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https://mathoverflow.net/questions/43388 | 30 | One out of the almost endless supply of identities discovered by Ramanujan
is the following:
$$ \sqrt[3]{\rule{0pt}{2ex}\sqrt[3]{2}-1} = \sqrt[3]{\frac19} - \sqrt[3]{\frac29} + \sqrt[3]{\frac49}, $$
which has the following interpretation in algebraic number theory: the fundamental unit
$\sqrt[3]{2}-1$ of the pure cubic number field $K = {\mathbb Q}(\sqrt[3]{2})$ becomes a cube in the extension $L = K(\sqrt[3]{3})$.
Are there more examples of this kind in Ramanujan's work?
| https://mathoverflow.net/users/3503 | Ramanujan and algebraic number theory | $$(7 \sqrt[3]{20} - 19)^{1/6} = \ \sqrt[3]{\frac{5}{3}}
- \sqrt[3]{\frac{2}{3}},$$
$$\left( \frac{3 + 2 \sqrt[4]{5}}{3 - 2 \sqrt[4]{5}}
\right)^{1/4}= \ \ \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1},$$
$$\left(\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}\right)^{1/2}
= \ \ (1 + \sqrt[5]{2} + \sqrt[5]{8})^{1/5} = \ \
\sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}},$$
and so on. Many of these were submitted by Ramanujan as problems to the
Journal of the Indian Mathematical Society. See the following link:
[jims.ps](https://faculty.math.illinois.edu/%7Eberndt/jims.ps "Bruce C. Berndt, Youn-Seo Choi, and Soon-Yi Kang: The problems submitted by Ramanujan to the Journal of the Indian Mathematical Society") for more precise references.
Quote: "although Ramanujan never used the term unit, and
probably did not formally know what a unit was,
he evidently realized their fundamental properties.
He then recognized that taking certain powers of units
often led to elegant identities."
| 17 | https://mathoverflow.net/users/nan | 43456 | 27,629 |
https://mathoverflow.net/questions/43066 | 8 | Let $A$ be a Banach algebra and let $X$ be an $A$-bimodule. Is there a notion of (relative) injectivity for $X$ which would imply that $\mathcal{H}^n(A,X)$ vanishes for all $n\ge 1$? Here $\mathcal{H}^n(A,X)$ denotes the continuous Hochschild cohomology of $A$ with coefficients in $X$ ($X$ can be assumed to be dual $A$-bimodule).
I expected there is such a notion, but after reading books by Helemskii, Runde and a few other sources on cohomology of Banach algebras I can't seem to find a general statement of this type, even though versions of projectivity and injectivity are discussed there.
| https://mathoverflow.net/users/nan | Injectivity for bimodules and Hochschild cohomology | A couple of people have encouraged me to post this as an answer, so here goes. I am currently without a copy of Helemskii's Pink Book so I can't give chapter-and-verse references as I would have liked. Everything that follows should be somewhere in there, although perhaps expressed slightly differently, and probably slightly better. Certainly what follows is too wordy, but I haven't had time to work out a condensed version.
To recap: Piotr is asking about the continuous Hochschild cohomology groups ${\mathcal H}^n(A,X)$ where $A$ is a Banach algebra and $X$ a Banach $A$-bimodule. To simplify the discussion slightly, I shall assume that $A$ has an identity element (which is indeed the case if $A$ is one of the usual convolution-type algebras associated to a discrete group) and that $X$ is unit-linked, i.e. that the identity of $A$ acts as the identity operator on $X$.
Conceptual/abstract POV (Helemskian)
------------------------------------
As in the classical theory of Cartan-Eilenberg vintage, (continuous) Hochschild cohomology can be expressed in terms of relative Ext. One way to approach this, as Helemskii does, is to introduce the enveloping algebra $A^e$ of a unital Banach algebra $A$.
This has underlying Banach space $A\hat{\otimes} A$ (projective tensor product) and has multiplication defined by $(a\otimes b)\cdot (c\otimes d) = (ab\otimes dc)$.
(The definition is slightly different for the non-unital case, and the artificial dichotomy that arises in places in the Pink Book is something that vexes some of us. But I digress...)
The purpose of doing this is as follows: every Banach $A$-bimodule $X$ becomes a left Banach $A^e$-module via $(a\otimes b)\cdot x = axb$; and conversely, every left Banach $A^e$-module becomes a Banach $A$-bimodule via the same formula. Now, taking as read the definition of relative Ext that is given in the Pink Book, we have
$$ {\mathcal H}^n(A,X) \cong \operatorname{Ext}\_{A^e}^n (A,X) $$
This is an isomorphism of seminormed spaces for each $n$ (I guess it would be more precise to say an isomorphism of seminormed-space-valued $\delta$-functors or some such high-falutin' phrase)
Now, recall that if $B$ is a Banach algebra, then a left Banach $B$-module $X$ is said to be (relatively) $B$-injective if it satisfies the following:
whenever $N$ is a left Banach $B$-module and $M$ is a closed $B$-submodule of $N$ which is complemented as a Banach subspace, then each continuous linear $B$-module map $M\to X$ has a continuous linear extension to a $B$-module map $N\to X$.
Moreover, if $X$ is relatively $B$-injective then $\operatorname{Ext}\_B^n(\cdot,X)=0$ for each $n\geq 1$. (The converse also holds, in fact.) Therefore:
>
> if $X$ is relatively $A^e$-injective, then ${\mathcal H}^n(A,X)=0$ for all $n\geq 1$.
>
>
>
The get-your-hands-dirty approach (Johnsonite)
----------------------------------------------
We continue to suppose that $A$ has an identity element. Now let $E$ be any Banach space and equip $V\_E :={\mathcal L}(A\hat\otimes A, E)$ with the following natural $A$-bimodule structure:
$$ (b\cdot T \cdot a)(c\otimes d) = T(ac\otimes db) \quad\quad(T\in V\_E). $$
**Claim:** ${\mathcal H}^n(A,V\_E)=0$ for all $n\geq 1$.
This is most easily proved by proving something stronger:
**Exercise:** Let $\delta: {\mathcal C}^n(A,V\_E)\to {\mathcal C}^{n+1}(A,V\_E)$ denote the Hochschild coboundary operator. Define $\sigma: {\mathcal C}^{n+1}(A,V\_E) \to {\mathcal C}^n(A,V\_E)$ by
$$ [\sigma\psi(a\_1,\dots,a\_n)]{(c\otimes d)} = [\psi(d,a\_1,\dots, a\_n)]{(c\otimes 1)}. $$
Then $\delta\sigma(\psi)+\sigma\delta(\psi) =\psi$ for every $\psi\in\mathcal C^k(A,V\_E)$.
We now observe the following: if $V$ is any Banach $A$-bimodule, and it can be written as $V\cong X\oplus Y$ where $X$ and $Y$ are closed $A$-sub-bimodules of $V$, then
$$ {\mathcal H}^n(A,V) \cong {\mathcal H}^n(A,X) \oplus {\mathcal H}^n(A,Y) \quad\hbox{for all $n$.} $$
One can check this directly or appeal to the long exact sequence of Hochschild cohomology (which is a special case of the one for relative Ext).
Finally, for each Banach $A$-bimodule $X$ there is a canonical $A$-bimodule map $J:X\to V\_X$ which is defined by $[J(x)](c\otimes d) = dxc$. Therefore:
>
> If there exists an $A$-bimodule map $P:V\_X\to X$ such that $PJ$ is the identity, then ${\mathcal H}^n(A, X)=0$ for all $n\geq 1$.
>
>
>
Clowns to the left of me, jokers to the right
---------------------------------------------
As may be apparent to anyone who's read this far: the two conditions we have obtained on $X$, each of which implies that Hochschild cohomology with coefficients in $X$ vanishes, are one and the same condition. [The calculations in the second version actually show that $V\_E$ is $A$-bi-injective - meaning the same as $A^e$-injective. This relied on $A$ having an identity element! Then, knowing that a complemented submodule of an injective module is injective, we see that the second condition implies $X$ is $A$-bi-injective.] The nice thing about the direct approach is that it gives one explicit formulas one can try even in settings where the coefficient module is not bi-injective (see for instance [my first excuse for a paper](http://arxiv.org/abs/math.FA/0606364) ). Personally I think it is good to have both points of view.
It should lastly be noted that almost none of the above actually used analysis - everything is taken care of by working in a particular category with a particular tensor product. So what I have just written out is no more than was known at the time of Cartan-Eilenberg.
| 4 | https://mathoverflow.net/users/763 | 43457 | 27,630 |
https://mathoverflow.net/questions/43423 | 3 | Dudley's theorem (1966) states that if $(X, d)$ is a metric space and if $X$ is separable and $\mu$, $\mu\_i$ are Borel probability measures then $\mu\_i \to \mu$ narrowly iff $d\_{\text{BL}}(\mu\_i, \mu) \to 0$ where $d\_{\text{BL}}$ is the bounded Lipschitz metric.
**Definitions:**
$(\mu\_i)$ converges narrowly to $\mu$ (where all measures are Borel probability measures) if
$$\int f \, d\mu\_i \to \int f \, d\mu \text{ for all $f$ bounded and continuous on $X$}$$
The bounded Lipschitz metric is a metric on the space $\text{BL}(X,d) := \{f : X \to \mathbb{R} : f \text{ is bounded and Lipschitz} \}$. Then define
$$d\_\text{BL}(\mu, \nu) := \sup \left \{ \left | \int f \, d\mu - \int f \, d\nu\, \right | : f \in \text{BL}(X,d), \|f\|\_\text{BL} \leq 1 \right \}$$
where $\|f\|\_{\text{BL}}$ is the sum of the Lipschitz-norm and the $\infty$-norm.
The proof uses Arzela-Ascoli, but I wonder what would be a counterexample if $X$ isn't separable? From right-to-left still works.
| https://mathoverflow.net/users/5295 | Does Dudley's theorem hold for nonseparable metric spaces? | Let $X$ be a set with $2$-valued measurable cardinal. (Real-valued measurable can also be done, but with some more complications, so I do not do that now.) Give it the discrete metric. Let $\mu$ be a countably-additive measure on the Borel sets (i.e., the power set) with values $0$ and $1$ such that each singleton has measure $0$ but $\mu(X) = 1$. There is a net $\mu\_i$ of point-masses converging to $\mu$ narrowly, but not in the BL metric.
A **point-mass** is a measure that assigns measure $1$ to a certain singleton, and measure $0$ to the complement. As long as our net of point-masses is eventually outside each set of measure $0$, we have convergence to $\mu$ in the narrow topology. But any point-mass $\mu\_i$ at the point $a\_i$ is far away from $\mu$ in the BL topology, since the indicator function of the singleton $a\_i$ is a BL function with norm $2$.
Another note. For any bounded function $f \colon X \to \mathbb{R}$, there is a set $F\subseteq X$ with $\mu(F)=1$ and $f$ is constant on $F$; the constant value there is the integral $\int f d\mu$.
| 5 | https://mathoverflow.net/users/454 | 43460 | 27,631 |
https://mathoverflow.net/questions/43462 | 23 | By Borel's theorem, for any sequence of real numbers $a\_n,$ there is a $C^{\infty}$-function
$f:\mathbb{R}\to\mathbb{R}$ whose Taylor series at 0 is $\sum a\_nx^n.$ In particular, there are $C^{\infty}$-functions whose Taylor series at a point has 0 radius of convergence.
Motivated by this I have the following question. Is there a $C^{\infty}$-function
$f:\mathbb{R}\to\mathbb{R}$ whose Taylor series has 0 radius of convergence at every
point in $\mathbb{R}?$ I realize that this might sound like a homework problem, but, well,
it's not.
| https://mathoverflow.net/users/8257 | Existence of a smooth function with nowhere converging Taylor series at every point | S.S. Kim and K.H. Kwon gave an explicit example of a *monotone* smooth but nowhere analytic function ([link](http://www.jstor.org/pss/2589322)), which is an anti-derivative of the function
$$\psi(x)=\sum\limits\_{k=1}^{\infty} \frac{1}{k!}\phi(2^k(x-[x])),$$
where
$$\phi(x) = \begin{cases} \exp{\left(-\frac{1}{x^2}-\frac{1}{(x-1)^2}\right)},\qquad & 0 < x < 1,\ \\\ \\\ 0, & \mbox{otherwise.} \end{cases}$$
In fact, the set of smooth but nowhere analytic functions on $\mathbb R$ is of the second category in $C^{\infty}(\mathbb R)$ (just like the set of all continuous but nowhere differentiable functions is of the second category in $C(\mathbb R)$). See a one page note by R. Darst ["*Most infinitely differentiable functions are nowhere analytic*"](http://books.google.co.uk/books?id=k86bShPCxfUC&pg=PA597&lpg=PA597&dq=%2522Most+infinitely+differentiable+functions+are+nowhere+analytic%2522&source=bl&ots=5LvFhmFVwR&sig=U5IxIiA6VBo_W1oPcuIAPqrrbHQ&hl=en&ei=Ff7ETL3cNcufOpW5xYUM&sa=X&oi=book_result&ct=result&resnum=7&ved=0CDcQ6AEwBg#v=onepage&q=%2522Most%2520infinitely%2520differentiable%2520functions%2520are%2520nowhere%2520analytic%2522&f=false).
**Edit.** Kim and Kwon mention in their paper that the first concrete example of smooth
but nowhere analytic function dates back to A Pringsheim ([*"Zur Theorie der Taylor'schen Reihe und der analytischen Functionen mit beschränktem Existenzbereich."*](http://www.ams.org/mathscinet-getitem?mr=1510771) *Math. Ann.* 42 (**1893**), no. 2, 153–184.)
| 20 | https://mathoverflow.net/users/5371 | 43468 | 27,635 |
https://mathoverflow.net/questions/43466 | 8 | Wikipedia claims that the group of units of Z24 (1,5,7,11,13,17,19,23), which all have order 2, and are isomorphic to (Z/2Z)^3 have an important connection to Monstrous Moonshine theory, however, I cannot find any other reference besides Wikipedia that claims this --- It was recommended on sci.math that I pose this question here.
Perhaps it's a mistake? And he meant that the primes of the Monster, which continue to 71, are what are considered in Moonshine.
Paul Hjelmstad, B.M, B.A.
[**Edit** (PLC): Here is the relevant passage from wikipedia:]
>
> 24 is the highest number $n$ with the property that every element of the group of units $(\mathbb{Z}/n\mathbb{Z})^{\times}$ of the commutative ring $\mathbb{Z}/n\mathbb{Z}$, apart from the identity element, has order $2$; thus the multiplicative group $(\mathbb{Z}/24\mathbb{Z})^{\times} = \{1,5,7,11,13,17,19,23\}$ is isomorphic to the additive group $(\mathbb{Z}/2\mathbb{Z})^3$. This fact plays a role in monstrous moonshine.
>
>
>
| https://mathoverflow.net/users/10350 | Monstrous Moonshine | $\newcommand{\Q}{\mathbf Q} \newcommand{\Z}{\mathbf Z}$
I don't know about monstrous moonshine, but $(\Z/24\Z)^\times$ is the group of automorphisms of the maximal elementary abelian $2$-extension $\Q\_2\left(\root2\of{\Q\_2^\times}\right)=\Q\_2(\root2\of5, \root2\of3, \root2\of2)=\Q\_2(\zeta\_{24})$ of $\Q\_2$. See for example Lemma 8 of Lecture 19 of my [course](http://arxiv.org/abs/0903.2615) on Local arithmetic.
| 4 | https://mathoverflow.net/users/2821 | 43476 | 27,640 |
https://mathoverflow.net/questions/43454 | 0 | The usual action of $fg$ on $u⊗v$ , where $f,g$ are elements in the Universal Enveloping Algebra $U(G)$ of a Lie algebra $G$ and $u,v$ are elements of a representation $V$ of $G$, is given by $fg(u⊗v)=fgu⊗v+fu⊗gv+gu⊗fv+u⊗fgv$, using the comultiplication, right?
How to state this fact for $V^{\otimes n}$, i.e. $fg$ acting on $u⊗v$, where $u=\otimes\_{i=1}^{n-k} u\_i$ and $v=\otimes\_{i=1}^k v\_i$, for each $k=1,...,n-1$ ?
Thanks,
| https://mathoverflow.net/users/40886 | How to define the action of $U(G)$ in this situation? | There is nothing deep here. The coproduct $\Delta : U(\mathfrak g) \to U(\mathfrak g)\otimes U(\mathfrak g)$ simply implements Leibniz's product rule: if $v\_1\in V\_1$ and $v\_2\in V\_2$, then $x\in \mathfrak g$ acts on $V\_1\otimes V\_2$ by $x: v\_1\otimes v\_2 \otimes (xv\_1)\otimes v\_2 + v\_1\otimes (xv\_2)$. Extending this, on a large tensor product we have:
$$ x(v\_1 \otimes v\_2 \otimes \cdots \otimes v\_n) = xv\_1 \otimes v\_2 \otimes \cdots \otimes v\_n + v\_1 \otimes xv\_2 \otimes \cdots \otimes v\_n + \cdots + v\_1 \otimes v\_2 \otimes \cdots \otimes xv\_n$$
If $x,y\in \mathfrak g$, then $xy \in U(\mathfrak g)$ acts via the composition $x\circ y$. For example,
$$ xy \left( \bigotimes\_{k=1}^n v\_k\right) = \sum\_{k=1}^n (xy\text{ acts on }v\_k) + \sum\_{j\neq k}(x\text{ acts on }v\_j\text{ and }y\text{ acts on }v\_k)$$
Note that in general, for $f\in U(\mathfrak g)$, we do not have $f(u\otimes v) = fu \otimes v + u\otimes fv$. For example, the constant $1$ is an element of $U(\mathfrak g)$, and $1(u\otimes v) = u\otimes v \neq 2u\otimes v = 1u\otimes v + u\otimes 1v$. More generally, not all differential operators are derivations: certainly you do not expect $\frac{\partial^2}{\partial x^2}$ to satisfy a Leibniz rule!
| 3 | https://mathoverflow.net/users/78 | 43485 | 27,644 |
https://mathoverflow.net/questions/43483 | 0 | I have a sequence of matrices $\lbrace A\_i \rbrace\_{i=1}^N$ and I want to select a column from each of these matrices so that the following sum is minimized:
$\sum\_{i=1}^N || A\_{i} \vec{x\_{i}}- A\_{i+1} \vec{x}\_{i+1} ||\_2^2$
$\vec{x}\_i$ is a binary vector which selects a column of $A\_i$. Formally: $x\_{ij} \in \lbrace 0,1 \rbrace$ for $\forall i, j$ and $\sum\_j x\_{ij} = 1$ for $\forall i$.
How can I tackle this problem? Any hints or resources?
| https://mathoverflow.net/users/5223 | How to solve this integer programming problem? | I think I found a solution using Dijkstra's shortest path algorithm. I would appreciate if anybody could check my solution.
Construct a graph as follows:
1. Create a starting node $s$ and connect it to each column of $A\_1$. The cost of these connections are all the same and equal to some arbitrary constant.
2. Create a terminal node $t$ and connect it to each column of $A\_N$. The cost of these connections are all the same and equal to some arbitrary constant.
3. Connect the $j^{th}$ column of $A\_i$ to the $k^{th}$ column of $A\_{i+1}$ with a cost of $|| A\_i \vec{y\_i} - A\_{i+1}\vec{y}\_{i+1}||\_2^2$ where $\vec{y\_i}$ is all zeros except its $j^{th}$ element, and $\vec{y}\_{i+1}$ is all zeros except its $k^{th}$ element.
4. Compute the shortest path between $s$ and $t$. Then, it is trivial to determine $\vec{x\_i}$s from the selected nodes in the solution.
| 1 | https://mathoverflow.net/users/5223 | 43486 | 27,645 |
https://mathoverflow.net/questions/43489 | 11 | Hi! This is my first post on Math Overflow. I have two equations: $a(3a-1) + b(3b-1) = c(3c-1)$ and $a(3a-1) - b(3b-1) = d(3d-1)$. I'm trying to find properties of $a$ and $b$ that lead to solutions, where $a, b, c, d \in \mathbb{N}$. I'm having trouble applying any of the techniques in my abstract algebra book, as they mostly only apply to linear Diophantine equations.
So far, I only really have managed to deduce the following things:
$2b(3b-1) + d(3d-1) = a(3a-1) +b(3b-1)$
$2b(3b-1) = c(3c-1) - d(3d-1)$
$2b(3b-1) = (c-d)(3(c+d)-1)$
Any ideas on where to go from here would be greatly appreciated. Thanks!
| https://mathoverflow.net/users/10313 | Analysis of a quadratic diophantine equation | One thing to do is to try to express these in terms of squares. Note that
$$12x(3x-1)=36x^2-12x=(6x-1)^2-1$$
so that your equations become
$$a\_1^2+b\_1^2=c\_1^2+1$$
and
$$a\_1^2-b\_1^2=d\_1^2-1$$
where $a\_1=6a-1$ etc. Then the variables $a\_1$ etc are constrained to be
congruent to $5$ modulo $6$.
Homogenizing these gives
$$X^2+Y^2=Z^2+T^2$$
and
$$X^2-Y^2=Z^2-T^2.$$
Searching for rational solutions of your equation is essentially looking
for rational points on the intersection of these two quadrics in
$\mathbf{P}^3$. In general the intersection of two quadrics in
$\mathbf{P}^3$ is an elliptic curve, so it looks like your
problem will boil down to something like finding the integer points on
an elliptic curve.
**Added**
There's a blunder in the above: I must thank Fedor for noticing
that the second equation should be
$$X^2-Y^2=W^2-T^2.$$
So the variety is the intersection of two quadrics in
$\mathbf{P}^4$. Hartshorne mentions in passing that in general
this construction gives a del Pezzo surface. Del Pezzo surfaces are rational
so there should be a birational parametrizion (in terms of two
affine parameters) of the **rational** solutions to the original
pair of equations.
| 5 | https://mathoverflow.net/users/4213 | 43490 | 27,647 |
https://mathoverflow.net/questions/43459 | 0 | Hi,
I'm attempting to implement a QR factorization with column pivoting so that the returned R matrix has decreasing diagonal elements (that is, $r\_{i,i} \leq r\_{i-1,i-1}$ for all $i\geq 2$). Mathematically, it would involve finding the matrix $P$ so that $AP=QR$, or $A=QRP^T$.
I'm using Gram-Schmidt to compute QR.
One source I found was this: <http://www.mathworks.de/matlabcentral/newsreader/view_thread/250632>
Quote from the answer:
"During the iteration #k, it is easy to show that if we pick among the set of remaining vectors (i.e., not yet included in the span) the vector that has the *largest* orthogonal component to the current subspace (generated by k first vectors), then the diagonal of R must decrease."
Sounds simple enough, but HOW do I determine which vector of the remaining ones that have the largest orthogonal component to the subspace that's already been found?
Thanks a lot in advance for any help on this matter!
Best regards
Hallgeir
| https://mathoverflow.net/users/10309 | QR factorization: How to get decreasing r_ii | At each step $k$, choose the column of the "reduced" working matrix $A(k:n,k:n)$ with largest Euclidean norm and bring it in front with a permutation. Notice that $r\_{11}$ is the Euclidean norm of the first column...
| 0 | https://mathoverflow.net/users/1898 | 43492 | 27,648 |
https://mathoverflow.net/questions/43464 | 16 | I'm an undergrad who is taking a Complex Analysis Course mainly for its applications in number theory.
So I would like to ask some guidelines about which theorems/concepts should I focus on in order to develop a narrower path for self study.
In addition, it would be helpful to know if there is a book that does a good job showing off how the
complex analysis machinery can be used effectively in number theory,
or at least one with a good amount of well-developed examples in order to provide a wide background of the tools that complex analysis gives in number theory.
| https://mathoverflow.net/users/3937 | Complex Analysis applications toward Number Theory | This question is like asking how abstract algebra is useful in number theory: lots of it is used in certain areas of the subject so there's no tidy answer. You probably won't be using Morera's theorem directly in number theory, but most of single-variable complex analysis is needed if you want to understand basic ideas in analytic number theory. A few topics you should pay attention to are: the residue theorem, the argument principle, the maximum modulus principle, infinite product factorizations (esp. the Hadamard factorization theorem), the Fourier transform and Fourier inversion, the Gamma function (know its poles and their residues), and elliptic functions. Basically pay attention to the whole course! There really isn't a whole lot in a first course on complex variables where one can say "that you should ignore if you are interested in number theory".
If you want to be careful and not just wave your hands, you need to know conditions that guarantee the convergence of series and products of analytic functions (and that the limit is analytic), the existence of a logarithm of an analytic function (it's not the composite of the three letters "log" and your function), that let you reorder terms in series and products, that justify termwise integration, and of course the workhorse of analysis: how to make good estimates.
| 23 | https://mathoverflow.net/users/3272 | 43498 | 27,653 |
https://mathoverflow.net/questions/43478 | 60 | A subset of ℝ is [meagre](http://en.wikipedia.org/wiki/Meagre_set) if it is a countable union of nowhere dense subsets (a set is nowhere dense if every open interval contains an open subinterval that misses the set).
Any countable set is meagre. The Cantor set is nowhere dense, so it's meagre. A countable union of meagre sets is meagre (e.g. all rational translates of the Cantor set).
There can also be meagre sets of positive measure, like "fat Cantor sets". To form a fat Cantor set, you start with a closed interval, then remove some open interval from the middle of it, then remove some open intervals from the remaining intervals, and so on. The result is nowhere dense because you removed open intervals all over the place. If the sizes of the intervals you remove get small fast, then the result has positive measure.
So does meagreness have any connection at all to measure? Specifically, are all measure zero sets meagre?
| https://mathoverflow.net/users/1 | Is there a measure zero set which isn't meagre? | On the relation between null sets and meagre sets, you can also look at
[this article](http://www.artsci.kyushu-u.ac.jp/~ssaito/eng/maths/duality.pdf).
Two theorems mentioned in this note (both classical and not due to the author):
1. (As already mentioned above) There exist a meagre $F\_\sigma$ subset $A$ and a null $G\_\delta$ subset $B$ of $\mathbb R$ that satisfy $A\cap B=\emptyset$ and $A\cup B=\mathbb R$.
2. (The Erdős-Sierpiński Duality Theorem) Assume that the Continuum Hypothesis
holds. Then there exists an involution (bijection of order two) $f:\mathbb R\to\mathbb R$ such that $f[A]$
is meagre if and only if $A$ is null, and $f[A]$ is null if and only if $A$ is meagre for every subset $A$ of $\mathbb R$.
While (1) says that the ideals of null, respectively meager sets are "orthogonal", (2) says that assuming CH they behave identically. But it is well known that this duality between measure and category fails dramatically once we take a more abstract point of view: Shelah proved that you need large cardinals to construct a model of set theory (ZF, no axiom of choice) where every set of reals is Lebesgue measurable, but no large cardinals are necessary to construct a model where every set of reals has the Baire property (the corresponding notion to measurability for category).
| 35 | https://mathoverflow.net/users/7743 | 43502 | 27,655 |
https://mathoverflow.net/questions/43499 | 2 | Let R be a local ring, I an ideal, M a finitely generated module and $N=\cap \_nI^nM$. Then the Krull intersection theorem states that $N=IN$. Now if R is a local ring of characteristic $p>0$, for each $e\geq 0$ let $I^e$ denote the ideal generated by the $p^e$-th power of the elements of $I$. let $N=\cap \_eI^eM$. Is it ture that $N=I^0N$?
| https://mathoverflow.net/users/5775 | A question arising from the Krull intersection theorem. | First Krull's theorem is for Noetherian (not necessarily local) rings. Let $n\ge 1$. If $I$ is generated by $r$ elements $x\_1, \dots, x\_r$, then the usual $n$-power $I^n$ of $I$ is contained in your $I^e$ if $n/r \ge p^e$ (any element of $I^n$ is a combination of $x\_1^{a\_1}...x\_r^{a\_r}$ with $a\_1+\dots + a\_r=n$, so $\max\_i{a\_r} \ge n/r$). Therefore your $N$ is equal to the usual $\cap\_n I^nM$ and the answer to your question is yes.
| 5 | https://mathoverflow.net/users/3485 | 43503 | 27,656 |
https://mathoverflow.net/questions/43438 | 5 | This is something of a follow-up question to [this one](https://mathoverflow.net/questions/43255/); I hope people won't think this is a duplicate. At least in my head, it seems like a distinct enough question to merit a fresh start.
All my schemes will be finite type over an algebraically closed field $k$. Let $X\to S$ be a flat family of affine schemes over smooth affine base. Let's say for now that each fiber and the whole family have rational singularities, and thus are Cohen-Macaulay. Assume, furthermore, that $X$ has an action of the group scheme $T=(\mathbb{G}\_m)\_S$; this is the same data as a grading on $k[X]$ such that $k[S]$ has degree 0.
Now, we can take the schematic fixed points $X^T$ of this family, which is a subscheme of $X$ whose points over any ring are invariant points of $X$. Concretely, this is the vanishing set of the ideal generated by all functions of non-zero degree.
>
> Must the morphism $X^T\to S$ be flat? If not, are there stricter hypotheses than I gave above which would assure it is?
>
>
>
For example, consider the family $$X=\mathrm{Spec}[x,y,z,a\_0,\dots, a\_{n-1}]/(xy=z^n+a\_{n-1}z^{n-1}+\cdots + a\_0)$$ where $S=\mathrm{Spec}[a\_0,\dots, a\_{n-1}]$ with $x$ having degree 1, $y$ degree $-1$ and $z,a\_i$ having degree 0. In this case $$X^T=\mathrm{Spec}[z,a\_0,\dots, a\_{n-1}]/(z^n+a\_{n-1}z^{n-1}+\cdots + a\_0=0),$$ which is, of course, flat over $S$, even though the number of closed points in a fiber (the number of roots of $z^n+a\_{n-1}z^{n-1}+\cdots + a\_0$) varies from $n$ to $1$.
| https://mathoverflow.net/users/66 | Under what hypotheses are schematic fixed points of a flat deformation themselves flat? | Here is a counterexample. Let $\mathbb G\_{\rm m}$ act on $\mathbb A^2$ by $t\cdot(x,y) = (tx,t^{-1}y)$, and let $f\colon \mathbb A^2 \to \mathbb A^1$ be defined by $f(x,y) = xy$.
I am positive that when $X$ is smooth over $Y$, the fixed point scheme is also smooth; but I doubt that one can say much more, in general.
[Edit] Here is a variant. Let $\mathbb G\_{\rm m}$ act on $\mathbb A^4$ by $t\cdot(x,y, z, w) = (tx,t^{-1}y,tz,t^{-1}w)$, and let $f\colon \mathbb A^4 \to \mathbb A^1$ be defined by $f(x,y) = xy + zw$.
| 8 | https://mathoverflow.net/users/4790 | 43523 | 27,665 |
https://mathoverflow.net/questions/43514 | 15 | Let us have a symmetric matrix $C \in \mathbb{R}^{n\times n}$ having non-negative values. Suppose that we have the eigenvalue decomposition for this particular matrix such that
$$C e\_i = \lambda\_i e\_i$$
where $e\_i$ are the eigenvectors and $\lambda\_i \geq 0$ are the corresponding eigenvalues. In matrix form,
$$CE = ES$$ where $E$ is the matrix involving eigenvectors as columns and $S$ is the diagonal matrix involving eigenvalues on the diagonal entries.
Now, we delete $k^\text{th}$ row and column of the matrix and form a new matrix $\tilde{C} \in \mathbb{R}^{(n-1) \times (n-1)}$ and we want to find its eigenvalues and eigenvectors such that
$$\tilde{C} \tilde{E} = \tilde{E} \tilde{S}$$
Instead of computing them from scratch, I wonder if there exists an analytical way to find the eigenvectors and eigenvalues iteratively using $E$ and $S$. In other words, is there a link between $E$, $S$ and $\tilde{E}$, $\tilde{S}$?
For example, for a 3-by-3 matrix $C$ where
$$C = \left[\begin{array}{ccc}a & b & c \\\\
b & d & e \\\\
c & e & f\end{array}\right]$$
if I delete the third row and column, then I get
$$C = \left[\begin{array}{cc}a & b \\\\
b & d \end{array}\right]$$
I know that the number of dimensions of the eigenvectors are one less and we have one less eigenvalues but may there be a projection of the others onto some bases?
**P.S.(1)** I've read the answers to the question ["How does eigenvalues and eigenvectors change if the original matrix changes slightly"](https://mathoverflow.net/questions/20492/how-does-eigenvalues-and-eigenvectors-change-if-the-original-matrix-changes-sligh), but I couldn't find a connection with this question. Sorry if I couldn't get a point and created a duplicate question.
**P.S.(2)** If you wonder why I'm asking this question, here it is: I'm computing the eigenvalues and eigenvectors of the covariance matrix of some samples. Let each sample have 3 dimensions and let $D \in \mathbb{R}^{k\times n}$ be the data matrix where each row is a sample and we have $k$ samples. Then the covariance matrix is $C = D^T D$. If we have 3 dimensions and the columns of $D$ are $d\_1$, $d\_2$ and $d\_3$, then we have
$$ C = \left[\begin{array}{ccc}d\_1^T d\_1 & d\_1^T d\_2 & d\_1^T d\_3 \\\\
d\_2^T d\_1 & d\_2^T d\_2 & d\_2^T d\_3 \\\\
d\_3^T d\_1 & d\_3^T d\_2 & d\_3^T d\_3\end{array}\right]$$
I'm interested in the decomposition when we have some dimensions missing in the data. If, as in the example, the third dimension is missing, then we have
$$ \tilde{C} = \left[\begin{array}{cc}d\_1^T d\_1 & d\_1^T d\_2 \\\\
d\_2^T d\_1 & d\_2^T d\_2 \end{array}\right]$$
I wondered whether there's a way to compute the eigenvectors and eigenvalues of the missing data's covariance matrix using the ones that we computed from the full data in an iterative manner instead of computing it from scratch.
Kind regards and thanks for any ideas.
| https://mathoverflow.net/users/5287 | How do eigenvectors and eigenvalues change when we remove a row/column pair of a matrix? | Let $A$ be a symmetric matrix, with eigenvalues $\lambda\_1 \leq \lambda\_2 \leq \cdots \leq \lambda\_n$. Let $B$ be the matrix obtained by deleting the $k$-th row and column from $A$, with eigenvalues $\mu\_1 \leq \mu\_2 \leq \cdots \leq \mu\_{n-1}$. Then
$$\lambda\_1 \leq \mu\_1 \leq \lambda\_2 \leq \mu\_2 \leq \lambda\_3 \leq \cdots \leq \lambda\_{n-1} \leq \mu\_{n-1} \leq \lambda\_{n}.$$
This is a special case of [Cauchy's interlacing theorem](https://en.wikipedia.org/wiki/Min-max_theorem#Cauchy_interlacing_theorem). The operator $P$ in the wikipedia article should be taken to be the projection on the coordinate vectors other than the $k$-th one.
| 20 | https://mathoverflow.net/users/297 | 43526 | 27,666 |
https://mathoverflow.net/questions/43521 | 7 | I was listening to a talk about ultraproducts and one result there suggested, that every finitely generated group can be written as a colimit of residually finite groups (over a directed system).
As I don't see a trivial proof, I expect it to be false (it is a statement about all groups). But I don't see a counterexample.
| https://mathoverflow.net/users/3969 | Is every finitely generated group colimit of residually finite groups | The answer is: This does not hold in general.
If the group in question is finitely generated, then the maps into the colimit will eventually be surjective. If the group in question is also finitely presented, then eventually, all relations will hold in the groups in the colimit diagram. Hence, you can split the surjection and see the colimit group as a subgroup of one of the groups in the diagram.
Hence, your assertion would imply that every finitely presented group *is* residually finite (since this property passes to subgroups), which of course is wrong. Higman's group is the classical counterexample:
$$H:= \langle a,b,c,d \mid ba = a^2b, cb=b^2c, dc=c^2d, ad=d^2a \rangle.$$
It has no finite quotients and is finitely presented.
| 12 | https://mathoverflow.net/users/8176 | 43527 | 27,667 |
https://mathoverflow.net/questions/43529 | 6 | Let $K$ be a perfect field. In what follows, an algebraic group $G/K$ is by definition a group scheme of finite type over $K$.
The following seems to be well-known:
**Theorem:** Let $G/K$ be a connected smooth algebraic group. Then there is a connected smooth *affine* normal closed subgroup $N$ of $G$, an abelian variety $A/K$ and a homomorphism $G\to A$ with kernel $N$ such that the sequence
$$0\to N\to G\to A\to 0$$
is exact for the fppf-topology (say).
Can someone give me a proper reference or a hint why this is true? (Checking the literature I find on the one hand plenty of references treating affine algebraic groups, and on the other hand references containing the theory of abelian varieties, but I was surprised not to find a reference containing a proof of this Theorem about the "mixed case".)
| https://mathoverflow.net/users/8680 | Decomposition of an algebraic group in an affine and a proper part | This question is answered in the [Wikipedia page for algebraic groups](http://en.wikipedia.org/wiki/Algebraic_group). The article says that this is a difficult result of Chevalley, and it has a link to a [modern write-up by Brian Conrad](http://math.stanford.edu/~conrad/papers/chev.pdf) of Chevalley's result. So that is surely a proper reference. No particular hint leaps out at me for "why" it is true, but I can say something about what the proof is really saying. The subgroup $N$ appears as the common kernel of all algebraic homomorphisms from $G$ to all abelian varieties. So it is a functorial construction and $N$ is actually a characteristic subgroup, not just a normal subgroup. Relatively early on it is shown that there are no non-trivial algebraic homomorphisms from an affine group to an abelian variety, in fact not even any non-trivial algebraic morphisms that don't have to be homomorphisms.
It is also relatively quick to show that $G/N$ is an abelian variety. The really hard part is to show that the kernel is affine. You might as well let $G = N$ and you might as well let $K$ be algebraically closed. The hard theorem is that if $G$ does not have any non-trivial homomorphisms to an abelian variety, then it is affine. It is important to remember that $G$ is affine if and only if it is linear, i.e., an algebraic subgroup of $\text{GL}(V)$ for some finite-dimensional vector space $V$. This vector space $V$ is the most difficult construction of the paper.
| 13 | https://mathoverflow.net/users/1450 | 43534 | 27,673 |
https://mathoverflow.net/questions/43537 | 6 | Suppose $G$ is a discrete group and $H \leq G$ a subgroup of finite index. If $H$ has Kazhdan property (T), does it follow that $G$ has property (T)? (I've read somewhere that (T) is preserved by exact sequences, so if $N$ is normal and $G/N$ is finite, then the fact above holds ; here, however, we do not assume $H$ to be normal)
| https://mathoverflow.net/users/1121 | Property (T) and subgroups of finite index | Yes. For groups $H\subset G$, with H a lattice, H has (T) iff G has (T). When both groups are discrete being a lattice is the same as being finite index.
Almost every thing you ever need to know about Property (T) can be found here
<http://perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf>
| 8 | https://mathoverflow.net/users/5732 | 43540 | 27,674 |
https://mathoverflow.net/questions/43550 | 13 | This question was suggested when trying to find an explicit example of a continuous function with compact support in $\mathbb{R}$ whose Fourier transform is not integrable. The existence of such a function was proved by an abstract argument in this [MO question](https://mathoverflow.net/questions/3764/does-there-exist-a-continuous-function-of-compact-support-with-fourier-transform), but no explicit example was given. It is clear that such a function cannot be differentiable everywhere.
The functions $(1-x^2)^{\alpha}\\_{+}$ for $0<\alpha<1$ look like reasonable candidates, since they have a singularity at $\xi=\pm1$ and converge as $\alpha\to0$ to the characteristic function of $[-1,1]$, whose Fourier transform is not integrable. However $\hat f\_{\alpha}(\xi)=O(|\xi|^{-(1+\alpha)})$ as $|\xi|\to\infty$, and hence is integrable.
For any $\alpha,\beta>0$, the function
$$g\_{\alpha,\beta}(x)=(1-\beta\log(1-x^2))^{-\alpha}\hbox{ if }|x|<1,\quad 0 \hbox{ if } |x|\ge1$$
is more singular than any $f\_{\alpha}$. So my questions are:
* Is $\hat g\_{\alpha,\beta}$ is integrable?
* What is the asymptotic behaviour of $\hat g\_{\alpha,\beta}$ at infinity?
I suspect, based on numerical calculations, that the answer to the first question is no, at least for sufficiently small $\alpha$ and $\beta$.
| https://mathoverflow.net/users/1168 | Is the Fourier transform of 1/(1-log(1-x^2)) (supported in [-1,1]) integrable? | If one performs a smooth dyadic decomposition of $g\_{\alpha,\beta}$ around the singularities $x = \pm 1$ (i.e. using smooth partitions of unity to decompose $g\_{\alpha,\beta}$ into pieces that are localised in the region $1-|x| \sim 2^{-n}$ for $n \geq 0$), and then takes the Fourier transform of these pieces, one soon arrives at the conclusion that $\hat g\_{\alpha,\beta}$ decays like $\frac{1}{|\xi| \log^\alpha |\xi|}$ (times something like $\sin \xi$) as $\xi \to \infty$ (and one can then back up this intuition with stationary phase, or use some localised form of the Plancherel theorem for an $L^2$ averaged result, which should be enough for the application at hand). So absolute integrability should fail for $\alpha \leq 1$.
| 14 | https://mathoverflow.net/users/766 | 43557 | 27,682 |
https://mathoverflow.net/questions/43554 | 9 | Is there a compass and straightedge construction of parallel lines in hyperbolic geometry?
That is, given a line and a point not on the line, construct a line parallel to the given line.
| https://mathoverflow.net/users/10327 | Is there a compass and straightedge construction of parallel lines in hyperbolic geometry? | The quickest way to get you started is to refer you to my article, reference [5] (a pdf) on
<http://en.wikipedia.org/wiki/Squaring_the_circle>
and then to the fourth edition (2008) of Marvin Jay Greenberg's book, which is reference [6].
I'm guessing what you want is Bolyai's construction, given a line and a point off the line, of the two rays through the point that are asymptotic to the line, one in each direction. When I wrote the article, I relied on an earlier edition of Marvin's book, along with *The Foundations of Geometry and the Non-Euclidean Plane* by George E. Martin, which has a nice little section at the very end. There is also, now, *Geometry: Euclid and Beyond* by Robin Hartshorne.
The most complete reference I know on constructions is in Russian, by Smogorshevski, other very helpful books by Kagan and by Nestorovich. Of course, at this point I have my own versions of it all.
| 14 | https://mathoverflow.net/users/3324 | 43558 | 27,683 |
https://mathoverflow.net/questions/43181 | 1 | If $X\sim \Gamma(a,\sigma\_x^2)$ and $Y\sim \Gamma(b,\sigma\_y^2)$. What will be the probability density function of R? Where $R=\frac{X+C}{X+Y}$, here $C$ is a positive constant, $\Gamma(.,.)$ denotes standard gamma probability density function and '$\sim$' represents 'distributed as'. X and Y are independent random variables.
| https://mathoverflow.net/users/8576 | what will be the distribution of ratio of correlated gamma distributed random variables? | First of all, since $R=(X+C)/(X+Y)$ (and $X$ and $Y$ are independent gamma variables), then the valid range of $R$ is a priori $(0,\infty)$. The density of $(R,S)$ is given by $f\_{R,S} (r,s) = f\_X (sr - C)f\_Y (s - sr + C)s$, where $f\_X$ and $f\_Y$ are the densities of $X$ and $Y$ (see the remark below). This leads to $sr-C > 0$ and $s-sr+C > 0$. Thus, $s > C/r$ and $s(r-1) < C$. Hence, if $r > 1$, then $C/r < s < C/(r-1)$, while if $0 < r < 1$, then $C/r < s < \infty$. So, for $0 < r < 1$ you would use $f\_R (r) = \int\_{C/r}^\infty {f\_{R,S} (r,s)\,{\rm d}s}$, and for $r > 1$, $f\_R (r) = \int\_{C/r}^{C/(r-1)} {f\_{R,S} (r,s)\,{\rm d}s}$. In general, a nice expression for $f\_R (r)$ is not likely to be found. [Minor point: in view of the title, note that $X+Y$ is, in general, not gamma distributed.]
Remark: As Didier observed, a factor of $s$ was missing in the original expression for $f\_{R,S}(r,s)$ (now fixed). The density $f\_{R,S}(r,s)$ is found as follows. $(X,Y)$ has density $f\_{X,Y}(x,y)=f\_X (x) f\_Y (y)$. Noting that $X=SR-C$ and $Y=S-SR+C$, it follows by the standard formula for transformation of rv's that $f\_{R,S} (r,s)$ is given by $f\_{X,Y}(sr-C,s-sr+C)$ times $|J(r,s)|$, where $J(r,s)$ is given by the determinant $\frac{{\partial (sr - C)}}{{\partial r}}\frac{{\partial (s - sr + C)}}{{\partial s}} - \frac{{\partial (sr - C)}}{{\partial s}}\frac{{\partial (s - sr + C)}}{{\partial r}} = s$. Thus, $f\_{R,S} (r,s)$ is $s$ times the original expression.
| 1 | https://mathoverflow.net/users/10227 | 43559 | 27,684 |
https://mathoverflow.net/questions/43535 | 10 | Let $k$ be a field (of char. not $2$) and $X\_k=\text{Spec} (k[x\_1,\cdots,x\_n]/(x\_1^2+\cdots +x\_n^2-1))$. Do we know the Chow groups $A\_i (X\_k)$? I could not find any references, even for $X\_{\mathbb R}$.
What (I think) I know: the K-groups were computed by [Swan](http://www.jstor.org/pss/1971371), so we know the total Chow group up to torsions. In codimension $1$ (i.e., class groups) I am fairly certain the answers are known.
| https://mathoverflow.net/users/2083 | Do we know the Chow groups of spheres? | The book
[The Algebraic and Geometric Theory of Quadratic Forms.](http://www.math.jussieu.fr/~karpenko/publ/Kniga.pdf) by R. Elman, N. Karpenko and A. Merkurjev (American Mathematical Society Colloquium Publications, 56., American Mathematical Society, Providence, RI, 2008. 435 pp.)
contains a lot of information. I was just reading
B. Totaro, The automorphism group of an affine quadric, Math. Proc. Cambridge Philos. Soc. (2007) vol. 143 (1) pp. 1-8
and he is referring to this book for information on Chow groups of spheres.
| 6 | https://mathoverflow.net/users/8176 | 43562 | 27,686 |
https://mathoverflow.net/questions/43569 | 6 | Birkhoff's theorem states:
*The set of $n \times n$ doubly stochastic matrices is a convex set whose extreme points are the permutation matrices*
This theorem seems to be commonly attributed to Birkhoff (perhaps also von Neumann). But I recall listening to a talk by Harold Kuhn, where he said that this theorem should actually be attributed to some $P$ where $P \in \{$Jacobi, Dénes Kőnig, Jenő Egerváry, Somebody else?$\}$.
**Question**: Does anybody know whom Kuhn might have meant, and to whom this theorem should really be attributed?
I would be very happy to learn the connection (also, yes, am embarrassed that despite listening carefully during the talk, I have still forgotten!)
| https://mathoverflow.net/users/8430 | Birkhoff's theorem about doubly stochastic matrices | See the [Wikipedia page](http://en.wikipedia.org/wiki/Birkhoff_polytope) for the Birkhoff polytope. It says that equivalent results were obtained by Steinitz in 1894 and by Kőnig in 1916.
| 9 | https://mathoverflow.net/users/1450 | 43572 | 27,692 |
https://mathoverflow.net/questions/43481 | 4 | Last week, George Lowther provided a rather sophisticated counter-example of a continuous process $\{W(t):t \geq 0\}$ with $W(0)=0$ and $W(t)-W(s) \sim {\rm N}(0,t-s)$ for all $0 \leq s < t$, yet not a Brownian motion; see [link text](https://mathoverflow.net/questions/43015/the-conditions-in-the-definition-of-brownian-motion).
Apparently, his approach relied heavily on special properties of BM.
Now, what about the analogue for a Poisson process: Can you find an example of a càdlàg (right-continuous with left limits) process $\{X(t):t \geq 0\}$ with $X(0)=0$ and $X(t)-X(s) \sim {\rm Poi}(t-s)$ for all $0 \leq s < t$, yet not being a Poisson process?
Bonus question: If the last question turns out to be too easy to answer (etc.), then what about the general Lévy process case? That is, given a Lévy process $X$ (defined below) with law $\mu\_t$ at time $t>0$, does there exist a càdlàg process $\tilde X$ with $\tilde X(0) = 0$ and $\tilde X(t)-\tilde X(s) \sim \mu\_{t-s}$ for all $0 \leq s < t$, which is yet not identical in law to $X$ (hence not a Lévy process)? [Here, assume that $X$ is non-deterministic, equivalently, $\mu\_t$ is not a $\delta$-distribution.]
Definition: A stochastic process $X=\{X(t):t \geq 0\}$ is a *Lévy process* (say, real-valued) if the following conditions are satisfied: (1) $X(0)=0$ a.s.; (2) $X$ has independent increments; (3) $X$ has stationary increments; (4) $X$ is stochastically continuous; (5) Almost surely, the function $t \mapsto X(t)$ is right-continuous (for $t \geq 0$) and has left limits (for $t>0$). [In fact, condition (4) is implied by (1), (3), and (5).]
PS: you are still welcome to try and find a simpler counter-example for the Brownian motion case.
| https://mathoverflow.net/users/10227 | The conditions in the definition of Poisson process (and a Lévy process generalization) | You cannot define a Lévy process by the individual distributions of its increments, except in the trivial case of a deterministic process *X**t* − *X*0 = *bt* with constant *b*. In fact, you can't identify it by the n-dimensional marginals for any n.
>
> 1) Let *X* be a nondeterministic Lévy process with *X*0 = 0 and *n* be any positive integer. Then, there is a cadlag process *Y* with a different distribution to *X*, but such that (*Y**t*1,*Y**t*2,…,*Y**t**n*) has the same distribution as (*X**t*1,*X**t*2,…,*X**t*n) for all times *t*1,*t*2,…,*t**n*.
>
>
>
Taking *n* = 2 will give a process whose increments have the same distribution as for *X*.
The idea (as in my answer to [this related question](https://mathoverflow.net/questions/43015/the-conditions-in-the-definition-of-brownian-motion)) is to reduce it to the finite-time case. So, fix a set of times 0 = *t*0 < *t*1 < *t*2 < … < *t**m* for some *m* > 1.
We can look at the distribution of *X* conditioned on the ℝ*m*-valued random variable *U* ≡ (*X**t*1,*X**t*2,…,*X**t**m*). By the Markov property, it will consist of a set of independent processes on the intervals [*t**k*−1,*t**k*] and [*t**m*,∞), where the distribution of {*X**t* }*t* ∈[*t**k*−1,*t**k*] only depends on (*X**t**k*−1,*X**t**k*) and the distribution of {*X**t* }*t* ∈[*t**m*,∞) only depends on *X**t**m*. By the [disintegration theorem](http://en.wikipedia.org/wiki/Disintegration_theorem), the process *X* can be built by first constructing the random variable *U*, then constructing *X* to have the correct probabilities conditional on *U*. Doing this, the distribution of *X* at any one time only depends on the values of at most two elements of *U* (corresponding to *X**t**k*−1,*X**t**k*). The distribution of *X* at any set of *n* times depends on the values of at most 2*n* values of *U*.
Choosing *m* > 2*n*, the idea is to replace *U* by a differently distributed ℝ*m*-valued random variable for which any 2*n* elements still have the same distribution as for *U*. We can apply a small bump to the distribution of *U* in such a way that the *m* − 1 dimensional marginals are unchanged. To do this, we can use the following.
>
> 2) Let *U* be an ℝ*m*-valued random variable with probability measure μ. Suppose that there exist (non-trival) measures μ1,μ2,…,μ*m* on the reals such that μ1(*A*1)μ2(*A*2)…μ*m*(*A**m*) ≤ μ(*A*1×*A*2×…×*A**m*) for all Borel subsets *A*1,*A*2,…,*A**m* ⊆ ℝ.
> Then, there is an ℝ*m*-valued random variable *V* with a different distribution to *U*, but with the same *m* − 1 dimensional marginal distributions.
>
>
>
By 'non-trivial' I mean that μ*k* is a non-zero measure and does not consist of a single atom.
By changing the distribution of *U* in this way, we construct a new cadlag process with a different distribution to *X*, but with the same *n* dimensional marginals.
Proving (2) is easy enough. As μ*k* are non-trivial, there will be measurable functions ƒ*k* on the reals, uniformly bounded by 1 and such that μ*k*(ƒ*k*) = 0 and μ*k*(|ƒ*k*|) > 0. Replacing μ*k* by the signed measure ƒ*k*·μ*k*, we can assume that μ*k*(ℝ) = 0.
Then
$$
\mu\_V = \mu + \mu\_1\times\mu\_2\times\cdots\times\mu\_m
$$
is a probability measure different from μ. Choosing *V* with this distribution gives
$$
{\mathbb E}[f(V)]=\mu\_V(f)=\mu(f)={\mathbb E}[f(U)]
$$
for any function ƒ: ℝ*m* → ℝ+ independent of one of the dimensions. So, *V* has the same *m* − 1 dimensional marginals as *U*.
To apply (2) to *U* = (*X**t*1,*X**t*2,…,*X**t**m*), consider the following cases.
1. *X* is continuous. In this case, *X* is just a Brownian motion (up to multiplication by a constant and addition of a constant drift). So, *U* is joint-normal with nondegenerate covariance matrix. Its probability density is continuous and strictly positive so, in (2), we can take μ*k* to be a multiple of the uniform measure on [0,1].
2. *X* is a Poisson process. In this case, we can take μ*k* to be a multiple of the (discrete) uniform distribution on {2*k*,2*k* + 1} and, as *X* can take any increasing nonnegative integer-valued path on the times *t**k*, this satisfies the hypothesis of (2).
3. If *X* is any non-continuous Lévy process, case 2 can be used to change the distribution of its jump times without affecting the *n* dimensional marginals: Let ν be its jump measure, and *A* be a Borel set such that ν(*A*) is finite and nonzero. Then, *X* decomposes as the sum of its jumps in *A* (which occur according to a Poisson process of rate ν(*A*)) and an independent Lévy process. In this way, we can reduce to the case where *X* is a Lévy process whose jumps occur at a finite rate, with arrival times given by a Poisson process.
In that case, let *N**t* be the Poisson process counting the number of jumps in intervals [0,*t*]. Also, let *Z**k* be the *k*'th jump of *X*. Then, *N* and the *Z**k* are all independent and,
$$
X\_t=\sum\_{k=1}^{N\_t}Z\_k.
$$
As above, the Poisson process *N* can be replaced by a differently distributed cadlag process which has the same *n* dimensional marginals. This will not affect the *n* dimensional marginals of *X* but, as its jump times no longer occur according to a Poisson process, *X* will no longer be a Lévy process.
| 6 | https://mathoverflow.net/users/1004 | 43585 | 27,701 |
https://mathoverflow.net/questions/43571 | 3 | Consider the space $K$ of all immersions of $S^1$ into $\mathbb R^3$.
The set of knots with self-intersection is a discriminant in $K$ and divide it into "chambers".
Let $f$ be a knot with $n$ double points. Everybody know that neighbourhood of $f$ in $K$ looks like $\mathbb R^n$, the origin be f, and hyperplanes $x\_i=0$ be a knots with self-intersection. (In other words $f$ can be considered as intersection of n planes modelling dicriminant, there are $2^n$ chambers adjacent to $f$, looks like octants in $\mathbb R^n$).
My question: what is it mean? Does there exist a continuous map of neighboorhod $f\in U$ into $\mathbb R^n$ with above prescribed properties?
We can define "tangent" vector for $f$ as piecewise-smooth vector field along $f$. So, does there exist a map from "tangent" vector space of $f$ to $\mathbb R^n$ with above properties? Can we define subspace which is "tangent" to discriminant?
| https://mathoverflow.net/users/4298 | Discriminant locus in knot space | If $X$ is an infinite-dimensional manifold (say a Hilbert or a Frechet manifold), and $A \subset X$, we say $A$ has co-dimension strictly larger than $n$ if for all $n$-dimensional manifolds $N$, the space of smooth maps $f : N \to X$ ( $Map(N,X)$) has as an open and dense subspace maps which are disjoint from $A$.
Basically, this definition is motivated by the truth of the above statement in the finite-dimensional case -- see a differential topology text like Guillemin and Pollack, or Milnor's "Topology from a Differentiable Viewpoint".
edit, continued from the comments above: The notion of co-dimension used by Vassiliev is very much in the spirit of the above comments. When you have a map $D^n \to X$ such that its restriction to $S^{n-1}$ can be perturbed to be disjoint from $A$, yet no small perturbation of the original map can be made to be disjoint from $A$, that's when you are in the co-dimension $n$ setting. A map $D^n \to X$ can't be a neighbourhood of a point in its image since $X$ isn't finite-dimensional, but you can consider such maps to be a "slice" of a neighbourhood of a point in the image. By that I mean you have a map $f : V \to X$ whose image *is* open in $X$, and you decompose $V$ into an $n$-dimensional subspace and a complementary subspace. The the image of the $n$-dimensional subspace under $f$ will be your "resolutions" of your knot-with-multiple (double/triple, etc) points. The complementary subspace will consist of perturbations of the knot complementary to the resolution bump functions.
| 2 | https://mathoverflow.net/users/1465 | 43591 | 27,705 |
https://mathoverflow.net/questions/43237 | 9 | What role do generalized geometries (in terms of Dirac structures, for instance, symplectic, Poisson, complex, and generalized complex structures in the sense of Hitchin, Cavalcanti, and Gualtieri) play in string theory?
EDIT: More generally, what role to Dirac structures (subbundles of the generalized tangent bundle $TM \bigoplus T^\*M$ which are maximally isotropic to the natural pairing and closed under the Courant bracket) play?
| https://mathoverflow.net/users/6527 | Role for generalized geometries in string theory | Let me add something to what David and Urs have written already, since the way those two answers are shaping up, perhaps what I'm about to say does not get mentioned.
One of the most interesting applications of generalised geometry in string theory is in the study of *supersymmetric flux compactifications*. Ten-dimensional superstring theories have a well-defined limit (=the effective theory of massless states) which corresponds to ten-dimensional supergravity theories. One way to view these theories is as variational problems for certain geometric PDEs which generalise the Einstein-Maxwell equations. The dynamical variables consist of a lorentzian ten-dimensional metric and some extra fields, depending on the theory in question. One set of of fields common to the ten- and eleven-dimensional supergravity theories are $p$-forms obeying possibly nonlinear versions of Maxwell equations. In the Physics literature these $p$-form fields are called *fluxes* and the geometric data consisting of the lorentzian manifold, the fluxes and any other fields all subject to the field equations are known as *supergravity backgrounds*.
These supergravity backgrounds are actually not just lorentzian manifolds, but in fact they are spin and part of the baggage of the supergravity theory is a connection (depending on the fluxes,...) on the spinor bundle, which defines a notion of parallel transport. Parallel spinor fields are known as *(supergravity) Killing spinors* and backgrounds admitting Killing spinors are called *supersymmetric*.
One way to make contact with the 4-dimensional physics of everyday experience is to demand that the ten-dimensional geometry be of the form $M \times K$, where $M$ is a four-dimensional lorentzian spacetime (usually a lorentzian spaceform: Minkowski, de Sitter or anti de Sitter spacetimes) and $K$ a compact six-dimesional riemannian manifold, known as the *compactification manifold*.
When all fields, except the metric, are set to zero, the connection agrees with the spin lift of the Levi-Civita connection and supersymmetric backgrounds of this type are lorentzian Ricci-flat manifolds admitting parallel spinor fields. If we demand that they be metrically a product $M \times K$ as described above, then a typical solution is $M$ being Minkowski spacetime and $K$ a six-dimensional manifold admitting parallel spinors; that is, a Calabi-Yau manifold, by which I mean simply a manifold with holonomy contained in $SU(3)$. This result, which today seems quite unassuming, was revolutionary when it was first discovered in the 1985 paper of [Candelas, Horowitz, Strominger and Witten](http://inspirebeta.net/record/16270). That paper is responsible for the interest of physicists in Calabi-Yau manifolds and ensuing *rapprochement* between physicists and algebraic geometers, the fruits of which we're still reaping today.
But Calabi-Yau compactifications are in fact very special from the physics point of view: since most of the fields in the theory (especially the fluxes) have been turned off. Generalised geometry enters in the search for more realistic "flux compactifications". One of the fields which all ten-dimensional supergravity theories have in common is the $B$-*field* (also called Kalb-Ramond field). One of Hitchin's motivations for the introduction of generalised geometry was to give a natural geometric meaning to the $B$-field. For example, the automorphism group of the Courant algebroid $T \oplus T^\*$ is the semidirect product of the group of diffeomorphisms and $B$-field transformations.
More generally, I think that it is still true that all *known* supersymmetric flux compactifications $M \times K$ (even allowing for warped metrics) of ten-dimensional supergravity theories are such that $K$ is a generalised Calabi-Yau manifold. The fluxes turn out to be related to the pure spinors in the definition of a GCY structure. There are many papers on this subject and perhaps a good starting point is [this review by Mariana Graña](http://arxiv.org/abs/hep-th/0509003).
There are other uses of generalised geometry in string theory, e.g., the so-called doubled field theory formalism, as in [this recent paper of Chris Hull and Barton Zwiebach](http://arxiv.org/abs/0908.1792).
| 10 | https://mathoverflow.net/users/394 | 43603 | 27,714 |
https://mathoverflow.net/questions/43586 | 15 | The line bundle $O(-1)$ on a projective space or $O(-\rho)$ on a flag variety has a property that all its cohomology vanish. Is there a story behind such sheaves?
Here are more precise questions. Let $X$ be a smooth complex projective surface (say, a nice one like Del Pezzo or K3). Does there always exist a coherent locally free sheaf $M$ whose derived global sections vanish? Can one describe all such sheaves? Is there a coarse moduli space of such sheaves?
| https://mathoverflow.net/users/5301 | Sheaves without global sections | The bundles with no derived global sections (more generally the objects $F$ of the derived category $D^b(coh X)$ such that $Ext^\bullet(O\_X,F) = 0$) form the left orthogonal complement to the structure sheaf $O\_X$. It is denoted $O\_X^\perp$. This is quite an interesting subcategory of the derived category.
For example, if $O\_X$ itself has no higher cohomology (i.e. it is exceptional) then there is a semiorthogonal decomposition $D^b(coh X) =< O\_X^\perp, O\_X >$. Then every object can be split into components with respect to this decomposition and so many questions about $D^b(coh X)$ can be reduced to $O\_X^\perp$ which is smaller. Further, if you have an object $E$ in $O\_X^\perp$ which has no higher self-exts (like $O(-1)$ on $P^2$), you can continue simplifying your category --- considering a semiorthogonal decomposition $O\_X^\perp = < E^\perp, E >$. For example if $X = P^2$ and $E = O(-1)$ then $E^\perp$ is generated by $O(-2)$, so there is a semiorthogonal decomposition $D^b(coh P^2) = < O\_X(-2), O\_X(-1), O\_X >$ also known as a full exceptional collection on $P^2$. It allows a reduction of many problems about $D^b(coh P^2)$ to linear algebra.
Another interesting question is when $O\_X$ is spherical (i.e. its cohomology algebra is isomorphic to the cohomology of a topological sphere). This holds for example for K3 surfaces. Then there is a so called spherical twist functor for which $O\_X^\perp$ is the fixed subcategory.
Thus, as you see, the importance of the category $O\_X^\perp$ depends on the properties of the sheaf $O\_X$.
| 43 | https://mathoverflow.net/users/4428 | 43626 | 27,731 |
https://mathoverflow.net/questions/43627 | 2 | It is known that
If $f:U→ R$ is a real-valued convex function defined on a convex open set in the Euclidean space $R^n$, a vector v in that space is called a subgradient at a point $x\_0$ in $U$ if for any $x$ in U one has
$f(x)-f(x\_0)\geq v\cdot(x-x\_0)$
What if for function $f$, at any $x\_0$ I can find $v$, such that $f(x)-f(x\_0)\geq v\cdot(x-x\_0)$ for any $x$, does this show that $f$ is convex?
| https://mathoverflow.net/users/10331 | Can subgradient infer convexity? | Yes. Let $x, y \in U$. Let $z = \lambda x + (1 - \lambda) y$, for $\lambda \in [0,1]$. Let $v\_z$ be a subgradient for $z$.
Then $$f(x) \geq f(z) + v\_z \cdot (x - z) = f(z) + v\_z \cdot \left(x - ( \lambda x + (1 - \lambda) y)\right) $$
$$= f(z) + (1 - \lambda) v\_z \cdot (x - y).$$
Similarly,
$$f(y) \geq f(z) - \lambda v\_z \cdot (x - y).$$
Multiplying the first inequality by $\lambda$ and the second by $1 - \lambda$ and adding the two, we obtain
$$\lambda f(x) + (1 - \lambda)f(y) \geq f(z),$$
proving that $f$ is convex.
| 2 | https://mathoverflow.net/users/9716 | 43628 | 27,732 |
https://mathoverflow.net/questions/40365 | 7 | In answer to Pete L. Clark's question [Must a ring which admits a Euclidean quadratic form be Euclidean?](https://mathoverflow.net/questions/39510/) on Euclidean quadratic forms, I also gave an example in six or fewer variables, repeated below. Pete's Euclidean property (in the case of positive definite integral quadratic forms) is simply that for any point $\vec x \in \mathbf Q^n$ but $\vec x \notin \mathbf Z^n,$ we require that there be at least one $\vec y \in \mathbf Z^n$ such that $$ q(\vec x - \vec y) < 1. $$
The question on the example with 7 variables was a big success, see [Verifying an example in the Geometry of Numbers and Quadratic Forms](https://mathoverflow.net/questions/40195/)
Could some kind soul please verify the example(s) below. Note how very symmetric this one is, I have little doubt that the "worst" point(s) must occur on the main diagonal $x\_1 = x\_2 = \cdots x\_n.$ Indeed, I think that for any point the orthogonal projection onto the main diagonal has worse "Euclidean" value.
For six or fewer variables we can use one of the easiest constructions, include all mixed terms so that the Gram matrix becomes
$$ P\_6 \; \; = \; \;
\left( \begin{array}{cccccc}
1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\\
\frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\
\frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} \\\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} \\\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1
\end{array}
\right) . $$
Then the worst $\vec x$ is either
$$ \vec x = \left( \frac{3}{7}, \frac{3}{7}, \frac{3}{7} , \frac{3}{7}, \frac{3}{7}, \frac{3}{7} \right) $$
or
$$ \vec x = \left( \frac{4}{7}, \frac{4}{7}, \frac{4}{7} , \frac{4}{7}, \frac{4}{7}, \frac{4}{7} \right) $$
with ``Euclidean minimum'' $\frac{6}{7}.$
This construction is much easier to figure out. In dimension $ n$ we have determinant $\frac{n +1}{2^n}$ and characteristic polynomial
$$ \left( \frac{1}{2^n} \right) \left(2 x - (n+1) \right) \left(2 x - 1 \right)^{n-1}. $$ For even $n $ the worst $\vec x$ has either all entries $\frac{n}{2(n + 1)}$ or $\frac{n + 2}{2(n + 1)}$ with a Euclidean minimum of $\frac{n^2 + 2 n}{8 (n+1)}.$
For odd $n $ the worst $\vec x$ has all entries $\frac{1}{2}$ with a Euclidean minimum of $\frac{n+1}{8}.$
The formulas $\frac{n^2 + 2 n}{8 (n+1)}$ and $\frac{n+1}{8}$ show that this recipe fails (just barely) for $n=7$ and more obviously for larger $n.$ I don't believe there are any examples with $n \geq 9$ and I have my doubts that there can be any for $n=8.$ I did try half of Gosset's root lattice for $E\_8,$ see
<http://en.wikipedia.org/wiki/E8_lattice>
but it does not seem to work to have any of the squared terms with a coefficient other than $1,$ in all likelihood as soon as $n \geq 4.$
| https://mathoverflow.net/users/3324 | Verifying my other example in the Geometry of Numbers and Quadratic Forms | It's strange that I didn't see this question before, since Will and I have started thinking about these issues off-site. Anyway, recently I found (in the sense of located, not discovered) the answer to this and a bit more.
First, the MAGMA computational system has a built in command to compute the Euclidean minimum of (the lattice associated to) a positive definite integral quadratic form, namely ${ \tt CoveringRadius }$. In the examples I tried, this computation was almost instantaneous up until about $6$ variables, relatively quick for $7$ and $8$ variables, and didn't terminate when run on my local server starting in $9$ variables. Here is how it goes for Will's six dimensional lattice:
>
> $>$ F61 := MatrixRing(IntegerRing(), 6) ! [2,1,1,1,1,1, 1,2,1,1,1,1, 1,1,2,1,1,1\
> , 1,1,1,2,1,1, 1,1,1,1,2,1, 1,1,1,1,1,2];
>
> $>$ L61 := LatticeWithGram(F61);
>
> $>$ L61;
>
> Standard Lattice of rank 6 and degree 6
>
> Determinant: 7
>
> Factored Determinant: 7
>
> Inner Product Matrix:
>
> [2 1 1 1 1 1]
>
> [1 2 1 1 1 1]
>
> [1 1 2 1 1 1]
>
> [1 1 1 2 1 1]
>
> [1 1 1 1 2 1]
>
> [1 1 1 1 1 2]
>
> $>$ (1/2)\*CoveringRadius(L61);
>
> 6/7
>
> $>$ #GenusRepresentatives(L61);
>
> 1
>
> $>$ IsIsomorphic(L61,Lattice("A",6));
>
> true
>
>
>
Some comments: you can see that the Gram matrix is twice the one Will gives. This is because MAGMA's lattice package likes matrices with integral entries. Thus we compute half the covering radius rather than the covering radius to take care of this, and the answer is indeed what Will said it would be: $\frac{6}{7}$. (In particular, this *is* a Euclidean quadratic form.)
The second calculation shows that this lattice has class number one in the sense of quadratic forms, i.e., its genus has only one class. Will, Jon Hanke and I believe that any Euclidean quadratic form should have class number one, but we haven't (yet!) been able to prove this, so this was an interesting data point for that. (I computed nearly $60$ other examples late last week.)
The last calculation shows that Will's lattice (when rescaled by multiplication by $2$) is isomorphic to
the $A\_6$ root lattice. The covering radii for all the root lattices are well known to the experts (though not to me): for instance, the covering radius of $A\_n$ is given on p. 109 of Conway and Sloane's *Sphere Packings, Lattices and Groups* and the formula agrees with the one given in the question above: in particular it is asymptotic to $\frac{n}{8}$. (Of course, in their book they explain what is going on geometrically, which is a much more desirable answer than just running a software package.)
| 4 | https://mathoverflow.net/users/1149 | 43639 | 27,737 |
https://mathoverflow.net/questions/43634 | 5 | I am looking for an example (or definition) of a *quantum probability experiment* (if there is such a thing). Ideally it should have these properties:
1. Be purely mathematical; no mention of physics or other empirical sciences;
2. in the example, all variables should be replaced by constants that are as small or simple as possible without collapsing back into classical probability;
3. it should state what are the analogues of the ingredients of a *classical probability experiment*: sample space $\Omega$, outcome $\omega\in\Omega$, event $E\subseteq\Omega$, probability $\mathbb P(E)$, random variable $X:\Omega\rightarrow V$ where $V=\mathbb R$ or something else.
4. show explicitly how $\sigma$-additivity or finite additivity $\mathbb P(\sqcup\_i A\_i)=\sum\_i \mathbb P(A\_i)$ fails, or is replaced by some other rule, as the case may be. Use of measure theory is welcome.
5. be short enough to fit in an MO answer.
I looked at Greg Kuperberg's draft article [here](http://math.ucdavis.edu/~greg/intro.pdf) and answer [here](https://mathoverflow.net/questions/14803/are-there-nonequivalent-randomnesses/42879#42879), and while I had trouble extracting what I am describing above, it made me hopeful that such a thing might be possible.
| https://mathoverflow.net/users/4600 | Quantum probability experiment? | There is an old example due to Kochen and Specker of sort-of this point, which later was replaced by a much better example due to Bell. (Actually, the Kochen-Specker construction is related to an earlier, stronger result of Andrew Gleason.) Bell's example, after some tidying up, was actually tested in a physical experiment by Alain Aspect, and by many other people since. In the Kochen-Specker thought experiment, finite additivity does not directly fail. Rather, you can exhibit a set of Booleans which cannot generate a traditional Boolean algebra that is consistent with the quantum interpretation. In other words, you cannot have "underlying determinism" that is consistent with natural embeddings of classical probability into quantum probability.
Consider the quantum probability space $M\_3 = M\_3(\mathbb{C})$ of $3 \times 3$ matrices. For each line $L \in \mathbb{C}^3$, there is a corresponding Boolean event given as an operator by the orthonormal projection onto $L$. When Boolean events commute, they generate a commutative algebra, and thus a classical probability space, and one has traditional additivity. So if you have an orthonormal frame of lines, you get a copy of the Boolean algebra of events on three points. However, these Boolean algebras are inconsistent, even just using lines in $\mathbb{R}^3$. There exists a collection of 31 lines in $\mathbb{R}^3$ that do not admit a boolean function $f$ such that exactly one of $f(L\_1)$, $f(L\_2)$, and $f(L\_3)$ is true for every orthogonal frame $\{L\_1,L\_2,L\_3\}$. (Following Gleason, Kochen and Specker found 117 lines, but this was simplified to 31 by Conway and Kochen.) The set of lines is complicated, but there is a simpler construction of Peres in $\mathbb{R}^4$, which then establishes the contradiction for $M\_4$. Namely, you should take the 12 diagonals of the regular 24-cell and the 12 diagonals of the dual 24-cell. Then it is an exercise to check that these 24 lines have the same inconsistency with respect to orthonormal frames.
At another level, the contradiction that you seek is not possible without an extra structure on classical or quantum probability. Classical probability makes a category whose objects are Boolean algebras or $L^\infty$ algebras, and whose morphisms are, say, stochastic maps. (Actually you want contravariant stochastic maps. An map from algebra $A$ to algebra $B$ yields a map of states from $B$ to $A$, and stochastic maps are usually defined on states.) Quantum probability does also, where the objects are von Neumann algebras and the morphisms are quantum stochastic maps, by definition completely positive, normal, unital maps. However, any such category embeds in the classical probability category or even the category **Set**. This formal result has been considered important by people who want to make quantum probability look classical. (E.g., it is essentially Bohm's point.) It means that if you have a single quantum system, you can always build a countability additive classical probability model "behind" it.
---
However, classical probability is also a tensor category to model joint systems, and so is quantum probability. Bell's theorem is that there does not exist a tensor embedding (speaking loosely; I'd have to think about how to rigorously define what it excludes) of quantum probability into classical probability. In physics terminology, a non-tensor-preserving construction such as that of Bohm is called "non-local".
Here is a description of an optimized form of Bell's construction called the CHSH inequality. Let $a\_1, a\_2 \in A$ and $b\_1, b\_2 \in B$ be four Booleans or, more conveniently, four $\pm 1$-valued Bernoulli random variables. Then their correlation matrix $E[a\_j b\_k]$ in $A \otimes B$ satisfies this elementary inequality in classical probability:
$$E[a\_1 b\_1] + E[a\_1 b\_2] + E[a\_2 b\_1] - E[a\_2 b\_2] \le 2.$$
To prove this inequality, consider that each term is $\pm 1$ if you plug in four values of the variables; if the first three term are 1 then all four are equal and the last term is $-1$. (You can easily make them unbiased so that $E[a\_j b\_k]$ really is the correlation matrix.) However, if you take $A \cong M\_2$ (the qubit system) and $B \cong M\_2$, then it is easy to find four such random variables and a state on $A \otimes B$ such that instead,
$$E[a\_1 b\_1] + E[a\_1 b\_2] + E[a\_2 b\_1] - E[a\_2 b\_2] = 2\sqrt{2}.$$
(This is the maximum possible value in quantum probability.) This is a classical impossible set of correlations. You can repeatedly interrogate system $A$ by randomly measuring either $a\_1$ or $a\_2$, and the same for $B$. The correlation makes you think that $A$ and $B$ are in communication with each other, even when it cannot be true. This is what was actually demonstrated in Aspect's experiment. His experiment directly violated the CHSH inequality, in a protocol in which there wasn't enough time for light to travel from $A$ to $B$.
Here are a few more comments about the violation demonstrated in Aspect-type experiments. What is actually measured is
$$E = E[(-1)^{(j+1)(k+1)} a\_j b\_k],$$
where $j$ and $k$ are also random variables taking values in $\{1,2\}$. If $A$ and $B$ (or Alice and Bob) could communicate, then they could easily make $E > \frac12$, or even as close to 1 as they please, because they would both know the pair $(j,k)$ and they could pick a favorable pair of answers. But in the experiment, they are separated, and Alice is only told $j$ and Bob is only told $k$. In this case, they must separately provide $a\_j$ and $b\_k$, and the CHSH inequality applies if $A$ and $B$ are classical.
This is a somewhat formal and blasé summary of the illusion of telepathy in quantum probability that was first constructed by Bell. I have a livelier description of the same formulas [in a colloquium talk that I gave at Berkeley](http://www.math.ucdavis.edu/~greg/berkeley.pdf).
| 10 | https://mathoverflow.net/users/1450 | 43644 | 27,739 |
https://mathoverflow.net/questions/43642 | 5 | Let $G$ be an affine algebraic group defined over $\mathbf Z$. The kernel of the natural homomorphism $G(\mathbf Z/p^2\mathbf Z)\to G(\mathbf Z/p\mathbf Z)$, if abelian, is a group which comes along with the conjugation action of $G(\mathbf Z/p\mathbf Z)$.
In the case where $G$ is a classical group, this kernel is isomorphic (as a set with $G(\mathbf Z/p\mathbf Z)$-action) to the Lie algebra $\mathfrak g(\mathbf Z/p\mathbf Z)$ of $G(\mathbf Z/p\mathbf Z)$ (which comes with the adjoint action). It seems that this should be the case in general.
Does anyone know of a reference for this kind of thing?
| https://mathoverflow.net/users/9672 | kernel of G(Z/p^2 Z)->G(Z/pZ) is the lie algebra of G over Z/pZ? | Take a look at Waterhouse's book - Introduction to affine group schemes. I think Theorem 12.2 is what you're looking for.
| 8 | https://mathoverflow.net/users/1328 | 43645 | 27,740 |
https://mathoverflow.net/questions/43594 | 7 | While looking over the first chapter of
1) *Quantum Fields and Strings: A Course For Mathematicians* (P. Deligne, P. Etingof, D.S. Freed, L. Jeffrey, D. Kazhdan, J. Morgan, D.R. Morrison and E. Witten, eds.,), 2 vols., American Mathematical Society, Providence, 1999.
I wondered whether there would be any use to developing a theory of super-von Neumann algebras, mimicking the usual theory. Not knowing whether or not this would be a sterile or trivial exercise, I never tried.
I have always wondered, though:
>
> Would there be any benefit in developing a theory of super-von Neumann algebras, and if so what would the benefit likely be? Particularly, could such a theory tell us anything useful about ordinary von Neumann algebras we otherwise couldn't easily obtain?
>
>
>
Perhaps this is a trivial question, but I'm curious if anyone with broader knowledge can shed some light on this.
Of course, the dream is that looking at something like this would miraculously unveil something cool like a canonical time-evolution on $II\_{1}$-factors.
(This is another candidate for the 'dumb question' tag!)
| https://mathoverflow.net/users/6269 | Would a supersymmetric theory of von Neumann algebras be useful? | A von Neumann algebra is an associative Banach algebra over $\mathbb{C}$, which also has an anti-linear anti-involution \* such that $||a^\*a|| = ||a||^2$, and which also has a predual as a Banach space. In context, you can think of it as a non-commutative algebra with a certain semisimple-like property and certain fairly strong analytic closure properties.
Now, you can have a non-commutative superalgebra, but this is a somewhat thin combination, because the associativity axiom of an algebra (and in fact every axiom for a von Neumann algebra) does not use the switching map $v \otimes w \mapsto w \otimes v$ or its superized version $v \otimes w \mapsto (-1)^{(\deg v)(\deg w)} w \otimes v$. A supercommutative algebra is not usually a commutative algebra, a Lie superalgebra is not usually a Lie algebra, and a Hopf superalgebra is not usually a Hopf algebra; all of these objects have axioms that use the switching map. But an associative superalgebra is an associative algebra and a von Neumann superalgebra is a von Neumann algebra.
On the other hand, in quantum physics one is often interested in a classical limit which is commutative, or in the supersymmetry context, supercommutative. It is an interesting fact that you can make a commutative von Neumann algebra, which is then a model of classical probability. But you can't make a nontrivially supercommutative von Neumann algebra, because it doesn't have the semisimple-like properties of a von Neumann algebra. However, von Neumann algebra axioms really are necessary for the quantum probability model. So the conventional thing to do is to embed the supercommutative algebra that exists in a theory such as supersymmetry in a von Neumann algebra, even though it is not a von Neumann subalgebra. Or, you could say that supersymmetry (if you accept it) and quantum probability are two ultimately different reasons that classical probability has to be changed. Supersymmetry can be viewed as more of a geometric reason than a probabilistic reason.
| 4 | https://mathoverflow.net/users/1450 | 43648 | 27,741 |
https://mathoverflow.net/questions/43625 | 9 | Could you tell me what is the name and/or reference for the following theorem:
>
> Let $M$ be a metric space. Then any continuous function $f:M\to\mathbb R$ can be a be uniformly approximated by a locally Lipschitz functions.
>
>
>
| https://mathoverflow.net/users/10330 | Approximation by locally Lipschitz functions | Actually the uniform density of locally Lipschitz functions is quite an immediate consequence of the paracompactness of metric spaces (Stone's theorem), and of the fact that, of course, metric spaces admit locally Lipschitz partitions of unity. Note that this way you also have the general result for Banach-valued functions, that is, with a given Banach space as a codomain.
A close result is that *uniformly continuous* ($\mathbb{R}$-valued) functions on a convex set of a normed space can be uniformly approximated by (uniformly) Lipschitz functions. In this case, an explicit approximation for a function $f$ is obtained just taking $f\_k:=$ the infimum of all $k$-Lipschitz functions above $f.$ Then $f\_k$ is k-Lipschitz and $f\_k\to f$ uniformly as $k\to \infty$ (moreover, the uniform distance of $f$ and $f\_k$ can be evaluated in terms of the modulus of continuity of $f$), without need of Stone's theorem. I think that variant of this construction should work for locally Lipschitz approximation of continuous functions (always in the scalar-valued case).
| 12 | https://mathoverflow.net/users/6101 | 43653 | 27,743 |
https://mathoverflow.net/questions/43647 | 11 | Let me recall subj:
If $s>0$, $A$ and $B$ are two subsets of $\mathbb{S}^{n}$, $|A|=|B|$ ($|\cdot|$ stands for the Lebesgue measure on the sphere) and $B$ is a cup $B=\{ (x\_1,x\_2,\dots,x\_n)\in \mathbb{S}^n, x\_n\leq t \}$ (for some $t\in [-1,1]$), then $|A\_s|\geq |B\_s|$, where $A\_s$ means $s$-neighborhood of the set.
It leads to measure concentration inequalities for the sphere and so has numerous applications. So I guess that Levy's initial proof was simplified, maybe not once. What is the easiest proof of the inequality and where to read it?
| https://mathoverflow.net/users/4312 | Levy's isoperimetric inequality for sphere | The shortest and most amazing proof (in my opinion) is by Steiner symmetrization around half of a great circle. Given $A$, and given a half great circle $\gamma$, rotate the sphere so that $\gamma$ is a meridian arc. Then for each latitude sphere $H$, you can replace $A \cap H$ by the spherical cap in $H$ centered at $H \cap \gamma$. Let $A'$ be the result. Then it is not hard to show that $|A'\_s| \le |A\_s|$ for all $s > 0$; in fact even each $|A'\_s \cap H| \le |A\_s \cap H|$. And you can show that you can pick a sequence of half great circles such that $A$ converges to $B$ under symmetrization, and that some of the inequalities are strict unless $A$ is congruent to $B$.
Of course this is just an outline, but it is an accurate summary (I hope) of the Steiner symmetrization argument. It also works in Euclidean or hyperbolic space using a line rather than half of a line.
| 11 | https://mathoverflow.net/users/1450 | 43654 | 27,744 |
https://mathoverflow.net/questions/43629 | 4 | Assume that a positively graded ring R is generated in degree 1. Is it true that, if R is Cohen-Macaulay, then there exists a regular sequence **x** of elements of degree 1 so that R/**x** is zero dimensional?
I tend to believe that it holds, but could not find a reference. Maybe some extra condition should be imposed?
| https://mathoverflow.net/users/10332 | Regular sequence of elements of degree 1 for a homogeneous Cohen-Macaulay ring | You want to assume also that the residue field is infinite. Then a minimal reduction of the irrelevant ideal will be a system of parameters, each linear, and will be a maximal regular sequence.
| 6 | https://mathoverflow.net/users/460 | 43657 | 27,747 |
https://mathoverflow.net/questions/43564 | 6 | This is a sequel to my [earlier question](https://mathoverflow.net/questions/26094/a-topologically-mixing-subshift-with-multiple-measures-of-maximal-entropy), where I asked for an example of a shift space that is mixing but not intrinsically ergodic -- that is, it has multiple measures of maximal entropy (MMEs). Steve Huntsman's [answer](https://mathoverflow.net/questions/26094/a-topologically-mixing-subshift-with-multiple-measures-of-maximal-entropy/26151#26151) referred me to a paper by Haydn that gives such an example: however, in that example the two MMEs are supported on disjoint compact subsets of the shift space, and so in some sense the shift can be viewed as two intrinsically ergodic shifts that have been "glued together".
Is there an example of a transitive shift with multiple MMEs that are all fully supported? More precisely, does anybody know of a transitive shift space $X\subset \{0,1,\dots,p-1\}^\mathbb{Z}$ for which there are two distinct ergodic measures $\mu\_1, \mu\_2$ such that
1. $h\_{\mu\_1}(\sigma) = h\_{\mu\_2}(\sigma) = h\_\mathrm{top}(X,\sigma)$;
2. $\mu\_i(U)>0$ for every open set $U\subset X$ and $i=1,2$?
| https://mathoverflow.net/users/5701 | Transitive shifts with multiple fully supported MMEs | My recollection is that in one of the last chapters of the book by Denker-Grillenberger-Sigmund, there is a theorem which says something like: given $n$ ergodic measure-preserving transformations with entropy strictly less than $\log d$, there is a minimal subshift of the full shift on $d$ symbols which has precisely these invariant measures. This would certainly be fine if $n=2$ and two systems with identical entropy are given. Going to the Springer website, I think it's chapter 31, but unfortunately I can't get through the paywall, so I could be quite wrong...
That said, if it's not important that the topological entropy of the subshift is positive, then all that's needed is an example of a minimal subshift with zero topological entropy which is not uniquely ergodic. An example of this appears to be constructed in Section 3 of the 1984 paper "Toeplitz minimal flows which are not uniquely ergodic" by Susan Williams.
| 4 | https://mathoverflow.net/users/1840 | 43660 | 27,750 |
https://mathoverflow.net/questions/43426 | 6 | In the book "The Geometry of Schemes" of Eisenbud and Harris, page 26, it is said that the scheme theoretic closure of a closed subscheme Z of an open subscheme U is the closed subscheme of X defined by the sheaf of ideals consisting of regular functions whose restrictions to U vanish on Z.
I cannot verify this assertion when the open immersion of U in X is not quasi-compact (I mean I cannot prove that this sheaf of ideals is quasi-coherent).
Am I missing something here ?
| https://mathoverflow.net/users/3333 | Scheme theoretic closure of a locallly closed subscheme | It seems indeed that example 2.10 in the Stacks project morphisms of schemes chapter provides a counter example, where the sheaf of ideals of regular functions whose restrictions to U vanish on Z is not quasi-coherent, because if it were part (3) of lemma 4.3 would be fulfilled. Thank you again for this hint, and if I am not mistaken, I have answered my own question, and an errata should be added for this book on page 26, where the open immersion of U in X should be supposed quasi-compact.
| 2 | https://mathoverflow.net/users/3333 | 43671 | 27,755 |
https://mathoverflow.net/questions/43681 | 44 | In grad school I learned the isomorphism between de Rham cohomology and singular cohomology from a course that used Warner's book *Foundations of Differentiable Manifolds and Lie Groups*. One thing that I remember being puzzled by, and which I felt was never answered during the course even though I asked the professor about it, was what the theorem could be used for. More specifically, what I was hoping to see was an application of the de Rham theorem to proving a result that was "elementary" (meaning that it could be understood, and seen to be interesting, by someone who had not already studied the material in that course).
Is there a good motivating problem of this type for the de Rham theorem?
To give you a better idea of what exactly I'm asking for, here's what I consider to be a good motivating problem for the Lebesgue integral. It is Exercise 10 in Chapter 2 of Rudin's *Real and Complex Analysis*. If $\lbrace f\_n\rbrace$ is a sequence of continuous functions on $[0,1]$ such that $0\le f\_n \le 1$ and such that $f\_n(x)\to 0$ as $n\to\infty$ for every $x\in[0,1]$, then $$\lim\_{n\to\infty}\int\_0^1 f\_n(x)\thinspace dx = 0.$$
This problem makes perfect sense to someone who only knows about the Riemann integral, but is rather tricky to prove if you're not allowed to use any measure theory.
If it turns out that there are lots of answers then I might make this community wiki, but I'll hold off for now.
| https://mathoverflow.net/users/3106 | Motivating the de Rham theorem | Here is a really "trivial" application. Since a volume form (say from a Riemannian metric) for a compact manifold $M$ is clearly closed (it has top degree) and not exact (by Stoke's Theorem), it follows that the cohomology is non-trivial, so $M$ cannot be contractible.
| 43 | https://mathoverflow.net/users/7311 | 43685 | 27,763 |
https://mathoverflow.net/questions/43680 | 4 | Could you give me an example of a complete metric space wiht covering dimension $> n$ all of which compact subsets have covering dimension $\le n$?
| https://mathoverflow.net/users/10330 | Example in dimension theory | **A guess**
In $l^2$ Hilbert space, consider the set $E$ of points with all coordinates rational. Erdös ([reference](http://www.jstor.org/pss/1968851)) showed that $E$ has topological dimension $1$. (In separable metric space, all notions of topological dimension coincide.)
Does this $E$ have the property that every compact subset is zero-dimensional? This space (and thus any subset of it) is totally disconnected, and isn't it the case that for compact (metric) spaces, this implies zero-dimensinal?
| 6 | https://mathoverflow.net/users/454 | 43688 | 27,765 |
https://mathoverflow.net/questions/43631 | 10 | Actually I am not sure this is a legitimate question on MO. In April and June of this year Serre gave two talks on the same title "linear representations and the number of points mod p", one in ETH Number theory Days Zurich, another during Prof. Gross's birthday conference in Boston. Unfortunately I was in neither of those, nor could I find another reference about this talk online.
In the proof of Weil conjecture for curves, say an elliptic curve $E$ over $\mathbf{Q}$ has good reduction mod $p$, then the characteristic polynomial of the Frobenius operator on the Tate module for $E$ mod $p$ will essentially give us the number of points on the curve in $\mathbf{F}\_{p^n}$. So I would say this is an obvious example of relations between *two*-dimensional representation and number of points. But I wonder if Serre has more. E.g. (tangentially related) [Here](http://www.math.harvard.edu/conferences/gross_10/panel/panel.pdf) Mazur was interested in the **Chebyshev bias** (which is usually a quite analytic phenomenon) among the number of points corresponding to different $p$ (which, by the way, has few references also), and I'd like to know if the representation side could shed some light on this.
Of course since I didn't attend the talk, Serre could be talking about totally different things. I would greatly appreciate it if anyone attended the talk/have seen such notes/heard about this circle of ideas share some comments on this. Thanks!
| https://mathoverflow.net/users/1877 | Looking for reference on Serre's talk "linear rep and number of points mod p" | At the Dick Gross conference, Serre went over what he called "missing exercises from SGA 4.5". Basically he used the relation between the number of points mod $p$ on a variety and eigenvalues of Frobenius at $p$ to make statements that were less obvious in one situation, but pretty clear in the other (one side being points mod $p$, the other side being statements about functions on topological groups). Here's the dictionary: Let $X$ be a separated finite type scheme over $\mathbf{Z}$ and let $N\_X(p^e)=\mathrm{card}(X(\mathbf{F}\_{p^e}))$. On the one hand, one has that
* $N\_X(p^e)=\sum (-1)^i\mathrm{Tr}\left(\mathrm{Fr}\_p^e|H^i\_c(\overline{X},\mathbf{Q}\_\ell)\right)$ for all $p\geq p\_0$ and all $e\geq1$.
On the other hand,
* let $G=\mathrm{Gal}(\mathbf{Q}\_S/\mathbf{Q})$, where $\mathbf{Q}\_S$ is the maximal extension of $\mathbf{Q}$ unramified outside a finite set of primes $S$. Consider the virtual character $a$ of $\sum(-1)^iH^i\_c\left(\overline{X},\mathbf{Q}\_\ell\right)$.
Here are some situations he looked at:
1. If $N\_X(p)=N\_{X^\prime}(p)$ for a set of primes of density 1 then $N\_X(p^e)=N\_{X^\prime}(p^e)$ for all $p\geq p\_0$ and all $e\geq1$. (Under the dictionary, this is simply the statement that if $G$ is a topological group, $K$ is a topological field, and $a, a^\prime$ are two continuous functions $G\rightarrow K$ that agree on a dense subset, then they are equal.)
2. Suppose $|N\_X(p)-N\_{X^\prime}(p)|$ is bounded for a set of primes of density 1. Then
(i) $|N\_X(p^e)-N\_{X^\prime}(p^e)|$ has the same bound for all $p\geq p\_0$ and all $e\geq1$;
(ii) Base changing to a suitable finite extension of $\mathbf{Q}$, the value becomes constant.
3. In the special case of 2. when the bound is equal to 1, then either the difference is a constant, a quadratic character, or the negative of a quadratic character.
4. Let $B(X):=\sum\dim H^i\_c(X(\mathbf{C}),\mathbf{Q})$. Suppose $N\_X(p)\neq N\_X(p^\prime)$ for an infinite set of $p$. Then,
the set of such $p$ has density $\geq\frac{1}{(B(X)+B(X^\prime))^2}$.
Here the group theory statement is the following: let $G$ be a compact group (and set its total Haar measure to 1) and let $K$ be a locally compact field of characteristic $0$. If $\rho\_i:G\rightarrow \mathrm{GL}(n\_i,K)$ are two continuous linear representations and $a:=\mathrm{Tr}\rho\_1-\mathrm{Tr}\rho\_2$, then either
(i) $a=0$, or
(ii) {$g\in G:a(g)\neq0$} has volume $\geq (n\_1+n\_2)^{-2}$.
There's a little bit more he covered, and I've also left out the proofs. But that should give you a good idea of what his talk was about.
| 14 | https://mathoverflow.net/users/1021 | 43713 | 27,783 |
https://mathoverflow.net/questions/43611 | 11 | I posted this on Stack Exchange and got a lot of interest, but no answer.
A recent [Missouri State problem](http://people.missouristate.edu/lesreid/POW12_0910.html) stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction. That got me trying to decompose the plane into closed or open intervals. The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right. That gets the whole plane except the lattice points. This seems like it must be a standard problem, but I couldn't find it on the web. **Question:** So can the plane be decomposed into unit open intervals? closed intervals?
| https://mathoverflow.net/users/7408 | Decomposing the plane into intervals | Conway and Croft show it can be done for closed intervals and cannot
be done for open intervals in the paper:
[Covering a sphere with congruent great-circle arcs.
Proc. Cambridge Philos. Soc. 60, 1964, pp787–800](https://doi.org/10.1017/S0305004100038263).
| 10 | https://mathoverflow.net/users/3634 | 43715 | 27,785 |
https://mathoverflow.net/questions/42508 | 6 | Recall: We present an operad (with $S\_n$-action) in $R-Mod$ for any commutative ring $R$ not of characteristic 2 generated by a single element in degree $2$ satisfying the following identities:
* $\theta+\theta\tau=0$
* $\theta(1,\theta)+\theta(1,\theta)\sigma + \theta(1,\theta)\sigma^2$.
where $\tau$ and $\sigma$ are 2-cycles and 3-cycles respectively.
However, in characteristic $2$, this fails to characterize Lie algebras in the obvious way, since the first equation says that $\theta$ is skew-symmetric (and hence symmetric in characteristic 2).
The proper axiom to include is that $[x,x]=0$, i.e. that $[-,-]$ is alternating rather than skew-symmetric. Can we present this relation operadically? It seems like on the face of it, we can't, but I'd be happy to be surprised.
| https://mathoverflow.net/users/1353 | Repairing the Lie operad in characterstic 2? | In fact, several monads can naturally be associated to an operad $P$ and this might be used to answer your question.
In the usual setting, one considers a generalized symmetric algebra $S(P,X) = \bigoplus\_n (P(n)\otimes X^{\otimes n})\_{\Sigma\_n}$ where we form coinvariants under the action of the symmetric groups $\Sigma\_n$. But we can also take invariants instead of coinvariants and form another functor $\Gamma(P,X) = \bigoplus\_n (P(n)\otimes X^{\otimes n})^{\Sigma\_n}$ associated to $P$. The image of the norm map from coinvariants to invariants still gives another functor $\Lambda(P,X)$ associated to $P$.
Under the assumption $P(0) = 0$, we have a monad structure on $\Lambda(P): X\mapsto\Lambda(P,X)$ and $\Gamma(P): X\mapsto\Gamma(P,X)$ inherited from the operadic composition structure of $P$. See (1.2.12-1.2.17) in
<http://math.univ-lille1.fr/~fresse/PartitionHomology.pdf>
(ref.: <http://www.ams.org/mathscinet-getitem?mr=2005g:18015>)
For the operad $P = Lie$, the algebra category associated to $\Gamma(Lie)$ can be identified with the category of $p$-restricted Lie algebras (where $p$ is the cateristic of the ground ring), while the algebra category associated to $\Lambda(Lie)$ can be identified with the category of Lie algebras equipped with an alternating Lie bracket.
| 12 | https://mathoverflow.net/users/10354 | 43723 | 27,790 |
https://mathoverflow.net/questions/43711 | 5 | Let me try and put the question in context. I am studying certain subsets of the tangent bundle of a sphere. I also have a regular CW complex which is a deformation retract of such a subset. Hence I have a description of the cells and the information that tells me which cell is in the boundary of which other cell. Fortunately this cell complex is a homotopy colimit of a diagram of spaces. As a result I can compute the cohomology groups (but not the product).
All my examples concerning $S^1$ and $S^2$ show that these subsets have the homotopy type of wedge of copies of $S^1$ and $S^2$ respectively. Hence I am trying to prove that this is the case in all dimensions. In this process the only thing I was able to prove that there is a retraction from these subsets to the underlying sphere.
So I would like to know about various methods to show that a space is a wedge of spheres.
I understand that this question might sound vague and the information too little.
| https://mathoverflow.net/users/7494 | How to show that a space has the homotopy type of wedge of spheres ? | To follow up a bit on Mikael's answer, the notion of non-pure shellability is probably more relevant to your situation. Shellable simplicial complexes are wedges of spheres of equal dimension, but non-purity allows different dimensional spheres. You should look at papers by Michelle Wachs and Anders Bjorner if you're interested. However, this will require finding a simplicial decomposition of your space, which may be a challenge.
Added: Since this is now the accepted answer, I figure I should give the precise references. Both papers are on JSTOR (follow the links).
Björner, Anders; Wachs, Michelle L.
Shellable nonpure complexes and posets. I.
Trans. Amer. Math. Soc. 348 (1996), no. 4, 1299–1327.
<http://www.jstor.org/stable/i311403>
Björner, Anders; Wachs, Michelle L.
Shellable nonpure complexes and posets. II.
Trans. Amer. Math. Soc. 349 (1997), no. 10, 3945–3975.
<http://www.jstor.org/stable/i311413>
| 4 | https://mathoverflow.net/users/4042 | 43727 | 27,791 |
https://mathoverflow.net/questions/43721 | 26 | Is an arbitrary union of non-trivial closed balls in the Euclidean space $\mathbb{R}^n$ Lebesgue measurable? If so, is it a Borel set?
@George
I still have two questions concerning your sketch of proof.
First, how can you guarantee each of the open balls in the countable union has radius greater than or equal to 1?
Second, I don't know how to use convexity to prove $\mu (B') \leq (1+\epsilon)^{N}\mu(B)$
| https://mathoverflow.net/users/6018 | Is arbitrary union of closed balls in $\mathbb{R}^n$ Lebesgue measurable? | No, in dimension $N>1$, it does not have to be Borel measurable. E.g., in 2 dimensions, consider, a non Borel measurable subset of the reals $S$, and let $A$ be the union of closed unit balls centered at points $(x,0)$ for all $x\in S$. The intersection of $A$ with $\mathbb{R}\times \{1\}$ is the non-Borel set $S \times \{1\}$, so $A$ is not Borel.
On the other hand, for $N=1$, any union of non-trivial closed intervals is Borel-measurable. If $A$ is such a union and $B$ is the union of the open interiors, then it can be seen that $A$ is just the union of $B$ with (at most countably many) endpoints of connected components of $B$.
---
Lebesgue measurability *does* hold, however. Faisal posted a link for this as I was typing my answer, but I think its still worth giving a brief sketch of the proof I was starting to type (Edit: added more detail, as requested).
1. Reduce the problem to that of balls with at least some positive radius $r$ and within some bounded region. To do this, suppose that $S$ is the set of closed balls and $S\_r$ denotes the balls of radius at least $r$ and with center no further than $r$ from the origin. Then,
$$
\cup S=\bigcup\_{n=1}^\infty\left(\cup S\_{1/n}\right).
$$
As the measurable sets are closed under countable unions, it is enough to show that $\cup S\_r$ is Lebesgue measurable for each $r>0$. So, we can assume that all balls are of radius at least $r$ and are within some bounded distance of the origin.
2. Let $A$ be the union of the closed balls, and $B\subseteq A$ be the union of their interiors. This is open so, by second countability, is a union of countably many open balls of radius at least $r$. Also, $A$ lies between $B$ and its closure $\bar B$.
3. Show that the boundary $\bar B\setminus B$ of $B$ has zero measure. If we scale up the radius of each of the countable sequence of open balls used to obtain $B$ by a factor $1+\epsilon$ to get the new set $B^\prime$ then $\mu(B^\prime)\le(1+\epsilon)^N\mu(B)$. Showing this is the tricky part, but it does follow from convexity of the balls: If the balls have radius $r\_k$ and centres $x\_k$, then consider the sets
$$
B\_t=\bigcup\_{k=1}^\infty B(r\_k,tx\_k)
$$
for real $t$, so that $B\_1=B$. The function $t\mapsto\mu(B\_t)$ is increasing in $t\ge0$\*. Also, $B^\prime= (1+\epsilon)B\_{1/(1+\epsilon)}$ giving,
$$
\mu(B^\prime)=(1+\epsilon)^{N}\mu(B\_{1/(1+\epsilon)})\le(1+\epsilon)^{N}\mu(B)
$$
as claimed. As $\bar B\subseteq B^\prime$ we get $\mu(\bar B\setminus B)\le((1+\epsilon)^N-1)\mu(B)$ which can be made as small as we like by choosing ε small.
\* Edit: in my initial response, I was thinking that [this answer](https://mathoverflow.net/questions/27032/pushing-convex-bodies-together) is enough to prove that $\mu(B\_t)$ is increasing in $t$. However, as Mizar points out in the comments below, this is not clear. Actually, I don't think we can reduce it to that case. However, the result is still true, by the [Kneser-Poulson conjecture](http://www.openproblemgarden.org/op/kneser_poulsen_conjecture). This states that if the centres of set of unit balls in Euclidean space are all moved apart, then the measure of their union increases. Although only a conjecture, it has been proved for continuous motions, which applies in our case. Also, expressing each ball of radius greater than some arbitrarily small $r > 0$ as a union of balls of radius $r$, then it still applies in our case for balls of non equal radii.
---
Edit: having seen Faisal's explanation, the proof I outline here is completely different to his. The result Faisal quotes is a bit more general as it applies to convex sets with nonempty interior, rather than just balls. However, the proof given above also works for symmetric convex sets with nonempty interiors. As every convex set with nonempty interior is a union of (translates of) symmetric ones, this implies the same result
| 22 | https://mathoverflow.net/users/1004 | 43739 | 27,799 |
https://mathoverflow.net/questions/43726 | 22 | Is there someone who can give me some hints/references to the proof of this fact?
| https://mathoverflow.net/users/9401 | The free group $F_2$ has index 12 in SL(2,$\mathbb{Z}$) | To elaborate on Qiaochu's answer. The subgroup generated by the two matrices
$$\left[ \begin{array}{cc} 1 & 2 \\\ 0 & 1 \end{array} \right]$$
and
$$\left[ \begin{array}{cc} 1 & 0 \\\ 2 & 1 \end{array} \right]$$
is the Sanov subgroup. It consists, by an exercise in Kargapolov-Merzlyakov, of matrices of the form
$$\left[ \begin{array}{cc} 4k+1 & 2l \\\ 2m & 4n+1 \end{array} \right]$$ and det=1. The congruence subgroup $\Gamma(2)$ consists of matrices of the form $$\left[ \begin{array}{cc} 2k+1 & 2l \\\ 2m & 2n+1 \end{array} \right]$$ and det=1. Those matrices from $\Gamma(2)$ and not in the Sanov subgroup have the form $$\left[ \begin{array}{cc} 4k+3 & 2l \\\ 2m & 4n+3 \end{array} \right].$$ Taking the product of any two such matrices gives us a matrix from the Sanov subgroup. So the Sanov subgroup has index 2 in $\Gamma(2)$, and index 12 in $SL\_2(\mathbb Z)$.
| 24 | https://mathoverflow.net/users/nan | 43740 | 27,800 |
https://mathoverflow.net/questions/43759 | 5 | I've noted, that the following fact can be proven in a few lines using $C^\*$-algebra theory. I wonder if it has a simple elementary proof or not. Probably you can give me a reference.
>
> Suppose $X$ is a compact Hausdorff space, $V\subset X$ is a closed subset, $f\colon V\to \mathbb{R}$ is a continuous function. Then there exists a continuous function $g\colon X\to \mathbb{R}$, whose restriction on $V$ is $f$.
>
>
>
C\*-algebraic proof is the following.
First note, that it is enough to find $g\colon X\to \mathbb{C}$ (then we can take its real part). Consider restriction map $\phi\colon C(X)\to C(V)$. It is a \* -homomorphism, and therefore its image $\phi(C(X))$ is a commutative $C^\*$-algebra. Moreover $\phi(C(X))$ separates points of $V$ (because $C(X)$ does). By Stone-Weierstrass theorem $\phi(C(X))=C(V)$.
| https://mathoverflow.net/users/8134 | Every real-valued continuous function on a closed set of compact Hausdorff space has an extension. | I think this is a special case of the [Tietze extension theorem](https://en.wikipedia.org/wiki/Tietze_extension_theorem)
(since any compact Hausdorff space is normal). (Here is [one proof](https://planetmath.org/proofoftietzeextensiontheorem).)
| 6 | https://mathoverflow.net/users/2874 | 43760 | 27,813 |
https://mathoverflow.net/questions/43754 | 7 | Recently, I am looking into a non-convex optimization problem whose points satisfying KKT conditions can be obtained. Then the problem becomes how to decide whether the KKT conditions are sufficient for the global optimality of the solution. It is said on Wikipedia([link text](https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions#Sufficient_conditions)) that for "invex function" the KKT is sufficient for optimality. By doing some searching, I was led to a concept called "invexity" and the book "Invexity and Optimization"
([http://www.springerlink.com/content/978-3-540-78561-3#section=155972&page=1](https://link.springer.com/book/10.1007/978-3-540-78562-0?from=SL)).
But, why this "invexity" research, which was first proposed in 1980's, is only confined in a small group of people. And the book is cited for only 4 times.
In my current understanding, invexity is a generalization of convexity, and has some very good properties as in convexity, which should be very attractive and should have drawn lots of people's attention.
Is it because this concept is not interesting, not useful? Or it is being studied under other names?
| https://mathoverflow.net/users/10331 | Some questions about Invexity | There are a lot of generalizations or variations of convexity, such as quasi-convexity, pseudo-convexity, semilocal convexity, semilocal quasi-convexity, semilocal pseudo-convexity, strict versions of these, strong versions of these, etc. There is a reason for the existence of each term, in that each makes the hypotheses of some theorem tighter or has some other benefit (such as invexity making the KKT conditions not just necessary but sufficient). I agree with you that invexity ought to be better known, but it may have gotten lost in all the other generalizations/variations out there.
| 4 | https://mathoverflow.net/users/9716 | 43764 | 27,816 |
https://mathoverflow.net/questions/43581 | 4 | I'm not sure how long this [iterative](https://mathoverflow.net/questions/43438/) [questions](https://mathoverflow.net/questions/43255/) can go on, but let me try again. Let's say $X$ is a Cohen-Macaulay scheme with an action of $\mathbb{G}\_m$ (i.e. if $X$ is affine, a grading on the coordinate ring). Are the schematic fixed points $X^{\mathbb{G}\_m}$ of $X$ Cohen-Macaulay?
| https://mathoverflow.net/users/66 | Are schematic fixed-points of a Cohen-Macaulay scheme Cohen-Macaulay? | Here is a counterexample. Consider the action of $\mathbb G\_{\rm m}$ on $\mathbb A^4$ defined by $t \cdot(x,y,z,w) = (x, y, tz, t^{-1}w)$, and let $X$ be the invariant closed subscheme with ideal $(xy, y^2 + zw)$; this is a complete intersection, hence it is Cohen-Macaulay. The fixed point subscheme is obtained by intersecting with the fixed point subscheme in $\mathbb A^4$, which is given by $z = w = 0$; hence it is the subscheme of $\mathbb A^2$ given by $xy = y^2 = 0$, which is of course the canonical example of a non Cohen-Macaulay scheme.
Developing this idea a little, one can show that any kind of horrible singularity can appear in the fixed point subscheme of a $\mathbb G\_{\rm m}$-action on a complete intersection variety.
| 8 | https://mathoverflow.net/users/4790 | 43766 | 27,818 |
https://mathoverflow.net/questions/43768 | 53 | I am currently trying to learn a bit about Grothendieck-Riemann-Roch...
To try to get a better feeling for it, I am looking for examples of nice applications of GRR applied to a proper morphism $X \to Y$ where $Y$ is *not* a point. I already I know of a fair number of nice applications of HRR, i.e. GRR when $Y$ is a point. I've read through some of the relevant sections of Fulton's Intersection Theory book, but I've only found applications of HRR there, though it's very possible that I overlooked something.
I am also interested in seeing worked-out, explicit, concrete examples, with explicit Chow/cohomology classes.
Thanks much!
| https://mathoverflow.net/users/83 | Applications of Grothendieck-Riemann-Roch? | Check out Harris & Morrison's "Moduli of Curves", section 3E. There is a wealth of examples of applications of GRR coming from moduli theory, in which one applies it to projection from the universal family or some fibered power of the universal family. The basic idea in these cases is that both the base space and the total space are rather complicated beasts but the fibers of the morphisms are usually quite tractable, since they are just the gadgets you are trying to parametrize.
For more examples in the same vein, you could read the classic "Towards an enumerative geometry of the moduli space of curves" by David Mumford.
| 33 | https://mathoverflow.net/users/1310 | 43770 | 27,819 |
https://mathoverflow.net/questions/43769 | 14 | Let $P(x,y), Q(x,y)$ be polynomials of two variables over an algebraically closed field $k$. Suppose that the map $(x,y) \mapsto (P(x,y),Q(x,y))$ is not a dominant map from $k^2$ to $k^2$. Does this mean that one has $P(x,y) = R(T(x,y))$ and $Q(x,y) = S(T(x,y))$ for some polynomials $R,S,T$?
For fixed $x$, it seems to me that the curve $\{ (P(x,y), Q(x,y)): y \in k \}$ either degenerates to a point or is a curve of genus zero, which ought to kill off the problem, but I don't have enough of an understanding of the genus zero plane curves to do this.
A closely related question: if $P(x,y)-c$ is reducible for all $c \in k$ (or maybe all but finitely many $c$), does this mean that $P(x,y) = R(T(x,y))$ for some polynomials $R, T$, with $T$ having strictly smaller degree than $P$? I played around a bit with some Galois theory but was only able to convince myself that this question was more or less equivalent to the original question.
| https://mathoverflow.net/users/766 | Non-dominant polynomial maps in the plane | Yes, both results are true. For the first, as you say, the image of the map is either a point, or dominates an affine curve $C$ of geometric genus 0. By standard results, it factors through the normalization $X$ of $C$; the curve $X$ is a smooth affine curve of genus 0, so it is the complement of a finite subset of $\mathrm P^1$. This finite subset cannot contain more than one point, because otherwise $X$ would have non-constant invertible regular function, which would pull back to non-constant invertible regular functions on $\mathbb A^2$, which don't exist. Hence $X = \mathbb A^1$, and you can factor your map through $\mathbb A^1$, which gives what you want.
For the second question, Quim's argument works.
| 14 | https://mathoverflow.net/users/4790 | 43775 | 27,822 |
https://mathoverflow.net/questions/38430 | 0 | I asked a question before where I wanted a simple example where regularity up to the boundary fails for a linear elliptic PDE. I was presented an example with $\Omega = B(0,1) \backslash \{0\}$ (ball minus a point) which is nice but I would like something less pathalogical. I would like an example where my domain is at least Lipschitz (so a rectangle is an example).
In the case where we subtract off a point, a boundary value problem doesn't really even make sense in a weak formulation to begin with.
More precisely,
**Question:** I would like an example where an elliptic PDE with smooth coefficients satisfies $Lu = f$ in $\Omega$ for smooth $f$ (or zero) and $u = 0$ on $\partial \Omega$ (which is of Lipschitz class) but where $u$ somehow fails to be "regular" up to the boundary (I'm being vague on purpose here as any failure of regularity will do for the most part).
My guesses have been try try looking at the upper right quadrant $[\{ (x,y) : x, y > 0\}]$ but nothing has come from this so far. Any suggestions/ideas are welcome and appreciated.
| https://mathoverflow.net/users/8755 | Failure of regularity up to the boundary for a linear elliptic PDE | You might want to consider the function $v(z)=\Im(z^{\pi/\alpha})=r^{\pi/\alpha}\sin(\pi\theta/\alpha)$ in the sector $0<\theta<\alpha$ of the complex plane. It is harmonic in the interior, continuous up to the boundary and vanishes there, but has a "singularity" at $0$ if $\alpha$ is not of the form $\pi/n$. More precisely, the derivative of order $k$ along the ray $\theta=\alpha/2$ explodes at $0$ when $k>\pi/\alpha$ and $\pi/\alpha$ is not an integer.
More generally, you can take $v(z)=\Im(F(z))$, where $F$ is [edit: the inverse of] a [Schwarz-Cristoffel mapping](http://en.wikipedia.org/wiki/Schwarz%25E2%2580%2593Christoffel_mapping), taking the interior $\Omega$ of a polygon in the complex plane to the upper half plane $\Im w >0$.
| 5 | https://mathoverflow.net/users/6451 | 43779 | 27,824 |
https://mathoverflow.net/questions/43673 | 3 | Consider an arbitary positive semidefinite operator ρ, acting on ℂA ⊗ ℂB ⊗ ℂC, for A,B,C finite. Also, let P be an orthogonal projector on ℂB ⊗ ℂC . For the sake of concision, I will write R = 1ℂA ⊗ P ; this of course is also an orthogonal projector. Consider the completely positive transformation
>
> M(ρ) = (1 − R) ρ (1 − R) + R ρ R .
>
>
>
As R is an orthogonal projector, it is easy to show that || M(ρ) ||2 ≤ || ρ ||2 . This is because we may represent ρ as matrix in a basis consisting of the eigenvectors of R; if we divide ρ into block according to rows/columns representing vectors in the image or the kernel of R, the effect of the map M is to set the non-diagonal blocks to zero.
I am interested in how the map M may similarly affect the operator 2 norm of reduced operators on ℂA ⊗ ℂB. So I would like to know:
Is it also true that || trC( M(ρ) ) ||2 ≤ || trC(ρ) ||2 — where trC is the trace operator acting on ℂC, taken in tensor product with 1ℂA ⊗ 1ℂB ?
| https://mathoverflow.net/users/3723 | Bounds on operator 2-norms on partial traces of linearly related operators | Okay, it turns out in retrospect that the problem is trivial. The answer is "no": such a bound does not hold in general.
A simple counterexample is yielded by taking A=1 (so that we effectively deal with ℂB ⊗ ℂC throughout), B=C=2, and taking
>
> P = ½ **ψ****ψ\*** where **ψ** = **e**1 ⊗ **e**2 − **e**2 ⊗ **e**1
>
>
>
for standard basis vectors **e**j for ℂ2. Note that P is the projector onto the antisymmetric subspace of ℂB ⊗ ℂC. The map M may then be re-presented as
>
> M(ρ) = ½ ρ + ½ UρU\* where U = 1ℂB ⊗ 1ℂC − 2P.
>
>
>
The operator U is unitary, and has the effect of 'swapping' the two spaces B and C; that is, for all tensor products **α** ⊗ **β** , we have U(**α** ⊗ **β**) = **β** ⊗ **α** . We may then construct an operator ρ for which the desired bound does not hold, by taking a tensor product of an operator with low 2-norm with one of high 2-norm, e.g.
>
> ρ = 1ℂB ⊗ **e**1**e**1\*.
>
>
>
We then have trC(ρ) = 1ℂB , which has a 2-norm of $\sqrt 2$ ; and trC(UρU\*) = 2 **e**1**e**1\*, which has a 2-norm of $2$. By the convexity of the 2-norm, we may then show that
|| trC( M(ρ) ) ||2 > || trC(ρ) ||2 for this choice of P and ρ. A similar construction can be made for any B=C>1, and letting P be the projector onto the antisymmetric space of ℂB ⊗ ℂC .
I'm interested now in what upper bounds may be obtained for || trC( M(ρ) ) ||2 − || trC(ρ) ||2 , or related quantities, in the case that P is a rank-1 projector on ℂB ⊗ ℂC . If anyone can show such an interesting such bound, I may 'accept' it; but for the meantime, this answers my original question.
| 1 | https://mathoverflow.net/users/3723 | 43783 | 27,827 |
https://mathoverflow.net/questions/43796 | 3 | Let $V$ be a geometrically irreducible and reduced scheme defined over the real numbers, and let $K = K(V)$ be its function field.
1. If $V$ does not have any real points, is it true that $K$ is not formally real? It seems this is a theorem due to (E.) Artin but I cannot find a modern reference and my German needs a little work.
2. If $V$ does have real points, is $K$ necessarily formally real?
Thanks for the help.
| https://mathoverflow.net/users/7531 | Function Fields of Real Varieties | The theorem you want is due to Serge Lang, from the following paper:
>
> The theory of real places.
> Ann. of Math. (2) 57, (1953). 378–391.
>
>
>
The statement is almost, but not quite, what you suggest. To see the problem, a good example to consider is the affine plane curve over $\mathbb{R}$ defined by $\mathbb{R}[x,y]/(y^2+x^2+x^4)$. This defines a geometrically integral curve over $\mathbb{R}$ with non-formally real fraction field but possessing an $\mathbb{R}$-point, namely $(0,0)$. The key is that $(0,0)$ is the only $\mathbb{R}$-point on this curve and (thus!) it is a singular point.
So the correct result is that the function field of an integral affine variety $V\_{/\mathbb{R}}$ is formally real iff $V$ admits a nonsingular $\mathbb{R}$-point. (Note that projective real algebraic varieties are also affine(!!).) Probably you could extend this to finite-type integral schemes without any trouble.
I also looked in *Bochnak, Coste and Roy*, following Thierry Zell's suggestion, but only found "half" of this result, namely the Artin-Lang Homomorphism Theorem. It seems likely though that I just didn't look hard enough: perhaps someone can enlighten me.
| 6 | https://mathoverflow.net/users/1149 | 43801 | 27,834 |
https://mathoverflow.net/questions/43799 | 4 | A minimal complex is a CW complex whose only cells are the homology cells.
Is there some sort of criterion on CW complexes about existence of minimal complexes?
Actually I am working on a problem of understanding homotopy type of certain spaces
(see: [How to show that a space has the homotopy type of wedge of spheres ?](https://mathoverflow.net/questions/43711/how-to-show-that-a-space-has-the-homotopy-type-of-wedge-of-spheres))
My hope was to use discrete Morse theory (acyclic matching of face poset to be precise) and find the minimal complex. But then I don't know if the existence of the minimal complex is always guaranteed.
| https://mathoverflow.net/users/7494 | Discrete Morse theory and existence of minimal complex | This is not exactly what you asked, but it's certainly not the case that every CW complex has a discrete vector field where the Morse complex has trivial differential. In particular this would imply that chain complex is simple-homotopy equivalent to a chain complex with no differential. However, simple-homotopy equivalence is well-known not to generate homotopy equivalence. In particular, the Whitehead torsion is an obstruction which lives in the Whitehead group of the fundamental group. Marshall Cohen's book on Whitehead torsion is the canonical place to learn about this.
| 5 | https://mathoverflow.net/users/9417 | 43803 | 27,836 |
https://mathoverflow.net/questions/43805 | 16 | Is there a *finitely generated* nontrivial group $G$ such that $G \cong G \times G$?
Here are some properties which such a group $G$ has to satisfy:
* $G$ is not abelian (otherwise $G$ is a noetherian $\mathbb{Z}$-module, and the composition of the first projection $G \times G \to G$ with an isomorphism $G \cong G \times G$ will be bijective, i.e. $G$ is trivial).
* $G$ is perfect (apply the first observation to $G/G'$)
| https://mathoverflow.net/users/2841 | When is $G$ isomorphic to $G \times G$? | Yes. Some Googling turns up [J. M. Tyrer Jones, "Direct products and the Hopf property," *J. Austral. Math. Soc.* **17** (1974), 174-196](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=4894728).
| 20 | https://mathoverflow.net/users/290 | 43810 | 27,840 |
https://mathoverflow.net/questions/43795 | 8 | Let $X$ be a finite CW complex. Swan's theorem provide an equivalence
$$
{\rm Vec}(X)\xrightarrow\sim{\rm ProjMod}(\mathop{\rm hom}\nolimits\_{\rm Top}(X,\mathbb{R}))
$$
between the category of **finite dimensional** vector bundles over $X$ and the category of finitely generated projective modules over the ring of continuous functions from $X$ to the reals. This isomorphism behaves well with the monoidal structure $\oplus$.
There is an intermediate step in this construction: The category ${\rm Vec}(X)$ of finite dimensional vector bundles over $X$ is equivalent to locally free modules of finite rank over the sheaf $C\_X(-)=\mathop{\rm hom}\nolimits\_{\rm Top}(-,\mathbb{R})$ on $X$.
The category ${\rm Cov}(X)$ of covers of $X$ is equivalent to the category of locally constant sheaves **of sets** on $X$. Is it possible to formulate this analogously to the above correspondence? So maybe locally constant sheaves are somehow special modules over $C\_X(-)$ and this category possibly corresponds to some special modules over $C\_X(X)$. Maybe this is also compatible with disjoint unions of coverings and sums of the corresponding modules. Maybe it is also necessary to require that the covering is **regular**.
(The **bold** things are edits made, partially based on the answers below.)
| https://mathoverflow.net/users/2625 | Swan-like theorem and covering spaces | If I can take only the finite covers, then yes, I think. (After all, Swan's theorem is a characterization of finite-dimensional vector bundles, not all vector bundles.) This is easier to do over $\mathbb{C}$ than over $\mathbb{R}$. In addition to the entire sheaf $C\_X(-)$, let $C(X) = C\_X(X)$ be the algebra of global continuous functions.
Since the module $M$ is locally free, what you want to do is to choose a basis for $C(U) = C\_X(U)$ for enough open sets $U$, and such that the bases agree when you restrict to smaller open sets. You could just ask for this directly, but there is an indirect algebraic condition that comes to the same thing. Namely, you can ask for $M$ to not only be finitely generated and projective, but also a semisimple commutative algebra over $C(X)$. This gives you the unordered basis in each fiber.
Over $\mathbb{R}$, it's not quite enough to require that $M$ be a semisimple algebra, because you could end up creating $C(Y,\mathbb{C})$ for a finite cover $Y$ of $X$. So, you could also impose the condition that $f^2 + g^2 = 0$ has no non-trivial solutions in $M$.
| 6 | https://mathoverflow.net/users/1450 | 43813 | 27,842 |
https://mathoverflow.net/questions/43812 | 6 | My question is much more specific than the title:
>
> Given a symmetric
> distribution $\Xi$ on $\mathbb R^2$, when is it possible to construct a sequence $\xi\_1,\xi\_2,\dots$ of random variables such that the joint distribution of any two of them is $\Xi$?
>
>
>
For example, if the pdf of $\Xi$ is decomposable: $p\_\Xi(x,y) = p(x) p(y)$, then one can just take a sequence of independent r.v.'s.
(To construct certain counterexample related to fractional Brownian motion) I am particularly interested in the pdf $p\_\Xi(x,y) = \frac{(a+1)(a+2)}2 |x-y|^{a}1\_{[0,1]}(x)1\_{[0,1]}(y)$, $a\in(-1,0)$.
| https://mathoverflow.net/users/8146 | When is it possible to construct a joint law from its two-dimensional marginals? | I recently came upon this question in the context of distributions taking values in a finite set, but since yours take values in the compact interval $[0,1]$ I don't think much will go wrong applying the answer to your setting.
Certainly a sufficient condition is that you can construct an exchangeable sequence $\chi\_1,\chi\_2,\ldots$ for which the marginal of $\chi\_1$ and $\chi\_2$ is $\Xi$, or equivalently that $\Xi$ is a mixture of i.i.d. distributions per de Finetti's theorem. It turns out this is also necessary.
To see this suppose that $\Xi$ satisfies the condition you give on its two-variable marginals. Then for any finite $n$ there is a finite sequence of random variables $\xi\_1,\ldots,\xi\_n$ (just take the first $n$ variables of your given sequence) all of whose two-variable marginals are $\Xi$. We can construct a new sequence of random variables $\chi\_1,\ldots,\chi\_n$ by randomly permuting the $\xi\_1,\ldots,\xi\_n$. By linearity the marginal of any two of these will still be $\Xi$.
But the distribution of the $\chi\_1,\ldots,\chi\_n$ is invariant under arbitrary permutations, by symmetry. Using a diagonalization and compactness argument, we get from the existence of such a sequence $\chi\_1,\ldots,\chi\_n$ for all finite $n$ the existence of an exchangeable sequence $\chi\_1,\chi\_2,\ldots$ whose two-variable marginal distribution is $\Xi$.
| 10 | https://mathoverflow.net/users/5963 | 43822 | 27,848 |
https://mathoverflow.net/questions/43833 | 32 | Can anyone give an example of a Markov process which is not a strong Markov process? The Markov property and strong Markov property are typically introduced as distinct concepts (for example in Oksendal's book on stochastic analysis), but I've never seen a process which satisfies one but not the other.
Many thanks
-Simon
| https://mathoverflow.net/users/9564 | A Markov process which is not a strong markov process? | Consider the following continuous Markov process X, starting from position x
1. if x = 0 then Xt = 0 for all times.
2. if x ≠ 0 then X is a standard Brownian motion starting from x.
This is not strong Markov (look at times at which it hits zero).
| 17 | https://mathoverflow.net/users/1004 | 43841 | 27,860 |
https://mathoverflow.net/questions/43844 | 5 | Are there algorithms / theorems to find an acyclic matching on the Hasse diagram of a poset.
I am particularly interested in the face poset of a regular CW complex.
Also, how to decide if the given acyclic matching is perfect (impossible to add one more vertex).
Is there some structure on the set of all acyclic matchings ?
| https://mathoverflow.net/users/7494 | Algorithm to find an acyclic matching on a poset | You can start with a small matching and try to improve it. This paper of Patricia Hersch might be useful in this direction:
Hersh, Patricia(1-IN)
On optimizing discrete Morse functions. (English summary)
Adv. in Appl. Math. 35 (2005), no. 3, 294–322.
As for the last question, Chari and Joswig have a paper analyzing the space of all acyclic matchings. I don't know that the analysis is useful for the first two questions, though.
Chari, Manoj K.(1-LAS); Joswig, Michael(D-TUB-IM)
Complexes of discrete Morse functions. (English summary)
Discrete Math. 302 (2005), no. 1-3, 39–51.
| 5 | https://mathoverflow.net/users/4042 | 43847 | 27,863 |
https://mathoverflow.net/questions/43849 | 2 | I’m having some problems in ensuring the non-negativity of **KLD**!
I know that **KLD** is always positive and I went over the proof. However, it doesn’t seem to work for me. In some cases I’m getting negative results. Here is how I’m using **KLD**:
$${\rm KLD}( P(x) || Q(x) ) = \sum P(x) \log \left( \frac{P(x)}{Q(x)} \right) \, ,$$
where the Log is in base 2, and $P(x)$ and $Q(x)$ are two different distributions for all $x \in X$.
For example, $P(x) = {\rm Frequency}(x)/{\text Total Size}$; just a normal PMF! The same thing for $Q(x)$.
Note that the Total\_Size of $P$ might be different from that of the $Q$ distribution.
Could you please let me know if I’m missing something? Are there any special conditions that I have to take into consideration to avoid having negative results?
| https://mathoverflow.net/users/10378 | How to ensure the non-negativity of Kullback-Leibler Divergence KLD Metric (Relative Entropy)? | "The K-L divergence is only defined if P and Q both sum to 1 and if Q(i) > 0 for any i such that P(i) > 0."
I suspect that the second condition is your problem. Say that you have x which appears in P but not Q -- in this case you're probably adding zero contribution to the sum in your code so that you don't have to divide by zero or take the logarithm of zero, but this is effectively throwing out mass from P and you get a negative number for the divergence.
<http://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence>
| 5 | https://mathoverflow.net/users/9501 | 43851 | 27,864 |
https://mathoverflow.net/questions/43850 | 1 | I'm a little bit suprised at the moment, so i'll ask here if I see this wrong:
Given a sheaf of algebras $R$ ( e.g. maximal order or Azumaya) on a smooth projective scheme $X$ with generic point $p$. Asumme $M$ and $N$ are two left $R$-modules, coherent and torsion free over $O\_X$, such that $M\_p$ and $N\_p$ are simple $R\_p$-modules. Now given a nontrivial R-morphism $f: M \rightarrow N$. I think that f is automatically injective in this case.
Injectivity can be check on the open subsets $U \subset X$, there we have $f\_U : M(U) \rightarrow N(U)$. We also have the injections $i\_{U,p} :M(U)\rightarrow M\_p$ and $j\_{U,p}: N(U)\rightarrow N\_p$ and the induced map $f\_p: M\_p \rightarrow N\_p$, with the property $j\_{U,p}\circ f\_U=f\_p\circ i\_{U,p} (1)$.
Assume $f\_p=0$, then we have $j\_{U,p}\circ f\_U=0$ by $(1)$ for every open subset $U$. But since $j\_{U,p}$ is injective, this would imply that $f\_U=0$ for every $U$ so $f=0$, which contradicts $f\neq 0$. So $f\_p: M\_p \rightarrow N\_p$ is a nontrivial map between simple modules, so it is an isomorphism, especially injective. This implies that $f\_U$ must be injective for every open subset $U \subset X$ because of $(1)$. So f is injective.
What do you think? Am I missing something?
| https://mathoverflow.net/users/3233 | Is every nontrivial morphism already injective in this case? | I think you are right. Another way to prove this is the following:
Let $K=\ker [f:M\to N]$ and $I={\rm im}[f:M\to N]$.
Since $f$ is non-trivial, $I\neq 0$ and since it is torsion-free (as a subsheaf of $N$), $I\_p\neq 0$. Then $K\_p\subsetneq M\_p$, so $K\_p=0$, but then $K$ is a torsion-sheaf and hence $0$ since it is a subsheaf of the torsion-free sheaf $M$.
| 4 | https://mathoverflow.net/users/10076 | 43854 | 27,867 |
https://mathoverflow.net/questions/43836 | 3 | Hi!
I started to read the chapter 31 in Jechs book about proper forcing. Unfortunately it is written in a rather sketchy way and I do have some issues in proving a lemma about two equivalent definitions of $(M,P)$-genericity. Jech gives the following definition:
Let $(P,<)$ be a fixed notion of forcing, $\lambda > 2^{|P|}$ and let $M \prec (H\_{\lambda}, \in, <, ..)$, where $H\_{\lambda}$ is the set of all sets whose transitive closure has size less than $\lambda$, and $' <'$ is a well-ordering of $H\_{\lambda}$. Then a condition $q \in P$ is $(M,P)$-generic if for every maximal antichain $A \in M$, the set $A \cap M$ is predense below $q$.
He moves on to state a small lemma, saying that its proof is a routine verification (that leads us to the assertion that I am not an old hand):
Assume the same situation as in our definition above. Then the following are equivalent:
(i) $q$ is $(M,P)$-generic.
(ii) $ q \Vdash \dot{G} \cap M$ is a filter on $P$ generic over $M$
The (ii) $\Rightarrow$ (i) direction is quite easy but I couldn't find a proof for (i) $\Rightarrow$ (ii).
Thank you
| https://mathoverflow.net/users/4753 | Equivalent definitions of $(M,P)$-genericity | Let $G$ be a $(V,P)$-generic filter containing $q$. We want to show that $G \cap M$ is a filter on $P$ generic over $M$. $G \cap M$ will be countable so it typically won't be closed upwards in $P$, the relevant point is that it's closed upwards in $P \cap M$. So it's not hard to see that it's a filter on $P \cap M$. Next we need to see it's generic over $M$.
Let $A \in M$ be a maximal antichain in $P$. Let $p$ be the sole element of $A \cap G$. Let $r$ extend both $p$ and $q$, which is possible because they're both in $G$ and $G$'s a filter. By (i), $A \cap M$ is predense below $q$, and $r \leq q$, so there's some $p' \in A \cap M$ compatible with $r$. But since $A$ is a maximal antichain, there's precisely one element of $A$ compatible with $r$, and we already know $p \in A$ is compatible with $r$ (in fact it's above $r$), so $p' = p$. Therefore $p \in A \cap G \cap M$, meaning $G \cap M$ meets $A$. Since $A$ was arbitrary, $G \cap M$ meets every maximal antichain belonging to $M$, hence it's generic over $M$.
Just as genericity of a filter can be defined in terms of dense sets, open dense sets, predense sets, or maximal antichains, it can so be done for genericity of a condition. See Lemma 1.2 in my notes below; it gives three equivalent conditions for $(M,P)$-genericity. The first and third are just like your (i) and (ii) but with antichains replaced with dense sets. The second one is also in terms of dense sets but slightly different from your (i) and (ii), and it's another useful way of looking at $(M,P)$-genericity.
<http://math.berkeley.edu/~akgupta/PFA.pdf>
| 4 | https://mathoverflow.net/users/7521 | 43859 | 27,871 |
https://mathoverflow.net/questions/43110 | 12 |
>
> Assume $\Gamma$ acts by isometries on a separable Hilbert space $H$, and
> $$\operatorname{diam} H/\Gamma\le 1.$$
> Is it true that $H/\Gamma$ is compact?
>
>
>
---
**Stupid example.** Assume the action of $\Gamma$ on $H=\ell\_2$ is generated by coordinate translations $x\_n\mapsto x\_n+\epsilon\_n$. Then
$$\operatorname{diam} H/\Gamma=\tfrac12\cdot\sqrt{\sum\_{n=1}^\infty\epsilon\_n^2}.$$
Thus, if $\operatorname{diam} H/\Gamma\le 1$ then $H/\Gamma$ is a quotient of Hilbert cube, and has to be compact.
| https://mathoverflow.net/users/1441 | Cobounded ⇒ cocompact? | The answer is "NO". To show this let us use the following:
>
> **Lemma.** Let $L$ be a lattice in $\mathbb R^q$ ($q$ is any positive integer).
> Assume $$\operatorname{diam} \mathbb R^q/L>1000.$$
> Then there is a midpoint $m$ of two points in $L$ such that $|m-x|>1$ for any $x\in L$.
>
>
>
Modulo Lemma one can construct an action of parallel translations the following way:
Let us construct inductively a sequence of lattices $L\_q$ on $\mathbb R^q$ such that $\mathop{diam} \mathbb R^q/L\_q<1000$ and such that $|x|>1$ for any $x\in L$.
Start with standard $L\_1=\mathbb Z$ in $\mathbb R$.
To construct $L\_{q}$ take
$$L\_{q}'=L\_{q-1}\times \mathbb Z\subset \mathbb R^{q-1}\times\mathbb R = \mathbb R^{q}.$$
If $\mathop{diam} \mathbb R^q/L'\_q < 1000$ set $L\_q = L'\_q$.
Othewise pass to the minimal lattice which contains $L'\_q$ and the midpoint provided by the Lemma.
Applying this construction finitely many times you get a lattice $L\_q$ with $\mathop{diam} \mathbb R^q/L\_q<1000$.
Continue the process, we get lattice $L\_\infty$ in $H$ which is a $1000$-net, its fundamental doamin contains a ball of radius 1; i.e. $H/L\_\infty$ is not compact.
*Proof of Lemma.*
For $z\in\mathbb R^q$, denote by $\rho(z)$ the minimal distance to a point in $L$.
Take a point $z\in\mathbb R^q$ which maximize distance to $L$.
So $\rho(z)\ge 1000$. Then there is a couple of points $x,y\in L$ such that
$\angle xzy\ge\pi/2$ and $|x-z|=|x-z|=\rho(z)$.
Let $m$ be the midpoint for $x$ and $y$.
Then
$$|z-m|\le \frac{\rho(z)}{\sqrt{2}}$$ and therefore the distance from $m$ to any point of $L$ is at least $1000{\cdot}(1-\tfrac1{\sqrt{2}})>1$. $\square$
| 4 | https://mathoverflow.net/users/1441 | 43862 | 27,873 |
https://mathoverflow.net/questions/43861 | 13 | Apart from J. B Nation's [Notes on Lattice Theory](https://math.hawaii.edu/%7Ejb/), is there any other (mostly introductory) material on Lattices available online?
**NB**: The last update of Nation's notes was 2017, as of Feb 2023.
| https://mathoverflow.net/users/1662 | Online introduction to Lattice Theory? | For something brief to begin with see the [notes](http://www.rasmusen.org/GI/lattice.theory.notes.txt) by Eric Rasmusen, the introductions to lattice theory by [Zukowski](http://mizar.org/JFM/pdf/lattices.pdf) and [Wang](http://www.math.s.chiba-u.ac.jp/%7Ewang/research/coin/lattice.pdf)
[An essay on history](http://www.ams.org/notices/199711/comm-rota.pdf), somewhat from a personal view, by Giancarlo Rota is also nice.
| 6 | https://mathoverflow.net/users/2149 | 43869 | 27,878 |
https://mathoverflow.net/questions/43855 | 0 | Let $C\_c^\infty(\mathbb R^n)$ be the functions from $\mathbb R^n$ to $\mathbb R$ with compact support, further let $X$ be a separable Hilbert space with a fixed orthonormal basis $(e\_n)\_n$. Define the cylindrical functions:
\begin{align\*}
\text{Cyl}(X) := \{f : X \to &\mathbb R : \text{there exists } d \in \mathbb{N} \text{ and } \phi \in C\_c^\infty(\mathbb R^n) \text{ such that }\\\
&f(x) = \phi(\langle x, e\_1 \rangle, \ldots , \langle x, e\_d \rangle ) \text{ for all } x \in X \}
\end{align\*}
Now, if $\langle . , . \rangle$ is the inner product on $X$ define
$$\langle x, y \rangle\_\omega := \sum\_n \frac{1}{n^2} \langle x, e\_n \rangle \langle e\_n, y \rangle$$
Now, why is every $f \in \text{Cyl}(X)$ continuous with respect to $\langle . , . \rangle\_\omega$? Sure, it is Lipschitz and continuous with respect to the weak topology (because it is with respect to the strong topology). Further I know that on bounded sets, the topology induced by $\langle . , . \rangle\_\omega$ is the same as the weak topology. However, $f^{-1}(A)$ does not have to be bounded. What am I missing, I'm sure it is easy.
| https://mathoverflow.net/users/5295 | Continuity of cylindrical functions. | For each $j$, the inequality $$|\langle x , e\_j\rangle| = j \sqrt{\langle x , e\_j\rangle\langle e\_j , x\rangle\over j^2}\leq j\sqrt{\langle x , x\rangle\_\omega}$$ shows that the map $x\mapsto \langle x , e\_j\rangle$ is continuous from $(X , \langle\cdot ,\cdot\rangle\_\omega)$ to $\mathbb R$. Thus, for fixed $d$ the map $x\mapsto (\langle x , e\_1\rangle, \dots,\langle x , e\_d\rangle)$ is continuous from $(X , \langle\cdot ,\cdot\rangle\_\omega) $ into ${\mathbb R}^d$. Composing with the smooth map $\phi:{\mathbb R}^d\to {\mathbb R}$ gives you a continuous function from $(X , \langle\cdot ,\cdot\rangle\_\omega)$ into $\mathbb R$ again.
| 1 | https://mathoverflow.net/users/nan | 43871 | 27,880 |
https://mathoverflow.net/questions/43846 | 23 | My question is how should one think of p-adic L functions? I know they have been constructed classically by interpolating values of complex L-functions. Recently I have seen people think about them in terms of Euler systems. But we know only a few Euler systems and there are lot of p-adic L functions.
In case of elliptic curves(at least over $\mathbb{Q}$) complex L-functions give information about the Galois representations. Should the p-adic L-function give some information about some p-adic Galois representation? It seems to be the case in case of cyclotomic fields where we think of the cyclotomic character as a 1-dimensional representation.
I apologize in advance if my questions are vague. I am just starting to learn about the subject.
| https://mathoverflow.net/users/2081 | P-adic L functions | There are three way to obtain $p$-adic L-functions. The big dream is that one can do all of them for a large class of $p$-adic Galois representations $V$. To study them one starts best to look at the cases $\mathbb{Q}(1)$ for the classical Kubota-Leopoldt $p$-adic $L$-functions or the Tate-module of an elliptic curve etc.
Let $K\_{\infty}=\mathbb{Q}(\mu\_{p^{\infty}})$ be the union of all cyclotomic fields of roots of unity of $p$-power order. Let $G$ be its Galois group, which is isomorphic to $\mathbb{Z}\_p^{\times}$.
* Attached to $V$ there is a complex $L$-function and there are conjectures saying that certain values are algebraic and satisfy to certain congruences modulo powers of $p$, e.g. Kummer congruences. So in some cases, one can show the algebraicity and the congruences. So the values fit together to a $p$-adic analytic function. But the better way of presenting the $p$-adic $L$-function is by constructing a measure on the Galois group $G$ with values in $\mathbb{C}\_p$. One can then evaluate the $p$-adic $L$-function on characters of the group $G$. This way the $p$-adic $L$-function resembles a lot its complex counterpart as they are described in Tate's thesis. See Lang's Cyclotomic Fields or Washington or Mazur-Tate-Teitelbaum for instance.
* On the algebraic side, we have a Selmer group or a class group that we watch growing in the tower $K\_{\infty}/\mathbb{Q}$. The characteristic series of the dual of this Selmer group as a $\Lambda$-module is a sort of a generating function for this growth. Like zeta-functions for varieties over finite fields. These characteristic series are in fact power-series, but they are defined up to a unit (as they are generators of some ideal). Greenberg's paper give a good introduction to this side.
* The Euler system (if we are lucky to be in one of the few cases where we have one) is a system of norm-compatible cohomology classes. In particular they give an element in $H^1(K\_n, V)$ for each intermediate field $K\_n$. But there should be an element over sufficiently many abelian extensions. The norm-compatibility is involves a factor that looks like an Euler factor of the complex $L$-function. There is a general map, called the Coleman map or the logarithme élargi or whatever, from the inverse limit of the $H^1(K\_{n,p}, V)$ to a ring of power-series. The image of the Euler system under this map should be the analytically defined $p$-adic $L$-function. Typically one shows that they satisfy the same interpolation property.
In some sense the Euler system is the bridge between the analytic and the algebraic world. Under the Coleman map it links to the analytic side. In the other direction, one can form derivative classes out of the cohomology classes. These derived classes can be analysed locally and they can be used to bound the Selmer group and hence the characteristic series. That is how one can prove the main conjecture in some cases in one direction. Probably a good place to start is Coates-Sujatha.
The $p$-adic $L$-function of an elliptic curve is conjectured to satisfy a $p$-adic Birch and Swinnerton-Dyer formula. (Mazur-Tate-Teitelbaum and Bernardi-Perrin-Riou in the supersingular case). On the algebraic side instead, we know almost that the characteristic series satisfies this formula. The order of vanishing is known to be at least as large as the rank and if they agree then the leading term has the desired shape involving the Tate-Shafarevich group; of course only up to a $p$-adic unit.
In the geometric case, say an elliptic curve over a function field $K$ of a curve over a finite field $k$, the complex and the $p$-adic function are the same ($p\neq\text{char}(k)$), since they are both just a polynomial with integer coefficients. Tate's Bourbaki talk on BSD shows how one can use the tower $K\_{\infty} = \bar{k} \cdot K$ to prove a good deal about BSD. Iwasawa theory tries to mimic this.
So I believe that $p$-adic $L$-functions are just as nice and interesting as their complex counterparts. Even if they seem more mysterious and the definition is less straight forward, we sometimes know more about them. Now I stop otherwise I am going to write a book about it here...
| 26 | https://mathoverflow.net/users/5015 | 43885 | 27,888 |
https://mathoverflow.net/questions/43743 | 8 | An alternative title is: When can I homotope a continuous map to a smooth immersion?
I have a simple topology problem but it's outside my area of expertise and I worry may be rather subtle. Any help would be appreciated.
The set-up is the following:
Let $M$ be some (closed say) $n$ dimensional manifold and suppose that $\Sigma\_1$ and $\Sigma\_2$ are two closed submanifolds of $M$ of dimension $k$. Note that $\Sigma\_1$ and $\Sigma\_2$ are allowed to intersect (in my situation they are also embedded but I don't believe this is effects anything). Suppose in addition that $\Sigma\_1$ and $\Sigma\_2$ are homologous. If $k\leq n-2$ I would like a compact manifold with boundary $\Gamma$ with $\partial \Gamma=\gamma\_1\cup \gamma\_2$ and a smooth immersion $F:\Gamma\to M$ so that $F(\gamma\_i)=\Sigma\_i$. In other, words the homology between $\Sigma\_1$ and $\Sigma\_2$ can be realized by a smooth immersion.
I believe by approximation arguments one can always get a smooth such $F$ without restriction on $k$ but it need not be an immersion (especially if $k=n-1$). My gut is that when you have $k\leq n-2$ since the dimension of the image of $F$ is codimension one you have enough room to perturb it to be an immersion. That is that $F$ is homotopic rel boundary to our desired immersion.
Unfortunately, I don't know enough to formalize this and all my intuition comes from considering curves and domains in $\mathbb{R}^3$ so I'm afraid there may be obstructions in general.
References would be greatly appreciated.
Thanks!
Edit:
As suspected, the question is somewhat subtle . To make it tractable lets assume that $M$ is a $C^\infty$ domain in $\mathbb{R}^3$ (so is a fairly simple three-manifold with boundary) and that the $\Sigma\_i$ are curves. This is where my intuition says that there should be such a smooth immersion.
| https://mathoverflow.net/users/26801 | Realizing a homology by a smooth immersion | There is a general strategy for these kind of problems, which sometimes helps (the ''h-principle''): separate the homotopical and smooth aspects of the problem. Setup: $f:N \to M$ a map of smooth manifolds, $dim (N) < dim (M)$, $f|\_{\partial N}$ is an immersion.
Step 1: if your $f$ is going to be homotopic to an immersion $g$, then there should exist a bundle monomorphism $\phi:TN \to f^\* TM$, extending the derivative of $f|\_{\partial}$ to all of $N$. Otherwise, the hypothetical immersion won't exist, because there is no candidate for its derivative. Constructing this monomorphism is a purely bundle theoretic problem. Whether this can be successfully attacked depends very much on the concrete situation -- maybe it is trivial, maybe it is very hard.
In any case -- if there is no candidate for an immersion in sight -- it is be easier to construct a bundle map than an immersion.
Step 2: if you succeed in solving the homotopy problem, you are done! This is the content of the Smale-Hirsch-Theorem: once $\phi$ exists, $f$ will be homotopic to an immersion $g$ and the derivative is homotopic (as a bundle mono) to $\phi$. There is a relative statement as well; you have to extend the immersion $f\_{\partial M}$ to an immersion of a collar of $\partial M$ in $M$. This is not trivial since the collar has one dimension
more. If $dim(N)=dim(M)$, then the result still holds (in your situation, where $N$ is a connected cobordism with nonempty boundary), but you cannot fix $f$ on the boundary.
Your intuition in the low-dimensional case seems to be true: if $N$ is a surface and $M$ a domain in $R^3$, then both are parallelizable.
References: Eliashberg-Michachev: Introduction to the h-principle, Haefliger: Lectures on the theorem of Gromov, Ponaru: Homotopy theory and differentiable singularities. Adachi: Immersions and embeddings.
| 7 | https://mathoverflow.net/users/9928 | 43899 | 27,897 |
https://mathoverflow.net/questions/43888 | 0 | Given a commutative Artin algebra $A$ over an algebraically closed field $k$ one has a decomposition $A=A\_1\oplus\ldots\oplus A\_n$ into local Artin subalgebras, see for example *Atiyah-McDonald, Introduction To Commutative Algebra, Theorem 8.7*. The subalgebras $A\_i$ are uniquely determined up to the isomorphism.
The question is as follows. Are the inclusions $A\_i\subset A$ uniquely determined as well?
They should be, but I cannot find an accurate proof.
UPD: So, the inclusions are not necessarily unique. But may there exist an infinite number of inclusions? Or the number of inclusions is necessarily finite?
Motivation: If there is a finite number of ways for the embedding $A\_i\to A$ then the connected group of unity $(Aut A)^{\circ}$ of the automorphism group of algebra $A$ stabilizes the subalgebra $A\_i$.
| https://mathoverflow.net/users/10386 | local Artin algebras | Yes, the decomposition is unique. The uniqueness of inclusions is a moot point because rings may have nontrivial endomorphisms.
The proof goes like this: consider decompositions of $1$ into the sums of orthogonal idempotents $1=\sum\_i p\_i$. Orthogonality means that $p\_ip\_j=0$ whenever $i\neq j$. From general nonsense (commutativity will be needed) you can find unique maximal decomposition and then $A\_i = p\_iA$.
| 4 | https://mathoverflow.net/users/5301 | 43902 | 27,899 |
https://mathoverflow.net/questions/43864 | 11 | **Could one describe the subsets of the integers closed under the binary operation Ax+By
where A and B are arbitrary fixed integers ?** That is, describe the subsets S
of the integers such that if $x,y\in S$ then $Ax+By\in S$. Or just the minimal such subsets
containing 1.
Do I guess correctly that this question belongs to additive combinatorics ?
| https://mathoverflow.net/users/4745 | describe subsets of the integers closed under the binary operation Ax+By | I think the problem is pretty much solved in a series of papers by Klarner et al;
David A Klarner and Karel Post, Some fascinating integer sequences.
A collection of contributions in honour of Jack van Lint.
Discrete Math. 106/107 (1992), 303–309, MR 93i:11031
D G Hoffman and D A Klarner, Sets of integers closed under affine operators—the finite basis theorem.
Pacific J. Math. 83 (1979), no. 1, 135–144, MR 83e:10080
D G Hoffman and D A Klarner, Sets of integers closed under affine operators—the closure of finite sets.
Pacific J. Math. 78 (1978), no. 2, 337–344, MR 80i:10075
| 6 | https://mathoverflow.net/users/3684 | 43927 | 27,913 |
https://mathoverflow.net/questions/43923 | 23 | The following problem is homework of a sort -- but homework I can't do!
The following problem is in Problem 1.F in *Van Lint and Wilson*:
>
> Let $G$ be a graph where every vertex
> has degree $d$. Suppose that $G$ has
> no loops, multiple edges, $3$-cycles
> or $4$-cycles. Then $G$ has at least
> $d^2+1$ vertices. When can equality
> occur?
>
>
>
I assigned the lower bound early on in my graph theory course. Solutions for $d=2$ and $d=3$ are easy to find. Then, last week, when I covered eigenvalue methods, I had people use them to show that there were no solutions for $d=4$, $5$, $6$, $8$, $9$ or $10$. (Problem 2 [here](http://www.math.lsa.umich.edu/~speyer/PSet6.pdf).) I can go beyond this and show that the only possible values are $d \in \{ 2,3,7,57 \}$, and I wrote this up in a [handout](http://www.math.lsa.umich.edu/~speyer/Solution6.pdf) for my students.
Does anyone know if the last two exist? I'd like to tell my class the complete story.
| https://mathoverflow.net/users/297 | Is there a 7-regular graph on 50 vertices with girth 5? What about 57-regular on 3250 vertices? | This is the [Moore graph](http://en.wikipedia.org/wiki/Moore_graph), which is a regular graph of degree $d$ with diameter $k$, with maximum possible nodes. A calculation shows that the number of nodes $n$ is at most
$$
1+d \sum\_{i=0}^{k-1} (d-1)^i
$$
and as you mentioned it can be shown by spectral techniques that the only possible values for $d$ are
$$ d = 2,3,7,57. $$
Example for $d=7$ is the [Hoffman–Singleton graph](http://en.wikipedia.org/wiki/Hoffman-Singleton_graph), but for the case $d=57$ it is still open. See Theorem 8.1.5 in the book "[Spectra of graphs](http://homepages.cwi.nl/~aeb/math/ipm.pdf)" by Brouwer and Haemers for reference.
| 26 | https://mathoverflow.net/users/4248 | 43928 | 27,914 |
https://mathoverflow.net/questions/43924 | 6 | This question is inspired by a [riddle](https://math.stackexchange.com/questions/8101/iterated-polynomial-problem) in math.stackexchange.
Let $P$ be a polynomial, and $O = \{P^{(n)}(0) : n \geq 0\}$ be its orbit under zero (viewed as a set). Suppose that $O$ contains infinitely many integers. Is it true that for some $n$, $P^{(n)}$ is a polynomial with integral coefficients?
We can ask the same question replacing integers with rationals.
EDIT: Nick and David gave simple counterexamples for the first question.
Still open:
1. In the setting of the original question, is it true that some composition power of $P$ takes integers to integers?
2. The original question with rationals.
| https://mathoverflow.net/users/7732 | When are infinitely many points in the orbit of a polynomial integers? | $P(x)= \frac{x(x+1)}{2} +1$.
It is easy to see that $P^{n+1}(0) > P^n(0)$ and $P$ maps the integers into the integers.
But I think (didn't check it, might be one of these facts which are obvious but wrong) that
$$P^{(n)}(x) = \frac{1}{2^{m}} x^{2^n}+....\notin \mathbb{Z} $$
where $m$ is probably $m=2^n+1$.
The right question to ask might be if $f$ maps the integers into the integers....
**Disregard the following part**, as it was pointed in the comments, it only works if for each $k$ we can find an $l$ and $n\_1,..., n\_k$ so that $f^{(n\_i)}(0)$ and $f^{(n\_i+l)}(0)$ are integers(or rational for the second question).
EDIT: P.S. The answer with the rationals turns out to be true, I think (my algebra is rusty):
Let $P$ be such a polynomial, and let $m$ be the degree of $P$. Then using the Lagrange interpolation formula, you can reconstruct $P(x)$ from $m+1$ distinct integer values of the type $P^{(k)}(0)$, and since all of these are rational, all the coefficients are rational. Actually this way one can prove the following Lemma:
| 7 | https://mathoverflow.net/users/9313 | 43929 | 27,915 |
https://mathoverflow.net/questions/42160 | 6 | Let A be finite commutative group say $(Z\_m)^h$. I will say that $S \subset A$ is an orbit if exist group $H$
which acts on A such that $S$ is an orbit of $H$.
Can one give a simple characterization of all orbits of $(Z\_m)^h$?
By action on $A$ I mean automorphisms of a group A.
| https://mathoverflow.net/users/4246 | Orbits in commutative groups. | The abelian group in question is the product of its Sylow-$p$ subgroups, which are preserved by automorphisms. Therefore the orbits in it are the products of orbits in the Sylow $p$-subgroups. Therefore, we may consider the case where $m=p^k$ for some prime $p$ and some natural number $k$.
I can answer this question for maximal orbits (orbits under the full automorphism group). I think the more general questions may not have a nice answer.
In $(Z\_{p^k})^h$, there are precisely $k+1$ orbits of the full automorphism group, represented by $e, pe, \ldots, p^ke$, where $e=(1,0,\ldots,0)$. The orbit of $p^i e$ consists of those vectors where the gcd of entries divides $p^i$, but not $p^{i+1}$ (except for $i=k$, where the orbit is just the element $0$).
For a general finite abelian group, this problem was solved more than a 100 years ago by Miller, and also discussed by Birkhoff and Baer. For the exact references, as well as a modern treatment see <http://arxiv.org/abs/1005.5222>.
| 3 | https://mathoverflow.net/users/9672 | 43935 | 27,919 |
https://mathoverflow.net/questions/43934 | 3 | I'd like to prove that two definitions of sheaves on a site are equivalent, but I'm having trouble proving one direction.
Let $C$ be a category with pullbacks. Let $(C,T)$ be a site defined through a collection $\Phi$ of covering families. (So that a subfunctor of $Hom(-,U)$ is a covering sieve in $T$ if and only if it contains a covering family in $\Phi$).
Let $F$ be a presheaf on $C$, taking values in some complete category.
Then it should be the case that the following are equivalent characterisations of $F$ being a sheaf.
1) For each $U\in C$, and each covering sieve $R \subset Hom(-,U)$, the map $F(U) \to {{\lim \atop \leftarrow} \atop {V\to U \in R}} F(V)$ is an isomorphism.
2) The following diagram is an equalizer for each $U$ in $C$, and covering family {$U\_\alpha \to U$} in $\Phi$:
$$ F(U) \to \prod\_\gamma F(U\_\gamma) {\rightarrow \atop \rightarrow} \prod\_{(\alpha,\beta)} F(U\_\alpha \times\_U U\_\beta)$$
**Question:** How do you prove $2) \implies 1)$?. (I've proved $1) \implies 2)$)
Here's my proof attempt of $2) \implies 1)$ so far:
$\bullet$ Suppose we have a covering family {$U\_\alpha \to U$} in $\Phi$,and that it generates a sieve $S$. I proved that the following is an equalizer:
$$ {{\lim \atop \leftarrow} \atop {V\to U \in S}} F(V) \to \prod\_\gamma F(U\_\gamma) {\rightarrow \atop \rightarrow} \prod\_{(\alpha,\beta)} F(U\_\alpha \times\_U U\_\beta)$$
(This is also what I used to prove $1) \implies 2)$).
$\bullet$ Let $R$ be an arbitrary covering sieve, which is a subfunctor of $Hom(-,U)$. Then it contains a covering family {$U\_\alpha \to U$}, and the diagram of $2)$ is an equalizer by assumption. If $S$ is the covering sieve generated by the covering family (so $S \subset R$), the diagram above is also an equalizer. By an isomorphism of equalizers, the map $F(U) \to {{\lim \atop \leftarrow} \atop {V\to U \in S}} F(V)$ is an isomorphism.
Because of the commutative diagram
$F(U) \to {{\lim \atop \leftarrow} \atop {V\to U \in R}} F(V)$
$\ \ \ \ \ \ \cong \searrow \ \ \ \ \ \downarrow$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\lim \atop \leftarrow} \atop {V\to U \in S}} F(V)$
we get that the map $F(U) \to {{\lim \atop \leftarrow} \atop {V\to U \in R}} F(V)$ is a split monic. From here, I'm not sure how to show that it's also epi.
So the crux of my issue is that I can prove the isomorphism in $1)$ only for covering sieves generated by covering families, but not for arbitrary covering sieves.
| https://mathoverflow.net/users/9109 | Equivalence of two definitions of sheaves on a site | <http://arxiv.org/abs/math/0412512> Prop 2.4.2 (sorry no time to type it myself right now).
| 3 | https://mathoverflow.net/users/4528 | 43948 | 27,928 |
https://mathoverflow.net/questions/43950 | 18 | The symbol $\Subset$ (occurring in places where $\subseteq$ could occur syntactically) comes up frequently in a paper I'm reading. The paper lives at the intersection of a few areas of math, and I don't even know where to begin looking for the meaning of a symbol whose latex code is "\Subset". Do you know what this usually denotes?
Edit: some context follows.
All the sets in question are subsets of $\hat{\mathbb{C}} = \mathbb{C}\cup\{\infty\}$.
Example 1. In a situation where $J$ is closed with empty interior, $U$ and $V$ are closed with $U\subsetneq V$, it is written "Note that $J \Subset U$ and, selecting a neighborhood $W \subset U$ of $J$ which is compactly contained in $V$, ..."
Example 2. In a situation where $R$ is a rational mapping, and where it is assumed that $B\subset \hat{\mathbb{C}}$ is such that $R(B)\Subset B$, it is written "Let $\Omega\_0 = \hat{\mathbb{C}}\setminus B$. Define $\Omega\_1 = R^{-1}(\Omega\_0)$. By the properties of $B$, we have $\Omega\_1\Subset\Omega\_0$. If we let $U\_0$ be any finite union of closed balls such that $\Omega\_1 \subset U\_0 \subset \Omega\_0$, ..."
In both cases I have paraphrased to simplify the notation, so I hope I have not introduced errors into it.
| https://mathoverflow.net/users/6649 | Meaning of $\Subset$ notation | In my experience $U \Subset V$ means that the closure of U is a compact subset of V. ${}{}$
| 24 | https://mathoverflow.net/users/10073 | 43953 | 27,932 |
https://mathoverflow.net/questions/43709 | 12 | Let $p\in M$ be a point in a closed riemannian manifold $M$. Recall that the *cut locus* of $p$ is the subset of $M$ consisting of all points that are connected to $p$ by at least 2 distance-minimizing geodesics.
I will start with a general question:
>
> Is it true that for generic metrics on $M$ and generic points $p$, the cut locus of $p$ is a $(n-1)$-dimensional polyhedron with generic singularities?
>
>
>
By "generic singularities" I mean that $M$ is a simple polyhedron. See for instance [this paper of Alexander and Bishop](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.40.6959).
This property is certainly not satisfied for some important specific metrics: for instance, if $M$ is a round sphere the cut locus is a point, no matter where $p$ is. If $M$ is a flat torus, we get a generic polyhedron for generic flat metrics. What about hyperbolic manifolds? So, this is my question:
>
> Let $M^n$ be a hyperbolic $n$-manifold. Is the cut locus of a generic point a $(n-1)$-polyhedron with generic singularities?
>
>
>
Of course I am mostly interested in the case $n=3$. In dimension $n=2$ one may also pick a generic hyperbolic metric.
**Edit**: In dimension 1, a simple polyhedron is a graph with vertices of valence 2 or 3. In dimension 2, it is a polyhedron such that the link of a point is either a circle, a circle with a diameter, or a circle with three radii.
In general, a $n$-dimensional compact polyhedron is *simple* if every point has a neighborhood which is the cone over the $(k-1)$-skeleton of the $(k+1)$-simplex, times a $(n-k)$-disc.
| https://mathoverflow.net/users/6205 | Is the cut locus of a generic point in a hyperbolic manifold a generic polyhedron? | The question you pose is stated as an open question (in the 3-dimensional hyperbolic case) in the following paper:
Díaz, Raquel; Ushijima, Akira
On the properness of some algebraic equations appearing in Fuchsian groups.
Topology Proc. 33 (2009), 81–106.
Quoting from the review on mathscinet:
[the paper] takes its motivation from the fact that, apparently, the statement about the genericity of Dirichlet fundamental polyhedra is open for $\mathbb{H}^3$. (According to the authors, the paper of T. Jørgensen and A. Marden [in Holomorphic functions and moduli, Vol. II (Berkeley, CA, 1986), 69--85, Springer, New York, 1988] has a gap which the present authors have so far been unable to fix.)
| 7 | https://mathoverflow.net/users/6206 | 43954 | 27,933 |
https://mathoverflow.net/questions/43962 | 11 | Let $G$ be a reductive group over an algebraic number field $k$. Denote with $k\_v$ a local field and with $A$ the ring of its adeles, let $G\_k$, $G\_{k\_v}$ resp. $G\_A$ be the group of its $k$- resp. $A$- points. What are necessary and sufficient conditions for a local representation $\pi\_v$ of $G\_{k\_v}$ to appear as $\otimes\_v$ $\pi\_v$ in the right regular representation of $G\_k \backslash G\_A$? What is a good reference to study this local to global process?
| https://mathoverflow.net/users/10400 | Local to Global principle for reductive groups | If $\pi\_v$ is supercuspidal, then (after making a twist if necessary) it should be possible to find an automorphic $\pi$ with $\pi\_v$ as the local factor at $v$. This kind of result is usually proved (although their are sometimes other possibilities) by an application of the simple trace formula. This method will also give good (although perhaps not complete) control of the ramification at other places. If you look at the paper of Deligne, Kazhdan, and Vigneras on Jacquet--Langlands for $GL\_n$, I think you will find an expose of the technique.
If $\pi\_v$ is not supercuspidal (or if one is not willing to make a twist), then this is not generally possible, just for cardinality reasons. The link that David Loeffler provides to the discussion of the $GL\_2$ case is somewhat indicative of the situation.
| 11 | https://mathoverflow.net/users/2874 | 43969 | 27,940 |
https://mathoverflow.net/questions/43979 | 16 | [Sturm's theorem](http://en.wikipedia.org/wiki/Sturm%27s_theorem) gives a way to compute the number of roots of a one-variable polynomial in an interval [a,b]. Is there a generalization to boxes in higher dimensions? Namely, let $P\_1,\dotsc,P\_n\in \mathbb{R}[X\_1,\dotsc,X\_n]$ be a collection of $n$ polynomials such that there are only finitely many roots of $P\_1=P\_2=\dotsb=P\_n=0$. I want to be able to compute the number of roots in $[a,b]^n$. I do not care if the roots are counted with or without multiplicity.
I would also be interested in upper bounds on the number of roots that is similar to [Descartes' rule of signs](http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs). The only work in this connection that I managed to find is by [Itenberg and Roy](http://www.ams.org/mathscinet-getitem?mr=1422527), who postulated a conjectural extension of Descartes' rule of signs, which however later was [shown to be false](http://www.ams.org/mathscinet-getitem?mr=1614424).
| https://mathoverflow.net/users/806 | Counting roots: multidimensional Sturm's theorem | To expand on David's answer, the bound given by Khovanskii's theorem is of the form $2^{\binom{N}{2}} (n+1)^N$ per quadrant (more or less). Incremental improvements on this bounds have been obtained <http://arxiv.org/abs/1010.2962> being the latest, but nothing revolutionary and we're nowhere near realistic bounds.
As far as I know, there have been no new attempts on a multivariable Descartes (something that would take signs into consideration) since Itenberg and Roy's paper, and it remains a major open problem in the area.
**Added Later:** As far as an *algorithm* is concerned, I don't think you can count without solving, in which case the standard one to use would be the Cylindrical Algebraic Decomposition pioneered by Collins. It is more or less based on Sturm and induction, but it is a *lot* more involved than the 1-dimensional case. For reference, I recommend the book by Basu Pollack and Roy [Algorithms in Real Algebraic Geometry](http://Cylindrical%20Algebraic%20Decomposition%20pioneered%20by%20Collins/) (free download).
(More on counting without solving: Marie-Françoise Roy mentions the problem of determining if a semi-algebraic set is non-empty *without* producing one point per connected components as one of the major open problems in algorithmic real algebraic geometry).
| 9 | https://mathoverflow.net/users/8212 | 43984 | 27,946 |
https://mathoverflow.net/questions/43965 | 16 | Given a $m \times n$ matrix $n>m$, I was trying to check if all its $m \times m$ minor vanish.
I remember hearing that one really does not need to check all possible minors in order to conclude that all of them would vanish.
If such a result is true, how many minors will do the job and which ones ?
I am wondering if it is even possible to calculate the value of all minors based on the value of a nicely chosen "generating subset" ?
---
**Edit:-** The question which I had asked does not have an affirmative answer as explained
by Steven Sam. But matrix minors do satisfy some relationships see the answer by Sheikraisinrollbank below. **If someone can modify the question to a more appropriate one (in light of Steven Sam and Sheikraisinrollbank answers ) please feel free to do so.**
I have often come across a situation (more so at present than ever before) where in order to answer a problem in my subject area I am led to questions which are totally different areas about which I have absolutely no familiarity. Most often these are quite basic and I would suppose well known to any one who works in those areas. It is natural that a person who is not familiar with a given field will end up asking for "a result of the following kind" rather than a precise question. For a person who is knowledgeable I understand the question may be irritating or look ill posed but mind you the hapless fellow is not a graduate student in the given field and please do not judge him accordingly. I think its desirable that if someone knows how to reformulate the question to something so that it becomes well posed or meaningful it should be done. Why not edit the question to something so that it becomes a valid well posed question, to something which is obviously much more interesting than which was originally posed ?
| https://mathoverflow.net/users/6766 | How many minors I need to check to conclude all minors will vanish ? | To counterbalance Steven Sam's answer some (b/c the OP's intuition is correct in a sense):
It's true that the right way to check that all m by m minors are zero in practice is Gaussian elimination. However, while the minors may be linearly independent, they satisfy quadratic relations ("Plucker relations", see for instance the wikipedia article on Grassmannians) that allow you to deduce some things. In the simplest non-trivial case of 2 by 4 matrices, writing $m\_{ij}$ for the $(i,j)$th $2$ by $2$ minor one has $$m\_{12}m\_{34}-m\_{13}m\_{24}+m\_{14}m\_{23}=0.$$ This might have some theoretical value for the OP's situation that Gaussian elimination does not. For instance, in this case it allows one to deduce that if $m\_{12}$ and $m\_{13}$ are both zero then either $m\_{14}$ or $m\_{23}$ is zero. But it's hard to know if this helps without knowing somewhat more about the motivating problem.
| 12 | https://mathoverflow.net/users/8552 | 43985 | 27,947 |
https://mathoverflow.net/questions/43986 | 32 | Let $V$ be an infinite dimensional topological vector space and consider the natural application $\iota\colon V\to V^{\*\*}$. The space $V$ is said to be reflexive if $\iota$ is an isomorphism.
Are there examples where $\iota$ fails to be an isomorphism but $V$ and $V^{\*\*}$ are nevertheless isomorphic?
Can one find an example where $V$ is a Banach space and the isomorphism is actually an isometry?
| https://mathoverflow.net/users/9871 | Are there non-reflexive vector spaces isomorphic to their bi-dual? | Yes, the James space.
This is a good question, and R. C. James is rightly praised for this example.
>
> MR0044024 (13,356d)
>
> James, Robert C.
>
> A non-reflexive Banach space isometric with its second conjugate space.
>
> Proc. Nat. Acad. Sci. U. S. A. 37, (1951). 174–177.
>
>
>
| 35 | https://mathoverflow.net/users/454 | 43987 | 27,948 |
https://mathoverflow.net/questions/43974 | 0 | The contravariant functor $C(-)$ given by
$$
\hom\_{Top}(-,\mathbb{R}):cCW\to Rng
$$
where $cCW$ is the category of compact CW complexes is injective on objects. What is known about surjectivity, faithfulness and fullness of this functor?
| https://mathoverflow.net/users/2625 | The functor of continuous functions from compact CW-spaces to the reals | Corollary 4.1.(i) in Johnstone's book Stone Spaces states that the category of realcompact spaces is dual to the full subcategory
of the category of commutative rings consisting of rings of the form C(X).
The functor C implements the duality.
The category of compact CW-complexes embeds into the category of realcompact
spaces as a full subcategory, hence the functor C is fully faithful.
| 5 | https://mathoverflow.net/users/402 | 43989 | 27,949 |
https://mathoverflow.net/questions/43996 | 1 | Given a finite subset $S$ not containing the identity element in a residually finite group $G$, does there always exist a normal subgroup of $G$ which has finite index (in $G$) and
which avoids $S$? (If $S$ is a singleton, this is of course the definition of a residually finite group.)
| https://mathoverflow.net/users/4556 | Finite subsets in residually finite groups | Yes, take the intersection of the normal subgroups $N\_1, N\_2, ..., N\_k$ of finite index avoiding elements $x\_1,x\_2,...,x\_k$ of your set. It is normal and of finite index (at most the product of indices of $N\_i$).
| 6 | https://mathoverflow.net/users/nan | 43998 | 27,954 |
https://mathoverflow.net/questions/44018 | 9 | I have just learned from mighty Wikipedia that the symmetric group of an infinite set is not a matrix group. Why?
I can see rep-theory reasons: sizes of minimal non-trivial, non-sign representations of $S\_n$ grow as $n$ grows. But I believe that there should really be an elementary linear algebra argument for this. Is there one?
| https://mathoverflow.net/users/5301 | Why is symmetric group not matrix? | **Edit:** My original idea doesn't work, but unknown's does. Here are the details.
Let $k$ be a field, which is WLOG algebraically closed. Let $V$ be a finite-dimensional representation over $k$ of dimension $n$. Then $S\_{\infty}$ contains $(\mathbb{Z}/p\mathbb{Z}))^{p^n}$ (in fact any finite group) as a subgroup, where $p \neq \text{char}(k)$. Let $g\_1, ... g\_{p^n}$ be its generators. Then by "elementary linear algebra" $V$ is a direct sum of $1$-dimensional irreps of $\langle g\_1 \rangle$ which $g\_2, ... g\_{p^n}$ must preserve, while still having order $p$. But there are only $p^n - 1$ nontrivial ways to do this; hence either one of the $g\_i$ acts as the identity or two of them are the same and $V$ cannot be faithful.
Of course, whether this is "elementary linear algebra" or representation theory is debatable, and I think irrelevant. All I did was find a sequence of finite groups such that the dimension of the smallest faithful representation goes to infinity and I could have done this any number of ways, e.g. I could have chosen $\text{PSL}\_2(\mathbb{F}\_q)$.
| 13 | https://mathoverflow.net/users/290 | 44020 | 27,965 |
https://mathoverflow.net/questions/43879 | 2 | I would like to know what are the formal power series $$f(t) = \sum\_a \omega\_a t^{-a}$$ over an algebraicially closed field of characteristic 2, with two properties: (1) The series represents a rational function, i.e. the coefficients satisfy a linear recursion, and (2) $\omega\_{2a} = \omega\_a^2$ for $a \ge 0$.
One family of solutions is $\omega\_a = p\_a(u\_1, \dots, u\_r)$ where $p\_a$ is the $a$-th power sum symmetric function in some finite subset of $F$, $p\_a = \sum\_{i = 1}^r u\_i^a$.
Are these (more or less) all the solutions?
| https://mathoverflow.net/users/10381 | some rational functions over a field of characteristic 2 | Kevin Buzzard gave the solution. Here it is with a little more detail:
Our assumptions include $\omega\_0 = \omega\_0^2$. Thus $\omega\_0 \in \{0, 1\}$.
The linear homogeneous recursion only kicks in eventually; say the $\omega\_a$ for $a \ge N$ satisfy such a recursion.
Let $v\_1, \dots, v\_m$ be the distinct roots of the characteristic polynomial of the linear recursion. Then there exist polynomials $h\_1, \dots, h\_m$ such that
$\omega\_a = \sum\_{i = 1} ^m h\_i(a) v\_i^a $ for $a \ge N$. Let $\alpha\_i$ be the constant term of
$h\_i$ for each $i$. Since the characteristic is $2$, we have $h\_i(2a) = \alpha\_i$ for all $a$.
For $a \ge N$,
\begin{equation}
\sum\_i \alpha\_i v\_i^{4a} = \omega\_{4 a} = \omega\_{2a}^2 = \sum\_i \alpha\_i^2 v\_i^{4a}.
\end{equation}
Because the characteristic of $F$ is $2$, each element has a unique $2^k$--th root for all $k \ge 1$; in particular all the $v\_i^4$ are distinct, so the displayed equation implies that $\alpha\_i^2 = \alpha\_i$ for all $i$, i.e. $\alpha\_i \in \{0, 1\}$. Let $u\_1, \dots, u\_d$ be the list of those $v\_j$ such that $\alpha\_j = 1$. Then we have $\omega\_{2a} = \sum\_i u\_i^{2a}$ for
$a \ge N$. For an arbitrary $a \ge 1$, chose $k$ such that $2^{k-1} a \ge N$. Then
$\omega\_a$ is the unique $2^k$--th root of $\omega\_{2^k a} = \sum\_i u\_i^{2^k a}$, namely
$\omega\_a = \sum\_i u\_i^a$.
Thus we have $\omega\_0 \in \{0, 1\}$ and $\omega\_a = p\_a(u\_1, \dots, u\_d)$ for $a \ge 1$.
THANKS, KEVIN !
| 0 | https://mathoverflow.net/users/10381 | 44026 | 27,968 |
https://mathoverflow.net/questions/44021 | 95 | This question is only motivated by curiosity; I don't know a lot about manifold topology.
Suppose $M$ is a compact topological manifold of dimension $n$. I'll assume $n$ is large, say $n\geq 4$. The question is: *Does there exist a simplicial complex which is homeomorphic to $M$?*
What I think I know is:
* If $M$ has a piecewise linear (PL) structure, then it is triangulable, i.e., homeomorphic to a simplicial complex.
* There is a well-developed technology ("[Kirby-Siebenmann invariant](https://en.wikipedia.org/wiki/Kirby-Siebenmann_invariant)") which tells you whether or not a topological manifold admits a PL-structure.
* There are exotic triangulations of manifolds which don't come from a PL structure. I think the usual example of this is to take a [homology sphere](https://en.wikipedia.org/wiki/Homology_sphere) $S$ (a manifold with the homology of a sphere, but not maybe not homeomorphic to a sphere), triangulate it, then suspend it a bunch of times. The resulting space $M$ is supposed to be homeomorphic to a sphere (so is a manifold). It visibly comes equipped with a triangulation coming from that of $S$, but has simplices whose link is not homemorphic to a sphere; so this triangulation can't come from a PL structure on $M$.
This leaves open the possibility that there are topological manifolds which do not admit *any* PL-structure but are still homeomorphic to some simplicial complex. Is this possible?
In other words, what's the difference (if any) between "triangulable" and "admits a PL structure"?
[This Wikipedia page on 4-manifolds](https://en.wikipedia.org/wiki/4-manifold) claims that the E8-manifold is a topological manifold which is not homeomorphic to any simplicial complex; but the only evidence given is the fact that its Kirby-Siebenmann invariant is non trivial, i.e., it doesn't admit a PL structure.
| https://mathoverflow.net/users/437 | Which manifolds are homeomorphic to simplicial complexes? | Galewski-Stern proved
<https://mathscinet.ams.org/mathscinet-getitem?mr=420637>
" It follows that every topological m-manifold, m≥7 (or m≥6 if ∂M=∅), can be triangulated if and only if there exists a PL homology 3-sphere H3 with Rohlin invariant one such that H3#H3 bounds a PL acyclic 4-manifold."
The Rohlin invariant is a Z/2 valued homomorphsim on the 3-dimensional homology cobordism group, $\Theta\_3\to Z/2$, so if it splits there exist non-triangulable manifodls in high dimensions.
| 47 | https://mathoverflow.net/users/3874 | 44039 | 27,977 |
https://mathoverflow.net/questions/44005 | 6 | I asked this on mathematics stack exchange and did not receive answer . I hope it is good manners to ask here. Thank you very much.
Let $X$ be integral scheme and $\mathcal K$ sheaf of rationnal functions on $X$. For any
point $y\in X$ different of generic point we know that fiber of $\mathcal K$ (defined as usual as $\mathcal K \_y / \mathcal m\_y \mathcal K\_y$) is zero. I'll be very gratefull if you explain intuitively why this is so, in language of restriction of $\mathcal K$ to reduced subscheme $Y=\overline{\{y \} }$. I have difficulty because many rationnal functions on $X$ can be restricted to nonzero rationnal functions on $Y$ . How is that compatible with fiber of $\mathcal K$ equals zero at $y$?
| https://mathoverflow.net/users/10408 | Intuition for rational functions | The non-classical aspect of this setup is that you're using a quasi-coherent sheaf that is not coherent, and beyond the coherent case one cannot expect information about a fiber (e.g., vanishing, 6 generators, etc.) to "propogate" to information in a neighborhoood (which would be the spirit behind the choice of word "coherent", I suppose). Computing the fiber of the field of all rational functions at a non-generic point likely has no classical counterpart, much as in number theory one doesn't ever try to reduce $\mathbf{Q}$ modulo 5, only $\mathbf{Z}\_{(5)}$ or its subrings.
| 10 | https://mathoverflow.net/users/3927 | 44040 | 27,978 |
https://mathoverflow.net/questions/44001 | 1 | This is a question from Jech's Set Theory (Ex. 17.12) which I'm reading at the moment and pretty much stuck on.
>
> If $D$ is a normal measure on $\kappa$
> and $\{ \aleph\_\alpha \colon
> > 2^{\aleph\_\alpha} \le
> > \aleph\_{\alpha+\beta}\} \in D$ (for
> some constant $\beta < \kappa$), then $2^\kappa
> > \le \aleph\_{\kappa + \beta}$
>
>
>
He gives the following hint: If $f$ is such that $f(\aleph\_\alpha) = \aleph\_{\alpha+\beta}$ for all $\alpha < \kappa$, then $[f]\_D = (\aleph \_{ \kappa+j(\beta)})^M$
I think that I am just confused about the whole representation in $M$ and how to use it to solve this problem. Hints, partial or complete solutions are most welcomed.
| https://mathoverflow.net/users/7206 | Normal measures and Elementary Embeddings | The question you've stated isn't the question in Jech, you've made a minor typo. Here's the actual problem:
>
> If $\beta < \kappa$ and {$\aleph \_{\alpha} : 2^{\aleph \_{\alpha}} \leq \aleph \_{\alpha + \beta}$} $\in D$ and $D$ is a normal measure on $\kappa$, then $2^{\aleph \_{\kappa}} \leq \aleph \_{\kappa + \beta}$
>
>
>
Note that since $\kappa$ is measurable, $\aleph \_{\kappa} = \kappa$.
Okay, now we know that a normal measure extends the club filter, and the set of cardinals below $\kappa$ is club in $\kappa$, hence it makes sense in the hint to define $f(\aleph \_{\alpha}) = \aleph \_{\alpha + \beta}$ without specifying how $f$ acts on non-cardinals. Following my comment, let $g(\aleph \_{\alpha}) = 2^{\aleph \_{\alpha}}$. Then $g \leq f$ almost everywhere, and so:
>
> $M \vDash [g] \leq [f]$
>
>
>
i.e.
>
> $M \vDash j(g)(\kappa) \leq j(f)(\kappa)$
>
>
>
i.e.
>
> $M \vDash 2^{\kappa} \leq \aleph \_{\kappa + j(\beta)}$
>
>
>
Since $\beta < \kappa$, $j(\beta) = \beta$. Thus there is an injection from $(2^{\kappa})^M$ to $\aleph \_{\kappa + \beta} ^M$. Since $P(\kappa) = P^M(\kappa)$, it means there's an injection from $2^{\kappa}$ to $\aleph \_{\kappa + \beta}^M$. Finally, $\aleph \_{\kappa + \beta} ^M \leq \aleph \_{\kappa + \beta}$ since $M \subseteq V$.
| 5 | https://mathoverflow.net/users/7521 | 44044 | 27,980 |
https://mathoverflow.net/questions/44033 | 6 | I was recently thinking about efficient algorithms for modular exponentiation. This led me to the (more interesting, in my opinion) question:
>
> Let $1 < a < n$ be an integer relatively prime to $n$. What is the order of ${\overline{a}}$ in $\mathbb{Z}/n\mathbb{Z}^\*$ (the multiplicative group of $\mathbb{Z}/n\mathbb{Z}$)?
>
>
>
I did some Google searching, but all I could find were the obvious facts that the order should divide the order of the group $\phi(n)$ and the exponent of the group $\lambda(n)$ (see [Carmichael function](http://en.wikipedia.org/wiki/Carmichael_function)). I asked several people if anything more could be said, but the answers were generally: "Some people study this. It is really hard." However, I couldn't find any other references.
>
> Is this a question that has been seriously considered? If so, what is known and does anyone have any good references?
>
>
>
I am happy to suppose that we know *a priori* the prime factorization of both $a$ and $n$. Even given this information, is there something precise that can be said?
Because this is a (potentially) open problem, it is possible that it should be a community wiki page, I am not entirely certain what the policy is there. If so, someone please wiki-hammer this, as I have not the power! It might also be deserving of the open-problem tag?
**Edit**: I do in fact have the power to make community wiki posts (which I discovered by checking the faq) just not to edit someone else's. Still, I would prefer that this be a "real" question unless that is inappropriate.
| https://mathoverflow.net/users/10204 | What is the order of a in (Z/nZ)*? | You seem to have been given some misinformation so I'll answer this question although I think it is elementary. You want to find the order of $a$ modulo $n$. The prime factorization of $a$ is largely irrelevant, the prime factorization of $n$ is crucial since otherwise you don't know the order of the group. Conversely, knowing the order of $a$ for many $a$'s will allow you to factor $n$. I'll assume you can factor $n$.
If $n$ is prime, then the group is cyclic, so any factor of $n-1$ is the order of some element. There isn't much more that can be said, you can't eyeball the order except in some obvious cases such as $a=\pm 1$. If you know a factorization of $n-1$, then you can run through the divisors of $n-1$ to find the order. If you don't know the factorization of $n-1$ then brute force is basically all you can do.
If $n$ is the power of a prime $p$, then if you can compute the order modulo $p$ (say $d$), it is easy to compute it modulo $n$ by finding the highest power of $p$ dividing $a^d-1$. This is an exercise which most number theory textbooks do when discussing primitive roots modulo prime powers.
In general, you get the order modulo $n$ by factoring $n$, and using the Chinese remainder theorem to reduce to the above cases.
| 7 | https://mathoverflow.net/users/2290 | 44046 | 27,981 |
https://mathoverflow.net/questions/44015 | 17 | Other than learning basic calculus, I don't really have an advanced background. I was curious to learn about Optimal Control (the theory that involves, bang-bang, Potryagin's Maximum Principle etc.) but any article that I start off with, mentions the following: "Consider a control system of the form..." and then goes on to defining partial differential equations. In short, I am lost.
Can someone suggest me a path I should take to learn more about Optimal Control from the very basics?
| https://mathoverflow.net/users/3560 | How do I approach Optimal Control? | My field is mathematical programming, and I tend to look at optimal control as just optimization with ODEs in the constraint set; that is, it is the optimization of dynamic systems. I would start by studying some optimization theory (not LPs but NLPs) and getting an intuitive feel for the motivations behind stationarity and optimality conditions -- that will lead naturally into optimal control theory.
I should mention there is another facet of optimal control, related to control systems. The systems considered are discrete time (as opposed to continuous in PMP) therefore it's difference equations instead of differential equations. Examples of optimal control laws in this latter sense are Linear Quadratic Regulators (LQRs), Linear Quadratic Gaussian (LQGs), Model Predictive Control (MPC). It is this latter type of optimal control that is actually applied in industry. The Pontryagin principle, while useful for analysis, is generally intractable for real-time application to nontrivial plants.
| 7 | https://mathoverflow.net/users/7851 | 44052 | 27,984 |
https://mathoverflow.net/questions/44051 | 4 | Let $S$ is a partition of a set $U$. Let $c$ is an ultrafilter on $U$.
Prove or disprove this conjecture:
At least one of the following is true:
* $\exists D\in S, C\in c:C\subseteq D$
or
* $\exists C\in c\forall D\in S: \mathrm{card}(C\cap D)\le 1$.
| https://mathoverflow.net/users/4086 | An ultrafilter and a partition | Take a partition of ${\mathbb N}^2$ into vertical lines $\{x\}\times {\mathbb N}$. In each vertical line take a non-principle ultrafilter $\omega\_x$. Now take the set of all sets $Y$ that intersect all but finitely many vertical lines by a subset from $\omega\_x$. Note that all complements of finite sets of ${\mathbb N}^2$ are in our set of sets, and that it is clearly a filter. Take any ultrafilter $\omega$ that contains that filter. It exists by the Zorn lemma. Clearly, the first option does not hold: none of the vertical lines is in $\omega$. Now suppose that for some $C\in\omega$, $C$ intersects each vertical line by at most 1 element. Then its complement intersects each vertical line by a subset that is either the whole line or the line without one element. That is impossible because we chose non-principle $\omega\_x$. Thus $\omega$ is a counterexample.
| 5 | https://mathoverflow.net/users/nan | 44055 | 27,986 |
https://mathoverflow.net/questions/44023 | 6 | Suppose we have a Grothendieck pretopology $\tau$ on a category C with fibered products. Now define a new Grothendieck pretopology $\tau'$ consisting of all families of morphisms refinable by $\tau$-covers. That is, the new covers are the families $\{V\_\beta \to X\}$ such that there exists some $\tau$-cover $\{U\_\alpha \to X\}$ and a factorisation $U\_\alpha \to V\_{\beta\_\alpha} \to X$ for each $\alpha$. This new set of families is also a Grothendieck pretopology and the question is: do they give the same topos? That is, is a presheaf a $\tau$-sheaf if and only if it is a $\tau'$-sheaf?
Edit: I could't read the relevant page in Elephant either, but Mike's answer lead me to the saturation section of <http://ncatlab.org/nlab/show/coverage> after which I worked out how to prove it myself. If someone explains to me how to typeset diagrams, I'll write up the answer.
| https://mathoverflow.net/users/8324 | Does adding "co"refinements to a Grothendieck pretopology change the topos? | The answer is yes. David Roberts had the right idea—adding those new covering families gives you a new pretopology which generates the same Grothendieck topology—but not because it's a sieve completion, rather because there is an additional saturation condition in the definition of Grothendieck topology (in addition to saying that it consists of sieves) which essentially gives you this property.
It's not hard to check that any presheaf which is a sheaf for your original pretopology must also be one for the new one you define. You can find it as C2.1.6 in the Elephant. Note what this does not say: it's not necessarily true that if you have just a pair of families with the same codomain one of which corefines the other, that a sheaf for one of them is necessarily a sheaf for the other. The proof uses the assumption that the first covering family is part of a pretopology, and in particular can be pulled back along any morphism to another covering family, for which your presheaf is *also* a sheaf.
| 4 | https://mathoverflow.net/users/49 | 44058 | 27,988 |
https://mathoverflow.net/questions/44060 | 53 | Let $(G,\cdot,T)$ and $(H,\star,S)$ be topological groups such that
$(G,T)$ is homeomorphic to $(H,S)$ and $(G,\cdot)$ is isomorphic to $(H,\star)$.
Does it follow that $(G,\cdot,T)$ and $(H,\star,S)$ are isomorphic as topological groups?
If no, what if they are both Hausdorff? What if they are both Hausdorff and two-sided complete?
| https://mathoverflow.net/users/nan | Does homeomorphic and isomorphic always imply homeomorphically isomorphic? | The 2-adic rationals $\mathbb{Q}\_2$ and the 3-adic rationals $\mathbb{Q}\_3$ are homeomorphic, because each one is a countable disjoint union of Cantor sets. They are also isomorphic as groups if you assume the axiom of choice, because they are both fields of characteristic 0 and therefore vector spaces over $\mathbb{Q}$ (of the same cardinal dimension). However, the 2-adic integers $\mathbb{Z}\_2$ are a compact subgroup of $\mathbb{Q}\_2$ in which every element is infinitely divisible by 3. On the other hand, in $\mathbb{Q}\_3$, any non-trivial sequence $x, x/3, x/9, \ldots$ is unbounded in the complete metric, and is therefore not contained in a compact subgroup.
---
Keith Conrad asks whether these is an example without the axiom of choice, and Jason De Vito asks whether there is an example using Lie groups. In fact, there is a cheap example using disconnected Lie groups. Let $G$ and $H$ be two connected Lie groups that are homeomorphic but not isomorphic. For instance, abelian $\mathbb{R}^3$, the universal cover $\widetilde{\text{SL}(2,\mathbb{R})}$, and the Heisenberg group of upper unitriangular, real $3 \times 3$ matrices are all homeomorphic, but not isomorphic. If $G'$ and $H'$ are $G$ and $H$ with the discrete topology, then $G' \times H$ and $G \times H'$ are explicitly isomorphic and explicitly homeomorphic. But they are not continuously isomorphic, because the connected component of the identity is $G$ for one of them but $H$ for the other one.
| 115 | https://mathoverflow.net/users/1450 | 44070 | 27,994 |
https://mathoverflow.net/questions/44042 | 19 | In a recent [answer](https://mathoverflow.net/questions/44005/intuition-for-rational-functions/44040#44040) to a recent [question](https://mathoverflow.net/questions/44005/intuition-for-rational-functions), [BCnrd](https://mathoverflow.net/users/3927/bcnrd) wrote
>
> [...] beyond the coherent case one cannot expect information about a fiber (e.g., vanishing, 6 generators, etc.) to "propogate" to information in a neighborhoood (which would be the spirit behind the choice of word "coherent", I suppose). [...]
>
>
>
Is that what motivates the adjective *coherent*? Is this documented somewhere?
| https://mathoverflow.net/users/1409 | What's coherent about coherent sheaves? | Looking at the paper of MALATIAN
"Faisceaux analytiques: étude du faisceau des rélations entre p fonctions holomorphes",
Séminaire Henry Cartan, tome 4 (1951-52), exp. n.15, p. 1-10
one finds the
**Definition 3**
>
> "On dit qu'un sous-faisceau analytique $\mathcal{F}$ de $\mathcal{O}\_E^q$ is $cohérent$ au point $x \in E$, s'il existe un voisinage ouvert $U$ de $x$ et un système fini d'elements $u\_i \in \mathcal{O}\_U^q$ jouissant de la propriété suivante: pour tout $y \in U$, le sous-module de $\mathcal{O}\_U^q$ engendré par les $u\_i$ est précisement $\mathcal{F}\_y$.
> On dit qu'un faisceau $\mathcal{F}$ est cohérent (tout court) s'il est coherent en tout point de $E$."
>
And, in the following page:
>
> "...En d'autre termes, cette condition exprime que le faisceau $induit$ par $\mathcal{F}$ sur l'ouvert $U$ est "engendré" par un sous-module de $\mathcal{O}\_U^q$."
>
Reading this, it seems that the original definition given by Cartan in its seminar is somehow related to the "coherent behaviour" of $\mathcal{F}$ as a subsheaf of $\mathcal{O}\_U^q$, in terms of generation of the stalks.
**EDIT.**
However, this is not the whole story. Loking at the introduction of the book of Grauert-Remmert, as Brian suggests, it appears that the word "coherent" was actually introduced by Cartan some years before, in the middle of the '40; in fact, he investigated the so-called "coherent systems of punctual modules" when studyng the Cousin's problem. But he does not mention this previous work in his Seminar, when he introduces coherent analytic sheaves.
Grauert-Remmert write that
>
> "coherence is, in a vague sense, a local principle of analytic continuation".
>
And Cartan himself, in its collected works, says
>
> "En gros on peut dire que, pour en $A$-faisceaux $\mathcal{F}$ cohérent en un point $a$ de $A$, la connaissance du module $\mathcal{F}\_a$ détermine les modules $\mathcal{F}\_x$ attachées aux points $x$ suffisamment voisins de $a$."
>
| 18 | https://mathoverflow.net/users/7460 | 44077 | 27,999 |
https://mathoverflow.net/questions/44065 | 2 | Is the wikipedia definition of Lipschitz Euclidean domain correct?
See: <http://en.wikipedia.org/wiki/Lipschitz_domain>
i was wondering what stops me just showing the condition holds for one point and then just scale and translate that function $h\_p$ for any point on the boundary... This doesn't seem right? What am I missing?
Thanks in advance.
| https://mathoverflow.net/users/2011 | Lipschitz smooth boundary definition | A domain of $\mathbb{R}^n$ with Lipschitz boundary is an open subset $\Omega\subset \mathbb{R}^n$, which is locally the sub-graph of a Lipschitz function w.r.to some choice of orthogonal coordinates. In other words, for any $p\in\partial \Omega$, up to an orthogonal change of coordinates, there is an open set
$V\subset\mathbb{R}^{n-1}$, and a Lipschitz function $\phi:V\to(a,b)$ such that $U:=V\times (a,b)$ is a nbd of $p$ and
$$\Omega\cap U=\{(x,t)\in U\ | \ t< \phi(x) \}.$$
This is equivalent to the definition given in the link, which is closer to the general definition of manifold with boundary (here the transition mappings are the bi-Lipschitz homeomorphisms).
In particular, for any point $p\in \partial\Omega$ there is a small cone
$C$ with vertex in the origin, and a nbd $U$ of $p$ such that for any $q\in U\cap\partial\Omega$ the cone $q+C$ is disjoint from $\Omega$, and the cone $q-C$ is included in $\bar \Omega.$ This is a third equivalent definition.
| 8 | https://mathoverflow.net/users/6101 | 44091 | 28,007 |
https://mathoverflow.net/questions/44096 | 3 | Given a directed cycle in the plane I need to walk it and detect whether it is clockwise or counterclockwise.
My first idea is to gather the sum of the turn angles, where a "left" turn is a negative angle, and a "right" turn is a positive angle. If I go with this one, I need a good way to calculate the angle between two vectors, and also which sign this angle should have (+/-).
Even if I had the right tools to calculate the (+/-) angle I feel like this could be done simpler.c
| https://mathoverflow.net/users/10422 | Detecting whether directed cycle is clockwise or counterclockwise | The orientation of a triangle (clockwise/counterclockwise) is the sign of the determinant
$\begin{bmatrix}
1&x\_1&y\_1\\\\
1&x\_2&y\_2\\\\
1&x\_3&y\_3
\end{bmatrix}$, where $(x\_1,y\_1), (x\_2,y\_2), (x\_3,y\_3)$ are the Cartesian coordinates of the three vertices of the triangle.
| 5 | https://mathoverflow.net/users/806 | 44098 | 28,010 |
https://mathoverflow.net/questions/44094 | 6 | Throughout, let $X$ be a *connected finite* CW-complex.
>
> **Question:** If $X$ is of dimension $n$. Is there some integer $n'$ (maybe depending only on $n$), such that all homotopy groups $\pi\_k(X)$ for $k \geq n'$ are finite?
>
>
>
For the spheres $S^n$, $n'=2n+1$ works by Freudenthal's Suspension Theorem and Serre's result that the stable homotopy groups in that range are finite. More generally, if $\pi\_1(X)=0$, then the Milnor-Moore theorem relates the rational homotopy groups to the rational homology of the loop space of $X$ and I believe that this can be used to get a similar conclusion. But what if $\pi\_1(X) \neq 0$?
**EDIT:** Igor Belegradek (besides answering the question) pointed out that what I stated in the last three lines is not correct.
| https://mathoverflow.net/users/8176 | Finiteness of higher homotopy groups of finite complexes | The answer is no in a very strong way even for simply-connected complexes. In rational homotopy theory there is a famous dichotomy between elliptic and hyperbolic spaces: a simply-connected finite complex is either elliptic or hyperbolic. Elliptic means that all but finitely many homotopy groups are finite. Hyperbolic means that the sum of ranks of first $k$ homotopy groups grows exponentially with $k$. In some sense most spaces are hyperbolic. If I remember correctly, $m$-fold connected sum of $S^2\times S^2$ with itself is hyperbolic if $m>1$. You can read more of this in the book "Rational homotopy theory" by Felix-Halperin-Thomas.
| 18 | https://mathoverflow.net/users/1573 | 44101 | 28,013 |
https://mathoverflow.net/questions/44105 | 0 |
>
> **Possible Duplicate:**
>
> [Order types of positive reals](https://mathoverflow.net/questions/25100/order-types-of-positive-reals)
>
>
>
The title is self-explanatory.
| https://mathoverflow.net/users/10424 | Is every countable well-order embeddable in \mathbb{R}? | The answer is yes. Choose an enumeration $\alpha\_1,\alpha\_2,\dots$ of your well-ordering and define a map by setting:
$$\alpha\_n \mapsto \sum\_{k : \alpha\_k < \alpha\_n} \frac1{2^k} \in \mathbb R.$$
| 5 | https://mathoverflow.net/users/8176 | 44106 | 28,015 |
https://mathoverflow.net/questions/44053 | 17 | This may seems to be an elementary question, but I found no answers on MO nor google.
I have always heard "polynomials are easier to handle with than integers". For example:
1. When $n$ is quite large, maybe 200 or more, it's relatively easier to factorize a polynomial $f$ of degeree $n$ than to factorize an integer with $n$ bytes.
2. When multiplying large integers, we see them as polynomials,use techniques such as FFT,intepolations to multiply polynomials,and then back to integers.
3.The zeta functions of $F[x]$ and $\mathbb{Z}$, and the former are easier to study than the latter.
Of course there are other examples, but because of my shortage of knowledge, I can only lise these above.
So my question is (as in the titile): Why are polynomials easier to handle with than integers? I ask this because contrary to our intuitives, polynomials are "more complex" objects than integers.
| https://mathoverflow.net/users/10416 | Why are polynomials easier to handle with than integers? | How Halloweeny can you get with your questions? The norm on $Z$ is Archimedean and the norm on $F[X]$ is non-Archimedean, and, in general, non-Archimedean maths is easier than Archimedean...
| 4 | https://mathoverflow.net/users/5301 | 44111 | 28,016 |
https://mathoverflow.net/questions/44093 | 13 | Throughout, let $X$ be a *connected finite* CW-complex. If the universal covering of $X$ is contractible, then $\pi\_n(X)=0$ for all $n \geq 2$. In this case $X$ is a model for $B\pi\_1(X)$.
I am wondering whether this is the only reason why higher homotopy groups vanish above a certain degree. More precisely:
>
> **Question:** Let $k \geq 3$ be an integer. Can it happen that $\pi\_n(X) = 0$ for all $n \geq k$ and $\pi\_{k-1}(X)\neq 0$?
>
>
>
| https://mathoverflow.net/users/8176 | Vanishing of higher homotopy groups of finite complexes | No, it cannot happen. In [a paper by McGibbon and Neisendorfer](https://doi.org/10.1007/BF02566349 "McGibbon, C.A., Neisendorfer, J.A. On the homotopy groups of a finite dimensional space. Commentarii Mathematici Helvetici 59, 253–257 (1984). zbMATH review at https://zbmath.org/?q=an:0538.55010"), it is proven that if *X* is a 1-connected space and its mod-p-homology is non-zero in some degree, but zero in all higher degrees, then the $\pi\_n X$ contain a subgroup of order p for infinitely many *n*. This can be applied to the universal cover of your *X*.
| 18 | https://mathoverflow.net/users/2039 | 44114 | 28,019 |
https://mathoverflow.net/questions/44081 | 1 | This is actually something in a paper but the author claimed it without proof.
Let x be a positive elment of norm 1 in a $C^\*-$algebra A, and Her(x) is the hereditary subalegbra generated by x. Given $\epsilon>0$,let $f\_\epsilon$ be thecontinuous function on R defined as follow:
$f\_\epsilon \equiv 0 \quad on \quad [-\infty,\epsilon/2]$
$f\_\epsilon \quad is \quad linear \quad on\quad [\epsilon/2,\epsilon]$
$f\_\epsilon \equiv 1 \quad on\quad [\epsilon, +\infty]$
So $f\_\epsilon$ increase to the identiy function on [0,1] when $\epsilon$ decrease to 0, and $Her(x)=\overline{\cup\_{\epsilon>0} f\_\epsilon(x)Af\_\epsilon(x)}$. Let p be a projection in Her(x). then how do we know that there must exist a $\epsilon$ such that $p\in \overline{f\_\epsilon(x)Af\_\epsilon(x)}$? Or more generally, Let A be the inductive limit of {$A\_n$}, and p is a projection in A,does it follow that p is actually in some $A\_n$?
| https://mathoverflow.net/users/9858 | Projection in Hereditary C* subalgebra | No. Let $A$ be the C$^\*$-algebra of compact operators on $\ell\_2$ and $x$ is the diagonal operator $(1,1/2,1/3,\ldots)$. Then, $f\_\epsilon(x)Af\_\epsilon(x)$ is a matrix algebra in the left upper corner. The rank one projection corresponding to any vector of infinite support does not belong to $\bigcup f\_\epsilon(x)Af\_\epsilon(x)$.
However, it is a standard fact that if $a$ is a positive element such that $\| p - a \| < 1/2$, then the spectrum of $a$ has a gap around $1/2$ and $q=\chi\_{[1/2,3/2]}(a)$ is a projection in $C^\*(a)$ such that $\| p - q \| < 1$, which implies that $p$ and $q$ are unitarily equivalent.
| 6 | https://mathoverflow.net/users/7591 | 44122 | 28,022 |
https://mathoverflow.net/questions/44119 | 4 | Hi there,
Consider linear endomorphisms ("endos") of a finite dimensional vector space.
How can those endos be characterized, for which said vector space has a basis with respect to which the endo has a matrix with only nonnegative entries?
Not all endos have this property: e.g., $x\mapsto -x$. More generally, necessary conditions can be derived from Perron-Frobenius theory.
Does anybody know this problem? Or its solution?
-DOT
| https://mathoverflow.net/users/10426 | For which linear endomorphisms can one find a basis such that the matrix is nonnegative? | This is related to the mathoverflow question [Perron-Frobenius "inverse eigenvalue problem"](https://mathoverflow.net/questions/35320/perron-frobenius-inverse-eigenvalue-problem). Doug Lind's theory characterizes when an endomorphism $E$ can be extended to an endomorphism $\bar E$ of some larger vector space having a basis such that the matrix for the extended endomorphism is nonnegative: this can be done if and only if it satisfies the Perron-Frobenius condition, that the spectral radius of $E$ is an eigenvalue of $E$. Even for an automorphism of $R^3$, the dimension necessary for an extension with this property can be arbitrarily large (as shown by examples of Lind).
For dimensions 1 and 2, it's easy to see the Perron-Frobenius condition is sufficient.
You can think of the question geometrically in terms of projective space. For simplicity, take $E$ to be an automorphism. There's a sub projective space $M$ (which might be a single point) that mapped isometrically to itself, consisting of linear combinations of eigenvectors (real and complex) whose eigenvalue has maximal absolute value. Almost everything else is contracted toward $M$. The question is whether there's a simplex $T$ in projective space that contains its image.
For low dimensions it should be possible to describe the specific inequalities the set of eigenvalues need to satisfy by making use of the relatively simple geometry of the orbits of linear maps. For larger dimensions, I believe this question is too hopelessly complicated to expect a complete concrete description or feasible general algorithm.
| 5 | https://mathoverflow.net/users/9062 | 44128 | 28,027 |
https://mathoverflow.net/questions/44102 | 116 | Ten years ago, when I studied in university, I had no idea about [definable numbers](https://en.wikipedia.org/wiki/Definable_number), but I came to this concept myself. My thoughts were as follows:
* All numbers are divided into two classes: those which can be unambiguously defined by a limited set of their properties (definable) and those such that for any limited set of their properties there is at least one other number which also satisfies all these properties (undefinable).
* It is evident that since the number of properties is countable, the set of definable numbers is countable. So the set of undefinable numbers forms a continuum.
* It is impossible to give an example of an undefinable number and one researcher cannot communicate an undefinable number to the other. Whatever number of properties he communicates there is always another number which satisfies all these properties so the researchers cannot be confident whether they are speaking about the same number.
* However there are probability based algorithms which give an undefinable number as a limit, for example, by throwing dice and writing consecutive numbers after the decimal point.
But the main question that bothered me was that the analysis course we received heavily relied on constructs such as "let $a$ be a number such that...", "for each $s$ in the interval..." etc. These seemed to heavily exploit the properties of definable numbers and as such one can expect the theorems of analysis to be correct only on the set of definable numbers. Even the definitions of arithmetic operations over reals assumed the numbers are definable. Unfortunately one cannot take an undefinable number to bring a counter-example just because there is no example of undefinable number. How can we know that all of those theorems of analysis are true for the whole continuum and not just for a countable subset?
| https://mathoverflow.net/users/10059 | Is the analysis as taught in universities in fact the analysis of definable numbers? | The concept of *definable* real number, although seemingly
easy to reason with at first, is actually laden with subtle
metamathematical dangers to which both your question and
the Wikipedia article to which you link fall prey. In
particular, the Wikipedia article contains a number of
fundamental errors and false claims about this concept. (**Update**, April 2018: The Wikipedia article, [Definable real numbers](https://en.wikipedia.org/wiki/Definable_real_number), is now basically repaired and includes a link to this answer.)
The naive treatment of definability goes something like
this: In many cases we can uniquely specify a real number,
such as $e$ or $\pi$, by providing an exact description of
that number, by providing a property that is satisfied by
that number and only that number. More generally, we can
uniquely specify a real number $r$ or other set-theoretic
object by providing a description $\varphi$, in the formal
language of set theory, say, such that $r$ is the only
object satisfying $\varphi(r)$.
The naive account continues by saying that since there are
only countably many such descriptions $\varphi$, but
uncountably many reals, there must be reals that we cannot
describe or define.
But this line of reasoning is flawed in a number of ways
and ultimately incorrect. The basic problem is that the
naive definition of definable number does not actually
succeed as a definition. One can see the kind of problem
that arises by considering ordinals, instead of reals. That
is, let us suppose we have defined the concept of definable
ordinal; following the same line of argument, we would seem
to be led to the conclusion that there are only countably
many definable ordinals, and that therefore some ordinals
are not definable and thus there should be a least ordinal
$\alpha$ that is not definable. But if the concept of
definable ordinal were a valid set-theoretic concept, then
this would constitute a definition of $\alpha$, making a
contradiction. In short, the collection of definable
ordinals either must exhaust all the ordinals, or else not
itself be definable.
The point is that the concept of definability is a
second-order concept, that only makes sense from an
outside-the-universe perspective. [Tarski's theorem on the
non-definability of
truth](http://en.wikipedia.org/wiki/Tarski%27s_undefinability_theorem)
shows that there is no first-order definition that allows
us a uniform treatment of saying that a particular
particular formula $\varphi$ is true at a point $r$ and
only at $r$. Thus, just knowing that there are only
countably many formulas does not actually provide us with
the function that maps a definition $\varphi$ to the object
that it defines. Lacking such an enumeration of the
definable objects, we cannot perform the diagonalization
necessary to produce the non-definable object.
This way of thinking can be made completely rigorous in the
following observations:
* If ZFC is consistent, then there is a model of ZFC in
which every real number and indeed every set-theoretic
object is definable. This is true in the minimal
transitive model of set theory, by observing that the
collection of definable objects in that model is closed
under the definable Skolem functions of $L$, and hence by
Condensation collapses back to the same model, showing
that in fact every object there was definable.
* More generally, if $M$ is
any model of ZFC+V=HOD, then the set $N$ of parameter-free
definable objects of $M$ is an elementary substructure of
$M$, since it is closed under the definable Skolem
functions provided by the axiom V=HOD, and thus every
object in $N$ is definable.
These models of set theory are *pointwise definable*,
meaning that every object in them is definable in them by a
formula. In particular, it is consistent with the axioms of
set theory that EVERY real number is definable, and indeed,
every set of reals, every topological space, every
set-theoretic object at all is definable in these models.
* The pointwise definable models of set theory are
exactly the prime models of the models of ZFC+V=HOD, and
they all arise exactly in the manner I described above, as
the collection of definable elements in a model of V=HOD.
In recent work (soon to be submitted for publication),
Jonas Reitz, David Linetsky and I have proved the following
theorem:
**Theorem.** Every countable model of ZFC and indeed of
GBC has a forcing extension in which every set and class is
definable without parameters.
In these pointwise definable models, every object is
uniquely specified as the unique object satisfying a
certain property. Although this is true, the models also
believe that the reals are uncountable and so on, since
they satisfy ZFC and this theory proves that. The models
are simply not able to assemble the definability function
that maps each definition to the object it defines.
And therefore neither are you able to do this in general.
The claims made in both in your question and the Wikipedia
page on the existence of non-definable numbers and objects,
are simply unwarranted. For all you know, our set-theoretic
universe is pointwise definable, and every object is
uniquely specified by a property.
---
**Update.** Since this question was recently bumped to the main page by an edit to the main question, I am taking this opportunity to add a link to my very recent paper ["Pointwise Definable Models of Set Theory", J. D. Hamkins, D. Linetsky, J. Reitz](https://arxiv.org/abs/1105.4597), which explains some of these definability issues more fully. The paper contains a generally accessible introduction, before the more technical material begins.
| 233 | https://mathoverflow.net/users/1946 | 44129 | 28,028 |
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