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https://mathoverflow.net/questions/47028 | 5 | Since I found out about it, I've always been interested in the Axiom of Determinacy rather than the Axiom of Choice. Along these lines, I've kept flipping back to <http://en.wikipedia.org/wiki/%CE%98_%28set_theory%29>, and occasionally looking on google, because I keep thinking ZF+AD should be able to prove non-obvious things about it, although I haven't found anything other than (I think) something saying it must be regular. So, I'm finally asking here.
$\Theta := \operatorname{sup}(\{\alpha \in \operatorname{Ord} : (\exists f \in \alpha^\mathbb{R})(\operatorname{Range}(f) = \alpha)\})$
What is known about $\Theta$ in ZF+AD? In particular, how big is it?
For example, is it known to be different from $\omega\_2$?
Is anything more known in ZF + AD + V=L($\mathbb{R}$) ?
| https://mathoverflow.net/users/nan | value of Theta in ZF+AD | Ricky,
A good reference for this question specifically and determinacy in general is Kanamori's book on large cardinals, "The Higher infinite". The last chapter is devoted to determinacy.
We know a huge deal about $\Theta$. For example, it is much larger than $\omega\_2$. It does not need to be regular, but it *is* regular if in addition we assume $V=L({\mathbb R})$.
The key fact to see that $\Theta$ is large is *Moschovakis's coding lemma* which, in its simplest version, says that:
>
> (Under AD) if there is a surjection $f:{\mathbb R}\to\alpha$, then there is a surjection $f:{\mathbb R}\to{\mathcal P}(\alpha)$.
>
>
>
Harvey Friedman used this to prove that $\Theta$ is a limit cardinal. This is easy; the point is that there is a definable bijection between ${\mathcal P}(\tau)$ and ${\mathcal P}(\tau\times\tau)$ for any infinite ordinal $\tau$. But if $\tau$ is a cardinal, there is a surjection from ${\mathcal P}(\tau\times\tau)$ onto $\tau^+$: If $A\subseteq\tau\times\tau$ codes a well-ordering, send it to its order type. Else, to 0.
With a bit more effort, you can check that $\Theta=\aleph\_\Theta$ and in fact it is limit of cardinals $\kappa$ such that $\kappa=\aleph\_\kappa$, and it is a limit of limits of these cardinals, etc.
In $L({\mathbb R})$, $\Theta$ is regular (Solovay was first to prove this). In fact, if $V=L(S,{\mathbb R})$ for $S$ a set of ordinals, or $V=L(A,{\mathbb R})$ for $A\subseteq{\mathbb R}$, then $\Theta$ is regular.
(A technical aside: If AD holds, it holds in $L({\mathcal P}({\mathbb R}))$. Woodin defined a strengthening of AD that is now called $AD^+$. It is open whether $AD^+$ is strictly weaker than AD, since all known models of AD are models of $AD^+$ and any current technique that gives us a model of one gives us a model of the other. If $L({\mathcal P}({\mathbb R}))$ is a model of $AD^+$, then it is either of the form $L(A,{\mathbb R})$ for some $A\subseteq{\mathbb R}$, or else it is a model of $AD\_{\mathbb R}$, the strengthening of AD where we allow reals (rather than integers) as moves of the games.)
However, ZF+AD does not suffice to prove that $\Theta$ is regular. If DC holds, the "obvious" diagonalization shows that ${\rm cf}(\Theta)>\omega$. But Solovay proved that $ZF+AD\_{\mathbb R}+{\rm cf}(\Theta)>\omega$ implies the consistency of $ZF+AD\_{\mathbb R}$, so by the incompleteness theorem, ZF+AD or even the stronger $ZF+AD\_{\mathbb R}$ cannot prove that ${\rm cf}(\Theta)>\omega$.
Nowadays we know much more. For example, $\Theta$ is a Woodin cardinal in the HOD of $L({\mathbb R})$, and the computation of the large cardinal strength of $\Theta$ in the HOD of the models of AD is a guiding principle of what is now known as *descriptive inner model theory*.
You may be interested in the slides of recent talks by Grigor Sargsyan on the core model induction (which should be available somewher online, or from him by email). You will see there that the large cardinal strength of AD assumptions is calibrated by the large cardinal character of $\Theta$ inside HOD, and this is associated with the length of the so called *Solovay sequence* which keeps track of how difficult it is to define the surjections $$f:{\mathbb R}\to\alpha$$ as $\alpha$ increases. This difficulty is related to the *Wadge degree* of sets of reals present in the AD model.
(I can be much more detailed, but this will require me to get significantly more technical. Let me know.)
| 12 | https://mathoverflow.net/users/6085 | 47033 | 29,732 |
https://mathoverflow.net/questions/47038 | 4 | I've been reading some wonderful [blog entries](http://terrytao.wordpress.com/2007/09/25/the-quantitative-behaviour-of-polynomial-orbits-on-nilmanifolds/) where Terry Tao and Ben Green prove some generalizations of [Weyl Equidstribution](http://en.wikipedia.org/wiki/Equidistribution_theorem) using a "higher" Fourier Analysis. Unfortunately, all the information I can find about nilmanifolds is embedded for some [difficult papers](http://arxiv.org/pdf/0709.3562v5) (at least for non-Harmonic analysts. also [this one](http://arxiv.org/pdf/math/0606088v2)).
---
Weyl Equidistribution says you can take remainders (mod 1) of $\{n \alpha\}$ for 0 < n < N and the odds of it lying in any interval approaches the uniform distribution. (I think it's even known how quickly it converges.)
You can tell this sequence apart from purely random numbers in other ways. The "gaps" in $\{ \{ n \alpha\}: 0 < n \leq N \}$ take up to three values while random numbers on the circle have [Poisson](http://en.wikipedia.org/wiki/Poisson_distribution) distributed gaps.
---
The only example of a Nilmanifold I'll mention is the **Heisenberg nilmanifold**.
$$ \left( \begin{array}{ccc}
1 & \mathbb{R} & \mathbb{R} \\\\
0 & 1 & \mathbb{R}\\\\
0 & 0 & 1
\end{array} \right) \mod
\left( \begin{array}{ccc}
1 & \mathbb{Z} & \mathbb{Z} \\\\
0 & 1 & \mathbb{Z}\\\\
0 & 0 & 1
\end{array} \right)
$$
This quotient space is the unit cube, but the quotient map is funny. I think it's $(x,y,z) \equiv ( \{ x\}, \{y\}, z - \lfloor z - x\lfloor y \rfloor \rfloor )$. It's not even clear to me the third coordinate lies in [0,1]. This can be extended to n by n matrices. **Are all Nilmanifolds quotients of the Heisenberg group in this way?** .
---
To prove equidistribution, you show the average value $e^{ i n \alpha }$ as 0 < n < N approaches 0 as N gets large. If you move around a circle enough, it's kinda of intuitive that your average location is in the center.
With these nilmanifolds (which the circle S1 is also an example) you can get equidistribution for "bracket polynomials" like $\alpha n \lfloor \beta n \rfloor $ mod 1.
**What are the analogues of the Fourier coefficients here?**
---
**Clarification**, I'm using "Heisenberg group" to mean the group of upper-triangular matrices with real entries above the diagonal. I guess I'm trying to ask if groups like
$$ \left( \begin{array}{cccc} 1 & \mathbb{R} & \mathbb{R} & \mathbb{R} \\\\
0 & 1 & \mathbb{R} & \mathbb{R} \\\\
0 & 0 & 1 & \mathbb{R} \\\\
0 & 0 & 0 & 1 \end{array} \right) \mod \left( \begin{array}{cccc} 1 & \mathbb{R} & \mathbb{Z} & \mathbb{Z} \\\\
0 & 1 & \mathbb{Z} & \mathbb{Z} \\\\
0 & 0 & 1 & \mathbb{Z} \\\\
0 & 0 & 0 & 1 \end{array} \right) $$
are nilmanifolds. [there's no typo] And if all of them look like that? It seems one way to build a nilpotent lie group abstractly is to take a lie group and quotient out $G^{(n)} = [G^{n-1},G]$.
| https://mathoverflow.net/users/1358 | Are all Nilmanifolds quotients of Heisenberg Group | The answer to the question in the title is emphatic no in higher dimensions. In dimension $3$ nilmanifolds (that are not tori) are indeed quotients of the Heisenberg group.
There are lots of nilmanifolds in each dimension $>2$, in fact nilpotent Lie algebras are not classified (and probably not classifiable as there are too many of them with no apparent structure), and as long as all structure constants of the Lie algebra are rational the corresponding Lie group has a torsion free lattice, whose quotient is a nilmanifold.
Topologically nilmanifolds are precisely the iterated principal circle bundles, e.g. in dimension $3$ any principal circle bundle over a $2$-torus is a nilmanifold. Such circle bundles are classified by the Euler number which can take any value in $\mathbb Z$, so there are countably many $3$-dimensional nilmanifolds. In dimensions $1$ and $2$ the only nilmanifolds are tori.
| 8 | https://mathoverflow.net/users/1573 | 47041 | 29,736 |
https://mathoverflow.net/questions/46770 | 7 | Regarding reals as functions from $\omega$ to $\omega$, let's say a real $f$ *eventually dominates* $g$ iff $(\exists n)(\forall m > n)[ f(m) > g(m)]$. Let's say that a (non-trivial separative) forcing poset $P$ *doesn't always add a dominating real* iff there is a generic extension by $P$ which doesn't contains a real that eventually dominates every real from the ground model. Let's say that $P$ *never adds a dominating real* iff every generic extension by $P$ doesn't contain any real that eventually dominates all the ground model's reals. I'm interested in combinatorial/order-theoretic conditions which may be necessary or sufficient for either of these notions.
* **$\omega$-closure** implies you add no reals, hence you add no dominating reals; Cohen forcing is not $\omega$-closed but it never adds a dominating real
* one can show that **separability** implies you never add a dominating real (by separability, I mean containing a countable dense subset); the Cohen forcing that adds uncountably many reals isn't separable but never adds a dominating real
* Hechler forcing **has size at most continuum** but always adds a dominating real; the Cohen forcing that adds more than continuum many reals (where "continuum" is "continuum as computed in the ground model" obviously) has size greater than continuum but never adds a dominating real
* Hechler forcing also has the **countable chain condition** yet adds a dominating real; the forcing that adds a function $\omega \_1 \to \omega \_1$ with countable partial functions doesn't have the ccc but it's $\omega$-closed hence adds no new reals and thus never adds any dominating reals.
My question:
**What are some combinatorial/order-theoretic conditions on a poset that are necessary and/or sufficient for the poset to never/not always add a dominating real?**
| https://mathoverflow.net/users/7521 | Characterizing forcings that don't add any dominating reals | Stefan's answer pointed me in the right direction, and then talking it over with prof. Leo Harrington we've got an answer:
A complete Boolean algebra $\mathbb{B}$ never adds a dominating real iff for any collection $\{ u \_{m,k} : m, k \in \omega \} \subset \mathbb{B}^+$ the following weaker form of weak $(\omega ,\omega )$-distributivity holds:
>
> $\prod \_{m \in \omega} \sum \_{k \in \omega} u \_{m,k} = \sum \_{f \in \omega ^{\omega}} \prod \_{n \in \omega} \sum \_{n < m < \omega} \sum \_{k < f(m)} u \_{m,k}$
>
>
>
For the reverse implication, suppose "weak weak $(\omega ,\omega )$-distributivity" holds, and for contradiction let $u \neq 0$ be the Boolean value of the sentence "there exists a dominating real," and let $\dot{g}$ be a name witnessing this, i.e. the sentence "$\dot{g}$ is a dominating real" has Boolean value $u$. Define $u \_{m,k} = || \dot{g} (m) = k ||$. Now if $G$ is any $\mathbb{B}$-generic filter containing $u$, noting that the left side of the distributivity identity is (at least) $u$, we know that the right side belongs to $G$. It's then not hard to see that:
>
> $(\exists f \in (\omega ^{\omega})^V )(\forall n \in \omega )(\exists m > n)(\exists k < f(m))(u \_{m,k} \in G)$
>
>
>
which is to say that there's a real $f$ in the ground model such that:
>
> $(\forall n)(\exists m > n)(\dot{g}^G (m) < f(m))$
>
>
>
so $f$ is not dominated by $\dot{g}$, contradiction.
For the forward implication, it should suffice to show it in the case where for each $m$, the set $\{ u \_{m,k} : k \in \omega \}$ is an antichain with least upper bound $u$ independent of $m$ (I haven't checked this detail personally). So let $\{ u \_{m,k}\}$ be such a collection for which the identity fails. Consider the name:
>
> $\dot{g} = \{ (u \_{m,k}, (m,k)) : m, k \in \omega \}$
>
>
>
It's not hard to see that the right side of the identity is at most $u$, so assuming the identity fails it's strictly less than $u$, so since $\mathbb{B}$ is separative there's a generic $G$ containing $u$ avoiding the right side of the identity. It's not hard to see from here that $\dot{g}^G$ will dominate all the ground model's reals.
---
I should add that if we think of $u \_{m,n}$ as saying "$\dot{g}(m) = n$" and replace the Boolean operations with the corresponding quantifiers, then the left side says "$\dot{g}$ is a real," and the right side says "$\dot{g}$ doesn't dominate every real in the ground model." This suggests how we can characterize forcings that don't add any unbounded reals, for example, namely the following identity holds:
>
> $\prod \_{m \in \omega} \sum \_{k \in \omega} u \_{m,k} = \sum \_{f \in \omega ^{\omega}} \prod \_{m \in \omega} \sum \_{k < f(m)} u \_{m,k}$
>
>
>
Forcings that don't add any reals are precisely those that satisfy the following identity:
>
> $\prod \_{m \in \omega} \sum \_{k \in \omega} u \_{m,k} = \sum \_{f \in \omega ^{\omega}} \prod \_{m \in \omega} u \_{m,f(m)}$
>
>
>
You can easily generalize this to talking about functions $\kappa \to \lambda$; the above two results so generalized are precisely Theorem 15.38 and Lemma 15.39 in Jech, "Set Theory".
| 7 | https://mathoverflow.net/users/7521 | 47050 | 29,740 |
https://mathoverflow.net/questions/47057 | 7 | On page 3 of *Introduction to Lattices and Order*, Davey and Priestley define an antichain in a poset $\langle P,\leq\rangle$ as a set of pairwise **incomparable** elements:
>
> The ordered set P is an antichain if $x\leq y$ in P only if $x=y$
>
>
>
Gratzer's definition is equivalent, but stated in a manner which is difficult to excerpt.
On page 53 of *Set Theory, an Introduction to Independence Proofs*, Kunen defines an antichain in $\langle P,\leq\rangle$ as a set of pairwise **incompatible** elements, saying that two elements $p$ and $q$:
>
> are *incompatible* ($p\bot q$) iff $\neg\exists r\in P(r\leq p\wedge r\leq q)$. An *antichain* in $P$ is a subset $A\subset P$ such that $\forall p,q\in A(p\neq q\rightarrow p\bot q)$.
>
>
>
So, given a three-element partially ordered set $\{0,a,b\}$ with $0\leq a$, $0\leq b$ the only (non-reflexive) related pairs in the partial order, it appears that $\{a,b\}$ is an antichain in the lattice sense but not in the forcing sense.
**Question**: is it in fact true that "antichain in a poset" means something different to set theorists than to lattice theorists?
| https://mathoverflow.net/users/2361 | Does "antichain" mean something different in set-forcing than in lattice theory? | Adam:
Yes, the notions are different, but I believe the ambiguity is older than forcing; doesn't Halmos use "antichain" for the forcing notion in his book on Boolean algebras?
Typically, when the need arises of distinguishing both notions, I've seen used (and used myself) "$A$ is a weak antichain" for "the elements of $A$ are pairwise incomparable", while "$A$ is a strong antichain" is reserved for the forcing version, "the elements of $A$ are pairwise incompatible."
Usually context suffices to know which version is used. In combinatorial contexts I would think using "antichain" for the "weak" version is more common. Certainly whenever forcing is used, it is the "strong", Boolean- (or forcing-)theoretic version that is used. In any paper where ambiguity could be an issue, I've seen at least a remark.
| 9 | https://mathoverflow.net/users/6085 | 47059 | 29,748 |
https://mathoverflow.net/questions/42461 | 8 | Background
----------
The origin of this question is a conversation I had with some friends a few years ago. At the time, Roger Federer and Tiger Woods were dominating professional tennis and golf, respectively, and we were comparing and contrasting the two. It occurred to me that there was a mathematical question that was relevant to our discussion; namely, the structure of golf tournaments vs. tennis tournaments.
For example, the Masters is a four-round tournament. After two rounds, roughly the bottom half of the field is sent home. The winner is the person with the lowest total score after four rounds. On the other hand, Wimbledon is a single-elimination tournament; the winner must defeat seven other players in head-to-head competition. From a structural standpoint, if you are the best player in the field, is it harder to win a tournament like the Master's or harder to win a tournament like Wimbledon?
---
The mathematics
---------------
More generally, how does a tournament's structure affect the likelihood that the best player will win?
This question is a bit fuzzy because there are some modeling issues involved that will affect the answer. Something like the following seems a reasonable place to start.
>
> Let $F\_1, F\_2, \ldots F\_n$ be independent normal distributions such that a random variable drawn from distribution $F\_i$ gives the performance by the $i$th best player in a particular round of competition. So we would have $\mu\_{F\_i} > \mu\_{F\_j}$ when $i < j$. Assuming that the $F\_i$'s have the same variance seems a reasonable starting point, too, as does the assumption that a given player's performances from round to round are independent.
>
>
> First question: What is the best way to model the $\mu\_{F\_i}$'s vs. the $\sigma\_{F\_i}$'s? The difference $\mu\_{F\_1} - \mu\_{F\_2}$ ought to be much larger than $\mu\_{F\_{99}} - \mu\_{F\_{100}}$, so maybe something like $\mu\_{F\_i} = 1/i$ would work, but I'm not sure what makes for a reasonable variance to go with this function.
>
>
>
---
For specificity's sake, let's assume three types of tournament structure: 1) that of the Masters, 2) that of Wimbledon, and 3) that of the World Cup (which has a round-robin stage before moving to a single-elimination stage).
>
> Second question: Given a satisfactory answer to the first question, what is the probability that Player 1 will win each of these three tournaments?
>
>
>
---
My reasoning so far
-------------------
It seems to me that the two most important factors involved that would prevent the best player from winning the tournament are
1. an unusually poor performance from the best player in a particular round, and
2. an unusually good performance from someone else in the field in a particular round.
There's not much that the tournament's structure could do to mitigate factor (1), although a single-elimination tournament would seem to be the most unforgiving. On the other hand, the structure of the tournament probably has a large effect on the impact of factor (2). For instance, an incredible performance from someone in two separate rounds of the Masters raises the bar quite a bit for the best player. On the other hand, in a tournament like Wimbledon two great performances might lead to two upsets of major players but doesn't provide any advantage in later rounds, and, for the best player to be negatively affected, he/she would have to be playing directly against the overperforming player. Also, if there are enough players around (like the early rounds of Wimbledon and all the way through the Masters) there is a high probability that someone in the field will turn in two great performances in two different rounds.
So, if you are the best player in the field it seems to me that contests in which you are essentially playing most of the field simultaneously, like the Masters, would be more difficult to win than single-elimination tournaments like Wimbledon, which in turn would be more difficult to win than those with a round-robin format in the early rounds and single-elimination in the later rounds, like the World Cup.
---
>
> Third question: Are there any known results that address this problem of the effect of tournament structure on the best player's chances of winning?
>
>
>
I would be happy to see critiques/comments on my modeling and my reasoning as well.
| https://mathoverflow.net/users/9716 | How does a tournament's structure affect the likelihood that the best player will win? | It turns out this problem has been studied extensively in the economics literature. The motivation is to create some sort of competition that will maximize the likelihood of the best candidate for a job or the best application for a grant actually being awarded the job or grant.
For example, "[The Predictive Power of Noisy Elimination Tournaments](http://www.cerge.cuni.cz/pdf/wp/Wp252.pdf)," by Dmitry Ryvkin, examines the effects of seeding under different numbers of players and some different performance probability distributions.
The paper "[Three Prominent Tournament Formats: Predictive Power and Costs](http://www.cerge-ei.cz/pdf/wp/Wp303.pdf)," (apparently published in *Management Science* under the title "[The Predictive Power of Three Prominent Tournament Formats](http://mansci.journal.informs.org/cgi/content/abstract/54/3/492)"), by Ryvkin and Andreas Ortmann, addresses my question exactly, though. They calculate the exact probability (under uniform, normal, and Pareto distributions for player performance) that the best player wins a round robin tournament, a binary elimination tournament, and a contest. (The last involves all players performing simultaneously at once; the winner is the player with the best performance.) By calculating these probabilities for specific values they show numerically that for all but small numbers of players in a tournament, the best player in the tournament has a higher probability of winning a round robin tournament than a binary elimination tournament and a higher probability of winning a binary elimination tournament than a contest.
Given these results, it does appear that (all other things being equal) the structure of golf tournaments makes them more difficult to win than tennis tournaments, and, consequently, that dominating professional golf is even more impressive than dominating professional tennis.
| 5 | https://mathoverflow.net/users/9716 | 47062 | 29,750 |
https://mathoverflow.net/questions/47048 | 8 | Suppose $f: \mathbb{C}^n \to \mathbb{C}^n$ is a proper polynomial mapping and $\gamma: [0,1] \to \mathbb{C}^n$ is a continuous path. Further, suppose $z\_0 \in \mathbb{C}^n$ satisfies $f(z\_0)=\gamma(0)$. Does there exist a (not necessarily unique) lift of $\gamma$ under $f$ based at $z\_0$? Is there a published reference for this fact?
| https://mathoverflow.net/users/7844 | Do proper polynomial mappings have a path-lifting property? | The answer is yes, here is a proof.
Let $k=2n$. The polynomial, regarded as a map $f:\mathbb R^k\to\mathbb R^k$, has the following properties:
(1) The pre-image of every point if finite, moreover the cardinality of a pre-image is uniformly bounded by some constant $N$ (by Bezout's theorem, see comments).
(2) The set $\Sigma$ of singularities of $f$ (i.e. points where $\det df=0$) is a union of finitely many smooth submanifolds of (real) codimension at least 2.
This (along with smoothness and properness) implies the path-lifting property.
First, observe the following facts:
(3) $f(\Sigma)$ is closed (because $\Sigma$ is closed and $f$ is proper).
(4) $\mathbb R^k\setminus f(\Sigma)$ is path connected. Moreover any path in $\mathbb R^k$ is a uniform limit of paths avoiding $f(\Sigma)$. This follows from the fact that $f(\Sigma)$ has Hausdorff dimension at most $k-2$.
(5) For every compact set $K\subset\mathbb R^k$ and every $\varepsilon>0$, there exists $\delta>0$ such that for every connected set $S\subset K$ of diameter greater than $\varepsilon$, the diameter of $f(S)$ is greater than $\delta$. Indeed, if this is not the case, there would be a sequence $S\_i$ of connected subsets of $K$ with diameters at least some $\varepsilon\_0>0$ and diameters of images going to zero. By choosing a subsequence, we may assume that the images converge to some $y\_0\in\mathbb R^k$.
Since $S\_i$ is connected and has diameter at least $\varepsilon\_0$, it contains a finite subset $P\_i$ of cardinality $N+1$ such that all points of $P\_i$ are separated away from one another by distance at least $\varepsilon\_0/2N$. A subsequence of $\{P\_i\}$ converge to a set $P$ of cardinality $N+1$, and all points of $P$ are mapped by $f$ to $x\_0$, contrary to (1).
Now, in order to lift a path $\gamma$, approximate it by paths $\gamma\_i\subset\mathbb R^k\setminus f(\Sigma)$. The restriction of $f$ to the pre-image of $R^k\setminus f(\Sigma)$ is a covering map to $R^k\setminus f(\Sigma)$ because it is a proper local homeomorphism. So there are lifts $\tilde\gamma\_i$ of $\gamma\_i$. Now it suffices to find a converging subsequence of $\{\tilde\gamma\_i\}$. To do this, it suffices to show that this sequence is equi-continuous. And this follows from (5): if you could find arbitrarily close pair of points on some $\gamma\_i$ with lifts bounded away from each other, the lift of the segment of $\gamma\_i$ between these points would have diameter bounded away from zero but the diameter of the image arbitrarily small, contrary to (5).
To make sure that the limit lift starts from $z\_0$, choose $\gamma\_i$'s so that their starting points have pre-images near $z\_0$ and start $\tilde\gamma\_i$'s from there.
| 9 | https://mathoverflow.net/users/4354 | 47067 | 29,754 |
https://mathoverflow.net/questions/33582 | 16 | This question concerns the characteristic $0$ representation theory of the symmetric group $S\_n$. I'm a topologist, not a representation theorist, so I apologize if I state it in an odd way.
First, a bit of background. The finite-dimensional irreducible representations of $S\_n$ are given by the [Specht modules](https://en.wikipedia.org/wiki/Specht_modules) $S^{\mu}$. Here $\mu$ is a partition of $n$, which is best visualized as a [Young diagram](https://en.wikipedia.org/wiki/Young_diagram). There are classical rules for restricting $S^{\mu}$ to $S\_{n-1}$ and inducing $S^{\mu}$ to $S\_{n+1}$ (these are known as branching rules). Namely, we have the following.
1. The restriction of $S^{\mu}$ to $S\_{n-1}$ is isomorphic to the direct sum of the $S\_{n-1}$-representations $S^{\mu'}$ as $\mu'$ goes over all ways of removing a box from the Young diagram for $\mu$ (while staying in the world of Young diagrams).
2. The induction of $S^{\mu}$ to $S\_{n+1}$ is isomorphic to the direct sum of the $S\_{n+1}$-representations $S^{\mu'}$ as $\mu'$ goes over all ways of adding a box to the Young diagram for $\mu$ (again while staying in the world of Young diagrams).
These two rules are equivalent by [Frobenius reciprocity](https://en.wikipedia.org/wiki/Frobenius_reciprocity).
There is a nice concrete proof of the restriction rule (I believe that it is due to Peel, though I first learned about it from James's book "The representation theory of the symmetric groups"). Assume that the rows of the Young diagram for $\mu$ from which we can remove a box are $r\_1 < \ldots < r\_k$, and denote by $\mu\_i$ the partition of $n-1$ obtained by removing a box from the $r\_i^{\text{th}}$ row of $\mu$. There is then a sequence
$$0=V\_0 \subset V\_1 \subset \cdots \subset V\_k = S^{\mu}$$
of $S\_{n-1}$-modules such that $V\_i/V\_{i-1} \cong S^{\mu\_i}$. In fact, recalling that $S^{\mu}$ is generated by elements corresponding to Young tableaux of shape $\mu$ (known as polytabloids), the module $V\_i$ is the subspace spanned by the polytabloids in which $n$ appears in a row between $1$ and $i$.
**Question** : Is there a similarly concrete proof of the induction rule (in particular, a proof which does not appeal to Frobenius reciprocity and the restriction rule)?
| https://mathoverflow.net/users/317 | Making the branching rule for the symmetric group concrete | Chapter 17 of James' book, The Representation Theory of the Symmetric Groups, is about modules which have Specht filtrations (where the field is arbitrary) and includes induction of the Specht modules as a special case. (See in particular Example 17.16.) It gives a concrete proof of the branching rule for induction without using restriction.
| 6 | https://mathoverflow.net/users/10487 | 47087 | 29,761 |
https://mathoverflow.net/questions/47086 | 3 | * Let $(\hat{M},\hat{g})$ be the conformal compactification of a space-time $(M,g)$. Let $I^+$ be the conformal null infinity and $J^{-}(I^+)$ be its causal past. Then the spacetime will be called "asymptotically strongly predictable" if there exists a subset $\hat{V}$ of $\hat{M}$ such that $(\hat{V},\hat{g})$ is globally hyperbolic and contains the closure in $\hat{M}$ of $M \cap J^{-}(I^+)$. Define as a "black hole" the region $B=M /\ J^{-}(I^+)$.
1. I think from the above it should imply that there cannot be a causal curve which starts in the region $B$ and enters $M \cap \hat{V}$. Is this correct and if yes then how does one prove it?
(In specific examples what happens is that such a curve cannot remain always causal if it has to do such a journey. But I guess this not a good way of thinking since the restriction is purely topological)
1. Under the above conditions when does it also follow that $(M \cap \hat{V},\hat{g})$ and/or $(M\cap \hat{V},g)$ is globally hyperbolic?
2. What is the meaning of "trace of a terminally indecomposable past set" ? Is there any special significance if a compact set happens to be the trace of a terminally indecomposable past set of a maximal future directed time-like curve? (Reading around I get the impression that such compact sets somehow specify physically reasonable initial conditions)
| https://mathoverflow.net/users/2678 | Some questions about causal structure of space-time. | 1st statement is false. You need to add "future directed", else a past causal curve can of course leave the black hole region, since time-reversed, it is just a white-hole. For the correct statement, the proof is immediate following the usual causal relations of Penrose and Kronheimer: if $\exists$ such a causal curve, there must $\exists p,q$ such that $p\in B$ and $q\in \hat{V}$ with $q \in J^+(p)$. Which is equivalent to $p \in J^-(q)$. Now using that the causal relations form an ordering, you have that
$$p \in J^-(q), q\in J^-(I^+) \implies p \in J^-(I^+) $$
contradicting the assumption that $p\in B$. (As a side remark, this is very, very basic stuff. If the question were just about this fact, I would've voted to close. You should read E. H. Kronheimer and R. Penrose (1967). On the structure of causal spaces. Mathematical Proceedings of the Cambridge Philosophical Society, 63.)
(Second side remark: I don't think the word topological means what you think it means. Pure topology is not enough to contrain causal structure, beyond some trivial things about Euler characteristic and completely compact space-times.)
---
For the second question (labeled 1 again by the Markdown software), it is important to remember that the causal structure of space-time is conformally invariant, and that global hyperbolicity is an **intrinsic notion** rather than extrinsic. To say it another way, you need to remember that
1. If $(V,g)$ is a globally hyperbolic space-time domain, then for any conformal change of metric $g\to \hat{g}$, $(V,\hat{g})$ is also globally hyperbolic. (This is immediate using the Cauchy surface definition, since conformal changes preserve causal relations; using the compact causal diamond definition, you need to remember that continuous functions attain its maxima and minima on a compact set.)
2. If $(V,g)$ is a globally hyperbolic space-time domain, and let $\phi$ be an isometric embedding of $(V,g)\to (M,\hat{g})$, then, the restricion $(\phi(V),\hat{g})$ is also globally hyperbolic.
So, $\hat V$ embeds into $\hat V \cap M$ isometrically by the identity map. And $(\hat V, \hat g)$ is globally hyperbolic, so by the above two points its image must be globally hyperbolic. (By the way, the above two observations are trivial consequences of the definitions for global hyperbolicity. You should be able to prove them yourself.)
---
For your third question (as a side remark: the question is actually pretty poorly posed. If you are going to ask about terminology, please at least provide a reference on which paper it is in which you found the phrase that is confusing you; my crystal ball tells me that you are asking about Christodoulou's 1999 CQG paper "On the global initial value problem and the issue of singularities", but most other people won't have a convenient divination device in their office.), it helps to note that the original phrase is about
>
> the trace of a terminally indecomposable past set on a space-like hypersurface.
>
>
>
And as such it means the same thing generally meant when you take the trace of *any* object onto a hypersurface: as a set you want to consider the intersection of whatever object you are talking about with the hypersurface, and you want to inherit any geometric structure by pushing forward with the restriction operator.
In context the condition that "the trace of a TIP on a space-like hypersurface is compact" means that the intersection of the TIP with the space-like hypersurface is a compact set in the induced topology of the hypersurface.
The intuition is that in Minkowski space (or any small perturbations of it), the only TIP associated to maximal time-like geodesics is the whole space-time. And in particular, its trace on any Cauchy hypersurface **cannot** be compact. (For maximal time-like curves, you can get particle horizons if the curves become asymptotically null, but it is easy to see that those TIPs correspond to a half-space in Minkowski space that lies below a null hyperplane, so the same conclusion holds.) In general, if you have a compactification of space-time with a future time-like infinity (the terminus of all maximally extended, complete future time-like curve), then the TIP associated to it must have non-compact trace on any Cauchy hypersurface.
So in fact, I think your parenthetical impression is completely wrong. That a TIP has compact trace should suggest to you that the maximal time-like curve your are looking at is incomplete, which should suggest to you that its terminus is a "singularity". (You should compare this picture with the usual picture in the Penrose singularity theorem.) Which is why Christodoulou formulated his version of weak cosmic censorship conjecture in terms of such sets. (It would also do you good to work through the paper of Geroch, Kronheimer and Penrose, "Ideal Points in Space-time", Proc. Roy. Soc. London (1972). )
| 5 | https://mathoverflow.net/users/3948 | 47091 | 29,764 |
https://mathoverflow.net/questions/47082 | 5 | Let $M$ be a complex manifold, and $\omega$ be a $(p,q)$-form. Then $d\omega$ is an element of $\Omega^{p+1,q}(M)\oplus\Omega^{p,q+1}(M)$, so that $d = \partial + \overline{\partial}$, where $\partial$ and $\overline{\partial}$ are the Dolbeault operators.
Now let $M$ be *almost* complex. It is commonly stated that $d = \partial + \overline{\partial}$ only holds for complex manifolds, and not for almost complex manifolds. But why is this? After extending $d$ to also be complex linear, if $\omega = \sum\_i f(z)dz^i$ is a $(0,1)$-form, I'd say that we would have $ d\omega = \sum\_i df\wedge dz\_i
= \sum\_{i,j} \frac{\partial f}{\partial z^j}dz^j\wedge dz^i + \frac{\partial f}{\partial \overline{z}^j}d\overline{z}^j\wedge dz^i,$
which clearly does not have a $(0,2)$-part. Why is this wrong?
On the other hand, let $X, Y$ be antiholomorphic tangent vectors, then $d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]) = -\omega([X,Y])$. Since $M$ is not nessecarily complex, $[X,Y]$ is not nessecarily also antiholomorphic, so that this term does not nessecarily vanish. But $d\omega$, being a 2-form, can only give a nonzero result if it has a $(0,2)$-part. So from this I can see that it has to have one, but I can't see why this contradicts the calculation of $d\omega$ above.
| https://mathoverflow.net/users/11031 | Exterior derivative on almost complex manifolds | In writing $\omega$ you used a symbol $dz$ which doesn't make sense unless there is a holomorphic coordinate. Your $dz$ should really be an element of a frame of (1,0) 1-forms, which need not be closed (as you have assumed).
| 15 | https://mathoverflow.net/users/1186 | 47092 | 29,765 |
https://mathoverflow.net/questions/46350 | 27 | It is well known that primitive recursion is not powerful enough
to express all functions, Ackermann function being probably the best
known example.
Now, in the logic courses (that I have had look at) one always proceeded from primitive recursion to mu-recursion. In computer science terms this basicly means we are jumping from a formalism where programs are quaranteed to halt to a Turing-complete formalism where halting is a non-computable property i.e. we can't say for every program if it will eventually halt.
I got curious if there is any hierarchy between primitive recursion and
mu-recursion. After a while I found a programming language called Charity. In Charity (according to Wikipedia)
all programs are quaranteed to stop, thus its not Turing-complete, but,
on the other hand, it is expressive enough to implement Ackermann function.
This suggests there is at least one level between mu-recursion and primitive recursion.
My question is: does there exists any other halt-for-sure formalisms that are more expressive than primitive recursion? Or, even better, does there exist some known hierarchies between mu-recursive and primitive recursive functions? I'm curious about how "much" we can compute with a formalism that guarantees halting.
| https://mathoverflow.net/users/nan | Between mu- and primitive recursion | I think it would also be reasonable to (explicitly) mention the functions definable by primitive recursion *in higher types* - instead of only defining functions $\mathbb{N}\to\mathbb{N}$ by primitive recursion, we may also define families of functions (i.e. functions $\mathbb{N}\to\mathbb{N^N}$) by primitive recursion (and, of course, even more complicated things).
As an example, let us first define an iteration function $g\colon\mathbb{N}\times\mathbb{N^N}\to\mathbb{N^N}$ by primitive recursion, by:
\begin{array}{l}
g(0,f) = i \qquad\qquad\text{(the identity function)}\\\\
g(n+1, f) = f\circ g(n, f)
\end{array}
We can then easily define a `curried' version of the Ackermann function as a function $A\colon\mathbb{N}\to\mathbb{N^N}$, by:
\begin{array}{l}
A(0) = S \quad\quad\text{(successor)}\\\\
A(n+1) = g(n, A(n))
\end{array}
This is quite popular with (some groups of) computer scientists, particularly those interested in functional programming, and some form of primitive recursion in higher types is behind the strength of the language Charity you mentioned (which also allows primitive recursion over other data structures than the natural numbers).
| 16 | https://mathoverflow.net/users/11035 | 47098 | 29,769 |
https://mathoverflow.net/questions/47103 | 19 | Is every field the field of fractions of an integral domain which is not itself a field?
What about the field of real numbers?
| https://mathoverflow.net/users/11030 | Is every field the field of fractions of an integral domain? | Every field $F$ of characteristic zero or of prime characteristic
but not algebraic over its prime field
is the field of fractions of a proper subring of $F$.
But no algebraic extension of $\mathbb F\_p$ is, since its only subrings are fields.
If $F$ is not an algebraic extension of some $\mathbb F\_p$
then $F$ contains a subring $A$ isomorphic to $\mathbb Z$
or $\mathbb F\_p[X]$. Each of these rings $A$ has a nontrivial
valuation $v$. The valuation $v$ can be prolonged to $F$.
Its valuation ring is a proper subring of $F$ whose quotient field
is $F$.
| 30 | https://mathoverflow.net/users/4213 | 47106 | 29,775 |
https://mathoverflow.net/questions/46998 | 23 | Roth first proved that any subset of the integers with positive density contains a three term arithmetic progression in 1953. Since then, many other proofs have emerged (I can think of eight off the top of my head).
A lot of attention has gone into the bounds in Roth's theorem, and in particular what kind of bounds different proofs get you (e.g. Fourier analysis gives log type bounds, regularity lemma arguments give Ackermann type bounds, etc.)
Also, some proofs are more amenable to generalisation (to longer arithmetic progressions) than others.
My question is
>
> If we are only concerned with brevity and directness (i.e. not with a deeper theoretical understanding or sharp quantitative bounds), what is the shortest proof of Roth's theorem?
>
>
>
Running through the arguments I know, it seems like the shortest one may be Roth's original one (with a couple of simplifications): show there is a large Fourier coefficient and deduce some sort of density increment argument and iterate it - a good exposition is in Tao and Vu, or Ben Green's notes at <http://people.maths.ox.ac.uk/greenbj/notes.html> . With all details fleshed out, this could probably be done in 8 pages or an hour lecture.
The odd thing is that this proof also gives fairly good quantitative bounds; if $r\_3(N)$ is the size of the largest subset of $\{1,...,N\}$ without three term arithmetic progressions, then even a crude version of this argument gives
$$ r\_3(N)=O\left(\frac{N}{\log\log N^c}\right)$$
whereas all Roth needs is $o(N)$. Hence my second question,
>
> Is it inevitable that the most direct and simple proofs would also lead to fairly good quantitative bounds?
>
>
>
Finally, an exercise in Tao and Vu mentions that Behrend's example of a lower bound for $r\_3(N)$ shows that simple pigeonhole type arguments couldn't be used to prove Roth's theorem. Hence,
>
> What other proof techniques wouldn't work with Roth's theorem?
>
>
>
| https://mathoverflow.net/users/385 | What is the shortest route to Roth's theorem? | There's a short-cut in Roth's approach if one only cares to get $o(N)$. Adolf Hildebrand told me so, and here is my shortest writeup.
**Notation:** Let $r(N),\rho(N)$ be the largest cardinality and density of a subset of $[N]$ that is free of 3-term APs, and let $\rho=\lim \rho(N)$, which must exist by [Fekete's subadditive lemma](http://seaneberhard.blogspot.hu/2013/01/feketes-lemma-and-sum-free-sets.html) since $r(N+M)\leq r(N)+r(M)$. Let $A\subset [N]$ witness $r(N)$, and let $S(x) = \sum\_{a\in A} e( a x)$ (where $e(ax)=\exp(2\pi i a x)$, naturally). Let $T(x)=\sum\_{n=1}^N e(nx)$.
**Lemmata:** Now $r(N)=|A|=I:=\int\_0^1 S(x)^2 S(-2x)dx$, since $A$ is 3-free. By Parseval, $|A|=\int\_0^1 |S(x)|^2dx$. By expanding $S(x)^2 T(-2x)$ and exchanging integration and summation, $(|A|/2)^2 \leq I\_0:=\int\_0^1 S(x)^2 T(-2x)$.
**Main Engine:** As long as $0< M\leq N$, $$\sup\_{x\in {\mathbb R}} |S(x)-\rho(M) T(x)| \leq N(\rho(M)-\rho(N)) + 10 M \sqrt{N}.$$ Proof is by circle method; set $E(x):=S(x)-\rho(M)T(x)$. Three steps are
* $|E(a/q)|\leq N(\rho(M)-\rho(N))+2Mq$, the hard one;
* $|E(a/q+\beta)| \leq N(\rho(M)-\rho(N))+2Mq +2\pi|\beta|NMq$;
* $|E(x)| \leq N(\rho(M)-\rho(N))+10M\sqrt{N}$.
**Main Lemma:** By Main Engine and Lemmata, $\Delta:=|I-\rho(M)I\_0|$ is at most $(N(\rho(M)-\rho(N))+10M\sqrt{N})|A|$. By Lemmata, $\Delta$ is at least $|A|(\rho(M)\rho(N)N/4-1)$. Therefore, $$\rho(N)\rho(M)\leq 4(\rho(M)-\rho(N))+50M/\sqrt{N}.$$
**Conclusion:** Let $N\to\infty$ and then $M\to\infty$ to get $\rho^2\leq 0$.
The step labeled "the hard one" in the Main Engine is tricky, and uses the fact that we can restrict $A$ to arithmetic progressions and still get a 3-free set. The length of the progressions we restrict to is connected to $M$.
I suppose the point is that a write-up can be almost arbitrarily short or extremely long, depending on what the reader can be trusted to fill in.
| 17 | https://mathoverflow.net/users/935 | 47117 | 29,781 |
https://mathoverflow.net/questions/47095 | 2 | Hi, if the Fourier series development of $g(t)$ (periodic, $C^\infty$) is
$$
g(t)=\sum\_{-\infty}^{+\infty}a\_n e^{in\omega t}
$$
does the series
$$
\sum\_{-\infty}^{+\infty}\frac{a\_n^2}{n^2}?
$$
converges toward something known like average $g^2$ or something like that?
| https://mathoverflow.net/users/9039 | Series of squared Fourier coefficients | Assume $a\_0=0$, which easily can be arranged by adding a constant to $g$.
Then the function
$$
h(t)=\frac1{i\omega } \sum\_{n}\frac{a\_n}ne^{in\omega t}
$$
is the primitive of $g$.
Let $h^\* (x)=\overline{h(-x)}$, then the sum you asked for equals the inner product
$$
\langle h,h^\*\rangle.
$$
| 3 | https://mathoverflow.net/users/nan | 47126 | 29,786 |
https://mathoverflow.net/questions/47122 | 3 | I am wondering if there are some results about the depth of a diffeomorphism on a manifold.
More precisely, $(M,f)$ be a diffeomorphism. For each compact invariant subset $E$, let $\Omega(f, E)$ be the nonwandering subset of $f$ relative to $E$. Let $\Omega\_1=\Omega(f,M)$, $\Omega\_{n+1}=\Omega(f,\Omega\_n)$, and $\Omega\_a=\cap\Omega\_b$ over $b < a$ for a limit cardinal $b$... etc
So my question is; are there some conditions under which the diffeo has a finite depth, that is, $\Omega\_{n+1}=\Omega\_n$ for some $n$?
There are examples of topological systems with countable depths.
I do not know what can happen in the smooth category.
I googled and found that the depths of circle maps or interval maps are less than 2.
Thanks!
---
To rpotrie: I am looking for sufficient conditions on the spaces (say, manifolds) and the maps (say, the regularity) such that $f$ has finite center depth. As rpotrie mentioned, Axiom A maps (hence all Anosov) always have center depth 1, the maps with $\Omega(f)$ hyperbolic have center depth less than 2.
For example partially hyperbolicity may not be a good candidate since the direct product $f\oplus g:M\times N\to M\times N$ has transfinite center depth if one of $f$ or $g$ has.
| https://mathoverflow.net/users/11028 | nonwandering set and Birkhoff center | There are some results which guaranty that the depth is $1$, but I don't know if they are quite what you are looking for.
First, in the case of Homeomorphisms and $C^1$-diffeomorphisms of a Manifold, the famous [Closing Lemma](http://www.scholarpedia.org/article/Pugh_closing_lemma) implies that generic homeomorphisms (resp. generic $C^1$-diffeomorphisms) verify that the set of periodic points is dense in the non-wandering set, and thus, the depth is one.
Also, when the map has some hyperbolic conditions, one knows that in surfaces the nonwandering set must be the closure of hyperbolic points. This is a Theorem by Newhouse and Palis which says that: If $f$ is a surface diffeomorphism and the nonwandering set is hyperbolic, then, it is the closure of its periodic points (and thus, the depth is one).
In higher dimensions, there are examples [by Dankner](http://www.jstor.org/pss/1971127) of diffeomorphisms whose nonwandering set is hyperbolic but the periodic points are not dense (in this hyperbolic case, this implies that the depth is $>1$). In this examples, the depth is exacly two, since for a hyperbolic set $\Lambda$, the periodic points are dense in $\Omega(f|\_\Lambda)$.
Other of this results, the only knowledge I have is that examples where the depth is bigger than one are extremely patological (although I don't believe it is even known for $C^r$ topology with $r\geq 2$ if having depth one is a generic property, but maybe it can be proved that the depth is finite generically).
| 4 | https://mathoverflow.net/users/5753 | 47127 | 29,787 |
https://mathoverflow.net/questions/47134 | 16 | It's an obvious and well-known fact that there is no uniform probability measure on a set of natural numbers (i.e. the one that gives the same probability to each singleton).
On a recent probability seminar one professor mentioned that this nonexistance is one of the main objections to measure-theoretic formulation of probability and that even Kolmogorov admitted the problem in it.
What are possible solutions (i.e. alternative definitions of probability) to this problem?
I'm looking either for an informative answer or a good reference.
| https://mathoverflow.net/users/6159 | "Uniform probability" on a set of naturals | The correct notion here is **amenability**. This means that there is a positive linear form $\phi$ on the space of bounded functions that is invariant under translations:
$$\phi(\tau\_nf)=\phi(f),\qquad\forall f\in\ell^\infty(\mathbb Z),n\in\mathbb Z,$$
where $\tau\_nf(m):=f(m+n)$. As mentioned by Qiaochu, it is not countably additive, but only finitely additive (when applied to characteristic functions). It satisfies in addition $\phi({\bf 1})=1$. Of course, if $f$ has a finite support,then $\phi(f)=0$. Therefore, we must forget about "counting".
Not every group is amenable. Free groups with two or more generators are not.
**Edit**. As mentionned by Nate, one does not have an explicit form of $\phi$, because its existence follows from the axiom of choice. In particular, it is not unique.
| 12 | https://mathoverflow.net/users/8799 | 47141 | 29,794 |
https://mathoverflow.net/questions/47139 | 5 | Excuse the possible naivete of this question. Since reading a nice survey article by Daniel Biss a few years ago, I'm always worried about what $P^1(R)$ is, for a ring $R$.
So that I stop worrying, I'm looking for an answer to the following question: For what (commutative, of course) rings $R$ is it true that $P^1(R)$ is naturally identifiable with the set of pairs $(a,b) \in R^2$ such that $(a,b)$ equals the unit ideal, modulo the natural action of $R^\times$?
| https://mathoverflow.net/users/3545 | When is the projective line the seminaive projective line? | This is equivalent to the property that every invertible (=rank-1 projective) $R$-module generated by two elements is free. Examples: semilocal rings, unique factorization domains, finite products of such rings.
| 10 | https://mathoverflow.net/users/7666 | 47143 | 29,796 |
https://mathoverflow.net/questions/45983 | 1 | If $$f:\mathscr{X} \to \mathscr{Y}$$ is a map between (possibly ineffective) orbifolds (in the sense of differentiable stacks, or orbifold groupoids), does it follow that $f$ induces a map between their underlying effective orbifolds? At first, I thought it should always be true, but now that I think about it, you might need $f$ to be open (in this case, I can prove it). Is this known? (by this I mean, is it known to always hold, even without this open assumption?) Note, this really has nothing to do with the differentiable structure, so you may as well ask this for proper etale topological stacks, in fact, I doubt properness plays a role.
| https://mathoverflow.net/users/4528 | When do maps of ineffective orbifolds descend to their effective part? | Not true:
Take *X* to be a non-trivial *Z*/2-gerbe on S^2, take *Y* to be a faithful vector bundle over *X*, and take *f* to be the inclusion of the zero section.
The effective quotient of *X* is S^2, and it has no map back to *X*.
In particular, it has no map to *Y* that's compatible with *f*.
| 2 | https://mathoverflow.net/users/5690 | 47147 | 29,798 |
https://mathoverflow.net/questions/47140 | 1 | Fibonacci numbers are defined by the recurrence relation
$f\_{n+2}=f\_{n+1}+f\_{n}$ and
Tribonacci numbers by
$f\_{n+3}=f\_{n+2}+f\_{n+1}+f\_{n}$
One can define, in general, K-Bonacci numbers as
$f\_{n+K}=f\_{n+K-1}+...+f\_{n+1}+f\_{n}$
(they show up naturally if you consider the problem of counting binary strings of length n which do not contain sequences of K adjacent zeroes).
The characteristic polynomial associated to K-Bonacci numbers is
$$P\_K(t):=t^K-(t^{K-1}+t^{K-2}+...+t+1)$$
By the way, he same polynomial turns up when trying to calculate the asymptotic growth rate via generating functions and, as $K \to +\infty$, the biggest real root approaches 2.
Question: **do these polynomials have a already a name?**
| https://mathoverflow.net/users/7979 | Characteristic polynomials for $K$-Bonacci numbers: what's their name? | The dominant root of such a polynomial is often referred to as a **multinacci number**. These numbers are known to be [Pisot numbers](http://mathworld.wolfram.com/PisotNumber.html) and, indeed, tend to 2.
| 4 | https://mathoverflow.net/users/8131 | 47154 | 29,802 |
https://mathoverflow.net/questions/47163 | 9 | For the univariate central limit theorem, the Berry-Esseen theorem gives a quantitative bound on the rate of convergence of distributions to the Normal distribution under Kolmogorov distance:
<https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem>
Are similar statements known for the multivariate version of the central limit theorem, that use some standard distance measure?
<https://en.wikipedia.org/wiki/Central_limit_theorem#Multidimensional_central_limit_theorem> ([current revisions](https://en.wikipedia.org/w/index.php?title=Central_limit_theorem&oldid=397609863#Multidimensional_central_limit_theorem))
This question is a re-post from
<https://math.stackexchange.com/questions/11596/quantitative-bounds-for-multivariate-central-limit-theorem>
Thanks,
| https://mathoverflow.net/users/11044 | Quantitative bounds for multivariate central limit theorem | There is a bunch of such statements which can be obtained by [Stein's method](https://en.wikipedia.org/wiki/Stein%27s_method).
You might be interested in the paper ["On the Rate of Convergence in the Multivariate CLT"](https://projecteuclid.org/euclid.aop/1176990448) by Gotze, which is specifically devoted to Berry-Esseen theorems in the multidimensional setting. Have a look also at the very recent book [*Normal Approximation by Stein's Method*](https://books.google.co.uk/books?id=5jMVpAbs9UkC&printsec=frontcover&dq=Normal+Approximation+by+Stein%2527s+Method&hl=en&ei=UFbsTPL-FcShOqrIzJEB&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCkQ6AEwAA#v=onepage&q&f=false) by Chen, Goldstein and Shao.
| 9 | https://mathoverflow.net/users/5371 | 47165 | 29,810 |
https://mathoverflow.net/questions/47185 | 19 | A *choice* function maps every set (in its domain) to an element of itself. This question concerns existence of an *anti-choice* function defined on the family of countable sets of reals. In an answer to [a question about uncountability proofs](https://mathoverflow.net/questions/46970/proofs-of-the-uncountability-of-the-reals) it was suggested that while Cantor's diagonal method furnishes a Borel function mapping each countable *sequence* $S$ of real numbers to a number not in the seqence, the same is not true for countable *sets* of reals. This seems surprising (but believable) to me and an important insight, if true. But why is it true? Put more provocatively (if this is fair, if not explain it as stated)
>
> Let $\mathcal{C}$ be the family of countable subsets of $\mathbb{R}$ .` Is it the case that for every Borel function $f:\mathcal{C} \rightarrow \mathbb{R}$ there is an $X \in \mathcal{C}$ such that $f(X) \in X$?
>
>
>
| https://mathoverflow.net/users/8008 | Why is there no Borel function mapping every countable set of reals outside itself? | The initial function you mention is the diagonalizing
function $d:\mathbb{R}^\omega\to\mathbb{R}$, for which one
ensures that $z=d(x\_0,x\_1,\ldots)$ is distinct
from every $x\_n$ simply by making the $n$-th digit of $z$
different from the $n$-th digit of $x\_n$ in some regular
way. Since the graph of this function is arithmetically
definable (needing to look only at the individual digits
of the input and output), it follows that $d$ is a Borel
function.
The point of your question, however, is that this function is not
well-defined on different enumerations of the same set---the resulting diagonal value will be different if you rearrange the input.
What would be desired is a function
$f:\mathbb{R}^\omega\to\mathbb{R}$ such that always
$f(x\_0,x\_1,\ldots)\neq x\_n$, but for which $f$ gives the
same value for different enumerations of the same countable
set. Unfortunately, there is no such function that is
Borel.
**Theorem.** There is no Borel function
$f:\mathbb{R}^\omega\to\mathbb{R}$ such that
* $f(x\_0,x\_1,\ldots)\neq x\_n$ for every sequence $\vec x$
and index $n$, and
* $f(x\_0,x\_1,\ldots)=f(y\_0,y\_1,\ldots)$, whenever
$\{\ x\_0,x\_1,\ldots\ \}=\{\ y\_0,y\_1,\ldots\ \}$.
Proof. The nonexistence of such a function is closely related in spirit, in the
context of Borel equivalence relation theory, to the
impossibility of a Borel reduction from the equivalence relation
$E\_{set}$ to $=$, and the argument below belongs to that
subject. The argument will use set-theoretic forcing, and
is an instance where forcing is used in order to make a
conclusion about the ground model $V$, rather than to prove an
independence result.
To begin, suppose that $f$ is a Borel function with the two
properties. Let $\mathbb{P}=\text{Coll}(\omega,\mathbb{R})$
be the forcing to collapse $\mathbb{R}$ to $\omega$. That
is, conditions in $\mathbb{P}$ are finite sequences of
reals, ordered by end-extension. The generic object will be
a countable enumeration consisting of all the reals of the
ground model $V$. Let $g$ and $h$ be mutually $V$-generic
for $\mathbb{P}$, and consider the corresponding forcing
extensions $V[g]$, $V[h]$ and their common extension
$V[g][h]$. Since $f$ was a Borel function, it has a Borel
code that may be re-interpreted in any of these universes.
Furthermore, the assertion that $f$ has the stated features
is a $\Pi^1\_1$ statement about this Borel code, and hence
absolute between $V$ and these larger universes. That is,
the re-interpreted function $f$ continues to have the
desired properties in $V[g][h]$. Since $g$ and $h$ both
enumerate the same set $\mathbb{R}^V$, it follows that
$f(g)=f(h)$ in $V[g][h]$. In particular, the value
$z=f(g)=f(h)$ is in both $V[g]$ and $V[h]$. But since $g$
and $h$ are mutually generic, it follows that $V[g]\cap
V[h]=V$, and so $z\in V$. But this contradicts the fact
that $f(g)$ should be a real not listed in $g$, since $g$
lists all the reals of $V$, including $z$. Contradiction!
QED
The practitioners of Borel equivalence relation theory have
a large bag of tools at their disposal---many arguments
proceed with one's choice of forcing or ergodic theory and
group actions or something else---and I expect similarly
that there is a forcing-free proof of the theorem above
(perhaps someone can post such an argument?). But to my way of thinking, the
forcing proof is fairly sharp.
Lastly, let me say that if there are sufficient large
cardinals, then projective truth is absolute from $V$ to
$V[g][h]$, and in this case, the same argument shows that
there can be no projective function $f$ with the two
properties. Since the Borel functions sit merely at the
doorstep of the projective hierarchy, this would be an
enormous expansion of the phenomenon.
| 28 | https://mathoverflow.net/users/1946 | 47191 | 29,823 |
https://mathoverflow.net/questions/47184 | 5 | Let X be a projective variety (say, irreducible) and E a vector bundle on X or rank r. Is it true that there exists a codimension 2 closed subset Z in X such that restriction of E(n) (for n large enough) to U = X - Z has a trivial sub-bundle of rank (r-1)? Is this written somewhere? What happens when X varies in a flat family over a base S?
UPD: corrected the question - an ample twist of E is supposed to have a trivial sub-bundle.
| https://mathoverflow.net/users/11051 | trivial subbundles | Stated like this it can't be true. Indeed if $X$ is normal then by Hartogs lemma each embedding $O\_{X-Z} \to E\_{|X-Z}$ (i.e. a section of $E\_{|X-Z}$) should extend to $O\_X \to E$. But a priori $E$ can be without global sections. For example $E = O(-1) \oplus O(-1)$ on $P^2$. Maybe you want to have a subbundle which is trivial up to a twist?
UPD. If a twist is allowed then the answer is yes. Indeed, after an appropriate twist we can assume that $E$ is generated by a vector space $V$ of global sections, that is a map $V\otimes O\_X \to E$ is surjective. Let $E' = Ker(V\otimes O\_X \to E)$. It is a vector bundle on $X$. Consider the Grassmannian $Gr(r-1,V)$ of vector subspaces of rank $r-1$. Let $U$ denote the tautological subbundle. On the product $X\times Gr(r-1,V)$ we have a composition
$$
U\boxtimes O \to V\otimes O\boxtimes O \to O\boxtimes E.
$$
Let $Y$ be its discriminant locus (the scheme of points where the rank is $\le r-2$).
The fiber of $Y$ over a point $x \in X$ consists of all $(r-1)$-dimensional subspaces which intersect with the subspace $E'\_x \subset V$ of codimension $r$. A simple parameter count shows that it has codimension 2 in the Grassmannian. Since this is true for all $x \in X$ we conclude that $codim Y = 2$. Hence for generic point $u \in Gr(r-1,V)$ the fiber $Y\_u \subset X$ also has codimension 2. But since the corresponding map $O^{r-1} \to E$ is an embedding out of $Y\_u$, we just take $Z = Y\_u$ and restrict the above map to $X - Z$.
| 6 | https://mathoverflow.net/users/4428 | 47196 | 29,826 |
https://mathoverflow.net/questions/47080 | 14 | I was reminded of this topic by some of the answers to [this question](https://mathoverflow.net/questions/46966), where it was noted that "typical" measure-preserving transformations are weak-mixing but not strong-mixing for several senses of "typical". As a result, it occurred to me that I do not know of any very natural, explicit examples of transformations which are weakly but not strongly mixing. So,
>
> What are some good examples of measure-preserving transformations which are weak-mixing but not strong-mixing?
>
>
>
To clarify "good": I'm particularly interested in examples where it can be proved in a concise and self-contained manner that weak mixing occurs and strong mixing does not, in examples which arise constructively, and in examples which arise directly from a continuous transformation of a compact metric space (as opposed to abstract measure-theoretic constructions).
Thanks in advance!
| https://mathoverflow.net/users/1840 | Examples of transformations which are weak-mixing but not strong-mixing | The Chacon transformation has a nice and fairly explicit description as a uniquely ergodic subshift: set $B\_0=0$ and set $B\_{k+1}=B\_kB\_k1B\_k$.
The subshift is the set of all infinite words, all of whose finite subwords occur as a someword of some $B\_k$.
From this it is easy to see why the lengths $n=|B\_k|$ fail to be good times for strong mixing. As mentioned previously Parry shows the weak mixingness.
| 9 | https://mathoverflow.net/users/11054 | 47198 | 29,827 |
https://mathoverflow.net/questions/47206 | 4 | does the finite dimensionlity of the first cohomology group ($ H^1 $) of the sheaf of meromorphic sections of a holomorphic line bundle on a compact riemann surface follow easily from the finite dimensionality of the cohomologies of the sheaf of holomorphic functions on the surface?
| https://mathoverflow.net/users/11058 | finite-dimensionality of cohomology groups on compact riemann surfaces | The derivation is indeed "easy", in the sense that no additional results from analysis are needed. Consider the sheaf map $\mathcal{O}(L) \to \mathcal{M}(L)$, it is injective and the cokernel is the sheaf $\mathcal{H}(L)$ of principal parts of meromorphic sections of $L$. The sheaf $\mathcal{H}$ is fine. Therefore you get from the long exact sequence associated with
$$0\to \mathcal{O}(L) \to \mathcal{M}(L) \to \mathcal{H}(L) \to 0$$
a short exact sequence
$$
H^1 (X;\mathcal{O}(L)) \to H^1 (X;\mathcal{M}(L)) \to H^1 (X;\mathcal{H}(L))=0
$$
and the finite-dimensionality of $H^1 (X;\mathcal{O}(L))$ implies the result.
Addendum: it is known that any line bundle admits a nonzero meromorphic section $f$. Multiplication by $f$ induces an isomorphism $\mathcal{M}\cong \mathcal{M}(L)$, so $dim (H^1 (X;\mathcal{M}(L)))$ does not depend on $L$. Since there is a line bundle $L$ with $H^1 (X;\mathcal{O}(L))=0$ (this happens if the degree of $L$ is a sufficiently large positive number). This show that $H^1 (X;\mathcal{M}(L))=0$ for ANY line bundle.
The proof that $H^1 (X;\mathcal{O})$ is finite-dimensional (which requires quite a bit of analysis) generalizes to general line bundles. Or one can use the result for trivial line bundles plus the existence of meromorphic sections, see the answer by Francesco.
The existence of meromorphic sections is not easy to establish; it follows from Riemann-Roch. There is also a more direct argument, using the analysis involved in the proof of Riemann-Roch.
| 6 | https://mathoverflow.net/users/9928 | 47228 | 29,845 |
https://mathoverflow.net/questions/38825 | 25 | In any physics book I've read the Lagrangian is introuced as as a functional whose critical points govern the dynamics of the system. It is then usually shown that a finite collection of non-interacting particles has a Lagrangian $\frac{1}{2}(m\_1\dot{x}\_1^2 + \cdots + m\_n \dot{x}\_n^2)$. It is then generally argued that $L=T-U$. I feel like something is missing here.
What exactly are the physical hypotheses that go into this? Can we have other forms of the Lagrangian? How do we know those are "right"? Do we always have to compare them to the form of the equations we derived previously? For example, the Lagrangian formalism seems to be justified usually in so far as it 'works' for a finite collection of particles. Then you can solve any dynamics problem involving a collection of particles.
I have been vague so let me try to be more precise in my question.
Is the principle of least action an experimental hypothesis? Is it always true that $L=T-U$? When we don't know what the Lagrangian is, do we have to just guess and hope it is compatible with the dynamical equations we had already? Or can we perhaps start with the ansatz of a Lagrangian in some cases?
I hope this is sufficiently precise.
| https://mathoverflow.net/users/8755 | What kind of Lagrangians can we have? | As a theoretical physicist who shifted to pure mathematics, I think to answer this question and as a clarification to previous posts, we should not forget the historical side of the evolution and origins of the terms involved in physics theory modeling.
The origin of the principle of stationary ('least' most of the time) action comes as a variational formulation using a functional $S[q(t)]:=\int\_{t\_1}^{t\_2}L(q^i (t),\dot{q}^i (t);t)dt$ for the Lagrange equations of the real motion of a system with generalized coordinates $q^i (t)$ (obtained this way they are called Euler-Lagrange equations):
$$\delta S[q\_{real}(t)]=0\Rightarrow \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i}-\frac{\partial L}{\partial q^i}=0$$
The original Lagrange equations of motion were obtained as a reformulation of Newton's second law of mechanics for the case of generalized coordinates (the remaining degrees of freedom after introducing constraints) as an easier to handle version of D'Alembert's principle. D'Alembert wrote the 'virtal work' principle as a way to state a general equation for statics, which in a way amounts to postulate that constraint forces do not exert work (as they should be internal and the weak 3rd law applies). Now given Newton's second law for a system of particles, D'Alembert reformulates constrained dynamics as statics introducing the inertial force (you can see details [at wikipedia](http://en.wikipedia.org/wiki/D%2527Alembert%2527s_principle)) $$\sum\_i (\vec{F}\_{i}^{ext}-\frac{d\vec{p}\_i}{dt})\cdot\delta \vec{r\_i}=0$$
Lagrange wanted to work only with generalized coordinates which changing variables in D'Alembert's principle led him to [the original Lagrange's equations of motion](http://en.wikipedia.org/wiki/Lagrange_equations) in terms of a kinetic generalised energy $T$ and generalized forces $Q\_i$; furthermore he absorbed those conservative forces which derived from a potential $Q\_i^c=-\partial U/\partial q\_i$ (or more generally those which even depended on speed or time but had the required form of the left-hand side) along with the kinetic energy, into a function $L=T-U$ now called the Lagrangian $$\frac{d}{dt}\frac{\partial T}{\partial \dot{q}^i}-\frac{\partial T}{\partial q^i}=Q\_i\Rightarrow \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i}-\frac{\partial L}{\partial q^i}=Q\_i^{no-c.}$$
That is the origin of the form of the lagrangian as $L=T-U$. Now, the non-conservative forces $Q\_i^{no-c.}$ appear because we are considering general 'open systems' which interchange energy with their enviroment, other systems as sources of energy, etc. To check this form in detail, first consdier that a closed system is always of the form $L=T-U$ because of a constructive reasoning as follows. For a free particle, at least locally, because of the homogeneity of space and time, and isotropy of space, we must have $L(q\_i,\dot{q}\_i;t)=T(v^2)$, that is a scalar which only depends on the particle and its speed (but no direction); since a free particle must still be free in another inertial system by def. the relation must be linear $L=a\cdot v^2$ (this is because any other Lagrangian giving the same equations of motion has to be of the form $L'=A\cdot L+\frac{d}{dt}\Omega(q;t)$ for which an inertial transformation $\vec{v}'=\vec{v}-\vec{\epsilon}$ must imply at first order $-2\vec{v}\cdot\vec{\epsilon}\partial L(v^2)/\partial (v^2)\propto d\Omega(q;t)/dt\Leftrightarrow \partial L/\partial (v^2)=0$); now the constant $a$ characterizes the particle, and must reduce to the free Newtonian second law ($p\_i=m\cdot v\_i =0$) therefore $a=m/2$. For a system of free particles without interactions the same applies summing over the individual kinetic energies $L\_{free}=T=\frac{1}{2}\sum\_i m\_i v\_i^2=\frac{1}{2}\sum\_{i,j} A\_{ij}\dot{q}^i\dot{q}^j$ (when $T$ is a cuadratic form the system is called 'natural' which happens when $\partial \vec{r}\_k(q^i(t),\dot{q}^i(t);t)/\partial t = 0$ for all particles $k$, common for the natural case of constraints not dependent on time). To add interactions between particles in a closed system, their degrees of freedom must be coupled, which just means that the relative energies are not lost to the exterior, and this is satisfactorily implemented by a potential function $U=U(q\_i,\dot{q}\_i)$. By this method we obtain a general framework for mechanics using Lagrange's equations: postulating different potentials we obtain different models of interactions which shall be contrasted with experiment to constrain the form of $U$ (this may be done to obtain just effective phenomenological models, nevertheless all fundamental interactions of physics seem to fit this framework very well). Finally, when part B of the system A+B is fixed (like considered external) its motion can be treated like already 'solved', which opens the previous closed system, separating some of the degrees of freedom $q\_B^i(t)$; therefore the Lagrangian decouples $L\_{open}=T\_A+T\_B(\dot{q}^i\_B(t))-U(q\_A,q\_B(t),\dot{q}\_A,\dot{q}\_B(t))$, in this case $T\_B=d\Omega(q\_B,t)/dt$ for some $\Omega$, so we can neglect that part since does not contribute to the equations of motion and therefore $L\_{open}=T\_A-U\_{open}(q\_A,\dot{q}\_A,t)$. This way $U$ creates the effect of non-conservative forces which are just those that add or subtract energy from the system without taking into account where this energy goes.
Therefore we can consider the Lagrange equation (without non-conservative forces) as the general more fundamental equation given individual interactions between the particles of the Universe, at least in principle. This is why the form $L=T-U$ is the usual approach in theoretical physics to deal with dynamics, since the very concept of instantaneous force at a distance à-la Newton is anti-natural (above all after special-general relativity) and the energetic local approach is not only mathematically better but philosophically more satisfactory. The [Hamiltonian approach](http://en.wikipedia.org/wiki/Hamiltonian_mechanics) can be taken also as a starting point, or as a Legendre transformation of Lagrange dynamics, but that way one makes a non-relativistic break down of the coordinates $x^i, t$ which is useful for non-relativistic quantum mechanics.
Now in general, physics is a theory of fields so we want to add these to the framework. This is easily done by considering them as continuous systems of degrees of freedom, providing generalizations of kinetic and potential energies of those.
Furthermore one wishes for a (special or general) relativistic invariant theory which forces the $Ldt$ to be a Lorentz scalar in order to get Lorentz-covariant ($SO(3,1)$) equations of motion (i.e. covariant as tensor equations in a pseudo-Riemannian manifold). This way the free case reasoned above leads us to $ds=Ldt$ (where $ds$ is the relativistic space-time interval or 'proper time' measured by the particle) which indeed reduces to $ds\approx \frac{1}{2}mv^2$ for speeds $v\ll c$, and therefore can be taken as a relativistic generalization of the framework. Nevertheless point-particles are idealizations and both the classical and quantum theories require a field-theoretic treatment of matter (in the classical case, particles appear as small density lumps, and in the quantum case particle behaviour arises for discrete-like energy levels of the quantum field states). In the wikipedia article about [Lagrangians](http://en.wikipedia.org/wiki/Lagrangian) different examples can be seen from point particles to fields, their kinetic terms and so on.
In any case, the traditional form $L=T-U$ has its roots in the very nature of mechanics and any modern field theory is created by building up possible $T-U$ functionals from the field's degrees of freedom, i.e. the fields $\phi\_a(x^\mu)$ and their "velocities" $\partial \phi\_a/\partial x^\nu$ (for relativistic invariance reasons time-like speed $\frac{\partial}{\partial t}$ is not enough and the whole $\partial\_\nu$ must be used). Hence one constructs kinetic terms like (Einstein's summation convenction) $T\_\phi=\frac{1}{2}\partial\_\mu\phi\_a\partial^\mu\phi\_a$ for scalar fields or $T\_A=F\_{\mu\nu}F^{\mu\nu}$ for vector fields ($F\_{\mu\nu}:=\partial\_\mu A\_\nu-\partial\_\nu A\_\mu$ is in this case the tensor constructed from the field used to define its kinetic energy because we want to build invariant 'speed-like' $\partial\_\mu$ scalars that are as well Gauge invariant which is another physical symmetry typically required besides Lorentz; furthermore, requiring gauge invariance for different Lie groups, tyipcally $SU(N)$, one forces the automatic appearance of fields and couplings between the matter fields responsible for the interactions, which is just working with connections and curvatures of fiber bundles). Since any term which makes the field equation of motion non-homogeneous is considered a source of perturbation, or force, one sees the corresponding couplings in the Lagrangian like a potential energy, giving as always $\mathcal{L}=T\_\phi+T\_A-U(A\_\mu,\phi)$. With this, the coupled physical equations of motion are deduced for the principle of stationary action, where now the integration must be over the whole space and an interval of time (i.e. $\mathcal{L}$ is a density), the field Euler-Lagrange equations of motion:
$$\partial\_\mu\frac{\partial\mathcal{L}}{\partial (\partial\_\mu\phi)}-\frac{\partial\mathcal{L}}{\partial\phi}=0$$
Finally these equations give the solutions of motion, that is the real field configurations at any point in space-time, which contribute the most in a quantum-theoretic framework, where one weighs complex transition amplitudes by [Feynman operational methods](http://en.wikipedia.org/wiki/Feynman_path_integral) using the action for each possible field-configuration ('motion'): $e^{\frac{i}{\hbar}S[\phi,A\_\mu]}$. As remarked in other comment, the semi-classical approximation $\hbar\rightarrow 0$ makes the classical solution of Euler-Lagrange the predominant one, hence deducing the principle of stationary action. (Also, in the non-relativistic hamiltonian quantum-mechanics Ehrenfest's theorem provides a classical Newton-like equation of motion for the average degrees of freedom)
Indeed as science makes progress we must see previous theories and models as limiting cases of new more precise theories, but in the beginning any physical theory must be constructed by physical insight, intuition and permanent comparison with experiment. Even though Feynman titled his thesis that way, he does not develop his approach to quantum theory starting from the action-principle, on the contrary he works with standard (hamiltonian) quantum mechanics and obtains a novel method for computing quantum probability amplitudes in which the classical action appears. You could start with quantum mechanics as an axiomatic system, develop Feynman's approach to quantum propagators of motion and deduce that classical approximate solutions of motion must obey, at first order of quantum corrections, the classical Euler-Lagrange equations of motion. At the same time, you need the input from the experimentally succsessful classical mechanics to get to quantum mechanics... and so forth. In the end at any moment of time in the history of science we are getting better and better mathematical structures that model reality to different degrees of precision; the important point is that a new one must contain an old one as an approximation in some regime, and explain even more. This way, theoretical physics tends to get better structures to encompass more phenomena from Nature in a simpler and simpler manner as it explains more effective laws with less theories (currently almost all the phenomena observed is explained by general relativity and quantum field theory, and unifying both is the neverending quest for the holy grail of physics).
| 14 | https://mathoverflow.net/users/10867 | 47240 | 29,854 |
https://mathoverflow.net/questions/47135 | 13 | Let $E$ be an elliptic curve over $\mathbb{Q}$ and $T$ the $p$-adic Tate module of $E$. Kato's Euler system, constructed in the paper "P-adic Hodge theory and values of zeta functions of modular forms" (Asterisque 295, 2004), gives rise to an element $\mathbf{z}\_{\rm Kato}$ lying in the Iwasawa cohomology $ H^1\_{\mathrm{Iw}}(\mathbb{Q}\_p, T)[\frac{1}{p}]$. In Theorem 12.5(4) of the paper, Kato shows that if the image of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ in $\mathrm{Aut}(T)$ contains $\mathrm{SL}\_2(\mathbb{Z}\_p)$, then in fact $\mathbf{z}\_{\rm Kato} \in H^1\_{\mathrm{Iw}}(\mathbb{Q}\_p, T)$.
Is it known if there are weaker conditions that are sufficient to ensure that $\mathbf{z}\_{\rm Kato}$ has this integrality property? Are there examples where it is genuinely non-integral, or is it conjectured that it should always be so?
I would be the last person to claim I understand Kato's argument, but it looks to me as if he only actually uses the weaker statement that the mod $p$ Galois representation $T/pT$ is irreducible. I'd be interested to know if this weaker condition is indeed sufficient, and whether anything is known in this direction if the weaker condition doesn't hold (i.e. if $E$ admits a $p$-isogeny).
(EDIT: Added more detail and references.)
| https://mathoverflow.net/users/2481 | Are Kato's zeta elements integral? | There are two issues. Let $H=H^1\_{\mathrm{Iw}}(\mathbb{Q},T)$ where $T=V\_{\mathbb{Z}\_p}(f)(1)$ and $f$ is the modular form associated to the isogeny class of $E$.
(1) What is $T$ ?
$T$ will correspond to the lattice $\Lambda$ in $\mathbb{C}$ generated by all modular symbols. This is because $V\_{2,\mathbb{Z}}(f)$ is the image of $V\_{2,\mathbb{Z}} (Y\_1(N)) = H\_1 \bigl( X\_1(N)(\mathbb{C}),\{\text{cusps}\},\mathbb{Z}\bigr)$ inside $H\_1(E(\mathbb{C}),\mathbb{Z})$. The lattice $\Lambda$ will contain the lattice $\Lambda\_0$ of the $X\_0$-optimal (strong Weil) curve $E\_0$, but it may be strictly larger if $E\_0$ has rational torsion points, e.g. for 11a. Note also that $\Lambda$ need not be the lattice of an elliptic curve in the isogeny class, but rather $\tfrac{1}{2}\Lambda\_E$ for some elliptic curve $E$. in case $E\_0[2] \subset E\_0(\mathbb{Q})$, e.g. for 17a. But can only happen for $p=2$ and for $p$ of additive reduction. So if $p$ is a prime of odd semi-stable reduction, then $T$ is the Tate-module of an elliptic curve $E\_\*$ which is an étale quotient of $E\_0$. So let $E$ be this curve for the rest of the answer.
(2) Are Kato's elements integral in $H$ ?
There are two kinds of them. The $z\_{\gamma}$ and the ${c,d}$ $z\_{m}$. (sorry I don't seem to be able to produce indices before the symbol in MathJax) The latter are in $H$, see 8.1 of Kato, but they depend on the choices of $c$ and $d$. They are useful for bounding the Selmer group as, for a fixed $c$ and $d$ they form an Euler system.
The $z\_{\gamma}$ instead is linked to the $p$-adic $L$-function and they are independent of the choices. They are obtained by dividing by $\mu(c,d)$, page 229 of Kato. So they need not be integral anymore. The appendix A in Delbourgo's book "Elliptic curves and big Galois representations" discusses this in detail. Kato shows that they are in $H\otimes \mathbb{Q}\_p$
Kato shows that $z\_{\gamma}$ is integral if $H$ is a free $\Lambda$-module of rank 1, e.g. as shown in 12.4.(3) if $T/pT$ is irreducible. In fact it is not hard to show that $H$ is free also if $E(\mathbb{Q})[p]$ is trivial.
Now if the curve admits an isogeny of degree $p$, one can show that for all curves $A$ in the isogeny class $H^1\_{\text{Iw}}(\mathbb{Q},T\_p A)$ is a free $\Lambda$-module of rank $1$, except for at most a single one of them (I mean up to non-$p$-isogenies of course). This exception - if present - will always be the minimal curve $E\_{\text{min}}$ in the class. Moreover if $E\_{\text{min}}$ is does not have a free $H^1\_{\text{Iw}}$, then there is an embedding of it into $\Lambda$ with image equal to the maximal ideal of $\Lambda$. One can now conclude from the interpolation property of the $p$-adic $L$-function that even if $E\_\*=E\_{\text{min}}$ then $z\_{\gamma}$ will be in $H$.
Using this one can prove that the $p$-adic $L$-function is integral and that $\mu\geq 0$, but it would not say anything about Greenberg's $\mu=0$ conjecture. Furthermore one gets a proof of the divisibility in the main conjecture as in 12.5.(3), if the $T/pT$ is reducible. However this conclusion can not be extended at present to all odd semi-stable primes, because there may be primes for which the Galois representation is not surjective, yet $T/pT$ is irreducible; because the Euler system method requires and element $\binom{1\ 1}{0\ 1}$ in the image of the Galois representation, see Hyp($K\_{\infty},T)$ in Rubin's book. So the integrality won't help, yet.
In summary, all zeta-elements of Kato are integral with respect to $T$ in the case of an elliptic curve. For Kato's divisibility on the other hand, the surjectivity of the representation to $\mathrm{GL}\_(\mathbb{Z}\_p)$ or its reducibility is still needed.
(edits: quite a few in the whole answer above, now that I konw the full answer to the question)
| 11 | https://mathoverflow.net/users/5015 | 47243 | 29,856 |
https://mathoverflow.net/questions/47207 | 3 | Hello,
If we look at the class of all vector spaces over some field, we can note two things:
1) this class should not have cardinality.
2) for two elements of this class, we should not want to be able to verify whether they are equal
But we can have some situations which have only one of the two features. For example:
* Look at "beasts" which assign to every vector space a morphism from it to its dual. Then the class of all such beasts should not have cardinality, but we can say whether two such beasts are equal or not.
My question is rather vague: Does (axiomatic) set theory think about the difference between these two features 1) and 2) ? Or, can someone share his way of thinking about the two features and their relation?
Edit: I apologize for being not very precise. In (2), I am talking from a (biased?) categorical point of view; there should be no meaning for two vector spaces to be equal; they can be known to be isomorphic, or given a specific isomorphism, but not equal (if we can or can not ask in a specific axiomatic system whether two vector spaces are equal does not really make a difference). In (\*), we can verify whether two beasts are equal just by checking, for every vector space, whether the morphisms from it to its dual provided by the two beasts are equal or not. In a comment it was told the collection of beasts does have a cardinality; Anyway, it was just an example. I would think of them as not having cardinality (probably one might provide some example in the same spirit for which there is no cardinality?), or would not care about their cardinality, just to stress the point.
Thank you,
Sasha
| https://mathoverflow.net/users/2095 | "classes" with no cardinality; "classes" with no equality notion | Classical axiomatic set theories (eg ZFC, NGB) are formulated in first-order logic with equality, so *any* things you can quantify over (i.e. that you can talk about as actual objects of the language), you can talk about equality of, as a basic given of the language.
In particular, in either ZFC or NGB, you can certainly talk about equality of vector spaces. In ZFC, you can’t talk about beasts as quantify-over-able objects (since they can only be represented as proper classes, not as sets); in NGB, you can, and so you get equality of them.
Cardinality is a bit slipperier: it’s generally considered as a defined rather than a basic notion, and the exact definitions used vary in ways your question will be sensitive to. Most often, an object called the “cardinality” is only specifically defined for sets[1]; for classes, “*C* and *D* have the same cardinality” is considered syntactic sugar for “there is a class-bijection between *C* and *D*”. So it’s not quite clear what it means to ask if a class “has cardinality”, but whatever it is depends heavily on having an equality relation on it, to be able to talk about bijections to/from it.
---
On the other hand, there are some more recent set/type theories in which equality isn’t given, or is a more flexible notion.
In some versions of the Calculus of Constructions, if I remember right, there is a universe of small sets (possibly multiple universes), and an arbitrary product of sets is again a set, possibly in some higher universe (this has to be formulated carefully to avoid inconsistency); and each set has equality on it, but there’s no equality on the universe(s). So there, vector spaces wouldn’t form a set, and wouldn’t have an equality relation; but beasts would form a set (a certain product of hom-sets) so would have an equality relation. (The C of C’s is a little out of what I know, so this may need correcting by someone more knowledgeable.)
Similarly, there are versions of Martin-Löf Type Theory with *identity types* which address this issue; roughly, identity types can represent something like an ordinary equality relation, but more generally they can also look like the sets/categories of (weak) ismorphisms in a (higher) category. So you can *define* an object to be *0-categorical*[1] if all its identity types are just truth-values; then an arbitrary product of 0-categorical types is again 0-categorical.
In this setup, the type of all vector spaces within some universe will have identity types, so equality of a sort, but not of the objectionable kind — “equality” of vector spaces will exactly be isomorphism between them. The type of beasts over this universe will now be 0-categorical: we will have equality of beasts in the simplest sense. (Also, in this foundation, all beasts will automatically respect isomorphism!)
---
[1]: The two main definitions of $\|X\|$ I know are “the least ordinal bijective to $X$” (elegant, but requires choice to be defined for all $X$), or “$\{ Y \in V\_\alpha\ |\ Y \cong X \}$, where $\alpha$ is minimal such that this is non-empty” (less transparent but more robust).
[2]: I first heard this definition from Voevodsky, though I’m pretty sure it had been considered by others before as well. He calls this property being a *set*, but I want to make unambiguous that it’s a restriction of *categorical dimension* not of *size*.
| 5 | https://mathoverflow.net/users/2273 | 47261 | 29,865 |
https://mathoverflow.net/questions/47259 | 4 | Comrades,
Let $S$ be a non-ruled, minimal (smooth projective complex algebraic) surface. Let $K$ be a canonical divisor of $S$ and $H$ a hyperplane section of $S$ (for your favorite embedding).
Suppose I know that $K^2 > 0$. Then supposedly $S$ being non-ruled forces $H.K > 0$. Is this correct or should it only say $H.K \geq 0$? Why is it correct?
This is at the beginning of chapter 9 in beauville's surfaces book. I can say for sure that $H.K \geq 0$ since a surface is ruled iff there exists a non-exceptional curve $C \subseteq S$ satisfying $C.K < 0$.
Thanks in advance.
edited, i said something stupid in original post.
| https://mathoverflow.net/users/11071 | Non-Ruled Minimal Surfaces | $K^2 >0$ and $S$ non ruled implies that $S$ is of general type. In particular $mK$ is effective for some $m \geq 1$.
Since $H$ is very ample it follows $mKH >0$, that is $KH >0$.
| 5 | https://mathoverflow.net/users/7460 | 47268 | 29,870 |
https://mathoverflow.net/questions/47017 | 3 | Which book or books do you recommend that cover advanced engineering topics and problem solving using matlab?
I already finished a very good introductory book and i want something more advanced.
Do you think it's better to read a book that covers several topics or search for the topics i'm interested in and then explore the methods applied on them?
I'm more interested in applied mathematics, algorithms, computation and engineering.
I study product design and engineering if this helps.
| https://mathoverflow.net/users/11019 | Matlab book recommendation | Here is a list of some books that you might find useful.
1. Introduction to Scientific Computing by Charles F. Van Loan
2. Matrix Computations by Golub and Van Loan (uses Matlab notation, but contains a wealth of material)
3. You might also benefit from trying to extend the methods in "Numerical Recipes" to Matlab. They have some information about that [here](http://www.nr.com/nr3_matlab.html)
4. Have a look at "Numerical Computing with Matlab" by Cleve Moler (the guy who created Matlab); [check it out here](http://www.mathworks.com/moler/chapters.html)
But really, the best way to learn more is to actually have an application or targeted problem at hand. Then, while attempting to solve your problem, you will gradually pick up the tricks of the trade. While doing that, perhaps people at the mathworks file exchange or over at stackoverflow might be able to help you further. Happy experimenting!
| 3 | https://mathoverflow.net/users/8430 | 47270 | 29,872 |
https://mathoverflow.net/questions/47271 | 4 | In many places, I have seen the slogan that "simplicial abelian group = chain complexes of abelian groups". These same sources usually tell me how to go in one direction. Namely, given a simplicial abelian group, one gets a chain complex whose $n^{\text{th}}$ term is the abelian group corresponding to the $n$-simplex and whose differential is the alternating sum of the face maps.
However, this process does not appear to me to be reversible. My question then is in what sense is the above slogan true? In other words, is there some procedure that takes a chain complex of abelian groups and produces a simplicial abelian group? If there is, in what sense is it an inverse to the above procedure?
| https://mathoverflow.net/users/11074 | From chain complex to simplicial abelian group | There are three standard functors from simplicial abelian groups to chain complexes. Let $C\_{\ast}A$ be the one that you described. Let $D\_{\ast}A$ be the sum of the images of all the degeneracy maps in $A$; then $D\_{\ast}A$ is a contractible subcomplex of $C\_{\ast}A$ and we define $N\_{\ast}A=C\_{\ast}A/D\_{\ast}A$, which is chain homotopy equivalent (but not isomorphic) to $C\_{\ast}A$. We also put $M\_{\ast}A=\{a\in A: d\_i(a)=0 \text{ for all } i>0\}$; this is a subcomplex of $C\_{\ast}A$, and in fact $C\_{\ast}A=M\_{\ast}A\oplus D\_{\ast}A$, so $M\_{\ast}A\simeq N\_{\ast}A$.
Now let $F(n)$ be the free simplicial abelian group on a generator in degree $n$, so the term in degree $k$ is the free abelian group on the set of nondecreasing maps $[n]\to [k]$. For any chain complex $U\_{\ast}$ we let $L\_nU\_{\ast}$ denote the group of chain maps from $N\_{\ast}F(n)$ to $U\_{\ast}$. This defines a functor $L$ that is inverse to $N\_\*$.
| 10 | https://mathoverflow.net/users/10366 | 47273 | 29,874 |
https://mathoverflow.net/questions/47153 | 0 | Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit.
My question is,
>
> Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$)?
>
>
>
The main issue I have is that it might take infinite time to reach Y. To see how things may go wrong if $f$ is not continuous, consider the following example:
* $\dot{x} = -1-x$, $\quad$ for $x<-1$,
* $\dot{x}=-x$, $\quad\quad$ for $-1 \le x \le 1$, $\quad$ and
* $\dot{x}=1-x$, $\quad$ for $x>1$.
In this case every $x$ in $[-1,1]$ converges to zero, and every point outside $[-1,1]$ converges to either $-1$ or $1$, but it takes infinite time to do so and it is not true that all points in $R$ converge to zero.
>
> Smoothness of $f$ might be enough to rule out this behavior; I am unable to construct a counterexample with $f$ continuous. Is it actually enough, and if so, why?
>
>
>
| https://mathoverflow.net/users/8460 | When is convergence transitive? | If the equation is defined in $\mathbb{R}$, then $f$ continuous is enough. It is clear that if a point of $X$ does not reach the set $Y$ in finite time, then $f$ vanishes in this limit point. Since every point of $Y$ converges to $Z$, you get that this limit point must thus belong to $Z$.
In higher dimensions, I would say that the same holds only that there is the subtelty that you don't have a priori a dynamical system defined (due to lack of uniqueness of solutions, the group condition fails) however, if $\dot x = f(x)$ integrates into a flow (which must then be a continuous flow, this will hold for example if $f$ is locally Lipchitz) an analogous argument works (now $f$ may not vanish at the omega-limit set of a trayectory, but if $x$ does not enter $Y$, then it will converge to a point in $Z$ anyway).
| 1 | https://mathoverflow.net/users/5753 | 47274 | 29,875 |
https://mathoverflow.net/questions/47212 | 16 | Let $f:X\to S$ be a universal homeomorphism of schemes. Assume $X(S')\neq\emptyset$ for some étale surjective $S'\to S$. Does $f$ have a section?
The answer is yes if $S$ is reduced, by descent. Indeed, note that if $S\_1$ is a reduced $S$-scheme then $X(S\_1)$ has at most one element. Apply this to $S\_1=S'\times\_S S'$.
Interesting special case: if $S$ has prime characteristic $p$, let $G$ be a finite locally free $S$-group scheme with connected (i.e. "infinitesimal") fibers, such as $\alpha\_p$ or $\mu\_p$. Is $H^1\_{\mathrm{et}}(S,G)$ trivial?
| https://mathoverflow.net/users/7666 | Universal homeomorphisms and the étale topology | I think that the following might work. Let $X\_0$ be a reduced scheme over a field $k$ of characteristic $p > 0$, and let $X$ be the product of $X\_0$ with the ring of dual numbers $k[\epsilon]$. Then $\mathcal O\_X = \mathcal O\_{X\_0} \oplus \epsilon\mathcal O\_{X\_0}$, and the $p^{\rm th}$ roots of 1 are those of the form $1 + \epsilon f$; hence the Zariski sheaf of $p^{\rm th}$ roots on 1 on $X$ is isomorphic to $\mathcal O\_{X\_0}$. Hence if $\mathrm H^1(X\_0, \mathcal O\_{X\_0}) ≠ 0$ there is a non-trivial $\mu\_p$-torsor on $X$ that is locally trivial in the Zariski topology, thus giving a counterexample.
| 6 | https://mathoverflow.net/users/4790 | 47277 | 29,877 |
https://mathoverflow.net/questions/47276 | 1 | I recently stumbled upon a paper (NON CANCELLATION FOR SMOOTH CONTRACTIBLE AFFINE THREEFOLDS) about the cancellation problem: If $X$ is a variety over $\mathbb{C}$ of dimension $d$ such that $X \times \mathbb{A}^n \cong \mathbb{A}^{n+d}$ when is $X \cong \mathbb{A}^d$?
Apparently when $d = 1,2$ the answer is always. The paper remarked that the case $d = 1$ is trivial, but I'm having trouble coming up with an argument. The paper does reference a paper (ON THE UNIQUENESS OF THE COEFFICIENT RING IN A POLYNOMIAL RING) which proves the more general cancellation problem -when does $X \times \mathbb{A}^n \cong Y \times \mathbb{A}^n$ imply $X \cong Y$- for curves (again the answer for curves is always).
However I wonder if there is a simple argument for the less general question. I can see that $X \times \mathbb{A}^n \cong \mathbb{A}^{n+1}$ implies that $X$ should be irreducible, affine and smooth. Seems like if you could show genus$(X) = 0$ you would be done, but I'm kinda stuck here.
Does someone have a simple argument for this?
| https://mathoverflow.net/users/7 | Cancellation problem for curves | The only smooth affine curve admitting a non-constant map from an affine space is $\mathbb A^1$. It must have genus 0, because of Lüroth's theorem, so it is $\mathbb P^1$ minus $d$ points for some $d$. But if $d$ were larger than 1 the curve would have a non-constant invertible function, which would pull back to a non-constant invertible function on an affine space, and this is impossible.
| 14 | https://mathoverflow.net/users/4790 | 47279 | 29,878 |
https://mathoverflow.net/questions/47278 | 7 | Hi all,
Suppose that $\mathcal{B}$ is a Boolean algebra. It there a way to extend $\mathcal{B}$ to a smallest Boolean algebra $\mathcal{B}'$ that contains an isomorphic copy of $\mathcal{B}$ and is countably complete, i.e. every countable subset of $\mathcal{B}'$ has a least upper bound in $\mathcal{B}'$? By "smallest" I mean that the inclusion $i: \mathcal{B} \hookrightarrow \mathcal{B}'$ has the obvious universal property, i.e. for every homomorphism $f$ from $\mathcal{B}$ to a countably complete Boolean algebra $\mathcal{C}$ there exists a unique homomorphism $g: \mathcal{B}' \to \mathcal{C}$ such that $g \circ i = f$ (it would be nice if $g$ turned out to commute with countable sups too). If no such $\mathcal{B}'$ exists, is there some other useful definition of "smallest" countably complete Boolean algebra containing $\mathcal{B}$?
If it makes any difference, I'm mostly interested in the special case where $\mathcal{B}$ is a direct limit of a sequence of finite Boolean algebras.
**Edit:** Thanks very much for the replies, it's a shame I can only mark one as the answer. It will take me a while to absorb the various references I've been given, so if I run into difficulty I'll bump the thread with an edit.
**Edit 2:** Bumping with followup question, please see my answer below.
| https://mathoverflow.net/users/7842 | Is there such a thing as the sigma-completion of a Boolean algebra? | The short answer is "yes", and it's a special case of a much, much more general theorem on relatively free algebraic constructions.
In other language, you are asking whether the underlying functor from countably complete Boolean algebras to Boolean algebras has a left adjoint. The more general question is whether, given a homomorphism $\phi: S \to T$ between two monads on $Set$, the evident underlying functor
$$Set^\phi: Set^T \to Set^S$$
from the category of $T$-algebras to the category of $S$-algebras has a left adjoint. For this I'll direct your attention to this [nLab article](http://ncatlab.org/nlab/show/Lawvere+theory#AdjCatAlgs).
Of course, we have to know that countably complete Boolean algebras can be described as algebras of a monad on $Set$, but this too follows from general theory. I'll refer you to another [nLab article](http://ncatlab.org/nlab/show/algebraic+theory) for this; the article is not complete but it should give the idea. The upshot is that for any algebraic theory with only a small set of operations of each arity, there is a corresponding monad on $Set$ whose algebras are the models of the theory. The general constructions go back to work in the sixties, due to Lawvere, Linton, and others.
**Edit:** I'll remark that had you said "complete" instead of "countably complete", then the answer would have been **no**. In fact, the underlying functor from complete Boolean algebras to sets has no left adjoint; this is mentioned for instance in Categories for the Working Mathematician. But in your case, the theory is generated by a *set* of operations and equations, and all is well.
| 17 | https://mathoverflow.net/users/2926 | 47282 | 29,880 |
https://mathoverflow.net/questions/46599 | 7 | This question is inspired by [this question](https://mathoverflow.net/questions/46597/why-does-the-grothendieck-group-k-0r-of-a-ring-not-depend-on-our-choice-of-us) about the dependence of K-theory on the order of multiplication in the ring. I did not think long about it, so maybe the answer really lies on the surface; but I do not know.
Let $G$ be a discrete group, $G^{op}$ its opposite group (i.e. the one with reversed multiplication). Let $H \subset G$ be a subgroup, such that $H=[H,H]=[G,G]$. Note that $H^{op} \subset G^{op}$ has the same properties. I denote the Quillen Plus-Construction of a space with fundamental group $G$ with respect to $H$ by $X \mapsto X^+\_H$
>
> **Question:** Is there a homotopy equivalence between $BG\_H^+$ and $B(G^{op})\_{H^{op}}^+$, such that the induced map on $\pi\_1$ is induced by identity $id: G \to G^{op}$.
>
>
>
Note that there is clearly a homotopy equivalence between $BG\_H^+$ and $B(G^{op})\_{H^{op}}^+$ such that the induced map on $\pi\_1$ is induced by the inverse $inv: G \to G^{op}$; but that is not the one I am looking for.
However, this shows that one could also ask:
>
> **Question:** Is there a homotopy equivalence between $BG\_H^+$ and itself, such that the induced map on the abelian group $\pi\_1(BG\_H^+) = G/H$ is the inversion.
>
>
>
Given the motivation, any good answer in the case $G=GL\_{\infty}(R)$ (for some ring) would be interesting too.
**EDIT:** Johannes Ebert suggested in a comment a strategy to give a negative answer to Question 2 in general. The Kan-Thurston construction gives a way of obtaining every (finite?) cell complex $X$ as the plus-construction on $BG$ with respect to *some* perfect subgroup $H$. It seems that the only missing piece is a finite cell-complex $X$ with $\pi\_1$ abelian and *no* $\pi\_1$-inversion-inducing self-homotopy equivalence. Then, $X = BG^+\_H$ for some group $G$ and some perfect subgroup $H \subset G$. However, since $\pi\_1 = G/H$ is abelian, we see that $[G,G] \subset H$ and hence, $[H,H] = [G,G]=H$ as required.
>
> **Question:** Can anybody give an example of such a space?
>
>
>
| https://mathoverflow.net/users/8176 | How commutative is Quillen's Plus-Construction? | To answer the final question, here is how you can construct a space with no self-equivalence inducing negation on (abelian) $\pi\_1$.
If $X \to Y$ is a homotopy equivalence, then for any basepoint it induces isomorphisms $\pi\_n(X) \to \pi\_n(Y)$ commuting with the action of $\pi\_1$. In particular, if we can construct abelian groups $A$ and $B$ with an action of $A$ on $B$ such that there is no isomorphism $\phi: B \to B$ such that ${}^a \phi(b) = \phi({}^{(-a)}b)$, then we can use these to construct the desired space; let $X$ be the homotopy orbit space $K(B,2) \times\_A EA$, which has $\pi\_1 = A$ and $\pi\_2 = B$ with the given action.
If $B$ is cyclic, all endomorphisms commute and so it suffices to construct an action so that ${}^a b = {}^{(-a)}b$ does not hold for all $a, b$.
"Minimal" examples include the action of $\mathbb{Z}/4$ on $\mathbb{Z}/5$ with generator $x \mapsto 2x$ (not isomorphic to $x \mapsto 3x$) and the action of $\mathbb{Z}/3$ on $\mathbb{Z}/7$ with generator $x \mapsto 2x$ (not isomorphic to $x \mapsto 4x$).
| 5 | https://mathoverflow.net/users/360 | 47285 | 29,883 |
https://mathoverflow.net/questions/47301 | 1 | Given two finite-dimensional Hilbert spaces $U, V,$ a linear transformation $T:U\to V$ contracts the inner product if for all $x,y \in U,$
$$\langle x,y \rangle\_U \ge \langle Tx, Ty\rangle\_V.$$
All unitary transformations satisfy this criterion; is there a larger class of linear transformations that do?
| https://mathoverflow.net/users/756 | Which linear transformations between f.d. Hilbert spaces contract the inner product? | Such a map will preserve orthogonality, and any such map must be a scalar multiple of an isometry. This is true in great generality, e.g. the map $T$ doesn't have to be linear, and $U$ and $V$ don't have to be finite-dimensional; see Theorem 1 in
>
> Chmieliński, *Linear mappings approximately preserving orthogonality.* J. Math. Anal. Appl. **304** (2005), no. 1, 158–169.
>
>
>
Consequently, a map $T \colon U \to V$ contracts the inner product if and only if $T = \alpha S$, where S is an isometry and $|\alpha| \leq 1$.
---
**Edit:** Here's a simple proof of the assertion that an orthogonality-preserving linear map between finite-dimensional inner product spaces is a scalar multiple of an isometry. Let $T \colon U \to V$ be such a map and fix an orthonormal basis $\{e\_1, \ldots, e\_n\}$ for $U$. Observe that $e\_i + e\_j \perp e\_i - e\_j$. Thus
$$ 0 = \langle T(e\_i + e\_j), T(e\_i - e\_j) \rangle = \langle Te\_i, Te\_i \rangle - \langle Te\_j, Te\_j \rangle. $$
So we may set $\alpha = \langle Te\_i, Te\_i \rangle$; this is a nonnegative constant independent of $i$. In particular, if $T$ kills one $e\_i$, it kills all the others. It follows that either $T=0$ or else $\{Te\_1, \ldots, Te\_n\}$ is an orthogonal basis for the range of $T$. In the latter situation, an easy computation yields
$$ \|Tx\|^2 = \sum\_i \frac{|\langle Tx, Te\_i \rangle|^2}{\langle Te\_i, Te\_i \rangle} = \sum\_i \frac{|\langle \sum\_j \langle x, e\_j \rangle Te\_j, Te\_i \rangle|^2}{\langle Te\_i, Te\_i \rangle} = \sum\_i |\langle x,e\_i \rangle|^2 \langle Te\_i, Te\_i \rangle = \alpha \|x\|^2 $$
for all $x \in U$. It follows that $\frac{1}{\sqrt{\alpha}}T$ is an isometry.
| 7 | https://mathoverflow.net/users/430 | 47303 | 29,893 |
https://mathoverflow.net/questions/29011 | 6 | At the risk of asking a stupid question I have the following problem.
Suppose I have a measure preserving dynamical system $(X, \mathcal{F}, \mu, T\_s)$, where
* $X$ is a set
* $\mathcal{F}$ is a sigma-algebra on $X$,
* $\mu$ is a probability measure on $X$,
* $T\_s:X \rightarrow X$, is a group of measure preserving transformations parametrized by $s \in \mathbb{R}$.
Suppose that this dynamical system is ergodic, so that for any $f \in L^1(\mu)$,
$\lim\_{t\rightarrow \infty}\frac{1}{2t}\int\_{-t}^t f(T\_s x) ds = \int f(x)d\mu(x)$.
Now let $B\_s$ be a real valued Wiener process such that $B\_0 = 0$, then I can define the following process:
$\frac{1}{t}\int\_{0}^t f(T\_{B\_s} x) ds$
Does anybody know how this process would behave as $t\rightarrow \infty$? Intuitively I would expect it to converge to a similar constant for a.e realisation of the brownian motion, but I can't find a convincing argument.
Thanks for your help.
| https://mathoverflow.net/users/4047 | Birkhoff ergodic theorem for dynamical systems driven by a Wiener process | Not a stupid question, but I think the answer is no.
The paper Random Ergodic Theorems with Universally Representative Sequences
by Lacey, Petersen, Wierdl and Rudolph gives a counterexample in the case where the system is being driven by a simple symmetric random walk (based on an application of Strassen's functional law of the iterated logarithm). I'm pretty sure the same technique would give a counterexample here.
The paper can be found online at: <http://www.numdam.org/item?id=AIHPB_1994__30_3_353_0>
| 5 | https://mathoverflow.net/users/11054 | 47307 | 29,896 |
https://mathoverflow.net/questions/47176 | 10 | I am wondering if there are necessary and sufficient conditions under which an one-dimensional subbundle of $TM$ has a nowhere vanishing vector field.
More precisely let $M$ be a compact smooth manifold.
a. When dose there exist a one-dimensional (smooth or continuous) subbundle $L\subset TM$?
b. If $L\subset TM$ is a continuous/smooth line subbundle of $TM$, does there exist a nowhere vanishing continuous/smooth section $X:M\to L$? If so, the euler characteristic of $L$ should be zero.
This is related to the partially hyperbolic system $f:M\to M$ and $TM=E^s\oplus E^c\oplus E^u$. I am curious if there is a 'center flow' if $\dim E^c=1$.
---
To Ryan: Am I right to say the following about your answers:
1. If there exists an 1-dimensional subbundle $L$ of $TM$, then $\chi(M)=0$. This is independent of the case whether $M$ is orientiable or not.
2. If $M$ is orientable, then there always exists an orientable 1-dimensional subbundle $L$ of $TM$.
Another question is, when is an 1-dimensional subbundle $L\subset TM$ orientable? Is it sufficient to assume that $M$ is orientable?
---
Thank you all. I did not formulate some questions properly. What I really mean is:
I. For a given line bundle $L\subset TM$, what is the obstruction for $L$ beging orientable? (or equivalently trivial according to Georges)
For example let $L\_{\mathbb{C}}$ be a complex line bundle over a complex manifold $M$, if the top Chern class $c\_1(L\_{\mathbb{C}})$ does not vanish, then $L\_{\mathbb{C}}$ can not be trivial. Is there some similar results in the real case?
II. Is there an example such that $M$ is orientable and has a non-orientable line bundle $L\subset TM$?
| https://mathoverflow.net/users/11028 | nowhere vanishing vector field on a manifold | A bundle is orientable if and only if its first Stiefel-Whitney class is 0 (one can see the first Stiefel-Whitney class as the function $w\_1: H\_1(M)\rightarrow \mathbb{Z}\_2$ which associate to a loop the sign of the determinant of the monodromy).
As mentionned by Ryan, if a line bundle is non-orientable then there is a two sheeted cover of $M$ which orients $L$ (the covering correspond exactly to the index two subgroup $ker(w\_1)$) this implies that if an oriented bundle admits a 1-dimensionnal sub-bundle the its Euler class has to be $0$ (regardless if the bundle is the tangent bundle or not).
Finally one can easily see that $T(T^2)$ is the sum of two non-trivial line bundle:
The canonical line bundle $\gamma$ on $\mathbb{R}P^1$ is non trivial but $\gamma\oplus\gamma^\*$ is (it is oriented, 2 dimensionnal and admit a section given by the trace map). Pulling back this bundle by the projection $T^2\rightarrow S^1$ you get a trivial bundle (hence the tangent bundle) on $T^2$ written as a sum of two non trivial line bundle ($w\_1$ is non zero on each summand).
| 5 | https://mathoverflow.net/users/10111 | 47318 | 29,903 |
https://mathoverflow.net/questions/47298 | 3 | I have a reasonably precise question which I hope is clear enough to get a nice answer. Let R be a Noetherian non-commutative ring which is finite as a module (and flat/free if it helps) over it's center Z(R) which we can assume has finite Krull dimension. One can also assume R integral over Z(R). By a two sided prime ideal, I mean a two sided ideal P where if $xRy$ is contained in P, then either x and/or y are in P. Consider now the abelian category of left modules and suppose we know that $Ext^m(R/P,M)=0$ for any M and $m>n$ for some n (that is independent of P). Now there aren't enough (two-sided) prime ideals in a general non-commutative ring, but there are a fair number in rings such as the one I describe above.
\**Question: Does it follow that for any finitely generated left module N, $Ext^m(N,M)=0$ for $m>n'$ which can depend on N \**. The issue I am having is that we don't have quite as effective a filtration sequence of any N, we only have a filtration such that $N\_i/N\_{i-1}=I/P$ for some left ideal and a prime P. That might be the end of the story, but there might be other tricks I am not aware of. Either way, I haven't been able to sort it out or find a good reference. If there is an extra hypothesis that helps the situation, I'd like to know about it.
| https://mathoverflow.net/users/6986 | Finite Homological Dimension of R/P for all P for module finite non-commutative rings | Yes. If we know that $\mathrm{Ext^{m}(R/P,M)}=0$ for all $m\gneq n$, then that tells us, in this situation, that the injective dimension of $M$ is less than or equal to $n$.
To see this, take an minimal injective resolution $I$ of $M$ (This is an injective resolution such that $ker(\partial\_{i})\leq\_{e}I^{i}$ for all $i$, i.e. you are doing the obvious thing and taking injective hulls at each step while constructing it). Let us define $E\_{P}$ to be the injective hull of a uniform left ideal of $R/P$ as an $R$ module. This construction does not depend on the choice of uniform left ideal if you take your ring to be left noetherian.
A result, which appears as Lemma 2.3 in
K.A. Brown, Fully bounded noetherian rings of finite injective dimension, Quart. J. Math. Oxford (2), 41, (1990) 1-13
tells us that $E\_{P}$ appears as a summand in the $i^{th}$ term of $I$ if and only if $\mathrm{Ext^{i}(R/P,M)}=0$ is not torsion as a left $R/P$ module. (Note that the bimodule structure of $R/P$ gives these $\mathrm{Ext}$ groups a left $R/P$ module structure.) Thus, if they all vanish for $m\geq n$, we have that for any $m\geq n$, $I^{m}$ contains no summands of the form $E\_{P}$ for any prime.
However, your assumptions on $R$ tell us that $R$ is a left fully bounded noetherian ring. In such a ring, we have that every indecomposable injective is of the form $E\_{P}$ for some prime $P$, and that every injective is a direct sum of indecomposable injectives. Thus $I^{m}=0$ for all $m\gneq n$, and $M$ has injective dimension less than or equal to $n$. In particular $\mathrm{Ext^{m}(N,M)}=0$ for any module $N$ and any $m\gneq n$.
| 7 | https://mathoverflow.net/users/3613 | 47323 | 29,906 |
https://mathoverflow.net/questions/45441 | 7 | There is a method of constructing representations of classical Lie algebras via Gelfand-Tsetlin bases. It has also been applied to Symmetric groups by Vershik and Okounkov. Does anybody know of any application of the method to complex representations of $GL\_n(\mathbb F\_q)$? Or, at least, any results in this directions, like what is the centralizer of $GL\_{n-1}$ in $\mathbb C[GL\_n]$?
| https://mathoverflow.net/users/6772 | Gelfand-Tsetlin bases for Lie groups over finite fields | My earlier comment was not at all well-focused. After more thought, I'm inclined to be pessimistic about using a Gelfand-Tsetlin approach here (even if it has some success for symmetric groups). Though of course it would be interesting to be proven wrong.
As Matt Davis reminds me, my offhand reference to Schur-Weyl duality is not helpful here since the work of Benson, Doty, and others deals mainly with the representations of various groups over fields of prime characteristic. (See especially Doty's papers on arXiv.) Irreducible representations of finite general linear groups over $\mathbb{C}$ are very difficult to construct directly and have very little in common with the finite dimensional representations of general linear groups or their Lie algebras in characteristic 0. Instead, the theory imitates more closely the infinite dimensional Harish-Chandra approach to Lie group representations in which parabolic induction is exploited together with a study of "discrete series".
J.A. Green's 1955 TAMS paper followed somewhat this pattern in developing combinatorially the *character* theory of finite general linear groups. But there is little insight here into constructing the elusive discrete series characters; instead orthogonality relations and the like are exploited. The best approach to an actual construction of discrete series representations was given in Lusztig's 1974 *Annals of Mathematics Studies* No. 81. Soon after that, Deligne and Lusztig pioneered a more sophisticated method for constructing generalized characters of arbitrary finite groups of Lie type. This has become the dominant influence in the subject, since Lusztig's earlier techniques don't go far enough beyond the finite general linear case.
| 3 | https://mathoverflow.net/users/4231 | 47336 | 29,913 |
https://mathoverflow.net/questions/47332 | 5 | Hi all,
We all know that the lie derivative of the metric tensor along a Killing Vector vanishes, by definition. I am trying to show that the Lie derivative of the Ricci tensor along a Killing vector also vanishes, and I am hoping to interpret it physically.
What might be a good direction to proceed? Thanks!
| https://mathoverflow.net/users/7780 | Killing vectors and Ricci Tensor | Recall that the definition of the Lie derivative of a tensor field $T$ with respect to a vector field $X$ is given by "dragging" $T$ with respect to the one-parameter (quasi) group $\phi\_t$ generated by $X$, i.e., computing $\phi\_t^\*(T)$, and differentiating wrt $t$ at $t = 0$. But to say that $X$ is a Killing field means that the $\phi\_t$ are (partial) isometries, and so not only preserve the metric tensor but also the Riemann curvature tensor and its contraction the Ricci tensor or any other tensor field that is defined canonically from the metric tensor and so preserved by isometries. Thus any such tensor field is preserved by dragging, i.e., $\phi\_t^\*(T)$ is constant in $t$ and so has a zero derivative.
Regarding the physical interpretation, let me try to answer a slightly different question. Recall that the Ricci tensor comes up as the Euler-Lagrange expression for the Einstein-Hilbert functional, and that the latter is invariant under the group of ALL diffeomorphisms. So it is natural to ask what the Noether Theorem (connecting one-parameter groups that preserve a Lagrangian to constants of the motion of the corresponding Euler-Lagrange equations) leads to in this case. The answer is that it gives the contracted Bianchi identity for the Ricci tensor. Perhaps this is what your question about physical significance was aiming at.
| 5 | https://mathoverflow.net/users/7311 | 47338 | 29,914 |
https://mathoverflow.net/questions/47342 | 4 | Assume $\mathcal{A}$ is a small cocomplete abelian $\otimes$-category (see [here](https://mathoverflow.net/questions/47079/line-bundles-in-abelian-otimes-categories) for a definition). Is there a cocontinuous, full, faithful, exact $\otimes$-functor $\mathcal{A} \to \text{Mod}(R)$ for some ring $R$?
See [here](https://mathoverflow.net/questions/32173/mitchells-embedding-theorem) for a related question. The reason why I'm asking is that I want to prove that a certain (perfectly natural!) construction works in such a category $\mathcal{A}$ and it seems to me that the axioms of $\mathcal{A}$ are not enough to check that a certain sequence is exact. In module categories there are no problems.
| https://mathoverflow.net/users/2841 | Tensor variant of Mitchell's embedding theorem | Let ${\bf 1}$ denote the unit object of ${\mathcal A}$. If $F: {\mathcal A} \rightarrow Mod(R)$ is a tensor functor, then $F( {\bf 1} ) \simeq R$. If $F$ is fully faithful, then you can recover $R$ as $End({\bf 1})$, and the functor $F$ is given by $A \mapsto Hom( {\bf 1},A)$.
This is rarely a fully faithful embedding. For example, if ${\mathcal A}$ is the category
of complex representations of a finite group, then $F$ annihilates every nontrivial irreducible representation.
| 12 | https://mathoverflow.net/users/7721 | 47344 | 29,917 |
https://mathoverflow.net/questions/47351 | 9 | I've come across the notion of Monodromy transformations while reading some aspects of variations of Hodge structures in context of Classical Mirror symmetry. I am having difficulty in grasping the concept of Monodromy transformations, probably due to lack of any good reference. My question is :
How to think of monodromy transformations, what are these intuitively and what are some examples of monodromy.
Please suggest any good references for Variations of Hodge structures and monodromy.
| https://mathoverflow.net/users/9534 | how to think of monodromy transformations | A local system is a sheaf of finite dimensional vector spaces that is locally isomorphic to the constant sheaf $k^n$. If $\gamma: [0,1] \to X$ is a continuous path in $X$, $\gamma^{-1}(L)$ is again local system on $[0,1]$ but one shows that any locally constant sheaf on $[0,1]$ is actually constant. So the fibers at 0 and 1 are canonically identified. This means that we have a map
$$
\gamma\_{\*} : L\_{\gamma(0)} \to L\_{\gamma(1)}
$$
Moreover this is
1. linear: $\gamma\_\* (v + \lambda w) = \gamma\_\* (v) + \lambda \gamma\_\* (w)$
2. invariant by homotopy: if $\gamma \sim \gamma'$, $\gamma\_\*v = \gamma'\_\*v$.
3. compatible with composition of homotopy classes of paths: $(\gamma')\_\*(\gamma\_\*x) = (\gamma'\gamma)\_\*x$.
So to any local system $L$ corresponds a representation $\pi\_1(X,x) \to GL(L\_x)$ of the fundamental group at $x$. This is the monodromy representation. You can rebuild $L$ from it: this is the sheaf of sections of $(\tilde{X}\times V)/\pi\_1(X,x) \to X$ where $\tilde{X}$ is the universal covering of $X$. We have sketched:
**Theorem**: If $X$ is connected, the functor "fiber at $x$" induces an equivalence of categories $LS(X) \to Rep(\pi\_1(X,x))$.
This is all very abstract so let us look at an example.
Consider $\mathcal{K}$, the trivial rank 2 vector bundle $\cal{O}\_{\mathbb{C}^\times}^2$ and connection
$$
\nabla \begin{pmatrix} f\_1 \cr f\_2 \end{pmatrix} =
d\begin{pmatrix} f\_1 \cr f\_2 \end{pmatrix} -
\begin{pmatrix}0 & 0 \cr 1 &0 \end{pmatrix} \begin{pmatrix} f\_1 \cr f\_2 \end{pmatrix} \frac{dz}{z}
= \begin{pmatrix} df\_1 \cr df\_2 - f\_1 \frac{dz}{z} \end{pmatrix}
$$
Horizontal sections are the solutions of $\nabla f = 0$. This is a system of two first order linear differential equations. On any simply connected $U$ we can chose a determination of the logarithm and the solution can be written
$$
\begin{pmatrix} f\_1 \cr f\_2 \end{pmatrix} =
\begin{pmatrix} A \cr A \log z + B \end{pmatrix}
$$
where $\log$ is any determination of the logarithm function. This means that the sections
$$
e\_1 = \begin{pmatrix} 1 \cr \log z \end{pmatrix} \qquad
e\_0 = \begin{pmatrix} 0 \cr 1\end{pmatrix}
$$
trivialize the sheaf of solutions on $U$. Covering $\mathbb{C}^\times$ by simply connected open sets we see that the solutions form a local system $L$.
When we turn once around 0 following the orientation, our determination $\log z$ changes to $\log z + 2\pi i$. So if $\gamma(t) = xe^{2\pi i t}$
$$
v = \begin{pmatrix} A \cr A\log x + B \end{pmatrix} \mapsto
\begin{pmatrix} A \cr A(\log x + 2\pi i) + B \end{pmatrix} = \gamma\_\*(v)
$$
The monodromy representation is
$$
\pi\_1(\mathbb{C}^\times,x) = \mathbb{Z} \to GL\_2(\mathbb{C}) \qquad
1 \mapsto \begin{pmatrix} 1 & 0 \cr 2\pi i & 1 \end{pmatrix}
$$
It tells you everything there is to know about our differential equation (because it has regular singularities). For example, the space of global solutions is identified with the space of invariant of the representation:
$$
\Gamma(\mathbb{C}^\times,L) = Hom(k\_{\mathbb{C}^\times},L) = Hom\_{\pi\_1(X,x)}(k,L\_x) = L\_x^{\pi\_1(X,x)}
$$
This is the 1 dimensional space generated by $e\_0$.
Another good example to work out is the equation $df = \alpha f\frac{dz}{z}$ (the monodromy can be quite different depending on $\alpha$).
Pierre Schapira's webpage has notes of a course on sheaves and algebraic topology focussing on local systems. Claire Voisin's book on Hodge theory is a good reference for variations of Hodge structures.
| 19 | https://mathoverflow.net/users/1985 | 47370 | 29,931 |
https://mathoverflow.net/questions/47369 | 30 | The proof that column rank = row rank for matrices over a field relies on the fact that the elements of a field commute. I'm looking for an easy example of a matrix over a ring for which column rank $\neq$ row rank. i.e. can one find a $2 \times 3$-(block)matrix with real $2\times 2$-matrices as elements, which has different column and row ranks?
| https://mathoverflow.net/users/6415 | Example for column rank $\neq$ row rank | Let $D$ be a skew field and consider the sets of $2\times 1$-matrices (columns) and $1\times 2$-matrices (lines) as left vector spaces over $D$. Let $a$ and $b$ be two non-commuting elements of $D$. Then $(a,ab)\in D(1,b)$, on the other hand $(b,ab)^{\rm T}\not\in D(1,a)^{\rm T}$.
In particular the matrix
$$
\left(\begin{array}{cc}
1 & b\\
a & ab
\end{array}
\right)
$$
is not invertible, but its transpose
$$
\left(\begin{array}{cc}
1 & a\\
b & ab
\end{array}
\right)
$$
is invertible.
| 33 | https://mathoverflow.net/users/4767 | 47374 | 29,933 |
https://mathoverflow.net/questions/47362 | 12 | Let $A$, $B$ be square matrices over infinite field (we identify them with linear operators on the vector space of columns). It is given that for all scalars $a,b$ the matrix $aA+bB$ is singular. Does it follow that there exist matrices $P$, $Q$ such that rank$(P)$+rank$(Q) > n$ but $PAQ=PBQ=0$?
If yes, is the same true for arbitrary subspaces of singular matrices? Well, the answer is no for antisymmetric matrices $3\times 3$... But how can subspaces of singular matrices be described (if they can)?
| https://mathoverflow.net/users/4312 | subspaces of singular matrices | Since the question in the new formulation is quite different, I am adding a new answer. Now the answer is positive, but the proof is not so simple, I will sketch the basic steps.
First of all, assume $A$ and $B$ are matrices of size $n$. Let $V$ and $W$ be $n$-dimensional vector spaces, so $A,B \in Hom(V,W)$. Then consider $P^1$ with coordinates $(x:y)$ and consider the morphism $V\otimes O(-1) \to W\otimes O$ given by $xA + yB$. Let $K$ be its kernel and $C$ its cokernel. Thus we have an exact sequence
$$
0 \to K \to V\otimes O(-1) \to W\otimes O \to C \to 0.
$$
The condition of singularity implies $r(K) = r(C) > 0$. Also from the exact sequence it follows that $d(K) = d(C) - n$. Now let us take $Q$ to be the induced map
$$
H^1(P^1,K(-1)) \to H^1(P^1,V\otimes O(-2)) = V
$$
and $P$ to be the induced map
$$
W = H^0(P^1,W\otimes O) \to H^0(P^1,C).
$$
Then one can check $Q$ is an embedding, $P$ is a surjection and that $PAQ = PBQ = 0$, so it remains to check that $\dim H^1(P^1,K(-1)) + \dim H^0(P^1,C) > n$. But this can be done like this. First,
$$
\dim H^0(P^1,C) \ge \chi(C) = r(C) + d(C).
$$
Further,
$$
H^1(P^1,K(-1)) \ge - \chi(K(-1)) = - (r(K) + d(K) - r(K)) = -d(K) = n - d(C).
$$
Summing up we see that
$$
\dim H^1(P^1,K(-1)) + \dim H^0(P^1,C) \ge r(C) + d(C) + n - d(C) = n + r(C) > n.
$$
| 9 | https://mathoverflow.net/users/4428 | 47375 | 29,934 |
https://mathoverflow.net/questions/47150 | 2 | Hello, I would like to know if this already has been researched.
There has been lot of research done, where logics are limited. They are often limited in the axioms or inference rules, which makes them weaker.
However, I am interested if someone has researched logics that are limited in arithmetic hierarchy. I am interested in a system that has only sentences of $\Pi^0\_2$.
Has someone worked that out?
Lucas
| https://mathoverflow.net/users/5917 | logics restricted in arithmetic hierarchy | $\Pi\_2$ statements can be modeled in the form of a "question and answer." Specifically, the statement $(\forall a \in A)(\exists b \in B)\phi(a,b)$ can be thought of as follows: $A$ is a set of questions, $B$ is a set of answers, and $\phi(a,b)$ determines whether $b$ is a correct answer to question $a$. It turns out that this scenario lends itself to interpreting Girard's Linear Logic. This is described in detail by Andreas Blass in [Questions and Answers — A Category Arising in Linear Logic, Complexity Theory, and Set Theory](http://www.math.lsa.umich.edu/~ablass/qa.pdf); in fact, Andreas Blass has [several papers on the subject](http://www.math.lsa.umich.edu/~ablass/ll.html).
| 2 | https://mathoverflow.net/users/2000 | 47376 | 29,935 |
https://mathoverflow.net/questions/47377 | 4 | Suppose we have $n$ lines in general position in the plane. Prove that there are at least $n-2$ ''small'' triangles. Here a "small" triangle is a triangle that is not contained in any larger triangle.
| https://mathoverflow.net/users/2623 | $n$ lines in general position; there are $n-2$ small triangles | It is well-known problem, but quite now I am unable to find a link on AoPS. For any line $a$ take all $n-1$ points, in which it meets other lines, and for any two consecutive points $B=a\cap b$, $C=a\cap c$ consider the triangle, formed by lines $a$, $b$, $c$ and draw a flower inside this triangle near the midpoint of its side $BC$. Totally, we get $(n-2)n$ flowers. On the other hand, in any part, which is not a triangle, we have at most two flowers (because any two flowers in the same part must lie on neighbouring sides of this part). Since we have $n(n+1)/2+1$ parts (simple induction), and $2n$ of them are unbounded (common sense), we get at most $3T+2(n(n+1)/2+1-2n-T)$ flowers, hence $(n-2)n\leq n^2-3n+2+T$, $T\geq n-2$, where $T$ is the number of triangular parts.
| 11 | https://mathoverflow.net/users/4312 | 47381 | 29,937 |
https://mathoverflow.net/questions/47350 | 24 | Sorry about the dumb title.
I'd like to understand Wick's theorem. More specifically, I have seen it pop up in several different contexts and I am really puzzled by the different statements of it that I have seen. My own background/interest is in moduli of curves, if that helps.
The first version, that is also the only one I have seen more than one time, is in the context of infinite wedge space. Here Wick's theorem is a formula about how to decompose any product of the fermionic operators $\psi\_k$ and their adjoints $\psi\_k^\ast$ as a sum over normally ordered products. This is for instance how it is explained in Kac-Raina.
A second version is in "Graphs on surfaces" by Lando and Zvonkin as Theorem 3.2.5. Here it is a statement about how to integrate a polynomial against a Gaussian measure on the real line. If $\langle f \rangle$ denotes the integral $\frac{1}{\sqrt{2\pi}}\int\_{\mathbb R} f(x) \exp(-x^2/2) dx$, then Wick's theorem states that if *f* is a product $f = f\_1 f\_2 \cdots f\_{2k}$ of linear polynomials, then $\langle f \rangle$ can be written as an explicit sum of products of pairs $\langle f\_if\_j\rangle$.
Now I can somehow believe that the two theorems above are talking about the same thing or that the second is a special case of the first. But what really got me scratching my head was the following statement from page 2 of Getzler & Kapranov's paper on modular operads (sorry for the lengthy quote):
> [...]
> As a model for this calculation, take the formula for the enumeration of graphs known in mathematical physics as Wick’s Theorem. Consider the asymptotic expansion of the integral $W(\xi,\hbar) = \log \int \exp \frac 1 \hbar \left( x \xi - \frac{x^2}{2} + \sum\_{2g-2+n > 0} \frac{a\_{g,n}\hbar^g x^n}{n!}\right) \frac{dx}{\sqrt{2\pi\hbar}}$
> considered as a power series in $\xi$ and $\hbar$. (The asymptotic expansion is independent of the domain of integration, provided it contains 0.) Let $\Gamma((g,n))$ be the set of isomorphism classes of connected graphs $G$, with a map $g$ from the vertices Vert(G) of $G$ to $\{0,1,2,...\}$ and having exactly $n$ legs numbered from 1 to $n$, such that
> $g = b\_1 + \sum\_{v\in \mathrm{Vert}(G)} g(v)$ where $b\_1$ is the first Betti number of the graph. If $v$ is a vertex of $G$, denote by $n(g)$ its valence, and let |Aut(G)| be the cardinality of the automorphism group of $G$. Wick’s Theorem states that
> $W \sim \frac 1 \hbar \left(\frac{\xi^2}{2} + \sum\_{2g-2+n>0} \frac{\hbar^g\xi^n}{n!} \sum\_{G\in \Gamma((g,n))} \frac{1}{|\mathrm{Aut}(G)|} \prod\_{v\in \mathrm{Vert}(G)} a\_{g(v),n(v)}\right)$.
I also heard a version of Wick's theorem at a talk of Rahul Pandharipandhe about two months ago, which I will not be able to state correctly here since I can't really make sense of my notes. In that version of Wick's theorem one studied an $n$-fold product of a variety with itself by interpreting it as a configuration space of $n$ "particles" moving on the variety. The goal was to simplify certain complicated products of cohomology classes given by diagonals (= particles coinciding) and Chern classes of the tangent/cotangent bundle at one of the "particles". This was all done pictorially, and one represented the diagonals as a line connecting the two points, which at least shows some connection with the Wick formalism since I think I have at one point seen these lines between particles also in the context of Feynman diagrams.
Can someone give a hint about how these Wick theorems fit together?
| https://mathoverflow.net/users/1310 | What's up with Wick's theorem? | Let's take for granted the Gaussian integration formula, which holds for both bosonic and fermionic integrals, if they are properly interpreted:
**Theoreom (Gauss, Wick):** Let $X$ be a vector space with a chosen volume form ${\rm d}x$, $f: X \to \mathbb C$ a polynomial, and $a: X^{\vee 2} \to \mathbb C$ a symmetric bilinear form with inverse $a^{-1}\in X^{\vee 2}$ such that the Gaussian measure $\exp(-\frac12 a\cdot x^{\otimes 2}){\rm d}x$ is defined. (So for example $X$ can be bosonic finite-dimensional and $a$ can have positive-definite real part; or $a$ can be invertible pure-imaginary and all integrals can be taken as conditionally convergent; or $X$ can have even-dimensional fermionic parts and the integral can be defined a la Berezin.) Then we have
$$ \int\_X \sum f^{(n)} \cdot \frac{x^{\otimes n}}{n!} \exp \left(-\frac12 a\cdot x^{\otimes 2}\right){\rm d}x = \sqrt{\det(2\pi a)} \sum f^{(2k)} \cdot \frac{(a^{-1})^{\otimes 2k}}{2^kk!} $$
Or, anyway, when $X$ is finite-dimensional Bosonic this is correct. In the fermionic case, it's off by some $\sqrt{-2\pi}$s, and $\det$ is Berezin's superdeterminant. Here $f^{(n)} : X^{\vee n} \to \mathbb C$ is the $n$th Taylor coefficient of $f$ at $0$; it's a symmetric tensor. In the fermionic case, some care must be taken with words like "symmetric", but I will ignore this subtlety.
**Proof:** Integrate by parts. $\Box$
Now the trick is to interpret the RHS combinatorially: you should draw each summand as a graph with one vertex, labeled $f^{(2k)}$, and $k$ self-loops, labeled $a^{-1}$; then if you count automorphisms correctly, the denominator of the summand is the number of automorphisms of the graph, and the numerator is the "evaluation" of the graph as a picture of tensor contractions. You can also draw the left hand summands combinatorially: a vertex with $n$ incoming strands corresponds to $f^{(n)}$, and the $n!$ counts the symmetries.
What would have happened if you had not just a single polynomial but a product? You can draw $\frac1{m!} f^{(m)} \otimes \frac1{n!} g^{(n)}$ as two vertices, one labeled $f$ with $m$ incoming strands and the other labeled $g$ with $n$ incoming strands, and the symmetries are correct, and using $\frac1{m!} f^{(m)} \otimes \frac1{n!} g^{(n)} = \binom{m+n}{m,n} (f^{(n)}\otimes g^{(m)})$ and the above theorem, the RHS now can be taken as a sum over all graphs (possibly disconnected) with two vertices, one labeled $f$ and the other labeled $g$, where the edges are still valued $a^{-1}$ and a labeled graph is weighted by its automorphism group.
Ok, so now let's try to calculate asymptotics of integrals, and I will henceforth ignore the "determinant" prefactors. My domain of integration will always be a "small" neighborhood of $0$. I'm interested in the $\hbar \to 0$ asymptotics of
$$ \int\_X \exp \frac1\hbar\left( - a\cdot \frac{x^{\otimes 2}}2 + \sum\_{n\geq 3} b^{(n)}\cdot \frac{x^{\otimes n}}{n!} \right) {\rm d}x $$
First rescale $x \mapsto \sqrt\hbar x$; this just rescales the overall integral by $\hbar^{\dim X/2}$, and I'm dropping those terms. Then the integral is $\exp\bigl( - a\cdot \frac{x^{\otimes 2}}2 + O(\hbar) \bigr)$, so keep the $O(1)$ part in the exponent but expand the $O(\hbar)$ part out:
$$ = \int\_X {\rm d}x \exp \left(-\frac12 a\cdot x^{\otimes 2}\right) \times \sum\_{m\geq 0} \frac1{m!} \left( \sum\_{n\geq 3} b^{(n)}\cdot \frac{x^{\otimes n}}{n!} \right)^m$$
Expanding the sum further, the $b$s still look like vertices with $n\geq 3$ incoming strands (and $n!$ symmetries), but now I get to have $m$ many of them, weighted by the $m!$ symmetries from permuting the vertices. So the sum is over all collections of trivalent-and-higher vertices, counted with symmetry, and an $n$-valent vertex is valued $\hbar^{\frac n 2 - 1}b^{(n)}$.
Then we integrate by connecting them up. All together, we get:
$$ = \sum\_{\text{graphs } \Gamma } \frac{ \hbar^{\#} \operatorname{ev} (\Gamma) }{ \lvert \operatorname{Aut} \Gamma \rvert } $$
Graphs can be disconnected, empty, have parallel edges and self loops, etc. We compute $\operatorname{ev}(\Gamma)$ by assigning $a^{-1}$ to each edge, $b^{(n)}$ to an $n$-valent vertex, and contracting tensors. The power on $\hbar$ is $-1$ for each vertex, and $\frac12$ for each half-edge (each part of an edge that arrives at a vertex), i.e. it's $-1$ for each vertex, $+1$ for each edge, i.e. it's the negative of the Euler characteristic of the graph.
Finally, let me get to the version from Getzler-Kapranov. Above, I used the fact that if $\star$ is a sum of connected things (counted with symmetry), then $\exp(\star)$ is a sum (counted with symmetry) over all possible collections of disjoint copies of $\star$. We've ended up with a sum-with-symmetry over disjoint things. Taking $\log$ gives the sum of connected things. For a connect graph, the negative Euler characteristic is precisely one less than the first Betti number.
**Exercise:** Redo the above calculations with $O(\hbar)$ corrections to the exponent of the initial integral, to end up with the precise Getzler-Kapranov result.
---
Finally, I should say one more thing about Feynman diagrams. Feynman and Dyson disagreed about the meaning of Feynman's diagrams: Feynman thought of them as pictures of particles interacting, and Dyson thought of them they way I do above: as graphs computing the asymptotics of integrals.
The point is the following. The most important integrals like those above that physicists care about come from particularly nice quantum field theories, where $X$ is a space of sections of some vector bundle (with some extra structure) on a Riemannian manifold, and $a$ is the Laplacian, and the $b^{(n)}$ are all "local". In this situation, $a^{-1}$ computes the "heat flow" for the Riemannian manifold, which is nothing more nor less than an integral over all paths connecting two endpoints of some "amplitude for a free particle to travel along this path". Then $\operatorname{ev}(\Gamma)$ can be interpreted as an integral over all embeddings of $\Gamma$ into your manifold of: the amplitude for your particle to travel along the edges, times an amplitude for an "interaction" at the vertices.
One good reference is the first chapter or two of Costello's recent book on QFT.
| 23 | https://mathoverflow.net/users/78 | 47385 | 29,939 |
https://mathoverflow.net/questions/47390 | 21 | **Main Question**: Does anyone know of a reference that can tell me which axioms of ZFC Quine's New Foundations prove, disprove, and leave undecided?
**Secondary Question**: I've read that diagonal arguments don't go through in NF and thus can't be used to prove that the reals are uncountable. Does NF manage to prove the uncountability of the reals by some other means or does that fact (normally rendered as "$P \_1(\mathbb{N}) < P(\mathbb{N})$" in order to make sense in NF) turn out to be undecidable in NF?
| https://mathoverflow.net/users/7521 | How much of ZFC does Quine's New Foundations prove? | Hi Amit,
Pairing (true in NF), Choice (false) and Infinity (true) are well documented. I would expect that [Thomas Forster](http://www.dpmms.cam.ac.uk/~tf/)'s book addresses if not outright answers most of your question; I suppose one would need to restate things like replacement appropriately to even make the question meaningful for some formulas. The book is "Set Theory with a Universal Set: Exploring an Untyped Universe" (Oxford Logic Guides), 1995, and you may enjoy reading it anyway.
Thomas is also interested in ZF, so even if the book doesn't completely answer your question, he may help guide you through the relevant literature if you email him directly.
As for the secondary question, quite a few basic ZF facts go through for NF when reformulated as you suggest (this is part of the reason why Forster, Randall Holmes, and other NF researchers, are interested in ZF, and why set theorists like Jensen and Solovay have thought about NF). One of these facts is Cantor's result. You may also be interested in Greg Kirmayer, "A refinement of Cantor’s theorem", Proceedings of the AMS 83 (4) (Dec., 1981), 774.
(Email me in a few days if this doesn't work out, and I'll go across the hall and ask Randall.)
| 19 | https://mathoverflow.net/users/6085 | 47392 | 29,944 |
https://mathoverflow.net/questions/47340 | 1 | Let $\mathbb R^4\_A$, $\mathbb R^4\_B$ be spacetime as seen by two inertial observers $A$, $B$ respectively, and call $f:\mathbb R^4\_A \to \mathbb R^4\_B$ the change of coordinates.
We assume that $f$ is a bijection that send straight lines in spacetime corresponding to the motion of free bodies to straight lines (this is to account for $A$ and $B$ being inertial).
In special relativity, by the constancy of light, one has that light cones are mapped to light cones by $f$, and Alexandrov theorem tell us that that is sufficient for $f$ to be affine.
I've read in a book (which title i can't recall at the moment, but i would guess that the following is a common statement) that in Galilean physics you still get that $f$ is linear (this time by the fundamental theorem of affine geometry) because it sends straight lines to straight lines.
That doesn't seem so obvious to me, since while in classical mechanics you assume that there's no restriction on the velocity of a body, i don't see why lines with t = const (which would correspond to bodies with infinite velocity) should be sent to straight lines unless one impose continuity or something else on $f$. Am i missing something?
| https://mathoverflow.net/users/10763 | Alexandrov's theorem analogue for Galilean kinematics | The exclusion of horizontal lines from the assumption of the theorem does not make a big difference.
If all non-horizontal lines are sent to lines, then all non-horizontal (2-dimensional) planes are sent to planes, because you can make a plane out of lines (even if one of the lines' directions is forbidden). Then all lines (including horizontal ones) are sent to lines, because every line can be obtained as an intersection of two non-horizontal planes.
| 3 | https://mathoverflow.net/users/4354 | 47406 | 29,951 |
https://mathoverflow.net/questions/47400 | 0 | as we all know that the slice theorem is very important in symplectic geometry , especially in the proof of marsten-sternberg-weinstein reduction theorem . so I wonder a similar question that does there is a similar theorem when the symplectic manifold have some sigular point ? and i need a original proof of the slice theorem 'Sur certains groupes de transformations de Lie' by Koszul , if you have the electric version of this paper,please send a copy to me [email protected].
| https://mathoverflow.net/users/4437 | the slice theorem in the symplectic manifold | I am only aware of results concerning singular symplectic reduction, i.e. when the Hamiltonian group action is not free and the quotient space is only a stratified symplectic space. The theorem is due to Sjamaar and Lerman ("Stratified symplectic spaces and reduction").
| 1 | https://mathoverflow.net/users/3509 | 47408 | 29,952 |
https://mathoverflow.net/questions/47399 | 21 | When one writes down the axioms of ZFC, or any other axiomatic theory for that matter, and making statements like "let x, y ..." doesn't this assume an understanding (and thus existence) of natural numbers implicitly? (Q1)
How is the reader to interpret statements such as existence of separate symbols, nevermind sets, without an intuitive notion of numbers?
Bourbaki talks about this within the framework of metamathematics, but then declares that the reader can read words, differentiate between different words etc. and that to assume otherwise is idiotic.
Is there an introduction to these circle of ideas & debates somewhere you'd recommend? (Q2)
| https://mathoverflow.net/users/6321 | Don't the axioms of set theory implicitly assume numbers? | This issue came up both in the logic course that I taught last semester and in the set theory course that I am teaching this semester.
What I told the students is that in order to do mathematical logic, we need a basic understanding of words over a finite alphabet. We cannot build a theory from less than that.
But if we understand finite strings, we basically have the natural numbers.
Some parts of mathematical logic assume some basic set theory, such as the completeness theorem of first order languages over uncountable alphabets (or just alphabets that are not recursively enumerable). But this can be avoided if you stick to sufficiently simple
alphabets (or even finite alphabets).
Similarly, you cannot do axiomatic set theory without a basic understanding of logic, which in turn requires a basic understanding of strings.
On the other hand, once you have built a sufficient theory of logic and set theory, you can use that in order to analyse mathematics. This is somewhat similar to the way that we learn mathematics: You learn to add natural numbers first, and then (usually something like 12 or more years after that) you learn about Peano Axioms that put everything on a solid foundation. I believe that this sort of circle cannot be avoided.
| 20 | https://mathoverflow.net/users/7743 | 47410 | 29,953 |
https://mathoverflow.net/questions/47396 | 2 | Suppose that $f:U\subset\mathbb{C}\to\mathbb{C}$, where $U$ is a region in the complex plane, is a holomorphic function.
Of course, if $c\in\mathbb{R}$ is a regular value for $\text{Re}(f(z))$ then $\text{Re}(f(z))^{-1}(c)$ is an union of differentiable curves in the plane.
**Question:**
If $c$ is not a regular value and $\text{Re}(f(z))^{-1}(c)$ have at least one cluster point is this set also a piece-wise differentiable curve ?
| https://mathoverflow.net/users/2386 | What $Re(f(z))=c$ can be if $f$ is a holomorphic function ? | Yes. The set of points in $\mathbb C$ having real part equal to $c$ form a line, i.e. a smooth simple curve $l$. The counterimage $f^{-1}(l)$ of any smooth simple curve $l$ via a holomorphic function $f$ is always piecewise smooth.
To prove the last sentence, take any point $z\_0\in U$ with $f(z\_0) \in l$. There are two local diffeomorphisms at $z\_0$ and $f(z\_0)$ that move $z\_0$ to $0$ and transform $f$ locally into $g(z) = z^n$. Since $f$ is not constant, we have $n>0$. Local diffeomorphisms send smooth curves to smooth curves. The counterimage along $g$ of a smooth curve passing through $0$ is the union of $n$ smooth curves exiting from 0. Therefore $f^{-1}(l)$ is a piecewise smooth curve.
| 5 | https://mathoverflow.net/users/6205 | 47411 | 29,954 |
https://mathoverflow.net/questions/47384 | 6 | What were the initial motivations of the use of the proper forcing.?
| https://mathoverflow.net/users/11094 | The history of Proper Forcing | I agree with Andres that this is a very ambitious question. Let me throw in a tiny little bit of information:
As far as I know, Jensen's construction of a model of CH without Souslin trees was one of the first uses of both countable support iteration and master conditions (conditions generic over
a countable elementary submodel of a sufficiently large initial part of the universe).
One of the first publications using proper forcing seems to be Shelah 100, with the hilarious title "Independence results" (JSL 45 (1980), 563-573).
This paper not only introduces proper forcing (without proofs of the the iteration theorem and so on), but also oracle-ccc forcing and oracle-proper forcing.
| 9 | https://mathoverflow.net/users/7743 | 47412 | 29,955 |
https://mathoverflow.net/questions/47428 | 10 | Let $R$ the polynomial ring in $n$ variables with complex coefficients and $I$ an ideal of $R$. Is it true that if $R/I$ is CM also $R/J$ is CM (where $J$ is the radical of $I$)?
Is there a relations between a resolution of $R/J$ and one of $R/I$? What if I suppose that $proj.dim(R/I)=2$?
| https://mathoverflow.net/users/4821 | CM for radical ideal | It is [not true](http://www.uni-due.de/~mat306/preprints/radical.pdf), but the example is not easy to find $I = (x\_2^2-x\_4x\_5,x\_1x\_3-x\_3x\_4, x\_3x\_4-x\_1x\_5)$!
| 11 | https://mathoverflow.net/users/2083 | 47434 | 29,966 |
https://mathoverflow.net/questions/46368 | 4 | The sequence A000140 is studied
<http://oeis.org/A000140>
(Kendall-Mann numbers: the maximum number of permutations on n letters having the same number of inversions )
and I am looking for a proof that
M(n)/M(n-1)=n+1/2 when n= infinity, M(n) - max element in row n.
If you have any ideas how to proove or disproove it, even the question is too hard, could you let me know in anyway.
Modelling with Pari GP shows
for n = 0 to 149 M(n)/M(n-1):
1.00000000, 1.00000000, 2.00000000, 3.00000000, 3.66666667, 4.59090909, 5.67326733, 6.69458988, 7.61939520, 8.57906801, 9.60953383, 10.6235009, 11.5884536, 12.5657349, 13.5817521, 14.5907723, 15.5704306, 16.5558579, 17.5656455, 18.5718445, 19.5585507, 20.5484134, 21.5549876, 22.5594838, 23.5501133, 24.5426559, 25.5473665, 26.5507683, 27.5438066, 28.5380914, 29.5416285, 30.5442887, 31.5389122, 32.5343930, 33.5371446, 34.5392804, 35.5350028, 36.5313400, 37.5335406, 38.5352923, 39.5318079, 40.5287792, 41.5305788, 42.5320411, 43.5291478, 44.5266018, 45.5281005, 46.5293394, 47.5268986, 48.5247283, 49.5259956, 50.5270586, 51.5249718, 52.5230999, 53.5241854, 54.5251073, 55.5233026, 56.5216716, 57.5226117, 58.5234188, 59.5218427, 60.5204088, 61.5212309, 62.5219434, 63.5205550, 64.5192845, 65.5200094, 66.5206430, 67.5194106, 68.5182772, 69.5189212, 70.5194882, 71.5183871, 72.5173696, 73.5179455, 74.5184560, 75.5174660, 76.5165477, 77.5170657, 78.5175276, 79.5166329, 80.5157998, 81.5162682, 82.5166882, 83.5158756, 84.5151165, 85.5155421, 86.5159256, 87.5151843, 88.5144896, 89.5148781, 90.5152297, 91.5145507, 92.5139127, 93.5142686, 94.5145921, 95.5139679, 96.5133798, 97.5137071, 98.5140058, 99.5134299, 100.512886, 101.513188, 102.513465, 103.512932, 104.512428, 105.512707, 106.512964, 107.512470, 108.512001, 109.512260, 110.512499, 111.512039, 112.511602, 113.511843, 114.512067, 115.511637, 116.511229, 117.511454, 118.511663, 119.511261, 120.510879, 121.511090, 122.511285, 123.510909, 124.510550, 125.510748, 126.510932, 127.510578, 128.510241, 129.510426, 130.510599, 131.510267, 132.509949, 133.510123, 134.510286, 135.509973, 136.509673, 137.509838, 138.509992, 139.509696, 140.509412, 141.509568, 142.509713, 143.509434, 144.509165, 145.509312, 146.509450, 147.509185, 148.508930,
n+0.5 for n = infinity
[email protected]
| https://mathoverflow.net/users/10903 | The property of Kendall-Mann numbers | It is known that
$$ \left| P\left( \frac{\mathrm{inv}(\pi)-\frac 12{n\choose 2}}{\sqrt{n(n-1)(2n+5)/72}}\leq x\right)-\Phi(x)\right| \leq \frac{C}{\sqrt{n}}, $$
where $\Phi(x)$ denotes the standard normal distribution. From this it is immediate that $M(n+1)/M(n)=n-\frac 12+o(1)$.
For some references, see <http://arxiv.org/PS_cache/math/pdf/0508/0508242v2.pdf>.
| 12 | https://mathoverflow.net/users/2807 | 47440 | 29,968 |
https://mathoverflow.net/questions/47442 | 28 | It's probably common knowledge that there are Diophantine equations which do not admit any solutions in the integers, but which admit solutions modulo $n$ for every $n$. This fact is stated, for example, in Dummit and Foote (p. 246 of the 3rd edition), where it is also claimed that an example is given by the equation
$$ 3x^3 + 4y^3 + 5z^3 = 0. $$
However, D&F say that it's "extremely hard to verify" that this equation has the desired property, and no reference is given as to where one can find such a verification.
So my question is: Does anyone know of a readable reference that proves this claim (either for the above equation or for others)? I haven't had much luck finding one.
| https://mathoverflow.net/users/430 | Diophantine equation with no integer solutions, but with solutions modulo every integer | It is actually quite straightforward to write down examples in one variable where this occurs. For example, the Diophantine equation $(x^2 - 2)(x^2 - 3)(x^2 - 6) = 0$ has this property: for any prime $p$, at least one of $2, 3, 6$ must be a quadratic residue, so there is a solution $\bmod p$, and by Hensel's lemma (which has to be applied slightly differently when $p = 2$) there is a solution $\bmod p^n$ for any $n$. We conclude by CRT. (**Edit:** As Fedor says, there are problems at $2$. We can correct this by using, for example, $(x^2 - 2)(x^2 - 17)(x^2 - 34)$.)
Hilbert wrote down a family of quartics with the same property. There are no (monic) cubics or quadratics with this property: if a monic polynomial $f(x) \in \mathbb{Z}[x]$ with $\deg f \le 3$ is irreducible over $\mathbb{Z}$ (which is equivalent to not having an integer solution), then by the Frobenius density theorem there are infinitely many primes $p$ such that $f(x)$ is irreducible $\bmod p$.
| 24 | https://mathoverflow.net/users/290 | 47443 | 29,970 |
https://mathoverflow.net/questions/44861 | 10 | It is possible that on a sphere $S^n$ there is a natural Riemannian metric in $R^(n+1)$. But it is not always possible for pseudo Riemann metric since the sum of two symmetric matrix which are not positive definite but may have rank different from the two matrix. So I wonder what is the sufficient and necessary condition for the dimension of a sphere which can be endowed with a Lorentz metric.
| https://mathoverflow.net/users/1964 | For which $b$ it is possible that $S^n$ can have a Lorentz metric? Why? | **A compact simply connected manifold carriez a Lorenz metric iff its Euler characteristic vanishes.**
Proof: If $\chi(M)=0$, $M$ carries a nowhere vanishing vector field $X$. Pick up a Riemannian metric $g$ on $M$ (using a partition of unity argument) and denote by $\eta$ the 1-form dual to $X$: $\eta(Y):=g(X,Y)$ for all $Y\in TM$. Then
$$g-2\frac{\eta\otimes\eta}{g(X,X)}$$
is a Lorenz metric.
Conversely, if $M$ has a Lorenz metric $h$ of signature $(n-1,1)$, pick again a Riemannian metric $g$ and consider the symmetric endomorphism $A$ of $TM$ defined by $h(.,.)=g(A.,.)$. The eigenspaces of $A$ corresponding to the unique negative eigenvalue define a line sub-bundle of $TM$ which is trivial if $M$ is simply connected, so $\chi(M)=0$.
Therefore, the answer to your question is: $S^n$ carries a Lorenz metric iff $n$ is odd.
| 31 | https://mathoverflow.net/users/10675 | 47446 | 29,973 |
https://mathoverflow.net/questions/44016 | 2 | Consider probability space W with pair of random variables having same distribution. On how much this variables distinct in terms of W symmetries? Namely, let's talk about automorphism as measure-preserving self-mapping of W defined almost everywhere. The following question must be well studied. When such random variables may be combined by some automorphism of W (up to measure-zero set, of course)? Sorry for bad english.
| https://mathoverflow.net/users/8906 | Random variables with same distribution | There are some obvious restrictions in the case when the base probability
space $W$ is allowed to have atoms. For instance, if $W$ consists of an atom
and a continuous part with equal masses $1/2$, then their indicator functions
have the same distribution, but an automorphism in question clearly does not
exist.
A slightly more involved counterexample is provided by the following pair of
random variables defined on the unit interval endowed with the usual Lebesgue
probability measure. One is the identity map, and the other one is $x\mapsto
2x$ (mod 1).
I will give a complete answer to your question in the situation when the base
probability space $(W,P)$ is a [Lebesgue
(standard) probability space](http://en.wikipedia.org/wiki/Standard_probability_space) (which is the only reasonable generality in
probability nowadays, although there is a lot of people here who are very fond
of discussing various exotic if not outright pathological measure spaces). The
*signature* of a measure space is the mass of its non-atomic part plus
the non-increasing sequence of the weights of its atoms. Rohlin (see the above
Wikipedia article for a reference to his 1949 article very appropriately called "On
the fundamental ideas of measure theory") proved that, up to isomorphism,
Lebesgue spaces are completely characterized by their signatures. In
particular, there is only one purely non-atomic Lebesgue measure space, which
is just the unit interval with the Lebesgue measure on it (whence the term).
It is less known that Rohlin also obtained a complete classification of
homomorphisms of Lebesgue spaces (equivalently, of their measurable
partitions, or of complete sub-$\sigma$-algebras). Signature of the quotient
measure and signatures of the conditional measures associated with the
homomorphism provide an obvious system of conjugacy invariants of such
homomorphisms. Rohlin proved that this system is, in fact, a complete system
of invariants. In the simplest purely non-atomic case it means that any
homomorphism of Lebesgue spaces with a purely non-atomic quotient space and
purely non-atomic conditional measures is conjugate to the coordinate
projection of the unit square onto the unit interval (both being endowed with
the canonical Lebesgue measures).
Applied to your original question, Rohlin's classification implies that two
random variables with the same distribution on a Lebesgue space are equivalent
in your sense if and only if for a.e. value taken by these variables the
corresponding conditional measure spaces are isomorphic, i.e., have the same
signature. In particular, the latter is the case if almost all conditional
measures are purely non-atomic.
| 5 | https://mathoverflow.net/users/8588 | 47451 | 29,976 |
https://mathoverflow.net/questions/47447 | 13 | In the questions [Is "semisimple" a dense condition among Lie algebras?](https://mathoverflow.net/questions/9661/is-semisimple-a-dense-condition-among-lie-algebras) and [What is the Zariski closure of the space of semisimple Lie algebras?](https://mathoverflow.net/questions/9719/what-is-the-zariski-closure-of-the-space-of-semisimple-lie-algebras), something equivalent to the following is mentioned: if you have a smoothly varying family of semisimple Lie algebras, all the Lie algebras in the family are isomorphic. e.g. the following quote:
"Because the Cartan classification of isomorphism classes of semisimples is discrete (no continuous families), connected components of the space of semisimples are always contained within isomorphism classes."
I can't see how this follows just from the discreteness of the classification. Can anyone explain why it's true or give a counterexample?
e.g. could you not have a $\mathbb{P}^1$ of semisimple Lie algebras which are generically isomorphic to $\mathfrak{d}\_7 \oplus \mathfrak{a}\_1$ say, but at one point you get $\mathfrak{e}\_8$, or something similar?
| https://mathoverflow.net/users/11108 | Deformations of semisimple Lie algebras | The experts should correct me if there is a fatal mistake in the argument, I am neither an algebraic geometer nor a Lie theorist. I am working over $\mathbb{C}$.
1. Let $\mathfrak{g}$ be a semisimple Lie algebra, $G$ be the adjoint group, $Aut(\mathfrak{g})$ be the automorphism group. Let
$Aut\_0 (\mathfrak{g})$ be the subgroup of automorphisms that preserve the decomposition of $\mathfrak{g}$ into simple ideals.
It has finite index in the whole automorphism group, since the decomposition into ideals is unique and an automorphism has to
map a simple ideal into a simple ideal. The group $Aut\_0(\mathfrak{g}$ is the product of the automorphism groups of the simple factors.
The automorphism group of any simple Lie algebra has the adjoint group as a finite index subgroup; the quotient is the automorphism group of
the Dynkin diagram. The upshot of this discussion is: $Aut(\mathfrak{g})$ has $G$ as a finite index subgroup. In particular, the dimension of
$Aut(\mathfrak{g})$ only depends on the dimension of $\mathfrak{g}$!
EDIT: there is a better proof of this step in the literature, e.g. in Procesi's book, page 301.
1. Let $V$ be the variety of all Lie algebra structures on $C^n$; it has an action of $GL\_n$ on it, the stabilizers are the
automorphism groups, the orbits are the isomorphism classes. Now let $\mathfrak{g} \in V$ be semisimple and let
$O \subset V$ be its $GL\_n$-orbit. Now use the orbit closure theorem (Borel, Linear
Algebraic Groups, page 53). It says that any $GL\_n$-orbit $O$ is open in its closure and that $\bar{O} \setminus O$ consists
of orbits of smaller dimension. Hence all Lie algebras in the closure of $O$ which are not isomorphic to $\mathfrak{g}$
must have a larger-dimensional automorphism group. By part 1, they are not semisimple.
| 15 | https://mathoverflow.net/users/9928 | 47457 | 29,981 |
https://mathoverflow.net/questions/47461 | 4 | I know this question is absolutely trivial, but having self-studied the subject I feel extremely unsure on the basics. Do not hesitate to downgrade the question, if you feel it deserves so.
Given a morphism of complex of sheaves $\varphi:\mathcal{F}^\bullet\to \mathcal{G}^\bullet$ on a topological space $X$ (eventually as nice as needed), we have two induced morphisms, one in local cohomology, $\mathcal{H}^\bullet(\varphi):\mathcal{H}^\bullet(\mathcal{F}^\bullet)\to \mathcal{H}^\bullet(\mathcal{G}^\bullet)$, and one in global cohomology $H^\bullet(\varphi):H^\bullet(X;\mathcal{F}^\bullet)\to H^\bullet(X,\mathcal{G}^\bullet)$.
The question is: does $\mathcal{H}^\bullet(\varphi)=0$ imply $H^\bullet(\varphi)=0$?
I guess so, since there is a spectral sequence with $E\_2$-page given by cohomology with cefficients in local cohomology abutting to global cohomology. But as I said, I do not trust myself too much on self-study..
| https://mathoverflow.net/users/8320 | Does trivial on local cohomology implies trivial on global cohomology? | The exact sequence
$0\rightarrow\mathrm{Z}/p\rightarrow\mathrm{Z}/p^2\rightarrow\mathrm{Z}/p\rightarrow0$
($p$ a prime say) gives a map $\mathrm{Z}/p\rightarrow\mathrm{Z}/p[1]$ in the
derived category of abelian groups (which can be realised as a map of
complexes). It clearly induces zero on cohomology. We can then take any
topological space $X$ and pull this back to $X$ giving a map (in the derived
category of sheaves on $X$) between constant sheaves again inducing zero in
(local) cohomology. The maps $H^i(X,\mathrm{Z}/p)\rightarrow
H^{i+1}(X,\mathrm{Z}/p)$ induced by this map are the Bockstein operations. Hence
we can let $X$ be any space with non-zero Bockstein map. A particular case for
$p=2$ are the real projective spaces of dimension at least $2$.
| 6 | https://mathoverflow.net/users/4008 | 47465 | 29,984 |
https://mathoverflow.net/questions/47181 | 3 | I'll ask you to consider a situation wherein one has a series of edges for a graph, $(e\_1, e\_2, ..., e\_N) \in E$, each with a specifiable length $(l\_1, l\_2, ..., l\_N) \in L$, and the goal is to insure that the connected graph has a unique topology in 3-space. More specifically, I'm interested in insuring that some graph with the connectivity of a polytope can only be drawn as the skeleton of that particular polytope - that there should be no crossed edges or knots possible for the specified edge lengths.
To provide a physical example:
I use a group of rods to represent the edges of the desired graph (with pencils or the like) and color/symbol-encode their ends to represent vertex-assignments. I want to choose rod lengths in such a way that if I hand them to a naive-constructor (i.e. a 3-year old or a computer-controlled robot), and tell him/her/it to connect the ends of the rods together that have the same color or symbol, after waiting an arbitrarily long time there will only be a unique geometry satisfying the connectivity constraints of the graph I originally had in mind.
Is there a known computational complexity for this problem? Is there even a solution in the general case, or in the case where we apply the restriction that the specified polytope is convex?
I appreciate any feedback!
EDIT 1: The edges of the graph must be straight lines in 3-space, they cannot be bent to accommodate a particular edge length.
EDIT 2: Does the problem become easier if one assumes some physical diameter for the edges?
| https://mathoverflow.net/users/11048 | Can we uniquely define a graph to have the topology of a polytope via proper edge length selection? | This seems like a question in rigidity theory. In particular it seems like part of what you want is conditions for **global rigidity** in 3 dimensions.
Let me write down some definitions and basic facts from the introduction of [this nice set of slides by Dylan Thurston](http://www.math.columbia.edu/~dpt/speaking/Rigidity/talk.pdf) and then post some references that might be helpful.
A *framework* is a graph and a map from its vertices into $d$-dimensional Euclidean space $\mathbb{E}^d$. A framework is *locally rigid* if every other framework in a small neighborhood with the same edge lengths is related to it by an isometry of $\mathbb{E}^d$. A framework is *globally rigid* if every other framework in $\mathbb{E}^d$ with the same edge lengths is related to it by an isometry of $\mathbb{E}^d$.
It turns out that checking global rigidity is NP hard, even in 1 dimension (Saxe 1979). However, if you're just interested in "generic" frameworks, i.e. those for which the edge lengths do not satisfy any polynomial relation, then work of Connelly and S. Gortler, A. Healy and D. Thurston characterizes these frameworks in any dimension with an efficient randomized algorithm. See [the paper of GHT](http://www.cs.harvard.edu/~sjg/papers/ggr.pdf) or the slides above. I must admit that I have not yet studied their work in any detail.
Since you are requiring that your frameworks are skeleta of polytopes, there may be extra structure which you can exploit. Let me just point you to [Cauchy's rigidity theorem](http://en.wikipedia.org/wiki/Cauchy%27s_theorem_%28geometry%29) which states that convex polyhedra are rigid if you force the faces to be rigid in addition to the edges. If you don't have this restriction on the faces, then there are nonrigid examples, e.g. the 1-skeleton of a cube can be sheared, also pointed out in sleepless in beantown's answer. If you do have the restriction on the faces, but you allow nonconvex polyhedra, then there are [flexible polyhedra](http://en.wikipedia.org/wiki/Flexible_polyhedra).
In addition to the links above, there are several surveys on the webpage of [Robert Connelly](http://www.math.cornell.edu/~connelly/) on various topics in rigidity theory.
| 4 | https://mathoverflow.net/users/353 | 47467 | 29,985 |
https://mathoverflow.net/questions/47462 | 14 | Does there exist polynomial $p(x)$ with integer coefficients such that $p(x) > 0$ for all real values of $x$, but for any integer $n > 0$ there exists integer $k$ such that $n$ divides $p(k)$? The trick with multiplying quadratic polynomials (as here [Diophantine equation with no integer solutions, but with solutions modulo every integer](https://mathoverflow.net/questions/47442/diophantine-equation-with-no-integer-solutions-but-with-solutions-modulo-every-i)) does not work, if I am not mistaken.
| https://mathoverflow.net/users/4312 | Positive polynomial having roots modulo any integer | $\newcommand{\Q}{\mathbf{Q}}\newcommand{\Z}{\mathbf{Z}}$
I now suspect the answer might be no! This isn't a complete answer but it might be an idea that turns into one.
So let me assume that such $p$ exists and let me go for a counterexample.
First I claim that if such $p$ exists, then a monic $p$ exists. For if $p$ works, then so does $Np$ for a large positive integer $N$, and if $p=cx^d+\ldots$ and $N=c^{d-1}$ then $Np=q(cx)$ with $q$ monic with integer coefficients, and $q$ also works. So WLOG $p$ is monic.
Now say $p$ factors as $p\_1p\_2\ldots p\_r$ in $\Q[x]$ with the $p\_i$ monic. By Gauss' Lemma the $p\_i$ are all in $\mathbf{Z}[x]$. Note
that $p$ must have positive degree so $r\geq 1$. Let $K$ be the splitting field of $p$ and let $K\_i$ be the field $\Q[x]/(p\_i)$. Now none of the $K\_i$ have any real places. Let $c$ denote any complex conjugation in $Gal(K/\Q)$. By Cebotarev density, there's a prime number $\ell$, unramified in $K$, and such that $Frob\_\ell$'s conjugacy class in $Gal(K/\Q)$ is that of $c$, and indeed there are infinitely many such $\ell$.
Certain primes cause me trouble, so let me get rid of them now: for each $K\_i$ let $S\_i$ be the index of $\Z[x]/(p\_i)$ in the full ring of integers of $K\_i$, and let $S$ be the product.
Now let $\ell$ be a prime as above, and such that $\ell$ doesn't divide $S$.
I claim that this $\ell$ is going to cause us problems. Say $p(k)$ is zero mod $\ell$. Then one of the $p\_i(k)$ is zero mod $\ell$, so the factorization of $p\_i$ mod $\ell$
has a linear factor, and so $\ell$ factors in the integers of $K\_i$ into a bunch of primes
one of which has degree 1. I now want to argue something like the following: the Frobenius element at $\ell$ corresponding to this prime is fixing a root of $p\_i$ so complex conjugation is fixing a root of $p\_i$ so $p$ has a real root! But the cricket has just started and I can't think straight. Can this be turned into a proof?
| 13 | https://mathoverflow.net/users/1384 | 47470 | 29,986 |
https://mathoverflow.net/questions/47468 | 2 | Hi,
I am reading the book "Topics in Optimal Transportation" by Cedric Villani.
The Brenier's theorem states (among other things) that there is a unique transport plan for the optimal transport with the quadratic cost if the measure $\mu$ (to be transported toward $\nu$) does not give mass to small sets.
(page 66)
A counter example, in the case of mu giving mass to small sets, is the case where $\mu$ and $\nu$ are measures on $\mathbb R^2$, concentrated on $\{(0,0),(1,1)\}$ and $\{(0,1),(1,0)\}$ respectively. In that case, there is no uniqueness. (page 67)
However, I am wondering why this is only the case when $\mu$ does not give mass to small sets. We can easily imagine a case where $\mu$ and $\nu$ are continuous but are made such that there is no uniqueness. For example, with an analogy with the example above, $\mu$ can spread its mass on a line between $(0,0)$ and $(1,1)$ with some thickness, and $\nu$ on a line between $(0,1)$ and $(1,0)$ with the same thickness (so the two lines are symmetric and don't have zero measure). In that case, I still see two possible transport plans...
Same thing if we consider 2 gaussians centered at $(0,0)$ and $(1,1)$ in the first measure and at $(0,1)$ and $(1,0)$ in the second one....
Any idea ?
Thanks
| https://mathoverflow.net/users/8646 | Brenier's theorem | Because there's no reason to ship a whole gaussian en masse to the same target gaussian. It's cheaper to send all the mass that is to one side of the diagonal (line joining (0,0) to (1,1)) to one gaussian and the rest of the mass to the other side. The same is true for the thickened line.
| 4 | https://mathoverflow.net/users/613 | 47476 | 29,990 |
https://mathoverflow.net/questions/47474 | 6 | Does anyone have an answer to the three-dimensional analogue of the 2009 Putnam Competition A1 problem, viz., if $f\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ satisfies $\sum\_{i=1}^8 f(a\_i) = 0$ whenever $a\_1, \ldots, a\_8$ are the vertices of a cube, must $f$ be identically zero?
A few thoughts: Since the cube is not self-dual above dimension $2$, the solution (well, at least, my solution) to the $2$-dimensional problem doesn’t generalize. To try to show that the answer to the $3$-dimensional problem is “no,” one might try letting $\Omega$ be the set of ordered pairs $(X, f\_X)$ where $X \subseteq \mathbb{R}^3$ is a subset and $f\_X\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ is a function that sums to zero over any eight points of $X$ that form the vertices of a cube. Ordering $\Omega$ in the usual way, we can invoke Zorn’s Lemma to obtain a maximal $(X, f\_X) \in \Omega$. If $X \neq \mathbb{R}^3$, then every $x \in \mathbb{R}^3 \setminus X$ must, in two “incompatible” ways, be the eighth vertex of a cube whose remaining vertices lie in $X$. But I don't see how a contradiction arises from this (and of course one could make the same argument in two dimensions, where a contradiction does not arise).
| https://mathoverflow.net/users/6521 | function that sums to zero over cube vertices | First prove that the sum of the values of your function at the vertices of any regular tetrahedron is zero. You can do this by considering a $2\times 2\times 2$ cube made out of $8$ smaller cubes that you can color in a checkerboard fashion. Sum all the vertices of the black cubes and subtract the sum of the vertices of all the white cubes.
Second you can show that the sum of the values at the vertices of a $x\times x\sqrt{2}$ rectangle is always zero. You need to consider two tetrahedra sharing an edge for this.
From here it is very similar to the Putnam problem you mention. (In fact some solutions apply in the same exact way.)
P.S. This shows that it is enough to have the information only for cubes whose sides are parallel to the axis.
| 7 | https://mathoverflow.net/users/2384 | 47477 | 29,991 |
https://mathoverflow.net/questions/47466 | 11 | How do I compute the compact cohomology of a hypersurface?
For example, let $f$ be a Newton polynomial of a polytope in $\mathbb{R}^n$ and let $X = (f=0)$
inside $(\mathbb{C}^\\*)^n$ (maybe there is some dependency on the coefficients of $f\;$?). Can you tell me anything about $H^\*\_c(X)$? Perhaps I should know better, but I don't.
Thanks!
| https://mathoverflow.net/users/1186 | How do I compute the compact cohomology of a hypersurface? | The classic reference is Danilov-Khovanskii's "Newton polyhedra and an algorithm for calculating Hodge-Deligne numbers". There is subsequent work by Cox, Batyrev, Malvyutov, etc. but they are mainly concerned with more general toric ambient spaces; if you want a hypersurface in the torus then this original paper should have all you need.
| 12 | https://mathoverflow.net/users/6107 | 47493 | 30,001 |
https://mathoverflow.net/questions/47484 | 4 | Genericity is still a little bit mysterious to me, although not as much as it used to be.
Here is a rough paraphrase of Theorem 3.5 of Kunen's *Set Theory: an Introduction to Independence Proofs*
>
> Let $M$ be a countable transitive model, $\langle\mathbb P,\leq\rangle\in|M|$ a partial order and $G\subseteq{\mathbb P}$ $M$-generic. For any $\phi$ it is the case that $M[G]\vDash\phi$ if and only if $(\exists p\in G) p\Vdash\phi$.
>
>
>
I will refer to the right-to-left direction of this implication as "things which are forced are true in the extension".
Assume that $\mathbb P$ is splitting and we've chosen some $\phi$ such that $M\nvDash\phi$ and there is some $p\in\mathbb P$ such that $p\Vdash\phi$. Let $\hat p$ be the principal $\mathbb P$-filter generated by $p$. Among the $M$-generic filters $G$, those for which $M[G]\vDash\phi$ are exactly those for which $\hat p\subseteq G$.
But what about $\hat p$ itself? Well, all the principal filters of $\mathbb P$ are sets of the ground model, so $\hat p\in|M|$. So forcing with $\hat p$ will give us back an extension model which is isomorphic to $M$ -- no new sets. So $M[\hat p]\nvDash\phi$.
So, if I haven't screwed up so far, we know that for generic filters "things which are forced are true in the extension", and for filters which happen to be elements of $|M|$ it is not necessarily the case that "things which are forced are true in the extension". This leaves a gap: filters which are not elements of $|M|$, yet are not $M$-generic either.
**Question:** does "things which are forced are true in the extension" hold for $\{any,all\}$ filters $F\supseteq\hat p$ such that $F\notin|M|$ yet $F$ is also not $M$-generic?
Phrased another way: does the right-to-left implication rely on the filter being generic, or only on the fact that it isn't an $M$-set?
I can see quite clearly why the left-to-right implication ("things which are true in the extension are forced") relies on the fact that the filter is generic, and couldn't possibly work without "intersects every dense set". I'm having a harder time seeing why genericity (rather than simply "not in $|M|$") is necessary in the other direction, though. Kunen uses it in his proof (the case where $\phi$ is $\tau\_1=\tau\_2$), but it's not as easy to see why (or if) the use is essential.
This isn't a homework question (or even close to any of them). Hope you'll trust me.
| https://mathoverflow.net/users/2361 | Is genericity essential to "things which are forced are true in the extension" or only to its converse? | There are some statements $\phi$ in the forcing language that are true in
$M[G]$ iff they are forced by a condition in the filter, no matter whether $G$ is generic or not.
The simplest is $p\in\Gamma$ where $\Gamma$ is the natural name for the generic filter.
(Assuming the forcing notion is separative.)
But I understand that this is not what you are looking for, since you are asking about statements not true in $M$ but true in the extension.
Now, the problem with filters that are not generic and also not in the ground model is that
the extension might not be a model of ZFC at all. Let me give an example:
Consider the forcing for adding a single Cohen real. Every function from $\omega$ to $2$ corresponds to a filter. $M$ is countable and transitive and the ordinals intersected with $M$ are some countable ordinal $\alpha$. Fix some function $f:\omega\to 2$ that codes a well-ordering of $\omega$ that is longer than $\alpha$.
Let $G$ be the corresponding filter. If $M[G]$ was a model of ZFC, then it would contain
the ordinal coded by $f$. But for every name $\sigma$ in $M$, the evaluation of $\sigma$ at
$G$ has rank at most the rank of $\sigma$, showing that $M[G]$ has the same ordinals as $M$.
This technology should be enough to give you all kinds of counter examples you might be looking for.
| 5 | https://mathoverflow.net/users/7743 | 47498 | 30,004 |
https://mathoverflow.net/questions/47492 | 18 | Consider a continuous irreducible representation of a compact Lie group on a finite-dimensional complex Hilbert space. There are three mutually exclusive options:
1) it's not isomorphic to its dual (in which case we call it 'complex')
2) it has a nondegenerate symmetric bilinear form (in which case we call it 'real')
3) it has a nondegenerate antisymmetric bilinear form (in which case we call it 'quaternionic')
It's 'real' in this sense iff it's the complexification of a representation on a real vector space, and it's 'quaternionic' in this sense iff it's the underlying complex representation of a representation on a quaternionic vector space.
Offhand, I know just **four compact Lie groups whose continuous irreducible representations on complex vector spaces are all either real or quaternionic in the above sense**:
1) the group Z/2
2) the trivial group
3) the group SU(2)
4) the group SO(3)
Note that I'm so desperate for examples that I'm including 0-dimensional compact Lie groups, i.e. finite groups!
1) is the group of unit-norm real numbers, 2) is a group covered by that, 3) is the group of unit-norm quaternions, and 4) is a group covered by *that*. This probably explains why these are all the examples I know. For 1), 2) and 4), all the continuous irreducible representations are in fact real.
**What are all the examples?**
| https://mathoverflow.net/users/2893 | Which groups have only real and quaternionic irreducible representations? | An irreducible representation is real or quaternionic precisely when its
character is real-valued. By the Peter-Weyl theorem all characters are
real-valued precisely when every element in the group is conjugate to its
inverse. When the group is connected a more precise answer is as follows: The
Weyl group (in its tautological representation) must contain multiplication by
$-1$ and this is true precisely when all indecomposable root system factors have
that property. I don't remember off hand which indecomposable root systems have
this property but it is of course well known (type A is out, type B/C is in,
type D depends on the parity of the rank).
**Addendum**: Found the relevant places in Bourbaki. All characters are real-valued precisely when the element he calls $w\_0$ is $-1$ (Ch. VIII,Prop. 7.5.11) and one can also read off if a given representation is real or quaternionic (loc. cit. Prop 12). From the tables in Chapter 6 one gets that $w\_0=-1$ precisely for $A\_1$, B/C, D for even rank, $E\_7$, $E\_8$, $F\_4$ and $G\_2$.
| 34 | https://mathoverflow.net/users/4008 | 47500 | 30,005 |
https://mathoverflow.net/questions/47433 | 6 | Questions:
1. What is the connection between representation theory of complex semisimple Lie groups and representations of (maybe "proper") Lorentz groups?
2. Why should one read Bargmann's paper on irred. unitary representations of Lorentz group if one wants to know unitary representation?
| https://mathoverflow.net/users/1930 | Representations of Lorentz group | Weyl's theorem states that any finite dimensional representation of a compact Lie group is completely reducible. The Lorentz group is not compact, but its maximal compact subgroup is $SU(2)$. This is why there is a 1-1 correspondence between the representations of the Lorentz group (algebra) and those of $SU(2)$ (respectively $su(2)$).
You can find more details about this relation in
* R. O. Wells, Jr. Differential analysis on complex manifolds. Published 1980 by Springer-Verlag in New York. I quote from page 173:
>
> **Proposition 3.1:** The mappings $r\_1$,
> $r\_2$ and $d$ in (3.7) are all
> bijective, i.e., there is a one-to-one
> correspondence between representations
> of $SL(2,\mathbb C)$, $sl(2,\mathbb
> > C)$, $SU(2)$ and $su(2)$.
>
>
>
The representations of $SU(2)$ and $su(2)$ are treated in most books on representation theory.
Indeed, Wigner's and Bargmann's articles are useful if you are interested in how the spin particles occur from representations of the Lorentz group:
* E. P. Wigner. On unitary representations of the inhomogeneous Lorentz group. Annals of Mathematics, (40):149{204, 1939.
* V Bargmann. On Unitary Ray Representations of Continuous Groups. Ann. of Math., 59:1{46,
* E. P. Wigner. Group Theory and its Application to Quantum Mechanics of Atomic Spectra. Academic Press, New York, 1959.
The main idea is that the wavefunctions should transform in wavefunctions at a Poincare transformation, and the transformation should be unitary. So, we need unitary representations of the Poincare group.
In order to classify the irreducible representations of a group, one can use the Casimir invariants. The Lie algebra of the $ISL(2,\mathbb C)$ group, $isl(2,\mathbb C )$ (isomorphic to the Poincare Lie algebra $so(1,3)$) has two Casimir invariants, namely $m^2=p^a p\_a$ and the squared angular momentum about the center of mass, $S^2=s(s+1)$, where the spin $s$ takes semi-integer values. Usually is considered that only the representations corresponding to $m^2\geq 0$ have physical meaning, the ones with $m^2<0$ being tachyonic. For the case $m^2>0$ $s$ is of the form $0,\frac 1 2, 1, \frac 3 2, \ldots \frac n 2 \ldots$. For the case $m^2=0$, $s$ can be $0,\pm\frac 1 2, \pm1, \pm\frac 3 2, \ldots\pm\frac n 2 \ldots$. In this last case there exists also representations with continuous spin, but no physical evidence support this kind of representations.
**Added.**
The nice relation between the representations of $SL(2,\mathbb C)$ and $SU(2)$ refers, as I stated, to the finite dimensional case. But what's the connection between the finite-dimensional and the infinite-dimensional representations? The infinite-dimensional reps of $SL(2,\mathbb C)$ which are of interest in quantum mechanics are spinor fields. That is, they are superpositions of sections in finite-dimensional complex vector bundles which are associated to $SL(2,\mathbb C)$. To construct such an associated finite-dimensional bundle, you start with a finite-dimensional representation. Strictly speaking, the things are more complicated for infinite-dimensional representations, but for quantum mechanical systems (with a finite number of particles), there is this nice connection between infinite-dimensional and finite-dimensional representations.
| 4 | https://mathoverflow.net/users/10095 | 47501 | 30,006 |
https://mathoverflow.net/questions/47168 | 9 | The following is inspired by [this](https://math.stackexchange.com/questions/11487/y-xn-what-is-exyy) recent question on math.stackexchange.
Two standard exercises in conditional expectation are to find ${\rm E}(X\_1|X\_1+X\_2)$ where:
1) $X\_i$, $i=1,2$, are independent ${\rm N}(0,\sigma\_i^2)$ rv's;
2) $X\_i$, $i=1,2$, are independent Poisson($\lambda\_i$) rv's.
The solutions are given by $\frac{{\sigma \_1^2 }}{{\sigma \_1^2 + \sigma \_2^2 }}(X\_1 + X\_2)$ and $\frac{{\lambda \_1 }}{{\lambda \_1 + \lambda \_2 }}(X\_1 + X\_2)$, respectively. A proof for case 1) is given on math.stackexchange. For case 2) we have
${\rm E}(X\_1|X\_1 + X\_2 = n) = \sum\limits\_{k = 0}^n {k{\rm P}(X\_1 = k|X\_1 + X\_2 = n)}.$
A straightforward calculation shows that the right-hand side sum is equal to
$\sum\limits\_{k = 0}^n {k{n \choose k}\bigg(\frac{{\lambda \_1 }}{{\lambda \_1 + \lambda \_2 }}\bigg)^k \bigg(\frac{{\lambda \_2 }}{{\lambda \_1 + \lambda \_2 }}\bigg)^{n - k} },$
which is the expectation of the binomial distribution with parameters $n$ and $\lambda\_1 / (\lambda\_1 + \lambda\_2)$, hence given by $ n \lambda\_1 / (\lambda\_1 + \lambda\_2)$. The result for case 2) is thus proved. In this context, what is common to the normal and Poisson distributions is that both are infinitely divisible (ID). More specifically,
the characteristic function of $X\_i \sim {\rm N}(0,\sigma\_i^2)$ is given by ${\rm E}[{\rm e}^{{\rm i}zX\_i} ] = {\rm e}^{\sigma \_i^2 ( - z^2 /2)}$, and that of $X\_i \sim {\rm Poisson}(\lambda\_i)$ by ${\rm E}[{\rm e}^{{\rm i}zX\_i} ] = {\rm e}^{\lambda \_i ({\rm e}^{{\rm i}z} - 1)}$. Now, consider integrable, ID, independent rv's $X\_i$, $i=1,2$, with characteristic functions of the form ${\rm E}[{\rm e}^{{\rm i}zX\_i} ] = {\rm e}^{c\_i \psi(z)}$, $c\_i > 0$ (loosely speaking, the characteristic function of an arbitrary ID rv is of that form). In view of the normal and Poisson examples considered above (the former requires somewhat tedious algebra for the solution), and the fact that many important rv's fall into the general category of integrable ID rv's (e.g., gamma rv's), it would be very useful to have the following result: ${\rm E}(X\_1 | X\_1 + X\_2) = \frac{{c \_1 }}{{c\_1 + c\_2 }}(X\_1 + X\_2)$. In fact, I have proved it recently.
Now to my questions: 1) Have you encountered this result before? 2) Can you provide a rigorous but simple proof of it?
3) Can you provide some intuition?
EDIT: 1) Here's another interesting example: if $X\_i \sim {\rm Gamma}(c\_i,\lambda)$, $i=1,2$, so that $X\_i$ has density $f\_{X\_i } (x) = \lambda ^{c\_i } {\rm e}^{ - \lambda x} x^{c\_i - 1} /\Gamma (c\_i )$, $x > 0$, then ${\rm E}(X\_1 | X\_1 + X\_2) = \frac{{c\_1 }}{{c\_1 + c\_2 }}(X\_1 + X\_2 )$. 2) It is very instructive to reformulate the result in terms of L\'evy processes: if $X = \{ X(t): t \geq 0 \}$ is an integrable L\'evy process, then ${\rm E}[X(s)|X(t)] = \frac{s}{t}X(t)$, $0 < s < t$.
EDIT: The "direct" solution for the gamma case considered above is now given
[here](https://math.stackexchange.com/questions/11673/ex-1-x-1-x-2-for-independent-gamma-random-variables). This shows, once more, the effectiveness of the general formula.
EDIT: A complete solution is given in my first (according to date) answer below.
EDIT: An important extension is considered in my second answer below.
| https://mathoverflow.net/users/10227 | $E(X_1 | X_1 + X_2)$, where $X_i$ are (integrable) independent infinitely divisible rv's "of the same type" | Regarding a "rigorous but simple proof" of the relation the OP is interested in, such a proof is, almost completely, already written in the original post.
To see this, consider independent integrable random variables $X$ and $Y$ and assume that their characteristic functions, defined for every real number $t$, are such that $E({\mathrm e}^{\mathrm{i}tX})=\mathrm{e}^{a\psi(t)}$ and $E(\mathrm{e}^{\mathrm{i}tY})=\mathrm{e}^{b\psi(t)}$ for a given function $\psi$ and given real numbers $a$ and $b$. Let $S=X+Y$. Now, to prove that $$
(a+b)E(X\vert S)=aS,
$$
it suffices to show that, for every real number $t$,
$$
(a+b)E(X\mathrm{e}^{\mathrm{i}tS})=aE(S\mathrm{e}^{\mathrm{i}tS}).
$$
Since both sides of the equality can be explicitly written in terms of $a$, $b$, the function $\psi$ and its derivative $\psi'$, the proof is, in a way and modulo some easy computations, already over.
For example,
$$
E(S\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})+E(Y\mathrm{e}^{\mathrm{i}tY})E({\mathrm e}^{\mathrm{i}tX}),
$$
because $X$ and $Y$ are independent. Here, both $E({\mathrm e}^{\mathrm{i}tX})$ and $E({\mathrm e}^{\mathrm{i}tY})$ are already known, and both $E(X{\mathrm e}^{\mathrm{i}tX})$ and $E(Y{\mathrm e}^{\mathrm{i}tY})$ are derivatives of the former with respect to $(\mathrm{i}t)$. Hence,
$$
E(S\mathrm{e}^{\mathrm{i}tS})=-\mathrm{i}(a+b)\psi'(t)\mathrm{e}^{(a+b)\psi(t)}.
$$
Likewise,
$$
E(X\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})=-\mathrm{i}a\psi'(t)\mathrm{e}^{(a+b)\psi(t)}.
$$
Comparing these two formulas, we are done.
(If this helps, one can note that the signs of $a$ and $b$ must be the same, in the sense that $ab>0$ or that $X$ or $Y$ must be $0$ with full probability.)
| 4 | https://mathoverflow.net/users/4661 | 47505 | 30,007 |
https://mathoverflow.net/questions/47404 | 16 | I am considering the following two isomorphisms:
First, if $X$ is a reasonably nice topological space, then $X$ has a normal covering space which is maximal with respect to the property of having an abelian group of deck transformations. The group of deck transformations of this covering space is isomorphic to the abelianizatin of the fundamental group of $X$, which can be identified with the singular homology group $H\_1(X,\mathbb{Z})$.
Second, if $K$ is a number field, class field theory gives an isomorphism between the Galois group of the maximal abelian unramified extension of $K$ (the Hilbert class field) and the ideal class group of $K$. The ideal class group can be identified with the sheaf cohomology group $H^1(\mathrm{Spec}(\mathcal{O}\_K),\mathbf{G}\_m)$.
Given the apparent similarity between these two theorems, is there some more general theorem which implies both of these results as special cases?
| https://mathoverflow.net/users/5263 | Connection between isomorphisms of algebraic topology and class field theory | As BCnrd says, the theorem you want is geometric class field theory. One version says that the abelianization of the fundamental group of a curve over an algebraically closed field is the fundamental group of its Jacobian. One can use this to derive class field theory for curves over finite fields. Over the complex numbers, this is a topological statement, since the Jacobian of a Riemann surface can be constructed by topological methods, such as $J(C)=H\_1(C;\mathbb R/\mathbb Z)$. Also, consider Weil's construction of the Jacobian by using Riemann-Roch to recognize a high symmetric power of a curve as a $\mathbb C\mathbb P^n$ bundle over the Jacobian. The projective space bundle is probably not a topological invariant, but the symmetric power is and it already has the right fundamental group. That has an extensive topological generalization, the [Dold-Thom theorem](https://ncatlab.org/nlab/show/Dold-Thom+theorem), that the homology of a reasonable space is the homotopy groups of its infinite symmetric power.
The key unifying ingredient is, as BCnrd says, the Jacobian, even though it is missing from both of your statements.
| 12 | https://mathoverflow.net/users/4639 | 47516 | 30,017 |
https://mathoverflow.net/questions/47504 | 6 | I'm trying to learn how to compute stalks of IC sheaves, and I was wondering about the following example:
Fix $n$. Let $X \subset \mathbb{C}^n$ be the variety cut out by the equation $x\_1 \cdots x\_n =0$, i.e. the coordinate hyperplanes. What are the stalks of $\mathrm{IC}(X)$ at the various points of $X$, in particular at the origin?
This seems like a natural toy example, but if the general answer is difficult, I'd be happy to know how to compute this for small $n$.
| https://mathoverflow.net/users/788 | Intersection Cohomology of Coordinate Hyperplanes | Let $Y$ denote the disjoint union of the coordinate hyperplanes in $\mathbb{C}^n,$ and let $f:Y \to X$ denote the corresponding resolution of singularities.
1) Show that $f\_{\ast}\mathbb{C}\_Y[n-1] \simeq IC\_X$ (consider, for example, the support conditions and the fact that both sheaves are isomorphic to $\mathbb{C}\_U[n-1]$ when restricted to the nonsingular open $U \subset X$).
Edit (some details added): Letting $U$ denote the complement of the set where any two coordinate planes intersect, $f$ is an isomorphism when restricted to $U.$ We therefore have that the restriction (i.e., pullback) of $f\_{\ast}\mathbb{C}\_Y[n-1]$ to $U$ coincides with $\mathbb{C}\_U[n-1]$ (by proper base change if you like).
In order to conclude that $f\_{\ast}\mathbb{C}\_Y[n-1] \simeq IC\_X,$ we now just need to check the support and cosupport conditions which uniquely define the intersection cohomology sheaf (together with the fact that its restriction to $U$ is the (shifted) constant sheaf). These conditions are similar to, but more restrictive than, the support and cosupport conditions for perverse sheaves.
I recommend looking at page 21 of the wonderful article by de Cataldo and Migliorini, which can be found at <http://arxiv.org/abs/0712.0349> for a statement of these support and cosupport conditions (and figure 1 on page 25 for a visual illustration of the definition).
Since the fibers of $f$ consist of a finite number of points, the cohomology of the fibers is non-zero only in degree zero. This shows that the first condition (the support condition) is satisfied.
For the second condition (the cosupport condition), you can either derive it from the support condition using Verdier duality and the properness of $f,$ or you can simply note that an open ball in $\mathbb{C}^{n-1}$ has non-zero compactly supported cohomology only in degree $2n-2.$
2) Now it's straightforward to compute any of the stalks since the fiber of $x \in X$ consists of anywhere between one point and n points, depending on how many hyperplanes $x$ lives inside of.
Alternatively, it is also possible to do this by using only basic definitions. To compute the stalk at $x,$ intersect a sufficiently small open ball around $x$ in $\mathbb{C}^n$ with $X$ and then calculate the intersection cohomology by considering intersection cochains (just like you would for singular cohomology, but now with a less restrictive notion of cochain).
| 6 | https://mathoverflow.net/users/916 | 47519 | 30,018 |
https://mathoverflow.net/questions/47138 | 7 | I'm wondering if a particular theory of second order arithmetic has been studied or is known to be equivalent to some other theory.
Consider the formulas generated by $\Pi^1\_1$ and $\Sigma^1\_1$ formulas by propositional combinations (the title refers to this, rather informally, as the difference hierarchy, since these formulas are essentially the difference between two $\Pi^1\_1$ formulas, the difference between two such, and so on). Let's call this class of formulas $\Delta$, for convenience.
My question is what the strength (either reverse mathematical or proof theoretic) of $\Delta-TI\_0$, the theory which allows transfinite induction along any well-ordering for $\Delta$ formulas, is. This has proof theoretic strength strictly greater than $\Pi^1\_2-TI\_0$ (Rathjen and Weierman's proof that $\Pi^1\_2-TI\_0$ implies well-foundedness of $\psi\Omega^{\Omega^n}$ goes through, and furthermore this theory can carry out the key induction internally), and strictly less than $\Pi^1\_1-CA\_0$ (the standard argument that $\Pi^1\_1-CA\_0$ proves the existence of a $\beta$-model of $\Pi^1\_\infty-TI\_0$ suffices).
| https://mathoverflow.net/users/8991 | Strength of Transfinite Induction on the Difference Hierarchy | I hope it's not too tacky to answer my own question now that I've had a few days and train rides to think about it.
The answer is that the proof theoretic ordinal of $\Delta-TI\_0$ is the Howard-Bachmann ordinal (the ordinal of $\Pi^1\_1-CA\_0^-$, $\Pi^1\_\infty-TI\_0$, $ID\_1$, and $KP\omega$).
The upper bound is easy to see ($\Delta-TI\_0$ is a subtheory of $\Pi^1\_\infty-TI\_0$). (In fact, each instance of $\Delta-TI\_0$ is provable in $\Pi^1\_1-CA\_0^-$, modulo some work to handle parameters in the formulas in $\Delta$.)
For the lower bound, consider the easy embedding of $ID\_1$ into the language of second order arithmetic, in which arithmetic formulas map to arithmetic formulas and the least fixed point in $ID\_1$ is mapped to a $\Pi^1\_1$ formula. Then all formulas of $ID\_1$ map to $\Delta$ formulas, so every axiom of $ID\_1$ becomes a theorem of $\Delta-TI\_0$ (the point here is that the only instances of induction axioms appearing in $ID\_1$ are induction on $\Delta$ formulas). In particular, the proof of well-foundedness below the Howard-Bachmann ordinal can be carried out, unchanged, in $\Delta-TI\_0$.
| 3 | https://mathoverflow.net/users/8991 | 47526 | 30,021 |
https://mathoverflow.net/questions/47509 | 14 | Brown representability states that any contravariant functor from the homotopy category $CW\_\*$ of pointed CW complexes to the category of pointed sets is representable if it turns coproducts into products and satisfies a type of Mayer-Vietoris gluability axiom, which I like to think of as a weak version of "the functor sends push-outs into pull-backs" as any representable functor must. The proof very much relies on the fact that CW complexes can be built up in a steady and predictable manner, as it uses Whitehead's theorem that a weak homotopy equivalence is automatically a homotopy equivalence. Namely, one shows that an element which is "universal" for the spheres is actually a "universal element" for this functor (in the sense of Yoneda's lemma).
Brown representability has many interesting consequences, e.g. that there is a "universal" principal $G$-bundle for pointed CW complexes (where $G$ is a topological group) or that the Eilenberg-Maclane spaces represent the cohomology functors. However, in the former case, it's actually true that the universal bundle exists for any topological space, not just CW complexes. I don't know whether the cohomology functors are representable on the category of all pointed topological spaces (even if one restricts to non-pathological ones: say Hausdorff, with nondegenerate basepoint), though I would imagine that a CW complex couldn't do it. This leads me to ask:
>
> Is there a version of Brown representability for arbitrary pointed topological spaces?
>
>
>
There is a version of it on the [nLab](http://ncatlab.org/nlab/show/Brown+representability+theorem) in more generality, but I don't know enough about categorical homotopy theory to understand anything.
Could someone perhaps translate some of that into the special case of topological spaces?
| https://mathoverflow.net/users/344 | Brown representability beyond CW complexes |
>
> Is there a version of Brown representability for arbitrary pointed topological spaces?
>
>
>
The answer is: No and Yes.
If you take a particular construction of a generalised cohomology theory that makes sense for all topological spaces then there is no guarantee that it will be representable in the homotopy category of topological spaces. For example, the various flavours of ordinary cohomology are only guaranteed to coincide for CW complexes (okay, and stuff with the homotopy type of such). That they disagree elsewhere shows that at least one of them can't be representable. Another example is K-theory. It's great that for compact Hausdorff spaces (and some others) that K-theory is exactly what you get when you take vector bundles and group complete, but it's not true for other spaces.
That's the "No". Now for the "Yes". The point about the "Yes" is that Brown representability is *so* good to have that when we move outside the realm of CW-complexes, we often *define* our cohomology theory to be (homotopy) homs into the representing object (found by restricting to CW-complexes). That is, we take our "natural construction" of whatever cohomology theory it is, use Brown representability to find the representing object for CW-complexes, and then define the extension of the theory to all topological spaces to be $[X,\underline{E}]$. Then it is representable, but by construction rather than by any fancy theory.
| 17 | https://mathoverflow.net/users/45 | 47529 | 30,024 |
https://mathoverflow.net/questions/47439 | 6 | A is an abelian variety over number field K, with simple good reduction at a finite field $\kappa$, can we deduce that $A$ itself is simple?
| https://mathoverflow.net/users/2008 | Can the simplicity of abelian varieities be implied by the reduction | Here is a slightly different proof. We have the following facts:
>
>
> (1) If $B, C$ are abelian varieties over $K$, then the Néron model of $B\times C$ is the product of the Néron models (this is a simple onsequence of the universal property of Néron models). In particular, if $B\times C$ has good reduction, then $B, C$ also have good reduction, and the reduction of $B\times C$ is the product of the reductions of $B$ and $C$.
>
>
>
>
>
> (2) If $A$ and $B$ are isogeneous abelian varieties over $K$ and if $A$ has good reduction, then $B$ has good reduction and any isogeny $A\to B$ induces an isogeny on the reductions.
>
>
>
Proof of (2): if $A\to B$ is an isogeny of degree $n$, then there exists an isogeny $B\to A$ such that the composition $A\to B\to A$ is the multiplication-by-$n$ map $[n]\_A: A\to A$ on $A$. On the reductions we have the maps $A\_k \to B\_k^0 \to A\_k$ whose composition is $[n]\_{A\_k}$. Consider a positive integer $\ell$ prime to $n$ and to $\mathrm{char}(k)$. As the restriction of $[n]\_{A\_k}$ to $A\_k[\ell]$ is an isomorphism, $A\_k[\ell]\to B^0\_k[\ell]$ is injective, so $B^0\_k[\ell]$ contains $(\mathbb Z/\ell \mathbb Z)^{2\dim B}$. Therefore $B\_k^0$ is an abelian variety and $B$ has good reduction. Finally $A\_k\to B\_k=B\_k^0$ is an isogeny because its kernel is contained in $A\_k[n]$.
This is a special case of Néron-Ogg-Shafarevich criterion (Serre-Tate: *Good reduction of abelian varieties*, §1).
Now to answer the original quesiton, if $A$ was not simple, then it is isogeneous to a product of abelian varieties $B\times C$. Then $B\times C$, $B, C$ have good reduction by (2) and (1), and $A\_k$ is isogeneous to $B\_k\times C\_k$. So $A\_k$ would not be simple.
| 5 | https://mathoverflow.net/users/3485 | 47539 | 30,031 |
https://mathoverflow.net/questions/47546 | 1 | Let $M$ be a centered parallelepiped, the intersection of $M$ and any plane $P$ that passes through the origin is a parallelogram or hexagon. Each parallelogram or hexagon has a cubic box that is the smallest box that can contain the parallelogram or hexagon. Denote the cubic box by $B(P)$. There exist planes $P\_{0}$ such that $B(P\_{0})$ is the smallest among all the boxes $B(P)$.
Is it true that there is always one of the planes $P\_{0}$ such that the cross-section of the centered parallelepiped $M$ by $P\_{0}$ is a parallelogram?
Thanks.
| https://mathoverflow.net/users/7738 | Is there always a parallelogram cross-section of parallelepiped contained in the smallest box | No, here is a counter-example (to revision 9).
Let $A$ be the linear map that sends the vector $(1,1,1)$ to $V:=(100,100,100)$ and is the identity on the orthogonal complement of this vector. Then any optimal cross-section is a hexagon whose plane intersects the six edges of $A(Q)$ separated from the vertices $V$ and $-V$. Indeed, any plane that avoids this configuration must intersect one of the edges adjacent to $V$. And the edges adjacent to $V$ are contained in the half-space $x+y+z\ge 100/3$, so any such section (and hence its minimal cube) has diameter at least $100/3\sqrt3\ge 10$. On the other hand, the intersection of $A(Q)$ with the plane $x+y+z=0$ is the same as the intersection of this plane with $Q$, so it fits in a unit cube (whose diameter is $2\sqrt3<10$). Hence no section of diameter $\ge 10$ can be optimal.
| 3 | https://mathoverflow.net/users/4354 | 47556 | 30,043 |
https://mathoverflow.net/questions/47557 | 5 | Hi Experts,
I have question regarding Kolmogorov's Superposition Theorem:
It is known that:
Let ${f(x\_1,x\_2,...,x\_m): \Re^m :=[0,1]^m \to \Re}$ be an arbitrary multivariate continuous function. From Kolmogorov’s Superposition Theorem we have the following representation:
${f(x\_1,x\_2,...,x\_m)= \sum\_{q=0}^{2m} \Phi\_q (\sum\_{p=1}^m \phi\_{p,q}(x\_p))}$
with continuous one-dimensional outer functions ${\Phi\_q}$ and inner functions ${\phi\_{p,q}}$. All these functions are defined on real line. The inner functions ${\phi}$ are independent of function ${f(x\_1,x\_2,...,x\_m)}$.
Question is:
Is it possible to find inner functions ${\phi\_p{(x\_p)}}$ which is independent of $q$, that satisfies the superposition theorem:
${f(x\_1,x\_2,...,x\_m)= \sum\_{q=0}^{2m} \Phi\_q (\sum\_{p=1}^m \phi\_p (x\_p))}$
Where ${\Phi\_q, \phi\_p, N}$ can be selected and defined where appropriate.
It is critical to our works on nonlinear control, and we look forward to your advises on possible solutions, tips, related documents,etc.
Thank You!
Wang Tao
| https://mathoverflow.net/users/11131 | Regarding Kolmogorov's Superposition Theorem | If I understand correctly, that doesn’t seem possible. If the inner functions are independent of $ q $, then the sum of the outer functions collapses to a single function $ \Phi $ with
$$
\Phi(\cdot) = \sum\_{q = 0}^{2 m} {\Phi\_{q}}(\cdot).
$$
Hence, the stronger form of the theorem that you’re looking for would be equivalent to:
>
> For every $ m \in \mathbb{N} $, there exist continuous functions $ \phi\_{1},\phi\_{2},\ldots,\phi\_{m}: [0,1] \to \mathbb{R} $ such that any continuous function $ f: [0,1]^{m} \to \mathbb{R} $ can be written as
> $$
> f(x\_{1},x\_{2},\ldots,x\_{m}) = {\Phi\_{f}} \left( \sum\_{p = 1}^{m} {\phi\_{p}}(x\_{p}) \right)
> $$
> for some continuous function $ \Phi\_{f}: \mathbb{R} \to \mathbb{R} $.
>
>
>
For each $ i \in \{ 1,\ldots,m \} $, take $ f\_{i} $ to be the $ i $-th projection function (i.e., $ {f\_{i}}(x\_{1},x\_{2},\ldots,x\_{m}) = x\_{i} $). It can be shown that this would imply the existence of a continuous function $ F: [0,1]^{m} \to \mathbb{R} $ that is also *one-to-one* (as all the coordinates can be recovered from it). However, a continuous function from an open set in $ \mathbb{R}^{m} $ to $ \mathbb{R} $ cannot be one-to-one for $ m > 1 $ (by Invariance of Domain), so we obtain a contradiction.
| 4 | https://mathoverflow.net/users/1229 | 47571 | 30,048 |
https://mathoverflow.net/questions/47458 | 20 | This was inspired by this recent [question](https://mathoverflow.net/questions/46970/proofs-of-the-uncountability-of-the-reals).
In my [answer](https://mathoverflow.net/questions/46970/proofs-of-the-uncountability-of-the-reals/47022#47022) there, I pointed out that, given $F:{\mathcal P}(X)\to X$, an argument dating back to Zermelo allows us to define a pair $(A,B)$ of distinct subsets of $X$ witnessing that $F$ is not injective. The pair is defined in terms of a well-ordering that is constructed using $F$.
Of course, the usual Cantor argument also shows that $F$ is not injective: One considers $$ A=\{F(Z)\mid F(Z)\notin Z\},$$ and argues that there must be a $B$ with $A\ne B$ and $F(A)=F(B)$.
My question is:
>
> Can we exhibit such a set $B$ (definably from $F$)?
>
>
>
| https://mathoverflow.net/users/6085 | Cantor's argument revisited | If I understood the OP correctly, the problem can be stated as follows :
**Problem 1.** Let $X$ be a set, let $F:{\cal P}(X) \to X$, and let $A$ be defined
as above: $$A=\lbrace F(Z) | Z\subseteq X, F(Z) \not\in Z\rbrace.$$ Find a definable $B$ (in terms of $F$) such that $B \neq A$ and $F(B)=F(A)$.
Now Problem 1 is equivalent to the simpler problem :
**Problem 2.** Let $Y$ be a set and let $Y'$ be a nonempty subset of $Y$. Find
a definable $y\_0$ (in terms of $Y$ and $Y'$) which is in $Y'$.
The interesting and nontrivial part of the equivalence is of course, to show that we can solve Problem 2 if we can solve Problem 1. Here is how. Let $Y$ and $Y'$ be as above. Take two elements $a,b$ and a countable set $W=\lbrace w\_k \rbrace\_{k \geq 0}$ outside of $Y$. Now define
$X$ to be the disjoint union of $\lbrace a,b \rbrace$ and $Y \times W$, and define
$F : {\cal P}(X) \to X$ by:
1. $F(\lbrace (y,w\_0) \rbrace)=a$, if $y\in {Y'} $,
2. $F(\lbrace (y,w\_{k+1}) \rbrace)=(y,w\_k)$ , for all $y\in Y$ and $k\ge0$,
3. $F(X)=a$, and
4. $F(Z)=b$ for all other subsets $Z$ of $X$ (thus $F(\emptyset)=b$).
Now, by construction, $A=X$, and any solution $B$ to Problem 1 is of the form $\lbrace (y,w\_0) \rbrace$ for some $y\in Y'$, thereby solving Problem 2.
| 13 | https://mathoverflow.net/users/2389 | 47576 | 30,050 |
https://mathoverflow.net/questions/47585 | 10 | Is it true that two random (w.r.t. Haar measure) rotations in $SO(3)$ generate a free group?
| https://mathoverflow.net/users/1121 | Random rotations in SO(3) and free group | Yes. Here's what should be a proof: the set of pairs of elements satisfying any particular relation is Zariski closed, hence has measure zero (to show that it is not $SO(3) \times SO(3)$ it suffices to know that at least one subgroup generated by two elements is free), and there are countably many relations.
| 20 | https://mathoverflow.net/users/290 | 47586 | 30,056 |
https://mathoverflow.net/questions/47554 | 6 | The following question is an attempt to find a lower bound for the value of a polynomial at integer points. It is something that I originally thought about while trying to understand how it would be possible to approach [this MO question](https://mathoverflow.net/questions/9731/polynomial-representing-all-nonnegative-integers) about polynomials representing the nonnegative integers, but does seem to be very interesting in its own right. I guess this would fall under the study of Diophantine approximation/geometry, which I am not at all expert on. So it could even be a known conjecture or theorem, or obviously false. Maybe someone on MO will be able to say?
As is well known, the [Thue-Siegel-Roth theorem](http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem) says that, for an irrational algebraic number $\alpha$, there are only finitely many pairs of integers $p,q$ satisfying the inequality $\vert p/q-\alpha\vert\le q^{-2-\epsilon}$. Here, $\epsilon$ is any fixed positive real number.
This is easily seen to be equivalent to the following lower bound on the growth of homogeneous and irreducible polynomials $f\in\mathbb{Q}[X,Y]$ of degree $d > 1$. For all but at most finitely many integer pairs $x,y$, the inequality
$$
\vert f(x,y)\vert\ge\vert y\vert^{d-2-\epsilon}\qquad\qquad{\rm(1)}
$$
holds. The argument is very simple. We can decompose $f(x,y)$ as $y^d\prod\_{i=1}^d(x/y-\alpha\_i)$ for distinct irrational algebraic numbers $\alpha\_i$. As $x/y-\alpha\_i$ can only be made arbitrarily small for at most one $i$ at a time, (1) is equivalent to applying the Thue-Siegel-Roth theorem to $x/y-\alpha\_i$.
Now for my question. Is there an extension of (1) to non-homogeneous polynomials $f$? Now, I realize that it cannot possibly carry directly across to the non-homogeneous case in the same form. For one thing, a simple change of variables allows us to replace $f$ by a polynomial of arbitrarily large degree, such as $\tilde f(x,y)=f(x+y^r,y)$, which would invalidate any inequality depending on the degree of $f$ and, similarly, the right hand side of (1) would change form under changes of variables. To guess how this can be fixed, we can look at [Siegel's theorem](http://en.wikipedia.org/wiki/Siegel%2527s_theorem_on_integral_points), which says that $f(x,y)=a$ has only finitely many integer solutions in $x,y$ whenever $f-a$ defines a curve over $\mathbb{Q}$ of genus $g$ at least one. It seems reasonable then, that a generalization of (1) should involve the genus $g$ of the curve defined by $f-a$, for typical rationals $a$, and not the degree.
So, to be precise, my question is whether there is an increasing and unbounded function $\phi\colon\mathbb{N}\to\mathbb{R}$ with the following property: If $f\in\mathbb{Q}[X,Y]$ is such that $f-a$ defines a curve of genus $g$ (for all but finitely many $a$), there exists a nonconstant polynomial $h\in\mathbb{Q}[X,Y]$ such that the inequality
$$
\vert f(x,y)\vert\ge \vert h(x,y)\vert^{\phi(g)-\epsilon}\qquad\qquad{\rm(2)}
$$
holds on $\mathbb{Z}\times\mathbb{Z}$ outside of a finite set (\*). We might even hope that $h(x,y)=y$ under a change of variables, but that seems like a bit much to ask. Comparing with the case where $f$ is homogeneous of degree d, the genus of $f-a$ for nonzero $a$ is given by $g=(d-1)(d-2)/2$ and we have $\phi(g)=(-1+\sqrt{1+8g})/2$, although I doubt that precise form would hold in the non-homogeneous case.
(\*) Edit: It is easy to construct polynomials which vanish or degenerate into something simpler on a given finite set of curves. E.g., $f=\prod\_{i=1}^n(X-a\_i)$ is zero on the curves $x=a\_i$. So I should say that (2) holds on $\mathbb{Z}\times\mathbb{Z}$ outside of a finite set of curves (of genus zero).
| https://mathoverflow.net/users/1004 | Bounding the growth of rational bivariate polynomials from below | What you are asking is for a strong effective form of Siegel's theorem. There are a number of conjectures in this direction, all considered very hard. For example, if $f=x^3-y^2$, what you are asking is essentially Hall's conjecture: <http://en.wikipedia.org/wiki/Hall%27s_conjecture> .
In general, it follows from Vojta's conjectures (LNM 1239). Nothing close is known. For certain classes of polynomials (general cubics, $y^m - f(x)$, a few others) you can get a lower bound of the type $\log \log \log \max\{x,y\}$ give or take a log or two, from the theory of linear forms in logarithms.
| 6 | https://mathoverflow.net/users/2290 | 47587 | 30,057 |
https://mathoverflow.net/questions/47596 | 14 | I am teaching a course on manifolds, and soon I will have to prove the Stokes' theorem which, of course, involves defining oriented manifolds. There are many ways to define an oriented manifold. My favorite way is by the reduction of the structure group of the tangent bundle. But this definition and a couple of other that I know give just one orientation for the point: $GL(V)/GL^{+}(V)$ is
${\mathbb Z}/2{\mathbb Z}$ when $\dim V \ge 1$, but when $\dim V=0$ then $GL(V)$ has only one element. Of course, it is possible to define two orientations of a point by convention. I would like to know if is there any way to define the orientation for smooth manifolds in a uniform way that would also yield two orientations of a point.
| https://mathoverflow.net/users/3635 | orientations for zero-dimensional manifolds | I think instead of reducing the structure group of the tangent bundle, you could consider reducing the structure group of its top exterior power, which is always a one-dimensional vector bundle, even in dimension zero. Then the set of orientations of each connected component is a torsor under $GL\_1(\mathbb{R})/GL\_1^+(\mathbb{R}) \cong \{ \pm 1 \}$.
| 16 | https://mathoverflow.net/users/121 | 47597 | 30,061 |
https://mathoverflow.net/questions/47590 | 13 | My question is about monoidal categories. To motivate it, let me first recall something about group objects.
Assume you define a group object in a category $C$ with products by an object $G$ together with morphisms $G \times G \to G, G \to G, \* \to G$, so that the diagrams commute which correspond to the group axioms. Then you want to generalize some elementary properties from usual groups ($C=Set$) to these group objects, but it is quite hard to write down the relevant diagrams. For example when you want to prove that there is a actually at most one unit $\* \to G$, a morphism $G \to G'$ between group objects which respects the multiplication already respects the inversion and the unit, left-inverse implies right-inverse, etc. But all this may be reduced to the case $C=Set$ by using the Yoneda-Lemma and the the definition of a group object as an object together with a factoriation of its hom-functor over the category of usual groups. Then these calculations are easy and you don't have to produce all these diagrams.
My question is: Is there a similar definition for a monoidal category? Specifically, I want to see a neat (diagram-free?) proof of Lemma 3.2.5 in [this note](http://www.mathematik.uni-muenchen.de/~pareigis/Vorlesungen/98SS/Quantum_Groups/LN3_2.PDF) (Pareigis' lectures on quantum groups), which is intuitive and does not come up with diagrams without motivating them. (For me, "We tried to prove it and finally this diagram worked" is no intuition.) The lemma implies for example that the endomorphism monoid of the unit object of a monoidal category is abelian, which is quite surprising (for me). This Lemma is just one example. It seems to me that monoidal categories are "weak monoids in the 2-category Cat", but I don't see yet if this description actually simplifies these diagrams.
| https://mathoverflow.net/users/2841 | Alternative definition of monoidal categories | The kind of thing you are looking for applies not just to monoidal categories, but to bicategories, and it is called the bicategorical Yoneda lemma. If $B$ is a small bicategory, one may form the strict 2-category $[B^{op}, Cat]$ consisting of weak 2-functors (aka homomorphisms), pseudonatural transformations, and modifications from $B^{op}$ to $Cat$. Then there is a Yoneda embedding
$$y: B \to [B^{op}, Cat]$$
sending an object $b$ to $\hom(-, b)$, and the mapping of $B$ onto its image is a bi-equivalence. The image however is a strict 2-category, and hence we get from this a coherence theorem which assures us that all definable diagrams (in say the free monoidal category generated by a discrete category) commute. This line of thinking was introduced by Street in his paper [Fibrations in Bicategories](http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1980__21_2/CTGDC_1980__21_2_111_0/CTGDC_1980__21_2_111_0.pdf) (for which there is also a [correction](http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1987__28_1/CTGDC_1987__28_1_53_0/CTGDC_1987__28_1_53_0.pdf)).
This is really the modern point of view on coherence in monoidal categories. An exposition which focuses on monoidal categories can be found (I believe) in Braided Tensor Categories by Joyal and Street.
It is true that monoidal categories can be described as weak monoids, but this by itself doesn't solve coherence issues.
| 9 | https://mathoverflow.net/users/2926 | 47598 | 30,062 |
https://mathoverflow.net/questions/47603 | -1 | Is it possible to express the functions $S(x)=x+1$ and $Pd(x)=x\dot{-}1$ in terms of the functions $f\_1$, $f\_2$, $f\_3$ and $f\_4$, where $f\_1(x)=0$ if $x$ is even or $1$ if $x$ is odd, $f\_2(x)=\mbox{quot}(x,2)$, $f\_3(x)=2x$ and $f\_4(x)=2x+1$? For example, $S(x)=f\_4(f\_2(x))$ if x is even. Is there a similar formula if $x$ is odd?
| https://mathoverflow.net/users/11147 | Other ways to define naturals | I assume that the problem is: does there exists a sequence $(u\_1,\dots,u\_n)\in{}\{1,2,3,4\}$ such that for all $x\in\mathbb{N}$, $S(x) = f\_{u\_n}(\dots (f\_{u\_1}(x))\dots)$ (and similarly for $Pd$)
Let’s prove by induction on $n$ that every such function either is of the form $f(x)=2^kx+l$ where $k\ge 0$ and $0\le l<2^k$ or $f(2)=f(3)$ or the image of $f$ contains only two elements.
If there is an $f\_1$ somewhere, then the image of $f$ contains only two elements, and this is preserved by composition. We suppose now that there is no $f\_1$.
* If $n=0$, $f(x) = x$ ok.
* If $f(2)=f(3)$ then $f\_i(f(2)) = f\_i(f(3))$
* If $f(x) = 2^kx+l$, $f\_3\circ f$ and $f\_4\circ f$ are of the form $2^{k+1}x+l'$ with $0\le l'< 2^{k+1}$, and $f\_2\circ f$ is also of the same form if $k\ge 1$, and if $k=0$ we have $f\_2(f(3))=f\_2(f(2))=1$.
This proves that neither $S$ nor $Pd$ can be written as a composition of the functions $f\_i$.
| 5 | https://mathoverflow.net/users/10217 | 47610 | 30,071 |
https://mathoverflow.net/questions/47620 | 14 | How difficult is it to know what $\pi\_1(Spec(\mathbb{Z}[1/(p\_1...p\_r)]))$ is? Is it independent of the choice of $p\_1,...,p\_r$? When is it known, and what is known about it?
| https://mathoverflow.net/users/5309 | Does $\pi_1(Spec(\mathbb{Z}[1/p]))$ depend on p? | The full etale fundamental groups in question are, I believe, complicated infinite profinite groups. (They are however "small" in the technical sense that they have only finitely many open normal subgroups of any given finite index, as follows from Hermite's finiteness theorem in algebraic number theory.)
The abelianization of $\pi\_1(\operatorname{Spec}(\mathbb{Z}[\frac{1}{p}])$ is the Galois group of the maximal abelian extension of $\mathbb{Q}$ which is ramified only at $p$ (and infinity). By Class Field Theory, this field is the direct limit of the ray class fields of conductor $p^n (\infty)$, i.e., the field generated by all $p$-power roots of unity. The Galois group is thus the inverse limit of the groups $(\mathbb{Z}/p^n \mathbb{Z})^{\times}$. When $p$ is odd, this is isomorphic to $\mathbb{Z}\_p \times \mu\_{p-1}$ (where the second factor is cyclic of order $p-1$). So yes, this depends on $p$!
| 17 | https://mathoverflow.net/users/1149 | 47623 | 30,075 |
https://mathoverflow.net/questions/47621 | 0 | If we have n homogeneous polynomials (over algebraically closed field) $f\_1\ldots , f\_n$ on variables $x\_0, \ldots , x\_n$
$$
f\_i(x\_0, \ldots , x\_n) = \sum\_{j\_0,\ldots , j\_n} a\_{i, j\_0, \ldots , j\_n} x\_0^{j\_0}\ldots x\_n^{j\_n}
$$
then the common condition is the condition when there are finitely many solutions of the system $f\_1=\ldots = f\_n=0$.
What is the codimension of non-common condition on coefficients $a\_{i, j\_0, \ldots , j\_n}$ ? Is it true that it is at least 2?
The origin of this question is of following: there is an article of Shakirov <http://arxiv.org/abs/0807.4539> where he is stating his formula on resultants. There is a Poisson's lemma: $Res(f\_0, \ldots , f\_n)=C\prod\_{i=1}^N f\_0(a\_i)$ where $a\_i$ are the common zeroes of $f\_1, \ldots , f\_n$ (if $f\_1, \ldots , f\_n$ are in common condition) and C is a non-zero element of the field. Then he is using this formula to state an equation on logarithms of resultants. But if the non-common condition is of codimension 1 then his proof fails: the element $C$ may be a rational function with zeroes and poles lying inside non-common condition.
I've wrote him and he said that the editor of his article (in Russian journal "Functional analysis") also is in misunderstanding with his proof.
The common-condition means that the factor $\Bbbk [x\_0, \ldots , x\_n]\_m/(f\_1, \ldots , f\_n)\_m$ (where the index $\_m$ means that it is a homogeneous part of degree $m$) is of maximal possible dimension when $m$ is sufficiently large. This gives us a system of inequalities (some determinants are non equal to zero). But this inequalities can still have a common factor...
If you don't understand my English I'm very-very sorry :)
| https://mathoverflow.net/users/11072 | Codimension of non-common condition is 2? | I think the answer is yes, this set has at least codimension $2$.
I suppose by solutions, you mean solutions in $\mathbb P^n$.
So, let $V\_i$ be the hypersurface defined in $\mathbb P^n\_{x\_0,\dots,x\_n}\times \mathbb P^N\_{a\_{i, j\_0,\dots,j\_n}}$ by $f\_i=0$.
I believe it is easy to prove that for these $V\_i$ it is true, that if $V=\cap V\_i$, then
$$
\dim V = N.
$$
The point is that each $V\_i$ cuts down the dimension by $1$.
I also claim that all the $V\_i$ and $V$ are irreducible. I think this should be easy to prove, but "I leave this to the reader" because I don't have time to think about it much. (I am fully aware that some reader might point out that this is actually not true, but let's go with assuming it for now. I might come back later to fix this part.)
Now, then the projection $V\to \mathbb P^N$ is a surjective, generically finite morphism between irreducible (projective) varieties of the same dimension. Let $Z\subset \mathbb P^N$ be the "non-common condition set" and $W\subset V$ the pre-image of $Z$ in $V$.
Then it follows, since $W$ is a proper subvariety that
$$\dim V > \dim W \geq \dim Z +1$$
(also since the restriction of the projection $W\to Z$ has positive dimensional fibers).
In particular ${\rm codim}\_{\mathbb P^N} Z\geq 2$.
| 2 | https://mathoverflow.net/users/10076 | 47632 | 30,082 |
https://mathoverflow.net/questions/47533 | 8 | Let $f:[0,1]\to [0,1]$ be a continuous function. Must it have a point $x$ that $f^{-1}(x)$ is at most countable?
Added: Must it have a point $x$ that $dim\_H(f^{-1}(x))=0$ ? ($dim\_H$ means the Hausdorff dimension)
| https://mathoverflow.net/users/4298 | Uncountable preimage of every point | A simple modification of the ideas of André Henriques, Sergei Ivanov and others shows that it is possible that all fibers have Hausdorff dimension $1$. For completeness I write down a complete proof.
Form a Cantor set as follows: let $I\_0 = [0,1/3]$, $I\_1=[2/3,1]$. If $I\_{i\_1,\ldots, i\_n}$ has been defined, let $I\_{i\_1,\ldots,i\_n,0}, I\_{i\_1,\ldots,i\_n,1}$ be the two intervals obtained by removing from $I\_{i\_1,\ldots, i\_n}$ a central open interval of length $1/(n+2)$ times the length of $I\_{i\_1,\ldots, i\_n}$. The Cantor set $C$ is then
$$
C := \bigcap\_{n=1}^\infty C\_n :=\bigcap\_{n=1}^\infty \bigcup\_{i\_1\ldots i\_n} I\_{i\_1\ldots i\_n}.
$$
(This is just like the construction of the usual Cantor set, except that the relative lengths of the removed intervals tend to $0$ rather than staying constant).
We define a continuous function $f:C\to [0,1]$, and then extend $f$ to all of $[0,1]$ in an arbitrary way.
Let $x$ be a point of $C$. Then $x$ is coded by a unique sequence $i\_1 i\_2\ldots\in \{0,1\}^{\mathbb{N}}$, and we define
$$
f(x) = \sum\_{n=1}^\infty i\_{n^2} 2^{-n}.
$$
It is very easy to check that $f$ is indeed continuous.
Now let $t=[0,1]$, and let $t =\sum\_{n=1}^\infty a\_n 2^{-n}$ be a binary expansion of $t$. Then $f^{-1}(t)$ consists of those points in $C$ whose code $i\_1 i\_2\ldots$ satisfies $i\_{n^2} =a\_n$. In other words, except for places corresponding to perfect squares, the other elements of the sequence are completely arbitrary.
It is a fairly easy exercise in the calculation of Hausdorff dimension to show that $f^{-1}(t)$ has dimension $1$. The easiest way to give a formal proof is perhaps to use the mass distribution principle (see e.g. Falconer's or Mattila's books): we define a measure $\mu$ on $f^{-1}(t)$ inductively as follows. We start by assigning a unit mass to $[0,1]$. After $k$ steps, we have assigned a mass to all intervals $I\_{i\_1\ldots i\_k}$, which is $0$ whenever $I\_{i\_1\ldots i\_k}\cap f^{-1}(t)=\varnothing$. Now, if $k+1=n^2$ for some $n$, then we specify that $\mu(I\_{i\_1\ldots i\_k a\_n}) = \mu(I\_{i\_1\ldots i\_k})$ and $\mu(I\_{i\_1\ldots i\_k (1-a\_n)})=0$. Otherwise, if $k+1$ is not a perfect square, then we specify that
$$
\mu(I\_{i\_1\ldots i\_k 0})=\mu(I\_{i\_1\ldots i\_k 1}) = \frac{1}{2}\mu(I\_{i\_1\ldots i\_k}).
$$
(In other words, at the stages where the next symbol isn't determined, split mass uniformly; otherwise, pass all the mass to the required interval of next level.)
In this way we have assigned a mass to each of the intervals in the construction of $C$, so that $\mu$ is a well defined measure, which is supported on $f^{-1}(t)$ by construction. Note that if $n^2\le k < (n+1)^2$, then $\mu(I\_{i\_1\ldots i\_k})=2^{k-n}$. Hence
$$
\lim\_{k\to\infty} \frac{\mu(I\_{i\_1\ldots i\_k})}{-\log(2^k)} = 1,
$$
for any sequence $i\_1 i\_2\ldots$, and since
$$
\lim\_{k\to\infty} \frac{\log|I\_{i\_1\ldots i\_k}|}{-\log(2^k)} = 1,
$$
it is easy to deduce that
$$
\lim\_{r\to 0} \frac{\log(\mu(B(x,r))}{\log r} = 1
$$
for $\mu$-almost every $x$ (indeed for all $x$ in the support of $\mu$). Since $\mu(f^{-1}(t))=1$, the mass distribution principle implies that $\dim\_H(f^{-1}(t))\ge 1$ (and hence it is exactly $1$).
Two concluding remarks. A straightforward modification of the argument shows that for any gauge function $\varphi(x)$ such that $\lim\_{x\to 0} \varphi(x)/x=+\infty$, there is a continuous function $f:[0,1]\to [0,1]$ such that every fiber $f^{-1}(t)$ has positive $\varphi$-dimensional Hausdorff measure.
Finally, these results are not really surprising, since continuity alone does not imply any Hausdorff dimension bounds. This is the same reason why space-filling curves exist - continuity does not prevent the dimension of the image from being as large as the ambient space, and likewise it does not prevent the dimension of the fibers from being as large as the ambient space.
| 10 | https://mathoverflow.net/users/11009 | 47633 | 30,083 |
https://mathoverflow.net/questions/44192 | 8 | This question came up when my supervisors, my colleague, and I were considering arithmetic progressions in sets of fractional dimension. In particular, we were interested in "extracting" Salem sets from other sets, and we came across the following question:
We define the *Fourier dimension* of $E \subseteq \mathbb{R}$, $\mathrm{dim}\_F(E)$, as the supremum of $\beta \in [0,1]$ such that for some probability measure $\mu$ supported on $E$,
$
|\widehat{\mu}(\xi)| \leq C|\xi|^{-\beta/2}.
$
**Suppose $E\_1,E\_2$ are two subsets of $\mathbb{R}$. What is the relationship, if any, between $\mathrm{dim}\_F(E\_1 + E\_2)$ and $\mathrm{dim}\_F(E\_1)$, $\mathrm{dim}\_F(E\_2)$?**
A quick calculation shows
$
\mathrm{dim}\_F(E\_1 + E\_2) \geq \min(\mathrm{dim}\_F(E\_1) + \mathrm{dim}\_F(E\_2),1),
$
but can we do any better than this? Or can strict inequality hold?
For some motivation, we may look to Hausdorff dimension instead (it is known that $\mathrm{dim}\_F(E) \leq \mathrm{dim}\_H(E)$; a set $E$ such that $\mathrm{dim}\_F(E) = \mathrm{dim}\_H(E)$ is called a *Salem set*). I believe Falconer gave an example of sets $E\_1$ and $E\_2$ such that $\mathrm{dim}\_H(E\_1) = 0 = \mathrm{dim}\_H(E\_2)$, yet $\mathrm{dim}\_H(E\_1+E\_2) = 1$.
| https://mathoverflow.net/users/7165 | Fourier dimension of the sum of sets | It is possible that $\dim\_F(E\_1)=\dim\_F(E\_2)=0$ yet $\dim\_F(E\_1+E\_2)=1$, so there is no inequality in the opposite direction.
In fact, Falconer's example of sets $E\_1, E\_2$ such that $\dim\_H(E\_1)=\dim\_H(E\_2)=0$ but $\dim\_H(E\_1+E\_2)=1$ already works. In Falconer's example, not only $E\_1+E\_2$ has dimension $1$, but in fact $E\_1+E\_2$ is an interval. Hence $\dim\_F(E\_1)=\dim\_F(E\_2)=0$ (since $\dim\_F$ is bounded above by $\dim\_H$) but $\dim\_F(E\_1+E\_2)=1$.
There are also examples which are not forced by the Hausdorff dimensions of the sets. Indeed, let $C$ be the ternary Cantor set. It is a classical result of Kahane and Salem that if $\mu$ is any measure supported on $C$, then $\widehat{\mu}(\xi)\nrightarrow 0$ as $|\xi|\to\infty$; in particular, $\dim\_F(C)=0$. Clearly, the same is true for any dilate $t C$ with $t\neq 0$.
On the other hand, it is well known that $C+C$ equals the interval $[0,2]$. Indeed, since $C$ and also $tC$ have Newhouse thickness equal to $1$, Newhouse's gap lemma (the endpoint version), guarantees that $C+tC$ has nonempty interior for all $t\neq 0$, and therefore also Fourier dimension $1$ (note that I'm not claiming that the indicator function of $C+tC$ has Fourier decay, just that there is a measure supported on (an interval in) $C+tC$ that does).
Hence, for all $t$ we have $\dim\_F(C)=\dim\_F(tC)=0$ but $\dim\_F(C+ tC)=1$. This shows that the opposite inequality fails in a rather dramatic function: it does not even hold typically in the sense of Marstrand's Theorem.
Morally speaking, there is no reason why $$\dim\_F(E\_1+E\_2)=\min(1,\dim\_F(E\_1)+\dim\_F(E\_2))$$ should hold in general. Leaving Hausdorff dimension considerations aside, if $E\_1$ or $E\_2$ are not Salem, this tells us that there are some resonances in the construction of the sets at a set of frequencies (possibly very sparse). These special frequencies will in general be lost in the sum $E\_1+E\_2$ (unless $E\_1$ and $E\_2$ also resonate to each other in some strong form), so one would expect that $\dim\_F(E\_1+E\_2) > \dim\_F(E\_1)+\dim\_F(E\_2)$. However, I suspect it is not trivial at all to give specific examples where $\dim\_H(E\_1+E\_2)<1$ (because proving Salemness or even some good decay of Fourier coefficients is usually hard).
On the other hand, equality $\dim\_F(E\_1+E\_2)=\min(1,\dim\_F(E\_1)+\dim\_F(E\_2))$ can certainly hold. Indeed, it is easy to see this is always the case when $E\_1$ and $E\_2$ are Salem and additionally one of them has coinciding Hausdorff and upper box-counting dimension (which is the case for all known constructions of Salem sets). Indeed, denoting upper box-counting dimension by $\dim\_B$, it is well known that $\dim\_H(A+B)\le \dim\_H(A)+\dim\_B(B)$, so in this case $\dim\_H(E\_1+E\_2)\le \dim\_H(E\_1)+\dim\_H(E\_2)$, and it follows from salem-ness and $\dim\_F\le \dim\_H$ that
$$
\dim\_F(E\_1+E\_2)\le \dim\_F(E\_1)+\dim\_F(E\_2).
$$
| 4 | https://mathoverflow.net/users/11009 | 47639 | 30,085 |
https://mathoverflow.net/questions/47637 | 0 | I am trying to find an explicit way to view global holomorphic sections of $\Omega^{1} \otimes \mathcal{O} (2)$ over $\mathbb{CP}^{2}$. I guess what I mean by "explicit" would be a formulation over an affine open $U\_i \subset \mathbb{CP}^{2}$. According to what I found in Okoneck, Schneider and Spindler, there is a 3-dimensional space of such sections, but I want this for a computation in differential geometry.
| https://mathoverflow.net/users/11156 | Global sections of $\Omega^{1} \otimes \mathcal{O} (2)$ over $\mathbb{CP}^{2}$ | If $x,y,z$ are coordinates on $P^2$ then the 3 sections of $\Omega(2)$ are given by $xdy - ydx$, $ydz-zdy$, and $zdx-xdz$.
| 4 | https://mathoverflow.net/users/4428 | 47645 | 30,088 |
https://mathoverflow.net/questions/47661 | 0 | Given a function $f(X\_1,\cdots,X\_n,Y)$ on random variables $\{X\_i\}$ and $Y$, which is continuous ,
I want to
show that $f$ concentrates around its expectation $\operatorname\*{E}[f]$, i.e., a formula like this:
$\Pr[|f(X\_1,\cdots,X\_n, Y)-\operatorname\*{E}[f(X\_1,\cdots,X\_n, Y)]|\geq t]\leq \exp(-\frac{t^2}{2c^2})$, where $c^2$ is the Lipschitz-type bound on $f$.
The case considered here is different from the traditional one which does not have the additional continous
random variable $Y$. However, we can still use the traditional way to show the concentration. By the
Law of total probability, it is equal to bound
$\operatorname\*{E}\_Y[\Pr[|f(X\_1,\cdots,X\_n, y)-\operatorname\*{E}[f(X\_1,\cdots,X\_n, y)]|\geq t|Y=y]] \qquad (1)$.
Given $y$ , if we have that
$|\operatorname\*{E}[f|X\_1,\cdots,X\_{i-1},X\_i=x\_i,y]-|\operatorname\*{E}[f|X\_1,\cdots,X\_{i-1},X\_i=x'\_i, y]\leq c\_i(y),$
for all $i$ with $1\leq i\leq n$ and any $x\_i,x'\_i$.
then by the stardard use of Azuma's inequality,
$\Pr[|f(X\_1,\cdots,X\_n, y)-\operatorname\*{E}[f(X\_1,\cdots,X\_n, y)]|\geq t|Y=y]\leq \exp(-\frac{t^2}{2\sum\_{i=1}^n c\_i^2(y)})$.
Thus, from (1),
$\Pr[|f(X\_1,\cdots,X\_n, Y)-\operatorname\*{E}[f(X\_1,\cdots,X\_n, Y)]|\geq t]\leq \operatorname\*{E}[\exp(-\frac{t^2}{2\sum\_{i=1}^n c\_i^2(Y)})] \qquad (2)$
My question is that can the above inequality be improved as:
$\Pr[|f(X\_1,\cdots,X\_n, Y)-\operatorname\*{E}[f(X\_1,\cdots,X\_n, Y)]|\geq t]\leq \exp(-\frac{t^2}{2\sum\_{i=1}^n \operatorname\*{E}[c\_i(Y)]^2}) \qquad (3)$.
P.S. I think that the Jassen's inequality (i.e., $\operatorname\*E[g(Z)]\geq g(\operatorname\*E[Z])$ for convex function
$g$) may be useful here, but I donot see the convexity of the right hand of inequality (2).
| https://mathoverflow.net/users/7279 | Concentration bound using Azuma's inequality and Law of total probability | See my comment above for some problem in your argument but anyhow (3) is wrong. If the $X\_i$-s are constants then the right hand side of (3) is 0, while the left hand side is not in general. If you don't like to use constant r.v.: if each of the $X\_i$ takes values in a small interval, the right hand side is arbitrarily close to zero while the left hand side not, in general (say, for the function $f(x\_1,...,x\_n,y)=y$ and any reasonable $Y$).
| 2 | https://mathoverflow.net/users/6921 | 47665 | 30,092 |
https://mathoverflow.net/questions/47681 | 2 | This question is very closely related to my other question [here](https://mathoverflow.net/questions/43348/is-there-a-characterization-of-free-groups-in-terms-of-the-unitary-dual).
Let $\Gamma$ be a countable discrete group and $\pi:\Gamma \rightarrow \mathcal{H}$ be a unitary representation of $\Gamma$. A map $b:\Gamma \rightarrow \mathcal{H}$ is an additive cocycle if $$b(gh)=\pi(g)b(h)+b(g)$$ for all $g,h\in G$.
>
> If we consider the collection of all $(b,\pi)$ where $\pi$ is a
> unitary representation of $\Gamma$ and
> $b$ is a cocycle as above, can this collection
> "see" if $\Gamma$ is a free
> group?
>
>
>
Again, I'm being deliberately vague here regarding the word "collection". Roughly, what I am asking is whether including these cocycles in my previous question would yield a structure that can detect freeness of the underlying group.
| https://mathoverflow.net/users/6269 | Is there an abstract characterization of freeness in terms of additive unitary cocycles? | I think your question is not precise enough. If you just look at this as a set or topological space, it is most likely not enough. If you add additional structure like sum and tensor product, then it is enough.
Maybe this is what you are looking for: If $\Gamma$ is free on the set $X$, then for any unitary representation $\pi \colon \Gamma \to U(H)$ any map $c \colon X \to H$ extends uniquely to a $1$-cocycle on $\Gamma$. Moreover, this happens if and only if $\Gamma$ is free. In other words, the linear space of 1-cocycles for $\pi$ on $\Gamma$ is precisely the space of $H$-valued functions on $X$.
This observation can be used to show that the first $\ell^2$-Betti number of $\Gamma$ is $|X|-1$. (Minus one, because you have to substract the inner 1-cocycles.) The first $\ell^2$-Betti number of $\Gamma$ precisely counts how many 1-cocycles with values in the left-regular representation $\lambda$ exist.
One does not even have to know this for all unitary representations $\pi$, it is enough to know it for the left-regular representation $\lambda$. Indeed, a group on $n$ generators is free if and only if its first $\ell^2$-Betti number equals $n-1$. This is sharp, as their are $n$-generated groups which are not free but the first $\ell^2$-Betti number exceed $n-1 - \varepsilon$, for example $\Gamma = \langle a,b \mid a^k \rangle$ for $k$ high enough. More interestingly, Denis Osin has constructed $n$-generated *torsion* groups with first $\ell^2$-Betti number higher than $n-1 - \varepsilon$.
| 3 | https://mathoverflow.net/users/8176 | 47683 | 30,100 |
https://mathoverflow.net/questions/47343 | 5 | In the context of ZFC, one normally uses von Neumann's definition of the ordinals. However, originially an ordinal was just the order-type of a well-ordered set (where "order-type of A" may for example be defined to be the equivalence class of all ordered sets that are order-isomorphic to A; this definition is of course no longer allowed in ZFC, but was common in pre-ZFC naive set theory).
I am now looking for a complete exposition of Burali-Forti paradox, with the original definition of ordinal. One can in a number of papers and books find expositions similar to the following one cited from Wikipedia:
"The "order types" (ordinal numbers) themselves are well-ordered in a natural way, and this well-ordering must have an order type Ω. It is easily shown in naïve set theory (and remains true in ZFC but not in New Foundations) that the order type of all ordinal numbers less than a fixed α is α itself. So the order type of all ordinal numbers less than Ω is Ω itself. But this means that Ω, being the order type of a proper initial segment of the ordinals, is strictly less than the order type of all the ordinals, but the latter is Ω itself by definition. This is a contradiction."
To complete this exposition, we need proofs for the following two facts:
* The ordinals are well-ordered under their natural ordering.
* The order type of all ordinal numbers less than a fixed α is α itself.
The book "Grundbegriffe der Mengenlehre" by Gerhard Hessenberg (1906) (which can be read online at <http://www.archive.org/stream/grundbegriffede00hessgoog#page/n79/mode/1up>) presents proofs for these facts, which to me however seem invalid (I do not understand why he may conclude "und umgekehrt ist jeder Zahl ν<μ ein Abschnitt in M eindeutig zugeordnet" on page 550).
I have found a complicated proof for the first fact, which however is based on induction (over the natural numbers) and three applications of the Axiom of Choice. It seems to me that the second fact may be proven using transfinite induction (transfinite induction, it seems to me, may only be used once one has established the first fact). So in principle, I think I can complete the Burali-Forti paradox as stated above, but this completed derivation would be very lengthy and involved.
So what I am actually looking for is a more concise or less involved complete derivation of the Burali-Forti paradox. Can anyone present such a derivation here, or point me to an existing one in the literature?
| https://mathoverflow.net/users/5199 | Looking for a complete exposition of the Burali-Forti paradox | I have now found a textbook that provides a complete proof of the Burali-Forti paradox without making use of von Neumann's definition of ordinals: "Basic Set Theory" by Azriel Levy. Before providing von Neumann's definition, he works just on the assumption that some "order types" can be defined such that the order type of two well-ordered sets is identicall iff they are order-isomorphic. Based only on this assumption, and, importantly for my concern, without using the Axiom of Foundation, he shows that the class of ordinals cannot be a set.
| 3 | https://mathoverflow.net/users/5199 | 47686 | 30,102 |
https://mathoverflow.net/questions/47219 | 3 | Let $\mathcal C$ be a category and let $\mathcal F:\mathcal C\to\mathcal C\textrm{at}$ be a strong bifunctor. Given another category $\mathcal D$, let $\triangle\_{\mathcal D}$ denote the constant functor $\mathcal C\to\mathcal C\textrm{at}$. Now define lax/strong limits and colimits as follows:
* A **lax limit** of $\mathcal F$ is a
category $\mathsf{lim}\mathcal F$
together with a natural equivalence
$$[\triangle\_{(-)},\mathcal
F] \cong \mathcal
C\textrm{at}(-,\mathsf{lim}\mathcal
F).$$ Here $[\triangle\_{(-)},\mathcal F]$ denotes the category
of lax natural transformations and modifications.
* A **lax colimit** of $\mathcal F$ is a
category $\mathsf{colim}\mathcal F$
together with a natural equivalence
$$[\mathcal
F,\triangle\_{(-)}] \cong \mathcal
C\textrm{at}(\mathsf{colim}\mathcal
F,-).$$
* We define **strong limits** and **strong colimts** by replacing lax natural
transormations with strong natural transfomations.
Now, if my calculations are correct, a lax colimit of such a functor $\mathcal F$
is given by the grothendieck construction $\mathcal C\int\mathcal F$ and a lax limit is given by the category of **strict** sections
$s:\mathcal C\to \mathcal C\int\mathcal F$, i.e. the category $\mathcal C\textrm{at}/\mathcal C(\operatorname{id}\_\mathcal C,\pi)$, where $\pi:\mathcal C\int\mathcal F\to\mathcal C$ is the opfibration corresponding to $\mathcal F$.
If we consider only the category of opcartesian sections, that is sections that map every morphism in $\mathcal C$
to an opcartesian morphism, we get a strong limit.
Now for the question:
**Is there an explicit description of the strong colimit of a functor $\mathcal F:\mathcal C\to\mathcal C\textrm{at}$?**
| https://mathoverflow.net/users/1261 | Strong colimits of categories. | Adapted from the papers by Fiore and by Porter that I referred to above:
To form the usual lax colimit of P we take the disjoint union of the $P\_i$ for each $i \in C$ and then adjoin new arrows to represent the action of P: for each $m \colon i \to j$ in C and each $X \in P\_i$ there is an arrow $X \to Pm(X)$. Then we quotient by a congruence that ensures that the assignment of $X \to Pm(X)$ to $X$ is lax natural. This lax transformation is the universal cone.
To form the pseudo colimit we simply make sure each $X \to Pm(X)$ is an isomorphism: adjoin a formal inverse along with it and add the requisite equations to the congruence.
| 3 | https://mathoverflow.net/users/4262 | 47689 | 30,105 |
https://mathoverflow.net/questions/47660 | 7 | Let $X$ be an algebraic variety and $X[n]$ be the Fulton-MacPherson compactification of the configuration space $F(X,n)$ introduced in the paper "A compactification of configuration spaces".
In this paper the authors give an explicit construction of the space $X[n]$ by a sequence of blow-ups, which is inductive. They assume that the space $X[n]$ is already constructed. They describe $X[n+1]$ as an explicit sequence of blow-ups of $X[n] \times X$.
As they mention in their paper their construction is not symmetric. For example, when $n=4$, if one starts by blowing up the small diagonal in $X^4$ and then blow-up proper transforms of the next larger diagonals, then the proper transform of succeeding diagonals will not be separated, so extra blow-ups are needed to get a smooth compactification.
I am interested in the case where $X$ is a smooth curve. I was wondering if there is a construction of the space $X[n]$ as an explicit and symmetric sequence of blow-ups of $X^n$.
I want to get a smooth space at each stage of the construction.
Question: Is there any such sequence?
| https://mathoverflow.net/users/5286 | Symmetric sequence of blow-ups for the Fulton-MacPherson compactification | The interesting question of to what extent "wonderful compactifications" like the Fulton-MacPherson space depend on the order of blowups was studied -- and I think, mostly resolved -- by Li Li in his thesis. The paper <http://arxiv.org/abs/math/0611412> gives "a condition on the order of blow-ups in the construction....such that each blow-up is along a nonsingular center."
| 4 | https://mathoverflow.net/users/5081 | 47692 | 30,107 |
https://mathoverflow.net/questions/47682 | 12 | Given a commutative, $\mathbb N$-graded ring, one can associate to it a scheme via the $Proj$ construction.
What happens if one tries to copy this procedure but instead of $\mathbb N$ with another indexing gadget (say commutative monoid) ?
Some thoughts about this:
Considering projective varieties is roughly the same as studying affine varieties equivariant under the multiplicative group.
So I would guess that replacing $\mathbb N$ by something else corresponds to replacing the multiplicative group by something else.
| https://mathoverflow.net/users/2837 | Proj for rings graded by different things then $\mathbb N$ ? | Weighted projective spaces $\mathbb{P}(a\_1,\ldots,a\_n)$ are examples where a grading other than the standard grading is used. In general you can study gradings coming from any finitely generated abelian group, and this grading gives rise to a torus action on the ring. The Proj you speak of is then a GIT-quotient of $Spec R$ by this torus action (if you are familiar with GIT, the choice of linearization of the action corresponds to a choice of irrelevant ideal). This GIT-quotient construction is completely analogous to the usual construction
$$
\mbox{Proj} k[x\_0,\ldots,x\_n]=\left(\mbox{Spec} k[x\_0,\ldots,x\_n]-V(x\_0,\ldots,x\_n)\right) // \mathbb{G}\_m
$$
The best reference I can give for this stuff is Chapter 1 of the book ["Cox rings"](http://arxiv.org/abs/1003.4229) by Arzhantsev, Derenthal, Hausen and Antonio Laface. Other nice references are
[The Homogeneous Coordinate Ring of a Toric Variety](http://arxiv.org/abs/alg-geom/9210008) by Cox, which deals with Toric varieties
[Lectures on invariant theory](http://books.google.co.uk/books?id=B9S_0iPvfrUC&printsec=frontcover&dq=dolgachev+invariant+theory&source=bl&ots=qxIPsqhMks&sig=59jcHL5Tn8AX6-N0H8EfixFSAco&hl=en&ei=NdLzTJCCMtGLhQeewJyxCA&sa=X&oi=book_result&ct=result&resnum=4&ved=0CCsQ6AEwAw#v=onepage&q&f=false) by Dolgachev, which is a nice introduction to quotients in algebraic geometry.
| 13 | https://mathoverflow.net/users/3996 | 47694 | 30,109 |
https://mathoverflow.net/questions/47655 | 5 | Hello,
this question is related to [Differential graded structures on free resolution?](https://mathoverflow.net/questions/40282/differential-graded-structures-on-free-resolution).
Given a regular local ring $S$ and $f\in{\mathfrak m}\_S\setminus\{0\}$, I am interested in studying $R$-modules through their $S$-free resolutions. More precisely, given an $R$-module $M$, any $S$-free resolution $F^{\ast}$ of $M$ admits homotopies $s\_n$ of respective degree $2n-1$ such that $s\_0$ equals the differential, $s\_1$ is a nullhomotopy for the multiplication with $f$ and for the higher $s\_n$ we have
$s\_0 s\_n + s\_1 s\_{n-1} + ... + s\_n s\_0 = 0$ for $n\geq 2$.
If $s\_1$ can be chosen in such a way that $s\_1^2=0$, we can consider $F^{\ast}$ as a dg-module over the Koszul dg-S-algebra of $S/f$.
Now I have two questions:
(1) What happens if the $s\_1$ can *not* be chosen such that $s\_1^2=0$? Can we still write the datum of higher homotopies $s\_n$ as a dg-module structure over some dg-resolution of $S/f$? A naive guess would be a free non-commutative dg-algebra generated by elements $s\_n$ of degree $2n-1$ such that $\text{d}(s\_n) = s\_1 s\_{n-1} + ... + s\_{n-1} s\_1$ for $n\geq 2$ and $\text{d}(s\_1) = f\cdot 1$. Is this studied anywhere?
(2) Can we do all this somehow functorially in $M$? I'm thinking of something like the canonical functor from R-mod into the derived category, which after identification with the homotopy category of projectives turns "projective resolution" into a functor. However, in the concrete example I'm struggling with extending a morphism between R-modules to a morphism of free resolutions respecting the chosen homotopies. Perhaps this is just some diagram chase, but I don't see it.
*Apart from technicalities, my goal is to study $S/f$-modules through dg-modules over an appropriate dg-$S$-algebra up to homotopy. If this is possible: do you know books or articles where it is treated?*
Thank you very much!
Hanno
| https://mathoverflow.net/users/3108 | Multiplicative Structures On Free Resolutions | You can always find a free resolution $F$, of $M$ over $S$, such that $F$ is a dg-module over the Koszul complex. It may not be minimal, but in many cases that's not an issue. This is the path taken by Avramov and Buchweitz in their paper "Homological algebra modulo a regular sequence with special attention to codimension 2". In particular these resolutions can be taken functorially.
Also see Avramov's book "Infinite free resolutions" whose main themes are how to put dg-algebra and dg-module structures on free resolutions, and how this helps to study these resolutions.
More recently Dyckerhoff and Polishchuk/Vaintrob have used dg-categorical methods to study the situation you're interested in. See their recent papers on the Arxiv.
| 8 | https://mathoverflow.net/users/3293 | 47697 | 30,111 |
https://mathoverflow.net/questions/47695 | 8 | The following question is for my own curiosity as I take some time to get reacquainted with group theory.
Let G be a semi-direct product of the groups N and K with multiplication defined by the automorphism $\phi$ from K to Aut(N). Let Fix($\phi$) be the set of all elements of N that are mapped to themselves by all elements of the range of $\phi$. Clearly every element of Fix($\phi$) commutes with all elements of K and every element of the kernel of $\phi$ commutes with every element of N.
If Fix($\phi$) is the trivial group in N and Ker($\phi$) is the trivial group in K, does that imply that the center of G is trivial? If so, could someone point me to a reference or proof. If not, then a counter example.
| https://mathoverflow.net/users/10596 | Centers of Semidirect Products | Suppose that $z=xy$ is in the centre where $x\in N$ and $y\in K$.
Then for all $u\in K$, $uxy=xyu$. But $uxy=\phi(u)(x)uy$ so that
$x=\phi(u)(x)$ (and $uy=yu$). As this is true for all $u\in K$
then by the assumption on Fix($\phi$), $x=1$. Therefore $z=y\in K$.
As $y$ commutes with all elements of $N$ then $y$ lies
in Ker($\phi$) and is trivial. So $z=1$ and the centre of $G$
is trivial.
| 7 | https://mathoverflow.net/users/4213 | 47698 | 30,112 |
https://mathoverflow.net/questions/47677 | 4 | Is there someone can tell me some papers or books about the basic material of Spectral flow? I want to know, what is spectral flow and how to use it to geometry.
| https://mathoverflow.net/users/3896 | An introduction paper or book to Spectral Flow | A good start could be to read this paper by Philips which was recommended to me, when I was looking for an overview:
[John Phillips, “Self-Adjoint Fredholm Operators And Spectral Flow”](https://www.cambridge.org/core/journals/canadian-mathematical-bulletin/article/selfadjoint-fredholm-operators-and-spectral-flow/CFA9F1362A48E77D9B67212CDCEFA902#).
| 5 | https://mathoverflow.net/users/11176 | 47715 | 30,121 |
https://mathoverflow.net/questions/47702 | 45 | In his 1967 paper *A convenient category of topological spaces*,
Norman Steenrod introduced the category *CGH* of **compactly generated Hausdorff spaces**
as a good replacement of the category *Top* topological spaces, in order to do homotopy theory.
The most important difference between *CGH* and *Top* is that in *CGH* there is a functorial homeomorphism $$\mathrm{map}(X,\mathrm{map}(Y,Z))\cong \mathrm{map}(X\times Y,Z),$$
a fact that is only true in *Top* under the extra assumption that $Y$ is locally compact.
---
But in more recent papers, I see that people use *CG**W**H* spaces instead of *CGH* spaces...
Why?
Could someone explain to me what goes wrong in *CGH* spaces
(please illustrate with an example),
and explain how the *"w"* fixes everything?
Also (following Jeff's comment), to whom should the *"w"* be attributed?
One more wish: can someone give me an example of a *CGWH* space that isn't *CGH*?
| https://mathoverflow.net/users/5690 | Why the "W" in CGWH (compactly generated weakly Hausdorff spaces)? | I *believe* that CGWH spaces were first used in a *systematic* way in the work of Lewis-May-Steinberger on spectra. It is certainly the case that Gaunce Lewis's (unpublished) thesis contains the best reference on CGWH spaces that I'm aware of. (I haven't looked at the McCord paper Andrey mentions. **Update:** Having looked at McCord's paper, it does indeed seem to be the one to introduce CGWH (the idea of which he attributes to J.C. Moore.))
As to why one might prefer to use CGWH spaces, I'm not precisely sure. But here is one possibility.
A key property of the category of CG spaces is that the product of a quotient map with a space is still a quotient map. In CGWH spaces, something even nicer is true: any *pullback* of a quotient map (along any map) is still a quotient map. (I don't know whether this nicer fact fails in CGH, but I suspect it does.)
Another nice fact about CGWH: regular monomorphisms are precisely the closed inclusions.("Regular monomorphism" means the monomorphism is an equalizer of some pair.) (I originally said here that regular epis in CGWH are precisely quotient maps, but on reflection I'm not sure this is true.)
| 24 | https://mathoverflow.net/users/437 | 47724 | 30,125 |
https://mathoverflow.net/questions/47611 | 16 | I am curious, what kind of exact formulas exist for the partition function $p(n)$?
I seem to remember an exact formula along the lines $p(n) = \sum\_k f(n, k)$, where $f(n, k)$ was some extremely messy transcendental function, and the approximation was so good that for large $n$ one could simply take the $k = 1$ term and truncate this to the nearest integer to get an exact formula.
Reviewing the literature, it seems that I misremembered Rademacher's exact formula, which *is* of the above type but which requires more than one term. I am curious if there are other exact formulas, particularly of the type I mentioned?
Also, if I am indeed wrong and no such formula has been proved, is some good reason why it would be naive to expect one?
Thanks.
| https://mathoverflow.net/users/1050 | Exact formulas for the partition function? | This doesn't really answer the question, so perhaps it would be better as a comment, but alas, I don't have the necessary reputation.
Following up on Thomas Bloom's reference to the work of Bringmann and Ono, there is a paper of Folsom and Masri (Mathematische Annalen, available here: <http://www.math.yale.edu/~alf8/Folsom-Masri-MathAnn07-10.pdf>) which considers the main term one would get in an asymptotic formula arising from BO's Poincare series formula. In particular, they also consider the problem of the error arising from truncating the infinite sum at $O(n^{1/2})$, obtaining power savings over the best known results of $O(n^{-1/2+\epsilon})$ if one truncates at $\lfloor \sqrt{n/6} \rfloor$.
| 4 | https://mathoverflow.net/users/2056 | 47725 | 30,126 |
https://mathoverflow.net/questions/47726 | 11 | Hello everybody! Recently in my research, I came across the Perron-Frobenius operator .. I would like to intuitive interpretation of this operator, ie, physical interpretations are possible, articles (can be physical).
I wonder how this operator originated.
| https://mathoverflow.net/users/nan | Ruelle Perron Frobenius Operator | The place to start is with the [Perron-Frobenius theorem](https://en.wikipedia.org/wiki/Perron-Frobenius_theorem), which (in its most basic form) says that a $d\times d$ matrix $A$ with only positive entries has exactly one positive eigenvector $\vec{v}$, which corresponds to the largest eigenvalue (which is real and positive). Furthermore, all the other eigenvalues are strictly smaller in absolute value, and the iterates of any non-negative vector under the matrices $A^n$ converge exponentially to the eigendirection spanned by $\vec{v}$. This can be interpreted very intuitively by looking at the action on the unit simplex $\Delta = \{\vec{w}\in \mathbb{R}^d \mid w\_i \geq 0, \|\vec{w}\|\_1 = 1\}$ given by $f(\vec{w}) = A\vec{w} / \|\vec{w}\|\_1$. This is discussed at more length in the books by [Brin and Stuck](http://books.google.com/books?id=isStwiHsoM4C&printsec=frontcover&dq=brin+stuck&hl=en&ei=1CL0TM7bJ4L6lwfq4Oi_DA&sa=X&oi=book_result&ct=book-thumbnail&resnum=1&ved=0CCwQ6wEwAA#v=onepage&q&f=false) and by [Katok and Hasselblatt](http://books.google.com/books?id=9nL7ZX8Djp4C&printsec=frontcover&dq=katok+hasselblatt&hl=en&ei=6yL0TIb-MIO0lQf2laD4DA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCwQ6AEwAA#v=onepage&q&f=false).
Once you understand that case, you have at least a general picture for what you're trying to prove in the more general setting, when $\vec{w} \in \mathbb{R}^d$ is replaced by a function $\phi$ in some Banach space $X$. The idea in a wide variety of settings is that you have a dynamical system which defines a natural action on spaces of functions, and you'd like to find a Banach space of functions on which the properties of that action (which is the Ruelle Perron-Frobenius operator) mimic those of the positive matrix $A$ above. The two cases are linked by the fact that if $A$ is a $0$-$1$ matrix that determines the admissible transitions for a mixing topological Markov chain on $d$ symbols, then the matrix $A$ actually ***is*** the Perron-Frobenius operator. (Note that the mixing property says that some power of $A$ is positive, so then you can apply the standard Perron-Frobenius theorem.)
| 12 | https://mathoverflow.net/users/5701 | 47728 | 30,128 |
https://mathoverflow.net/questions/47630 | 4 | Suppose $A$ and $B$ are two $n \times n$ real symmetric matrices, and $A$ is positive semidefinite. For what values of $k \in \mathbb R$ is matrix $kA-B$ positive semidefinite (we write as $kA-B \succeq 0$)?
If $A$ is positive definite, we may find an $n \times n$ nonsingular matrix $D$ such that $A = D^T D$. As a result, $kA-B \succeq 0$ is equivalent to $$kI \succeq (D^{-1})^T B D^{-1}$$ or $$k \geq \lambda\_{\max}((D^{-1})^TBD^{-1}))$$
But how to deal with the situation when $A$ is singular but still positive semidefinite?
I know for certain that in this case we must impose additional constraints on matrix $B$. In particular, let the columns of matrix $N$ consist of a basis of the null space of $A$, then we must have that $N^T B N \preceq 0$ (i.e., $N^T B N$ is negative semidefinite). But what is the lower bound on $k$?
Thanks.
| https://mathoverflow.net/users/7595 | For what values of $k$ is matrix $k A - B$ positive semidefinite? | You can make a orthogonal change of coordinates as follows. First choose an orthogonal basis of the range of $A$, and one of the kernel. This makes a basis of $\mathbb R^n$ in which $A$ is block-diagonal with a zero block:
$$A=\begin{pmatrix} A\_+ & 0 \\\\ 0 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} B\_1 & B\_2 \\\\ B\_2^T & B\_3 \end{pmatrix}.$$
Then a necessary condition is that $B\_3\le0$. Now change the basis of $\ker A$ by choosing orthogonal bases of the kernel and range of $B\_3$. Then $A$ is unchanged, whereas $B$ becomes
$$B==\begin{pmatrix} B\_1 & B\_{12} & B\_{13} \\\\ B\_{12}^T & B\_{22} & 0 \\\\ B\_{13}^T & 0 & 0 \end{pmatrix}.$$
By assumption, we have $B\_{22}< 0$. A new necessary condition appears, that $B\_{13}=0$. Now the necessary and sufficient condition over $k$ is that
$$k\ge\lambda\_{max}(S(B\_{1}-B\_{12}B\_{22}^{-T}B\_{12}^T)S),$$
where $S:=A\_+^{-1/2}$.
| 5 | https://mathoverflow.net/users/8799 | 47739 | 30,135 |
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