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https://mathoverflow.net/questions/38660 | 3 | A node on a curve is a singular point that locally looks like the intersection of two lines. I think the precise way to say this is that $p \in X$ is a (closed?) point on a scheme $X$ (of finite type over a field $k$?), then the completion of the local ring at $p$, $\widehat{\mathcal{O}}\_{X,p}$ should be isomorphic to $k[[x,y]]/(xy)$ (the completion for the intersection of two lines).
The first question is whether this is correct.
The second question is whether you get the n-dimensional version of a node by requiring
the completion of the local ring to look like intersection of $n+1$ coordinate planes: $k[[x\_0,x\_1,...,x\_n]]/(x\_0 x\_1\cdots x\_n)$.
The third question is: what exactly does it mean for such a singularity to be isolated? This
is easy to imagine over $\mathbb{C}$ (there is analytic neighborhood containing no other singularity), but how to say this in something like the Zariski topology?
| https://mathoverflow.net/users/9046 | Higher dimensional nodes | **[Edit: Over an algebraically closed field]** a node should be an isolated hypersurface singularity whose (projectivized) tangent cone is a nondegenerate quadric. This means that in local coordinates the equation has no linear part, and the quadratic part is nondegenerate.
For a curve your definition is equivalent to this one so it is correct. In higer dimensions it is very different. In fact your hypersurface has a non-isolated singularity (for instance, all points in the coordinate lines are singular).
For the last question, the singular points of an algebraic set (or variety, or scheme) are a Zariski closed subset, which splits in irreducible components. A singular point is isolated if it is one of the irreducible components (and it is not embedded, ie, it does not belong to any other irreducible component, but this has sense only in the scheme-theoretic case).
For the **non algebraically closed field** case, see BCnrd's comment and reference!
| 3 | https://mathoverflow.net/users/1939 | 38669 | 24,797 |
https://mathoverflow.net/questions/33628 | 20 | This is essentially the same as the closed question [Representation of rational numbers as the sum of 1/k](https://mathoverflow.net/questions/32956/representation-of-rational-numbers-as-the-sum-of-1-k) but I hope I can make a case for it as an MO-worthy question.
Ed Pegg, Jr., in his Math Games column for 19 July 2004 at the MAA website, <http://www.maa.org/editorial/mathgames/mathgames_07_19_04.html> writes, "Here is an interesting sequence of fractions that would likely would [sic] have fascinated Ahmes: $$1/2, 2/3, 4/5, 8/11, 14/17, 19/23, 24/29, 49/59, 65/71, 76/83, 61/157, 183/191, 260/269, 289/299.$$ $8/11 = 1/2 + 1/6 + 1/21 + 1/77$. This is the simplest Egyptian fraction that requires 4 parts.
$14/17 = 1/2 + 1/4 + 1/20 + 1/55 + 1/187$ requires 5 parts. 289/299 is the simplest fraction that requires 14 parts. One might think that this sort of thing was well known, but it isn't.... What is the simplest fraction that requires 15 parts, 16 parts, and beyond?"
Pegg never defines "simplest," but presumably it means smallest (positive) denominator and, among fractions with the same denominator, smallest (positive) numerator. So the general question would be, given $s$, what's the simplest rational that can be expressed as a sum of $s$ unit fractions, but not fewer?
In this form, it's probably an open, and maybe impossible, problem (that is, I don't think anyone will find a simple formula for the rational as a function of $s$), so let me ask a bit less. Has there been any advance beyond 14 since 2004? Are there any bounds in the literature (that is, bounds on the "complexity" of the rational as a function of $s$)?
I note that Pegg gives no source for his list of 14. The Online Encyclopedia of Integer Sequences does not recognize the sequence of numerators, nor the sequence of denominators. Before anyone suggests typing "Egyptian fractions" into Google, or looking at the Wikipedia article on that subject, I hope he or she will verify that the particular question I'm asking is in fact answerable by such means.
EDIT: As per the comments, it appears that only the first four terms in Pegg's list are correct, and that the current state of knowledge is $${1\over2},{2\over3},{4\over5},{8\over11},{16\over17},{77\over79},{732\over733}.$$
Also as per the comments, if we are after
$f(s)=\min\lbrace b:N(a,b)=s{\rm\ for\ some\ }a,1\le a\lt b\rbrace$ then
$f(s)\ge e^{Cn^2}$ for some $C>0$, and, conjecturally, $f(s)\ge e^{e^{Cn}}$ for some $C>0$.
At this point I will gladly settle for a calculation of $s(8)$.
| https://mathoverflow.net/users/3684 | What's the simplest rational not expressible as a sum of a given number of unit fractions? | $s(8) = \frac{27538}{27539}$.
I have made the code I used available at <http://crypt.org/hv/maths/least_eg-0.01.tar.gz>, with a README file at <http://crypt.org/hv/maths/least_eg-0.01-README>.
Update: those links no longer available, code is available via github at <https://github.com/hvds/seq/tree/master/least_eg>.
The package includes both PARI/GP code and C code using the GNU GMP library to calculate the results, as well as a synopsis of the results for each denominator from 2 to 27539 which may be of use for further analysis.
I estimate the PARI code would have taken about a CPU-year to find the result; the C code runs over 20 times faster on my machine, and I don't understand why the difference is so great. (I'd appreciate email if someone can explain.)
| 16 | https://mathoverflow.net/users/6089 | 38675 | 24,800 |
https://mathoverflow.net/questions/38674 | 1 | Is this graph transformation G\_1 to G\_2 efficiently computable?
1. All vertices in G\_1 are unique edges in G\_2
2. Adjacent vertices in G\_1 are adjacent edges in G\_2
The inverse transformation (edges to vertices) is trivial.
I can encode it as CSP, but I am not sure solving is tractable:
adjacent V1,V2 in G\_1 to (V1',V1''), (V2',V2'') in G\_2, must share a vertex, constraints: V1'=V2' / V1'=V2'' / V1''=V2' / V1''=V2'' and some disequations.
References, search terms will be apreciated. Thank you.
| https://mathoverflow.net/users/9222 | Is the graph transformation vertices to edges efficiently computable? | This is only possible if $G\_1$ is the [line graph](http://en.wikipedia.org/wiki/Line_graph) of $G\_2$. Not all graphs are line graphs.
>
> A graph $G$ is the line graph of some other graph, if and only if it is possible to find a collection of cliques in $G$, partitioning the edges of $G$, such that each vertex of $G$ belongs to at most two of the cliques.
>
>
>
[Here](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V0F-45FCW4T-37&_user=10&_coverDate=10%2F31%2F1973&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=bde3a2a5cadbc74f8878ccdfc7ac7989&searchtype=a) you can find am algorithm to recover the original graph from it's line graph.
| 2 | https://mathoverflow.net/users/2384 | 38677 | 24,802 |
https://mathoverflow.net/questions/38680 | 19 | I'm fearful about putting this forward, because it seems the answer should be elementary. Certainly, the Weak Approximation Theorem allows every system of simultaneous inequalities among archimedean absolute values to be satisfied. But equality combined with inequality?
| https://mathoverflow.net/users/4994 | Can an algebraic number on the unit circle have a conjugate with absolute value different from 1? | Yes. Take
$$
\alpha=\sqrt{2-\sqrt{2}}+i\sqrt{\sqrt{2}-1}.
$$
Neither of the conjugates
$$
\sqrt{2+\sqrt{2}}\pm \sqrt{\sqrt{2}+1}
$$
have absolute value 1.
It is impossible, however, if $\mathbb{Q}(\alpha)/\mathbb{Q}$ is abelian, since then all automorphisms commute with complex conjugation.
This was all stolen from Washington's Cyclotomic Fields book.
| 32 | https://mathoverflow.net/users/35575 | 38683 | 24,805 |
https://mathoverflow.net/questions/38670 | 3 | Let $X$ be a quasi-projective variety, $Y$ a projective variety, and $f:X \rightarrow Y$ an open immersion. If $\mathcal{F}$ is a locally free coherent sheaf, what can be said about $f\_\ast \mathcal{F}$? Is it coherent? Is it torsion free? Is it reflexive?
| https://mathoverflow.net/users/9220 | Is the pushforward of a locally free sheaf by an open immersion coherent? | About your new question:
Let $Y$ be a projective variety and let $X\subset Y$ be an open subset with complement the closed subset $S:=Y\setminus X$. Call $f:X\hookrightarrow Y$ the inclusion.
Let $\mathcal F$ be an algebraic coherent sheaf without torsion on $X$.
**Theorem (Serre-Grothendieck)** Suppose that $Y$ is normal and that $S$ has codimension $\geq 2$. Then the sheaf $f\_\ast \mathcal F$ is coherent.
Serre, *Prolongement de faisceaux analytiques cohérents*, Ann.Inst.Fourier **16** (1966), 363-374
| 7 | https://mathoverflow.net/users/450 | 38687 | 24,809 |
https://mathoverflow.net/questions/38659 | 11 | In popular science books and articles, I keep running into the claim that the total energy of the Universe is zero, *"because the positive energy of matter is cancelled out by the negative energy of the gravitational field"*.
But I can't find anything concrete to substantiate this claim.
As a first check, I did a calculation to compute the gravitational potential energy of a sphere of uniform density of radius R using Newton's Laws and threw in $E=m{c}^2$ for energy of the sphere, and it was by no means obvious that the answer is zero !
So, my questions:
1. What is the basis for the claim – does one require general relativity, or can one get it from Newtonian gravity ?
2. What conditions do you require in the model, in order for this to work ?
3. Could someone please refer me to a good paper about this ?
| https://mathoverflow.net/users/8528 | Total energy of the universe | In fact, two very well-known mathematicians, Schoen and Yau, in a much quoted paper, proved the long standing conjecture that the ADM mass is always POSITIVE (except for flat space). Here is the reference and abstract:
Commun. math. Phys. 65, 45--76 (1979)
On the Proof of the Positive Mass Conjecture in General Relativity
Richard Schoen and Shing-Tung Yau
Abstract. Let M be a space-time whose local mass density is non-negative
everywhere. Then we prove that the total mass of M as viewed from spatial
infinity (the ADM mass) must be positive unless M is the fiat Minkowski
space-time. (So far we are making the reasonable assumption of the existence of
a maximal spacelike hypersurface. We will treat this topic separately.) We can
generalize our result to admit wormholes in the initial-data set. In fact, we
show that the total mass associated with each asymptotic regime is non-
negative with equality only if the space-time :is fiat.
| 18 | https://mathoverflow.net/users/7311 | 38690 | 24,812 |
https://mathoverflow.net/questions/38632 | 40 | I asked this question on the new Theoretical Computer Science "overflow" site, and commenters suggested I ask it here. That question is [here](https://cstheory.stackexchange.com/questions/1160/projective-plane-of-order-12), and it contains additional links, which I doubt I can embed here because I don't have enough reputation. Anyway, here goes:
**Objective**: Settle the conjecture that there is no projective plane of order 12.
In 1989, using computer search on a Cray, Lam proved that no projective plane of order 10 exists. Now that God's Number for Rubik's Cube has been determined after just a few weeks of massive brute force search (plus clever math of symmetry), it seems to me that this longstanding open problem might be within reach. I'm hoping this question can serve as a sanity check.
The Cube was solved by reducing the total problem size to "only" 2,217,093,120 distinct tests, which could be run in parallel.
Questions:
1. There have been special cases of nonexistence shown (again, by computer search). Does anyone know, if we remove those and exhaustively (cleverly?) search the rest, if the problem size is on the order of the Cube search? (Maybe too much to hope for that someone knows this....)
2. Any partial information in this vein?
| https://mathoverflow.net/users/9197 | Projective Plane of Order 12 | I am actually not aware of many results on planes of order 12 in the vein of what Lam et. al. did (I list the few I know of below). There seems to be a plethora of papers proving restrictions on the collineation group of a hypothetical such plane, but I am not aware of how any of these could be used to settle the existence problem.
Moreover, I am quite skeptical that disproving the existence of planes of order 12 by a computer search would help for the general theory much. Though it certainly would be nice to know, and if one actually found a plane of order 12, that would be quite exciting; but it's hard to gain deep insights from these combinatorial brute force searches.
Extending the approach by Lam et. al. to planes of order 12 is in principle possible. But probably still not feasible with today's computers, as the search space is a *lot* bigger than for order 10. Anyway, here are some reasons why I think that, and at the same time a sketch of things that would have to be done. But my personal belief is that one will need some substantially new ideas to make progress on this. Then again, only by actually trying to do it can one be sure... :)
From here on, I'll assume you are familiar with Lam's ["The Search for a Finite Projective Plane of Order 10"](http://www.cecm.sfu.ca/organics/papers/lam/) and the notation used within.
A crucial point was the reduction of the (non-)existence to the value of certain weight enumerator coefficients $w\_0$ to $w\_{n^2+n+1}$ (a good exposition can be found in ["On the existence of a projective plane of order 10"](http://www.ams.org/mathscinet-getitem?mr=313089) by MacWilliams, Sloane and Thompson). But the real breakthrough was when [Assmus and Mattson proved](http://www.dtic.mil/cgi-bin/GetTRDoc?AD=AD718114&Location=U2&doc=GetTRDoc.pdf) that one only needs to know $w\_{12},w\_{15},w\_{16}$ to determine all others. I'll refer to these as *essential* weight enumerator coefficients.
Some steps towards this for order 12 have been executed in ["Ternary and binary codes for a plane of order $12$"](http://www.ams.org/mathscinet-getitem?mr=734976) by Hall and Wilkinson. Yet many nice properties and theorems will be hard to recover for order 12. E.g. for orders of the form $8m+2$, one knows the $\mathbb{F}\_2$-rank of the incidence matrix. Not so for order 12, where working with a ternary code is in some ways more "natural." In particular, the $\mathbb{F}\_3$-rank of the incidence matrix is known, but, alas, working with a ternary code means losing the natural identification of codewords with point sets, so tons of new machinery would be needed to exploit the ternary code. Thus I'll focus on the binary code case here.
Anyway, let's assume we reduced the number of essential weight enumerator coefficients as much as we can (Hall and Wilkinson pushed it down to 16; remember, for $n=10$ we had only 3). We must compute the essential coefficients.
According to Lam, for $n=10$ and the case $w\_{12}$, they estimated, using a Monte-Carlo method (before doing it) that $4\times 10^{11}$ configuration had to be checked. I don't have a good means to compute a good estimate for $n=12$, but for that there are 16 coefficients to determine, and I'd hazard to guess that some of them are much, much harder than the three cases for $n=10$ put together. Several orders of magnitude. However, this is just gut feeling.
So let's assume we had somehow managed to overcome this and had computed all essential weight enumerator coefficients. We then would have the full weight enumerator at hand (and no projective plane arose as a byproduct of our search). Now, the hard part starts (corresponding roughly to the second half of Lam's paper), the one that took them 2 years for $n=10$: We have to somehow derive a contradiction (or construct a plane). A lot of ground work needs to be done (extending stuff from $n=10$), before one can even start writing code...
Ah well. To anybody who wants to try out this strategy on $n=12$, I would recommend to first try reproducing the $n=10$ result -- with modern computers it should be possible to do this much, much quicker than it took Lam et. al. originally (this verification might already interest some people on its own). Actually, at the very start, try it with even smaller examples ($n=6,8$), then go up.
UPDATE 2023-08-10: I just learned that the $n=10$ result has been reproduced not once but twice in the meantime:
1. in 2010/2011 by Dominique J. Roy in their [master thesis "Confirmation of the non-existence of a projective plane of order 10"](https://doi.org/10.22215/etd/2011-09202) at Carleton University, using special purpose code. They also give a more detailed overview of what is involved in the whole endeavour.
2. And then again more recently with help of a standard SAT solver (!), in an effort culminating in: Bright, C., Cheung, K. K. H., Stevens, B., Kotsireas, I., & Ganesh, V. (2021). *A SAT-based Resolution of Lam’s Problem.* Proceedings of the AAAI Conference on Artificial Intelligence, 35(5), 3669-3676. [DOI](https://doi.org/10.1609/aaai.v35i5.16483).
| 69 | https://mathoverflow.net/users/8338 | 38707 | 24,824 |
https://mathoverflow.net/questions/38699 | 5 | Take an equator on the two sphere $S^2$ and parametrize it by arc-length obtaining a closed loop $\alpha: S^1 \to S^2$. The curve $(\alpha,\alpha'):S^1 \to T^1S^2$ in the unit tangent bundle of $S^2$ is homotopically non-trivial.
However if you consider the concatenation $\beta$ of two copies of $\alpha$, you can take one copy and turn it about a diameter passing through two points of $\alpha$ so that it is now a copy of $\alpha$ but traversed in the opposite sense. In other words, the curve $(\beta,\beta') \in T^1S^2$ is homotopically trivial.
Does this occur on other compact orientable surfaces?
On the torus the answer is no.
What about on the double-torus (which is a hyperbolic surface)?
Also, does anybody have references for the above statements about the sphere and the torus? In particular, are they correct?
I've come across these problems while trying to picture the lift of a general geodesic flow to the universal covering space of $T^1M$ where $M$ is a compact orientable surface (in particular, what do homotopically trivial closed geodesics look like?).
| https://mathoverflow.net/users/7631 | What immersed closed curves on the double-torus are non-trivial when lifted to the unit tangent bundle? | If $M$ is not $S^2$ or $RP^2$, then $\pi\_1(T^1M)$ does not have elements of finite order
(in particular a double non-contractible loop is also non-contractible). Indeed, consider the long exact sequence of our fibration $E=T^1M\to M$:
$$
\dots\to \pi\_2(M)\to \pi\_1(F)\to\pi\_1(E)\to\pi\_1(M)\to\dots
$$
where $F$ is a fiber (a circle).
Note that $\pi\_2(M)=0$ (since the universal cover of $M$ is the plane), hence the arrow $\pi\_1(F)\to\pi\_1(E)$ is injective. Hence the kernel of the arrow $\pi\_1(E)\to\pi\_1(M)$ is isomorphic to $\pi\_1(F)$ which is $\mathbb Z$. So this kernel does not contain elements of finite order. And if a non-kernel element has a finite order, then so does its image in $\pi\_1(M)$. But $\pi\_1(M)$ has no elements of finite order (e.g. due to existence of a nonpositively curved metric on $M$).
| 8 | https://mathoverflow.net/users/4354 | 38713 | 24,829 |
https://mathoverflow.net/questions/38648 | 7 | I'm trying to get a better grasp of iterated forcing, and I ran across the following problem:
0) Let $P\_\alpha$ be posets in a c.t.m. $M$, $\alpha<\beta$, and for each $\alpha$ let $G\_\alpha$ be $P\_\alpha$ generic. Let $P$ be the finite support iteration of the $P\_\alpha$. Then is there necessarily a $G$ which is $P$-generic such that for all $\alpha$, $G\_\alpha\in M[G]$?
My guess for how to do this would be transfinite induction, but I run into some problems when I try that. Specifically, the successor step makes sense, but the limit step and the base case $(\beta=2)$ don't.
[For what it's worth, here's the successor step. Suppose the answer to (0) is "yes" for all $\beta<\delta$, $\delta\ge 2$. Let $P\_\alpha$ be posets for $\alpha<\delta+1$, let $G\_\alpha$ be $P\_\alpha$-generic, let $P$ be their finite support iteration, and let $P'$ be the finite support iteration of the $P\_\alpha$ for $\alpha<\delta$. Then $P$ is equivalent (in fact, isomorphic) to $P'\times P\_\delta$. By the induction hypothesis, we have some $G'$ which is $P'$-generic such that for all $\alpha<\delta$, $G\_\alpha\in M[G']$. Applying the induction hypothesis again (since $\delta>1$) we have some $G$ which is $P'\times P\_\delta$-generic such that $G', G\_\delta\in M[G]$. But then $M[G']\subseteq M[G]$, so $G\_\alpha\in M[G]$ for all $\alpha<\delta$.]
Related questions:
1) What happens if we restrict to c.c.c. posets?
2) What happens if we specify $\beta=2$?
3) What happens if we try to generalize to other kinds of supports?
I have a feeling I'm missing something obvious, but I've thought about this for a while now with no success.
| https://mathoverflow.net/users/8133 | A question about iterated forcing | I'm not sure I really understand the question, because of the mixture of iteration and product, but I believe the following negative answer is independent of such issues because it uses $\beta=2$ and takes both factors (or iterands) to be Cohen forcing (so the product forcing is equivalent to the iteration). Recall that a real (regarded, as usual in such contexts, as an element of $2^\omega$) is Cohen-generic over $M$ if and only if it belongs to every dense $G\_\delta$ set coded in $M$. Since $M$ is countable, the set of Cohen-generic reals over $M$ is comeager. Now choose some real $z$ that codes a well-ordering of $\omega$ longer than the height of $M$. Since both $\{x\in 2^\omega:x\text{ Cohen over }M\}$ and its translate $\{x\in 2^\omega:x\oplus z\text{ Cohen over }M\}$ (where $\oplus$ is pointwise addition mod 2) are comeager, they have an element $x$ in common. Take $G\_0$ to be the generic filter coding $x$ and take $G\_1$ to be the generic filter coding $x\oplus z$. No generic extension of $M$ (by $P\_0\times P\_1$ or by any other forcing) can contain both of these $G\_\alpha$'s. [Proof: Such a model would contain both $x$ and $x\oplus z$, and would therefore contain $z$. Being a model of ZF, it would have to contain the ordinal isomorphic to the ordering of $\omega$ coded by $z$. But the ordinals of a forcing extension of $M$ are the same as those of $M$, so this contradicts the choice of $z$.]
| 10 | https://mathoverflow.net/users/6794 | 38718 | 24,831 |
https://mathoverflow.net/questions/38666 | 7 | Fix a countable transitive model $M$ of ZFC.
In my answer to [this question](https://mathoverflow.net/questions/38648/a-question-about-iterated-forcing) I indicated that there are forcing iterations
$((Q\_\alpha:\alpha\leq\omega),(\dot P\_\alpha:\alpha<\omega))$ in $M$ and sequences
$(G\_\alpha:\alpha<\omega)$ of filters such that the following happens:
Each $G\_\alpha$ is a filter in the evaluation of $\dot P\_\alpha$ with respect to the filter $G\_0\*\dots\*G\_{\alpha-1}$ and $G\_\alpha$ is generic over $M[G\_0,\dots,G\_{\alpha-1}]$
(call such a sequence $(G\_\alpha:\alpha<\omega)$ a sequence of generics),
but there is no $Q\_\omega$-generic filter over $M$ whose
$\alpha$-th projection is $G\_\alpha$ for all $\alpha<\omega$.
An example can be obtained as follows:
Take the countable support iteration of Sacks forcing
(or any other nontrivial $\omega^\omega$-bounding proper forcing notion) of length $\omega$
(i.e., the supports are actually everything, but this doesn't matter).
This forcing adds no Cohen real.
Compare this to the finite support iteration of the same forcing notions.
This iteration does add a Cohen real. The Cohen real is coded by the sequence of generics
and hence this sequence of generics does not come from a generic filter for the countable support iteration mentioned before. This sequence of generics is not even contained in
a forcing extension obtained using the countable support iteration.
Now here are two questions:
1) Is there an example of a sequence of generics (of length $\omega$) that cannot come from
any iteration of the $\dot P\_\alpha$?
I am asking here for iterations where the finite initial segments are as usual (just plain iteration) and we choose whatever ideal for the supports, including all finite subsets of the index set.
But I am open to more general forms of iteration. For example take a large forcing notion
$Q$ along with commuting complete embeddings of the $Q\_\alpha$, $\alpha<\omega$.
This would be an iteration of the $\dot P\_\alpha$, too, the most general one that I can think of right now.
2) Is there an example of a sequence of generics over $M$ that is not contained in any countable transitive extension of $M$ with the same ordinals as $M$ that is a model of ZFC?
Obviously, a positive answer to 2) solves 1) as well.
| https://mathoverflow.net/users/7743 | A sequence of generic filters that does not come from an iteration | There are a number of interesting things to say.
The answer to your first question is yes. Suppose that $M$ is a countable transitive model of set theory and we have a forcing iteration $P\_\omega$ in $M$ of length $\omega$, forcing with, say, Cohen forcing $Q\_n$ at stage $n$. Let $z$ be any real that cannot be added by forcing over $M$, such as a real that codes all the ordinals of $M$. This real cannot exist in any extension of $M$ to a model of ZFC with the same ordinals. Now, suppose that $G$ is any $M$-generic filter for the iteration, with $G\_n$ being the stage $n$ generic filter. Let $H\_n$ be the filter that results from $G\_n$ by changing the first bit so as to agree with $z(n)$. That is, we change a single bit at each stage. The resulting sequence $\langle H\_n | n\in\omega\rangle$ will be generic at every stage, since only finitely many bits are changed by a given stage, but the whole sequence computes $z$, which cannot be added by forcing.
Second, a similar phenomenon occurs even just with 2-step product forcing:
**Theorem.** If $M$ is a countable transitive model of ZFC, then there are two $M$-generic Cohen reals $c$ and $d$ such that $M[c]$ and $M[d]$ have no common extension to model of ZFC with the same ordinals.
The proof is to build $c$ and $d$ in stages. Fix a real $z$ which cannot exist in any extension of $M$ with the same ordinals, and enumerate the dense sets of $M$ by $D\_0, D\_1$ and so on. Build $c$ and $d$ in zig-zag fashion: first provide $c\_0$ meeting $D\_0$, and $d\_0$ all $0$s of the same length as $c\_0$, followed by the first digit of $z$. Now extend $d\_0$ to $d\_1$ meeting the dense set, adding all $0$s to $c\_0$ making $c\_1$ of the same length, and adding one more bit of $z$. And so on. The point is that we ensure that each of $c$ and $d$ is $M$-generic, but together, they reveal the coding points of $z$. So no model extending $M$ with the same ordinals can have both $c$ and $d$, for then it would have $z$.
Third, this is essentially the only kind of obstacle, for there is a positive result here. The following theorem is proved in a paper with G. Fuchs, myself and J. Reitz on the topic of [set-theoretic geology](http://arxiv.org/abs/1107.4776):
**Theorem.** If $M$ is a countable transitive model of set theory, and $M[G\_n]$ is a sequence of generic extensions of $M$ by forcing $G\_n\subset P\_n\in M$ of bounded size in $M$, such that the extensions are finitely amalgamble, in the sense that any finitely many of the $M[G\_n]$ have a common forcing extension $M[H]$, then there is a single forcing extension $M[H]$ containing all $M[G\_n]$.
I'll try to post a proof sketch later, but the main idea is to perform a very large combination of forcing, and then perform surgey so as to replace certain coordinate generics with $G\_n$, in such a way that the resulting extension can see only finite fragments of the sequence $\langle G\_n | n\lt\omega\rangle$, without being able to construct the whole sequence.
A special case of this theorem answers your second question, in a sense, for if we have a sequence of extensions $M\subset M[G\_0]\subset M[G\_1]\subset\cdots$, then these extensions are finitely amalgamable, and so there is indeed a common extension $M[H]$ containing every $M[G\_n]$. This extension, however, is not an $\omega$-iteration of the forcing notions in your iteration, and in general we cannot expect that the sequence $\langle G\_n | n\lt\omega\rangle$ is in $M[H]$, for the reasons described above.
| 8 | https://mathoverflow.net/users/1946 | 38720 | 24,832 |
https://mathoverflow.net/questions/38651 | 1 | Say given elliptic curve $ \{ (x,y) | y^2 = (x^2-1)(x^2-k^2) \}$, what is the right form of the K$\ddot{a}$hler form and how to compute the K$\ddot{a}$hler moduli of this elliptic curve? Thank you.
| https://mathoverflow.net/users/1790 | compute the Kähler moduli of an elliptic curve | The curve you wrote in equations lies in C^2, while the "elliptic curve" of your text is presumably a compact projective variety -- meaning you imagine making your equations homogeneous (or even quasi-homogeneous) and considering the closure of the set of points described by your equation in a (quasi-)projective plane. Not every "homogenization" will lead to an elliptic curve (Calabi-Yau) upon compactification, so you have to do this correctly (as noted by Kevin Buzzard above).
Having said that, the answer is that every projective variety is also Kahler: just restrict e.g. the Fubini-Study(-like) Kahler form. In plain English, since a Kahler form on a complex curve is just a volume form, the volume of the compact curve inside projective space gives you your answer.
| 5 | https://mathoverflow.net/users/1186 | 38723 | 24,835 |
https://mathoverflow.net/questions/38716 | 8 | The $\mathbb{Z}\_2$ topological degree of a (non-constant) polynomial in one variable, clearly, coincides with its degree as a polynomial, $\mod 2$.
Consider further a polynomial self-mapping $F$ on $\mathbb{R}^2$, and assume it is a proper map (in case, even more generally a map in higher dimension)
>
> Is there a short way to decide what's
> the parity of the topological degree of $F$?
>
>
>
E.g. it's odd if $F$ is an odd map, or more generally, if $F$ can be transformed into an odd map by a proper homotopy. Actually: is there a short way to understand if a polynomial map is a proper map? What about the case of a gradient map (I mean, the gradient of a polynomial $f\in\mathbb{R}[x,y]$)?
I'm somehow confident that there may be a simple criterion known, at least in $\mathbb{R}^2.$ After all, what is required is just a one-bit information (well this argument doesn't convince me either).
| https://mathoverflow.net/users/6101 | Topological degree of polynomial maps. | Here is an idea, I'm not 100% confident that it makes sense in all cases, but I'll try it anyway. Assuming that $f(\mathbb{R}^2)$ is 2-dimensional, the degree mod 2 of your map is the cardinality of the preimage of a generic point. If your components have degree respectively *d* and *e*, then Bezout gives you a preimage size of *de*.
That's projective solutions. In generic cases, you would expect no solutions at infinity. Unfortunately as David points out below, you may have an *odd* number of solutions at infinity.
So in the case when there's nothing at infinity, we have solutions over the complexes, but non-real solutions will come in complex conjugate pairs, so in that nice case the degree mod 2 is given by *de*.
| 3 | https://mathoverflow.net/users/8212 | 38729 | 24,839 |
https://mathoverflow.net/questions/38719 | 3 | If I have an arbitrary positive monotonically decreasing function $f(x), x \in [0,\infty]$, is there an 'efficient' method for finding $y$ in:
$r = \int\limits\_0^y f(x) dx $
for a known $r \in [0, \int\limits\_0^\infty f(x) dx]$. By efficient I guess I mean more efficient than doing numerical integration until one lands in within a given distance from $r$.
I particularly care about the cases where the integral of $f$ has no closed form.
---
Best answer so far (rewrite of below), doubt it gets much better than this:
Start with $s\_0 = r$, $y\_0 = 0$, then
$y\_{i+1} = y\_i + \frac{s\_i}{f(y\_i)} $
$s\_{i+1} = s\_i - \int\limits\_{y\_i}^{y\_{i+1}}f(x)dx$
| https://mathoverflow.net/users/9199 | Numerical Solution to Inverse Integral (Pseudo Random Number Generation) | I guess the only non-trivial thing about the problem is that:
$$
x f(0) \geq \int\_0^{x} f(t) dt \geq x f(x).
$$
So you start by computing the integral
$$
r\_1 = \int\_0^{y\_1} f(t) dt,\quad y\_1 = \frac{r}{f(0)}.
$$
Then replace $r$ by $r - r\_1$. I think under reasonable assumptions this should converge pretty quickly (and always lower bound). It should be noted that I only use one of the inqualities, one can probably optimize it by using the other one.
| 3 | https://mathoverflow.net/users/3983 | 38731 | 24,840 |
https://mathoverflow.net/questions/38717 | 12 | Let's start with the following random example: If $F$ is a presheaf, then for every chain of open subsets $U \subseteq V \subseteq W$, the morphisms $F(W) \to F(V) \to F(U)$ and $F(W) \to F(U)$ coindice. But this may be an evil ([nlab link](http://ncatlab.org/nlab/show/evil)) definition, especially when the values of $F$ are categories (which occurs in my current research). So we should just impose that these morphisms are equivalent. But then we arrive at compatibility conditions between these equivalences, which are, again, equalities, and may be evil. So what is the consequence of this: Should every mathematical theory take place in a $\infty$-category? Or is 'real' mathematics basically evil? I know this question is quite imprecise, but currently I just don't see where and why this process of increasing the "depth" of category theory should end. Anyway, I have to admit that my knowledge of higher category theory is very, very rudimental.
Here is a evil question, which is part of my confusion: Let $Ring$ denote the category of rings, ring homomorphisms and $AbCat$ denote the category of abelian categories, functors. Then $Ring \to AbCat, A \to Mod\_A$ should be a functor, right? If $A \to B$ is a ring homomorphism, just tensor with $B$ over $A$. But then functoriality is only satisfied up to equivalence ($2$-isomorphism) in $AbCat$. To get a honest functor, we might mod out these equivalences to get a $1$-category $\tilde{AbCat}$. But I don't think that this is the most natural way to handle this. Or we may define quasi-functors (cf. the presheaf example). Anyway, may we *think* of it as a usual functor, without turning into troubles? Or is it important, in practice, to have this higher category theoretic point of view? Or is it possible to turn this functor into a honest functor, by choosing the tensor products $M \otimes\_A B$ carefully? I think the latter is interesting, although I know that nobody really cares about such evil equalities like $(M \otimes\_A B) \otimes\_B C = M \otimes\_A C$.
| https://mathoverflow.net/users/2841 | evil properties, higher category theory and well-chosen tensor products | There are a lot of questions here, but I'll try to answer them all.
>
> Should every mathematical theory take place in a ∞-category? Or is 'real' mathematics basically evil?
>
>
>
I would say that all mathematics should take place in its natural context. Sometimes you have things that are sets where equality makes sense, like an ordinary presheaf, and then you work in a 1-category. Sometimes you have things where only isomorphism makes sense, like a presheaf of categories, and then you work in a 2-category. Etc.
It is true that any n-category for finite n can be considered a special case of an ∞-category with only identity cells above n, so in this degenerate sense all n-categories are ∞-categories, and thus one might say that "all mathematics takes place in an ∞-category" — at least if one believes that all mathematics takes place in an n-category for some n! But even that is not clear, e.g. some mathematics naturally takes place in other categorical structures, such as a double category or a proarrow equipment. Some mathematics uses no category theory at all (at least as far as anyone has noticed so far), and so it would be a stretch to say that it takes place in any sort of category.
>
> Anyway, may we *think* of it as a usual functor, without turning into troubles? Or is it important, in practice, to have this higher category theoretic point of view? Or is it possible to turn this functor into a honest functor, by choosing the tensor products $M\otimes\_A B$ carefully?
>
>
>
I would say qualified yes, yes, and yes, respectively. You can think of it as a usual functor as long as doing so doesn't cause you to think that it behaves in any way that a pseudofunctor doesn't! Which is sort of a vacuous statement, but the point is that pseudofunctors really shouldn't be a very scary concept (as opposed to a technical definition, which might be a bit complicated, though cf. Harry's comment) — they really are just like ordinary functors, except that you're dealing with things (e.g. categories) for which it doesn't really make sense to ask morphisms to be equal, only isomorphic.
On the other hand, the "higher category theoretic" fact that pseudofunctors are *not* all strict functors *is* very important. I believe that Benabou, the inventor of bicategories, once said that the important thing about bicategories is not that they themselves are "weak," but that the *morphisms* between them are weak. In particular, although every bicategory is equivalent to a strict 2-category, not every pseudofunctor between bicategories is equivalent to a strict functor.
But on the third hard, it *is* true that any pseudofunctor with values in the 2-category Cat *is* equivalent to a strict functor. In the language of fibrations, this says that any fibration is equivalent to a split one. Tyler mentioned one construction of an equivalent strict functor in the case of modules and tensor products. There is also a general construction which, applied to the case of modules, will replace $Mod\_A$ by a category whose objects are pairs (M,φ) where M is an R-module and φ:R→A is a ring homomorphism. We regard such a pair as a formal representative of $M\otimes\_R A$ and define morphisms between them accordingly, to get a category eequivalent to $Mod\_A$. Now the extension-of-scalars functor $\psi\_!:Mod\_A \to Mod\_B$ is represented by the functor taking a pair (M,φ) to (M,ψφ), which is strictly functorial since composition of ring homomorphisms is so.
| 13 | https://mathoverflow.net/users/49 | 38734 | 24,841 |
https://mathoverflow.net/questions/38733 | 2 | I'm reading up on [maximal sets](http://en.wikipedia.org/wiki/Maximal_set) and the word "coinfinite" pops up in the first sentence. I tried searching on Wolfram Mathworld as well as Google, but nothing concrete has come up. What does it mean and in what context can it be used?
| https://mathoverflow.net/users/5837 | What does coinfinite mean? | Perhaps it means the complement is infinite.
Certainly "cofinite" and "cocountable" are used this way.
| 10 | https://mathoverflow.net/users/454 | 38735 | 24,842 |
https://mathoverflow.net/questions/38724 | 16 | For $d=3$, vertex coordinates of a [regular simplex](http://en.wikipedia.org/wiki/Simplex) have a simple expression since vertices correspond to four vertices of a cube. Is there a simple expression for higher dimensions? In particular I'm interested in $d=2^n-1$, integer $n$.
**Edit**: by coordinates I mean points in $\mathbb{R}^d$. Every $d$-simplex has a simple expression for coordinates in $\mathbb{R}^{d+1}$, as Mariano shows below
| https://mathoverflow.net/users/7655 | coordinates of vertices of regular simplex | It is known that there is a regular simplex of side length $\sqrt{(d+1)/2}$ whose vertices are vertices of the cube $[-1,1]^d$ in $\Bbb{R}^d$ if and only if there exists a [Hadamard matrix](http://en.wikipedia.org/wiki/Hadamard_matrix) of order $d+1$; this is a square matrix of $\pm 1$-entries with pairwise orthogonal columns.
In particular, there exist Hadamard matrices of order $2^n$, one of which can be constructed using the recursive Sylvester's construction as explained on the above linked [wikipedia page](http://en.wikipedia.org/wiki/Hadamard_matrix#Sylvester.27s_construction):
Let $H\_0=[1]$ and $H\_{n+1}=\left[\array{H\_n & H\_n \\\\ H\_n & -H\_n}\right].$
Note that the first column of $H\_n$ consists only of ones. Delete it to obtain $2^n$ row vectors in $\Bbb{R}^{2^n-1}$. These are the coordinates of a regular simplex.
| 13 | https://mathoverflow.net/users/932 | 38736 | 24,843 |
https://mathoverflow.net/questions/38748 | 2 | Let $f: X \to Y$ be a fibration of pointed Kan complexes, and let $F$ be the fiber.
Question: How do you prove that the following diagram of homotopy groups commutes?:
$\pi\_n(Y) \to \pi\_{n-1}(\Omega Y)$
$\ \ \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow$
$\pi\_{n-1}(F) \to \pi\_{n-2}(\Omega F)$
Admittedly, I don't know for certain that it commutes, but it looks like it should.
All the arrows are boundary maps ($\delta$) from long exact sequences of homotopy groups for a fibration. The horizontal maps are isomorphisms.
The definition of $\delta$ that I'm using is that for for a fibration $X \to Y$, $\alpha \in \pi\_n(Y)$, $\delta ([\alpha]) = [\beta d^0]$, where $\beta:\Delta^n \to X$ fits into the following diagram:
$ \Lambda^n\_0 \to^\* X$
$\ \downarrow \ \ \ \ \ \downarrow$
$\Delta^n \to^\alpha Y$
So far, I've attempted to chase elements around this diagram and use prismatic arguments, but I haven't found one that works.
$\Omega F \to \Omega X \to \Omega Y$
$\ \ \downarrow\ \ \ \ \ \ \downarrow\ \ \ \ \ \ \ \downarrow$
$PF \to PX \to PY$
$\ \ \downarrow\ \ \ \ \ \ \downarrow\ \ \ \ \ \ \ \downarrow$
$\ \ F \to \ \ X \to \ \ Y$
It feels like a useful fact that $PX \to PY\times\_Y X$ is a fibration, which is true since pointed simplicial sets form a simplicial model category.
Edit: The path and loop spaces I'm using are defined by the following:
For pointed simplicial sets X and Y, define the simplicial set $hom\_\*(X,Y)$ to have n- simplices $hom\_{sSet\_\*}(X \wedge \Delta^n\_+,Y)$, where $\Delta^n\_+$ is the standard n-simplex with a disjoint basepoint.
Then I'm using $PX=hom\_\*(\Delta^1 , X)$
and
$\Omega X=hom\_\*(\Delta^1/\partial \Delta^1,X)$
| https://mathoverflow.net/users/9109 | Commutativity of a diagram of boundary morphisms from the long exact sequence of homotopy groups of a fibration and its loop spaces | The boundary map of homotopy groups is induced by a map $\partial : \Omega Y \to F$
of spaces; then the commutativity follows from the naturality of the
isomorphism $\pi\_n \circ \Omega \cong \pi\_{n+1}$.
| 1 | https://mathoverflow.net/users/3634 | 38761 | 24,855 |
https://mathoverflow.net/questions/38763 | 22 | Is $L^p(\mathbb{R}) \setminus 0$ contractible? My intuition says that the answer is yes, but I'm afraid that this is based on thinking of this as somehow similar to a limit of $\mathbb{R}^n \setminus 0$ as n approaches $\infty$, which is of course nonsense. In any case, every contraction I've tried ends up making some function pass through $0$.
| https://mathoverflow.net/users/9189 | Is $L^p(\mathbb{R})$ minus the zero function contractible? | Here is something really cheap and dirty. Let $p<+\infty$. Take $g=\frac{1}{1+x^2}$. Then $f(x,t)=e^{-(1+|x|)t/(1-t)}f(x)$ ($0\le t\le 1$) is a continuous contraction of $L^p\setminus\{g\}$ to $0$. (the reason is that your only chance to hit $g$ is to start with it because $g(x)e^{s(1+|x|)}$ is not in $L^p$ for $s>0$).
Let's make it more interesting without making it more abstract. Can we find a uniformly continuous (both in space and time, as usual) contraction of the unit ball in $L^p$ without the center to a point?
| 22 | https://mathoverflow.net/users/1131 | 38769 | 24,860 |
https://mathoverflow.net/questions/38756 | 1 | The Harmonic series is well known and its divergence was proven back in the middle ages.
I've taken an introductory course in model theory so I know a bit about RCF and some properties of it. We did not explore it thoroughly though and haven't seen many interesting examples.
However, I do know that we can take some real closed field which is large enough (i.e. has cofinality $>\aleph\_0$) and then the harmonic series will possibly converge.
My question if we take some $\mathcal{F}$ to be a model of RCF in which $\mathbb{R}$ is embedded and that the type $p(x) = \{ x > n | n\in\mathbb{N}\}$ is realized,
$$x = \sum\_{n \in \mathbb{N}^+} \frac{1}{n}$$ then $\forall y\in\mathbb{R}(x>y)$ then obviously $x$ is an upper-bound for the real numbers in the field we've chosen. However since $x$ is a non-Archimedean number, it is also clear that $x-1$ is an upper bound of the real numbers in $\mathcal{F}$.
This is the part where I get confused. What is $x$ and what is the conditions required for it to exist in the model?
| https://mathoverflow.net/users/7206 | Convergence of the harmonic series in larger fields | Real closed fields are not complete (unless they are isomorphic to the reals), so the fact that some increasing sequence is bounded does not imply that it has a supremum.
If x is the sum of the harmonic series, then we seem to get x=1+ 1/3 + ...+ 1/2+1/4+...>1/2+1/4...+1/2+1/4+..=x/2+x/2 = x, suggesting that x does not exist in any real closed field.
| 1 | https://mathoverflow.net/users/51 | 38770 | 24,861 |
https://mathoverflow.net/questions/38752 | 23 | I have not studied category theory in extreme depth, so perhaps this question is a little naive, but I have always wondered if analysis could be taught naturally using categories. I ask this because it seems like a quite a lot of topological and group theoretic concepts can be defined most succinctly using categorical concepts, and the categorical definitions are more revealing. So my question is: (1) Is it possible/beneficial to teach analysis using category theory? and (2) Are there any good textbooks that use this method?
| https://mathoverflow.net/users/6856 | Analysis from a categorical perspective | I hesitate to let this out, but there's always this cute little note that I learned from another MO answer (I don't know which one): [https://www.maths.ed.ac.uk/~tl/glasgowpssl/banach.pdf](https://www.maths.ed.ac.uk/%7Etl/glasgowpssl/banach.pdf). Maybe this will satisfy your curiosity, but I maintain that it takes a warped mind to identify such a categorical formulation of integration as the "right" way to think about integrals.
---
The advantage of categorical thinking in my view is that it helps to organize computations and arguments involving several different kinds of structures at the same time. For instance, (co)homology is all about capturing useful invariants associated to a complicated structure (e.g. a geometric object) in a much simpler structure (e.g. an abelian group). When we want to determine how the invariants behave under certain operations on the complicated structure (e.g. products, (co)limits) it helps to have a theory already set up to tell us what will happen to the simpler structure. That's where category theory comes into its own, and instances of this paradigm are so ubiquitous in algebra and topology that category theory has taken on a life of its own. It seems that people working in those areas have found it convenient to build categorical constructions into the foundations of their work in order to emphasize generality (one can treat algebraic varieties and solutions to diophantine equations on virtually the same footing), keep track of different notions of equivalence (e.g. homotopy versus homeomorphism), build new kinds of spaces (e.g. groupoids), and to achieve many other aims.
In many kinds of analysis, this kind of abstraction isn't necessary because there's often only one structure to keep track of: $\mathbb{R}$. When you think about it, analysis is only possible because we are willing to seriously overburden $\mathbb{R}$. Take, for example, the expression "$\frac{d}{dt}\int\_X f\_t(x) d\mu(x)$" and consider all of the different ways real numbers are being used. It is used as a geometric object (odds are X is built out of some construction involving the real numbers or a subspace thereof), a way to give $X$ additional structure (it wouldn't hurt to guess that $\mu$ is a real valued measure), a parameter ($t$), and a reference system ($f$ probably takes values in $\mathbb{R}$ or something related to it). In algebraic geometry, one would probably take each of these roles seriously and understand what kind of structure they are meant to bring to the problem. But part of the power and flexibility of analysis is that we can sweep these considerations under the rug and ultimately reduce most complications to considerations involving the real numbers.
All that being said, the tools of category theory and homological algebra actually have started to make their way into analysis. Because of the fact that analysts generally consider problems tied to certain very specific kinds of structure, they have historically focused on providing the sharpest and most detailed solutions to their problems rather than extracting the crude, qualitative invariants for which cohomological thinking is most appropriate. However, as analysts have become more and more attuned to the deep relationships between functional analysis and geometry, they have turned to ideas from category theory to help keep things organized. K-theory and K-homology have become indispensable tools in operator theory; there is even a bivariant functor $KK(-,-) $ from the category of C*-algebras to the category of abelian groups relating the two constructions, and many deep theorems can be subsumed in the assertion that there is a category whose objects are C*-algebras and whose morphism spaces are given by $KK(A,B)$. Cyclic homology and cohomology has also become extremely relevant to the interface between analysis and topology.
So ultimately I think it all comes down to what kinds of subtleties are most relevant in a given problem. There is just something fundamentally different about the kind of thinking required to estimate the propagation speed of the solution operator for a nonlinear PDE compared to the kind of thinking required to relate the fixed point theory in characteristic 0 of a linear group acting on a variety to that in characteristic p.
| 34 | https://mathoverflow.net/users/4362 | 38777 | 24,866 |
https://mathoverflow.net/questions/38780 | 45 | An integral domain $R$ is said to be Euclidean if it admits some Euclidean norm: i.e., a function $N: R \rightarrow \mathbb{N} = \mathbb{Z}^{\geq 0}$ such that: for all $x, y \in R$ with $N(y) > 0$, either $y$ divides $x$ or there exists $q \in R$ such that $N(x-qy) < N(y)$. A well-known "descent" argument shows that any Euclidean domain is a PID. In fact, the argument that a Euclidean domain is necessarily a UFD is a little more direct and elementary than the argument that shows that a PID is a UFD (because, in the latter case, one needs some kind of ideal-theoretic argument to show the existence of factorizations into irreducible elements). Because of this, Euclidean domains are a familiar staple of undergraduate algebra.
A lot of texts seem to emphasize the fact that a PID need not be a Euclidean domain. In order to show this, one has to show not only that some particular norm (and often there is a preferred norm in sight, see below) is not Euclidean, but that there is no Euclidean norm whatsoever. In general this is a very delicate question: for instance, the proof of the most standard example -- that the ring of integers of $\mathbb{Q}(\sqrt{-19})$ is a PID but does not admit *any* Euclidean norm -- is already rather intricate.
My question is this:
>
> Given a ring $R$ that we already know is a PID, why do we care whether or not it admits some Euclidean norm?
>
>
>
Note that in contrast, many domains admit natural norms. A class of domains which I have been thinking about recently are the infinite domains satisfying (FN): the quotient by every nonzero ideal is finite. In this case, the map $0 \mapsto 0$, $x \in R \setminus \{0\} \mapsto \# R/(x)$ is a multiplicative norm, which I call **canonical**. For instance, the usual absolute value on $\mathbb{Z}$ is the canonical norm, as is the norm on any ring of integers in a number field that you meet in an algebraic number theory course.
I have recently realized that I care quite a bit about whether certain specific norms on integral domains are Euclidean. (This has come up in my work on quadratic forms and the Davenport-Cassels theorem.) There is some very natural algebra and discrete geometry here. But why do I care if some crazy Euclidean norm exists?
Here are three reasons that one might care about this:
1. If a domain admits an "effective" Euclidean norm, one can give effective algorithms for linear algebra over that ring, whereas the structure theory of modules over an arbitrary PID is not *a priori* algorithmic in nature.
2. (in algebraic K-theory): If $R$ is Euclidean, $\operatorname{SK}\_1(R) = 0$, but there exists a PID with nonvanishing $\operatorname{SK}\_1$. (Thanks to Charles Rezk for giving the precise result based on my vague allusion to it.)
3. In algebraic number theory, there has been a lot of work towards proving the conjecture that if $K$ is a number field which is *not* $\mathbb{Q}(\sqrt{D})$ for $D = -19, -43, -67, -163$, then the ring $\mathbb{Z}\_K$ of integers of $K$ is a PID iff it is Euclidean (for some crazy norm). In particular, disproving this would disprove the generalized Riemann hypothesis.
Comments on 1: There is something to this, but I somehow doubt that it's such a big deal. For instance, the ring of integers of $\mathbb{Q}(\sqrt{-19})$ is not Euclidean, but I'm pretty sure that there are algorithms for modules over it. In particular, it seems to me that for algorithmic purposes, having a Dedekind-Hasse norm is just as good as a Euclidean norm, and every PID has a Dedekind-Hasse norm. In fact, for every PID which satisfies (FN), the canonical norm is a Dedekind-Hasse norm. (See p. 27 of [http://alpha.math.uga.edu/~pete/factorization2010.pdf](http://alpha.math.uga.edu/%7Epete/factorization2010.pdf) for this.)
Comments on 3: if I knew more about this result, I might appreciate it better. It does seem to involve some interesting geometry of numbers. But this convinces me why I should be interested in the special case of rings of integers in number fields, which, as a number theorist, I am already convinced are more worthy of scrutiny from every possible angle than an arbitrary domain.
If there are other good reasons to care, I'd certainly like to know.
| https://mathoverflow.net/users/1149 | Why do we care whether a PID admits some crazy Euclidean norm? | There are a lot of results in elementary number theory that can be proved with the quadratic reciprocity law. In such a proof you usually have to invert some Jacobi symbol $(a/b)$ and then reduce the numerator modulo the denominator. For number fields that are not Euclidean with respect to some simple map you have a problem if you want to follow this route (the same goes for applications of quadratic and higher residues to cryptography, although this is mostly a theoretical business). In principle, Dedekind-Hasse will also do the trick in some cases.
If the ring of integers you're interested in is not Euclidean for the canonical norm, the first idea is to modify it. You could give prime ideals a different weight (weighted norms), or allow division chains in which the norm does not necessarily get smaller in every step (k-stage Euclidean rings), or try some version of Dedekind-Hasse. But if (given a pair $(a,b)$ of elements in a ring) you want to make the norm of $ka-bq$ small, you need more than just the knowledge that a suitable $k$ exists: you need a method for finding $k$ (in addition to finding $q$), perhaps by showing that you can select it from a finite set of elements with bounded norm or something similar.
**Edit.** The Euclidean algorithm is closely related to continued fractions, and the latter are routinely used for doing calculations of units and ideal class groups of real quadratic number fields. For number fields that admit a Euclidean algorithm, something similar can be done: Hurwitz and Mathews worked out a theory of continued fractions over the Gaussian integers, and people like Arwin, Trinks, Degel, Lakein, Stein etc. generalized this to complex Euclidean number fields and used it for computing units and class numbers. I am not aware of too many recent contributions in this direction, but a short search has at least revealed D. Fried, Reduction theory over quadratic imaginary fields, T. Number Theory 2005.
| 28 | https://mathoverflow.net/users/3503 | 38789 | 24,871 |
https://mathoverflow.net/questions/38751 | 19 | I'm seeking a function which is Hölder continuous but does not belong to any Sobolev space.
**Question:** More precisely, I'm searching for a function $u$ which is in $C^{0,\gamma}(\Omega)$ for $\gamma \in (0,1)$ and $\Omega$ a bounded set such that $u \notin W\_{loc}^{1,p}(\Omega)$ for any $1 \leq p \leq \infty$. Take $\Omega$ to be bounded, open.
My first guess is to do a construction with a Weierstrass function. I know this is differentiable 'nowhere' but that doesn't convince me it isn't weakly differentiable in some bizarre way. Hopefully someone knows of an explicit example.
| https://mathoverflow.net/users/8755 | A Hölder continuous function which does not belong to any Sobolev space | Your guess is indeed right. Following a similar idea gives you the Takagi or [blancmange function](http://en.wikipedia.org/wiki/Blancmange_curve).
It is even quasi-Lipschitz (it has a modulus of continuity $\omega(t)=ct(|\log(t)|+1)$ for a suitable constant $c>0$), thus it's Hoelder of any positive exponent less than 1. It is not even BV in any open interval, thus $W^{1,p} \_ {loc}$ for no $p\geq1$.
*Rmk 1.* The above example is for dimension 1: but of course it holds in any dimension a fortiori.
*Rmk 2.* To get an example with a more classical flavor, actually a Weierstrass function, replace $s(x)$ with $\cos(x)$. I'd say that the resulting Fourier series defines a function with the same features, by the same reasons (the function $\cos(x)$ works better than $\sin(x),$ in view of point 2 below.)
*Rmk 3.* Once you know that the Weierstrass function $f(x):=\sum\_{k=0}^\infty 2^{-k}\cos(2^k x)$ is nowhere differentiable, you also have that it is BV on no open interval, for BV on an interval would imply differentiability a.e. there. However, for your needs it seems more direct just showing it has infinite variation on any interval.
*Details.*
1. To prove that the Takagi function $f(x)$ admits the above modulus of continuity, recall that that $f$ is characterized as the fixed point of the affine contraction $T:C\_b(\mathbb{R})\to C\_b(\mathbb{R})$ such that $(Tf)(x)=\frac{1}{2}f(2x)+s(x),$ for all $x\in\mathbb{R}$, where $s(x)$ is the distance function from the integers (a zig-zag piecewise 1-periodic function). Just find a $c$ such that the subset of $C\_b(\mathbb{R})$ of functions that admit $\omega$ as modulus of continuity is a $T$-invariant set. The latter subset is obviously closed and non-empty, so the fixed point is there. (The above illustrated a standard general technique to prove properties of objects found by means of the contraction principle).
2. Proving that $f$ is not of bounded variation on $[0,1]$ (hence in no open interval, due to the self-similarity encoded in the fixed point equation), requires a small computation on the partial sum $f\_n$ of the series defining $f$. Let
$$f\_n(x):=\sum\_{k=0}^{n-1}\, 2^{-k} s(2^k x).$$
First note that the derivative of $f\_n$ only takes integer values, which of course come as a result of the sum of $n$ terms $\pm 1$ (with all the $2^n$ possible signs). In particular, for any $n\in\mathbb{N}$ the function $f\_{2n}$ has ${2n \choose n} $ flat intervals of lenght $2^{-2n}$ within the unit interval $I$, and has derivative larger than $2$ in absolute value elsewhere in $I$. Thus, for the subsequent odd index $2n+1,$ the function $f\_{2n+1}$ has
${2n \choose n}$ local maxima in $I$ (located in the mid-points of the above intervals). Moreover, passing to $f\_{2n+1}$ each maximum point contributes to the increment of the total variation with $2^{-2n}$, while the total variation remains unchanged passing from $2n+1$ to the next even index $2n+2$. The conclusion is that, for any $n$, the total variation of $f\_n$ on $I$ is
$$V(f\_n;I)=\sum\_{0\leq k < n/2}{2k\choose k}2^{-2k} =O\big(\sqrt{n}\big),$$
since by the classical asymptotics for the central binomial coefficient, ${2k \choose k}=\frac{4^k}{\sqrt{\pi k}}(1+o(1)),\, k\to\infty.$ So actually $V(f\_n;I)$ diverges. Yet this would not be sufficient to conclude that $V(f,I)=\infty,$ as the total variation is only lower semicontinuous with respect to the uniform convergence. However, the discrete variation on a given subdivision $P:=\{t\_0 < \dots < t\_r \}$
$$V(f\_n; P\, )=\sum\_{i=0}^{r-1}\, \big|f\_n(t\_{i+1})-f(t\_i)\big|$$
does of course pass to the limit under even pointwise convergence. Now the point is that, for the binary subdivision $P\_m:=\{ k2^{-m} \, : \, 0 \le k \le 2^m \},$ we have $V(f\_n;I)=V(f\_n;P\_m)$ as soon as $n \geq m$. So for all $m$ letting $n\to\infty$
$$V(f;P\_m)=\lim\_{n\to\infty }V(f\_n;P\_m)=V(f\_m;P\_m)$$ and
$$V(f;I)=\sup\_{m\in\mathbb{N}}V(f;P\_m)=\infty,$$
as we wished to show.
| 27 | https://mathoverflow.net/users/6101 | 38791 | 24,872 |
https://mathoverflow.net/questions/38767 | 8 | I heard that De Giorgi-Nash-Moser type regularity arguments fail for elliptic systems, but do not know where to start looking for more substantial information. Why does the regularity fail? Is there some cases where the Moser iteration can be successfully applied to elliptic systems?
| https://mathoverflow.net/users/824 | Moser iteration for elliptic systems | Hi.
The point is not the ellipticity. In fact the Argument of De Giorgi, Moser and Nash was designed for elliptic problems. The point is that solutions $u: \Omega\to\mathbb{R}^N$ of elliptic problems with $N>1$ just aren't $C^{1,\alpha}$ any more in general. This is no problem with the method, it's intrinsic. The famous counterexample is by De Giorgi himself.
See Giusti - The direct method of variational calculus for more details to this topic. The counterexample itself can be found as example 6.3 in this book. The paper from De Giorgi is "Un esempio di estremal discontinue per un problema variazionale di tipo ellittico", Boll. U.M.I., 4 (1968), 135-137
| 12 | https://mathoverflow.net/users/3041 | 38797 | 24,873 |
https://mathoverflow.net/questions/38657 | 7 | Let W be a finite word on a two symbol alphabet {0,1}; let us say that W is maximal if it is the last item in the list of all its cyclic permutation (ordered lexicographically).
So, for instance:
{0,1} are the maximal words of length 1;
{00, 10, 11} are the maximal words of length 2;
{000, 100, 110, 111} are the maximal words of length 3;
{0000, 1000, 1010, 1100, 1110, 1111} are the maximal words of length 4;
{00000, 10000, 10100, 11000, 11010, 11100, 11110, 11111} are the maximal words of length 5; ...
und so weiter.
Let
k(n):= number of maximal words of length n
Is there some formula for k(n)?
| https://mathoverflow.net/users/7979 | Minimal words of length n | For aperiodic (sometimes also called, full period) strings, the term you are looking for is Lyndon words. These are the (unique lexicographically-least) representative of a full-period necklace (as stated in the comments, a necklace is the equivalence class under cyclic rotation). The number $k(n)$ you ask for is exactly the number of necklaces, and again, as stated in the comments, it is given by
$k(n)=\frac{1}{n}\sum\_{d|n}\phi(d)2^{n/d}$. You can check out a proof for this in S.W.Golomb's book "Shift Register Sequences" (in the 1967 edition, start looking at around page 171 and look for the cycles of $PCR\_n$).
| 6 | https://mathoverflow.net/users/9044 | 38798 | 24,874 |
https://mathoverflow.net/questions/38795 | 1 | Define the Borel sigma-algebra on $\mathbb{R}^n$ as the smallest sigma-algebra containing all $n$-rectangles
$(a\_1, b\_1) \times \cdots \times (a\_n, b\_n)$.
Is it true that the Borel sigma algebra contains all sets of the form $A\_1 \times \cdots \times A\_n$, where each $A\_i$ is some Borel set in $\mathbb{R}$?
I thought this would be trivially true, but I had a lot of trouble trying to prove it, and I'm not even sure its true anymore.
If this is a well-known result, could you please refer me to a text where it has been (dis)proved ?
| https://mathoverflow.net/users/8528 | Borel Sets on $\mathbb{R}^n$ | A way to prove it:
1/ any set of the form $A\_1 \times \mathbb R \ldots \times \mathbb R$, where $A\_1$ is Borel, or more generally a "Borel rectangle" with only one slice not equal to the whole space, is in the Borel sigma-algebra (this is essentially a 1-dimensional Borel set, and those are generated by open intervals).
2/ any product $A\_1 \times \ldots \times A\_n$ (with each $A\_i$ Borel) is a finite intersection of sets of the above form.
Not sure I should have answered this, it may be a homework problem... I'd have just written a comment but I'm not reputable enough to do so :)
Any standard reference on measure theory will provide a proof of the result you're asking about (say, Dudley's book).
| 8 | https://mathoverflow.net/users/8923 | 38799 | 24,875 |
https://mathoverflow.net/questions/38792 | 5 | I suppose this question is probably elementary for experts, but I'd like to present my arguments, about which I have some doubts, and see if they are correct, or if corrections and improvements are possible.
The setting is as follows: $k$ is the base field of characteristic zero, $G$ a connected semisimple $k$-group, and $Rep(G)$ the Tannakian category of finite-dimensional algebraic $k$-representations of $G$, with the canonical fiber functor in $k$-vector spaces, whose objects are called $k$-representations for short. Unless otherwise stated, reductive $k$-groups are connected.
The motivation is as follows. For $H$ semisimple $k$-subgroup of $G$, one has the restriction functor, $Rep(G)\rightarrow Rep(H)$, sending a $k$-representation $V$ of $G$ to the restriction $V$ as a $k$-representation of $H$. What kind of irreducible $k$-representation of $G$ remains irreducible viewed in $Rep(H)$?
Recall that a reductive $k$-group is *$k$-isotropic* if it contains a $k$-split $k$-torus, and *$k$-anisotropic* otherwise.
*Fact* For $H\subset G$ a semisimple $k$-subgroup, $H$ extends to a parabolic $k$-subgroup $H\subset P\subsetneq G$ if and only if $Z(H,G)$ the centralizer of $H$ in $G$ is $k$-isotropic.
(from this one also sees that if $L$ is the Levi $k$-subgroup of a $k$-parabolic $P$, then its connected center $C(L)$ is $k$-isotropic.)
*Claim* Let $H$ be a semisimple $k$-subgroup of $G$ as above, such that $Z(H,G)$ is $k$-anisotropic, then for any irreducible $k$-representation $(\rho,V)$ of $G$, its restriction to $H$ is irreducible as an algebraic $k$-representation of $H$. Conversely, if the restriction functor $Rep(G)\rightarrow Rep(H)$ respects irreducibility, with $H\subset G$ a semisimple $k$-subgroup, then $Z(H,G)$ is $k$-anisotropic.
*Sketch of proof* To prove the first part, assume that for some irreducible $(\rho,V)$, the restriction to $H$ is not irreducible. Then in $Rep(H)$ one has a non-trivial splitting $V=V\_1\oplus V\_2$. Define $F\_0(V)=V$, $F\_1=V\_1$, $F\_2=0$ etc, one gets a non-trivial decreasing filtration on $V$. $V$ generates a full Tannakian subcategory, which is of the form $Rep(G')$, equipped with the non-trivial filtration generated by $F(V)$. By Tannaka duality, $Rep(G')\rightarrow Rep(G)$ corresponds to an epimorphism $G\rightarrow G'$. $G'$ is thus semisimple. The non-trivial filtration on $Rep(G')$ corresponds to a cocharacter defined over $k$, which is equivalently characterized by $k$-parabolic $P'$ of $G'$, and $P'$ lifts to a $k$-parabolic $P$ of $G$. One checks easily that $P$ contains $H$, because $H$ preserves the filtration generated by $F(V)$. This shows that $Z(H,G)$ is $k$-isotropic.
Conversely, when $Z(H,G)$ is $k$-isotropic, $H$ extends to a non-trivial $k$-parabolic $H\subset P\subsetneq G$. This gives a filtration on $Rep(G)$, preserved by $P$ and $H$. In particular, there exists at least one irreducible $k$-representation $(\rho,V)$ of $G$ on which $F(V)$ is non-trivial, and then the restriction of $\rho$ to $H$ splits non-trivially.
Here I use the notion of filtration on $Rep(G)$, which means for each $V\in Rep(G)$ one has a finite separated exhaustive decreasing filtration $F(V)$, moving functorially: it respects the tensor products and direct sums in the filtered sense, and is strict with respect to all exact sequences in $Rep(G)$. To see a filtration on $Rep(G')$ extends to a filtration on $Rep(G)$ for an epimorphism $G\rightarrow G'$ as above, it suffices to transfer to the Lie algebra side: $LieG=LieG'\oplus Lie G''$ for some semi-simple $k$-subgroup $G''$ of $G$, then use the fact that $Rep(LieG)$ equals the "exterior tensor product" of $Rep(LieG')$ with $Rep(LieG'')$, and pass equivalently to the $k$-group side, as $k$ is of characteristic zero. In this way the filtration on $Rep(G')$, together with the trivial filtration on $Rep(G'')$, gives a filtration on $Rep(G)$ by tensorial construction.
I would like to know if these arguments make sense. If so, is there any other elementary proof, essentially different (modulo the Tannakian duality). Moreover, what if one allows reductive $k$-subgroups? Does that imply the claim that over $\mathbb{R}$, if one takes a pair of compact groups, say $SO\_3\subset SO\_4$, every irreducible representation of $SO\_4$ remains irreducible when restricted to $SO\_3$? and does it have anything to do with the branching rule? I would be grateful if further references, like expository articles, are mentioned concerning branching rules for reductive $k$-groups, even in the case of non-algebraically base fields (I guess one might do something from the algebraically closed case through Galois descent, but I'm quite lost when doing this for reductive $k$-groups.)
Thanks a lot.
| https://mathoverflow.net/users/9246 | When does an irreducible representation remain irreducible after restriction to a semi-simple subgroup? | One method for computing branching rules in favorable situations is to use the Littelmann path model---this has a wiki page
<http://en.wikipedia.org/wiki/Littelmann_path_model>
In this situation (of semisimple $G$ and with $H$ the Levi subgroup of a parabolic) irreducibles essentially never remain irreducible.
Edit in response to the comment below:
Another situation that sometimes occurs (e.g. $SO\_n \subseteq SL\_n$) is that $H$ is the fixed point set of an involution $\phi$ of $G$. In this case, given any irreducible $G$ module $V$ we can construct another $G$-module $V^\phi$ by twisting by $\phi$. If $V$ and $V^\phi$ are isomorphic as $G$-modules, then we obtain an involution of $V$ as an $H$-module whose $\pm 1$ eigenspaces are $H$-submodules. This allows one to prove that certain $G$-modules are not irreducible as $H$-modules.
| 4 | https://mathoverflow.net/users/8552 | 38801 | 24,876 |
https://mathoverflow.net/questions/38054 | 9 | I'm seeking a simple example of where elliptic (preferably linear) boundary regularity fails due to a simple kink in the domain.
So far my gueses were to look at $-\Delta u = f$ on $[0,2\pi] \times [0,2\pi]$ with $0$ Dirichlet boundary conditions and choose an $f$ which was far from $0$. This hasn't seem to produce any results (I was checking regularity directly by the method of Fourier series).
So more precisely, I would like an example where
1) $Lu = f$ in $\Omega \subset \mathbb{R}^n$ with $f$ smooth
2) $L$ is elliptic and $u = 0$ on $\partial \Omega$
3) $\Omega$ is *not smooth* and consequently $u$ is not smooth up to the boundary.
| https://mathoverflow.net/users/8755 | A simple example where elliptic boundary regularity fails due to a kink in the domain | This is the same idea as timur's answer but with more details and less generality. A frequent test problem in numerical analysis is the Poisson equation $-\Delta u = 1$ on the L-shaped domain
$\Omega = ([-1,1] \times [-1,1]) \setminus ([-1,0] \times [-1,0])$
with homogeneous Dirichlet boundary conditions: $u = 0$ on $\partial\Omega$. The solution has a singularity at the origin: it is continuous but not differentiable. More precisely, close to the origin we have
$u(r,\theta) \approx r^{2/3} \sin \frac{2\theta+\pi}{3}$
in polar coordinates, according to equation (1.6) in <http://eprints.ma.man.ac.uk/894/02/covered/MIMS_ep2007_156_Sample_Chapter.pdf> (sample chapter from Elman, Silvester and Wathen, Finite Elements and Fast Iterative Solvers, Oxford University Press, 2005).
**Added:** I don't know the details and I don't have time to do the necessary computations, but I think that you can solve the PDE by converting the Laplacian to polar coordinates and applying separation of variables. I imagine that you get that
$u(r,\theta) = r^{2n/3} \sin \frac{2n}{3} (\theta + \frac{1}{2}\pi)$
with $n$ a positive integer satisfies the boundary conditions at $r=0$ and $\theta=-\pi/2$ and $\theta=\pi$ (*as Dorian comments below, these are all harmonic functions, so there must be something else*). Then take a linear combination of those to match the conditions on the rest of the boundary of the L-shaped domain. Close to the origin, the $n=1$ term dominates. Perhaps somebody else can confirm / amend?
| 7 | https://mathoverflow.net/users/2610 | 38803 | 24,878 |
https://mathoverflow.net/questions/38800 | 0 | Hello,
I am wondering whether it is possible to convert the following integral equation to a partial differential equation.
[Integral Equation here http://ima.epfl.ch/~lechen/images/integralEq.jpg](http://ima.epfl.ch/~lechen/images/integralEq.jpg)
where $J\_0(t,x)$ is some given nonnegative function and $\nu>0$ is a constant. It is clear $t\ge 0$.
The aim is to solve this equation. To convert it to PDE is just one possible way to solve it, since latter we can use the hopefully the fundamental solutions.
My current solution is
[PDE http://ima.epfl.ch/~lechen/images/PDE.jpg](http://ima.epfl.ch/~lechen/images/PDE.jpg)
But I am not sure whether it is right or not.
Thanks for any comments or hints!
| https://mathoverflow.net/users/36814 | From an integral equation to a differential equation | Of course. When $\nu=1$, if you apply the operator $\partial\_t-\partial^2\_{xx}$ to the last integral you obtain precisely $f(t,x)$ so the equation is
$$ f\_t - f\_{xx} = (\partial\_t-\partial^2\_{xx}) J\_0^2 + f.$$
EDIT: you seem to know already the answer, so I stop here :) You edited your question when I was writing my answer...
By the way, if you want to solve the PDE just set $ f(t,x) = e^{t} g(t,x) $ and the equation in $g$ is a homogeneous heat equation. This sounds like some textbook exercise, I musr say
| 3 | https://mathoverflow.net/users/7294 | 38809 | 24,882 |
https://mathoverflow.net/questions/38811 | 2 | Hello,
For almost one year, I am searching for the Green's function for wave equation in R² or R³ with some boundary conditions. As far as I know, when the boundaries permit the method of images, we can get the Green's function. But this requirement is too strong.
What we would like to have is that: the wave is confined in a convex, sufficiently smooth domain. On the boundary, either Dirichlet or Neumann's conditions can be put. To impose these conditions is just to avoid the diffraction problem, which can be too much complicated for us.
During this searching, I encountered books by Prof. Melrose, Prof. Michael E. Taylor and also the formidable three volumns by Prof. Hormander. I still feel hopeless in finding that.
Thanks in advance for any comments!
Best!
| https://mathoverflow.net/users/36814 | Green's function for wave equations in R² or R³ | What do you mean by *getting the Green's function* ? If you mean in closed form, then this is hopeless for most domains.
Otherwise, the proper way to express the solution of
$$u\_{tt}=\Delta u,\qquad u(0)=u\_0,\qquad u\_t(0)=u\_1$$
with homogeneous boundary conditions BC (say Dirichlet or Neumann)
is to use the Laplace transform $\hat u$ of $u$ as an auxiliary function:
$$\hat u(z):=\int\_0^{+\infty}\exp(-sz)u(s)ds.$$
For each $z$ of positive real part, $\hat u(z)$ solves the elliptic problem
$$(-\Delta+z^2)w=u\_1+zu\_0.$$
The above problem, with BC, is well-posed, for every $z$ away from the imaginary axis, and the map $z\mapsto w$ is holomorphic. One recovers $u$ through a Cauchy integral along an appropriate contour in the complex plane. This amounts to express the Green's function of the wave equation as a Cauchy integral in terms of the Green's functions of the elliptic problems parametrized by $z$. This expression may be used to analyze the singularities of the Green's function.
| 4 | https://mathoverflow.net/users/8799 | 38817 | 24,887 |
https://mathoverflow.net/questions/38819 | 2 | Let $I$ be an ideal of a commutative ring $R$. $M$ be an $R$-module. In *Local cohomology: an algebraic introduction with geometric applications* of Brodmann M. P., Sharp R. Y we have
$$D\_I(M)=\mathop {\lim }\limits\_{\begin{subarray}{c}
\longrightarrow \\
\end{subarray}} Hom\_R(I^{n},M) $$
called ideal transform of $M$ respect to $I$ or $I-transform$ of $M$.
I heard someone talked about Deligne's formula but i can not find it.
Can anyone help me to find it?
I think that, it is
$$D\_I(M)=\mathop {\lim }\limits\_{\begin{subarray}{c}
\longrightarrow \\
\end{subarray}} M\_a $$
where $M\_a$ is the localization of $M$ with respect to multiplicative systems of powers of a single $M$-regular element $a$ in $I$.
| https://mathoverflow.net/users/9141 | ideal transform | I've seen the following called Deligne's formula (it is in Hartshorne, Algebraic Geometry, Chapter III, Exercise 3.7), and I think essentially answers your question.
It says that if $Z = V(I)$ (the closed subset of $X = \text{Spec} R$) and $U = X \setminus Z$, then
$$
\Gamma(U, \widetilde{M}) = \lim\_{\to} \text{ Hom}\_R(I^n, M)
$$
| 3 | https://mathoverflow.net/users/3521 | 38826 | 24,890 |
https://mathoverflow.net/questions/38829 | 2 | I learnt from a paper that "Let cov(K) be the least cardinal k such that a perfect Polish space can be expressed as a union of k meager sets. (It does not matter which perfect Polish space is used to define cov(K) ) " . I don't know why cov(K)'s are equal for different perfect Polish spaces. I want to use this result in my own paper, but I fail to find any books contain this result, could anybody kindly to help me ?
| https://mathoverflow.net/users/9252 | Covering number of the meager ideal | This should be in standard texts on descriptive set theory, like Moschovakis's "Descriptive Set Theory" or Kechris's "Classical Descriptive Set Theory". The basic idea is that any two perfect Polish spaces become homeomorphic after you remove (at most) countably many points. Countably many points constitute a meager set and therefore don't affect cov(K).
| 4 | https://mathoverflow.net/users/6794 | 38831 | 24,894 |
https://mathoverflow.net/questions/38771 | 39 | This question was motivated by the comments to [Dual of Zorn's Lemma?](https://mathoverflow.net/questions/38754/dual-of-zorns-lemma)
Let's denote by the Dual Schroeder-Bernstein theorem (DSB) the statement
>
> For any sets $A$ and $B$, if there are surjections from $A$ onto $B$ and from $B$ onto $A$, then there is a bijection between them.
>
>
>
In set theory without choice, assume that the Dual Schroeder-Bernstein theorem holds. Does it follow that choice must hold as well?
I strongly suspect this is open, though I would be glad to be proven wrong in this regard. In all models of ZF without choice that I have examined, DSB fails. This really does not say much, as there are plenty of models I have not looked at. In any case, I don't see how to even formlate an approach to show the consistency of DSB without AC.
The only reference I know for this is Bernhard Banaschewski, Gregory H. Moore, *The dual Cantor-Bernstein theorem and the partition principle*, Notre Dame J. Formal Logic **31 (3)**, (1990), 375–381. In this paper it is shown that a strengthening of DSB does imply AC, namely, that whenever there are surjections $f:A\to B$ and $g:B\to A$, then there is a bijection $h:A\to B$ contained in $f\cup g^{-1}$. (Note that the usual Schroeder-Bernstein theorem holds -without needing choice- in this fashion.)
The *partition principle* is the statement that whenever there is a surjection from $A$ onto $B$, then there is an injection from $B$ into $A$. As far as I know, it is open whether this implies choice, or whether DSB implies the partition principle. Clearly, the reverse implications hold.
If you are interested in natural examples of failures of DSB in some of the usual models, Benjamin Miller wrote a nice note on this, available at his [page](http://wwwmath.uni-muenster.de:8013/persdb/show_perspage.php?id=674).
---
**Added Sep. 21. [Edited Aug. 14, 2012]** It may be worthwhile to point out what is known, beyond the Banaschewski-Moore result mentioned above.
Assume DSB, and suppose $x$ is equipotent with $x\sqcup x$. Then, if there is a surjection from $x$ onto a set $y$, we also have an injection from $y$ into $x$. (So we have a weak version of the partition principle.) This *idemmultiple hypothesis* that $x\sqcup x$ is equipotent to $x$, for all infinite sets $x$, is strictly weaker than choice, as shown in Gershon Sageev, *An independence result concerning the axiom of choice*, Ann. Math. Logic 8 (1975), 1–184, MR0366668 (51 #2915).
Also, as indicated in Arturo Magidin's answer (and the links in the comments), H. Rubin proved that DSB implies that any infinite set contains a countable subset.
| https://mathoverflow.net/users/6085 | Dual Schroeder-Bernstein theorem | This is only a partial answer because I'm having trouble reconstructing something I *think* I figured out seven years ago...
It would seem the Dual Cantor-Bernstein implies Countable Choice. In a [post in sci.math](http://groups.google.com/group/sci.math/msg/28543d2b17d8f4ab) in March 2003 discussing the dual of Cantor-Bernstein, Herman Rubin essentially points out that if the dual of Cantor-Bernstein holds, then every infinite set has a denumerable subset; this is equivalent, I believe, to Countable Choice.
Let $U$ be an infinite set. Let $A$ be the set of all $n$-tuples of elements of $U$ with $n\gt 0$ and even, and let $B$ be the set of all $n$-tuples of $U$ with $n$ odd. There are surjections from $A$ onto $B$ (delete the first element of the tuple) and from $B$ onto $A$ (for the $1$-tuples, map to a fixed element of $A$; for the rest, delete the first element of the tuple). If we assume the dual of Cantor-Bernstein holds, then there exists a one-to-one function from $f\colon B\to A$ (in fact, a bijection). Rubin writes that "a 1-1 mapping from $B$ to $A$ quickly gives a countable subset of $U$", but right now I'm not quite seeing it...
| 7 | https://mathoverflow.net/users/3959 | 38833 | 24,896 |
https://mathoverflow.net/questions/38522 | 3 | This question is geared towards the experts, so I will only briefly gloss the definitions. Everything I say is in the category of finite-dimensional smooth manifolds, and whenever I say "$\mathbb Z$-graded" I'm implicitly using the Koszul or "super" rule for signs.
Recall that a vector bundle $A\to X$ determines a $\mathbb Z$-graded commutative algebra whose degree-$k$ part is $\Gamma(\wedge^k A^\*)$. A **Lie algebroid** is a vector bundle $A\to X$ along with a degree-one square-zero derivation (Koszul rule!) on $\Gamma(\wedge^k A^\*)$, i.e. it makes $\Gamma(\wedge^k A^\*)$ into a complex and in fact a differential graded algebra. **Lie algebroid cohomology** is the *homology* of this complex (closed sections mod exact sections). (The "co", as far as I can tell, comes from the typical example: for any $X$, the tangent bundle ${\rm T}X \to X$ is a Lie algebroid via the de Rham $d$, and the homology of the complex $\Gamma(\wedge^k {\rm T}^\* X)$ is de Rham *cohomology* of $X$. Or maybe the etymology is different. The functor from Lie algebroids to their cohomology is contravarian, so maybe that's it.) A morphism of Lie algebroids is a morphism of vector bundles that induces a morphism of the corresponding complexes.
A **VBLA** is a "vector bundle in the category of Lie algebroids". I.e. it is a pair of Lie algebroids $D \to B$ and $A\to X$, a morphism of algebroids $\{D\to B\} \to \{A\to X\}$, and maps $0: \{A\to X\} \to \{D\to B\}$ and $+ : \{D\to B\} \times\_{\{A\to X\}} \{D\to B\} \to \{D\to B\}$ satisfying the conditions you would think of if you write down the words "vector bundle" in categorical language. (Equivalently, the induced maps on manifolds $B \to X$ and $D\to A$ are vector bundles, and there's some compatibility conditions.) For details, peruse the papers by Mehta and Garcia-Saz, or by Mackenzie.
In particular, a VBLA is a morphism of algebroids, and so induces a map on cohomology. My question is:
>
> What "geometric" conditions on the VBLA are equivalent to the map on cohomology being an isomorphism? For example, does every VBLA induce an isomorphism on cohomology? If not, is there a way to check whether the cohomologies are isomorphic without computing the full cohomologies of both?
>
>
>
I'm realizing that I don't know enough examples (that I can actually compute) to gather up intuition. And the answer must be known by the experts (I'm guessing that the answer is known much more generally than just for VBLAs?). As with any mathoverflow question, I'm perfectly happy with an answer consisting entirely of a good reference.
| https://mathoverflow.net/users/78 | When does a VBLA induce an isomorphism on Lie algebroid cohomology? | The short answer is that a VBLA always induces an injection on cohomology, but in general it isn't an isomorphism.
The less short answer is that the complex for $D \to B$ decomposes as a direct sum of a bunch of subcomplexes, the first of which is the complex for $A \to M$. Thus, you have an isomorphism if and only if all of the higher subcomplexes are acyclic.
I'll give more detail. First of all, it helps to take the point of view of graded geometry. Here, a vector bundle $A \to X$ is associated to a graded manifold $A[1]$, whose algebra of "functions" is $\Gamma(\wedge^k A^\*)$. Then the differential (let's call it $d\_A$) can be thought of as a degree 1 vector field satisfying the equation $[d\_A,d\_A] = 0$. From this point of view, a VBLA corresponds to a vector bundle in the category of graded manifolds $D[1] \to A[1]$, where the differential $d\_D$ is a *linear* vector field over the base vector field $d\_A$ (so as a derivation, it preserves linear functions). This means you can impose a grading on the "functions" on $D[1]$ (different from the homological grading), which is preserved by $d\_D$. The decomposition via this grading is the one that I alluded to above.
You can go even further by using the product structure. As an algebra, the "functions" on $D[1]$ is, up to some closure, generated by the functions coming from $A[1]$ and the linear functions. I think you can use this fact to conclude that the cohomology of $D$ is isomorphic to the cohomology of $A$ if and only if the subcomplex of linear functions is acyclic. In our paper on VB-algebroids, Gracia-Saz and I studied the linear subcomplex in terms of Lie algebroid superrepresentations, and we gave a classification of those satisfying a regularity assumption. So take a look at that---we didn't directly address the question, but it should be fairly easy to use the classification data to describe which ones are acyclic.
| 5 | https://mathoverflow.net/users/9251 | 38837 | 24,899 |
https://mathoverflow.net/questions/38782 | 3 | After thinking on Joel's answer at [Computable nonstandard models for weak systems of arithemtic](https://mathoverflow.net/questions/38160/computable-nonstandard-models-for-weak-systems-of-arithemtic) for a few days, I do not see how to develop enough tuple machinery in I-Sigma\_0 (PA with induction restricted to Sigma\_0 formulas) to prove the necessary result.
Does I-Sigma\_0 prove that "for any number d, there is a number c coding the set of Turing machine programs less than d that halt on input 0 with output 0 in at most d steps", with a coding such that determining whether a standard number n is in the set is computable from (=,0,S,+,\*)?
With the coding Joel suggested, which is code(set) := $\displaystyle\Pi\_{n \in set} \; p\_n$, I don't even see a way for I-Sigma\_0 to prove "For all n, there is a number coding the set of numbers less than n", and I can't think of a better coding either.
| https://mathoverflow.net/users/nan | Tuple machinery in I-Sigma_0 | As I recall, McAloon's method for proving that there are no computable nonstandard models of $I\Delta\_0$ was to show that there are initial segments that are nonstandard models of PA. The usual Tennenbaum tricks can then be used to show that addition and multiplication are not computable.
Additional Comment--
Here are references for McAloon's paper and the paper of Wilmers that proves a similar result for
$IE\_1$ the fragment of $I\Delta\_0$ where you only have induction for formulas with bounded existential quantifiers
McAloon, Kenneth, On the complexity of models of arithmetic.
J. Symbolic Logic 47 (1982), no. 2, 403--415.
Wilmers, George Bounded existential induction. J. Symbolic Logic 50 (1985), no. 1, 72--90.
| 2 | https://mathoverflow.net/users/5849 | 38858 | 24,911 |
https://mathoverflow.net/questions/38846 | 3 | Let $X$ be a topological space, and $Homeo(X)$ the group of self-homeomorphisms of $X$.
(1) What is the exact meaning of: $H^\*(X)$ is a an $A\_\infty$-module over $Homeo(X)$?
(2) Does $H\_\*(X)$ also have an $A\_\infty$-module structure? Is it the same as that of $H^\*(X)$?
Added later:
Jeff Giansiracusa gave a nice answer to (1). But his answer uses the ring structure in cohomology, leaving (2) open: Is there an $A\_\infty$ structure on homology as well?
| https://mathoverflow.net/users/7867 | $A_{\infty}$ structure of (co)homology of a space | Your category $X^X$ is just the group of homeomorphisms of $X$. This group certainly acts on the homology and cohomology, making them strict modules. But since the group of homeomorphisms is actually acting on $X$, it gives automorphisms of the rational homotopy type. The rational homotopy type of $X$ can be encoded in an $A\_\infty$ algebra structure on the rational cohomology ring (technically, it is a $C\_\infty$ structure, which is a special kind of $A\_\infty$ structure). Thus the group of homeomorphisms of $X$ gives homotopy self-equivalences of the $A\_\infty$ algebra $H^\*(X)$. That is, a homeomorphism $\phi: X \to X$ gives an $A\_\infty$ map $H^\*(X) \to H^\*(X)$ that is an equivalence. Note that such a map contains potentially more information than simply an automorphism of $H^\*(X)$ as an ordinary ring.
| 6 | https://mathoverflow.net/users/4910 | 38873 | 24,921 |
https://mathoverflow.net/questions/38620 | 1 | I know that if $\Lambda$ is a stochastic positive linear map, i.e., $\Lambda(I) = I$, it is true that
`\[ \|\Lambda(B)\| \leq \| B \| \]`
For any operator $B$, where $\|\cdot\|$ is the standard operator norm $\|B\| := \max\_{|v|= 1} |Bv|$. Is it true for any other $p$-norm? Specifically, I want to prove it for the $2$-norm
`\[ \|B\|^2_2 := \operatorname{tr} (B^*B)\]`
also known as Hilbert-Schmidt norm, and I'm only interested in self-adjoint operators.
Naturaly, this question only makes sense if these operators have well-defined norms, so $\Lambda$ can be taken to act in this subalgebra.
It would be nice if the infinite-dimensional case could be done, but the main focus is on the finite-dimensional case.
| https://mathoverflow.net/users/9211 | Norm inequality for stochastic maps | Positivity is not enough (complete positivity is).
Indeed, in $M\_2(\mathbb{C})$, let $\Lambda\left(\left[\begin{array}{cc}a&b \\\\ c&d\end{array}\right]\right)=\left[\begin{array}{cc}a&0 \\\\ 0&a\end{array}\right]$. Then $\Lambda$ is positive and $\Lambda(I)=I$; but if $B=\left[\begin{array}{cc}1&0 \\\\ 0&0\end{array}\right]$ we have, for any $1\leq p<\infty$,
$\|B\|\_p=1$, $\|\Lambda(B)\|\_p=2^{1/p}$.
For a unital completely positive map, Stinespring leads to an easy proof that the inequality holds for $p=2$. For other $p$ it looks trickier, but I think it might work too.
| 1 | https://mathoverflow.net/users/3698 | 38875 | 24,922 |
https://mathoverflow.net/questions/38868 | 5 | Question : Does there exist a surjective function $F$ that maps $\mathbb{R}^n\_+$ to $\mathbb{R}^n$ (where $\mathbb{R}^n\_+$ denotes the set of vectors of length $n$ with only positive entries). The answer is yes by considering the function $F$:
$$(x\_1,\ldots,x\_n)\to(\log x\_1,\ldots,\log x\_n)$$
It is easy to see that $F$ is surjective.
Now my real question is if there is such a function $F$ whose components are restricted to be polynomials of degree at most $n$ (and $F$ maps $\mathbb{R}^n\_+$ to $\mathbb{R}^n$ and is surjective). For example:
$$(x\_1,x\_2,x\_3,x\_4)\to(1-x\_1,x\_1\cdot x\_2+x\_3,x\_3^2+x\_2-x\_4,5x\_4)$$
has polynomial components, but is not surjective as the restriction $x\_i>0$ implies we cannot get a $1$ in the first coordinate.
Any help or advice is greatly appreciated. I think the answer is no but am not sure how to prove it.
| https://mathoverflow.net/users/nan | Vector valued functions | Complete Version
----------------
This is motivated by the complex cubing map described below, which generalizes to the cubing map on quaternions:
For $n > 1$, the map analogous to cubing a complex number or a quaternion is
f\_3: $ (x\_1, x\_2, \dots, x\_n) \rightarrow
(x\_1^3 -3 x\_1( x\_2^2 + x\_3^2 + \dots + x\_n^2),
-x\_2^3 + 3 x\_2 x\_1^2,
\dots, - x\_n^3 + 3 x\_n x\_1^2)$.
This map takes any plane through the $x\_1$ axis to itself, and acts like the complex cubing map in such a plane.
Therefore, the image of the cone $C\_{60}$ within a 60 degree angle of the $x\_1$ axis is all of $\mathbb R^n$.
There are linear transformations that take the positive orthant to nearly all of a half-space, in particular, there are linear images that contain $C\_{60}$.
Composing, we get a surjective homogeneous degree 3 polynomial map from the positive orthant to all of $\mathbb R^n$.
The dimension 1 case is impossible with any degree polynomial, as discussed in Richard Palais's response, and the degree 2 cases cannot be done with degree < 3, as discussed below.
NEW VERSION
-----------
>
> earlier version at end.
>
>
>
For n = 2, if we consider a homogeneous quadratic polynomial, it acts on lines through the origin as a rational map of degree 2. This is either a double cover map of $S^1$ to itself, equivalent to the complex map $z \rightarrow z^2$ , or it folds the circle in half to an interval. The only chance it has to be surjective is the double cover case. The image of any interval under the complex squaring map may be surjective on the circle of lines, but when lifted to its double cover, the circle of rays, its image is an interval, since there are lines which are only hit once. Therefore, no homogeneous degree 2 polynomial is surjective.
Adding a linear or constant term doesn't change the limiting action on rays unless the homogeneous part is degenerate and is 0 on some line, as for instance
$(x,y) \rightarrow (0, y^2)$ (up to linear transformations in the domain and range).
Adding lower degree terms obviously can't make this surjective. So, for $n=1$ or $2$, the answer is **NO**.
There are degree 3 polynomial maps of $\mathbb R^2$ to itself that take the positive quadrant to the whole plane: first, compress the plane to fatten the quadrant to make an angle more than 120 degrees, then cube it as a complex number.
In even dimension $2m \ge 4$, a solution in complex coordinates, as noted by fedja in comments, is
$(z\_1, \dots, z\_m) \rightarrow (z\_1^4, \dots, z\_m^4)$, and a solution of degree 3 is obtained by performing the degree 3 map above coordinatewise.
For odd dimensions $2m+1 \ge 7$, a solution of degree 6 can be obtained as the degree 3 solution on $2m$ coordinates followed by the complex squaring map on the last coordinate with any of the first $2m$. The degree of the composition is 6.
This leaves dimensions 3 and 5 unanswered. Maybe with a little more cleverness?...
Earlier Version
---------------
>
> Note, added: I was hasty reading the question, and didn't pay attention to degree restrictions. The comment of fedja to the question took this into account, and answered it as well as this. The missing cases at the moment seems to be $n = 3, 5, 7$.
>
>
>
In the complex plane, the map $z \rightarrow z^4$ maps the positive quadrant surjectively to $\mathbb C$. Expressed in terms of real and imaginary parts, this satisfies your question for dimension $n = 2$.
You can get a positive answer for any $n \ge 2$ by composing copies of this function applied to pairs of coordinates (the pairs can overlap).
For dimension $n = 1$, it is impossible
| 13 | https://mathoverflow.net/users/9062 | 38876 | 24,923 |
https://mathoverflow.net/questions/38865 | 4 | Let $p:E \to B$ be a fiber bundle over a triangulated base $B$ with fiber $F$, $\sigma$ simplex in $B$, $\sigma \mapsto H\_{\*}(p^{-1}(\sigma)) \simeq H\_{\*}(F)$ the obvious map and let $\mathcal{S}$ be the category of simplices in $B$ with inclusions.
Then $\sigma \hookrightarrow \tau$ in $\mathcal{S}$ gives us a map $\mathcal{S} \to H\_{\*}(p^{-1}(\sigma)) \to H\_{\*}(p^{-1}(\tau))$. Ie, a morphism in $\mathcal{S}$ gives us an element of $End(H\_{\*}(F))$
What I'd like to do in this set-up is now construct a map $H\_{\*}(\Omega B) \to End(H\_{\*}(F))$ using something like the monodromy representation.
(1) Does this map exist? I'd really love to see a construction.
(2) If the answer to (1) is "yes", is this then a map of $A\_{\infty}$-algebras?
Details would be most welcome - this kind of thing is hard to track down in the literature...
| https://mathoverflow.net/users/7867 | Homology of bundles over a triangulated base and $A_\infty$-algebras | I think one can do something like the following. Let $M = map([0,1], B)$ and $e:M \to B$ be evaluation at 0: this is a Hurewicz fibration and a homotopy equivalence. Now form the pullback fibration $e^\*E \to M$, and consider the composite $\bar{E} := e^\*E \to M \to B$. This is a fibration fibrewise homotopy equivalent to your original one, and a point in the fibre $\bar{F}\_b$ over $b \in B$ is a path $\gamma$ from $b$ to a $b\_1$ and point in $p^{-1}(b\_1)$. There is an evident $A\_\infty$ action of the $A\_\infty$ space $\Omega\_b B$ on this fibre by composing $\gamma$ with loops at $b$.
Thus there is an $A\_\infty$ map $\Omega\_b B \to End(\bar{F}\_b)$, which should give you what you want.
Doing the usual Moore loop tricks, one can find an equivalent fibration with an actual action of the grouplike monoid of Moore loops $\Lambda\_b B$.
| 2 | https://mathoverflow.net/users/318 | 38886 | 24,931 |
https://mathoverflow.net/questions/38911 | 6 | I am wondering where I might look to see what has been done in terms of Calculus of Functors for more general weak equivalences and Model Categories.
I am at least aware of some of the extended definitions of the main concepts in Calculus of Functors to weak equivalence such as homotopy limits, but I was wondering if a document existed that worked through the basic of COF in this setting.
I am aware also of Lurie's work [here.](http://www.math.harvard.edu/~lurie/papers/GoodwillieI.pdf)(Thanks Harry for pointing this out.)
I appreciate your help.
| https://mathoverflow.net/users/348 | A reference for Calculus of Functors for Model Categories | In Calculus III and its predecessors I studied functors from Top to Top and a few related cases. The ideas clearly generalize to functors $C\to D$ between model categories satisfying some pretty weak axioms, but I did not try to find the right axioms, and I don't think anyone has ever written anything definitive about that.
Is that what you are asking about?
The paper mentioned by Tilson takes the ideas in a somewhat different direction, I think: it's about treating the categories of functors themselves as model categories and finding Quillen adjoint pairs that refine my ideas.
| 8 | https://mathoverflow.net/users/6666 | 38957 | 24,980 |
https://mathoverflow.net/questions/38793 | 6 | Let $AbCat$ denote the $2$-category of abelian categories with additive functors. Is the forgetful functor $AbCat \to Cat$ representable; i.e. is there an abelian category $T$ such that for every abelian category $A$, the category $Hom(T,A)$ is naturally isomorphic to (the category underlying) $A$?
This would be nice, because then it is possible to reconstruct an abelian category which has only a universal property.
I try to work out the structure of $T$: The isomorphism $Hom(T,T) \cong T$ maps $1\_T$ to some object $x \in T$. If $A$ is arbitrary, then the isomorphism $Hom(T,A) \cong A$ is given by mapping a functor $F : T \to A$ to $F(x)$ and mapping a natural transformation $\eta : F \Rightarrow G$ to the morphism $\eta(x) : F(x) \to G(x)$.
Let $T'$ be the smallest full abelian subcategory of $T$, which contains $x$ (Existence: Construct inductively subcategories $T\_{n+3k}$, which contain direct sums $(n=0)$, kernels $(n=1)$ and cokernels $(n=2)$ from $T\_{n+3k-1}$.). Then $T'$ has the same universal property as $T$. Thus $T'=T$, and we see that $T$ is generated by $x$. Now we have to find such a $T$, which has no additional relations. In particular, we have to ensure that a natural transformation between additive functors on $T$ is determined by the morphism at $x$, although the functors don't have to be exact.
| https://mathoverflow.net/users/2841 | Classifying functors of abelian categories | If we require an isomorphism (as opposed to an equivalence) the answer is no, for reasons having little to do with the (interesting aspect of) the question.
Assume there exists an abelian category $T$ and an object $x$ of $T$ such that for every abelian category $A$ and object $a$ there is a unique functor $F\_a:T\to A$ sending $x$ to $a$. Clearly, $x$ is nonzero, so the object $x^2$ has a nontrivial automorphism $\sigma$. Consider the functor $\Phi:T\to T$ defined as follows:
$\Phi$=identity on objects,
for a map $f:u\to v$, put:
$\Phi(f)=f$ if $u\neq x^2\neq v$ or $u=x^2=v$,
$\Phi(f)=\sigma f$ if $u\neq x^2$ and $v=x^2$,
$\Phi(f)=f\sigma^{-1}$ if $u=x^2$ and $v\neq x^2$.
Now $\Phi$ is indeed a functor (case-by-case inspection) sending $x$ to $x$. So it should be equal to the identity functor but isn't.
| 3 | https://mathoverflow.net/users/7666 | 38963 | 24,985 |
https://mathoverflow.net/questions/38855 | 2 | Hi,
I have a (symmetric) matrix $M$ that represents the distance between each pair of nodes. For example,
```
A B C D E F G H I J K L
A 0 20 20 20 40 60 60 60 100 120 120 120
B 20 0 20 20 60 80 80 80 120 140 140 140
C 20 20 0 20 60 80 80 80 120 140 140 140
D 20 20 20 0 60 80 80 80 120 140 140 140
E 40 60 60 60 0 20 20 20 60 80 80 80
F 60 80 80 80 20 0 20 20 40 60 60 60
G 60 80 80 80 20 20 0 20 60 80 80 80
H 60 80 80 80 20 20 20 0 60 80 80 80
I 100 120 120 120 60 40 60 60 0 20 20 20
J 120 140 140 140 80 60 80 80 20 0 20 20
K 120 140 140 140 80 60 80 80 20 20 0 20
L 120 140 140 140 80 60 80 80 20 20 20 0
```
Is there a method to extract clusters from $M$ (if needed, the number of clusters can be fixed), such that each cluster contains nodes with small distances between them. In the example, the clusters would be (A, B, C, D), (E, F, G, H) and (I, J, K, L).
Thanks a lot.
| https://mathoverflow.net/users/8515 | Nodes clusters with a distance matrix | The best answer is **distance threshold**. Look at the problem as defining an undirected graph based on a given distance matrix and trying to create subcomponents of the graph by selectively deleting edges based on the weight of each edge as defined by the distance.
If you make the assumption that the underlying metric space (which you have not clearly defined in this case) is Euclidean (linear and therefore homogeneous throughout, I believe), then one simple way of creating subsets of your elements from the distance matrix which you have available is the following.
(*edit*: No assumtion of homogeneity of the underlying metric space is actually required for this approach of selectively deleting edges.)
>
> Given a distance matrix $M$ with elements $M\_{i,j}$ describing a distance between the elements $i$ and $j$ where $i\in S$ and $j\in S$, and $S$ is the set of items you are attempting to separate into subsets (hopefully disjunct)
>
>
> define a **binary matrix** which describes whether element $i$ is in the same subset as element $j$ by using a **distance threshold**, $d$.
>
>
> $D$ = the matrix with elements
>
>
> $D\_{i,j}=0$ if $M\_{i,j}\ge d$ or if $i=j$, and
>
>
> $D\_{i,j}=1$ if $M\_{i,j} < d$ and $i \ne j$
>
>
> use this binary matrix to draw a graph, and call each separate **connected component** of the graph a separate **cluster**
>
>
>
If it is possible to separate the elements of your set $S$ into disjunct partitions, there is a minimum distance $d\_{min}$ that you can use.
For example, setting the distance threshold to $40$ allows the partitioning of $S$ into disjunct (non-intersecting) subsets:
```
0 1 1 1 0 0 0 0 0 0 0 0
1 0 1 1 0 0 0 0 0 0 0 0
1 1 0 1 0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 1 0 1 1 0 0 0 0
0 0 0 0 1 1 0 1 0 0 0 0
0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1 1 1
0 0 0 0 0 0 0 0 1 0 1 1
0 0 0 0 0 0 0 0 1 1 0 1
0 0 0 0 0 0 0 0 1 1 1 0
```
Now, drawing the graph structure of this binary matrix leads to a three component graph composed of these three graphs (which all happen to be $K\_4$, the complete graph on 4 vertices) with vertex sets composed of
* {A, B, C, D}
* {E, F, G, H}
* {I, J, K, L}
---
However, the binary matrix for the undirected graph "in the same cluster" using distance $d<50$ as the threshold results in
```
0 1 1 1 1 0 0 0 0 0 0 0
1 0 1 1 0 0 0 0 0 0 0 0
1 1 0 1 0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0 0 0 0
1 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 1 0 1 1 1 0 0 0
0 0 0 0 1 1 0 1 0 0 0 0
0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 1 1 1
0 0 0 0 0 0 0 0 1 0 1 1
0 0 0 0 0 0 0 0 1 1 0 1
0 0 0 0 0 0 0 0 1 1 1 0
```
A distance threhold of 50 does not break the dataset into disjunct sets, as when you follow the linkages all of the elements are effectively connected by certain bridging elements. If you draw this as a graph structure, you will see that there are subgraphs
* {B,C,D}
* {F,G,H}
* {J,K,L}
* with bridging element A connecting {B,C,D} and E
* and bridging element F connects to bridging element I,
* etc.
---
For the example distance matrix which you have given, setting the threshold $20 \le d \lt 40$ will give you what looks like a correct result. Setting $d$ too low results in more isolated components to the graph, setting $d$ too high leads to larger components.
>
> There is no guarantee that there will be a threhold $d\_{min}$ which will separate such a set into disjunct subsets.
>
>
>
For example, if you set the distances between A and E to zero, and the distances between F and I to zero, there is no threshold which will separate the sets using only the distance matrix.
You may have to "manually" adjust the **distance threshold** to get the best separation of the set into disjunct subsets, if such a partitioning exists.
| 1 | https://mathoverflow.net/users/8676 | 38967 | 24,988 |
https://mathoverflow.net/questions/38973 | 24 | First some context. In most algebraic number theory textbooks, the notion of
discriminant and different of an extension of number fields $L/K$, or rather, of the corresponding extension $B/A$ of their rings of algebraic integers is defined.
The discriminant, an ideal of $A$, is the ideal generated by the discriminant of the quadratic form $\text{tr}(xy)$ on $B$. The different, an ideal of $B$, is the inverse of
the fractional ideal $c$ of $L$ defined by $c=\{x \in L, \text{tr}(xy) \in A \ \forall y \in B\}$. The norm of the different is the discriminant.
Now the discriminant makes sense in a much more general context, say for any extension of (commutative) rings $B/A$ that is finite projective, since the trace map $\operatorname{tr}$ makes sense in this context. My question is: is there a standard definition of the different
in this context? if so, where can I find it in the literature, if possible with the basic results about it?
I am pretty sure the answer to the first question is yes, but I have not been able
to find a reference. The problem when I try to use google or MathSciNet seems to be that
"different" is not a very discriminant name: almost every paper in mathematics contains it.
Let me propose an answer to my own question: we could define the different of
$B/A$ by the Fitting ideal of the universal $B$-modules of differentials $\Omega\_{B/A}$. The fact that it gives the correct definition in the number field cases is [Serre, Local Fields, chapter III, Prop. 14], and moreover it behaves well under base change. This definition may very possibly be a remembrance of something I had heard in an earlier life. But even if it is the correct definition, I'd like to know a reference where it is stated.
| https://mathoverflow.net/users/9317 | Discriminant and Different | In chapter 8 (entitled "Traces, Complementary Modules, and Differents") of the book Residues and Duality for Projective Algebraic Varieties by Kunz, he gives exactly the definition you propose and proves some basic results about its properties.
| 16 | https://mathoverflow.net/users/397 | 38974 | 24,992 |
https://mathoverflow.net/questions/38978 | 8 | **Problem**
I am interested in the random recurrence relation of the form $x\_{n+1}=\alpha x\_n \pm \beta x\_{n-1}$ where $\alpha$, $\beta$ are known constants and the $\pm$ sign is chosen with equal probability.
In particular I would like to investigate the limit of $|x\_n|^{1/n}$ as $n\rightarrow\infty$. (i.e the exponential growth rate)
**Motivation**
I was considering the problem of finding $\lim\_{n\rightarrow\infty}|x\_n|^{\frac{1}{n}}$ where $x\_n$ is the solution to $x\_{n+1}=2x\_n \pm x\_{n-1}$, $x\_0=x\_1=1$. All numerical evidence suggests that $\lim\_{n\rightarrow\infty}|x\_n|^{\frac{1}{n}}\approx 1.91$ almost surely but I am having difficulty even proving that the sequence almost surely convergences.
**Survey**
The solution to the recurrence relation of the from $x\_{n+1}=x\_n \pm \beta x\_{n-1}$ has been show to have expontential growth almost surely. See <http://en.wikipedia.org/wiki/Embree%E2%80%93Trefethen_constant>
Also there has been work on the random fibonacci sequence $x\_{n+1}=x\_n \pm x\_{n-1}$ for example Viswanath(2000), "Random Fibonacci sequences and the number 1.13198824"
| https://mathoverflow.net/users/2011 | Random linear recurrence relations | If $(x\_n)$ solves the recursion you are interested in, then the sequence $(y\_n)$ of general term $y\_n=x\_n/\alpha^n$ is a random Fibonacci sequence such as in the Embree-Trefethen page you linked to.
| 9 | https://mathoverflow.net/users/4661 | 38980 | 24,996 |
https://mathoverflow.net/questions/38943 | 6 | Let $P\in{\mathbb R}[X]$ be a monic polynomial with roots on the unit circle. For the problem below, we may assume wlog that the roots are simple and distinct from $\pm1$. It can be shown that there exists a matrix $M\in{\bf SO}\_n({\mathbb R})$, whose characteristic polynomial is $P$ (an *orthogonal companion matrix* of $P$, in short OCM). See for instance Exercise 99 on my list <http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf> .
Regretfully, this exercise uses the square root of Hermitian positive definite matrices, which cannot be computed in finitely many operations.
Does there exist a construction of an OCM that uses only finitely many elementary operations (including the square root of complex numbers) ?
Thanks to the reduction to Hessenberg form, which can be done in finite time and which preserves the orthogonal group, we may restrict our attention to a Hessenberg orthogonal matrix $M$. It writes
$$\left( \begin{array}{ccccc}
c\_1 & s\_1c\_2 & s\_1s\_2c\_3 & s\_1s\_2s\_3c\_4 & \ldots \\\\
-s\_1 & c\_1c\_2 & c\_1s\_2c\_3 & c\_1s\_2s\_3c\_4 & \ldots \\\\
0 & -s\_2 & c\_2c\_3 & s\_2s\_3c\_4 & \ldots \\\\
0 & 0 & -s\_3 & c\_3c\_4 & \ldots \\\\
0 & 0 & 0 & -s\_4 & \ldots
\end{array} \right)$$
where $(c\_j,s\_j)$ are cosine/sine pairs.
| https://mathoverflow.net/users/8799 | An orthogonal companion matrix | I'd do this in three steps:
1. Find any $2n \times 2n$ matrix $A$ whose eigenvalues are $e^{\pm i \theta}$.
2. Find a positive definite quadratic form preserved by $A$. In equations, we want $A P A^T = P$.
3. Find an orthonormal basis for $P$, using the [Gram-Schimdt algorithm](http://en.wikipedia.org/wiki/Gram-Schmidt). In equations, we want $S P S^T = \mathrm{Id}$.
Then $S A S^{-1}$ is orthogonal and has the required eigenvalues.
---
I can think of two ways to do step 2. The first is more purely algebraic, the second I think would be much easier to implement.
**Algebra:** Let $f(x) = \prod\_{j=1}^{n} (x-e^{i \theta\_j}) (x - e^{-i \theta\_j}) = \prod (x^2 - 2 \cos \theta\_j + 1)$ be your characteristic polynomial. Let $V$ be the ring $\mathbb{R}[x]/f(x)$. Note $1$, $x$, ..., $x^{2n-1}$ is a basis for this ring, in which multiplication can be written down algebraically in terms of the coefficients of $f$. Also, multiplication by $x$ has the desired eigenvalues, so that accomplishes part 1.
For $y \in T$, let $T(y)$ be the trace of multiplication by $y$. Also, let $y \mapsto \overline{y}$ be the automorphism of $V$ induced by $x \mapsto x^{-1}$. Again, both of these can be written down, in the monomial basis, algebraically in terms of the coefficients of $f$.
Then $\langle y,z \rangle = T(y\*\overline{z})$ is the desired positive definite quadratic form. Namely, observe that the ring $V$ is isomorphic to $\mathbb{C}^{\oplus n}$. In terms of this isomorphism, $\langle (z\_1, \ldots, z\_n), (z\_1, \ldots, z\_n) \rangle = 2 \sum |z\_i|^2$.
**In practice:** The condition that $APA^T = P$ is a linear condition on $P$. Let $W$ be the subspace of the vector space of symmetric matrices where this condition is satisfied; finding $W$ is just algebra. Now, our goal is to find a positive definite element of $W$. For large $N$, $(1/N) \left( \mathrm{Id} + AA^T + A^2 (A^{T})^2 + \cdots + A^{N-1} (A^T)^{N-1} \right)$ is positive definite and is near $W$. I would guess that the orthogonal projection of this matrix onto $W$ would probably be positive definite for large $N$.
| 3 | https://mathoverflow.net/users/297 | 38999 | 25,006 |
https://mathoverflow.net/questions/38996 | 2 | Let $G$ be a group acting properly by biholomorphisms on a complex manifold $X$, so $X // G$ is a complex orbifold. Let the *holomorphic Picard group* $Pic\_{hol}(X//G)$ be the group of isomorphism classes of $G$-equivariant holomorphic line bundles on $X$, under tensor product. This is naturally isomorphic to the group $H^1(X/G; \mathcal{O}^\times)$.
The first Chern class furnishes a map
$$c\_1 : Pic\_{hol}(X//G) \to H^2(X//G;\mathbb{Z})$$
to the second integral cohomology of the obifold (where cohomology is taken in the orbifold sense: it is not the integral cohomology of the actual quotient $X/G$). This is the connecting homomorphism for the exponential sequence $\mathbb{Z} \to \mathcal{O} \to \mathcal{O}^\times$ of sheaves on $X//G$.
I am interested in conditions on the orbifold so that this map is *injective*, and am happy to suppose that $H^1(X//G;\mathbb{Z})=0$. (Please do not tell me that the condition I want is that $H^1(X//G;\mathcal{O})=0$: I know this, and want conditions for it to hold.)
| https://mathoverflow.net/users/318 | Picard group of complex orbifolds | I guess that for reductive $G$ one has $H^1(X//G,O) = H^1(X,O)^G$ (the $G$-invariants). So, the condition you want is $H^1(X,O)^G = 0$. For nonreductive $G$ there is a spectral sequence $H^q(G,H^p(X,O)) \Rightarrow H^{p+q}(X//G,O)$ so the condition is
$$
H^1(G,H^0(X,O)) = 0,
\qquad
Ker(H^1(X,O)^G \to H^2(G,H^0(X,O))) = 0.
$$
| 3 | https://mathoverflow.net/users/4428 | 39008 | 25,012 |
https://mathoverflow.net/questions/38890 | 15 | These are questions on D. Quillen's 1978 paper *Homotopy properties of the poset of nontrivial p-subgroups of a group*.
Let $G$ be a finite group, $p$ a prime number, $\mathcal S(G)$ the poset of non-trivial $p$-subgroups of $G$, and $\mathcal A (G)$ the poset of non-trivial elementary Abelian $p$-subgroups of $G$, both ordered by inclusion.
>
> Question One: Is $\mathcal S(G)$ homotopic or weakly homotopic to $\mathcal A (G)$? (Whichever is true is a theorem of Quillen.) If the latter, can someone give a specific example showing that the two posets are not homotopic?
>
>
>
Quillen also proved that for $G$ solvable, $\mathcal A (G)$ is contractible if and only if $G$ has a non-trivial normal $p$-subgroup.
**Conjecture (Quillen):** $\mathcal A (G)$ is contractible if and only if $G$ has a non-trivial normal $p$-subgroup.
>
> Question Two: Is this still open (I know it was a few years ago)? What are lines of attack on this problem? Have attempts to prove it led to any *a priori* unrelated work?
>
>
>
| https://mathoverflow.net/users/7867 | Status of Quillen's conjecture on elementary abelian p-groups | To answer your first question, the inclusion $A(G)\to S(G)$ is a homotopy equivalence. This is an application of Quillen's "Theorem A" (aka his "fiber lemma"). See Prop. 2.1 in his paper. To apply the fiber lemma, you just need to show that the fibers, which are those points mapping below a particular P-subgroup in S, are contractible. This means you need to show that the elementary abelian p-subgroups of a P-group form a contractible poset. Well, that's done by a conical contraction: just multiply each subgroup by the maximal elementary abelian subgroup of the center (to slide it up above this characteristic subgroup), and then slide it down to this subgroup.
The Quillen Conjecture is definitely still open. Grodal showed some connections with higher derived limits (he shows that certain strengthenings of Quillen's conjecture are equivalent to statements about higher limits), and there's an old approach using finite topological spaces due to Richard Stong. There's been some interesting developments lately due to Shareshian (Hypergraph matching complexes and Quillen complexes of symmetric groups. J. Combin. Theory Ser. A 106 (2004), no. 2, 299--314) and others, but mostly they're about figuring out more specific information in particular cases, rather than attacks on the full conjecture.
In the early 90's, Aschbacher and Smith made a lot of progress (see their 1993 Annals paper). They proved the conjecture for groups not containing certain matrix groups as subnormal subgroups.
I don't know that anyone really has ideas for how to prove it in general.
| 15 | https://mathoverflow.net/users/4042 | 39012 | 25,015 |
https://mathoverflow.net/questions/39002 | 4 | Suppose $F(x)$ is a convex objective function on $n\times n$ matrices, and I need to numerically optimize $F$ with the condition that $x$ has spectral radius less than $1$. This might be too hard, so an approximation would be needed. Has this problem been studied before?
Motivation: Boltzmann machines are hard to evaluate when spectral radius of the weight matrix is large, especially if it's above $1$ so best fit to data subject to this constraint would give a useful model.
Example: Let $X=\{1,-1\}^d$ and $\hat{X}$ some list of $\{1,-1\}$ $d$-tuples. Find $$\max\_A \sum\_{x\in \hat{X}} \mathbf{x}'A\mathbf{x} - |\hat{X}|\log \sum\_{x\in X} \exp(\mathbf{x}'A\mathbf{x})$$
Where $A$ is symmetric real-valued $d\times d$ matrix with spectral radius < 1. This needs to be done in time polynomial in $d$ and linear in $|\hat{X}|$. When spectral radius is <1, belief propagation gives a reasonably accurate way to approximate gradient of this objective in $O(|\hat{X}|d^2)$ time
| https://mathoverflow.net/users/7655 | Optimizing over matrices with spectral radius <1? | If your matrices are symmetric, the set of matrices with spectral radius $\le 1$ is convex, and can be modelled using a linear matrix inequality (LMI), see e.g. [page 147 in Lectures on Modern Convex Optimization](http://books.google.com/books?id=M3MqpEJ3jzQC&lpg=PR1&ots=O2nXBEkTUU&dq=modern%2520convex%2520optimization&pg=PA147#v=onepage) by Ben-Tal and Nemirovski. If you wanted to *minimize* a convex objective that is also semidefinite-representable, you could in principle formulate and solve your problem as a semidefinite programming problem. However, *maximizing* a convex objective over a convex set is a much more difficult problem.
| 5 | https://mathoverflow.net/users/1184 | 39013 | 25,016 |
https://mathoverflow.net/questions/39011 | 3 | Let $J\_1$ be the Bessel function of the first kind and let $H\_1(x) = \frac{J\_1(|x|)}{|x|}$ for $n = 1$. Define the operator $Tf(x) = (f \* H\_1)(x)$ from $L^2$ to $L^2$.
Since the $H\_1$-function is the Fourier transform of something it must be in $L^2$, so we have a Hilbert-Schmidt operator which is in this case self-adjoint and compact, so the spectral theorem applies which for example says all the eigenvalues form a countable set, are real and go to zero.
What I am now interested in is the largest eigenvalue of $T$. What theorem or method could I try to obtain this? (I don't need a full solution, just a hint would suffice).
| https://mathoverflow.net/users/5295 | Eigenvalues convolution-type operator | In Fourier space, the operator $T$ is given by
$$\hat{(T f)}(\xi)=c\sqrt{1-\xi^2}\hat f(\xi)$$
for some constant $c\ne0$ ($c$ dependson the normalization chosen for thr Fourier transform.) If $\lambda$ is an eigenvalue and $f$ is a corresponding eigenfunction, then for almost all $\xi$
$$\lambda \hat f(\xi)=c\sqrt{1-\xi^2}\hat f(\xi).$$
It follows that the only eigenvalue is $\lambda=0$, and the coresponding eigenfunctions are all $L^2$ functions whose Fourier transform vanish almost everywhere on $[-1,1]$.
| 2 | https://mathoverflow.net/users/1168 | 39022 | 25,022 |
https://mathoverflow.net/questions/38294 | 10 | The generation function of the Gromow-Witten invariants (with descendants) of the point is known to be Kontsevich-Witten tau-function of KdV, partition functions of $P^1$ and equivariant $P^1$ are known to be tau-functions of extended Toda and 2-Toda respectively. Are there any other manifolds (except of orbifolds made of mentioned previously manifolds ) for which generation functions of GW invariants are identified with tau-functions of some integrable hierarchy?
| https://mathoverflow.net/users/3840 | Gromov-Witten and integrability. | Short answer: essentially, the point and $P^1$ are the only spaces where the GW generating function is a tau-function. However, you mentioned two variations on this spaces: equivariant orbifold versions. And there are other variations that go a bit further -- twisted and relative invariants, and, a little wilder, Landau-Ginzburg theory. But in all cases, everything seems to be a reduction of either the KP hierarchy or the 2-Today hierarchy.
Twisted Invariants
------------------
Let $L\_i$ be some line bundles on $X$, and let $B$ be the total space of $\oplus L\_i$. The GW theory of $B$ doesn't make sense since it's not compact, and so neither will $\overline{\mathcal{M}}\_{g,n}(B, \beta)$, but we can take a $\mathbb{C}^\*$ on $B$ that fixes $X$ and just acts on the fibers, and then use localization on
$\overline{\mathcal{M}}\_{g,n}(B, \beta)$.
The fixed point locus will be $\overline{\mathcal{M}}\_{g,n}(X, \beta)$, but the invariants will be different because of extra terms coming from the euler class of the normal bundle. We call and so it will make sense to integrate, and we define these to be $GW(B)$, and call them twisted invariants of $X$.
A Grothendieck-Riemann-Roch calculation begun by Mumford and extended and expanded by Faber-Pandharipande and Coates-Givental will reduce $GW(B)$ to $GW(X)$, but in a messy way that would for instance change the invariants, and certainly change around the integrable hierarchy.
So, for instance, in the case of a point, we'll just get integrals over $\overline{\mathcal{M}}\_{g,n}$, the normal bundle will essentially be copies of the Hodge bundle, and so we'll get $\lambda$ classes appearing in our integrals along with the usual $\psi$ classes: these are called Hodge integrals.
In the case that there is just one line bundle (so $B=\mathbb{C}$, the integral will be linear in the $\lambda$ classes, and will be the integral appearing in the ELSV formula. [Kazarian](http://arxiv.org/abs/0809.3263) shows that the resulting GW theory satisfies the KP hierarchy.
The case with three line bundles ($B=\mathbb{C}^3$) is the case covered by the topological vertex. As far as I know, there is no known integrable hierarchy satisfying the whole thing. But [Zhou](http://arxiv.org/abs/math/0310408) has shown that certain specializations lead to KP and 2-Toda type equations.
Similarly, we can twist the GW theory of $\mathbb{P}^1$ and hope to get integrable hierarchies here. [Brini](http://arxiv.org/abs/1002.0582) has made some progress here for the direct sum of two line bundles, so that $B$ is a three-fold. In particular, for the resolved conifold (the total space of $\mathcal{O}(-1)\oplus\mathcal{O}(-1)$, he gets connections with the Ablowitz-Ladik hierarchy, apparently some reduction of 2-Toda.
Relative Invariants
-------------------
Another variation we can play is to use relative invariants, working relative to a divisor.
In dimension zero this doesn't work, but we can work relative to 0 and $\infty$ on $\mathbb{P}^1$. Okounkov and Pandharipande have shown that this flavor also satisfies some 2-Toda type hierarchy.
In the paper of Zhou mentioned above, he also sets up certain relative generating functions for toric varieties that satisfy KP and 2-Toda hierarchies.
Landau-Ginzburg
---------------
In another direction, if we move slightly away from GW theory we can get more interesting examples. Fan, Jarvis and Ruan have recently finished rigorously constructing a Landau-Ginzburg A-model -- essentially, a Gromov-Witten theory for hypersurface singularities.
These theories have a central charge that acts as a dimension. The central charge 0 case has an ADE classification. Witten conjectured that the theory of these should satisfy the corresponding Kac-Wakimoto / Drinfeld-Sokolov reduction of the KP hierarchy. Note that for $A\_1$, this is the Gromov-Witten theory of a point, and the usual KdV hierarchy.
In the $A\_r$ case, the analytic machinery of FJR is not needed to define the theory, and it goes under the name $r$-spin curves -- essentially, we're doing integrals over the moduli space of orbifold curves with an chosen $r$-th root of the canonical bundle. Witten's conjecture was proven here by Faber, Shadrin and Zvonkine. FJRW have recently proven the D and E cases.
| 13 | https://mathoverflow.net/users/1102 | 39023 | 25,023 |
https://mathoverflow.net/questions/38966 | 85 | What is sheaf cohomology intuitively?
For local systems it is ordinary cohomology with twisted coefficients. But what
if the sheaf in question is far from being constant?
Can one still understand sheaf cohomology in some "geometric" way?
For example I would be very interested in the case of coherent $\mathcal{O}\_X$-Modules.
Or even just line bundles.
| https://mathoverflow.net/users/2837 | What is sheaf cohomology intuitively? | One way to think about $H^1(A)$ is to use the long exact sequence not as a property of cohomology, but outright as a definition. That is, given an exact sequence of sheaves,
$$ 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$
then $H^1(A)$ is measuring the obstruction of global sections to be exact:
$$ 0\rightarrow \Gamma(A)\rightarrow \Gamma(B)\rightarrow \Gamma(C)\rightarrow H^1(A)$$
In words, $H^1$ is `measuring the failure of $\Gamma$ to preserve surjectivity'.
If you want this idea to actually *define* $H^1(A)$, you have to be careful to chose $B$ so that $H^1(B)=0$. But, as far as intuition goes, this works pretty well for me.
A demonstrative example of this, at least for me, is to think about the the complex variety $\mathbb{P}^1$, with $\mathcal{O}$ the structure sheaf and $\mathbb{C}\_p$ the skyscraper sheaf at a point. Then there is a surjective map (it is surjective because it is surjective on stalks):
$$ \mathcal{O}\rightarrow\mathbb{C}\_p$$
which has kernel $\mathcal{O}(-1)$, the twisted structure sheaf. This whole short exact sequence can be twisted by $(-1)$, noting that twisting a skyscraper sheaf $\mathbb{C}\_p$ gives an isomorphic sheaf $\mathbb{C}\_p(-1)$ (which I identify with the original sheaf):
$$0\rightarrow \mathcal{O}(-2)\rightarrow \mathcal{O}(-1)\rightarrow \mathbb{C}\_p\rightarrow 0$$
On global sections, we then get
$$ 0\rightarrow \Gamma(\mathcal{O}(-2))\rightarrow \Gamma(\mathcal{O}(-1))\rightarrow \Gamma(\mathbb{C}\_p)\rightarrow H^1(\mathcal{O}(-2))$$
We know that $\Gamma(\mathcal{O}(-1))=0$ and $\Gamma(\mathbb{C}\_p)=\mathbb{C}$, so the middle arrow is no longer surjective. Hence, $H^1(\mathcal{O}(-2))$ must contain at least $\mathbb{C}$ (in fact, it is exactly $\mathbb{C}$, since $H^1(\mathcal{O}(-1))=0$).
Higher cohomology may be also thought of this way: $H^{i+1}$ measures the failure of $H^i$ to preserve surjective maps. However, I don't find this very useful for thinking about higher cohomology, since it would need that I somehow understand lower cohomology much better.
| 33 | https://mathoverflow.net/users/750 | 39027 | 25,026 |
https://mathoverflow.net/questions/38582 | 0 | I'm calculating the roots of the function
\begin{equation}
R(x) = \sum\_{k=1}^{\infty}\frac{\mu(k)}{k}li(x^{1/k})
\end{equation}
This function seems to have a largest and smallest positive root. Can anyone tell me if the roots of $R(x)$ have any significance for the prime counting function?
| https://mathoverflow.net/users/2011 | Do the roots of R(x) have any significance for the prime counting function? | Not that I know of. If you have not already, see the paper by Folkmar Bornemann that describes a method for finding the roots of R(x) (see link below). It's a very interesting method.
Best regards,
Tom
[Paper](http://www-m3.ma.tum.de/m3old/bornemann/challengebook/AppendixD/waldvogel_problem_solution.pdf)
| 1 | https://mathoverflow.net/users/8955 | 39032 | 25,030 |
https://mathoverflow.net/questions/39033 | 0 | It's possible, that equation $\sum\_{n} n!^s=1+2\sum\_n (2n+1)!^s$ is correct for all $s \in \mathbb{R}$ with which sum $\sum\_{n} n!^s$ is convergent?
I'm looking for closed formula of that sum and correctness of that equation is very important to me. Thanks for help.
| https://mathoverflow.net/users/9067 | Correctness of equation for $\sum_{n} n!^s$ | Assuming that your sums run over the nonnegative integers,
$$\sum\_{n \ge 0} n!^s$$
is convergent if and only if $s < 0$. (Apply the ratio test.)
It's easy to evaluate both sides at $s = -1$. In that case, the left-hand side is just
$$ \sum\_{n \ge 0} {1 \over n!} = e$$
and the left-hand side is $1 + 2 \sinh 1$, where $\sinh$ is the hyperbolic sine. In particular, the sum $\sum\_{n \ge 0} 1/(2n+1)!$ is greater than 1 (the $n=1$ term is 1), and so $1 + 2\sum\_{n \ge 0} 1/(2n+1)! > 3$. So the identity you want to prove is false.
| 6 | https://mathoverflow.net/users/143 | 39034 | 25,031 |
https://mathoverflow.net/questions/39029 | 3 | [(Medeen, et all, 1998)"](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.34.6043) show that Maximum Likelihood estimate is admissible for multinomial distribution under squared error. On other hand, James and Stein showed that arithmetic average is not an admissible estimator of Gaussian location parameter in 3 dimensions. But since maximum likelihood estimates of multinomial parameters are averages of observed counts, which become normally distributed for large sample sizes, why doesn't Stein effect happen here?
$\hat{p}$ is an inadmissible estimator of $\theta$ if there's an estimator that is no worse for every $\theta$ and better for at least one
| https://mathoverflow.net/users/7655 | Why doesn't Stein effect happen for multinomial distributions? | This is not an answer, but maybe worth thinking about (and I cannot yet leave comments). My intuition about the Stein phenomenon is that while the individual coordinates of the Gaussian random variable are independent, the loss function involves all of the location parameters jointly. Stein type estimators take this into account and by doing so outperform the MLE, making it inadmissible.
In the case of the multinomial parameters, they inherently have dependence via the sum-to-one constraint as a probability vector and you take this into account when averaging over possible parameter values. So a question related to yours, which may shed some light on it, is whether or not the MLE is admissible for a Gaussian location vector $\mu$ under the restriction that $\|\mu\| = c$ for some positive constant $c$.
UPDATE:
"Admissibility and complete class results for the multinomial estimation problem...", Ighodaro, Thomas & Brown (Journal of Mult. Analysis '82) shows the MLE for the multinomial parameter becomes inadmissible if you remove the vertices of the simplex from the action space. It is a property of the risk behavior of the MLE at these extremal points that makes it admissible, then. Since the corresponding Gaussian problem has no such extremal points, this may constitute an explanation to your question.
| 6 | https://mathoverflow.net/users/8719 | 39040 | 25,036 |
https://mathoverflow.net/questions/39045 | 0 | Let $H = L\_2(S)$ be the complex Hilbert space over $S$ with the counting measure. (There might be another term for this concept, but) I define a continuous linear operator $L$ on $H$ with matrix representation $A$ to be Frobenius-finite if and only if $\displaystyle\sum\_{(i,j) \: \in \: S \times S} |a\_{i,j}|^2 < \infty$
1.
Is Frobenius-finiteness invariant under unitary similarity?
2.
If yes, is there a categorical (i.e., not using the concept of matrix representation) characterization of when a continuous linear operator is Frobenius-finite?
3.
Is there another term for what I am calling Frobenius-finite?
| https://mathoverflow.net/users/nan | "Frobenius-finite" linear operators on a Hilbert Space | Since $\sum\_{(i,j)∈S×S}|a\_{i ,j}|^2 =$ Trace$(A^\*A)$ the answers to 1. and 2. are both affirmative, and as has already been said, the answer to 3. is "Hilbert-Schmidt."
| 4 | https://mathoverflow.net/users/7311 | 39054 | 25,046 |
https://mathoverflow.net/questions/39042 | 2 | I should start by saying that I have not studied field theory in depth, so if this question is totally off base, I apologize. Something I noticed as I studied group theory is many concepts that were very difficult to define directly had simple and elegant categorical definitions. For example, the direct definition of the free group is rather long and arduous, whereas the categorical definition, i.e. any function $S\to G$, where $G$ is a group factors through a homomorphism from the free group generated by $S$ to $G$, is quite simple. However, for the most part, it seems to me that categorical methods are most easily used on infinite groups, and in particular, infinite abelian groups. Despite this limitation, categorical methods seemed so natural that I couldn't help but wonder if they can be applied to field theory with similar results. So my question is: (1) is it beneficial to study infinite field theory in the generality that category theory necessitates, and (2) are there any good books that use this approach.
| https://mathoverflow.net/users/6856 | Infinite Field Theory and Category Theory | I think Mike Skirvin's comment above should be expanded into an answer.
There are no homomorphisms at all between fields of different characteristic. Hence one has to look at the category of fields of a fixed characteristic $p$.
An elementary fact about fields is that they have no nontrivial ideals.
It follows that all homomorphisms between fields are 1-1.
This implies that there are no free fields of any characteristic $p$
(except for the free field of char $p$ over the empty set of generators).
Finally, as Mike Skirvin pointed out in his comment, there are in general
no products of fields, even of a fixed characteristic.
I think this sufficiently explains why categorial constructions are not very
useful in field theory.
| 4 | https://mathoverflow.net/users/7743 | 39057 | 25,047 |
https://mathoverflow.net/questions/39056 | 28 | Can someone please tell me some introductory book on symplectic geometry? I have no prior idea of the subject but I do know about Lagrangian and Hamiltonian dynamics (at the level of Landau-Lifshitz Vol. 1). Thanks in advance. :-)
| https://mathoverflow.net/users/9292 | Book on symplectic geometry | If you are physically inclined, V.I.Arnold's *Mathematical methods of classical mechanics* provides a masterful short introduction to symplectic geometry, followed by a wealth of its applications to classical mechanics. The exposition is much more systematic than vol 1 of Landau and Lifschitz and, while mathematically sophisticated, it is also very lucid, demonstrating the interaction between physical ideas and mathematical concepts that support them. (It is also worth mentioning that Arnold was largely responsible for the reawakening of interest to symplectic geometry at the end of 1960s and pioneered the study of symplectic topology. Some of these developments were brand new when the book was first published in 1974 and are briefly discussed in the appendices).
In addition to the notes by Cannas da Silva mentioned by Dick Palais, here are further two advanced books covering somewhat different territory:
>
> Michèle Audin, *Torus actions on symplectic manifolds* (2nd edition)**A**
>
>
> Dusa McDuff and Dietmar Salamon, *Introduction to symplectic topology*
>
>
>
---
**A** In her book, Michèle Audin herself recommends
>
> Paulette Libermann and Charles-Michel Marle, *Symplectic geometry and analytical mechanics*
>
>
>
as a wonderful introduction to symplectic geometry.
| 26 | https://mathoverflow.net/users/5740 | 39060 | 25,050 |
https://mathoverflow.net/questions/39073 | 5 | Typically, when defining the functor category $\mathcal{D}^\mathcal{C}$, where objects are functors $\mathcal{C}\rightarrow\mathcal{D}$ and the morphisms between such objects $F,G$ are the natural transformations $F\rightarrow G$ with the obvious composition and identities, one requires the category $\mathcal{C}$ to be *small*, i.e. that its objects form a set. $\mathcal{D}$ can be any category.
I've always thought that the reason for the smallness requirement on $\mathcal{C}$ is that "surely you'll get into trouble with the morphisms $F\rightarrow G$ in $\mathcal{D}^\mathcal{C}$ potentially not forming a set otherwise", and I've never given it much thought.
Now, does anyone know of an *example* which fixes a large category $\mathcal{C}$, a category $\mathcal{D}$ and two functors $F,G:\mathcal{C}\rightarrow\mathcal{D}$ such that that the collection of natural transformations $F\rightarrow G$ does not form a set? I've always thought in the back of my head that this should be easily doable with $\mathcal{C}=\mathcal{D}=\mathrm{Set}$ and some simple functors $F,G:\mathcal{C}\rightarrow\mathcal{D}$ by somehow cooking up at least one natural transformation $F\rightarrow G$ for each object of $\mathcal{C}$, i.e. for each set. Maybe my set theory fu is just weak.
I could of course just be missing something basic here. I'm asking because the books I've checked tend to just state $\mathcal{C}$ must be small. Apologies if this is all trivial!
| https://mathoverflow.net/users/8415 | Example of what goes wrong with the functor category $D^C$ if $C$ is not small? | Hi,
let C be a discrete category (i.e. the only morphisms in C are the identity morphisms) and let D be the category consisting of two objects 0 and 1 and (apart from the identity morphisms) two parallel arrows $0\rightrightarrows 1$. Consider the constant functors F with value 0 and G with value 1. A natural transformation $F\to G$ corresponds to the choice of an element in a set of two elements for every object of C so that the conglomerate of all natural transformations $F\to G$ can be identified with the power conglomerate of Ob(C). This is not a set unless Ob(C) is a set.
| 10 | https://mathoverflow.net/users/2308 | 39075 | 25,056 |
https://mathoverflow.net/questions/39074 | 0 | i am currently reading the Probabilistic robotics book where the filters are discussed.
Such filters as kalman filter or particle filters.
Now I can understand one thing while reading about the Kalman filter.
First I want to say that I could successfully understand about Bayes filtering.
I've read some of theory of random processes and I can understand it.
Let me give you some details about the problem, if you think it is not sufficient I will give more. Who knows about that book I write here the page: it is about Kalman filter at page 41.
Let me not to explaing the whole problem, I hope the reader could be able to understand it as it is related with Kalman filtering. However I will write more if it is needed.
1. The state transition probability $p(x\_t|u\_t, x\_{t-1})$ must be a linear function in its arguments with added Gaussian noise. This is expressed by the following equation:
$$x\_t = A\_t x\_{t-1} + B\_t u\_t + \epsilon\_t (1)$$
$x\_t$ and $x\_{t-1}$ are state vectors, and $u\_t$ is a control vector at time $t$.
$\epsilon\_t$ is a gaussian noise.
Also it is given the definition of multivariant normal distribution:
$$p(x) = det\left(2\pi\Sigma\right)^{-1/2}exp\left(-1/2(x-\mu)^T\Sigma^{-1}(x-\mu)\right) (2)$$
Where the $\mu$ and $\Sigma$ are the mean and covariance.
The $x\_t$ and $u\_t$ are of the form:
$$x\_t = \left(x\_{1,t}, x\_{2,t}, ..., x\_{n,t}\right)^T$$
$$u\_t = \left(u\_{1,t}, u\_{2,t}, ..., u\_{m,t}\right)^T$$
So it is said that in order to obatain the state transition probability $p(x\_t|u\_t, x\_{t-1})$ you need to plugg the equation $(1)$ to the multivariant normal distribution (2).
Ok, I got that, but it is also written there that (here is the problem):
The mean of the posterior state is given by: $A\_tx\_{t-1}+B\_tu\_t$ and the covariance by $R\_t$
What I can't understand is that how does the mean in the multivariant normal distribution equation $(2)$ is calculated to $A\_tx\_{t-1}+B\_tu\_t$?
The whole formula affter plugging $(1)$ to $(2)$ become:
$$p(x\_t|u\_t, x\_{t-1}) = det\left(2\pi R\_t\right)^{-1/2}exp\left(-1/2(x\_t-A\_tx\_{t-1}-B\_tu\_t)^TR\_t^{-1}(x\_t-A\_tx\_{t-1}-B\_tu\_t)\right) $$
I think that the mean of $x\_t$ should be calculated like this:
$$E{x\_t} = E(A\_tx\_{t-1}+B\_tu\_t+\epsilon\_t)$$
$$E{x\_t} = A\_tE(x\_{t-1})+E(B\_tu\_t)$$
And I can't understand anyway how does the
$Ex\_t$ is equal to $A\_tx\_{t-1}+B\_tu\_t$
If you see any mistakes in my reasoning please tell me. If you need more details please comment this question for that. Thank you very much! Hope you help!
EDIT 1:
From book it is said that Kalman Filter (KF) is an implementation of Bayes Filter(BF).
I understood BF. Actually it calculates belief $bel(x\_t)$ at time $t$ from belief at time $t-1$.
BF algorithm:
```
For all $x_t$:
```
$\hat{bel}(x\_t) = \int{p(x\_t|u\_t, x\_{t-1})bel(x\_{t-1})dx\_{t-1}}$
$bel(x\_t) = \etap(z\_t|x\_t)\hat{bel}(x\_t)$
```
end for
return $bel(x_t)$
```
So about KF:
$$\hat{bel}(x\_t) = \int{p(x\_t|x\_{t-1}, u\_t)bel(x\_{t-1})dx\_{t-1}}$$
where $bel(x\_{t-1})$ is represented by mean $\mu\_{t-1}$ and covariance $\Sigma\_{t-1}$
The state transition probability $p(x\_t|x\_{t-1}, u\_t)$ is given as a normal distribution over $x\_t$ with mean $A\_tx\_{t-1}+B\_tu\_t$ and covariance $R\_t$.
If $x\_t$ can't be observed directly, so then what is $E(x\_t)$?
| https://mathoverflow.net/users/3195 | kalman filter: understanding the mathematical part | Do you really mean $Ex\_t$? That's the unconditional mean without looking at any of the data, which is a constant. Normally, the Kalman filter tells you how to compute the conditional mean based on the data you have at a particular moment in time.
I'm not familiar with the book, but I assume that you mean what the [Wikipedia page](http://en.wikipedia.org/wiki/Kalman_filter) calls the predicted state estimate.
You don't say anything about measurement error. Do you observe $x\_t$ exactly? By assumption, since you choose $u\_t$ you know it at time $t-1$. Then the formula is just telling you that
$$
E(x\_t | u\_t, x\_{t-1}) = A E(x\_{t-1} | u\_t, x\_{t-1}) + B E(u\_t | u\_t, x\_{t-1}) + E(\epsilon\_t | u\_t, x\_{t-1}).
$$
Since $E(x\_{t-1} | u\_t, x\_{t-1}) = x\_{t-1}$ and $E(u\_t | u\_t, x\_{t-1}) = u\_t$, and the noise is independent,
then
$$
E(x\_t | u\_t, x\_{t-1}) = A x\_{t-1} + B u\_t.
$$
If you don't observe $x\_t$ exactly, then you observe it with some error, given by
$$
z\_t = H\_t x\_t + \nu\_t,
$$
where $z\_t$ is your observation at time $t$ and $\nu\_t$ is again Gaussian white noise, independent of everything else.
Now, at time $t-1$ you only know $z\_1, \ldots, z\_{t-1}$, so the best you can do is
$$
E(x\_t | u\_t, z\_1, \ldots, z\_{t-1}),
$$
which I'll write as
$$
E\_{t-1} (x\_t),
$$
since it's the conditional mean given all information at time $t-1$ (including $u\_t$, which you choose at time $t-1$.
Using the definition,
$$
E\_{t-1}(x\_t) = A E\_{t-1}(x\_{t-1}) + B E\_{t-1}(u\_t) + E\_{t-1}(\epsilon\_t),
$$
but the most this simplifies is to
$$
E\_{t-1}(x\_t) = A E\_{t-1}(x\_{t-1}) + B u\_t.
$$
With measurement error, the Kalman filter by design just works by calculating the two conditional means $E\_{t-1}(x\_t)$ and $E\_{t-1}(x\_{t-1})$, and no other conditional or unconditional means.
| 4 | https://mathoverflow.net/users/3711 | 39082 | 25,059 |
https://mathoverflow.net/questions/39077 | 3 | I am not sure if this is a proper place for my question, but:
can anybody recommend any good online latex editor?
---
Anyone interested in this would probably benefit from the answers to this question on the tex.SE site: <https://tex.stackexchange.com/questions/3/compiling-documents-online>
| https://mathoverflow.net/users/3840 | Online latex editor | Have a look at [Scribtex](https://www.scribtex.com/).
| 4 | https://mathoverflow.net/users/3356 | 39087 | 25,063 |
https://mathoverflow.net/questions/39061 | 32 | According to Wikipedia, the [Bohr-Mollerup Theorem](https://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem) (discussed previously on MO [here](https://mathoverflow.net/questions/23229/importance-of-log-convexity-of-the-gamma-function)) was first published in a textbook. It says the authors did that instead of writing a paper because they didn't think the theorem was new.
What other examples are there of significant theorems that first saw the light of day in a textbook? (I'm assuming Wikipedia is right about Bohr-Mollerup.)
I recognize that the word "significant" is imprecise; I have in mind theorems that mathematicians have picked up on and used in their own work, but I'm open to other interpretations.
| https://mathoverflow.net/users/3684 | Theorems first published in textbooks? | I recall that, and Wikipedia independently confirms that L'Hôpital's rule first appeared in a textbook, apparently the first textbook on differential calculus: [*Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes*](https://en.wikipedia.org/wiki/Analyse_des_Infiniment_Petits_pour_l%27Intelligence_des_Lignes_Courbes) published by Guillaume de l'Hôpital and made up of content mostly provided by Johann Bernoulli, who was on retainer to l'Hôpital, more or less, for this purpose.
| 26 | https://mathoverflow.net/users/7505 | 39101 | 25,074 |
https://mathoverflow.net/questions/39096 | 18 | I have written a paper, which includes an appendix discussing how to obtain numerical evidence for the result of the paper. Now the computation essentially works as follows:
* Create a large tridiagonal matrix.
* Compute its eigenvalues.
* Compute the difference of consecutive eigenvalues, and output it.
The implementation of such an algorithm is rather straightforward, but in order to look at large matrices, I started using algorithms from a package called [LAPACK](http://en.wikipedia.org/wiki/LAPACK), which turned out to be faster then regular algorithms provided by Matlab. (I'm no specialist, so not exactly sure what happens).
I am curious if one should provide the source code for such a computation, and if yes in what form. I cam up with the following options:
* Pseudocode (as above)
* Simplified matlab, that works with any installation of matlab, but is too slow to actually do the computations.
* The real code, which most people will not be able to get to run without some effort.
I am also curious if one should include some sort of source code in the paper, and if yes, in what form? Or what people have done in such a case...
The simplified code is available at:
<http://math.rice.edu/~hk7/ftp/matlab_code/SkewSpecDense.m>
I have not put the real code online, because it requires external packages, and I am not sure how easy it is to install them...
| https://mathoverflow.net/users/3983 | How to distribute the source of programs used in a paper? | My preference is detailed pseudocode, at a high-enough level of abstraction to allow understanding the algorithm.
Of course, as pointed out by Ryan Budney's comment, it depends strongly on what the journal requirements are and in which journal you publish. However, I feel strongly that the complete code-set which you use should be available from some resource, either through the journal article's publsher, or from your own website, your academic website, or via Arxiv.
If the pseudo-code is detailed enough to allow reimplementing the algorithm straightforwardly by another mathematician, then that should be sufficient.
If the pseudo-code has to leave out certain details which are germane to the computation, then the interpreted code which implements the algorithm in a numerical computational package (such as Maple, Matlab, Sage, or [Octave](http://www.gnu.org/software/octave/) or [Scilab](http://en.wikipedia.org/wiki/Scilab) ([download link](http://www.scilab.org/) ) which are free open source software packages capable of running code similar to or equivalent to matlab) should be provided.
Why not provide both? -- If you can provide a link to your own webpage for the paper, or for its supporting supplemental materials, I don't see why you couldn't *provide both* the interpreted code and the compileable C or C++ code on your webpage, unless there are copyright issues involved such as if you did not write all of the code yourself and do not have the right to release all of the code source. I am a supporter of free open-source software and the [Gnu](http://gnu.org) organization's [GPL licensing](http://www.gnu.org/licenses/gpl.html), which would allow others to benefit from your code and to contribute back to it via incremental improvements.
I suggest that you **specify which version** of software package, operating system, compiler, and/or library you used in running your program or in creating the binary application from your code. This is necessary because different versions of Octave (2.3 vs. 3.0) or Matlab (R10, R13, etc.) or any software package may implement or include different routines and may not be capable of correctly running your software program.
I would recommend that if particular packages are necessary in order to run the interpreted code in Octave or Matlab that you list which packages they are. In the same vein, if your C or C++ code requires particular libraries such as LAPACK or BLAS, make sure to list them in a text file or in a header file. If you know how to use the [make](http://en.wikipedia.org/wiki/GNU_make) program, you can create a makefile to help others in compiling your software.
The **make** program, the Gnu compiler collection, and many other development tools are all standard parts of Gnu/Linux distributions, such as Debian.
My preference is detailed pseudocode, at a high-enough level of abstraction to allow understanding the algorithm.
| 9 | https://mathoverflow.net/users/8676 | 39102 | 25,075 |
https://mathoverflow.net/questions/39100 | 8 | I am interested in the following problem: I have a finite field $F\_q$, two positive integers
$n>m$ and elements $a\_1,...,a\_m\in F\_q$. How many of the polynomials
$x^n+a\_1x^{n-1}+...+a\_mx^{n-m}+c\_{m+1}x^{n-m-1}+...+c\_n,c\_i\in F\_q$ are irreducible?
What are the best known estimates, esp. for $q$ fixed and $m,n\to\infty$?
| https://mathoverflow.net/users/9304 | Number of irreducible polynomials with some coefficients fixed over a finite field | This is similar to counting irreducibles in arithmetic progressions modulo $x^m$ (once you replace $x$ by $1/x$). You can turn the problem into counting rational points on a curve (coming from a "cyclotomic function field" in the sense of Carlitz) over $F\_{q^n}$ and get an estimate $q^n/n + O(gq^{n/2})$, where $g$ is the genus of the curve. Unfortunately $g$ grows like $mq^m$ so you only get good estimates for $m$ small and nothing when $m$ gets close to $n/2$. There are plenty of papers on this (e.g. by S. Cohen). There is also some experimental work by Panario et al.. Mathscinet should help you locate these.
| 10 | https://mathoverflow.net/users/2290 | 39108 | 25,080 |
https://mathoverflow.net/questions/38727 | 14 | This is a question I wonder a little about every now and then.
It is immediate, using forcing, that if there is a transitive set model of set theory, then there are continuum many.
>
> Can one prove a weak version of this without using the forcing machinery?
>
>
> (Perhaps in the presence of reasonable large cardinal assumptions?)
>
>
>
Here are some specific versions:
1. Suppose we know there are two transitive models of set theory. Can we prove there are infinitely many?
2. Suppose we know there are two transitive models of set theory. Can we prove there are two with the same height?
3. Suppose we know there is an uncountable model. Are there continuum many?
I'm going to leave "reasonable" loose, but we do not want to assume much. For example, if there is a transitive model of "there is a measurable cardinal" then one can (easily) check that there are continuum many countable transitive models of set theory. There are also a few more or less obvious observations in the same spirit that follow from $\Sigma^1\_2$ absoluteness.
Also: if there is a countable transitive model $M$ of set theory, there is a comeager set of reals $C$ and a measure 1 set $R$ such that if $x$ is in $C\cup R$, then $M[x]$ is a model of set theory. Any weakening of this or something similar in spirit that can be established without forcing would also be welcome.
Now: I do not think I want something where we do forcing in disguise. So I am not sure presenting forcing as some variant of Bairwise compactness or that sort of thing would be appropriate here.
Of course, any references you think I should be aware of are more than welcome.
| https://mathoverflow.net/users/6085 | Number of transitive models of set theory | Suppose you could prove, in ZFC, without forcing, the statement
(A) If there are two transitive models of ZFC, then there is a third.
Then you could also prove, in ZFC, without forcing, the statement
(B) If there are two transitive models of ZFC, then there is a transitive model of ZFC + $V\neq L$.
[Proof: Work in ZFC and assume there are two transitive models of ZFC. By (A) there is a third. If one of them satisfies $V\neq L$ we're done, so assume all three satisfy $V=L$ and are therefore of the form $L\_\xi$ for some ordinals $\xi$. Let $\alpha<\beta<\gamma$ be the first three ordinals occurring as heights of transitive models of ZFC (+ $V=L$). Since $L\_\gamma$ sees both $L\_\alpha$ and $L\_\beta$, it must, by (A) again, see another transitive model of ZFC. By minimality of $\alpha<\beta<\gamma$, $L\_\gamma$ can't see another model of the form $L\_\xi$, so it must see a model of ZFC + $V\neq L$. (I've tacitly used that the notion of "transitive model of ZFC" is absolute between the real world and transitive models of ZFC.)]
What does this have to do with the question? I claim that a non-forcing proof of (B) would be significant news --- it would say that people could have deduced the consistency of $V\neq L$ from highly plausible assumptions before Cohen. So I feel reasonably confident in saying that no ZFC proof of (B) without forcing is known. Therefore, by the argument above, no ZFC proof of (A) without forcing is known.
[Concerning "highly plausible," note that the existence of lots of transitive models of ZFC follows from assumptions just slightly beyond ZFC itself. My favorite such extension of ZFC is to add a satisfaction predicate for formulas in the language of ZFC, add axioms saying this predicate obeys the usual recursive definition of satisfaction, and allow this new predicate in the replacement scheme of ZFC. With these assumptions, you can apply a L"owenheim-Skolem argument to get lots of elementary submodels of the universe.]
Note that the same discussion goes through if, in (A) we replace the conclusion (that there exists a third model) with the statement that there are two transitive models of ZFC of the same height.
| 17 | https://mathoverflow.net/users/6794 | 39112 | 25,084 |
https://mathoverflow.net/questions/39113 | 2 | It's well known that in a Hilbert space, good inequalities exist concerning the norm due to the existence of inner product.Now let X be a general Banach algebra, are there good inequalities concerning the norm? To be precise, let's consider an example, let X be a commutative Banach algebra with identity I,is the following claim ture or not(especially when X is infinite dimension)?
Either for every element b in X with norm 1, we have the norm of b^2 is also 1, or inf ||b^2||=0, with b running over all elements in X with norm 1.
P.S.This problem is derived from a question concerning the existence of a nilpotent element in X, in other words, the linear span of all the multiplicative linear functionals may not equal to the dual space of X.
| https://mathoverflow.net/users/9305 | Are there good inequalities on the norm? | The way it's formulated, the claim can fail in the finite-dimensional case. For example, consider $\ell^1(\mathbb{Z}\_p)$. Then if we take an element $a$ of norm 1, $\sum\_{k=1}^p|a\_k|=1$. This implies that there is $k$ with $|a\_k|\geq1/p$. Then $\|a^2\|\geq1/p^2$ (it's likely that a sharper inequality can be found, but that's not necessary to answer your question).
Edit: on the suggestion of Yemon, we now know how to provide an infinite dimensional counterexample. So let $A\_0$ be the algebra $\mathbb{C}^2$ with the norm $\|(\lambda,\mu)\|\_1=|\lambda|+|\mu|$. As mentioned in the first paragraph, this algebra has the property that if $\|a\|=1$, then $\|a^2\|\geq1/2$, and this bound is achieved. And now construct $A=\ell^\infty(\mathbb{N},A\_0)$ with the supremum norm. This one is infinite-dimensional, and it still has the same lower-bound-for-the-square property.
| 2 | https://mathoverflow.net/users/3698 | 39118 | 25,089 |
https://mathoverflow.net/questions/30140 | 20 | There are many situations which arise where one might consider different Model categories with the same underlying category. For example in (left) Bousfield localization you start with a model category M and construct a new model category structure on M with the same cofibrations, but with more weak equivalences and fewer fibrations. In these settings the identity adjoint equivalence often serves as a Quillen pair relating the Model categories.
One classic example is when the underlying category is a the category of functors from a (small) category C into a (nice) model category M. In that case, we have two naturally occuring model structures: the projective and the injective model structures. In some cases, when the source category C is a Reedy category, there is a third model structure on D=Fun(C,M) known as the Reedy model structure. All three of these model category structures are Quillen equivalent. In fact the weak equivalences are exactly the same. They are the level-wise weak equivalences.
In some sense which I'm trying to make precise, the injective and projective model structures on on opposite sides of the spectrum. The Reedy model structure is something like a mix of these two. In fact the model structures with a fixed set of weak equivalences should form a poset. For example we can just look at the set of cofibrations. Afterall the cofibrations and the weak equivalences determine the model structure if it exists. Then the model category structures with a fixed set of weak equivalences are ordered by inclusion (of sets of cofibrations). In the above example we have the ordering: $$Proj \subseteq Reedy \subseteq Inj$$
So this raises the question, what is known about different model structures on a category with a fixed set of weak equivalences? **Is there always a maximal/minimal model structure?** If not, are there some conditions which ensure its existence? **Are there properties which characterize these model structures?** Since the weak equivalences are always the same, then the identity functor should induce an isomorphism on homotopy categories. Thus if two such model structure are comparable (so that the identity adjoint equivalence is a Quillen pair) then they are Quillen equivalent. So if there is a minimal or maximal model structure, they all these model structures are Quillen equivalent via a zig-zag of Quillen equivalences. Is this the case?
| https://mathoverflow.net/users/184 | How many model categories have the same weak equivalences? | Tibor Beke has some comments on this too.
<http://faculty.uml.edu/tbeke/cofib.pdf>
| 7 | https://mathoverflow.net/users/3075 | 39126 | 25,094 |
https://mathoverflow.net/questions/36377 | 9 | To a morphism of sets $f\colon E\to B$ with finite fibers, one may assign a function $$|f^{-1}|\colon B\to{\mathbb N}$$ sending an element $b\in B$ to the cardinality of the fiber $f^{-1}(b)$.
---
To a proper morphism of manifolds imbedded in Euclidean space $f\colon E\to B$ one may assign a function $$|f^{-1}|\colon B\to{\mathbb R}$$ by sending an element $b\in B$ to the volume of the fiber $f^{-1}(b)$. ... Or we could assign the dimension of the fiber (valuing in ${\mathbb N}$), or we could assign the number of points in the fiber (valuing in ${\mathbb N}\cup${$\infty$}).
---
If $M$ is a commutative monoid (thought of as a set with addition operation), let $Set\_{/M}$ denote the category of sets equipped with a map to $M$. Then to a morphism $f\colon E\to B$ in $Set{/M}$, where $f$ has finite fibers, one may assign a function $$|f^{-1}|\colon B\to M$$ sending an element $b\in B$ to the sum of the elements in the fiber. ... Or we could use a possibly non-commutative monoid $M$ but replace Sets-over-$M$ with Sequences-over-$M$.
---
To a discrete op-fibration of categories $f\colon E\to B$ one can assign a functor $$|f^{-1}|\colon B\to Set$$ sending $b$ to its fiber.
---
---
Question: What do all these have in common? More specifically, where can I find some category-theoretic way to understand situations of this type? The "type" here seems to be something like: a "finite type" morphism in a concrete category with "valuation" can be converted into a "valuation" on the base.
One might call it "integration along the fiber" or "gysin" or "sheaf-to-function correspondence." What is the generalized setup? What is the notion of "valuation" or "measure" supposed to be?
| https://mathoverflow.net/users/2811 | Bundle-to-function correspondence | This answer comes by private correspondence from Mathieu Anel. I record it here, with some minor clean-up, because it's exactly what I was looking for. --David Spivak
---
Here are some thoughts about what I've understood of your question.
We suppose that $f:E\to B$ is a (kind of) fibration.
If there exists a moduli object $M$ for fibrations, $f$ is classified by a function $[f]:B\to M$ (I've changed your notation).
Now I assume you want the map $f\mapsto [f]$ compatible with addition and products of fibers like $[f\times\_Bg]=[f][g]$.
Remark : $M$ has a monoidal (or even a rig) structure iff the fibrations have a monoidal (or rig) structure.
for example, discrete and Grothendieck fibrations are stable by disjoint sums and pull-backs and their moduli objects ($Sets$ and $Cat$) are rigs.
other example, the direct sum and tensor product of vector bundles.
in each case $f\mapsto [f]$ is always compatible with the operations (it's tautological).
Now compose with any rig morphism $g:M\to R$ where $R$ is any rig (e.g. a ring) and $g[f]$ is a "measure." And, $[f]$ is the "universal measure."
Examples:
1. $M$ = moduli of (finite) sets $M\to {\mathbb N}$ = cardinality
2. $M$ = moduli for (finite dimensional) vector bundles $M\to {\mathbb N}$ = dimension (this one is compatible only with tensor product)
3. $M$ = moduli of compact riemannian manifolds $M\to {\mathbb R}$ = volume
4. K-theory : $M$ is anything that is a rig, and $g:M\to \Pi\_0(M)^+$ (where $^+$ is the additive group completion)
5. $M$ = moduli for l-adic sheaves, $M\to R$ = trace of the frobenius operator
| 5 | https://mathoverflow.net/users/2811 | 39130 | 25,096 |
https://mathoverflow.net/questions/39121 | 1 | I remember hearing somewhere that strongly irreducible Heegaard surfaces in hyperbolic 3-manifolds "look like" fibers.
I know that Otal's result about short geodesics in hyperbolic mapping tori being unlinked with respect to fibers has an analogue in the setting of strongly irreducible Heegaard surfaces.
Can someone tell me some other similarities between strongly irreducible Heegaard surfaces and fibers in hyperbolic 3-manifolds? Or why someone would say that strongly irreducible Heegaard surfaces look like fibers? Any references would be greatly appreciated.
| https://mathoverflow.net/users/4325 | Why do strongly irreducible Heegaard surfaces look like fibers? | Something more basic is true. Strongly irreducible Heegaard splittings act a lot like incompressible surfaces (of which fibers are a special case). Here are two results as evidence.
First, suppose that the three-manifold is equipped with a triangulation. Then Haken showed that incompressible surfaces can be normalized. That is, isotoped to intersect every tetrahedron in a collection of standard disks; normal triangles and normal quads. Rubinstein (and Stocking) showed that strongly irreducible splittings can be "almost normalized"; in every tetrahedron it has normal disks and in at most one tetrahedron there is an almost normal annulus or octagon. Michelle Stocking's thesis is a standard reference for this material. Many people (Rubinstein, Hass, Bachman, Scott, ...) will say in public that the above is a PL version of an analytic truth: incompressible surfaces can be isotoped to be minimal surfaces of index zero while strongly irreducible surfaces can be made minimal of index one.
Second, suppose that $S, T$ are surfaces, with $T$ incompressible. If $S$ is also incompressible then it is an exercise in innermost disks to show that, after isotoping $S$ to meet $T$ minimally, all curves of intersection are essential on both surfaces. If $S$ is instead strongly irreducible then a one-parameter sweep-out argument, followed by an innermost disk argument shows that there is some position of $S$ where all curves of intersection are essential on both surfaces.
Ok, third (I couldn't resist), suppose that the ambient three-manifold is hyperbolic. Minsky's approach to the ending lamination conjecture says that the geometry "around" the incompressible surface or a strongly irreducible splitting can be modelled using "blocks" based on the four-holed sphere or once-holed torus, where the blocks are glued to each other vertically using pairs of pants and horizontally using solid tori (Margulis tubes).
| 5 | https://mathoverflow.net/users/1650 | 39140 | 25,105 |
https://mathoverflow.net/questions/39148 | 12 | Suppose $A$ is an $m\times n$ real matrix and we need to find $\left\|A\right\|\_p$ for $p \notin \{ 1, 2, \infty \}$. What is the most efficient way to compute $\left\|A\right\|\_p$?
Here's one naive approach I can think of. Sample random points $\left\|x\right\|$ on the unit hypersphere , computing $\left\|Ax\right\|\_p$ for each such and take the maximum. What I would like to know is the runtime of this approach for the "average" A, and how we can optimize this for special classes of matrices (like Diagonal, Orthonormal, etc.)?
| https://mathoverflow.net/users/6495 | Efficiently computing a matrix's induced p-norm | On the negative side, there is a [result](http://arxiv.org/abs/0908.1397) by myself and Julien Hendrickx that the matrix $p$-norm is NP-hard to approximate whenever $p$ is not $1,2,$ or $\infty$.
On the positive side, the [M.S. thesis of Daureen Steinberg](http://www2.isye.gatech.edu/~nemirovs/Daureen.pdf) has an efficient algorithm for computing the $p$-norm of a nonnegative matrix (see Remark 3.4 on page 48).
| 10 | https://mathoverflow.net/users/9316 | 39151 | 25,111 |
https://mathoverflow.net/questions/39141 | 8 | Let $\Omega\subset \mathbb{R}^n$ open and bounded. The Poincaré inequality
$$\|u\|\_p \le C \|\nabla u\|\_p$$
($\|\cdot\|\_p$ denotes the usual $L^p(\Omega)$-norm; the Lebesgue measure shall be used here)
is valid in the following cases:
* $u$ is zero on $\partial\Omega$ (in the $W^{1,p}$-sense)
* $u$ has an average value of zero (this implies an estimate of $\|u-\bar u\|\_p$, but not of $\|u\|\_p$ for general $u$)
* $\mu := |\{u=0\}| > 0$ (Lebesgue measure), with a constant that blows up as $\mu\to 0$.
Of course the inequality cannot hold in general (take $u$ to be a constant function), but the only obstruction seems to be that $u$ can be far away from zero. Therefore it should be true that the inequality holds whenever $u$ is zero somewhere in $\Omega$ (in a suitable sense, for example, zero is contained in the essential range of $u$).
So the question is: Is there a version of the inequality for this case?
| https://mathoverflow.net/users/8794 | Version of the Poincaré Inequality | The Poincaré inequality need not hold in this case. The region where the function is near zero might be too small to force the integral of the gradient to be large enough to control the integral of the function.
For an explicit counterexample, let $$\Omega = \{(x,y) \in \mathbb{R}^2 : 0 < x < 1, 0 < y < x^2\}$$ be the region under the graph of a parabola, and take $p=2$. For $\epsilon > 0$, let $$u\_\epsilon(x,y) = \begin{cases} x/\epsilon, & x < \epsilon \\\ 1, & x \ge \epsilon \end{cases}.$$ $0$ is in the essential range of each $u\_\epsilon$, but one can easily verify $||\nabla u\_\epsilon||\_2^2 = \epsilon/3 \to 0$ as $\epsilon \to 0$, whereas $||u\_\epsilon||\_2^2 \to |\Omega| = 1/3$.
**Edit:** To address Piero's comment, the irregularity of the domain is not the main issue here. For another counterexample, let $\Omega$ be the unit ball in $\mathbb{R}^d$, $d \ge 3$, and take
$$u\_\epsilon(x) = \begin{cases} \frac{|x|^2}{\epsilon^2}, & |x| < \epsilon \\\ 1, & |x| \ge \epsilon \end{cases}.$$ Using polar coordinates, one easily computes $||\nabla u\_\epsilon||\_2^2 \sim \epsilon^{d-2}$ while again $||u\_\epsilon||\_2^2 \to |\Omega|$.
What's really the problem is that the set where $u$ vanishes has zero capacity. If you can control from below the capacity of the set where $u$ vanishes, then you can get a Poincaré inequality.
Indeed, the following theorem may be what the asker wants: for "reasonable" $\Omega$ (Lipschitz suffices), if $\mathrm{Cap}(\{u = 0\}) \ge \delta$, then $||u||\_2^2 \le \frac{C}{\delta} ||\nabla u||\_2^2$. The precise statement (for any value of $p$) can be found in section 4.5 of William P. Ziemer's *Weakly Differentiable Functions* (along with the definition of capacity).
Note in particular that $\mathrm{Cap}(E) \ge m(E)$, so it's enough to control the Lebesgue measure. This is your bullet point 3.
| 13 | https://mathoverflow.net/users/4832 | 39154 | 25,113 |
https://mathoverflow.net/questions/39157 | 3 | Often when computing in category theory, one has to show that some square is cartesian. Depending on the number of maps involved, and their arrangement, it's somewhat difficult to write down exactly all of the relationships between the various squares.
Take for instance the following problem (please don't answer it, I've already proven it):
We have a commutative diagram
$$\begin{matrix}C\_3 &\to& C\_2 &\to& C\_1\\
\downarrow&&\downarrow&&\downarrow\\
D\_3&\to &D\_2&\to& D\_1
\end{matrix}$$
Label the right vertical map $p$, and assume we further have a commutative diagram
$$\begin{matrix}\ast &\to& C\_1\\
\downarrow&&\downarrow\\
D\_3&\to& D\_1
\end{matrix}$$
Where map on the top of the square is called $x$ (it classifies a point $x$, and the left vertical map is called $\sigma$. The bottom map and right map are the same maps from the first diagram $D\_3\to D\_1$ is simply the composite of the two bottom maps in the first diagram.
We would like to show that the canonical map between fibers $$C\_3\times\_{D\_3\times\_{D\_1} C\_1} \{\sigma,x\}\to C\_2\times\_{C\_1} \{x\}\times\_{D\_2\times\_{D\_1} \{px\}}\{e\}$$ is a pullback of the canonical map $C\_3\to C\_2\times\_{D\_2} D\_3$ (if you really care to know, this is to show that the first map is a trivial fibration, since we knew at the time that the second map was a trivial fibration).
Note: The vertex $\{e\}$ is the image of $\sigma$ in $D\_2$.
The diagram I drew to realize this was a sheet of four cartesian squares with a cartesian cube attached at the top left square. This is before even taking fibers.
Another example: If anyone's following Ravi Vakil's notes for his schemes course, there is a similar, albeit substantially less complicated problem to prove that the "magic square" is cartesian. I did a similar computation, which ended again with the surprising cartesianness ultimately arising from a cartesian cube attached to a cartesian sheet.
Then the question: Is there a more efficient way to verify claims like this (i.e. without drawing out "every single possible pullback"?
| https://mathoverflow.net/users/1353 | Efficiently computing with pullbacks and pushouts | Try doing it in sets; then you can usually write it out "syntactically". It looks like you've got a map
$$ \{c\_3\in C\_3|p(c\_3)=\sigma, fg(c\_3)=x\} \to \{c\_2 \in C\_2| f(c\_2)=x \}$$
sending $c\_3\mapsto g(c\_3)$
and you want to know it's the pullback of the map
$$ \{c\_3\in C\_3\} \to \{(c\_2,d\_3)| p(c\_2)=g(d\_3)\}$$
sending $c\_3\mapsto (g(c\_3),p(c\_3))$.
I'm writing "$p$" for all vertical maps, "$g$" for $C\_3\to C\_2$ and $D\_3\to D\_2$, and "$f$" for $C\_2\to C\_1$ and $D\_2\to D\_1$.
Having done this, your claim looks false: it looks like the pullback set is just $\{c\_3\in C\_3|gf(c\_3)=x\}$. But quite possibly I've misinterpreted your diagrams.
The point of this exercise is that, if the set theoretic argument holds, then you can prove the version in the category $C$ using the Yoneda lemma; apply $\mathrm{Hom}\_{\mathcal{C}}(T,{-})$ to everything, for arbitrary objects $T$, and observe you get a pullback of sets, therefore you must have a pullback in $\mathcal{C}$.
| 5 | https://mathoverflow.net/users/437 | 39166 | 25,117 |
https://mathoverflow.net/questions/39155 | 0 | Let $k$ be a field of char $0$ and let $\mathbb{Z}$ act on $\mathbb{A}^1\_k$ by the action induced by $G\to\mathrm{Aut}\_k(k[X]), n\mapsto X+n$. It is rather easy to show that the orbit space $\mathbb{A}^1\_k/\mathbb{Z}$ is just $\mathrm{Spec}(k)$. At least to me this is surprising at the moment since dividing out the analogue action of $\mathbb{Z}$ on the topological spaces $\mathbb{R}$ resp. $\mathbb{C}$ gives the sphere $S^1$ (up to homotopy).
**Q.:** Is that simply a pathological example of a categorical quotient or is there a geometric explanation why this is the right orbit space?
| https://mathoverflow.net/users/2146 | Geometric explanation of an orbit space: Integer action on the affine line | Your observation $\mathbb{A}^1/\mathbb{Z}=Spec(k)$ refers to the fact that there are no non-constant, $\mathbb{Z}$-invariant **polynomial** functions.
On the other hand there are plenty $\mathbb{Z}$-invariant **continues** functions on $\mathbb{A}^1(\mathbb{R})$, hence you get a non-trivial quoteint.
Also note that $\mathbb{Z}$ is no group-scheme on the nose. Taking
the group ring yields $k[\mathbb{Z}] = k[t,t^{-1}]$ which is a torus.
So to get a group-scheme $G$ with $G(k)=\mathbb{Z}$ you need to consider
an infinite union of copies of $Spec(k)$, which would be not finitely generated.
There is no good way to take quotients of schemes by infinite, discrete groups in general.
For example there is no way to produce curves as quoteints of the upper half plane $\mathbb{H}/\Gamma$, or abelian varietis as quotients $\mathbb{C}^n/ \Lambda$, in the realm of scheme theory.
| 2 | https://mathoverflow.net/users/5714 | 39167 | 25,118 |
https://mathoverflow.net/questions/39159 | 7 | Let A be a convex compact set in the plane (with a piecewise smooth boundary, say). We want to `inflate' it in such a way that the diameter does not increase.
More accurately, we are looking for all sets C such that
a) A is a subset of C;
b) diam(A)=diam(C)
Let now B is the largest possible set C which satisfies these two properties.
By `largest' I mean either that it m(B) = max m(C), where m is the Lebesgue measure; or that B actually contains any C with these properties. Let us call B the *isodiametric hull* of A.
The simplest example of A is of course the square: here B is the superscribed disc, and it is the isodiametric hull of A in the strong sense.
Another example is the equilateral triangle, for which B is the [Reuleaux triangle](http://mathworld.wolfram.com/ReuleauxTriangle.html). Similarly, for any regular 2n-gon we have the disc, and for any regular (2n+1)-gon its isodiametric hull is a Reuleaux polygon.
The first non-trivial example that comes to mind is an isosceles triangle that isn't equilateral. It is clear that the hull is always a set of constant diameter but how does one actually obtain it? It seems that its boundary - a curve of constant width - is not a finite union of circular arcs.
I wonder if all this is well known (being such a natural question!). In particular, does the isodiametric hull of a set always exists in the strong sense?
**Added:** of course, if there is no IDH in the strong sense, B may not be unique. Its area *is* unique, though. How does one find it?
| https://mathoverflow.net/users/8131 | Isodiametric hull |
>
> For any shape that is not constant width, there are many different maximal elements of the same diameter containing it.
>
>
>
Constant width shapes are maximal. They are regions swept out by line segments of length $D$ swept by their midpoints moving perpendicularly along curves perpendicular to along smooth curves (with cusps) whose tangent line turns 180 degrees
Any shape that is not constant width has some projections that do not have maximal diameter.
For each projection that has maximal diameter, there is a diameter as above perpendicular to the line of projection.
Any method of interpolating a positively turning curve between the existing diameters will create a maximal element of the set of shapes of the given diameter. There are always infinitely many ways to do this unless the set is already maximal --- any one solution for an interpolating curve can be perturbed anywhere it is not constrained.
This observation can be applied, for exmaple, to the case of squares mentioned in the question. For example, you can begin enlargement of a square by adding an arc of a circle centered at one corner and passing through an opposite corner, then take the convex hull. More systematically: the square only defines two diameters. There are many Legendrian sections of
tangent line bundle of plane $\to \mathbb {RP|^2}$ (cf. [Is the sphere the only surface all of whose projections are circles? Or: Can we deduce a spherical Earth by observing that its shadows on the Moon are always circular?](https://mathoverflow.net/questions/39127/is-the-sphere-the-only-surface-all-of-whose-projections-are-circles-or-can-we-d/39133#39133)) interpolating these two elements.
The total area is locally given by an easy formula, but globally it seems like a highly irregular function, since it depends on how the sweeping diagonal overlaps itself. I doubt if there is a good theory of a maximal area shape of the same diameter
containing the given one.
| 5 | https://mathoverflow.net/users/9062 | 39178 | 25,126 |
https://mathoverflow.net/questions/39165 | 3 | Consider the game "Ruler", which is defined as follows. We start with finitely many coins in a line. A move in this game consists of turning over any number of coins, but they must be consecutive, and the rightmost coin must be turned from heads to tails. Then the position in this game where a coin in the $n$th position is heads and all others are tails has Sprague-Grundy value given by
$$ g(n) = mex \{ 0, g(n-1) , g(n-1) \oplus g(n-2), \cdots, g(n-1) \oplus \cdots \oplus g(1) \} $$
From here, according to page I-31 of [Ferguson's game theory notes](http://www.math.ucla.edu/~tom/Game_Theory/comb.pdf), "it is easy to show" that $g(n)$ is the largest power of two dividing $n$.
Except it's not easy. But it's a nice fact and I'd like to be able to present a proof of it to my students.
*Fair Game* by Guy and *Winning Ways*, the two other sources I've seen this in, both state this without proof. It seems that it might be closely related to results about Gray codes - the partial sums $g(1), g(1) \oplus g(2), g(1) \oplus g(2) \oplus g(3), \cdots$ are a binary Gray code for the integers.
| https://mathoverflow.net/users/143 | Sprague-Grundy sequence for the ruler game | As in all combinatorial game theory problems, we want to use strong induction on $n$. So suppose we know it for $1, ..., n-1$, and let's prove it for $n$.
Write $n$ in the form $2^k(2x+1)$. Then the highest power of $2$ dividing $n-i$ for $i<2^k$ is the same as the highest power of $2$ dividing $i$, so it's pretty easy to see that by the time we hit $g(n-1)\oplus \cdots \oplus g(n-2^k+1)$ we have hit every nimber from $0$ to $2^k-1$ (by Gray codes, if you like).
Now we just have to show that the nimber $2^k$ never shows up in that mex. The key idea, I think, is that the order of the highest nonzero bit in the binary representation of $g(n-1)\oplus \cdots \oplus g(n-i)$ never decreases as $i$ increases. This is because of the fact that for any subsequence $g(a), g(a+1), ..., g(b)$ of the largest-power-of-two sequence, the largest value only occurs once (if it occurred twice, there would be a larger power of two halfway in between). Now we just notice that for $i=2^k$, we have $n-i = 2^k(2x+1)-2^k = 2^{k+1}x$, so $g(n-1)\oplus \cdots \oplus g(n-i) \ge 2^{k+1}$ for all $i$ greater than or equal to $2^k$.
Thus $2^k$ does not show up in the mex, while every smaller nimber does. The induction is done :)
| 5 | https://mathoverflow.net/users/2363 | 39182 | 25,129 |
https://mathoverflow.net/questions/39172 | 3 | Let $k$ be an algebraically closed field, and $V$ a normal (irreducible) affine variety over $k$. Does there necessarily exist a closed immersion $V \hookrightarrow \mathbb{A}^n$ of $V$ into affine space such that the closure of $V$ in projective space $\mathbb{P}^n$ is normal?
| https://mathoverflow.net/users/5094 | Making the projective closure of a normal affine variety normal | Yes:
Take a closure $\bar X$ of $X$ in some projective space. We can write $\bar X= X\cup D$, with $D$ ample. Normalize $\bar X$ to get new projective variety $\pi:\tilde X\to \bar X$. The preimage $\pi^{-1}D$ is ample with complement $X$ because $\pi$ is finite. So $\tilde X$ can be re-embedded in another projective space $\mathbb{P}^N$ so that $\pi^{-1}D$ is set theoretically the intersection of $\tilde X$ with a hyperplane $H$. Under the embedding of $X\subset \mathbb{A}^N=\mathbb{P}^N-H$, the closure is $\tilde X$ which is normal.
| 7 | https://mathoverflow.net/users/4144 | 39183 | 25,130 |
https://mathoverflow.net/questions/37161 | 16 | The von Neumann-Halperin [vN,H] theorem shows that iterating a fixed product of projection operators converges to the projector onto the intersection subspace of the individual projectors. A good bound on the rate of convergence using the concept of the Friedrichs number has recently been shown [BGM].
A generalization of this result due to Amemiya and Ando [AA] to the product of random sequences of projection operators drawn from a fixed set also shows convergence to the projector onto the intersection subspace.
My question is: are there any known bounds on the convergence rate for the latter problem analogous to the earlier one? In my application I'm only interested in the case of finite-dimensional Hilbert spaces.
[vN] J. von Neumann, Functional operators, Annals of Mathematics Studies No. 22, Princeton University Press (1950)
[H] I. Halperin, The product of projection operators, Acta. Sci. Math. (Szeged) 23 (1962), 96-99.
[BGM] C. Badea, S. Grivaux, and V. M¨uller. A generalization of the Friedrichs angle
and the method of alternating projections. Comptes Rendus Mathematique, 348(1–2):53–56, (2010).
[AA] I. Amemiya and T. Ando, Convergence of random products of contractions in Hilbert space, Acta. Sci. Math. (Szeged) 26 (1965), 239-244.
| https://mathoverflow.net/users/8629 | Random products of projections: bounds on convergence rate? | If you only care about the bound having the correct form, and don't mind obtaining constants that are much worse than the actual asymptotic convergence, then all you have to do is apply [BGM] to a subsequence. Specifically, let $k$ be the number of projections from which you sample, and let $p\_0, p\_1, \ldots, p\_{k-1}, p\_k = p\_0$ be a particular circular ordering of them. Given a random sequence $X\_i$ of projections, consider the initial segment $S(n)$ of $n$ projections, and define $L(n)$ such that $L(n) \ge 1$ if and only if $(p\_0, p\_1)$ occurs consecutively in $S(n)$, such that $L(n) \ge 2$ if and only if the consecutive pair $(p\_1,p\_2)$ occurs somewhere after $(p\_0, p\_1)$ in $S(n)$, such that $L(n) \ge 3$ if and only if that is somewhere followed by $(p\_2, p\_3)$, and so forth. For large values of $n$, the random variable $L(n)$ is tightly concentrated around a value close to $n/k^2$, and the convergence of, say, the segment $S(2k^2n)$ will, with high probability, be at least as good as the fixed cyclic ordering of length $n$.
The one technical lemma to prove is that you cannot lose by replacing each $p\_i$ in the fixed sequence by a product of projections that both starts and ends with $p\_i$.
| 6 | https://mathoverflow.net/users/7936 | 39189 | 25,135 |
https://mathoverflow.net/questions/38378 | 5 | Thinking about exotic 7-spheres, one can look at the maps $\cdots \rightarrow \Omega^2Diff(D^4, rel \space \partial) \rightarrow \Omega Diff(D^5, rel \space \partial) \rightarrow Diff(D^6, rel \space \partial)$. There are then homomorphisms $\cdots \rightarrow\pi\_2Diff(D^4, rel \space \partial) \rightarrow \pi\_1 Diff(D^5, rel \space \partial) \rightarrow \pi\_0Diff(D^6, rel \space \partial)$.
The map $\pi\_1 Diff(D^5, rel \space \partial) \rightarrow \pi\_0Diff(D^6, rel \space \partial)$ is onto by an appeal to a well-known theorem of Jean Cerf, so "$\pi\_1$ detects the exotic 7-sphere". But his theorem needs dimension at least 5, hence only applies to the right-most map.
Question: Can we lift up to $\pi\_2$? What can be said about the map $\pi\_2Diff(D^4, rel \space \partial) \rightarrow \pi\_1 Diff(D^5, rel \space \partial)$?
| https://mathoverflow.net/users/7867 | Exotic spheres detected in higher homotopy | My answer here is to just point to Ryan Budney's comments above - they seem to cover all that is known at present (ie, very little).
| 1 | https://mathoverflow.net/users/7867 | 39190 | 25,136 |
https://mathoverflow.net/questions/39037 | 0 | Hey guys,
I have a slightly imprecise question. I would like say something about a whole set of binary strings evaluated by a binary function by just looking at some type of average. The easiest example I can think of is probably a binary function $f: \{0,1\}^n \rightarrow \{0,1\}$ that is linear with $f(0) = 0$. Now in order to count the number of assignments resulting in $1$ I can do the following:
$1/2^n \* \sum\_{x \in \{0,1\}^n} f(x) = f(1/2^n \* \sum\_{x \in \{0,1\}^n} x) = f(1/2 e)$
where $e = (1,\dots, 1)$ is the all-one vector and thus $f(1/2 e) \* 2^n$ gives me the answer i am looking for.
I vaguely recall that I have seen something like this beforehand and I guess that there is something like a whole theory about this type of combinatorial argument out there. It is also somehow about inferring the structure of the boolean function by evaluating it at non-boolean inputs. For example, I think that this is part of the idea of the algebraization as a barrier to showing P != NP where one of the oracles get enhanced power by not only being able to evaluate a certain function at 0/1 assignments but also any other point contained in $[0,1]^n$.
I would really appreciate any pointers or references or just names for what I am actually looking for.
Thanks a lot,
Alberto
| https://mathoverflow.net/users/8994 | boolean functions and averaging / counting | There's a canonical way to extend a boolean function to the unit n-cube: you replace the boolean arguments by real numbers that are the probabilities of independent events, and the new output is the probability of the compound event defined by the original boolean function. For example, take the boolean function $(a,b,c)\mapsto (a\wedge b) \vee c$. If $a$, $b$, and $c$ are independent events with respective probabilities $p$, $q$, and $r$, then you can check that the probability of the event $(a\wedge b) \vee c$ is $pq +r -pqr$. I wrote a somewhat long post about this here: [Finding minimal or canonical expressions for Boolean truth tables](https://mathoverflow.net/questions/4930/finding-minimal-or-canonical-expressions-for-boolean-truth-tables/4951#4951)
I bring this up because if $f$ is an n-ary boolean function and $f^\\*$ is the probability version, then $f^\\*(1/2,...,1/2)$ is precisely the average value of $f$ over all possible boolean inputs. (That's obvious: if you flip a fair coin to determine the truth value for each of the n arguments of $f$, then by definition, $f^\\*(1/2,...,1/2)$ is the probability of $f$ evaluating to "true," but that probability is also clearly the average value of $f$ in this case.) That seems to be what you're looking for. (Unfortunately, *computing* $f^\*$ for any reasonably complicated function $f$ is pretty much intractable: indeed, there is no polynomial-time algorithm unless P = #P, by the simple fact stated above.)
| 0 | https://mathoverflow.net/users/302 | 39197 | 25,140 |
https://mathoverflow.net/questions/39201 | 5 | Let bigset(X) and rel(X,Y) be otherwise arbitrary formulas in the language of second-order arithmetic with the indicated variables free, and thmemberof(Z,x,X) be the formula asserting that X is the xth member of the sequence of sets coded by Z. Does it follow that second-order arithmetic proves
$((\exists X)(bigset(X)) \; \land \; (\forall X)(bigset(X) \implies (\exists Y)(bigset(Y) \; \land \; rel(X,Y)))) \implies$
$(\exists Z)(\forall x)(\exists X)(\exists Y)($
$bigset(X) \; \land \; bigset(Y) \; \land \; thmemberof(Z,x,X) \; \land \; thmemberof(Z,\operatorname{S}(x),Y) \; \land \; rel(X,Y))$
?
By second-order arithmetic, I mean Robinson arithmetic + full comprehension + the second-order induction axiom.
| https://mathoverflow.net/users/nan | Does second-order arithmetic prove every expressible instance of Dependent Choice? | Carl has pointed out that my previous answer missed a clause in the theorem I cited.
Simpson's book, Subsystems of Second Order Arithmetic, does address this in section VII.6. He shows that dependent choice for $\Sigma^1\_2$ formulas is equivalent to $\Delta^1\_2$ comprehension plus $\Sigma^1\_2$ induction (Theorem VII.6.9).
However even regular (non-dependent) $\Sigma^1\_3$ choice is independent of full comprehension; he attributes this result to Feferman and Levy, and cites Theorem 8 of Levy's "Definability in axiomatic set theory, II".
The result I mentioned before, that $\Sigma^1\_k$ dependent choice is equivalent to $\Delta^1\_k$ comprehension plus $\Sigma^1\_k$ induction for $k\geq 2$, holds for $k\geq 3$ requires the additional assumption that the universe is constructible from from some set of integers.
Your statement of dependent choice is a bit more complicated than necessary; you can fold bigset into rel (take $rel'(X,Y)$ to hold if either $\neg bigset(X)$ and $bigset(Y)$, or if $bigset(X)$, $bigset(Y)$, and $rel(X,Y)$). Conversely, it's a bit simpler than the version Simpson uses (which I believe is standard), in which $rel$ can depend on the parameter $x$ as well.
| 9 | https://mathoverflow.net/users/8991 | 39202 | 25,143 |
https://mathoverflow.net/questions/39207 | 2 | Is anybody know a solution of this problem?
(Sorry, correct question is [here](https://mathoverflow.net/questions/39210/solve-in-positive-integers-n-mm1).)
| https://mathoverflow.net/users/5712 | Solve in positive integers $n!=m^2$ | Bertrand's postulate (<http://en.wikipedia.org/wiki/Bertrand%27s_postulate>).
| 11 | https://mathoverflow.net/users/6153 | 39209 | 25,148 |
https://mathoverflow.net/questions/39211 | 2 | Let $f : X \to Y$ be an open faithfully flat morphism of schemes. In the text I'm reading ([Angelo Vistoli's notes on descent](http://homepage.sns.it/vistoli/descent.pdf)) it is claimed that then every point $x \in X$ admits an open neighorhood $U$ such that $f(U)$ is open and the morphism $U \to f(U)$ is quasi-compact. The latter is one of the possible definitons of a fpqc morphism. However, I don't understand at all why $U \to f(U)$ should be quasi-compact.
This is needed to prove that the fpqc topology is finer than the fppf topology.
| https://mathoverflow.net/users/2841 | open faithfully flat morphisms are fpqc | [Edit] (add some details).
Replacing $Y$ with an affine open neighborhood $V$ of $f(x)$ (and $X$ with $f^{-1}(V)$), one can suppose that $Y$ is affine. Cover $X$ by affine open subsets {$U\_i$}$\_i$. As $Y$ is quasi-compact, a finite number of the $f(U\_i)$ cover $Y$. If necessarily, we can add one more $U\_i$ so $x$ belong to one of these $U\_i$'s. The union $U$ of these (finitely many) $U\_i$ is quasi-compact, and we have $f(U)=Y$, $x\in U$. The morphism $f|\_U : U\to Y$ is a morphism from a quasi-compact scheme to an affine scheme, so it is quasi-compact because for any affine open subset $V$ of $Y$, $(f|\_U)^{-1}(V)\cap U\_i= V\times\_Y U\_i$ is affine.
| 8 | https://mathoverflow.net/users/3485 | 39216 | 25,150 |
https://mathoverflow.net/questions/39129 | -1 | I am not a specialist in maths, so I thank you very much for any help you can give me.
Consider two circles C1, C2.
Q1: Find the points that are in the intersection of C1 and C2, this is easy !
Q2: Find two points p1 and p2, such that (p1 \in C1) and (p2 \in C2), and (distance(p1, p2)= D).
Is it possible to solve this problem ?
Now I want to generalize it to more than two circles and to arbitrary geometric predicates (or patterns)
Consider the circles C1, C2, ..., Cn
Q3: Find the points (p1 \in C1), (p2 \in C2), ... , (pn \in Cn) such that (Some\_Geometric\_Predicate(p1, ..., pn)= true).
Have you encountered this problem before ? are they any references that speak about such kind of problems ?
Thank you in advance for your help.
| https://mathoverflow.net/users/9307 | Points in circles that form a given geometric pattern | Permit me to reformulate a specific version of Q3 that
Ellipsissi posed in the comments:
>
> **P1**. Given three non-intersecting circles
> $\{C\_1,C\_2,C\_3\}$,
> find all triples
> $\{p\_1,p\_2,p\_3\}$ with $p\_i \in C\_i$
> such that $\triangle p\_1 p\_2 p\_3$ is
> similar to a given triangle $T$.
>
>
>
This differs from the posed question in (a) the non-intersecting condition,
and (b) not demanding that a specific angle be realized at
a specific corner $p\_i$.
The form above is analogous to this problem:
>
> **P2**. Given a plane curve $\gamma$,
> find all triples
> $\{p\_1,p\_2,p\_3\}$ of points on $\gamma$
> such that $\triangle p\_1 p\_2 p\_3$ is
> similar to a given triangle $T$.
>
>
>
Much is known about P2, under various restrictions on $\gamma$.
For example, if $\gamma$ is a smooth Jordan curve,
then I believe it is almost completely understood now,
through recent work of
Benjamin Matschke, and of
Jason Cantarella, Elizabeth Denne, and John McCleary.
See especially Cantarella's fascinating [web pages on the topic](http://www.jasoncantarella.com/webpage/index.php?title=Square_Peg_problem#Results_on_Inscribed_Triangles).
So, here is a high-level plan for P1. Connect the three circles by thin
corridors to form a plane curve $\gamma$.
Solve P2, and discard solutions with points on the corridors, or more than one point
on one $C\_i$.
The efficacy of this plan depends on the degree to which
P2 is completely solved in its various guises.
**References**
1.
M. J. Nielsen. "Triangles inscribed in simple closed curves,"
*Geometriae Dedicata* **43**: 291-297 (1992).
2.
Benjamin Matschke.
"On the Square Peg Problem and some Relatives."
[arXiv](http://arxiv.org/abs/1001.0186) (2009)
3.
Wikipedia article on the [Inscribed Square Problem](http://en.wikipedia.org/wiki/Inscribed_square_problem), with triangles discussed under "Variants."
| 1 | https://mathoverflow.net/users/6094 | 39217 | 25,151 |
https://mathoverflow.net/questions/39212 | 3 | Let $C$ be a small category. Consider the class of all Grothendieck topologies on $C$, it is a preorder with the relation "finer". Does this preorder has all infima and suprema?
(For example, how do you prove that there is always the canonical topology?)
| https://mathoverflow.net/users/2841 | Infima and Suprema of Grothendieck topologies | Yes. In fact, Grothendieck topologies on any small category constitute a locale.
See Proposition 3.2.13 in Borceux's [Handbook of Categorical Algebra 3](http://gen.lib.rus.ec/get?nametype=orig&md5=C1C3ADB25F2F79762F44CE46C69DF2BC).
| 6 | https://mathoverflow.net/users/402 | 39218 | 25,152 |
https://mathoverflow.net/questions/39224 | 86 | [Zipf's law](http://en.wikipedia.org/wiki/Zipf%27s_law) is the empirical observation that in many real-life populations of n objects, the $k^{th}$ largest object has size proportional to $1/k$, at least for $k$ significantly smaller than $n$ (and one also sometimes needs to assume $k$ somewhat larger than 1). It is a special case of a power law distribution (in which $1/k$ is replaced with $1/k^\alpha$ for some exponent $\alpha$), but the remarkable thing is that in many empirical cases (e.g. frequencies of words, or sizes of cities), the exponent is very close to 1.
My question is: is there a "natural" random process (e.g. a birth-death process) that one can rigorously demonstrate (or at least conjecture) to generate populations of n non-negative quantities $X\_1,\ldots,X\_n$ (with n large but possibly variable) that obey Zipf's law on average with exponent 1? There are plenty of natural ways to generate processes that have power law tails (e.g. consider n positive quantities $X\_1,\ldots,X\_n$ evolving by iid copies of log-Brownian motion), but I don't see how to ensure the exponent is 1 without artificially setting the parameters to force this.
Ideally, such processes should be at least somewhat plausible as models for an empirical situation in which Zipf's law is observed to hold, such as city sizes, but I would be happy with any non-artificial example of a process.
One obstruction here is the exponent one property is not invariant with respect to taking powers: if $X\_1,\ldots,X\_n$ obeys Zipf's law with exponent one, then for any fixed $\beta>0$, $X\_1^\beta,\ldots,X\_n^\beta$ obeys the power law with a different exponent $\beta$. So whatever random process one would propose for Zipf's law must somehow be quite different from its powers.
| https://mathoverflow.net/users/766 | Is there a natural random process that is rigorously known to produce Zipf's law? | I'm not sure if this is an "answer" to your question, but I recall seeing somewhere that someone had shown that if you create a document by selecting the characters a...z plus a space character with uniform frequency then the "words" of such a document have a frequency distribution that follows Zipf's Law. (A little anecdote: when I was an undergraduate, I took a course on "Inductive Logic" given by Zipf. I recall being rather annoyed because he spent a lot of the time lecturing about "his" law and having us form groups that as part of our class work collected statistics to test it :-)
(Added Remarks) I recalled that when we tested Zipf's Laws for city populations back then (more than 50 years ago !) the results were quite good---i.e., the population of the n-th city was pretty close to $1/n$ times the population of the first for many countries. I decided to see if that was still so. For the US it pretty much is:
<http://www.infoplease.com/ipa/A0763098.html#axzz0zuwyduxq>
However, for China, it is WAY off---not even close:
<http://en.wikipedia.org/wiki/List_of_cities_in_the_People%27s_Republic_of_China_by_population>
Of course the population of Chinese cities has been changing rapidly due to migrations into them from the countryside, and perhaps Zipf's Law pertains only to stable situations when things are in equilibrium.
| 41 | https://mathoverflow.net/users/7311 | 39232 | 25,159 |
https://mathoverflow.net/questions/39243 | 1 | Consider a cycle of length $(2n+2)$. Now we quadrangulate this cycle into $n$ quadrants. We want to enumerate the number of quadrangulations, and we denote this number by $q\_n$. Now we triangulate this quadrangulation by triangulating each quadrant. We denote the number of triangulations $t\_n$. It is clear that $t\_n = 2^nq\_n$.
Do I count some triangulations multiple times? That is, will some triangulation be the result of multiple different quadrangulations?
| https://mathoverflow.net/users/1539 | Enumerating triangulations of quadrangulations in cycles | No triangulation occurs multiple times. At least one of the triangles $T$ of the triangulation has two edges on the boundary. There is a unique quadrant (I would say quadrilateral) $Q$ made up of two triangles of the triangulation, one of which is $T$. Remove the three edges of $T$, obtaining two quadrangulated cycles (possibly degenerate), and induct.
Note that not every triangulation of a $(2n+2)$-cycle is obtained by bisecting the quadrilaterals of a quadrangulation. Also, it is well-known that the number of quadrangulations of a $(2n+2)$-gon is $\frac{1}{2n+1}\binom{3n}{n}$.
| 4 | https://mathoverflow.net/users/2807 | 39246 | 25,166 |
https://mathoverflow.net/questions/39240 | 4 | In the case of the circle I can hardly make any conclusions from the integral $(1)$, most of the theorems come from geometrical considerations. It's not clear how to prove periodicity from this integral or derive the addition theorem.
$$\arcsin(y) = \displaystyle \int\_{0}^{y} \frac{\mathrm dy}{\sqrt{1 - y^2}} (1)$$
In the case of the lemniscate $(2)$ we (Fagnano) can derive a doubling theorem by trying out substitutions in analogy with those which rationalize the first integral. (I learned this from Siegel - Topics in Complex Function Theory). One result (of the sort I wish I could find more) which does come nicely from this integral is via the substitution $y \mapsto iy$ we notice it has double periodicity.
$$\text{sl}^{-1}(y) = \displaystyle \int\_{0}^{y} \frac{\mathrm dy}{\sqrt{1 - y^4}} (2)$$
Euler and others were able to produce theorems about the elliptic integral $(3)$ by analogy with the lemniscate (I read this in Stillwell - Mathematics and its History) - Just as ideas from $\arcsin$ helped to produce theorems about the lemniscate integral. Still, these theorems are very hard earned and it appears that you have to be a master like Euler to derive them.
$$F(y,k) = \displaystyle\int\_0^y\frac{\mathrm dy}{\sqrt{(1-k^2 y^2)(1-y^2)}} (3)$$
---
What I would really like to know is, can we derive more results about the functions from these integrals - maybe using integral manipulation techniques I just don't know about?
Another question that has been bothering me deeply is the integrands (which I believe are called invariant differentials of the lie groups for the algebraic curve) - What sort of coincidence is it that allows the integrands to be of this particular form?
Thank you!
| https://mathoverflow.net/users/4361 | What is the advantage of inverting elliptic integrals? | Some (self biased) links to get you started:
**Expository:**
[Ziegler - talk on the AGM at the Technion's math-club](http://www.technion.ac.il/~tamarzr/agm-talk.pdf)
[Cox: The arithmetic-geometric mean of Gauss.. Enseign. Math. (2) 30 (1984), no. 3-4.](http://retro.seals.ch/cntmng?type=pdf&rid=ensmat-001%3A1984%3A30%3A%3A87&subp=hires)
[Ritzenthaler: AGM for elliptic curves.](http://iml.univ-mrs.fr/~ritzenth/cours/AGM-ec.pdf)
**Textbook:**
[Borwein & Borwein: Pi and the AGM via Amazon (w/ look inside)](http://rads.stackoverflow.com/amzn/click/047131515X)
**Research:**
[Donagi & Livne:The arithmetic-geometric mean and isogenies for curves of higher genus](http://arxiv.org/abs/alg-geom/9712027)
[Dolgachev & Lehavi: On isogenous principally polarized abelian surfaces](http://arxiv.org/abs/0710.1298)
[Lehavi & Ritzenthaler: Formulas for the arithmetic geometric mean of curves of genus 3](http://arxiv.org/abs/math/0403182)
[Humbert: Sur les fonctions abéliennes singulières (Troisième Mémoire)](http://math-tenere.ujf-grenoble.fr/JMPA/feuilleter.php?id=JMPA_1901_5_7)
---
And a couple of non online sources:
Richelot, De transformatione integralium Abelianorum primi ordinis comentatio.
J. reine angew. Math. 16 (1837) 221–341.
J.-B. Bost and J.-F. Mestre, Moyenne arithmético-géométrique et périodes des courbes de genre 1 et 2, Gaz. Math. 38 (1988), 36–64.
| 4 | https://mathoverflow.net/users/404 | 39248 | 25,168 |
https://mathoverflow.net/questions/39230 | 5 | Given a regular local ring $(R,m)$ and a finitely generated $R$-algebra $S$, which is free as an $R$-module. Let $M$ be a left $S$-module of finite length, $\ell\_S(M)=r<\infty$.
Under what conditions is $\ell\_R(M)<\infty$? If this is the case, can we compute $\ell\_R(M)$ in terms of $\ell\_S(M)$?
For example if $S=M\_n(R)$, then i think we have $\ell\_R(M)=n\ell\_S(M)$.
If $S$ is commutative and local, with maximal ideal n, then according to Liu one has the following: $\ell\_R(M)=[S/n:R/m]l\_S(M)$.
Are there general formulas for length and "restriction of scalars"? I'm especially interested in the case when $S$ is a maximal $R$-order in a division algebra. Literature tips are also appreciated.
| https://mathoverflow.net/users/3233 | Length of a module over different rings | Let $\{V\_i\}$ be representatives from each of the isomorphism classes of simple left $S$-modules. For any finite length module ${}\_S M$, let $\ell\_S(M; V\_i)$ denote the number of times that $V\_i$ occurs in a composition series for $M$. Then the following formula holds (where almost all $\ell(M;V\_i)$ are zero because $M$ has finite length, but any single $\ell\_R(V\_i)$ could be infinite, and $\infty \cdot 0 = 0$):$$\ell\_R(M) = \sum \ell\_R(V\_i) \cdot \ell\_S(M; V\_i).$$
Thus, a finite length $S$-module $M$ has finite $R$-length if and only if every simple $S$-module that occurs in $M$ has finite $R$-length. This makes it clear that *every* finite length left $S$-module has finite $R$-length if and only if all simple left $S$-modules have finite $R$-length. The above discussion is true with no requirements on the ring $S$.
Now let's assume that $S$ is finitely generated as an $R$-module. (I'm not sure if this is what you meant by "finitely generated as an $R$-algebra," but it's probably true in the cases that you're studying if you're looking at maximal orders.) I'll prove that *$S$ has finitely many simple modules, each of which has finite $R$-length.* Let $J(A)$ denote the Jacobson radical of any ring $A$. Then $\mathfrak{m} = J(R) \subseteq J(S)$ because $S$ is a module-finite $R$-algebra; for a proof of this fact, see Lam's *A First Course in Noncommutative Rings*, Proposition 5.7. Now the simple $S$-modules are the same as the simple $S/J(S)$ modules (see Proposition 4.8 of the same text). But $S/\mathfrak{m} S \twoheadrightarrow S/J(S)$. Because $S$ is module-finite over $R$, $S/\mathfrak{m} S$ has finite $R$-length. Thus $S/J(S)$ has finite $R$-length. So $S/J(S)$ is artinian (the terminology here is that $S$ is a *semilocal* ring) and thus has finitely many simple modules, each of which has finite $R$-length. The same must be true for the simple $S$-modules
The formula above also verifies your formula in the case of a matrix ring. In case $S = \mathbb{M}\_n(R)$, $S$ has a unique simple left module $V = (R/\mathfrak{m} \ \cdots \ R/\mathfrak{m})^T$, the set of all column vectors of length $n$ with entries in $R/\mathfrak{m}$. (At least one way to see that this is the only simple $S$-module is through Morita theory: the Morita equivalence between $R$-$\operatorname{Mod}$ and $S$-$\operatorname{Mod}$ sends the unique simple $R$-module $R/\mathfrak{m}$ to $V$, so that $V$ must be the unique simple $S$-module.) Since $\ell\_R(V) = n$, the formula above reduces to $\ell\_R(M) = n \cdot \ell\_S(M)$.
(If something above doesn't make sense, you may want to review [Jordan-Hoelder theory](http://en.wikipedia.org/wiki/Composition_series#For_modules).)
| 5 | https://mathoverflow.net/users/778 | 39257 | 25,171 |
https://mathoverflow.net/questions/39210 | 35 | Does anybody know a solution to this problem? (Sorry, I've missed one summand in the [previous post](https://mathoverflow.net/questions/39207/solve-in-positive-integers-n-m2).)
| https://mathoverflow.net/users/5712 | Solve in positive integers: $n!=m(m+1)$ | I'm pretty sure this is open. As suggested from Brocard's problem, it is interesting to investigate the Diophantine equations $$n!=P(m)$$ for polynomials $P$. You can see the paper ["On polynomial-factorial diophantine equations"](http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/home.html), by D. Berend, J.E. Harmse where they make some advances on the problem and prove that this equation has finitely many solutions for many classes of polynomials (irreducible, with an irreducible factor of large degree or with an irreducible factor to a large power).
So by their results it is known that the equation $n!=m^r(m+1)$ has finitely many solutions if $r\geq 4$. But for $r\in \{1,2,3\}$ the problem is open.
| 38 | https://mathoverflow.net/users/2384 | 39260 | 25,173 |
https://mathoverflow.net/questions/37394 | 4 | Let $k$ be an algebraically closed field and $A$ a finitely generated $k$-algebra, together with a specified surjective morphism $\phi \colon k[x\_1, \dotsc, x\_n] \to A$. For $f \in A$, define $\mathrm{deg}(f)$ to be the minimum of $\mathrm{deg}(g)$, where $g$ ranges over all polynomials in $k[x\_1, \dotsc, x\_n]$ such that $\phi(g) = f$. [Note: by $\mathrm{deg}(g)$, I mean the degree of the highest-degree monomial, where $\mathrm{deg}(x\_1^{i\_1} \dotsm x\_n^{i\_n}) = i\_1 + \dotsb + i\_n$.] If it is helpful, we can assume $A$ is an integral domain, even integrally closed if necessary.
>
> Let $u \in A^\*$ be a unit such that $\mathrm{deg}(u) > 0$, or equivalently, $u \not\in k^\*$. Is it necessarily true that $\deg(u^n) \to \infty$ as $n \to \infty$?
>
>
>
Thoughts: If we have a monomial order that respects degree (such as grlex or grevlex, but not lex), and take a Groebner basis of $\ker(\phi)$, then we see that powers of $u$ remain predictable as long as their leading terms fall outside the ideal generated by the leading terms of the groebner basis (aka, the initial ideal).
Motivation: I'm trying to prove a classical theorem using model theory, and the proof I have in mind would require the above to be true.
| https://mathoverflow.net/users/5094 | Lower bounds on the degrees of representatives of $u^n$ as $n \to \infty$ | If $u \in A$ and $deg(u^t), t = 0,1,2, \dots$ is bounded by $d$, then the powers of $u$ lie
in $\varphi(V)$, where $V$ is the finite-dimensional $k$-vector space spanned by the
monomials in $k[x\_1, \dots , x\_n]$ of degree $\leq d$. Hence the powers of $u$ are linearly
dependent over $k$, so $u$ is algebraic over $k$. If $A$ is a domain, this implies that
$u \in k$, since $k$ is algebraically closed.
| 3 | https://mathoverflow.net/users/9347 | 39274 | 25,183 |
https://mathoverflow.net/questions/39194 | 7 | Let $M\_n$ be the set of $n$-by-$n$ matrices with complex entries, viewed as a variety over $k=\mathbb{C}$. Equip $M\_n$ with the conjugation action of $\mathrm{GL}(n)=\mathrm{GL}(n,\mathbb{C})$. Consider $A:=\mathrm{Mor}^{\mathrm{GL}(n)}(M\_n \oplus M\_n, M\_n)$, the set of $\mathrm{GL}(n)$-equivariant maps (of algebraic varieties) $M\_n \oplus M\_n \to M\_n$, where $GL(n)$ acts diagonally on $M\_n \oplus M\_n$. Then the (standard) algebra structure of $M\_n$ (with multiplication given by matrix multiplication) induces an algebra structure on $A$.
The following maps belong to $A$:
1. $M\_n \oplus M\_n \to M\_n \colon (A,B) \mapsto A$;
2. $M\_n \oplus M\_n \to M\_n \colon (A,B) \mapsto B$;
3. $M\_n \oplus M\_n \to M\_n \colon (A,B) \mapsto f(A,B)I\_n$, where $I\_n$ is the identity matrix and $f$ is an element of the ring of invariants $k[M\_n \oplus M\_n]^{\mathrm{GL}(n)}$.
Do they generate $A$ as an algebra?
| https://mathoverflow.net/users/nan | Generators for the algebra of GL(n)-equivariant maps from M_n + M_n to M_n | The answer is affirmative not only in the case of 2 matrices, but also in the case of any number of matrices; in fact, an analogous statement is true for quiver representations (in characteristic 0).
The original question can be restated as follows.
>
> Let $P$ be the space of polynomial functions of 2 $n\times n$ matrices, with the adjoint action of $GL\_n$ and the ring of invariants $I.$ Consider the space $\text{Hom}\_{GL\_n}(M\_n,P)$ as an $I$-module. Is it true that it is generated by the products of matrices?
>
>
>
For the case of any number of generic matrices $A\_1,\ldots,A\_k,$ Procesi proved that over a field $k$ of characteristic 0, $I$ is spanned by the traces of the products of matrices. Formally, consider words in the free monoid with $k$ generators, substitute the generic matrices, and take a trace.
Procesi, C. *The invariant theory of n×n matrices*. Advances in Math. 19 (1976), no. 3, 306–381
The statement follows by adjoining an extra generic matrix $A\_0$ and converting an $M\_n$-space into a $GL\_n$-invariant forming a product with $A\_0$ and taking the trace, then undoing the trace of the term in the trace polynomial from Procesi's theorem containing $A\_0.$
---
Here is a vast generalization due to Le Bruyn and Procesi. Given a finite quiver $Q$ and a dimension vector $\alpha,$ consider the corresponding representation space $R(Q,\alpha)$ with the action of the algebraic group $GL(\alpha)$ and the space $P$ of polynomial functions on $R.$ (If the quiver consists of a single vertex with $k$ loops and $\alpha=n$ then the representation space is given by $k$ generic $n\times n$ matrices with the simultaneous conjugation action by $GL\_n.$) Then, over a field of characteristic 0, the algebra $I$ of polynomial invariants is spanned by the traces of matrix products over oriented cycles in $Q$ and for any pair of vertices $(i,j)$ of $Q,$ the space $\text{Hom}\_{GL(\alpha)}(\text{Hom}\_k(V\_i,V\_j),P)$ is generated as an $I$-module by the products over oriented paths connecting $i$ with $j.$
Lieven Le Bruyn, Claudio Procesi, *Semisimple representations of quivers*.
Trans. Amer. Math. Soc. 317 (1990), no. 2, 585–598
| 5 | https://mathoverflow.net/users/5740 | 39291 | 25,190 |
https://mathoverflow.net/questions/39308 | 7 | I've been working on a paper with some collaborators. Barring a breakthrough on some unresolved questions, the math content is finalized. I would like to give a talk on the results, but I am unclear on the etiquette. Should I check with my collaborators before presenting the material?
In general, how careful should I be discussing and presenting unpublished material without clearing it collaborators?
| https://mathoverflow.net/users/750 | Should I check with collaborators before presenting unpublished material? | Check with your collaborators on this one. Opinions vary.
| 20 | https://mathoverflow.net/users/2620 | 39309 | 25,199 |
https://mathoverflow.net/questions/39312 | 4 | **Motivation**: I want to see how the 3-dimensional Weisfeiler-Lehman algorithm (see [Logical complexity of graphs](http://arxiv.org/PS_cache/arxiv/pdf/1003/1003.4865v1.pdf), p. 14) distinguishes between two non-isomorphic [strongly regular graphs srg(v,k,λ,μ)](http://en.wikipedia.org/wiki/Strongly_regular_graph) in a specific example.
>
> **Question**: What are the smallest non-isomorphic strongly regular graphs
> with the same v,k,λ,μ?
>
>
>
| https://mathoverflow.net/users/2672 | Smallest non-isomorphic strongly regular graphs | This page <http://www.maths.gla.ac.uk/~es/srgraphs.html> lists some strongly regular graphs on few vertices, and gives two (16,6,2,2) graphs (which I didn't check but I presume they're non-isomorphic). I imagine they're the smallest possible but I haven't checked: <http://www.maths.gla.ac.uk/~es/16.vertices>
| 6 | https://mathoverflow.net/users/4580 | 39319 | 25,203 |
https://mathoverflow.net/questions/39316 | 4 | Let's view the category of algebraic spaces as a full subcategory of the category of "spaces" over the opposite category of commutative rings, that is, the category of sheaves on $CRing^{op}$ in the étale topology. Is there any sort of interesting model structure on this category, or a suitable enlargement of it (perhaps by looking at simplicial sheaves or by changing the topology), capturing the theory of formally smooth morphisms (as fibrations)? As a bonus, is there any way to describe the interesting spaces of this category, say algebraic spaces, as a category of fibrant or cofibrant objects?
If there is some obvious failure that I'm overlooking, is there any way to rescue the idea? Is there any homotopical content in the definition of formally smooth morphisms by a lifting property?
| https://mathoverflow.net/users/1353 | Model category with formally smooth morphisms as fibrations? | This addresses just the last question "Is there any homotopical content...". It would belong into a comment but doesn't fit.
Mathieu Anel shows in a [very recommendable article](http://arxiv.org/abs/0902.1130) how two classes of maps in the opposite of a locally presentable category, one having the (left/right) lifting property w.r.t. the other, yield factorization systems - every map of the category factors as a map of the left class followed by a map of the right class. To a factorization system he associates a Grothendieck topology (you can just read the two pages addressing this, the article is very readable). Now a category of presheaves over a site is a model category with weak equivalences those morphisms which become isomorphisms after applying sheafification, cofibrations the monomorphisms and fibrations those with the lifting property. This already gives a homotopical content to a lifting system, but yet not as suggested in your question.
Now if the topology on a site arose via a lifting system, I would guess that the fibrations of the model structure on presheaves can be described in terms of maps from the right class (I am thinking of something like: a map of sheaves is a fibration if every pullback of it into the realm of affines gives a map from the right class). Or maybe one can find a new model structure on presheaves with the same weak equivalences involving both the left and the right class from the site. I guess this is known to someone, but not to me...
| 2 | https://mathoverflow.net/users/733 | 39320 | 25,204 |
https://mathoverflow.net/questions/39343 | 2 | I tried to find a solution to this in the web but couldn't.
Can you tell me if the following sentence is correct or else give me a counterexample?
$G$ is $4$-colorable if and only if each sub-graph $G'$ in $G$ is not isomorphic to $K\_5$.
At first glance it seems to be related to the four color theorem but it is not exactly a planar graph (e.g. $3\times 3$ complete bipartite graph) so it is not identical to FCT.
Any ideas?
| https://mathoverflow.net/users/9369 | A conjectured criterion for 4-colorable graphs | This is false. In fact, there are graphs that contain no K3's but have arbitrarily high chromatic number.
See this wiki article for one such construction <http://en.wikipedia.org/wiki/Mycielskian>
| 13 | https://mathoverflow.net/users/3655 | 39345 | 25,220 |
https://mathoverflow.net/questions/39294 | 9 | There are a couple of beautiful results in finite group theory that look trivial, at least on a first glance, but require non-trivial facts to prove. I am basically interested in whether these results actually have relatively easier proofs to the ones I will outline below. *More specifically, I am interested in whether these results may be proven without the aid of the powerful subnormality theory.*
The first, and perhaps most beautiful result, is due to Horosevskii:
>
> Let $\sigma\in Aut(G)$ where Aut($G$) denotes the automorphism group of a non-trivial finite group $G$. Then the order of $\sigma$ in Aut($G$) in less than $\left|G\right|$.
>
>
>
The second result, though perhaps more specialized, actually is an important ingredient in the purely group-theoretic proof of Burnside's $p^aq^b$-theorem:
>
> Let $t$ be an involution in a finite group $G$, and assume that $t\not\in {\bf{O}}\_2(G)$, where ${\bf{O}}\_2(G)$ denotes the 2-core of $G$. Then there exists an element $x\in G$ of odd prime order such that $txt=x^{-1}$.
>
>
>
(I think that this result is due to Matsuyama but please do not take my word for this because I am not completely certain. For definiteness, I will refer to this result as the "result on involutions".)
Now the only proof I know of the second result (on involutions) relies on Baer's theorem in subnormality theory. For reference, Baer's theorem states that:
>
> Let $H$ be a subgroup of a finite group $G$. Then $H\subseteq {\bf{F}}(G)$ if and only if $\left\langle H,H^x \right\rangle$ is nilpotent for all $x\in G$.
>
>
>
(Here, ${\bf{F}}(G)$ denotes the Fitting subgroup of $G$ and $H^x$ denotes the conjugate of $H$ by $x$; that is, $H^x = \{x^{-1}hx|h\in H\}$.)
The proof (at least the one I know) of the result on involutions uses Baer's theorem, but actually, it really only uses a very special case of Baer's theorem: the subgroup $H$ in the statement of Baer's theorem is chosen to be $\{1,t\}$ in the proof, where $t$ is the involution quoted in the result. (The proof also uses a very easy fact about dihedral groups.)
This leads me to wonder whether there exists a "Baer-free" proof of the result on involutions. More specifically, my question is:
* Question 1: *Does there exist a proof of the result on involutions independent of subnormality theory?*
It would be interesting if such a proof existed since Baer's theorem relies on the non-trivial Wielandt "zipper lemma" (which I will quote at the end of my question) - and the zipper lemma really looks unrelated to the result on involutions.
Now Horosevskii's theorem relies on another non-trivial consequence of subnormality theory: Lucchini's theorem. For reference, Lucchini's theorem states that:
>
> Let $A$ be a cyclic proper subgroup of a finite group $G$, and let $K=\mbox{core}\_G(A)$. Then $\left|A:K\right|<\left|G:A\right|$, and in particular, if $\left|A\right|\geq \left|G:A\right|$, then $K>1$.
>
>
>
What is amazing, at least to me, is that a fact that one might consider fundamental about automorphisms (Horosevskii's theorem) relies on a result (Lucchini's theorem) that at least on a first glance seems very specialized. My second question is nearly identical to my first:
* Question 2: Does there exist a proof of Horosevskii's theorem independent of subnormality theory?
(What follows is not really subsumed in my question, but it may be useful for answering my question.)
Succinctly, the proof of Baer's theorem (at least the proof I know) relies on the amazing Wielandt "zipper lemma" which states:
>
> Suppose that $S\subseteq G$, where $G$ is a finite group, and assume that $S$ is subnormal in $H$ for every proper subgroup $H$ of $G$ that contains $S$. If $S$ is not subnormal in $G$, then there is a unique maximal subgroup of $G$ that conains $S$.
>
>
>
Likewise, the proof of Lucchini's theorem relies on Zenkov's theorem on intersections of abelian subgroups, the proof of which in turn relies on Baer's theorem. Zenkov's theorem states:
>
> Let $A$ and $B$ be abelian subgroups of a finite group $G$ and let $M$ be a minimal member of the set $\{A\cap B^g|g\in G\}$. Then $M\subseteq {\bf{F}}(G)$.
>
>
>
(Note that "$B^g$" denotes the conjugate of $B$ by $g$; that is, $B^g=\{g^{-1}bg|b\in B\}$.)
(Note: I am well aware that there may be other theorems due to Baer, Lucchini and Horosevskii that are referred to as "Baer's theorem", "Lucchini's theorem" or "Horosevskii's theorem". However, I hope that by stating the results above, no confusion arises. For proofs of the theorems quoted, please see the book *Finite Group Theory* by I. Martin Isaacs. More specifically, please see Theorem 2.9., Chapter 2, Section 2A, page 50 for the statement and proof of the Wielandt "Zipper Lemma", Theorem 2.12., Chapter 2, Section 2B, page 55 for the statement and proof of Baer's theorem, Theorem 2.18., Chapter 2, Section 2D, page 61 for the statement and proof of Zenkov's theorem, and Theorem 2.20., Chapter 2, Section 2D, page 63 for the statement and proof of Lucchini's Theorem.)
Thanks!
| https://mathoverflow.net/users/4842 | Non-trivial consequences of Baer's theorem and Lucchini's theorem in subnormality theory | **Baer–Suzuki:** The subgroup generated by {x,x^g} is a p-group for all g in G if and only if x is contained in the p-core of G.
Baer's proof emphasized commutators, rather than subnormality. In some sense these are the same thing, but perhaps it will feel different enough for you. Baer's presentation is given in textbook form in Huppert's Endliche Gruppen as III.6.15, page 298. See also IX.7.8 in Huppert–Blackburn, Finite Groups, Vol 2, p. 500. Baer's original paper is:
* Baer, Reinhold. "Engelsche elemente Noetherscher Gruppen."
Math. Ann. 133 (1957), 256–270.
MR[86815](http://www.ams.org/mathscinet-getitem?mr=86815)
DOI:[10.1007/BF02547953](http://dx.doi.org/10.1007/BF02547953)
Suzuki's proof is given in Gorenstein's Finite Groups, 3.8.2, p. 105. It also avoids subnormality, rather using ideas about fusion of p-elements, and is probably how Bender thought of it.
---
Subnormality is a pretty critical idea, and many of Bender's insights use subnormality, so I would not suggest avoiding subnormality. Other characterizations of the Fitting subgroup in terms of subnormality are given in Huppert's textbooks. In particular, the Fitting subgroup as the elements that centralize chief factors is a very important viewpoint. It generalizes to F-subnormality in the finite soluble world, and Bender's p\*-nilpotency in the finite insoluble world. Kegel and Carter have a number of nice papers that explore subnormality in ways that have heavily influenced both the soluble and the insoluble worlds.
Robinson's group theory textbook (and Lennox–Stonehewer MR[902857](http://www.ams.org/mathscinet-getitem?mr=902587)) have a good description of subnormality in the infinite case.
Wielandt's collected works contains several good textbook style presentations of subnormality that are not properly contained in any other works that I have found. They avoid assuming finiteness, and tend to have very interesting relationships between perfect subgroups and subnormality, that complement Bender's work.
---
One very nice thing about Isaacs's FGT is how it exposes you to important techniques. It does not try to "optimize" the presentation either by using the bare minimum of tools or by using the most general tool here-to-fore created. It just uses some nice results in a realistic way that more people should know. Suzuki's textbooks also have this nice property, though they are not as easy to quote from.
| 4 | https://mathoverflow.net/users/3710 | 39349 | 25,224 |
https://mathoverflow.net/questions/39326 | 15 | I've been looking at unit fractions, and found a paper by Erdős ["Some properties of partial sums of the harmonic series"](http://www.ams.org/journals/bull/1946-52-04/S0002-9904-1946-08550-X/S0002-9904-1946-08550-X.pdf) that proves a few things, and gives a reference for the following theorem:
$$\sum\_{k=0}^n \frac{1}{m+kd}$$ is not an integer.
The source is:
>
> Cf. T. Nagell, *Eine Eigenschaft gewissen Summen*, Skrifter Oslo, no. 13 (1923) pp. 10-15.
>
>
>
>
> ---
>
>
>
Question
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Although I would like to find this source, I've checked with my university library and it seems pretty out of reach. What I'm really hoping for is a source that's more recent or even written in English.
Finding this specific source isn't everything, I'll be fine with pointers to places with similar results.
| https://mathoverflow.net/users/1150 | Unit fraction, equally spaced denominators not integer | You can cite H. Belbachir and A. Khelladi, On a sum involving powers of reciprocals of an arithmetic progression, Ann. Math. Inform. 34 (2007) 29-31, MR2385421, where a more general result is given. If you are OK with Russian, there is Z. D. Gorskaya, On an arithmetic property of a harmonic sum, Ukrain. Mat. Z. 6 (1954) 375-384, MR0069200 (16,998j).
Nagell wrote a nice introductory number theory textbook, which was republished by the American Math. Society. Maybe the result is in it.
EDIT: I have had a look at the Belbachir and Khelladi paper, at <http://www.kurims.kyoto-u.ac.jp/EMIS/journals/AMI/2007/ami2007-belbachir.pdf>, and I find that it rests heavily on the Shorey and Tijdeman paper cited in Gjergji Zaimi's answer.
FURTHER EDIT: I think that Erdős himself proves the result in the paper "Egy Kürschák-féle elemi számelméleti tétel általánosítása", [KöMaL](https://www.komal.hu/verseny/korabbi.e.shtml) 39 (1932) 17-24. This is freely available, with a summary in German at the end, at [http://www.renyi.hu/~p\_erdos/1932-02.pdf](http://www.renyi.hu/%7Ep_erdos/1932-02.pdf). It would seem that in 1932 Erdős was unaware of Nagell's work.
| 5 | https://mathoverflow.net/users/3684 | 39350 | 25,225 |
https://mathoverflow.net/questions/39292 | 14 | If $p$ is a polynomial with real coefficients and p(x)>0 on [0,1], then $p(x)=\sum c\_{i,j} x^i(1-x)^j$ with $c\_{i,j}$ positive. I know this is true but but I need a proof/reference. Thanks!
| https://mathoverflow.net/users/9354 | Polynomial positive on an interval | See "Polynomials that are positive on an interval" by myself and Bruce Reznick, Trans. Amer. Math. Soc. 352 (2000), 4677-4692. We give a brief history of this problem along along with a bound for the minimum $m$ so that $p$ can be written $p = \sum\_{i=1}^m c\_i x^i (1-x)^{m-i}$ with $c\_i \geq 0$. The bound depends on the minimum of $p$ on $[0,1]$ and the size of the coefficients. The best reference for this (without the bound) is probably P\'olya-Szego.
| 18 | https://mathoverflow.net/users/9372 | 39354 | 25,229 |
https://mathoverflow.net/questions/39163 | 8 | In an [article](http://www.cut-the-knot.org/blue/weight1.shtml) describing the twelve balls weighing problem, the author mentions a solution that involves the finite projective plane of order 3, discovered by Rick Wilson. Does anyone know what this solution could have been?
| https://mathoverflow.net/users/491 | 12 balls weighing puzzle | Will Orrick is right, the problem is solved by exhibiting a matrix $3\times 12$ with entries in $\{-1,0,1\}$ where all columns are pairwise independent and the row sums are zero, as mentioned in Wilson's book.
In general you can solve the $\frac{3^r-1}{2}-1$ coin problem using $r$ weighings. You need to use one of the [generator matrices](http://en.wikipedia.org/wiki/Generator_matrix) of the simplex code, so the columns are given by the points in the projective space $P(r-1,3)$ (you throw out the point at infinity) and by induction show that the choices can be made to arrange the zero sum rows. For the case of 12 coins it suffices to consider the projective plane, and you get a [constant weight code](http://en.wikipedia.org/wiki/Constant-weight_code) and thus end up weighing groups of 4 coins each time.
| 9 | https://mathoverflow.net/users/2384 | 39358 | 25,232 |
https://mathoverflow.net/questions/39332 | 2 | I can't find a single solid explanation of how to implement this -- whitepapers too detailed/confusing. Closest I came to an answer was this:
<http://www.hep.ucl.ac.uk/~bino/libbpm/doc/pro/html/gsl__linalg_8c-source.html>
see:
[1] gsl\_linalg\_bidiag\_decomp
[2] gsl\_linalg\_SV\_decomp (which calls [1])
Which perform it, but really, really obfusticate the process underneath.
| https://mathoverflow.net/users/9729 | Bidiagonalization and SVD of matrix | **Golub-Kahan Bidiagonalisation**
In this process householder reflectors are applied alternatively on the left and then the right. The $i^{\text{th}}$ left reflector introduces zeros below the diagonal in the $i^{\text{th}}$ column. The $i^{\text{th}}$ right reflector introduces zeros to the right of the first super-diagonal in the $i^{\text{th}}$ row.
In software packages I suspect they use a mixture of this Golub-Kahan bidiagonalisation and a process called Lawson-Hanson-Chan (LHC) bidiagonalisation depending on the size of the matrix.
**Computing the SVD**
The first phase of computing the SVD is bidiagonalising the matrix. Then the SVD of the bidiagonal matrix is determined by a process very similar to the QR algorithm. This process is described in Golub and Kahn, ["Calculating the singular values and pseudo-inverse of a matrix"](http://www.amath.washington.edu/~narc/win08/papers/golub-kahan.pdf) (1960).
Since this paper there have been some alterations to provide better accuracy when the singular values are small, see Demmel and Kahan: "Accurate singular values of bidiagonal matrices" (1990)
Having been lectured by N. Trefethen on this very subject he briefly mentioned a divide-and-conquer type algorithm was now state-of-the-art though I don't know much about the details.
| 3 | https://mathoverflow.net/users/2011 | 39364 | 25,236 |
https://mathoverflow.net/questions/39359 | 2 | (base theory = ZFC)
Are any Hamel bases for the vector space $\mathbb{R}^{\omega}$ in the
1. analytical hierarchy?
2. projective hierarchy?
In any of the above cases where the answer is not simply "no", is anything known about what levels they are or can be in?
My knowledge of descriptive set theory is basically just what's on wikipedia, so I probably won't know other theorems even if they are proved in every textbook on the subject. However, I suspect the answers will be
"1. no; 2. none are below $\Delta^1\_n$, if V=L then they are in $\Delta^1\_n$, if projective determinacy then no"
with n a small natural number explicitly known but not to me.
| https://mathoverflow.net/users/nan | Descriptive complexity of Hamel bases of R^ω | A projective Hamel basis under V=L should be easy: Take a $\Delta^1\_2$ wellordering of $\mathbb R^\omega$ and prove the existence of a Hamel basis using this wellordering. That will give you a projective Hamel basis low ($\Delta^1\_2$?) in the projective hierarchy.
Negative results are often proved by constructing sets without the Baire property or nonmeasurable sets (which then tells that the constructed set is at least $\Delta\_2^1$) from your assumption. My guess would be that your suggested answer is true with $n=2$,
but I don't quite see how to construct a "weird" set from the Hamel basis yet.
By the way, the question what a basis for $\mathbb R$ over $\mathbb Q$ looks like has been studied quite a bit.
| 4 | https://mathoverflow.net/users/7743 | 39367 | 25,238 |
https://mathoverflow.net/questions/39373 | 2 | Does n!m!=t! have infinitely many solutions in positive interger besides trivial ones? (n=0 m=1 etc)
Can't work this one out. thanks.
| https://mathoverflow.net/users/9382 | does n!m! = t! have infinitely many solutions? besides trivial ones | $(n!)!=(n!-1)!\cdot n!$ - is it trivial or not?
| 9 | https://mathoverflow.net/users/1888 | 39374 | 25,242 |
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