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https://mathoverflow.net/questions/38004 | 3 | Is there a classification theorem for "finitely generated abelian groups with a distinguished element"? If it helps, you can restrict this to the cases where the order of each element divides the order of the distinguished element.
My idea would be to try to use this for a classification theorem for rings with a finitely generated additive group, where the distinguished element is 1.
| https://mathoverflow.net/users/nan | slight extension to classification of finitely generated abelian groups | You say that we may assume that the order of every element divides the order of the distinguished element. That's the easy case. If the distinguished element generates a finite cyclic group $C$ of order $n$ and every element has order dividing $n$, then in fact $C$ is a direct summand. ("For the ring $\mathbb Z/n$ the free module of rank one is injective.")
| 3 | https://mathoverflow.net/users/6666 | 38012 | 24,427 |
https://mathoverflow.net/questions/38007 | 7 | I've been working through the heat-equation proof of the Atiyah-Singer index theorem. My question is what is the motivation for the definition of the index of an operator? I know there is the isomorphism between the homotopy classes of maps from a compact topological space to the space of Fredholm operators and the first K-group of the topological space given by the index map, but is there a simpler example of why we want to study the index?
| https://mathoverflow.net/users/1132 | Index of an Operator | Usually one really wants to know the dimension of the kernel of an operator rather than its index. The problem is that the dimension of the kernel is not a continuous function of the operator, so is very hard to compute in terms of topological data. The key point about the index is that it is a continuous function of the operator, which makes it a lot easier to compute. And it's a modification of the kernel, so with some luck one can compute the kernel from it.
| 16 | https://mathoverflow.net/users/51 | 38017 | 24,431 |
https://mathoverflow.net/questions/37849 | 5 | Is the (algebraic) span a finite set of vectors in a Hausdorff topological vector space over a complete field always closed?
I suspect yes, but I can't come up with a proof, and it seems like locally convex might be needed to get this.
| https://mathoverflow.net/users/nan | Closedness of finite-dimensional subspaces | This holds indeed for complete fields: see Theorem 2, Section I.2.3, of Bourbaki's "Espaces Vectoriels Topologiques".
Here is the argument.
Let $K$ be a (not necessarily commutative) field equipped with a complete nontrivial [absolute value](http://en.wikipedia.org/wiki/Absolute_value#Fields) $x\mapsto|x|$, let $n$ be a positive integer, let $\tau$ be a Hausdorff vector space topology on $K^n$, and let $\pi$ be the product topology on $K^n$.
**THEOREM** $\tau=\pi$.
**REMINDER** A topological group $G$ is Hausdorff iff {1} is closed. [Proof: {1} is closed $\Rightarrow$ the diagonal of $G\times G$ is closed (because it's the inverse image of {1} under $(x,y)\mapsto xy^{-1}$) $\Rightarrow$ $G$ is Hausdorff.]
**LEMMA** The Theorem holds for $n=1$.
**The Lemma implies the Theorem.** We argue by induction on $n$. The continuity of the identity from $K^n\_\pi$ to $K^n\_\tau$ (obvious notation) is clear (and doesn't use the Lemma). To prove the continuity of the identity from $K^n\_\tau$ to $K^n\_\pi$, it suffices to prove the continuity of an arbitrary nonzero linear form $f$ from $K^n\_\tau$ to $K\_\pi$. By induction hypothesis, the kernel of $f$ is closed, and the Theorem follows from the Reminder and the Lemma.
**Proof of the Lemma.** We'll use several times the fact that $K^\times$ contains elements of arbitrary large and arbitrary small absolute value. As already observed, we have $\tau\subset\pi$. If $x$ is in $K^\times$, write $B\_x$ for the open ball of radius $|x|$ and center 0 in $K$. Let $a$ be in $K^\times$, and let $\tau\_0$ be the set of those $U$ such that $0\in U\in\tau$.
It suffices to check that $B\_a$ contains some $U$ in $\tau\_0$.
We can find a $b$ in $K^\times$ and a $V$ in $\tau\_0$ such that $a$ is not in $B\_bV$, and then a $c$ in $K$ with $|c|>1$ and a $W$ in $\tau\_0$ such that $a$ is not in $B\_cW$. Then $U:=c^{-1}W$ does the job.
| 7 | https://mathoverflow.net/users/461 | 38030 | 24,439 |
https://mathoverflow.net/questions/37946 | 3 | Let $X\_t\in\mathbb{R}$ be an Ito diffusion process given by $$ dX\_t=a(b-X\_t)dt+\sigma dW\_t$$, then the characteristic operator of $X\_t$ is given by $$L=a(b-x)\frac{\partial}{\partial x}+\frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2}$$
(more details about the characteristic operator can be found here <http://en.wikipedia.org/wiki/It%C5%8D_diffusion>). Where in this case $a>0$, $b\in\mathbb{R}$, and $\sigma>0$.
Now I assume that $\tau$ is a random variable with density function $f(t)=\lambda e^{-\lambda t}\chi(t)\_{[0,\infty)}$, $\alpha\_t$ is a random variable taking only two values $1$ and $2$.
$$\alpha\_{t}=1\chi\_{(0,\tau)}(t)+2\chi\_{[\tau,\infty)}(t) $$
Then I change $b=b(\alpha\_{t}$), i.e, b is random and taking two values $b(1)\in\mathbb{R}$ and $b(2)\in\mathbb{R}$
My question is : how we find the characteristic of Ito diffusion $X\_t$ given by $$dX\_t=a[b(\alpha\_t)-X\_t]dt+\sigma dW\_t $$ ?
Thanks for your time and consideration
| https://mathoverflow.net/users/9059 | Characteristic operator | The wikipedia page cited in the question provides most of the answer: to get your operator compute
\begin{equation}
\lim\_{\delta \rightarrow 0} \frac{ {\mathbb E}[f(X\_\delta)] -f(x)}{\delta}
\end{equation}
The difference between your problem and the case covered in the wikipedia article is that $f$ in the above display is a function of $x$ only. However, your problem has an additional state variable (the binary variable that takes one of the values $1$ or $2$ depending on $\alpha$). So, the correct limit to study is:
\begin{equation}
\lim\_{\delta \rightarrow 0} \frac{ {\mathbb E}[f(X\_\delta,\alpha\_t)] -f(x,1)}{\delta}.
\end{equation}
Thus, you don't have one function, but two functions $f(x,1)$ and $f(x,2)$ and two PDEs that these functions satisfy.
It is implicitly assumed that $\tau$ is independent of the dynamics of $X$ before $\tau$. Furthermore, before $\tau$ the dynamics of $X$ are governed by the first SDE given in the question. One can use these to write the above expectation in two pieces: one piece over the set $\{\delta < \tau\}$ the other over $\{\delta < \tau\}^c$. Once this is done, the usual use of Ito's formula gives:
$$
L\_1 f(x,1)=-\lambda f(x,2)~~~ (\*)
$$
and
$$
L\_2 f(x,2) = 0.
$$
where
$$
L\_i = a(b\_i - x) \frac{\partial}{\partial x} + \frac{1}{2} \sigma^2\frac{\partial^2}{\partial x^2}
$$
Further details:
\begin{align\*}
{\mathbb E}[ f(X\_\delta,\alpha\_\delta) ]&=
{\mathbb E}[ f(X\_\delta,\alpha\_\delta) 1\_{\{ \tau > \delta\}} ] +
{\mathbb E}[ f(X\_\delta,\alpha\_\delta) 1\_{\{ \tau \le \delta\} }]\\\\
&\approx (1-\lambda \delta){\mathbb E}[ f(X^1\_\delta,\alpha\_\delta)] +
\lambda \delta f(x,2),
\end{align\*}
where $X^1$ is a process that is independent of $\tau$ with dynamics determined by $L\_1$.
Here you use several things: 1) $P( \tau < \delta) \approx \delta \lambda$ 2) if a jump occurs before $\delta$, you can ignore what happens between $\tau$ and $\delta$ (the contribution of this part is in the order of $\delta^2$ and when divided by $\delta$ and
$\delta$ is let go to $0$, it disappears).
To get (\*) from the previous display: use Ito's formula on the first expectation, subtract $f(x,1)$, divide by $\delta$ and let $\delta \rightarrow 0$. $f(x,2)$ is a function of what happens after $\tau$; after $\tau$ the stochastic process is a simple diffusion with generator $L\_2$: this is why (\*\*) holds.
| 2 | https://mathoverflow.net/users/3370 | 38041 | 24,446 |
https://mathoverflow.net/questions/38043 | 1 | Given the definition of subsets and equality of sets:
* A $\subset$ B, if x $\epsilon$ A $\rightarrow$ x $\epsilon$ B for every set x.
* A = B, if A $\subset$ B and B $\subset$ A
Why is it impossible to decide whether two circular sets I = {I} and J = {J} are equal.
I mean, the way is see it is that I is not an element of J, since only J is an element of J, so the two circular sets are not equal.
What's wrong in my reasoning?
| https://mathoverflow.net/users/9081 | Equality of two circular sets | In ZF minus the axiom of foundation there is no way of proving
that all "circular" sets are equal. You could take a model of set theory
allowing ur-elements and replace some or all of these by circular sets.
If you accept the popular alternative to Foundation, Peter Aczel's
[Anti-Foundation axiom](http://en.wikipedia.org/wiki/Anti-Foundation_Axiom), then there is just one circular set.
Effectively the AFA decrees that two sets with the same membership digraph
on its transitive closure are equal.
| 5 | https://mathoverflow.net/users/4213 | 38045 | 24,448 |
https://mathoverflow.net/questions/37812 | 4 | I have been studying of late about formation of naked singularities in certain collapse scenarios in Einstein's theory. It seems to me that the canonical paper to read about how such a formation is established is the 1984 paper by Christodoulou in Communications in Mathematical Physics. ( <http://www.ams.org/mathscinet-getitem?mr=742192> )
I was wondering if there is a reference which gives a more modern rewriting of the proof in that paper which say sort of highlights the generic technique of the proof which the reader can take away from there for other scenarios.
Somehow even the most recent books on Einstein's theory like the otherwise brilliant book by Choquet-Bruhat also doesn't dwell on techniques of testing in a collapse scenario whether the curvature singularity is naked or not.
I haven't seen till now any generic method or algorithm for testing this.
It would be great if someone can give me some references/explanations along these lines.
| https://mathoverflow.net/users/2678 | Christodoulou's paper on naked singularities in inhomogeneous dust collapse | Your question is very broad, and so I'll just give some very broad answers too.
**Collapse scenarios** Are you absolutely sure you want to restrict yourself to dust collapse? In the case of a spherically symmetric scalar field, there is also <http://www.ams.org/mathscinet-getitem?mr=1307898> (and [this paper](http://www.jstor.org/stable/121023) which shows that naked singularities in the scalar field model are unstable).
There's some problem of the interpretation of the dust collapse as a physical formation. Here I'll quote Demetri from [his book on vacuum collapse due to incoming gravitational waves](http://arxiv.org/abs/0805.3880)
>
> With the above remarks in mind the author turned to the study of the
> gravitational collapse of an inhomogeneous dust ball. In this case, the
> initial state is still spherically symmetric, but the density is a function of the
> distance from the center of the ball. The corresponding spherically symmetric
> solution had already been obtained in closed form by Tolman in 1934, in
> comoving coordinates, but its causal structure had not been investigated. This
> required integrating the equations for the radial null geodesics. A very different
> picture from the one found by Oppenheimer and Snyder emerged from this
> study. The initial density being assumed a decreasing function of the distance
> from the center, so that the central density is higher than the mean density,
> it was found that as long as the collapse proceeds from an initial state of low
> compactness, the central density becomes infinite before a black hole has a
> chance to form, thus invalidating the neglect of pressure and casting doubt on
> the predictions of the model from this point on, in particular on the prediction
> that a black hole eventually forms.
>
>
>
Essentially the same comment was made by Hájíček in his MathReviews on the dust collapse paper.
**Modern re-writes** Part of the reason that there are no modern re-writes of the proof is because of what I mentioned above, that the violation of weak cosmic censorship is unphysical (so it didn't attract that much attention). On the other hand, the basic idea behind the proof is not too hard (it is in the implementation of the analysis that is difficult). Simply speaking, in spherical symmetry, there is a lot of qualitative information that can be extracted without knowing too much details of the matter model involved (we must have matter as for vacuum, spherically symmetric space-times, we have Birkhoff's theorem). Perhaps a best modern reference is [Mihalis Dafermos' paper in CQG](http://www.ams.org/mathscinet-getitem?mr=2145241). The most important part is that under symmetry conditions and some mild assumptions, future null infinity must be complete; in the spherically symmetric space-times, future null infinity is characterised by the area-radius $r\to\infty$ along out-going null geodesics. Thus the basic idea is to show that there exists out-going null geodesics, such that if you travel along the future direction the area radius increases without bound, and that if you travel along the past direction you will hit the singularity.
Now, in the case of the Tolman dust which was studied by Christodoulou, since the exterior of the dust cloud is glued to a Schwarzschild solution, there is a dichotomy: either the out-going null geodesic escapes the dust region before hitting the apparent horizon, or the null geodesic hits the apparent horizon first. In spherical symmetry, it is a general fact that the apparent horizon consists of space-like or expanding null portions. So once a null geodesic passes the apparent horizon it can no longer escape to infinity. On the other hand, inside the Schwarzschild region the apparent horizon agrees with the event horizon, so once the null geodesic escapes from dust without hitting the apparent horizon, it will remain in the domain of outer communications and escape to infinity.
So to demonstrate existence of naked singularity, it suffices to show that a null geodesic emanating from the first point of singularity (it doesn't make sense to consider later points, as they will no longer belong to the Cauchy development of an initial data set) will escape the dust cloud before hitting the apparent horizon. To do so one needs to estimate the size of the solution of an ODE. In spherical symmetry, the apparent horizon is characterised by $2m/r = 1$, where $m$ is the Hawking mass and $r$ is the area radius. Now the Hawking mass satisfies ordinary differential equations (with source) in spherical symmetry (see, for example, Equation 3 [here](http://williewong.wordpress.com/2010/01/28/the-hoop-conjecture-of-kip-thorne-and-spherically-symmetric-space-times/); or see Section 3 of [this paper](http://dx.doi.org/10.1007/BF00375144).) So to estimate the size of the Hawking mass at the matching boundary, it is necessary to estimate the source terms in the ODE. This will lead to equation chasing using the Einstein equations and the matter-field equations. The rest is just being clever (in deciding in which order to estimate the various quantities) and doing the hard work of computation.
**General techniques** The reason there are no references for general techniques in testing whether a singularity is naked or not is very simply, *there exists no such techniques*. In spherical symmetry there is the simple description I posted above, but at the end of the day the estimates strongly depend on the structure of the matter equation (its separability and what not), so can only be really dealt with in a case-by-case basis. Of course, *if* you are given an explicit solution of the equations, testing whether the singularity is naked is often a simple computation re-writing the solution in some sort of null coordinates. The problem is that to prove genericity or to study singularities without reference to a explicit solution, one needs analytical estimates which, like I said, depends on which equations you are studying. In fact, if you could come up with a usable, generic method of testing whether a singularity is clothed, you would be half-way there resolving the general weak cosmic censorship conjecture.
**Some last comments** The general issue of weak cosmic censorship is a wide open one. The problem is that the statement contains the word "generic" in it (generic initial data sets lead to blah blah blah). So while there has been quite a lot of work going into constructing solutions and verifying that those solutions contain naked singularities, these works say nothing about weak cosmic censorship. (Explicit solutions tend to be non-generic in the space of solutions, except in the case you have strong rigidity theorems like Birkhoff's theorem for spherically symmetric Einstein-vacuum/electro-vacuum space-times or the No Hair Theorem for four-dimensional stationary axi-symmetric solutions.) The only real progress to weak cosmic censorship have all been due to Christodoulou (most of the physics papers are lacking in rigour). (Interestingly, there has been more developments in strong cosmic censorship, which, despite the name, has relatively little to do with weak cosmic censorship.)
Most recently, the focus in the community seems to be that the next model to consider for cosmic censorship should be the Einstein-Vlasov model (in spherical symmetry). (Well, it has been under consideration for around 10 years now and still prohibitingly hard.) For general solutions without symmetry assumptions, there has been essentially zero work in the field. There was an [attempt](http://stacks.iop.org/0264-9381/16/i=12A/a=302) to reformulate the conjecture into something mathematically tractable, but not has been done with the reformulation.
| 8 | https://mathoverflow.net/users/3948 | 38055 | 24,453 |
https://mathoverflow.net/questions/38049 | 37 | I know this is a dangerous topic which could attract many cranks and nutters, but:
According to Wikipedia [*and probably his own website, but I have a hard time seeing exactly what he's claiming*] Louis de Branges has claimed, numerous times, to have proved the Riemann Hypothesis; but clearly few people believe him. His website is:
[http://www.math.purdue.edu/~branges/site/Papers](http://www.math.purdue.edu/%7Ebranges/site/Papers)
but I find his papers difficult to follow. However, whether or not you believe him, his arguments presumably should prove *something*, even if not the full RH.
So, my question is:
***Are there any theorems related to the Riemann Hypothesis and similar problems, arising from his work, which have been fully accepted by the mathematical community and published (or at least submitted)?***
| https://mathoverflow.net/users/6651 | What, exactly, has Louis de Branges proved about the Riemann Hypothesis? | The paper by Conrey and Li "A note on some positivity conditions related to zeta and L-functions"
<https://arxiv.org/abs/math/9812166>
discusses some of the problems with de Branges's argument. They describe a (correct) theorem about entire functions due to de Branges, which has a corollary that certain positivity conditions would imply the Riemann hypothesis. However Conrey and Li show that these positivity conditions are not satisfied in the case of the Riemann hypothesis.
So the answer is that de Branges has proved theorems in this area that are accepted, and his work on the Riemann hypothesis has been checked and found to contain a serious gap. (At least the version of several years ago has a gap; I think he may have produced updated versions, but at some point people lose interest in checking every new version.)
Update: there is a more recent [paper by Lagarias](http://www.springerlink.com/content/l7n6670q03k54150/) discussing de Branges's work. *Lagarias, Jeffrey C.*, Hilbert spaces of entire functions and Dirichlet $L$-functions, Cartier, Pierre (ed.) et al., Frontiers in number theory, physics, and geometry I. On random matrices, zeta functions, and dynamical systems. Papers from the meeting, Les Houches, France, March 9–21, 2003. Berlin: Springer (ISBN 978-3-540-23189-9/hbk). 365-377 (2006). [ZBL1121.11057](https://zbmath.org/?q=an:1121.11057), [MR2261101](https://mathscinet.ams.org/mathscinet-getitem?mr=2261101). [Author's website](https://dept.math.lsa.umich.edu/%7Elagarias//doc/debranges-houches.pdf) and [Wayback Machine](https://web.archive.org/web/20210508190901/https://dept.math.lsa.umich.edu/%7Elagarias//doc/debranges-houches.pdf).
| 49 | https://mathoverflow.net/users/51 | 38057 | 24,454 |
https://mathoverflow.net/questions/38037 | 3 | This context of this question is Rutten's [Universal Coalgebra](http://en.wikipedia.org/wiki/F-coalgebra), used for modelling systems. I'm interested in finding a description of a functor between different types of coalgebras corresponding to finding a certain subcoalgebra.
An $F+1$-coalgebra $\langle S,\alpha:S\to F(S)+1\rangle$ can be thought of as a system whose transition shape is given by functor F plus the possibility of an error/termination, given by the +1. Assume that $F$ is a so-called Kripke polynomial functor: $F::=Id ~|~ B ~|~ F+F ~|~ F\times F ~|~ F^ A ~|~ \mathcal{P}\_\omega F$, thus it preserves pullbacks.
A subcoalgebra of $\langle S,\alpha:S\to F(S)+1\rangle$ is coalgebra $\langle S',\alpha':S'\to F(S')+1\rangle$, where $S'\subseteq S$ and $\alpha'$ is $\alpha'$ restricted to $S'$, such that its range falls within $F(S')+1$.
I want to find the maximal subcoalgebra of this coalgebra which corresponds to an $F$-coalgebra. In terms of my application, this means I'm looking for the subset of states $S'\subseteq S$ that do not lead to the error state.
Clearly, I can take the pullback of the functions $\alpha:S\to F(S)+1$ and $inl:F(S)\to F(S)+1$ to get a set $S\_0\subseteq S$ which do not lead to an error in the first step. Iterating this process for functors $F^i+1$ seems to lead to progressively smaller subsets $S\_i$ of $S$ each avoiding the error state for $i$ steps. What I'm lacking is a coherent description of the process.
Is whether there is a more universal description of this construction in terms of limits or colimits, or at least, some known approaches to the problem?
I asked this question on the computer science theory overflow ([here](https://cstheory.stackexchange.com/questions/901/maximal-subcoalgebras-of-an-f1-coalgebra-corresponding-to-an-f-coalgebra)), but am re-asking it here as I received no feedback.
| https://mathoverflow.net/users/2620 | Maximal subcoalgebras of an $F+1$-coalgebra corresponding to an $F$-coalgebra | I think the construction you're looking for can be seen as a right adjoint, and hence the details of the construction can be seen as coming from general transfinite constructions of adjoints.
$\newcommand{\inl}{\mathrm{inl}} \newcommand{\Coalg}{\mathbf{Coalg}}$
There's a functor $\inl^\* : F$-$\Coalg \longrightarrow (F+1)$-$\Coalg$; it embeds $F$-coalgebras asthe full subcategory of "error-free" $F+1$-coalgebras, and is induced by the natural transformation $\inl : F \rightarrow F+1$ in an obvious-once-you-write-down-the-diagram way.
Now, if I'm understanding right, the construction you're looking at, the "error-free core" of an $F+1$ coalgebra, is the right adjoint to this.
Moreover, I *think* there should be theorems that show automagically why this can be computed by the construction you give, as an $\omega$-long limit of pullbacks — but I'm not sure exactly where, I'm afraid. It's almost certainly deducible from the Kelly "Unified treatment of transfinite constructions" paper, well-described by Tom Leinster [here](https://mathoverflow.net/questions/19906/are-monads-monadic/20345#20345); the constructions of that have a very similar flavour.
Possibly relevant well-known constructions to compare (in Kelly and elsewhere): the construction of an algebraically-(co)free (co)monad on an endofunctor; the construction of the free $T$-algebra on a $T$-graph; the free $S$-algebra on a $T$-algebra, given a monad map $S \to T$.
| 2 | https://mathoverflow.net/users/2273 | 38060 | 24,455 |
https://mathoverflow.net/questions/38047 | 13 | The ordinary notions of limit and colimit are universal solutions to a problem, specifically, finding terminal/initial objects in slice/coslice categories. In the context of homotopy right Kan extensions (it's not hard to show that the theory of homotopy limits reduces to this case (the same holds for left Kan extensions/colimits)), we say that given a functor $f:C\to C'$ between small categories, and a functor $F:C\to A$ where $A$ is a cofibrantly-generated model category, that a natural transformation of functors $\alpha: H\to Ran\_f F$ exhibits $H$ as the homotopy right Kan extension of $F$ if there exists an injectively fibrant replacement $G$ of $F$ such that the composite $H\to Ran\_fF \to Ran\_fG$ is a weak equivalence.
When $A$ is combinatorial, we can also simply define a homotopy right Kan extension functor along $f$ to be $Ran\_f (Q (-))$, where $Q$ is a functorial injectively fibrant replacement functor.
This is easy enough to define, but why is this the definition? Why would we want to take fibrant/cofibrant replacements and consider their ordinary Kan extensions/limits/etc? I suspect that it has to do with the fact that homotopy is an honest equivalence relation on arrows from a cofibrant object into a fibrant object, but I would appreciate an actual explanation.
| https://mathoverflow.net/users/1353 | What is the "universal problem" that motivates the definition of homotopy limits/colimits (and more generally "derived" functors)? | There is a notion of homotopical Kan extension defined for "homotopical categories" (cats with a class of weak equivalences satisfying the 2-out-of-6 property). I cannot go into much detail without making a bazillion definitions, but the reference is Dwyer, Kan, Hirschorn, and Smith's "Homotopy Limit Functors on Model Categories and Homotopical Categories". Homotopical kan extensions and homotopy limit functors are defined, and model categories are demonstrated to be homotopically complete and cocomplete, and of course, fibrant and cofibrant replacement play a pretty big role in the proof. I think of the book as MacLane's chapter on Kan extensions generalized to the homotopical case.
| 7 | https://mathoverflow.net/users/2468 | 38062 | 24,457 |
https://mathoverflow.net/questions/38059 | 9 | Is there a way of talking about continuity and smoothness for set valued functions? More precisely, consider $M$ and $N$ topological/smooth manifolds, and let $f$ a function that associates to each point $p\in M$ a subset $f(p) \subset N$ (I haven't made any assumptions on what target sets are allowed, but feel free to discuss cases where some restrictions are required). Is there a meaningful/canonical way of saying that $f$ is continuous or smooth?
For my particular application in mind, $M$ is a smooth manifold, and $f$ associates to each $p\in M$ an open, convex cone inside $T\_pM$.
**Edit**: I should clarify that a convex cone $K$ in some real vector space $V$ is a subset such that
(a) conic: for any $v\in K$ and $r\in \mathbb{R}\_+$ $\implies rv \in K$.
(b) convex: for any $v,w\in K$ and $a,b \in \mathbb{R}\_+$ $\implies av + bw \in K$
It is open if $K$ is an open subset of $V$, so in particular open cones do not contain the origin.
| https://mathoverflow.net/users/3948 | Notion of smoothness for set-valued functions | My idea is that if we want to compare $f(p)$ and $f(q)$ for nearby points $p$ and $q$, then we need to be able to put $f(p)$ and $f(q)$ into the same space. To do this, I'm going to assume that $M$ is a finite-dimensional Riemannian manifold, so that we can make use of a connection $\nabla$ on $M$.
For all $p$, let $f(p)$ be an open cone of $T\_p M$. Let $U\_p M$ denote the unit sphere in $T\_p M$, and define $$g(p) = f(p) \cap U\_p M.$$ Since $f(p)$ is a cone, it is the linear span of $g(p)$. Thus, any smoothness on $g$ will apply to $f$ as well.
Let $w \in T\_p M$, and define the covariant derivative of $g(p)$ in the direction $w$ by $$\nabla\_w g(p) := \{ \nabla\_w v \}\_{v \in g(p)}.$$
Let $\gamma$ be a smooth curve on $M$ with $\gamma(0) = p$ and $\dot \gamma(0) = w$. Define the parallel transport of $g(p)$ along $\gamma(t)$ by $$\nabla\_{\dot\gamma(t)} g(p) := \{ \nabla\_{\dot\gamma(t)} v \}\_{v \in g(p)}.$$ That is, the parallel transport of the set $g(p)$ is given by transporting each vector in $g(p)$ along the curve $\gamma(t)$.
Now, both $\nabla\_{\dot\gamma(t)} g(p)$ and $g(\gamma(t))$ are open subsets of the unit tangent space $U\_{\gamma(t)} M$ at the point $\gamma(t)$. If the set-valued function $g$ is to be smooth, then these two sets should be comparable.
Let $\operatorname{Vol}\_q$ denote the (finite) volume measure on the unit sphere $U\_q M$ at the point $q \in M$, and let $\Delta$ denote the symmetric difference of two sets. Let us say that $g$ is smooth at $p$ in the direction $w$ if $$\operatorname{Vol}\_{\gamma(t)} \left( \nabla\_{\dot\gamma(t)} g(p) ~\Delta~ g(\gamma(t)) \right) = O(t)$$
for all smooth functions $\gamma$ with $\gamma(0) = p$ and $\dot \gamma(0) = w$. If $g$ is smooth at all points in all directions, then we shall say it is smooth on $M$. Consequently, we shall say that $f$ is smooth if $g$ is smooth.
| 4 | https://mathoverflow.net/users/238 | 38072 | 24,462 |
https://mathoverflow.net/questions/38064 | 7 | This question is somewhat related to [Differential inclusions for distributions](https://mathoverflow.net/questions/37524/differential-inclusions-for-distributions) but I am asking for something rather more specific, so I hope it is alright to leave this as a separate, new question.
Let $M$ be a smooth manifold, then given a one-form $\omega$ on $M$, Frobenius' theorem gives a simple way to test whether the distribution defined by the kernel of $\omega$ in $TM$ is integrable.
Now let $F\subset T^\*M$ be a subset of the cotangent bundle such that the restriction of the canonical projection $\pi$ is onto, and such that $\pi^{-1}(p)$ is an open convex cone for every $p$ (I don't know whether this condition will affect the answer; I'm just including the information on what I know). I am interested in knowing conditions which will guarantee that there exists (locally) an integrable distribution in $F$.
Trvially some restrictions must apply. One may imagine that $F$ is in some sense not continuous, such that for any $\omega\_p\in F\_p$, there does *not* exist any smooth extension $\omega$ to any neighborhood of $p$. An example would be taking $M$ to be $\mathbb{R}^2$, and $F\_p$ to be the first quadrant for all $p\neq 0$, and the second quadrant for $p = 0$.
So one specific question is: is this lack of freedom the only difficulty? Is the following statement true?
>
> Suppose $F$ has the property that, for any $p$, and any $\omega\_p \in F\_p$, there exists some smooth one-form $\omega$ that is a section of $F$, such that $\omega |\_p = \omega\_p$, and $d\omega |\_p = 0$, then for any $q\in M$, we can find an open neighborhood $U$ of $q$ and some one-form $\eta$ over $U$ such that $\eta$ is a section of $F|\_U$ and $d\eta = 0$.
>
>
>
Feel free to ask for clarifications.
| https://mathoverflow.net/users/3948 | Hypersurfaces orthogonal to a cone | There are a global obstrutions in some cases when local solutions exist, even in the case that the cone is an open subset of the cotangent bundle. For example: let $M^3$ be the tangent line bundle to any Riemannian surface $N^2$. The Levi-Civita connection defines a 2-plane field, where curves tangent to these planes represent transportation of a tangent line that is parallel under the connection. This is not integrable except where the Gaussian curvature is 0. Let's relax the condition, allowing the open cone of motions of lines where they drift from parallelism by a rate say < .1 degree per unit of arc length. By the h-principle, there are local solutions --- actually, you can get local solutions by parallel transport of lines along geodesics from a central point, up to a small radius.
However: note that for any global solution, the leaves of the foliation are covering spaces of $N^2$. If the surface is a sphere, this would mean they project homeomorphically, so each leaf would give a global line field --- but we all know you can't comb a billiard ball, so this is impossible. It's also known that if the surface has negative Euler characteristic, it can't be done, by a theory developed by John Milnor and John Wood. A circle bundle over a surface other than $S^2$ admits a foliation transverse to the fibers if and only if the Euler class of the bundle has absolute value that does not exceed -Euler characteristic of base. For the tangent sphere bundle, this just barely works: some significant examples are the Anosov foliations of the geodesic flow for any metric of negative curvature. For the tangent line bundle, it doesn't work, since its Euler class is doubled from the tangent sphere bundle.
Quite a bit more is known about existence and non-existence of foliations, but I think this answers your specific question.
| 11 | https://mathoverflow.net/users/9062 | 38078 | 24,466 |
https://mathoverflow.net/questions/37127 | 2 | I've found (as have others), that for some analytic functions, a Padé approximant of it has an infinite convergence radius, whereas its associated Taylor series has a finite convergence radius. $f(x)=\sqrt{1+x^2}$ appears to be one such function. My questions are:
1) Is there any function where the Taylor series has the largest convergence radius of all associated Padé approximants? If so, is the Taylor series radius strictly larger, or only equal to the convergence radius of other Padé approximants (i.e. excluding the Taylor series itself)?
2) If not, is there any function that is analytic everywhere, and yet for which there is no (limit of) Padé approximant(s) that has an infinite convergence radius?
It would be both very cool and very useful if there is always a (limit of) Padé approximant(s) that has an infinite convergence radius for any function that is analytic everywhere, though I haven't the slightest how one checks/analyzes convergence of Padé approximants if the degrees of numerator and denominator both approach infinity. :)
One extra question, if there is always such a Padé approximant:
3) Is there always a numerically stable method of computing this approximant up to a finite order?
| https://mathoverflow.net/users/8864 | Can Convergence Radii of Padé Approximants Always Be Made Infinite? | Don't be lured into thinking Pade approximates are 'nice'. Here is why:
**Notation:**
Let $[L/M]\_f(z)$ be the Pade approximation $\frac{p\_n(z)}{q\_m(z)}$ to $f$ where $\text{deg}(p\_n)\leq n$ and $\text{deg}(q\_m)\leq m$, $q\_m(0)=1$.
**Parital Answer:**
1) The partial theta function $h\_q(z)$ with $q=e^{iz}$ and $z/2\pi$ real and irrational is a counterexample by Lubinsky and Saff. It is analytic in $|z|<1$ but there is no subsequence of pade approximants $[L/M]\_f(z)$ with $M$ fixed that converges for all $|z|<1$. See Pade Approximants by Baker and Braves Morris, p279.
2) A theorem by Wallin shows that there is an entire function such that the approximant sequence $[L/L]\_f(z)$, $L\in\mathbb{N}$ is unbounded in $z\neq0$. (I cannot remember a reference for this.) However, I don't have an answer if you consider pade sequences $[L/M]\_f(z)$ where $L$ and $M$ are two independent sequences.
3) For special sequences of Pade approximations (i.e. $[L/L]\_f(z)$ -the diagonal sequence) there are numerically stable ways to calculate the Pade approximant. Not sure in general.
| 4 | https://mathoverflow.net/users/2011 | 38087 | 24,469 |
https://mathoverflow.net/questions/38085 | 12 | Let $R$ be a ring, and $R\text{-Mod}$ its category of all left modules. There is a "forgetful" functor $\operatorname{Forget}: R\text{-Mod} \to \text{AbGp}$, which is additive, continuous, and cocontinuous (in particular, exact). Since $R\text{-Mod}$ is both complete and cocomplete, $\operatorname{Forget}$ has both a left adjoint $\operatorname{Free}: \text{AbGp} \to R\text{-Mod}$ and a right adjoint $\operatorname{Cofree}: \text{AbGp} \to R\text{-Mod}$.
You can see what these functors are explicitly. Let me write $\_R R\_{\mathbb Z}$ for "$R$ as a left module" and $\_{\mathbb Z} R \_R$ for "$R$ as a right module".
The $\operatorname{Forget}$ functor is (isomorphic to) the functor $\operatorname{Hom}\_R({\_R R\_{\mathbb Z}},-)$ — this description makes it clearly continuous, and its left adjoint is $\operatorname{Free} \cong {\_R R\_{\mathbb Z}} \otimes\_\mathbb Z (-)$. But we also have $\operatorname{Forget} \cong {\_{\mathbb Z} R \_R}\otimes\_R (-)$, whence its right adjoint is $\operatorname{Cofree} \cong \operatorname{Hom}\_{\mathbb Z}({\_{\mathbb Z} R \_R},-)$.
I feel like I have some positive amount of experience with free modules. (I would say, given the above, that the correct definition of "free module" is "object in the essential image of $\operatorname{Free}$", although what's actually used is "object of the form $\operatorname{Free}(\mathbb Z^{\oplus \kappa})$ for some cardinal $\kappa$.) But I hardly ever come across the essential image of $\operatorname{Cofree}$, or indeed the cofree functor at all. (Again, maybe the "standard" definition of "cofree module" is "module isomorphic to $\operatorname{Cofree}((\mathbb Q/\mathbb Z)^{\times \kappa})$," or something.) The functors are not the same: when $ R = \mathbb Z/2$, then $\operatorname{Free}(\mathbb Z) = \mathbb Z/2$, whereas $\operatorname{Cofree}(\mathbb Z) = 0$. If you would rather replace $\mathbb Z$ by a field throughout, then they are still not the same when $R$ is infinite-dimensional (for example).
So: Do people use cofree modules? If so, how? If not, why not? Are free modules just a lot nicer than cofree ones, and if so, how?
| https://mathoverflow.net/users/78 | Why not _co_free modules? | This construction is used frequently (at least, I use it frequently in my work).
For example, it appears in the usual proof that module categories have enough injectives.
(In this case one studies $Cofree(\mathbb Q/\mathbb Z)$, as you anticipated.)
If we generalize slightly, and replace $\mathbb Z$ by the group ring $k[H]$ and $R$ by
the group ring $k[G]$ (with $H$ being a subgroup of $G$), then
$Hom\_{k[H]}(k[G],\text{--})$ is precisely the functor of induction from $H$-representations to $G$-representations, and the adjointness you note is a form of Frobenius reciprocity.
If $R$ is a Hecke algebra (over $\mathbb Z$) on a space of weight $k$-cuspforms of some level,
then $Cofree(\mathbb Z)$ is the space of modular forms of weight $k$ with coefficients in $\mathbb Z$. (This technical relationship between Hecke operators and the space of modular
forms on which they operate is used frequently by number theorists working on the arithmetic of
modular forms.)
There are lots of other contexts in which this functor (and its variants, replacing
$\mathbb Z$ by other rings) appear, but maybe I've said enough for now.
| 8 | https://mathoverflow.net/users/2874 | 38091 | 24,471 |
https://mathoverflow.net/questions/38089 | 19 | The following is some version of Tannaka-Krein theory, and is reasonably well-known:
>
> Let $G$ be a group (in Set is all I care about for now), and $G\text{-Rep}$ the category of all $G$-modules (over some field $\mathbb K$, say). It is a fairly structured category (complete, cocomplete, abelian, $\mathbb K$-enriched, ...) and in particular carries a symmetric tensor product $\otimes$. The forgetful functor $\operatorname{Forget}: G\text{-Rep} \to \text{Vect}$ respects all of this structure, and in particular is (symmetric) monoidal. Let $\operatorname{End}\_\otimes(\operatorname{Forget})$ denote the monoid of monoidal natural transformations of $\operatorname{Forget}$. Then it is a group, and there is a canonical isomorphism $\operatorname{End}\_\otimes(\operatorname{Forget}) \cong G$.
>
>
>
The following is probably also reasonably well-known, but I don't know it myself:
>
> Let $G$, etc., be as above, but suppose that we have forgotten what $G$ the category $G\text{-Rep}$ came from, and in particular forgot, at least momentarily, the data of the forgetful functor. We can nevertheless recover it, because in fact $\operatorname{Forget}$ is the unique-up-to-isomorphism ADJECTIVES functor $G\text{-Rep} \to \text{Vect}$.
>
>
>
My question is: what are the words that should go in place of "ADJECTIVES" above? Certainly "linear, continuous, cocontinuous, monoidal" are all reasonable words, although my intuition has been that I can drop "cocontinuous" from the list. But even with all these words, I don't see how to prove the uniqueness. If I had to guess, I would guess that the latter claim is a result of Deligne's, although I don't read French well enough to skim a bunch of his papers and find it. Any pointers to the literature?
| https://mathoverflow.net/users/78 | What's so special about the forgetful functor from G-rep to Vect? | If $G$ is an affine algebraic group (for example a finite group), then the category of $k$-linear cocontinuous symmetric monoidal functors from $\mathsf{Rep}(G)$ to $\mathsf{Vect}\_k$ is equivalent to the category of $G$-torsors over $k$. In particular, not every such functor needs to be isomorphic to the identity. For example, if $k'$ is finite Galois extension of k with Galois group $G$, then the functor $F(V) = (V \otimes\_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.
| 35 | https://mathoverflow.net/users/7721 | 38092 | 24,472 |
https://mathoverflow.net/questions/38086 | 20 | Consider $n$ circles with variable radii $r\_1,\ldots, r\_n$ that pack inside a fixed circle of unit radius. In other words, all $n$ variable-radius circles are contained in the unit radius circle and their interiors have empty intersections. The tangency graph of a packing comprises $n+1$ vertices, one for each circle, and edges between vertices if the corresponding circles are tangent.
>
> Conjecture: in a packing that
> maximizes $r\_1+\cdots +r\_n$,
> the corresponding tangency graph is
> planar and triangulated.
>
>
>
This conjecture looks like it might be related to the Koebe-Andreev-Thurston circle packing theorem. The latter states that for every planar triangulated graph there is a corresponding circle packing of the kind described and that this packing is unique up to conformal transformations. While it may turn out that the KAT theorem can provide some insights on proving the conjecture, I believe that something else is going on. For instance, the radius-sum objective function is not conformally invariant.
I have good numerical evidence in support of this conjecture. The optimum configurations I've found up to $n=20$ all have triangulated graphs. I'm posting this on MO because I also have something that looks like it may be "close" to a proof. Perhaps someone can close the gap or convince me that the gap is actually a bottomless chasm -- either would be helpful!
Here is my proof strategy:
1. Use convexity to show that an optimal configuration maximizes the number of edges in the tangency graph.
2. Use Euler's theorem to show that
a tangency graph that maximizes the
number of edges is triangulated.
This is a constrained optimization problem in $\mathbb{R}^{3n}$. Consider the constraint that applies to circles 1 and 2: $(x\_1-x\_2)^2+(y\_1-y\_2)^2 \ge (r\_1+r\_2)^2$. This type of constraint is called "reverse convex" (the feasible region is the complement of an open convex set). Feasible regions in reverse convex problems (intersection of open set complements) can be quite complex -- they may not even be connected. On the other hand, they have a very nice property when we are maximizing a convex function: an optimum can always be found at a "vertex" of the feasible region. In a reverse convex problem in $\mathbb{R}^{N}$, a vertex is a point of the feasible region where at least $N$ of the constraints are equalities. We can think of reverse convex problems as generalizing linear programming in a way that inherits all the nice local properties.
The existence of a global optimizer requires that the feasible region is non-empty and compact. This is not an issue for the circle packing problem since we can let the radii range over all the real numbers and add reverse convex constraints $r\_1\ge 0,\ldots,r\_n\ge 0$.
The alert reader will already have realized that not all of the constraints in the circle packing problem are reverse convex! The constraints that apply to the fixed unit circle have the wrong sense of the inequality, e.g. $x\_1^2 + y\_1^2 \le (1-r\_1)^2$. One can try to fix this problem by replacing the fixed unit circle with a regular $M$-gon and taking the limit (in some sense) of large $M$. This has two nice consequences. First, the optimization is now truly reverse convex (half-plane constraint for every side of the polygon) and so there is an optimizer where exactly $3n$ constraints are active (at their equality value). To see the second nice feature we have to do some counting.
The tangency graph has one new feature when the fixed circle is replaced by a regular $M$-gon: it is no longer simple because it may have doubled edges between the variable-radius circles and the polygon (whenever a circle is tangent to adjacent polygon edges). Let the number of circles with double tangencies be $D$. If $E$ and $F$ are the number of edges and faces of the graph, and $\tilde{E}$ and $\tilde{F}$ are these quantities when the doubled edges are merged into single edges, then $E=\tilde{E}+D$ and $F=\tilde{F}+D$. Since our graph has $n+1$ vertices, and reverse convex programming tells us there is an optimum with $E=3n$ tangencies, Euler's theorem gives $n+1-3n+F=2$, or $F=2n+1$. We therefore have the following formulas for the "reduced graph" after merging doubled edges: $\tilde{E}=3n-D$, $\tilde{F}=2n+1-D$. The reduced graph is simple and planar and satisfies $2\tilde{E}\ge 3\tilde{F}$ where equality implies that the graph is triangulated. Using our formulas this inequality becomes $D\ge 3$.
The result of this analysis is that optimum configurations in the $M$-gon have at least 3 circles with double tangency, and that the reduced graph is triangulated when this minimum holds. The number 3 is interesting. I believe it corresponds to the fact that the conformal transformations are fixed by specifying 3 points on the boundary of the region (the $M$-gon) where the circles are mapped. Sacrificing the symmetry of the fixed circle paid off because it allowed the optimization problem to have discrete solutions (whose existence follow from reverse convex programming).
There are two gaps in the proof. How do we take results for the $M$-gon and by some limiting process prove a theorem about the circle? Second, how do we prove $D=3$? Optimal configurations with $D>3$ become more unlikely as $M$ becomes large because in that case more than the minimum number of active constraints arise from double tangencies. After all, the pair of constraints at a double tangency become degenerate at $M=\infty$.
I believe the conjecture is true for the class of objective functions $r\_1^p+\cdots + r\_n^p$ where $1\le p < 2$. The case $p\ge 2$ is uninteresting because the optima degenerate into a single circle that completely fills the fixed circle, the rest having zero radius.
| https://mathoverflow.net/users/40739 | A circle packing conjecture | You have an elaborate set of ideas, and I haven't thought through all of what you outlined, but here's a suggestion:
Oded Schramm generalized the circle packing theory to include arbitrary convex shapes, and showed they work in much the same way. (The famous case of packing squares is one instance included in this generalization). His theory even allows the shape to be a function of position and size, but that generality seems unnecessary here. The suggestion: consider a set of regular N-gons packed inside an N-gon. At corners, they will touch at more than one point, but it is a connected set, so there is a well-defined adjacency graph with out doubled edges, and no room for extra disks to try to hide in the corners. The reverse convex constraints become piecewise linear. I think the limiting process should be straightforward to analyze.
| 19 | https://mathoverflow.net/users/9062 | 38096 | 24,475 |
https://mathoverflow.net/questions/38066 | 16 | André Weil's likening his research to the quest to decipher the Rosetta Stone (see this [letter](http://www.ams.org/notices/200503/fea-weil.pdf) to his sister) continues to inspire contemporary mathematicians, such as Edward Frenkel in [Gauge Theory and Langlands Duality](http://arxiv.org/abs/0906.2747).
Remember that Weil's three 'languages' were: the 'Riemannian' theory of algebraic functions; the 'Galoisian' theory of algebraic functions over a Galois field; the 'arithmetic' theory of algebraic numbers. His rationale was the desire to bridge the gap between the arithmetic
and the Riemannian, using the 'Galoisian' curve-over-finite-field column as the best intermediary, so as to transfer constructions from one side to the other. (See also 'De la métaphysique aux mathématiques' 1960, in volume II of his Collected Works.)
That fitted rather neatly with demotic Egyptian mediating between priestly Egyptian
(hieroglyphs) and ordinary Greek on the real Rosetta Stone. But just as one might have expressed that text in
a range of other contemporary languages - Sanskrit, Aramaic, Old Latin, why should
there not be other columns in Weil's story? Frankel himself adds a fourth column (p. 11)
'Quantum Physics'.
So now the questions:
>
>
> Are there other candidate languages for Weil's stone? Might there be a further language for which we would need intermediaries back to the arithmetic? Could there be a meta-viewpoint which determines all possible such languages.
>
>
Presumably the possession of a zeta function is too weak a condition as that would allow the language of
[dynamical systems](http://www.ams.org/notices/200208/fea-ruelle.pdf).
| https://mathoverflow.net/users/447 | Which languages could appear on Weil's Rosetta Stone? | All the 12 or more approaches to geometry over the field with one element are tentatives to create such intermediate languages. But you seemed to ask more about a pre-existing area of it's own which may serve as a bridge - in this direction there are
$\bullet$ Alexandru Buium's theory of [Arithmetic Differential Equations](http://books.google.com/books?id=aqwYKFjW5nwC&printsec=frontcover&dq=buium+differential&source=bl&ots=Ari-GIDIJQ&sig=yUCY-rxrmID_YDxo0EYJAJIsU0o&hl=de&ei=GuSHTOiXIZSSOKLr7aUO&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBUQ6AEwAA#v=onepage&q&f=false) which brings the theory of differential equations into play as an intermediate language.
$\bullet$ I remember Alexandru Buium saying that there is also a theory of *difference* equations, different from his, but don't know more about this.
$\bullet$ Shai Haran uses probability theory as an intermediate language in his book ["Mysteries of the Real Prime"](http://books.google.com/books?id=bsUkkEraBvAC&dq=shai+haran&source=bl&ots=T6L2sJwl1N&sig=eisso9vgJb2PnxqPvgwoSLlvUcY&hl=de&ei=OOaHTKayJ5KLOPeqvdwO&sa=X&oi=book_result&ct=result&resnum=3&ved=0CCMQ6AEwAg). It connects to the quantum theory column, but I don't know whether it's in the same way that Frenkel suggested.
$\bullet$ Homotopy theory might be a future candidate for a column of its own, on the one hand via motivic homotopy theory as it is getting available over more and more general base schemes with more and more general coefficients and thus moving towards arithmetic. And on the other hand possibly via ring spectra which may serve as the base deeper than the integers which is hoped to trigger the translation process one day...
$\bullet$ You write that the possession of Zeta functions is too weak to make the theory of dynamical systems an intermediate language. But there is certainly much more connecting dynamical systems with Arithmetic, as shown in [Deninger's work](http://www.emis.de/journals/DMJDMV/xvol-icm/00/Deninger.MAN.html).
$\bullet$ The works of Bost, [Connes, Marcolli](http://arxiv.org/abs/math/0501424), [Meyer](http://arxiv.org/abs/math/0412277v2), [Laca](http://www.msri.org/publications/ln/msri/2000/simplec/laca/1/index.html) and others connect Arithmetic to the Theory of Operator Algebras, and via those again to dynamical systems, quantum physics and Thermodynamics
| 12 | https://mathoverflow.net/users/733 | 38097 | 24,476 |
https://mathoverflow.net/questions/38082 | 6 | Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume $V$? Or more generally a hyperbolic $n$-manifold of volume $V$?
| https://mathoverflow.net/users/4325 | Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V? | A couple of things are true:
1. If you have any Riemannian manifold of bounded infinitesimal geometry (curvature pinched above and below), its thick part, where the injectivity radius $> \epsilon$, can be triangulated with a number of simplices bounded by a constant times volume, where the constant depends on the curvature bounds and the dimension. I don't personally know the constant even for hyperbolic 3-manifolds, but I think there are people who can produce explicit bounds. This is basically a consequence of the compactness of the set of manifolds of bounded infinitesimal geometry and injectivity radius bounded below, together with the fact that all smooth manifolds admit a smooth triangulation, and that any smooth triangulation of a closed subset can be extended.
1. For hyperbolic 3-manifolds, if you allow "spun triangulations" where some tetrahedra are allowed to have missing vertices that spiral infinitely around a short closed geodesic, then there is a similar bound, the number is less than some constant times volume. To do it: first triangulate the thick part leaving a boundary torus, then make cones on the boundary triangles that spiral around a short geodesic.
The answers are the same whether you're asking for a geodesic triangulation of a hyperbolic manifold, or any smooth triangulation.
| 19 | https://mathoverflow.net/users/9062 | 38099 | 24,477 |
https://mathoverflow.net/questions/38080 | 9 | Fill in the blank, please :)
>
> Let $\mathcal C$ be a complete and cocomplete abelian category. A **generator** in $\mathcal C$ is an object $X \in \mathcal C$ such that every object $Y \in \mathcal C$ is a *colimit* of a (small) diagram made entirely of $X$s; in this way, $X$ knows everything there is to know about $\mathcal C$. When $\mathcal C$ is the category of all modules of some ring $R$, then an example of a generator is $R$, thought of as an $R$-module. A **cogenerator** in $\mathcal C$ is an object $X \in \mathcal C$ such that every object $Y \in \mathcal C$ is a *limit* of a (small) diagram made entirely of $X$s; in this way, $X$ knows everything there is to know about $\mathcal C$. When $\mathcal C$ is the category of all modules of some ring $R$, then an example of a cogenerator is `________`.
>
>
>
The only examples I know are: when $R = \mathbb Z$, an example of a cogenerator is the rational circle $\mathbb Q / \mathbb Z$; when $R = \mathbb K$ is a field, an example of a cogenerator is $\mathbb K$. But by some version of the Law of Small Numbers, these examples are not enough for me to see how to (or, in fact, whether it is possible to) generalize.
| https://mathoverflow.net/users/78 | What are examples of cogenerators in R-mod? | (This is closely related to Hailong's comment above.)
You can say (albeit rather abstractly) what *any* cogenerator must look like. The following can be found in T.Y. Lam's *Lectures on Modules and Rings*, Theorem 19.10. Let $\{V\_i\}$ be a complete set of simple right $R$-modules, with injective hulls $E(V\_i)$. Then $U\_0 = \bigoplus E(V\_i)$ is a cogenerator, called the *canonical cogenerator* of $\mathrm{Mod}\_R$, and any module $U\_R$ is a cogenerator for $\mathrm{Mod}\_R$ if and only if $U\_0$ can be embedded in $U$.
Note that if $R$ is right noetherian, then the direct sum of injective right $R$-modules is again injective. Hence $U\_0$ is injective, and in this case $U\_R$ is a cogenerator if and only if $U\_0$ embeds in $U$, if and only if $U\_0$ is a direct summand of $U$.
Referring to your examples of $R$ above: If $R = \mathbb{Z}$ then $U\_0 = \mathbb{Q}/\mathbb{Z}$. If $R = \mathbb{K}$ then $U\_0 = \mathbb{K}$. Both of these rings are noetherian, so the previous paragraph applies. (In particular, every nonzero $\mathbb{K}$-vector space is a cogenerator).
| 7 | https://mathoverflow.net/users/778 | 38102 | 24,479 |
https://mathoverflow.net/questions/38084 | 14 | Say a monoid $M$ *has infinite products* if, for any (possibly infinite) sequence $(m\_1,m\_2,\ldots)$ of elements of $M$, there exists an element $m\_1m\_2\cdots\in M$, satisfying some good properties. First, if the sequence is finite, it should coincide with the usual product on $M$. Second, concatenation of sequences results in multiplication of their products and is associative. Third, identities can be "thrown out," as can consecutive inverses ($m\_{i+1}=m\_i^{-1}$). (Other good properties to include?)
Another way to phrase this is: "$M$ is closed under small ordinal colimits." That is, if $M$ is considered as a one-object category, then for any small ordinal $[\kappa]$ and functor $m\colon[\kappa]\to M$, the colimit of $m$ exists in $M$.
Example: let ${\mathbb N}^+$ denote the monoid with underlying set ${\mathbb N}\cup ${$\infty$} and whose operation on a sequence $m=(m\_1,m\_2,\ldots)$ is given by addition if $m$ has only finitely many non-zero elements, and by $\infty$ otherwise.
Now suppose that $M$ is any monoid and I want to replace it by a monoid that has infinite products. I'm hoping there are two ways to do this. One would be to add colimits freely, and the other would be to add a single "$\infty$" element that served as a catch-all (as in the example above).
Q: Do these both exist (functorially in $M$)? If so, can you describe them in elementary terms? For example, I'm worried about sequences like $1-1+1-1+\cdots$. So in a good answer I'd hope to see what happens with such infinite sums.
| https://mathoverflow.net/users/2811 | Monoids with infinite products | If I understand the question, the short answer is "yes, you can freely and functorially adjoin infinite products to monoids". The basic idea is that algebraic theories can accommodate arbitrary arities (bounded above by some cardinal), and one can discuss relative free-forgetful adjunctions between categories of algebras in great generality.
At first pass, let me just focus on the purely monoid-like aspects for now, because those are easier to visualize. Once the story for that is clear, one can work in inverses.
As a warm-up, let's recall that one way of defining an ordinary monoid $M$ is as an algebra of the terminal nonpermutative operad. In plainer English, this means we a single operation
$$\mu\_n: M^n \to M$$
for each finite ordinal $n$, so that
$$\mu\_{n\_1 + \ldots + n\_k} = \mu\_k(\mu\_{n\_1} \times \ldots \times \mu\_{n\_k})$$
(generalized associativity equation).
Now let's generalize operads so as to allow operations of countably infinite arity. By "arities", I will really mean countable ordinals. Given an ordinal $k \lt \aleph\_1$ and ordinals $n\_j \lt \aleph\_1$ for $j \lt k$, you can concatenate the $n\_j$ to get a new ordinal $\sum\_j n\_j \lt \aleph\_1$. Concatenation is associative in an evident sense. Now define an $\omega$-monoid to be a set $M$ equipped with operations
$$\mu\_k: \hom(k, M) \to M,$$
one for each $k \lt \aleph\_1$, such that $\mu\_{\sum\_j n\_j} = \mu\_k (\prod\_j \mu\_{n\_j})$. (I am not completely certain we have to go all the way up to $\aleph\_1$, but if not it will be some suitable initial segment. Let's just say $\aleph\_1$ for now.) This condition can be interpreted in any category with countable products, such as $Set$.
The free $\omega$-monoid on a set $X$ will be $\sum\_{k \lt \aleph\_1} \hom(k, X)$. We get in this way a monad $T$ for a free-forgetful adjunction between $\omega$-monoids and sets.
There is a general bit of nonsense that for any morphism of monads $\phi: S \to T$ on $Set$, there is a forgetful functor $Alg\_T \to Alg\_S$, and this forgetful functor has a left adjoint. This follows from an adjoint functor theorem, although if I'm not mistaken, in this particular scenario a more direct construction is available: if $S$ is the monad for ordinary monoids and $T$ is as above, the evident inclusion $S \to T$ induces a forgetful functor
$$\omega-Mon \to Mon$$
which has a left adjoint $L$ described by the type of coequalizer familiar from tensor products:
$$L(M) = coeq((\mu\_T \circ \phi) M, T\theta: TSM \stackrel{\to}{\to} T M)$$
where $\mu\_T: TT \to T$ is the monad multiplication and $\theta: S M \to M$ is the structure of $M$ as $S$-algebra. This left adjoint $L$ would correspond to what I think you were asking for with "freely adjoined colimits", and the left adjoint means we indeed have a functorial construction.
If you want to work inverses in, you can do that too. Long story short: for any set of formal operation symbols of arbitrary arity, subject to any set of well-formed equations you jolly well please, you can form a monad whose algebras are precisely the models of for the corresponding algebraic theory. So: together with operations of countable arity as above, subject to generalized associativity equations, you can certainly toss in an unary inversion operation as well. I leave it to you to decide what, in addition to associativity, are the sensible equations to impose on inversion $i$, but it seems to me you might want to impose only
$$\mu\_2(id \times i) = id = \mu\_2(i \times id)$$
and stop there. (Operations involving infinitely many instances of inversion are still permissible, but the equations would enforce only finitely many cancellations at a time.) You get in this way a monad for "$\omega$-groups", and again the forgetful functor from $\omega$-groups to groups admits a left adjoint, constructed in a way analogous to the above.
| 4 | https://mathoverflow.net/users/2926 | 38111 | 24,483 |
https://mathoverflow.net/questions/38118 | 2 | Hi all:
I'm wondering if there is a simple formula for this.
Simple Example:
```
x=cos(pt);
y=cos(qt);
where p,q are integers.
```
Question: How many intersection points are there?
0) only need to consider (p,q) are relatively prime.
1) Firstly, I thought it was just basic counting:
Let N(p,q) be the number of crossing.
```
If p is odd, q is odd: N(p,q) = p(q-1) + (p-1)q = 2pq - p - q.
If p is even, q is odd: N(p,q) = (p-1)(q-1)/2.
```
But later I found some special cases p=2,q=5,where the intersection becomes unusual and the formula falls down.
Thanks
| https://mathoverflow.net/users/8744 | Formula for number of intersection points for Lissajous curve? | EDIT: with both functions switched to cosine I get $$ \frac{(p-1)(q-1)}{2} $$ for both odd-odd and for even-odd.
ORIGINAL, both functins sine: For $p$ even and $q$ odd I get $$ 2 p q - p - q $$ which is the same as you have when both are odd.
Examples, I had to count over a few times,
(p,q,count) :
(2,1,1); (2,3,7); (2,5,13); (2,7,19);
(4,1,3); (4,3,17); (4,5,31); (4,7,45);
(6,1,5); (6,5,49); (6,7,71); (6,11,115);
Then for both odd I get $$ \frac{(p-1)(q-1)}{2} $$
(1,q,0);
(3,q, q-1);
(5,q, 2 q - 2);
(7,q, 3 q - 3);
However for both odd the figure is traced twice, there is no way around that, look on the website you supplied and slowly vary the maximum of $\Theta.$ So it is also reasonable to claim that both segments are doubled and what appears to be a single intersection counts as four as far as the parameter values are concerned. If you prefer that, the formula for both odd becomes
$$ 2 p q - 2 p - 2 q + 2. $$
| 1 | https://mathoverflow.net/users/3324 | 38124 | 24,491 |
https://mathoverflow.net/questions/38126 | 5 | Let $p>2$ be a prime, $C\_p$ be the additive group of integers mod $p$. Then the multiplicative group $\{1,...,p-1\}$ of units in the field $Z/pZ$ is cyclic of order $p-1$, it acts on $C\_p$ by left multiplication. Let $G\_p$ be the corresponding semi-direct product of order $p(p-1)$. Question: does this group admit a presentation with 2 defining relations for some (all) $p > 3$? For $p=3$, $G\_p$ is the symmetric group $S\_3$ and it admits the balanced presentation $\langle a,b \mid a^2=1, aba=b^2\rangle$. (The question is attributed to Alex Lubotzky.) For example, what if $p=5$? Does $G\_5$, a cyclic-by-cyclic group of order 20, admit a presentation with 2 generators and 2 defining relations?
| https://mathoverflow.net/users/nan | a balanced presentation of a cyclic-by-cyclic group? | G5 has a presentation on generators a,b and relations ba=aab, abbabb=1.
| 7 | https://mathoverflow.net/users/3710 | 38135 | 24,498 |
https://mathoverflow.net/questions/38137 | 5 | I am forced to know the etale fundamental group of the grassmannian over the rational field. I searched it but couldn't find any hint. I am wondering whether there are some positive results or recipe to compute it.
Thank you!
| https://mathoverflow.net/users/9096 | Etale $\pi_1$ of Grassmannian | The answer is that any Grassmannian is geometrically simply connected, so the etale fundamental group over $\mathbb{Q}$ is simply [**edit**: !!] the absolute Galois group $\operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\mathbb{Q}$.
In more detail: let $X$ be a geometrically integral variety defined over $\mathbb{Q}$, let
$\overline{X}$ be its basechange to an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $\mathfrak{g}\_{\mathbb{Q}} = \operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ be the absolute Galois group of $\mathbb{Q}$.
1) Via a choice of geometric points (which we suppress), we get a short exact sequence of profinite groups
$1 \rightarrow \pi\_1(\overline{X}) \rightarrow \pi\_1(X) \rightarrow \mathfrak{g}\_{\mathbb{Q}} \rightarrow 1$.
2) Let $K$ be an algebraically closed field of characteristic $0$. Suppose that *either* $X$ is complete or that $X$ is nonsingular [in our application, both hold]. Then the natural map
$\pi\_1(\overline{X}) \rightarrow \pi\_1(\overline{X} \otimes K)$ is an isomorphism.
[Comment: if instead of $\mathbb{Q}$, our ground field was a field $k$ of positive characteristic and $K$ is an algebraically closed field containing $\overline{k}$, this map is still an isomorphism for complete varieties but not necessarily for all smooth varieties and indeed, not even for the affine line!]
3) Take $K = \mathbb{C}$. Then the analytification functor induces an isomorphism from the
profinite completion of the topological fundamental group of $X(\mathbb{C})$ (with the $\mathbb{C}$-analytic topology) to the etale fundamental group $\pi\_1(X \otimes \mathbb{C})$.
4) If $X/\mathbb{Q}$ is any Grassmannian, then it is nonsingular, complete and its analytification is the usual complex Grassmannian, which is simply connected: see e.g.
<http://books.google.com/books?id=WHjO9K6xEm4C&pg=PA748&lpg=PA748&dq=simply+connected+Grassmannian&source=bl&ots=waYTv_whVx&sig=ErlPHYKL5FdQPUdBlIrIgYQGhbE&hl=en&ei=9FiITJuNNYjW9ASP_p3fDg&sa=X&oi=book_result&ct=result&resnum=6&ved=0CC8Q6AEwBTgK#v=onepage&q&f=false>
Therefore putting the previous parts together the geometric fundamental group $\pi\_1(\overline{X})$ is trivial, so we naturally have $\pi\_1(X) \cong \mathfrak{g}\_{\mathbb{Q}}$.
References for these facts may be found, for instance, in Chapter 5 of Szamuely's *Galois Groups and Fundamental Groups*.
| 15 | https://mathoverflow.net/users/1149 | 38142 | 24,501 |
https://mathoverflow.net/questions/38098 | 12 | Often when people write about the geometrization conjecture they assume (for simplicity) that the manifold is orientable. I never seriously thought of non-orientable 3-manifolds, but while reading Morgan-Tian's [paper](https://arxiv.org/abs/0809.4040) I realized that they prove the geometrization for every compact 3-manifold $M$ with no embedded two-sided projective planes, which was news to me. There is a lot of notation in their paper so I ask
**Question 1.** Is the above a correct interpretation of what is proved in Morgan-Tian's paper?
Note: one way to rule out two-sided projective planes is to assume that $\pi\_1(M)$ has no 2-torsion (because then a two-sided projective plane would lift to the orientation cover where it cannot exist). In particular, this gives the geometrization conjecture when $M$ is aspherical (in which case $\pi\_1(M)$ is torsion free).
**Question 2.** What is the status of the geometrization conjecture for manifolds that contain two-sided projective planes?
Note: on the last two pages of Scott's wonderful survey ["The Geometries of 3-manifolds"](http://www.math.lsa.umich.edu/~pscott/8geoms.pdf) he describes a version of the geometrization conjecture that makes sense in the presence of two-sided projective planes.
UPDATE:
1. Looks like my reading of Morgan-Tian was hasty, and I no longer think they prove the geometrization for non-orientable manifolds that contain no two-sided projective planes. They only prove it for manifolds that become extinct in finite time under Ricci flow.
2. As we discussed with Ryan in comments the geometrization for manifolds that contain two-sided projective planes reduces to geometrization of certain non-orientable orbifolds with isolated singular points. However, in contrast with what Ryan says, I was unable to find the geometrization for such orbifolds in the literature. Again, lots of particular cases are known, but I could not find it claimed (let alone proved) in full generality.
| https://mathoverflow.net/users/1573 | Geometrization for 3-manifolds that contain two-sided projective planes | Most 3-manifold topologists tend to hypothesize away 2-sided projective planes. If a 3-manifold contains a 2-sided projective plane, then it must be non-orientable, and the preimage of the projective plane in the orientable double cover must be essential. Cutting along a maximal collection of disjoint essential embedded 2-spheres in the orientable cover (which may be made equivariant by [Meeks-Yau](http://www.ams.org/mathscinet-getitem?mr=595203)), and capping off with spheres, one obtains an irreducible 3-manifold with an orientation reversing involution. Corresponding to projective planes will be spheres coned off which are acted on by the involution as the suspension on the antipodal map. If the components of this 3-manifold are atoroidal, then this is covered by a result of [Dinkelbach and Leeb](https://arxiv.org/abs/0801.0803%20) for spherical, hyperbolic, and $S^2\times R$ metrics.
The case of the other homogeneous metrics was taken care of before by Meeks and Scott.
As far as I can tell, the toroidal case may still be open in general. One may argue that the involution preserves the JSJ tori, so it preserves the decomposition into geometric pieces. But I'm not sure it is now completely proven that the action on each geometric piece is standard, so that the quotient is geometric on each JSJ piece, because the results quoted above only cover involutions on closed geometric manifolds.
A [theorem of Hatcher](http://www.ams.org/mathscinet-getitem?mr=420620) implies that it is isotopic to such an involution, but I'm not sure his theorem implies that the isotopy is achieved by a path of involutions (with fixed points).
Kleiner and Lott are working on a proof of the orbifold theorem using Ricci flow, and I think their argument ought to give a unified argument for all of these things.
For aspherical non-orientable 3-manifolds, I think the proof via Ricci flow still works, even though Perelman didn't formulate it this way. Or Thurston's original proof should work, since these are Haken manifolds.
| 12 | https://mathoverflow.net/users/1345 | 38146 | 24,502 |
https://mathoverflow.net/questions/38145 | 4 | A self-affine tile is a compact set $T$ in $\mathbb R^n$ of positive Lebesgue measure for which there is an $n\times n$ expanding matrix $A$ (i.e. all its eigenvalues have modulus greater than 1) such that the affinely inflated copy $A(T)$ of $T$ can be perfectly tiled with essentially disjoint translates of $T$.
Thus we have $$ A(T) = \cup\_{i=1}^m (T+d\_i); \mathcal D= d\_1,d\_2,\dots,d\_m $$
where $|det(A)| =|\mathcal D|= m$
Results of Kenyon (Projecting the one-dimensional Sierpinski gasket Projecting the one-dimensional Sierpinski gasket.
Israel J. Math. 97 (1997), 221--238.) and Lagarias and Wang (Self-affine tiles in $R^n$. Adv. Math. 121 (1996), no. 1, 21--49) tells that such sets always can be used to give a translational tiling of $R^n$ and has boundary of measure zero and has nonempty interiors.
Thus in one dimension we can think of them as a union of intervals (possibly infinitely many ).
My question is :-
Is there a characterization of self-affine tiles in $\mathbb R$ which are union of finitely many intervals ?
| https://mathoverflow.net/users/6766 | A question about self-affine tiles | The classification is given in section 5 of ["Integral Self-Affine Tiles in $\mathbb R^n$ I. Standard and Nonstandard Digit Sets"](http://jlms.oxfordjournals.org/content/54/1/161.abstract) by Lagarias and Wang (Theorem 5.2 and corollary 5.2a). Their result builds on the previous paper by A. M. Odlyzko, "Non-negative digit sets in positional number systems".
| 2 | https://mathoverflow.net/users/2384 | 38153 | 24,507 |
https://mathoverflow.net/questions/38155 | 0 | Can every solution of a homogeneous linear system be approximated by a solution in rational numbers?
In mathematical terms: Let $$Ax=0$$ be a homogeneous linear system in $n$ determinates for an $m\times n$-matrix $A$ (possibly $m>n$) with integer entries (say all entries $1,0,-1$ for simplicity).
Given a solution $x\in {\Bbb R}^n$ and $\epsilon>0$, do there exist solutions in ${\Bbb Q}^n$ within distance $< \epsilon$ from $x$?
I am sure this kind of question has been considered somewhere. However, as a topologist, I have no idea where to look this up. Apart from answers also hints to literature about this genre of questions would be appreciated.
| https://mathoverflow.net/users/39082 | Rational solutions of homogeneous equations | Since A has integer entries, putting it in reduced row-echelon form shows that the solution-space is spanned by vectors with rational coordinates. Rational multiples of the spanning vectors are then dense in the solution-space, so vectors with rational coordinates are also dense in the solution-space. Therefore every solution of a homogeneous linear system can be approximated by a solution in the rational numbers. QED
| 4 | https://mathoverflow.net/users/nan | 38156 | 24,509 |
https://mathoverflow.net/questions/38160 | 15 | By Tennenbaum's theorem, PA itself does not have any computable nonstandard models. The integer polynomials which are 0 or have a positive leading coefficient form a computable nonstandard model of Robinson arithmetic, which also happens to make the order relation total. Since Presburger arithmetic is decidable, we can add axioms giving it a nonstandard number and work through Henkin's proof of the completeness theorem to get a computable nonstandard model of Presburger arithmetic. (There's probably a simpler way to get one, though.)
Is any system strictly weaker than PA known to have no computable nonstandard models?
What other systems weaker than PA are known to have computable nonstandard models?
.
possible examples of either include:
I-Delta-0, I-Delta-0(exp), I-Sigma-1
Elementary Function Arithmetic
Elementary Recursive Arithmetic, Primitive Recursive Arithmetic
Robinson arithmetic + Euclidean division, Robinson arithmetic + Euclidean division + order relation is total
| https://mathoverflow.net/users/nan | Computable nonstandard models for weak systems of arithemtic | One of the usual ways of proving Tennenbaum's theorem also
applies to many of the theories on your list, and so they
can have no computable nonstandard models.
The proof I have in mind is the following, which I also
explained in [this MO
answer](https://mathoverflow.net/questions/12426/is-there-a-computable-model-of-zfc/12434#12434).
Let $A$ be the set of Turing machine programs that halt on
input $0$ with output $0$, and let $B$ be the set of
programs that halt on input $0$ with output $1$. These sets
are disjoint and [computably
inseparable](http://en.wikipedia.org/wiki/Recursively_inseparable_sets),
meaning that there is no computable $C$ containing $A$ and
disjoint from $B$. Now, suppose that $M$ is a nonstandard
model of arithmetic. Let $d$ be a nonstandard natural
number, and inside $M$, consider the set of programs below
$d$ that halt on input $0$ in at most $d$ steps with output
$0$. In $M$, this is a (nonstandard) finite list, and so
there is a nonstandard number $c$ coding this list of
programs. Now, let $C$ be the set of standard programs $p$
that $M$ thinks appear on the list coded by $c$. This
includes every program in $A$, since all such programs halt
in a standard finite time with output $0$, and hence $M$
will agree that they halt before time $d$. Second, for a
similar reason, this set includes no programs in $B$, since
those programs halt in finite time with output $1$, and $M$
will see that. Finally, the set $C$ is computable from the
operations of $M$, since we need only perform the decoding
procedure to see if a given number $p$ is on the list coded
by $c$. For example, we might use the coding that would
require us merely to check whether $M$ thinks that the
$p^{th}$ binary digit of $c$ is $1$ or not. If the
operations of $M$ were computable, then this would be a
computable procedure, in contradiction to the fact that $A$
and $B$ are computably inseparable. QED
Now, we haven't really used much of PA in this argument.
Any theory $T$ that is able to perform basic Goedel coding
and simulate Turing machine computations will be sufficient
for the argument. This includes any $I\Sigma\_n$, even
$I\Sigma\_0$, since the operation of a Turing machine is
inductively iterating a very trivial process. So none of
the stronger theories on your list have computable
nonstandard models.
Meanwhile, however, I am unsure about the very weakest
theories on your list, but this argument reduces the
question to: can the given theory prove that for any number
$d$, there is a number $c$ coding the list of Turing
machine programs less than $d$ that halt on input $0$ with
output $0$ in at most $d$ steps?
This is a comparatively simple statement in arithmetic, and
any theory proving it will not have computable nonstandard
models.
| 11 | https://mathoverflow.net/users/1946 | 38162 | 24,511 |
https://mathoverflow.net/questions/38151 | 2 | How many different rectangles (in terms of area) can fit in a 20-unit-wide square? The rectangles can be squares, and their dimensions are integers.
| https://mathoverflow.net/users/9107 | How many different rectangles (in terms of area) can fit in a 20-unit-wide square? | If you're looking for the number of different areas realizable by fitting rectangles in a 20x20 square *with (integer-length) edges parallel to the coordinate axes*, the answer is the number of elements in {$ \{ x \times y | x,y \in \{ 1..20 \} \} $}. In Haskell, `length . List.nub . sort $ [x*y| x<-[1..20] , y<-[1..20]]` evaluates to 152.
| 0 | https://mathoverflow.net/users/5790 | 38172 | 24,519 |
https://mathoverflow.net/questions/38161 | 62 | Obviously there exists a list of the finite simple groups, but why should it be a nice list, one that you can write down?
[Solomon's AMS article](https://www.ams.org/notices/199502/solomon.pdf) goes some way toward a historical / technical explanation of how work on the proof proceeded. But, though I would like someday to attain some appreciation of the mathematics used in the proof, I'm hoping that there is some plausibility argument out there to convince the non-expert (like me!) that a classification ought to be feasible at all. A few possible lines of thought come to mind:
* Groups have very simple axioms. So perhaps they should be easy to classify. This seems like not a very convincing argument, but perhaps there is some way to make it more convincing.
* Lie groups have a nice classification, and many tools are available for their study and that of their finite analogues. And in fact, it turns out that almost all finite simple nonabelian groups fall under this heading. Is it somehow clear *a priori* that these should be essentially all the examples? What sort of plausibility arguments might lead one to believe this?
* If there are *not* currently any good heuristic arguments to convince a non-expert that a classification *should* be possible, then will this always be the case? Or will we someday understand things better...
There is probably a model-theoretic way to formalize this question. As a total guess, it might be something along the lines of "Do the finite simple groups have a finitely axiomatizable first-order theory?", except probably "finitely axiomatizable first-order theory" doesn't really capture the idea of a classification. If someone could point me towards how to formalize the idea of "classifiable", or "feasibly classifiable", I'd appreciate it.[FSGs up to order SE](https://math.stackexchange.com/questions/1423/number-of-finite-simple-groups-of-given-order-is-at-most-2-is-a-classificati)[FSGs up to order MO](https://mathoverflow.net/questions/34424/number-of-finite-simple-groups-of-given-order-is-at-most-2-is-a-classification)
**EDIT:**
To clarify, what I'd like is an argument that finite simple groups should be classifiable which does not boil down to an outline of the actual classification proof. Joseph O'Rourke asked on StackExchange [Why are there only a finite number of sporadic simple groups?](https://math.stackexchange.com/questions/2427/why-are-there-only-a-finite-number-of-sporadic-simple-groups). There, Jack Schmidt pointed out the work of Michler towards a uniform construction of the sporadic groups, as reviewed [here](http://www.ams.org/bull/2009-46-01/S0273.../S0273-0979-08-01215-9.pdf). Following the citation trail, one finds [a 1976 lecture by Brauer](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-1/issue-1/Blocks-of-characters-and-structure-of-finite-groups/bams/1183542332.full) in which he says that he's not sure whether there are finitely many or infinitely many sporadic groups, and which he concludes with some historical notes that describe a back-and-forth over the decades: at times it was believed there were infinitely many sporadic groups, and at times that there were only finitely many. So it appears that the answer to my question is **no**-- at least up to 1976, there was no evidence apart from the classification program as a whole to suggest that there should be only finitely many sporadic groups.
So I'd like to refocus my question: are such lines of argument developing today, or likely to develop in the (near? distant?) future? And has there been further clarification of what exactly is meant by a classification? (Is this too drastic a change? should I start a new thread?)
| https://mathoverflow.net/users/2362 | Heuristic argument that finite simple groups _ought_ to be "classifiable"? | It is unlikely that there is any easy reason why a classification is possible, unless someone comes up with a completely new way to classify groups. One problem, as least with the current methods of classification via centralizers of involutions, is that every simple group has to be tested to see if it leads to new simple groups containing it in the centralizer of an involution. For example, when the baby monster was discovered, it had a double cover, which was a potential centralizer of an involution in a larger simple group, which turned out to be the monster. The monster happens to have no double cover so the process stopped there, but without checking every finite simple group there seems no obvious reason why one cannot have an infinite chain of larger and larger sporadic groups, each of which has a double cover that is a centralizer of an involution in the next one. Because of this problem (among others), it was unclear until quite late in the classification whether there would be a finite or infinite number of sporadics.
Any easy way to get around this has been overlooked by about a hundred finite group theorists.
| 67 | https://mathoverflow.net/users/51 | 38174 | 24,521 |
https://mathoverflow.net/questions/37651 | 73 | I'm looking for explicit examples of Riemannian surfaces (two-dimensional Riemannian manifolds $(M,g)$) for which the distance function d(x,y) can be given explicitly in terms of local coordinates of x,y, assuming that x and y are sufficiently close. By "explicit", I mean things like a closed form description in terms of special functions, by implicitly solving a transcendental equation or (at worst) by solving an ODE, as opposed to having to solve a variational problem or a PDE such as the eikonal equation, or an inverse problem for an ODE, or to sum an asymptotic series.
The only examples of this that I know of are the constant curvature surfaces, which can be locally modeled either by the Euclidean plane ${\bf R}^2$, the sphere ${\bf S}^2$, or the hyperbolic plane ${\bf H}^2$, for which we have classical formulae for the distance function.
But I don't know of any other examples. For instance, the distance functions on the surface of the solid ellipsoid or solid torus in ${\bf R}^3$ look quite unpleasant already to write down explicitly. Presumably Zoll surfaces would be the next thing to try, but I don't know of any tractable explicit examples of Zoll surfaces that are not already constant curvature.
| https://mathoverflow.net/users/766 | Riemannian surfaces with an explicit distance function? | I'll briefly spell out what others have pointed to concerning geodesics on surfaces of revolution (or more generally, surfaces with a 1-parameter group of symmetries), because it's nice and not as widely understood as it should be.
Geodesics on surfaces of revolution conserve angular momentum about the central axis, so the geodesic flow splits into 2-dimensional surfaces having constant energy (~length) and angular momentum (The more general principle is that the inner product of the tangent to a geodesic with any infinitesimal isometry of a Riemannian manifold is constant). The surfaces are generically toruses. The shadow of these toruses on the surface of revolution is an annulus, a component of a set of $r \ge r\_0$, where on each point with $r > r\_0$ there are two vectors having the given angular momentum, but they merge at the boundary, both becoming tangent to the boundary of the annulus. If you sketch the picture, you will *see* the torus. The geodesics correspond to the physical phenomenon of the pattern of string or thread mechnically but passively wound around a cylinder. As string builds up in the middle, geodesics start to oscillate back and forth in a sinusoidal pattern, further amplifying the bulge in the middle.
To find the geodesic from point x to point y, you need to know which angular momentum will take you from x to y. For any two meridian circles and any choice of angular momentum, the geodesics of given angular momentum map one circle to the other by a rotation. Both the angle of rotation of the map and the length of the particular family of geodesics traversing the annulus is given by an integral over an interval cutting across the annulus, since the the slope of the vector field at all intervening points is known. I have an aversion to actual symbolic computation so I won't give you example formulas, but I believe this should meet your criterion for explicitness.
But to take a step back: this question, asking for an explicit formula, has an unstated (and probably unintended) connotation that is worth examining: this use of language implicitly suggests that non-symbolic forms are less worthy. I don't know the background motivation for the question, but an alternative question for some purposes would be to give example of surfaces where you can *exhibit* the distance function. Communication of mathematics is biassed toward symbolic forms. However, for many people and many purposes, some kind of graphical representation of the distance function, and/or diagrams or explanations of why it is what it is as well as a striaghtforward method for computing it, would often be better than a symbolic answer.
The geodesic flow of course is an ordinary differential equation. It is a vector field on the 3-manifold of unit-length tangent vectors to the surface, defined by very easy equations: the vectors are tangent to the surface, and their derivative (= the 2nd derivative of a geodesic arc) is normal to the surface. The solutions may not always have a nice symbolic form, but they always have a nice and easy-to-compute geometric form. Finding the distance involves the implicit function theorem, but this is easy and intuitive. One could, for instance, easily draw a parametric surface that is the graph of distance as a function of position directly from solutions to the ODE (which no doubt sometimes even have reasonable symbolic representations). Both the ODE for the geodesic flow and the inverse function to give distance as a function of position are easy to compute numerically, and easy to understand qualitatively.
| 53 | https://mathoverflow.net/users/9062 | 38183 | 24,525 |
https://mathoverflow.net/questions/38191 | -2 | Consider a markov chain matrix P of size n x n (n states).
P is known to be:
1- Not irreducible (i.e. there exist at least a pair of states i, j such that we cannot go from i to j)
2- Not all states are recurrent.
3- Aperiodic (the return to some states can occur at irregular times).
4- there are at least two absorbent states i,j (P\_i,i = P\_j,j = 1)
It is true that limit when n goes to infinity of P^n converges? Is this result well known or is the proof simple?
Thanks.
| https://mathoverflow.net/users/8021 | Convergence of a markov matrix | Suppose the answer is Yes. Then suppose we add two more states $i\ne j$ with $P\_{i,j}=1$ and $P\_{j,i}=1$, and no other state can go to states $i$ or $j$. Then for the new matrix the assumptions are still satisfied, but now the answer is No. Therefore the answer must be No.
| 0 | https://mathoverflow.net/users/4600 | 38194 | 24,530 |
https://mathoverflow.net/questions/38139 | 2 | I am having hard time to convert following set of differential equation to state space equation. I am a biologist and my math skills fall short as I don't know where to start. Any suggestion or feedback is highly appreciated. Thanks in advance.
**Updates**:
Based on comments I have updated the question (which is not a good practice, my sincere apologies to everyone),
I am modelling biological system with a cascade of signal transduction steps with feedback loops. Each level in the cascade has corresponding equation $\dot{y\_i}$ or $dy\_i/dt$ given by
\begin{equation\*}
\frac{dy\_i}{dt}= g\_i(v\_i, y\_i)+ c\_i + e\_i
\end{equation\*}
For example
\begin{equation\*}
\frac{dy\_0}{dt}= g\_0(a\_{00}y\_0+a\_{02}y\_2, y\_0)+ c\_0
\end{equation\*}
\begin{equation\*}
\frac{dy\_1}{dt}= g\_1(a\_{10}y\_0+a\_{11}y\_1+a\_{12}y\_2, y\_1)\\
\end{equation\*}
\begin{equation\*}
\frac{dy\_2}{dt}= g\_2(a\_{20}y\_0+a\_{21}y\_1+a\_{22}y\_2+a\_{23}y\_3+a\_{24}y\_4+a\_{25}y\_5, y\_2)+e\_2
\end{equation\*}
\begin{equation\*}
\frac{dy\_3}{dt}= g\_3(a\_{32}y\_2+a\_{33}y\_3, y\_3)
\end{equation\*}
\begin{equation\*}
\frac{dy\_4}{dt}= g\_4(a\_{42}y\_2+a\_{44}y\_4, y\_4)
\end{equation\*}
\begin{equation\*}
\frac{dy\_5}{dt}= g\_5(a\_{25}x\_2-a\_{25}x\_5)
\end{equation\*}
and so on. $y\_i$ represents state variable and only $y\_5$ can be observed ("output") others are hidden. $e\_2$ is input variable. $a\_{jk}$ represents parameters. Function $g\_i$ is given by multiplication of $h\_i$ and $r\_i$
\begin{equation\*}
g\_i({v\_i}, y\_i)= h\_i ({v\_i})\cdot r\_i({v\_i}, y\_i)\\
\end{equation\*}
where function $h\_i$ and $r\_i$ are given by
\begin{equation\*}
h\_i({v\_i}) = \begin{cases} \frac{{v\_i}}{1+\frac{{v\_i}}{S\_i}(1-exp(-{v\_i}/S\_i))} & \mbox{when } {v\_i}> 0, \cr {v\_i} & \mbox{when } {v\_i}\leq 0, \end{cases}
\end{equation\*}
\begin{equation\*}
r\_i({v\_i}, y) = \begin{cases} 1-exp(\frac{{v\_i}^2S\_i}{{v\_i}(\varepsilon -y)^2}) & \mbox{when } y<\varepsilon\ \&\ {v\_i}< 0, \cr 1 & \mbox{otherwise.} \end{cases}
\end{equation\*}
where
Function $h\_i$ puts physiologically relevant soft upper limit.
Function $r\_i$ ensures non-negative ligand concentration.
$\dot v$ is corresponding linear rate.
$S\_i$ is max size of the pool.
$\varepsilon$ is a small positive constant.
**Update-2:**
Ok, I want to know if I am doing right thing here.
I need state space equation for my system of differential equation with nonlinear function. Technically I can have two different versions
**A non-linear version**
\begin{equation\*}
\frac{dy\_i}{dt}= g\_i(v\_i, y\_i)+ c\_i + e\_i
\end{equation\*}
**A linear version**
\begin{equation\*}
\frac{dy\_i}{dt}= v\_i+ c\_i + e\_i
\end{equation\*}
For linear version I can write the state equation as
$y'= Ay+c+e$ ([see the matrix equation as Image](http://farm5.static.flickr.com/4150/4973551912_6b8a7c4d49.jpg))
Taking this to next step can I write like following?
$y'= g(Ay)+c+e$
| https://mathoverflow.net/users/9097 | Converting an ODE system to State space formulation | Ok, let me give this a shot:
* Because your system is nonlinear, I'm assuming you want the nonlinear state-space form. You can easily get the linear form by doing a Taylor series expansion on it around some equilibrium point.
* The fact that most of the states are not measurable is not a big problem. You can estimate them using your output variables (subject to observability conditions), using an state observer such as a Kalman filter, Moving Horizon Estimator (MHE) or a Luenberger observer. Also note for a nonlinear system, only local observability can be checked.
* Because you have conditional statements, I don't believe you'll be able to write the above as a single state-space system. You have 4 conditions, but they can be reduced to 3 disjunctions, so you'll need 3 state-space systems and a conditional switching equation that "activates" the correct state-space system depending on the values of $v$ and $y$. This is known as a hybrid (or switched) system.
* As to the treatment of a hybrid system, perhaps you could clarify what the purpose is of getting your model into state-space form. Is it for simulation reasons? Do you need it in order to do analysis (i.e. controllability, observability)? Or do you need to do optimization? If it is the last case, you can write the logic as a disjunctive program, which will allow you generate a very efficient mixed-integer programming (MIP) problem.
Anyway, this is one way of writing your state-space system:
$$
\begin{align}
\frac{dy\_{i}}{dt} &= g\_{i}^{m}(v\_{i},y\_{i}) + p\_{i} c\_{i} + q\_{i} e\_{i},\quad i=0,\ldots,N-1\\
v\_{i} &= \sum\_{j=0}^{N-1} a\_{ij}y\_{j},\quad i=0,\ldots,N-1
\end{align}
$$
where $p\_{i},q\_{i} \in \{0,1\}$ = coefficients, $N$ = number of states, and $m \in \{1,2,3\}$ = modes of the system. In addition, you will need a switching function $T(m,v\_{i},y\_{i}) = 0$ to select the appropriate modes based on the current states of your system. This can be done programmatically through `IF-THEN-ELSE` clauses (or via integer variables in an optimization problem).
* For $m = 1$ (where $v\_{i} < 0, y\_{i} < \varepsilon$):
$$g\_{i}^{1}(v\_{i},y\_{i}) = v\_{i} \left[1-\exp\left(\frac{{v\_i}^2S\_i}{{v\_i}(\varepsilon -y\_{i})^2}\right)\right]$$
* For $m = 2$ (where $v\_{i} \leq 0, y\_{i} \geq \varepsilon$):
$$g\_{i}^{2}(v\_{i},y\_{i}) = v\_{i}$$
* For $m = 3$ (where $v\_{i} > 0, y\_{i} \in \mathbb{R}$):
$$g\_{i}^{3}(v\_{i},y\_{i}) = \frac{{v\_i}}{1+\frac{{v\_i}}{S\_i}(1-\exp(-{v\_i}/S\_i))}$$
| 2 | https://mathoverflow.net/users/7851 | 38197 | 24,531 |
https://mathoverflow.net/questions/38188 | 6 | Is there an easy example of a (closed) hyperbolic 3-manifold that fibers over the circle but contains some totally geodesic surface?
(Of course such manifolds exist if the 'Virtually Fibered Conjecture' were correct, since a geodesic surface lifts to the fibered cover. But is there something more eplicit?)
| https://mathoverflow.net/users/39082 | Totally geodesic surfaces in fibered 3-manifolds | There are many specific known examples. Here is one construction:
Start with the 3-torus $T^3$, parametrize in the standard way as $R^3/Z^3$. It fibers over the circle in many ways.
Let $a$, $b$ and $c$ be three disjoint circles, coming form lines parallel to the x, y and z axes.
For most fibrations, these three circles are transverse to the fibers.
Form a branched cover of the torus with two-fold branching over all preimages of these 3 circles. The resulting manifold has a hyperbolic structure that can be constructed from right-angled hyperbolic dodecahedra, and is commensurable with the 4-fold branched cover of $S^3$ over the Borromean rings. You can think of it this way: you can take a unit cube as fundamental domain for the torus, and arrange that a, b and c lie on faces of the cube, each bisecting a pari of (glued together) opposite facce. This induces a subdividision of the boundary of the cube into what look like rectangles, but are really pentagons.
The map (x,y,z) -> x+y+z gives a fibration over the torus, also works for any branched cover as described. The preimage of any face of the cube is an extended face plane of a dodecahedron, and is always a totally geodesic immersed surface, but it splits into two embedded surfaces for suitable branched covers of $T^3$ (perhaps the one you first come up with.)
The tiling of hyperbolic space by right-angled dodecahedra has a cameo appearance in the video "Not Knot" we made at the Geometry Center, available together with "Outside In" on DVD from [AKPeters](http://akpeters.com/searchresults.asp?type=quick&keywords=Not+Knot&btnSubmit.x=0&btnSubmit.y=0). In the 1984 Scientific American Article *The Mathematics of three-dimensional manifolds* that Jeff Weeks and I wrote, a manifold in this family (constructed from right-angled hyperbolic dodecahedra and having the properties you asked for) was described as the configuration space of a mechanical linkage. I don't think these particular properties were pointed out in Scientific American.
This and other examples that are counterintuitive at first were a good part of my motivation when I raised the question whether all hyperbolic 3-manifolds virtually fiber over the circle, which at the time was a radical idea.
| 24 | https://mathoverflow.net/users/9062 | 38206 | 24,537 |
https://mathoverflow.net/questions/38193 | 34 | For simplicity, let me pick a particular instance of Gödel's Second Incompleteness
Theorem:
ZFC (Zermelo-Fraenkel Set Theory plus the Axiom of Choice, the usual foundation of mathematics) does not prove Con(ZFC), where Con(ZFC) is a formula that expresses that
ZFC is consistent.
(Here ZFC can be replaced by any other sufficiently good, sufficiently strong set of axioms,
but this is not the issue here.)
This theorem has been interpreted by many as saying "we can never know whether mathematics is consistent" and has encouraged many people to try and prove that ZFC (or even PA) is in fact inconsistent. I think a mainstream opinion in mathematics (at least among mathematician who think about foundations) is that we believe that there is no problem with
ZFC, we just can't prove the consistency of it.
A comment that comes up every now and then (also on mathoverflow), which I tend to agree with, is this:
(\*) "What do we gain if we could prove the consistency of (say ZFC) inside ZFC? If ZFC were inconsistent, it would prove its consistency just as well."
In other words, there is no point in proving the consistency of mathematics by a mathematical proof, since if mathematics were flawed, it would prove anything, for instance its own non-flawedness.
Hence such a proof would not actually improve our trust in mathematics (or ZFC, following the particular instance).
Now here is my question: Does the observation (\*) imply that the only advantage of the Second Incompleteness Theorem over the first one is that we now have a specific sentence
(in this case Con(ZFC)) that is undecidable, which can be used to prove theorems like
"the existence of an inaccessible cardinal is not provable in ZFC"?
In other words, does this reduce the Second Incompleteness Theorem to a mere technicality
without any philosophical implication that goes beyond the First Incompleteness Theorem
(which states that there is some sentence $\phi$ such that neither $\phi$ nor $\neg\phi$ follow from ZFC)?
| https://mathoverflow.net/users/7743 | Interpretation of the Second Incompleteness Theorem | For the philosophical point encapsulated in (\*) in the question, it seems that corollaries of the second incompleteness theorem are more relevant than the theorem itself. If we had doubts about the consistency of ZFC, then a proof of Con(ZFC) carried out in ZFC would indeed be of little use. But a proof of Con(ZFC) carried out in a more reliable system, like Peano arithmetic or primitive recursive arithmetic, would (before Gödel) have been useful, and I think this is what Hilbert was hoping for. Gödel's second incompleteness theorem tells us that this sort of thing can't happen (unless even the more reliable system is inconsistent).
| 32 | https://mathoverflow.net/users/6794 | 38210 | 24,539 |
https://mathoverflow.net/questions/38219 | 17 | As I have been studying algebraic topology, something that I found puzzling was the existence of finite homotopy groups. For instance, $\pi\_{4}(S^{2})\cong\pi\_{5}(S^{4})\cong\mathbb{Z}/2\mathbb{Z}$. I was wondering if there was any kind of intuitive reason for why this might be true, and if there are spaces $X$ such that $\pi\_{1}(X)$ is finite. Speaking very roughly, it would seem that a finite, nontrivial fundamental group means that if you repeat a closed path enough times, it can be contracted to a point, something which I find rather hard to visualize. So the question is: Is there any intuitive reason for the existence of finite homotopy groups?
| https://mathoverflow.net/users/6856 | Intuition on finite homotopy groups | The simplest (to understand) case of finite $\pi\_1$ is the group $SO\_3$. This can be illustrated using an arm or a belt! $SO\_3$ is the group of rotations in space and a based loop in $SO\_3$ can be thought of as a description of the motion of an object in such a way that it ends up back where it started. By attaching a strip of paper to the object, it's possible to see this path in space. For example, taking a belt and holding one end fixed whilst moving the other, or moving your hand (your arm forms the "strip").
So: hold your hand out in front of you, this is easiest if you do it palm-up. Keeping it palm-up, rotate it *under* your arm back to where you started. Your arm is now twisted (hopefully not too badly). Continue moving your hand **in the same direction** and with your palm facing up but this time over the top of your arm. When your hand gets back to where it started, your arm is now magically untwisted! So **two** times round the loop gets you back to an untwisted state, thus $2\gamma = 0$. But $\gamma \ne 0$ as evidenced by your twisted arm at the half-way stage.
If you find this difficult to do, here's an alternative way using a belt. Take a belt and twist it once (that is, hold it out straight and imagine an axis along its length, then twist one end all the way around). Now try to straighten it *without* twisting either end (though you can move either end in space). Can't be done. But if you twist the belt *twice* then it can.
(There's some funky you-tube videos showing the arm twists. If you get really good at it, you should do it with a beaker of water.)
| 24 | https://mathoverflow.net/users/45 | 38220 | 24,543 |
https://mathoverflow.net/questions/38141 | 7 | Suppose $K$ is a field endowed with a non-archimedian absolute value. Assume $K$ has characteristic $p>0$ and that $[K:K^p] < \infty$. Let $L$ be the completion of $K$ with respect to this absolute value. Is it always true that $[L:L^p] = [K:K^p]$?
| https://mathoverflow.net/users/9099 | Degree of $[K:K^p]$ and Completion | In fact, BCnrd's comment says that for dvr's, only for non-excellent one's $[L:L^p]\ne [K:K^p]$ can happen. Actually, suppose $d=[K:K^p]$ is finite. Consider the canonical map $K\otimes\_{K^p} L^p \to L$. The source is a $L^p$-vector space of dimension $d$, its image is therefore closed ($L^p$ is complete) and contains $K$, so the map is surjective. Therefore $[L:L^p]=d$ if and only if the above map is an isomorphism, or equivalently if $K$ and $L^p$ are linearly disjoint over $K^p$. Extracting $p$-th root in all these fields, this is also equivalent to $K^{1/p}$ and $L$ linearly disjoint over $K$. Which is also equivalent to $L$ separable (i.e. geometrically reduced) over $K$. This is the definition for the dvr of $K$ to be excellent.
Example of non-excellent dvr with finite $[K : K^p]$: Let $s$ be an element of $\mathbb F\_p((t))$ transcendent over $\mathbb F\_p(t)$, let $K=\mathbb F\_p(t,s)⊂\mathbb F\_p((t))$ be endowed with the $t$-adic valuation. Then $[K:K^p]=p^2$ and $[L:L^p]=p$. If we take any number of algebraically independent elements $s\_1,...,s\_n$ (instead of just one $s$) over $\mathbb F\_p(t)$, then we have $[K:K^p]=p^{n+1}$.
| 8 | https://mathoverflow.net/users/3485 | 38229 | 24,550 |
https://mathoverflow.net/questions/37739 | 8 | For a large enough $n$, and a parameter $ m $ I'm looking for a subset of the prime numbers in the range $[n,2n]$ with a unique structure. I am looking for a prime $p$ and a set of $m$ positive (not necessary different) integers: $\Delta\_1,\Delta\_2,\ldots,\Delta\_m$ such that the following set consists only of prime numbers in the range $[n,2n]$:
$$\{{ p+\sum\_{i=1}^{m} a\_i \cdot \Delta\_i} \mid a\_i \in \{0,1\} \}$$
For example, for $n=10$ and $m=2$ there is such a set with parameters:
$$p=11,\quad \Delta\_1 = 2,\quad \Delta\_2 = 6$$
And the set is:
$$\{11,13,17,19\}
$$
The question is how large can $m$ be asymptotically?
Note that an arithmetic progression of $m+1$ primes is a cube with all $m$ deltas having the same value.
I've managed to prove that $m$ could be $\frac{\log\log(n)}{2}$ though I'm sure my bound is far from tight.
| https://mathoverflow.net/users/9017 | Boolean Cube of Primes | The argument that gives you cubes in dense sets shows roughly speaking (via repeated applications of Cauchy-Schwarz) that the number of k-dimensional cubes in a set of density delta is at least something around $\delta^{2^k}n^{k+1}$, which is the number you would get in a random set. (I am in fact giving the result for the integers mod n, but one can think of a subset of the first n integers of density δ as a subset of $\mathbb{Z}\_{2n}$ of density δ/2. So this says that the best dimension should be that k for which $\delta^{2^k}$ is around $n^{-(k+1)}.$ Taking logs twice, that says (ignoring constants) that $k+\log\log(1/\delta)=\log k+\log\log n,$ or roughly $k=\log\log n - \log\log(1/\delta).$ If $\delta=1/\log n,$ then that second term is not making much difference.
The worst case is for random sets. Although the primes are not random, one can make them more random, and denser, by applying the so-called W-trick (roughly speaking, you restrict to an arithmetic progression that contains no multiples of small primes, thereby increasing the chances that a number in that progression is prime). But this does not increase the density by enough to have a significant effect on the estimate you might get for k. If you're interested in $2^k$, then the question becomes more delicate and the W-trick might get you a useful, if smallish, improvement. But basically it looks to me as though your $\log\log n$ estimate for the dimension should be right, up to a constant.
| 2 | https://mathoverflow.net/users/1459 | 38233 | 24,553 |
https://mathoverflow.net/questions/38187 | 1 | Is there a complete "infinite mixture of gaussians representation" for densities? What I mean is, is there, for any reasonably big class of densities $\phi(x)$ I can come up with a function $c(\mu, \tau)$ such that:
$\displaystyle\phi(x) = \int d\mu d\tau\; c(\mu, \tau) \exp\left(-\frac{\tau (x-\mu)^2}{2}\right)$??
($\tau = \frac{1}{\sigma}$ is the inverse of the variance and the normalization of the gaussian was absorbed in $c(\mu,\tau)$)
I'm using the word "complete" in the sense theoretical physicists talk about the completeness of a basis for a vector space. I'm not sure if it's an adequate use of the term for mathematicians (probably not).
If yes, does it generalize for multivariate distributions?
Also, is there a known "inverse integral transform" for this representation? Something like:
$\tilde{\phi}(\Sigma, \mu) = \int dx \; \tilde{\mathcal{N}}(\Sigma, \mu | x) \phi(x)$
$\phi(x) = \int d\Sigma \mathcal{N}(x | \Sigma, \mu) \tilde{\phi}(\Sigma,\mu)$
where $\Sigma$ and $\mu$ are the covariance matrix and the vector of means.
| https://mathoverflow.net/users/757 | Completeness of an "infinite mixture of gaussians" representation | How about this: the set of finite mixtures of (non-degenerate) Gaussians is weakly dense in the space of probability measures on $\mathbb{R}$. Proof: the set of finite mixtures of constants is certainly weakly dense. But a constant can be approximated as a Gaussian with small variance.
| 1 | https://mathoverflow.net/users/4832 | 38242 | 24,559 |
https://mathoverflow.net/questions/36673 | 2 | Shishikura (1991) proved that the Hausdorff Dimension of the boundary of the Mandelbrot set equals 2, in [this paper](http://arxiv.org/abs/math/9201282), but I can't figure out one thing : can we say **all open subsets** of this boundary has dimension 2 ? Are there any references ? Thanks.
| https://mathoverflow.net/users/8779 | Hausdorff dimension of subsets of the Mandelbot set. | Yes - and it is indeed in Shishikura's paper, as in your comment.
The "open subsets" of the boundary with respect to which topology? That of the ambient space, or the relative topology of the set itself? Since the latter topology is extremely wild, it is much clearer to use the ambient topology, as Shishikura does.
| 2 | https://mathoverflow.net/users/3993 | 38247 | 24,562 |
https://mathoverflow.net/questions/38262 | 5 | (This is related to my question at [Computable nonstandard models for weak systems of arithemtic](https://mathoverflow.net/questions/38160/computable-nonstandard-models-for-weak-systems-of-arithemtic) )
Is there a nontrivial computable discrete ordered ring with Euclidean division that is not isomorphic to Z?
If so, what other first-order properties could it share with Z?
.
Possible properties include:
all numbers with rational square roots are perfect squares
Lagrange 4-square theorem
prime elements are unbounded
.
Could it satisfy all Pi\_1 properties satisfied by Z?
(bounded quantifiers referring to the absolute value being bounded)
| https://mathoverflow.net/users/nan | Computable rings similar to Z | Berarducci and Otero in "A Recursive Nonstandard Model of Normal Open Induction" (Journal of Symbolic Logic v61, 1996)
give a discretely ordered ring $R$ with recursive operations having the following properties:
1. $R$ is integrally closed in its quotient field. (So elements with "rational" square roots are perfect squares.)
2. The prime elements of $R$ are cofinal
3. $R$ satisfies the induction axioms for quantifier-free formulas
In an earlier paper of Otero (Journal of Symbolic Logic, vol 55, 1990) Otero proves that every model of Open Induction extends to one in which Lagrange's Four-Square Theorem holds. Whether this can be done effectively I don't know.
As for the general question "What properties can a recursively presented discretely ordered ring share with $\mathbb{Z}$": Let's say that a discretely ordered ring $R$ is "diophantine correct" if it satisfies all universal sentences that hold in $\mathbb{Z}$. Assuming that the language of rings has signature $(+, -, \times , \le,0 ,1)$, diophantine correctness amounts to the requirement that any system of polynomial equations and inequalities that are solvable in $R$ is solvable in the standard integers. Incidentally, models of open induction satisfy a weaker property: Any system of equations solvable in some (at least one) model of open induction has a p-adic solution.
The question whether a nonstandard diophantine correct model of open induction can be effectively constructed was raised by Adamowic and Morales-Luna in "A Recursive Model for Arithmetic with Weak Induction", (Journal of Symbolic Logic v50, 1985). I believe that this question is still open. I also believe that the models constructed in the Otero-Berarducci paper are in fact diophantine correct, but the proof of this seems to bump up against open problems in number theory. This is all discussed in an article on diophantine correct open induction by Sidney Raffer in "Set theory, Arithmetic, Philosophy: Essays in Memory of Stanley Tennenbaum (edited by J. Kennedy and R. Kossak), Cambridge University Press. (To appear.)
This is a response to a question from the comments: It is too long to fit there.
Proof of the Euclidean Division Theorem from the axioms of Open Induction:
The problem is to show that if $A$ is a model of open induction and if $x,y$ are elements of $A$ with $y>0$ and $x \ge 0 $ then there are unique elements $r, q$ of $A$ such that $x=yq+r$ and $0 \le r < q$. (The statement actually holds for all $x$ and the proof for $x < 0 $ is similar.)
1. First show that for every $x\in A$ with $x\ge 0$, there is some $q\in A$ such that $yq\le x < y(q+1)$. (Suppose, by way of by contradiction, that $x \ge 0$ and there is no such $q$. Let $S$ be the subset of $A$ defined by the quantifier-free formula $\sigma(q): q \ge 0 \wedge yq\le x$. Show that $S$ contains 0 and is closed under successor. By induction $S$ contains every positive element of $A$. But this is impossible because $x+1$ cannot be in $S$.)
2. Next, given $y > 0$ and $x \ge 0 $ choose $q$ (as in Part 1) such that $yq \le x < y(q+1)$. Put $r=x-yq$. Using that fact that $A$ is discretely ordered it is follows easily that $r$ and $q$ are the unique elements of $A$ satisfying $x=yq+r$ and $0 \le r < q$.
| 6 | https://mathoverflow.net/users/5229 | 38265 | 24,576 |
https://mathoverflow.net/questions/38264 | 3 | I'm seeking a function which belongs to $W^{1,p}(\Omega)$ for $p < n$ which is *not differentiable a.e*. There is a standard theorem which shows that if $p > n$ then in fact any function in $W^{1,p}$ is differentiable a.e. I would like an example where a weak derivative exists, lies in $W^{1,p}$ for $p < n$ but fails to be differentiable a.e.
| https://mathoverflow.net/users/8755 | A function in $W^{1,p}(\Omega)$ for $1 < p < n$ which is not differentiable a.e | Pick a number $\alpha$ with $0<\alpha< n/p-1$ and a smooth, nonnegative function $g(r)$ defined for $r>0$ with $g(r)=r^{-\alpha}$ when $r$ is small, $g(r)=0$ when $r$ is large. Then $x\mapsto g(|x|)$ belongs to $W^{1,p}(\mathbb{R}^n)$. Write $$f(x)=\sum\_{i=1}^\infty 2^{-i}g(|x-q\_i|)$$ where $(q\_i)$ is a dense sequence in $\mathbb{R}^n$. The series converges in $W^{1,p}(\mathbb{R}^n)$ and the sum is unbounded in any neighbourhood of any $q\_i$, hence unbounded in any nonempty open set. Differentiability is rather hard to achieve under those circumstances.
(Edit: Need $g\ge0$.)
| 3 | https://mathoverflow.net/users/802 | 38273 | 24,582 |
https://mathoverflow.net/questions/38283 | 5 | I am trying to compute the asymptotic growth-rate in a specific combinatorial problem depending on a parameter $w$, using the Transfer-Matrix method. This amounts to computing the largest eigenvalue of the corresponding matrix.
For small values of $w$, the corresponding matrix is small and I can use the so-called power method — start with some vector, and multiply it by the matrix over and over, and under certain conditions you'll get the eigenvector corresponding to the largest eigenvalue. However, for the values of $w$ I am interested in, the matrix becomes too large, and thus the vector becomes too large, say, $n > 10^7$ entries or so, and it can't be contained in the computer's memory anymore and I need extra programming tricks or a very powerful computer.
As for the matrix itself, I don't need to store it in memory — I can access it as a black box, i.e., given $i, j$ I can return $A\_{ij}$ via a simple computation. Also, the matrix has only $0$ and $1$ entries, and I believe it to be sparse (i.e., only around $\log n$ of the entries are $1$'s, $n$ being the number of rows/columns). However, the matrix is **not** symmetric.
Is there some method more space-effective for computation of eigenvalues for a case like this?
| https://mathoverflow.net/users/9136 | Computing the largest eigenvalue of a very large sparse matrix | You could use the Arnoldi Iteration algorithm. This algorithm only requires the matrix A for matrix-vector multiplication. I'm expecting that you will be able to black-box the function v→Av. What you generate is an upper Hessenberg matrix H whose eigenvalues whose can be computed cheaply (by a direct method or Rayleigh quotient iteration) and which approximate the eigenvalues of A. Arnoldi Iteration will give the best approximation to the dominant eigenvalue so I suspect you won't have to do many iterations before you have a good estimate.
An excellent introduction to this is: "Numerical Linear Algebra" by Trefethen and Bau. (p250)
The basic algorithm can be found here: <http://en.wikipedia.org/wiki/Arnoldi_iteration>
Now the only thing that is required to make this a fully functional algorithm is a termination condition. Since you don't seem to need the dominant eigenvalue to a high degree of accuracy I would not worry and just stop when the dominant eigenvalue estimate doesn't change too much.
If you have Matlab you can always use the built in function eigs(Afun,n,...) where Afun is the black-box function handle that computes Av.
| 4 | https://mathoverflow.net/users/2011 | 38288 | 24,592 |
https://mathoverflow.net/questions/38238 | 43 | Recently I learned that there is a useful analogue of mathematical induction over $\mathbb{R}$ (more precisely, over intervals of the form $[a,\infty)$ or $[a,b]$). It turns out that this is an old idea: it goes back to Khinchin and Perron, but has for some reason never quite caught on and thus keeps getting rediscovered. A nice recent paper is
<http://alpha.math.uga.edu/~pete/Kalantari07.pdf>
[As a side note, I gave a talk two days ago in the VIGRE Graduate Seminar at UGA -- essentially, a colloquium for graduate students, with talks which are supposed to be accessible to first and second year students -- about induction over the real numbers with applications to snappy proofs of essentially all the basic "hard theorems" of honors calculus / elementary real analysis. It was a smashing success: the students both enjoyed stretching their minds around this new form of induction and appreciated the application to consolidated proofs of theorems which were, in their recent memeory, not so quick or easy to prove.]
The way I like to state the principle is a little different from Kalantari's approach. [I actually got confused by Kalantari's "axioms for induction" when I first saw them and was under the impression that they were wrong. In fact *I* was wrong, and he quickly responded to my email on the subject, setting me straight and, kindly, mentioning that others had made the same mistake.] Here's my preferred version:
Let $(X,\leq)$ be a totally ordered set with a least element, which we may as well call $0$.
Say that $X$ **has infima** if...every nonempty subset $S$ of $X$ has an infimum.
Say that $S \subset X$ is an **inductive subset** if all of the following hold:
(POI1) $0 \in S$.
(POI2) For all $x \in S$, if there exists $z \in X$ such that $x < z$ -- in other words, if $x$ is not a maximum element of $X$ -- then there exists $y > x$ such that the interval $[x,y]$ is contained in $S$.
(POI3) For all $x \in X$, if $[0,x) \subset S$, then $x \in S$.
Finally, we say that $X$ satisfies the **principle of ordered induction** if the only inductive subset of $X$ is $X$ itself.
Theorem: For a totally ordered set $X$ with a minimum element, TFAE:
(i) $X$ has infima.
(ii) $X$ satisfies the principle of ordered induction.
The proof is straightforward. Applying this to half-closed intervals as above gives real induction. Also, applying it to a well-ordered set recovers transfinite induction exactly as it is usually stated -- i.e., with an extra axiom for "limit elements", even though one could formally combine (POI2) and (POI3) into a single case.
This issue came up on Tuesday on our sister site math.SE -- someone asked whether there was such a thing as real induction -- and I answered it, **yes**, as above. Then someone commented on my answer: what about generalizations to partially ordered sets?
There is a well-known principle of induction on partially ordered sets satisfying the descending chain condition, or equivalently, in which every nonempty subset has a minimum. This is called, by mathematicians of various stripes, well-founded induction or Noetherian induction. (As far as I can see it should be called **Artinian induction**. ~~Anyone want to address that?~~ **Edit**: I am satisfied by Dave Anderson's comment below.)
Note that a partially ordered set with DCC need not have a minimum element, which is certainly necessary for the above setup in totally ordered sets. But this isn't a big deal: if $(X,\leq)$ is a poset satisfying DCC, the poset $X\_0$ obtained by adjoining a minimum element $0$ still satisfies DCC, and nothing is lost here.
[**Edit**: as Francois Dorais points out, adding a minimum is not enough; still a partially ordered set satisfying DCC need not have infima. So what I am asking about really is different from Noetherian induction.]
After a little thought I was optimistic that there should be a version of induction partially ordered sets with a minimum element. I even thought that the right definition of inductive subset should be essentially the one given above, with (POI2) modified slightly to
(POI2'): for every $x \in S$ and $z \in X$ with $x < z$, there exists $y \in (x,z]$ such that the entire interval $[x,y]$ is contained in $S$.
And then I tried to prove that any poset with a minimum and having infima satisfies the principle of ordered induction. And I couldn't. Eventually I found the following counterexample: let $A$ be an infinite set and $X = 2^A$ be its power set, partially ordered by inclusion. Of course $X$ has infima: take the intersection. Let $S$ be the collection of all finite subsets of $A$. Then $S$ satisfies (POI1), (POI2') and (POI3) but is proper.
>
> Neither can I think of some small modification of (POI2') which evades this example. I still think there should be *some kind* of principle of induction in partially ordered sets with infima, but I don't know what to do. Can anyone state such a principle which recovers ~~as special cases~~ as a special case the principle of ordered induction ~~and the principle of Noetherian induction~~?
>
>
>
| https://mathoverflow.net/users/1149 | A principle of mathematical induction for partially ordered sets with infima? | Something very close to François' conditions achieves the
desired if-and-only-if version of the theorem for partial
orders, providing an induction-like characterization of the complete partial orders, just as Pete's theorem characterizes the complete total orders.
Suppose that $(P,\lt)$ is a partial order. We say that it
is *complete* if every subset $A$ has a least upper bound
$sup(A)$ and a greatest lower bound $inf(A)$. This implies
that $(P,\lt)$ has a least and greatest element, and in
this case, it is easy to see that completeness is equivalent to the assertion that
every set has a greatest lower bound (the least upper bound of a set with upper bounds is the greatest lower bound of its upper bounds). The complete partial
orders are exactly the [complete
lattices](http://en.wikipedia.org/wiki/Complete_lattice).
For the purpose of this question, let us define that
$S\subset P$ is *inductive* if the following occur:
* (1) $S$ is downward closed: $y\lt x\in S\to y\in S$;
* (2) $S$ has no largest element, except possibly the largest element of $P$;
* (3) If $d=sup(A)$ for some $A\subset S$, then $d\in S$.
In (2), by a *largest* element, I mean an element that is
larger than all other elements (in distinction with
*maximal* element, a weaker concept). Conditions (2) and
(3) are both slightly weaker than François' conditions. The
difference in (2) is that we no longer assume that a
condition $x\in S$ can be extended in any particular
direction, and the difference in (3) is that we are not
assuming here that $P$ is complete, but only that $S$
contains the suprema of its subsets, when these suprema exist.
Lastly, define that a partial order $(P,\lt)$ has
*partial-order induction* if the only inductive set is all
of $P$.
**Theorem.** For any partial order $(P,\lt)$, the
following are equivalent:
* $(P,\lt)$ is complete.
* $(P,\lt)$ has least and greatest elements and satisfies partial-order induction.
Proof. For the forward implication, suppose $(P,\lt)$ is
complete. It follows that $(P,\lt)$ has least and greatest
elements, since these are the sup and inf of the emptyset.
Suppose that $S\subset P$ is inductive. By (3) we know
$sup(S)\in S$, which would make it the largest element of
$S$, contrary to (2), unless $sup(S)$ is largest in $P$, in which case
$S=P$ by (1). So $(P,\lt)$ has partial-order induction.
Conversely, suppose that $(P,\lt)$ has least and greatest
elements and satisfies partial-order induction. We want to
prove completeness. Suppose $B\subset P$ is nonempty and
has no greatest lower bound. Let $S$ be the set of lower
bounds of $B$. This is closed downwards, and so (1) holds.
Since $B$ has no greatest lower bound, it follows that $S$
has no largest element, and so (2) holds. Finally, if
$A\subset S$ and $sup(A)$ exists in $P$, then it remains a
lower bound of $B$, since every element of $B$ is an upper
bound of $A$ and $sup(A)$ is the least upper bound of $A$.
Thus, (3) holds and so $S$ is inductive, contrary to the
fact that it contains no elements of $B$. Thus, $B$ must
have a greatest lower bound after all, and so $(P,\lt)$ is
complete. QED
Note that when $(P,\lt)$ is a total order, then these
concepts reduce to the conditions Pete mentions in the
question (adding a least and greatest element if
necessary). Thus, this theorem seems to be the desired
generalization.
(I have deleted my other answer to the question, as it was
based on a misunderstanding of the question.)
| 15 | https://mathoverflow.net/users/1946 | 38292 | 24,595 |
https://mathoverflow.net/questions/38026 | 15 | Let $G = S\_n$ (the permutation group on $n$ elements).
Let $A\subset G$ such that $A$ generates $G$.
Is there an $n$-cycle $g$ in $G$ that can be expressed as
$g = a\_1 a\_2 ... a\_k$
where $a\_i\in A \cup A^{-1}$
and $k\leq c\_1 n^{c\_2}$, where $c\_1$ and $c\_2$ are constants?
What about $2$-cycles, or elements of any other particular form?
| https://mathoverflow.net/users/398 | Generating n-cycles | That all elements in a symmetric group with a specified arbitrary generating set can be reached in a polynomial amount of steps is a known open problem, and has been investigated for some time now. This long-standing conjecture has been proven for most choices of generating sets. Heuristically one expects this to be true for the reasons mentioned by Colin Reid above. That is, if certain conjectures are true, then every Cayley graph has a Hamiltonian cycle and so is expected to have exponentially many such cycles. So one expects to be able to reach any element of the group pretty fast.
For the symmetric group the problem of determining tight bounds on the diameter of its Cayley graph has been studied (and partially resolved) by L. Babai and coauthors. In the paper ["On the diameter of cayley graphs of the symmetric group"](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WHS-4D7CYNS-GC&_user=10&_coverDate=09%2F30%2F1988&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=fd830a2a697193261a6cacc86268a0c5&searchtype=a) it is proved that one can reach any elements using words of length at most $e^{\sqrt{n\log n}(1+o(1))}$ for any generating set, while the optimal bound should be $O(n^{c})$. [Here](http://portal.acm.org/citation.cfm?id=982956&dl=GUIDE&coll=GUIDE&CFID=101320652&CFTOKEN=30527812) and [here](http://portal.acm.org/citation.cfm?id=1070584) more partial results are proved, showing evidence that polynomial bounds are in fact the true asymptotic. I will remark again that one expects this behavior for all nonabelian finite simple groups too.
These papers are a good survey of what is currently known about this problem, I don't know if restricting your target to specific conjugacy classes of elements (such as n-cycles) makes the problem easier so I will think about it a bit more.
| 9 | https://mathoverflow.net/users/2384 | 38296 | 24,597 |
https://mathoverflow.net/questions/38304 | 1 | Let $F$ be a field, $n$ be a positive integer. Denote by $h\_{F}(n)$ the maximal dimension of a subspace $X\subset F^n$ such that $(x,y)=0$ for any two (not necessary distinct) vectors $x,y\in F^n$, where $(x,y)=x\_1y\_1+\dots+x\_ny\_n$ for $x=(x\_1,\dots,x\_n)$, $y=(y\_1,\dots,y\_n)$. For example, $h\_{\mathbb{R}}(n)=0$ for any $n$.
It is clear that $h\_F(n)$ is non-decreasing by $n$, that $h\_F(n)\leq n/2$ (such $X$ is contained in $X^{\perp}$, hence $\dim(X)\leq \dim(X^{\perp})=n-\dim(X)$) and that always $h\_{F}(n+k)\geq h\_F(n)+h\_F(k)$. It allows to get $h\_{F}(n)=[n/2]$ for $F=\mathbb{C}$ or $F=\mathbb{F}\_p$ with prime $p=4k+1$ or $p=2$.
Now take $p=4k+3$ and $F=\mathbb{F}\_p$.
It is easy to get $h\_{F}(2)=0$, $h\_{F}(3)=1$, $h\_{F}(4)=2$ (take span of $(a,b,c,0)$ and $(0,-c,b,a)$ with $a^2+b^2+c^2=0$). Hence $h\_{F}(4n)=2n$, $h\_{F}(4n+1)=2n$, $h\_{F}(4n+3)=2n+1$. But what about $h\_F(4n+2)$? If it equals $2n+1$ for some $n$, then also for all greater $n$. But does there always exist such $n$ and if it does exist, how to find it as a function of $p$? I managed only to observe by hands that $h\_{F}(6)=2$ for $p=3$.
| https://mathoverflow.net/users/4312 | maximal number of mutually orthogonal vectors | This is the question of finding maximal [isotropic subspaces](http://en.wikipedia.org/wiki/Isotropic_quadratic_form) of
an inner-product space. The results for finite fields of odd characteristic
are well-known and can be found in Serre's *Course in Arithmetic*.
Let's consider the quadratic form $Q=x\_1^2+\cdots+x\_n^2$.
When $p\equiv 3$ (mod $4$) the dimension of the maximum isotropic
subspace is $2k$ for $n=4k$, $4k+1$, $4k+2$ and $2k+1$ for $n=4k+3$.
To see that it isn't $2k+1$ for $n=4k+2$ we argue as follows.
If a nonsingular quadratic form has a dimension $r$ isotropic subspace then $V$
it has a subspace $W$ or dimension $2r$ on which the form restricts to
$y\_1 y\_2+y\_3y\_4+\cdots+\cdots y\_{2r-1}y\_{2r}$. If $r=n/2$ the discriminant
of this form is $(-1)^r$. But this is not the same as the discriminant of $Q$
(namely $1$) modulo squares if $r$ is odd (since $p\equiv 3$ (mod $4$).
| 2 | https://mathoverflow.net/users/4213 | 38306 | 24,603 |
https://mathoverflow.net/questions/38280 | 1 | Let $f$ and $g$ be two discrete signals. I want to find a monotone function h such that
$h=argmin\_{h}\sum\_{n\in[0,N]}{(f(n)-h(g(n)))^2}$
I don't really care about finding the global optimum, I just want a good fit. What would be a good representation of f to achieve that? Thanks!
| https://mathoverflow.net/users/180 | Finding an optimal monotone function? | Here is another try:
Assume w.l.o.g. that values of $g(n)$ are in increasing order, i.e. $g(0) \le g(1) \le \cdots\le g(n)$. Moreover, assume the first $n\_1$ values in that sequence are equal, then the next $n\_2$ values are equal, etc. and that there are $m$ distinct values
i.e.
$$g(0)=g(1)=\cdots=g(n\_1-1) < g(n\_1)=\cdots=g(n\_1+n\_2-1) < \cdots < g(n\_1+\cdots+n\_{m-1}) + \cdots g(n\_1+\cdots+n\_m-1).$$We have to decide the following $m$ values of $h$: $$h\_k = h(g(k)), \quad k=1,2,\ldots m$$
Say we are looking for an increasing $h$ (you can look for a decreasing $h$ in a similar way and take the best of the two solutions). Denote by $H\_r(t)$ the value of the solution of the problem limited to the first $r$ groups, where $t$ is an additional upper bound $t$ on the values taken by $h$, i.e.
$$H\_r(t) = \min\_{h\_1 \le h\_2 \le \cdots \le h\_r \le t} \quad \sum\_{k=1}^r \quad \sum\_{i=n\_1+\cdots+n\_{r-1}}^{n\_1+\cdots+n\_r-1} (f(i) - h\_k)^2$$
When $r=1$, the value of $H\_1(t)$ can now be easily computed: it is easy to check that the optimal $h\_1$ is equal to the average $\bar{f\_1}=\frac{1}{n\_1}(f(0)+f(1)+ \ldots+ f(n\_1-1))$ if it is less than $t$, and to $t$ otherwise, and that $$H\_1(t)=constant + n\_1 (\min (t-\bar{f\_1},0))^2 .$$From that we can successively compute the values of $H\_2(t)$, $H\_3(t)$, etc. and end up with the final solution when $r=n$, using the recurrence $$H\_{r}(t) = \min\_{s \le t}\ H\_{r-1}(s) + constant + n\_s(s-\bar{f\_r})^2$$ (this is similar to dynamic programming).
| 1 | https://mathoverflow.net/users/1184 | 38313 | 24,606 |
https://mathoverflow.net/questions/38305 | 4 | Hi all,
I would like to know if there is a way to compute some measure of similarity between two ordinary graphs with weighted edges. Graphs do not share vertices and can differ in number of vertices and edges.
Any hints, suggestions and thoughts are highly appreciated.
Best,
Jozef
| https://mathoverflow.net/users/9139 | Similarity of weighted graphs | If you view the weights as edge lengths then you can view each graph as a metric space, and then use the [Gromov-Hausdorff](http://en.wikipedia.org/wiki/Gromov%E2%80%93Hausdorff_convergence) distance between the two metric spaces. This may not be at all suitable for your application but it has been very useful in my own research.
| 9 | https://mathoverflow.net/users/3401 | 38316 | 24,608 |
https://mathoverflow.net/questions/38307 | 23 | At some point during my research I was confronted with this problem, but I did not dedicate serious time to it. Anyway it stayed in the back of my mind and I'm still interested in hints for it. Application: asymptotic properties of Schroedinger equations, scattering.
You have two convex (compact, smooth, everything) disjoint sets in the plane. Consider a ray starting in the complement of the two sets and bouncing on the boundary of the sets in the usual way, with the ingoing and outgoing rays forming equal angles with the normal to the boundary. Q.: does it always exist a trapped ray which never leaves a ball containing the two sets? This can happen of course if the ray keeps bouncing forever between the two bodies. A trivial example is obtained if the sets have two parallel sides, and the ray is chosen perpendicular to both. Less trivial examples (even strictly convex) can be constructed by choosing a trajectory first, and then joining the dots (i.e., the turning points of the trajectory) with convex curves; with some work and some adjustments in the trajectory, you can produce plenty of examples.
But, is this always the case? given two arbitrary bodies, does it always exist a trapped ray?
EDIT (see Pietro's comment): I mean, another trapped ray besides the 'trivial' trapped ray bouncing between the closest points of the two sets (a general version of the trivial case mentioned above of two parallel sides).
EDIT 2 (quick summary of the discussion for the benefit of future readers): the answer is yes for smooth boundaries and large (in particular, with nonempty interior) sets of initial points. A continuity argument is enough to prove this. If the boundary is non smooth problems may arise. E.g. for two polygons with a couple of facing parallel sides, the only trapped ray is the periodic one.
PS in retrospect, the question was quite elementary, but I really enjoyed to discuss it here :)
| https://mathoverflow.net/users/7294 | Trapped rays bouncing between two convex bodies | Yes, there is always a trapped ray. The simplest way to see it is to find the path between the two bodies that minimizes length. It is necessarily perpendicular to both surfaces.
EDIT: I see the question was edited to ask for more than this trivial answer, so the new answer: there is a unique trapped ray from any starting point, but it is not trapped in backward time unless it is on the shortest path between the bodies. One can find it by minimizing distance of a zig-zag path alternately touching the two bodies a finite number of times, then passing to a limit.
Here is a generalization: suppose you have a collection of smooth disjoint convex shapes $\{S\_i\}$ in the plane arranged in a way that no straight line intersects more than two. Then, for any doubly infinite sequence of indices $ \dots, i\_{-1}, i\_{0}, i\_{1}, \dots $ such that $i\_j \ne i\_{j+1}$, there is a unique trajectory that intersects the shapes in that order, starting with $S\_{i\_1}$ in the positive direction and $S\_{i\_0}$ going backward. If the sequence is periodic, you can find the trajectory just as for the case of two objects. For the infinite case, you can take limits.
Even if the shapes are not convex, as long as they are smooth the trajectories still exist, but they are not necessarily unique. If you want to say something about the case when the obstacles are not smooth, you can extend the rule to make it a non-deterministic dynamical system, where a ray hitting a corner has choices which way to go.
This kind of system is classical dynamical systems, which has been well-understood since early last century. Perhaps someone more knowledgable will supply appropriate references.
It is a limiting special case of the theory of the geodesic flow on surfaces of negative curvature.
In response to a comment, here is some more detail (that doesn't itself fit into a comment).
The question was about stability and how to prove convergence under the limiting process.
To prove existence, you don't need stability: just take a sequence of longer and longer rays, and choose a convergent subsequence. This exists because of compactness of the set of possible initial directions.
To prove uniqueness: this follows from the hyperbolicity of the flow. Think of the convex obstacles as trick mirrors that make you look skinny, cylinders with a convex cross-section. The convexity implies that reflected rays diverge at least as fast as they would from a flat mirror. Successive reflected images of the two mirrors in each other get thinner and thinner, so they narrow down to a unique point. (In the three-dimensional picture, they're also narrowing vertically, just at the relatively slow rate at which images shrink with distance in Euclidean space rather than at the exponential rate resulting from mirrors that are convex to 2nd order).
One way to formalize the discussion above is by use of triangle comparison theorems. Double the complement of the convex bodies to make a surface. The surface can be smoothly approximated by a surface of nonpositive curvature if it's comforting, but that's not technically necessary; the (intuitively obvious) statement about image sizes above become cases of the [Toponogov comparison theorem](http://en.wikipedia.org/wiki/Theorem_of_Toponogov).
| 34 | https://mathoverflow.net/users/9062 | 38320 | 24,612 |
https://mathoverflow.net/questions/38324 | 26 | In 1977, [Henry Pogorzelski](http://en.wikipedia.org/wiki/Henry_Pogorzelski) published what some believed was a claimed proof of Goldbach's Conjecture in [Crelle's Journal (292, 1977, 1-12)](http://www.deepdyve.com/lp/de-gruyter/goldbach-conjecture-frA6e4f0DS). His argument has not been accepted as a proof of Goldbach's Conjecture, but as far as I know it has not been shown that his argument is incorrect.
Pogorzelski's argument is said to depend on the "Consistency Hypothesis," the "Extended Wittgenstein Thesis," and "Church's Thesis." Pogorzelski has a Ph.D. in mathematics (his advisor was Raymond Smullyan).
Daniel Shanks says in *Solved and Unsolved Problems in Number Theory* (fourth edition, 1993) that: "It seems unlikely that (most) number-theorists will accept this as a proof [of Goldbach's Conjecture] but perhaps we should wait for the dust to settle before we attempt a final assessment." ([page 222](http://tinyurl.com/37jl3px))
Did Pogorzelski claim to present a proof of Goldbach's Conjecture? If so, and this claimed proof has not been disproven after 33 years, I am curious why this would be the case, given that Shanks considers it important enough to mention in his book.
| https://mathoverflow.net/users/2594 | Did Pogorzelski claim to have a proof of Goldbach's Conjecture? | In the 1970's Pogorzelski published a sequence of four papers in Crelle concerning the Goldbach Conjecture (and various generalizations and abstractions):
>
> MR0347566 (50 #69) Pogorzelski, H. A. On the Goldbach conjecture and the consistency of general recursive arithmetic. Collection of articles dedicated to Helmut Hasse on his seventy-fifth birthday, II. J. Reine Angew. Math. 268/269 (1974), 1--16.
>
>
> MR0505402 (58 #21554) Pogorzelski, H. A. Dirichlet theorems and prime number hypotheses of a conditional Goldbach theorem. J. Reine Angew. Math. 286/287 (1976), 33--45.
>
>
> MR0434999 (55 #7961) Pogorzelski, H. A. Semisemiological structure of the prime numbers and conditional Goldbach theorems. J. Reine Angew. Math. 290 (1977), 77--92.
>
>
> MR0538046 (58 #27414) Pogorzelski, H. A. Goldbach conjecture. J. Reine Angew. Math. 292 (1977), 1--12.
>
>
>
The last paper is the one referred to in the question. (Edit: as I have been writing this, the question has been edited to make this reference explicit, which is good.)
I think that describing Pogorzelski's last paper as a purported proof of the Goldbach Conjecture is a mischaracterization. Rather what he shows is an implication: three statements which are not known to be true imply Goldbach. (I don't pretend to understand these three statements. The only one that I recognize at all is Church's Thesis, but although I think I know what that means, it does not denote to me a precise mathematical conjecture, so I am for sure out of my depth here.)
So far as I can see, Pogorzelski himself never claimed that his 1977 paper is a proof of Goldbach. Indeed, he worked for many years thereafter on the problem and published nearly two thousand pages of further work. Specifically, in 1982 he published the first of a proposed seven volume series, *Foundations of a semiological theory of numbers*, whose ultimate goal is to prove Goldbach by showing that a disproof is impossible in a certain formal system. Volume 1 is 608 pages. Volume 2 (743 pages) appeared in 1985. Volume 3 (522) pages appeared in 1988. MathSciNet does not list any further volumes.
To summarize, his programme for proving Goldbach seems to be as yet unfinished (and, of course, may well be unfinishable), but none of the reviews I read -- some of which are written by leading mathematicians -- raised any mathematical objections to the work that has been published.
| 35 | https://mathoverflow.net/users/1149 | 38328 | 24,618 |
https://mathoverflow.net/questions/24131 | 19 | The group of $n\times n$ matrices with integer entries and determinant equal to 1, $SL(n,Z)$, is a finitely generated group (in fact, it is generated by 2 matrices). I am interested to know if the semigroup of the matrices in $SL(n,Z)$ where all the entries are nonnegative is also finitely generated. This is true at least in dimension $n=2$.
| https://mathoverflow.net/users/3960 | Is the semigroup of nonnegative integer matrices with determinant 1 finitely generated? | The question has been amply answered, but perhaps it's worth explaining the geometry of the situation. The positive monoid acts by projective transformations on the $n-1$-simplex (the view of the positve orthant as seen from the origin, and the partially order on the monoid is the order by inclusion of the image under this action. In $R^n$, the image of the simplex spanned by the origin together with the basis vectors cannot contain any interior lattice points. This implies that in the projective action, no proper image can contain its barycenter. This, with its ramifications, gives a strong limitation on images of the entire simplex under the positive monoid.
On the other hand, it's not hard to prove that for $n \ge 3$, the set of images of any face are dense. In fact, using elementary and well-known theory, there is a positive density of $(n-1)$-tuples of lattice points that extend to a basis for $Z^n$, and any such $n$-tuple of positive elements can be extended to a positive basis.
This implies that there is no finite set of generators, because a single image of the $n-1$-simplex can contain at most simplices near the barycenter over a small range of angles. Here's a picture for the 2-dimensional case:
[alt text http://dl.dropbox.com/u/5390048/MatrixSemigroup.jpg](http://dl.dropbox.com/u/5390048/MatrixSemigroup.jpg)
The line segment is the projection of the pair of lattice points
$L\_1=(285, 684, 112)$ and $L\_2 = (764, 318, 949)$ found, by pseudorandom selection from lattice cubes that projected near the targeted segment, subject to the condition that they extend to a free basis.
A third element of a basis can be transformed to be project arbitrarily close to any desired point on the line through the image of $L\_1$ and $L\_2$ by adding $N\*L1 + M\*L2$, for suitable $N$ and $M$.
Moreover, this characterize the closure (with respect to the Hausdorff topology) of the set of images of the triangle under the positive monoid: it consists of all triangles spanned by positive free bases, together with set of all line segments contained in the triangle.
| 16 | https://mathoverflow.net/users/9062 | 38332 | 24,622 |
https://mathoverflow.net/questions/38119 | 16 | I wrote a research paper "A mathematical model of the Mafia game" ([arXiv:1009.1031](http://arxiv.org/abs/1009.1031) [math.PR]). However, I do not know where to publish it. As an undergraduate studying majorly physics, I have little knowledge of mathematical journals. Moreover, its not easy for me to classify its subject.
The paper itself is generally "using applied mathematics (and physicist's approach) to model a party game (psychology)". Just to give some of its features:
* Uses pure death process and gives its closed-form solution
* Is elementary and didactic (a tricky use of generating functions, Wallis formula, ...)
* Somehow "cool" topic
I thought about [The American Mathematical Monthly](http://www.maa.org/pubs/monthly.html). Besides the obvious advantages I see two drawbacks:
* In AMM there are usually an old results in a nice/didactic/cool form or short findings
* It won't reach to psychologists who might investigate the Mafia game experimentally
Can you recommend me any proper journal? (popular mathematics, mathematics and psychology, ...?)
Any other advice are appreciated as well.
| https://mathoverflow.net/users/9093 | Where to publish a paper on the Mafia game? | I have only glanced at your paper, but one possibility is to submit it to [*The Mathematical Intelligencer*](http://www.springer.com/mathematics/journal/283?detailsPage=editorialBoard), in particular to Michael Kleber, who edits the "Mathematical Entertainments" column. This is a great place for serious mathematical analysis of problems with a recreational flavor. The *Intelligencer* has a wide readership, which is what you want for your article.
By the way, contrary to some other commentators, I don't see any reason why you necessarily need to find a mentor just because you're an undergraduate. Your paper seems to be well-written enough. Of course it doesn't hurt to find a mentor but what matters is the quality of your writing and not your employment status.
Also, regarding KConrad's comment that only existing players of the game will find it interesting, I don't believe that this is true. First of all, as you note, Mafia is in fact a pretty well-known game. Secondly, the subtlety of the game means that there is a lot of interesting mathematics buried in it. If your article can draw more mathematicians into studying it, that would be a very good thing.
| 18 | https://mathoverflow.net/users/3106 | 38337 | 24,623 |
https://mathoverflow.net/questions/38323 | 10 | I try to understand some of the topology of the space of pointed non-compact hyperbolic surfaces (with the pointed Gromov-Hausdorff topology). It is known that the fundamental
group of a non-compact surface is a free group, so I am interested in free Fuchsian groups
(discrete, free groups of direct isometries of the hyperbolic plane, not necessarily finitely generated).
Call ``ideal polygon'' any domain of the hyperbolic plane that is an intersection of half-planes limited by geodesics that are pairwise disjoint.
Is it true that any free fuchsian group has a fundamental domain that is an ideal polygon?
I think that I can manage do do it by hand for the simplest examples (e.g. covering groups of an hyperbolic punctured tori, or a hyperbolic trice punctured sphere) and the result seems plausible, but I feel that either true or false it shall be well-known. Any reference on this, or more generally on uniformization of non-compact, possibly infinite genus hyperbolic surfaces would be welcome.
| https://mathoverflow.net/users/4961 | fundamental domains for free fuchsian group. | Yes, this is true. Topologically, one may find a locally finite collection of properly embedded arcs in a connected surface whose complement is homeomorphic to $R^2$. Then make each of these arcs geodesic in the hyperbolic metric. The complement will be the fundamental domain of the type you want.
Addendum: I'll add some comments on one way to obtain these properly embedded arcs in the infinite topology case. Ian Richards gave a [classification of connected surfaces](http://www.jstor.org/stable/1993768). In Theorem 3 of that paper, he explains how to construct all surfaces. A planar surface $\Sigma\cong S^2-X$ is obtained by removing a totally disconnected compact set $X\subset S^2$ from $S^2$. As explained in Prop. 5 of the paper, one may consider the totally disconnected set $X\subset S^2$ to be a subset of the Cantor set, and therefore a subset of the interval (including the endpoints) $X\subset I\subset S^2$. Then the properly embedded arcs $I\cap (S^2-X)$ give a decomposition of $\Sigma$ into $R^2$.
If the surface is non-planar, then one removes from $S^2-X$ a properly embedded countable collection of disks $D\_1,D\_2,\ldots$, and makes identifications of their boundaries. We may assume after an isotopy that these disks are all centered on $I$, and that the identifications either identifies antipodal points, or identifies two disks which are adjacent along a component of $I-X$ with a $\pi$ twist. The complement $U=S^2-(I\cup\_i D\_i)$ is again homemorphic to $R^2$. If we identify antipodal points of $D\_j$, then this identifies two arcs in the boundary of $U$ to obtain an open Mobius strip. We add two arcs connecting antipodal points of $D\_j$ to a point $x\in X$ at the end of the interval of $I-X$ which intersects $D\_j$, which forms a single arc after identification of antipodal points of $D\_j$, and cuts the Mobius strip back up into $R^2$. If adjacent disks $D\_i, D\_{i+1}$ are identified, then the complement $U$ gives a punctured torus. We add 4 arcs connecting these points to $x$ (again, $x\in X$ is at the end of the interval of $I-X$ containing $D\_i$), cutting the surface into $R^2$ again. Continuing in this fashion inductively, we get a locally finite collection of arcs cutting the surface up into $R^2$.
| 12 | https://mathoverflow.net/users/1345 | 38338 | 24,624 |
https://mathoverflow.net/questions/38315 | 3 | A pair $(X,O\_X)$ is a ringed space if $X$ is a topological space and $O\_X$ is a sheaf of rings. If every stalk $O\_{X,x}$ is a local ring, then we say that $(X,O\_X)$ is a locally ringed space.
In the case of $X$ being an abstract algebraic variety, not necessarily irreducible, and $O\_X$ its sheaf of regular functions:
1) Is it possible to have a ringed space which is not a locally ringed one? And if $X$ is irreducible? My guess is NO for the first question and YES for the second.
Correct me if I am wrong. My reasoning goes as follows, since stalks are local we can work on the affine case. All irreducible algebraic varieties are quotient of $K[x\_1,\ldots,x\_n]$ by a prime ideal and prime ideals always have a unique maximal ideal, corresponding to a point, so the quotient on the point is a field. This seems to work for $K=\mathbb{C}$. In the irreducible case I am not sure of what happens, but for the complex case you can have two lines crossing and I get each of them may work independently giving two local ring ideals? Pure speculative...
2) Is this true for non-algebraically closed fields or fields with positive characteristic?
3) Could you please provide with any example of non-locally ringed spaces? In varieties or schemes, if it is possible.
4) If $X$ is a variety/scheme, is there any example of morphism of ringed spaces which are locally ringed but which is not a morphism of locally ringed spaces? (i.e. for $f\colon X\rightarrow Y$ there is a sheaf morphism $f^\sharp\colon O\_Y \rightarrow f\_\* O\_X$ which is not a morphism of local rings on the stalks).
These questions are a bit vague but hopefully you understand what I mean.
| https://mathoverflow.net/users/1887 | Ringed and locally ringed spaces | 1. No, all schemes are locally ringed spaces (cf. Robin Chapman's comment above). In fact, schemes are often defined by using locally ringed spaces.
2. It depends on what you mean by "this". If you have a pair of lines crossing, the local ring at the crossing has zero divisors, but it's still local.
3. Take a point, and set its ring of functions to be the ring of rational integers (see Robin Chapman's comment again). It's not a scheme. There are more interesting examples in, e.g., work by Kato and Nakayama on log analytic spaces.
4. Yes. Hartshorne produces one as example 2.3.2 in his *Algebraic Geometry*: The spectrum of a discrete valuation ring $R$ is a locally ringed space made of an open dense point (whose stalk is the fraction field of $R$) and a closed point (whose stalk is $R$). There is a ringed space map from the spectrum of $R$ to itself that takes both points to the closed point, where the restriction to the closed stalk is identity, and the restriction to the open dense point is defined by the inclusion of $R$ into its fraction field.
| 8 | https://mathoverflow.net/users/121 | 38339 | 24,625 |
https://mathoverflow.net/questions/38344 | 11 | Of course, no continuous real valued non-constant function can attain only rational or irrational values, but can there be a pair of nowhere-constant continuous functions f and g such that for all x, at least one of f(x) and g(x) is rational? Or maybe a countable collection of continuous functions, {f1, f2...} such that for all x there is n such that fn(x) is rational?
Thanks
| https://mathoverflow.net/users/4903 | Can there be two continuous real-valued functions such that at least one has rational values for all x? | If you allow the functions to be constant on some intervals, then there are some easy examples, and Ricky has provided one.
But if you rule that out, then there can be no examples, even with countably many functions. To see this,
suppose that $f\_n$ is a list of countably many continuous functions which are never constant on an interval. Enumerate the pairs $(r,n)$ of
rational numbers $r$ and natural numbers $n$ in a countable list
$\langle (r\_0,n\_0), (r\_1,n\_1),\ldots\rangle$. Let $C\_0$ be any closed
interval. If the closed interval $C\_i$ is defined, consider
the function $f\_{n\_i}$ and the rational value $r\_i$. Since $f\_{n\_i}$ is not constant value $r\_i$ on $C\_i$, we may
shrink the interval to $C\_{i+1}\subset C\_i$ such that
$f\_{n\_i}$ on $C\_{i+1}$ is bounded away from $r\_i$. By
compactness, there is some $x\in C\_i$ for all $i$. Thus,
$f\_n(x)$ is not $r$ for any rational number $r$.
| 16 | https://mathoverflow.net/users/1946 | 38346 | 24,628 |
https://mathoverflow.net/questions/37976 | 5 | Given any subspace $A\subset X$ of a topological space with Lebesgue dimension $\le N$.
Let $\bar{A}$ denote the closure of $A$. Assume, that the pair $(\bar{A},A)$ satisfies the Z-set condition, i.e. there is a homotopy $H:\bar{A}\times [0;1]\rightarrow \bar{A}$, such that $H\_1=id$ , Image$(H\_t)\subset A$ for all $t<1$.
Does this imply, that the Lebesgue dimension of $\bar{A}$ is at most the Lebesgue dimension of $A$?
| https://mathoverflow.net/users/3969 | Lebesgue dimension of closures satisfying the Z-set condition | No for general topological spaces, yes for metrizable ones (and I believe the argument can be generalized to all normal spaces).
Bad example: $X=\{a,b,c\}$ with open sets $\emptyset$, $X$, $\{a\}$, $\{a,b\}$, $\{a,c\}$. Let $A=\{a\}$, then $\bar A=X$. The homotopy is given by $H\_1=id$, $H\_t\equiv a$ for $t<1$. The dimension of $A$ is 0 but the dimension of $\bar A$ is 1.
On the positive side, let me begin with a quick and dirty proof in the case when $\bar A$ is a compact metric space. Let $\{U\_i\}$ be an open covering of $\bar A$. We need to find a refined covering of multiplicity at most $N+1$ where $N=\dim A$. It suffices to find a continuous map $f:\bar A\to A$ and an open covering $\{V\_j\}$ of $A$ such that $\{f^{-1}(V\_j)\}$ is a refinement of $\{U\_i\}$. Indeed, in this case we can find a refinement of $\{V\_j\}$ of multiplicity at most $N+1$ and its $f$-preimage is the desired refinement of $\{U\_i\}$.
In the compact case, let $\{V\_j\}$ be the covering by $(\rho/3)$-balls where $\rho$ is the Lebesgue number of the covering $\{U\_i\}$. Then, for some $t$ sufficiently close to 1, the map $f=H\_t$ satisfies the desired property: the preimage of every $V\_j$ has diameter less than $\rho$ and hence is contained in some of the sets $U\_i$. Indeed, suppose the contrary. Then there is a sequence $t\_k\to 1$ and sequences $x\_k,y\_k\in \bar A$ such that $|x\_ky\_k|\ge\rho$ but $|H\_{t\_k}(x\_k)H\_{t\_k}(y\_k)|<2\rho/3$. Due to compactness we may assume that $x\_k$ and $y\_k$ converge to some $x,y\in\bar A$. Then $|xy|\ge\rho$ but $|H\_1(x)H\_1(y)|\le 2\rho/3$, a contradiction.
In the general metric space case, let $\rho(x)$ denote the local Lebesgue number of $\{U\_i\}$ at $x$, that is the supremum of $\rho$ such that the ball $B\_\rho(x)$ is contained in one of the set $U\_i$. Note that $x\mapsto\rho(x)$ is a positive 1-Lipschitz function on $\bar A$. It is easy to construct a continuous function $u:\bar A\to[0,1)$ such that the distance from $x$ to $H\_t(x)$ is less that $\rho(x)/10$ for all $x\in\bar A$ and all $t>u(x)$. Then the map $f:\bar A\to A$ given by $f(x)=H\_{u(x)}(x)$ and the covering of $A$ by the balls of the form $B\_{\rho(x)/10}(x)$, $x\in A$, will do the job.
| 4 | https://mathoverflow.net/users/4354 | 38352 | 24,630 |
https://mathoverflow.net/questions/37993 | 21 | Let P be an arbitrary probability space.
I would like to find a compact topological group $G$ so that the Haar probability measure on $G$ admits a measurable map to the probability space $P$.
By a measurable map, I mean a function which lifts measurable sets to measurable sets of the same measure. That is, $f : Q \longrightarrow P$ is measurable when for all measurable $E \subseteq P,$ the set $f^{-1}(E)$ is measurable and satisfies
$$ \mu\_Q (f^{-1}(E)) = \mu\_P (E). $$
I imagine that some enormous product space $(\mathbb{R} / \mathbb{Z})^{\kappa}$ will do. Anyone see a nice way to make this work?
Edit:
My hunch that some power of the circle group will work is based on the game twenty questions. One player thinks of an object (a trampoline), and the other players ask a sequence of at most twenty yes/no questions (Can it fly? Is it legal to drive one on the highway? Would it hurt to swallow?). A strategy for twenty questions would consist of an enormous decision tree that says what question to ask next.
Here we're trying to guess a point of $P$ using at most $\kappa$ measurable yes/no questions. The answers come from the coordinates of $[0,1]^{\kappa}$ which may be interpreted according to the probability of a yes. The function $f$ could be built out of a suitable decision tree.
Perhaps a decision tree could be constructed out of a well-ordering of the sigma-algebra of measurable sets.
Update:
Thanks for all the help!
Here's the story:
What I really want is to perform convolution in $L^2(P)$. I'll admit, $P$ is not a group, so I guess it's okay to switch to $L^2(G)$ provided that $G$ has $P$ as a factor space. But it turns out even this was too greedy.
Let $S \subseteq G$ be a dense subset with a measure structure inherited from $G$. Now we may perform convolution in $L^2(S)$, even though $S$ is not a group! The continuous functions are dense in $L^2(S)$, so it will suffice to convolve two of them. But any continuous function on $S$ extends uniquely to one on $G$. So we convolve in $L^2(G)$ and then restrict the result to $L^2(S)$.
Because of this, George Lowther's weaker result will suffice for my purposes. After all, a subset of full outer measure is certainly dense. I will accept his answer unless a full answer to the original question materializes.
| https://mathoverflow.net/users/9068 | Is every probability space a factor space of the Haar Measure on some group? | It is possible to find the following: A compact abelian group G with Haar measure $\mu\_G$, a subset $S\subseteq G$ of full outer Haar measure and a measurable function $f\colon S\to X$ with $\mu\_P(E)=\mu\_S(f^{-1}(E))$ for measurable $E\subseteq P$. In fact, as you mention, G can be taken to be a large enough product of the circle group.
Here, I am implicitly referring to the sigma algebra $\mathcal{B}(S)\equiv\{S\cap E\colon E\in\mathcal{B}(G)\}$ and $\mu\_S$ is the restriction of the Haar measure to $\mathcal{B}(S)$, $\mu\_S(S\cap E)=\mu\_G(E)$. Ideally, we would like to enlarge S so that it is actually of full measure, then it could be enlarged to all of G. I'm not sure if this is possible though (and suspect that it is not possible in ZFC). The problem is that if $f\colon P\to Q$ is a measure preserving map of probability spaces then $f(P)$ will be of full outer measure, but need not be measurable. If $f^{-1}\colon\mathcal{B}\_Q\to\mathcal{B}\_P$ is an onto map of their sigma algebras then P and Q are "almost isomorphic" probability spaces. If, however, $f(P)$ is not measurable then f does not have a right inverse (even up to zero probability sets), unless we restrict to the subset $f(P)\subseteq Q$.
In the following, I write 2={0,1}, so that, for a set I, 2^I is the set of {0,1}-valued functions from I.
Letting $\pi\_i\colon2^I\to\{0,1\}$ be the projection onto the i'th coordinate, the sets of the form $\pi\_i^{-1}(S)$ generate a sigma algebra, which I will denote by $\mathcal{E}\_I$.
First, we can reduce the problem to that of measures on $2^I$.
>
> Step 1: Given a collection $\{A\_i\}\_{i\in I}$ of sets generating the sigma algebra on P, construct a probability measure $\mu\_I$ on $2^I$ and a measure preserving map $f\colon P\to 2^I$ such that $f^{-1}\colon\mathcal{E}\_I\to\mathcal{B}\_P$ is onto.
>
>
>
Then, f will have a right inverse $g\colon f(P)\to P$ which is automatically measurable and measure preserving.
To construct f, define $f(p)\in 2^I$ by $f(p)(i)=1$ if $p\in A\_i$ and =0 otherwise. It can be checked that $\mu\_I(S)=\mu(f^{-1}(S))$ for $S\in\mathcal{E}\_I$ satisfies the required properties.
Now, I will use $G^I=(\mathbb{R/Z})^I$, which is a compact abelian group with Haar measure $\mu\_{G^I}$, which is the product of the uniform measure on the circle $G=\mathbb{R/Z}$.
>
> Step 2: Construct a measure preserving map $f\colon G^I\to 2^I$.
>
>
>
Once this map is constructed, putting it together with step 1 gives what I claimed.
For any $J\subseteq I$, use $\pi\_J\colon 2^I\to2^J$ and $\rho\colon G^I\to G^J$ to denote restriction to J. Also use $\mu\_J(S)=\mu\_I(\pi\_J^{-1}(S))$ for the induced measure on $2^J$.
Zorn's lemma guarantees the existence of a subset $J\subset I$ and measure preserving map $f\colon G\_J\to 2^J$ which is maximal in the following sense: for $J\subseteq K\subseteq I$ and measure preserving $g\colon G^K\to2^K$ with $\pi\_J\circ g=f\circ\rho\_J$ then K=J.
Then, J=I and we have constructed the required map. If not, choose any $k\in I\setminus J$, $K=J\cup\{k\}$ and define $h\_k\colon G^K\to 2$ as follows (I let $A\_k=\pi^{-1}\_k(1)\subseteq2^I$).
$$
h\_k(x)=\begin{cases}
1,&\textrm{if }0\le x\_k\le\mu\_I\(A\_k\mid\mathcal{E}\_J\)\\_{f(x)}\\\\
0,&\textrm{otherwise}.
\end{cases}
$$
Defining $g\colon G^K\to2^K$ by $g(x)\_i=f(x)\_i$ for $i\in J$ and $g(x)\_k=h\_k(x)$ gives a measure preserving map extending f, and contradicting the maximality of J.
| 9 | https://mathoverflow.net/users/1004 | 38362 | 24,636 |
https://mathoverflow.net/questions/38347 | 14 | Framed functions arose in the work of K. Igusa defining cohomology invariants for smooth manifold bundles (Igusa-Klein torsion). In the late 80's, he proved a strong connectivity result about the "space of framed functions" using Morse theory and conjectured that this space was, in fact, contractible. Jacob Lurie recently verified this conjecture using the language of higher category theory in the course of his proof of the Cobordism Hypothesis.
(1) Is there any progress on a more geometric proof of this? The relevant work of Igusa is very geometric and Lurie himself points out that a more direct proof should be found. Perhaps more realistically: is there a geometric heuristic for Lurie's result?
(2) Does anyone know of applications of framed functions beyond defining Igusa-Klein torsion and proving the Cobordism Hypothesis? Pointers into the literature on this topic would be much appreciated.
EDIT: A framed function on $M$ is a pair $(f,\xi)$ of a smooth function $f$ on $M$ having at most $A2$ singularities in the interior of $M$ and no singular points on the boundary, and an orthonormal framing $\xi$ of the negative eigenspace of the Hessian of $f$ at each critical point.
| https://mathoverflow.net/users/7867 | The space of framed functions | It is possible to prove the contractibility directly (and thereby bypass the obstruction theory arguments sketched at the end of section 3 of my paper). The statement itself is an example of an h-principle: namely, one can show fairly easily that the framed function space of ${\mathbb R}^{n}$ is contractible, so the general result hinges on knowing that the framed function spaces also satisfy some sort of "local to global" principle for arbitrary manifolds. (If I remember correctly, Igusa told me that he and Eliashberg sketched out a proof along these lines, but never had the motivation to write it up.)
| 15 | https://mathoverflow.net/users/7721 | 38368 | 24,641 |
https://mathoverflow.net/questions/38367 | 8 | Notation: Let$M$ be a smooth, closed manifold, $S$ any submanifold of $M$, $Diff(M)$ the group of diffeomorphisms of $M$ and $Imb(S, M)$ the group of smooth imbeddings of $S$ into $M$.
A classical result of R. Palais from the 1960 paper *Local triviality of the restriction map for embeddings* says that the map $Diff(M)$ $\rightarrow$ $Imb(S, M)$ given by restriction is a fibration.
I feel like I've heard during numerous teas that there are various refinements and generalizations of this due to J. Cerf and (possibly) others.
(1) Can anyone summarize what else is known in this direction beyond the theorem of Palais?
(2) Is there a way to see Palais' result easily? [added: from the responses it sounds like the original paper is still a great way to see this result. But see the answers of Randal-Williams and Palais for an alternate route.]
| https://mathoverflow.net/users/7867 | Restrictions of Diffeomorphisms | It's not clear what you mean by "various refinements and generalizations". Cerf has a huge paper published by IHES "Topologie de certains espaces de plongements" which goes into many related details. In a way it's more of a ground-up collection of basic information on the topology of function spaces.
Regarding your 2nd question, if instead of demanding a fibre bundle you ask for a Serre fibration, the proof is relatively simple. It's just the isotopy extension theorem with parameters, and the proof is pretty much verbatim Hirsch's proof of isotopy extension in his "Differential Topology" text plus the observation that solutions depend smoothly on the initial conditions.
Regarding your 2nd question, yes of course. Palais's paper is quite nice. If you haven't had a look at it, you might as well try -- it's only 7 pages long. If you want to discover the proof on your own I'd start with the case $S$ a finite set. Then move up to $S$ a positive-dimensional submanifold. You'll want to be comfortable with things like the proof of the tubular neighbourhood theorem, the concept of injectivity radius, etc.
| 9 | https://mathoverflow.net/users/1465 | 38372 | 24,645 |
https://mathoverflow.net/questions/38348 | 2 | Fix $N>0$. Let $b\_i=(b\_{i,1}, b\_{i,2}, b\_{i,3}, b\_{i,4})$, $i=1,\ldots, m$, be distinct 4-tuples of integers with with all $0\leq b\_{i,j}< N$. (The zero tuple is disallowed.)
Define $w\_i=(\prod\_{j=1}^4 (z-z\_j)^{b\_{i,j}})^{\frac1{N}}$.
Consider $w\_i$ as an element of the following vector space: the algebraic closure of $\mathbb{C}(z)$ over the field $\mathbb{C}(z)$.
I believe that Puiseux series can be used to show the linear independence of the $w\_i$. Are there any other approaches that might show the linear independence of the $w\_i$?
More generally, are there any other general techniques for proving linear independence of functions in the given vector space?
| https://mathoverflow.net/users/5399 | Linear independence in the algebraic closure of $\mathbb{C}(z)$ | You can also use Galois theory or monodromy. Take a minimal linear dependence relation and apply the automorphimsm of the algebraic closure that fixes $(z-z\_j)^{1/N},j>1$ and fot $j=1$ multiplies the function by an N-th root of unity, thus getting a new relation and you can produce a shorter relation from those two.
| 1 | https://mathoverflow.net/users/2290 | 38375 | 24,648 |
https://mathoverflow.net/questions/38376 | 4 | $\mathbf{n}$ is nilpotent Lie algebra with $N$ being the corresponding *algebraic* Lie group. Now one neat feature of this setting is that you can take the exponential map to be identity. In other words you can define a group structure on $\mathbf{n}$ using the Campbell-Hausdorff formula. I have the following questions (they might be very easy; I just don't know):
1. If $\pi: \mathbf{n} \rightarrow \text{End}(V)$ is a representation of the nilpotent Lie algebra then does $x \in N$ acts as $\exp (\pi (x))$ on $V$ ?
2. If $\mathbf{m}$ is a subalgebra of $\mathbf{n}$ then you get a corresponding group $M \subset N$. Do we have: $N/M = \mathbf{n}/\mathbf{m}$ as sets?
**EDIT:** Victor's comments are to the point so in order to clear up the confusion I added that $N$ is algebraic.
| https://mathoverflow.net/users/8811 | Nilpotent Lie algebras and unipotent Lie groups | I will suppose that $\mathfrak n$, etc is finite-dimensional.
If by "the corresponding Lie group" you mean "the corresponding connected simply-connected Lie group", then what you say is correct (in spite of Victor Prostak's comment). Let $\{\mathfrak n,[,]\}$ be a Lie algebra over a field $k$ of characteristic $0$. Then the BCH formula truncates, and in particular converges. In general, BCH defines a "partial group" structure in its domain of convergence (as for "typical uses", in finite dimensions over $\mathbb R$, BCH has positive radious of convergence, and [by gluing together such patches one can prove the Lie III theorem](https://mathoverflow.net/questions/8784/why-is-lies-third-theorem-difficult)), and so for nilpotent $\mathfrak n$ it defines a group structure $\{\mathfrak n, \operatorname{BCH}\}$ on the set of points in $\mathfrak n$. In fact, when $\{\mathfrak n,[,]\}$ is nilpotent, the fact that BCH truncates means that it is polynomial, and so $\{\mathfrak n, \operatorname{BCH}\}$ is affine algebraic over $k$. In particular, when $k = \mathbb R$, then BCH defines a real algebraic group $\{\mathfrak n, \operatorname{BCH}\}$, and in particular a Lie group, and it is straightforward to check that $\operatorname{Lie}(\{\mathfrak n, \operatorname{BCH}\}) = \{\mathfrak n,[,]\}$. Since as a topological space $\mathfrak n$ is connected and simply connected, $\{\mathfrak n, \operatorname{BCH}\}$ is the connected simply-connected group corresponding to $\{\mathfrak n,[,]\}$. This I think you already know, but I thought it best to spell it out for other readers.
You ask two questions.
(1.) Yes, if I'm understanding correctly. Let $\pi : \{\mathfrak n,[,]\} \to \operatorname{End}(V)$ be a Lie algebra homomorphism, and $\dim V < \infty$. Since $\{\mathfrak n, \operatorname{BCH}\}$ is connected and simply connected, $\pi$ determines a group homomorphism $\Pi: \{\mathfrak n, \operatorname{BCH}\} \to \operatorname{GL}(V)$. Given $x\in \mathfrak n$, yes, $\Pi(x) = \exp(\pi (x))$.
There's nothing special about unipotent groups here. Let $\mathfrak g$ be a (finite-dimensional) Lie algebra (over $\mathbb R$) and $G$ the corresponding connected simply-connected Lie group, and let $\exp : \mathfrak g \to G$ be the exponential map (in your example, $\exp: \{\mathfrak n,[,]\} \to \{\mathfrak n, \operatorname{BCH}\}$ is an isomorphism of spaces, so we call it the identity). Let $H$ be any other Lie group and $\mathfrak h = \operatorname{Lie}(H)$, and let $\exp: \mathfrak h \to H$ be the exponential map. Then there is a canonical bijection between Lie algebra homomorphism $\mathfrak g \to \mathfrak h$ and Lie group homomorphisms $G \to H$, and the bijection intertwines the two exponential maps ($\Phi\circ \exp = \exp \circ \phi$). One proof is uses the following description of $\exp$: we can identify $\mathfrak g \hookrightarrow \Gamma({\rm T}G)$ as left-invariant vector fields, so that $x\in \mathfrak g$ determines an ODE on $G$, and it turns out that this ODE has solutions for all times, and $\exp(x)$ is the image of $e\in G$ under the flow-by-time-$1$ map.
(2.) No. I think what you are asking is the following. Let $\{\mathfrak n,[,]\}$ be a (finite dimensional) nilpotent Lie algebra (over $\mathbb R$) and $\{\mathfrak m,[,]\}$ a subalgebra. Then we have connected simply-connected groups $\{\mathfrak n,\operatorname{BCH}\}$ and $\{\mathfrak m,\operatorname{BCH}\}$, and the latter is canonically a subgroup of the former. Then I think what you are asking is whether the cosets of $\{\mathfrak m,\operatorname{BCH}\}$ in $\{\mathfrak n,\operatorname{BCH}\}$ are the same as the cosets of the vector space $\{\mathfrak m,+\}$ in the vector space $\{\mathfrak n,+\}$.
Consider the three-dimensional Heisenberg algebra, which is generated by two elements $x,y$, with the condition that $[x,y]$ is central (then $x,y,[x,y]$ form a basis). This has the matrix representation:
$$ x = \left( \begin{array}{ccc} 0 & 1 & 0 \\ & 0 & 0 \\ & & 0 \end{array} \right), \quad y = \left( \begin{array}{ccc} 0 & 0 & 0 \\ & 0 & 1 \\ & & 0 \end{array} \right), \quad [x,y] = \left( \begin{array}{ccc} 0 & 0 & 1 \\ & 0 & 0 \\ & & 0 \end{array} \right)$$
Then $\operatorname{BCH}(ax,y) = ax + y + \frac a 2 [x,y]$. In particular, letting $\mathfrak n$ be the Heisenberg algebra and $\mathfrak m$ the one-dimensional subalgebra spanned by $x$, we see that the (right, say) cosets of $\{\mathfrak m,\operatorname{BCH}\}$ do not agree with the cosets of $\{\mathfrak m,+\}$: in particular, the cosets through $y$ are different (one is a line parallel to the span of $x$, the other is a line parallel to the span of $x + \frac12 [x,y]$).
Note that as Emerton points out, if $\{\mathfrak m,[,]\}$ is a Lie ideal in $\{\mathfrak n,[,]\}$, then the answer is yes. Let $y \in \mathfrak n$. Then the coset of $\{\mathfrak m,\operatorname{BCH}\}$ through $y$ consists of elements of the form $x + y + l$, where $x \in \mathfrak m$ and $l$ is a Lie polynomial with at least one $x$, and hence (since $\mathfrak m$ is an ideal), it is in $\mathfrak m$. I.e. the coset is precisely $y + \mathfrak m$.
| 8 | https://mathoverflow.net/users/78 | 38388 | 24,653 |
https://mathoverflow.net/questions/38385 | 6 | Let R and S be commutative rings with a 1 different from zero. Let m and n be positive integers. Assume the ring of m-by-m matrices over R is isomorphic to the ring of n-by-n matrices over S. Does it follow that R is isomorphic to S? Does it follow that m = n? Does either of those follow from the other? I'm interested in both where R,S are finite and where R,S are infinite. (although the second question is trivial in the former case)
| https://mathoverflow.net/users/nan | Exotic isomorphism of matrix rings | Yes and yes. Let $T=M\_m(R)=M\_n(S)$.
The center of $T$ is isomorphic to both $R$ and $S$.
The $1\times m$ matrices over $R$ form an $(R,T)$-bimodule and the $n\times 1$ matrices over $S$ form a $(T,S)$-bimodule. Tensor these over $T$ to get an $(R,S)$-bimodule. As an $S$-module the direct sum of $m$ copies of this is free of rank $n$. For a nonzero commutative ring this implies that $m$ divides $n$. (Tensor with a residue field to get a vector space of dimension $\frac{n}{m}$.) Likewise, looking at it as an $R$-module, $n$ divides $m$.
| 8 | https://mathoverflow.net/users/6666 | 38391 | 24,654 |
https://mathoverflow.net/questions/37721 | 4 | The topology of a closed surface can be constructed
by identifying edges of a [fundamental polygon](http://en.wikipedia.org/wiki/Fundamental_polygon) of an
even number $2n$ of edges.
Labeling the edges and using $\pm 1$ exponents to indicate
direction,
the construction can be specified by a string of $2n$ symbols:
$a b a^{-1} b^{-1}$ for the torus, $a a b b$ for the Klein bottle, etc.
My question is: Does it make sense to have an infinite
number of boundary identifications, i.e., does it
define some topological object?
More specifically:
1. Does an infinite string of symbols representing pairwise
identifications correspond to some surface?
For example, the generalization of a non-orientable genus-$n$ surface as $n \rightarrow \infty$:
$$a\_1 a\_1 a\_2 a\_2 a\_3 a\_3 \cdots a\_i a\_i \cdots \;.$$
2. Does it make sense to have an uncountable number of pairwise point identification
around a circle?
For example, parametrize the circle circumference from 0 to 1
and identify points with complementary binary representations:
$$.011100100011\ldots \leftrightarrow
.100011011100\ldots \;.$$
These extensions may be nonsensical, in which case I apologize for the distraction!
But if something along these lines has been studied, I'd appreciate a reference. Thanks!
**Edit.** What was nonsensical was my bit-complement example, as pointed out by both
Victor and Sergei. I'll leave it so their remarks make sense. I intended a more patternless
pairing.
| https://mathoverflow.net/users/6094 | Fundamental polygons with infinite pairwise identifications | Identification spaces with infinitely complicated identifications do occur naturally in various contexts. Here are a few more remarks:
1. If you define an involution on a countable set of intervals on the boundary of a surface, as in Victor Protsak's remark, and extend the equivalence relation to finite equivalence classes of the closure (i.e., vertices), you can construct arbitrary noncompact surfaces. The classification of these is known, and rather interesting. In the orientable case: The first invariant is the set of *ends* of the space, meaning, the inverse limit of the set of components of complements of compact sets. I think this is the same as the closure of the equivalence relation extended to the rest of the circle, minus the surface, but I haven't thought it through properly. The set of ends can be an arbitrary closed homemorphism type of compact totally disconnected space. I believe it's known that the number of types has the cardinality of the continuum; it's easy to see it's at least $\omega$ and at most $2^\omega$. The second invariant is a closed subset of the set of ends, that is the limit of "handles", that is, pairs of curves that intersect in one point. Finally, if there are only finitely many handles, you need the genus.
2. There is a structure called a geodesic lamination in the disk: a closed collection of disjoint hyperbolic lines ending at infinity. Given a lamination, you can identify the endpoints. This kind of identification arises in multiple ways. First, it describes the structure of the cut locus" for a curve in the plane. For which pairs of points is there a circle whose interior is contained on the inside, but touches only those two points? Second, if you identify all lines to points, the resulting space is still homeomorphic to the plane. They give explanations for the structure of Julia sets and limit sets of Kleinian groups, etc.
More generally, one can represent identifications by collections of lines that are allowed to cross, but the theory becomes more complicated. In the case of Julia sets, the identification is in some sense isometric. (Actually, the involution is nontrivial only on a set of measure 0 with dense complement, but it is a limit of isometric involution on finite collections of intervals.) If you identify the circle using two geodesic laminations, one on the inside and one on the outside, you typically get $S^2$, and the circle becomes a space-filing curve: a sort of Hamiltonian path that is a limit of simple curves.
3. For a dynamical system, it is natural to look at the set of all orbits. This is an identification space that is somtimes "nice" but more often is (in Ivanov's words) weird: non-Hausdorff and failing other separation properties. Nonetheless, these quotient spaces often carry structure that is helpful to understanding the dynamics. A simple example: the flow in the plane minus the origin $\phi\_t(x,y) = (\exp(t) x, \exp(-t) y) of a linear differential equation. The orbit space is a 1-dimensional manifold, because you can take a cross-section near any point that meets any orbit at most once, so it embeds in the quotient space. However, the quotient is not Hausdorff.
| 10 | https://mathoverflow.net/users/9062 | 38394 | 24,657 |
https://mathoverflow.net/questions/38380 | 4 | When solving the heat equation on say $\mathbb{R}$ (or $[0,2\pi]$, whichever is easier to talk about) we are posing Cauchy data on the surface $t=0$. My understanding is that $t=$constant are precisely the characteristic surfaces of the heat equation.
I realize this question may seem elementary to those who know the answer, but I'm confused as to how this makes sense. I understand how to solve the heat equation using Fourier series (on $[0,2\pi]$) or the fundamental solution on $\mathbb{R}$ but I thought we were not able to pose Cauchy data on characteristic surfaces? Is the point here that we are not solving the equation using the characteristics? Shouldn't $u\_t$ be automatically specified in terms of what $u\_{xx}(t=0,x)$ is?
| https://mathoverflow.net/users/8755 | Posing Cauchy data for the heat equation: $t=0$ a characteristic surface? | The concept of a non-characteristic surface for a PDE or a system of PDE's is useful primarily for only establishing the existence and uniqueness of real analytic or formal power series solutions to the initial value problem using the Cauchy-Kovalevsky theorem.
The generalization of this to the smooth category is the class of hyperbolic PDE's, where you need the initial hypersurface to be more than characteristic. It has to be space-like.
Parabolic equations are a set of PDE's for which the initial value problem in time is well-posed only in the smooth category and not in the real analytic category. Indeed, if you try to apply the Cauchy-Kovalevsky theorem, the $t = c$ hypersurface is characteristic. From the point of view of this theorem, the initial value problem for the standard heat operator is well-posed only for hypersurfaces that are noncharacteristic with respect to the space-like Laplacian. The time derivative is lower order and does not even appear in the symbol.
However, in any useful application of parabolic equations, you want smooth solutions to the initial value problem in time. The explicit formula for the heat kernel shows that the solution is not necessarily real analytic in time. For parabolic equations, the study of solutions that are real analytic or have a power series expansion in the time variable is of little interest. You want to use a weaker category of solutions.
| 8 | https://mathoverflow.net/users/613 | 38403 | 24,662 |
https://mathoverflow.net/questions/38414 | 26 | A friend of mine recently asked me if I knew any simple, conceptual argument (even one that is perhaps only heuristic) to show that if a triangulated manifold has a non-vanishing vector field, then Euler's formula (the alternating sum of the number of faces of given dimensions) vanishes. I didn't see how to get started, but it seems like a good MO question.
| https://mathoverflow.net/users/7311 | Euler Characteristic of a manifold with non-vanishing vector field, | Consider a straight simplex $\Delta^n$ in $\mathbb R^n$ and take a generic constant vector field $v$ (transversal to the faces of $\Delta^n$). Choose all faces of $\Delta^n$ such that the field moves the center of the face inside the simplex. Then the alternating sum of the numbers of these simplices (signed by the parity of the dimension) is zero.
Now, if you have a fine enough triangulation of $M$ and a vector field transversal to all faces, we can apply the above reasoning to the whole manifold.
**Edited**. There was an explanation here with a mistake (spotted by Sergei) of why each simplex contributes zero, but the statement is correct. The new proof is a follows: $(-1)^{n-1}+(-1)^n=0$.
Proof. Let us say that $v$ is the sunlight. Then it enlightens a part of the simplex $\Delta^{n+1}$. Consider the shade from $\Delta^n$ on some plane below the simplex. The shade is an convex set. It is naturally decomposed into simplices, so the sum of simplices over this shade is $(-1)^{n-1}$ (because the simplices in the boundary of this convex set do not contribute). And we also get $(-1)^n$ for $\Delta^n$.
| 22 | https://mathoverflow.net/users/943 | 38417 | 24,670 |
https://mathoverflow.net/questions/38421 | 2 | Hello,
do you know any papers or books that use algebraic geometry in evolutionary game theory ?
| https://mathoverflow.net/users/9166 | Applications of Algebraic Geometry in Evolutionary Game Theory | Perhaps Datta's [papers](http://math.berkeley.edu/~datta/) and his thesis [Algebraic Methods in Game Theory](http://math.berkeley.edu/~datta/thesis.pdf) could be a good place to start.
| 2 | https://mathoverflow.net/users/2149 | 38423 | 24,673 |
https://mathoverflow.net/questions/38185 | 3 | Given a Lebesgue measurable set A with strictly positive measure, can we find an open interval (a,b) such that x belongs to A for almost every x in (a,b)?
Thanks in advance for any comments!
| https://mathoverflow.net/users/36814 | About measurable sets and intervals | The usual Cantor set constructed by removing 1/3 at each step is nowhere dense but has measure 0. However, there exist nowhere dense sets which have positive measure. The trick is to try to remove less, for instance you remove 1/4 from each side of [0,1] during the first step then 1/16 from each pieces etc...
The resulting set is the fat Cantor set: it is nowhere dense and it has positive measure.
| 4 | https://mathoverflow.net/users/3859 | 38425 | 24,674 |
https://mathoverflow.net/questions/38308 | 3 | hi,
assume that I have a function $q$ which is a Fourier Multiplier of order zero, i.e.
$$
\left|\left( \frac{d}{dx}\right)^nq(x)\right|\lesssim \left(\frac{1}{1+|x|}\right)^n\quad \mbox{for all }n\geq 0.
$$
Can I conclude that $q$ is the Fourier Transform of a finite Radon Measure?
If not, what are the conditions on $q$?
Actually, what I really want to know is when is the Fourier Multiplication Operator defined by
$q$ bounded on $L^p$? The Mihlin Multiplier theorem gives an affirmative answer for $1<p<\infty$. How about the boundary cases $1$ and $\infty$?
| https://mathoverflow.net/users/6035 | fourier transform of radon measure | No, such function doesn't need to be a Fourier transform of a finite measure (and, thereby, doesn't need to be a multiplier in $L^1$ or $L^\infty$). This is well-known and the most classical counterexample is just a smoothed Heaviside function $q$ that is $0$ on $(-\infty,1]$, $1$ on $[1,+\infty)$ and whatever you want (just keep it $C^\infty$) in between. One reason it is not a Fourier transform of a measure is that on one hand, such measure should simultaneously have and have not point masses by Wiener's formula
$$
\sum\_x|\mu(\{x\})|^2=\lim\_{|I|\to+\infty}|I|^{-1}\int\_I |\widehat \mu(y)|^2\ dy
$$
(usually it is written for the interval $[-T,T]$, but the truth is that the limit can be taken over any sequence of intervals whose lengths tend to infnity). The existence of this limit is quite a restrictive condition on the absolute value of a bounded function that wants to be a Fourier transform of a finite measure.
The full description of Fourier transforms of measures seems beyond reach (not in the sense that it is hard to figure out what happens for each particular function or whether a given condition is necessary or sufficient, but in the sense that there are no easily checkable conditions that would be exactly equivalent to that property).
| 5 | https://mathoverflow.net/users/1131 | 38428 | 24,676 |
https://mathoverflow.net/questions/35531 | 15 | For a category $\mathcal{C}$, let $\mathcal{C}-Set$ denote the category of functors $\mathcal{C}\to{\bf Set}$. Recall that given a functor $F\colon\mathcal{B}\to\mathcal{C}$, the ``composition with $F$" functor is denoted $F^\*\colon\mathcal{C}-Set\to\mathcal{B}-Set.$ It has a left and a right adjoint, $F\_!$ and $F\_\*$. I call these functors the pullback, the left pushforward, and the right pushforward.
Let $p\colon A\to B$ be a function of sets, thought of as a functor $P\colon[1]\to{\bf Set}$, where $[1]$ is the "free-arrow category," $[1]="\bullet\to\bullet$." Suppose one wants to find the image of $p$, but he or she can only use pull-backs, left pushforwards, and right pushforwards to manufacture it. In other words, suppose one wants to find a zigzag of functors $[1]=:C\_0\leftarrow C\_1\rightarrow C\_2\leftarrow C\_3\rightarrow\cdots\rightarrow C\_n=[0]$ such that if we perform a pullback along all leftward functors and either a left pushforward or a right pushforward along rightward functors, then the end result will be the image set $im(p)$ of $p$ (considered as a functor $[0]\to{\bf Set}$).
This can be done. To do it, I used a sequence of the form $$[1]\leftarrow C\_1\rightarrow C\_2\rightarrow [0].$$ If the functors are denoted (left to right) by $F,G,$ and $H$, I found that $H\_! \circ G\_\*\circ F^\ast (P)=im(P)$.
I'm not going to bore you with the details of $C\_1, C\_2$ and $F,G,H$.
Here's the question. I've seen things like $H\_! \circ G\_\*\circ F^\ast$ before in the context of polynomial functors. Unfortunately, I don't know enough about them to know if there's a connection. Is there?
I also don't know if I can get the whole epi-mono factorization somehow. I haven't worked that long at it, but suppose I want not to end up with the set $im(p)$ but instead the maps $A\to im(f)\to B$. Can I achieve that by use of pullbacks and pushforwards as above (with $C\_n=[2]$ now)? Is there any rhyme or reason to such constructions?
Thanks.
| https://mathoverflow.net/users/2811 | First: upper-star, then: lower-star, finally: lower-shriek | First of all: yes, there's certainly a connection. See <http://ncatlab.org/nlab/show/polynomial+functor>. If the base category is $Set$, the composite
$$Set/W \stackrel{f^\ast}{\to} Set/X \stackrel{g\_\ast}{\to} Set/Y \stackrel{h\_!}{\to} Set/Z$$
first takes a $W$-indexed set $S\_w$ to an $X$-indexed set $T\_x = S\_{f(x)}$, then takes this to the $Y$-indexed set $U\_y = \prod\_{x: g(x) = y} T\_x$, then takes this to the $Z$-indexed set $V\_z = \sum\_{y: h(y) = z} U\_y$. Putting this together, the composite is a family of polynomials, each a sum of monomial terms
$$P(\ldots, S\_w, \ldots) = (z \mapsto \sum\_{y \in h^{-1}(z)} \prod\_{x \in g^{-1}(y)} S\_{f(x)})$$
I'll give a quick example. Suppose we want to express the free monoid functor
$$F(S) = \sum\_{n \geq 0} S^n$$
in this form. Then we take $W = 1$, $X = \mathbb{N} \times \mathbb{N}$, $Y = \mathbb{N}$, $Z = 1$. There's only one choice for $f$ and $h$, and $g$ is rigged so that the fiber over $n \in \mathbb{N}$ is an $n$-element set: $g(m, n) = m + n + 1$. One can easily check this works.
As for the other question: it would have been nice if you had "bored" us! Because I don't see how to reconstruct what you did. What I have to get the image is a zig-zag of length 4
$$Set^{[1]} \stackrel{F^\ast}{\to} Set^{C\_1} \stackrel{G\_\ast}{\to} Set^{C\_2} \stackrel{H^\ast}{\to} Set^{C\_3} \stackrel{J\_!}{\to} Set^{[0]}$$
where $C\_1$ is the generic cospan $a \to c \leftarrow b$, $C\_2$ is the generic commutative square, $C\_3$ is the generic span $a \leftarrow d \to b$, and then $G$ and $H$ are the evident inclusion functors, and $F$ takes each arrow of the generic span to the arrow of $[1]$. Then $F^\ast$ takes $p: A \to B$ to the cospan consisting of two copies of $p$; hitting this with $G\_\ast$ takes this cospan to the pullback square (pulling back $p$ against itself); hitting this with $H^\*$ restricts the pullback square to the span consisting of the pullback projections; finally, hitting this with $J\_!$ takes this span to its colimit = pushout, which is the same as the coequalizer of the pullback projections (because they have a common right inverse). (Based on his comment, I'm guessing that some guy on the street was doing more or less the same thing.)
Could you tell us what you had in mind?
| 7 | https://mathoverflow.net/users/2926 | 38433 | 24,680 |
https://mathoverflow.net/questions/38434 | 1 | Suppose $A$ is a positive definite matrix and $B$ is a non-symmetric
matrix with all positive principal minors.
Is their product $AB$ a matrix with all positive principal minors?
I believe the answer is yes, and I have been trying to find a proof but got stuck along the way. The wiki page for minor gives a corollary to the Cauchy-Binet formula which I think may be of use:
<http://en.wikipedia.org/wiki/Minor_%28linear_algebra%29#Applications>
Thank you,
Alex
| https://mathoverflow.net/users/7154 | Principal Minors of Matrix Product | This isn't true even if $A$ and $B$ are both symmetric and positive definite. For example, let $$A=\begin{pmatrix} 1 & 2\\2 & 5\end{pmatrix}, \quad B=\begin{pmatrix} 1 & -2\\-2 & 5\end{pmatrix},\quad\text{then}\quad AB=\begin{pmatrix} -3 & 8\\-8 & 21\end{pmatrix}.$$
| 6 | https://mathoverflow.net/users/5740 | 38436 | 24,681 |
https://mathoverflow.net/questions/38437 | 31 | There is a question someone (I'm hazy as to who) told me years ago. I found it fascinating for a time, but then I forgot about it, and I'm out of touch with any subsequent developments. Can anyone better identify the problem or fill in the history, and say whether it's still unsolved? It's a challenging question if I've gotten it right. Here it is:
Suppose you have some kind of machine with two buttons, evidently designed by people with poor instinct for UI. The machine has many states in which the buttons do different things.
Here are the assumptions:
1. There is no periodic quotient of the state space: no way to label states by an n-cycle so that both buttons advance the label by 1 mod n.
2. It is not reversible: there are situations when two states merge into one.
3. It's ergodic: you can get from any state to any other state by some sequence of buttons.
Now suppose its dinky little LCD is faded or broken, so you can't actually tell what the state it's in. Is there necessarily a universal reset code, a sequence that will get you to a known state no matter where you start?
(Formally, this is a finite state automaton, or an action of the free 2-generator semigroup on a finite set, and asks whether some element acts as a constant map).
| https://mathoverflow.net/users/9062 | Is there a reset sequence? | I believe you are referring to the [Road coloring theorem](http://en.wikipedia.org/wiki/Road_coloring_problem). It was solved [in this preprint](http://arxiv.org/abs/0709.0099).
| 17 | https://mathoverflow.net/users/121 | 38440 | 24,682 |
https://mathoverflow.net/questions/38454 | 9 | In a bonus exercise last year, we were asked to compute the completion in general of such a stalk on a smooth manifold of dimension $n$ (it is isomorphic to the ring of formal power series over $\mathbb{R}$ in $n$ unknowns). It's clear that this is a bad case to work with, since smooth manifolds admit bump functions (which allows us to prove that there exists a nonzero element in the intersection of all (finite) powers of the maximal ideal), and therefore the completion contains very little data.
However, what kind of "stuff" does this technique allow one to do in the analytic/holomorphic cases? Similarly, in the algebraic case, we can often look at the henselization for the same information, but I still am not really sure why one would want to do so in the first place. That is, what geometric idea corresponds to the idea of "completion" (including henselization and strict henselization depending on the context) in the same way that localization at a prime corresponds to taking stalks geometrically?
| https://mathoverflow.net/users/1353 | What information does the completion of a stalk at its maximal ideal give us in the holomorphic, analytic, or algebraic cases? | First I think you are a little bit unfair when you say that the completion of
the ring of germs of $C^\infty$-functions "contains very little". Mapping a
function into that completion gives you the Taylor series of function which
contains a lot of data about the function even though you certainly miss a
significant amount of information...
Anyway, you are certainly right looking at the (strict) Henselisation of a local
ring is often better than looking at the completion. This is mainly because the
Henselisation is a direct limit of very well-behaved extension rings and hence
many things which are defined over the Henselisation will be defined over one of
these extension rings. However, if you look at the local rings of analytic
spaces they already are Henselian yet you still use the completion even in those
case. The reason for that is that constructing elements of the completion can be
done by a step by step by step procedure, constructing one term of a Taylor
expansion at the time. In classical complex analysis one then usually performs a
closer analysis and shows that the resulting power series is actually
convergent but the first step is still important.
From that point of view the Artin approximation (and generalisations) gives a
very general criterion for when certain processes automatically give convergent
power series provided that it gives any power series at all. Note that there are
many more classical results which allow you to pass to the completion without
using something like the approximation theorem. One such is that the completion
(of a Noetherian local ring) gives a faithfully flat extensions which in
particular allows you to check many equalities by passing to the completion.
However, if we back down to an algebraic scheme over $\mathbb C$ then you get
several local rings; the local ring of a point, its Henselisation, the ring of
germs of analytic functions and the completion. The ring of germs of analytic
functions doesn't make algebraic sense so Henselisation and completion are
algebraic substitutes (from "both sides"). The Henselisation is closer to the
original ring which is an advantage but is often more difficulat to work with,
while the completion is easier to work with but it is more difficult to get back
to the original ring. The approximation theorems should be seen as a very
powerful way of getting back to the Henselisation and then one can work at
getting from the Henselisation to the original or one can stay at the
Henselisation (this is what the étale topology is all about). (This is the
optimistic view on the approximation theorems, rather than the pessimistic one
that they say that you never need to pass to the completion as everything
already lies in the Henselisation.)
**Added**: To add some specific examples. On is the proof that a regular local ring is a UFD (let us assume that it contains a copy of its residue field to simplify). This one shows by first showing that the ring is UFD it its completion is. For the completion which then is a power series ring over the residue field one can use the Weierstrass preparation theorem to show that one can reduce to the polynomial ring in the last variable over the power series ring in all but the last. This is a UFD by induction and the fact that the polynomial ring over a UFD is a UFD ("Gauss lemma"). Here the Henselisation does not appear at all.
Another (far more sophisticated) example is the one which I guess was one of Artin's motivation for the approximation theorem to begin with. Here one wants to show that some functor is representable by an algebraic space. This means constructing a universal element over some suitable base. The first step is to use deformation theory to show that the functor (for a fixed point over some field) is prorepresentable. This is done exactly by showing that it is representable over the category of local Artinian rings for which a fixed power of the maximal ideal is zero. This is done by induction over the fixed power (and hence can be said to do a Taylor expansion one going from one order to the next). The end result (if everything works!) is a formal deformation over some complete local ring. Then one uses further properties of the functor to show that this formal deformation is given by an actual element of the functor (this typically uses Grothendieck's GAGA-type results for formal schemes). Then one uses the approximation theorem to show that the element comes from the Henselisation of something of finite type (there is an extra trickyness in that the complete ring is not known beforehand to descend as some such Henselisation). One then stops there and uses the universality to get gluing data for an algebraic space.
Note that there are situations when this doesn't work. A particular class of
examples arise when one is dealing with differential equations: There are
differential equations (with coefficients in the local ring of a smooth variety)
which has no solution in the Henselisation but does have a solution in germs of
analytic functions and there are differential equations that have formal power
series solutions but no convergent ones.
As for this setup for a general locally ringed space I do not know of any
general results except for the situation where a problem can be reduced to a
commutative algebra type problem concerning the local rings of the space which
then can be solved by forgetting about the space altogether.
| 14 | https://mathoverflow.net/users/4008 | 38459 | 24,692 |
https://mathoverflow.net/questions/38450 | 8 | S is uncountable := |$\mathbb{N}$| < |S|
S is noncountable := |S| $\not\leq |\mathbb{N}|$
(X,$T$) is a nice space := (X,$T$) is a compact Hausdorff space without isolated points
Does [ ZF / ZF + Countable Choice ] prove that every nice space is [ uncountable / noncountable ] ?
If not, is it known to prove that the statement implies some choice principle?
What if the spaces are additionally assumed to be metrizable?
Now, that's basically 12 questions, so I certainly don't anticipate answers for all of them.
If it matters, the one I'm most curious about is "Does ZF prove that every nice space is noncountable?".
| https://mathoverflow.net/users/nan | Compact Hausdorff spaces without isolated points in ZF | One of the usual ways of proving in ZFC that every compact
Hausdorff space $X$ without isolated points ([perfect
space](http://en.wikipedia.org/wiki/Perfect_space)) is
uncountable is by proving that there is a copy of the
Cantor space $2^\omega$ inside it, as follows. Pick two
points and separate them with neighborhoods $U\_0, U\_1$
having disjoint closures. Inside each of these
neighborhoods, pick two points and separating neighborhoods
$U\_{00},U\_{01}\subset U\_0$ and $U\_{10},U\_{11}\subset U\_1$
having disjoint closures inside those neighborhoods, and so
on proceeding inductively. Every infinite binary sequence
$s\in 2^{\omega}$ determines a unique nesting sequence of
these sets, which must be nonempty. And so we have
continuum many points in $X$, so it is uncountable.
This proof, however, makes several uses of the axiom of
choice. First, we have the choices involved with picking
the points to be separated, and second, the choices
involved with picking the separating neighborhoods.
Although there are only countably many choices being made
here, this is an instance of [Dependent
Choice](http://en.wikipedia.org/wiki/Axiom_of_dependent_choice),
a stronger principle than mere countable choice, since the
choices are being made in succession. Finally, third, a
subtle point, we have the choices involved in picking for
each binary sequence a single point from the intersection
of the corresponding nested neighborhoods. After all, there
could be many points in that intersection.
With some additional assumptions on $X$, however, we can get around
these uses of choice, and thereby obtain answers to some of your questions. For example, if we only aim to prove
that $X$ is noncountable, rather than uncountable, then we
may assume towards contradiction that $X$ is countable, which provides for us a
canonical way of picking points from the space. (In the
case of the first use of choice, it would suffice if $X$
were separable, since we could just pick points from a
fixed countable dense set.) If $X$ were a metric space,
then we have a canonical way to pick neighborhoods of any
given point. Also, by making these neighborhoods shrink to $0$ as
the construction proceeds, we ensure that the intersection
of the nested sets contains a single point.
Thus, this argument shows in ZF, without any choice, that
every compact Hausdorff metric space having no isolated
points is noncountable. More generally, it shows, again without any choice, that every separable compact Hausdorff metric space without isolated points has uncountable size at least continuum.
If we have Dependent Choice, then we can prove that every
compact Hausdorff metric space is uncountable of size at
least continuum, since DC allows us to overcome the first
two uses of choice (picking the points and the
neighborhoods), and by shrinking the neighborhoods we avoid
the need for choice in the last step.
A clever person may be able to
improve these arguments to cover additional cases.
Meanwhile, let me mention an interesting example on the other side of the question. This example illustrates that several of the usual equivalent formulations of compactness are no longer equivalent in the non-AC context. Namely, it is consistent with ZF that there is an infinite but [Dedekind finite](http://en.wikipedia.org/wiki/Dedekind-finite) set $D$ of real numbers. That is, $D$ is infinite, but has no countably infinite subset. It follows that $D$ has at most finitely many isolated points, since otherwise we could enumerate the rational intervals and find these isolated points, thereby enumerating a countably infinite subset of $D$, which is impossible. Let us simply omit these finitely many isolated points and thereby assume without loss of generality that $D$ is an infinite Dedekind-finite set of reals having no isolated points. Since $D$ is Dedekind-finite, every sequence in $D$ has only finitely many values and hence has a convergent (constant) subsequence. Thus, $D$ is a sequentially compact set of reals. In other words, $D$ is a sequentially compact metrizable space with no isolated points. However, $D$ is not uncountable in the sense you mentioned, since we don't even have $|\mathbb{N}|\leq D$, as there is no countably infinite subset of $D$. Nevertheless, $D$ is noncountable.
| 11 | https://mathoverflow.net/users/1946 | 38467 | 24,697 |
https://mathoverflow.net/questions/38451 | 13 | Let $W:[0,1]\rightarrow\mathbb R$ be standard Brownian motion with $W(0)=0$.
Let $F\_n$ denote the collection of all the $2^n$ many piecewise linear continuous functions $f:[0,1]\rightarrow\mathbb R$ such that $f(0)=0$ and $f$ is linear with slope $\pm \sqrt{n}$ on the intervals $[\frac in,\frac{i+1}n]$ for $0\le i<n$.
Let $\psi\_n$ denote a uniformly randomly chosen element of $F\_n$, i.e., $\mathbb P(\psi\_n=f)=2^{-n}$ for each $f\in F\_n$.
Let $\phi\_n$ denote a uniformly randomly chosen element of
$$
\text{arg min}\_{f\in F\_n}\left(\sup\_{0\le x\le 1}|W(x)-f(x)|\right).
$$
In other words, $\phi\_n$ is an element of $F\_n$ that minimizes the sup-norm distance to $W$. More simply, we can say that $\phi\_n$ is a *nearest walk* to Brownian motion.
*Question:* Do $\phi\_n$ and $\psi\_n$ have the same distribution? In other words, is $\mathbb P(\phi\_n=f)=2^{-n}$ for all $f\in F\_n$? I am mostly interested in the case $n\rightarrow\infty$, but will accept a rigorous answer for $n=2$.
*Motivation*: Donsker's Theorem says that $\psi\_n$ converges weakly to $W$, whereas it is clear that almost surely, $\phi\_n$ converges to $W$ uniformly.
EDIT: Here is a [follow-up question](https://mathoverflow.net/questions/38481/is-the-nearest-walk-to-brownian-motion-approximately-uniform).
| https://mathoverflow.net/users/4600 | Is the nearest walk to Brownian motion uniform? | My understanding of the question is this:
There is a function $V\_n$ that maps the $L^\infty([0,1])$ to the set $F\_n$ of $2^n$ piecewise linear functions as defined. $V\_n$ gives the Voronoi subdivision, taking each point of the big set to the nearest neighbor of the smaller set.
I believe you're asking whether $V\_n$ pushes Brownian measure on $L^\infty$ to uniform measure(?)
This doees not seem plausible, starting with $F\_2$, because unlike random walks and Brownian motion, this process has a memory effect.
Think of the provisional grouping of paths into an upper group and a lower group, based on behavior in the interval $[0,1/2]$. The two distributions of positions at time 1/2 are skewed, because of how Gaussian tails decay rapidly. The provisionally upper group has mean $\sqrt 2/2$, but more of them are less than $\sqrt 2/2$ than greater than $\sqrt 2/2$. This means that those that start provisionally up have less than probability .5 to continue upward, more than probability .5 to go down or to switch to down-up, and low probability of switching to down-down. In other words, I believe the Voronoi measure is biased toward paths that end at 0, rather than at $\pm \sqrt 2$.
It should be possible for someone to compute the exact distribution in this case.
| 7 | https://mathoverflow.net/users/9062 | 38469 | 24,698 |
https://mathoverflow.net/questions/38435 | 6 | Alright, so [a similar question was recently asked](https://mathoverflow.net/questions/38026/generating-n-cycles) about the theoretical bound for generating certain permutations in polynomial time. I had been thinking about a related problem in algorithms (with applications to a specific problem in graph theory - namely, discrete moves of sets of points among the vertices of a graph) and H A Helfgott's question inspired me to ask here.
Suppose I have some "black box" that spits out permutations $\rho\_i \in S\_n$. I know the following things about the permutations it spits out:
* $\rho\_i$ is of cycle type $(k\_i,1,1,\cdots,1)$.
* This "black box" is fast in $n$ (linear in $n$ or so, maybe plus a few log terms).
* If I run this black box long enough, it will spit out all of the $k$-cycles in some subgroup $H \subseteq S\_n$. I don't know what $H$ is *a priori*, although I can tell you (based on other constraints of the general problem) if $H \subseteq A\_n$.
Let $G \subseteq S\_n$ be the group generated by the $\rho\_i$. (Note that $G$ may not in fact be either $H$ or $S\_n$.)
I'd like to test if $A\_n \subseteq G$.
1. Is there a computationally efficient test to see if the $\rho\_i$ act primitively on $[1,n]$? I want to say that if they act transitively and if the $k\_i$ do not all share some nontrivial factor, they act primitively, but I am not sure of this.
2. Assuming that the answer to (1) is yes, I can guarantee that the natural action of $G$ on $[1,n]$ is transitive and primitive. Does this guarantee that $G = A\_n$? If not, what computationally non-intensive criterion do I need to add to guarantee that $G = A\_n$?
Note: right now my algorithm for solving this problem is somewhere in that scary, scary territory beyond $O(n!)$ (yeah, *that's* how I'm testing to see if the darn thing is the alternating group), so any polynomial-time algorithm here would be super-awesome.
| https://mathoverflow.net/users/8345 | Testing permutations to see if they generate $S_n$ | The answer to (1) is yes, primitivity can be checked in O(n^3) time and practical computer implementations have been widely available for decades. See Butler's [Fundamental Algorithms for Permutation Groups](http://dx.doi.org/10.1007/3-540-54955-2) p.76 for this and various related algorithms (such as testing transitivity) explained in a friendly manner. Holt et al.'s [Handbook of Computational Group Theory](http://www.ams.org/mathscinet-getitem?mr=2129747) also contains this material in textbook form. [GAP](http://www.gap-system.org/) contains open-source implementations of most of the algorithms mentioned (for instance IsTransitive and IsPrimitive would be useful).
The answer to (2) is usually yes, since proper primitive groups do not tend to contain many k-cycles. You are just looking for what are called "giant tests" that can be applied to your restricted setting. Some old theorems of Jordan can be used for this in ways that are described in Seress's [Permutation group algorithms](http://www.ams.org/mathscinet-getitem?mr=1970241), especially 10.2.1 and 10.2.2. These are refined to give probabilistic runtime estimates, some of which you could probably use if your black box had a (vaguely) known probability distribution. See also Holt's textbook and GAP's DoSnAnGiantTest.
Section 3.3 of [Dixon–Mortimer's book](http://www.ams.org/mathscinet-getitem?mr=1409812) also contains results which can be used (with other results) to rule out "small" k. In some ways this is done in section 5.3 and 5.4: if k is smaller than √n, then G contains the alternating group.
Be careful about your guarantee in (2). In particular, can you tell if H is transitive and primitive from the graph? If you are only gathering information from the group G, then be careful that G may not be transitive even if H is (where G is generated by all of the elements of H that happen to be cycles).
| 5 | https://mathoverflow.net/users/3710 | 38474 | 24,702 |
https://mathoverflow.net/questions/27609 | 22 | A countable discrete group $\Gamma$ is said to be exact if it admits an amenable action on some compact space.
So clearly amenable groups are exact, but large familes of non-amenable groups are as well.
For many of the families that I know of (ex. linear groups, hyperbolic groups) that are exact, they also satisfy the von Neumann conjecture (i.e. that if they are non-amenable then they have subgroup isomorphic to a free group.)
So my questions is:
Are there examples of exact groups that are non-amenable and do not contain free subgroups?
| https://mathoverflow.net/users/5732 | An example of a non-amenable exact group without free subgroups. | I did not check the details, but most probably Gromov's random groups can be made torsion as well (and thus will not contain $F\_2$). Just impose the relations $u^{n\_u}$ on the steps with even numbers and do Gromov's construction of embedding the next graph of an expander on steps with odd numbers. Here $u$ runs over all words of the free group (more precisely, $u\_k$ is the smallest length word that has infinite order in the group number $k-1$, and $n\_{u\_{k}}>>1$, odd, is chosen after the word $u\_k$ is determined). Thus there are non-exact groups without free subgroups. If one does only the even steps of this construction, then the resulting group will be torsion and non-amenable (it will have the previous group as its factor) and most probably exact, although I am not as sure about it as about the non-exact example above. So the answer to the original question is most probably "yes". The way to prove exactness can be through finiteness of asymptotic dimension.
| 12 | https://mathoverflow.net/users/nan | 38476 | 24,703 |
https://mathoverflow.net/questions/38487 | 1 | I remember the following problem back from my undergraduate days:
>
> Suppose that $f\in C^1(\mathbb{R}^n)$ is a map such that for all *p*, we have $df(p)\in SO(n)$. Then, $df$ is a constant rotation, or in other words, $f$ is an affine rotation.
>
>
>
There is a clever proof of this fact using local inversion to prove that at any point, your map $f$ must be locally an orientation-preserving isometry.
I can't think of any similar-looking result, so my question is the following:
>
> *Is this exercise just an isolated fact, or is it a special case of a more general phenomenon?*
>
>
>
Given how often we build undergraduate-level problems from special cases, I am now curious.
| https://mathoverflow.net/users/8212 | Generalisation of a multivariable calc problem | There is also a generalization in a slightly different direction. The simplest version I know is that if $f:\mathbb R^n\to \mathbb R$ and $|\nabla f|=1$ everywhere (which is true for each coordinate mapping in your original problem), then $f$ is linear. The solution is to go along the gradient accent/descent curves and to note that they are straight lines because otherwise you have too big increments over the straight shortcuts. Once you know that, you note that if you have any point $p$ with $f(p)=y$, then any other point $q$ with $f(q)=y$ is on the plane perpendicular to the gradient line through $p$ (otherwise we can go far enough on that gradient line and then take a shortcut to $q$ from there). Since the level sets are nice locally, we conclude that they are planes, after which everything becomes obvious.
Unfortunately, I do not remember now what other theorems are there about what can and what cannot be an absolute value of a gradient and how rigid is the set of solutions if it is non-empty.
| 6 | https://mathoverflow.net/users/1131 | 38493 | 24,712 |
https://mathoverflow.net/questions/38492 | 2 | I would like to show that for $s \in \mathbb{R}$ and a nonnegative integer $k$
$$
\triangle^k ((1+n)^s) \lesssim (1+|n|)^{s-k}
$$
where $\triangle$ is the discrete derivative, i.e. $\triangle^1 ((1+n)^s) = (2+n)^s - (1+n)^s$.
---
This is easy when $s \in \mathbb{Z}$, and in the continuous analogue because
$$
\partial\_x^k (1+x)^s = (s)(s-1)\cdots (s-k+1) (1+x)^{s-k}
$$
---
I think that you can use the generalized binomial theorem to prove this, but I was wondering if there was anything more straightforward, e.g. some kind of convexity argument to use the continuous case.
---
Note: I wasn't sure about the tags, feel free to re-tag as appropriate.
| https://mathoverflow.net/users/1540 | Elementary proof of bounds on discrete derivative applied to $(1+n)^s$ | There's a generalization of the mean value theorem that can be
applied here, namely that if $f$ is smooth enough then for each $x$
there is $y$ such that
$$\Delta^kf(x)=f^{(k)}(y)$$
and $x < y < x+k$.
**Added** see
[Is it true that all the "irrational power" functions are almost polynomial ?](https://mathoverflow.net/questions/18054/is-it-true-that-all-the-irrational-power-functions-are-almost-polynomial/18056#18056) .
| 3 | https://mathoverflow.net/users/4213 | 38494 | 24,713 |
https://mathoverflow.net/questions/38475 | 11 | Does any additive, covariant functor preserve direct sum?
| https://mathoverflow.net/users/9141 | Additive, covariant functor preserve direct sum? | I'm not sure what is being asked, but I'll try rephrasing what I suspect the question is: if $A$ and $B$ are enriched in abelian groups, and if $F: A \to B$ is a functor such that all the maps $\hom(a, a') \to \hom(F(a), F(a'))$ are homomorphisms, then is it true that $F$ takes finite sums (coproducts) in $A$ to finite sums in $B$? The answer to that question is 'yes'. José Figueroa-O'Farrill has already given some hints, but I think they deserve to be amplified.
The result is standard in homological algebra textbooks (I believe you can find it in Hilton and Stammbach for instance). It helps to observe first that in $Ab$-enriched categories $A$, or even in categories enriched in commutative monoids, finite sums in $A$ are the same as [biproducts](http://ncatlab.org/nlab/show/biproduct), which roughly speaking are simultaneously products and coproducts (in a compatible way).
Suppose $a + b$ is a coproduct in $A$. Then by definition there are coproduct inclusions $i\_a: a \to a + b$, $i\_b: b \to a + b$. Also by definition, a map $f: a + b \to c$ is uniquely determined by the pair of maps $f\_a := f \circ i\_a$, $f\_b = f \circ i\_b$. In particular, we have a map $p\_a: a + b \to a$ determined by the pair $1\_a: a \to a$, $0: b \to a$, where $0$ denotes the zero element in the abelian group $\hom(b, a)$. Similarly, we have a map $p\_b: a + b \to b$. By definition, we have the equations
$$p\_a i\_a = id\_a \qquad p\_a i\_b = 0 \qquad p\_b i\_a = 0 \qquad p\_b i\_b = id\_b$$
Note also that $i\_a p\_a + i\_b p\_b = 1\_{a+b}$, because each side is a map $f$ satisfying the equations $f i\_a = i\_a$, $f i\_b = i\_b$, and there is only one such map (by the universal property of coproducts).
Now, the crucial point is that whenever you have maps $i\_a: a \to c$, $i\_b: b \to c$, $p\_a: c \to a$, $p\_b: c \to b$ satisfying the four equations above and a fifth equation $i\_a p\_a + i\_b p\_b = 1\_c$, then $c$ is simultaneously a product and coproduct of $a$ and $b$; in this situation we call $c$ a biproduct. I invite you to write out the proof. (Hint: given $g: d \to a$ and $h: d \to b$, define $\langle g, h \rangle: d \to c$ to be $i\_a g + i\_b h$. Show this is the unique map such that $p\_a \langle g, h \rangle = g$ and $p\_b \langle g, h \rangle = h$, so that $c$ satisfies the universal property of a product. By dualizing this argument, $c$ also satisfies the universal property of a coproduct.)
But any $Ab$-enriched functor preserves these five equations since it preserves composition, identities, zero elements, and addition. Therefore $Ab$-enriched functors $F: A \to B$ take biproducts in $A$ to biproducts in $B$. That is the same as saying it takes coproducts in $A$ to coproducts in $B$.
Just to round out the story: there are converses to these statements as follows. If a category $A$ has biproducts (see the nLab page cited above), then $A$ is automatically enriched in commutative monoids: the sum of two morphisms $f, g \in hom(a, b)$ is given by the composite
$$a \stackrel{\Delta\_a}{\to} a \oplus a \stackrel{f \oplus g}{\to} b \oplus b \stackrel{\nabla\_b}{\to} b$$
where $\oplus$ denotes biproducts, and the diagonal and codiagonal are defined using the the biproduct structure. (The zero element in $\hom(a, b)$ is the composite $a \to 0 \to b$ factoring through the zero object.) And if $A$ and $B$ have biproducts and $F: A \to B$ preserves them, then $F$ is an enriched functor: preserves the addition and zero carried by the hom-sets.
| 19 | https://mathoverflow.net/users/2926 | 38496 | 24,714 |
https://mathoverflow.net/questions/38217 | 4 | I'm creating a system that will allow people to rate images.
My idea is to use an Elo Rating system (<http://en.wikipedia.org/wiki/Elo_rating_system>) for each image and then use crowdsourcing to have people say if an individual image is better than another i.e
Is A better than B
This will be used to updated the Elo rating of A and B, eventually I would end up ranking all the images from supposedly the best to worse.
For this I have two questions
1. Is this the correct use of Elo or should I be looking at another rating scheme.
2. If the ELO rating is correct and I have 100 images how many "matches" do I need before I can confidently look at the ranking ?
| https://mathoverflow.net/users/9120 | Elo Rating System Help with the Maths around number of matches | By the way, Elo ratings are named after Élő Árpád, so only the first letter should be capitalized.
1) An implicit assumption of the Elo rating system is that if you know the true advantage of A over B, and of B over C, then you know the true advantage of A over C. I don't think that should hold for the preferences for images, so I don't think the Elo system will fit well. The consequence is that the ratings should change based on which matches you set up. If the Elo ratings were a good fit, then it should not matter much which opponent you choose for a particular image. In the Elo system, if A and B are equally matched, and C is preferred 2:1 over A, then the preference for C over B is precisely 2:1. I can imagine that this would not be the case at all, and that you could manipulate the rating of C by choosing whether it is paired against A rather than against B.
Imagine a martial arts tournament. Perhaps an Elo rating would be appropriate if karate practitioners spar against each other, all trying to strike each other from a distance. However, very different skills would be used if they faced someone using judo who attempted to close to apply choke holds and throws, and there should be little reason to suppose that the variations between the karate students would predict how well they would do when facing someone using a completely different strategy. Back to images: Perhaps an Elo rating would work comparing images of different models of cars against each other for wallpaper, but would comparisons between cars predict how people compare an image of a car against an image of food or a person? Some people might almost always choose the image of food over a car as wallpaper, and others might do the reverse, which is only consistent with an Elo rating if the preferences between car images can't be strong.
2) Different implementations of the Elo system use different parameters. However, in general, if the Elo system is appropriate, then an individual's rating in a large population behaves as a random walk in a potential well centered at the true rating. One consequence is that when a rating is close to the true rating, there is an exponential decay of the influence of perturbations. Another is that there is a stable distribution which is roughly normal about the true rating. I looked at these for Elo ratings in backgammon for a nonmathematical audience in [this article](http://www.bkgm.com/articles/Zare/Ratings.html), originally published in the online magazine GammonVillage.
| 10 | https://mathoverflow.net/users/2954 | 38500 | 24,715 |
https://mathoverflow.net/questions/38498 | 8 | A fundamental construction in a first course on manifolds is to build a smooth function $\psi\colon \mathbb{R} \to \mathbb{R}$ with the property that for some $0<\delta<\epsilon$ we have
* $\psi(x)=1$ for $|x|\leq\delta$;
* $\psi(x)=0$ for $|x|\geq\epsilon$.
Using this "bump function", one can do all sorts of "gluing" tricks: for example, if $f\colon \mathbb{R}^n \to \mathbb{R}^n$ is any smooth map and $\epsilon>0$ is such that $|f(x)|<\epsilon$ for all $|x|\leq\delta$, then we can build a smooth map $F\colon \mathbb{R}^n\to \mathbb{R}^n$ such that
* $F(x) = f(x)$ for $|x|\leq\delta$;
* $F(x) = x$ for $|x|\geq\epsilon$.
However, the map we obtain in this manner does not necessarily preserve all the nice properties of $f$. For example, if $f$ is a local diffeomorphism, it does not immediately follow from the above construction that $F$ can also be taken to be a local diffeomorphism.
Intuitively, it seems clear that this is (usually) the case:
* For linear maps with $n=1$, it's just a matter of drawing a smooth curve that starts on the line $y=\lambda x$ and goes to the line $y=x$ without ever having a horizontal tangent. (Of course we must take $\lambda>0$ for this.)
* For linear maps with $n=2$, the image of a small ball is an ellipse, so by smoothly deforming the ellipse into a ball, rotating so that the map is a multiple of the identity, and then using the trick from $n=1$ to make the eigenvalues equal to $1$, we can find a smooth homotopy $f\_t$ such that $f\_0=f$ and $f\_1$ is the identity, and furthermore, setting $F(x) = f\_t(x)$ for $|x| = \delta + t(\epsilon - \delta)$ makes $F$ a diffeomorphism.
* Since linear maps approximate arbitrary maps, the above procedure ought to generalise. (Modulo the restriction that $Df(0)$ should have positive determinant.)
I expect that there's a general result along these lines, and that it is quite standard and well-known. But I don't know it (and I'd rather re-invent the wheel as few times as possible). Can someone help me out with a statement of a general theorem, and ideally a reference?
**Edit**: Since that was kind of rambling, here's the specific question. Let $B(r)$ denote the (open) ball of radius $r$ in $\mathbb{R}^n$ centred at the origin. Suppose $f\colon B(\delta) \to \mathbb{R}^n$ is a diffeomorphism onto its image, and suppose $\overline{f(B(\delta))} \subset B(\epsilon)$. Let $\delta' < \delta$. Does there necessarily exist a diffeomorphism $F\colon \mathbb{R}^n \to \mathbb{R}^n$ that agrees with $f$ on $B(\delta')$ and is the identity map outside of $B(\epsilon)$?
I think we need to require that $Df(0)$ have positive determinant. Are there any other obstructions? If there are, can they be removed (for a given $f$) by decreasing $\delta$ so that $f$ is close to being a linear map? Is there a general theorem from which all this follows?
| https://mathoverflow.net/users/5701 | Gluing two diffeomorphisms together | The group $GL(n,R)$ has an affine structure, coming from the coefficients of the group.
It has an easily described and fairly large convex neighborhoods of the identity, satisfying that for no vector $V \ne 0$ is $A(V)$ negative real multiple of $V$ (*i.e.*, no negative real eigenvalues).
The first derivative of a convex combinations of two diffeomorphisms, the first derivative is the convex combination of derivatives, plus a correction term of the difference of derivatives applied to the gradient of the convex coefficient. Convex combinations of more terms are similar. In any case, if you bound the maximal gradient of a partition of unity, you can get an explicit bound for collections of first derivatives of local diffeomorphisms guaranteeing that there compostion is diffeomorphism.
I think a bigger issue to watch for is the difficulty in practice to give an actually valid atlas for an abstract manifold that doesn't start life already embedded in $\mathbb R^m$. For two overlapping charts, an approximate diffeomorphism is fine. However, when three charts overlap, in principle the overlaps must satisfy a cocycle condition. In practice, there aren't many good choices of classes of gluing maps that are simply described and have reasonable group laws; note in particular that convex combinations disrupt relationships of compositions.
One solution is, to define how an intersection of $m$ charts patch together, work in the product of the collection of local coordinates, $\mathbb R^{nm}$. Build the graph of the relation among all the charts at once, rather than doing it pair by pair. Consistency is straightforward to arrange using partitions of unity on the graphs. Even for an overlap of two charts, rather than specify a diffeomorphism which may be hard to invert explicitly, specify the graph of a diffeomorphism, which is trivial to invert.
For the specific question of extending a diffeomorphism defined on a small ball: there is a standard technique that works if you use one of Goodwillie's suggested modifications.
The set of differentiable embeddings of a ball in $\R^n$ retracts to the set of linear embeddings by conjugating with a family of contractions, $\phi\_t(x) = \phi((1-t) x)/(1-t)$.
The set of linear maps retracts to orthogonal maps, using a parametrized Gramm-Schmidt process. Every orhtogonal map is homotopic to either the identity, or some chosen reflection. Then, apply the isotopy extension theorem to make all of $\R^n$ move along with the moving embedding. You can compress everything that happens into the unit ball.
You could also do this in another way: For any neighborhood $U$ of the identity in $GL(n,R)$, any diffeomorphism that is isotopic to the identity can be expressed as the composition of elements in the small ball (and conversely). You can make a sequence of overlapping annular coordinate charts, where in the main part of each one you use an initial segment of the composition. The overlap maps is now in a small neighborhood where the convex combination principle works.
One more remark: in the particular case of dimension 2, if you define a lift of the derivatives to the universal cover of SL(2,R), then there is an averaging process for derviatives that always works, using complex logarithmic coordinates for the plane minus the origin. In the particular case of dimension 3, the universal cover of $SO(3)$ is the group of unit quaternions, homeomorphic to $S^3$, so there is an averaging procedure in that group (and by quadratic form manipulation, in $SL(3,R)$ which works much better than averaging in the obvious coordinates. Among other things, this is very useful for motion control. We used to get workable mathematical specifications for segments of the video *Outside In*. [``We'' primarily means the actual implementers---I especially recall discussions about the difficulties and solutions with my son Nathaniel Thurston].
| 11 | https://mathoverflow.net/users/9062 | 38502 | 24,716 |
https://mathoverflow.net/questions/38497 | 5 | Hey everyone, I was reading about obstruction theory, here's a bit of a summary. Take a cellular space $X$ and a fibre bundle $p:E \to X$ with fiber $F$; consider the problem of extending a section $s$, defined on the $(n-1)$-skeleton over to the $n$-skeleton. We work cell by cell pulling back the bundle via the characteristic map and the section via the restriction of the Char. Map to the boundary of our $n$-cell, since the cell is contractible, the bundle is isomorphic to $D^n \times F$ so the section defines a map from $S^{n-1} \to D^n \times F$ i.e. an element of $\pi\_{n-1}(D^n \times F) \cong \pi\_{n-1}(F)$. Define the obstruction cochain as the element in $C^n(X^n,\pi\_{n-1}(F))$ taking each $n$-cell to the element in the $(n-1)$-homotopy group constructed before.
Here's what's bothering me, in Steenrod's book (The topology of Fibre Bundles) he proves that this cochain is actually a cocycle in a really weird way, it looks to me as if he makes no distinction between the homological boundary and the topological boundary of a cell. Roughly he writes the following composition:
$ C\_{q+1}(X) \stackrel{\partial\_\*}\to Z\_q(X) \stackrel{hurewicz}\to \pi\_q(X^q) \stackrel{f=(p\_2\circ \sigma)\_\*}\to \pi\_q(F) $
And claims that this composition is the value of the obstruction cochain in an $n+1$ cell, how might one verify this?
Not being happy with this proof i went and looked at the one Kirk and Davis' book (Lecture notes on algebraic topology) and found it too complex (I know, I dont like anything sorry).
What I was wondering is if there was a way to prove this affirmation (the obstruction cochain is a cocycle) directly, i.e. denoting the cochain by $\Theta$ doing something like:
$\delta \Theta (e) = \Theta( \partial e) = \Theta (\sum [w\_i;e]w\_i) = \sum [w\_i;e]\Theta(w\_i) = \dots = 0$
(where $e$ is a $(n+1)$-cell, $w\_i$ is a $n$-cell and $[w\_i;e]$ is their incidence number, so that the third term is $\Theta$ evaluated on the cellular (homological) boundary).
Any help on the subject or a good refference is very very much appreciated!
Thanks and have a great week.
| https://mathoverflow.net/users/9187 | Obstruction Cocycles | You should have a look at the paper given in the answer to [my earlier question on obstruction theory](https://mathoverflow.net/questions/31147/obstruction-theory-for-non-simple-spaces). It gives a very nice and direct proof that the obstruction cochain is a cocycle that also works in the case of non-simple spaces (a setting that most modern treatments gloss over). It's written in simplicial rather than cellular language, but I imagine the techniques could be carried over with a bit of effort.
| 3 | https://mathoverflow.net/users/396 | 38505 | 24,717 |
https://mathoverflow.net/questions/38506 | -2 | In an algorithm book once the first example was how to compute a multiplication in a loop (only that, so I just remembered, and wanted to do it programmatically but with all operations)
Multiplication was simple, say 10 \* 4:
base = 0, x = 10, y = 4:
```
While y != 0:
base + x = 10, y--
base + x = 20, y--
base + x = 30, y--
base + x = 40, y-- (y is now 0)
Result of base: 40 (correct)
```
As the opposite of taking away is giving, I assumed the same was for division and multiplication (doubling, halving), but: `"base - x.."` clearly does not give anything useful in an iteration..
Is this impossible? Am I looking at an algorithm to divide completely the wrong way (not like my mult)?
| https://mathoverflow.net/users/9191 | Mul + div using only add/sub ? | Division is repeated subtraction. Take 11 \div 3 as an example.
11 - 3 = 8.
Increment the quotient by 1.
Since 8 > 3, keep going.
8 - 3 = 5.
Increment the quotient by 1.
Since 5 > 3, keep going.
5 - 3 = 2.
Increment the quotient by 1.
Since 2 < 3, set the remainder to 2 and terminate.
I'll leave it to you to figure out the code.
| 0 | https://mathoverflow.net/users/8871 | 38509 | 24,719 |
https://mathoverflow.net/questions/38527 | 25 | Here is a rather pathetic question. In a [comment](http://gowers.wordpress.com/2010/08/21/icm2010-ngo-laudatio/#comment-9720) on Tim Gower's weblog, I tentatively stated that the fundamental lemma was necessary for the [work of Skinner and Urban](http://www.mathunion.org/ICM/ICM2006.2/Main/icm2006.2.0473.0500.ocr.pdf) relating ranks of Selmer groups of elliptic curves to the vanishing of their $p$-adic $L$-functions. Now, I believe it is correct that some endoscopic version of transfer from a unitary group to a general linear group is necessary for the construction of their $\Lambda$-adic representations. However, having a really poor understanding of the actual techniques, I don't know which version is crucial. That is to say, it's entirely likely that some earlier special case is sufficient for Skinner-Urban. Could I trouble some expert to give a brief outline of the situation?
The pathetic part of this is that the journalist I mentioned in the comment will call in about 4 hours, so it would be nice to know before that. Of course I shouldn't have agreed to speak about something I know so little about, but it was hard to refuse under the circumstances. Oh, in case you're worried that I'm going to discuss Skinner-Urban with the fellow, don't. I just want to bone up on the background.
---
Added:
For people who like the idea of linguistic diversity in mathematics, I am including a link to a [report](http://math.postech.ac.kr/~minhyong/kmsfields.pdf) written (with Sugwoo Shin) for the Korean Mathematical Society that expands on the comment to the journalist.
| https://mathoverflow.net/users/1826 | The fundamental lemma and the conjecture of Birch and Swinnerton-Dyer | Dear Minhyong,
My understanding, based on recalling talks of Skinner and also briefly looking over the ICM paper that you linked to is that, yes, they do rely on the fundamental lemma, namely, the fundamental lemma for unitary groups as proved by Laumon and Ngo. Unfortunately, I'm not sufficiently educated in their work, or in the field of unitary Shimura varieties in general,
to be sure whether their Conjecture 4.1.1 is actually a theorem at this point or not. Certainly related results have been proved by Sug Woo Shin and by Sophie Morel, both relying
on Laumon--Ngo. But whether these results address the precise unitary groups considered by Skinner--Urban, I'm not sure. (Just looking briefly at what Skinner and Urban write, it looks as if they are considering quasi-split groups, or at least groups that are not anisotropic,
so that their Shimura varieties are non-compact; thus the work of Morel (who considers the non-compact case) may be more directly relevant than that of Shin (who considers the compact case); but both Morel and Shin's work come with certain technical restrictions, so that the precise requirements of Conjecture 4.1.1 (in particular, *precise* local-global compatibility away from the residue characteristic) may not follow in the full generality considered by Skinner and Urban from the work of either.)
I'm sorry that I can't say more, but I do think that it's safe to say that the fundamental lemma is a crucial (albeit highly technical) ingredient in their program.
At the risk of adding unsolicited and unwanted advice:
You could also mention the complete proof of Sato--Tate for elliptic curves over totally real fields (with no requirement that the $j$-invariant be non-integral, unlike in the original work of Clozel--Harris-Shepherd-Barron-Taylor), which follows from the work of CHSBT + that of Shin (the work on compact unitary Shimura varieties mentioned above, which as I already remarked requires Laumon--Ngo as an ingredient).
This is a pretty nice Diophantine statement (technical for a journalist obviously, but presumably one can convey the gist) with the fundamental lemma forming one of the pillars
that supports it.
| 20 | https://mathoverflow.net/users/2874 | 38530 | 24,730 |
https://mathoverflow.net/questions/38447 | 3 | What is an example of an action of a *linearly reductive* group variety acting on an affine variety with the property that there exists a closed orbit that is not separable?
To be more precisely, let's work over a fixed algebraically closed field $k$. Suppose that we are given an affine variety $X$ and a group variety $G$ acting on $X$. Given a closed point $x \in X$, we define the **orbit** $\operatorname{O}(x)$ to be the image of the map $G \to X$ given by $g \mapsto g x$ . We say that the orbit is **separable** if the natural map
$$
G \to \operatorname{O}(x)
$$
given by $g \mapsto gx$ is separable.
This question is only interesting in characteristic $p>0$. In this case, the condition that $G$ is **linearly reductive** is very strong: it implies $G$ is the product of a multiplicative torus and a finite group of order prime-to-$p$.
| https://mathoverflow.net/users/5337 | What is a closed orbit that is not separable? (ANSWERED) | BCnrd writes: "This happens for every G of positive dimension, since all one needs is a non-smooth subgroup scheme (such as kernel of Frobenius). Indeed, if H is a closed k-subgroup scheme of G then let X=G/H equipped with the natural left G-action (so X is smooth, since G is smooth, regardless of how "bad" H may be). Then the orbit map through x=1 is the natural surjection G→X which is not separable precisely when H is not smooth. The simplest example is G=GL(1) acting on X=G via $t\cdot x=t^px$, whose orbit map through $x=1$ is $t\mapsto t^p$."
| 2 | https://mathoverflow.net/users/66 | 38533 | 24,732 |
https://mathoverflow.net/questions/38544 | 3 | Let $U (\mathbf{R})$ be the standard unipotent subgroup of $SL(3, \mathbf{R})$. So $U(\mathbf{R})$ is the group of 3 by 3 upper triangular matrices with 1s on the diagonal. I am interested in the quotient space $U (\mathbf{R})/ U (\mathbf{Z})$. I think it is just the cube, $[0,1]^3$, but I am having difficulty writing the actual map. The only non-obvious part is the (1,3)-entry where the addition is not straightforward. Also is the generalization that one gets the $\frac{1}{2}(n^2 - n)$ dimensional cube for the standard unipotent subgroup of $SL(n, \mathbf{R})$ correct?
---
**@ Bill**
Your answer helped clear a lot of confusion but it also created a number of new questions:
1. So basically $U(\mathbf{R})$ has three copies of $\mathbf{R}$ however they way they are knit together is not the simple direct sum. Is that right?
2. If I pick an element $u \in U$ is the group generated by $u$ isomorphic to $\mathbf{G}\_a$ ?
3. Is the space $U (\mathbf{R})/ U (\mathbf{Z})$ compact? The Haar measure on $SL(3, \mathbf{R})$ gives a measure on $U(\mathbf{R})$. Then presumably I get a measure on $U (\mathbf{R})/ U (\mathbf{Z})$ what is the volume of the quotient with respect to this measure? Is there a place I can find these calculations?
Ultimately I am interested in the arithmetic quotient $SO(3)\backslash SL(3, \mathbf{R}) / SL(3, \mathbf{Z})$ and $U (\mathbf{R})/ U (\mathbf{Z})$ is supposed to be its 'base' and the end result should be an algebraic surface (hence a 4-dimensional real manifold I think).
| https://mathoverflow.net/users/8811 | Quotients of unipotent groups | The group $U(\mathbb Z)$ is nilpotent, but not isomorphic to $\mathbb Z^3$.
The quotient space is not the torus $T^3$ (= cube with opposite sides identified).
The fundamental domain which glues together to make the quotient can be described, but I'm not sure it's what you really want or need: there are easier ways to understand it.
The element $c $ with the 1,3 entry equal to 1 generates the center of $U(\mathbb Z)$, with quotient $\mathbb Z^2$, and it is the commutator of the other two elementary matrices $a$ and $b$. Geometrically, the
quotient has a map to a torus $T^2$ (factoring out by your troublesome entry) with fiber (preimage of any point) a circle, but the circles form a fiber bundle that is not trivial.
You can map all this into R^3 as a group of affine transformations, and visualize
how it works: $c$ acts as a vertical translation.
The element $a$ translates in the $x$ direction by one unit, but at the same time shears planes perpendicular to the x axis so the y-axis develops a slope of -1/2. The element $b$
is similar, translating in the $y$ direction, taking vertical lines to vertical lines and shearing the $x$ axis to have slope 1/2. Now look at what happens when you do the commutator
$a^{-1} b^{-1} a b$. This group is also called the Heisenberg group.
[alt text http://dl.dropbox.com/u/5390048/AffineGeometries.jpg](http://dl.dropbox.com/u/5390048/AffineGeometries.jpg)
In higher dimensions, the picture is similar but more elaborate, with iterated circle bundles over previous stuff, exhibiting additional shearing. (There is more than one reasonable way to choose coordinates, so don't be surprised if you don't get the exact same description when you work through details).
You can find pictures and descriptions in my book on Three-dimensional Geometry and Topology (among other places). It's the basic example for one of the 8 geometries for 3-manifolds. I first learned this picture by reading Poincare.
---
**Addendum**
In response to the added questions:
Keivan Karai and Max have addressed most of it, in comments to the question.
It's true, there are many sources to read about these things. Many people have studied the geometry of $SO(3)\backslash SL(3,\mathbb R)/SL(3,\mathbb Z)$, it's challenging to come to grips with it completely, but a vast amount is known.
You can interpret $SO(3)\backslash SL(3, \mathbb R)$ fairly concretely as the space of positive definite quadratic forms of determinant 1 on $\mathbb R^3$, and it has a natural metric in which a path of quadratic forms has arc length equal to the total strain it takes to do the associated deformations --- you can imagine an object, perhaps a ball, made of a highly viscous and inelastic material, perhaps modeling clay, that you can squeeze into different shapes that don't spring back. You're squeezing the ball into different ellipsoid shapes via linear transformations, and keeping track of how much energy it takes. (use the $L^2$ norm of the first derivative, as measured in using the current shape).
In 3 dimensions, quadratic forms have 6 coefficients, but the condition that the determinant is 1 brings it down to 5. To think about $SO(3)\backslash SL(3,\mathbb R)/SL(3,\mathbb Z)$, imagine a 3-torus with a flat metric, that is, $\mathbb R / \mathbb Z^3$ for some action of $\mathbb Z$.
The set of possible shapes for the torus, up to isometry is how you can interpret it.
(This is beccause $SL(3,Z)$ is the group of affine automorphisms of the torus, up to translations, so it acts on the set of Euclidean metrics on a torus with fixed basis --- in the double coset space, we're forgetting the basis).
The Heisenberg group gives the small scale geometry of the quotient when you go off to infinity in certain directions --- that is, if you have a torus with one loop $x$ of size $l\_1 < \epsilon$ and another loop $y$ size $l\_2 < l\_1 \epsilon$, then in the moduli space for toruses, the subgroup of $SL(3,\mathbb Z)$ generated by low-energy loops is isomorphic to the Heisenberg group. By making $\epsilon \rightarrow 0$ the geometric size of this cross-section 3-manifold ( where the shortest and second shortest lengths are held constant)
get arbitrarily small. This is *part* of the picture of the space near infinity, but to get a fuller picture, you need to examine the global structure of how different behaviors at infinity connect to one another.
There's a lot more needed before you have a very good picture of what's known about
$SL(3,Z)$ or of this locally symmetric space --- a lot is known from a number of different points of view, and it's challenging to fully understand. This isn't the place for an exposition, even if I knew all the answers (which I don't).
| 16 | https://mathoverflow.net/users/9062 | 38545 | 24,736 |
https://mathoverflow.net/questions/38539 | 2 | Let $G$ be a (discrete, say) group and $\mathbb K$ a field. The *regular representation* $G^{\mathbb K}$ is the vector space of all functions $G \to \mathbb K$. It is a (left, say) $G$-module: given $g\in G$ and $f: G \to \mathbb K$, the action is $g\cdot f: x \mapsto f(xg)$. Then $G^{\mathbb K}$ is a commutative algebra object in $G\text{-rep}\_{\mathbb K}$, the symmetric monoidal category of $\mathbb K$-valued $G$-representations, under pointwise multiplication $f\_1f\_2: x \to f\_1(x)f\_2(x)$.
But the pointwise product is not necessarily the only commutative algebra (in $G\text{-rep}$) structure that can be put on $G^{\mathbb K}$. For example, when $\mathbb K = \mathbb R$ and $G = \mathbb Z/2$, as an algebra $G^{\mathbb K} \cong \mathbb R[\epsilon]/(\epsilon^2 = 1)$, with the $G$-action corresponding to conjugation $\epsilon \mapsto -\epsilon$. The same $G$-module supports the algebra structure $\mathbb R[\epsilon]/(\epsilon^2 = -1) = \mathbb C$, which is patently a different algebra.
My question is whether there are any examples with $\mathbb K = \mathbb C$? I.e.:
>
> Does there exist a group $G$ so that there is a commutative algebra object in $G\text{-rep}\_{\mathbb C}$ that is isomorphic to $G^{\mathbb C}$ as a representation but not as an algebra?
>
>
>
I believe that any such group must be rather large: in particular, I'm sure that it cannot be finite.
| https://mathoverflow.net/users/78 | Are there other algebra structures on the regular representation of a group? | **Summary** Yes, there are many such examples, even for finite groups.
**Construction** Let $W<GL(V)$ be a complex reflection group, $A=\mathbb{C}[V]$ be the algebra of polynomial functions on $V$ and $A^W$ be the subalgebra of $W$-invariants. Then by the [Chevalley–Shephard–Todd theorem](http://en.wikipedia.org/wiki/Chevalley%E2%80%93Shephard%E2%80%93Todd_theorem), $A^W$ is a polynomial algebra and $A$ is a free $A^W$-module. This may be viewed as a deformation of $\mathbb{C}^W$ as follows. For any $z\in\text{Spec}A^W,$ consider the fiber $A\_z=A/zA.$ By the freeness property, each $A\_z$ carries the regular representation of $W.$ For a regular $z,$ corresponding to a $W$-orbit of a regular point in $V,$ the algebra $A\_z$ may be identified with the algebra of functions on the orbit, which consists of $|W|$ points; in particular, $A\_z\cong \mathbb{C}^W$ as a $W$-algebra. But for values of $z$ corresponding to singular orbits, algebras $A\_z$ are not reduced. The most singular fiber is $A\_0=\mathbb{C}[V]/(\mathbb{C}[V]^W\_{+})$ and is a graded nilpotent Frobenius algebra, with one-dimensional socle and radical, called the *covariant algebra* of $W.$
**Example** Let $W=\mathbb{Z}\_n$ acting on the one-dimensional vector space with coordinate $x$ by the $n$th roots of unity. Then regular fibers $\mathbb{C}[x]/(x^n-a)$ with $a\ne 0$ are semisimple and isomorphic to $\mathbb{C}^{\mathbb{Z}\_n}$ (explicitly, $x$ is mapped to $b\sum\_k \zeta^{-k}\delta\_k,$ where $b^n=a$), whereas the singular fiber $\mathbb{C}[x]/(x^n)$ is a graded nilpotent algebra.
| 6 | https://mathoverflow.net/users/5740 | 38548 | 24,738 |
https://mathoverflow.net/questions/38541 | 1 | I am desperately searching for a paper I once have read, but cannot find anymore. It was about a counting function f(n) - the only thing I remember is that f's expression contained n! and 2n - and the question in the title was something like "What does f(n) count?". The paper compared three or so different approaches, and was written as a trialog.
>
> Can someone point me to this paper?
>
>
>
If you know about other papers on this question - "What does f(n) count?", with f(n) any counting-but-what function, I'd be happy to learn about them.
| https://mathoverflow.net/users/2672 | "What does f(n) count?" - in search for a paper | Could it be [*The answer is $2^n·n$! What's the question?*](http://dx.doi.org/10.2307/2589493) by Gary Gordon (Amer. Math. Monthly **106** (1999), no. 7, 636–645)?
| 10 | https://mathoverflow.net/users/989 | 38552 | 24,740 |
https://mathoverflow.net/questions/38549 | 10 | I am trying to understand what exactly Ngo's support theorem says about how the cohomology of compactified Jacobians varies over the versal deformation of a curve with plane singularities.
That is, let $C$ be a compact complex curve with singularities of embedding dimension 2. Let $\mathcal{C} \to \mathbf{V}$ be a versal deformation, i.e., there is a point $0 \in \mathbf{V}$ over which the fibre is $C$, and any deformation of $C$ pulls back from this family. Let $\pi: \overline{\mathcal{J}} \to \mathbf{V}$ be the relative compactified Jacobian of the family; I want to understand the family of cohomologies $\mathrm{R}\pi\_\* \mathbb{Q}$. As pointed out in the [few](http://www.institut.math.jussieu.fr/projets/fa/bpFiles/Ngo.pdf) [references](http://arxiv.org/abs/1009.1862) I've [browsed](http://arxiv.org/abs/0712.0349), this is a situtation to which Ngo's support theorem applies: there is a relative action of the (uncompactified but generalized) Jacobian $\mathcal{J}$, and the locus $\mathcal{V}\_h$ where the affine part of $\mathcal{J}$ is of dimension $h$ is of precisely codimension $h$ (by an old result of Teissier). However, said references go on to treat a more famous example instead -- the Hitchin fibration.
Unfortunately my competence with the machinery of perverse sheaves is so poor that I can barely understand the statement of the support theorem. So, I apologize if the following questions are overly trivial or confused.
0.) Is the application of the support theorem to the situation I describe above treated explicitly in the literature?
1.) Does the support theorem imply that $\mathrm{R} \pi\_\* \mathbb{Q}$ is the IC-sheaf on $\mathbf{V}$ determined by the local system of cohomologies of Jacobians on the locus $\mathbf{V}\_{\mathrm{reg}} \subset \mathbf{V}$ of smooth curves? If not, does it specify what other local systems on what other loci must be added?
2.) Does the knowledge of the Picard-Lefshetz monodromy around nodal degenerations determine (explicitly) the local system of cohomologies of Jacobians on $\mathbf{V}\_{\mathrm{reg}}$, and, if so, does it allow an in principle calculation of the cohomology of the central fibre? Can it be done in practice?
3.) Does it follow that the cohomology of the compactified Jacobian of the central fibre is determined by the topology of the singularity of the discriminant (the complement of the locus of smooth curves)? Or -- perhaps it is necessary to know how monodromies around different nodes interact -- the stratification of the discriminant by the (closures of the) loci with a fixed number of nodes?
4.) Any other comments on how this (or anything else) would be used to compute the cohomology of the compactified Jacobian are welcome.
| https://mathoverflow.net/users/4707 | Ngo's support theorem & the Jacobian over the versal deformation of a planar curve | Very partial answer - I don't think I can comment yet...
I found it helpful to rephrase the statement of the support theorem like this:
Let $R\pi\_\ast \mathbb Q = \bigoplus \_i IC\_{Z\_i}(L\_i)[d\_i]$ be the decomposition of the pushforward sheaf. If Z is one of the $Z\_i$ appearing in the sum above, then $Z$ is the support of a direct summand of $R^{2d}\pi\_\ast \mathbb Q$ (i.e. there is extra stuff appearing in the top cohomology of the fibres). Such a summand only will occur when there are extra irreducible components in the fibres of $\pi$.
In particular, if the fibres of $\pi$ are irreducible, then the only possible support appearing in the direct sum is the whole of V. This means that the pushforward is the IC extension of the local system over the locus where $\pi$ is smooth.
I think this means that in your situation, the answer to (1) is yes (as compactified Jacobians are irreducible for such singularities). I would be interested to find out the answers to (2) and (3).
| 6 | https://mathoverflow.net/users/7762 | 38557 | 24,744 |
https://mathoverflow.net/questions/1988 | 19 | There is supposed to be a strong analogy between the arithmetic of number fields and the arithmetic of elliptic curves. One facet of this analogy is given by the class number formula for the leading term of the Dedekind Zeta function of K on the one hand and the conjectural formula for the leading term of the L function associated to an elliptic curve defined over a number field on the other. One can look at the terms that show up in these two expressions and more or less 'pair' them off as analogous quantities.
One of these pairs consists of the respective regulators. The regulator of a number field K is defined by taking a basis for the free part of the units of the ring of integers and then using the embeddings of K into C, taking logs of absolute values, etc. The regulator of an elliptic curve is defined by taking a basis for the free part of the K points of the curve and then computing the determinant of a symmetric matrix built out of this basis using a height pairing.
My question is: is there some way to view the number field regulator as coming from some kind of symmetric pairing on the units of K? Alternatively, just give some reasoning why these constructions appear so different.
| https://mathoverflow.net/users/493 | Regulators of Number fields and Elliptic Curves | Let me first give you a heuristic "reason", why the regulator in the class number formula looks different from the regulator in the Birch and Swinnerton-Dyer conjecture. It is often more convenient (and more canonical) to combine the regulator and the torsion term in the Birch and Swinnerton-Dyer conjecture: one chooses a free subgroup $A$ of the Mordell-Weil group of $E(K)$ with finite index and looks at the quantity $R(A)/[E(K):A]^2$, where $R(A)$ is the absolute value of the determinant of the Néron-Tate height pairing on a basis of $A$. As you can easily check, the square in the denominator insures that this quantity is independent of the choice of $A$. Now, in the class number formula, the torsion term is replaced by the number of roots of unity in $K$ and that term is not squared. So for such a canonical formulation to be possible, it is reasonable to expect that not the regulator of the number field should be defined through a symmetric pairing on the units, but its square. Then you could make the same definition as in the elliptic curves case, and it would be independent of the choice of finite index subgroup.
Now, that we have established this, there are several ways to bring the two situations closer together.
You could do the naïve thing: take the matrix $M=(\log|u\_i|\_{v\_j})$, where $u\_i$ runs over a basis of the free part of the units (or more generally $S$-units for any set of places $S$ which includes the Archimedean ones), and where $v\_j$ runs over all but one Archimedean place (or more generally all but one place in $S$), v\_0, say. The absolute values have to be suitably normalised (see e.g. Tate's book on Stark's conjecture). Now take the symmetric matrix $MM^{tr}$ and define this to be the matrix of a new pairing on the units. In other words, you would define your symmetric pairing as
$$
(u\_1,u\_2) = \sum\_{v\in S\backslash\{v\_0\}} \log|u\_1|\_v\log|u\_2|\_v.
$$
Then, it's clear that you have a symmetric pairing and that the determinant of that pairing with respect to any basis on the free part of the units is $R(K)^2$. As I mentioned above, the square was expected.
Depending on what you want to do, this pairing might not be the best one to consider. For example if now $F/K$ is a finite extension and you consider the analogous pairing on the $S$-units of $F$ and restrict it to $K$, then it's not clear how to compare it to the pairing on $K$. In the elliptic curves case by contrast, the former is $[F:K]$ times the latter. To fix this, we can make the following definition:
$$
(u\_1,u\_2) = \sum\_{v\in S} \frac{1}{e\_vf\_v}\log|u\_1|\_v\log|u\_2|\_v,
$$
where $e$ and $f$ are the absolute ramification index and residue field degree respectively. With this definition, the compatibility upon restriction to subfields is the same as in the elliptic curves case. On the other hand, the relationship with the actual regulator is slightly less obvious. It is however quite easy to show (and I have done it in <http://arxiv.org/abs/0904.2416>, Lemma 2.12, in case you are interested in the details) that the determinant of this pairing is given by
$$
\frac{\sum e\_vf\_v}{\prod e\_vf\_v}R(K)^2,
$$
with both the sum and the product again ranging over the places in $S$.
So I guess, the moral is that one shouldn't seek analogies between the BSD-formula and the class number formula but rather between the BSD and the square of the class number formula (note that also sha, which is supposed to be the elliptic curve analogue of the class number, has square order whenever it is finite). There is also a corresponding heuristic on the analytic side.
| 16 | https://mathoverflow.net/users/35416 | 38558 | 24,745 |
https://mathoverflow.net/questions/38556 | 2 | Hi,
consider the following random walk on the lattice $\{0,\dots,n\}^2$. It starts at $(0,0)$ and then move either up or right, with probability respectively $p$ and $1-p$. Once it reaches the right border (respectively the up border), it goes up (respectively it goes right) to $(n,n)$. What properties do we know about this random walk ? Especially can we say anything about the times when it crosses the main diagonal ?
| https://mathoverflow.net/users/9203 | Random walk on a two-dimensional uniform grid | Turn your square one fourth turn to the left and project it down.
To give some details: consider a simple random walk $Y$ on $\mathbb{Z}$ that is constrained to go left when yours goes up, and to go right when yours goes right. When your walk is on the boundary, then draw randomly independently the direction of $Y$. Since $Y$ is a simple random walk, all the results you can dream of are available. The only issue is that the two walks decorelate as soon as yours hit the boundary, but this is no problem for your question on the diagonal. Indeed, if $Z$ is the walk that is equal to $Y$ for steps $\leq n$, and equals the projection of $Y$ to $[-n+k,n-k]$ at steps $n+k$, then up to a $\sqrt{2}$ factor $Z$ is the projection on the second diagonal of your random walk, and it hits $0$ at time $t$ (meaning that your walk hits the diagonal at that time) if and only if $Y$ does.
| 4 | https://mathoverflow.net/users/4961 | 38564 | 24,751 |
https://mathoverflow.net/questions/38577 | 1 | If a group G has a subgroup H of finite index which is torsion free, then does it satisfy $H\_\ast (G,Q) = H\_ \ast (H,Q)$? (probably it is very well known...)
| https://mathoverflow.net/users/9205 | a small question about group homology | No, it is not true. For example, let $\mathbf{Z}/2$ act on $\mathbf{Z}$ by inversion, and $G$ be the semidirect product. Then $\mathbf{Z}$ is a torsion-free finite index normal subgroup of $G$, but one easily computes that the rational cohomology of $G$ is trivial, by the Hochschild-Leray-Serre spectral sequence for the extension.
In general, if $H$ is a normal subgroup of finite index, then $H^\*(G;\mathbb{Q}) \cong H^\*(H;\mathbb{Q})^{G/H}$, for the action of $G/H$ on $H$ by outer automorphisms.
If the subgroup is of finite index but not normal, the most one can say is that $H^\*(G;\mathbb{Q}) \to H^\*(H;\mathbb{Q})$ is split injective, which is proved using the transfer.
| 6 | https://mathoverflow.net/users/318 | 38584 | 24,759 |
https://mathoverflow.net/questions/38586 | 7 | The $n$-th Mersenne number $M\_n$ is defined as
$$M\_n=2^n-1$$
A great deal of research focuses on Mersenne primes. What is known in the opposite direction about Mersenne numbers with only small factors (i.e. smooth numbers)? In particular, if we let $P\_n$ denote the largest prime factor of $M\_n$, are any results known of the form
$$\liminf\_{n\rightarrow \infty}\frac{P\_n}{f(n)}= 1$$
for some function $f$?
I've only come across two (fairly distant) bounds so far. If we consider even-valued $n$, then $M\_n=M\_{n/2}(M\_{n/2}+2)$, so:
$$\liminf\_{n\rightarrow \infty}\frac{P\_n}{2^{n/2}}\leq 1$$
In the other direction, [1] shows that $P\_n\geq 2n+1$ for $n>12$, so
$$\liminf\_{n\rightarrow \infty}\frac{P\_n}{2n}\geq 1$$
[1] A. Schinzel, On primitive prime factors of $a^n-b^n$, Proc. Cambridge Philos. Soc. 58
(1962), 555-562.
| https://mathoverflow.net/users/8938 | Smoothness in Mersenne numbers? | I can give you a slightly better upper bound. Recall that $2^n - 1 = \prod\_{d | n} \Phi\_d(2)$ where $\Phi\_d$ is a cyclotomic polynomial. Now,
$$\Phi\_d(2) = \prod\_{(k, d) = 1} (2 - \zeta\_d^k) \le 3^{\varphi(d)}$$
so that in particular the largest prime factor of $2^n - 1$ is at most $3^{\varphi(n)}$. By taking $n$ to be a product of the first $k$ primes and letting $k$ tend to infinity we have $\liminf\_{n \to \infty} \frac{\varphi(n)}{n} = 0$, hence
$$\liminf\_{n \to \infty} \frac{P\_n}{c^n} = 0$$
for any $c > 1$. In fact if $n$ is the product of the first $k$ primes then we should expect something like $3^{\varphi(n)} \approx 3^{ \frac{n}{\log k} }$ but this doesn't seem like a big improvement to me.
| 3 | https://mathoverflow.net/users/290 | 38591 | 24,764 |
https://mathoverflow.net/questions/38551 | 2 | Consider two countable families of objects, given as unions of finite subfamilies:
$F^k = \bigcup\_{n \in \mathbb{N}} F^k\_n$, *k* = 1,2.
Let there be a bijection $f: F^1 \rightarrow F^2$ such that $x \in F^1\_n \Leftrightarrow f(x) \in F^2\_n$ for all *n* $\in \mathbb{N}$.
This means: The two families have the same counting functions $f^1(n) = f^2(n)$ for all *n* $\in \mathbb{N}$ with $f^k(n) = |F^k\_n|$.
This may be *by sheer accident*, or it may be because the two families are in some sense *essentially the same*.
>
> Can the notions of "by accident" and
> "essentially the same" be distinguished in this context, and how?
>
>
>
"Essentially the same" might mean: "there is a *computable* bijection" and "by accident" might mean: "there is *no* computable bijection". Or might category theoretical notions lead further?
>
> Are there known examples of two
> families as above that have the same counting
> functions by accident (in the meaning just mentioned or another one)?
>
>
>
PS: More sensible tags are welcome!
| https://mathoverflow.net/users/2672 | Equivalence of families of objects with the same counting function | If you adopt Philippe Nadeau's proposal to define "essentially the same" in terms of isomorphism of species, then a standard example of "accidental" is the following. For any finite set $S$, there are just as many linear orderings of S as there are permutations of $S$ (i.e., bijections from $S$ to itself), namely $|S|!$. But the two species are not isomorphic. There is a distinguished permutation, namely the identity, but there is no distinguished linear ordering (provided $|S|>1$).
Some people feel that these two are, nevertheless, in some sense the same. (In fact, some people indiscriminately use the word "permutation" for both.) That feeling can be formalized in the observation that, if you choose one linear ordering of $S$, then it determines a natural bijection between the linear orderings and the permutations: Given a permutation, apply it to the chosen linear ordering. So people who want to think of these two species as essentially the same would want a weakening of "isomorphism of species" that would allow this sort of "isomorphism modulo an arbitrary choice." Unfortunately, it's not clear how big an object could reasonably be arbitrarily chosen --- presumably not a whole isomorphism. So I don't see a good way to make this intuition precise; as a result, I'm inclined to stick with "isomorphism of species" and to accept that linear orderings and permutations are not essentially the same.
| 7 | https://mathoverflow.net/users/6794 | 38592 | 24,765 |
https://mathoverflow.net/questions/38570 | 7 | Let $L$ be the constructible universe and $x,y \in L$ such that $x \subseteq y$. Is then $x \in D(y)$, i.e. $x$ a definable subset of $y$?
If this is not true, do we at least have the following: If $x,y \in L$, then $\rho(x) \leq \rho(y)$? Here $\rho$ denotes the $L$-rank.
And if this is also false, how do you prove, that for $x \in L$, we have $L \cap P(x) \in L$? So I need this to prove the power set axiom in $L$.
| https://mathoverflow.net/users/2841 | Inclusions of constructible sets are *not* definable | If you just want to prove that $L$ satisfies the power set axiom, you don't need condensation; Kunen's comment is correct. Given $x\in L$, to show that $L\cap P(x)\in L$, first consider an arbitrary $y\in P(x)\cap L$ and observe that, since $y\in L$, there is a first ordinal $\alpha(y)$ such that $y\in L\_{\alpha(y)}$. By replacement (in the real world), there is an ordinal $\beta$ that is larger than all these $\alpha(y)$ for all $y\in P(x)\cap L$. So $P(x)\cap L\subseteq L\_\beta$. But then $P(x)\cap L$ is first-order definable over $L\_\beta$, by the definition $\{y\in L\_\beta:L\_\beta\models y\subseteq x\}$, and is therefore an element of $L\_{\beta+1}$.
Note that this proof, unlike the one given by Francois, tells you nothing about how far you must go in the $L$ hierarchy before you find $P(x)\cap L$; it provides no information about the size of $\beta$. To get such information, condensation is needed.
Concerning the first two parts of the original question, the answer is no for both in general. Take $y$ to be $\omega$ and note that it has only countably many definable subsets and only countably many subsets of lower or equal $L$-rank, but it may have (and will have if, for example, $V=L$) uncountably many subsets in $L$.
| 7 | https://mathoverflow.net/users/6794 | 38594 | 24,766 |
https://mathoverflow.net/questions/38599 | 2 | Let X be a variety and $E$ an ample vector bundle on $X$. Let $G=G(r+1,E)$ be the Grassmann bundle over $X$ whose fiber over $x\in X$ is the Grassmannian of the $r+1$-dimensional subspaces of $E\_x$. Let $U$ denote the universal subbundle on $G$. Under which hypothesis is the dual of $U$ ample on $G$?
| https://mathoverflow.net/users/33841 | Ampleness of the universal subbundle | I think this is essentially never true, again by restricting to a fiber over $x\in X$. The problem is that (somewhat counterintuitively) the universal quotient bundle on $Gr(k,n)$ is not ample, and for the same reason, neither is the dual of the universal sub. (Except of course when $k=1$!) See Examples 6.1.5 and 6.1.6 in Lazarsfeld's *Positivity in Algebraic Geometry II*.
| 5 | https://mathoverflow.net/users/5081 | 38602 | 24,768 |
https://mathoverflow.net/questions/38529 | 4 | What is a quotient of an affine scheme that is not a universal quotient? Let's recall some terminology.
Suppose that $k$ is an algebraically closed field and $G$ is a reductive group acting on an affine scheme $X$. Theorem 1.1 of [Geometric Invariant Theory](http://books.google.com/books?id=dFlv3zn_2-gC&printsec=frontcover&dq=geometric+invariant+theory&hl=en&ei=SpaNTInmIY2MnQfwmdDUCw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCgQ6AEwAA#v=onepage&q&f=false) states that the uniform categorical quotient $X//G$ of $X$ exists.
In other words, $X \to X//G$ is universal with respect to $G$-invariant morphisms out of $X$ and this property persists under base change by a *flat* morphism $T \to X//G$.
When $\text{char}(k)=0$, the theorem states that $X \to X//G$ is a universal categorical quotient, so that the universal property persists under base change by an *arbitrary* morphism $T \to X//G$.
What is an example where $X \to X//G$ is not a universal quotient?
I'd be particularly interested in the case where the stabilizers of the action on $X$ are all linearly reductive.
| https://mathoverflow.net/users/5337 | Uniform Quotient vs Universal Quotient | Here is an example, which is in some sense the simplest one. Suppose that $k$ has characteristic $p > 0$; set $X := \mathop{\rm Spec} k[x,y]$. Let $G$ be a cyclic group of order $p$ acting via $(x,y) \mapsto (x, x+y)$. The ring of invariants is $k[u,v] := k[x, y^p - x^{p-1}y]$. Consider the point $\mathop{\rm Spec} k = \mathop{\rm Spec} k[u,v]/(u,v)$ of $X/G = \mathop{\rm Spec} k[u,v]$; the inverse image $Y$ in $X$ is $\mathop{\rm Spec} k[x,y]/(x, y^p)$; it is immediate to check that the action of $G$ on $Y$ is trivial, so $Y/G = Y \neq \mathop{\rm Spec} k$.
If you want an example with a connected group, embed $G$ into $\mathrm{GL}\_n$ and take the induced action.
I don't know any example with linearly reductive stabilizers, and I suspect that they don't exist.
| 3 | https://mathoverflow.net/users/4790 | 38604 | 24,770 |
https://mathoverflow.net/questions/38019 | 27 | It was asked in the Putnam exam of 1969, to list all sets which can be the range of polynomials in two variables with real coefficients. Surprisingly, the set $(0,\infty )$ can be the range of such polynomials. These don't attain their global infimum although they are bounded below. But is it also possible that such polynomials with range $(0,\infty )$ also have a non zero gradient everywhere?
| https://mathoverflow.net/users/9075 | Zeros of Gradient of Positive Polynomials. | $(1+x+x^2y)^2+x^2$
| 52 | https://mathoverflow.net/users/1131 | 38634 | 24,776 |
https://mathoverflow.net/questions/34724 | 16 | ### Overview
For integers n ≥ 1, let T(n) = {0,1,...,n}n and B(n)= {0,1}n. Note that |T(n)|=(n+1)n and |B(n)| = 2n.
A certain set S(n) ⊂ T(n), defined below, contains B(n). The question is about the growth rate of |S(n)|. Does it grow exponentially, like |B(n)|, so that |S(n)| ~ cn for some c, or does it grow superexponentially, so that cn/|S(n)| approaches 0 for all c> 0?
### Definition
The set S(n) is defined as follows: an n-tuple t = (t1,t2,...,tn) ∈ T(n) is in S(n) if and only if ti+j ≤ j whenever 1 ≤ j< t i. For example, if t ∈ T(10) with t 4=5, t 5 can be at most 1, t 6 can be at most 2, , t 7 can be at most 3, and t 8 at most 4, but there is no restriction (at least not due to the value of t 4) on t 9 or t 10; t 9 and t 10 can have any values in {1,...,10}.
### Alternate formulation (counting triangles)
The elements of S(n) can be put into one-to-one correspondence with certain configurations of n right isosceles triangles, so that |S(n)| counts the number of such configurations.
For integers k>0 (size) and v≥0 (vertical position), let Δ k,v be the triangle with vertices (0,v), (k,k+v), and (k,v). (Δ0,v is the degenerate triangle with all three vertices at (0,v).)
Now associate with an n-tuple t = (t1,t2,...,tn) ∈ T(n) the set Dt = $\lbrace\Delta\_{t\_k,k}:1\le k \le n\rbrace$. (That's "\lbrace\Delta\_{t\_k,k}:1\le k \le n\rbrace," if you can't read it.) The set D t contains n isosceles right triangles that extend to the right of the y-axis, one triangle at each of the points (0,k) for 1 ≤ k ≤ n.
The tuple t is in S(n) if and only if the triangles in D t have disjoint interiors. (This isn't hard to show, and if it is, I've probably made a mistake in my definitions, so let me know.) Thus |S(n)| counts the number of ways one can arrange n isosceles right triangles of various sizes (between size zero and size n) at n consecutive integer points on the y-axis so the triangle can extend to the right and up without overlapping. Triangles of the same size are indistiguishable for the purpose of counting the number of arrangements. (It may help to think of right isosceles pennants attached at an acute-angle corner to a flagpole in a stiff wind.)
### Question
Does |S(n)| grow exponentially with n, or faster?
### Calculations
If I’ve counted correctly, the first few terms of the sequence {|S(n)|} beginning with n=1 are 2, 8, 38, 184, 904, and 4384. This sequence (and some sequences resulting from minor variations of the problem) fails to match anything in the Online Encyclopedia of Integer Sequence.
Links to similar counting problems mentioned or solved in the literature would help.
Thanks!
| https://mathoverflow.net/users/8201 | Counting certain arrangements of n triangles. Does the count grow superexponentially? | Your sequence is bounded by $(125+\epsilon)^n$. Obviously, this isn't close to a good bound, but it answers the question.
We start by bounding a different question: Let $\Gamma\_n$ be the convex hull of $(0,0)$, $(0,n)$ and $(n,n)$. (So $\Gamma$ is rotated $180^{\circ}$ with respect to your $\Delta$.) Let $q\_n$ be the number of ways to pack non-overlapping triangles into $\Gamma\_n$.
Given any packing of triangles in $\Gamma\_n$, which uses at least one triangle, let the largest triangle be of size $k$ and have a vertex at $(0,r)$. (If there is more than one largest triangle, make an arbitrary choice; this will just lead to a larger bound in the end.) So all the other triangles must fit into one of two trapezoids: one with base $r$ and height $k$ and the other with base $n-r$ and height $k$. In any case, these two trapezoids fit into translations of $\Gamma\_r$ and $\Gamma\_{n-r}$. So we obtain the inequality
$$q\_n \leq \sum\_{r=1}^{n-1} q\_r q\_{n-r} + 1,$$
where the $+1$ is because we have to remember the possibility that there might be no triangles in the packing. If we take $q\_0=0$ for convenience, we get that $\sum q\_n z^n$ is term by term dominated by the solution of
$$Q(z) = Q(z)^2 + \frac{z}{1-z}.$$
Solving the quadratic,
$$Q(z) = \frac{1}{2} \left( 1 - \sqrt{1-\frac{4z}{1-z}} \right).$$
Notice that $Q(z)$ has radius of convergence $1/5$ so $q\_n \leq (5+\epsilon)^n$.
I previously had an argument here that didn't work, so here is something even more sloppy. All the triangles you are considering fit inside $\Gamma\_{3n}$. So your quantity is bounded by $q\_{3n}$, and hence by $(125+\epsilon)^n$.
I suspect that $5^n$ may be pretty close to the right rate of growth, especially given Roland Bacher's computation.
| 6 | https://mathoverflow.net/users/297 | 38637 | 24,778 |
https://mathoverflow.net/questions/38581 | 5 | Let
\begin{equation}
R(x) = \sum\_{k=1}^{\infty}\frac{\mu(k)}{k}li(x^{1/k})
\end{equation}
where $\mu$ is the Mobius function and
\begin{equation}
li(x) = \int\_0^x \frac{dt}{\log t}
\end{equation}
Is there a proof in the literature of
\begin{equation}
\pi(x)=R(x)-\sum\_{\rho}R(x^{\rho})
\end{equation}
where $\pi$ is prime counting function and the sum is over all complex zeros of $\zeta(s)$. The literature seems to treat it as fact while stating no proof is available - a strange situation.
Thanks in advance.
| https://mathoverflow.net/users/2011 | Proof in the literature of an equality involving the prime counting function | Stopple, A Primer of Analytic Number Theory, proves a theorem which looks something like the one under discussion. On page 248, he has $$\pi(x)=R(x)+\sum\_{\rho}R(x^{\rho})+\sum\_{n=1}^{\infty}{\mu(n)\over n}\int\_{x^{1/n}}^{\infty}{dt\over t(t^2-1)\log t}$$
You say that the literature treats your formula as a fact, but you give no citation. Where in the literature do you find your formula?
| 2 | https://mathoverflow.net/users/3684 | 38646 | 24,785 |
https://mathoverflow.net/questions/38633 | 2 | This may not be precise enough for MO, but I'll give it a go.
Let $M$ be a symmetric closed monoidal model category with unit $u\in Ob(M)$. We define the vertices of an object $A$ to be points $x\in \operatorname{Hom}\_M(u,A)$. We define the fibre of $f:B\to A$ over a vertex of $x$ of $A$ to be the pullback of $f$ by the map $x:u\to A$ classifying $x$.
It can be shown in the category of simplicial sets with the cartesian monoidal structure (and unit $\Delta^0$) that a Kan fibration is acyclic if and only if all of its fibres are contractible (I'm pretty sure that a similar statement holds for CGWH spaces in the Quillen model structure). However, the proof uses specific properties of the category of simplicial sets to prove this fact. Is there some general additional structure on $M$ that we would need to prove this statement in more generality?
Edit: It's at least clear that we must require the monoidal unit to be trivially fibrant.
| https://mathoverflow.net/users/1353 | In what generality does the following statement hold: A fibration is acyclic if and only if all fibres are contractible fibrant objects. | Consider sSet×sSet with the induced structure from sSet. The monoidal unit is (1,1) (which is trivially fibrant, as is the unit in any cartesian monoidal model category), and so an object of the form (X,0) has no vertices at all (with your definition). Hence any map (X,0)→(Y,0) has (vacuously) all its fibers contractible, but is not in general acyclic.
I think what you'd need is some sort of "well-pointedness" of M, which will probably be a very special property in general.
| 3 | https://mathoverflow.net/users/49 | 38653 | 24,788 |
https://mathoverflow.net/questions/38622 | 3 | Let $Dom$ be a uniform space, and $\hspace{.04 in}f$ be a continuous function from $Dom$ to itself satisfying:
1. For all non-empty open subsets $U$ and $V$ of $Dom$, there exists a natural
number $n$ and a member $x$ of $U$ such that $f^n(x)$ is a member of $V$.
2. The periodic points of $f$ are dense in $Dom$.
Does it follow that $f$ satisfies (3)?
$\;\;$3.$\:\:$There exists an entourage $E$ of $Dom$ such that for all members $x$ of
$\quad \;\;$ $Dom$ and all neighborhoods $U$ of $x$, there exists a member $y$ of $U$ and
$\quad \;\;$ a natural number $n$ such that $\:\langle \hspace{.05 in}f^n(x)\hspace{.02 in},\hspace{.03 in}f^n(\hspace{.03 in}y)\rangle\:$ is not a member of $E$.
According to [this paper](http://pb.math.univ.gda.pl/chaos/pdf/on_intervals.pdf), the implication holds in metric spaces.
| https://mathoverflow.net/users/nan | Chaos in uniform spaces | I believe the following works. Follow the proof in the reference supplied by Matthew Daws: choose two distinct periodic orbits and choose a compatible pseudometric $\rho$ such that all points in those orbits are at least $1$ apart under this pseudometric.
The proof establishes that the entourage $\lbrace(x,y):\rho(x,y)<\frac18\rbrace$ is as required.
| 2 | https://mathoverflow.net/users/5903 | 38667 | 24,795 |
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