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https://mathoverflow.net/questions/41492 | 9 | It seems to me that the results in [this important paper of Kahn-Markovich](http://arxiv.org/abs/0910.5501) imply the following fact. Let $M^3$ be any closed hyperbolic 3-manifold. For every $\epsilon > 0$ there is a natural number $R(\epsilon)$ such that for every $R>R(\epsilon)>0$ there is a closed geodesic whose lenght is between $R-\epsilon$ and $R+\epsilon$.
That is, the more $R$ grows, the more the spectrum of all geodesics having length more than $R$ becomes uniformly crowded, without holes. Am I right that this is a consequence of their results? If so, is there a more direct way to prove this? Does this property generalize to closed hyperbolic manifolds with arbitrary dimension $n$?
| https://mathoverflow.net/users/6205 | Is the spectrum of closed geodesics in a closed hyperbolic 3-manifold asymptotically homogeneously dense? | I think this should just follow from the exponential mixing of the geodesic flow (due to Pollicott).
Exponential mixing says that there is a constant $q$ such that if you have two smooth functions $f$ and $h$ on the unit tangent bundle, and $g\_t$ is the geodesic flow, then there is a constant $C$ depending on $f$ and $g$ (some function of some Sobolev norm) such that
$\Big| \int\_{T^1M} (g\_t^\*f)h - (\int\_{T^1M} f)(\int\_{T^1M}h ) \Big | \leq C e^{-qt}$
If you take a vector $v$ and then let $f = h$ be a function with integral $1$ supported on an $\epsilon$-neighborhood of $v$, then you find that there is some constant $T(\epsilon)$ depending only on $M$ and $\epsilon$ such that for any $T \geq T(\epsilon)$, there is a closed path whose length is within $\epsilon$ of $T$ and which is a geodesic except at the basepoint, where it is broken at an angle of $\pi - \epsilon$. If $\epsilon$ is small, this path will be close to a bona fide geodesic, which should give you what you want.
I played fast and loose there with the constants, but that's the idea, I think.
| 14 | https://mathoverflow.net/users/1335 | 41521 | 26,504 |
https://mathoverflow.net/questions/41525 | 16 | When talking about the Eilenberg-Maclane space $K(G,n)$, we usually restrict our attention to the situation where $G$ is abelian. In that case, we get $\Omega K(G,n)=K(G,n-1)$, so we can call $K(G,n)$ a *delooping* of $K(G,n-1)$.
Since $\pi\_n$ is always abelian for $n>1$, it only makes sense to talk about $K(G,1)=BG$ for $G$ nonabelian anyways. So there definitely shouldn't be delooping of this space, because then it would have $\pi\_2=G$, which is impossible. From the previous paragraph, it seems like we should therefore be able to say that the nonabelianity of $G$ (i.e., the nontriviality of the commutator $[G,G]$) is the obstruction to delooping $BG$. But this isn't very satisfying, because I can't quite see what's going on with the actual space.
All of which motivates my (slightly open-ended/up-to-interpretation) question:
**How should I think about delooping? Is it nothing more than thing like "for the space $X$ that we care about, it just so happens that we've got $Y$ with $\Omega Y\simeq X$", or is there a definite way to measure obstructions? In the cases where a delooping exists, is there an explicit method for its construction?**
| https://mathoverflow.net/users/303 | How should I think about delooping? | One possible answer: Stasheff proved that a (connected) space $X$ is (homotopy equivalent to) a loopspace if and only if $X$ is an algebra over the $A\_\infty$ operad (or rather I should say *an* $A\_\infty$ operad).
See for instance [this article](http://www.ams.org/notices/200406/what-is.pdf).
| 10 | https://mathoverflow.net/users/83 | 41526 | 26,505 |
https://mathoverflow.net/questions/41516 | 0 | Let $X$ a smooth manifold. Is the pullback morphism
$\Omega^\bullet(X)\to\Omega^\bullet(X\times \mathbb{R}^n)$ an acyclic cofibration of differential graded commutative algebras? I guess so, and even that this should be the basic example to have in mind, but being no expert, I don't trust myself too much.
| https://mathoverflow.net/users/8320 | Cofibrations of differential graded commutative algebras | It depends completely on what you mean by cofibrations. The choice is
not quite simple to make as the homotopy category of real commutative dga's is
anti-equivalent to "real homotopy" which would suggest that the cofibrations
should correspond very roughly to fibrations of spaces (judging from your
example this looks like the notion you are searching for). Then the proper notion
would seem to be a pseudofree extension algebra (i.e., the extension algebra
forgetting the differential) should be a polynomial algebra over the base. In
that case the map $\Omega^\bullet(X)\rightarrow\Omega^\bullet(X\times\mathbb
R^n)$ is not a cofibration. I find it difficult to imagine an interesting model structure on
commutative dga's which would make it a cofibration.
| 1 | https://mathoverflow.net/users/4008 | 41528 | 26,507 |
https://mathoverflow.net/questions/41522 | 1 | Let $L=L\_1 \cup ... \cup L\_n$ be the union of $n$ distinct lines through the origin in $\mathbb{R}^{3}$. I'd like a convincing argument that $\mathbb{R}^{3} \setminus L$ is homotopy equivalent to a wedge of $n$ circles (if that is true). In fact, I especially care about the case $n=2$.
I know this sounds like a homework problem, but I have other purposes in mind (this space naturally showed up as the fibre in a certain fibration) and I don't find typical text-book
explanations of such problems very convincing, so I would appreciate a clear answer.
| https://mathoverflow.net/users/6254 | Complement of lines and wedges of spheres | First, deformation retract $\mathbb{R}^3$ minus $L$ to $S^2$ minus $2n$ points (you can do this since you've removed the origin). Stereographically project from one of the punctures, and you've got $\mathbb{R}^2$ minus $2n-1$ points. Choose a point away from the punctures and draw disjoint based loops around each of the remaining holes. Now deformation retract to those loops and you've got a wedge of $2n-1$ circles.
| 6 | https://mathoverflow.net/users/353 | 41529 | 26,508 |
https://mathoverflow.net/questions/41470 | 5 | I'm looking for such pathological examples of uniform spaces which are not metrizable, but whose underlying topology is metrizable. Willard in his General Topology text constructs such a uniformity using ordinals. I am asking for examples which do not rely on ordinals.
EDIT: Below is an example by Daniel Tausk using families of pseudometrics. I forgot to mention that, if possible, I would like an example that uses the definition through a diagonal uniformity, not the definition through pseudometrics nor the definition through uniform covers. See [Cryptomorphisms](https://mathoverflow.net/questions/15731/cryptomorphisms) for some elaboration on this phenomenon.
| https://mathoverflow.net/users/6249 | Nonmetrizable uniformities with metrizable topologies | The family of **all** neighbourhoods of the diagonal of $\mathbb{R}$ (with its normal topology) is a uniformity without a countable basis but it generates the normal topology.
The normal topology of $\mathbb{R}$ is the topology generated by the usual distance.
To see that the open sets in the plane that contain the diagonal generate a uniformity let $U$ be such an open set. For each $x$ there is $r\_x$ such that $B(x,r\_x)^2\subseteq U$. Now let $V$ be the union of the squares $B(x,\frac13r\_x)^2$; this is again an open set and it is not hard to show that $V\circ V\subseteq U$.
If $U$ is an open set in the plane that contains the diagonal then the vertical section $U[x]$ is open in $\mathbb{R}$, so this uniformity generates at best a subtopology of the normal topology and hence certainly not the discrete topology. To see that it generates the normal topology consider, given $x$ and $r$, the open set $U\_{x,r}=B(x,r)^2\cup(\mathbb{R}\setminus\lbrace x\rbrace)^2$; then $U\_{x,r}[x]=B(x,r)$.
The uniformity does not have a countable base: if $\langle U\_n:n\in\mathbb{N}\rangle$ is a sequence of neighbourhoods of the diagonal then you can use diagonalisation to produce a neighbourhood such that $U\_n\not\subseteq U$ for all $n$: make sure that the square $[n,n+1]^2$ contains a point in $U\_n$ but not in $U$.
| 5 | https://mathoverflow.net/users/5903 | 41531 | 26,509 |
https://mathoverflow.net/questions/41483 | 2 | I've been reading about reverse mathematics (mostly on wikipedia), and I had been thinking that I understood how to prove the equivalences to WKL0 and ACA0 mentioned in the its article. However, I now realize that my idea of how WKL0 can prove that every continuous real valued function on [0,1] is bounded. My idea would have started "since f is continuous, there f is locally bounded near each point, so there is an open cover of [0,1] such that f is bounded on each member of the cover", but I can't figure out how to express "f is bounded on the interval with rational endpoints (q,r)" as a Sigma\_1 property, and I can't figure out how to get around this issue, either.
How does WKL0 prove that every continuous real valued function on [0,1] is bounded?
| https://mathoverflow.net/users/nan | Proving boundedness of continuous images of [0,1] in WKL0 | Here is a sketch of how the proof might go.
If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S\_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $S\_n\ne\emptyset$ for all $n\in\mathbb N$. According to the definition of continuous function the sets $f^{-1}[(-n,n)]$ are represented as $\Sigma\_1$ sets of (endpoints of) rational intervals ($\Sigma\_1$ definable in the model). So the closed dyadic rational intervals intersecting $S\_n$ for $n\in\mathbb N$ can be represented as an infinite $\Pi\_1$ (or by a standard trick, equivalently $\Delta\_1$) tree, which by Weak König's Lemma must have a path, so $\cap\_n S\_n\ne\emptyset$ which is absurd.
| 2 | https://mathoverflow.net/users/4600 | 41533 | 26,511 |
https://mathoverflow.net/questions/7492 | 18 | Inspired by [this thread](https://mathoverflow.net/questions/7439/algebraic-varieties-which-are-also-manifolds), which concludes that a non-singular variety over the complex numbers is naturally a smooth manifold, does anyone know conditions that imply that the topological space underlying a complex variety is a topological manifold without necessarily implying it is smooth?
| https://mathoverflow.net/users/1231 | Algebraic varieties which are topological manifolds | The answer from Dmitri motivates this partial answer from the topological side of the question.
It is [a theorem of Mark Goresky](http://www.math.ias.edu/~goresky/pdf/triangulations.pdf) and others that every stratified space, and in particular every complex variety $V$, has a smooth triangulation. Moreover, I would bet (although I don't know that Goresky's paper has it) that the associated piecewise linear structure is unique. This means that the PL homeomorphism type of the link of a singular point $p$ of $V$ is a local invariant. I don't know how to compute this local invariant in general, but there must be some way to do it from the local ring at $p$. There can't be a simple calculation of this invariant that is fully general. As a special case, $V$ can be the cone of a projective variety $X$. If so, then the link at the cone point $p$ is the total space of the tautological bundle on $X$. $X$ and therefore the link can be all sorts of things. If $p$ is an isolated singularity, then the type of this link is obtained by "intersecting with a small sphere", as Dmitri says.
The variety $V$ is a PL manifold if and only if the link of every vertex is a PL sphere. This is the case for the Brieskorn examples.
On the other hand, [a theorem of Edwards](http://www.ams.org/journals/proc/1999-127-10/S0002-9939-99-04860-1/S0002-9939-99-04860-1.pdf) (or maybe Cannon and Edwards) says that a polyhedron is a topological $n$-manifold (for $n \ge 3$) if and only if the link of every vertex is simply connected and the link of every point is a homology $(n-1)$-sphere. In particular, the link of a simplex which is not a point does not have to be simply connected! For example, if $\Gamma \subseteq \text{SU}(2)$ is the binary icosahedral group, then $\mathbb{C}^2/\Gamma$ is not a manifold, because the link of the singular point is the Poincaré homology sphere. But $(\mathbb{C}^2 / \Gamma) \times \mathbb{C}$ is a topological manifold, even though it is not a PL manifold.
So for the question as stated, you would want to combine Goresky's theorem with Edwards' theorem, and with a method to compute the topology of the link of a singular point. On the other hand, whether a variety $V$ is a PL manifold could be a more natural question than whether it is a topological manifold.
---
At least in the case of isolated singularities, the possible topology of the link of a singular point has been studied in the language of complex analytic geometry rather than complex algebraic geometry. I found [this paper](http://www.scichina.com:8081/sciAe/fileup/PDF/06ya1441.pdf) by Xiaojun Huang on this topic. The link of the singular point is in general a strictly pseudoconvex CR manifold. This is a certain kind of odd-dimensional analogue of a complex manifold and you could study it with algebraic geometry tools. (I think that strict pseudoconvexity also makes it a contact manifold?) But the analytic style seems to be more popular, maybe because a CR manifold is not a scheme.
Sometimes, for instance in the case of a Brieskorn-Pham variety, such a CR manifold has a circle action whose quotient is a complex algebraic variety. At a smooth point, this quotient is just the usual Hopf fibration from $S^{2n-1}$ to $\mathbb{C}P^{n-1}$. In the famous Brieskorn examples, the link is a topological sphere with a circle action, but the circle action yields a non-trivial Seifert fibration over an orbifold-type complex variety. On the other hand, I don't think that this circle action always exists.
| 14 | https://mathoverflow.net/users/1450 | 41537 | 26,512 |
https://mathoverflow.net/questions/41535 | 17 | Suppose that I have $n$ homogeneous polynomials $f\_1, \dots, f\_n \in \mathbb{C}[x\_1, \dots, x\_m]$ and that $n < m$. Is there a well known method or algorithm to determine if these polynomials are algebraically independent?
As far as I know the Jacobian criterion works only for the case where $n=m$.
| https://mathoverflow.net/users/9899 | How to show a set of polynomials is algebraically independent? | The polynomials are algebraically independent if and only if
$$
df\_1 \wedge df\_2 \wedge \cdots \wedge df\_n
$$
is not identically zero. In other words, you have only to check that
one of the maximal minors of the matrix $\left( \frac{\partial f\_i}{\partial x\_j} \right)$ is nonzero.
| 21 | https://mathoverflow.net/users/605 | 41539 | 26,514 |
https://mathoverflow.net/questions/40403 | 2 | I'm having a hard time finding some references on series solutions for "nonlinear" ODE's, the most I could find was a small excerpt on Wikipedia.
<https://en.wikipedia.org/wiki/Power_series_solution_of_differential_equations>
Most books just say something along the lines of ... and the method is applicable to nonlinear ODE's. But none I've seen go into detail let alone an example. Can anyone suggest me a good book or reference (in particular for 2nd order nonlinear ODEs)?
Thanks
| https://mathoverflow.net/users/9404 | Power series solutions for nonlinear ordinary differential equations - references | Nonlinear differential equations is hard to find good references on-partly due to the difficulty of the subject and partly due to the highly specialized nature of most of the research problems connected with them. But a lot of these problems are really problems of numerical approximation-so I think you'll have greater luck if you begin searching the literature on THAT,math.
A very good book to start with that has a lot of great material on this is Atkinson and Kan's *Theoretical Numerical Analysis*. Not only is it terrifically written and comprehensive with lots of examples,it's one of the most scholarly texts I've ever seen with **complete and opiniated** references. I think you'll find this book's references will give you a great deal of direction for further study on nonlinear solution of ODE's.
An older book that has a lot of nice material on power series and other numerical methods for ODE's is Einar Hille's *Lectures On Ordinary Differential Equations*. Why most of Hille's texts-which are all wonderful-are out of print mystifies me.
That should help you get started,especially the Atkinson/Han book. Good hunting!
| 1 | https://mathoverflow.net/users/3546 | 41549 | 26,520 |
https://mathoverflow.net/questions/41502 | 10 | Let $(A,\alpha, G)$ be a $C^\*$-dynamical system, where $G$ is a discrete group. Let $\Gamma$ be a subgroup of $G$, then we can form two universal crossed products $A\rtimes\_\alpha \Gamma$ and $A\rtimes\_\alpha G$.
Question 1: Is the canonical map $A\rtimes\_\alpha \Gamma \to A\rtimes\_\alpha G$ injective?
Question 2: What about reduced crossed products $A\rtimes\_{\alpha,r} \Gamma \to A\rtimes\_{\alpha,r} G$?
For the universal case, I guess it's wrong even though I can not find a counterexample. But if we let $\alpha$ be the trivial action of $G$, then we only need to look at $A\otimes\_{max} C^\* (\Gamma) \subseteq A\otimes\_{max} C^\* (G) $.
For the reduced case, I guess it's should be the case. Just follows from the facts that the left regular representaion of $G$, restricted to $\Gamma$, is a multiple of the left regular representation of $\Gamma$ and $A\rtimes\_{\alpha,r} \Gamma = C^\*(\pi(A), 1\otimes \lambda(G))$, where $\pi: A\subseteq B(H) \rightarrow B(H\otimes l^2(G))$ and $\lambda$ is the left regular representaion of $G$. Do I make any mistake?
| https://mathoverflow.net/users/9401 | Given a C-star dynamical system and a subgroup of the acting group, is the corresponding map on crossed product algebras necessarily an injection | The universal case is also injective. What one needs to show is that any covariant representation of $(A,\Gamma)$ extends to a covariant representation of $(A,G)$. This can be shown by using induced representation.
| 7 | https://mathoverflow.net/users/7591 | 41556 | 26,523 |
https://mathoverflow.net/questions/41559 | 4 | Let $X$ be a smooth projective variety and let $D$ be an effective divisor on $X$. Is there a natural way to describe the tangent space to $|D|$ (or $|D|^\vee$, of course) at a divisor $D'$? Preferably as some sort of cohomology group, again ideally on $X$. I would prefer to avoid using the fact that the linear system is a projective space, and do things naturally.
| https://mathoverflow.net/users/622 | Tangent Space to a Linear System | The linear system $|D|$ is the projectivization of $H^0(X,\mathcal O(D)).$ The divisor $D'$ corresponds to a line $\ell\_{D'} \subset H^0(X,\mathcal O(D)).$ (This line consists of all the sections whose zero loci are equal to $D'$.) The space $Hom(\ell\_{D'},H^0(X,\mathcal O(D))/\ell\_{D'})$
is a vector space of the same dimension as $|D|$, and is naturally isomorphic to the
tangent space of $|D|$ at $D'$.
(Here is am using the general fact that if $V$ is an vector space, and $\ell \subset V$
a line through the origin, then the tangent space to the projectivization $\mathbb P(V)$
at the point corresponding to the line $\ell$ is identified with
$Hom(\ell,V/\ell)$.)
[Thanks to Georges Elencwajg for correcting an earlier misstatement here.]
One can say a little more; before doing so, it's convenient to note that $D$ can be any divisor in the linear system, and so it is no loss of generality to set $D = D'$; this
eases the notation somewhat. We also fix a section of $\mathcal O(D)$ cutting out $D$,
i.e. a basis of $\ell\_D$, which gives an identification
$\ell\_D = k = H^0(X,\mathcal O).$; this allows us to rewrite $Hom(\ell\_D,H^0(X,\mathcal O(D))/\ell\_D)$
simply as
$H^0(X,\mathcal O(D))/\ell\_D$.
Our choice of basis for $\ell\_D$ gives a short exact sequence
$$0 \to \mathcal O \to \mathcal O(D) \to \mathcal O(D)\\_{| D} \to 0,$$
and taking global sections gives
$$0 \to \ell\_{D} \to H^0(X,\mathcal O(D)) \to H^0(D, \mathcal O(D)\\_{| D}),$$
and hence an injection
$$H^0(X,\mathcal O(D))/\ell\_{D} \hookrightarrow H^0(D,\mathcal O(D)\\_{| D}).$$
But this is not going to be an isomorphism in general, I guess.
Indeed,
the cokernel embeds into $H^1(X,\mathcal O)$, which is the tangent space to Pic $X$,
while $H^0(D,\mathcal O(D)\\_{|D})$ is the tangent space to the Hilbert scheme of $X$
at $D$. [Note: To see this, observe that our choice of section cutting out $D$ corresponds to
a choice of isomorphism $\mathcal O(D) \cong \mathcal I\_D^{-1},$ and it is
$(\mathcal I\_{D}^{-1})\\_{| D}$ that is canonically the normal bundle to $D$.] The map
$H^0(D,\mathcal O(D)\_{|D}) \to H^1(X,\mathcal O)$ then measures the extent to which
the deformations of $D$ in $X$ fill up the component of the Picard scheme
containing $D$. I imagine that if $D$ is sufficiently positive then this map is surjective;
at least when $X$ is a surface, this is the main result of Mumford's "Lectures on curves on an algebraic surface" (if I am remembering correctly).
| 9 | https://mathoverflow.net/users/2874 | 41560 | 26,525 |
https://mathoverflow.net/questions/39808 | 7 | The following essentially implies the equivalence of Anantharaman-Delaroche's compact approximation property (page 337 of [Link](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-171/issue-2/Amenable-correspondences-and-approximation-properties-for-von-Neumann-algebras/pjm/1102368918.full)) and the Haagerup approximation property.
Let $M$ be a type ${II}\_{1}$ factor with trace $\tau$. Let $\Omega$ denote the standard unit cyclic trace vector in $L^{2}(M)$ (associated to the element $1\in M$). If $\phi:M \rightarrow M$ is a normal completely positive map, we naturally associate an operator $T\_{\phi}\in B(L^{2}(M))$ extending $T\_{\phi}(x\Omega)=\phi(x)\Omega$.
If the map $x\mapsto \phi(x) \Omega$ is a compact linear map from $M$ with the operator norm into $L^{2}(M)$, is the operator $T\_{\phi}$ compact?
| https://mathoverflow.net/users/6269 | Are the compact and Haagerup approximation properties equivalent? | *(I removed my first answer as it contained an egregious mistake, pointed out by Yemon; here's a second attempt)*
I think that $T\_\phi$ may fail to be compact.
Fix a sequence of projections $\{p\_k\}$ in $M$, pairwise orthogonal, with $\tau(p\_k)=2^{-k}$ ($\tau$ the trace in $M$) and sum 1. Now define
$$
\phi:M\to M,\ \ \ \mbox{ given by } \phi(x)=\sum\_k 2^k\tau(xp\_k)p\_k.
$$
(the series converges strongly because all of its terms are positive and any partial sum is bounded by $\|x\|$). This map is ucp (it is an infinite sum of cp), and since it commutes with $\tau$, it is normal. Let us also define the maps
$$
\phi\_n=\sum\_{k=1}^n 2^k\tau(xp\_k)p\_k.
$$
The maps $x\mapsto \phi\_n(x)\Omega$ are all finite-rank.
If $\|x\|\leq1$, we have, using that the set $\{p\_k\}$ is orthogonal in $L^2(M)$,
$$
\left\|\phi(x)-\phi\_n(x)\right\|\_2^2=\left\|\sum\_{k>n}2^k\tau(xp\_k)p\_k\right\|\_2^2
=\sum\_{k>n}|\tau(xp\_k)|^2\leq\sum\_{k>n}\tau(p\_k)^2\leq\frac1{3\times 4^n}
$$
This shows that $\|\phi-\phi\_n\|<4^{-n}$ in $B(L^2(M))$: so the map $x\mapsto \phi(x)\Omega$ is compact.
Now consider the orthonormal set $\{2^kp\_k\}$ in the unit ball of $L^2(M)$. Since $\phi(p\_k)=p\_k$, we get that $T\_\phi(2^kp\_k)=2^kp\_k$; so the range of $T\_\phi$ contains an orthonormal set: $T\_\phi$ is not compact.
| 4 | https://mathoverflow.net/users/3698 | 41562 | 26,526 |
https://mathoverflow.net/questions/41561 | 10 | Liouville's theorem states that all bounded holomorphic functions on $\mathbb{C}^n$ are constant.
I'm wondering which connected complex manifolds have this property ?
Connected compact complex manifolds have it since all holomorphic functions there are constant.
There are no simply connected proper open subsets of $\mathbb{C}$ satisfying this because of the Riemann Mapping Theorem. But are there some other open subsets satisfying it ?
In higher dimension there are some, such as the Fatou–Bieberbach domains (open subsets of $\mathbb{C}^n$ biholomorphic to it).
I would be interested in references on this property on complex manifolds.
Incidentally, is it true that any complex manifold where all holomorphic functions are bounded is compact ?
| https://mathoverflow.net/users/9152 | Complex manifolds where bounded holomorphic functions are constant | We have that there are no non-constant bounded functions on $\mathbb C^\*=\mathbb
C\setminus\{0\}$. The easiest way to see that is to notice that such a function
has a removable singularity at the origin and hence comes from a bounded
function on $\mathbb C$ (which incidentally has a removable singularity at
$\infty$ and hence extends to the Riemann sphere and therefore is constant).
As for the higher-dimensional problem it is no doubt hopeless to get anything
like a complete description: There are as you say domains in $\mathbb C^n$ (and
there are many of those), the compact manifolds but also compact manifolds minus
a codimension $2$ closed analytic subspace, the blowing up of some space that
has the property, any product of two manifolds with the property and so on.
Complex manifolds in higher dimension are simply too varied.
Finally, there are non-compact manifolds with only constant holomorphic
functions. If $L\rightarrow X$ is an analytic line bundle over a complex
manifold $X$ and $f\colon L\rightarrow\mathbb C$ is a holomorphic function, then
we may Taylorexpand $f$ along the zero section of $L$: First we just look at the
restriction of $f$ to the zero section which gives a function on $X$. Then we
may take any local section of $L$, think of that as a tangent vector at the
zero section and take the derivative of $f$ along this tangent vector. This
glues together to give the first derivative as a global section of $L^{-1}$ and
similarly the $n$'th derivative of $f$ along the zero section will be a section
of $L^{-n}$. If $X$ is compact and $L^{-n}$ for $n>0$ only has the zero section
as global section, then $f$ is constant along the zero section and all its
higher derivatives along it are zero so that $f$ is constant in a neighbourhood
of the zero section and hence constant. There are lots of such examples $(X,L)$.
| 14 | https://mathoverflow.net/users/4008 | 41565 | 26,527 |
https://mathoverflow.net/questions/41546 | 2 | Hi
Consider Poisson equation with Neumann boundary condition but the right hand side of boundary condition is in term of the unknown function $u$.
How we can solve it?
$\Delta u(x) = f(x)\quad in~ \Omega$
$\frac{\partial u(x)}{\partial n }=g(u(x))\quad on~\partial \Omega$
where n is outward normal vector.
For special case let $g=\sqrt u$.
| https://mathoverflow.net/users/9900 | Poisson equation with special Neumann BC | Set $G$ a primitive of $g$. Then the solution is a critical point of the functional $E:H^1(\Omega)\rightarrow{\mathbb R}$ defined by
$$E[u]:=\int\_\Omega \left(\frac12|\nabla u|^2+fu\right)dx-\int\_{\partial\Omega}G(u)ds.$$
If $G(\pm\infty)=-\infty$, you may look for a minimum of $E$ over $H^1(\Omega)$. When $g$ is a decreasing function, $G$ is concave and $E$ is strictly convex: your solution is unique.
| 1 | https://mathoverflow.net/users/8799 | 41570 | 26,531 |
https://mathoverflow.net/questions/41575 | 1 | Given a smooth projective surface $S$ over an algebraically closed field, a sheaf rings or algebras $R$ on $S$ and a simple left $R$-module $M$, i.e. $Hom\_R(M,M)=k$.Then we have $Hom\_R(M,M(-i))=H^{0}(S,\mathcal{H}om\_R(M,M)\otimes O(-i))=0$ for $i>0$.
Now given some $a\in k, a\neq 0$. Then $Hom\_R(M,M(-i))=0$ implies multiplication with $a$ doesn't give a global morphism $M\rightarrow M(-i)$. But what is the reason for this? Is this because as a constant $a$ doesn't have any zeroes or poles?
| https://mathoverflow.net/users/3233 | Why is multiplication with a scalar no global morphism? | The element $a \in k$ has degree zero so it gives a global morphism $M \to M$. These are the only global morphisms because of your simplicity assumption.
If you take instead an element of degree $i > 0$, multiplication with it gives a global morphism $M(-i) \to M$.
Think about $S=\mathbb{P}^2$ and $M=\mathcal{O}$ if you want to convince yourself with a basic example.
| 3 | https://mathoverflow.net/users/7460 | 41578 | 26,534 |
https://mathoverflow.net/questions/41576 | 4 | Let $k$ be an algebraic closure of a finite field of characteristic $p$. Fix an integer $l\neq p$. For a separated $k$-scheme $X$ of finite type, we define the (compactly supported) Euler characteristic of $X$ to be $$e(X) =\sum\_i (-1)^i \dim\_\mathbf{Q\_l} H^i\_c(X,\mathbf{Q}\_l).$$ Here $H^i\_c(-,\mathbf{Q}\_l)$ denotes the $l$-adic cohomology with compact support.
For example, if $X$ is smooth and projective over $k$, we have that $e(X)$ equals the degree of the top Chern class of $X$.
Let $X$ and $Y$ be separated $k$-schemes of finite type. Let $\pi:X\longrightarrow Y$ be a finite etale morphism of degree $d$.
**Question.** Is it true that $e(X) = d \cdot e(Y)$?
Let $M$ and $N$ be separated $\mathbf{C}$-schemes of finite type. Let $\pi:M\longrightarrow N$ be a finite etale morphism of degree $d$. Then $e\_c(M) = d \cdot e\_c(N)$. To prove this, we may and do assume that $M$ and $N$ are connected and that $\pi$ is Galois. Let $G$ be the Galois group. Let $K\_0(\mathbf{Q}[G])$ be the Grothendieck group of finitely generated $\mathbf{Q}[G]$-modules. Since the action of $G$ is free, a nontrivial element $g\in G$ has no fixed points. By the Lefschetz trace formula (see the paper by Deligne-Lusztig), we have that $$\sum (-1)^i \textrm{Tr}(H^i\_c(g)) = 0.$$ Therefore, by character theory or some result in *loc. cit*, we have that the class of $H^\cdot\_c(M,\mathbf{Q})$ in $K\_0(\mathbf{Q}[G])$, defined to be the alternating sum of the classes of $H^i\_c(M,\mathbf{Q})$, is an integer multiple of the regular representation. (Here $H^i\_c(-,\mathbf{Q})$ denotes the cohomology with compact support and coefficients in $\mathbf{Q}$ on the category of para-compact Hausdorff spaces.) The result then follows from an easy computation.
**Question.** The same proof works to answer my above question positively when the cover $\pi:X\longrightarrow Y$ above is *tame*. In particular, if $p>d$. But what about the *wild* case?
| https://mathoverflow.net/users/4333 | Behaviour of euler characteristics in characteristic p for finite etale covers | It's not true in general. Over a field of characteristic $p>0$, the map $f:\mathbf{A}^1\to\mathbf{A}^1$ defined by $f(z)=z^p+z$ is etale because its derivative is $1$. The degree of $f$ is $p$, and the Euler characteristic of $\mathbf{A}^1$ is $1$, but $1\neq 1\times p$.
| 5 | https://mathoverflow.net/users/1114 | 41579 | 26,535 |
https://mathoverflow.net/questions/41557 | 11 | Let $s(x)$ is the length of continued fraction expansion of $x$, and let $l(x)$ be the sum of partial quotients. I can prove that for any rational $\alpha$ ratios $\frac{s(\alpha x)}{s(x)}$ and $\frac{l(\alpha x)}{l(x)}$ (for all rational $x$) are bounded with some constants depending on $\alpha$ only.
Is this result new?
| https://mathoverflow.net/users/5712 | Lengths of continued fractions for the numbers with fixed ratio | It definitely is not new for the length, and I am nearly sure that is not for height either.
See, for example,
Labhalla, Salah; Lombardi, Henri
Transformation homographique appliqu´ee `a un d´eveloppement en fraction continue fini ou
infini. (French) [Fractional linear transformations applied to finite and infinite continued
fractions]
Acta Arith. 73 (1995), no. 1, 29–41.
| 7 | https://mathoverflow.net/users/4312 | 41589 | 26,543 |
https://mathoverflow.net/questions/41598 | 5 | This is related to my [recent question](https://mathoverflow.net/questions/41597/transfinite-induction-a-theorem-of-pedersen-and-chains-of-subalgebras-of-bh) and would provide a natural positive answer to Question 2. I am sure this must be known to experts.
>
> **Question:** Is there a monotone injection $(\omega\_1,<) \to ([0,1],<)$ ?
>
>
>
| https://mathoverflow.net/users/8176 | Monotone injection of an ordinal into $[0,1]$ | No, because you could use it to construct an injective map $\omega\_1\to\mathbb{Q}$, mapping $\alpha<\omega\_1$ to some rational number between $\alpha$ and $\alpha+1$.
| 18 | https://mathoverflow.net/users/9342 | 41601 | 26,550 |
https://mathoverflow.net/questions/41572 | 1 | Let $X$ be a scheme, $S$ a $K3$ surface and $F$ a flat family of coherent sheaves on $S$ parametrized by $X$. Let us assume that for every $x\in X$ $F\_x$ is locally free, has fixed Chern classes and satisfies $h^1(F\_x)=h^2(F\_x)=0$. Does there exist a scheme $A$ together with a morphism $p:A\to X$ such that $p^{-1}(x)$ is isomorphic to $Aut(F\_x)$ for every $x\in X$? (by $F\_x$ I mean $F|\_{X\_x}$ with $X\_x$ the fiber of the projection $X\times S\to X$ over the point $x$)
| https://mathoverflow.net/users/33841 | Families of sheaves and automorphisms | There are standard ways of constructing this kind of objects, but I can't immediately think of a reference, so here it goes:
Let $p:Y\to X$ be a projective map (in your case, $Y=S\times X$), let $F$ and $G$ be coherent sheaves on $Y$ with $G$ flat over $X$. Then there is a scheme $H$ of finite type over $X$ whose fiber over
$x$ is the space $Hom(F\_x,G\_x)$. (More concretely, $H$ represents certain natural functor.)
In your case, $F=G$, and your $A$ is an open subscheme of $H$ consisting of invertible homomorphisms, i.e., automorphisms.
Sketch of construction of $H$: Write $F$ as the cokernel of a map
$$d:O(-n\_1)^{N\_1}\to O(-n\_2)^{N\_2}$$
for $n\_1,n\_2\gg 0$.
Morphisms from $O(-n)$ to $G$ correspond to sections of $p\_\*(G(n))$; by assumptions,
it is a vector bundle if $n\gg0$. Let $G\_n$ be the total space of this vector bundle. Then $d$ induces a map
$$(G\_{n\_2})^{N\_2}\to (G\_{n\_1})^{N\_1},$$
and $H$ is the preimage of the zero section.
| 2 | https://mathoverflow.net/users/2653 | 41602 | 26,551 |
https://mathoverflow.net/questions/41606 | 0 | Hello!
*I recently started (it's purely self-education) reading a **"Mathematical programming and optimizations"** book, did a vast part of the exercises related to the theoretical part and at one moment I got the following question about convex sets:*
I'm almost sure this statement is correct, but unfortunately, couldn't find something similiar on the internet and I tried to prove it, but I couldn't.
Assume we have some set $S \subset \mathbb{R\_n}$ and for this set: $S = \overline{S}$ *(set closure equals the set itself)*.
*Now, there exists only one projection of arbitrary point $y$ which doesn't belong to the set* $S :$
$\forall y \in \mathbb{R\_n}, \space y \notin S: \space \exists ! \space p = \pi\_S(y) $
This should mean that $S$ is a ***convex*** set.
Could someone please point me if I'm wrong *(or right, but with some limitations for this statement)* and help me proving it if I'm right.
*I also understand that this question be a too "basic" to post here, but I've just started educating myself in this sphere and hope that sometimes I'll get smart enough to ask really bright questions :)*
Thank you.
| https://mathoverflow.net/users/6256 | Convex sets and projections | I presume what you want to prove is the following. Let $S$ be a
nonempty closed subset of $\mathbb{R}^n$. Then if there is a point $y\in\mathbb{R}^n$
and there are at least two points $p$ and $q$ in $S$ with Euclidean distance $d$ from $y$
(where $d$ is the distance of $y$ from $S$), then $S$ is not convex. To see
this, note that the midpoint $r$ of the line segment $pq$ is closer to $y$
than $p$ of $q$ is, and so cannot lie in $S$. Hence $S$ isn't convex.
| 4 | https://mathoverflow.net/users/4213 | 41608 | 26,554 |
https://mathoverflow.net/questions/41604 | 8 | Lets say I have a [geometric distribution](http://en.wikipedia.org/wiki/Geometric_distribution) (of the number X of Bernoulli trials needed to get a success) with parameter `p` (success probability of a trial).
Assume I randomly sample `n` elements from this distribution.
My problem is: what is the expected *maximum* element of such a sample (it should depend on `n` I guess)? Hopefully it makes sense...
For `n=1`, e.g. if I only pick a single element from the distribution, the answer would be the mean `1 / p` of the distribution. For samples of larger cardinalities?
| https://mathoverflow.net/users/7083 | What is the expected maximum out of a sample (size N) from a geometric distribution? | Here's an answer: $\sum\_{i=1}^n \binom{n}{i} (-1)^{i+1} \frac{1}{1-q^i}$, where $q = 1-p$. Maybe someone else can help you simplify that further.
The argument: Let $Y = \max\{X\_1, X\_2, \ldots, X\_n\}$, where the $X\_i$ are $n$ sampled values from the geometric distribution.
Then $E[Y] = \sum\_{k=0}^{\infty} P(Y>k) = \sum\_{k=0}^{\infty} P(X\_1 > k \text{ or } X\_2 > k \text{ or } \cdots \text{ or } X\_n > k)$.
Applying the principle of inclusion-exclusion and the fact that the $X\_i$ are iid, we have
$P(X\_1 > k \text{ or } X\_2 > k \text{ or } \cdots X\_n > k)=$
$=\binom{n}{1} P(X\_1 > k) - \binom{n}{2} P(X\_1 > k)^2 \pm \cdots (-1)^{n+1} \binom{n}{n} P(X\_1 > k)^n$.
Since $P(X\_1 > k) = q^k$, this sum becomes
$\sum\_{i=1}^n \binom{n}{i} (-1)^{i+1} q^{ki}$.
Thus
$E[Y] = \sum\_{k=0}^{\infty} \sum\_{i=1}^n \binom{n}{i} (-1)^{i+1} q^{ki} = \sum\_{i=1}^n \binom{n}{i} (-1)^{i+1} \sum\_{k=0}^{\infty} (q^i)^k$
$= \sum\_{i=1}^n \binom{n}{i} (-1)^{i+1} \frac{1}{1-q^i}.$
| 13 | https://mathoverflow.net/users/9716 | 41614 | 26,556 |
https://mathoverflow.net/questions/41597 | 11 | This post is closely related to [this one](https://mathoverflow.net/questions/24440/must-we-close-weakly-to-apply-the-spectral-theorem). (In fact I copied some of its content.)
Let $H$ be an infinite dimensional separable complex Hilbert space. All $C^{\star}$-subalgebras of $B(H)$ are assumed to be non-degenerate.
The spectral projections of a self-adjoint element $T$ of $B(H)$ lie in the weakly closed algebra generated by $T$. In the early 1970s Pedersen [proved](http://blms.oxfordjournals.org/cgi/reprint/4/2/171) that if a $C^{\star}$-subalgebra $A$ of $B(H)$ contains all of the spectral projections of each of its self-adjoint elements, then $A$ is weakly closed, i.e. $A=A''$. Now, concerning the consequences of this result, Pedersen says in his book:
"For any $C^\star$-subalgebra $A$ of $B(H)$ define $a(A)$ as the smallest $C^{\star}$-subalgebra of $B(H)$ containing all spectral projections of each self-adjoint element in $A$. [...] Note that [...] a transfinite (but countable) application of the operation $a$ will produce $A''$."
>
> **Question 1:** What is the reasoning here?
>
>
>
I see that $\omega\_1$ applications of the operation $a$ produce $A''$ and also that each element of $A''$ appears at the $\alpha$-th application for some $\alpha < \omega\_1$ (This is due to the fact that the closure in the norm-topology is a sequential closure and hence, only countably many elements play a role.); but why is $a^{\alpha}(A)=A''$ for some $\alpha < \omega\_1$?
Maybe there is something deeper behind:
>
> **Question 2:** Is there an $\omega\_1$-chain of unital subalgebras of $B(H)$?
>
>
>
Here $\omega\_1$-chain means an $\omega\_1$-index family $(A\_{\alpha})\_{\alpha< \omega\_1}$ of subalgebras of $B(H)$ such that for all $\beta < \omega\_1$ we have
$$\overline{\cup\_{\alpha< \beta}A\_{\alpha}}^{\|.\|} \subsetneq A\_{\beta}.$$
| https://mathoverflow.net/users/8176 | Transfinite induction, a theorem of Pedersen, and chains of subalgebras of $B(H)$ | The answer to Question 2 is yes; such a chain can be found using only diagonal operators with respect to some fixed basis $(e\_n)\_{n\in{\mathbb N}}$ for $H$. As a starting point, you can find an $\omega\_1$-sequence of subsets $S\_\alpha\subset{\mathbb N}$ such that if $\alpha<\beta$, $S\_\beta\setminus S\_\alpha$ is finite and $S\_\alpha\setminus S\_\beta$ is infinite (that is, the sequence $(S\_\alpha)$ is strictly decreasing "modulo finite sets"). This is easy to do because by diagonalizing, given a countable decreasing sequence of infinite subsets of ${\mathbb N}$, you can always find another infinite set which is strictly contained in all of them modulo finite sets. Now let $A\_\alpha$ denote the algebra of diagonal operators such that the eigenvalues of the eigenvectors $e\_n$ for $n\in S\_\alpha$ form a convergent sequence. These are closed because a uniform limit of convergent sequences is convergent. These are nested because if $\alpha<\beta$, then all but finitely many elements of $S\_\beta$ are contained in $S\_\alpha$ so if the $S\_\alpha$-eigenvalues converge, so do the $S\_\beta$-eigenvalues.
Here's a more conceptual explanation of this construction. The algebra of diagonal operators is naturally isomorphic to the algebra $C\_b({\mathbb N})$ of bounded continuous functions on $\mathbb N$, which is in turn naturally isomorphic to the algebra $C(\beta{\mathbb N})$ of all continuous functions on the Stone-Cech compactification of $\mathbb N$. Each set $S\_\alpha\subset{\mathbb N}$ has a closure $\overline{S\_\alpha}\subset\beta{\mathbb N}$, and the statement that the $S\_\alpha$ are decreasing modulo finite sets says exactly that the sequence of closed sets $C\_\alpha=\overline{S\_\alpha}\setminus S\_\alpha$ is actually decreasing. The subalgebra $A\_\alpha$ is then exactly the set of continuous functions on $\beta{\mathbb N}$ which are constant on the set $C\_\alpha$.
| 7 | https://mathoverflow.net/users/75 | 41617 | 26,558 |
https://mathoverflow.net/questions/41569 | 2 | A neutral Tannakian category over a field $k$ is a rigid $k$-linear abelian tensor category
$\mathcal{C}$ whose unit $1$ satisfies $\mathrm{End}(1) \simeq k$, and is
moreover equipped with an exact faithful tensor functor $\omega : \mathcal{C} \rightarrow
\mathrm{Vect}\_k$ into the category of finite dimensional $k$-vector spaces.
Question:- Why the condition $\mathrm{End}(1) \simeq k$ is necessary to get equivalence with the category of finite dimensional representations of some affine group scheme?
Thanks in advance!!
| https://mathoverflow.net/users/8141 | The condition End(1) = k in Tannakian Categories | If you say the other axioms correctly, then the condition on $\operatorname{End}(1)$ is redundant. Indeed, the word "tensor functor" implies that $\omega: 1 \mapsto k$, and the word "faithful" implies that $\operatorname{End}(1) \hookrightarrow \operatorname{End}(\omega(1))$. What you should include that you don't on your list is that $\omega$ be $k$-linear. You should also demand that $\mathcal C$ be a nontrivial category; then you cannot have $\operatorname{End}(1) = 0$, as $\operatorname{End}(1)$ acts on all other homsets via the $1$ action and in particular $\operatorname{id}\in \operatorname{End}(1)$ acts as the identity on all other homsets. With all of this, it follows that $\operatorname{End}(1) = k$.
Conversely, you can see the condition that $\operatorname{End}(1) = k$ as being a "nontriviality" condition. It is necessary only to assure that $\mathcal C \neq 0$. In particular, no group has zero representation theory, as every group has a trivial representation on $k$.
If you believed in "the empty group", then you would not need this restriction: the zero category is the category of representations of the zero ring, which is "the group ring of the empty group".
| 6 | https://mathoverflow.net/users/78 | 41623 | 26,562 |
https://mathoverflow.net/questions/41612 | 5 | Hi, I have a general question about the relationship between exponential sums and differential equations. In particular, I have been trying to read Katz' work on the subject (his book and his lecture notes) but I am having trouble understanding the big idea and getting confused with all of the algebraic geometry background. Can someone explain the general picture, or direct me towards more elementary works introducing the subject?
| https://mathoverflow.net/users/9769 | Exponential sums and differential equations | In simplistic terms, exponential sums arise in characteristic p geometry because curves have Artin-Schreier coverings. The appearance of (complete) exponential sums in that context is a reflection in cohomology of that kind of Galois covering, and can be understood pretty much in Weil's terms and technology. (Appendix in Basic Number Theory? Unfortunately I have the first edition that doesn't have such conveniences.) And linear differential equations on curves have been understood geometrically since Riemann's time, admittedly in varying languages. The case corresponding to finite coverings (i.e. curve morphisms) is that of finite monodromy.
So far so good? Nick Katz's papers have got easier to read as the years go by (at least 10% per decade I think, as he has moved away from the Grothendieckising style). But are probably still hard going.
| 2 | https://mathoverflow.net/users/6153 | 41627 | 26,565 |
https://mathoverflow.net/questions/41624 | 7 | Recall that $\operatorname{Sch}$ can be identified with the subcategory of (Zariski-)locally representable (by an affine) étale sheaves on $\operatorname{CRing^{op}}$. In this case, $\operatorname{Spec}(-):\operatorname{CRing^{op}}\to \operatorname{Sch}$ is simply the co-Yoneda embedding $R\mapsto \operatorname{Hom}(R,-)$ (which makes sense since the étale topology is subcanonical and $\operatorname{Hom}(R,-)$ is obviously locally affine).
Is there a nice "sheaf-theoretic" description of $\operatorname{Proj}:\operatorname{(Gr\_{{\mathbf Z}\_{\geq 0}}CRing)^{op}}\to \operatorname{Sch}$ (I've only seen $\operatorname{Proj}$ for nonnegative integral grading. If we can use more exotic gradings, I guess I'd be interested in that too). Hoping for it to be as nice as $\operatorname{Spec}$ seems like a bit of a pipe dream, but I'm wondering if there is a nicer way to describe it than Hartshorne's approach, which feels rather arbitrary.
Edit: To clarify, I'm looking for a construction $\operatorname{Proj}:\operatorname{(Gr\_{{\mathbf Z}\_{\geq 0}}CRing)^{op}}\to \operatorname{Sh}\_{\acute{et}}(\operatorname{CRing^{op}})$, which we can then see lands in $\operatorname{Sch}$ by showing that we can cover it with Zariski-open affines.
| https://mathoverflow.net/users/1353 | A sheaf-theoretic version of the proj construction? | Perhaps I've not understanding your question, but it sounds like you're asking "What is the functor of points of a Proj?"
The answer, of course, is the functor that sends a scheme $X$ to the set of line bundles $L$ on that scheme equipped with a graded map of $R$ to $\Gamma(X;\oplus\_{n\geq 0} L^n)$ whose image generates $L^n$ as a $\mathcal{O}\_X$-module. The affine open sets are given by the subset where a positive degree element $r$ of $R$ is sent to non-vanishing section; these open sets are easily seen to be the Spec of the degree 0 part of $R\_r$, the localization of $R$ at $r$.
**EDIT:** Let me incorporate some of the things I said below: if you want to just work with rings, then replace "line bundle" with "invertible projective" and $\Gamma(X;\oplus\_{n\geq 0} L^n)$ with "tensor algebra over A."
If you're OK with sheafifying, you can do something simpler, which is assume that your line bundle is trivial (**but not with a fixed trivialization**), i.e. that your invertible projective is free of rank 1 (**but not canonically isomorphic to A**).
A graded map from $R$ to the tensor algebra of a rank 1 free module is the same as a map from $R\to A$, after you pick an isomorphism of that module to $A$. However, you have to identify maps that come from the same map to the tensor algebra under different isomorphisms (i.e. mod out by the action of the units of A), and throw out maps where the images of the degree 1 elements don't generate A (this is why you get a Proj-ish thing).
| 7 | https://mathoverflow.net/users/66 | 41630 | 26,567 |
https://mathoverflow.net/questions/41616 | 16 | The classifying space $BG=|Nerve(G)|$ of an arbitrary topological group $G$ does not necessarily have the homotopy type of a CW-complex but the fundamental group should still be accessible. What is $\pi\_{1}(BG)$? A reference on this would be great. My initial guess: $\pi\_{1}(BG)$ is the quotient group $\pi\_{0}(G)$ for arbitrary $G$
Motivation: There is a natural way to make $\pi\_1$ a functor to topological groups. I am interested in relating the topologies of $G$ and $\pi\_{1}(BG)$ but the topology on $\pi\_{1}(X)$ is boring (discrete) when $X$ is a CW-complex.
| https://mathoverflow.net/users/5801 | What is π_1(BG) for an arbitrary topological group $G$? | If $G$ is homeomorphic to a Cantor set (e.g. $G=\mathbb Z\_p$), then $BG$ contains a copy of the Hawaiian earrings in it. To see this, take a sequence of points of $G$ that converges to the identity element: you'll get a corresponding sequence of 1-cells in $BG$ that converge to the the degenerate 1-cell. The fundamental group of the Hawaiian earrings is a rather wild object, and looks nothing like the free group that you might naively expect.
If, on the other hand, if you agreed to redefine $BG$ to be the **fat** geometric realization of the simplicial space $NG$, then you would get $\pi\_1(BG)\cong\pi\_0(G)$, as desired.
I would even bet that the above isomorphism respects the natural topologies that are present on both sides.
| 32 | https://mathoverflow.net/users/5690 | 41632 | 26,569 |
https://mathoverflow.net/questions/41638 | 6 | Hello, I was looking for an answer to the following question:
Consider an algebraically closed field $K$ and a map $K \rightarrow K^n$ given by $a \mapsto (p\_1 (a) , \ldots p\_n (a))$ for $p\_i \in K[t]$. Is the image of this map an algebraic set?
Certainly this is false for $K^2 \rightarrow K^n$, for example $(x,y) \mapsto (x, xy)$ but I feel like polynomials in $1$ variable aren't complicated enough to give this bad behavior.
Thanks!
| https://mathoverflow.net/users/18 | Regular Morphism From Affine Line | Dear Damien, let's show that your morphism $f: \mathbb A^1\_K \to \mathbb A^n\_K $ is proper, hence closed, hence certainly has closed image.
For that it is enough to prove that each $f\_i:\mathbb A^1\_K \to \mathbb A^1\_K$ is proper. But this follows from the stronger property that $f\_i$ is finite or dually that the ring morphism $K[T] \to K[T]: T\to p\_i(T)$ is finite. This is elementary: it follows, for example, from the fact that $T$ is (tautologically) integral over $K[p\_i (T)]$.
Note that in this proof you needn't assume that the field $K$ is algebraically closed.
**Edit:** As BCnrd remarks, this proof only works if all polynomials $p\_i(T)$ are non-constant. Let me modify the proof to take his judicious comments into account. If all polynomials are constant, your morphism is not proper but its image is clearly closed. If at least one polynomial is non-constant, say the first, then the argument above proves that the corresponding morphism $f\_1: \mathbb A^1\_K \to \mathbb A^1\_K $
is finite.The obvious closed immersion $j: \mathbb A^1\_K \to \mathbb A^n\_K $ (last $n-1$ coordinates given by the other polynomials) is finite and the composition, which is your morphism $f=j\circ f\_1: \mathbb A^1\_K \to \mathbb A^n\_K $ , is thus also finite.
| 5 | https://mathoverflow.net/users/450 | 41667 | 26,588 |
https://mathoverflow.net/questions/41660 | 5 | Given an operator $f\colon R^m\to R^n$, can one always find a non-zero vector
$x\in \{ 0,1 \}^m$ such that $\|f(x)\|/\|x\|\ge0.01\|f\|$? (Here I denote by
$\|\cdot\|$ both the Euclidean norms in $R^m$ and $R^n$ and the induced
operator norm.) The answer may well be negative -- any examples?
---
In case the answer to the question above is ``no'' (or unknown), would it
help to assume that the matrix of $f$ with respect to the standard
orthonormal bases of $R^m$ and $R^n$ has all its elements equal to $0$ or
$1$?
---
As I see it, this is basically a question in the geometry of numbers, and I
would expect the answer should be known.
| https://mathoverflow.net/users/9924 | Is the operator norm always attained on a $\{0,1\}$-vector? | The answer is no. First, to understand the question, WLOG $f$ is symmetric and positive definite; a general $f$ has a polar decomposition $f = os$ and the orthogonal factor $o$ has no effect on any of the norms in question. Then, WLOG $f$ is a rank 1 projection. The second and subsequent eigenvalues of $f$ do not increase $||f||$, but they could increase $||f(x)||$ for some specific $x$. So in summary, we can assume that $f = vv^T$ for some vector $v$. The question is whether $v$ must always make a small angle with some binary vector.
Let
$$v = (1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\ldots,\frac{1}{\sqrt{m}}).$$
If $w$ is a binary vector of weight $k$, then $|v \cdot w|$ is maximized when the non-zero entries of $w$ are at the beginning. However,
$$||w|| = \sqrt{k} \qquad ||v|| = \Theta(\sqrt{\log m}) \qquad |v \cdot w| = O(\sqrt{k}).$$
This means that the angle between $w$ and $v$ is large, and therefore $||f(w)|| = ||v (v \cdot w)||$ is small compared to $||f||\;||w|| = ||v||^2 ||w||$.
The same proof works if $\{0,1\}$ is replaced by $\{-1,0,1\}$, or indeed by any finite subset of $\mathbb{R}$. On the other hand, there is a variation of the question with a positive answer.
Similar to Pietro Majer's remark, you can interpret the question as a comparison between two norms on $\mathbb{R}^m$. One is the $\ell^2$ norm, and the other is the norm whose unit ball is a polytope whose vertices are at the points in $S = \{0,1\}^m$ and its negative. By the theory of spherical packings on a sphere, for any $c < 1$, there exists a set $S$ of exponential size in $m$ such that the two norms are equal up to a factor of $c$. This is then a positive answer for that sample set of vector, even for constants close to 1. But such a set (coming from the centers of a sphere covering of the sphere) has to be fairly complicated, and I don't know if there are explicit asymptotic examples.
| 9 | https://mathoverflow.net/users/1450 | 41669 | 26,589 |
https://mathoverflow.net/questions/41666 | 15 | Given odd positive integer $n$ and a monic polynomial $f(x)=(x-x\_1)\dots (x-x\_n)$ with $n$ distinct real roots. Is it always true that $\sum f'(x\_i) > 0$? I may prove it for $n=3$ and $n=5$ and it looks plausible.
| https://mathoverflow.net/users/4312 | sum of derivatives in roots of a polynomial of odd degree | If I'm not mistaken this is basically the same question as [this](http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366673&sid=0f77743cc628257f609bb649b7ecfc84#p366673) question from the international mathematical olympiad in 1971. The statement is only true for 3 and 5 variables showing that there is no obvious generalization to [Schur's inequality](http://en.wikipedia.org/wiki/Schur%27s_inequality) in many variables.
| 19 | https://mathoverflow.net/users/2384 | 41674 | 26,590 |
https://mathoverflow.net/questions/41619 | 13 | A nice property of $\mathbb P^n$ is:
>
> Property 1: Two subvarieties $U,V$ such that $\operatorname{dim} U +\operatorname{dim} V \geq n$ always intersect.
>
>
>
(for example, any 2 curves in $\mathbb P^2$ intersect)
There are other smooth varieties $X$ when Properties 1 holds. For example, a sufficient condition is that the ranks of $\text{CH}^i\_{num}(X)$ are $1$ for $i\leq n/2$. Here $n = \operatorname{dim} X$ and $\text{CH}^i\_{num}(X)$ is the Chow group of codimension $i$ modulo [numerical equivalences](http://en.wikipedia.org/wiki/Adequate_equivalence_relation).
My question is whether some converse is true:
>
> Question: Let $X$ be a smooth projective variety satisfying Property 1. Does that impose some upper bounds on the ranks of $\text{CH}^i\_{num}(X)$ for $i\leq n/2$?
>
>
>
Let's assume we are over $\mathbb C$, but I am also interested in results over any ground fields.
One can ask the same questions for the ranks of $\text{CH}^i\_{hom}(X)$ (I think they are conjectured to be the same). The baby case is $i=1$, where the question asks if Property 1 tells us something about the rank of the Neron-Severi group of $X$.
I am aware that the question is a little vague (upper bound as function of what?), but that was because of my ignorance, so comments to improve the question are welcome.
| https://mathoverflow.net/users/2083 | Intersection of subvarieties versus ranks of Chow groups modulo numerical equivalences | This is a very interesting question and I guess that a general answer is unknown, already in the case $i=1$. Let me just make the following
**Remark.** There exists no upper bound on $\textrm{rank } NS(X)$ which is independent on the dimension.
In fact, let us consider a complex Abelian variety $X$ of dimension $g$ such that $End\_{\mathbb{Q}}(X)$ is a totally real number field of degree $g$ over $\mathbb{Q}$. These
varieties do exist and the general one is simple, see [Birkenhake - Lange, Chapters 5 and 9].
Therefore it is known that
$\rho(X) =\textrm{rank } \textrm{NS}(X) = \textrm{rank } End^s\_{\mathbb{Q}}(X)=g$,
where $End^s\_{\mathbb{Q}}(X)$ denotes the subgroups of elements in $End\_{\mathbb{Q}}(X)$ which are symmetric with respect to the Rosati involution.
On the other hand, in a simple Abelian variety any effective divisor is ample, so two effective divisors always intersect and $X$ satisfies Property 1.
| 7 | https://mathoverflow.net/users/7460 | 41678 | 26,592 |
https://mathoverflow.net/questions/41675 | 11 | Hi... I am wondering how 'eigenvalues' that don't lie in my Hilbert space combine into producing the spectral measure. I study probability and I am quite ignorant in the field of spectral analysis of operators on Hilbert spaces so please go easy on me :), yet i tried reading parts of the classical Simon and Reeds "Functional analysis" volume 1, and other books. I feel i am very far from an answer. At least, now i can formulate my question.
The mathematical setting is the following: I consider a general (possibly unbounded) operator $A$ on a Hilbert space H with scalar product $(. , .)$, say $H = L^2( \mathbb{R} )$. $D(A)$ will be a dense domain. I do understand that the spectrum $\sigma(A)$ is defined as the complementary of the resolvent set, and can be broken into continuous, residual and point spectra.
In the case of self-adjoint operators (hence closable), a very abstract version the Von-Neumann spectral theorem asserts that $A$ can be diagonalized using a spectral decomposition of the identity. The full setting would look like (cf "Quantum physics for mathematicians" by Leon Takhtajan):
* There are a spectrum-indexed family of projectors $P\_\lambda(.)$ $\lambda \in \sigma(A)$. These are the "spectral projectors" and reduce in finite dimension to $P\_\lambda(x) = \sum\_{ \mu \leq \lambda }(x, e\_\mu) e\_\mu$, $e\_\mu$ being the orthonormal diagonalizing basis. And when $\lambda$ is in the point spectrum: $dP\_\lambda(x) = (x, e\_\lambda) e\_\lambda$
* The image of spectral projectors grows with respect to the spectral parameters so that the following identity is true:
$$ \forall f \in H, P\_\lambda \circ P\_\mu(f) = P\_{min(\lambda,\mu)}(f)$$
* The behavior regarding the $\lambda$ parameter is such that:
$$ \forall (f,g) \in H^2, (P\_\lambda(f), g) = \mu\_{f,g}(]-\infty; \lambda])$$ with $\mu\_{f,g}$ a measure. Because of this property, $P\_.(f)$ can be seen as a measure on H itself whatever that means.
* The spectral decomposition of the identity (trivial in finite dimension):
$$ \forall f \in H, f = \int\_{-\infty}^\infty dP\_\lambda(f) $$
* The spectral decomposition of our operator:
$$ \forall f \in H, Af = \int\_{-\infty}^\infty \lambda d P\_\lambda(f) $$
This last one being the generalization of the very basic linear algebra identity valid for hermitian matrices:
$$ \forall x \in \mathbb{R}^n, A(x) = \sum\_{ \lambda \in \sigma(A) } \lambda (x, e\_\lambda) e\_\lambda $$
It is well known that at $\lambda$ an eigenvalue (in the point spectrum), the spectral measure has a Dirac, as we find ourselves in same situation as the finite dimensional case. I am interested in "generalized eigenfunctions" that are functions not necessarily in $L^2$, but that verify still $ Af = \lambda f$ for a certain $\lambda$ in the general spectrum. I am now including two classical examples.
In the case of the "position" operator:
$$ D(A) = (\{ f \in H, \int x^2 f(x)^2 dx < \infty \}) $$
$$ (Af)(x) = x f(x) $$
The spectrum is well known and only continuous: $\sigma(A) = \mathbb{R}$. It is obvious that the operator is already diagonal, and that matter is reflected by the fact that the "generalized eigenfunctions" are Dirac distributions:
$$ (A \delta\_\lambda)(x) = x \delta\_\lambda(x) = \lambda \delta\_\lambda(x)$$
$$ \forall f \in H, Af = \int\_{\mathbb{R}} \lambda f(\lambda) \delta\_\lambda(.) $$
In the case of $A = -\Delta$:
$$ D(A) = (\{ f \in H, \int f''(x)^2 dx < \infty \}) $$
The spectrum is well known and only continuous: $\sigma(A) = \mathbb{R}^+$. The operator is diagonalized using the Fourier transform $\mathcal{F}$, and that matter is reflected by the fact that the "generalized eigenfunctions" are complex unitary characters $e^{i\sqrt{\lambda}x}$ and $e^{-i\sqrt{\lambda}x}$. The spectral theorem takes a simple shape thanks to the Fourier transform. Indeed:
$$ \forall f \in H, \mathcal{F}(Af)(k) = k^2 \mathcal{F}(f)(k) $$
Then:
$$ \forall f \in H, (Af)(x) = \frac{1}{2 \pi} \int\_{\mathbb{R}} k^2 e^{-i k x} \mathcal{F}(f)(k) dk = \frac{1}{2 \pi} \int\_{\mathbb{R}^+} \lambda ( e^{-i \sqrt{\lambda} x}\mathcal{F}(f)(\sqrt{\lambda}) - e^{i \sqrt{\lambda} x}\mathcal{F}(f)(-\sqrt{\lambda}) ) d\lambda$$
This last way of writing the operator 'diagonalization' shows that the spectral measure is a superposition of the two types of 'waves' (positively propagating $e^{i\sqrt{\lambda}x}$ and negatively propagating $e^{-i\sqrt{\lambda}x}$) with a weight given by the Fourier transform of f, whatever that really means.
In those two cases, we see that those "generalized eigenfunctions" (Diracs and unitary complex characters) combine in a special way in order to produce the spectral measure. I read somewhere a sentence that left me puzzled: `those eigenfunctions combine into a Schwartz kernel`. I think i read that in "Quantum physics for mathematicians" by L. Takhtajan. Now i get the feeling that fully diagonalizing a self-adjoint operator can be a very hard task. Can you also provide me for other references than those i used to far?
In the end, my question could be formulated as the following, even if i am not satisfied with it, as it is still too vague:
Suppose that by some mean, i know all or some "generalized eigenfunctions". Then can i express the spectral measure in terms of those eigenfunctions? If so, how?
Side questions:
* It seems natural to ask my generalized eigenfunctions to be a complete orthonormal system, whose cardinal is the cardinal of the spectrum at least (if no multiplicities). It then makes me feel that those functions (or distributions) will lie in a non-separable space as all orthonormal systems is separable spaces are countable. For the same reason, an operator's point spectrum is always countable. This makes me think that generalized eigenfunctions have to be looked for in a very big space with a special topology.
* Why some "generalized eigenfunctions" count and others don't? I am thinking of the Laplacian case, as any $f(x) = e^{zx}, z \in \mathbb{C}$ verifies $f'' = z^2 f$. But clearly, only those with imaginary $z$ count. There are also the harmonic polynomials. Is this related to the fact of being unitary?
I would very much appreciate any references or hints. And i am sorry if my question is not stated in the proper terms of spectral analysis/operator theory.
Phew that was long to type...
Cheers
Reda
| https://mathoverflow.net/users/9419 | How "generalized eigenvalues" combine into producing the spectral measure? | Wouldn't Theorem 4.2 in [here](http://mathserver.neu.edu/~king_chris/GenEf.pdf) answer your question?
**Theorem 4.2.** *(Generalized Eigenfunctions, by Mustafa Kesir)* Let $\mathcal H$ be a Hilbert Space. Given a self-adjoint operator $A$ in $\mathcal H$ and a Hilbert–Schmidt rigging of $\mathcal H$, there exists a complete system of generalized eigenfunctions of $A$.
| 4 | https://mathoverflow.net/users/3698 | 41681 | 26,593 |
https://mathoverflow.net/questions/41679 | 1 | 3 players are playing a game where they get to pick independently without knowing the other players picks one of 2 prizes (A,B) and the payout is (a,b) for the two prizes, divided by how many people picked the specific prize.
For example, if the prizes are (3,1) and 2 people picks A and 1 person picks B, the 2 people get 3/2 = 1.5 each and the third person gets 1 for himself.
In a zero-sum setting, if the prizes are (3,1), it would always be best to pick the first location, since you are guaranteed a prize of at least 1. If prizes are (2,1), this is also true, since no other player can beat that strategy. But, if prizes are (3,2), and two players use the strategy, the third player can do better by always picking the 2nd prize.
Is it possible to find an equilibrium for this particular problem, and does it generalize to further players and prizes? Does it fit into some standard theory?
| https://mathoverflow.net/users/9926 | Simple(?) game theory | This looks like a homework exercise on mixed strategy Nash equilibria.
My ASCII art skills are a little rusty, but let's write up a "nice" litte 2x2x2 cube:
```
+--------------+-----------------+
C plays 3 / / / |
+--------------+----------------+ |
C plays 2 / / /| /
+---------------+---------------+ | /|
B plays 3 | (3/2,3/2,2) | (1,3,1) | |/ /
+---------------+---------------+-/ /
B plays 2 | (3,1,1) | (2/3,2/3,2/3) | |/
+---------------+---------------+-/
A plays 3 A plays 2
```
Yes I know it's ugly as sin. Just sketch it on a piece of paper instead and fill it out with the payoffs. And leave a spot somewhere for the (A:3,B:2,C:3) corner.
Now let $p\_X$ denote the propability that player $X$ plays $3$ and let $U\_{X,y}$ denote the expeted payoff for player $X$ playing $y$
Because of the symmetries in this game, in a mixed strategy equilibrium $A$, $B$ and $C$ will all play the same mixed strategy, i.e. $p\_A = p\_B = p\_C$.
$$
U\_{C,3} = p\_Ap\_B\cdot 1 + p\_A(1-p\_B)\cdot\tfrac{3}{2} + (1-p\_A)p\_B\cdot\tfrac{3}{2} +
(1-p\_A)(1-p\_B)\cdot 3 $$
$$ = p\_A^2 + \tfrac{3}{2}p\_A - \tfrac{3}{2}p\_A^2 + \tfrac{3}{2}p\_A - \tfrac{3}{2}p\_A^2 + 3
-6p\_A + 3p\_A^2 = p\_A^2 -3p\_A + 3$$
$$
U\_{C,2} = p\_Ap\_B \cdot 2 + p\_A(1-p\_B)\cdot 1 + (1-p\_A)p\_B\cdot 1 + (1-p\_A)(1-p\_B)\tfrac{2}{3} $$
$$ = \tfrac{2}{3}p\_A^2 + \tfrac{2}{3}p\_A + \tfrac{2}{3}$$
Then we just solve $U\_{C,3} = U\_{C,2}$ for $p\_A$ and we get $\tfrac{1}{3}p\_A^2 - \tfrac{11}{3}p\_A + \tfrac{7}{3} = 0$ which has solutions $p\_A = \frac{11 \pm \sqrt{93}}{2}$. Only the root corresponding to minus gives $p\_A \in [0,1]$, so we conclude that the players should play $3$ with a probability of approximately $0.678$.
PS: I've given up on `align`. It just doesn't work. Ever.
PPS: There's almost certainly a mistake. This was a pretty quick and dirty computation.
| 2 | https://mathoverflow.net/users/8842 | 41687 | 26,597 |
https://mathoverflow.net/questions/41649 | 4 | I've been searching google and scholar google, but i only have come upon orderings and Hermitian forms on \*-fields.
Has real algebraic geometry been carried over to \*-rings? \*-rings are rings with an involution. For example are there Positivstellensätze and characterizations of sums of squares by total positivity etc.
If anyone has references, it would be of great help. Books, papers or online resources are all great. thanks in advance.
| https://mathoverflow.net/users/nan | Doing Real Algebraic Geometry on *-Rings | Konrad Schmüdgen has set up a programme to develop analogues of the basic results in Real Algebraic Geometry in a setup of $\star$-algebras. See [arxiv.org/abs/0709.3170](http://arxiv.org/abs/0709.3170) and for example [arxiv.org/abs/0903.2708](http://arxiv.org/abs/0903.2708). There is also interesting recent work by Jaka Cimpric on that topic, [arxiv.org/abs/0807.5020](http://arxiv.org/abs/0807.5020). Looking up the references in those articles will give you a lot more interesting sources of information.
There is also a reformulation of the Connes embedding conjecture in terms of sums of squares, which is due to Hadwin, Radulescu and Klep-Schweighofer in different variations.
| 1 | https://mathoverflow.net/users/8176 | 41690 | 26,599 |
https://mathoverflow.net/questions/41676 | 7 | I'm reading stalling's article "the augmented ideal in group ring" in Ann. Math. Studies 84(R. H. Fox memorial volume)
In his final remark, he says that Milnor's link invariant could be interpreted by using Spectral sequence.(see Milnor, Isotopy of links, Algebraic geometry and topology, Princeton press)
Are there anybody who knows about further advances in this story??
| https://mathoverflow.net/users/7776 | Milnor's isotopy invariant using spectral sequence? | Have you read the Wikipedia page on Massey products?
<http://en.wikipedia.org/wiki/Massey_product>
It mentions Massey products are differentials in the Atiyah-Hirzebruch spectral sequence for a K-theory with local coefficients. The Atiyah-Hirzebruch spectral sequence is to a (co)homology theory what cellular (co)homology is for standard singular (co)homology.
| 4 | https://mathoverflow.net/users/1465 | 41692 | 26,601 |
https://mathoverflow.net/questions/41543 | 10 | I cannot find any answer to that apparently simple problem :
On a square lattice, a path is given by a sequence of relative moves in {"move forward", "turn right" and "turn left"}.
Is there a rule that characterizes if a path is self-avoiding (or not) ?
Edit : Let me precise the kind of rule I am looking for:
Let's imagine a walk described, this time, by absolute moves (N=move North, S=move South, E=move East, W=move West), then the presence of a loop in the sequence is characterized by a subsequence for which nb(N) = nb(S) and nb(E)=nb(W). That's a simple rule.
**Is there such a rule in the case of a sequence of relative moves ? Or do we have to translate the sequence to absolute moves ? Thanks.**
Example (to make myself clear):
here is a walk (or part of a walk), written in absolute moves {North, East, West, South}:
EENWNNWSSS
=> We immediately know it is a loop, without having to draw anything or keep track of the positions visited, because nb(N)=nb(S) and nb(E)=nb(W).
Now here is the same walk written in relative moves {Forward, Turn right, Turn left}:
FFLFLFRFFLFLFFF
=> Without drawing anything, nor converting to absolute moves. Is there a rule that allows to say it is a loop ?
| https://mathoverflow.net/users/8779 | How to characterize a Self-avoiding path. | I'm only now starting to understand the question. Let me formulate it slightly differently, but I think it's easy to translate between this and your formulation. I'll imagine a little machine with three possible operations: take a step (which is always in the direction of an arrow on the machine's head, so would be your "forwards"), turn to the left (which changes the direction of the arrow but doesn't change the square the machine is on), and turn to the right.
If you're prepared to characterize self-avoidance with a statement like "There is no subsequence such that ...," as you seem to be, then it is enough to characterize loops. This is easy to do if we convert into absolute moves, as you say, so the idea is not to do that. It is quite hard to say precisely what it means not to keep track of the direction the machine is pointing in, but here is an attempt.
Now there are various ways that a sequence of moves can be simplified. For instance, RRR=L, LLL=R, and RL=LR="do nothing", so we can always rewrite so that there are at most two turns in a row, when they have to be the same. Also, FRRF can be simplified to RR, so if we ever have RR or LL we can cancel Fs on either side until we have three turns in a row and then cancel those.
After all this cancellation, we may assume that we never have two turns in a row. But that is not all the cancellation that can be done. For instance, $FRF^kRF$ cancels to $RF^kR$. So if we ever have $RF^kR$ we can cancel Fs on both sides until we get two turns in a row and go back to the previous method for getting rid of that.
So now we may assume that the sequence consists of Fs with isolated Rs and Ls that have to alternate. That won't be a loop unless the sequence is null.
I think this can be done without thinking about what the absolute direction is, but one could perhaps argue that the cancellation rules are working mod 4 and keeping track of the direction in an implicit way. But I'm not sure that's right, since we can do the cancellation in the middle of the sequence without caring about the absolute direction at all. So perhaps this counts as a sort of answer to your question.
| 12 | https://mathoverflow.net/users/1459 | 41694 | 26,602 |
https://mathoverflow.net/questions/41693 | 8 | I expect this to be a small understanding problem and not a real interesting question.
In [Gabriel's thesis](http://ncatlab.org/nlab/show/Des+Cat%25C3%25A9gories+Ab%25C3%25A9liennes) you find a proof of the theorem that every small abelian category $C$ admits a faithful exact functor to the category of abelian groups. The proof goes like this: Consider the abelian category $Sex(C,Ab)$, which is the category of left-exact functors $C \to Ab$. By the way, does anybody know why Gabriel has chosen this notation? Why not "Lex" for left-exact or "Gex" for the french "exacte a gauche"? Anyway, it can be shown that $Sex(C,Ab)$ is a (nowadays called) Grothendieck category and has thus enough injective objects. Besides, every injective object is an exact functor $C \to Ab$. If we embed the generator $U=\sum\_{X \in C} Hom(X,-)$ into an injective object, we are done.
Now Gabriels claims that $U$ is actually a projective generator. I doubt that this is used in the proof, nevertheless it is interesting. By Yoneda the functor $F \mapsto Hom(U,F)$ is isomorphic to $F \mapsto \prod\_{X \in C} F(X)$, thus we have to prove that every epimorphism in $Sex(C,Ab)$ is actually pointwise an epimorphism of abelian groups. This is not clear to me since a cokernel in $Sex(C,Ab)$ is defined by the universal left-exact functor associated to the pointwise-defined cokernel (cf. Prop. 5 in II.2). The vanishing does not seem to imply the vanishing of the pointwise-defined cokernel (cf. Lemme 3 b).
Note that Grothendieck has supervised this thesis. Although there are many many typos, I don't think that such a statement will just be wrong. What am I missing?
| https://mathoverflow.net/users/2841 | projective generator in the category of left-exact functors | The object $U$ is a projective generator in the category of additive functors $C\to Ab$. It is also a generator of the category of left exact functors $C\to Ab$. It is not projective in the latter category, though.
Generally, left exact functors are similar to sheaves (on the category opprosite to $C$; epimorphisms in the latter category, i.e., monomorphisms in $C$, play the role of the coverings). There are no projective sheaves, generally speaking. The object $U$ is like the direct sum of the constant (pre)sheaves on all open subsets, extended by zero to the outside. It is a projective presheaf. It is not supposed to be a projective sheaf.
Update: I have been asked to provide a specific counterexample. It suffices to present an example of a morphism of left exact functors whose object-wise cokernel is not left exact. Let us find such a morphism of functors among morphisms of representable functors. So we want a morphism $X\to Y$ in an abelian category such that the object-wise cokernel of the morphism $Hom(Y,{-})\to Hom(X,{-})$ is not left exact. This means that there is a commutative square of morphisms $X\to Y$, $A\to B$, $X\to A$, $Y\to B$ such $A\to B$ is a monomorphism and the morphism $X\to A$ does not factor through the morphism $X\to Y$. Start with any nonsemisimple abelian category and choose a nonsplit monomorphism $X\to Y$. Set $A=X$ and $B=Y$.
| 5 | https://mathoverflow.net/users/2106 | 41699 | 26,606 |
https://mathoverflow.net/questions/41682 | 2 | We know the fact that $K\_0(-)$ and $K\_1(-)$ are continuous under inductive sequence of $C^\*$-algebras (in fact inductive system), i.e. $K\_0(\varinjlim A\_n)=\varinjlim K\_0(A\_n)$ similar for $K\_1(-)$. In fact it is also true that $M\_k(\lim\_{\rightarrow} A\_n)=\varinjlim M\_k(A\_n)$ for $k\in \mathbb N$.
Q1: Does $\widetilde{(\varinjlim A\_n)}$ coincide with $\varinjlim\tilde{(A\_n)}$? In fact this is a claim in someone's book, but without a proof. If we let $(X,\lambda\_n)$ be the inductive limit of $\tilde{A\_1}\rightarrow \tilde{A\_2}\rightarrow~\cdots$, then by universal property we get a unique morphism $\lambda: X\rightarrow \widetilde{\varinjlim A\_n}$. How can we show $\lambda$ is injective? NB morphisms need not be unital, even though $C^\*$-algebras are unital.
Q2: Can we find any other continuous functors?
What about the universal group $C^{\star}$-algebras, tensor product of $C^{\star}$-algebras, cross product of $C^{\star}$-algebras and so on?
Q3: Do we know any functor which is not continuous?
| https://mathoverflow.net/users/9401 | Continuity of functors under inductive sequence of $C^*$-algebras. | The answer to Q1 is a standard fact, which can be found in the standard text books like Bruce Blackadar's book on $C^{\star}$-algebras. The answer to Q2 is "yes" and, again, looking in a text book helps.
For Q3, I think that it is worth noting that the Algebraic K-theory of $C^{\star}$-algebras is very interesting but not continuous, in fact the natural map
$$K^{alg}\_i({\mathbb C}) = \lim\_n \ K^{alg}\_i(M\_n{\mathbb C}) \to K^{alg}\_i(\lim\_n \ M\_n{\mathbb C}) = K^{alg}\_i(\mathcal K)$$
is only an isomorphism if $i=0$ (in fact, it is the zero-map if $i \neq 0$). The comparison map is much better behaved for commutative algebras if $i \leq 0$, but still not known to be continuous.
| 3 | https://mathoverflow.net/users/8176 | 41700 | 26,607 |
https://mathoverflow.net/questions/40800 | 25 | Barry Mazur and I have come across the question below, motivated by (but independent
of) issues regarding the Leopoldt conjecture.
Suppose that $\mathbf{C}$ is the complex numbers.
Let $H$ be a finite set, and
let $S$ be a subset of $H$. The vector space $X = \mathbf{C}^{H}$
has a canonical basis consisting of the generators $[h]$ for $h \in H$. Let
$X\_S$ denote the subspace generated by $[h]$ for $h \in S$.
Suppose we consider a subspace $U$ of $X$, and ask for the
dimension of $U \cap X\_S$. The answer will depend on $U$, but
we expect the codimension of the intersection will *generically*
be the sum of the codimensions of each space. In other words, let
$G(X,U)$ denote the set of vector spaces $V$
inside $X$ which
are abstractly isomorphic to $U$ - that is, $\dim(V) = \dim(U)$.
Then
$$\min \{ \dim(V \cap X\_S), \ | \ V \in G(X,U)\} = \min\{0,\dim(U) + |S| - |H|\}\}.$$
Indeed, $G(X,U)$ is a Grassmannian, and equality holds on an open set in $G(X,U)$.
Suppose we now assume that $H$ is a *group*. We continue to assume that $S$ is
a subset of $H$ (not necessarily a subgroup),
and that $X\_S$ is the subspace generated by $[h]$ for $h \in S$.
The vector space $X$ now has extra structure --- it has a left and right action of $H$,
indeed, $X \simeq \mathbf{C}[H]$ is just the regular representation of $H$.
Suppose that $U$ is a subspace of $X$, and let us additionally assume that
$H.U = U$, that is, $U$ is a representation of $H$, and the inclusion $U
\subseteq X$ is an inclusion of left $X = \mathbf{C}[H]$-modules. We now suppose
that $G\_H(X,U)$ is the set of left $X$-modules $V$ inside $X$ which are abstractly
isomorphic to $U$ as representations of $H$ --- this is contained in but generally
much smaller than the space of vector subspaces isomorphic to $U$. The question is: can one compute
$$\delta(H,S,U) := \min \{ \dim(V \cap X\_S), \ | \ V \in G\_H(X,U)\}.$$
That is: what is the expected dimension of this intersection *given* the structure
of $U$ as an $H$-module?
$G\_H(X,U)$ can be identified with a product of Grassmannians, namely
$G(v\_i,u\_i)$ where $V\_i$ are the irreducible representations of $H$,
$v\_i = \dim(V\_i)$, and $u\_i = \dim \mathrm{Hom}\_H(V\_i,U)$. A lower bound for
$\delta(H,S,U)$ is given by $\min\{0,\dim(U) + |S| - |H|\}$, but this is
not optimal in general.
Remark: If $u\_i \in \{0,v\_i\}$ for all $i$, then $G\_H(X,U)$ consists of a single
point.
Example: Let $H = \langle \sigma \rangle$ be cyclic of order four. Let $U$
consist of the subspace generated by $[1] + [\sigma^2]$ and $[\sigma] + [\sigma^3]$.
Then $G\_H(X,U) = \{U\}$ is a point. If $S = \{[1],[\sigma^2]\}$ then
$\dim X\_S \cap U = 1$, even though $\dim(U) + |S| -|H| = 0$.
The most general question is: Can one compute $\delta(H,S,U)$ in a nice way?
A more specific question is: Can one compute $\delta(H,S,U)$ when $H$ has an
element $c$ of order two and $U = X^{c = 1}$, that is, the elements of $X$
which $c$ acts by $1$ on the right. $u\_i = \dim \mathrm{Hom}\_H(U,V\_i)$ is equal to
$\dim(V\_i,c = 1)$ in this case.
An even more specific question is: Can one compute $\delta(H,S,U)$ when
$H = S\_4$ (with representations of dimension $1$, $1$, $2$, $3$ and $3$, and
$U = X^{(12) = 1}$ has corresponding multiplicities $1$, $0$, $1$, $1$, and $2$?
What about $H = D\_{8}$ or $D\_{10}$, and $c$ is a reflection?
Example: Suppose that $U = X^{c = 1}$, and suppose that $c$ is central in $H$.
This is exactly the condition on $c$ to ensure that $G\_H(X,U)$ is a point. Then
$$\delta(H,S,U) = \frac{1}{2} | \{ s \in S \ | \ cs \in S\}|.$$
This generalizes the previous Example, where $c = \sigma^2$.
---
Let me make the following remark. As I mentioned in the question, One can obtain the complete answer when $c$ is central. In this case, one obtains generic intersections that are "larger" than what one expects from the linear algebra, at least when $|G| > 2$. It follows that a similar thing happens whenever $G$ admits a quotient $G/H$ of order $> 2$ where $c$ is central. This explains why one would expect degenerate answers in dihedral groups $D\_{2n}$ of order divisible by $4$. Having done $D\_3 = S\_3$ by hand, the next "interesting" case is $D\_5$. If I understand Greg's answer correctly, the generic intersection is always as small as possible for $D\_5$ and $D\_7$ and any $S$. Thus the most optimistic conjecture is that this is always the case providing that $|G^{ab}| \le 2$, for example, if $G = S\_4$.
| https://mathoverflow.net/users/nan | How does one compute invariants of certain Grassmannians inside the regular representation? | This is not a complete solution, but it is a way to compute things in sort-of a nice way. The module $U$ has a complementary module $U^\perp$, and the kernel of a generic module map $F:X \to U^\perp$ is isomorphic to $U$. Then you can look at the restriction $F:X\_S \to U^\perp$ and try to compute its generic rank. This puts things as close as possible to linear algebra, because $F$ comes from a linear family of linear maps. The first question is whether there is any anomalous intersection, which amounts to asking whether $F$ has maximal rank, which amounts to computing the full minors of $F$. Each maximal minor simply corresponds to restricting to $X\_T$ for a subset $T \subseteq S$ such that $|T| = \dim U^\perp$. This is simply the statement that
$$\dim (X\_S \cap V) - \dim (X\_T \cap V) \le \dim X\_S - \dim X\_T,$$
which is true solely because $X\_T$ is a subspace of $X\_S$.
You ask about certain special cases in which $U$ comes from a permutation representation $H/c$, and $U^\perp$ is thus induced from the sign representation $c = -1$. In such a case, the matrix of $F$ has a simple form in which each entry is a single variable, and the variables are negated or repeated across entries. For example, if $H = S\_3$ and $c$ is a transposition, then I get
$$F = \begin{pmatrix}
x & -y & -x & z & y & -z \\
y & -x & -z & x & z & -y \\
z & -z & -y & y & x & -x
\end{pmatrix}$$
for all of $X$. We are interested in determinants and ranks of minors of this matrix, and evidently we can remove all of the minus signs without changing any answers. This is explained by the fact that $c$ is complemented by the alternating group and we can tensor $X$ by the sign representation to switch $U$ with $U^\perp$. For the same reason, we can switch $U$ and $U^\perp$ for the dihedral cases as well, because $c$ is complemented by a subgroup of index 2. Then $F$ becomes a generic module map between two permutation representations of $H$; each matrix entry of such an $F$ is a single variable.
It is easy to check that for the example of $S\_3$ and $c$ a transposition, every full minor has the property that one of three variables ($x$, $y$, or $z$) is a transversal. This is a sufficient condition for that minor to be generically non-zero. So, there are no anomalous generic intersections for this $H$ and this $U$.
If $c$ is central, then the matrix of $F$ has repeated columns. Obviously, if a minor of $F$ has repeated columns, then it vanishes. But this is not the only way to get an anomalous generic intersection. For instance, if $H = D\_4$ and $c$ is a reflection, then (if you switch from $U = X^{c=1}$ to $U = X^{c=-1}$ as I argued you can), you get the matrix
$$F = \begin{pmatrix} x & y & z & w & x & y & z & w \\
w & x & y & z & y & z & w & x \\
z & w & x & y & z & w & x & y \\
y & z & w & x & w & x & y & z \end{pmatrix}$$
The odd-numbered columns are a vanishing minor. The reason is that the vector
$$(1, 0, 1, 0,-1, 0, -1, 0)$$
is always in $V$.
So, although I certainly don't have a complete solution, I can find examples of $S$ where there isn't an anomalous intersection (because one of the variables is a transversal), and examples where there is an anomalous intersection (because every admissible $V$ contains the same vector in $X\_S$). And, this formulation of the problem makes it easier to check a given $S$ by computer.
---
In other words, the question is essentially a matroid problem. The set $F(H)$ is a spanning set in the module $U^\perp$, and you are interested in the matroid structure of $F(H)$ for a generic $H$, because a circuit or linearly dependent subset $F(S)$ gives you an $S$ such that $V \cap X\_S$ is larger than expected. I wrote a short program in Sage to compute this matroid for $D\_n$, where the element $c$ is a reflection, using a convenient matroid package written by David Joyner.
```
n = 6 # For the dihedral group D_n
gen = primes(100,200) # Generic parameters
# http://boxen.math.washington.edu/home/wdj/sagefiles/matroid_class.sage
load 'matroid_class.sage'
R.<x,y> = PolynomialRing(ZZ,2)
names = [x^k for k in range(n)] + [x^k*y for k in xrange(n)]
a = list(gen)[:n]
F = matrix(QQ,[[a[(i-j)%n] for i in range(n)] + [a[(i+j)%n] for i in range(n)]
for j in range(n)])
for S in vector_matroid(F).circuits():
if len(S) <= n: print [names[k] for k in S]
```
For $D\_4$ and $D\_6$, the program found various matroid circuits of size 4 and 6, respectively. For $D\_5$ and $D\_7$, it did not find any non-trivial circuits. I do not understand their full structure, but there they are. For $D\_8$, it eventually reporteed that there are 362 non-trivial circuits of size 8 and none that are smaller. The first one listed is $S = \{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$, if the group is
$$D\_8 = \langle x,y | x^8 = y^2 = xyxy = 1 \rangle$$ and $y=c$.
Although this package is convenient, it is not efficient. With a more efficient code, it should be possible to find all of the non-trivial circuits in $D\_{10}$ and maybe $S\_4$.
| 7 | https://mathoverflow.net/users/1450 | 41707 | 26,611 |
https://mathoverflow.net/questions/41658 | 3 | The $k$th Chebyshev polynomial is denoted by $T\_k$ where
$T\_k(x) = \cos(k\cos^{-1}(x))$
I was wondering where this notation came from. It has been suggested that it comes from Tschebyscheff (the Russian name for Chebyshev) but does anyone know the first use of this notation or verify this is the reason?
| https://mathoverflow.net/users/2011 | Where does the Chebyshev polynomial notation come from? | Great Soviet mathematician [N.I. Akhiezer](http://en.wikipedia.org/wiki/Naum_Akhiezer) mentions in his survey article *"Чебышевское направление в теории функций"* (*"Function Theory According to Chebyshev"*) that the notation
$$T\_n(x)=\frac{1}{2^{n-1}} \cos{(n\arccos x)}$$
was first introduced by S. Bernstein.
I think that the first published paper on the Chebyshev polynomials by Bernstein was *"О наилучшем приближении непрерывных функций посредством ~~полиномов~~ многочленов данной степени"* which appeared in "Сообщения Харьковского математического общества", series 2, vol. 13 (**1912**), pp. 49-194. The paper is in Russian as you may guess.
In his paper, Bernstein refers to the Chebyshev polynomials as *trigonometric polynomials* which probably might explain the letter *T* in the notation.
---
English translation of Akhiezer's survey article is contained in [*Mathematics of the 19th Century*](http://rads.stackoverflow.com/amzn/click/0817658459) edited by A.N. Kolmogorov.
**Edit added.** I don't know if there is an English translation of the original paper by Bernstein. [This source](http://www.archive.org/details/sovietmenofscien00turk) refers to the paper as *"The optimum approximation to continuous functions by polynomials of a given power"*, Reports of the Kharkov Mathematical Society, Second Series, 1912, 13, #2-3. The original paper can be also found in volume 1 of S.N. Bernstein Collected Works (*С.Н. Бернштейн, Собрание сочинений (Том 1. Конструктивная теория функций [1905-1930])*, Москва, 1952).
| 12 | https://mathoverflow.net/users/5371 | 41717 | 26,615 |
https://mathoverflow.net/questions/41722 | 7 | Is every balanced pre-abelian category abelian? That is, given an additive category $\mathcal{A}$ in which cokernels and kernels exists, such that every morphism, which is a mono- and an epimorphism, is an isomorphism; does it follow that $\mathcal{A}$ is abelian? Note that it would suffice to prove that the canonical morphism $coim(f) \to im(f)$, where $f$ is an arbitrary morphism, is a mono- and an epimorphism.
Note that the usual examples for non-abelian categories somehow suggest this (filtered modules, topological abelian groups). See also this [related question](https://mathoverflow.net/questions/41353/abelian-categories-vs-additive-categories).
After a google search I have found the following theorem (in "Basic homological algebra" by M. Scott Osborne, Cor. 7.18): If $\mathcal{A}$ is a balanced pre-abelian category with a separating class of projectives and a coseparating class of injectives, then $\mathcal{A}$ is abelian. Ok then this does not seem to be true in general. Does anybody know an example?
| https://mathoverflow.net/users/2841 | Is every balanced pre-abelian category abelian? | Let $B$ be the abelian category of 3-term sequences of vector spaces and linear maps $V^{(1)}\to V^{(2)}\to V^{(3)}$ (the composition can be nonzero). There are 6 indecomposable objects in this category; denote them by $E\_1$, $E\_2$, $E\_3$, $E\_{12}$, $E\_{23}$, and $E\_{123}$. Here $dim E^{(i)}\_J=1$ for $i\in J$ and $0$ otherwise. Let $A\subset B$ be the full additive subcategory whose objects are the direct sums of all the indecomposables except $E\_{12}$ and $f$ be the morphism $E\_3\to E\_{123}$.
Then one has $Coker\_B f=E\_{12}$, $Coker\_A f=E\_1$ and $Im\_A f=E\_{23}$, while $Coim\_A f=E\_3$ and therefore $Coker\_A(Coim\_A f\to Im\_A f)=E\_2\ne0$. It is easy to see that any morphism in $A$ has a kernel and cokernel. Besides, any morphism in $A$ with zero kernel and cokernel has also zero kernel and cokernel in $B$, hence is an isomorphism.
Further discussion can be found in my paper <http://arxiv.org/abs/1006.4343> , Example A.5(7) (pages 59-60 in version 2).
| 12 | https://mathoverflow.net/users/2106 | 41726 | 26,620 |
https://mathoverflow.net/questions/41736 | -4 | Is there a way to calculate the number of non-repeating digits that precede the periodic repeating portion of a decimal expansion? For example:
1/6 = 0.1666.... (there is 1 non repeating digit) \*\*(Correction)
1/12 = 0.08333... (there are 2 non repeating digits)
7/12 = 0.58333....(there are 2 non repeating digits)
1/96 = 0.01041666..(there are 5 non repeating digits)
Do any forumulas exist for predicting the maximum length n, of the number of non repeating digits preceding the repeating portion?
I know that if the denominator of a fraction is n, the maximum length of the repeating periodic portion is n-1. Must also the length of the preceding portion before the cycle be n-1?
Thank you!
| https://mathoverflow.net/users/9934 | How do you calculate/prove the length of n, the number of non-repeating digits preceeding a periodic sequence of a fractional repeating decimal | When one writes an irreducible fraction $m/n$ as a periodic digit number all one does is to write
$m/n=\frac{a}{999...9000.00}$
So the number of digits before the period is the maximum of the power of $2$ and $5$ in $n$,
i.e. wirting $n=2^\alpha 5^\beta k$ with $k$ relatively prime to $10$, the number of digits before the
period is $\max\{\alpha, \beta \}$.
I think that this will follow for free from the following lemma, whose proof is trivial:
Lemma: if gcd$(k,10) =1$ then $k$ has a multiple of the form $999...9$.
| 6 | https://mathoverflow.net/users/9498 | 41743 | 26,630 |
https://mathoverflow.net/questions/41744 | 3 | A group $G$ is nilpotent if and only if there is a $c\gt 0$ such that the $(c+1)$st term of the lower central series is trivial. A group $G$ is solvable if and only if there is a $c\gt 0$ such that the $c$th term of the derived series is trivial.
Is there some similar criterion for supersolvability, or at least one which is purely commutator-theoretic?
| https://mathoverflow.net/users/3959 | Is there a commutator-theoretic criterion for supersolvability of a group? | Nilpotent groups of a given class and solvable groups of a given class form varieties, each of these varieties is defined by commutator identities. Supersolvable groups of any class do not form a variety of groups. Moreover it is not a union of varieties because there exists a supersolvable group which generates a variety where not all groups are supersolvable. So there is no definition in terms of identities - commutator or not.
| 11 | https://mathoverflow.net/users/nan | 41746 | 26,631 |
https://mathoverflow.net/questions/41745 | 5 | Let $U$ be the set of all non-null $n \times 1$ vectors $\mathbf{\mathrm{u}}$, where $u\_i \in \lbrace-1, 0, 1\rbrace$. Let $\mathbf{\mathrm{x}}$ be an $n \times 1$ vector in $\mathbf{R}^n$. Let $\mathbf{\mathrm{u\_x}}$ be the element of $U$ that is closest in angle to $\mathbf{\mathrm{x}}$. Then for any $\mathbf{\mathrm{x}}$, the maximum possible angle between $\mathbf{\mathrm{u\_x}}$ and $\mathbf{\mathrm{x}}$ is ...
For $n = 2$, I get $\pi / 8$, obviously, but what is the general expression for larger $n$? I tried to think of this as a nearest-lattice-point problem on the (hyper-)sphere with a latitude-longitude lattice, but didn't get very far.
Thanks.
| https://mathoverflow.net/users/9939 | Maximum distance to nearest-lattice-point on (hyper-)sphere with unit lat-lon lattice. | Given your initial answer for $\mathbf R^2,$ to move to $\mathbf R^3$ consider the vectors $(x,y,z)$ such that $x,y,z \geq 0$ that are equiangular between the plane vectors $(1,0,0)$ and $(1,1,0).$ These make up the plane
$$ y = (\sqrt 2 - 1) x $$ with arbitrary $z \geq 0.$ In order to get the same angle with
$(1,0,1)$ we get the ray
$$ (\sqrt 2 - 1) x = y = z. $$ In order to get the same angle with
$(1,1,1)$ we get the ray
$$ y = (\sqrt 2 - 1) x, \; \; z = (\sqrt 3 - \sqrt 2) x. $$
Note that $ (\sqrt 2 - 1) = 0.4142... $ while $ (\sqrt 3 - \sqrt 2) = 0.317837...$ So in $\mathbf R^3$ the latter comes first while increasing $z,$, and best vector is $(1, \; \;\sqrt 2 - 1, \; \; \sqrt 3 - \sqrt 2) $ where you can work out the angle.
The same process gets you from $\mathbf R^3$ to $\mathbf R^4,$ take this answer for $\mathbf R^3$ and increase the fourth coordinate until you have an equal angle with $(1,1,1,1).$
And so on.
EDIT: I get it. In $\mathbf R^n$ the optimal vector is
$$ V = (1, \; \;\sqrt 2 - 1, \; \; \sqrt 3 - \sqrt 2, \ldots, \sqrt n - \sqrt {n-1})$$ with equal angles to
$A\_1 = (1,0,\ldots,0),$ $A\_2 = (1,1,0,\ldots,0),$ $A\_3 = (1,1,1,0,\ldots,0), \ldots,$ $A\_n = (1,1,1,\ldots,1).$
EDIT 2: Note that the entries of $V$ are strictly decreasing. As a result, if instead we consider $B\_k$ with $k$ entries set to $1$ and the other $n-k$ set to $0,$ then
$$ | A\_k | \; = \; | B\_k | $$
but
$$ V \cdot A\_k > V \cdot B\_k .$$ Therefore the angle between $V$ and $B\_k$ is larger than the angle between $V$ and $A\_k,$ and the angle we actually constructed is the best we can get away with.
| 3 | https://mathoverflow.net/users/3324 | 41748 | 26,633 |
https://mathoverflow.net/questions/41609 | 22 | Any number less than 1 can be expressed in base g as $\sum \_{k=1}^\infty {\frac {D\_k}{g^k}}$, where $D\_k$ is the value of the $k^{th}$ digit. If we were interested in only the non-zero digits of this number, we could equivilantly express it as $\sum \_{k=1}^\infty {\frac {C\_k}{g^{Z(k)}}}$, where $Z(k)$ is the position of the $k^{th}$ non-zero digit base $g$ and $C\_k$ is the value of that digit (i.e. $C\_k = D\_{Z(k)}$).
Now, consider all the numbers of this form $(\sum \_{k=1}^\infty {\frac {C\_k}{g^{Z(k)}}})$ where the function $Z(k)$ eventually dominates any polynomial. Is there a proof that any number of this form is transcendental?
So far, I have found a paper demostrating this result for the case $g=2$; it can be found here: <http://www.escholarship.org/uc/item/44t5s388?display=all>
| https://mathoverflow.net/users/9712 | Have all numbers with "sufficiently many zeros" been proven transcendental? | I don't know of a paper proving the result, but I can prove it for you now. In fact, the methods in the paper you link generalize to an arbitrary base $g\gt2$. The authors of the paper don't seem to think that it generalizes quite so easily, as in the Open Problems section they state that "For bases $b\gt2$ there is the problem of having more than two possible digits. What kinds of bounds might be placed on counts of 1's and 2's for ternary expansions of algebraic numbers?". Hopefully I have not made any major mistakes...
[Edit: A paper by Bugeaud, *On the b-ary expansion of an algebraic number*, available from his [homepage](http://www-irma.u-strasbg.fr/~bugeaud/publi.html) gives lower bounds on the number of nonzero digits in an irrational algebraic number. There, he references the paper linked in the question, saying "Apparently, their approach does not extend to a base $b$ with $b\ge3$". However, he has just [responded to this question](https://mathoverflow.net/questions/41609/have-all-numbers-with-sufficiently-many-zeros-been-proven-transcendental/41794#41794), agreeing that the method does indeed generalize. So I'm more confident about my proof now.]
Use $\#(x,N)$ to denote the number of nonzero base-$g$ digits in the expansion of $x$, up to and including the $N$'th digit after the 'decimal' point, then what you are asking for is implied by the following.
>
> If $x$ is irrational and satisfies a rational polynomial of degree $D$ then $\#(x,N)\ge cN^{1/D}$ for a positive constant $c$ and all $N$.
>
>
>
First, I'll introduce some notation similar to that used in the linked paper. Use $r\_1(n)$ to denote the $n$'th base-$g$ digit of $x$, so that $0\le r\_1(n)\le g-1$ and
$$
x=\sum\_nr\_1(n)g^{-n}.
$$
It's enough to consider $1\le x\lt2$, so I'll do that throughout. Then $r\_1(n)=0$ for $n\lt0$ and $r\_1(0)=1$. Also use $r\_d(n)$ to denote
$$
r\_d(n)=\sum\_{p\_1+p\_2+\cdots+p\_d=n}r\_1(p\_1)r\_1(p\_2)\cdots r\_1(p\_d)=\sum\_{j+k=n}r\_1(j)r\_{d-1}(k)
$$
This satisfies the inequalities $r\_d(n)\ge r\_1(0)r\_{d-1}(n)=r\_{d-1}(n)$ and
$$
\sum\_{n\le N}r\_d(n)\le(g-1)^d\#(x,N)^d\le(g-1)^d(N+1)^d.\qquad\qquad{\rm(1)}
$$
Also, raising $x$ to the $d$'th power gives
$$
x^d=\sum\_nr\_d(n)g^{-n},
$$
which differs from the base $g$ expansion of $x^d$ only because $r\_d(n)$ can exceed $g$. We also introduce notation for the expansion of $x^d$ with the digits shifted to the left $R$ places and truncated to leave the fractional part,
$$
T\_d(R)=\sum\_{n\ge1}r\_d(R+n)g^{-n},
$$
so that $g^Rx^d-T\_d(R)$ is an integer. This can also be bounded, using (1),
$$
\begin{array}{rl}
\displaystyle T\_d(R)&\displaystyle\le\sum\_{n\ge1}(g-1)^d(R+n+1)^dg^{-n}\\ &\displaystyle\le\sum\_{n\ge1}(g-1)^d(R+1)^d(n+1)^dg^{-n}\\
&\displaystyle\le C\_d(R+1)^d
\end{array}
$$
where $C\_d=\sum\_{n\ge1}(g-1)^d(n+1)^dg^{-d}$ is a constant independent of $R$.
Now suppose that $x$ satisfies an integer polynomial of degree $D\gt1$,
$$
A\_Dx^D+A\_{D-1}x^{D-1}+\cdots+A\_1x+A\_0=0
$$
with $A\_D\gt0$. It follows that
$$
T(R)\equiv\sum\_{d=1}^D A\_dT\_d(R)
$$
is an integer for each $R$.
The following is similar to Theorem 3.1 in the linked paper.
>
> **Lemma 1**: For all sufficiently large $N$, there exists $n\in(N/(D+1),N)$ with $r\_1(n)\gt0$.
>
>
>
*Proof:* This is a consequence of [Liouville's theorem](http://en.wikipedia.org/w/index.php?title=Liouville_number&oldid=385130821#Liouville_numbers_and_transcendence) for rational approximation. If the statement was false then setting $m=\lfloor N/(D+1)\rfloor$, $p=\sum\_{n=0}^mr\_1(n)g^{-n}$, $q=g^{m}$ gives infinitely many approximations $\vert x-p/q\vert=q^{-D}o(1)$ as $N$ increases, contradicting Liouville's theorem.
In Lemma 1, Roth's theorem could have been used to reduce the $D+1$ term to $2+\epsilon$. In fact, Ridout's theorem as discussed in the comments can be used to reduce it even further to $1+\epsilon$. This isn't needed here, so I just used the more elementary Liouville's theorem.
Lemma 6.1 from the linked paper generalizes to base $b$, and puts upper bounds on the number of times at which $T(n)$ can be nonzero.
>
> **Lemma 2**: For large enough $N$, setting $K=\lceil 2D\log\_g N\rceil$ gives
> $$
> \sum\_{1\le R\le N-K}T\_d(R) < (g-1)^{d-1}\#(x,N)^d+1
> $$
> for $1\le d\le D$ and so,
> $$
> \sum\_{1\le R\le N-K}\vert T(R)\vert\le\sum\_{d=1}^D\vert A\_d\vert ((g-1)^{d-1}\#(x,N)^D+1)
> $$
>
>
>
*Proof:* Using similar inequalities to the proof used in the linked paper,
$$
\begin{array}{rl}
\displaystyle\sum\_{1\le R\le N-K}T\_d(R) &\displaystyle=\sum\_{m\ge1}g^{-m}\sum\_{R\le N-K}r\_d(R+m)\\
&\displaystyle\le\sum\_{m=1}^Kg^{-m}\sum\_{R\le N}r\_d(R)+g^{-K}\sum\_{m > K}g^{K-m}\sum\_{R\le N-K}r\_d(R+m)\\
&\displaystyle \le \frac{1}{g-1}\sum\_{R\le N}r\_d(R)+g^{-K}\sum\_{K\le R\le N}T\_d(R)\\
&\displaystyle\le(g-1)^{d-1}\#(x,N)^d+g^{-K}NC\_d(N+1)^d.
\end{array}
$$
The final term is bounded by $C\_d(N+1)^{d+1}/N^{2D}$ which will be less than 1 for $N$ large.
Lemma 6.2 also generalizes, which gives blocks where $T(R)$ is nonzero.
>
> **Lemma 3**: Let $R\_1\lt R\_2$ be positive integers with $r\_{D-1}(R)=0$ for all $R\in(R\_0,R\_1]$ and $T(R\_1)\gt0$. Then $T(R)\gt0$ for all $R\in[R\_0,R\_1]$.
>
>
>
*Proof:*
We have the following relation for $T$,
$$
T(R-1)=\frac{1}{g}T(R)+\frac{1}{g}\sum\_{d=1}^D A\_dr\_d(R).
$$
As $r\_d(n)\ge r\_{d-1}(n)$, the hypothesis implies that $r\_d(R)=0$ for all $1\le d\le D-1$ and $R\in(R\_0,R\_1]$. Therefore,
$$
T(R-1)=\frac{1}{g}T(R)+\frac{1}{g}A\_Dr\_D(R)\ge \frac{1}{g}T(R).
$$
Assuming inductively that $T(R)\gt0$ gives $T(R-1)\gt0$.
Putting this together gives the result (Theorem 7.2 in the linked paper).
>
> **Theorem 4**: There is a constant $c$ such that, for all sufficiently large $N$
> $$
> \#(x,N)>cN^{1/D}
> $$
>
>
>
*Proof*:
Suppose not, then for any $\delta\gt0$, there are infinitely many $N$ with $\#(x,N)\lt\delta N^{1/D}$ and, using (1),
$$
\sum\_{n\le N}r\_{D-1}(n)\le \delta N^{1-1/D}\qquad\qquad{\rm(2)}
$$
In particular, the proportion of integers $R$ with $r\_{D-1}(R)\gt0$ goes to $0$. Let $0=R\_1\lt R\_2\lt\cdots\lt R\_M\le N$ be those integers in the range $[0,N]$ with $r\_{D-1}(R\_k)\gt0$ and set $R\_{M+1}=N$. Then (2) gives $M+1\le\delta N^{1-1/D}$, and $r\_d(R)=0$ for $d\le D-1$ and $R$ in any of the ranges $(R\_i,R\_{i+1})$. So, $T\_d(R-1)=g^{R-R\_{i+1}}T\_d(R\_{i+1}-1)$.
Fixing $\epsilon\gt0$ and letting $I$ denote the numbers $i$ with $R\_{i+1}-R\_i\gt\epsilon N^{1/D}$ gives
$$
\sum\_{i\in I}(R\_{i+1}-R\_i)\ge N - (M+1)\epsilon N^{1/D}\ge N(1-\epsilon \delta).
$$
So, the intervals $(R\_i,R\_{i+1})$ larger than $\epsilon N^{1/D}$ cover most of the interval $[0,N]$, as long as $\epsilon\delta$ is small enough.
If $R$ is in the range $(R\_i,R\_{i+1}-D\log\_gN)$ and $r\_D(R)\gt0$ then $T(R-1)\gt0$:
$$
T(R-1)\ge \frac{1}{g}A\_D-g^{R-R\_{i+1}}\sum\_{d=1}^{D-1}\vert A\_d\vert T\_d(R\_{i+1}-1)
\ge\frac1g-N^{-D}\sum\_{d=1}^{D-1}\vert A\_d\vert C\_d R\_{i+1}^d
$$
which is positive, so long as $N$ is chosen large enough.
Assuming that $N$ is large enough, by Lemma 1, for each $i$ in $I$, there is
$$
j\in\left(\frac{1}{D+1}(R\_{i+1}-R\_i-D\log\_gN),R\_{i+1}-R\_i-D\log\_gN\right)
$$
with $r\_1(j)\gt0$. Then, $r\_D(R\_i+j)\ge r\_{D-1}(R\_i)r\_1(j)$ is positive, so $T(R\_i+j-1)\gt0$. Lemma 3 implies that $T(R\_i+j)$ is positive for all $0\le j\lt(R\_{i+1}-R\_i-D\log\_gN)/(D+1)$.
$$
\sum\_{1\le n< N-2D\log\_g N}\vert T(n)\vert\ge\frac{1}{D+1}\sum\_{i\in I}(R\_{i+1}-R\_i-2D\log\_gN)
\ge\frac{N(1-\epsilon\delta)}{D+1}-2\delta N^{1-1/D}\log\_gN
$$
This contradicts Lemma 2, which gives, for $N$ large,
$$
\sum\_{1\le n< N-2D\log\_g N}\vert T(n)\vert =O(\delta^D N).
$$
| 22 | https://mathoverflow.net/users/1004 | 41760 | 26,638 |
https://mathoverflow.net/questions/41757 | 0 |
>
> **Possible Duplicate:**
>
> [Eigenvalues of sum of anti-commuting matrices](https://mathoverflow.net/questions/41755/eigenvalues-of-sum-of-anti-commuting-matrices)
>
>
>
I know two anti-commuting (nxn)-matrices A and B, n -even. I know also that +-a and +-b are real eigenvalues among all eigenvalues of A and B respectively. How to show that the matrix A+B has also at least two real eigenvalues of the form +-\sqrt{a^2+b^2} ? (I know also that (A+B)(A+B)^t=(A+B)^t(A+B)=(a^2+b^2)I, where I is (nxn)-identity matrix and AA^t=A^tA=a^2I, BB^t=B^tB=b^2I and ^t denotes transposition).
| https://mathoverflow.net/users/9941 | Eigenvalues of sum of two anti-commuting matrices | Suppose for simplicity's sak that $A$ and $B$ are diagonalizable over $\mathbb{R}$
and are non-singular.
Let $V\_a$ be the $a$-eigenspace of $A$. Then by anti-commutativity,
we find $BV\_a\subseteq V\_{-a}$ etc. As $A$ and $B^2$ commute then there is
an eigenvector $v\in V\_a$ with $B^2v=b^2 v$ for some $v$. If we let
$w=bv+Bv$ then $Bw=bw$ so $b$ is real (assuming $B$ has real eigenvectors).
On the space $W$ spanned by $v$ and $w=b^{-1}Bv$ the linear transformation
$A+B$ has matrix
$$\left(\begin{array}{rr}
a&b\\\\
b&-a\\\\
\end{array}\right)$$
which has an eigenvectors with eigenvalues $\pm\sqrt{a^2+b^2}$.
| 1 | https://mathoverflow.net/users/4213 | 41761 | 26,639 |
https://mathoverflow.net/questions/41711 | 1 | One really stupid, trivial question: A Quasi coherent sheaf $F$ on an affine group scheme(Spec R) is simply an R-module. What happens in case R is a Hopf algebra? Will the Q.coherent sheaf $F$ be an algebra in this case?
| https://mathoverflow.net/users/9492 | Quasi coherent sheaf | No. Consider the case of the trivial group scheme over a field $k$ (so $R=k$). In this situation, a quasi-coherent sheaf is just a $k$-vector space. As Lennart Meier said in a comment, you need additional structure to get an algebra, e.g., a multiplication map.
**Added:** If you just want a "pointwise" multiplication operation on your module $M$, you should ask for an $R$-linear map $M \otimes M \to M$. This does not use the group law $m: G \times G \to G$ (i.e., the coalgebra structure on $R$). It sounds like you might be looking for some kind of convolution product that uses the group law, for example, a map $F \boxtimes F \to m^\*F$ on $G \times G$. Again, this is not a condition, but an extra structure. If we return to the case where $G$ is trivial, we see that the only conditions on a $k$-vector space that endow it with a canonical algebra structure are those conditions that imply the vector space is zero.
| 1 | https://mathoverflow.net/users/121 | 41762 | 26,640 |
https://mathoverflow.net/questions/41689 | 11 | In Bar-Natan's "Knots at Lunch" seminar at the University of Toronto, we are currently discussing [a talk by Alekseev at Montpellier](http://katlas.math.toronto.edu/drorbn/dbnvp/Alekseev-1006-1.php) about Rouvière's expansion of the [Duflo isomorphism](http://math.univ-lyon1.fr/~calaque/LectureNotes/LectETH.pdf) to the setting of symmetric spaces.
We understand the definition of a symmetric space, and we know that people have written books about them; but we I don't understand in what sense symmetric spaces are useful and interesting mathematical objects. In particular:
>
> Are there significant (analytic? geometric? algebraic?) techniques which work for symmetric spaces, but not for more general classes of homogenous spaces?
>
What I'm trying to understand (at least vaguely) is the role of symmetric spaces in the Kashiwara-Vergne picture, and the conceptual reason one might expect Duflo's isomorphism to generalize to this specific class of mathematical objects. There must be a conceptual explanation why symmetric spaces are the natural class of objects to consider in such contexts.
A closely related question is [THIS](https://mathoverflow.net/questions/20777/hermitian-symmetric-spaces-vs-hermitian-homogeneous-spaces).
| https://mathoverflow.net/users/2051 | Why symmetric spaces? | The algebra of invariant differential operators on a symmetric space is commutative, and this is certainly not true for an arbitrary homogeneous space. While it is not true that the commutativity of the algebra $D^{G}$ of $G$ invariant differential operators on a homogeneous space $X = G/H$ implies that $X$ is symmetric, if $G$ is reductive it is true that $D^{G}$ is commutative if and only if $G/H$ is weakly symmetric in the sense defined by Selberg (see E.B. Vinberg's survey in Russ. Math. Surveys 56(1)).
| 9 | https://mathoverflow.net/users/9471 | 41769 | 26,645 |
https://mathoverflow.net/questions/41776 | 0 | Let $C$ be a complete category, and let $P$ be presheaf on $C$. Let $X\to P$ and $Z\to P$ be objects of $\mathcal{Y}\downarrow P$, where $\mathcal{Y}$ is the Yoneda embedding. Is the pullback $X\times\_P Y$ in $Psh(C)$ representable? This is obvious when $P$ is representable, but I am not sure if it's true otherwise.
If it is true, does it still hold if $C$ is only finitely complete?
| https://mathoverflow.net/users/1353 | Pulling back representables over a presheaf | No. Take $C$ to be the terminal category, so that the category of presheaves is just $Set$. There is just one representable: the terminal set $1$. Let $P$ be a 2-element set, with elements $a: 1 \to P$ and $b: 1 \to P$. Then the pullback of these two morphisms is the empty set, which is not representable.
| 3 | https://mathoverflow.net/users/2926 | 41779 | 26,651 |
https://mathoverflow.net/questions/41758 | 6 | A *pinching* over $M\_n({\mathbb C})$ is an endomorphism $T$ where the $(i,j)$-entry of $T(M)$ is given either by $0$ or by $m\_{ij}$, depending on the pair $(i,j)$. Let us say that a pinching is symmetric if the rule is the same for $(i,j)$ and $(j,i)$ whenever $j\ne i$.
R. Bhatia has shown that the pinching $M\mapsto D(M):={\rm diag}(m\_{11},\ldots,m\_{nn})$ is a contraction for every unitarily invariant norm $\|\cdot\|$. In other words,
$$\|D(M)\|\le\|M\|,\qquad\forall M\in M\_n({\mathbb C}).$$
Remark that the map $D$ sends the cone of positive definite Hermitian matrices $HPD\_n$ into itself.
It is not difficult to extend Bhatia's result to block pinching $\Delta$, in which $\Delta(M)$ is block diagonal, made from diagonal blocks of $M$. Again $\Delta$ sends $HPD\_n$ into itself. On an other hand, it is known that some non-block diagonal pinching are not contracting and do not preserve $HPD\_n$. For instance, the linear map
$$M=\left( \begin{array}{ccc}
a & b & c \\\\
d & e & f \\\\ g & h & k \end{array} \right)\mapsto B(M)=\left( \begin{array}{ccc}
a & b & 0 \\\\
d & e & f \\\\ 0 & h & k \end{array} \right).$$
The operator norm of $B$ (when $M\_n({\mathbb C})$ is endowed with a unitarily invariant norm) is larger than $1$ (Bhatia), and there exists $H\in HPD\_n$ such that $B(M)$ is even not semi-positive definite.
**Question**. Are the following three properties equivalent ?
1. The pinching $T$ is symmetric and contracting for every unitarily invariant norm of $M\_n({\mathbb C})$.
2. The pinching $T$ sends $HPD\_n$ into itself.
3. The pinching $T$ is block diagonal.
| https://mathoverflow.net/users/8799 | Pinching and positive definite matrices | R. Bhatia proved (in [Amer. Math. Monthly 107](http://www.ams.org/mathscinet-getitem?mr=1786234)) that the operator $D\_k$ taking $M$ to its $k$-th diagonal ($M\_{ij}$, $j-i=k$) contracts any unitarily invariant norm, so 1 implies neither of 2, 3. Still, it is an interesting question to characterize those "contractive pinchings".
EDIT (partly answering the modified question): Consider a symmetric pinching, $M\mapsto P(M)=(p\_{ij}M\_{ij})\_{ij}$, $p\_{ij}=p\_{ji}\in\{O,1\}$. Property (2) implies $p\_{ii}=1$ for all $i$, since a positive definite matrix has positive diagonal elements. Property (1) doesn't imply the same thing, since all $p\_{ij}$ might be zero (as by Guillaume's comment) [EDIT: a slightly less trivial example is $p\_{11}=p\_{22}=0$, $p\_{12}=p\_{21}=1$, and $p\_{ij}=\delta\_{i,j}$ if $i>2$ or $j>2$ ]. So let us assume $p\_{ii}=1$ as part of the hypothesis. Then both (1) and (2) imply that the relation between indices $\{(i,j): p\_{ij}=1\}$ is reflexive and symmetric. If it is not transitive (i.e. if it is not an equivalence relation, or equivalently (3) doesn't hold), then there are three distinct indices $i,j,k$ with $p\_{ij}=p\_{jk}=1$ but $p\_{ik}=0$. Let $I=\{i,j,k\}$. Then,
considering your $B$ example, $P$ doesn't preserve positive definiteness, since the $I \times I$ principal minor of $P(M)$ can be negative for a positive definite $M$. Hence (2) implies (3), and they are equivalent. Similiarly, restricting $P$ to matrices supported on $I\times I$, if $B$ doesn't contract some "natural" unitarily invariant norm on $3\times 3$ matrices,the same must be true of $P$. But it is easily seen that $B$ doesn't contract the *trace norm* (sum of singular values), since the all ones $3\times3$ matrix $E$ has trace norm $3$ and $B(E)$ has trace norm $1+2\sqrt{2}$ (cf Bhatia).
Hence (1),(2),(3) are equivalent, if in (1) one assumes $p\_{ii}=1$ for all $i$ (i.e. the symmetric pinching also preserves the diagonal).
| 3 | https://mathoverflow.net/users/6451 | 41782 | 26,653 |
https://mathoverflow.net/questions/41587 | 4 | The following is classical theorem of Ore and Ryser, generalising famous Hall marriage theorem.
Assume that $n$ guys and $m$ girls live in a town, some guys like some girls. Three statements are equivalent:
(i) each guy may get one wife and one mistress (they do differ, but he should like both) so that all wifes are different and all mistresses are also different.
(ii) after removing any $k$ edges ($k=1,2,\dots$) in corresponding bipartite graph, at least $n-k/2$ guys may get wifes.
(iii) for each $k$ guys, (number of girls liked by at least one of them) plus (number of girls liked by at least two of them) is not less then $2k$.
It is easy to see that (ii) and (iii) are equivalent, and (i) implies both, so the interesting part is that (ii) implies (i).
One may look at this problem as follows. Consider the set $E$ of edges of our graph, call its subset independent, if no two edges of the subset have common endpoint. This is independence system, call it $M$, and its rank function $\rho$. Then consider new independent system $M\cup M$, in which independent set is a union of two sets, independent in $M$. Then (i) means that rank of $M\cup M$ equals $2n$ (it obviously can not be more), and (ii) means that
$$
2 \rho(E\setminus A)+|A|\geq 2n
$$
for any $A\subset E$. If $M$ was matroid, it would finish the proof by matroid union rank formula, but alas $M$ is (in general) not a matroid, but it is intersection of two matroids, one of them corresponds to "independent set of edges is the set of edges with different guys-endpoints", and another one to "independent set of edges is the set of edges with different girls-endpoints". Also, projection of $M$ onto guys and onto girls are ("transversal") matroids.
Now finally the question. Is there any general weaker, then being a matroid, condition on independence system, which is enjoyed by our $M$, and which shares the matroid union rank formula?
| https://mathoverflow.net/users/4312 | union of matroid intersection | The concept of "strongly base orderable" matroids seems to fit the bill, see for example
<http://lemon.cs.elte.hu/egres/open/Base_orderable_matroid>
In particular, partition matroids are strongly base orderable and a "union of intersections" theorem was proven by Davies and McDiarmid.
| 2 | https://mathoverflow.net/users/4020 | 41786 | 26,656 |
https://mathoverflow.net/questions/41784 | 28 | Consider the equation $x^2=x\_0$ in the symmetric group $S\_n$, where $x\_0\in S\_n$ is fixed. Is it true that for each integer $n\geq 0$, the maximal number of solutions (the number of square roots of $x\_0$) is attained when $x\_0$ is the identity permutation? How far may it be generalized?
| https://mathoverflow.net/users/4312 | Roots of permutations | $\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\SL}{SL}$
The maximum of the function counting square roots is attained at $x\_0=1$ and this statement generalises quite well.
Let $s(\chi)$ denote the Frobenius-Schur indicator of the irreducible character $\chi$. For the definition, see the edit below. One has $s(\chi)=1$ if the representation of $\chi$ can be realised over $\mathbb{R}$, $s(\chi)=-1$ if $\chi$ is real-valued but the corresponding representation is not realisable over $\mathbb{R}$, and $s(\chi)=0$ if $\chi$ is not real-valued. Then, the number of square roots of an element $g$ in any group is equal to
$$\sum\_\chi s(\chi)\chi(g),$$
where the sum runs over all irreducible characters of the group. See below for a reference and a quick proof of this identity.
It follows from the usual theory of representations of $S\_n$ that in this special case all Frobenius-Schur indicators are $1$, so the number of square roots of $x\_0$ is just $\sum\_\chi \chi(x\_0)$. This proves that the maximal number of solutions is indeed attained by $x\_0 = 1$, since each character value attains its maximum there. This generalises immediately to all groups for which every representation is either realisable over $\mathbb{R}$ or has non-real character, in other words has no symplectic (or sometimes called quaternionic) representations. That includes all abelian groups, all alternating groups, all dihedral groups, $\GL\_n(\mathbb{F}\_q)$ for all $n\in \mathbb{Z}\_{\geq 1}$ and all prime powers $q$ (see [1, Ch. III, 12.6]), and many more.
[1] A. Zelevinsky, Representations of Finite Classical Groups, Lecture Notes in Mathematics, Vol. 869, Springer-Verlag, New York/Berlin, 1981.
---
*Edit*: One reference I have found for the identity expressing the number of square roots in terms of Frobenius-Schur indicators is Eugene Wigner, American Journal of Mathematics Vol. 63, No. 1 (Jan., 1941), pp. 57-63, ["On representations of certain finite groups"](http://www.jstor.org/stable/2371276). Once you get used to the notation, you will recognise it in displayed formula (11). Since the notation is really heavy going, I will supply a quick proof here:
**Claim:** If $n(g)$ is the number of square roots of an element $g$ of a finite group $G$, then we have
$$n(g) = \sum\_\chi s(\chi)\chi(g),$$
where the sum runs over all characters of $G$, and $s(\chi)$ denotes the Frobenius-Schur indicator of $\chi$, defined as $s(\chi)=\frac{1}{|G|}\sum\_{h\in G}\chi(h^2)$.
**Proof:** It is clear that $n(g)$ is a class function, so it is a linear combination of the irreducible characters of $G$. The coefficient of $\chi$ in this linear combination can be recovered as the inner products of $n$ with $\chi$. We can write $n(g) = \sum\_h \delta\_{g,h^2}$ (here $\delta$ is the usual Kronecker delta), so we obtain
$$
\begin{align\*}
\left< n,\chi \right> &= \frac{1}{|G|}\sum\_{g\in G}n(g)\chi(g) = \frac{1}{|G|}\sum\_{g\in G}\sum\_{h\in G}\delta\_{g,h^2}\chi(g)=\\\\
&=\frac{1}{|G|}\sum\_{h\in G}\sum\_{g\in G}\delta\_{g,h^2}\chi(g) = \frac{1}{|G|}\sum\_{h\in G}\chi(h^2),
\end{align\*}$$
as claimed.
---
*Edit 2*: I got curious and ran a little experiment. The proof above applies to all finite groups that have no symplectic representations. So the natural question is: what happens for those that do? Among the groups of size $\leq 150$, there are 1911 groups that have a symplectic representation, and for 1675 of them, the square root counting function does not attain its maximum at the identity! There are several curious questions that suggest themselves: is there a similar (representation-theoretic?) 2-line criterion that singles out those 300-odd groups that satisfy the conclusion but not the assumptions of the above proof? What happens for the others? Can we find a complete characterisation of the groups whose square root counting functions is maximised by the identity? Following Pete's suggestion, I have started two follow-up questions on this business: [one on square roots](https://mathoverflow.net/questions/42646/square-roots-of-elements-in-a-finite-group-and-representation-theory) and [one on $n$-th roots](https://mathoverflow.net/questions/42653/number-of-n-th-roots-of-elements-in-a-finite-group-and-higher-frobenius-schur-ind).
| 47 | https://mathoverflow.net/users/35416 | 41788 | 26,658 |
https://mathoverflow.net/questions/31848 | 5 | Despite the title, this is probably actually a question in linear algebra or algebraic geometry. Let me write the question(s) first, before I explain the background.
**Problems**
Let $h^{\mu\nu}\_{ij}$ represent a map from $\mathbb{R}^4\otimes\mathbb{R}^4$ to $\mathbb{R}^2\otimes\mathbb{R}^2$ (here $\mu\nu$ are indices in the $\mathbb{R}^4$ directions, and $ij$ are in $\mathbb{R}^2$ directions). We shall assume that $h$ is symmetric swapping the $\mu,\nu$ indices and also symmetric swapping the $i,j$ indices. Then for any $\xi\_\mu$ in $\mathbb{R}^4$, the object $H(\xi):= h^{\mu\nu}\_{ij}\xi\_\mu\xi\_\nu$ is a symmetric bilinear form on $\mathbb{R}^2$. We say that $\xi$ is characteristic if $H(\xi)$ is degenerate. In other words, $\xi$ is characteristic if $\det(H(\xi)) = 0$.
Since $H(\xi)$ is quadratic in $\xi$, the determinant is an 8th degree homogeneous polynomial in $\xi$. Furthermore, by definition if $\xi$ is characteristic, so is $-\xi$. Observe also that in general the characteristic set will have multiple sheets.
*Question 1, very specific*
Does there exist an $h$ such that the characteristic surface is given by $\xi\_1^4 + \xi\_2^4 + \xi\_3^4 - \xi\_4^4 = 0$?
*Question 2, slightly more general*
In general are there any obstructions to having a sheet of the characteristic surface described by the zero set of an irreducible (over the reals) polynomial of degree strictly higher than 2?
*Question 3, even more general*
What if we relax the condition on $h$ so that it is a map from $\mathbb{R}^m\otimes\mathbb{R}^m$ to $\mathbb{R}^d\otimes\mathbb{R}^d$ with the same symmetric properties. Define $H(\xi)$ analogously. Can a sheet of the characteristic surface have algebraic degree more than 2?
I'm particularly interested in concrete examples.
**Motivation**
This comes from the study of hyperbolic systems partial differential equations. Recall that a second degree partial differential equation
$$ h^{\mu\nu}\_{ij} \partial\_\mu\partial\_\nu u^i = 0 $$
is said to be strictly hyperbolic in the direction of $e\_\mu$ if the characteristic polynomial (a polynomial in $t$) $\det(H(x\_\mu - te\_\mu))$ is hyperbolic for any fixed $x\_\mu$ linearly independent from $e\_\mu$ and that the roots are distinct (it is enough that the second condition only holds for all by finitely many $x\_\mu$ modulo $e\_\mu$).
The classical examples for strictly hyperbolic systems (wave equation, crystal optics, etc) all have the sheets of the characteristic surfaces being linearly transformed versions of the standard quadratic double cone: in other words there exists a basis of $\mathbb{R}^m$ such that a sheet is given by $\sum\_{i = 1}^{m-1} e\_i^2 - e\_m^2 = 0$.
I am guessing that for strictly hyperbolic systems in fact all sheets must be of this form due to homogeneity (though please let me know if I am wrong).
So my question is: is it possible for a non-strictly hyperbolic system (but one still hyperbolic) where some of the sheets have higher multiplicity to not come from "the square of a quadratic sheet" but from a genuinely quartic or higher polynomial?
**Postscript**
Please do let me know if you need any clarification on my question. Thanks.
**Update**
I struck out question 1 for the following reason: in view of my motivation from hyperbolic polynomials arising from second order PDEs, the answer is negative. The argument is thus: for a hyperbolic system of PDEs, the time-like direction $\xi\_4$ should have its corresponding $h^{44}\_{ij}$ negative definite, whereas the space-like directions $\xi\_1,\xi\_2,\xi\_3$ should have their corresponding $h^{aa}\_{ij}$ positive definite. A simple computation shows that the coefficient to the $\xi\_a^4$ term in $\det H(\xi)$ must be $\det h^{aa}\_{ij}$. If the target space is two dimensional, both positive definite and negative definite matrices have positive determinants. So for any hyperbolic polynomial arising from a second order system of PDEs, the coefficients for $\xi\_a^4$ must be positive.
| https://mathoverflow.net/users/3948 | Characteristic surface for systems of PDE | There is a natural example where an irreducible component of the symbol, a hyperbolic polynomial, is of degree higher than $2$. It occurs in compressible magnetohydrodynamics (MHD). It is a coupling of the Euler system of compressible, inviscid gas dynamics, with Maxwell's equations, when the magnetic part dominates the EM field. It consists of $8$ conservation laws, governing mass density ($1$), momentum ($3$), energy ($1$) and magnetic field ($3$). When linearizing around a uniform state, we obtain a linear, first-order hyperbolic system. The determinant of the symbol splits into the square of a linear factor (pure transport), a quadratic factor (Alfvén mode) and an irreducible factor of degree $4$. The roots of the latter are the velocities in direction $\xi$ of the fast/slow backward/forward waves.
You can also get a complex structure in nonlinear models of EM fields. See [here](http://www.umpa.ens-lyon.fr/~serre/DPF/LTT2.pdf).
| 2 | https://mathoverflow.net/users/8799 | 41796 | 26,662 |
https://mathoverflow.net/questions/41795 | 12 | Every matrix $A\in SL\_2(\mathbb{Z})$ induces a self homeomorphism of $S^1\times S^1=\mathbb{R}^2/\mathbb{Z}^2$. For different matrices these homeomorphisms are not homotopic, as the induced map on $\pi\_1(S\_1\times S\_1)$ is given by $A$ (w.r.t the induced basis).
So I am wondering, whether a similar construction also works for other spheres than $S^1$. So is there any non-obvious self-homeomorphism of $S^2\times S^2$ (I know i can use degree $-1$ maps in any choice of coordinates / flip the coordinates /compose such maps).
| https://mathoverflow.net/users/3969 | Self homeomorphisms of $S^2\times S^2$ | The matrix $\text{SL}(2,\mathbb{Z})$ acts on $H^n(S^n \times S^n)$; one interpretation of your question is whether this action lifts to $\text{Diff}(S^n \times S^n)$. There is a simple reason that it doesn't when $n$ is even: The intersection form on $H^n(S^n \times S^n)$ is symmetric rather than anti-symmetric, and any diffeomorphism has to preserve this form. This more or less nails down everything, in the weaker category of homotopy self-equivalences; in this setting you can't do anything other than exchange the spheres or apply degree $-1$ maps. (But the full homotopy structure of the $\text{Diff}(S^n \times S^n)$ could be much more complicated.)
When $n$ is odd things are much more complicated. If $n=3$ or $n=7$, you can use multiplication in the unit quaternions or the unit octonions to lift the matrix
$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$
and its transpose. These matrices generate $\text{SL}(2,\mathbb{Z})$. On the other hand, for any other value of $n$, there is no diffeomorphism, nor even any homotopy equivalence, that realizes this matrix. Because, if you composed such a map with projection onto one of the factors, it would turn $S^n$ into an [H-space](http://en.wikipedia.org/wiki/H-space).
I don't know of a way to prove more than that just by citation for some other large, odd value of $n$. In other words, I know that you can't get all of $\text{SL}(2,\mathbb{Z})$, but I don't know how big of a group you can get.
| 12 | https://mathoverflow.net/users/1450 | 41807 | 26,670 |
https://mathoverflow.net/questions/41742 | 3 | Let $\pi\in S\_n$. I recently needed to understand the permutations $\rho$ such that $\rho\not\leq\pi$ in Bruhat order. Since there are $O(n!)$ of those I really wanted a description of the $O(n^2)$ minimal such.
I have a satisfying (to me) answer now, and so I am asking whether this question is addressed in the literature.
*My answer:* It is easy to prove that the minimal $\rho$ are biGrassmannian, i.e. of the form
$$1...r\ \ a+1...b\ \ r+1...a\ \ b+1...$$
for some $(r,a,b)$.
In $\pi$'s permutation matrix, make a diagram by crossing out strictly North and West of each $1$. Let the **co-essential boxes** be the NW corners of the remaining regions, except for the region containing the SE corner. (The usual diagram comes from crossing out *weakly* South and East, and Fulton's "essential set" is the SE corners of what remains.) For each such box, let $r$ be the number of $1$s weakly NW of it, and $(r+b-a,a)$ its position, i.e. use those to define $(r,a,b)$. Then the biGrassmannian above is a minimal $\rho$, and they all arise this way, corresponding to the co-essential boxes.
| https://mathoverflow.net/users/391 | Reference for: the Bruhat-minimal permutations not less than a fixed permutation pi? | Vic Reiner, Alex Yong, and I spell this out in Sections 4.1 and 4.2 of our paper on the cohomology rings of Schubert varieties: <http://arxiv.org/abs/0809.2981>
This is not really original to us: for type A we refer back to Lascoux and Schutzenberger's paper Trellis et bases des groupes de Coxeter, Elect J. Combin. 3, no. 2 R27, though perhaps they don't state things exactly in this form.
Note for Allen: The rest of our paper might not be as irrelevant to you as it might seem at first glance. Jim Carrell started a line of work back in the 80s relating cohomology rings of Schubert varieties to their local equations at the identity. Interestingly, their strongest results are only for type A.
| 3 | https://mathoverflow.net/users/3077 | 41814 | 26,673 |
https://mathoverflow.net/questions/41802 | 4 | Let $\sigma\_k(n)$ denote the sum of the k-th powers of the divisors of n. For any real value of k, we can find a sequence of numbers $s\_k$ that has increasing values of n at which $\sigma\_k(n)$ attains a new maximum. For k=0, that sequence is called the highly composite numbers, which is [A002182](http://oeis.org/classic/A002182) in the OEIS database. From numerical experimentation, it appears that if i > j >= 0, then $s\_j$ is a subsequence of $s\_i$. Is this a known result? If not, any ideas on how to prove it?
| https://mathoverflow.net/users/2360 | A hierarchy of k-highly composite numbers | It appears you got an answer while I was typing my comment. But let me expand the picture a bit. Ramanujan initiated a technique for finding certain subsequences of yours that possess a parametrization, where the highly composite numbers do not. First, see
two articles,
<http://en.wikipedia.org/wiki/Superior_highly_composite_number>
<http://en.wikipedia.org/wiki/Colossally_abundant_number>
The recipe is very simple, with one detail that is not usually pointed out: to define the sequence of "superior" numbers, note that for any $\varepsilon > 0,$ we can prove
$$ \lim\_{n \rightarrow \infty} \frac{\sigma\_k(n)}{n^{k + \varepsilon}} = 0. $$
As a result, for any fixed $\varepsilon > 0,$ the ratio attains a maximum at a finite number of integers (usually just one). If the maximum is attained at more than one integer, take the largest of these, and call that $S\_\varepsilon.$ As $\varepsilon$ decreases $S\_\varepsilon$ stays the same until $\varepsilon$ passes below the next of a sequence of threshold values, and then at the threshold is multiplied by a certain prime. So we get an increasing sequence $S\_\varepsilon,$ each a multiple of all before it, and each setting a new record for $\sigma\_k.$
Now, about the recipe, the construction, if you work through it carefully, gives a specific prime factorization for $S\_\varepsilon,$ where the exponent for any prime depends on $\varepsilon$ in a way that involves the greatest integer function $\lfloor \bullet \rfloor. $
However, as $\varepsilon$ decreases, the relative sizes of the exponents for two primes, say 2 and 3, depend upon the value of $k.$ And these sequences are a subsequence of yours. So, despite that fact that there are so many more numbers of a given size in one of your sequences, what this says to me is that we should not expect too much agreement for differing $k.$ At the same time, there are papers on the difficult Operations Research aspects of finding large highly composite numbers, say all those between two consecutive "superior" numbers. Thus the first disagreement (showing one is not a subsequence of the other) of two of your sequences may occur at a huge number, as you found.
Well, for $k=1$ this was written up by Alaoglu and Erdos. It was later discovered that Ramanujan did write this up as well, but the journal declined to publish the full length article owing to shortages of paper in 1915.
Contemporary authors working in this area include J.-L. Nicolas and G. Robin. Perhaps the most dramatic use of these numbers is that the Riemann hypothesis is equivalent to the statement that for $n > 5040,$ $$ \sigma\_1(n) < e^\gamma n \log \log n, $$ and that any possible counterexample must occur at a colossally abundant number
| 3 | https://mathoverflow.net/users/3324 | 41818 | 26,676 |
https://mathoverflow.net/questions/41820 | 1 | Let $X$ be a scheme. An open atlas for $X$ is a jointly epimorphic family of Zariski-open immersions $\{X\_i\to X\}$ where each $X\_i$ is an affine scheme.
A morphism $X\to S$ of schemes is called representable by an affine if for any map $Y\to S$ where $Y$ is affine, the pullback $X\times\_S Y$ is itself affine.
Then the question:
Given a scheme $S$, does there exist an open atlas for $S$ consisting only of morphisms representable by an affine?
| https://mathoverflow.net/users/1353 | Does every scheme admit an open atlas consisting only of morphisms representable by an affine? | A morphism $X \to Y$ is representable by an affine iff it is an affine morphism (preimages of open affines are open affine), since the latter is stable under base change. Now an open immersion $U \to X$ is affine iff $U \cap V$ is open affine in $X$ for every open affine $V$ in $X$. Thus $X$ has an atlas consisting of such maps iff every two open affines intersect in an open affine, i.e. iff the diagonal $\Delta : X \to X \times\_{\mathbb{Z}} X$ is affine (which is the case when $X$ is separated).
| 5 | https://mathoverflow.net/users/2841 | 41822 | 26,679 |
https://mathoverflow.net/questions/34213 | 6 | In all instances of convex optimization I know of, the dimension of the vector space is defined beforehand. Is there any work on convex optimization over a vector space of *varying* dimension?
For example,
$$\begin{array}{ll} \text{minimize} & k\\ \text{subject to} & x\_1 + \cdots + x\_k = 10\\ & 1 \leq x\_i \leq 2\end{array}$$
| https://mathoverflow.net/users/3609 | Convex optimization over vector space of varying dimension | This reminds me of the compressed sensing literature. Suppose that you know some upper bound for k, let that be K. Then, you can try to solve
$\min||{\bf x}||\_0$
subject to
${\bf 1}^T\_K{\bf x}=10$ and ${\bf x}\_i\in[1,2], \; \forall i\in\{1,\ldots,K\}$. The 0-norm counts the number of nonzero elements in ${\bf x}$.
This is by no means a convex problem, but there exist strong approximation schemes such as the $\ell\_1$ minimization for the more general problem
$\min||{\bf x}||\_0, \;{\bf A}{\bf x}={\bf b},\; {\bf x}\in\mathbb{R}^{K\times 1}$, where ${\bf A}$ is fat. If you googlescholar compressed sensing you might find some interesting references.
| 2 | https://mathoverflow.net/users/2763 | 41824 | 26,681 |
https://mathoverflow.net/questions/41833 | 7 | If $X$ is an algebraic stack fibered in sets (and therefore essentially a sheaf), is it an algebraic space? It seems conceivable that at least when $X$ is Deligne-Mumford, it is actually an algebraic space. However, when $X$ is an Artin stack, it is only required to have an atlas of smooth maps of affines, so it seems less likely that this would be the case.
| https://mathoverflow.net/users/1353 | Is every (Artin/DM) algebraic stack fibered in sets an algebraic space? | Yes. The criterion for an Artin stack to be Deligne-Mumford is that it should have unramified diagonal (this is somewhere in Laumon Moret Bailly, I don't have it here). If the stack is fibered in sets, the diagonal is a monomorphism, and a monomorphism is certainly unramified.
| 12 | https://mathoverflow.net/users/4790 | 41834 | 26,686 |
https://mathoverflow.net/questions/41830 | 3 | Let $\varphi : A \to B$ be an isogeny between 2 abelian varieties of dimension $g$. Are there known conditions for the $\ker\varphi$ so that this induces an isomorphism between $A$ and $B$? For example, if $\ker\varphi \cong (\mathbb{Z}/n\mathbb{Z})^{2g}$, then $A \cong B$, because $\varphi$ factors through the multiplication map $A \xrightarrow{\times n} A$ followed by an isomoprhism $A \to B$. I wonder if there are other cases that induce isomorphisms.
| https://mathoverflow.net/users/5197 | isogenies between abelian varieties that induce isomorphisms? | Kevin's comment is right on the money, but here it is in more detail: I will give a general criterion for an isogeny $\varphi: A \rightarrow B$ of abelian varieties to induce an isomorphism upon passage to the kernel.
Let me work over an unnamed algebraically closed field. Suppose that $A = B$ and $\eta \in \operatorname{End}(A)$ is a surjective endomorphism of $A$. (N.B.: If $A$ is simple -- i.e., contains no proper nontrivial subvariety -- then any nonzero endomorphism is surjective. In
particular this holds for all elliptic curves.) Then $\eta$ is also an isogeny: i.e., its kernel is a finite subgroup scheme, say $K$ and -- essentially, by the first isomorphism theorem for groups, as Kevin says -- it follows that there is an induced isomorphism
$A/K \stackrel{\sim}{\rightarrow} A$.
This condition is also necessary: if $\varphi: A \rightarrow B$ is an isogeny such that $B \cong A$, then composing with this isomorphism gives a surjective endomorphism of $A$ and the resulting map factors through an isomorphism $A/(\operatorname{ker}(\varphi)) \rightarrow B$. Thus all examples arise from a surjective endomorphism of $A$ as above, well-defined up to isomorphisms on the source and target.
| 3 | https://mathoverflow.net/users/1149 | 41841 | 26,687 |
https://mathoverflow.net/questions/41836 | 28 | Nakayama's lemma is as follows:
>
> Let $A$ be a ring, and $\frak{a}$ an ideal such that $\frak{a}$ is contained in every maximal ideal. Let $M$ be a finitely generated $A$-module. Then if $\frak{a}$$M=M$, we have that $M = 0$.
>
>
>
Most proofs of this result that I've seen in books use some non-trivial linear algebra results (like Cramer's rule), and I had come to believe that these were certainly necessary. However, in Lang's Algebraic Number Theory book, I came across a quick proof using only the definitions and induction. I felt initially like something must be wrong--I thought perhaps the proof is simpler because Lang is assuming throughout that all rings are integral domains, but he doesn't use this in the proof he gives, as far as I can see.
Here is the proof, verbatim: We do induction on the number of generators of $M$. Say M is generated by $w\_1, \cdots, w\_m$. There exists an expression $$w\_1 = a\_1w\_1 + \cdots + a\_mw\_m$$ with $a\_i \in \frak{a}$. Hence $$(1-a\_1)w\_1 = a\_2w\_2 + \cdots +a\_mw\_m$$ If $(1-a\_1)$ is not a unit in A, then it is contained in a maximal ideal $\frak{p}$. Since $a\_1 \in \frak{p}$ by hypothesis, we have a contradiction. Hence $1-a\_1$ is a unit, and dividing by it shows that $M$ can be generated by $m-1$ elements, thereby concluding the proof.
Is the fact that $A$ is assumed to be a domain being smuggled in here in some way that I missed? Or is this really an elementary proof of Nakayama's lemma, in full generality?
| https://mathoverflow.net/users/9960 | Elementary proof of Nakayama's lemma? | There are various forms of the Nakayama lemma. Here is a rather general one; note that it does *not* involve maximal ideals and is a constructive theorem (Atiyah-MacDonald, Commutative Algebra, Prop. 2.4 ff).
>
> Let $M$ be a finitely generated $A$-module, $\mathfrak{a} \subseteq A$ be an ideal and $\phi \in End\_A(M)$ such that $\phi(M) \subseteq \mathfrak{a} M$. Then there is an equation of the form
> $\phi^n + r\_1 \phi^{n-1} + ... + r\_n = 0$,
> where the $r\_i$ are in $\mathfrak{a}$.
>
>
>
The proof uses the equality $adj(X) \cdot X = \text{det}(X)$ for quadratic matrices over a ring. I call this an elementary linear algebra fact. Of course, there you only prove it for fields but using function fields implies the result for general rings. If we take $\phi=\text{id}\_M$, we get the following form:
>
> Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$. Then there exists some $r \in A$ such that $rM = 0$ and $r \equiv 1$ mod $\mathfrak{a}$.
>
>
>
In particular, we get:
>
> Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$ and $\mathfrak{a}$ lies in every maximal ideal of $A$. Then $M=0$.
>
>
>
Observe that this argument uses Zorn's lemma (namely that every non-unit is contained in a maximal ideal) and is thus nonconstructive. Which is of course not surprising since without Zorn's lemma it is consistent that there are nontrivial rings without any maximal ideals at all. This should convince you that the first form of the Nakayama lemma is the most easy and elementary one. The last form has another short proof, which is standard and given in the question above.
Here is another short well-known proof for the last form, which also works if $A$ is noncommutative (then we have to replace "maximal ideal" by "maximal left ideal"): Assume $M \neq 0$. Since $M$ is finitely generated, an application of Zorn's lemma shows that $M$ has a maximal proper submodule $N$. Then $M/N$ is simple, thus isomorphic to $A/\mathfrak{m}$ for some maximal left ideal $\mathfrak{m}$. Then $N = \mathfrak{m} M = M$, contradiction.
By the way, I don't know if the first form is true if $A$ is noncommutative. The theory of determinants is not really prosperous over noncommutative rings. Hints?
In many texts about algebraic geometry only the last form of the Nakayama lemma is needed. But the first one is stronger and is used in many results in commutative algebra.
| 34 | https://mathoverflow.net/users/2841 | 41842 | 26,688 |
https://mathoverflow.net/questions/41839 | 8 | Say that I have the set $[n] = \{1,2,...,n\}$ and a collection $\mathcal{C} = \{ S\_1, S\_2, ..., S\_k \}$ of subsets of $[n]$. Say that $\mathcal{C}$ is *valid* if it is closed under the superset operation; i.e., if $(S \in \mathcal{C} \wedge S \subseteq S' \subseteq [n]) \implies S' \in \mathcal{C}$. How many valid collections $\mathcal{C}$ are there, as a function of $n$?
Without the requirement to be closed under superset, the question is easier. There are $2^n$ subsets of $[n]$, and so there are $2^{2^n}$ ways to choose which of them belong to the collection. But not all collections are valid; for instance, if $n=2$, the valid collections are
$\mathcal{C} = \{ \emptyset, \{ 1 \} , \{ 2 \}, \{ 1,2 \} \}$,
$\mathcal{C} = \{ \{ 1 \} , \{ 2 \}, \{ 1,2 \} \}$,
$\mathcal{C} = \{ \{ 1 \}, \{ 1,2 \} \}$,
$\mathcal{C} = \{ \{ 2 \}, \{ 1,2 \} \}$,
$\mathcal{C} = \{ \{ 1,2 \} \}$, and
$\mathcal{C} = \{ \}$.
So rather than the answer being $2^{2^2} = 16$, there are only 6 valid collections.
Thank you in advance.
| https://mathoverflow.net/users/9961 | How many collections of subsets of {1,2,...,n} are closed under the superset operation? | Such a set is uniquely determined by its minimal members which form an *antichain*. The first few values and some links are [A000372](http://www.research.att.com/~njas/sequences/A000372) in the OEIS
| 17 | https://mathoverflow.net/users/8008 | 41845 | 26,689 |
https://mathoverflow.net/questions/41844 | 5 | Please consider a central, ordinary 2-sphere $S\_1$, of some radius $r\_1$, and a second ordinary sphere, $S\_2$, of radius $r\_2$, where $r\_2 \leq r\_1$.
My question concerns optimal values for the number of spheres of type $S\_2$ that can be packed in three-dimensional space so that they are non-overlapping and tangent to $S\_1$. Is there an analytical result for optimal packing a function of the ${r\_2 \over r\_1}$, or are there subsets of cases that are solved with methods beyond something like simulating annealing? Does it simplify the problem to apply the further restriction that $r\_2 << r\_1$?
I've been having trouble finding answers with a literature search, particularly for the latter situation where $r\_2 << r\_1$, and I appreciate everyone's time.
| https://mathoverflow.net/users/6588 | Optimal packing of spheres tangent to a central sphere | I believe what you are seeking is sometimes known as a
*spherical packing* or a *spherical code*. Here is
[the MathWorld article](http://mathworld.wolfram.com/SphericalCode.html) on the topic.
Here is [Neil Sloane's webpage](http://neilsloane.com/packings/index.html) on the topic, including "putative
optimal packings" up to $n=130$ spheres.
See also [integer sequence A126195](https://oeis.org/A126196) for the
"Conjectured values for maximal number of solid spheres of radius 1 that can be rolled all in touch with and on the outside surface of a sphere of radius $n$."
There are bounds, but no "analytical result" of the type for which you might be hoping.
**Tangential Addendum**.
There is an interesting variant of the sphere-packing kissing number
that seems little known, and for which the bounds are wide enough to invite further
work. It is called *Hornich's Problem*: What is the fewest number of non-overlapping
closed unit balls that can radially hide a unit ball $B$, in the sense that every ray from the
center of $B$ intersects at least one of the surrounding balls?
In $\mathbb{R}^3$, it is known that at least 30 balls are needed, and 42 suffice.
So rather different than the kissing number 12!
Described [on p.117](https://books.google.com/books?id=WehCspo0Qa0C&pg=PA117) of
*Research Problems in Discrete Geometry*,
Brass, Peter, Moser, William O. J., Pach, János,
2005.
([Springer book link](https://www.springer.com/us/book/9780387238159).)
| 5 | https://mathoverflow.net/users/6094 | 41847 | 26,690 |
https://mathoverflow.net/questions/41840 | 15 | Let $(X,d)$ be a separable metric space. Can $X$ always be covered by a sequence of balls $B(x\_i,r\_i) (i=1,2,\dots)$ s.t. radii $r\_i$ tend to 0?
| https://mathoverflow.net/users/4312 | covering a separable metric space by small balls | The answer is no for the Banach space $c\_0$. Suppose $B(x\_i,r\_i)$ is a sequence of balls with $r\_i\to 0$ and WLOG $x\_i$ is supported in $[1,N\_i]$ with
$N\_1<N\_2<...$. Consider a point $x$ in $c\_0$ whose $N\_i+1$ coordinate is $2 r\_i$.
I think the answer is no for any separable Banach space: IIRC, for any separable Banach space $X$ and any increasing sequence $E\_n$ of finite dimensional subspaces and any sequence of positive $r\_n\to 0$, there is a vector $x$ in $X$ s.t. the distance from $x$ to $E\_n$ is larger than $r\_n$ (in fact, even equal to $2r\_n$ if $r\_n$ is decreasing).
ADDED 10/12/10: It is not hard to check what I said in the second paragraph of my answer, from which it follows that the answer is no for any infinite dimensional Banach space. Is the answer no for any infinite dimensional linear metric space?
Can you characterize the metric spaces for which the answer is yes? I suspect that the reason Fedor is interested in the property is that a modification of the proof of the Vitali covering theorem yields that if $X$ is covered by such a sequence of balls, then there are DISJOINT balls $B(y\_n,t\_n)$ with $t\_n\to 0$ s.t. $B(y\_n,5t\_n)$ covers $X$.
| 14 | https://mathoverflow.net/users/2554 | 41852 | 26,693 |
https://mathoverflow.net/questions/41862 | 5 | Finite groups are solvable if they have no nontrivial perfect subgroup. But I am sure that for infinite groups, the two notions diverge. Is there standard terminology for groups with no perfect subgroups?
| https://mathoverflow.net/users/3634 | Groups with no perfect subgroups -- terminology? | In the infinite case, there is a close notion of "locally indicable group", i.e. a group where every finitely generated subgroup maps onto $\mathbb Z$ (see, for example, [this paper](https://www.jstor.org/stable/2699673)). Locally indicable groups are left (right) orderable, hence important. Note that in that notion, not all subgroups are considered but only finitely generated, and "non-perfect" is replaced by a stronger property "maps onto $\mathbb Z$". But in the finite case all subgroups are finitely generated, and "maps onto $\mathbb Z$" is an infinite analog of "maps onto a finite cyclic group" (= non-perfect). So "locally indicable" is possibly the infinite analog of the property you consider.
Update: The groups without perfect subgroups are called hypoabelian. See [this text.](https://planetmath.org/descendingseries)
| 11 | https://mathoverflow.net/users/nan | 41863 | 26,699 |
https://mathoverflow.net/questions/41867 | 2 | Let $X$ be a finite-dimensional smooth manifold, $\mathcal C^\infty(X)$ its algebra of smooth functions, $V\to X$ a finite-dimensional smooth vector bundle, and $\Gamma(V)$ the space of smooth sections of $V$. In particular, $\Gamma(V)$ is a $\mathcal C^\infty(X)$-module. I am interested in $\mathcal C^\infty(X)$-submodules $D \subseteq \Gamma(V)$.
>
> Is $D$ necessarily finitely-generated as a $\mathcal C^\infty(X)$-module?
>
>
>
If $X$ is not compact (or maybe even if it is?), then $\mathcal C^\infty(X)$ is not Noetherian. So it is not true that submodules of arbitrary finitely-generated modules are finitely generated. So I expect that the answer to my question is "no", but I'm having trouble coming up with a counterexample.
Actually, what I really want is for $D$ to receive a ($\mathcal C^\infty$-linear) surjection from $\Gamma(W)$ for some finite-dimensional vector bundle $W$. If $X$ is not compact, then I think it is still the case (using partitions of unity) that $\Gamma(W)$ is globally finitely-generated (the idea is to find a cover for which each open intersects only finitely many others in the cover, and then to double up the generators). But if it isn't, the actual question I want to ask is the one with the word "locally" sprinkled in all the necessary places.
| https://mathoverflow.net/users/78 | Is a submodule of the sheaf of sections of a smooth vector bundle necessarily finitely generated? | The module of all sections of $V$ that vanish to infinite order at a given point of the manifold will not be finitely generated (unless the bundle has rank zero or the manifold has dimension zero).
| 11 | https://mathoverflow.net/users/6666 | 41869 | 26,703 |
https://mathoverflow.net/questions/41868 | 5 | There is at least one result saying that the Mandelbrot set is undecidable, and there might be more, but I think it (or they all) use real computation rather than Turing machines. This makes some sense, as $\mathbb{C}$ is connected, so the only decidable subsets of it are $\{\}$ and $\mathbb{C}$ itself. However, I've been reading about reverse mathematics, and I was wondering if the set is computable in the representations used. The Mandelbrot set is easily seen to be the complement of an effectively open set, and I'd guess asking whether it is separably closed would run into the open problem of are the hyperbolic components dense. This still leaves located closed, so the question follows:
Let $f : \mathbb{Q}[i] \to \mathbb{R} \hspace{.05 in}$ be given by $f(q) := \operatorname{min}(\{d(q,z) : z\in (\operatorname{Mandelbrot set})\})$. Is $f$ computable?
| https://mathoverflow.net/users/nan | Is the distance function from a point to the Mandelbrot set computable? | Following [Giusto and Simpson, *Located sets and reverse mathematics*, J. Symbolic Logic Volume 65, Issue 3 (2000), 1451-1480], the Mandelbrot set $M$ is *located* if the distance function $f:\mathbb C\rightarrow\mathbb R$, $f(x)=d(x,M)$, exists in the model under consideration, which I assume you take to be the model containing only computable objects.
Such locatedness seems to be the same as Conjecture 4 of [Hertling, *Is the Mandelbrot set computable?*, Math. Logic Quarterly, 51(1):5-18, 2005]: The function $f:\mathbb C\rightarrow \mathbb R$ is computable.
(Here $f$ is *computable* if you can compute the value $f(x)$ with any desired precision (in terms of the distance $d$) when you know $x$ with sufficient precision. More precisely, the algorithm that computes $f$ might promise to output a rational interval of length at most $2^{-n}$ containing $f(x)$ once it is told an interval of length $2^{-{g(n)}}$ containing $x$, where $g$ is a computable function on $\mathbb N$ of the algorithm's choosing.)
Now, what happens if we replace $\mathbb C$ by $\mathbb Q[i]$? Could there be a way to compute $f(q)$ using a representation of $q$ as a rational, but nevertheless no way to approximate $f(x)$ given an arbitrary $x$, due to a lack of a useful modulus of continuity in $x\mapsto d(x,M)$?
No, because $|d(x,M)-d(q,M)|\le d(q,x)$.
Conclusion
==========
The question is equivalent to a known-to-be-open question.
| 5 | https://mathoverflow.net/users/4600 | 41871 | 26,704 |
https://mathoverflow.net/questions/41874 | 2 | A) Given a non-constant polynomial $q\in\mathbb{Z}[\alpha\_1,\alpha\_2,\ldots,\alpha\_n],$ if we pick random $\omega\_i\in\mathbb{F}$ (a finite field) uniformly and independently across $1\leq i\leq n,$ then, we know that $q(\omega\_1,\omega\_2,\ldots,\omega\_n)\neq 0$ with high probability (i.e. the probability goes to 1 as $|\mathbb{F}|\rightarrow\infty$).
B) Given another polynomial $r\in\mathbb{Z}[\alpha\_1,\alpha\_2,\ldots,\alpha\_n],$ I am interested in determining if there exists a field $\mathbb{F}$ and a choice of $\omega\_i\in\mathbb{F}$ which simultaneously satisfy $q(\omega\_1,\omega\_2,\ldots,\omega\_n)\neq 0$ and $r(\omega\_1,\omega\_2,\ldots,\omega\_n)= 0.$ Is there a theorem that gives necessary or sufficient conditions for this to happen? Is it true that if it happens over some field, then it happens over all sufficiently large finite fields?
Is it true that if there is a point which satisfies $r=0$ and $q\neq 0,$ then "most" of the points satisfying $r=0$ also satisfy $q\neq 0,$ in similar spirit to the result A which is the case of $r$ being the zero polynomial?
I am interested only in solutions over finite fields and not over their algebraic closures.
Thanks a lot.
| https://mathoverflow.net/users/7576 | Polynomials over Z evaluated with finite field arguments | This should work out for you, with a bit more geometry and perhaps a little trickery. You are asking for points over finite fields avoiding a given hypersurface, in affine space. It is conceptually a little easier to think of points in a projective space avoiding both a hypersurface and the hyperplane at infinity, because the typical estimates refer immediately to projective varieties. Your (A) works because the points on the hypersurface have main term which is the size of the finite field (often written q but your notation clashes) to the power of its dimension, which is n - 1. The error term is smaller, naturally, in its power, with implied constant that comes from Betti numbers that are fixed by the choice of q (your polynomial). The points on the hyperplane at infinity are easy to count.
For your (B) there will have to be some conditions at least as strong as r not dividing q. One trick that can be useful is to introduce a further variable y for which one requires yq(x) - 1 = 0: then that's a second equation to add to r(x) = 0 (rather than the inequation on q). The general theory applied to the simultaneous equations should then produce adequate numbers of solutions of the kind you want, over all large finite fields. Some care is needed to discuss the possibility that the simultaneous equations produce a singular variety over *Z*. When q and r are in "general position" this should go through easily. The old Lang-Weil paper on counting points might have enough to deal with it.
| 3 | https://mathoverflow.net/users/6153 | 41879 | 26,707 |
https://mathoverflow.net/questions/41883 | 9 | Let $G$ be an abelian group, $A$ a trivial $G$-module. We know that $\text{Ext}(G,A)$ classifies abelian extensions of $G$ by $A$, whereas $H^2(G,A)$ classifies central extensions of $G$ by $A$. So we have a canonical inclusion $\text{Ext}(G,A)\hookrightarrow H^2(G,A)$. Is there some naturally arising exact sequence/spectral sequence which realizes this injection?
Usually this kind of thing can be explained by constructing a clever short exact sequence, but here I have no idea how one might compare $R^1\text{Hom}\_\mathbb{Z}(G,\underline{\quad})$ with $R^2\text{Hom}\_G(\mathbb{Z},\underline{\quad})$.
| https://mathoverflow.net/users/5513 | Injection of Ext into H^2 | You get a description from the universal coefficient theorem which gives a (split) exact sequence
$$
0\to \mathrm{Ext}(H\_1(G),A) \to H^2(G,A) \to \mathrm{Hom}(H\_2(G),A) \to 0
$$
and the fact that $H\_1(G)=G$. We have that $H\_2(G)=\Lambda^2G$ and the map $H^2(G,A) \to \mathrm{Hom}(H\_2(G),A)$ associates to an extension its commutator map.
| 14 | https://mathoverflow.net/users/4008 | 41892 | 26,715 |
https://mathoverflow.net/questions/41902 | 4 | Suppose $\dot{x}=f(x)$ is a dynamical system, with $x$ in $R^n$, and $f:R^n \to R^n$ sufficiently smooth (for example, Lipschitz-continuous).
Assume that $x\_e$ is an *unstable* equilibrium point of the system. Even if $x\_e$ is unstable, the union of all trajectories having $x\_e$ as a limit point could be larger than just the singleton {$x\_e$}. But does this set always have zero measure?
This might be a basic result, but any pointer would be appreciated.
| https://mathoverflow.net/users/8460 | Measure of the stable set in a dynamical system | Take for example the equation
$\dot x = \lambda x$
$\dot y = y^2$
For $\lambda < 0$. The origin is an unstable equilibrium point, however, its stable manifold is the whole lower semiplane (including the $y=0$ axis).
If there is one eigenvalue of real part bigger than $0$ for the derivative in the equilibrium point, it is then true that the set of points converging to $x\_e$ has zero measure. See for example [here](http://www.informaworld.com/smpp/content~db=all~content=a904090891~frm=abslink) (proposition 4.1).
The proof is simple and based in the existence of center stable manifolds, this implies (toghether with the eigenvalue with positive real part) that the local stable set of the equilibrium has zero measure, after that, it is done since the stable set can be written as a countable union of sets diffeomorphic to this one.
| 4 | https://mathoverflow.net/users/5753 | 41903 | 26,721 |
https://mathoverflow.net/questions/41900 | 3 | Let M be a 2n-dimensional closed symplectic manifold.
Then is there a constant c such that , for any real 2-dimensional embedded J-holomorphic
disk u, the symplectic volume of u is bounded by c ?
If not, is there any result about a condition which makes the statement above to be true?
I really thank you for your any comment.
| https://mathoverflow.net/users/11705 | symplectic volume of embedded J-holomorphic disk | Unless I'm misunderstanding what you're asking, the answer is surely no...consider for instance $\mathbb{C}P^n$ with its standard symplectic and complex structures. This admits embedded $J$-holomorphic curves of arbitrarily large area (take a high-degree curve in a plane $\mathbb{C}P^2\subset \mathbb{C}P^n$), and restricting to a disc within any of these curves would give you a $J$-holomorphic disc $u$ of arbitrarily large area (which I assume is what you mean by the symplectic volume of $u$).
On more general symplectic manifolds $(M,\omega)$, the h-principle gives you immersed symplectic spheres in every homology class $A$ with $\int\_{A}\omega>0$; these spheres can be taken embedded if $\dim M\geq 6$ and embedded away from finitely many transverse double points if $\dim M=4$. In either case you could construct an almost complex structure $J$ on $M$ with respect to which an arbitrarily large-area subdisk of the surface is embedded and $J$-holomorphic. (This is admittedly a little weaker than the first example, since here we're choosing $J$ after we choose the surface--so all it shows is that for any $C$ there is a pair $(u,J)$ where $u$ is a $J$-holomorphic disc of area larger than $C$, with $J$ possibly depending on $C$.)
| 4 | https://mathoverflow.net/users/424 | 41907 | 26,723 |
https://mathoverflow.net/questions/41912 | 5 | In responding to
[Fast computation of multiplicative inverse modulo q](https://mathoverflow.net/questions/40997/fast-computation-of-multiplicative-inverse-modulo-q/41631#41631)
I mentioned an algorithm for computing the inverse of $a \mod p$ different from the extended Euclidean algorithm, hoping that someone could tell me how its speed stacks up against other algorithms. Since no one did, I'm asking directly if someone can tell me.
To compute the inverse of a modulo p, you can run the Euclidean algorithm starting with $p^2$ and $ap+1$, comparing the size of each remainder with $p$. The first remainder less than $p$ that appears will be an inverse for $a \mod p$.
This will always take either the same number of steps to reach the inverse as it takes to reach $\gcd(a,p)=1$ using the Euclidean algorithm with $a$ and $p$, or else one additional step, depending on whether the least positive residue of an inverse of $a$ is greater than $p/2$ or less than $p/2$.
Thus, it requires approximately the same number of computations as the first half of the extended Euclidean algorithm (albeit with bigger numbers initially), excepting an extra comparison with $p$ at each step.
Question: How does the speed of this algorithm compare to others?
Aside: Pedagogically, this is nice since the second half of the extended Euclidean algorithm is the one my students tend to mess up. However, assuming our ultimate goal is for students to understand why they're doing what they're doing, perhaps the extended Euclidean algorithm is preferable.
| https://mathoverflow.net/users/5373 | Computation of inverses modulo p followup | Here is a tidy way to solve $ax+by=gcd(a,b)$. Start with the matrix
$$\left(\begin{array}{ccc}
1&0&a\\
0&1&b\\
\end{array}\right)
$$
Suppose $a\ge b$. Then replace row 1 by row 1 minus $t$ times row 2, where $t=\lfloor a/b\rfloor$. Repeat this operation until the last entry in one of the two rows is zero. If the other row is $x,y,d$ then $ax+by=d$ and $d=gcd(a,b)$. This is very simple to program, and avoids the back-substitutions that students find confusing.
Most of the speedups of the Euclidean Algorithm described in Knuth's Art of Computer Programming v2 work here also.
| 3 | https://mathoverflow.net/users/5229 | 41925 | 26,730 |
https://mathoverflow.net/questions/41915 | 1 | Hello, i have NP hard problem. Let imagine I have found some polynomial algorithm that find ONLY one of many existing solutions of that problem, but at least one solution (if present in the probem). Is that algorithm considered as solution of NP=P question (if that algorithm transformed to mathematical proof)?
Thanks for answers
| https://mathoverflow.net/users/9992 | NP-Hard solution question | For concreteness, let's pick an NP-hard problem to talk about. Given a graph $G$, the 3-colouring problem asks: "can the vertices of $G$ be painted by three colours such that for any edge $uv$, $u$ and $v$ get different colours?" This is a decision problem --- its possible answers are "yes" or "no" --- but a "yes" answer can be *certified* by a proper 3-colouring.
Say you had a polynomial-time algorithm that found, for any input graph, a proper 3-colouring if one exists. Then your algorithm would solve the 3-colouring problem: it answers "yes" or "no" correctly, and even gives a nice certificate (or *witness*) of a "yes" answer. This would be enough to show that P=NP. It is not necessary to find all possible 3-colourings (indeed, there may be exponentially many of them).
Now, if you had some sort of "partial algorithm," which solves an NP-hard problem only for some specific instances, then this is not enough. For example, the 3-colouring problem can be easily solved for bipartite graphs, split graphs, and more. The reason for this is that the restriction of an NP-hard problem is not necessarily NP-hard.
Finally, just to elaborate on Jim's answer: many popular descriptions of NP-hard problems, like Travelling Salesman, don't sound like decision problems. But they are, really: they can be retranslated as a series of questions with yes or no answers (e.g. "does there exist a travelling salesman route of length at most $x$?").
| 9 | https://mathoverflow.net/users/785 | 41930 | 26,735 |
https://mathoverflow.net/questions/41932 | 10 | Does anyone know more examples of two nonisomorphic connected graphs with the same chromatic symmetric function? The only pair I know is the one in Stanley's paper on c.s.f.'s [<http://math.mit.edu/~rstan/pubs/pubfiles/100.pdf>; p.5 of the PDF file]. I would especially like to have an example in which at least one of the two graphs is triangle-free.
| https://mathoverflow.net/users/8604 | Graphs with the same chromatic symmetric function | I don't think there are any other published examples. I think your best bet is to look at the literature on "chromatically equivalent graphs" (graphs with the same chromatic polynomial) and do your own computations to find examples. (I assume that you wrote some code to compute the chromatic symmetric function when you investigated trees.)
| 4 | https://mathoverflow.net/users/3106 | 41938 | 26,738 |
https://mathoverflow.net/questions/41895 | 5 | Let $G$ be a reductive algebraic group over $\mathbb R$ and $K$ a maximal compact subgroup. Then we refer to the conjugacy class in $G$ of some $k \in K$ as an elliptic conjugacy class.
**Question:** Can one characterizes those conjugacy classes in $G$ which contain an elliptic conjugacy class in their closure?
(For $G = GL\_n(\mathbb R)$ they are characterized by the fact that all eigenvalues are of modulus one, if I a not mistaken.)
| https://mathoverflow.net/users/9927 | Conjugacy classes with elliptic limit points | For g in G, write g=gsgu as its Jordan decomposition into semisimple and unipotent parts. I claim that the closure of the conjugacy class of g contains an elliptic element if and only if gs is elliptic.
Let us first suppose that gs is not elliptic. Choose an embedding of G into GLn(ℂ). Then by our assumption, gs has an eigenvalue of norm greater than one, let λ be the absolute value of such an eigenvalue. Suppose for want of contradiction that the conjugacy class of gs contained an elliptic element a in its closure. WLOG a is in the special unitary group SUn. Let h be in the conjugacy class of gs. Then h has an eigenvalue of absolute value λ. Letting v be an eigenvector, we see that |(h-a)v| is at least (λ-1)|v|, so |h-a|≥λ-1, a contradiction.
Now suppose that gs is elliptic. We may replace G by the centraliser of gs is G, which is also reductive. So WLOG, gs is central in G. Now the Zariski closure of the group generated by gu is a one-dimensional unipotent subgroup of G. Let E be a non-zero element in its lie algebra. This is a nilpotent element. Then by the Jacobson-Morozov theorem, we can extend E to a sl2 triple E,F,H in Lie(G). Now consider conjugation by elements of the form exp(tH) with t real. This shows that gs is in the closure of the conjugacy class of g, and we're done.
| 2 | https://mathoverflow.net/users/425 | 41941 | 26,739 |
https://mathoverflow.net/questions/41940 | 7 | Let $\rho\_p : \mbox{Gal}(\overline{\mathbb{Q}}\_p / {\mathbb{Q}\_p}) \to \mbox{GL}\_n(\mathbb{Q}\_p)$ be a de Rham $p$-adic representation. Can one find a representation $\rho : \mbox{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \mbox{GL}\_n(\mathbb{Q}\_p)$ such that $\rho$ is geometric (in the sense of Fontaine-Mazur) and such that the restriction of $\rho$ to $\mbox{Gal}(\overline{\mathbb{Q}}\_p / {\mathbb{Q}\_p})$ is $\rho\_p$ ?
| https://mathoverflow.net/users/10001 | local to global Galois representation | No. There are uncountably many unramified representations from the local Galois group to
$\mathbb{Q}^{\times}\_p$, since Frobenius can be sent to anything in $\mathbb{Z}^{\times}\_p$.
However, there are only countably many global representations of this form, since, by class field theory, they all factor through the Galois group of a (finite) cyclotomic field.
The general picture is pretty much the same --- the local Galois representations form large $p$-adic analytic families, yet, assuming the Fontaine-Mazur conjecture, there are only countably many geometric representations.
| 7 | https://mathoverflow.net/users/nan | 41942 | 26,740 |
https://mathoverflow.net/questions/41922 | 18 | Can someone give an explicit example of a group with two generators $a$, $b$, such that $a^2 = b^3 = 1$ and $a b$ has infinite order, but which is not isomorphic to the free product of $\mathbb{Z}\_2$ and $\mathbb{Z}\_3$?
| https://mathoverflow.net/users/2926 | Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't. | It is straightforward to calculate that the commutator subgroup $G' = D$ of $G = \langle a,b \mid a^2, b^3 \rangle$ is a free group on the generators $x=bab^{-1}a$, $y=b^{-1}aba$, where $|G:D|=6$.
Now $(ab)^6$ is equal to the commutator $x^{-1}yxy^{-1}$, which lies in $D'$ but not in $D''$, so if we add any nontrivial element of $D''$ as an extra relator of $G$, then we will get an example with the required property.
| 18 | https://mathoverflow.net/users/35840 | 41944 | 26,741 |
https://mathoverflow.net/questions/41934 | 5 | I am trying to understand spin structures and am looking at the specific case of complex projective space (viewed as the quotient $SU(N)/U(N-1)$) and more generally the Grassmannians (viewed as the quotient $SU(N)/(U(N-k) \times U(k))$. My questions are as follows:
(1) For what values of $N$ does complex projective $N$-space have a spin structure?
(2) For these values, is the canonical spinor bundle equivarient with respect to the $U(N-1)$ action?
(3) If so, what is the associated representation of $U(N-1)$?
(4) All of the above for the Grassmannians?
| https://mathoverflow.net/users/1648 | Spin structures on the Grassmannians | I think there is a slight mistake in the formulation of the question. $\mathbb{CP}^n$ is the homogeneous space $U(n+1)/(U(n) \times U(1))=SU(n+1)/G$ with $G= SU(n+1) \cap (U(n) \times U(1))$. The right formulation of question (2) is: is the spin structure on $\mathbb{CP}^n$ (for odd $n$, there is unique spin structure on $\mathbb{CP}^n$, see Charles Siegel's answer) $U(n+1)$-equivariant?
The answer is no, for a very elementary reason: if the spin structure were $U(n+1)$-equivariant, then it certainly were $U(n)$-equivariant,
where $U(n)$ embeds into the product in the standard way. But the $U(n)$-action on $\mathbb{CP}^n$ has a fixed point and it is not too hard to see that the $U(n)$-representation on the tangent space to that fixed point is isomorphic to the standard representation of $U(n)$ on $\mathbb{C}^n$. So if the spin structure were equivariant, then the fixed-point representation has to be spin, which is of course wrong.
You can ask the same question for spheres (is the spin structure on $S^n$ $SO(n+1)$-equivariant), and the answer is again no. But the spin structure on $S^n$ is $Spin(n+1)$-equivariant; likewise the spin structure on $\mathbb{CP}^n$ will be equivariant under the double cover of $U(n)$.
What you can guess from these two examples is that the question has something to do with double covers (alias central extensions of your group by $\mathbb{Z}/2$). Here is the precise relation: $M$ a spin manifold, $s$ a spin structure (viewed as a double cover of the frame bundle of $M$), $G$ a topological group acting on M by diffeomorphisms. The spin structure defines a new group $G'$ and a surjective homomorphism $p:G' \to G$ with kernel. $G'$ consists of pairs $(f,t)$, $f \in G$ and $t$ is an isomorphism of spin structures $f^\* s \to s$. The spin structure is equivariant under $G'$, and it is $G$-equivariant iff there is $q:G \to G'$, $pq=\operatorname{id}$. If $G$ is a simply-connected topological group, this is always the case, but otherwise not in general.
This discussion implies that the spin structure on $\mathbb{CP}^n$ is indeed $SU(n+1)$-equivariant, if it exists. Grassmannians and other homogeneous spaces can be dealt with in the same way.
| 9 | https://mathoverflow.net/users/9928 | 41949 | 26,745 |
https://mathoverflow.net/questions/41827 | 3 | I'm reading a paper and here the authors say that a connected 4-manifold with zero rational top homology has a homotopy type of 3-dimensional CW-structure. I can't figure out how it can be done.
| https://mathoverflow.net/users/6569 | a CW-complex homotopic to a manifold | For this to be true, you need to assume that your $4$-manifold $M$ is not a compact nonorientable manifold. Otherwise, you would have $H\_4(M;\mathbb{Q}) = 0$ but $H\_4(M;\mathbb{Z}/2) \neq 0$, so there is no hope that your manifold is homotopy equivalent to a $3$-dimensional CW-complex.
Assuming this, your conditions imply that your $4$-manifold is not compact (otherwise the 4th homology group would be $\mathbb{Q}$). It is a general fact that smooth noncompact $n$-manifolds are homotopy equivalent to $(n−1)$-dimensional CW complexes. For details, see Mohan Ramachandran's answer to my question
[here](https://mathoverflow.net/questions/18454).
| 4 | https://mathoverflow.net/users/317 | 41964 | 26,750 |
https://mathoverflow.net/questions/41973 | 3 | In the paper "On the consistency strength of projective uniformization" Woodin proves a lemma "Assume $M$ is a model of ZFC that is $\Sigma^{1}\_{3}$-absolute. Then $M\vDash\forall x\in\mathbb{R}\,[x^{\sharp} \mathrm{\ exists}]$." He then goes on to say after the proof "It in fact now follows by a theorem of Martin-Solovay [6] that $\Sigma^{1}\_{3}$-absoluteness is equivalent to the existence of $S^{\sharp}$ for every set $S$."
[6] Martin, D. A. and Solovay, R. M., A basis theorem for $\Sigma^{1}\_{3}$ sets of reals, Ann. of Math. 89 (1969), 138-160.
When I read this article of Martin and Solovay I have trouble seeing the connection with the assertion that $S^{\sharp}$ exists for all sets $S$. I was wondering if anyone could clarify this for me.
| https://mathoverflow.net/users/7909 | Question about Woodin's paper "On the consistency strength of projective uniformization" | Rupert, the Martin-Solovay paper shows the absoluteness result follows form the existence of measurable cardinals. The measurables are used in the construction of certain trees (now called Martin-Solovay trees), where some ordinals are chosen by means of the measure to serve as witnesses (of ill-foundedness of some branches, say). There are standard arguments to tighten up this approach so indiscernibles (coming from sharps) suffice for this. (This is closely related to how $\Pi^1\_1$-determinacy can be established form sharps rather than measurables.)
I suspect Martin's draft of a book on determinacy has the details, but I do not know whether you have access to it (and I do not have my copy handy at the moment to double-check). In any case, Ralf Schindler and I have a joint paper where we explain in detail how $\Sigma^1\_3$-absoluteness is equivalent to the existence of sharps. The paper is "Projective well-orderings of the reals", Arch. Math. Logic (2006) 45:783–793. It is available at my webpage, see Theorem 3.
| 7 | https://mathoverflow.net/users/6085 | 41974 | 26,752 |
https://mathoverflow.net/questions/41939 | 21 | A box contains n balls coloured 1 to n. Each time you pick two balls from the bin - the first ball and the second ball, both uniformly at random and you paint the second ball with the colour of the first. Then, you put both balls back into the box. What is the expected number of times this needs to be done so that all balls in the box have the same colour?
Answer (Spoiler put through rot13.com): Gur fdhner bs gur dhnagvgl gung vf bar yrff guna a.
Someone asked me this puzzle some four years back. I thought about it on and off but I have not been able to solve it. I was told the answer though and I suspect there may be an elegant solution.
Thanks.
| https://mathoverflow.net/users/7576 | A balls-and-colours problem | It can probably be done by looking at the sum of squares of sizes of color clusters and then constructing an appropriate martingale. But here's a somewhat elegant solution: reverse the time!
Let's formulate the question like that. Let $F$ be the set of functions from $\{1,\ldots,n\}$ to $\{1,\ldots,n\}$ that are almost identity, i.e., $f(i)=i$ except for a single value $j$. Then if $f\_t$ is a sequence of i.i.d. uniformly from $F$, and
$$g\_t=f\_1 \circ f\_2 \circ \ldots \circ f\_t$$
then you can define $\tau= \min \{ t | g\_t \verb"is constant"\}$. The question is then to calculate $\mathbb{E}(\tau)$.
Now, one can also define the sequence
$$h\_t=f\_t \circ f\_{t-1} \circ \ldots \circ f\_1$$
That is, the difference is that while $g\_{t+1}=g\_t \circ f\_{t+1}$, here we have $h\_{t+1}=f\_{t+1} \circ h\_t$. This is the time reversal of the original process.
Obviously, $h\_t$ and $g\_t$ have the same distribution so
$$\mathbb{P}(h\_t \verb"is constant")=\mathbb{P}(g\_t \verb"is constant")$$
and so if we define $\sigma=\min \{ t | h\_t \verb"is constant"\}$ then $\sigma$ and $\tau$ have the same distribution and in particular the same expectation.
Now calculating the expectation of $\sigma$ is straightforward: if the range of $h\_t$ has $k$ distinct values, then with probability $k(k-1)/n(n-1)$ this number decreases by 1 and otherwise it stays the same. Hence $\sigma$ is the sum of geometric distributions with parameters $k(k-1)/n(n-1)$ and its expectation is
$$\mathbb{E}(\sigma)=\sum\_{k=2}^n \frac{n(n-1)}{k(k-1)}= n(n-1)\sum\_{k=2}^n \frac1{k-1} - \frac1{k} = n(n-1)(1-\frac1{n}) = (n-1)^2 .$$
| 29 | https://mathoverflow.net/users/1061 | 41985 | 26,759 |
https://mathoverflow.net/questions/41978 | 32 | I was wondering if anyone could offer some intuition for why [Alexander duality](http://en.wikipedia.org/wiki/Alexander_duality) holds. Of course, the proof is easy enough to check, and it is also easy to work out many examples by hand. However, I don't have any feeling for why it is true.
To give you an example of what I am looking for, when I think of Poincare duality I think of the picture in terms of triangulations and dual triangulations. Is there any picture like that for Alexander duality? Is there at least maybe some kind of obvious bilinear pairing between the two sides of it or something?
| https://mathoverflow.net/users/10019 | Intuition behind Alexander duality | Let $M$ be a closed orientable $n$-manifold containing the compact set $X$.
Given an $n-q-1$-cocyle on $X$ (I am choosing this degree just to match with the notation
of the Wikipedia article to which you linked), we extend it to some small open neighbourhood $U$ of $X$.
By Lefschetz--Poincare duality on the open manifold $U$, we can convert this $n-q-1$-cocylce
into a Borel--Moore cycle (i.e. a locally-finite cycle made up of infinitely many simplices)
on $U$ of degree $q+1$. Throwing away those simplices lying in $U \setminus X$,
we obtain a usual (i.e. finitely supported) cycle giving a class in $H\_{q+1}(U,U\setminus X) = H\_{q+1}(M,M\setminus X)$ (the isomorphism holding via excision).
Alexander duality for an arbitrary manifold then states that
the map $H^{n-q-1}(X) \to H\_{q+1}(M,M \setminus X)$ is an isomorphism. (If $X$ is very pathological, then we should be careful in how define the left-hand side, to be sure
that every cochain actually extends to some neighbourhood of $X$.)
Now if $M = S^{n+1}$, then $H^i(S^{n+1})$ is almost always zero, and so we may use the boundary map for the long exact sequence of a pair to
identify $H\_{q+1}(S^{n+1}, S^{n+1}\setminus X)$ with $H\_{q}(S^{n+1}\setminus X)$ modulo worrying about reduced vs. usual
homology/cohomology (to deal with the fact that $H^i(S^{n+1})$ is *non-zero* at the extremal
points $i = 0$ or $n$).
So, in short: we take a cocycle on $X$, expand it slightly to a cocyle on $U$,
represent this by a Borel--Moore cycle of the appropriate degree, throw away those simplices lying entirely outside $X$, so that it is now a chain with boundary lying outside $X$, and finally *take this boundary*, which is now a cycle
in $S^{n+1} \setminus X$.
(I found [these notes of Jesper Moller](http://www.math.ku.dk/~moller/f03/algtop/notes/homology.pdf) helpful in understanding the general structure of Alexander duality.)
One last thing: it might help to think this through in the case of a circle embedded in $S^2$. We should thicken the circle up slightly to an embedded strip. If we then take our cohomology class to be the generator of $H^1(S^1)$, the corresponding Borel--Moore cycle is just a longitudinal ray of the strip (i.e. if the strip is $S^1 \times I$, where $I$ is an open
interval, then the Borel--More cycle is just $\{\text{point}\} \times I$).
If we cut $I$ down to a closed subinterval $I'$ and then take its boundary, we get a pair
of points, which you can see intuitively will lie one in each of the components
of the complement of the $S^1$ in $S^2$.
More rigorously, Alexander duality will show that these two points generate the reduced $H^0$ of the complement of the $S^1$, and this is how Alexander duality proves the Jordan curve theorem. Hopefully the above sketch supplies some geometric intuition to this argument.
| 24 | https://mathoverflow.net/users/2874 | 41986 | 26,760 |
https://mathoverflow.net/questions/41970 | 7 | Let $X$ be an algebraic variety. Consider an exact sequence
$$0\to A\to B\to C\to 0$$
of vector bundles on $X$. I have seen in different papers the following type resolution of wedge product of $C$ (or $A$)
$$0\to S^kA\to S^{k-1}A\otimes B\to S^{k-2}A\otimes \wedge^2B\to \cdots\to \wedge^kB\to \wedge^k C\to 0.$$
Question: does this resolution come from certain geometric context? Is there a proof which involves certain geometric aspects, for example, using projective bundles associated to the vector bundles?
| https://mathoverflow.net/users/2348 | How to resolve a wedge product of vector bundles | The resolution in question is the $k$th graded piece of the symmetric algebra of the complex $A \rightarrow B$, where $A$ is considered to lie in even degrees (say degree $2$) and $B$ in odd degrees (say degree $1$). (Also, a kind of Koszul complex.)
The way I think of this: forgetting the differential $A \rightarrow B$ for a minute, we just have a graded vector bundle $A \oplus B$ with $A$ in degree $2$, $B$ in degree $1$. Since the symmetric algebra construction $\mathcal{S}$ takes sums to tensor products, we have $\mathcal{S}(A \oplus B) \simeq \mathcal{S}(A) \otimes \mathcal{S}(B)$. Since $A$ is in degree $2$, $S(A)=S(A)$ (commutative) and since $B$ is in degree $1$, $\mathcal{S}(B) \simeq \bigwedge(A)$ (the Koszul sign rule for graded tensor products makes this super-commutative).
Now you have to bring in the differential. To do this, it is useful to use the bialgera structure on the symmetric algebra construction, where the coproduct comes from applying $\mathcal{S}$ to the diagonal map $X \rightarrow X \oplus X$. To see how to get the differential out of this: given $S^{k-i}A \otimes \bigwedge^{i} B$, comultiply in the first factor to end up in $S^{k-i-1}A \otimes A \otimes \bigwedge^{i}B$, now apply the map $A \rightarrow B$ to the middle factor of $A$ to end up in $S^{k-i-1}A \otimes B \otimes \bigwedge^{i}B$, and finally, multiply the last two factors together to end up in $S^{k-i-1}A \otimes \bigwedge^{i+1}B$
This is a standard construction and can be found for instance in the paper of Akin, Buchsbaum, Weyman on Schur complexes, or in the book of Weyman on Cohomology of vector bundles and syzygies. If people have other references, I'd be glad to know about them, since the above mentioned ones are from the point of view commutative algebra (which is for me non-optimal).
While the above construction is certainly useful in geometry, I'm afraid the above is not geometric in the way you were looking for.
| 10 | https://mathoverflow.net/users/4659 | 41990 | 26,764 |
https://mathoverflow.net/questions/41979 | 33 | $\newcommand{\Spec}{\mathrm{Spec}\,}$Let $X=\Spec A$ be a variety over $k$, then we have the definition of the tangent bundle $\hom\_k(\Spec k[\varepsilon]/(\varepsilon^2),X)$ (note that this has the structure of a variety). On the other hand, we have the definition of a tangent sheaf $\hom\_{\mathscr{O}\_X}(\Omega\_{X/k},\mathscr{O}\_X)$. What is the relationship between the two? Also, when $X$ is an arbitrary scheme (not necessarily affine), then does the relationship still hold?
| https://mathoverflow.net/users/9035 | Relationship between Tangent bundle and Tangent sheaf | You can always apply the "vector bundle" construction to $\Omega:=\Omega\_{X/k}$ (locally free or not). What you get is a scheme $T=\mathrm{Spec\ Sym}(\Omega)\to X$ which deserves to be called "tangent bundle" (albeit not locally trivial); in particular its $k$-points are what you want and, more generally, if, say, $Z=\mathrm{Spec}\ C$ is an affine $k$-scheme, then $T(C)$ is just $\mathrm{Hom}\_k (\mathrm{Spec}\ C[\varepsilon]/\varepsilon^2, X)$.
On the other hand consider the ${\cal O}\_X$-dual ${\cal T}:={\Omega}^\vee$. For every $X$-scheme $y:Y\to X$ there is a canonical map $\Gamma(Y,y^\*\mathcal{T})\to \mathrm{Hom}\_X(Y,T)$. If $Y$ is an open subset of $X$ this is bijective. But if $y$ is a point of $X$, then the LHS is $\Omega^\vee \otimes \kappa(y)$ while the RHS is the $\kappa(y)$-dual of $\Omega\otimes \kappa(y)$. Clearly the image consists of those tangent vectors at $y$ which extend to vector fields in a neighbourhood. The computation when $X$ is the union of the two axes in the plane is a good exercise; if $y$ is the origin the above map is zero.
[EDIT] after seeing Unknown's answer (BTW, there are some problems with TeX there). The above argument shows that the "tangent bundle" is always a scheme, if you define it right. Another way of seeing this is that it's just an instance of Weil restriction: if $R$ is a finite-dimensional $k$-algebra you can define the functor $\underline{\mathrm{Hom}}\_k (\mathrm{Spec}(R),X)$ in a similar way. This is always an algebraic space, and it is a scheme if $X$ is quasiprojective. But it is also a scheme if $R$ is *local*, which is the case here with $R=D\_1(k)$. The reason is that if you cover $X$ by affines $U\_i$, every morphism frome a local scheme to $X$ factors through one of the $U\_i$'s, so we can construct the Weil restriction of $X$ by gluing those of the $U\_i$'s.
| 31 | https://mathoverflow.net/users/7666 | 41992 | 26,766 |
https://mathoverflow.net/questions/41984 | 29 | The "textbook" example of a smooth bijection between smooth manifolds that is not a diffeomorphism is the map $\mathbb{R} \rightarrow \mathbb{R}$ sending $x \mapsto x^3$. However, in this example, the source and target manifolds *are* diffeomorphic -- just not by the given map. Is there an example of a smooth bijection $X \rightarrow Y$ of smooth manifolds where $X,Y$ are not diffeomorphic at all? (and if so, what?)
(For instance, is it possible to arrange a smooth bijection from a sphere to an exotic sphere, failing to be a diffeomorphism because of the existence of critical points? or do homeomorphisms between different smooth structures on spheres fail to be everywhere smooth in some catastrophic way?)
| https://mathoverflow.net/users/379 | Smooth bijection between non-diffeomorphic smooth manifolds? | Every smooth manifold has a smooth triangulation, which yields a pseudofunctor from the category of smooth manifolds to the category of PL manifolds. (There is no actual functor; that would be crazy.) If two smooth manifolds are PL isomorphic, then the answer is yes. You can start with the PL isomorphism, and then build a homeomorphism that follows it and that has the property that all derivatives vanish in all directions perpendicular to every simplex. You can build the homeomorphism by induction from the $k$-skeleton to the $(k+1)$-skeleton using bump functions.
The PL Poincaré conjecture is true in dimensions other than 4, so all exotic spheres in the same dimension $n \ge 5$ are PL homeomorphic. (High-dimensional examples of exotic spheres start in dimension 7, it was calculated.) In dimension 4, by contrast, every PL manifold has a unique smooth structure, and it is not known whether there are any exotic spheres.
On the other hand, if the manifolds are homeomorphic but not even PL homeomorphic, then I don't know. It is known that every manifold of dimension $n \ge 5$ has a unique Lipschitz structure, but I do not know a Lipschitz version of the above argument. On the positive side, passing from smooth to Lipschitz is an actual functor, so the answer to a modified question, is there a Lipschitz-smooth homeomorphism, is yes, and you can even make it bi-Lipschitz.
| 22 | https://mathoverflow.net/users/1450 | 41993 | 26,767 |
https://mathoverflow.net/questions/42006 | 12 | Some simple questions, for which I know no precise reference (and would be deeply grateful for a nice one!):
1. Is it true that the category of (pure) polarized Hodge structures is abelian semi-simple, whereas the whole category of pure Hodge structures is not?
2. Should one only consider those morphisms of polarized Hodge structures that respect polarizations in order to obtain an abelian category?
3. Is it true that all pure Hodge structures 'that come from geometry' (for example, the graded pieces of the weight filtration of the singular cohomology of varieties and motives) are polarized?
| https://mathoverflow.net/users/2191 | On polarized (pure) Hodge structures | Fortunately, these questions are easy to answer. First of all, it helps to distinguish
between polarizable Hodge structures and polarized structures. For polarizable, we merely
require that a polarization exists, but it is not fixed. Let Hodge structure
mean pure rational Hodge structure below.
Then
* The category of polarizable pure Hodge structures is abelian and semisimple (morphisms are not required to respect polarizations). This is essentially proved in Theorie de Hodge II.
* The category of arbitrary Hodge structures is abelian but not semisimple.
To see the nonsemisimplicity, we can use a theorem of Oort-Zarhin [Endmorphism algebras of
complex tori, Math Ann 1995], that
any finite dimensional $\mathbb{Q}$-algebra is the endomorphism algebra of some complex torus, and therefore
of some Hodge structure.
* All pure Hodge structures of geometric origin are polarizable. So this is a very reasonable condition to impose. For $Gr\_WH^\*(X)$, this is explained for example in Beilinson's *Notes on absolute Hodge cohomology* (although this was already implicit in Deligne's construction).
| 21 | https://mathoverflow.net/users/4144 | 42008 | 26,775 |
https://mathoverflow.net/questions/41972 | 13 | Let $\theta \not\in \mathbb{Q}$. We know that $(n\theta)\_{n \geq 1}$ is equidistributed modulo 1.
Let $\epsilon\_n = \mathrm{sign}\bigl(\sin(n\pi \theta)\bigr)$ and $S\_N= \sum\_{n=1}^N \epsilon\_n$.
I'm looking for a "good" asymptotic bound for $|S\_N|$ (not $|S\_N|\leq N$ obviously).
It looks like for any $x>0$, we should have $S\_N =o(n^x)$, or even better, that $(S\_N)$ is bounded, but is it?
| https://mathoverflow.net/users/3958 | Consequence of equidistribution or not? | No, you cannot put any better bound than SN = *o*(N). There is a general technique, using the [Baire category theorem](http://en.wikipedia.org/w/index.php?title=Baire_category_theorem&oldid=379733059) of proving the existence of counterexamples to problems like this (which I discovered while trying to find a counterexample to a question by David Speyer, [link](https://mathoverflow.net/questions/35902/does-weyls-inequality-prove-equidistribution/35976#35976)). I see that Helge's answer is also pointing towards the same result.
First, for θ irrational,
$$
S\_N/N=\frac{1}{N}\sum\_{n=1}^N1\_{\{0< n\theta/2 <1/2{\rm\ (mod\ 1)}\}}-\frac{1}{N}\sum\_{n=1}^N1\_{\{1/2< n\theta/2 <1{\rm\ (mod\ 1)}\}}
$$
By [Weyl's equidistribution theorem](http://en.wikipedia.org/wiki/Weyl%2527s_equidistribution_theorem), both sides on the right hand side tend to 1/2 and SN / N → 0, so SN = *o*(N).
It is not possible to do better than this. In fact, if f: ℕ → ℝ+ is any function satisfying liminf f(N) / N = 0 then there will be an uncountable dense set of irrational θ for which limsup SN / f(N) = ∞. In particular, using f(n) = nx for x < 1 rules out bounds such as Sn = *O*(nx).
In fact, we can find a set of such θ as an intersection of countably many open dense subsets of ℝ, so the Baire category theorem shows the existence of uncountably many counterexamples.
Let u(x) = 1{0≤[x/2]<1/2} - 1{1/2≤[x/2]<1} where [x] is the fractional part of x, and SN(θ) = Σn≤N u(nθ). Let UK be the set
$$
U\_K=\left\{\theta\in\mathbb{R}\colon S\_n(\theta)>Kf(n){\rm\ for\ some\ }n\ge K\right\}.
$$
This contains a dense open subset of ℝ. In fact, if θ = 2p/q for q odd then, for 1 ≤ n < q, u((q-n)θ) = -u(nθ). So, Sq-1(θ) = 0 and Sq(θ) = 1. Then, by periodicity of [nθ/2], Snq (θ) = n and Sn(θ) increases linearly. So, Sn(θ) > Kf(n) for infinitely many n, and θ ∈ UK. By right continuity of u, (θ,θ+ε) ⊆ UK for small enough ε. This shows that (2p/q,2p/q+ε) is contained in the interior of UK and, as such 2p/q are dense, the interior of UK is a dense open subset of ℝ. The Baire category theorem implies that
$$
U\equiv\bigcap\_{K=1}^\infty U\_K
$$
is an uncountable dense subset of ℝ and, by construction, for any θ ∈ U, limsup Sn(θ) / f(n) > K for each K.
---
The further question was asked in the comment: are there *any* irrational θ for which SN = *O*(Nx) for x < 1. The answer is yes. In fact this holds for [almost every](http://en.wikipedia.org/wiki/Almost_everywhere) θ and every x > 1/2.
The idea is to consider rational approximations to θ, |θ/2 - p/q| ≤ q-2. Then, there will be an integer 1 ≤ a < q such that |1/2 - [ap/q]| ≤ 1/(2q). So, |1/2-[aθ/2]| ≤ 1/q. With u() as above, it follows that u(nθ) + u((n+a)θ) = 0 unless -2/q ≤ nθ ≤ 2/q (mod 1). So, there is a lot of cancellation in SN(θ),
$$
\begin{array}
\displaystyle
\vert S\_N(\theta)\vert &\displaystyle \le a +\sum\_{n=1}^N1\_{\{-2/q\le n\theta\le 2/q{\rm\ (mod\ 1)}\}}\\\\
&\displaystyle\le 2q +\sum\_{n=0}^{\lfloor N/q\rfloor}\sum\_{m=1}^q1\_{\{-2/q\le nq\theta+m\theta\le 2/q{\rm\ (mod\ 1}\}}\\\\
&\displaystyle\le 2q+\sum\_{n=0}^{\lfloor N/q\rfloor}\sum\_{m=1}^q1\_{\{-4/q\le nq\theta+2mp/q\le 4/q{\rm\ (mod\ 1)}\}}
\end{array}
$$
The points 2mp/q (mod 1) are equally spaced. If q is odd then they have spacing 1/q and no more than 9 of them can lie in an interval of length 8/q. If q is even then the spacing is 2/q and no more than 5 can lie in such an interval. In either case, the final sum over m above is bounded by 10=5\*2.
$$
\vert S\_N(\theta)\vert\le 2q+10N/q.
$$
If θ has [irrationality measure](http://mathworld.wolfram.com/IrrationalityMeasure.html) less than α then, for large enough N, the rational approximation p/q can be chosen such that N1/2 ≤ q ≤N(α-1)/2,
$$
\vert S\_N(\theta)\vert\le 2N^{(\alpha-1)/2}+10N^{1/2}.
$$
In particular, if θ has irrationality measure 2 then $S\_N=O(N^x)$ for every $x>1/2$. But, almost every real number has irrationality measure 2.
| 9 | https://mathoverflow.net/users/1004 | 42011 | 26,777 |
https://mathoverflow.net/questions/42024 | 0 | hello,
it is often said that a **conformal mapping** preserves the Laplace equetion in 2D.
However, if this is true for the **cartesian coordinates (x,y)**, where the laplacian is:
$$
\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=0
$$
is it true for the **cylindrical coordinates (r,z)** where:
$$
\frac{\partial^2 \phi}{\partial r^2}+\frac{\partial^2 \phi}{\partial z^2}+\frac{1}{r}\frac{\partial \phi}{\partial r}=0?
$$
More precisely, if you have a function $\phi(r,z)$ verifying the previous equation, is it also verified by $\psi(u,v)$ is $u+i v=f(z+ir)$ and f is analytic/holomorphic?
| https://mathoverflow.net/users/9039 | Invariance of the cylindrical Laplace equation under conformal transform | **No**. Take $\phi=r^2-2z^2$ and $f(z+ir)=(z+ir)^2$, thus $u=r^2-z^2$, $v=2rz$. Then $\phi(u,v)=r^4+z^4-10r^2z^2$ is not cylindrical-harmonic.
| 1 | https://mathoverflow.net/users/8799 | 42027 | 26,784 |
https://mathoverflow.net/questions/41998 | 12 | There are various notions of exact categories ([nlab](http://ncatlab.org/nlab/show/exact+category)). In a lecture I've seen the following definition of an exact category, which is basically (exact) = (abelian) − (additive):
A category $C$ is called **exact** if a) it contains a zero object, b) every morphism has a kernel and a cokernel, c) the canonical morphism $\operatorname{coim}(f) \to \operatorname{im}(f)$ is an isomorphism for every morphism $f$.
So for example, the category of pointed sets is an exact category in this sense. I think also the category of pointed compactly generated hausdorff spaces is an example.
**Questions**: 1) Which theorems and constructions of homological algebra carry over from abelian categories to exact categories in the above sense? 2) Where can I find literature about these categories? I can only find some about the other definitions.
| https://mathoverflow.net/users/2841 | Exact categories which are not additive | These categories are called Puppe-exact or p-exact categories. See paragraph 1.1, of *Jordan-Hölder, modularity and distributivity in non-commutative algebra*, by Francis Borceux and Marco Grandis (JPAA **208** (2007), 665-689 doi:[10.1016/j.jpaa.2006.03.004](https://doi.org/10.1016/j.jpaa.2006.03.004)), for non-abelian examples. And see the papers of Marco Grandis (e.g. *On the categorical foundations of homological and homotopical algebra* ([Numdam](http://www.numdam.org/item/CTGDC_1992__33_2_135_0/))) and Mitchell's book “Theory of categories” for homological results in this context (as a general rule, all homological lemmas non involving direct products hold).
| 12 | https://mathoverflow.net/users/10033 | 42034 | 26,788 |
https://mathoverflow.net/questions/41950 | 2 | I am wondering about approximation and idealization. Specifically, I am wondering if anyone has seen some work on the following. In the semantics of programming languages we find Domains as a place to talk about iteration and approximation. We can define a Scott Topology on the Domain and now our Domain-maps are continuous maps. Next,we can see our Domains as categories and turn the continuous Domain-maps into continuous functors. If we push the idea further, we have continuous functors and a notion of approximation which is now over categories. Lambek ponders the existence of *the* category of Sets. What about *approximations* to the category of sets. For that matter, what might it look like to approximate any well-known category like that of manifolds or FDVec?
I realize the question is vague. I wish I had a more concrete question, but my grasp of the material is too weak.
Naturally, any thoughts on this would be most appreciated.
| https://mathoverflow.net/users/10007 | approximating categories with continuous functors | As it happens, I just saw a paper about this very subject today -- Martin Hyland's ["Some Reaons for Generalizing Domain Theory"](http://www.dpmms.cam.ac.uk/~martin/Research/Publications/2010/srfg10.pdf), which is concerned with precisely the generalization you suggest, in order to clarify the semantics of concurrency. This paper was apparently inspired by some work by Cattani and Winskel, but I found their work to be more categorically sophisticated than I could easily digest, and this one to be at about the level I can presently cope with.
EDIT: I also found a paper by Winskel in which he discusses the intuitions underlying his ideas, ["Events, Causality, and Symmetry"](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.157.6016&rep=rep1&type=pdf). This seems quite accessible to me.
| 3 | https://mathoverflow.net/users/1610 | 42035 | 26,789 |
https://mathoverflow.net/questions/42010 | 12 | In the next days I have to give a talk in which I need to explain some of the usual singularities of pairs that one meets when dealing with the minimal model program: KLT, DLT and LC pairs.
In particular I would like to give an intuition (also to non specialists) of the reason why an LC pair is much more difficult to treat than a DLT pair, for example.
In the same context I would also like to give an intuition of what the LC centers of a pair are and why they are "special" subvarieties for a pair.
Note that I'm always considering normal (possibly non-smooth) projective varieties.
Do you have any ideas?
| https://mathoverflow.net/users/6430 | Singularities of pairs | For references, probably you already know things like Kollar-Mori and Kollar's "Singularities of pairs", for LC-centers and subadjunction, Kollar has some notes on that as well. There's some sections on this in the recent book by Hacon-Kovacs too that I've looked at recently.
With regards to your specific questions I have some comments but probably VA will have better comments:
**1. an LC pair is much more difficult to treat than a DLT pair**
The archetypical LC pair is probably $(\mathbb{A^2}, Div(x) + Div(y))$, the two coordinate axes (EDIT: this example is also DLT, and doesn't really represent the most poorly-behaved properties of LC-pairs). This is a simple normal crossings (SNC) pair and we understand how they behave pretty well with respect to numerous operations. On the other hand, if you have a pair $(X, D)$ where the pair is KLT, then even if the singularities of $D$ are bad, the fact that $(X, D)$ is klt means that you can perturb the coefficients of $D$ in many ways. This gives one a great amount of flexibility in numerous situations.
DLT, by Szabo's criterion, is a combination of KLT and SNC with coefficients $\leq 1$. Basically, a DLT pair is LC and where it is not KLT, it is SNC (simply normal crossings), which we understand. Dually, where it's not SNC, it's KLT, and we can perturb things and do those other tricks.
**2. what the LC centers of a pair are and why they are "special" subvarieties for a pair.**
The most basic example of a LC-center of a pair $(X, D)$ is a prime divisor $D\_i$ which is a component of $D$ and such that the $D\_i$-coefficient of $D$ is $1$. For example, if $X$ is smooth, and $D$ is a prime divisor, then $D$ is a LC-center of $(X, D)$. Why is this nice? This lets us relate the canonical divisor $K\_X + D$ of $(X, D)$ and the canonical divisor of $D$.
Explicitly, in this smooth case, $(K\_X + D)|\_D = K\_D$. Because of this, you can translate many properties of the pair $(X, D)$ to things on $D$ (look up extension theorems for example, or adjunction/inversion of adjunction, variants of this hold without the smooth hypothesis). This is very very useful for induction on dimension. LC-centers are a way to generalize this to higher codimensional subvarieties.
By definition, LC-centers are exactly the images of divisors $D\_i'$ where $D\_i'$ is a component of $D'$ with coefficient $1$, and $(X', D')$ is a pair obtained by taking a log resolution $\pi$ of $(X, D)$ and setting $D' = \pi^\*(K\_X + D) - K\_{X'}$.
In particular, if $W$ is an LC-center of a pair $(X, D)$, then one has $$(K\_X + D)|\_W = K\_W + (\text{some correction terms}).$$ Again, many properties of the pair $(X, D)$ can be translated to properties of $(W, (\text{correction terms}) )$, this isn't trivial to show.
One should note that even if you have a 1-dimensional LC-center, then there can still be a correction factor (just not in the smooth example I described above).
| 9 | https://mathoverflow.net/users/3521 | 42042 | 26,793 |
https://mathoverflow.net/questions/42016 | 23 | I am looking for algorithms on how to find rational points on an [elliptic curve](http://en.wikipedia.org/wiki/Elliptic_curve) $$y^2 = x^3 + a x + b$$ where $a$ and $b$ are integers. Any sort of ideas on how to proceed are welcome. For example, how to find solutions in which the numerators and denominators are bounded, or how to find solutions with a randomized algorithm. Anything better than brute force is interesting.
Background: a student worked on the [Mordell-Weil theorem](http://en.wikipedia.org/wiki/Mordell-Weil_theorem) and illustrated it on some simple examples of elliptic curves. She looked for rational points by brute force (I really mean *brute*, by enumerating all possibilities and trying them). As a continuation of the project she is now interested in smarter algorithms for finding rational points. A cursory search on Math Reviews did not find much.
| https://mathoverflow.net/users/1176 | Algorithms for finding rational points on an elliptic curve? | A good reference to get started from the algorithmic point of view is Chapter 3 of Cremona's [*Algorithms for Modular Elliptic Curves*](http://www.warwick.ac.uk/~masgaj/book/fulltext/index.html). It contains a good deal of pseudocode which explains how Cremona's C++ package mwrank computes rational points on elliptic curves.
Some of the more intricate details, such as second descents are left to Cremona's papers [here](http://www.warwick.ac.uk/~masgaj/papers/index.html). Given an elliptic curve with coefficients that aren't too big, your best bet to quickly find the points you're looking for will probably be to use mwrank as included in [Sage](http://www.warwick.ac.uk/~masgaj/papers/index.html).
As has been explained to me in the comments. Sage is not the only way to get access to mwrank and the other programs that make up Cremona's elliptic curve library (eclib), but it is arguably the easiest way to get it, and it contains much more elliptic curve functionality, such as the method E.analytic\_rank() which if run on elliptic curve of reasonably sized conductor, will return an integer that is proBably the analytic rank of the curve.
| 10 | https://mathoverflow.net/users/4872 | 42044 | 26,794 |
https://mathoverflow.net/questions/42038 | 3 | Is there example of hypersurface $X \subset \mathbb{P}^n$ satisfying
1. $X$ is of degree 2. (I mean, the Poincare dual PD(X) is 2u, where u is a generator of
$H^2(\mathbb{P}^n, \mathbb{Z})$.
2. Some odd betti number is non-zero.
Thank you for any comment.
| https://mathoverflow.net/users/11705 | Hypersurface of complex projective space | As noticed by Sasha in his comment, the answer is **no**.
The following proof also shows that this result cannot be generalized for higher values of the degree.
For any smooth complex hypersurface of degree $d$, say $X\_d \subset \mathbb{P}^{n+1}$, by standard arguments involving Lefschetz theorem we have $$H^k(X\_d)= H^k(\mathbb{P}^n) \quad \mathrm{for} \; k \neq n.$$
In particular, all the odd Betti numbers are zero, except possibly the middle Betti number
when $n$ is odd. On the other hand, the Euler-Poincare characteristic of $X\_d$ is equal to $$\chi(X\_d)= \langle c\_n(T\_{X\_d}), [X\_d] \rangle =\frac{1}{d}[(1-d)^{n+2}-1]+n+2,$$
so for $n$ odd and $d=2$ the middle cohomology group must be zero too. Notice that for $n$ odd and $d >2$ one always has a non-zero middle Betti number. For instence, if $X \subset \mathbb{P}^4$ is a smooth cubic hypersurface, then $b\_3(X)=10$.
A good reference for these results is Dimca's book *Singularities and topology of hypersurfaces*, Chapter 5, which also considers the case of hypersurfaces in $\mathbb{P}^{n+1}$ with isolated singularities.
| 13 | https://mathoverflow.net/users/7460 | 42048 | 26,796 |
https://mathoverflow.net/questions/42040 | 6 | In the decompositions I encountered so far, we all had orthogonal set of bases. For example in Singular Value Decomposition, we had orthogonal singular right and left vectors, in [discrete] cosine transform (or [discrete] fourier transform) we had again orthogonal bases.
To describe any vector $x \in \mathbb{C}^N$, we need to have $N$ independent set of basis vectors but independent doesn't necessarily mean orthogonal. My intentions behind selecting orthogonal vectors are as follows:
* The solution is not unique for $x$ if the basis are not orthogonal.
* It is easy to find the solution numerically by projecting $x$ onto each vector and this solution doesn't depend on the order of the bases. Otherwise, it depends on the order.
* If we are talking about some set of vectors, they might be correlated in the original space, but uncorrelated in the transformd space which might be important when analyzing the data, in dimensionality reduction or compression.
I'm trying to understand the big picture. Do you think that I am right with these? Do you have any suggestions, what is the main reason for selecting orthogonal bases?
| https://mathoverflow.net/users/5287 | Why do we want to have orthogonal bases in decompositions? | Your first point, non-uniqueness, is definitely false. One of the basic facts in linear algebra is precisely that for any fixed set of basis vectors (we don't even have to work on a vector space endowed with an inner product, so orthogonality doesn't come in at all), a given vector has a unique decomposition.
For if that were not true, let the basis vectors be $e\_1, \ldots, e\_n$, then you have two sets of numbers $a\_1, \ldots a\_n$ and $b\_1, \ldots, b\_n$ such that
$$ a\_1 e\_1 + \cdots + a\_n e\_n = x = b\_1 e\_1 + \cdots + b\_n e\_n \implies (a\_1 - b\_1) e\_1 + \cdots + (a\_n - b\_n) e\_n = 0$$
if the sets $a\_\*$ and $b\_\*$ are not identical, this implies that $e\_1\ldots e\_n$ are linearly dependent, contradicting the assumption that they form a basis.
The second point, however, is a biggie. Without an inner product you cannot define an orthgonal projection. Now, generally, this is not too much of a problem. Given the basis vectors $e\_1,\ldots, e\_n$, finding the coordinates $a\_1,\ldots, a\_n$ of a given vector $x$ in this basis is just solving a linear system of equations, which actually is not too hard numerically, in *finite dimensional systems*. In infinite dimensional systems this inversion of the transformation matrix business gets slightly tricky.
The key is to note that without using the orthogonal projection, you *cannot* answer the question "what is the length of $x$ in the direction of $e\_1$?" without knowing the entire set of basis vectors. (Remember that without using the orthogonal projection, you need to solve a linear system of equations to extract the coordinates; if you only specify one of the basis vectors, you do not have a complete system of equations, and the solution is underdetermined. I suspect this is what you had in the back of your mind for the first point.) This is actually a very fundamental fact in geometry, regarding coordinate systems. (I've once heard this described as the "second fundamental mistake in multivariable calculus" made by many students.)
Using the orthogonal projection/the inner product, you can answer the question, as long as you allow only orthgonal completions of the basis starting from the known vector. This fact is immensely useful when dealing with infinite dimensional systems.
I also don't quite like your formulation of the third point. A vector is a vector is a vector. It is independent of the coordinate representation. So I'd expect that if two vectors are correlated (assuming correlation is an intrinsic property), they better stay correlated without regard to choice of bases. What is more reasonable to say is that two vectors maybe uncorrelated in reality, but not obviously so when presented in one particular coordinate system, whereas the lack of correlation is immediately obvious when the vectors are written in another basis. But this observation has rather little to do with orthogonality. It only has some relation to orthogonality if you define "correlation" by some inner product (say, in some presentations of Quantum Mechanics). But then you are just saying that orthogonality of two vectors are not necessarily obvious, except when they are.
---
My personal philosophy is more one of practicality: the various properties of orthogonal bases won't make solving the problem harder. So unless you are in a situation where those properties don't make solving the problem easier, and some other basis does (like what Ryan Budney described), there's no harm in prefering an orthogonal basis. Furthermore, as Dick Palais observed above, one case where an orthogonal bases really falls out naturally at you is the case of the spectral theorem. The spectral theorem is, in some sense, the correct version of your point 3, that in certain situations, there *is* a set of basis vectors that are mathematically special. And this set happens to always be orthogonal.
---
**Edit** A little more about correlation. This is what I like to tell students when studying linear algebra. A vector is a vector. It is an object, not a bunch of numbers. When I hand you a rectangular box and ask you how tall the box is, the answer depends on which side is "up". This is similar to how you should think of the coordinate values of a vector inside a basis: it is obtained by a bunch of measurements. (Picking which side is "up" and measuring the height in that direction, however, in a non-orthogonal system, will require knowing all the basis vectors. See my earlier point.)
The point is that to quantitatively study science, and to perform numerical analysis, you can only work with numbers, not physical objects. So you have to work with measurements. And in your case, the correlation you are speaking of is correlation between the measurements of (I suppse) different "properties" of some object. And since what and how you measure depends on which basis you choose, the correlation between the data will also depend on which basis you choose. If you pick properties of an object that are correlated, then your data obtained from the measurements will also be correlated. The PCA you speak of is a way to disentangle that. It may be difficult to determine whether two properties of an object is correlated. Maybe the presence of a correlation is what you want to detect. The PCA tells you that, if there were in fact two independent properties of an object, but you just chose a bad set of measurements so that the properties you measure do not "line up" with the independent properties (that the properties you measure have a little bit of each), you can figure it out with a suitable transformation of the data at the end of the day. So you don't need to worry about choosing the best set of properties to measure.
| 11 | https://mathoverflow.net/users/3948 | 42053 | 26,798 |
https://mathoverflow.net/questions/42037 | 8 | Suppose I have $k$ $n$-dimensional polytopes $P\_1,\ldots,P\_k$, each explicitly specified as the intersection of a collection of hyperplanes. If there was a point $p \in \mathbb{R}^n$ that lay in the intersection of all of these polytopes ($p \in P\_1 \cap \ldots \cap P\_k$), I could efficiently find it by solving a linear program. Unfortunately, I have no guarantee that my polytopes have non-empty intersection. However, someone has promised me that there exists a point $p$ that lies in at least $2/3$ of my polytopes: that is, there exist indices $i\_1,\ldots,i\_{2k/3}$ such that:
$$p \in P\_{i\_1} \cap \ldots \cap P\_{i\_{2k/3}}$$
Does there exist an efficient algorithm (running in time polynomial in k and n) that can find such a point?
| https://mathoverflow.net/users/3027 | Finding a point that lies in a majority of polytopes | This problem is clearly in NP (guess which polytopes) and becomes NP-complete if we replace $2/3$ with $1/2$ and make it a decision problem, dropping the promise that such a $p$ exists. In particular we can reduce integer programming feasibility to this problem.
Say we wish to determine whether $Ax = b$ has a solution $x$ which is a zero-one vector. We can define $k = 2n$ polytopes $P\_i^j = \{x \mid x\_i = j, Ax = b, 0\leq x\leq 1\}$ for integers $1\leq i\leq n$ and $0\leq j\leq 1$. A zero-one solution of $Ax=b$ is the same as a point which lies in $l = n = k/2$ of these polytopes.
I imagine there is a simple reduction showing that the problem is still hard if we use the fraction $2/3$ (or any other fixed fraction) instead of $1/2$. But I'm guessing that you just gave $2/3$ as an example so I haven't thought about it. Similarly I am guessing that the problem is still hard if you somehow know that such a $p$ exists and merely want to find one, but I haven't thought of how to show this either.
| 4 | https://mathoverflow.net/users/5963 | 42064 | 26,804 |
https://mathoverflow.net/questions/42065 | 3 | I need the following result for an example in a paper I'm writing. It's easy enough to prove, but I'd prefer to just give a reference. Does anyone know one?
Fix $1 \leq k \leq n$. Define $X\_{n,k}$ to be the following poset. The elements of $X\_{n,k}$ are ordered sequences $\omega = (x\_1,\ldots,x\_m)$, where the $x\_i$ are distinct elements of the $n$-element set $\{1,\ldots,n\}$ and $m \geq k$. The order relation is that $\omega\_1 \leq \omega\_2$ if $\omega\_1$ is a subsequence of $\omega\_2$. The theorem then is that the geometric realization of $X\_{n,k}$ is $(n-1-k)$-connected.
| https://mathoverflow.net/users/317 | Posets of finite sequences are highly connected | If I understand your question correctly, an answer should appear in the paper "On lexicographically shellable posets" of Anders Bj\"orner and Michelle Wachs, in Transactions of the AMS 277, pp. 323-341.
| 3 | https://mathoverflow.net/users/36466 | 42074 | 26,808 |
https://mathoverflow.net/questions/42056 | 1 | Is there a simple rule to check whether a pretzel link P(n\_1,...,n\_k) is a trivial link?
I am interested in the 2-component case but every information would be helpful.
| https://mathoverflow.net/users/5001 | Trivial pretzel links | A pretzel link has as many components as the pretzel link with coefficients reduced $(\mod 2)$. So it will have two components if and only if there are precisely two even coefficients.
A conceptual classification is given by considering the 2-fold branched cover, or the $\pi$-orbifold, obtained by taking the orbifold quotient of the 2-fold branched cover. For pretzel links, the 2-fold branched cover is a connect sum of Seifert fibered spaces (this is more generally true for Montesinos links). Then a 2-component link is trivial if and only if the two-fold branched cover is $S^2\times S^1$. A theorem of [Bonahon and Siebenmann](http://www.ams.org/mathscinet-getitem?mr=827297) classifies Seifert fibered orbifolds, so in principal you can classify pretzel links from their classification.
| 7 | https://mathoverflow.net/users/1345 | 42075 | 26,809 |
https://mathoverflow.net/questions/42022 | 15 | Let $A$ be a C\*-algebra. Suppose that $P:A \rightarrow A$ is a contractive completely positive projection. Does the range $P(A)$ is completely order isomorphic to a $C^\*$-algebra?
In the case where the answer is false:
Is it true in supposing, in addition, that the range $P(A)$ is an operator system, $A$ unital and $P(1)=1$?
| https://mathoverflow.net/users/5210 | Range of completely positive projection | Yes. $P(A)$ is abstractly a C$^\ast$-algebra $B$ under the Choi-Effros product: $a\circ b := P(ab)$; and $P(A)$ is completely order isomorphic to $B$. See (the proof of) Theorem 3.1 in [M.-D. Choi and E. G. Effros; Injectivity and operator spaces. J. Functional Analysis 24 (1977), 156-–209]. Moreover, there is a surjective $\ast$-homomorphism from $C^\ast(P(A))$ onto the C$^\ast$-algebra $B$.
| 21 | https://mathoverflow.net/users/7591 | 42090 | 26,814 |
https://mathoverflow.net/questions/42072 | 2 | Everybody knows that a square matrix $A$ has the same eigenvalues as
$A^T$. And it is clear that if $A^T=BAB^{-1}$ then $B$ maps eigenvectors
of $A$ to those of $A^T$. But I have not found any discussion of the
benefits of knowing $B$. Perhaps it is unusual to know $B$ exactly,
without having analyzed $A$ completely. But I have some examples where
this is known.
It seems at least to be nontrivial information. For example, with $A$ real
and $B$ real and
symmetric, you know an indefinite inner product
$\langle x,y\rangle = x^HBy$
with respect to which $A$ is symmetric. This gives certain orthogonality
relations (for example non-real eigenvalues have null eigenvectors) but it doesn't
seem to lead to anything quantitative about eigenvalues.
The worst case may be for symmetric $A$, $A^T=IAI^{-1}$ tells you nothing.
So the question is
>
> Have examples been studied, where knowing $B$ in $A^T=BAB^{-1}$ helped to find eigenvalues of $A$?
>
>
>
| https://mathoverflow.net/users/6998 | Does knowing a conjugation of A to A^T determine eigenvalues of A? | Generally, over the algebraically closed field, say over $\mathbb{C}$, $A^T$ and $A$ are always conjugate, because they have the same Jordan form. So such $B$ always exists. But the knowledge of the concrete $B$ satisfying to this relation may clarify what the eigenspaces for $A$ are, because ${B^T}B$ always commutes with your A. As previously was mentioned, this condition $A^T=BAB^{-1}$ doesn't determine the eigenvalues of A uniquely, so it may help only with eigenspaces. And the amount of information it gives you depends on this particular $B$. For $B=I$ it gives you nothing, as you mentioned above.
| 2 | https://mathoverflow.net/users/10045 | 42092 | 26,816 |
https://mathoverflow.net/questions/42085 | 0 | How to estimate the probability of co-occurrence of the positive integers $c\_i$ and $d\_i$, $1 \leq i \leq t$ drawn from the uniform range $1$ to $2^k-1$, such that $\Sigma^t\_{i=1} c^2\_i = \Sigma^t\_{i=1} c\_i\times d\_i$ and $\forall i, c\_i\neq d\_i$ hold?
$\forall i, c\_i$ and $d\_i$'s are drawn from the same distribution, but each pair $(c\_i, d\_i, \forall i \neq j)$ is drawn from different distribution.
To simplify the problem for the case $t=2$, it is the probability of co-occurrence of four positive integers $c\_1, c\_2, d\_1$ and $d\_2$ such that $c^2\_1 + c^2\_2 = c\_1\times d\_1 + c\_2\times d\_2 $ and $c\_1\neq d\_1$ and $c\_2\neq d\_2$ hold. The pairs $(c\_1,d\_1)$ and $(c\_2,d\_2)$ are chosen from two different distributions of the same range $1$ to $2^k-1$.
| https://mathoverflow.net/users/10041 | Estimating the probability of co-occurrence of a set of positive integers | An approach for the case $t=2$. There are $N^4$ ways of choosing $a,b,c,d$, all between 0 and $N-1$. You want to know how many of them result in $a^2+b^2=ac+bd$. Let's take the case where $\gcd(a,b)=1$ and $a\lt b$. Then you must have $c=a+br$ and $d=b-ar$ for some $r$. This $r$ can't exceed $b/a$, nor can it exceed $(N-a)/b$. So, given $a,b$, the number of pairs $c,d$ is just the smaller of $b/a$ and $(N-a)/b$, suitably rounded. Now you have to sum that over all pairs $a,b$ with $\gcd(a,b)=1$ and $a\lt b$. Then you have to work out what modifications you need to handle the case $\gcd(a,b)=e\gt1$.
EDIT: Fixing $t$ but letting $N$ go to infinity, the probability must go to zero, since, when you've picked all the variables but one, there's only one way to choose that last variable (unless its partner is zero), so the probability should never exceed $2/N$. But maybe I misunderstand the question.
| 0 | https://mathoverflow.net/users/3684 | 42108 | 26,822 |
https://mathoverflow.net/questions/42113 | 3 | Let $r,s,n$ be positive integers with $r < s < n$. Let $U = \{1,\ldots,n\}$.
>
> Let $S$ contain $s$-element subsets of
> $U$ (of our choosing). What is that smallest we can make $S$ such that every
> $r$-element subset of $U$ is a subset
> of some element in $S$?
>
>
>
I'm curious what the best known upper and lower bounds are on the smallest we can make $S$, and I am especially interested in the case where $r << s << n$. I am more worried about asymptotic behavior than exact bounds.
This can also be viewed as a special case of a set-cover problem that has a lot of symmetry. I'm afraid/hoping there's something simple from graph theory that solves my problem.
EDIT: Thanks to @Gerhard, I see this is a well known problem called covering numbers / covering designs.
| https://mathoverflow.net/users/8574 | A small collection of large subsets covering all small subsets. | What you are looking for is an $(r,s,n)$-covering design. A good starting point might be the [La Jolla Covering Repository](http://www.ccrwest.org/cover.html) or in fact, any book on design theory. In general, the smallest possible size for your covering design is given by $\binom{n}{r}/\binom{s}{r}$ in which case every $r$ -element subset is contained in exactly one $s$-element subset and the design is then called a $(r,s,n)$-Steiner system. There are necessary conditions for Steiner systems to exist such as $\binom{s-i}{r-i} | \binom{n-i}{r-i}$ for all $i$. Quite a lot is known about the existence of $(2,3,n)$-Steiner systems, and $(3,4,n)$-Steiner systems, but in general their existence is an open question.
| 9 | https://mathoverflow.net/users/9044 | 42115 | 26,825 |
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