkq4jK0h2fTxjUgWcH1_KS-uC8A&cad=rja) of Birte Feix and
[this work](http://arxiv.org/abs/math/0011256) of Dima Kaledin.
| 15 | https://mathoverflow.net/users/439 | 46755 | 29,575 |
https://mathoverflow.net/questions/46748 | 21 | This is not actually a question asked by me. But since I do not know the answer, I would love to know if someone here could answer it.
| https://mathoverflow.net/users/4760 | Is every locally connected subset of Euclidean space R^n locally path connected ? | No. There are locally connected subsets of $\mathbb{R}^2$ which are totally path disconnected. See my answer to this old MO question "[Can you explicitly write R2 as a disjoint union of two totally path disconnected sets?](https://mathoverflow.net/questions/156/can-you-explicitly-write-r2-as-a-disjoint-union-of-two-totally-path-disconnected)". Also, Gerald Edgar's response to the same question says that such sets cannot be totally disconnected, although he does not mention local connectedness. In fact, the sets given by my answer are locally connected, so provide a counterexample to your question.
As in the linked question: Let $S$ be a subset of the reals such that $S\cap[a,b]$ and $S^{\rm c}\cap[a,b]$ cannot be written as a countable union of closed sets for any $a < b$ (e.g., this [explicit example of a non-Borel set](https://planetmath.org/alebesguemeasurablebutnonborelset)). Then, $A=(S\times\mathbb{Q})\cup(S^{\rm c}\times\mathbb{Q}^{\rm c})$ and $B=(S\times\mathbb{Q}^{\rm c})\cup(S^{\rm c}\times\mathbb{Q})$ partition the plane into a pair of locally connected and totally pathwise disconnected sets.
That they are totally pathwise disconnected is proven in my answer to the linked question. Let us show that $A\cap U$ is connected for any nonempty 'open rectangle' $U=(x\_0,x\_1)\times(y\_0,y\_1)$. If not, there would be nonempty disjoint open sets $V,W\subset U$ with $A\cap(V\cup W)=A\cap U$. If $\pi(x,y)=x$ is the projection onto the *x*-axis then $\pi(V)\cup\pi(W)=\pi(V\cup W)=(x\_0,x\_1)$ is connected. So, we can find a nontrivial closed interval $[a,b]\subseteq\pi(V)\cap\pi(W)$.
Now, for every $x\in S^{\rm c}\cap[a,b]$, the line segment $\lbrace x\rbrace\times(y\_0,y\_1)$ intersects with both $V$ and $W$ and, by connectedness of line segments, it will intersect with $U\setminus(V\cup W)$. Hence, there is a $q\in\mathbb{R}$ with $(x,q)\in U\setminus(V\cup W)\subset B$. So, $q\in\mathbb{Q}$. For each rational $q$, let $S\_q$ be the (closed) set of $x\in[a,b]$ such that $(x,q)$ is in $U\setminus(V\cup W)$.
Then, $S^{\rm c}\cap[a,b]=\bigcup\_{q\in\mathbb{Q}}S\_q$ is a countable union of closed sets, giving the required contradiction. Hence $A$ is locally connected and, exchanging $S$ and $S^{\rm c}$ in the argument above, so is $B$.
| 27 | https://mathoverflow.net/users/1004 | 46758 | 29,576 |
https://mathoverflow.net/questions/46757 | 3 | Suslin's problem is:
>
> Given a complete dense linear order without endpoints, if it has the ccc must it be isomorphic to $\mathbb{R}$?
>
>
>
The answer is that it's independent of ZFC. The related question:
>
> Given a complete dense linear order without endpoints, if it's separable must it be isomorphic to $\mathbb{R}$?
>
>
>
has a positive answer under ZFC. Now consider the following analogous questions:
>
> Given a non-trivial separative forcing poset, if it has the ccc must it have size at most continuum?
>
>
>
The answer to this is no, for example the Cohen forcing that adds more than continuum-many reals is ccc but has size greater than continuum. So what about:
>
> **Given a non-trivial separative forcing poset, if it's separable (i.e. has a countable dense set) must it have size at most continuum?**
>
>
>
| https://mathoverflow.net/users/7521 | A problem about posets similar to Suslin's problem | Amit:
If ${\mathbb P}$ is a non-trivial separative partial order, and it is countable, an easy argument (back-and-forth) shows that it is forcing isomorphic to Cohen forcing. (This is an exercise in Chapter VII of Kunen's book, I believe, and it can be found in a few other sources, such as the appropriate chapter of the "Handbook of Boolean Algebras".)
It follows that if ${\mathbb P}$ is non-trivial, separative, and admits a countable dense subset, then ${\mathbb P}$ is again Cohen forcing. Now, ${\mathbb P}$ embeds into its Boolean completion which must then coincide with the Boolean completion of Cohen (Borel sets/meager) and therefore has size ${\mathfrak c}=|{\mathbb R}|$. This gives ${\mathfrak c}$ as an upper bound for $|{\mathbb P}|$.
Here is a cute application: Fix $\epsilon>0$, and let ${\mathbb P}^\epsilon$ be the collection of open subsets $p$ of ${\mathbb R}$ that are a finite union of intervals with rational end-points and such that the sum of their lengths (the Lebesgue measure of $p$) is less than $\epsilon$. The generic object gives us an open set that covers the ground model reals, and has measure $\epsilon$. Since ${\mathbb P}^\epsilon$ is countable, it is Cohen forcing. Since $\epsilon$ is arbitrary, it follows that adding a Cohen real makes the set of ground model reals have measure zero.
---
**Edit:** As Andreas Blass pointed out, one can actually prove the upper bound $|{\mathbb P}|\le{\mathfrak c}$ directly. This actually gives us a way of *proving* the characterization of Cohen forcing.
| 9 | https://mathoverflow.net/users/6085 | 46766 | 29,581 |
https://mathoverflow.net/questions/46626 | 4 | Let $c\_0$ be the Banach space of doubly infinite sequences $$\lbrace
a\_n: -\infty\lt n\lt \infty, \lim\_{|n|\to \infty} a\_n=0 \rbrace.$$ Let $T$ be the space of $2\pi$ periodic functions integrable on $[0,2\pi]$.
Let $$S=\lbrace \lbrace a\_n\rbrace \in c\_0: a\_n=\hat{f}(n) \forall n \mbox{ for some function } f\in T\rbrace,$$
where $\hat{f}(n)$ denotes the $n$-th Fourier coefficient of $f$, i.e. $$\hat{f}(n)=\frac{1}{2\pi}\int\_{-\pi}^\pi f(x)e^{-inx}\,dx.$$
When I was a graduate student, I was told that no known characterizations of $S$ were known. Is this still true?
| https://mathoverflow.net/users/10583 | Characterizations of a linear subspace associated with Fourier series | This is to summarize what were discussed in the comments, so the title will not be listed as unanswered.
The linear subspace $S$ of $c\_0(\mathbb{Z})$ is equal to the convolution product of two copies of $\ell^2(\mathbb{Z})$.
More precisely, $\lbrace a\_n \rbrace$ is in $S$ if and only if there exist two sequences $\lbrace b\_n \rbrace$ and $\lbrace c\_n \rbrace$ in $\ell^2(\mathbb{Z})$ such that $$a\_n=\sum\_{k=-\infty}^\infty b\_k c\_{n-k} $$ for all $n$.
This follows since every function in $L^1[0,2\pi]$ is a product of two functions in $L^2[0,2\pi]$, and that for any functions $f,g$ in $L^2[0,2\pi]$ one has, by Parseval identity,
$$\frac{1}{2\pi}\int\_{-\pi}^\pi f(x)g(x)e^{-inx}dx=\frac{1}{2\pi}\int\_{-\pi}^\pi f(x)\overline{h(x)}e^{-inx}dx=\sum\_{k=-\infty}^\infty \hat{f}(k) \hat{g}(n-k)$$
where $h(x)=\overline{g(x)}$.
(One also uses that the mapping that maps each $f$ in $L^2[0,2\pi]$ to its Fourier coefficient sequence in $\ell^2(\mathbb{Z})$ is a surjective isomorphic isometry.)
| 2 | https://mathoverflow.net/users/10583 | 46773 | 29,584 |
https://mathoverflow.net/questions/46742 | 2 | $A$ a Cohen-Macaulay ring (not necessarily local), $M$ a Cohen-Macaulay $A$-module. Then does it necessarily follow that $\mbox{ann}(M)$ is height-unmixed?
| https://mathoverflow.net/users/5292 | CM module is height-unmixed? | This is true if $A$ is local but fails in general.
First, a counterexample. Let $A=\mathbb Z[X]$ and $M= A/p\oplus A/q$ with $p=(2)$, $q=(3,X)$. Since any maximal ideal $m$ of $A$ can not contain both $2$ and $3$, when you localize at $m$ only one of the summands can survive at most, so $M\_m$ will be CM. The annihilator is $p\cap q$ which is mixed.
Now assume $A$ is local and $I = \text{ann} (M)$ . We need to show that all the associated prime of $I$ have same height. Any associated prime $p$ of $I$ would be an associated prime of $M$, as $M$ is a faithful $A/I$ module. But then $\dim A/p = \dim M$ (see Bruns-Herzog, Theorem 2.1.2). Now, as $A$ is also CM, $\text{height}(p) = \dim A - \dim A/p =\dim A-\dim M$. QED
| 5 | https://mathoverflow.net/users/2083 | 46783 | 29,592 |
https://mathoverflow.net/questions/46769 | 17 | Define the *length* of a set of arithmetic progressions
of natural numbers
$A=\lbrace A\_1, A\_2, \ldots \rbrace$
to be $\min\_i | A\_i |$: the length of the shortest sequence
among all the progressions.
Say that $A$ *exactly covers* a set $S$
if $\bigcup\_i A\_i = S$.
Let $P'$ be the primes excluding 2.
>
> What is the longest set of arithmetic progressions
> that exactly covers the primes $P'$?
>
>
>
In other words, I want to maximize the length of
a set of such arithmetic progressions.
Call this maximum $L\_{\max}$.
$L\_{\max} \ge 2$ because
$$
P' \;=\;
\lbrace 3,5 \rbrace
\cup
\lbrace 7,11 \rbrace
\cup
\cdots
\cup
\lbrace 521,523 \rbrace
\cup
\cdots
$$
Perhaps it is possible that
$$P' \;=\;
\lbrace 3, 11, 19 \rbrace
\cup
\lbrace 5, 17, 29,41,53 \rbrace
\cup
\lbrace 7,19,31,43 \rbrace
\cup
\cdots
\;,$$
but I cannot get far with sequences of length $\ge 3$.
(I know Green-Tao establishes that there are arbitrarily
long arithmetic progressions in $P$, but I don't know
if that helps with my question.)
I am number-theoretically naïve,
and apologize if this question is nonsensical or trivial.
In any case, I appreciate the tutoring!
**Addendum**. Although my question should be revised (as Idoneal suggests)
in light of George Lowther's proof that 3
cannot be in a progression of length 4, George has shown that it is likely that $L\_{\max}=3$
but certification requires resolving an open problem. So I've added the *open-problem* tag.
Thanks for everyone's interest!
| https://mathoverflow.net/users/6094 | Covering the primes by arithmetic progressions | Despite the comments to the question (including mine), this is a bit easier than it seems at first sight. We can show that $L\_{\max}=2$ or $3$. Almost certainly we have $L\_\max=3$. However, determining which of these is actually the case seems to be beyond current technology, according to this MO answer "[Are all primes in a PAP-3?](https://mathoverflow.net/questions/34197/are-all-primes-in-a-pap-3/34298#34298)".
Showing that, $L\_{\max} < 4$ is easy. That is, not every odd prime is contained in an arithmetic progression of primes of length 4. More specifically, 3 is not contained in an arithmetic progression of length 4. Suppose that $\lbrace x, x+d, x+2d, x+3d \rbrace$ was such a progression for $d > 0$. Then $x\not=2$, otherwise we would have $x=2,d=1$, but $x+2d=4$ is not prime. So, $x=3$. But, then, $x+3d=3(1+d)$ is not prime.
Edit: Looking at $\tilde L\_\max \equiv \max\_A\liminf\_i \vert A\_i\vert$ might be more interesting. I expect that this is infinite but, again, showing that $\tilde L\_\max > 2$ appears to be beyond current means.
| 10 | https://mathoverflow.net/users/1004 | 46784 | 29,593 |
https://mathoverflow.net/questions/46523 | 7 | If $X$ is a smooth, projective variety over $\mathbb{F}\_q$, the [Weil conjectures](http://en.wikipedia.org/wiki/Weil_conjectures) tell us:
$$\prod \mathrm{det} (I - TF|\_{H^i\_c(X)})^{(-1)^{i+1}} = \mathrm{exp}\left(\sum\_{m=1}^{\infty} \frac{N\_m}{m} T^m \right)$$
here, $T$ is a formal variable, $H^i\_c(X)$ is an appropriate cohomology theory, $F$ is the Frobenius automorphism, and $N\_m$ is the number of $\mathbb{F}\_{q^m}$ points of $X$.
I would like to replace $\mathbb{F}\_q$ with $\mathbb{C}((z))$, on the pretext that both have absolute Galois group $\hat{\mathbb{Z}}$. I am thinking of $\mathbb{C}((z))$ as the ring of functions on a very small punctured disc, and of a variety over $\mathbb{C}((z))$ as a family over this punctured disc. I will also conflate the Frobenius automorphism with the monodromy action.
>
>
> >
> > Let $X$ be a variety over $\mathbb{C}((t))$; interpret the LHS of the equation above by understanding $F$ as the monodromy action. In what, if any, sense does the number $N\_m$ count points over the field $\mathbb{C}((t^{1/m}))$ ?
> >
> >
> >
>
>
>
Note that if one naively takes the cardinality of the set of these points, one would often find $N\_m = \mathrm{\infty}$.
>
>
> >
> > Is there any structure on the set of $\mathbb{C}((t))$ points which would allow me to take an Euler number?
> >
> >
> >
>
>
>
Finally, let me view $\overline{\mathbb{C}((t))}$ as the field over which tropical geometry happens. Making the modifications appropriate to discuss the non-projective case, let me take $X$ to be an affine variety.
>
>
> >
> > Can I "count" $\mathbb{C}((t^{1/m}))$ points of $X$ in terms of "counting" $\frac{1}{m} \mathbb{Z}$ points of $\mathrm{Trop}(X)$?
> >
> >
> >
>
>
>
| https://mathoverflow.net/users/4707 | Zeta function of monodromy and counting points over C((t)) | Hi Vivek. You should have a look to the following papers :
>
> J. Nicaise and J. Sebag (2007).
> *Motivic Serre invariants, ramification, and the analytic Milnor
> fiber*. Inventiones mathematicae,
> **168** (1), p. 133-173.
>
>
> J. Nicaise (2009). *A trace formula
> for rigid varieties, and motivic Weil
> generating series for formal schemes*.
> Math. Ann. **343**, p. 285–349.
>
>
>
They prove a trace formula along the lines you suspect: Let $X$ be a proper $\mathbf C((t))$-variety, let $S(X)$ be its motivic Serre invariant: this is an element of the Grothendieck group of complex varieties modulo the ideal generated by $\mathbf L-1$, where $\mathbf L$ is the class of the affine line; it is computed as the special fiber of any weak Néron model of~$X$ over $\mathbf C[[t]]$. The Euler characteristic of $S(X)$ is well-defined
and $\chi(S(X))$ equals the (alternate sum) of the traces of the action of the monodromy on the étale cohomology of $\bar X$.
| 9 | https://mathoverflow.net/users/10696 | 46789 | 29,594 |
https://mathoverflow.net/questions/46785 | 12 | Hello,
I'd love to learn more about the field of additive combinatorics. From what I've understand, there's a book by Tao and Vu out on the subject, and it looks fun, but I think I lack the prerequisites. Right now, I've had basic real analyis (Rudin), read the first volume of Stanley's "Enumerative combinatorics", and some algebra (some graduate). I have no experience of probability theory whatsoever, or functional analysis or ergodic theory. So I'm curious, from my background, what would be needed to reach the level where I can read and understand the book of Tao and Vu? Are there any certain books to reach that level which you can recommend?
Best regards,
CM
| https://mathoverflow.net/users/10984 | A learning roadmap for Additive combinatorics. | Some portions of their book should be accessible without too much background. Take a look at their sections on additive geometry, graph-theoretic methods, and algebraic methods, for example. For the bulk of the book, though, knowing some probability theory will make a big difference.
A recent book that I like and you might find more accessible is Alfred Geroldinger-Imre Z. Ruzsa, "Combinatorial Number Theory and Additive Group Theory", Birkhäuser, 2009.
From their Foreword:
>
> This book collects the material delivered in the 2008 edition of the *DocCourse in Combinatorics and Geometry* which was devoted to the topic of *Additive Combinatorics*.
>
>
> The two first parts, which form the bulk of the volume, contain the two main advanced courses, *Additive Group Theory and Non-unique Factorizations*, by Alfred Geroldinger, and *Sumsets and Structure*, by Imre Z. Ruzsa.
>
>
> The first part focusses on the interplay between zero-sum problems, arising from the Erdős–Ginzburg–Ziv theorem, and nonuniqueness of factorizations in monoids and integral domains.
>
>
> The second part deals with structural set addition. It aims at describing the structure of sets in a commutative group from the knowledge of some properties of its sumset.
>
>
> The third part of the volume collects some of the seminars which accompanied the main courses and covers several aspects of contemporary methods and problems in Additive Combinatorics.
>
>
>
I would recommend that you work through the second part, and see how you find the material. It should be accessible.
You may also want to take a look at Ben Green's [notes](http://www.dpmms.cam.ac.uk/%7Ebjg23/notes.html) on the structure theory of Set Addition.
Let me add: If you are mainly interested in classical additive combinatorics, as it applies to the natural numbers, then I strongly recommend Melvyn Nathanson, "Additive number theory. Vol II: Inverse problems and the geometry of sumsets", Springer, GTM 165, 1996.
| 7 | https://mathoverflow.net/users/6085 | 46790 | 29,595 |
https://mathoverflow.net/questions/46794 | 1 | Let $A$ be a commutative ring with identity. If $A$ is a ring with only a finite set of prime ideals $p\_1...p\_n$ and moreover $\prod\_{i=1}^n p\_i^{k\_i}=0$ for some k\_i. Is $A$ then isomorphic to $\prod\_{i=1}^nA\_{(p\_i)}$?
| https://mathoverflow.net/users/10988 | Generalization of the Structure theorem for artinian rings? | No. Let $A$ be a DVR. It has two prime ideals: the maximal ideal $p\_1=\mathfrak m\subset A$ and $p\_2=(0)\subset A$. So, $p\_1p\_2=0$, but $A$ is not a product (of two local rings).
| 8 | https://mathoverflow.net/users/10076 | 46796 | 29,599 |
https://mathoverflow.net/questions/46779 | 1 | Let $X,Y$ be smooth varieties over a field $k$ (which in my case is perfect of finite characteristic $p$; we may also assume that $X,Y$ are connected); $s:Y\to X$ is a finite morphism of degree $d$. Then the graph of $s$ could be considered both as a finite correspondence from $Y$ to $X$ and as a finite correspondence from $X$ to $Y$ (in the sense of Voevodsky, or in the sense of the 'classical' intersection theory). Now compose these correspondences to get a correspondence from $X$ to $X$ (using intersection theory); is this true that the result is $d\cdot id\_{X}$? I would say yes, since the composite correspondce seems to have only the diagonal component as a cycle in $X\times X$, and it would be very strange if its multiplicity would be anything else but d. Yet I would be glad both for any explanations/references here as well as for pointing out any subtle points (since I don't know much about intersection theory). Does one require any extra conditions here? It would be ok for me to replace $X$ and $Y$ by some open subvarieties, but I don't want to demand $s$ to be generically etale.
| https://mathoverflow.net/users/2191 | A composition of a finite morphisms with the transpose correspondence: is it the multiplication by the degree? | Let $f:Y \to X\times Y$ be the graph of $s$ and $f^T:Y \to Y\times X$ the transpose of the graph. By definition you should take the fiber product of $(1\times f^T):X\times Y \to X\times Y\times X$ and $(f\times 1):Y\times X \to X\times Y\times X$. It is easy to see that the fiber product is $Y$. Then we should take the composition of the resulting map $Y \to X\times Y\times X$ with the projection $X\times Y\times X \to X\times X$. It is easy to see that this map factors as $Y \stackrel{s}\to X \stackrel{\Delta}\to X\times X$. This is (by definition) the composition of the correspondences. It remains to note that the pushforward of the fundamental cycle of $Y$ along $s$ is $d$ times the fundametal cycle of $X$, and the the diagonal gives the identity map.
| 1 | https://mathoverflow.net/users/4428 | 46803 | 29,604 |
https://mathoverflow.net/questions/46787 | 24 | A well known theorem in algebraic topology relates the (co)homology of the Thom space $X^\mu$ of a orientable vector bundle $\mu$ of dimension $n$ over a space $X$ to the (co)homology of $X$ itself: $H\_\ast(X^\mu) \cong H\_{\ast-n}(X)$ and $H^\ast(X^\mu) \cong H^{\ast-n}(X)$.
This isomorphism can be proven in many ways: Bott & Tu has an inductive proof using good covers for manifolds and I learned on MathOverflow that one can use a relative Serre spectral sequence. However, I believe that there should also be a proof using stable homotopy theory, in the case of homology by directly constructing a isomorphism of spectra $X^\mu \wedge H\mathbb{Z} \to X\_+ \wedge \Sigma^{-n} H\mathbb{Z}$, where $X^\mu$ denotes the Thom spectrum, $H\mathbb{Z}$ the Eilenberg-Mac Lane spectrum for $\mathbb{Z}$ and $X\_+$ the suspension spectrum of $X$ with a disjoint basepoint added.
Is there an explicit construction of such a map implementing the Thom isomorphism on the level of spectra? I am interested in such a construction for both homology and cohomology. If so, is there a similar construction for generalized (co)homology theories? I would also be interested in references.
| https://mathoverflow.net/users/798 | Is there a map of spectra implementing the Thom isomorphism? | There is a construction for both Thom isomorphisms, homological and cohomological, via classical stable homotopy theory. You find the details in Rudyaks book "On Thom spectra, orientability, and cobordism", chapter V, §1. The Thom class is a map $X^{\mu} \to\Sigma^{n} H \mathbb{Z}$. Moreover, there is a map of spectra $X^{\mu} \to X\_+ \wedge X^{\mu}$ which is induced from the map of vector bundles $\mu \to \mathbb{R}^0 \times \mu$ over the diagonal map $X \to X \times X$. Here is the definition of the homological Thom isomorphism; the cohomological one is in the same spirit. Consider the composition
$X^{\mu} \wedge H \mathbb{Z} \to X\_+ \wedge X^{\mu} \wedge H\mathbb{Z} \to X\_+ \wedge \Sigma^n H \mathbb{Z} \wedge H \mathbb{Z} \to X\_+ \wedge \Sigma^n H \mathbb{Z} $. On homotopy groups, it induces a map lowering the degree by $n$ (there is a sign mistake in your question that confused me for some minutes).
It is clear that this works for orientations with respect to other ring spectra as well.
| 23 | https://mathoverflow.net/users/9928 | 46809 | 29,609 |
https://mathoverflow.net/questions/46804 | 37 | I was writing up some notes on harmonic analysis and I thought of a question that
I felt I should know the answer to but didn't, and I hope someone here can help me.
Suppose I have a compact Riemannian manifold $M$ on which a compact Lie group $G$
acts isometrically and transitively---so you can think of $M$ as $G/K$ for some closed
subgroup $K$ of $G$. Then the real Hilbert space $H = L^2(M, \mathbb{R})$ is an orthogonal representation space of $G$ and hence splits as an orthogonal direct sum of finite
dimensional irreducible sub-representations. On the other hand, the Laplacian $L$
of $M$ is a self-adjoint operator on $H$, so $H$ is also the orthogonal direct sum
of its eigenspaces---which are also finite dimensional. My question is, when do these
two orthogonal decompositions of $H$ coincide? Put slightly differently, since $L$
commutes with the action of $G$, each eigenspace of $L$ is a finite dimensional
subrepresentation of $H$ and so a direct sum of irreducibles, and I would like to
know conditions under which each eigenspace is in fact irreducible. For example, this
is true for the circle acting on itself and for $SO(3)$ acting on $S^2$ (where we
get the harmonic polynomials of various degrees). Is it perhaps always true for the
case of a symmetric space? Of course a standard reference in addition to the answer
would be most welcome.
| https://mathoverflow.net/users/7311 | When are the eigenspaces of the Laplacian on a compact homogeneous space irreducible representations? | The Peter-Weyl theorem tells you that $L^2(G)$ is isomorphic to $\bigoplus\_{\pi}\pi\otimes\pi^\*$ as $G\times G$ representation, where $\pi$ runs through all irreducible unitary representations.
It follows that
$$
L^2(G/K)\cong L^2(G)^K\cong\bigoplus\_\pi \pi\otimes(\pi^\*)^K.
$$
So, the first thing you absolutely need, is a multiplicity one property, which says that $\dim\pi^K\le 1$ for every $\pi$.
This is already a rare property, but known to be true for, say $G=SO(n)$ and $K=SO(n-1)$, see Zhelobenko's book for this.
But, the Laplacian may have the same eigenvalue on different representations.
For this you need highest weight theory (see for instance the book by Broecker and tom Dieck): Assume $G$ to be connected. The irreducible representations are parametrized by highest weights and the Laplace eigenvalue depends on the value of a quadratic form on the space of weights.
So, in each case you need to identify those weights with $K$-invariants and consider the values of the quadratic form, which in the case of a simple group should be the Killing form. I guess that in the above cases it might actually be true.
| 23 | https://mathoverflow.net/users/nan | 46814 | 29,611 |
https://mathoverflow.net/questions/46815 | 13 |
>
> Question: Assuming finiteness of the Tate-Shafarevich group, is there an algorithm to determine whether a curve $C$ defined over a number field $K$ has infinitely many $K$-rational points?
>
>
>
I believe that this is (a) true and (b) sufficiently important that it has been carefully explained somewhere, but I don't know a reference. Any help from the MO community would be very much appreciated!
P.S. To make the question precise, $C$ is specified, say, by its function field (a simple extension of $K(x)$), and all abelian varieties over all number fields have finite Sha. (And if the algorithm takes $10^{10}$ years for C: y=3x+1 over the rationals, I don't care.)
| https://mathoverflow.net/users/3132 | Which curves have infinitely many rational points | I believe the following is an algorithm, albeit a horrible one.
First, as the OP surely knows, it comes down entirely to curves of genus one. Indeed, if the genus is at least $2$ then by Faltings' Theorem there are only finitely many $K$-rational points, whereas if the genus is zero, there are infinitely many rational points iff the curve is isomorphic over $K$ to the projective line iff a certain Hilbert symbol vanishes. This is all very well understood.
Step 1: If for an elliptic curve $E\_{/K}$ the group $Sha(K,E)$ is finite, then there is an algorithm to compute the Mordell-Weil group $E(K)$.
Indeed, it's enough to know that there exists some prime number $p$ such that $Sha(K,E)[p] = 0$. Then the weak Mordell-Weil group $E(K)/pE(K)$ is isomorphic to the $p$-Selmer group, which is known to be (in principle!) effectively computable. Since the torsion subgroup is well-known to be effectively computable, knowing $E(K)/pE(K)$ gives us the Mordell-Weil rank, and if you know the rank then by enough searching you can find a basis for the free part of the Mordell-Weil group.
[**Added**: You don't actually need to know an explicit value of such a prime number $p$. You can compute the $p$-Selmer group for any value of $p$ you want and you can set up a program that given infinite time will compute $E(K)/pE(K)$. By running these programs on enough primes simultaneously, in finite time you will find a prime $p$ such that $E(K)/pE(K) = \operatorname{Sel}(K,E)[p]$.]
Step 2: Suppose that $C\_{/K}$ is a genus one curve over $K$. One may effectively decide (Hensel's Lemma, Weil bounds...) whether or not $C$ has points over every completion of $K$. If not, then certainly $C(K)$ is empty and hence finite.
Step 3: Next compute the Mordell-Weil group of the Jacobian elliptic curve of $C$ using Step 1. If this group is finite, then $C(K)$ is finite -- possibly empty.
Step 4: Suppose that $C$ has points everywhere locally and the Jacobian $E$ has positive rank. Then $C$ represents an element of $Sha(K,E)[n]$ for some $n \in \mathbb{Z}$. Since we can effectively compute the weak Mordell-Weil and Selmer groups of $E$, we can compute $Sha(K,E)[n]$. If it happens to be trivial then $C$ is necessarily isomorphic to $E$ so has infinitely many rational points.
Step 5: Finally, suppose that $Sha(K,E)[n]$ is nontrivial. Thus the question is whether $C$ represents a nontrivial element of this group. But one can compute defining systems of equations for each of the curves $C\_i$ representing the elements of this group (I am pretty sure, anyway; if this is the sticking point, let me know and I'll think about it more). Now one can do the following ridiculous thing: search for an isomorphism between $C$ and $C\_i$ by trying all possible maps. We know that $C$ is isomorphic to one of these curves -- possibly $C\_1 = E$ -- so eventually we will find it!
[**Added**: the explicit geometric realization of elements of the $n$-Selmer group is discussed in [this important paper](http://www.mathe2.uni-bayreuth.de/stoll/papers/Explicit-Descent-II-2006-11-20.pdf).]
| 16 | https://mathoverflow.net/users/1149 | 46819 | 29,613 |
https://mathoverflow.net/questions/36085 | 11 | A Hausdorff space $(X,\tau)$ is said to be **minimal Hausdorff** if for each topology $\tau' \subseteq \tau$ with $\tau' \neq \tau$ the space $(X,\tau')$ is not Hausdorff.
Every compact Hausdorff space is minimal Hausdorff.
I would like to know:
1) Is every minimal Hausdorff space compact?
2) Does every Hausdorff topology contain a minimal Hausdorff topology?
Many thanks!
| https://mathoverflow.net/users/8628 | Minimal Hausdorff | The answer to both questions is no - see 7.5 in Porter and Woods book, "Extensions and Absolutes of Hausdorff Spaces", Springer-Verlag, 1988. The space of rational numbers with the usual topology has no coarser minimal Hausdorff topology. Every Hausdorff space can be embedded in a minimal Hausdorff space; in particular, if you start with a Hausdorff space X that is not Tychonoff and embed it in a minimal Hausdorff space Y, then Y can not be compact Hausdorff.
| 12 | https://mathoverflow.net/users/10995 | 46825 | 29,619 |
https://mathoverflow.net/questions/46802 | 5 | Let $M,B$ be $R$-modules, and suppose we're given an n-extension $E\_1\to\dots\to E\_n$ of $B$ by $M$, that is, an exact sequence $$0\to M\to E\_1\to\dots\to E\_n \to B\to 0.$$
A morphism of $n$-extensions of $Y$ by $X$ is defined to be a hammock
$$\begin{matrix}
&&A\_1&\to&A\_2&\to&A\_3&\to&\ldots&\to &A\_{n-2}&\to &A\_{n-1}&\to& A\_{n}&&\\
&\nearrow&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&\searrow&\\
X&&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&&Y\\
&\searrow&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&\nearrow&\\
&&B\_1&\to&B\_2&\to&B\_3&\to&\ldots&\to &B\_{n-2}&\to &B\_{n-1}&\to& B\_{n}&&\end{matrix}$$
This detetermines a category $n\operatorname{-ext}(Y,X)$. Further, we're given an almost-monoidal sum on this category given by taking the direct sum of exact sequences, then composing the first and last maps with the diagonal and codiagonal respectively. Taking connected components, we're left with a set $ext^n(Y,X)$, and the sum reduces to an actual sum on the connected components, which turns $ext^n(Y,X)$ into an abelian group (and therefore an $R$-module).
It's well known that these functors, called Yoneda's Ext functors are isomorphic to the Ext functors $\pi\_n(\underline{\operatorname{Hom}}(sM,sN))$ where $\underline{\operatorname{Hom}}(sM,sN)$ is the homotopy function complex between the constant simplicial $R$-modules $sM$ and $sN$ (obtained by means of cofibrant replacement of $sM$ in the projective model structure, fibrant replacement of $sN$ in the injective model structure, or by any form of Dwyer-Kan simplicial localization (specifically the hammock localization)).
In a [recent answer](https://mathoverflow.net/questions/45534/formally-smooth-morphisms-the-cotangent-complex-andre-quillen-cohomology-and-r/45987#45987), Charles Rezk mentioned that we can compute this in the case $n=1$ as $\pi\_0(\underline{\operatorname{Hom}}(sM,sN[1]))$, where $sN[1]$ is the simplicial $R$-module with homotopy concentrated in degree $n$ equal to $N$. That is, these are exactly the maps $M\to N[1]$ in the derived category.
It was also mentioned that for the case $n=1$, there exists a universal exact sequence in the derived category: $$0\to N\to C\to N[1]\to 0$$ where $C$ is weakly contractible such that every extension of $N$ by $M$ arises as $$N\to M\times^h\_{N[1]} C\to QM,$$ which is $\pi\_0$-short exact. (And QM is a cofibrant replacement of $M$, for obvious reasons).
Questions:
Why can we get $\pi\_1$ of the function complex by looking at maps into $N[1]$? At least on the face of it, it seems like we would want to look at maps into $N[-1]$, that is, look at maps into the "loop space", not the "suspension" (scare quotes because these are the loop and suspension functors in $sR\operatorname{-Mod}$, not in $sSet$.
What is this this "universal extension" $$0\to N\to C\to N[1]\to 0$$ at the level of simplicial modules?
Is there a similar "universal extension" for $n>1$? If so, what does it (one of its representatives) look like at the level of simplicial modules?
Given an $n$-extension $\sigma$ of $B$ by $M$, how can we produce a morphism in the derived category $B\to M[n]$ that generates an $n$-extension in the same connected component of $n\operatorname{-ext}(B,M)$?
Lastly, since Yoneda's construction of Ext looks suspiciously like Dwyer and Kan's hammock localization, and there is homotopy theory involved in the other construction, I was wondering if there was any connection between the two. That is, I was wondering if there is another construction of Ext using DK-localization directly that shows why Yoneda's construction works.
| https://mathoverflow.net/users/1353 | Simplicial "universal extensions", the hammock localization, and Ext | For your first question:
>
> Given an $n$-extension $\sigma$ of $B$ by $M$, how can we produce a morphism in the
> derived category $M\to N[n]$ that generates an n-extension in the same connected
> component
> of $n{-}ext(B,M)$?
>
>
>
Do you mean a morphism $B\to M[n]$? In which case, isn't this a the standard construction? Let $E$ be the extension of $B$ by $M$, let $P$ be a projective resolution of $B$, and cover the identity map of $B$ by a map of chain complexes $P\to E$. The resulting map $P\_n\to M$ can be thought of as a morphism $B\to M[n]$ in the derived category, and it's the one that represents the corresponding component of $n{-}ext(B,M)$.
As for your second question, there's certainly some connection. What you've described is a category $C=n{-}ext(X,Y)$, whose objects are $n$-extensions of $Y$ by $X$. Let $W=C$, and perform the hammock construction $L\_H(C,W)$ of the pair $(C,W)$; then $L\_H(C,W)$ is a simplicially eniched category, which represents an $\infty$-groupoid since you inverted everything. The ext group is the set of equivalence classes of objects in $L\_H(C,W)$.
Dwyer and Kan showed that if $C=W$, then the simplicial category $L\_H(C,W)$ and the simplicial nerve $NC$ of $C$ represent the same $\infty$-groupoid. So $\mathrm{Ext}^n(X,Y)=\pi\_0 (NC)$.
What is the homotopy type of $NC$? It is supposed to be equivalent to the space of maps $\mathrm{map}(X,Y[n])$ in the $\infty$-category associated to the derived category.
I'm unaware of a reference where this is proved, however.
But take a look at Stefan Schwede's very nice paper on [An exact sequence interpretation of the Lie bracket in Hochschild cohomology](http://www.math.uni-bonn.de/~schwede/Ext-Crelle.pdf), which shows that $\pi\_1(NC)=\mathrm{Ext}^{n-1}(X,Y)$, and gets some interesting results from that. (He's looking at $A$-bimodule extensions of $A$ by $A$ (i.e., Hochschild cohomology), and he extracts the Gerstenhaber-algebra structure on the ext-groups by means of homotopical constructions involving the $NC$.
| 7 | https://mathoverflow.net/users/437 | 46829 | 29,621 |
https://mathoverflow.net/questions/46834 | 1 | Given a flat and projective morphism $f:X\rightarrow Y$ of noetherian schemes over some algebraically closed $k$ and $F$, $G$ coherent $O\_X$-modules, flat over $Y$.
Then the base change theroem for the relative $Ext$ sheaves reads:
Let $y\in Y$ and assume $\tau^i(y): \mathcal{E}xt\_f^i(F,G)\otimes k(y)\rightarrow Ext\_{X\_y}^i(F\_y,G\_y)$ is surjective. Then:
i) There is a neighbourhood $U$ of $y$ s.t. $\tau^i(y')$ is an isomorphism for all $y' \in U$
ii) $\tau^{i-1}(y)$ is surjective if and only if $\mathcal{E}xt\_f^i(F,G)$ is locally free in a neighbourhood of $y$
So now assume, we are in the following situation: $Y$ is a smooth projective surface and $X$ is the product of $Y$ with another projective smooth surface and $f$ is the projection. We have $Ext\_{X\_y}^i(F\_y,G\_y)=0$ for all $i\geq3$ and all $y\in Y$. Furthermore we have: $Ext\_{X\_y}^i(F\_y,G\_y)=0$ for $i=0,1,2$ and all $y\in Y\setminus \{y\_0\}$ for a fixed $y\_0 \in Y$. Finally $Ext\_{X\_{y\_0}}^0(F\_{y\_0},G\_{y\_0})=Ext\_{X\_{y\_0}}^2(F\_{y\_0},G\_{y\_0})=k^s$ and $Ext\_{X\_{y\_0}}^1(F\_{y\_0},G\_{y\_0})=k^{2s}$ for some $s\geq1$.
The claim is that we get (with an application of the base change theorem):
(a)$\mathcal{E}xt\_f^i(F,G)=0$ for $i=0,1$ and (b)$\mathcal{E}xt\_f^2(F,G)\otimes k(y\_0)=k^s$
So, since $Ext\_{X\_y}^3(F\_y,G\_y)=0$ for all y, we have by i) that $\tau^3(y)$ is surjective for all y, so it is an isomorphism for all $y\in Y$ and $\mathcal{E}xt\_f^3(F,G)=0$, which this is locally free on $Y$. So $\tau^2(y)$ is an isomorphism for all $y\in Y$ by ii). So we get (b).
But how about (a). We have that $\tau^1(y)$ is surjective for all $y\in Y\setminus y\_0$. But what about $\tau^1(y\_0)$. What can be said about $\mathcal{E}xt\_f^2(F,G)$? Is it localy free somewhere? How do we get the vanishing of $\mathcal{E}xt^1\_f(F,G)$ with the base change theorem?
Background: I'm still working on the Example [http://books.google.de/books?id=\_mYV1q0RVzIC&printsec=frontcover&dq=%22geometry+of+moduli+spaces+of+sheaves%22&source=bl&ots=ZGL9LDSVjV&sig=p9l1BZ-UOXZdlxakuC3k15sTtBk&hl=de&ei=flHpTOHyPI3oOaXj7IcK&sa=X&oi=book\_result&ct=result&resnum=2&sqi=2&ved=0CCMQ6AEwAQ#v=onepage&q=base%20change&f=false](http://books.google.de/books?id=_mYV1q0RVzIC&printsec=frontcover&dq=%22geometry%20of%20moduli%20spaces%20of%20sheaves%22&source=bl&ots=ZGL9LDSVjV&sig=p9l1BZ-UOXZdlxakuC3k15sTtBk&hl=de&ei=flHpTOHyPI3oOaXj7IcK&sa=X&oi=book_result&ct=result&resnum=2&sqi=2&ved=0CCMQ6AEwAQ#v=onepage&q=base%20change&f=false) at the bottom of page 169.
Edit: Sasha gave an answer that uses derived categories. But i'm also interested if one can prove this, just by using i) and ii) in the base change theorem.
| https://mathoverflow.net/users/3233 | Application of the base change theorem | It is always useful to look from the derived category point of view. What you want is the object of $D(Y)$ which is $Rf\_\*R{\mathcal H}om(F,G)$ --- the sheaves ${\mathcal E}xt^i\_f$ are its sheaf cohomology. The base change tells you that $Li^\*Rf\_\*R{\mathcal H}om(F,G) \cong RHom(F\_{y\_0},G\_{y\_0})$, where $i$ is the embedding of the point $y\_0$. Since you know $Ext^2(F\_{y\_0},G\_{y\_0}) = k^s$, you have a map $RHom(F\_{y\_0},G\_{y\_0}) \to k^s[-2]$. It gives by adjunction a map $Rf\_\*R{\mathcal H}om(F,G) \to O\_{y\_0}^s[-2]$. You want to prove that it is an isomorphism. Include this morphism into exact triangle
$$
Rf\_\*R{\mathcal H}om(F,G) \to O\_{y\_0}^s[-2] \to C.
$$
You want to check that $C = 0$. You already know that $C$ is zero on the complement of $y\_0$. Then it suffices to check that $Li^\* C = 0$. Apply $Li^\*$ to the triangle, you will get
$$
RHom(F\_{y\_0},G\_{y\_0}) \to Li^\* O\_{y\_0}^s[-2] \to Li^\* C.
$$
Note that $Li^\* O\_{y\_0} = Li^\* i\_\* k$ has cohomology $k$, $k^2$ and $k$ in degrees $0$, $-1$ and $-2$. Hence you see that the first two terms of the sequence have the same cohomology. So, it remains to check that the map induces their isomorphism. This is the tricky part.
To do this, you need more information about the situation. In your case, remember that $L\_1i^\* O\_{y\_0} = T\_{y\_0}^\* Y$, and since $Y$ is the moduli space of sheaves $G$, it is isomorphic to $Ext^1(G\_{y\_0},G\_{y\_0})^\*$. So, you have to check that $Ext^\bullet(F\_{y\_0},G\_{y\_0})$ as a module over $Ext^\bullet(G\_{y\_0},G\_{y\_0})$ is free, which is evident in your example.
| 4 | https://mathoverflow.net/users/4428 | 46850 | 29,629 |
https://mathoverflow.net/questions/46844 | 1 | The Hodge star operator $\ast$ acts on the differential forms of a differential manifold sending $\Omega^{k}$ to $\Omega^{N-k}$. If the manifold is complex, then for $p+q=k$, does $\ast$ map $\Omega^{p,q}$ into some $\Omega^{a,b}$, where $a+b=N-k$.
| https://mathoverflow.net/users/1867 | Does the Hodge star operator respect complex structure? | As Abtan requested, I'm converting my comments to an answer:
Suppose that $X$ is an $N$ (complex) dimensional complex manifold endowed with a Hermitean metric, or equivalently a Riemannian metric g satisfying $g(JX,JY)=g(X,Y)$, where $J$ is the complex structure.
Let $\*$ denote the $\mathbb{C}$-antilinear extension of the Hodge star operator to complex
valued forms (some people -- including me -- prefer to write this as $\overline{\*}$ as Spiro points out in the comments). Then as one finds on page 82 of Griffiths and Harris,
$$\*\Omega^{pq}\subset \Omega^{N-q,N-p}$$
where I'm following the notation in the question and writing $\Omega^{pq}$ for the
space of $C^\infty$ forms of type $(p,q)$.
| 5 | https://mathoverflow.net/users/4144 | 46853 | 29,631 |
https://mathoverflow.net/questions/46854 | 15 | Let $f: \mathbb R^n \rightarrow \mathbb R^n$, where $n> 1$ be a bijective map such that the image of every line is a line.
Is $f$ continuous?
I think it is, but the proof isn't immediately obvious to me.
Related to [this question on math.SE](https://math.stackexchange.com/q/11232/3638).
Feel free to retag.
| https://mathoverflow.net/users/10876 | Continuity in terms of lines | This is called the fundamental theorem of affine geometry. Let $f : E \to E'$ be a map between affine spaces over a field $K$. Suppose that
1. $f$ is bijective;
2. $\dim E=\dim E'\ge 2$;
3. If $a, b, c\in E$ are aligned, then so are $f(a), f(b), f(c)$.
Then $f$ is semi-affine: fix some $a\_0\in E$, then there exists a field automorphism $\sigma$ of $K$ such that the map $h: v\mapsto f(a\_0+v)-f(a\_0)$ (which goes from the vector space attached to ${E}$ to that attached to $E'$) is additive and $h(\lambda v)=\sigma(\lambda)h(v)$ for all $v$ and all $\lambda \in K$. I don't have an URL for this theorem, I find it in Jean Fresnel: *Méthodes Modernes en Géométrie*, Exercise 3.5.7. But I think it is in any standard textbook on affine geometry.
When $K=\mathbb R$, it is known that $K$ has no non-trivial field automorphism. So your $f$ is an affine function, hence continuous. If $K=\mathbb C$, as pointed out by Kevin in above comments, take any non-trivial automorphism of $\mathbb C$, then you get a semi-affine map $\mathbb C^n \to \mathbb C^n$ which will not be affine, even not continuous (if $\sigma$ is not the conjugation).
| 22 | https://mathoverflow.net/users/3485 | 46860 | 29,635 |
https://mathoverflow.net/questions/46855 | 9 | I know how to generate all Abelian groups of order n, but how would I generate the others? I can't seem to find anything about this.
By "generate", I mean produce the Cayley tables for all groups of order n.
| https://mathoverflow.net/users/866 | How to generate all finite groups of order n? | See
Hans~Ulrich Besche, Bettina Eick, and E.A. O'Brien.
A millennium project: constructing small groups.
*Internat. J. Algebra Comput.*, 12:623-644, 2002.
for a description of the construction of groups of order up to 2000. (I believe they narrowly failed to achieve this before the end of the year 2000.) In fact they did not construct the groups of order 1024 individually, but it is known that there are $49\,487\,365\,422$ groups of that order. The remaining $423\,164\,062$ groups of order up to 2000 (of which $408\,641\,062$ have order 1536) are available as libraries in GAP and Magma.
I would guess that 2048 is the smallest number such that the exact number of groups of that order is unknown. It is known that, for $p$ prime, the number of groups of order $p^n$ grows as $p^{\frac{2}{27}n^3+O(n^{8/3})}$: see <http://en.wikipedia.org/wiki/P-group>.
| 18 | https://mathoverflow.net/users/35840 | 46861 | 29,636 |
https://mathoverflow.net/questions/46866 | 52 | The falsity of the following conjecture would be a nice counter-intuitive fact.
Given a square sheet of perimeter $P$, when folding it along origami moves, you end up with some polygonal flat figure with perimeter $P'$.
**Napkin conjecture**: You always have $P' \leq P$.
In other words, you cannot increase the perimeter using any finite sequence of origami folds.
**Question 1**: Intuition tells us it is true (how in hell can it increase?). Yet, I think I read somewhere that there was some weird folding (perhaps called "mountain urchin"?) which strictly increases the perimeter. Is this true?
*Note 1*: I am not even sure that the initial sheet's squareness is required.
I cannot find any reference on the internet. Maybe the name has changed; I heard about this 20 years ago.
The second question is about generalizing the conjecture.
**Question 2**: With the idea of generalizing the conjecture to continuous folds or bends (using some average shadow as a perimeter), I stumble on how you can mathematically define bending a sheet. Alternatively, how do you say "a sheet is untearable" in mathematical terms?
*Note 2*: It might also be a matter of physics about how much we idealize bending mathematically.
| https://mathoverflow.net/users/3005 | Is the "Napkin conjecture" open? (origami) | There is a general version of this question which is known as *"the rumpled dollar problem"*. It was posed by V.I. Arnold at his seminar in 1956. It appears as the very first problem in [*"Arnold's Problems"*](http://books.google.co.uk/books?id=LqvLnu8c3ToC&printsec=frontcover&dq=arnold%2527s+problems&source=bl&ots=l0ULwBNvsc&sig=mLNHb1upSiddEtfrHRjbjvJPfzE&hl=en&ei=6ZPpTIvqLcuaOozXvZEK&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBYQ6AEwAA#v=onepage&q&f=false):
>
> Is it possible to increase the perimeter of
> a rectangle by a sequence of foldings and unfoldings?
>
>
>
According to the same source (p. 182),
>
> Alexei Tarasov has shown that a rectangle admits a realizable folding with arbitrarily large perimeter. A realizable folding means that it could be realized in such a way as if the rectangle were made of infinitely thin but absolutely nontensile paper. Thus, a folding is a map $f:B\to\mathbb R^2$ which is isometric on every polygon of some subdivision of the rectangle $B$. Moreover, the folding $f$ is realizable as a piecewise isometric homotopy which, in turn, can be approximated by some isotopy of space (which corresponds to the impossibility of self-intersection of a paper sheet during the folding process).
>
>
>
Have a look at
* A. Tarasov, *Solution of Arnold’s “folded rouble” problem.* (in Russian) Chebyshevskii Sb. 5 (2004), 174–187.
* I. Yashenko, *[Make your dollar bigger now!!!](http://link.springer.com/article/10.1007/BF03025296)* Math. Intelligencer 20 (1998), no. 2, 38–40.
---
A history of the problem is also briefly discussed in Tabachnikov's [review](http://www.math.psu.edu/tabachni/prints/Arnoldpr.pdf) of "Arnold's Problems":
>
> It is interesting that the problem was solved by origami practitioners way before it was posed (at least, in 1797, in the Japanese origami book “Senbazuru Orikata”).
>
>
>
| 47 | https://mathoverflow.net/users/5371 | 46867 | 29,638 |
https://mathoverflow.net/questions/46863 | 6 | Assuming the existence of enough large cardinals (I'm not sure whether I mean in the original V or in L(R), do whatever is standard), is the partially ordered class of cardinals order-isomorphic to something simpler?
If so, what is the weakest large cardinal assumption that gives this result?
Is L(R) ≠ L sufficient?
My best guess would be that it is order-isomorphic to $\omega \cup (\operatorname{Ord} \times \operatorname{Ord})$, with all elements of $\omega$ being smaller than all elements of $\operatorname{Ord} \times \operatorname{Ord}$, and $\operatorname{Ord} \times \operatorname{Ord}$ having the product partial order.
| https://mathoverflow.net/users/nan | What does the partially ordered class of cardinals look like in L(R)? | Ricky:
I assume you mean to ask your question in $L({\mathbb R})$.
In general, without choice, the ordering of cardinals tends to be rather pathological, although we do not yet know by how much. Here is an example: It is open whether ZF proves that, if there is no infinite set all of whose members are pairwise incomparable (cardinality-wise) then choice holds.
On the other hand, if choice fails, then for every finite $n$ there are $n$ pairwise incomparable sets.
Assuming enough large cardinals (in $V$ or, equivalently, determinacy in $L({\mathbb R})$), the ordering of cardinals in $L({\mathbb R})$ is *much more complicated* than you suggest.
To give you an idea of how little we know: We do not know yet whether there are infinitely many successors of $|{\mathbb R}|$.
For an example of the immense complexity present in the ordering, recall that $E\_0$ is the equivalence relation on $2^{\mathbb N}$ given by $x E\_0 y$ iff $\exists n\forall m\ge n (x(m)=y(m))$. Then:
>
> $|2^{\mathbb N}/E\_0|$ is a successor of $|{\mathbb R}|$, and above it but still below $|{\mathcal P}({\mathbb R})|$, you can embed the partial order of Borel subsets of ${\mathbb R}$ under containment.
>
>
>
In fact, you can realize these cardinals by taking suitable quotients of the reals by Borel equivalence relations. All these relations can be taken to be *countable* (i.e., each class has countably many members) and their definitions trace back to free measure preserving actions of $F\_2$ (the free group in two generators) on Polish spaces.
I take this as a strong indication that there is no sense in which we may have a "reasonable" description of the whole partial ordering. But really, we are far from being able to say much. Ketchersid and I have some recent results at the very bottom of the ordering, see "A trichotomy theorem in natural models of $AD^+$," to appear in the Proceedings of Boise Extravaganza in Set Theory, and also our forthcoming paper on "$G\_0$-dichotomies in natural models of $AD^+$." You may also want to take a look at a paper by Woodin that deals with a slightly more demanding version of determinacy, "The cardinals below $|[\omega\_1]^{\lt\omega\_1}|$," Annals of Pure and Applied Logic 140 (2006) 161–232.
Even if we restrict ourselves to better understood classes (specific collections of Borel sets), the ordering is rather elaborate and far from well understood. For a nice subclass for which there *is* a very decent picture, see Alex Andretta, Greg Hjorth, and Itay Neeman, "Effective cardinals of boldface pointclasses," J. of Mathematical Logic, vol. 7 (2007) 35–82.
---
I didn't address this above, but I should probably say something:
The assumption that there are enough large cardinals ensures that the theory of $L({\mathbb R})$ is invariant under forcing, so it is in a sense as canonical as we can hope.
Under weaker assumptions, the partial ordering may vary greatly. For example, $L({\mathbb R})\ne L$ does not suffice to preclude choice in $L({\mathbb R})$.
Assuming you do not have choice, it is open (even for $L({\mathbb R})$) whether well-foundedness of the partial ordering of cardinalities must fail. This is believed to be the case, and it is certainly so in all reasonable cases I have checked. It is easy to give examples by forcing over $L$ where the $L({\mathbb R})$ of one extension and the $L({\mathbb R})$ of another have non-equivalent partial orderings of cardinalities. (For example, by replicating or excluding the behavior mentioned above of quotients by free actions.)
Sometimes we have some form of "control", for example, if $L({\mathbb R})$ is a kind of Solovay model. But it already takes effort to show that in "nice" situations there are no infinite Dedekind finite subsets of ${\mathbb R}$ in $L({\mathbb R})$. In my view, the "right" version of these questions is under large cardinals, so we have canonicity, but already without it there are many difficulties and possibilities that may be interesting to explore.
| 7 | https://mathoverflow.net/users/6085 | 46868 | 29,639 |
https://mathoverflow.net/questions/46871 | 6 | Background/motivation: I'm investigating the construction of models for a first-order modal system (S5) as products of classical models. Since ultraproducts are all classical models and I need non-classical ones as well, I need to look at reduced products where the filter is not an ultrafilter. This leads me to ask about filters in general:
J.L. Bell & A.B. Slomson, in *Models and Ultraproducts* (p. 116), state and prove:
>
> Lemma 1.17. Let I be a countable set.
> Then the collections of non-principal,
> $\omega$-incomplete, uniform, and
> regular ultrafilters on I all
> coincide.
>
>
>
Suppose I alter their definitions slightly so the above properties are all defined for filters in general, then modify the lemma to assert that it holds for filters in general. Would that be true? Can anyone supply a reference to a proof or disproof? Thanks.
| https://mathoverflow.net/users/8224 | Which properties of ultrafilters on countable sets hold for filters in general? | If you use literally the definitions in Bell and Slomson, only changing "ultrafilter" to "filter," and if, as in the lemma you cited, you're interested only in flters on a countable set, then I believe non-principal is equivalent to $\omega$-incomplete, while "regular" is strictly stronger and "uniform" is strictly weaker. Unfortunately, I don't have time right now to check this carefully, so I hope someone will object loudly if I've messed it up.
Now that I have a bit more time, let me add the counterexamples that justify "strictly". Partition $\omega$ into two infinite pieces $A$ and $B$. The principal filter $F\_0$ generated by $A$ is uniform; that establishes the second "strictly" above. For the first, let $U$ be a nonprincipal ultrafilter that contains $B$, and let $F\_1=U\cap F\_0$. Then $F\_1$ is nonprincipal (the intersection of all the sets in it is $A$, which isn't in it), but it is not regular. (A function $f$ as in Bell and Slomson's definition of "regular" on page 114 would have to send each $a\in A$ to a finite set $f(a)$ that contains all elements $j$ of $\omega$, a contradiction.)
| 3 | https://mathoverflow.net/users/6794 | 46876 | 29,641 |
https://mathoverflow.net/questions/46874 | 9 | The intuitive idea is that the sphere connected the two manifolds is not contractible, which implies the (n-1)th homotopy group is not zero. Another argument, which I am not totally understand, uses the fact that the universal covering of an aspherical manifold has only one end. I am wondering if someone here could clarify these for me or give me a new argument.
| https://mathoverflow.net/users/4760 | How to prove the connected sum of two closed aspherical n-manfolds (n >2) is not asperical? | If $N$ is an open contractible manifold of dimension at least two, then it has one end. For instance, you can compute the zeroth reduced homology of $N$ minus a closed ball $B$ by using the long exact sequence for $(N,B)$ by using excision and the fact that $N$ is contractible.
If $M$ is a closed aspherical manifold of dimension at least three and $M$ is a connected sum of $M\_1$ and $M\_2$, then the fundamental group of $M$ is the free product of the the fundamental groups of $M\_1$ and $M\_2$.
If $M\_1$ and $M\_2$ are closed aspherical manifolds of dimension at least three, then their fundamental groups are nontrivial, and so $\pi\_{1}(M)$ would be a nontrivial free product. But the universal universal cover of $M$ is one ended, and one ended groups do not decompose as nontrivial free products.
| 20 | https://mathoverflow.net/users/7021 | 46878 | 29,642 |
https://mathoverflow.net/questions/46862 | 1 | Consider the 1-torus $\mathbb{T}$. Let $k$ be a smooth function on $\mathbb{T}^2$ and $K$ be the integral operator on $L^2(\mathbb{T})$ with kernel $k$. One can show that $K$ is of trace class, hence $|K|^{1/2}$ is a Hilbert Schmidt operator=integral operator. But what is the kernel of $|K|^{1/2}$?
| https://mathoverflow.net/users/9401 | Square root of integral operator | It seems to me that you are looking for a formula for the kernel of $|K|^{1/2}$. But, as Gerald, mentioned, such a formula (in the case where the space $\mathbb{T}$ is replaced by a finite set) would give you a formula for the entries of the square root of an arbitrary positive matrix. And I don't think such a thing exists (or, at least, I don't think it is known).
| 4 | https://mathoverflow.net/users/3698 | 46898 | 29,656 |
https://mathoverflow.net/questions/46907 | 37 | I attended a talk given by W. Hugh Woodin regarding the Ultimate L axiom and I wanted to verify my current understanding of what the search for this axiom means. I find it to be a fascinating topic but the details are so far beyond my grasp.
Given the language of set theory, one can write down a multitude of first-order sentences. By Godel's Incompleteness Theorem, it is known that from the ZFC axioms one can only derive the truth-values of a (small) fragment of these sentences.
In the past, it was hoped (by Godel, among others) that the Large Cardinal Axiom hierarchy would provide an infinite ladder of axioms of increasing strength such that any first-order sentence in the language of set theory would be either provable or refutable from ZFC + LCA for some suitable LCA.
However, it is now known (?) that the LCA hierarchy (pictorially represented as the vertical spine of the set-theoretic universe V) is not enough to settle all such questions. In particular, there is an additional horizontal "degree of freedom" due to Cohen forcing: for instance, when it comes to CH, it is known (or merely believed?) that both CH and ~CH are consistent with the LCA hierarchy.
Now, let a "completion of ZFC" be an assignment of truth-values to every first-order sentence in the language of set theory, such that a sentence is true whenever ZFC proves that sentence; moreover, for the other sentences (i.e. those which are undecidable in ZFC) the assignment of truth-values must be consistent.
My understanding of Ultimate L is that it picks out a unique completion of ZFC as being the "correct" one; that is, even though Cohen forcing allows us to have models (and therefore completions) of both ZFC + CH and also of ZFC + ~CH, Ultimate L eliminates the horizontal ambiguity and provides us with a unique completion of ZFC in which the truth-values of first-order sentences only depend on the vertical LCA hierarchy.
Is my understanding correct? And how do we know that there are (infinitely) many different completions of ZFC in the first place? Could it be that there is no way to consistently assign truth-values to all first-order sentences, i.e. that no completion exists?
Also, how would we know that Ultimate L + LCA picks out a unique completion (as opposed to a class of completions)? And would it be a valid completion (does consistency of ZFC + Ultimate L follow from Con ZFC)?
I would appreciate answers to any of the above questions, as I can't find anything on this topic in the literature. Thank you!
| https://mathoverflow.net/users/7154 | Completion of ZFC | I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version:
>
> There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\mathbb R})}\cap V\_\theta\models\varphi$.
>
>
>
(At least, this is the version of the axiom that was stated during Woodin's plenary talk at the 2010 ICM, which should be accessible from this [link](http://player.bitgravity.com/debug/embedcode.php?ap=true&video=http%253A//bitcast-a.bitgravity.com/highbrow/livearchive40009/26aug-13.45to14.45.flv). See also the slides for [this talk](http://www.phil.upenn.edu/WSTPM/Woodin)--Thanks to John Stillwell for the link.)
I do not think you will find much about it in the current literature, but Woodin has written a long manuscript ("Suitable extender models") that should probably provide us with the standard reference once it is published.
As stated, this is really an infinite list of axioms (one for each $\varphi$). The statement is very technical, and it may be a bit difficult to see what its connection is with Woodin's program of searching for nice inner models for supercompactness. (That was the topic of his recent series of talks at Luminy; I wrote notes on them and they can be found [here](http://andrescaicedo.wordpress.com/2010/10/19/luminy-hugh-woodin-ultimate-l-i/).)
Keeping the discussion at an informal level (which makes what follows not entirely correct), what is going on is the following:
Gödel defined $L$, the constructible universe. It is an inner model of set theory, and it can be analyzed in great detail. In a sense (guided by specific technical results), we feel there is only one model $L$, although of course by the incompleteness theorems we cannot expect to prove all its properties within any particular formal framework. Think of the natural numbers for an analogue: Although no formal theory can prove all their properties, most mathematicians would agree that there is only one "true" set of natural numbers (up to isomorphism). This "completeness" of $L$ is a very desirable feature of a model, but we feel $L$ is too far from the actual universe of sets, in that no significant large cardinals can belong to it.
The inner model program attempts to build $L$-like models that allow the presence of large cardinals and therefore are closer to what we could think of as the "true universe of sets"; again, the goal is to build certain canonical inner models that are unique in a sense (similar to the uniqueness of ${\mathbb N}$ or of $L$), and that (if there are "traces" of large cardinals in the universe $V$) contain large cardinals.
The program has been very successful, but progress is slow. One of the key reasons for this slow development is that the models that are obtained very precisely correspond to specific large cardinals, so that, for example, $L[\mu]$, the canonical $L$-like model for one measurable cardinal, does not allow the existence of even two measurable cardinals ($L$ itself does not even allow one).
Currently, the inner model program reaches far beyond a measurable, but far below a supercompact cardinal. Woodin began an approach with the goal of studying the coarse structure of the inner models for supercompactness. This would be the first step towards the construction of the corresponding $L$-like models. (The second step requires the introduction of so-called fine-structural considerations, and it is traditionally significantly more elaborate than the coarse step.)
The results reported in the talks I linked to above indicate that, if the construction of this model is successful, we will actually have built the "ultimate version of $L$", in that the model we would obtain not only accommodates a supercompact cardinal but, in essence, all large cardinals of the universe.
If we succeed in building such a model, then it makes sense to ask how far it is from the actual universe of sets.
A reasonable position (which Woodin seems to be advocating) is that it makes no sense to distinguish between two theories of sets if each one can interpret the other, because then anything that can be accomplished with one can just as well be accomplished with the other, and differences in presentation would just be linguistic rather than mathematical. One could also argue that of two theories, if one interprets the other but not vice versa, then the "richer" one would be preferable. Of course, one would have to argue for reasons why one would consider the richer theory "true" to begin with. This is a multiverse view of set theory (different in details from other multiverse approaches, such as [Hamkins's](http://jdh.hamkins.org/the-set-theoretic-multiverse/)) and rather different from the traditional view of a distinguished true universe.
Our current understanding of set theory gives us great confidence in the large cardinal hierarchy. $\mathsf{ZFC}$ is incomplete, and so is any theory we can describe. However, there seems to be a linear ordering of strengthenings of $\mathsf{ZFC}$, provided by the large cardinal axioms. Moreover, this is not an arbitrary ordering, but in fact most extensions of $\mathsf{ZFC}$ that have been studied are mutually interpretable with an extension of $\mathsf{ZFC}$ by large cardinals (and those for which this is not known are expected to follow the same pattern, our current ignorance being solely a consequence of the present state of the inner model program).
So, for example, we can begin with the $L$-like model for, say, a Woodin cardinal, and obtain from it a model of a certain fragment of determinacy while, beginning with this amount of determinacy, we can proceed to build the inner model for a Woodin cardinal. Semantically, we are explaining how to pass from a model of one theory to a model of the other. But we can also describe the process as establishing the mutual interpretability of both theories. Of course, if we begin with the $L$-like model for two Woodin cardinals, we can still interpret the other theory just as before, but that theory may not be strong enough to recover the model with two Woodin cardinals.
From this point of view, a reasonable "ultimate theory" of the universe of sets would be obtained if we can describe "ultimate $L$" and provide evidence that any extension of $\mathsf{ZFC}$ attainable by the means we can currently foresee would be interpretable from the theory of "ultimate $L$".
The ultimate $L$ list of axioms is designed to accomplish precisely this result.
Part of the point is that we expect $L$-like models to cohere with one another in a certain sense, so we can order them. We, in fact, expect that this order can be traced to the complexity of certain *iteration strategies* which, ultimately, can be described by sets of reals. Our current understanding suggests that these sets of reals ought to be [universally Baire](http://en.wikipedia.org/wiki/Universally_Baire_set). Finally, we expect that the models of the form $$HOD^{L(A,{\mathbb R})}\cap V\_\theta$$ as above, *are* $L$-like models, and that these are all the models we need to consider. The fact that when $A=\emptyset$ we indeed obtain an $L$-like model in the presence of large cardinals, is a significant result of Steel, and it can be generalized as far as our current techniques allow.
The $\Omega$-conjecture, formulated by Woodin a few years ago, would be ultimately responsible for the $HOD^{L(A,{\mathbb R})}\cap V\_\theta$ models being all the $L$-like models we need. (Though I do not quite see that formally the "ultimate $L$" list of axioms supersedes the $\Omega$-conjecture). Also, if there is a nice $L$-like model for a supercompact, then the results mentioned earlier suggest we have coherence for all these Hod-models.
The theory of the universe that "ultimate $L$" provides us with is essentially the theory of a very rich $L$-like model. It will not be a complete theory, by the incompleteness theorems, but any theory $T$ whose consistency we can establish by, say, [forcing](http://en.wikipedia.org/wiki/Forcing_%2528set_theory%2529) from large cardinals would be interpretable from it, so "ultimate $L$" is all we need, in a sense, to study $T$. Similarly, only adding large cardinal axioms would give us a stronger theory (but then, this strengthening would be immediately absorbed into the "ultimate $L$" framework).
It is in this sense that Woodin says that the "axiom" would give us a complete picture of the universe of sets. It would also be reasonable to say that this is the "correct" way of going about completing $\mathsf{ZFC}$, since any extension can be interpreted from this one.
[Note I am not advocating for the correctness of Woodin's viewpoint, or saying that it is my own. I feel I do not understand many of the technical issues at the moment to make a strong stance. As others, I am awaiting the release of the "suitable extender models" manuscript. Let me close with the disclaimer that, in case the technical details in what I have mentioned are incorrect, the mistakes are mine.]
---
**Edit**: (Jan. 10, 2011) Here is a [link](http://math.berkeley.edu/~steel/talks/phila2010.pdf) to slides of a talk by John Steel. Both Woodin's slides linked to above, and Steel's are for talks at the *Workshop on Set Theory and the Philosophy of Mathematics*, held at the University of Pennsylvania, Oct. 15-17, 2010. Hugh's talk was on Friday the 15th, John's was on Sunday. John's slides are a very elegant presentation of the motivations and mathematics behind the formulation of Ultimate $L$.
(Jul. 26, 2013) Woodin's paper has appeared, in two parts:
>
> W. Hugh Woodin. *Suitable extender models I*, J. Math. Log., **10 (1-2)**, (2010), 101–339. [MR2802084 (2012g:03135)](http://www.ams.org/mathscinet-getitem?mr=2802084),
>
>
>
and
>
> W. Hugh Woodin. *Suitable extender models II: beyond $\omega$-huge*, J. Math. Log., **11 (2)**, (2011), 115–436. [MR2914848](http://www.ams.org/mathscinet-getitem?mr=2914848).
>
>
>
He is also working on a manuscript covering the beginning of the fine structure theory of these models. I will add a link once it becomes available.
John Steel has a nice set of slides discussing in some detail the multiverse view mentioned above: [Gödel's program](http://math.berkeley.edu/~steel/talks/stanford2013copy.pdf), CSLI meeting, Stanford, June 1, 2013.
For more on why one may want to accept large cardinals as a standard feature of the universe of sets, see [here](https://mathoverflow.net/a/44185/6085).
---
**Edit** (May 17, 2017): W. Hugh Woodin has written a highly accessible survey describing the current state of knowledge regarding Ultimate $L$. For now, see here (I hope to update more substantially if I find the time):
>
> [MR3632568](http://www.ams.org/mathscinet-getitem?mr=3632568). Woodin, W. Hugh. *[In search of Ultimate-L. The 19th Midrasha Mathematicae lectures](https://www.cambridge.org/core/journals/bulletin-of-symbolic-logic/article/in-search-of-ultimatel-the-19th-midrasha-mathematicae-lectures/BDBD4F26AC73B54AA7CD76CD5808B957)*. Bull. Symb. Log. 23 (2017), no. 1, 1–109.
>
>
>
| 54 | https://mathoverflow.net/users/6085 | 46920 | 29,668 |
https://mathoverflow.net/questions/46888 | 6 | Let $P:=P(a\_1,\dots,a\_n)$ be a Pretzel link ( <https://en.wikipedia.org/wiki/Pretzel_link> ).
For every permutation $\sigma\in S\_n$ we can consider the link $$\sigma P:=P(a\_{\sigma(1)},\dots,a\_{\sigma(n)})$$. If $\sigma=(12\dots n)^k$ for some k then $P$ and $\sigma P$ are equivalent.
My questions are:
* Are there any results related to the general case?
* Which invariants can be used to distinguish pretzel links which differ only by a permutation of their coefficients?
Note that $\sigma P$ is a mutant of $P$ for every $\sigma\in S\_n$. So this problem is related to a more general problem i.e. how to distinguish mutant knots (or links).
| https://mathoverflow.net/users/5001 | How to distinguish Pretzel links with the same coefficients? | [Richard Bedient](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-112/issue-2/Double-branched-covers-and-pretzel-knots/pjm/1102709604.full) has proved in 1984 that two Pretzel knots $P$ and $\sigma P$ are equivalent if and only if $\sigma$ is a cyclic permutation, an order reversing permutation, or a composition of both. He computes a particular group constructed from the fundamental group of the complement to distinguish every non-equivalent pair of knots.
Finding invariants is non-obvious since mutant knots have the same Jones (and HOMFLY) polynomial and hyperbolic volume. These particular mutant knots also have the same branched double covering (a Seifert manifold fibering over $S^2$ with singular fibers of order $a\_1, \ldots, a\_n$). In fact, Bedient was interested in finding families of non-equivalent knots sharing the same branched double covering.
**Edit.** I missed some quite restricting hypothesis on the coefficients. The theorem holds if the $a\_i$'s are distinct and all strictly greater than $1$. Moreover we must have $\sum\_i 1/a\_i <1$.
As noted by Ian, such a theorem cannot hold in general. For instance, a consecutive pair $(+1,-1)$ can be moved everywhere.
| 4 | https://mathoverflow.net/users/6205 | 46925 | 29,671 |
https://mathoverflow.net/questions/46856 | 7 | Is the subgroup of $GL(2,\mathbb Z)$ generated by the matrices
$$ \left( \begin{array}{cc}
1 & 1 \\\
1 & 0 \end{array} \right) \ \ \text{and} \ \
\left( \begin{array}{cc}
2 & 1 \\\
1 & 0 \end{array} \right)
$$
free of exponential growth? More generally, how does one find all the relations between two matrices?
I am sure this is well known, so any relevant references will be appreciated.
My motivation comes from dynamical systems where these matrices specify two automorphisms of the 2-torus; I am interested in studying the orbits of their joint action.
| https://mathoverflow.net/users/8131 | A free subgroup of GL(2,Z)? | I interpret the question "how does one find all the relations between the matrices" as "find a set of defining relations for the group generated by the two matrices".
To do that we need a presentation of $GL(2,\mathbb{Z})$. I found one in the paper:
T. Brady, Automatic structures on Aut$F\_2)$, *Arch. Math.* 63, 97-102 (1994).
$\langle p,s,u \mid p^2=s^2=(sp)^4=(upsp)^2=(ups)^3=(us)^2=1 \rangle$
where $p= \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$, $s= \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$, $u= \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$.
Denoting your two matrices by
$a= \left(\begin{array}{cc}1&1\\1&0\end{array}\right)$,
$b= \left(\begin{array}{cc}2&1\\1&0\end{array}\right)$
we have $a=up$, $b=u^2p$.
Putting $H = \langle a,b \rangle$ and using coset enumeration in Magma, it turns out that $H={\rm GL}(2,\mathbb{Z})$. So your two matrices actually generate all of GL$(2,\mathbb{Z})$.
In fact, denoting $a^{-1}$ and $b^{-1}$ by $A$ and $B$, we have
$p=aBa$, $u=a^2Ba$, $s=abaBAbabA$.
Using the modified coset enumeration algorithm, we can compute a presentation of $H$, which came out as the not particularly enlightening
$H = \langle a,b \mid (aBa)^2, (AbaBA)^2, aBabABAbAbABAbAbABAbAbAB, (abABabaB)^3 \rangle$.
| 13 | https://mathoverflow.net/users/35840 | 46929 | 29,673 |
https://mathoverflow.net/questions/46926 | 2 | In Partial Differential Equation by Lawerence Evan p284 there is this theorem stated:
Let $U$ be a bounded open subset of $\mathbb{R}^n$ with $C^1$ boundary. Suppose $u\in W^{k,p}$ then if $k>n/p$ we have
$
u\in C^{\alpha, \gamma}(\overline{U})
$
where $\alpha = k-\left[n/p\right]-1$ and $\gamma = \left[n/p\right]+1-n/p$ if $n/p$ is not an integer and any $0<\gamma<1$ if $n/p\in\mathbb{N}$.
I have two questions:
1. Does this result extend to $U$ being an open subset with only Lipschitz boundary?
2.Does the result also holds $k\not\in\mathbb{N}$? The author doesn't mention anyway that $k$ should be an integer but I just wanted to check.
Thank you in advance.
| https://mathoverflow.net/users/2011 | General Sobolev Inequalities | 1- The Sobolev embedding is proved first in the case $U=\mathbb R^n$, and then for a general $U$ by using a right inverse $j$ of the restriction operator $W^{s,p}(\mathbb R^n)\rightarrow W^{k,p}(U)$. When $U$ is a half-space, a convenient $j$ is Babitch's extension. When the boundary is smooth, use an atlas to reduce to the previous case.
The situation is however not easy if the boundary is only Lipschitz. If $U$ is a (hyper-)rectangle, then $j$ can be build by applying $n-1$ times the Babitch operator. Thus the Sobolev embedding holds true in this case, whatever $\alpha$ is. Still true if the boundary is smooth, up to some kinks with angle $\frac{\pi}{m}$ with $m$ an integer. It is less clear for more general domains with Lipschitz boundary.
2- Sobolev embeddings hold even if $k$ is not an integer. The definition of the Sobolev space is then a bit more involved. What really matters is that $\beta:=k-\frac{n}{p}$ is not an integer. Then $\alpha=[\beta]$ and $\gamma=\beta-\alpha$. And of course regularity of the boundary.
| 1 | https://mathoverflow.net/users/8799 | 46931 | 29,674 |
https://mathoverflow.net/questions/46932 | 1 | Each book on algebraic geometry write I^2 when it deal with nongsingular varieties, here I
is a ideal sheaf. But no one give the definition. I guess it's the sheafification. It's right?
Thanks.
| https://mathoverflow.net/users/3525 | What is the definition of product of ideal sheaves? | It is the image of the map $I \otimes I \to O \otimes O = O$.
| 2 | https://mathoverflow.net/users/4428 | 46933 | 29,675 |
https://mathoverflow.net/questions/46936 | 10 | Exist simply connected CW complexes $X$, $Y$ and a mapping $f:X\to Y$ with the property that the reduced suspension $\Sigma f:\Sigma X\to\Sigma Y$ is a homotopy equivalence but $f$ is not?
| https://mathoverflow.net/users/10997 | Is a map a homotopy equivalence if its suspension is so? | Whitehead's Theorem (it is Corollary 4.33 in Allen Hatcher's [book](http://www.math.cornell.edu/~hatcher/#ATI)) says that a map between simply connected CW-complexes is a homotopy equivalence if and only if the induced map on homology (with $\mathbb Z$-coefficients) is an isomorphism. If $\Sigma f : \Sigma X \to \Sigma Y$ is a homotopy equivalence, then this is clearly the case, since suspension just shifts dimensions and the spaces are connected, so that there is no problem in dimension $0$.
It would be interesting to have an argument which does not use all the machinery that goes into Whitehead's Theorem, since your assumption is rather strong.
| 17 | https://mathoverflow.net/users/8176 | 46937 | 29,677 |
https://mathoverflow.net/questions/46934 | 6 | I would like to show that any Zariski-closed subsemigroup of $SL\_n(\mathbb{C})$ is a group. If I understand correctly, this is consequence 1.2.A of <http://www.heldermann-verlag.de/jlt/jlt03/BOSLAT.PDF> .
Is there a more elementary proof? For $SL\_2(\mathbb{C})$, the result is quite easy to show directly, or using the Hilbert basis theorem, .
| https://mathoverflow.net/users/408 | Zariski-closed subsemigroups of SL_n(C) are groups | It is quite elementary. Let $S$ be the semi-group in question. Then for any $g \in S$, the set
$g^kS$ for $k=1,2, \dots$ is a decreasing sequence of closed sets, hence it has to stabilize. So, $g^kS=g^{k+1}S$ implies that $gS=S$. Hence $S$ is closed with respect to taking inverse, and therefore is a group.
| 24 | https://mathoverflow.net/users/3635 | 46943 | 29,682 |
https://mathoverflow.net/questions/46942 | 13 | Away from the hyperelliptic locus, the moduli of curves immerses
in the moduli of principally polarized abelian varieties. The
ambient space has a riemannian metric, so one can ask about the
second fundamental form, the first-order deviation of the
submanifold from being totally geodesic. What is this second
fundamental form? Is anything known about it?
I think one could translate this into the language of algebraic
geometry by using the Serre-Tate formal coordinates, which exist
at each point of $A\_g$. With respect to these coordinates, $M\_g$
is not linear; what is its quadratic approximation? One could
interpret this as a version of the Schottky problem, which
suggests that existing solutions to it might be applicable.
| https://mathoverflow.net/users/4639 | What is the second fundamental form of moduli space? | The following papers might be useful:
$(1)$ E. Colombo- G. Pirola- A. Tortora
"Hodge-Gaussian maps"
Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 30 (2001), no. 1, 125–146.
$(2)$ E. Colombo - P. Frediani
"Siegel metric and curvature of the moduli space of curves"
Trans. Amer. Math. Soc. 362 (2010), no. 3, 1231–1246.
| 3 | https://mathoverflow.net/users/7460 | 46947 | 29,685 |
https://mathoverflow.net/questions/46949 | 15 | **Background**
One of my friends told me the following story: A child must walk from his home at point A = (1,0) to his school at point B = (0,1). The laws in his country state that you can only walk parallel to the horizontal and vertical axis. No matter how he tries to get to school, he finds that he must walk at least 2 miles. He is very frustrated that he cannot walk diagonally. He doesn't want to get in trouble, so he puts up with this silly law until his senior year of high school. At this point he takes calculus, and learns about limits, and so he decides that each day he will walk a new "zig-zag" path around the line, in such a way that his sequence of paths approaches the line from A to B uniformly. On the 30th day he is pulled over by a policeman for walking diagonally.
The point of the story is to get you to think about the notion of the length of a curve. Here we have an instance of a sequence of polygonal paths which approach a curve uniformly, but the lengths of the polygonal approximations do not converge to the length of the curve.
Our usual definition of the arclength of a curve involves approximating it with polygonal secant segments. My question is why this is the "best" definition. Or to make this precise, how do we know that any other sequence of polygonal approximations to the curve will not have a shorter limit? **EDIT**: In the case of a straight line, this is clear, but my precise question below is about answering this for more general curves.
**Precise Question**:
Let $C:[0,1] \to \mathbb{R}^n$ be a rectifiable curve (or feel free to add as many smoothness requirements as you like), and let $P\_n: [0,1] \to \mathbb{R}^n$ be a sequence of piecewise linear curves which converge uniformly to $C$. Is it true that limsup{Length($P\_n$)} $\geq$ Arclength(C)? **EDIT** I goofed: I meant to ask whether or not liminf{Length($P\_n$)} $\geq$ Arclength(C).
Hopefully this question is not too elementary; my analysis skills are almost pitifully weak. I strongly suspect that the answer to this precise question is "Yes", because otherwise I think that the usual definition of arc length is incorrect. I have enough faith in mathematics to believe that we have found the right definition, but I would still like to see a proof of this fact.
| https://mathoverflow.net/users/1106 | Justifying the definition of arclength | Yes it is true and quite elementary too. The definition of arclength you gave is also known as the (classical) total variation, and you can define it for any curve $\gamma:[a,b]\to M$ in any metric space $(M,d)$ as
$\mathrm{V}(\gamma,[a,b]):=\sup\_P \mathrm{V}(\gamma, P)$
where the supremum is taken over all subdivisions *à la Riemann* $P:=\{a=t\_0 < t\_1 < \dots < t\_n=b\}$ of the interval $[a,b],$ and where we denote
$\mathrm{V}(\gamma, P):=\sum\_{i=0}^{n-1}d\left(\gamma(t\_{i+1}),\gamma(t\_i)\right),$
the variation w.r.to $P$. Clearly, for any $P$, the variation wrto $P$ is a continuous functional on curves, even wrto point-wise convergence. The total variation over $[a,b]$ is a supremum of continuous functionals, and as such, what one can say is: it is (sequentially) upper semicontinuous, that is, the inequality you wrote holds for any sequence $\gamma\_n$ point-wise convergent to $\gamma$ :
$$\limsup\_{n\to\infty}V(\gamma\_n,[a,b])\geq V(\gamma, [a,b]).$$
(just because for all $P$ we have $\limsup\_{n\to\infty}V(\gamma\_n,[a,b])\geq
\limsup\_{n\to\infty}V(\gamma\_n,P)= V(\gamma, P)$: so we have it taking the $\sup\_P$ over all $P$.).
**Rmk.** The relevance of the notion of upper semi-continuity is linked to existence results for minima. For instance, it is true that in any compact metric space, if two points are joined by a curve of finite length, then there is also a curve of minimal length connecting them. It's a consequence of the Ascoli-Arzelà compactness theorem.
| 5 | https://mathoverflow.net/users/6101 | 46953 | 29,686 |
https://mathoverflow.net/questions/46957 | 9 | My question: Is the natural filtration of a right-continuous process also right-continuous?
I would say yes, but don't know where to start proving it.
Thanks for your help/ideas!
| https://mathoverflow.net/users/11011 | Right-continuity of natural filtrations | Right continuity fails even for canonical continuous processes.
The natural filtration on $C([0,\infty))$ is not right continuous.
For example, the event $\{\omega: {d^+\over dt}\ \omega\_t\mbox{ exists at }t=0\}$ belongs
to ${\cal F}\_{0+}$ but not ${\cal F}\_0$. In words, you can tell whether
the function $\omega\_t$ has a right derivative at $t=0$ with an infinitesimal peek *beyond* time 0,
but you cannot tell just from the value of the function $\omega\_t$ *at* time 0.
Right continuous filtrations are nicer to work with, and since it fails for the natural filtration,
we often use the right continuous version instead. Fortunately, many of the nice
properties of right continuous processes carry over even with this enlarged filtration. For example,
Brownian motion is still Markov with respect to ${\cal F}\_{t+}$ which leads to interesting results like Blumenthal's zero-one law.
| 9 | https://mathoverflow.net/users/nan | 46962 | 29,691 |
https://mathoverflow.net/questions/46900 | 42 | The complex irreps of a finite group come in three types: self-dual by a
symmetric form, self-dual by a symplectic form, and not self-dual at all.
In the first two cases, the character is real-valued, and in the third
it is sometimes only complex-valued. The cases can be distinguished by
the value of the Schur indicator $\frac{1}{|G|} \sum\_g \chi(g^2)$,
necessarily $1$, $-1$, or $0$. They correspond to the cases that
the representation is the complexification of a real one, the forgetful
version of a quaternionic representation, or neither.
A conjugacy class $[g]$ is called "real" if all characters take real
values on it, or equivalently, if $g\sim g^{-1}$. I vaguely recall the
number of real conjugacy classes being equal to the number of real irreps.
1. Do I remember that correctly?
2. Can one split the real conjugacy classes into two types,
"symmetric" vs. "symplectic"?
With #1 now granted, a criterion for a "good answer" would be that the number of symmetric real conjugacy classes should equal the number of symmetrically self-dual irreps.
(I don't have any application in mind; it's just bothered me off and on
for a long time.)
| https://mathoverflow.net/users/391 | Are there "real" vs. "quaternionic" conjugacy classes in finite groups? | It's a great question! Disappointingly, I think the answer to (2) is **No** :
The only restriction on a `good' division into "symmetric" vs. "symplectic" conjugacy classes that I can see is that it should be intrinsic, depending only on $G$ and the class up to isomorphism. (You don't just want to split the self-dual classes randomly, right?) This means that the division must be preserved by all outer automorphisms of $G$, and this is what I'll use to construct a counterexample. Let me know if I got this wrong.
**The group**
My $G$ is $C\_{11}\rtimes (C\_4\times C\_2\times C\_2)$, with $C\_2\times C\_2\times C\_2$ acting trivially on $C\_{11}=\langle x\rangle$, and the generator of $C\_4$ acting by $x\mapsto x^{-1}$. In Magma, this is G:=SmallGroup(176,35), and it has a huge group of outer automorphisms $C\_5\times((C\_2\times C\_2\times C\_2)\rtimes S\_4)$, Magma's OuterFPGroup(AutomorphismGroup(G)). The reason for $C\_5$ is that $x$ is only conjugate to $x,x^{-1}$ in $C\_{11}\triangleleft G$, but there there are 5 pairs of possible generators like that in $C\_{11}$, indistinguishable from each other; the other factor of $Out\ G$ is $Aut(C\_2\times C\_2\times C\_4)$, all of these guys commute with the action.
**The representations**
The group has 28 orthogonal, 20 symplectic and 8 non-self-dual representations, according to Magma.
**The conjugacy classes**
There are 1+7+8+5+35=56 conjugacy classes, of elements of order 1,2,4,11,22 respectively. The elements of order 4 are (clearly) not conjugate to their inverses, so these 8 classes account for the 8 non-self-dual representations. We are interested in splitting the other 48 classes into two groups, 28 'orthogonal' and 20 'symplectic'.
**The catch**
The problem is that the way $Out\ G$ acts on the 35 classes of elements of order 22, it has two orbits according to Magma - one with 30 classes and one with 5. (I think I can see that these numbers must be multiples of 5 without Magma's help, but I don't see the full splitting at the moment; I can insert the Magma code if you guys want it.) Anyway, if I am correct, these 30 classes are indistinguishable from one another, so they must all be either 'orthogonal' or 'symplectic'. So a canonical splitting into 28 and 20 cannot exist.
---
**Edit**: However, as Jack Schmidt points out (see comment below), it *is* possible to predict the number of symplectic representations for this group!
| 22 | https://mathoverflow.net/users/3132 | 46964 | 29,693 |
https://mathoverflow.net/questions/46966 | 7 | Consider the circle map $\times d:x\mapsto dx \mod 1$. The lebesgue measure is the only absolutely continuous invariant probability measure, but this map has many other invariant measures. Of course, one can take barycentric combinations of invariant measures to get a new one, so let us restrict to the extremal points, namely ergodic measures.
One can consider a uniform measure on any periodic orbit. There are also singular, atomless invariant measures. For example, the "uniform" measure on the usual middle-third Cantor set is invariant under $\times 3$. All this is pretty explicit and I'm fine with it. But I also heard about a thermodynamical formalism that yield many invariant measures; Bowen's lecture note are on my desk but it does not seem to answer my question, which is the following: what do these measures look like? What is their support? I guess that we cannot answer this for all invariant measures, but maybe for some of them less trivial than the atomic and easy Cantor ones.
Other question: Cantor measures are easily constructed for all $\times d$, $d>2$, but I cannot really get one for $d=2$. Am I clumsy or is there something special to this case?
| https://mathoverflow.net/users/4961 | What do singular, atomless invariant measures of $\times d$ look like? | If you do the Cantor measure construction for d=2, you just get Lebesgue measure... so it's a little bit special.
There are lots of fully supported invariant measures for the map $\times d$: the thermodynamic formalism that you mention gives you a whole zoo of them. In particular, if $\phi\colon [0,1]\to \mathbb{R}$ is any Hölder continuous function, then there is a unique "equilibrium state" for $\phi$, which is a probability measure $\mu\_\phi$. This measure can be shown to have a certain Gibbs property, which in particular implies that it has full support -- it gives positive measure to every open set in $[0,1]$.
The easiest of these to think about are the Bernoulli measures. Given a sequence $x\_1 x\_2 \cdots x\_n$ where $x\_i \in \{0,1,\dots,d-1\}$, consider the interval
$$
C(x\_1 x\_2 \cdots x\_n) = [x\_1 d^{-1} + x\_2 d^{-2} + \cdots + x\_n d^{-n}, x\_1 d^{-1} + x\_2 d^{-2} + \cdots + (x\_n+1) d^{-n}].
$$
These generate the $\sigma$-algebra of Borel sets, so given any probability vector $\mathbf{p} = (p\_1,\dots,p\_d)$, we can define an invariant measure $\mu\_{\mathbf{p}}$ by
$$
\mu\_{\mathbf{p}}(C(x\_1 \cdots x\_n)) = p\_{x\_1} p\_{x\_2} \cdots p\_{x\_n}.
$$
These are equilibrium states for potential functions that are constant on the $d$ intervals $C(x\_1)$. Potential functions that are constant on intervals $C(x\_1 \cdots x\_n)$ for some $n$ yield Markov measures as equilibrium states, and these can also be described quite explicitly.
All these measures are invariant under the $\times d$ map, fully supported on the interval, ergodic, and non-atomic.
See [this answer](https://mathoverflow.net/questions/37010/product-measure-only-possible-measure/37012#37012) for a little bit more on Markov measures, and [this one](https://mathoverflow.net/questions/42201/what-are-the-zero-entropy-invariant-measures-for-an-anosov-geodesic-flow/42214#42214) for some other invariant measures that are ergodic, non-atomic, and fully supported, but have zero entropy. It's worth pointing out that pretty much anything you say about invariant measures for the shift map $\sigma\colon \Sigma\_d^+ \to \Sigma\_d^+$ (here $\Sigma\_d^+ = \{0,\dots,d-1\}^{\mathbb{N}}$) can be translated into a statement about invariant measures for the $\times d$ map, since the two are topologically conjugate on a total probability set -- that is, a set that is given full weight by every invariant measure.
| 3 | https://mathoverflow.net/users/5701 | 46969 | 29,694 |
https://mathoverflow.net/questions/46976 | 0 | a short question: Is every 3-Sasakian manifold a Sasaki-Einstein manifold? If not, do you have an example? If yes, how can I prove this?
Thanks and best regards
| https://mathoverflow.net/users/7028 | Is every 3-Sasakian a Sasakian-Einstein manifold? | Yes, because a manifold is Sasaki-Einstein if and only if its metric cone is Ricci-flat Kähler, whereas the cone of a 3-sasakian manifold is hyperkähler.
See, for instance, Bär's "[Real Killing spinors and holonomy](https://doi.org/10.1007/BF02102106)" published in CMP.
| 3 | https://mathoverflow.net/users/394 | 46978 | 29,699 |
https://mathoverflow.net/questions/46970 | 96 | Recently, I learnt in my analysis class the proof of the uncountability of the reals via the [Nested Interval Theorem](http://personal.bgsu.edu/%7Ecarother/cantor/Nested.html) ([Wayback Machine](http://web.archive.org/web/20180402194030/http://personal.bgsu.edu/%7Ecarother/cantor/Nested.html)). At first, I was excited to see a variant proof (as it did not use the diagonal argument explicitly). However, as time passed, I began to see that the proof was just the old one veiled under new terminology. So, till now I believe that any proof of the uncountability of the reals must use Cantor's diagonal argument.
Is my belief justified?
Thank you.
| https://mathoverflow.net/users/5627 | Proofs of the uncountability of the reals | Mathematics isn't yet ready to prove results of the form, "Every proof of Theorem T *must use* Argument A." Think closely about how you might try to prove something like that. You would need to set up some *plausible system for mathematics* in which Cantor's diagonal argument is blocked and *the reals are countable*. Nobody has any idea how to do that.
The best you can hope for is to look at each proof on a case-by-case basis and decide, subjectively, whether it is "essentially the diagonal argument in disguise." If you're lucky, you'll run into one that your intuition tells you is a fundamentally different proof, and that will settle the question to your satisfaction. But if that doesn't happen, then the most you'll be able to say is that every *known* proof seems *to you* to be the same. As explained above, you won't be able to conclude definitively that every possible argument *must* use diagonalization.
**ADDENDUM** (August 2020). Normann and Sanders have a [very interesting paper](https://arxiv.org/abs/2007.07560) that sheds new light on the uncountability of $\mathbb R$. In particular they study two specific formulations of the uncountability of $\mathbb R$:
$\mathsf{NIN}$: For any $Y:[0,1] \to \mathbb{N}$, there exist $x,y \in [0,1]$ such that $x\ne\_{\mathbb{R}} y$ and $Y(x) =\_{\mathbb{N}} Y(y)$.
$\mathsf{NBI}$: For any $Y[0,1] \to \mathbb{N}$, either there exist $x,y \in [0,1]$ such that $x\ne\_{\mathbb{R}} y$ and $Y(x) =\_{\mathbb{N}} Y(y)$, or there exists $N\in\mathbb{N}$ such that $(\forall x\in [0,1])(Y(x) \ne N)$.
One of their results is that a system called ${\mathsf Z}\_2^\omega$ does not prove $\mathsf{NIN}$. Their model of $\neg\mathsf{NIN}$ can therefore be interpreted as a situation where the reals are countable! Nevertheless we are still far from showing that Cantor's diagonal argument is needed to prove that the reals are uncountable. A further caveat is that Normann and Sanders argue that the unprovability of $\mathsf{NIN}$ in ${\mathsf Z}\_2^\omega$—which might at first sight suggest that $\mathsf{NIN}$ is a strong axiom—is an artificial result, and that the proper framework for studying $\mathsf{NIN}$ and $\mathsf{NBI}$ is what they call a “non-normal scale,” in which $\mathsf{NIN}$ and $\mathsf{NBI}$ are very weak. In particular their paper gives lots of examples of statements that imply $\mathsf{NIN}$ and $\mathsf{NBI}$. I suspect, though, that you'll probably feel that the proofs of those other statements smuggle in Cantor's diagonal argument one way or another.
**ADDENDUM** (December 2022).
I just listened to an [amazing talk by Andrej Bauer](https://www.youtube.com/watch?v=4CBFUojXoq4), reporting on joint work with James Hanson.
If you start listening around [14:53](https://youtu.be/4CBFUojXoq4?t=893), you'll see how, in the context of intuitionistic logic, one can formulate precisely the question of whether there is a proof of the uncountability of the reals that doesn't use diagonalization. Bauer and Hanson don't answer this question, but they construct something they call a "parameterized realizability topos" in which *the Dedekind reals are countable*. In particular, this shows that higher-order intuitionistic logic (in which one cannot formulate the usual diagonalization argument) cannot show the reals are uncountable. Now, you could still justifiably claim that this whole line of research does not really address the original question, which I presume tacitly assumes classical logic; nevertheless, this still comes closer than anything else I've seen.
| 78 | https://mathoverflow.net/users/3106 | 46979 | 29,700 |
https://mathoverflow.net/questions/46971 | 2 | When I took model theory is an undergraduate, early on we wrestled with trying to state the fundamental theorem of arithmetic in the first order language of arithmetic. The problem was that we needed and unbounded and variable number of quantifiers Depending on each natural numbers factorization in the statement. I.e. For 6 we we need 2 quantifiers but for 110 needs 3. We stopped there because it illuatrated one of the shortcomings of the first order setting. We can quantify over a fixed number of primes and make the statement, but it will be false in general for any fixed prime, obviously.
I was recently refreshing my mind about non-standard models of arithmetic I found out that a form of unique factorization into primes does exist in these setting, but you may have non-standard primes in the factorization. So this means there is some first order sentence capturing the essence of FTA. I want to know what this sentence is. I am certain it will have to use some sort of recursive (=> definable In arithmetic) coding but the details I am unsure of. Can anyone assist me or point me onthe correct direction?
Thank you
| https://mathoverflow.net/users/10579 | FTA in first order setting | Possibly you will find [this](http://people.cs.uchicago.edu/~laci/REU09/tr9.pdf) somewhat useful.
| 2 | https://mathoverflow.net/users/2926 | 46982 | 29,702 |
https://mathoverflow.net/questions/46732 | 10 | This question grew out of this [post](https://mathoverflow.net/questions/46714/profinite-topologies).
>
> **Question:** Is there a finitely generated, infinite, residually finite group such that every finite index subgroup has $p$-power index for a fixed prime $p$?
>
>
>
The $p$-adic integers $\mathbb Z\_p$ give an example of a non-finitely generated such group. This is not entirely obvious, but follows from a result of Jean-Pierre Serre, which states that every finite index subgroup of $\mathbb Z\_p$ is closed.
**EDIT:** André Henriques has pointed out that Rostislav Grigorchuk constucted a finitely generated, infinite, residually finite $2$-group. All finite quotients of this group have to be $2$-groups. Colin Reid asked in a comment whether there is a torsionfree example. So let me take the freedom to extend my question:
>
> **Question:** Is there a torsionfree example?
>
>
>
| https://mathoverflow.net/users/8176 | Finitely generated, infinite, residually finite groups whose finite quotients are $p$-groups. | Yes, there are torsion free examples. I do not know who constructed them first, but some examples can be found in papers by Grigorchuk and his co-authors. For example Bartoldi and Grigorchuk proved that a certain Fabrykowski-Gupta group $\Gamma $ has the following properties (see Propositions 6.4 and 6.5 in arXiv:math/9911206):
(a) It is a subgroup of the automorphism group of a rooted tree.
(b) It is virtually torsion free.
(c) It satisfies the following 'congruence property': every finite index subgroup of $\Gamma $ contains a level stabilizer (i.e., the stabilizer of a level of the tree) and the index of every level stabilizer is a power of 3.
By (a) $\Gamma $ is residually finite. By (b) there is a torsion free subgroup $K$ of finite index in $\Gamma $. Since every finite index normal subgroup of $K$ contains a finite index normal subgroup of $\Gamma $, (c) implies that every finite quotient of $K$ is a 3-group.
| 11 | https://mathoverflow.net/users/10251 | 46999 | 29,710 |
https://mathoverflow.net/questions/46613 | 0 | For any universal cover p of the wedge S1 V S1 is it true that the two actions of π1(X, x\_0) on the fiber p^-1(x0) given by lifting loops at x0 and the action given by restricting deck transformations to the fiber coincide. I'm not sure but I think you need for π1(X , x0) to be abelian no?
| https://mathoverflow.net/users/10951 | wedge sum deck transformation | It's indeed true that the action by lifting loops and the action by deck transformations agree exactly when the fundamental group is abelian. This statement is a bit vague, so let me be precise.
Let $X$ be a space with universal cover $Y\stackrel{p}{\longrightarrow} X$, and choose basepoints $x\_0 \in X$, $y\_0 \in Y$. Then we can identify the fiber $p^{-1} (x\_0)$ with $\pi\_1 (X, x\_0)$ using path lifting: a loop $\gamma$ at $x\_0$ lifts to a path $\tilde{\gamma}$ starting at $y\_0$, and we identify $[\gamma]$ with $\tilde{\gamma} (0)$. Now the action of $\pi\_1 (X, x\_0)$ on the fiber, via lifting, corresponds to right multiplication in $\pi\_1 (X, x\_0)$. On the other hand, the group of deck transformations of $Y$ is also isomorphic to $\pi\_1 (X, x\_0)$, by sending a loop $\gamma$ to the deck transformation taking $y\_0$ to $\tilde{\gamma} (0)$. Now under the identifications of both the deck transformations and the fiber with $\pi\_1 (X, x\_0)$, the action of deck transformations on the fiber corresponds to left multiplication in $\pi\_1 (X, x\_0)$.
Checking these statements is a worthwile exercise. In some sense, this is easier to think about if you initially just think of actions as functions that assign group elements to bijections, and then you can later worry about left versus right.
In any event, a group G is abelian if and only if for every element $g\in G$, left multiplication by g and right multiplication by g are the same function $G\to G$.
I'll also note that any left action of a group $G$ on a set $S$ can be converted into a right action by setting $s\cdot g = g^{-1} \cdot s$, but the above discussion shows that if you convert the left action of $\pi\_1 (X, x\_0)$ on the fiber (via deck transformations) into a right action, you definitely do not get the lifting action.
| 3 | https://mathoverflow.net/users/4042 | 47012 | 29,718 |
https://mathoverflow.net/questions/47005 | 11 |
>
> Let $R$ be a commutative regular local ring. Is it true that for every $\mathfrak p \in \mathrm{Spec}(R)$ there is a finitely generated $R$-module $M$ such that $\mathrm{projdim}(M)=\mathrm{ht}(\mathfrak p)$ and $\mathrm{Ass}(M)=\{\mathfrak p\}$?
>
>
> Or is there some family of commutative noetherian rings where is this true?
>
>
>
I know that this holds if $R$ is commutative regular local of Krull dimension $\leq 4$ (up to dimension 3 it was easy, because factor rings $R/\mathfrak p$ where always CM, so we can take $M=R/\mathfrak p$. Dimension 4 was harder and in dimension 5 or more I don't know).
Is this an easy/hard/hopeless/already open/similar to something problem?
Thanks,
David
| https://mathoverflow.net/users/10931 | Homologically nice commutative rings | My feeling is that this is really about set-theoretic complete intersections.
Let $X={\rm Spec} A$ be a noetherian affine scheme such that every irreducible subscheme of $X$ is a set-theoretic complete intersection. In other words, for any prime $\mathfrak p\subset A$, there exist a set of elements $x\_1,\dots,x\_t\in \mathfrak p$ such that $t={\rm ht} (\mathfrak p)$ and the $x\_1,\dots,x\_t$ generate a $\mathfrak p$-primary ideal with $\mathfrak p= \sqrt{( x\_1,\dots, x\_t)}$, or equivalently the zero set $Z(x\_1,\dots,x\_t)=\overline{\{\mathfrak p\}}$.
In this case, take $M=A/(x\_1,\dots,x\_t)$ has the property that ${\rm Ass} (M)=\{\mathfrak p\}$.
If in addition $A$ is CM, then so is $M$ and then its projective dimension satisfies
that
$$
{\rm pd} (M)=\dim A-\dim M= {\rm ht} (\mathfrak p).
$$
I suppose the next step is to look at an affine scheme with an irreducible subscheme that is not a set-theoretic complete intersection and see what happens there.
(**Addendum**)
Regarding the case when there exists an irreducible subscheme that is not a set-theoretic complete intersection, one may mention, that in general, (still assuming that $A$ is CM, which follows if it is regular),
$$
{\rm pd} (M)=\dim A-{\rm depth}\_A M.
$$
If furthermore ${\rm Ass} (M)=\{\mathfrak p\}$, then it follows that
$$
{\rm depth}\_A M \leq \dim M = \dim A - {\rm ht} (\mathfrak p),
$$
So ${\rm pd} (M)\geq {\rm ht} (\mathfrak p)$ with equality iff $M$ is CM.
In other words, your desired condition is to find a CM module whose only associated prime is $\mathfrak p$.
At least for modules generated by a single element this seems to be pretty close to $\overline{\{\mathfrak p\}}$ being a set-theoretic complete intersection as
for an ideal $\mathfrak q\subseteq A$ in a noetherian ring the following holds:
$$
\mathfrak q \text{ is $\mathfrak p$-primary} \Leftrightarrow {\rm Ass}(A/\mathfrak q)=\{\mathfrak p\}.
$$
One way to ensure that $A/\mathfrak q$ is CM is to make sure that $\mathfrak q$ has the right number of generators and in order to have the condition on the associated primes one would need that $\mathfrak q$ is $\mathfrak p$-primary. Of course, I am not claiming that this is the only way to produce such modules, but this seems to be the obvious way.
| 3 | https://mathoverflow.net/users/10076 | 47018 | 29,722 |
https://mathoverflow.net/questions/47015 | 14 | A polygon $P\_k$ divided by $k-2$ diagonals into triangles is called a polygonal triangulation. These are the vertices of the triangulation graph $\mathcal P\_k$. Two vertices are connected by an edge if one triangulation is obtained from another by the *diagonal flip*, i.e. we take two triangles of the triangulation that share a side, and in their union (where that side is a diagonal), replace that diagonal by the other diagonal. Sleator, Tarjan, and Thurston proved that the diameter of the
triangulation graph ${\mathcal P}\_k$ is bounded above by $2k-10$. Hence the problem of finding a shortest path in that graph between two triangulations is in NP.
**Question 1.** Is it in P?
**Question 2.** What is known about the complexity of finding the shortest path in the triangulation graph of other surfaces?
**Update.** I have posted a followup [question.](https://mathoverflow.net/questions/47035/poincare-conjecture-and-the-graph-of-triangulations)
| https://mathoverflow.net/users/nan | The flip graph of triangulations | Just to add a little to Joseph's nice answer, for part 1 of your question: although the problem of computing the flip distance in polynomial time is wide open for triangulations of convex polygons, it can be solved in polynomial time for triangulations of certain highly nonconvex point sets (such as the intersection of the integer lattice with a convex set): see my paper "Happy Endings for Flip Graphs", SoCG 2007 and [JoCG 2010](https://doi.org/10.20382/jocg.v1i1a2).
| 9 | https://mathoverflow.net/users/440 | 47026 | 29,728 |
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