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https://mathoverflow.net/questions/39376 | 5 | My question is a very simple one.
What ways are there to generalize terms such as cardinality (or, more generally, the concept of finiteness) to abstract (and not concretizable) categories?
I have seen two ways to do so: One is to take a terminal object $\mathbf{T}$ and to count the numer of morphisms from $\mathbf{T}$ to an object $\mathbf{A}$ and define this number as the cardinality of $\mathbf{A}$. Clearly, this will coincide with the cardinality as defined in the category of sets as the number of morphisms from $\{1\}$ to another set is the cardinality of the letter. The downside of this approach is that it does not always work well: Even if a category has a terminal object, there is no garuantee that there are morphisms from this object to any other object and, thus, if we define cardinality in this way, it might give very misleading results: For instance, we can have a concrete category in which the number of morphisms from the terminal object does not reflect the finitness of an object that might be obtained by applying the forgetful functor to it.
Another way is to define an object as finite if it has a finite number of endomorphisms. Again, this coincides with the notional of finiteness in the category of sets. However, one can think of a category in which an object $\mathbf{A}$ only has the identity morphism as an endomorphism but an infinite number of morphisms exist between $\mathbf{A}$ and another object $\mathbf{B}$ (for which, again, only the identity morphism is an endomorphism). This creates a scenario in which it is hardly justified to think $\mathbf{A}$ and $\mathbf{B}$ as finite.
Are there other ways to define finitenes? In particular, is there a way to define finitness for a category such that, if this category is concretizable, it can be turned into a concrete category in which only the finite objects (abstractly defined) can have a finite image under the forgetful functor and this can be achieved for any abstractly defined finite object?
---
EDIT: In the comments, it was asked what categories I have in mind. Here is an answer to this:
The categories I have in mind are only required to have the following: There must exist an object $\mathbf{A}$ such that all finite, non-empty powers of $\mathbf{A}$ are also in the category (the product is of course defined in the usual category-theoretic way and the point of all this is to work with a generalized definition of clones in these categories). Apart from this condition, however, the category can have any possible structure. Although it would be nice to have a general definition of cardinality, all I am really interested in is to distinguish between finite and non-finite objects. The definition should satisfy the three conditions below.
1. Of course, it should coincide with the usual definition of finiteness if my category is Set.
2. If $\mathbf{A}$ and another object $\mathbf{B}$ is finite, then the number of morphisms from $\mathbf{B}$ to $\mathbf{A}$ should be finite (it would therefore be enough to have a notion of finiteness "with respect to $\mathbf{A}$", if such a thing makes sense to have).
3. The definition should give me as many finite objects as the "best" concretezation. By this, I mean the following: If my category is concretizable, then you should not be able to get more finite objects by defining finiteness over the image under the forgetful functor.
| https://mathoverflow.net/users/8590 | Finiteness and cardinality in abstract categories | If your category has a monoidal structure, you can ask for an object to be dualizable. This is a generalization of finite-dimensionality for vector spaces.
| 5 | https://mathoverflow.net/users/121 | 39377 | 25,243 |
https://mathoverflow.net/questions/38818 | 0 | Let $R$ be a commutative ring, $I$ is an ideal of $R$, $M$ is an $R$-module.
$$IM\supset I^2M\supset I^3M\supset\cdots$$
What is $\mathop {\lim }\limits\_{\begin{subarray}{c}
\longrightarrow \\
\end{subarray}}I^nM$ ?
| https://mathoverflow.net/users/9141 | What is lim⟶ I^n M? | I'll interpret this question by agglomerating information given in the comments: We choose an element $x \in I$, and want to know the structure of the direct limit of $\{ I^nM\}\_{n \geq 0}$, when the maps $I^nM \to I^{n+1}M$ are given by multiplication by $x$.
I think the answer is that we can't say very much at all in general without somewhat more information, and the assumption that $x$ is $M$-regular is not enough. For example, if $R = \mathbf{Q}[t]$ and $M=R$, then choosing $x = t$ yields a direct limit that is the zero module, while choosing $x=1$ yields a direct limit that is $M$ (which is $IM$ in this case). For more general rings $R$, you can get intermediate results by setting $x$ to be something that has an interesting fixed subspace, e.g., an idempotent.
| 1 | https://mathoverflow.net/users/121 | 39381 | 25,245 |
https://mathoverflow.net/questions/39390 | 5 | A projective variety $X$ is convex, if for any $f:\mathbb{P}^1 \to X$, the group $H^1(\mathbb{P}^1, f^\*(T\_X))$ vanishes. A big group of examples of convex varieties is made of homogeneous varieties. An homogeneous variety is the quotient variety $G/P$ of a Lie group $G$ by a parabolic subgroup $P$ of it.
My question is: is there an easy example of a projective variety, possibly smooth, that is convex but that does not admit such a description?
In the lectures: <http://www.math.uic.edu/~coskun/utah-notes.pdf>, the author asks whether every rationally connected smooth convex projective variety must also be homogeneous, thus suggesting that it should be rather easy to find such an example in the realm of non-rationally connected varieties.
| https://mathoverflow.net/users/9391 | Convex varieties that are not homogeneous | Of course every variety containing no rational curves is convex, by default. For a less trivial example, take the product of one such variety (for example, an abelian variety) with a homogeneous variety.
| 7 | https://mathoverflow.net/users/4790 | 39391 | 25,251 |
https://mathoverflow.net/questions/39371 | 3 | Given two closed disks of unit radius, such that center of one lies on the circumference of the other, let M denote their union. We want to place the maximum number of points in M such that their pairwise distance is strictly greater than 1. We can show that we cannot place 10 points, and we have examples where we can place 8. Is it possible to place 9 points? We have not been able to prove that 9 points is impossible. Any suggestions welcome. (I cannot seem to place image tags since this is my first question.. but the images showing why 10 points are impossible and a configuration showing 8 points are available at:
<http://www.freeimagehosting.net/image.php?3c080eeea5.jpg>, and
<http://www.freeimagehosting.net/image.php?7ff64600b3.jpg>
respectively.
| https://mathoverflow.net/users/9380 | Maximum number of points in two disks | Granted that there can be at most 5 in each disc, and a maximum of 2 in the intersection, the case of 2 in the intersection can only be from a total of at most 8. So look at the case of 1 point in the intersection. This reduces to showing at most 3 points possible in certain differences of discs. Now this might well follow from area considerations. It is anyway starting to look plausible in a picture.
| 1 | https://mathoverflow.net/users/6153 | 39394 | 25,254 |
https://mathoverflow.net/questions/39402 | 3 | Please consider the (presumably infinite) Euler product over the twin primes:
$$ f(z) = \prod\_{p\in\mathbb{P}}^{\infty} \Big( 1 - \frac{1}{(p(p+2))^ z} \Big) $$ (in which $p(p+2)$ is a divisor of $4((p-1)!+1) + p$ ).
The Euler Product is a product of a corresponding Dirichlet series. Which one is that?
Thanks in advance,
Max
Edit Update: error fixed.
| https://mathoverflow.net/users/93724 | Asociated sum series of the Euler Product over the Twin Primes? | In this paper: <http://arxiv.org/abs/0902.4352>, the authors discuss the analytic properties of the related Dirichlet series
$$ D\_{2r}(s) = \sum\_{n=1}^\infty \frac{\Lambda(n)\Lambda(n+2r)}{n^s}$$
where $\Lambda(\cdot)$ is von Mangoldt's function. See section $2$.
| 5 | https://mathoverflow.net/users/3659 | 39404 | 25,259 |
https://mathoverflow.net/questions/39409 | 6 | Suppose $f:[0,1]^2\to\mathbb{R}$, $(t,x)\mapsto f(t,x)$, is such that for each $t\in[0,1]$ $f(t,\cdot)$ is Lebesgue measurable on $[0,1]$, and for each $x\in[0,1]$ $f(\cdot,x)$ is continuous everywhere on $[0,1]\ni t$.
**1.** Does this imply that $f(t,x)$ is measurable on $[0,1]^2$?
**2.** Does this imply that the function $g(x)=\min\limits\_{t\in[0,1]}f(t,x)$ is measurable on $[0,1]$?
| https://mathoverflow.net/users/9397 | On measurable functions of two variables | 1. Yes, by continuity in the $t$ variable $f(t,x)=\lim\_n f(\lfloor n t\rfloor/n,x)$, which expresses $f$ as the pointwise limit of a sequence of measurable functions.
2. Yes, by continuity in the $t$ variable we have $\min\_{t\in[0,1]} f(t,x)=\min\_{t\in[0,1]\cap {\mathbb Q}} f(t,x)$, where $\mathbb Q$ means the rational numbers.
| 7 | https://mathoverflow.net/users/nan | 39419 | 25,264 |
https://mathoverflow.net/questions/39415 | 7 | I think one of the most interesting results in Elementary Group Theory is the so-called "[Burnside's Lemma](http://en.wikipedia.org/wiki/Burnside%27s_lemma)", counting the numbers of orbits of a (finite) group action.
I wonder if there is any (interesting) application in Elementary Geometry (I mean Euclidean, hyperbolic or elliptic geometry).
Searching on Google, I've found the article "[Applying Burnside’s lemma to a one-dimensional
Escher problem](http://users.wpi.edu/~bservat/strippat.pdf)" by T. Pisanski, but it sounds to me rather a combinatorial result.
| https://mathoverflow.net/users/47274 | Burnside's Lemma and Geometry | Burnside Lemma can be used as a first step to classify all finite subgroups of $\mathrm{SO}(3)$: it gives you that there are at most $3$ orbits in the action of any finite group $G$ on the set of intersections between axes of elements of $G$ and the unit sphere.
| 7 | https://mathoverflow.net/users/4961 | 39421 | 25,266 |
https://mathoverflow.net/questions/39408 | 27 | I've met tall people. That is: people taller than the average. Every now and then we encounter *really* tall people, even taller than the average of tall people i.e. taller than the average of those who are taller than the average. Meybe you've met someone who's even taller than the average of those who are taller than the average of those who are taller than the average... And so on.
So, take a quantity $X$ that we suppose normally distributed (caveat, I have no deep knowledge of probability theory), i.e. it's described by a gaussian distribution that we suppose standardized and call $f(x)$.
Now, define:
$M\_0:= \int\_{-\infty}^{\infty}f(x)dx=1$
$\mu\_0:=\int\_{-\infty}^{\infty}xf(x)dx=0$
and, inductively,
$M\_{n+1}:= \int\_{\mu\_n}^{\infty}f(x)dx$
$\mu\_{n+1}:=\frac{1}{M\_n}\int\_{\mu\_n}^{\infty}xf(x)dx$
I think this describes the situation in which your $X$ (e.g. height) has the value $\mu\_n$ precisely when you're as $X$ as the average of those who are more $X$ than the average of those who are more $X$ than...... (n times). If not, please explain why.
So my questions:
1. How does the sequence $\mu\_n$ behave asymptotically? Does it converge?
2. If yes, is there a nice expression for the limit?
3. Is there even a reasonably explicit expression ("closed form") for $\mu\_n$ as a function of $n$?
| https://mathoverflow.net/users/4721 | "Are you taller than the average of those who are taller than the average?" | As in Nate's answer, we are interested in iterating the function
$$G(y) := \frac{ \int\_{y}^{\infty} x e^{- x^2} dx}{\int\_{y}^{\infty} e^{- x^2} }.$$
The numerator is $e^{-y^2}/2$ (elementary). The denominator is $e^{-y^2}/2 \cdot y^{-1} \left( 1-(1/2) y^{-2} + O(y^{-4}) \right)$ (see [Wikipedia](http://en.wikipedia.org/wiki/Error_function#Asymptotic_expansion)). So $G(y) = y + (1/2) y^{-1} + O(y^{-3})$.
Set $z\_n = \mu\_n^2$. Then
$$z\_{n+1} = (\mu\_n+\mu\_n^{-1}/2 + O(\mu\_n^{-3}))^2 = \mu\_n^2 + 1 + O(\mu\_{n}^{-2}) = z\_n + 1 + O(z\_n^{-1}).$$
So $z\_n \approx n$ and we see that $\mu\_n \to \infty$ like $\sqrt{n}$.
I haven't checked the details, but I think you should be able to get something like $\mu\_n = n^{1/2} + O(1)$.
| 26 | https://mathoverflow.net/users/297 | 39427 | 25,269 |
https://mathoverflow.net/questions/39435 | 7 | You're playing pinball. When you first shoot a ball it randomly comes down through 1 of 3 gates. When you go through an unlit gate, it lights up. Similarly, a lit gate will go out. What is the expected number of balls you have to throw for all 3 gates to light up?
For example, ball A could go through gate 2, B through gate 3, and C through gate 1. This scenario took 3 rolls and has probability `1/27`.
I've put serious thought into this question twice over the last couple of years but my answer gets more and more complicated until my brain explodes.
**Follow up**
Douglas hit the nail on the head. For kicks, here's the Python script I used as a reality check for both the 2 and 3 gate cases.
```
from random import randint
def pinball(gates):
trials = []
for trial in range(10000):
state = [False for g in range(gates)]
balls = 0
while not all(state):
gate = randint(0, len(state) - 1)
state[gate] = not state[gate]
balls += 1
trials.append(1.0 * balls)
print sum(trials) / len(trials)
pinball(2)
pinball(3)
```
| https://mathoverflow.net/users/9402 | Expected number of pinballs to light up all 3 channels | This is the average time it takes for a random walk on the 1-skeleton of a cube to reach the opposite vertex. There are more general theories for such values, but you can determine this particular one with a simple set of linear equations. Let $T\_i$ be the expected time from when $i$ lights are lit. You want to determine $T\_0$.
$T\_0 = 1 + T\_1$
$T\_1 = 1 + T\_0/3 + 2T\_2/3$
$T\_2 = 1+ 2T\_1/3 + 0$
which has the solution
$\{T\_0=10, T\_1=9, T\_2=7\}$.
| 22 | https://mathoverflow.net/users/2954 | 39437 | 25,271 |
https://mathoverflow.net/questions/39447 | 3 | I'm working with Clifford algebras, of which the first few are $C\_0 = \mathbb{R}$, $C\_1 = \mathbb{C}$, $C\_2 = \mathbb{H}$, $C\_3= \mathbb{H}^2$, $C\_4 = M\_{2,2}(\mathbb{H})$, $C\_5= M\_{4,4}(\mathbb{C})$, $C\_6=M\_{8,8}(\mathbb{R})$, $C\_7 = M\_{8,8}(\mathbb{R})^2$, $C\_8 = M\_{16,16}(\mathbb{R})$, etc. (The operations on the product of two algebras are componentwise.)
The notes I'm working with claim without proof that the minimum dimensions of representations are 1, 2, 4, 4, 8, 8, 8, 8, 16. (Sorry I probably could've formatted this better.) I assume that they mean faithful representations, and that they're representations as $\mathbb{R}$-algebras (and not just as rings), i.e. they're $\mathbb{R}$-linear maps $C\_k \rightarrow End(V)$ (or equivalently maps $C\_k \otimes\_\mathbb{R} V \rightarrow V$, I think). I have no idea why these numbers are right, though. The first few I can see by ad hoc arguments, but that's not really that satisfying. I know no representation theory beyond a few definitions, so this might be really obvious and I just don't know the right theorems/arguments?
| https://mathoverflow.net/users/303 | Is there an easy way to find the minimum dimensions of representations for these R-algebras? | Yes. These can be proven by some very simple principles.
* The smallest representation of a division ring is the ring itself (this covers the first 3), since it has no left or right ideals.
* When you take $n\times n$ matrices in a division ring $D$, the smallest irrep is the obvious one $D^n$. This is just doing a computation of all the left ideals in matrices.
* And, of course, if you take the sum of two rings, the dimension of the smallest representation is just minimum of the two summands.
| 4 | https://mathoverflow.net/users/66 | 39456 | 25,282 |
https://mathoverflow.net/questions/39386 | 7 | Take the set $A\_n=\{a\_1,...,a\_n\}$. Let $S\_n$ be the set of subset-sums of $A\_n$. (The subset-sum of the empty set is assumed to be zero.) Assume that there are $2^n$ unique members of $S\_n$. How many possible sortings are there of set $S\_n$?
For instance, if $n=2$, we have $S\_2=\{0,a\_1,a\_2,a\_1+a\_2\}$. The number of possible sortings of $S\_2$ is 8:
$$\left\{ \begin{matrix} 0<a\_1<a\_2<a\_1+a\_2,\\ 0<a\_2<a\_1<a\_1+a\_2,\\ a\_2<0<a\_1+a\_2<a\_1, \\ a\_1<0<a\_1+a\_2<a\_2, \\ a\_2<a\_1+a\_2<0<a\_1, \\ a\_1<a\_1+a\_2<0<a\_2, \\ a\_1+a\_2<a\_1<a\_2<0, \\ a\_1+a\_2<a\_2<a\_1<0 \end{matrix} \right\}.$$
| https://mathoverflow.net/users/7089 | Number of unique sortings of subset-sums | As David Speyer points out in a comment, this is equivalent to the number of regions resulting when $\mathbb{R}^n$ is divided by hyperplanes of the form $\sum\_{i \in I}a\_i = \sum\_{i \in J}a\_i$ for all disjoint pairs of subsets $I,J \subseteq [n]$. Dual to this description, it is the number of ways that the $3^n-1$ nonzero vectors in $\{-1,0,1\}^n$ can be divided into "positive" and "negative" by a hyperplane (in general position) passing through the origin, which makes it easy to see that the answer is indeed $8$ for $n=2$.
As I mentioned in a comment, the total is always $2^n$ times what you get when making the assumption $a\_i > 0$ for all $i$. There is another symmetry that can also be exploited: making the assumption $a\_1 < a\_2 < \cdots < a\_n$ reduces the total by a further factor of $n!$, which gives a known sequence:
<http://www.oeis.org/A009997>
<http://arxiv.org/abs/math.CO/9809134>
Various key words are "coherent boolean term order", "coherent generalized term order", and "additive antisymmetric comparative probability order". It doesn't look like anyone knows the values beyond $n=7$. You'll want to check Maclagan's reference to Fine and Gill 1976 to see if they give any asymptotics.
Including the $2^nn!$ symmetries gives these values:
1. $2$
2. $8$
3. $96$
4. $5$ $376$
5. $1$ $981$ $440$
6. $5$ $722$ $536$ $960$
7. $138$ $430$ $238$ $607$ $360$
| 9 | https://mathoverflow.net/users/7936 | 39459 | 25,285 |
https://mathoverflow.net/questions/39460 | 8 | Suppose $f: A \to B$ and $g: B \to A$ are injections of rings
(*commutative with identity*). Must $A$ and $B$ be isomorphic as
rings?
According to [this
question](https://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold), this answer should be "no", but can someone give an example?
Thanks!
| https://mathoverflow.net/users/18 | Schroeder-Bernstein for Rings | Hey Damien, I think the following should work $\mathbb{C}$ and $\mathbb{C}(x)$. There is only one uncountable algebraically closed field of each cardinality in characteristic 0 and the algebraic closure of the right hand guy should have cardinality the continuum it should be isomorphic to $\mathbb{C}$. Probably, this assumes the axiom of choice though.
| 5 | https://mathoverflow.net/users/6986 | 39462 | 25,286 |
https://mathoverflow.net/questions/39454 | 1 | Hello
I'm trying to answer this question, but am completely stuck.
Argue that in analyzing the error in a stationery linear relaxation scheme applied to $Au=f$, it is sufficient to consider $Au=0$ with arbitrary initial guess, (say $v\_0$).
Any ideas?
I'm not even sure what the author is trying to say, is he saying when studying the error produced by the relaxation method (eg. the jacobi method) it is sufficient to study the error of the associated homogenous system $Au=0$ with arbitrary initial guesses (say $v\_0$). But in this case the error will simply be $e = -v\_0$. (since the error is defined as $e=u-v\_0$)
Maybe the residual equation plays a part, $Ae=r$ ... ???
Any help, would be much appreciated.
Incidently stationary linear relaxation schemes (eg. gauss seidel and jacobi) have the form
$v^{(k+1)}=(v^{k}+Br^{(0)})$. Not sure if this is usefull. here $r=f-Av$ is the residual vector, and $v^{(k)}$ is the kth iterate of the scheme
| https://mathoverflow.net/users/9404 | Relaxation Scheme for $Au=f$ error analysis | Relaxation method is a part of the theory of iterative methods for the calculation of approximate solutions to linear systems. It depends on a real or complex parameter $\omega$, the *relaxation parameter*. I must point out that Jacobi method is not a relaxation method, whereas Gauss-Seidel is one, corresponding to $\omega=1$. A necessary condition for the convergence is $|\omega-1|<1$. The choice $\omega\in(1,2)$ is called *over-relaxation*.
It is true that in the error analysis, one only needs to consider the case $f=0$. Because of linearity, it is the same to solve $Au=f$, starting from $u\_0$, or to solve $Av=0$ (obvious solution) starting from $v\_0=u\_0-A^{-1}f$. This is why the convergence theory treats only the homogeneous case.
It is not correct to write the iteration in the form $v^{k+1}=v^k+Br^{(0)}$. Such a formula would prevent convergence (obvious). Instead, the iteration writes $v^{k+1}=L\_\omega v^k+Bf$. The convergence occurs precisely when the spectral radius of $L\_\omega$ is less than $1$. The theory emphasises on the search of the optimal parameter $\omega^\*$, for which this spectral radius is minimal, that is, the convergence is the fastest.
For details, see for instance my book *Matrices; Theory and Applications*, GTM **216**. Springer-Verlag (2002).
| 2 | https://mathoverflow.net/users/8799 | 39468 | 25,290 |
https://mathoverflow.net/questions/38906 | 2 | Recall the related notions of [Lie groupoid](http://ncatlab.org/nlab/show/Lie+groupoid), [Lie algebroid](http://ncatlab.org/nlab/show/Lie+algebroid), [generalized morphism of Lie groupoids](http://ncatlab.org/nlab/show/Lie+groupoid#2CatOfGrpds), and [cohomology of Lie algebroid](http://ncatlab.org/nlab/show/Lie+algebroid#the_cealgebra_of_a_vector_bundle_with_anchor_19). Henceforth, I will drop the word "Lie" for all those things listed above, because I want to reuse it: there is a functor "Lie" from the 1-category of groupoids to the 1-category of algebroids. There is also a (contravariant) functor "cohomology" from the 1-category of algebroids to the 1-category of graded commutative algebras. However, their composition does not extend to generalized morphisms (which are really 1-morphisms in a 2-category of groupoids).
In particular, the algebroids for equivalent groupoids need not have isomorphic cohomology. A good example is as follows. For any manifold $M$, there is a "pair" groupoid $M\times M \rightrightarrows M$ with object the points in $M$ and a unique morphism between each pair of points. In fact, this groupoid is equivalent to the groupoid $\{\text{pt}\}$ with one object and one morphisms. But $\operatorname{Lie}(M\times M \rightrightarrows M) = {\rm T}M$ is the tangent groupoid, and the cohomology of this algebroid is the de Rham cohomology of $M$, which need not be trivial.
My question is:
>
> If $G\_1,G\_2$ are two equivalent groupoids, and if both are source-simply-connected, does it follow that the cohomologies of the algebroids $\operatorname{Lie}(G\_1)$ and $\operatorname{Lie}(G\_2)$ are isomorphic?
>
>
>
There is a converse question, for which I am less optimistic, and that I haven't thought much about myself:
>
> If $G\_1,G\_2$ are source-simply-connected groupoids and the cohomologies of $\operatorname{Lie}(G\_1)$, $\operatorname{Lie}(G\_2)$ are isomorphic, does it follow that $G\_1,G\_2$ are equivalent?
>
>
>
My motivation is the following. There ought to be (but there is not, although almost) a "Lie III theorem" that says that the categories of algebroids and of source-simply-connected groupoids are equivalent. Groupoids present stacks, and the question becomes what "stack-like" thing algebroids present. If the answers to both questions are "yes", then the "stack-like thing" presented by an algebroid just *is* its cohomology. But probably the answers are not both "yes" — even answers "yes, no" means that, well, the cohomology doesn't entirely determine the stack, but it is an invariant.
| https://mathoverflow.net/users/78 | Is the cohomology of the corresponding Lie algebroid an invariant under equivalence of source-simply-connected Lie groupoids? | I may be misinterpreting what you said (in particular, I don't know what you mean by "source simply connected"), but it sounds like you basically answered your own question in the negative: By employing the pair construction $M \times M \rightrightarrows M$, it suffices to find two simply connected manifolds with nonisomorphic cohomology. The standard examples are the point and the sphere $S^2$.
| 1 | https://mathoverflow.net/users/121 | 39476 | 25,295 |
https://mathoverflow.net/questions/39452 | 34 | Friedman [1] conjectured
>
> Every theorem published in the Annals of Mathematics whose
> statement involves only finitary mathematical objects (i.e., what logicians
> call an arithmetical statement) can be proved in [EFA](http://www.math.ohio-state.edu/~friedman/pdf/GodelLect060202.pdf). EFA is the weak
> fragment of Peano Arithmetic based on the usual quantifier free axioms for
> 0,1,+,x,exp, together with the scheme of induction for all formulas in the
> language all of whose quantifiers are bounded. This has not even been
> carefully established for Peano Arithmetic. It is widely believed to be
> true for Peano Arithmetic, and I think that in every case where a logician
> has taken the time to learn the proofs, that logician also sees how to
> prove the theorem in Peano Arithmetic. However, there are some proofs which
> are very difficult to understand for all but a few people that have
> appeared in the Annals of Mathematics - e.g., Wiles' proof of FLT.
>
>
>
Have there been any serious challenges to this or the weaker conjecture with Peano arithmetic in place of exponential function arithmetic?
[1] <http://cs.nyu.edu/pipermail/fom/1999-April/003014.html>
| https://mathoverflow.net/users/6043 | Status of Harvey Friedman's grand conjecture? | To Mark Sapir:
The conjecture says "can be proved in EFA". If it "was not proved in EFA" then that does not count. However, I am still interested if it "was not proved in EFA". Since EFA can still develop some theory of recursive functions, the fact that recursive functions are mentioned, or even used, does not imply that the proof is outside EFA.
It is also true that EFA is fully capable of proving recursive unsolvability theorems.
The only proofs we have that given mathematical statements are not provable in EFA is to show that the mathematical statements inherently give rise to functions that are not bounded by any finite height exponential stack. Is this definitely the case here? Or is there a conjecture to that effect?
Harvey Friedman
| 50 | https://mathoverflow.net/users/9411 | 39487 | 25,302 |
https://mathoverflow.net/questions/39484 | 1 | Suppose a Hilbert space W can be written as the direct sum (not necessarily orthogonal) of the closed subspaces H and V, where H is assumed to be of finite dimension. Define a new inner product via
||h+v||^2:=q(h)+|v|^2,
where |.| denotes the original norm on the Hilbert space and q is a positive definite quadratic form on H (one can assume w.l.o.g. q=|.|^2).
QUESTION: Are |.| and ||.|| equivalent?
||.||^2 is easily seen to be dominated by 2|.|^2, but I don't know about the other direction. (Also notice that the question is obviously true if V and H were orthogonal!)
| https://mathoverflow.net/users/3509 | Is the metric obtained by altering the metric of a Hilbert space on a finite-dimensional subspace equivalent to the original one? | Thanks to Bill Johnson!
My question is easily answered by a direct application of the closed graph theorem (one shows that the diagonal is closed in the mixed norms). Unfortunately, I did not have this one as an exercise in my functional analysis class!
| 0 | https://mathoverflow.net/users/3509 | 39495 | 25,306 |
https://mathoverflow.net/questions/39471 | 2 | I am puzzled about the following question:
Let C be a smooth, projective, absolutely irreducible curve defined over GF(q) and let g denote the genus of C. O is a rational point on C, and the divisor D is defined as D = (2g-1)O.
Question:
Whether there exist rational points $P\_1,P\_2,\cdots,P\_n~(n>g)$ on C such that for any g rational points $P\_{i\_1},P\_{i\_2},\cdots,P\_{i\_g}$ of them, $l(D-\sum^g\_{j=1}P\_{i\_j})=0$ (i.e. the dimension of $L(D-\sum^g\_{j=1}P\_{i\_j})$).
I know it is easy to find g such points. But the existence of further points really confused me.
Thank you for your help!
| https://mathoverflow.net/users/9407 | Existence question on rational points on a curve | Damiano's comments are spot-on and maybe you should clarify your hypotheses. Here is an inductive argument which shows that, if you have enough rational points on the curve, you can grow your set.
First, by using the linear system $|(2g-1)O|$ you can embed your curve in $\mathbb{P}^{g-1}$ as a curve of degree $2g-1$ and you want $n$ points such that no $g$ of them are in a hyperplane. Suppose you have $n$ such points and that your curve has more than $n + g{n \choose g-1}$ rational points. I claim you can choose a further point $P\_{n+1}$ and have $n+1$ points, no $g$ in a hyperplane. Indeed, for each subset of $g-1$ points of your $n$ points, there are at most $g$ other points on the intersection of the curve and the hyperplane spanned by them. If $P\_{n+1}$ is in none of these hyperplanes, you are done. My hypothesis ensures that there is at least one such point.
| 1 | https://mathoverflow.net/users/2290 | 39497 | 25,308 |
https://mathoverflow.net/questions/39490 | 2 | In most basic abstract algebra courses, the free group is directly constructed, a process that I find rather unwieldy. An alternate method of characterizing the free group is by means of its universal property: for any function $f:S\to G$, an arbitrary group, there is a function $g:S\to F\_{S}$ and a unique homomorphism $\varphi: F\_{S}\to G$ such that $f=\varphi g$. Of course, a direct construction of the free group is necessary to show that any group actually satisfies this definition. I was wondering what happened when the definition was dualized. In other words, let $P\_{S}$ be the group such that for any function $f:G\to S$, there is a function $g:P\_{S}\to S$ and a unique homomorphism $\varphi:G\to P\_{S}$ such that $f=g\varphi$. It would seem, in light of Cayley's theorem, that $P\_{S}$ is just the set of permutations on $S$, but I'm not sure of this. Does anyone know what $P\_{S}$ is?
| https://mathoverflow.net/users/6856 | Dualizing the definition of a free group | As has been noted in the comments, your definition of "free group on $S$" is not quite right. The map $g\colon S\to F\_S$ is fixed, and is part of the "free group" (that is, the free group on $S$ is the pair $(F\_S,g)$, with $g\colon S\to F\_S$ a set-theoretic map). The universal property is that for every set map $f\colon S\to G$ into an arbitrary group, there exists a unique homomorphism $\varphi\colon F\_S\to G$ such that $g = \varphi f$. But $f$ is not allowed to depend on $g$.
It is not hard to see that no such "cofree group" can exist on sets with more than one element. Suppose that you have a set $S$ with more than one element, and a "cofree group" on $S$, $C\_S$, together with a set-theoretic map $f\colon C\_S\to S$ such that for every group $G$ and every set-theoretic map $g:G\to S$, there exists a unique homomorphism $\varphi\colon G\to C\_S$ such that $f = g\varphi$. Let $a\in S$ be different from $f(e)$; then the map $g\colon G\to S$ with $g(x)=a$ cannot factor through $C\_S$.
As for the case $S=\{s\_0\}$, uniqueness of $\varphi$ forces $C\_S$ to be the trivial group, because both the zero map and the identity on $C\_S$ would satisfy the universal property relative to $f$.
The free group construction is the left adjoint of the underlying set functor. In general, left adjoints respect colimits and right adjoints respect limits; that is why the underlying set of a product of groups is the set-theoretic product of the underlying sets (underlying set is the right adjoint, so it respects limits like the product), and why the free group on the disjoint union of two sets is the free product of the free groups on the two sets (disjoint union being the coproduct in $Sets$, free product the coproduct in $Groups$, and coproduct being a colimit). As James Borger notes, if you had a dual of the free group construction, it would be the right adjoint of the underlying set functor and would therefore have to respect colimits. So that the underlying set of a free product of groups would have to be the disjoint union of the underlying sets of the groups. This does not occur, so no such object can exist.
P.S. As has also been pointed out in the comments, the universal property is nice and all, and can prove uniqueness, but in general one needs either very high-power categorical/universal algebraic theorems to deduce existence, or one must actually *construct* the objects in some way. In the case of free groups, while there are many constructions (e.g., as a "big direct product"; see the reference I'm about to give), it is via words or other equivalent constructions (e.g., the fundamental group of a bouqet of circles) that one can get a better handle of them. But if you like universal constructions (nothing wrong with that!) I recommend taking a look at George Bergman's [An Invitation to General Algebra and Universal Constructions](http://math.berkeley.edu/~gbergman/245/). It has three different constructions of the free group in Chapter 2.
| 9 | https://mathoverflow.net/users/3959 | 39501 | 25,309 |
https://mathoverflow.net/questions/39500 | 0 | As I understand it, an open cover of a Base Space and associated holomorphic transition functions on the intersection are sufficient data to define (up to isomorphism perhaps) a holomorphic (complex) line bundle.
So if we cover S1 with open sets U and V which intersect in open sets P and Q such that P and Q have empty intersection (you will note I am avoiding Latex here. Basically we cover S1 with a pair of horseshoes). We define the UV- transition function to be +1 on P and -1 on Q (locally constant thus holomorphic) we have the set up for an (infinite) Mobius band.
What I am strugling to define are the corresponding trivialisations from U or V to UxC or VxC
that correspond to this transition function.
Even Grifiths & Harris has some "hand waving" about identifying points which does not help. I just want to see the explicit map on the fibres over U and V.
Any suggestions?
Thanks
Noel R
| https://mathoverflow.net/users/9418 | Trivialisation of Mobius Line Bundle | There appears to be some confusion in the question: The circle $S^1$ is not a complex manifold, so it does not admit a meaningful notion of holomorphic line bundle. If you try to construct a complex line bundle on the circle, you will find that it is automatically a trivial line bundle. If you want to construct a Möbius band as a line bundle, you should take a real line bundle, with the transition functions you specified.
I would recommend a more topological text than Griffiths and Harris for information about the construction you are trying to do. Perhaps Hatcher's book on vector bundles and K-Theory will help. It is available on his web page.
| 0 | https://mathoverflow.net/users/121 | 39509 | 25,315 |
https://mathoverflow.net/questions/39491 | 3 | Consider a planar point process $X$ and call $N\_A = \text{Card}\big( X \cap A\big)$ the number of points inside the subset $A \subset \mathbb{R}^2$. If one knows the law of $(N\_{A\_1}, \ldots, N\_{A\_r})$ for any sets $A\_1, \ldots, A\_r$, then the process is completely characterized. I recently learned that it in fact suffices to know $f(A)=P(N\_A=0)$ (called the void-probability function) for any set $A$ in order to completely characterize the law of $X$.
Intuitively, I do not understand why such a result is true. Indeed, the knowledge of the function $f$ brings some information in the correlation structure of the process $X$: nevertheless, I still fail to understand how the function $f$ can encode the whole correlation structure of the process. Any thoughts on this ?
| https://mathoverflow.net/users/1590 | a point process is characterized by its void probabilities | This is only true for **simple** point processes (no duplicate points).
By the inclusion-exclusion principle, $f$ determines the joint distribution of several (disjoint) sets being empty or occupied. If the process is simple this allows recovering the law of $(N\_{A\_1},\dots,N\_{A\_r})$ as a limit over finer partitions.
| 7 | https://mathoverflow.net/users/9422 | 39517 | 25,319 |
https://mathoverflow.net/questions/39515 | 2 | Hi All,
I am learning Differential Equations, and came across a specific problem of Dirichlet BVP, which says that:
Given x'' = f(x'), x(0) = 0 = x(1), If f(0) $\neq $ 0 and f has two zeros of opposite sign (say, $r^+$ $\gt$ 0 and $r^−$ $\lt$ 0) then all solutions to Dirichlet BVP have derivatives satisfying
$r^−$ $\lt$ x'(t) $\lt$ $r^+$ , $\forall$t $\epsilon$ [0, 1].
Is this true? How can be this shown?
Also, can someone please tell me how to establish apriori bounds on the derivative of solutions on [0, 1] for the BVP x'' = [(x')$^2$ − 1]$^n$ ?
| https://mathoverflow.net/users/8245 | Specific Dirichlet BVP solutions | Say that $f$ is of class ${\mathcal C}^2$. Set $y:=x'$ and differentiate. You get $y''=f'(y)y'$. This is a linear ODE in $y'$, if we think of $f'(y)$ as a given function $g(t)$. Since $y$ is not $\equiv0$ (because $f(0)\ne0$), Cauchy-Lipschitz tells you that $y'=f(y)$ does not vanish over $[0,1]$. In particular, $y$ is strictly monotonous, say increasing. Because $\int\_0^1y(t)dt=x(1)-x(0)=0$, $y$ vanishes at some $t\_0$. Then $x''(t\_0)=f(0)$ shows that $f(0)>0$ and therefore ($r\_\pm$ are consecutive zeroes of $f$), $f$ remains $>0$ over $(r\_-,r\_+)$. Since $f(y)$ does not vanish, $y$ does not take the values $r\_\pm$. But $y$ takes the value $0$ (at $t\_0$). Therefore $y=x'$ remains in $(r\_-,r\_+)$.
| 2 | https://mathoverflow.net/users/8799 | 39518 | 25,320 |
https://mathoverflow.net/questions/39508 | 43 | When I was learning about spectral sequences, one of the most helpful sources I found was Ravi Vakil's notes [here](http://math.stanford.edu/~vakil/0708-216/216ss.pdf). These notes are very down-to-earth and give a kind of minimum knowledge needed about spectral sequences in order to use them.
Does anyone know of a similar source for derived categories? Something that concentrates on showing how these things are used, without developing the entire theory, or necessarily even giving complete, rigorous definitions?
| https://mathoverflow.net/users/5094 | A down-to-earth introduction to the uses of derived categories | I would suggest "Fourier-Mukai transforms in algebraic geometry" by Daniel Huybrechts.
| 29 | https://mathoverflow.net/users/4428 | 39532 | 25,330 |
https://mathoverflow.net/questions/39540 | 36 | The formula $\mathcal{L}\_X\omega = i\_Xd\omega + d(i\_X \omega)$ is sometimes attributed to Henri Cartan (e.g. Peter Petersen; Riemannian Geometry 2nd ed.; p.380) and sometimes to his father Élie (e.g. Berline, Getzler, Vergne; Heat Kernels and Dirac Operators, p.17), and often just to "Cartan" (e.g. <https://en.wikipedia.org/wiki/Lie_derivative> ).
Who is right? Reference?
| https://mathoverflow.net/users/9161 | Is "Cartan's magic formula" due to Élie or Henri? | Élie for sure. The formula is derived in *Les systèmes differentiels extérieurs et leur applications géométriques* which was probably written before Henri was born. BTW, here is a very short proof that Chern showed me long ago.
* The exterior derivative is an anti-derivation of the exterior algebra
and so is the interior product with a vector field while the Lie
derivative is a derivation. (These are all trivial to check.)
* Also, the anti-commutator of two anti-derivations is a derivation. Hence
both sides of the "magic formula" are derivations.
* It is obvious that
two derivations are equal if they agree on 0-forms $f$ and exact
1-forms $df$, since (locally) they generate the exterior algebra.
* Finally it is trivial that both sides of the magic formula agree on
such forms.
| 67 | https://mathoverflow.net/users/7311 | 39541 | 25,335 |
https://mathoverflow.net/questions/39545 | 3 | **Motivation:**
For the sake of concreteness, I'll state a very particular context, but my question is a little more general. I'm trying to find a function $\gamma\colon [0,\delta) \to [0,\delta')$ that satisfies the following functional equation:
$$
\gamma(y) + \gamma(y)^{1+\varepsilon} - y = \gamma(y - y^{1+\varepsilon}).
$$
Here $0<\varepsilon<1$, and I'm most interested in the behaviour of $\gamma$ for very small values of $y$. I've persuaded myself that $\gamma(y) = y^{1/(1+\varepsilon)}(1+o(1))$, but I'd really like an exact solution, which I imagine would have to come in the form of a power series $\gamma(y) = \sum\_n c\_n y^{a\_n}$. Due to the form of the equation, though, the exponents $a\_n$ aren't going to be integers, and unless I'm mistaken, they won't even all be integer multiples of some fixed $a\_0$, so I can't get back to "regular" power series by doing a simple change of coordinates. This motivates my question...
**Question:**
Can anyone suggest a reference on dealing with power series where the exponents take non-integer values (and are not all integer multiples of some fixed exponent)? Or suggest a paper where such power series are used (for any purpose)? Ideally I'd like to see how similar functional equations have been solved, but any references at all would be appreciated.
| https://mathoverflow.net/users/5701 | Power series with non-integer exponents | The series Σane-λnz are called (generalized) Dirichlet series, and are special cases of the Laplace transform of a discrete measure. For t=e-z you get the power series with fractional exponents you are asking about. See Widder's book "The Laplace transform" for more details.
| 7 | https://mathoverflow.net/users/51 | 39546 | 25,339 |
https://mathoverflow.net/questions/39537 | 12 | Let $M$ be a compact simply connected R. manifold. Let $x$ be a base point and let $\gamma$ be a smooth loop in $M$ starting and ending at $x$.
>
> Is there a base point preserving retraction of $\gamma$ to $x$ such that every point on $\gamma$ travels a distance at most, say, $2diam(M)$?
>
>
>
I think the answer is yes. The scheme I have in mind is the following
Consider the cut locus $CL(x)$ of $x$. By perturbing the metric and the curve we can assume that $CL(x)$ is triangulable and $\gamma$ intersects it at finitely many points. Then slide these points off the $CL(x)$. Then the new curve does not intersect $CL(x)$ and can be contracted to $x$ following the geodesics.
This is probably too complicated...
>
> Is there a simple proof? A reference?
>
>
>
Remark: 1) I would be happy with any estimate (not necessarily $2diam(M)$) which is independent of $\gamma$.
2) If there is a good curve shortening procedure that is not base-point-preserving please share it as well.
| https://mathoverflow.net/users/2029 | Effective contraction of a loop. Reference or a simple proof? | Yes there is a bound independent of $\gamma$. Fix a fine triangulation of $M$ (say, with simplices 10 times smaller than the injectivity radius of the metric). For each vertex $q$ of the triangulation, fix a shortest path $s\_q$ connecting $q$ to the marked point $p$. It is easy to deform any loop $\gamma$, via a short homotopy, into a path in the 1-skeleton of the triangulation. Then, for every vertex occuring on this path, "pull" a small piece of the loop to $p$ along the corresponding $s\_q$. Now the loop is a product of "elementary" loops of the form $s\_q^{-1}[qq']s\_{q'}$ where $[qq']$ is an edge of the triangulation. Contract these "elementary" loops one-by-one. Since there are only finitely many different elementary loops, and the same elementary loops work for all $\gamma$'s, we have a uniform upper bound (depending on $M$) for the length of the homotopy.
On the other hand, there is no bound for the homotopy length in terms of the diameter. In addition to the great Bill Thurston's answer, let me mention another construction. For every $\varepsilon>0$ there is a metric on $S^3$ with sectional curvature $\le 1$ and diameter $\le\varepsilon$ (it was first constructed in Gromov "Almost flat manifolds, see also Buser and Gromoll "On the almost negatively curved $3$-sphere"). In this metric, take a geodesic starting from $p$ of length $2\varepsilon$ and connect its endpoint back to $p$ by a shortest path. Since the curvature is bounded above by 1, any contraction of this curve will have a trajectory longer than $\pi$ (I think this fact is called Klingenberg's lemma but I am not sure).
| 11 | https://mathoverflow.net/users/4354 | 39556 | 25,346 |
https://mathoverflow.net/questions/39548 | 1 | Suppose I wanted to express a number $N$ as a difference of squares. For large $N$ this is in general difficult, as finding $N=a^2-b^2$ leads to the factorization $N=(a+b)(a-b)$. Even if the problem is weakened to searching for $a\neq b$ with $a^2\equiv b^2\pmod N$ the problem is still hard (though not as hard), since enough congruences could be used to factor $N$ with Dixon's method or any of its modern versions (in particular, the number field sieve).
So I am curious about the difficulty of these weak versions of the problem. Are any of these easier than finding relations with the NFS?
1. *Weak form.* Given $N$, $k$, and a factoring oracle, find $k$ distinct nontrivial congruences $a^2\equiv b^2\pmod N$.
2. *Semi-weak form.* Given $N$, $k$, and the complete factorization of $N$, find $k$ distinct nontrivial congruences $a^2\equiv b^2\pmod N$.
3. *Strong form.* Given $N$, $k$, and a partial factorization of $N$, find $k$ distinct nontrivial congruences $a^2\equiv b^2\pmod N$.
| https://mathoverflow.net/users/6043 | Differences of squares | I think he is more interested in how many ways can one number be written as a difference of squares.
The solution to the semiweak form is actually simple, since $a+b$ and $a-b$ have the same parity.
If $N$ is odd, then as long as $N$ has at least 2k divisors (which can easely be seen from the factorisation of $N$) you can simply write $k$ different pairs $(d,N/d)$ with $d$ smaller than $N/d$ and solve
$$a-b= d \,;\, a+b= N/d \,.$$
If $N\equiv 2 \text{mod} 4$ there is no solution as noted above.
If $N\equiv 0 \text{mod} 4$ then you do the same trick for $N/4$: as long as $N/4$ has at least 2k divisors you can simply write $k$ different pairs $(d,N/4d)$ with $d$ smaller than $N/4d$ and solve
$$a-b= 2d \,;\, a+b= N/2d \,.$$
Edit: Same thing works for the strong form, as long as the partial factorisation of $N$ is "big" enough: i.e. it allows us to find k distinct pairs pairs $(d,N/d)$ of divisors of same parity with $d$ smaller than $N/d$.
| 4 | https://mathoverflow.net/users/9313 | 39558 | 25,348 |
https://mathoverflow.net/questions/39563 | 3 | $N$ the positive natural numbers has one infinity.
$Z$ the integers has 2 infinities.
What object would as "naturally" as possible have 3 infinities?
This probably can be answered in many ways. Yet for me the algebraic side would be more important than the topological one, though this does not exclude both.
What troubles me is that $Z$ is natural as being final in the category of rings) and moreover it is the completion ( in fractional sense) of $N$.
| https://mathoverflow.net/users/3005 | 3 directions of infinity ? | The obvious first answer: take three copies of $\mathbb{N}$ as total orders, then join them at the bottom element to get an unbounded poset with bottom. This of course isn't satisfactory as it doesn't give $\mathbb{Z}$ for two copies of $\mathbb{N}$. This strikes me as a sort of 'what about a 3-dimensional version of the complex numbers?' question, and could benefit from considering 'four infinities'...
| 4 | https://mathoverflow.net/users/4177 | 39565 | 25,352 |
https://mathoverflow.net/questions/39479 | 1 | How will the kalman filtering model look like in the case when I just receive some data and want to filter them from noise? The data is actually an acceleration of some object.
So the system must be like this:
$$x\_t = A\_tx\_{t-1} + B\_tu\_t + \epsilon\_t$$
$$z\_t = C\_tx\_t + \delta\_t$$
Where the $\epsilon\_t$ and $\delta\_t$ are the white noise. $x\_t$ is a state variable.
The problem is that I can't figure out what will the system look like in my case,
when I receive acceleration measurements (observations - $z\_t$) at each time period $\Delta t$.
I think I don't need the control vector $u\_t$ in my case, so the system will be:
$$x\_t = A\_tx\_{t-1} + \epsilon\_t$$
$$z\_t = C\_tx\_t + \delta\_t$$
I suppose, but not sure about this kind of filter system:
$$x\_t = x\_{t-1} + \epsilon\_t$$
$$z\_t = x\_t + \delta\_t$$
But it seems too simple.
How to make the first iteration in the Kalman filtering procedure?
**EDIT 1:**
here is the kalman filtering algorithm, taken from the book: probabilistic robotics.
Kalman\_filter($\mu\_{t-1}$, $\Sigma\_{t-1}$, $u\_t$, $z\_t$)
$$\bar{\mu}\_t = A\_t\mu\_{t-1} + B\_tu\_t$$
$$\bar{\Sigma}\_t = A\_t\Sigma\_{t-1}A\_t^T + R\_t$$
$$K\_t = \bar{\Sigma}\_tC\_t^T\left(C\_t\bar{\Sigma}\_tC\_t^T + Q\_t\right)^{-1}$$
$$\mu\_t = \bar{\mu\_t} + K\_t\left(z\_t - C\_t\bar{\mu}\_t\right)$$
$$\Sigma\_t = \left(I - K\_tC\_t\right)\bar{\Sigma}\_t$$
return $\mu\_t$, $\Sigma\_t$
The thing that I do not understand here is:
Here the data that is unknown is - $\Sigma\_0$, $\mu\_0$
I supppose that I can choose some data by myself for that values. But one more data that is unknown for me is: $R\_t$
It comes from:
The state transition probability is given by $p(x\_t|u\_t,x\_{t-1})$.
And we got: $$x\_t = A\_t\mu\_{t-1} + B\_tu\_t+\epsilon\_t$$ as one of the equations of the Kalman filter.
We also know the normal distribution:
$$p(x) = det\left(2\pi\Sigma\right)^{-1/2}exp\left(-1/2(x-\mu)^T\Sigma^{-1}(x-\mu)\right)$$
(I've already asked a question from which you can see where it comes from: [question](https://mathoverflow.net/questions/39074/kalman-filter-understanding-the-mathematical-part))
So we have:
$\mu\_t = A\_tx\_{t-1} + B\_tu\_t$(discussed in the question, the link is above)
And also $R\_t$ is a covariance of the posterior state.
Here we got the whole formula.
$$p(x\_t|u\_t, x\_{t-1}) = det\left(2\pi R\_t\right)^{-1/2}exp\left(-1/2(x\_t-A\_tx\_{t-1}-B\_tu\_t)^TR\_t^{-1}(x\_t-A\_tx\_{t-1}-B\_tu\_t)\right)$$
So, how should be the $R\_t$ value estimated? It depends on $t$.
If I set some value by myself to $\Sigma\_0$ and $\mu\_0$ then what should be done with $R\_t$ which appears in the Kalman filter algorithm listed above in this step:
$$\bar{\Sigma}\_t = A\_t\Sigma\_{t-1}A\_t^T + R\_t$$
?
Correct me please if I am wrong:
$$R\_t = cov\left(x\_t|x\_{t-1}, u\_t\right) = E\left[x\_t^2|x\_{t-1}, u\_t \right] - \left(E\left[x\_t|x\_{t-1},u\_t\right]\right)^2$$
$$R\_t = E\left[x\_t^2|x\_{t-1}, u\_t \right] - \left(A\_tx\_{t-1}+B\_tu\_t\right)^2$$
So how to calculate the $R\_t$? Should it be also set by user?
Actually $R\_t$ is a covariance of the noise $\epsilon\_t$ in equation:
$$x\_t = A\_tx\_{t-1} + B\_tu\_t + \epsilon\_t$$.
And it depends on $t$.
The same thing about the noise covariance of $\delta\_t$ in case of this equation of the Kalman filter:
$$z\_t = C\_tx\_t + \delta\_t$$
**EDIT 2:**
So as I understood four parameters should be selected by the user (tuned), they are:
$Q\_t$, $R\_t$, $\mu\_0$ and $\Sigma\_0$
Am I right?
| https://mathoverflow.net/users/3195 | Kalman filtering: 1D case | A few remarks on your problem:
* You have to assume something for your initial variance (not covariance in this case, since it's univariate). The same applies in the multivariate case -- you have to know something about $P\_{0|0}$. You do not *calculate* the initial variance.
* If you really have no idea what to choose for your initial variance, choose a large number. This is equivalent to saying "I don't know what's going on in the system, so I'm going to be conservative and assume the worst." As the Kalman filter iterates, it will generally converge and the variance will tend to decrease.
* Given a measurement $z\_{0}$, you can do the rest (Kalman gain, prediction etc.). In fact in the linear case, it is proven that the Kalman gain can be calculated off-line (see "Separation Principle" <http://en.wikipedia.org/wiki/Separation_principle>).
* If your filter is having trouble converging (very unlikely in this simple case), you can use something called a Re-iterative Kalman Filter (<http://tinyurl.com/2fokknm>). This Kalman filters iterates $n$ steps and uses the information collected to correct $x\_{0}$. At $n+1$, it uses to the corrected $x\_{0}$ and recursively calculates $x\_{n+1}$; thereafter the Kalman filter will usually converge rapidly.
Peter D. Joseph (a pioneer in the use of Kalman Filters in the 1960s) wrote a simple tutorial on the subject in which he gives the reader an intuitive understanding of what these filters do -- in it he motivates the subject through the derivation of a 1-D example. Unfortunately the webpage no longer exists; however I managed to find the original document in text format: <http://www.humintel.com/hajek/kalman.txt>
If you're willing to reformat it into $\LaTeX$, I think you'll find the document helpful.
| 2 | https://mathoverflow.net/users/7851 | 39566 | 25,353 |
https://mathoverflow.net/questions/39554 | 3 | Given $n$ points $p\_i=(x\_i,y\_i)$ on the [Euclidean] plane, and a positive real number $\rho$. Can we have a polynomial spline (e.g., natural cubic spline) passing through all these points, such that: (a) successive segments of the spline have are continuous and have equal 1st & 2nd derivative at the meeting point (E.g., if $S\_1(x)$ joins $p\_1-p\_2$ and $S\_2(x)$ joins $p\_2-p\_3$, then $S\_1''(x\_2)=S\_2''(x\_2)$.) and (b) the curvature of the spline is bounded above by $\rho$?
Note that natural polynomial splines obey (a) but it's hard to say anything about (b). I am also unaware of any means to bound the curvature of a spline, and a literature search online didn't turn up much of interest.
Here are 2 other variations of the question above that I am unable to answer:
(V1) If the spline needs to be closed, i.e. $p\_{n+1}=p\_1$, how, if at all, does the answer change?
(V2) If we allow any type of interpolation spline at all that obeys (a) and (b), do we have a solution?
FYI, this isn't a homework problem. I ran into this question when trying to write code for an engineering application.
| https://mathoverflow.net/users/6495 | Interpolation splines of bounded curvature | If there are no more constraints, then you can do it with arbitrarily low curvature with any reasonable class of splines.
If the points are say within a 10cm region, make huge loops 1km in diameter (or bigger if you want smaller curvature). If the spline construction is smooth, continuous, and invariant under similarity, then the curvature converges to 0.
If the curves are required to stay in a bounded region of the plane, then as the region gets smaller, not even arbitrary $C^2$ curves can thread through them with bounded curvature. Just imagine $3$ points at the corners of an equilateral triangle 1 micron on a side, and ask for the curve to be confined to a box 2 microns on a side. The curvature will be on the order of $\pi / $ micron.
Here are two copies of a set of four points threaded with Adobe Illustrator splines to illustrate the phenomenon. Note: I added extra knots in the big loops to make them look better, but this isn't necessary to construct examples. (The mathematical characterization of these splines is not relevant to the answer, and furthermore, I don't actually know):
[alt text http://dl.dropbox.com/u/5390048/splines.jpg](http://dl.dropbox.com/u/5390048/splines.jpg)
The design considerations for splines are much more subtle than minimizing curvature.
However, I'd like to mention that the earlier meaning of [splines](http://en.wikipedia.org/wiki/Flat_spline) had to do with thin splints of wood used in woodworking, e.g. boat building, to lay out curves for cutting. Unlike the usual mathematical splines, they have fixed length, and a reasonable
mathematical model is that they trace out curves that locally minimize total curvature subject to their constraints (lead weights called ducks because that's what they resembled).
It's easy to get examples of these traditional splines with multiple local minima: cut a strip of paper (good enough for this) and bring the ends closer without turning them. The strip pops to one side or the other, giving two local minima.
| 3 | https://mathoverflow.net/users/9062 | 39571 | 25,357 |
https://mathoverflow.net/questions/39580 | 6 | It would be nice to find out what is known about the following problem.
First let us consider a free group $F$ with two generators $a$ and $b$. We are interested in its elements that are
1. not equal to identity,
2. of form $c\_1 c\_2 \ldots c\_n d\_1^{-1} d\_2^{-1} \ldots d\_n^{-1}$, where all $c\_i$ and $d\_i$ are either equal to $a$ or to $b$.
Let us denote all these elements by $W$.
Let us consider the group $S\_k$ now. What is the shortest word from $W$ with the following property: whatever elements of $S\_k$ we substitute for $a$ and $b$ we get identity (in $S\_k$)?
The best bounds for the smallest $n$ I am aware of are $2^{O(k)}$ and $\Omega(k^2)$.
| https://mathoverflow.net/users/3448 | Relations in symmetric group | You are asking for the shortest balanced semigroup identity in $S\_k$. Some info can be found here: Pöschel, R.; Sapir, M. V.; Sauer, N. W.; Stone, M. G.; Volkov, M. V. Identities in full transformation semigroups. Algebra Universalis 31 (1994), no. 4, 580--588. But the bound there is exponential. I believe the lower bound should be exponential too, but I do not think there were any more recent papers on the subject. You may also try to read this paper. It is also relevant: Cherubini, Alessandra; Kisielewicz, Andrzej; Piochi, Brunetto, On the length of shortest 2-collapsing words. Discrete Math. Theor. Comput. Sci. 11 (2009), no. 1, 33--44.
| 7 | https://mathoverflow.net/users/nan | 39584 | 25,363 |
https://mathoverflow.net/questions/39498 | 1 | A $1+n$ dimensional semi-Riemannian metric is called "regularly sliced" if it can be written as,
$ds^2 = -N^2 (\theta^0)^2 + g\_{ij}\theta ^i \theta ^j$
where $N$ is called the ``Lapse Function" and the $\theta's$ are defined as follows,
$\theta ^0 = dt$
$\theta ^i = dx^i + \beta ^i dt$
where $\beta ^i$ is called the ``Shift vector"
I would like to know how this particular form of the metric is motivated.
Is there something canonical or natural about this?
I am aware of Yvonne-Costakis' proof that being regularly sliced is an iff condition for the manifold to be globally hyperbolic and their proofs which use such metrics to establish conditions about the future completeness of non-spacelike geodesics depending on the behaviour of the Lapse Function.
For this metric the Christoffel symbols are apparently of the form,
$\Gamma ^i \_{00} = Ng^{ij}\partial \_j N $
$\Gamma ^0 \_{ij} = \frac{1}{2N^2} (\partial \_0 g\_{ij} - g\_{hj}\partial \_i \beta ^h -g\_{ih} \partial \_j \beta ^h)$
$\Gamma ^0 \_{io} = \frac{\partial\_i N}{N}$
$\Gamma ^0 \_{00} = \frac{\partial \_0 N}{N}$
I would like to know if there is any slick way of deriving the above equations.
I had asked a similar question [here](https://math.stackexchange.com/questions/4858/a-particular-method-of-pulling-back-a-metric-on-a-submanifold).
| https://mathoverflow.net/users/2678 | Techniques of calculating Christoffel symbols for regularly sliced metric. | This form of a semi-Riemannian metric is beloved of both numerical and mathematical relativists, but especially the former. The starting point is usually a globally hyperbolic spacetime $(M,h)$. Then $M$ can be foliated by surfaces $\Sigma\_t$ of constant $t$, where $t$ is a global time coordinate i.e. a $C^1$ function $t:M\to\mathbb{R}$ such that the 1-form $dt$ is non-vanishing and everywhere timelike: $h^{-1}(dt,dt)<0$. Each $\Sigma\_t$ is a Riemannian 3-manifold with metric $g$ inherited from $(M,h)$. The most convenient way to write this metric is $g=h+n\otimes n$, where $n$ is the normal 1-form of $\Sigma\_t$ with unit length. The lapse and shift are *not* canonical or unique. They are defined with respect to a *choice* of future pointing timelike vector field $\vec{t}$, conventionally normalized by $dt(\vec{t})=1$. Then the lapse is $N=-n(\vec{t})$ and the shift 1-form is defined by $\beta(\vec{v})=g(\vec{t},\vec{v})$ for all tangent vectors $\vec{v}$. Note that $\beta(\vec{n})=0$. Following the integral curves of $\vec{t}$ allows one to construct a diffeomorphism relating the different $\Sigma\_t$. Then one can interpret the lapse as measuring the proper time between different slices $\Sigma\_t$, and $\beta$ as measuring the relative velocity of observers that follow the normal $\vec{n}$ and those that follow $\vec{t}$.
The Einstein equations can be decomposed with respect to a regular slicing, yielding evolution and constraint equations. Different gauge choices – that is, different choices of $\vec{t}$ - are used by relativists to produce different decompositions that have favourable mathematical properties, usually relating to the particular flavour of hyperbolicity of the evolution equations.
So lapse and shift certainly both carry meaning, but neither is canonically defined inasmuch as they depend upon the choice of $\vec{t}$. The regular sliced form of the metric is indeed natural, in that it arises as described above once one has identified a global time coordinate on the spacetime.
Added: This attempts to answer the first question(s). I don't think that there is a better response to the second question - a slick way of calculating the Christoffel symbols - than that given by jc. The alternatives are also standard: a direct coordinate calculation or calculation of the connection 1-forms, from which one can read off the Christoffel symbols.
Edit: (In response to Anirbit's queries.) First, about the normal 1-form $n$. This is a 1-form field on $M$ that is fixed and smooth once the (smooth) foliation by $\Sigma\_t$'s is specified. So there is no choice at this level. The same holds for the form of the 3-metric $g$ on $\Sigma\_t$: this is just the spacetime 4-metric $h$ restricted to tangent vectors to $\Sigma\_t$: $g=h+n\otimes n$ is just the most convenient way to wite this metric.
A key point here is that each $\Sigma\_t$ is a 3 dimensional Riemannian manifold, and so the metric $g$ has signature $(+,+,+)$ at each point. That is, at any point the metric $g$ has three positive eigenvalues - where the eigenvectors are tangent to $\Sigma$. This is natural when we take the 3-dim perspective. When we think of $\Sigma\_t$ as being embedded in $M$, and allow $g$ to act on any tangent vector of $M$, we'd say that the signature is $(0,+,+,+)$. The zero eigenvalue corresponds to $n$.
Now we step down one more dimension to consider a spatial 2-surface $S$ embedded in $\Sigma\_t$. A typical example is a 2-sphere in $\mathbb{R}^3$ considered as a slice $t=$constant of Minkowski spacetime. The normal $m$ is a spacelike vector orthogonal to $n$. Again, the metric of $S$ is inherited naturally (no choice!) and has the convenient form $g^\prime = h+n\otimes n - m\otimes m$. The minus sign arises to make sure we get the right signature $(0,0,+,+)$ for $g^\prime$: without this signature - and without the minus sign - we would not have a Riemannian metric for $S$. The source of the sign difference is that $n$ is timelike, but $m$ is spacelike.
| 6 | https://mathoverflow.net/users/10563 | 39585 | 25,364 |
https://mathoverflow.net/questions/39579 | 6 | I'm sorry if this is an inappropriate forum to ask this question on, for I fear it is pretty undergraduate-level one :) I was contemplating on the study of non-linear PDEs. Is it possible to reduce a non-linear PDE on $\mathbb{R}^n$ to a distribution or a 'good' PDE on a smooth manifold? It seems to me like a natural step, but I don't know anything about it yet :(
| https://mathoverflow.net/users/44739 | Studying non-linear PDEs with manifolds | The question is vague (Denis is right) but it does make some sense. When looking at explicit solutions of nonlinear PDEs one frequently encounters singularities that remind of projections from some higher dimensional manifold (e.g. in Burger's equation). But I think the OP does not have a clear picture. Locally, working 'on' a manifold instead than on $R^n$ is essentially like changing coordinates in the equation. So usually you do not resolve nonlinearities, you just modify the coefficients of the equation.
On the other hand, if you consider equations with values into a manifold, then sometimes a very difficult nonlinear term reveals to be a very simple geometric object, 'linear' in some sense. This does not happen by chance; your equation must have some geometric meaning. Best examples are harmonic maps (elliptic) and wave maps (hyperbolic).
Some years ago, Serge Alinhac tried to develop a whole theory of 'geometric blow up': singularities in the solution which can be resolved by choosing an appropriate coordinate sistem in the target space. But I guess the results where not too satisfying since he abandoned the effort.
| 3 | https://mathoverflow.net/users/7294 | 39588 | 25,367 |
https://mathoverflow.net/questions/39577 | 12 | Given a pair of [Koszul dual](https://mathoverflow.net/questions/329/what-is-koszul-duality) algebras, say $S^\*(V)$ and $\bigwedge^\*(V^\*)$ for some vector space $V$, one obtains a triangulated equivalence between their bounded derived categories of finitely-generated graded modules.
Given a pair of Koszul dual operads, say the Lie and commutative operads, what is the precise analogue of a derived equivalence between their categories of algebras?
| https://mathoverflow.net/users/361 | koszul duality and algebras over operads | The situation for graded modules over a pair of Koszul dual algebras is more complicated, actually. What the question says is true for Koszul algebras $A$ and $A^!$ provided that $A$ is Noetherian and $A^!$ is finite-dimensional (including the case of the symmetric and exterior algebras) but not otherwise. In general one can say that the unbounded derived categories of positively graded modules with finite-dimensional components over $A$ and $A^!$ are anti-equivalent. The subcategories of complexes of positively graded modules bounded separately in every grading in these unbounded derived categories are also anti-equivalent.
One can replace the contravariant anti-equivalence with a covariant equivalence by considering positively graded modules over one of the algebras and negatively graded modules over the other one (both algebras being considered as positively graded). In this case one does not have to require the components of the modules to be finite-dimensional.
With algebras over operads, the analogue of the equivalence for graded modules involves DG-algebras with an additional positive grading (there being only the ground field $k$ in the additional grading $0$ and nothing in the negative additional grading), with the additional grading preserved by the differential. The Koszul duality is an anti-equivalence between the localizations of the categories of DG-algebras of this kind, with every component of fixed additional grading being a bounded complex of finite-dimensional vector spaces, by quasi-isomorphisms. For some operads (e.g., for Lie and Com) one has to assume the field $k$ to have characteristic $0$, while for some others (e.g., Ass) one doesn't.
If one wishes to replace the contravariant anti-equivalence with a covariant equivalence in the case of algebras over operads, one has to consider algebras on one side of the equivalence and coalgebras on the other side. Then the boundedness and finite-dimensionality requirements can be dropped.
What I've described above is the homogeneous Koszul duality; the nonhomogeneous case (with ungraded modules or algebras without the additional grading) is more complicated, though also possible. See my answer to the question linked to from the question above.
References: 1. Beilinson, Ginzburg, Soergel "Koszul duality patterns in representation theory", 2. My preprint "Two kinds of derived categories, ...", arXiv:0905.2621, Appendix A.
| 12 | https://mathoverflow.net/users/2106 | 39600 | 25,375 |
https://mathoverflow.net/questions/39616 | -2 | Given a finite list $x\_i$ of $N$ positive reals, it seems that $\sum\_{i=1}^N x\_i = \sum\_{i=1}^N x\_i {}^{-1} \Rightarrow \sum\_{i=1}^N x\_i \geq N$. Can anyone give me a proof?
| https://mathoverflow.net/users/799 | Little conjecture about sums of reciprocals | This is Cauchy-Schwarz inequality. Set $a\_i=x\_i^{1/2}$ and $b\_i:=x\_i^{-1/2}$. Then
$$N=(a,b)\le\|a\|\cdot\|b\|,$$
with equality if and only if $a$ and $b$ are colinear vectors. With your assumption, the right-hand side is precisely $\sum x\_i$.
| 3 | https://mathoverflow.net/users/8799 | 39619 | 25,385 |
https://mathoverflow.net/questions/39614 | 2 | If we quotient $U(N)$ by $U(N-1)$ we get the odd dimensional sphere $S^{2N-1}$. (Here the quotient is in the sense of embedding $U(N-1)$ in the bottom right hand corner (with 1 as the (1,1) entry and zero everywhere else) and taking its orbits as the set of new objects.) If we quotient now by $U(1)$ (embedded on the diagonal) we get ${\mathbb CP}^{N-1}$.
More generally, if we quotient $U(N)$ by $U(N-k)$, for some $k < N$ (with an analagous embedding), and then quotient by $U(k)$ (embedded again on the diagonal) we get the $k$-Grassmannian $G\_k({\mathbb C}^N)$.
My question is: What is the object we obtain when we quotient by $U(N-k)$? As we saw, it is the sphere for $k=1$. However, I cannot identify it with a familar object for higher $k$.
Also, more generally, if $F$ is a generalised flag manifold of signature $(d\_1, \ldots ,d\_k)$, then quotienting $U(N)$ by
$$
U(N-d\_1) \times \cdots \times U(N-d\_k),
$$
and then by
$$
U(d\_1) \times \cdots \times U(d\_k),
$$
gives $F$. What is the object we get from the first quotienting?
| https://mathoverflow.net/users/1867 | Is the object we get when we quotient $U(N)$ by $U(N-k)$ familar? | Putting a name to the space, it's a complex Stiefel manifold. See <http://en.wikipedia.org/wiki/Stiefel_manifold>. (But I wasn't the first.)
| 3 | https://mathoverflow.net/users/6153 | 39624 | 25,388 |
https://mathoverflow.net/questions/39620 | 1 | Given a smooth and projective surface $S$ over an algebraically closed field $k$ and a sheaf of Azumaya algebras $R$, i.e. $R$ is a locally free $O\_S$-module of finite rank. Let $M$ be a coherent and torsion free $O\_S$-module, which is also a left $R$-module, such that generically $M\_\eta$ is a simple $R\_\eta$-module. Then we have $Hom\_R(M,M)=k$.
Now $M^\\*:=Hom\_{O\_S}(M,O\_S)$ is a right $R$-module and $M^{\\*\\*}$ is a left $R$-module. We have the canonical map $\iota: M \rightarrow M^{\\*\\*}$.
Is it true that $Hom\_R(M,M^{\\*\\*})$ just consists of the muliples of $\iota$, i.e. is it a one dimensional $k$-vector space?
I tried to use the sequence $0\rightarrow M\rightarrow M^{\\*\\*} \rightarrow Q\rightarrow 0$. Since $M$ is torsion free $Q$ has support in codimension 2. Then apply $Hom\_R(M, - )$, which is left exact, so we get, with $Hom\_R(M,M)=k$: $0\rightarrow k\rightarrow Hom\_R(M,M^{\\*\\*}) \rightarrow Hom\_R(M,Q)$. But here i am stuck.
Or is this assertion wrong, i.e. are there more morphisms? If it is right, can it be generalized to a bigger class of algebras $R$?
| https://mathoverflow.net/users/3233 | Morphisms of a simple sheaf over an algebra to its double dual | Any $R$-homomorphism (in fact any $\mathcal O\_S$-homomorphism) $M \to M^{\*\*}$ extends to a morphism $M^{\*\*}\to M^{\*\*}$ (as $M$ is locally free in codimension $1$ and $M^{\*\*}$ is the maximal extension from outside codimension $2$. This gives what you want. as $Hom\_R(M^{\*\*},M^{\*\*})=k$ for the same reason as it is true of $M$.
| 3 | https://mathoverflow.net/users/4008 | 39625 | 25,389 |
https://mathoverflow.net/questions/39621 | 8 | Given a discrete group G. Is there a nice criterion to decide, whether there is a compact Hausdorff $G$- space X, that contains the discrete space $G$ as a subspace, such that the stabilizer of every point in $X$ is (virtually) cyclic ?
For example the free group admits such a compactification (As well as any hyperbolic group I think). Is it possible to decide, whether $\mathbb{Z}^2$ admits such a compactification? .
| https://mathoverflow.net/users/3969 | Which groups have nice compactifications ? | There are **many** compactifications of particular groups.
For your example of $\mathbb Z^2$: one construction for a compactification is to first embed it as a subgroup of
$S^1 = \mathbb R / \mathbb Z$ by picking two rationally independent numbers for the images of the generators.
Now compactify $\mathbb Z^2$ by making large elements connverge toward their image points in $S^1$.
The stabilizer of any point is trivial.
The same method works to get a compactification associated with any action of $G$ on a compact space $X$. Just pick a point $x \in X$, and adjoin the closure of the orbit of $X$ at infinity in $G$. If the action has no fixed points in the cloure of the orbit, then stabilizers are trivial. It's easier to avoid all but cyclic stabilizers. To make actions with small stabilizers, you can take products of examples; point stabilizes in the product become intersections of stabilizers in the factors. There are many tricks, some of them useful, for making compactifications that are Hausdorff metric spaces.
There's an ultimate (but non-constructive and of large cardinality) compactification, the Stone-Cech compactification, which has trivial point stabilizers for any group,
| 11 | https://mathoverflow.net/users/9062 | 39636 | 25,392 |
https://mathoverflow.net/questions/39628 | 3 | Hi
I have a problem which i find hard to modelize.
Suppose i have an urn with $N$ marbles. Among these marbles, one is white and all the other ones are black. I draw $P$ marbles without replacement. If the probability of drawing one marble is uniform, then the hypergeometric distribution tells me that the probability $P\_W$ of having the white marble among the $P$ marbles is:
$P\_W=\frac{\binom{1}{1} \binom{N-1}{P-1}}{\binom{N}{P}}=\frac{P}{N}$
That was the easy case.
Now suppose we have different weights for each marble. One marble $i$ is assigned a weight $w\_i$, and the probability of drawing the marble $i$ with one and only one draw is $p\_i=\frac{w\_i}{\sum\_i w\_i}$
Now let's go back to ou $P$ draws. What is the probability $P\_W$ according to $P$,$N$,$w\_i$.
All the ideas are welcome even with limit cases such as $P << N$
Thanks
| https://mathoverflow.net/users/9444 | Sequence of p draws without replacement with biased probabilities | This probability is always bounded from below by the probability with replacement, which is
$1-(1-w)^k$ where $w$ is the probability to pick the white marble in a single draw, and $k$ is the number of draws (changed from $P$ in your question which is rather unorthodox choice).
The probability of drawing the white marble at the $i$-th stage is bounded from above by $w/(1-\sum\_{j=0}^i w\_j)$, where $w\_0,\ldots$ are ordered by descending weight, and sum to 1. So the probability of drawing the white marble can be bounded from above by
$$1-\prod\_{i=0}^k(1-\frac{w}{1-\sum\_{j=0}^i w\_j})$$
In particular, we get the same asymptotic bound when $w\_0 << k^{-2}$.
| 2 | https://mathoverflow.net/users/1061 | 39641 | 25,394 |
https://mathoverflow.net/questions/39604 | 18 | Here I refer to Hamkins' slides:
<http://lumiere.ens.fr/~dbonnay/files/talks/hamkins.pdf>
particularly, to the "Universe view simulated inside Multiverse", p. 22.
My question is: is it very unsound to ask if the Multiverse view could be simulated (in a similar sense) inside Universe?
If it is, why is it? If it is not, why should one prefer one view to the other?
| https://mathoverflow.net/users/6466 | Universe view vs. Multiverse view of Set Theory | **Update**.(Sep 6, 2011) My paper on the multiverse is now available at the math arxiv at [*The set-theoretic multiverse*](http://arxiv.org/abs/1108.4223), and gives a fuller account of the ideas in the slides mentioned in the question. The particular issue of the question arises in the discussion of the toy-model approach to formalization, discussed on page 23, and also at greater length in my paper [Set-theoretic geology, G. Fuchs, J.D. Hamkins, J. Reitz](http://arxiv.org/abs/1107.4776).
---
Thanks for the question; I'm glad you're interested in it.
The multiverse view in set theory is a philosophical position offered in contrast to the Universe view, an orthodox position, which asserts that there is a unique background set-theoretic context or universe in which all our mathematical activity takes place. On the Universe view, there are definitive final answers to the question of whether a given mathematical statement, such as the Continuum Hypothesis, is true or not, and we seek to find these answers. On the Universe view, the fact that such a statement is independent of ZFC or another weak theory is regarded as a distraction from the question of determining whether or not it is ultimately true. For example, many set theorists regard the accumulating regularity consequences of large cardinals for properties of sets of reals as indicating that the large cardinal hierarchy is on the right track towards the final set-theoretic truth.
A paradox for the universe view, which I mention in the slides to which you link, is that the most powerful set-theoretic tools that have informed a half-century of research in set theory are most naturally understood as methods of constructing alternative set-theoretic universes. That is, from a given set-theoretic universe we can construct others, by means of forcing, ultrapowers, inner models, definability, large cardinal embeddings and so on. Indeed, we can often construct models of set theory to exhibit exact precise properties, and forcing especially has led to a staggering diversity of models.
The multiverse view takes these diverse models seriously, holding that there are diverse incompatible concepts of set, each giving rise to a set-theoretic universe in which they are instantiated. The set-theoretic tools provide a means of modifying any given concept of set to a closely related concept of set, whose resulting universes can be fruitfully compared in a single mathematical context. For example, we can understand the relationship between a ground model concept of set and that of its forcing extensions. Although the multiverse includes all the familiar models of set theory that we have built by forcing and other methods, it likely also includes universes arising from other set concepts that we have not yet imagined. There seems to be little reason that any two given concepts of set can be compared together in one set-theoretic context.
Now, the Universe view seems simulable inside the multiverse view by the idea of picking a single universe $V$, call it the *actual* universe if you like, and then restricting attention only to the universes that are somehow describable from the perspective of $V$.
But you ask about the other direction. There are a few ways to do it in a partial manner, but none of them seems fully satisfactory.
* First, one can mathematise the concept of multiverse, by just considering a multiverse as a collection of (set) models of set theory. For example, with Victoria Gitman (p. 44 of slides), we showed that if ZFC is consistent, then the set of all countable computably-saturated models of ZFC forms a multiverse satisfying all the multiverse axioms that I mention there (and others). This is just a straight theorem of ZFC. Since this multiverse does not include the set-theoretic background universe $V$ in which the collection was formed, however, we can recognize that it is not the full multiverse in which we are interested, and this is the sense in which we wouldn't really want to limit ourselves to that multiverse.
* Second, if one is interested only the *generic* multiverse---the part of the multiverse reachable by forcing, that is, by closing under forcing extensions and grounds in a zig-zag pattern---then one can formalize the whole set-up within ZFC. By describing exactly which forcing notions were used, one can index the models by their methods of construction. By this means, the concept of $\varphi$ is true throughout the generic multiverse of $V$'' is first-order expressible in $V$ in the language of set theory. This kind of analysis is a full answer to your question if you care only about the generic multiverse as opposed to the full multiverse.
But other researchers have cared about the full multiverse, and in general there is no satisfactory way to simulate it from the universe perspective. For example, the [Inner Model Hypothesis](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.109.7482&rep=rep1&type=pdf) of Sy Friedman (see also [Friedman, Welch, Woodin](http://www.maths.bris.ac.uk/~mapdw/imh.strengthpaper.17.09.06.pdf)), is described as the assertion that if the universe $V$ has an outer model with an inner model having a certain property, then there is already an inner model of $V$ with the property. Such a statement is explicitly appealing to the multiverse concept, but one cannot seem easily to formalize it within set theory. Instead, the official account of IMH backs off to the case where $V$ is a countable model of set theory, which takes it somewhat away from its initial multiverse sense.
So to finally come to an answer to your question, it doesn't seem possible to fully simulate the multiverse perspective from within the universe view, in a way that is fully satisfactory.
You ask, as a follow up, in this case why should we prefer one view to the other?
Well, these are philosophical positions on the fundamental nature of mathematical existence; this is a philosophical dispute rather than a mathematical one. Nevertheless, one's mathematical philosophy often suggests certain mathematical problems as interesting or solution methods as promising. The multiverse perspective naturally leads one to compare set-theoretic universes, and this led to the research on the modal logic of forcing and set-theoretic geology (mentioned in the slides). The Universe perpsective may lead elsewhere, perhaps towards an investigation of universes with highly structural features.
Surely the mathematicians' measure of a mathematical philosophy is the value of the mathematics to which it leads...
| 20 | https://mathoverflow.net/users/1946 | 39642 | 25,395 |
https://mathoverflow.net/questions/39645 | 2 | According the the introduction to Mazur's *Rational Isogenies of Prime Degree* the following question was open in 1978:
>
>
> >
> > Let $N$ be one of the integers 39, 65, 91, 125, or 169. Does the modular curve $X\_0(N)$ possess noncuspidal rational points?
> >
> >
> >
>
>
>
It seems likely that this should have been resolved in the past 32 years. Does anyone know of a reference for this?
| https://mathoverflow.net/users/4872 | For which composite $N$ does $X_0(N)$ possess a non-cuspidal rational point? | Let me make David Brown's answer more explicit.
Theorem ([Kenku, 1979](http://alpha.math.uga.edu/%7Epete/Kenku79.pdf)): $X\_0(39)(\mathbb{Q})$ consists entirely of the ($4$) cuspidal points.
Theorem ([Kenku, 1980](http://alpha.math.uga.edu/%7Epete/Kenku80.pdf)): $X\_0(65)(\mathbb{Q})$ and $X\_0(91)(\mathbb{Q})$ each consist entirely of the ($4$) cuspidal points.
Theorem ([Kenku, 1980](http://alpha.math.uga.edu/%7Epete/Kenku80b.pdf), see also [Kenku, 1980](http://alpha.math.uga.edu/%7Epete/Kenku81.pdf)): $X\_0(169)(\mathbb{Q})$ consists entirely of $2$ rational cuspidal points.
Theorem ([Kenku, 1981](http://alpha.math.uga.edu/%7Epete/Kenku81b.pdf)): $X\_0(125)(\mathbb{Q})$ consists entirely of $2$ rational cuspidal points.
| 6 | https://mathoverflow.net/users/1149 | 39661 | 25,408 |
https://mathoverflow.net/questions/39664 | 35 | Let $G=(V,E)$ be a graph and consider a random walk on it. Let $G'=(V',E')$ be a subgraph consisting of the vertices and edges that are visited by the random walk.
>
> Question 0: Is there a standard name for $G'$?
>
>
>
Intuitively $G'$ is a thin subgraph, so for instance, even when $G$ is transient, $G'$ can be recurrent.
>
> Question 1: Is there a counterexample? So, Is there a transient graph $G$ so that $G'$ is transient with positive probability?
>
>
>
I'm also curious to know what happens when one iterates this procedure, $G,G',G'',\dots$. Does it eventually look like a path graph?
>
> Question 2: What can one say about $G^{(n)}$ as $n\to \infty$?
>
>
>
| https://mathoverflow.net/users/2384 | Random walk inside a random walk inside... | Question 0: $G'$ is known as the *trace* of the random walk.
Question 1: $G'$ is always recurrent with probability one. This is [a result of Benjamini, Gurel-Gurevich, and Lyons](https://projecteuclid.org/euclid.aop/1175287760) from 2007.
Question 2: Since $G'$ is recurrent, with probability one we have $G^{(n)}=G'$ for all $n \geq 1$.
| 42 | https://mathoverflow.net/users/3401 | 39667 | 25,412 |
https://mathoverflow.net/questions/39670 | 1 | Question is the title. I suspect the answer is no, without some further conditions (clearly, normal is sufficient). Pointers to counterexamples would be appreciated, but not necessary.
| https://mathoverflow.net/users/4177 | Is a compactly generated Hausdorff space functionally Hausdorff? | There is an example at [PlanetMath](http://planetmath.org/?op=getobj&from=objects&id=5718) of a Hausdorff space which is not completely Hausdorff / functionally Hausdorff. On the other hand it is second-countable, hence first-countable and hence compactly generated.
| 4 | https://mathoverflow.net/users/290 | 39672 | 25,414 |
https://mathoverflow.net/questions/39618 | 5 | A bisymmetric matrix is a square matrix that is symmetric about both of its main diagonals.
If $A$ is a bisymmetric matrix and I'm interested in solving $Ax=b$.
Are there techniques used to exploit this structure when solving the system of linear equations?
Note: I'm looking for techniques which exploit more than just the fact that the matrix is symmetric.
| https://mathoverflow.net/users/2011 | Bisymmetric Matrix, solving set of linear equations. | The condition of symmetry about the antidiagonal says that $A$ commutes with reversal of coordinates. Call this operation $R$, so $R^2 = 1$ and $AR = RA$.
$R$ has a $+1$ eigenspace and a $-1$ eigenspace.
For any solution, you can project both $x$ and $b$ to the two eigenspaces, by averaging them with either their reversals or $-$ the reversals. You can get the induced action of $A$ on these (roughly if in odd dimension) half-size eigenspaces similarly. The two halves of $A$ are still symmetric, so you're left with the easier problem of solving two symmetric systems of equations in half the number of variables.
| 6 | https://mathoverflow.net/users/9062 | 39677 | 25,415 |
https://mathoverflow.net/questions/39673 | 1 | Consider the following integral,
$$
{1 \over 4\pi^{2}}\int\_{0}^{2\pi}\int\_{0}^{2\pi}
\sqrt{\, 9 -\sin^{2}\left(\theta\_{1} \over 2\right)
\sin^{2}\left(\theta\_{2} \over 2\right)\,}
\,{\rm d}\theta\_{1}\,d\theta\_{2}
$$
This integral comes up in computing the volume of $3$-dimensional special orthogonal matrices of Hessenberg form, i.e., the bottom left entry is $0$. Mathematica isn't able to produce close form solution. Numerically it's about $2.95$.
| https://mathoverflow.net/users/4923 | evaluating an integral related to the volume of Hessenberg orthogonal matrices | Assuming that
$$I=\int\_0^{2\pi} \int\_0^{2\pi}\sqrt{9-\sin^2 \frac{\theta\_1 }{2} \sin^2 \frac{\theta\_2 }{2}}\mathrm{d}\theta\_1 \mathrm{d}\theta\_2$$
is correct,
$$I=3\int\_0^{2\pi} \int\_0^{2\pi}\sqrt{1-\frac19 \sin^2 \frac{\theta\_1 }{2} \sin^2 \frac{\theta\_2 }{2}}\mathrm{d}\theta\_1 \mathrm{d}\theta\_2$$
then,
$$I=12\int\_0^{2\pi} \int\_0^{\frac{\pi}{2}}\sqrt{1-\frac19 \sin^2 \theta\_1 \sin^2 \frac{\theta\_2 }{2}}\mathrm{d}\theta\_1 \mathrm{d}\theta\_2$$
$$I=12\int\_0^{2\pi}E\left(\frac19 \sin^2 \frac{\theta\_2 }{2}\right)\mathrm{d}\theta\_2$$
($E(m)$ is the complete elliptic integral of the second kind, with parameter $m$; for the Maple people, what you have is $E(k)$ where $k^2=m$)
$$I=48\int\_0^{\frac{\pi}{2}}E\left(\frac19 \sin^2 \theta\_2\right)\mathrm{d}\theta\_2$$
and letting $m=\sin^2 \theta\_2$,
$$I=24\int\_0^1 \frac1{\sqrt{m}\sqrt{1-m}} E\left(\frac{m}{9}\right)\mathrm{d}m$$
which *Mathematica* evaluates to
$$12\pi^2 {}\_3 F\_2\left(-\frac12,\frac12,\frac12 ; 1,1 ; \frac19\right)$$
where ${}\_3 F\_2$ is a hypergeometric function; further "simplification" can be done using the formula [here](http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/04/02/01/06/0001/).
| 4 | https://mathoverflow.net/users/7934 | 39687 | 25,418 |
https://mathoverflow.net/questions/39692 | 4 | The sine function can take a complex argument. e.g. sin(x + iy)
But does it get used that way in any field? Either practical (e.g. electrical engineering) or in other fields of math? Naturally, I am not interested in trivial examples where the real or imaginary part of the argument is always zero.
I've asked elsewhere and the best anyone has come up with is that it can be a 2D solution of the Laplace equation. Anything more substantial? Any other interesting properties? Anyone stumbled upon it in deep in some analysis somewhere? Or is it really never used?
| https://mathoverflow.net/users/9462 | Are there any uses for complex sine? [sin z] | It turns up in a functional equation for the Gamma function, $\Gamma(s)\Gamma(1-s)={\pi\over\sin\pi s}$. From there one goes on to the functional equation for the Riemann zeta-function, $$\zeta(s)=2(2\pi)^{s-1}\Gamma(1-s)\sin(\pi s/2)\zeta(1-s)$$ The million dollar question is where are the (complex) zeros of $\zeta(s)$?
| 10 | https://mathoverflow.net/users/3684 | 39694 | 25,421 |
https://mathoverflow.net/questions/39686 | 20 | Having read a thread on a similar question on expository papers I'm reminded of reason #99 to drop my math PhD thingy, late c20th: I just couldn't blow up this paper to 4 pages. (OK, one half-page calculation was left to the expert reader (and other experts could guess), but...)
| https://mathoverflow.net/users/9161 | Which journals publish 1-page papers | There are plenty of examples [listed here](https://mathoverflow.net/questions/7330/which-math-paper-maximizes-the-ratio-importance-length).
| 12 | https://mathoverflow.net/users/121 | 39699 | 25,426 |
https://mathoverflow.net/questions/39702 | 0 | Hi.
While learning BVP, I came across a problem. It was mentioned that given the below left-focal BVP, a fixed–point problem can be formed for the system.
x'' = f (t, x), x (0) = 0, x(1) = 0.
where, f : [0, 1] × R → R is continuous and uniformly bounded. It was also mentioned that Schauder’s theorem can be applied to ensure the corresponding existence.
Can someone please help me in understanding this proof? Please let me know if the problem seems vague.
From this, can we infer that, in general, if
x'' = f (t, x), x(a) = x(b), x (a) = x (b) and f : [a, b] × R → R is continuous and uniformly bounded, then Schauder’s theorem can be applied to ensure that the system has at least one solution?
Regards,
Salil
| https://mathoverflow.net/users/8245 | Finding fixed points for BVP | *Fixed-point* is the name of a method, not of a problem (so far we speak of nonlinear differential equations). Its starts from the observation that, if your equation was $x''=-g(t)$, then the solution would be
$$x(t)=\int\_0^t(1-t)sg(s)ds+\int\_t^1t(1-s)g(s)ds,$$
which we write as
$$x(t)=\int\_0^1K(t,s)g(s)ds.$$
Therefore, your nonlinear problem is equivalent to the integral equation
$$x(t)=-\int\_0^1K(t,s)f(s,x(s))ds=:(Tx)(t).$$
This can be viewed as a fixed-point problem $Tx=x$. Hence the first idea of constructing a sequence of of 'approximate solutions' by $x^0\equiv0$ and then $x^{k+1}=Tx^k$, that is
$$x^{k+1}(t)=-\int\_0^1K(t,s)f(s,x^k(s))ds.$$
If $x^k$ converges pointwise boundedly to an $x$, then the convergence is uniform, thanks to the iteration, and $x$ is a fixed point. However, it is not always true that the sequence converges (it could oscillate, for instance), hence the second idea, which is of topological nature. Because $f$ is bounded, $T$ maps ${\mathcal C}(0,1)$ into a compact subset. In addition, $T$ is continuous. Then Schauder's fixed-point theorem ensures that $T$ admits a fixed point (not necessarily unique). Schauder's Theorem is a generalization of Brouwer's to infinite dimension (Banach spaces).
| 3 | https://mathoverflow.net/users/8799 | 39705 | 25,428 |
https://mathoverflow.net/questions/39704 | 0 | While learning differential equations, I was reading some notes, and it was mentioned that for Dirichlet BVP
$$x'' = f (t, x), \quad x(0) = 0 = x(1).$$ Suppose $f : [0, 1] \times \mathbb{R}\to \mathbb{R}$ is continuous and there is a constant $R > 0$ such that $f (t, R) \ge 0$, $f (t, −R)\leq 0$, for all $t \in [0, 1]$.
It can be shown that there is at least one solution, but the proof is missing.
Can someone please help me out with this?
How to find such an constant $R$, for some problem say, $$x'' = x^3 + t,\quad x(0) = 0 = x(1)$$ to show that this has at least one solution?
Regards,
Salil
| https://mathoverflow.net/users/8245 | Existence of solutions for differential equations | The assumption tells you that $x\_+\equiv R$ is a super-solution, and $x\_-\equiv -R$ a sub-solution, of your problem. This means that $x\_-''\ge f(t,x\_-)$, $x\_+''\le f(t,x\_+)$, and $x\_-\le0\le x\_+$ at $t=0,1$.
It is a general fact that when a BVP for a second-order differential equation admits a sub- and a super-solution $x\_\pm$, and if $x\_-\le x\_+$, then the problem admits a solution $x$ such that $x\_-\le x\le x\_+$.
Idea : Take $\lambda>0$ large enough that $x\mapsto g(t,x):=f(t,x)-\lambda x$ be non-increasing over $[-R,R]$, for every $t\in[0,1]$. Rewrite the problem as
$$x''-\lambda x=g(t,x).$$With $x(0)=x(1)=0$, this is equivalent to an integral equation
$$x(t)=-\int\_0^1K\_\lambda(t,s)g(s,x(s))ds=:(Tx)(t).$$
The kernel $K\_\lambda$ is bounded and continuous. Let us define the convex subset $C$ of ${\mathcal C}(0,1)$ by the conditions
$$x(0)=x(1)=0,\qquad x\_-\le x \le x\_+.$$
Because of the monotonicity of $g$, and because $Tx\_-\ge x\_-$, $Tx\_+\le x\_+$, $T$ maps $C$ into itself. In addition, $T(C)$ is relatively compact (a consequence of Ascoli-Arzela). By Schauder fixed point theorem, it admits a fixed point $\bar x$. This $\bar x$ is a solution of your problem.
| 3 | https://mathoverflow.net/users/8799 | 39710 | 25,432 |
https://mathoverflow.net/questions/39708 | 3 |
>
> Let $\mathcal{X} \to B$ be a flat family with some fibre $X\_b \to b$. Suppose I have a coherent sheaf $F\_b$ on $X\_b$. When does it spread out to a sheaf $\mathcal{F}$ on $\mathcal{X}$ flat over $B$?
>
>
>
What about a subscheme $z \subset X\_b$? Arbitrary diagrams of sheaves?
(I am only concerned with the case where everything is defined over $\mathbf{C}$, and moreover in the local case where $B$ is a disc and I am perfectly happy to take $\mathcal{X}$ to be affine. However $\mathcal{X}$ should not be assumed smooth, nor $X\_b$ to even be reduced.)
| https://mathoverflow.net/users/4707 | When do sheaves deform over a family? | Suppose that $B\_n$ it the $n^{\rm th}$ infinitesimal neighborhood of $b$ in $B$; that is, if $\frak m$ is the maximal ideal of $b$ in $B$, we set $B\_n := \mathop{\rm Spec} \mathcal O\_B/{\frak m}^{n+1}$. If $\mathcal F\_n$ is as extension of $\mathcal F\_b$ to $B\_n$, there is a canonically defined element of $({\frak m}^{n+2}/{\frak m}^{n+1})\otimes\_{\mathbb C}\mathop{\rm Ext}\_{\mathcal O\_{X\_b}}(\mathcal F, \mathcal F)$, called the *obstruction*; if this is zero, then the sheaf $\mathcal F\_n$ extends to $B\_{n+1}$. This depends on $\mathcal F\_n$, not only on $\mathcal F\_b$.
If this obstructions are always 0 (for example, if $\mathop{\rm Ext}\_{\mathcal O\_{X\_b}}(\mathcal F, \mathcal F) = 0$), then $\mathcal F\_b$ will extend to some étale neighborhood of $b$ in $B$. I don't think you can say much more in this generality.
There is a whole subject devoted to the study of this kind of problems (not only for sheaves, but for much more general objects), called *deformation theory*.
| 6 | https://mathoverflow.net/users/4790 | 39712 | 25,433 |
https://mathoverflow.net/questions/39690 | 3 | The counting class $\text{#P}$ and the related decision class $PP$ both involve counting the number of certificates to $NP$ problems.
Because $\text{#P}$ counts certificates, it seems obvious that $NP \subseteq \text{#P}$ and $co-NP \subseteq \text{#P}$. Furthermore, we can find any certificates using a brute force search, so $\text{#P} \subseteq EXP$.
The most interesting result that I found in Arora and Barak is Toda's Theorem which states that $PH \subseteq P^{\text{#SAT}}$, $\text{#SAT}$ being a $\text{#P}$-complete problem.
I'm wondering if there are any other results which relate $\text{#P}$ or $PP$ to other complexity classes, and what relationships are conjectured. For example, is it conjectured or known that $\text{#P} \subsetneq PSPACE$ ?
It occurred to me that because $\text{#P}$ is not a decision class some of these relationship may not be well-defined, but then again it seems natural enough to wonder how much time and space it takes to compute non-boolean (i.e. non-decision) functions of languages. Arora and Barak call Toda's Theorem a "big surprise", since as they put it $\text{#P}$ and $PH$ "both are natural generalizations of $NP$, but it seemed that their features ... are not directly comparable to each other". Given that result I'm hopeful that $\text{#P}$ can be related to other classes.
| https://mathoverflow.net/users/8981 | How is #P related to other complexity classes? | It is not known that $PP \subsetneq PSPACE$, but it is natural to conjecture it. What people really believe on this subject is murky territory. The fact that $P^{PP}$ contains the polynomial hierarchy is, to me, evidence that $P^{PP} = PSPACE$.
Let $a(n)$ be an unbounded time-constructible function. One example of a class which is "between" the polynomial hierarchy and $PSPACE$ is $\Sigma\_{a(n)} P$, the class of languages recognized by an alternating machine that makes $O(a(n))$ alternations on inputs of length $n$. It is not known for example if $\Sigma\_{a(n)} P \subseteq P^{PP}$ for any unbounded $a(n)$. If you were going to prove that $P^{PP} = PSPACE$, you would want to start by putting some "unbounded levels of the polynomial hierarchy" in $P^{PP}$ first. However, it may be helpful to know that by doing so, you would separate $PP$ from $LOGSPACE$. For all unbounded $a(n)$, we have $LOGSPACE \subsetneq \Sigma\_{a(n)} P$; this was proved in
>
> Lance Fortnow: Time-Space Tradeoffs for Satisfiability. J. Comput. Syst. Sci. 60(2): 337-353 (2000)
>
>
>
| 11 | https://mathoverflow.net/users/2618 | 39713 | 25,434 |
https://mathoverflow.net/questions/39475 | 6 | This is a followup of my previous question [Gromov-Witten and integrability](https://mathoverflow.net/questions/38294/gromov-witten-and-integrability). As I have learned from the answer (but guessed before), GW potentials of the point and $P^1$ (with different modifications) are, more or less, the only examples of the GW generation functions with established integrable properties. So what about higher genera curves? Are they really so complicated to establish integrability, at least for stable sector? What is the main problem with them?
| https://mathoverflow.net/users/3840 | Gromov-Witten and integrability 2. | Here's a sketch of my understanding of where the difficulty lies with higher genus curves. It got kind of long and vague, at parts, but hopefully it explains a few problems.
In Gromov-Witten theory, I'm aware of two or three general approaches to integrability currently. Certainly there's overlap among these approaches:
1. The quantumn cohomology of $X$ turns the cohomology of $X$ into a Frobenius manifold, and then Dubrovin has connections to various integrable hierarchies.
2. In some cases, the Gromov-Witten theory can be nicely expressed in terms of various Fock-spaces and the infinite wedges, and then you can get connections to the integrable hierarchies related to various infinite dimensional lie groups from the Kyoto school.
3. Matrix Models, and in particular the work of Eynard and Orantin. This seems to have mostly implemented in terms of the Topological vertex, which is rephrasing things in terms of combinatorics.
For higher genus curves, approach 1 is completely a no-go: there are no positive degree maps from a sphere to a higher genus curve, so the quantum cohomology is just the usual cohomology.
The GW-theory of higher genus curves is computed by Okounkov and Pandharipande very much in the general method of 2. The GW/Hurwitz correspondence shows that what they call the "stationary sector" (descendents of point classes only -- not the identity or odd cohomology classes. Is this the same as the stable sector?) is equivalent to Hurwitz theory, which is completely computable in terms of the symmetric group. There are nice ways of computing these characters, and this is where the connections to the infinite dimensional lie algebras and come up. However, this computation becomes much more complicated as we increase the genus. I'd like to explain how it's an entirely different beast for genus 2 or bigger.
The GW/Hurwitz correspondence turns insertions of point classes into ramification data, and hence as far as counting goes, into multiplying elements in the symmetric group. The ramification that shows up is nice and is easy to write in terms of the infinite wedge (free fermion) and I think can be written in terms of matrix model type stuff as well. One particular nice bit is that any given insertion only produces permutations with bounded supports: even if we let the degree get big (so considering symmetric groups $S\_d$ for $d$ large, we're only going to have a few nontrivial cycles in these permutations. Most points in $S\_d$ will be fixed.
For genus 0, we're only multiplying these special elements in the symmetric group, and this is why everything is so beautiful here.
For higher genus, one way to keep everything in terms of just the symmetric group is degenerate the curve by pinching off $g$ cycles, so that we again have a genus zero curve, but now with $g$ pairs of points identified. At each of these identified pairs of points, we're going to need to insert inverse permtuations, and we're going to have to sum over all permutations in $S\_d$ this way for each pair of points.
This is where things get ugly -- in higher genus, we have to consider these arbitrary permutations.
In genus one, things aren't all that bad: we only have two arbitrary permutations. So we can start with one of them, multiply in turn by the nice permutations that we know how to do, and then at end, instead of multiplying two arbitrary permutations, we only have to check that we have the same permutation, as the whole product has to be the identity. Essentially, we're taking the trace of some nice operator on the infinite wedge, and this why the quasi-modular forms show up here -- work of Bloch-Okounkov shows these are quasimodular forms.
Once we get to genus two though, we loose this as well. We really have to be able to multiply two arbitrary permutations of $S\_d$. I haven't read Mironov-Morozov's stuff on this closely at all, but seemed to recall them having some type of non-integrablity results for the general multiplication of three permutations, but couldn't find exactly this statement. The start of section three of [this paper](https://arxiv.org/abs/0904.4227) might touch on this, though.
The representation theory viewpoint is probably better for integrable systems, but I think it's similarly more difficult.
| 7 | https://mathoverflow.net/users/1102 | 39724 | 25,444 |
https://mathoverflow.net/questions/39751 | 11 | Last night I taught an algebra tutorial, and while writing out the multiplication table for the units of $\mathbb{Z}/5\mathbb{Z}$, a student remarked that it looked like a sudoku puzzle. I noted that it was similar, as the rows and columns all satisfy the sudoku condition, however the 2 by 2 sub-squares do not. The reason I gave is that multiplication is commutative in this group, and so the squares along the diagonal are symmetric. Afterwards I realized that there is an even simpler reason which excludes the possibility of nonabelian groups having multiplication tables which are sudoku squares: the identity element always messes up one of the sub-squares along the diagonal. This leads to the natural question: *do there exist associative sudoku squares of side length $n^2$*? (Meaning interpret the sudoku square as defining a binary operation on the integers $1$ through $n^2$. Is this binary operation ever associative? Of course I care about when $n > 1$ as $n = 1$ is clear.)
| https://mathoverflow.net/users/2677 | Do there exist associative sudoku squares? | If we are not obligated to keep the same order in rows and columns of the table, the answer is yes, and the group can be chosen to be commutative. For example
the group $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$,
the order of columns: $(1,1), (1,2), \dots, (1,n), (2,1), \dots, (n,n)$,
the order of rows: $(1,1), (2,1), \dots, (n,1), (1,2), \dots, (n,n)$.
Otherwise, the answer is no for any $n>1$, because in any table of this type both $a\cdot 1$ and $1\cdot a$ are in the same $n\times n$ square for some $a\in G$.
**Edit:** If this operation doesn't need to define a group, then the answer is no too. Here is the proof.
Suppose $G$ is $\{1,\dots,n^2\}$ endowed with some associative operation and suppose, that the table of this operation is a sudoku table. Take any $x\in G$. In $x$-th line of the table we can find element $x$. Therefore $xy=x$ for some $y\in G$. Then take $a\in G$, such that $a\neq y$ and $ay$ and $ya$ are in the same square of the table. We have $(xy)a=x(ya)$ and, therefore $xa=x(ya)$. Using, that there is only one place in $x$-th line of the table, where we can find $xa$, we see that $a=ya$. From this we have $a(ya)=(ay)a$, $aa=(ay)a$ and $a=ay$. Therefore $ay=a=ya$ and the table is not a sudoku table.
| 13 | https://mathoverflow.net/users/8134 | 39755 | 25,462 |
https://mathoverflow.net/questions/39578 | 21 | I received a request for [another](https://mathoverflow.net/questions/37963/lecture-notes-by-thurston-on-tiling) long-lost document:
>
> I am wondering if there is any way I
> might obtain a copy of
>
>
> The geometry of circles: Voronoi
> diagrams, Moebius transformations,
> convex hulls, Fortune's algorithm, the
> cut locus, and parametrization of
> shapes W.P. Thurston Technical Report,
> Princeton University, 1986.
>
>
> Is there a scanned version somewhere
> or might some library stock it?
>
>
>
Can anyone help out?
For context: I gave a series of presentations in a EECS course with Dave Dobkin, and I wrote these notes to go with it. It was fun material at the time, but the notes were only moderately distributed.
| https://mathoverflow.net/users/9062 | Missing document request | Here is the `.tex` not quite "as-is," but modified minimally so that it will compile: [`DTnotes.tex`](http://cs.smith.edu/~jorourke/MathOverflow/DTnotes.tex).
And here is `.pdf` produced by compiling that `.tex`: [`DTnotes.pdf`](http://cs.smith.edu/~jorourke/MathOverflow/DTnotes.pdf).
| 16 | https://mathoverflow.net/users/6094 | 39758 | 25,463 |
https://mathoverflow.net/questions/39756 | 17 | (ZFC)
Does there exist a function $f : \mathbb{R} \to \mathbb{R} \hspace{.1 in}$ such that for all $B$, if $B \subsetneq \mathbb{R}$ and $B$ is a nonempty Borel set, then $\lbrace x \in \mathbb{R} : f(x) \in B \rbrace$ is nonmeasurable?
| https://mathoverflow.net/users/nan | anti-measureable function | There are continuum many pairwise disjoint subsets of [0,1] each having Lebesgue outer measure 1. (By transfinite recursion of length continuum.) Assume that they are $\{A\_x:0\leq x\leq 1\}$ and their union is [0,1]. Then set $f:[0,1]\to[0,1]$ so that $f(y)=x$ is $y\in A\_x$. Now the inverse image of any $X\subseteq [0,1]$, $X\neq\emptyset, [0,1]$ has innermeasure 0, outer measure 1, hence $f^{-1}[X]$ is nonmeasurable.
| 21 | https://mathoverflow.net/users/6647 | 39759 | 25,464 |
https://mathoverflow.net/questions/21743 | 11 | I would like to find an example of principal ideal domain $R$, such that there exists a square matrix $A\in \mathfrak{M}\_n(R)$ with zero trace that is not a commutator (i.e. for all $B,C \in \mathfrak{M}\_n(R)$, $A\neq BC-CB$).
I know that such a PID (if it can be found) cannot be a field, or $\mathbb{Z}$.
| https://mathoverflow.net/users/3958 | Additive commutators and trace over a PID | It is not difficult to see that Rosset & Rosset's result for $2\times2$ matrices is equivalent to the surjectivity of the bilinear map $(X,Y)\mapsto X\times Y$ (called *vector product* when $A={\mathbb R}^3$) over $A^3$. For this, just search $B$ and $C$ such that $b\_{22}=c\_{22}=0$.
To prove it, let $Z=(a,b,c)\in A^3$ be given. One can choose a primitive vector $X=(x,y,z)$ such that $ax+by+cz=0$. By *primitive*, I mean that $gcd(x,y,z)=1$. Bézout tells that there exist a vector $U=(u,v,w)$ such that $ux+vy+wz=1$. Set $Y=Z\times U$. Then $Z=X\times Y$.
| 2 | https://mathoverflow.net/users/8799 | 39761 | 25,465 |
https://mathoverflow.net/questions/39762 | 7 | I'm trying to understand Vopěnka's Principle, which is a large cardinal axiom. One version of the principle is that there does not exist a proper class of directed graphs such that there are no homomorphisms between any two graphs in the class. This is a large cardinal axiom because it implies the existence of a proper class of measurable cardinals. If Vopěnka's principle is true, then it is not provable in ZFC, since it implies the consistency of ZFC. (It's falsity may be provable in ZFC.)
I presume that since it functions as a large cardinal axiom, then it must fail at small cardinals (i.e. cardinals whose existence is provable within ZFC, such as $\aleph\_\omega$). Is there an explicit construction for any such cardinal $\kappa$ there exists a set of graphs of size $\kappa$ such that Vopěnka's Principle fails for that set? In other words, there are no two homomorphisms between the two graphs in that set?
I can come up with a construction for $\aleph\_0$, but that's it. (For $\aleph\_0$, the set of directed cycle graphs with a prime number of vertices does the trick, I think.)
| https://mathoverflow.net/users/3711 | Vopenka's Principle at Small Cardinals | If $\kappa$ is almost huge (another large cardinal property), then for each family of size $\kappa$
of graphs of size $<\kappa$, one of the graphs embeds into another one from the family. See
[this post](http://www.cs.nyu.edu/pipermail/fom/2005-August/009023.html)
by Harvey Friedman.
Looking at $\kappa$-many graphs of size $<\kappa$ seems to be the right set analog to
a class of graphs (that are sets).
Could you clarify whether your family is of size $\kappa$, or the graphs?
---
Added after arsmath's comment: If you scroll down in Friedman's note, he says that
if $\kappa$ is Vopenka, then the set of extendible cardinals below $\kappa$ is stationary
in $\kappa$. Extendibility implies measurability (this should be in Kanamori's "The Higher Infinite") and hence there is a measurable cardinal below every Vopenka cardinal.
I think this qualifies as "no small cardinal is Vopenka".
| 6 | https://mathoverflow.net/users/7743 | 39767 | 25,467 |
https://mathoverflow.net/questions/39714 | 10 | The dual of an abelian category is again abelian, since the axioms are all preserved by the reversing of arrows. For example, the category of finite-dimensional vector spaces over a field is easily seen to be dual to itself, since we can just take linear duals of vector spaces. However, this is the only example where I know a 'concrete' description of both an abelian category and its dual, where by concrete I mean describing the category as a variant of, say, modules over a ring or sheaves on some space.
What are other examples of dual pairs of abelian categories which can both be described 'nicely'? For example, is there a concrete description of the dual of the category of abelian groups?
| https://mathoverflow.net/users/361 | Duals of Abelian Categories | A few (mostly trivial) examples (may be related to references in Tim Porter's answer?):
First of all, the self-duality of the category of vector spaces can be enriched to other self-dualities:
* The category of finite-dimensional representations of a group is self-dual.
* The category of finite-dimensional representations of a Lie algebra is self-dual.
(Of course, now we can generalize this to Hopf algebras.)
Another simple class of examples comes from taking the dual of some object, and asking what additional structure is necessary to reconstruct the original object. For instance:
* Let $V$ be a vector space. Consider $V^\*$. It has a natural topology, in which open subspaces are orthogonal complements of finite-dimensional subspaces of $V$. (This is
essentially weak topology if $V$ is viewed as a discrete space.) This gives a duality
between vector spaces and certain class of topological vector spaces that can be easily described explicitly. (A fancier way to view this: vector spaces are 'ind-finite-dimensional', so their dual are 'pro-finite-dimensional'.)
This can be modified to duality on Tate vector spaces, if you happen to like this sort of things.
There are also some examples in (algebraic) geometry (that are probably not as straightforward as the others).
* Category of perverse sheaves is self-dual.
* Category of holonomic $D$-modules is self-dual (this is related to the previous comment
by the Riemann-Hilbert).
Finally, there is a (somewhat cheap) trick to start with a duality on the derived category, then take the abelian category sitting inside it, and realize that it is dual to its image. For instance:
* Look at the derived category of constructible sheaves (say, on a scheme); it has Verdier's duality. Then the abelian subcategories corresponding to dual perversities are dual to each other. For instance, the category of constructible sheaves is dual to perverse sheaves for a certain perversity.
* Look at the derived category of coherent sheaves on a scheme; it has Serre's duality. Then the abelian subcategories corresponding to dual perversities are dual to each other. For instance, the category of coherent sheaves is dual to perverse sheaves for a certain perversity.
| 8 | https://mathoverflow.net/users/2653 | 39771 | 25,469 |
https://mathoverflow.net/questions/39626 | 32 | [This](https://mathoverflow.net/questions/39626/is-there-a-general-setting-for-self-reference) is a question about self-reference: Has anyone established an abstract framework, maybe a certain kind of formal language with some extra structure, which makes it possible to define what is a self-referential statement?
| https://mathoverflow.net/users/733 | Is there a general setting for self-reference? | I am not quite sure if it fits the bill but you can also check out:
[N. Yanofsky - A Universal Approach to Self-Referential Paradoxes, Incompleteness and Fixed Points](http://arxiv.org/abs/math/0305282)
| 11 | https://mathoverflow.net/users/2562 | 39776 | 25,471 |
https://mathoverflow.net/questions/39725 | 5 | I'm interested in an infinite dim'l Heisenberg group associated to the vector space $V = L\mathbb{C}/\mathbb{C}$ = {$f \colon S^1 \to \mathbb{C}$|$f$ smooth}/(const. maps). The group is $\mathbb{C}^\times \times V$ with group law
$(z,f)(z',g) = (zz' e^{\pi i (f,g)}, f+g)$
where $(f,g) = \int fdg$ is a symplectic form.
SOME BACKGROUND
There's a pairing $e(f,g) = e^{2\pi i (f,g)}$. The isotropic subspaces $W \subset V$ are the ones s.t. $e = 1$ on $W \times W$. General theory says for such $W$ you can construct a representation $F(W)$. Whenever you have a Lagrangian (= maximal isotropic) subspace $L \subset V$ you get up to equivalence a unique irreducible representation where $\mathbb{C}^\times$ acts by scalars.
(actually I've heard this for groups that are extensions by $U(1)$ and then then there is a unique *unitary* representation; but I'm guessing it works with $\mathbb{C}^\times$ too by just removing unitary)
One way to describe this representation is as continuous maps $\phi\colon V \to \mathbb{C}$ that satisfy $\phi(v + l) = e^{\pi i (v,\ l)}\phi(v)$ and $\int\_{V/L} |\phi|^2 dk < \infty$ where $dk$ is a Haar measure on $V/L$.
QUESTION
For $V = L\mathbb{C}/\mathbb{C}$, $z = e^{i\theta}$ it seems that $z^k$ for $k \in \mathbb{Z} - 0$ forms a basis. Also $L^\pm =$ the vector spaces spanned by the positive/negative powers of $z$ are Lagrangian. `Span' here doesn't mean finite linear combinations but linear combinations where the coefficients form maybe absolutely convergent series?
My question is what is an example of a $\phi \in F(L^+)$?
I think all such $\phi$ should be described as follows. Let $p\_\pm \colon V \to L^\pm$ be the projections. Then $\phi(v) = e^{i\pi (p\_-(v),\ p\_+(v))}\overline{\phi}(p\_i(V))$ where $\overline{\phi} \in L^2(L\_-;dk)$
Among my difficulties with answering this question is that $L\_-$ is still a really big space and I don't know what a Haar measure would be on this a space.
I should say, the answer to this question wont really help me in any research per se; I ask it because morally I feel better talking about $F(L)$ if I could write down at least one of its elements.
| https://mathoverflow.net/users/7 | What do representations of infinite-dimensional Heisenberg groups look like? | I'll give a description on the level of the polynomial Lie algebra, and then wave my hands about integrating and completing. As Victor Protsak mentioned in the comments, you can find a more precise treatment in section 9.5 of Pressley and Segal. There, the unitary representation arises from a choice of complex structure.
The Lie algebra of the Heisenberg group is (topologically) spanned by operators $\{ x\_k \}\_{k \neq 0}$ and $c$, where $x\_k$ describes the tangent vector corresponding to the basis vector $z^k$ in the group, and $c$ exponentiates to the central torus. The element $c$ is central in the Lie algebra, and the other generators obey the commutation relation $[x\_j, x\_k] = j\delta\_{j,-k} c$ (although you may need a factor of 1/2 with your choice of normalization). The Lie algebra of $L^+$ is topologically spanned by $x\_k$ for $k$ positive, and the corresponding statement holds for $L^-$ with $i$ negative.
The Fock representation is some completion of $\mathbb{C}[x\_{-1},x\_{-2},\dots,]$, i.e., ``finite energy'' elements are just finite sums of monomials in the generators of $L^-$. The action is given by the following rules:
1. When $k$ is negative, $x\_k$ acts by multiplication.
2. When $k$ is positive, $x\_k$ acts by $k\frac{\partial}{\partial x\_{-k}}$.
3. $c$ acts by the identity.
Exponentiating will give you a description of the action of group elements of the form $z^k$, and finite sums thereof, on finite energy elements of the representation. Modulo normalization, this will yield essentially the formula you gave in the background section. However, I don't think a Haar measure on $V/L$ exists.
Here is an explicit element: there is a distinguished vacuum vector $1$ (sometimes written $\Omega$ or $|0\rangle$), which is the unit element of the polynomial ring above. It is annihilated by all $x\_i$ for $i$ positive, so all $z^k$ act by identity on it when $k>0$. The action of $z^k$ for $k<0$ is by exponentiating $x\_k$ in the completion. Elements of the central torus act by ordinary scaling.
| 5 | https://mathoverflow.net/users/121 | 39785 | 25,477 |
https://mathoverflow.net/questions/39726 | 8 | I'm sure the following statement is well-known to experts:
Let $A$ be a dga. Let $perf(A)$ be the dg-category of perfect dg-modules over A. Then there is a quasi-isomorphism
$$C\_\bullet(perf(A)) \to C\_\bullet(A)$$
between their Hochschild chain-complexes.
Does anyone know a reference for it?
I'm aware of Keller's 2003 paper that I think gives the result for Hochschild cochain-complexes, but I need the chain version. Thanks!
| https://mathoverflow.net/users/2454 | Hochschild homology of dga's | I don't know a reference, but the assertion is easy to prove. The natural map goes in the opposite direction, $C\_\bullet(A)\to C\_\bullet(perf(A))$. It is quite simply induced by the embedding of the DG-category with a single object associated with $A$ into the DG-category $perf(A)$, sending the only object to the DG-module $A$ over $A$. To prove that it is a quasi-isomorphism, one can, e.g., interpret the Hochschild homology as the Tor of DG-bimodules, then use the fact that the derived categories of DG-bimodules over $A$ and $perf(A)$ are equivalent.
One can prove that what Caldararu-Tu call the Borel-Moore Hochschild homology are naturally isomorphic for a CDG-algebra $B$ and the DG-category $C$ of CDG-modules over $B$, projective and finitely generated as graded $B$-modules, in much the same way. I am finishing writing (or rather, editing now) a paper about this. It will be hopefully made public in a couple of weeks.
| 7 | https://mathoverflow.net/users/2106 | 39787 | 25,478 |
https://mathoverflow.net/questions/39782 | 6 | Let $Q$ be an acylic quiver. Let $E$ and $F$ be finite dimensional representations, with $E$ indecomposable. Suppose that, for some positive integer $r$, the representation $F$ injects into $E^{\oplus r}$. Suppose also that, for every vertex $v$ of $Q$, we have $\dim F\_v \leq \dim E\_v$.
Does it follow that $F$ injects into $E$?
There are results like this in the work of Derksen, Schofield and Weyman, but I can't find this particular statement. Thanks!
| https://mathoverflow.net/users/297 | A question about saturation of quivers | So I think you're asking if there is some kind of "saturation theorem" for generic rank.
The following is a counterexample. Let $Q$ be the ${\rm D}\_4$ quiver with vertices 1,2,3,4 (4 is the center) where the orientation is $1 \to 4$, $2 \to 4$ and $4 \to 3$. Let $E$ be the unique indecomposable (up to isomorphism) of dimension $(1,1,1,2)$ and take $F$ to be the unique representation of dimension $(0,0,0,2)$. Now the simple $S\_4$ injects into $E$ as the kernel of the map $E\_4 \to E\_3$, but $F$ does not inject into $E$. However, it does inject into $E \oplus E$.
| 6 | https://mathoverflow.net/users/321 | 39820 | 25,495 |
https://mathoverflow.net/questions/39831 | 10 | The string group $String(n)$ is by definition a 3-connected cover of $Spin(n)$. This definition determines the homotopy type of the string group.
[*In a previous version of this question I screwed up the definition and caused some confusion, see the comments below.*]
A common argument is saying that "the string group cannot be a Lie group because it has vanishing $\pi\_3$". This is obviously not a complete argument because $(\mathbb{R},+)$ is a nice Lie group with vanishing $\pi\_3$.
What is the correct statement about Lie group structures on the string group, and how does one prove it?
| https://mathoverflow.net/users/3473 | Why is the string group not a Lie group? | The result is that a compact, connected simple Lie group $G$ has $\pi\_3(G) = \mathbb{Z}$. Simple covering space or subgroups arguments should get you to $\mathrm{SO}(n)$ which is all that matters. For that matter start with the 1-connected $\mathrm{Spin}(n)$.
[OK, a short train ride later, now I'm home from work. To continue...]
The fibre of the 3-connected cover is a 2-type, and in the case of $\mathrm{Spin}(n)$ this is a $K(\mathbb{Z},2)$, so at the very least, $\mathrm{String}(n)$ can't be finite-dimensional. If one could construct a primitive[1] $PU(\mathcal{H})$-bundle on $\mathrm{Spin}(n)$ whose Dixmier-Douady classs was the generator $\langle -,[-,]\rangle \in H^3(\mathrm{Spin}(n),\mathbb{Z})$, then you would have an infinite-dimensional Lie group model for $\mathrm{String}(G)$ (here $\mathcal{H}$ is a infinite-dimensional separable Hilbert space, $PU(\mathcal{H})$ is then a smooth model for $K(\mathbb{Z},2)$, if we take the norm topology, making it a Banach Lie group).
([1] Primitive in the sense that for the group operations $G\times G\to G$ and $(-)^{-1}:G\to G$ there are bundle maps covering them.)
I don't know if this is possible or not, but I'm sure this idea has occurred to someone before, and since we haven't seen it, there might be a reason (well, I haven't seen it and everyone goes on about $\mathrm{String}\_G$ only being a topological group).
| 11 | https://mathoverflow.net/users/4177 | 39833 | 25,503 |
https://mathoverflow.net/questions/39804 | 4 | Let $P = a\_n(x) D\_x^n + a\_{n-1}(x) D\_x^{n-1} + \ldots + a\_0(x)$ be a linear ordinary differential operator with polynomial (or real analytic) coefficients $a\_j(x)$. Suppose that $a\_n(x)$ doesn't vanish on an interval $(a,b)$ and that $u$ is a [weak solution](http://en.wikipedia.org/wiki/Weak_solution) of $P u = 0$ on $(a,b)$. It can be concluded from the [elliptic regularity theorem](http://en.wikipedia.org/wiki/Elliptic_operator#Elliptic_regularity_theorem) that $u$ is in fact real analytic and a classical solution to the differential equation.
In other words, the operator $P$ is [analytically hypoelliptic](http://en.wikipedia.org/wiki/Hypoelliptic_operator).
However, appealing to the elliptic regularity theorem seems a bit much for the above case of an ordinary differential operator. Does anyone know a (canonical) reference for this (probably much more classical) case? I browsed through several books on differential equations but the closest I could find was in G. Folland's *Fourier analysis and its applications* where he mentions this fact (with just $C^\infty$ smoothness) on the top of page 344 without reference or proof.
*Update:* Thanks for the answers so far! While very helpful, I think I didn't make it clear what I was looking for. For an audience which may not be familiar with the theory of elliptic operators, what would be the proper way to (in one sentence plus a reference) justify that a weak solution in the case at hand is automatically real analytic? Currently, I have to make reference to a textbook on PDEs which treats elliptic regularity even though the problem for linear ODEs seems so much simpler.
Thank you!
| https://mathoverflow.net/users/359 | Analytic hypoellipticity of linear ordinary differential operators | The answers already given are quite complete and say it all, but maybe the OP was in search of a very simple explanation of what is happening. Let me try.
You will agree that it is not necessary to work in the full generality of an n-th order equation; if we can do it for the first order case, it is easy to generalize by induction. Moreover, we can assume that the highest order coefficient is equal to 1 (just divide by it). So essentially the question is: let $u$ be a weak solution of the equation
$$ u' + a(x) u = f(x) $$
on an open interval, with $ a(x) $ and $ f(x) $ analytic. Then we want to prove that $u$ is also analytic. A further reduction is possible: multiply both sides by $\exp (A(x)) $ where $A'=a$ and call $ v = e^A u $, $g=e^A f$. Then we are reduced to proving the same thing for the simple equation
$$ v ' = g. $$
Is $v$ analytic if $g$ is analytic? A further reduction is possible! just call $ w=v-G $,
where $G'=g$. Nice trick eh? we are reduced to the even simpler equation
$$ w' = 0 $$
and our serach will be over if we prove the fundamental fact that any weak solution of $w'=0$ must be a constant function. Notice that the same chain of arguments applies if the coefficients are $C^\infty$ (you get that $u$ is also $C^\infty$) and if $f$ is just $C^k$.
Now, in the greatest possible generality, if $w$ is any distribution on an open interval, with vanishing derivative, then $w$ must be a constant. This is proved in the following way: by definition, we know that $ w(\phi')=0 $ for any test function $\phi$. Fix a test function $\chi$ with $\int\chi=1$. Let $\psi$ be an arbitrary test function, define
$$ \psi\_1 = \psi-\chi \cdot \int \psi $$
and notice that $\int\psi\_1=0$. This means that $ \psi\_1 $ can be written as the derivative of another test function (guess which one?) $\psi\_1=\psi\_2'$ and hence
$$w(\psi\_1) = w(\psi\_2') = 0 $$
by assumption. This implies
$$ w(\psi) = w(\chi) \cdot \int \psi $$
which in the language of distributions means precisely that $w$ is equal to the constant $w(\chi)$ (notice that $\chi$ is fixed once and for all). That's it.
| 8 | https://mathoverflow.net/users/7294 | 39842 | 25,512 |
https://mathoverflow.net/questions/39848 | 15 | Given a finite group $G$, let $\{(1,1),(m\_1,n\_1),\ldots,(m\_r,n\_r)\}$ be the list of pairs $(m,n)$ in which $m$ is the order of some element, and $n$ is the number of elements with this order. The order of $G$ is thus $1+n\_1+\cdots+n\_r$, and the pair $(1,1)$ accounts for the neutral element.
Let $G,G'$ be two finite groups, with the same list. Is it true or not (I bet *not*) that $G$ and $G'$ are isomorphic ? If not, please provide a counter-exemple.
**Edit**. Nick's answer gives the correct terminology, of *conformal groups*. Ben's answer speaks of the refined notion of *almost conjugate subgroups*. Is there any other related notion ?
| https://mathoverflow.net/users/8799 | Finite groups with elements of the same order | There are easy examples that are $p$-groups. For instance, the mod 3 Heisenberg group is the nilpotent group with presentation
$\left < a,b,c \;\bigg |\, [a,b] = c, [a,c] = [b,c] = a^3 = b^3 = c^3 = 1 \right >$ has order 27, and all but the trivial element of order 3. This has the same order portrait as $C\_3^3$ where $C\_3 = \mathbb Z / 3\mathbb Z$ is the cyclic group of order 3.
| 32 | https://mathoverflow.net/users/9062 | 39850 | 25,519 |
https://mathoverflow.net/questions/39745 | 9 | **Definitions**
Suppose $A$ is a commutative algebra over $\mathbb{R}$ with unity. $\mathbb{R}$-linear map $\xi\colon A\to A$ is a [derivation](http://en.wikipedia.org/wiki/Derivation_(abstract_algebra)) of $A$ iff $\xi(ab)=a\xi(b)+\xi(a)b$ for any $a,b\in A$. If $\gamma\colon \mathbb{R}\to A$, $a\in A$, then we say, that $a=\frac{\partial}{\partial t}| \_ {t=\tau} \gamma(t)$ iff $h(a) =\frac{\partial}{\partial t}|\_{t=\tau} h(\gamma(t))$ for any $\mathbb{R}$-linear map $h\colon A\to\mathbb{R}$.
Suppose $\xi$ is a derivation of $A$. Then $\Phi\colon A\times\mathbb{R}\to A$ is it's flow iff $\Phi(a,0)=a$ for any $a\in A$ and
$$\frac{\partial}{\partial t} \Phi(a,t) = \xi \Phi(a,t).\tag{1}$$
**The question**
1. I like algebras $A$, such that any derivation of $A$ possesses a flow. Is there any simple sufficient condition for them?
2. Is there any simple condition for an algebra and it's derivation, from which it follows, that this derivation possesses a flow?
**Examples**
1. Algebra $C^\infty(M)$ of smooth functions on a [closed manifold](http://en.wikipedia.org/wiki/Closed_manifold) $M$ --- yes (if I haven't made a mistake), any derivation possesses a flow. This, I believe, can be checked using [Picard-Lindelof theorem](http://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem).
1'. Algebra $C^\infty(M)$ of smooth functions on a non-compact manifold without boundary --- no (see example 2), but a derivation possesses a flow if it preserves some function $H\in A$, such that for any $c\in\mathbb{R}$ subspace $\{x\mid H(x) < c \}$ of the topological space $M$ is compact.
2. Algebra $C^\infty((0,1))$ of smooth functions on an interval --- no, $\frac{\partial}{\partial x}$ does not possess a flow.
3. Algebra $C^\infty([0,1])$ of smooth functions on a segment --- yes, any derivation possesses a flow, but it's not always unique (for example, for $\frac{\partial}{\partial x}$ it is not). In order to prove this, one can consider an embedding of $[0,1]$ to some closed manifold $N$ and prolong any function from $[0,1]$ to $N$. Then use example 1.
4. Algebra $C^\infty(\mathbb{R})$ --- no, because it is isomorphic to the algebra from example 2.
5. Algebra $\mathbb{R}[x]$ --- no. In order to prove this one can consider derivation $\xi=x^2\frac{\partial}{\partial x}$ and manually solve equation (1) for $a=x$. Any solution locally should be of the form $\frac{x}{1+xt}$. It is not in $\mathbb{R}[x]$.
6. Algebra $\mathbb{R}[x,y]/(x^2+y^2-1)$ --- no. In order to prove this take $\xi = y (x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x})=\sin(\varphi)\frac{\partial}{\partial \varphi}$ and solve equation (1) manually (locally) in polar coordinates (take, for example, $a=y$). Check that the answer is not a polynomial.
7.(from Greg Muller's answer) Algebra $A$ is finite-dimensional --- yes, every derivation possesses the flow.
This question was already posted [here](https://math.stackexchange.com/questions/4963/in-which-commutative-algebras-does-any-derivation-possess-a-flow) on math.stackexchange.com, but it has received no answers even with a bounty. Any help is appreciated, both in the theme of question and in improving its wording.
| https://mathoverflow.net/users/8134 | In which commutative algebras does any derivation possess a flow? | If you really mean to get away with treating the algebra as an algebraic object -- no topology is given on this vector space -- then you don't even have the smooth compact manifold example. The given definition of $a=\gamma'(\tau)$ is rarely satisfied. There are too many linear maps from $A$ to $\mathbb R$. A function $\gamma:\mathbb R\to A$ cannot have a derivative in your sense at $\tau$ unless there is some $\epsilon$ such that the vectors $\gamma(t)$ for $|t-\tau|<\epsilon$ span a finite-dimensional vector subspace of $A$.
On the other hand, it is true that a smooth compact manifold (with its (sheaf of) smooth functions) can be reconstructed from the purely algebraic object which is the algebra of global functions, and also that the only derivations of that algebra are the ones arising from tangent vector fields. I don't how to describe (without introducing a topology on the algebra, or something like that) the relation between the derivations and the corresponding homomorphisms from the additive group of real numbers into the automorphism group of the algebra.
| 3 | https://mathoverflow.net/users/6666 | 39854 | 25,521 |
https://mathoverflow.net/questions/39858 | 3 | If $R=\mathbb{C}[x,y]$ is the polynomial ring in two variables $x$ and $y$ then we know that the localization of R at the multiplicative set $S=[1,x,x^2,x^3,...]$ is given by $R\_x=\mathbb{C}[x,x^{-1},y]$. Now, what will be the localization of $R$ at the prime ideal $(x)$. i.e. what will $R\_{(x)}$ be?
| https://mathoverflow.net/users/9492 | Localization of a polynomial ring at a prime ideal. | I doubt that there is a nice description which will satisfy you. As a $\mathbb{C}$-algebra, $R\_{(x)}$ is not finitely generated. Anyway every localization of a factorial domain at a principle prime ideal is a discrete valuation domain. In particular, $R\_{(x)}$ is such a domain with prime ideal $(x)$. Every nonzero element has a unique representation $x^n u$, where $u$ is a fraction, such that $x$ is coprime to both the numerator and the denominator of $u$. Probably an algebraic geometer would call this ring the local ring of the affine plane at the line $x=0$.
| 9 | https://mathoverflow.net/users/2841 | 39860 | 25,526 |
https://mathoverflow.net/questions/39818 | 14 | Since Ronnie Brown and his collaborators have come up with a general proof of the higher Van Kampen theorems, what impediments are there to using these to compute the unstable homotopy groups of spheres?
| https://mathoverflow.net/users/1353 | The higher Van Kampen Theorems and computation of the unstable homotopy groups of spheres | Here are some answers on the HHSvKT - I have been persuaded by a referee that we ought also to honour Seifert.
These theorems are about homotopy invariants of structured spaces, more particularly filtered spaces or n-cubes of spaces. For example the first theorem of this type
Brown, R. and Higgins, P.~J. On the connection between the second relative homotopy
groups of some related spaces. Proc. London Math. Soc. (3) \textbf{36}~(2) (1978) 193--212.
said that the fundamental crossed module functor from pairs of pointed spaces to crossed modules preserves certain colimits. This allows some calculations of homotopy 2-types and then you need further work to compute the 1st and 2nd homotopy group; of course these two homotopy groups are pale shadows of the 2-type.
As example calculations I mention
R. Brown ``Coproducts of crossed $P$-modules: applications to second homotopy groups and to the homology of groups'', {\em Topology} 23 (1984) 337-345.
(with C.D.WENSLEY), `Computation and homotopical applications of induced crossed modules', J. Symbolic Computation 35 (2003) 59-72.
In the second paper some computational group theory is used to **compute the 2-type**, and so 2nd homotopy groups as modules, for some mapping cones of maps $ Bf: BG \to BH$ where $f:G \to H$ is a morphism of groups.
For applications of the work with Loday I refer you for example to the bibliography on the nonabelian tensor product
<http://groupoids.org.uk/nonabtens.html>
which has 144 items (Dec. 2015: the topic has been taken up by group theorists, because of the relation to commutators) and also
Ellis, G.~J. and Mikhailov, R. A colimit of classifying spaces.
{Advances in Math.} (2010) arXiv: [math.GR] 0804.3581v1 1--16.
So in the tensor product work, we determine $\pi\_3 S(K(G,1))$ as the kernel of a morphism $\kappa: G \otimes G \to G$ (the commutator morphism!). In fact we compute the **3-type** of $SK(G,1)$ so you can work out the Whitehead product $\pi\_2 \times \pi\_2 \to \pi\_3$ and composition with the Hopf map $\pi\_2 \to \pi\_3$.
These theorems have connectivity conditions which means they are restricted in their applications, and don not solve all problems! There is still some interest in computing homotopy types of some complexes which cannot otherwise be computed. It is also of interest that the calculations are generally nonabelian ones.
So the aim is to make some aspects of higher homotopy theory more like the theory of the fundamental group(oid); this is why I coined the term `higher dimensional group theory' as indicating new structures underlying homotopy theory.
Even the 2-dimensional theorem on crossed modules seems little known or referred to! The proof is not so hard, but requires the notion of the homotopy double groupoid of a pair of pointed spaces. See also some recent presentations available on my [preprint page](http://groupoids.org.uk/brownpr.html).
Further comment: 11:12 24 Sept.
The HHSvKT's have two roles. One is to allow some calculations and understanding not previously possible. People concentrate on the homotopy groups of spheres but what about the homotopy types of more general complexes? One aim is to give another weapon in the armory of algebraic topology.
The crossed complex work applies nicely to filtered spaces. The new book (pdf downloadable) gives an account of algebraic topology on the border between homotopy and homology without using singular or simplicial homology, and allows for some calculations for example of homotopy classes of maps in the non simply connected case. It gets some homotopy groups as modules over the fundamental group.
I like the fact that the Relative Hurewicz Theorem is a consequence of a HHSvKT, and this suggested a triadic Hurewicz Theorem which is one consequence of the work with Loday. Another is determination of the critical group in the Barratt-Whitehead n-ad connectivity theorem - to get the result needs the apparatus of cat^n-groups and crossed n-cubes of groups (Ellis/Steiner).
The hope (expectation?) is also that these techniques will allow new developments in related fields - see for example work of Faria Martins and Picken in differential geometry.
Developments in algebraic topology have had over the decades wide implications, eventually in algebraic number theory. People could start by trying to understand and apply the 2-dim HHSvKT!
Edit Jan 11, 2014 Further to my last point, consider my answer on excision for relative $\pi\_2$: <https://math.stackexchange.com/questions/617018/failure-of-excision-for-pi-2/621723#621723>
See also the relevance to the [Blakers-Massey Theorem](http://www.ncatlab.org/nlab/show/Blakers-Massey+theorem) on this nlab link.
December 28, 2015 I mention also a presentation at CT2015 Aveiro on [A philosophy of modelling and computing homotopy types](http://groupoids.org.uk/pdffiles/aveiro-beamer-handout.pdf). Note that homotopy groups are but a pale "shadow on a wall" of a homotopy type. Note also that homotopy groups are defined only for a space with base point, i.e. a space with some structure. My work with Higgins and with Loday involves spaces with much more structure; that with Loday involves $n$-cubes of pointed spaces. As with any method, it is important to be aware of what it does, and what it does not, do. One aspect is that the work with Loday deals with nonabelian algebraic models, and obtains, when it applies, precise **colimit results in higher homotopy**. One inspiration was a 1949 Theorem of JHC Whitehead in "Combinatorial Homotopy II" on free crossed modules.
| 24 | https://mathoverflow.net/users/19949 | 39868 | 25,532 |
https://mathoverflow.net/questions/39792 | 6 | Let $X$ be a separable Banach space, and let $\mathbb P$ be a Radon probability measure on $X$ with zero mean and covariance operator $K : X^\* \to X$. Let $x$ be an $X$-valued random variable with distribution $\mathbb P$.
I would like a simple upper bound on the size of $\mathbb E \|x\|^2$ in terms of the operator norm $\|K\|$. In his textbook *The Concentration of Measure Phenomenon*, Ledoux proves that $$\mathbb E\|x\|^2 \le 4 \|K\|$$ as a consequence of a concentration inequality for a simpler example (where the vectors are sums of vectors with i.i.d. random coefficients $\eta\_i$ such that $|\eta\_i| \le 1$ a.s.). Because of the boundedness assumption, the argument doesn't quite work in this setting, though I'm certain I could generalize it if I needed to.
Nonetheless, I'm certain that the estimate I'm looking for is buried somewhere in the literature on concentration of measure (and in fact is probably due to Talagrand). Could you please point me in the right direction?
**Edit:** The inequality as I previously wrote it is incorrect. The correct inequality should be $$\operatorname{Var}(\|x\|) \le 4\|K\|,$$
implying $$\mathbb E\|x\|^2 \le 4\|K\| + \left( \mathbb E\|x\| \right)^2.$$ That is, the size of typical random element could be quite large (i.e. $\mathbb E\|x\| \gg 1$, as in Mark Meckes's example in the comments), but the deviation is only of the order $\|K\|^{1/2}$.
| https://mathoverflow.net/users/238 | The typical size of a random element in a Banach space | The inequality can't be true without additional assumptions. To see this, let $X = \ell\_2^n$ and let $x$ have a spherically symmetric distribution and let $R = \Vert x \Vert$. Then $R$ is an essentially arbitrary nonnegative random variable; indeed we could start by picking $R$ and defining $x=R\Theta$, where $\Theta$ is a uniform random vector in $S^{n-1}$ independent of $R$.
Now $\mathrm{Var}(\Vert x \Vert) = \mathrm{Var} (R)$, and $K = \frac{1}{n} \mathbb{E}(R^2) I\_n$, so $\Vert K \Vert = \frac{1}{n} \mathbb{E} (R^2)$. Thus for any fixed distribution of $R$, the inequality fails for sufficiently large $n$.
| 4 | https://mathoverflow.net/users/1044 | 39873 | 25,535 |
https://mathoverflow.net/questions/39882 | 39 | If $X$ and $Y$ are separable metric spaces, then the Borel $\sigma$-algebra $B(X \times Y)$ of the product is the $\sigma$-algebra generated by $B(X)\times B(Y)$. I am embarrassed to admit that I don't know the answers to:
Question 1. What is a counterexample when $X$ and $Y$ are non-separable?
Question 2. If $X$ is an uncountable discrete metric space, does
$B(X) \times B(X)$ generate the Borel $\sigma$-algebra on $X \times X$?
Question 3. If $X$ and $Y$ are metric spaces, with $X$ separable, does
$B(X) \times B(Y)$ generate the Borel $\sigma$-algebra on $X \times Y$?
| https://mathoverflow.net/users/2554 | Product of Borel sigma algebras | Q1. Discrete spaces with cardinal > c ... then the diagonal is a Borel set, but not in the product sigma-algebra.
This also answers Q2 (no)
but not Q3.
| 20 | https://mathoverflow.net/users/454 | 39883 | 25,541 |
https://mathoverflow.net/questions/39510 | 22 | The question is in the title, but employs some private terminology, so I had better explain.
Let $R$ be an integral domain with fraction field $K$, and write $R^{\bullet}$ for $R \setminus \{0\}$. For my purposes here, a **norm** on $R$ will be a function $| \ |: R^{\bullet} \rightarrow \mathbb{Z}^+$
such that for all $x,y \in R^{\bullet}$, $|xy| = |x||y|$ and $|x| = 1$ iff $x \in R^{\times}$. Also put $|0| = 0$.
[**Edit**: I forgot to mention that the norm map extends uniquely to a group homomorphism $K^{\bullet} \rightarrow \mathbb{Q}^{> 0}$.]
The norm is **Euclidean** if for all $x \in K \setminus R$, there exists $y \in R$ such that $|x-y| < 1$.
Now let $q(x) = q(x\_1,\ldots,x\_n)$ be a (nondegenerate) quadratic form over $R$, which I mean in the relatively naive sense of just an element $R[x\_1,\ldots,x\_n]$ which is homogeneous of degree $2$. Supposing that a norm $| \ |$ on $R$ has been fixed, I say that the quadratic form $q$ is **Euclidean** if for all $x \in K^n \setminus R^n$, there exists $y \in R^n$ such that $0 < |q(x-y)| < 1$.
(If $q$ is anistropic, then $x \in K^n \setminus R^n$, $y \in R^n$ implies that $x-y \neq 0$, so $|q(x-y)| \neq 0$, and the condition simplifies to $|q(x-y)| < 1$.)
As an example, the sum of $n$ squares form is Euclidean over $\mathbb{Z}$ iff $1 \leq n \leq 3$.
It is easy to see that the ring $R$ (with its fixed norm) is Euclidean iff the quadratic form $q(x) = x^2$ is Euclidean (iff the quadratic form $q(x,y) = xy$ is Euclidean), so the concept of a Euclidean quadratic form is indeed some kind of generalization of that of a Euclidean ring.
Conversely, the following is an obvious question that I have not been able to answer.
>
> Suppose that a normed ring $R$ admits some Euclidean quadratic form $q$. Is $R$ then necessarily a Euclidean domain? In other words, can there be any Euclidean quadratic forms if $q(x) = x^2$ is not Euclidean?
>
>
>
If the answer happens to be "no", I would of course like to know more: can this happen with an anisotropic form $q$? What can one say about a ring $R$ which admits a Euclidean quadratic form?
I have some suspicions that a domain which admits a Euclidean quadratic form is at least a PID. Indeed, let $q$ be a quadratic form over $R$. By an $R$-linear change of variables we may write it as $a\_1 x\_1^2 + \sum\_{j=2}^n a\_{1j} x\_1 x\_j + \sum\_{i=2}^n \sum\_{j=1}^n a\_{ij} x\_i x\_j$ with $a\_1 \neq 0$. Then if I take my vector $x$ to be $(x\_1,0,\ldots,0)$ for $x\_1 \in K \setminus R$, then the Euclidean condition implies the following: there exist $y\_1,z \in R$ such that $|a(x\_1-y\_1)^2 - z| < 1$. This is reminiscent of the Dedekind-Hasse property for a norm which is known to imply that $R$ is a PID. In fact, it is much stronger in that the $a \in R$ is fixed (and the $z$ is not arbitrary either). Unfortunately in place of an arbitrary element $x$ of $K$ we have a certain square, so this does not match up with the Dedekind-Hasse criterion...but it seems to me somewhat unlikely that a domain which is not a PID would satisfy it.
For more information on Euclidean forms, please feel free to consult
[http://alpha.math.uga.edu/~pete/ADCforms.pdf](http://alpha.math.uga.edu/%7Epete/ADCforms.pdf)
---
**Added**: a new draft which takes into account comments of J. Hanke and F. Lemmermeyer is available at
[http://alpha.math.uga.edu/~pete/ADCformsv2.pdf](http://alpha.math.uga.edu/%7Epete/ADCformsv2.pdf)
I warn that the new Section 2.1 showing that (primitive) binary Euclidean forms correspond to Euclidean ideal classes in quadratic orders in the sense of Lenstra -- as pointed out by Franz Lemmermeyer in his answer -- is rather miserably written at the moment, but at least it's there.
| https://mathoverflow.net/users/1149 | Must a ring which admits a Euclidean quadratic form be Euclidean? | Following Pete's request, I give the following as a second answer.
Take $R = {\mathbb Z}[\sqrt{34}]$ and $q(x,y) = x^2 - (3+\sqrt{34})xy+2y^2$; observe that the discriminant of $q$ is the fundamental unit $\varepsilon = 35 + 6 \sqrt{34}$ of $R$, and that its square root generates $L = K(\sqrt{2})$ since $2\varepsilon = (6+\sqrt{34})^2$. Then q is Euclidean over $R$ since the ring of integers in $L = K(\sqrt{2})$ is generated over $R$ by the roots of $q$, and since $L$ is Euclidean by results of J.-P. Cerri (see Simachew's
[A Survey On Euclidean Number Fields](http://www.algant.eu/documents/theses/simachew.pdf)). But $R$ is not principal ($L/K$ is an unramified quadratic extensions), so the answer to your question, if I am right, is negative.
| 9 | https://mathoverflow.net/users/3503 | 39884 | 25,542 |
https://mathoverflow.net/questions/39879 | 2 | Is there a nice trick for this? I would like to compute the eigenvalues more efficiently.
| https://mathoverflow.net/users/9501 | how to get nonzero eigenvalues of a large symmetric matrix with lots of duplicate rows | If you have a lot of duplicate rows (and you know what they are), you can reduce to a smaller matrix. I'll start with an example, because writing out the general case will be notationally annoying.
Let
$$A = \begin{pmatrix}
a & a & b & c \\
a & a & b & c \\
b & b & d & e \\
c & c & e & f
\end{pmatrix}$$
Then $A$ has a zero eigenvalue, and the other eigenvalues are the same as those of
$$A'=\begin{pmatrix}
2a & b \sqrt{2} & c \sqrt{2} \\
b \sqrt{2} & d & e \\
c \sqrt{2} & e & f
\end{pmatrix}$$
Proof: Let $v\_1$, $v\_2$, $v\_3$ be the orthonormal vectors $(1/\sqrt{2}, 1/\sqrt{2}, 0,0)$, $(0,0,1,0)$ and $(0,0,0,1)$. Complete to an orthonormal basis $(v\_1, v\_2, v\_3, w)$. Then $A$ annihilates $w$, and takes $\mathrm{Span}(v\_1, v\_2, v\_3)$ to itself by the matrix $A'$.
In general, if $I$ is a set of rows which are identical, then let $v\_I$ be the vector which is $1/\sqrt{|I|}$ on the coordinates in $I$ and $0$ elsewhere. The $v\_I$ are orthonormal, complete them to an orthonormal basis by adding vectors $w\_j$. Then $A$ will annihilate the $w\_j$ and will take $\mathrm{Span}(v\_I)$ to itself. The matrix of endomorphism of $\mathrm{Span}(v\_I)$ will have entries that look like $\sqrt{IJ} \cdot a\_{ij}$, with $i \in I$ and $j \in J$. So you ar reduced to computing the eigenvalues of this smaller matrix.
| 6 | https://mathoverflow.net/users/297 | 39889 | 25,545 |
https://mathoverflow.net/questions/39895 | 4 | Hi,
In the index of this book, under j, he references several 'jokes' found throughout the text. I can't find one on page 91 - anyone know what it is?
| https://mathoverflow.net/users/8867 | Jokes in Miles Reid's 'Undergraduate Algebraic Geometry' | Dear Robert, the joke on page 91 is that the ruled quadric depicted has "Central Electricity" written over it. It is an allusion to the cooling towers used by power plants. Here is the obligatory Wikipedia link:
<http://en.wikipedia.org/wiki/Cooling_tower>
| 1 | https://mathoverflow.net/users/450 | 39898 | 25,551 |
https://mathoverflow.net/questions/39881 | 7 | Hello,
i've been looking for a way to classify the non-trivial $p$-groups $G$ that live in an exact sequence of the form
$ 0 \rightarrow \mathbb{Z}/p\mathbb{Z} \rightarrow G \rightarrow (\mathbb{Z}/p\mathbb{Z})^{n-1} \rightarrow 0 $. Was this question settled before? Or is there any explicit computation of $H^2((\mathbb{Z}/p\mathbb{Z})^{n-1}, \mathbb{Z}/p\mathbb{Z})$?
Thanks!
| https://mathoverflow.net/users/3680 | Classification of $p$-groups of order $p^n$ with rank $n-1$ | Your group is such that $|G|=p^n$ and $|\Phi(G)|=p$. Since $(C\_p)^{n-1}$ is completely reducible, there is a subgroup $H$ of $G$ such that $G=HZ(G)$ and $H\cap Z(G)=\Phi(G)$. Thus $H$ is an extra-special group (possibly trivial), and we are taking the central product with the abelian group $Z(G)$, which is either of the form $(C\_p)^m$ or $(C\_{p^2})\times(C\_p)^m$. The first case is easy, since again, it is completely reducible, so we get a group of the form (extra-special) times (some copies of $C\_p$). The second case also gives (some group) times (some copies of $C\_p$). I believe the (some group) is uniquely determined by its order (that is the central product of either of the two non-abelian groups of order $p^3$ and $C\_{p^2}$ are isomorphic), but I haven't checked any cases but $p=2$.
Steve
| 4 | https://mathoverflow.net/users/1446 | 39900 | 25,552 |
https://mathoverflow.net/questions/39901 | 6 | I started studying the basics of category theory recently, and after seeing how a great deal of group theory could be described categorically, I began to wonder if it were possible to describe set theory, or set-theoretic concepts, without reference to elements, i.e., by only using sets and functions. For example, instead of saying $x\in S$, one could say that there is a map $f:0\to S$, where $0$ is a singleton. Similarly, one could describe the power set $\mathscr{P}(X)$ by saying that for any function $f:Y\to X$, there are functions $g:0\to \mathscr{P}(X)$ and $h:\mathscr{P}(X)\to X$ such that $f=hg$. Disjoint unions can be described as coproducts, and cartesian products can be described by means of a universal property. I was wondering if it were possible to describe all of naive set theory in this way, and if so, whether any attempts have been made to do so.
| https://mathoverflow.net/users/6856 | Set theory within the framework of category theory | Yes! Lawvere's [Elementary Theory of the Category of Sets](http://ncatlab.org/nlab/show/ETCS) (ETCS) is probably just what you want. It is a characterisation of Set up to equivalence from category theoretic principles alone. Todd Trimble explains a lot of it in some blog posting that have now migrated to the nLab, see the links [here](http://ncatlab.org/nlab/show/ETCS#todd_trimbles_exposition_of_etcs_4). This is what can be called a structural approach to foundations: one cares less about membership as how sets related to each other by functions.
It isn't the only one, Mike Shulman has [SEAR](http://ncatlab.org/nlab/show/SEAR) (Sets Elements And Relations) which takes the named entities as primitive. It relies a little more on dependent type theory. This isn't too scary, it just means you keep track of what sort of set your elements live in (so questions like, 'is $\pi = (sin:\mathbb{R} \to [0,1]$?', don't come up) Note that ETCS a priori isn't about elements, but they are recovered as in your question: maps $1 \to S$, but it is by the time one uses the axiom 'Set is well pointed' that this comes about. SEAR on the other hand has elements are primitive, but then turn out to be what you say once functions are defined and subsets an so on.
Both of these can be taken to include Choice or not (ETCS has Choice by default, but it isn't necessary). SEAR-C is what we call SEAR with Choice, and it is equivalent to ZFC, and SEAR is equiconsistent with ZF (but not equivalent).
ETCS is slightly weaker than SEAR, but this hasn't been explored too much.
One can also consider the variant on SEAR called [SEPS](http://ncatlab.org/nlab/show/SEAR#seps_using_pairs_and_subsets_instead_of_relations_49), which uses as primitives Sets, Elements, Pairing and Subsets (pairing is like canonical categorical product - stronger than the usual product in that it is specified. I wouldn't be surprised that if, after all's said and done, it's functorial). This is equivalent to SEAR
| 7 | https://mathoverflow.net/users/4177 | 39908 | 25,557 |
https://mathoverflow.net/questions/39885 | 9 | Following Gromov, take a metric space $(X,d)$ and consider $C(X)/\mathbb{R}$ the set of continuous functions to $\mathbb{R}$ with the topology of uniform convergence on compact sets after taking the quotient by constant functions (i.e. two functions are equivalent if the difference is a constant). Embed $X$ into this space by means of the map:
$x \mapsto f(x) = [d(x,\cdot)]$
A horofunction is an element of the closure of $f(X)$ that is not in $f(X)$.
In $\mathbb{R}^d$ all horofunctions are given by inner product with a unit vector.
In the upper-half plane model of the hyperbolic disc the function $h(z) = \log(\text{Im}(z))$ is a horofunction and all others can be obtained by composing with hyperbolic isometries.
What other spaces have well known horofunctions?
For example, are horofunctions of the model geometries $\text{Nil}$, $\text{Sol}$, and $\widetilde{\text{Sl}(2,R)}$ in dimension $3$ known relatively explicitly?
I know there is a general relationship between horofunctions and geodesics (Busseman functions) in non-positively curved spaces. However I'm interested in spaces where one can compute relatively explicitly (e.g. spaces where one knows what the distance function looks like).
| https://mathoverflow.net/users/7631 | What spaces have well known horofunctions? | Of course, the first example should be non-compact Riemannian symmetric spaces, where the Busemann (horofunctions in your terminology) functions are known in a pretty explicit form. I don't think there are other explicit examples.
More comments.
1) Geodesic rays always converge in this Busemann-Gromov compactification, and it has nothing to do with curvature.
2) I never understood why the definition of this compactification is formulated in terms of uniform convergence on compact sets instead of plain pointwise convergence (anyway, for Lipschitz functions the result is the same). Actually, this is a particular case of the general Constantinescu-Cornea compactification (see the book of Brelot "On Topologies and Boundaries in Potential Theory"), other examples of which are the Martin compactification in potential theory or Thurston compactification in the Teichmuller theory.
ADDED REFERENCES
In what concerns Busemann or horofunctions (I prefer to call them Busemann cocycles, because in invariant language they are cocycles, not functions) for symmetric spaces, there are several sources of various degree of explicitness.
(1) For the group $SL(d,\mathbb R)$ (and the associated symmetric space) the
Busemann cocycle essentially appears in Furstenberg's formula for the rate of
growth of random matrix products, see
MR0163345 (29 #648)
Furstenberg, Harry
Noncommuting random products.
Trans. Amer. Math. Soc. 108 1963 377--428.
From the geometrical point of view the main idea there is that if you want to find the linear rate of growth of $d(o,g\_1 g\_2\dots g\_n o)$, where $o$ is a reference point, and $g\_i$ is a stationary sequence of random isometries, then you can look at the increment $d(o,g\_1 g\_2\dots g\_n o)-d(o,g\_2 g\_3\dots g\_n o)=d(g\_1^{-1}o,g\_2g\_3\dots g\_n o)-d(o,g\_2 g\_3\dots g\_n o)$, which converges to the Busemann cocycle $\beta\_\gamma(o,g^{-1}o)$ provided $g\_2g\_3\dots g\_n o$ converges to a boundary point $\gamma$ in the Busemann compactification. The cocycle itself looks, roughly speaking, like $\log \|gv\|/\|v\|$, where $g$ is the matrix (or its exterior power) representing a point in the symmetric space, and $v$ is a vector representing the boundary point. There is also a lot about it in later papers by Guivarc'h.
(2) Busemann cocycles naturally appear in various works related to compactifications of symmetric spaces. Historically the first source is the monograph of Karpelevich
MR0231321 (37 #6876)
Karpelevic, F. I.
The geometry of geodesics and the eigenfunctions of the Beltrami-Laplace operator on symmetric spaces.
Trudy Moskov. Mat. Obšc. 14 48--185 (Russian); translated as Trans. Moscow Math.
Soc. 1965 1967 pp. 51--199. Amer. Math. Soc., Providence, R.I., 1967.
where he explicitly discusses pencils of convergent geodesics and introduces the associated horospheric coordinates. Later expositions are in two books on compactifications of symmetric spaces:
MR1633171 (2000c:31006)
Guivarc'h, Yves(F-RENNB-IM); Ji, Lizhen(1-MI); Taylor, J. C.(3-MGL)
Compactifications of symmetric spaces. (English summary)
Progress in Mathematics, 156. Birkhäuser Boston, Inc., Boston, MA, 1998. xiv+284 pp. ISBN: 0-8176-3899-7
and
MR2189882 (2007d:22030)
Borel, Armand(1-IASP); Ji, Lizhen(1-MI)
Compactifications of symmetric and locally symmetric spaces.
Mathematics: Theory \& Applications. Birkhäuser Boston, Inc., Boston, MA, 2006. xvi+479 pp. ISBN: 978-0-8176-3247-2; 0-8176-3247-6
| 5 | https://mathoverflow.net/users/8588 | 39913 | 25,561 |
https://mathoverflow.net/questions/39907 | 14 | Let $f:(0,1)\rightarrow(0,1)$ be a map with some regularity (${\mathcal C}^1$, ${\mathcal C}^2$, ${\mathcal C}^\infty$, analytic ?). We assume that $f(t)> t$ for every $t$, and that $f'> 0$.
Does there exist a vector field $X$ over $(0,1)$, whose flow at time $t=1$ is $f$ ?
If the answer is yes (as I bet), it will complete the existence proof in MO question [link text](https://mathoverflow.net/questions/34187/nth-root-of-a-b-mapsto-gm-am/39906#39906)
| https://mathoverflow.net/users/8799 | The vector field of a given flow | I assume your map is surjective, thus an increasing $C^k$ diffeo $f:(0,1)\to(0,1)$ (say $1\le k\le\infty).$ The latter, as a discrete dynamical system, turns out to be $C^k$ conjugated with the shift by translation $t\mapsto t+1$ on $\mathbb{R}$. Now if $h:(0,1)\to\mathbb{R}$ is such an (increasing) conjugation, define a field $X$ by taking for all $t\in (0,1)$
$$X(t)=\frac{1}{h^\prime(t)}.$$
Then $f$ is the flow of $X$ at time $1$, just because $u:=h^{-1}:\mathbb{R}\to(0,1)$ solves the autonomous ODE $u'=X(u).$ So the solution at time $1$ corresponding to the initial value $t$ at time $0$ is actually $u\left(h(t)+1\right)=f(t).$
*Rmk*. The general fact behind is: for a diffeo on a manifold, being a time-one map of a flow, is a property invariant by smooth conjugation -the flow transforms by conjugation, and its generator is the pull back of $X$. And the shift $x\mapsto x+1$ is, of course, the map at time 1 of a flow.
*Construction of the conjugation*. Note that here a conjugation is an increasing diffeo $h:(0,1)\to\mathbb{R}$ solving the linear functional equation $h\left(f(x)\right) = h(x)+1.$ A solution of class $C^k$ can be constructed fixing any smooth diffeo $h\_0:\left[\frac{1}{2},f(\frac{1}{2})\right]\to[0,1]$
with the convenient $k$-jets at the end-points of its domain, and extending it (uniquely) to a diffeo $h:(0,1)\to\mathbb{R}$ by means of the functional equation -the condition is that the $k$-jet of $h\_0\left(f(x)\right)$ at $x=\frac{1}{2}$ and the $k$-jet of $h\_0(x)+1$ at $x=f(\frac{1}{2})$ should coincide, in order that the glueing be $C^k$.
A maybe more clear way of achieving that, is: first fix any $C^k$-germ $H$ of local diffeo with $H(1/2)=0$ and $H'(0)>0$. Then take $h\_0$ as a smooth diffeo from a nbd of $J:=\left[\frac{1}{2},f(\frac{1}{2})\right]$ to a nbd of $[0,1]$, whose germs at $\frac{1}{2}$, and at $f(\frac{1}{2})$, are respectively $H$ and $H\circ f^{-1} + 1$. The extension of $h\_0$ from the two germs is of course immediate by using some cut-off function. Then we may proceed as said before, from ${h\_0}\_ { |J} $.
(Clearly, these naive constructions immediately fall, in the case of $C^\omega$ conjugation of $C^\omega$ diffeo's on open intervals with no fixed points.)
| 19 | https://mathoverflow.net/users/6101 | 39915 | 25,562 |
https://mathoverflow.net/questions/39923 | 21 | This should be a trivial question for mathematicians but not for typical physicists.
I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" and "residual" parts [I gather that no boundedness assumption is needed but I could be wrong]. I further know that the point spectrum coincides with the set of eigenvalues of the operator. It seems from the terminology that the point spectrum is a discrete set of isolated point and that the eigenvalues cannot form a continuum. But I haven't been able to find a clear statement in a math reference about this.
Actually, I'm mostly interested in self-adjoint operators on a Hilbert space; so a simpler version of my question would be: Can a self-adjoint operator have a continuous set of eigenvalues? And if yes, under what conditions do the eigenvalues have to be discrete?
I appreciate any help.
| https://mathoverflow.net/users/9504 | Can a self-adjoint operator have a continuous set of eigenvalues? | Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal. In a separable Hilbert space, any orthogonal set is countable. So a self-adjoint operator on separable Hilbert space has only countably many eigenvalues. (As noted, this does not mean the spectrum is countable.)
---
Proof for "Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal".
Let $\alpha \ne \beta$ be eigenvalues of the self-adjoint operator $A$. They are real. There exist nonzero eigenvectors $x, y$ so that $Ax=\alpha x, Ay=\beta y$. We must show $x \perp y$.
Compute
$$
\alpha\langle x,y\rangle =
\langle \alpha x, y\rangle =
\langle Ax, y\rangle =
\langle x, Ay\rangle =
\langle x, \beta y\rangle =
\beta\langle x,y\rangle
$$
Then
$$
(\alpha-\beta)\langle x,y\rangle = 0
$$
But $\alpha \ne \beta$, so $\alpha - \beta \ne 0$ and thus
$$
\langle x,y\rangle = 0,
$$
so $x \perp y$ as claimed.
| 21 | https://mathoverflow.net/users/454 | 39927 | 25,571 |
https://mathoverflow.net/questions/39928 | 13 | The Wiener process is defined by the three properties:
1. $W(0) = 0$,
2. $W(t)$ is almost surely continuous, and
3. $W(t)$ has independent increments with $W(t) - W(s) \sim N(0, t-s)$ (for $0 ≤ s < t$).
What would be an example of a process which satisfies 1) and 3), but not 2) ?
I am going to teach an introductory class on Brownian motion at advanced undergrad level.
Just wanted to make sure that all the conditions are mutually independent.
| https://mathoverflow.net/users/8528 | Wiener process related counterexample | This is not hard to find such an example. Let $P$ be Wiener measure on the space $\Omega = C([0,\infty))$ of continuous functions $t\mapsto \omega(t)$. Then the process $\omega(t)$ satisfies all three conditions of a Brownian motion.
Now let's define a new process $W(t)$ that is "almost" equal to $\omega(t)$, but where we deliberately wreck the sample path continuity.
Take any random time $T:\Omega\to [0,\infty)$ that has a continuous distribution on $(\Omega, P)$,
and let $W(t,\omega)=\omega(t)$ when $t\not=T(\omega)$, but $W(t,\omega)=\omega(t)+1$ otherwise. The process $W(t)$ still satisfies 1 and 3 but the sample path continuity fails at exactly at the time point $T(\omega)$ for each $\omega$.
There are many such random times $T$, for example you could use $T(\omega):=\inf [t>0: \omega(t)=1 ]$, i.e. the hitting time of 1.
| 13 | https://mathoverflow.net/users/nan | 39932 | 25,573 |
https://mathoverflow.net/questions/39828 | 120 | Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that.
So after this longish introduction, here goes: Many of us routinely use algebraic techniques in our research. Some of us study questions in abstract algebra for their own sake. However, historically, most algebraic concepts were introduced with a specific goal, which more often than not lies outside abstract algebra. Here are a few examples:
* Galois developed some basic notions in group theory in order to study polynomial equations. Ultimately, the concept of a normal subgroup and, by extension, the concept of a simple group was kicked off by Galois. It would never have occurred to anyone to define the notion of a simple group and to start classifying those beasts, had it not been for their use in solving polynomial equations.
* The theory of ideals, UFDs and PIDs was developed by Kummer and Dedekind to solve Diophantine equations. Now, people study all these concepts for their own sake.
* Cohomology was first introduced by topologists to assign discrete invariants to topological spaces. Later, geometers and number theorists started using the concept with great effect. Now, cohomology is part of what people call "commutative algebra" and it has a life of its own.
The list goes on and on. The axiom underlying my question is that you don't just invent an algebraic structure and study it for its own sake, if it hasn't appeared in front of you in some "real life situation" (whatever this means). Please feel free to dispute the axiom itself.
Now, the actual question. Suppose that you have some algebraic concept which has proved useful somewhere. You can think of a natural generalisation, which you personally consider interesting.
>
> How do you decide whether a generalisation (that you find natural) of an established algebraic concept is worth studying? How often does it happen (e.g., how often has it happened to you or to your colleagues or to people you have heard of) that you undertake a study of an algebraic concept and when you try to publish your results, people wonder "so what on earth is this for?" and don't find your results interesting? How convincing does the heuristic "well, X naturally generalises Y and we all know how useful Y is" sound to you?
>
>
>
Arguably, the most important motivation for studying a question in pure mathematics is curiosity. Now, you don't have to explain to your colleagues why you want to classify knots or to solve a Diophantine equation. But might you have to explain to someone, why you would want to study ideals if he doesn't know any of their applications (and if you are not interested in the applications yourself)? How do you motivate that you want to study some strange condition on some obscure groups?
Just to clarify this, I have absolutely no difficulties motivating myself and I know what curiosity means subjectively. But I would like to understand, how a consensus on such things is established in the mathematical community, since our understanding of this consensus ultimately reflects our choice of problems to study.
I could formulate this question much more widely about motivation in pure mathematics, but I would rather keep it focused on a particular area. But one broad question behind my specific one is
>
> How much would you subscribe to the statement that
> EDIT: "studying questions for the only reason that one finds them interesting is something established mathematicians do, while younger ones are better off studying questions that they know for sure the rest of the community also finds interesting"?
>
>
>
Sorry about this long post! I hope I have been able to more or less express myself. I am sure that this question is of relevance to lots of people here and I hope that it is phrased appropriately for MO.
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Edit: just to clarify, this question addresses the status quo and the prevalent consensus of the mathematical community on the issues concerned (if such a thing exists), rather than what you would like to be true.
---
Edit 2: I received some excellent answers that helped me clarify the situation, for which I am very grateful! I have chosen to accept Minhyong's answer, as that's the one that comes closest to giving examples of the sort I had in mind and also convincingly addresses the more general question at the end. But I am still very grateful to everyone who took the time to think about the question and I realise that for other people who find the question relevant, another answer might be "the correct one".
| https://mathoverflow.net/users/35416 | How do you decide whether a question in abstract algebra is worth studying? | Dear Alex,
It seems to me that the general question in the background of your query on algebra really is the better one to focus on, in that we can forget about irrelevant details. That is, as you've mentioned, one could be asking the question about motivation and decision in any kind of mathematics, or maybe even life in general. In that form, I can't see much useful to write other than the usual cliches: there are safer investments and riskier ones; most people stick to the former generically with occasional dabbling in the latter, and so on. This, I think, is true regardless of your status. Of course, going back to the corny financial analogy that Peter has kindly referred to, just *how* risky an investment is depends on how much money you have in the bank. We each just make decisions in as informed a manner as we can.
Having said this, I rather like the following example: [Kac-Moody algebras](http://en.wikipedia.org/wiki/Kac%E2%80%93Moody_algebra) could be considered 'idle' generalizations of finite-dimensional simple Lie algebras. One considers the construction of simple Lie algebras by generators and relations starting from a Cartan matrix. When a positive definiteness condition is dropped from the matrix, one arrives at general Kac-Moody algebras. I'm far from knowledgeable on these things, but I have the impression that the initial definition by Kac and Moody in 1968 really was somewhat just for the sake of it. Perhaps indeed, the main (implicit) justification was that the usual Lie algebras were such successful creatures. Other contributors here can describe with far more fluency than I just how dramatically the situation changed afterwards, accelerating especially in the 80's, as a consequence of the interaction with conformal field theory and string theory. But many of the real experts here seem to be rather young and perhaps regard vertex operator algebras and the like as being just so much bread and butter. However, when I started graduate school in the 1980's, this story of Kac-Moody algebras was still something of a marvel.
There must be at least a few other cases involving a rise of comparable magnitude.
Meanwhile, I do hope some expert will comment on this. I fear somewhat that my knowledge of this story is a bit of the fairy-tale version.
Added: In case someone knowledgeable reads this, it would also be nice to get a comment about further generalizations of Kac-Moody algebras. My vague memory is that some naive generalizations have not done so well so far, although I'm not sure what they are. Even if one believes it to be the purview of masters, it's still interesting to ask if there is a pattern to the kind of generalization that ends up being fruitful. Interesting, but probably hopeless.
Maybe I will add one more personal comment, in case it sheds some darkness on the question. I switched between several supervisors while working towards my Ph.D. The longest I stayed was with Igor Frenkel, a well-known expert on many structures of the Kac-Moody type. I received several personal tutorials on vertex operator algebras, where Frenkel expressed his strong belief that these were really fundamental structures, 'certainly more so than, say, Jordan algebras.' I stubbornly refused to share his faith, foolishly, as it turns out (so far).
Added again:
In view of Andrew L.'s question I thought I'd add a few more clarifying remarks.
I explained in the comment below what I meant with the story about vertex operator algebras.
Meanwhile, I can't genuinely regret the decision not to work on them because I quite
like the mathematics I do now, at least in my own small way. So I think what I had in mind was just
the platitude that most decisions in mathematics,
like those of life in general, are mixed: you might gain
some things and lose others.
To return briefly to the original question, maybe I do have some practical
remarks to add. It's obvious stuff, but no one seems to have written it so far on this page.
Of course, I'm not in a position to give anyone advice, and your question didn't really ask for it,
so you should read this with the usual reservations. (I feel, however, that what I write *is* an
answer to the original question, in some way.)
If you have a strong feeling about a structure or an idea, of course
keep thinking about it. But it may take a long time for your ideas
to mature, so keep other things going as well, enough to build up
a decent publication list. The part of work that belongs
to quotidian maintenance is part of the trade,
and probably a helpful routine for most people. If you go about it sensibly, it's really
not that hard either. As for the truly original
idea, I suspect it will be of interest to many people at some point, if
you keep at it long enough. Maybe the real difference between
starting mathematicians and established ones is the length of time
they can afford to invest in a strange idea before feeling
like they're running out of money. But by keeping a suitably interesting
business going on the side, even a young person can afford
to dream. Again, I suppose all this is obvious to you and many other people.
But it still is easy to forget in the helter-skelter of life.
By the way, I object a bit to how several people have described this question
of community interest as a two-state affair. Obviously, there are many different
degrees of interest, even in the work of very famous people.
| 30 | https://mathoverflow.net/users/1826 | 39938 | 25,577 |
https://mathoverflow.net/questions/39946 | 2 | The transitive closure of a *directed graph*, is another directed graph which encodes the reachability of nodes from other nodes. If $G$ is a graph, the edge $(v\_1,v\_2)$ is in it's transitive closure $G^{tc}$ iff there is a directed path from $v\_1$ to $v\_2$ in $G$.
A *multigraph* can have multiple edges between nodes.
The question is what would be natural definitions for the transitive closure of a multigraph?
An obvious answer would be the transitive closure of the induced graph (same graph with multiple edges between verices replaced with a single edge).
Are there already interesting graphs derivable from a multigraph which could earn the title of 'transitive closure'?
| https://mathoverflow.net/users/7412 | Transitive closure of multigraphs | Note that the term *transitive closure* comes from set theory. Every (simple) directed graph $G$ naturally defines a relation $R(G)$ on $V(G)$. The transitive closure $G'$ of $G$ is the (simple) directed graph $G'$ on $V(G)$ such that $R(G')$ is the transitive closure of $R(G)$. I am certainly not an expert, but I guess we need to generalize the notion of relation, where an element can be related to another element with multiplicity.
Anyway, after all that rambling another possible answer is for each $u, v \in V(G)$, put $n(u,v)$ directed edges from $u$ to $v$, where $n(u,v)$ is the maximum number of internally disjoint directed paths from $u$ to $v$.
| 1 | https://mathoverflow.net/users/2233 | 39948 | 25,581 |
https://mathoverflow.net/questions/39949 | 3 | (As suggested, this is a repost from [math.stackexchange.com](https://math.stackexchange.com/questions/4785/terminology-for-weighted-projective-spaces).)
For a sequence of positive integers $a\_1, \ldots, a\_n$ and a base ring $R$ there is a graded ring $R[x\_1,\ldots, x\_n]$ where $x\_i$ is in degree $a\_i$. We can then apply Proj and get a scheme, and this is usually called a weighted projective space; if all of the $a\_i$ are 1, then the resulting scheme really is projective space.
However, the way that this arises is as the quotient of $\mathbb{A}^n \setminus 0$ by an action of the multiplicative group, given by $(x\_1,\ldots,x\_n) \simeq (\lambda^{a\_1} x\_1, \ldots, \lambda^{a\_n} x\_n)$ for all $\lambda$. This is a "coarse" group quotient.
There is an alternative version where one instead takes the associated quotient stack/orbifold, and this has a number of nice properties (including possession of a line bundle $\mathcal{O}(1)$); this is true more generally of a graded ring.
It seems that most people associate "weighted projective space" with the scheme-theoretic notion. What is the appropriate terminology for the stack-theoretic version of this construction?
| https://mathoverflow.net/users/9509 | Terminology for weighted projective spaces | A google search for "[weighted projective stack](http://www.google.com/search?q=%22weighted+projective+stack%22)" currently yields 275 documents, many of which are quite legitimate.
| 6 | https://mathoverflow.net/users/121 | 39953 | 25,585 |
https://mathoverflow.net/questions/39597 | 35 | There was a recent question on intuitions about sheaf cohomology, and I answered in part by suggesting the "genetic" approach (how did cohomology in general arise?). For historical material specific to sheaf cohomology, what Houzel writes in the Kashiwara-Schapira book *Sheaves on Manifolds* for sheaf theory 1945-1958 should be adequate.
The question really is about the earlier period 1935-1938. According to nLab, cohomology with local coefficients was proposed by Reidemeister in 1938 (<http://ncatlab.org/nlab/show/history+of+cohomology+with+local+coefficients>). The other bookend comes from Massey's article in *History of Topology* edited by Ioan James, suggesting that from 1895 and the inception of homology, it took four decades for "dual homology groups" to get onto the serious agenda of topologists. It happens that 1935 was also the date of a big international topology conference in Stalin's Moscow, organised by Alexandrov. This might be taken as the moment at which cohomology was "up in the air".
Now de Rham's theorem is definitely somewhat earlier. Duality on manifolds is quite a bit earlier in a homology formulation.
It is apparently the case that *At the Moscow conference of 1935 both Kolmogorov and Alexander announced the definition of cohomology, which they had discovered independently of one another.* This is from <http://www.math.purdue.edu/~gottlieb/Bibliography/53.pdf> at p. 11, which then mentions the roles of Čech and Whitney in the next couple of years. This is fine as a narrative, as far as it goes. I have a few questions, though:
1) Is the axiomatic idea of cocycle as late as Eilenberg in the early 1940s?
2) What was the role of obstruction theory, which produces explicit cocycles?
Further, Weil has his own story. Present at the Moscow conference and in the USSR for a month or so after, his interest in cohomology was directed towards the integration of de Rham's approach into the theory. He comments in the notes to his works that he pretty much rebuffed Eilenberg's ideas. Bourbaki was going to write on "combinatorial topology" but the idea stalled (I suppose this is related). So I'd also like to understand better the following:
3) Should we be accepting the topologists' history of cohomology, if it means restricting attention to the "algebraic" theory, or should there be more differential topology as well as sheaf theory in the picture?
As said, restriction to a short period looks like a good idea to get some better grip on this chunk of history.
| https://mathoverflow.net/users/6153 | Timeline of cohomology (1935 to 1938) | What strikes me about the first fifty years of homology theory (from Poincaré to Eilenberg-Steenrod's book) is that the development was as much about stripping away unnecessary complication as about increasing sophistication. A famous example is singular homology, which was found very late, by Eilenberg. The construction as we know it presumably seemed too naive to Lefschetz, who misguidedly devised a theory of oriented simplices, and inadequate to those who were interested in general (not locally path connected) metric spaces.
I want to suggest that this process of stripping away is relevant to the introduction of cohomology and its product. (Cf. Dieudonné's "History of algebraic and differential topology", pp.78-81). I won't directly answer the questions, but will suggest that one of the motivations for cohomology came from an application of Pontryagin duality which was rendered obsolete by the new theory.
Alexander wrote up his Moscow conference talk, with improvements suggested by Cech, in a 1936 Annals paper (vol. 37 no. 3) ([JSTOR link](http://jstor.org/stable/1968484?seq=1)). In it, he proposes the cohomology ring ("connectivity ring") as a fundamental homological invariant of a space. In the introduction he hints at the line of thought that led him to the cohomology ring. The relation between cycles and differential forms is mentioned (without citation of de Rham), but what looks more surprising to modern eyes is the comment that the theory of cycles "has been very greatly perfected by Pontrjagin's cycles with real coefficients reduced modulo 1".
Pontryagin had recently developed his duality theory for locally compact abelian groups ([Annals, 1934](http://www.jstor.org/stable/1968438)) in order to apply it to Alexander duality (again, [Annals, 1934](http://www.jstor.org/stable/1968501)). If $K$ is a compact polyhedral complex in $\mathbb{R}^n$, there is a linking form which gives a pairing between $k$-cycles of $K$ and $(n-k-1)$-cycles of $\mathbb{R}^n-K$ and, in modern terms, induces an isomorphism of $H\_k(K)$ with $H^{n-k-1}(\mathbb{R}^n-K)$. Alexander's formulation equated the Betti numbers over a field (mod 2, initially - Dieudonné p. 57) of $K$ and its complement, but it was understood that the full homology groups of $K$ and $\mathbb{R}^n-K$ need not be isomorphic. Pontryagin showed that if one takes a Pontryagin-dual pair of metric abelian groups, say $\mathbb{Z}$ and $\mathbb{T}$, so that each is the character group of the other, then $H\_k(K;\mathbb{T})$ is Pontryagin-dual to $H\_{n-k-1}(\mathbb{R}^n-K;\mathbb{Z})$ via the linking form.
From Alexander's introduction:
>
> Now, if we use Pontrjagin's cycles, the $k$th connectivity [homology] group of a compact, metric space becomes a compact, metric group. Moreover, by a theorem of Pontrjagin, every such group may be identified with the character group of a countable, discrete group. This immediately suggests the advisability of regarding the discrete group, rather than its equivalent (though more complicated) metric character group as the $k$th invariant of a space.... One decided advantage of taking the discrete groups...as the fundamental connectivity groups of a space is that we can then take the product...of two elements of the same or different groups.
>
Guided by Pontryagin's generalisation of his own duality theorem, Alexander finds a simple construction that supersedes Pontryagin's as a basic invariant. (The universal coefficient theorem gives a modern perspective on why Pontryagin's choice of coefficient groups works. I must admit, his formulation of duality is very clean.)
Can anyone comment on Kolmogorov's route to cohomology?
ADDED. On obstruction theory: Charles Matthews's comments draw attention to a 1940 paper of Eilenberg. The MathSciNet review of that paper (by Hurewicz, whose homotopy groups, useful for obstruction theory, date from 1935-36) points me to its 1937 forerunner by Whitney, "The maps of an $n$-complex into an $n$-sphere" (Duke M.J. 3 (no.1), 51-55). This work, too, was presented at the Moscow conference in 1935. Though the topic is different, Whitney's introduction closely resembles Alexander's:
>
> The classes of maps of an $n$-complex into an $n$-sphere were classified by H. Hopf in 1932.
> Recently, Hurewicz [1935-6] has extended this theorem by replacing the sphere with more general spaces. Freudenthal [1935] and Steenrod have noted that the theorem and proof are simplified by using real numbers reduced mod 1 in place of integers as coefficients in the chains considered. We shall give here a statement of the theorem that seems most natural; the proof is quite simple.... The fundamental tool of the paper is the notion of "coboundary"; it has come into prominence in the last few years.
>
| 20 | https://mathoverflow.net/users/2356 | 39957 | 25,588 |
https://mathoverflow.net/questions/39403 | 26 | In the paper "Local Contractions and a Theorem of Poincare" Sternberg has mentioned the following question which was open when the paper was written:
Is the group of orientation-preserving homeomorphisms of the $n$-sphere arc-wise connected?
According to Sternberg's paper Kneser has shown this to be true for $n=2$. Does anyone know the current status of the problem?
| https://mathoverflow.net/users/3635 | connectivity of the group of orientation-preserving homeomorphisms of the sphere | This is known to be true for all $n$ as a consequence of the stable homeomorphism conjecture (SHC), itself closely related to the [annulus conjecture](http://en.wikipedia.org/wiki/Annulus_theorem).
The SHC says that any orientation preserving homeomorphism of $\mathbb{R}^n$ is *stable* i.e. a (finite) product of homeomorphisms each of which is the identity on some non-empty open set.
The same statement is then true for an homeomorphism $h$ of $\mathbb{S}^n$ : first, you can by composition with a (stable) homeomorphism of $\mathbb{S}^n$ assume that $h$ fixes the north pole $p$. Then by SHC $h$ restricted to $\mathbb{S}^n-p\simeq\mathbb{R}^n$ is a composition of homeomorphisms which are the identity on nonempty open sets. But an homeomorphism of $\mathbb{S}^n$ which is the identity on a non-empty open set is isotopic to the identity, by [Alexander's trick](http://en.wikipedia.org/wiki/Alexander%2527s_trick), hence the conclusion.
**PS** : in fact, the detailed story is somewhat complicated. What was known relatively early (due to [R. D. Anderson](http://www.ams.org/mathscinet-getitem?mr=98145), [G. M. Fisher](http://www.ams.org/mathscinet-getitem?mr=117712) around 1960) was that for any topological manifold $M$ (maybe non-compact, but paracompact), the group $H\_c(M)$ of homeomorphisms generated by those compactly supported in domains of topological charts $\mathbb{R}^n\simeq U\subset M$ is the smallest nontrivial normal subgroup of $H(M)$ of all homeomorphisms, and that it is simple. In particular, $H\_c(M)$ is arcwise connected. The proof is ingenious, but not very difficult.
The (much harder) methods and results of geometric topology in dimension $3$ (Bing, Moise,...) then allowed Fisher to prove that for a closed $3$-manifold $M$, $H\_c(M)$ is open in $H(M)$ and thus coincides with the identity component $H\_0(M)$. In particular $H(M)$ is locally arcwise connected, a not at all obvious fact -- later generalized by [Cernavskii](http://www.ams.org/mathscinet-getitem?mr=236948) and [Edwards-Kirby](http://www.ams.org/mathscinet-getitem?mr=283802), who proved the local contractibilty (hence local arcwise connectedness) of $H(M)$ in any dimension.
Fisher also considered the group of stable homeomorphisms $H\_s(M)$ (without the name) of a connected $M$ (otherwise the notion is empty). He managed to prove that it coincides with the group of orientation preserving ones $H\_+(M)$ for closed oriented $3$-manifolds $M$ admitting an orientation reversing homeomorphism. For $M=\mathbb{S}^3$, this implies that $H\_+=H\_0=H\_c$ : any orientation preseving homeomorphism of $\mathbb{S}^3$ is isotopic to the identity (note that for all $n$, $H\_s(\mathbb{S}^n)=H\_c(\mathbb{S}^n)$). This is the $n=3$ case of your question.
Then [M. Brown and H. Gluck](http://www.ams.org/mathscinet-getitem?mr=145497) named stable homeomorphisms, and studied stable structures on manifolds.
A puzzling aspect of the notion of stable homeomorphism is that it is very "contagious" : if $h\in H(M)$ coincides with an element $f$ of $H\_s(M)$ on a nonempty open set $U$, then $h$ is in $H\_s(M)$, since $h^{-1}f$ is the identity on $U$. So this is seen locally everywhere, like orientation preservation (to which it was eventually identified).
Brown and Gluck proved that SHC$\_n$ (SHC in dimension $n$) implies the annulus conjecture in dimension $n$ (AC$\_n$), and that AC$\_k$ in all dimensions $k\leq n$ imply SHC$\_n$. But this was still stuck at SHC$\_3$.
After that came [R. Kirby](http://www.ams.org/mathscinet-getitem?mr=242165) (and L. Siebenmann) in 1968, who proved (using the results of surgery by Wall et al), that SHC$\_n$ (hence AC$\_n$) is true in all dimensions $n> 4$.
But the remaining case SHC$\_4$ was only solved by F. Quinn in 1982 (after work of A. Casson and M. Freedman),
who proved AC$\_4$, hence SHC$\_4$ since the $n\leq 3$ cases were known. See the [survey](http://www.ams.org/mathscinet-getitem?mr=780581) by Edwards.
| 30 | https://mathoverflow.net/users/6451 | 39959 | 25,590 |
https://mathoverflow.net/questions/39960 | 13 | Consider the class of geodesic metrics $g$ on manifolds, that have the following
property: for each point $x$ there exists a neighbourhood $U\_x$ and
a smooth vector field $v\_x$ in $U\_x$ that vanishes at $x$ and whose flow (for small time) dilatates $g$ by a constant factor.
Let us call such metrics *dilatatable*.
An obvious example is provided by an Euclidean $\mathbb R^n$, the flow of the field
$\sum\_i x\_i \frac{\partial}{\partial x\_i}$ dilatates the Euclidean metric by
a constant factor. More generally one can take any Banach space.
I would like to make a guess about the structure of such metrics in general.
**Guess.** Suppose $g$ on $M^n$ is dilatatable. Then there exists a triangulation of $M^n$
such that the restriction of the metric $g$ to each simplex if flat with respect
to the flat structure on the simplex, and
$g$ is flat on the complement to the union of all co-dimension $2$ simplexes.
The **first question** is the following: was such class of metrics considered somewhere and
is this guess correct? Are there obvious counterexamples?
Second part of the question is about examples. It is not hard to construct an example
of such a metric, if we don't require $M^n$ to be a smooth manifold. Namely, we can take any polyhedral metric
on $M^n$, i.e. glue $M^n$ from a union of Euclidean simplexes (glue the boundaries by isometries).
Then for each point there is a conical neighbourhood, and obviously we can always scale this neighbourhood
by the radial field emanating from $x$. So now comes the
**Second question.** Take a topological manifold $M^n$ of dimension $n<7$ with such a polyhedral metric.
It is known then that such a manifold has a smooth structure (because a PL structure in dimension
up to $6$ always defines a unique smooth structure). Is it possible to chose this smooth structure
in such a way, that the polyhedral metric is dilatatable for the smooth structure?
The answer to this question is positive for $n=2$, but I don't know already what happen for $n=3$.
At the same time, there are non-trivial examples in higher dimensions, coming from complex geometry.
For example one can quotient some complex tori $\mathbb T^n$ by a finite group of isometries to get
$\mathbb CP^n$, the obtained polyheral metric on $\mathbb CP^n$ is dilatatable with respect
to the canonical complex (and hence smooth) structure on $\mathbb CP^n$.
| https://mathoverflow.net/users/943 | Geodesic metrics that admit dilatation at each point | Concerning the first question: you description is incomplete, even in the homogeneous case.
There are homogeneous geodesic metrics that admit smooth families of dilatations but are not made of flat Banach metrics. In particular, some [Carnot-Caratheodory metrics](http://en.wikipedia.org/wiki/Sub-Riemannian_manifold) are.
For example, consider the Heisenberg group $H$, which can be thought of as $\mathbb R^3$ equipped with the following group law:
$$
(x,y,z)\cdot(x',y',z') = (x+x',y+y',z+z'+x'y) .
$$
Observe that for every $t\in\mathbb R$, the map $\phi\_t:(x,y,z)\mapsto (e^tx,e^ty,e^{2t}z)$ is a group homomorphism, and these maps form a smooth 1-parameter group of diffeomorphisms (and hence a flow generated by a smooth vector field).
Consider a left-invariant two-dimensional distribution $V\subset TH$ spanned by left-invariant vector fields $X$ and $Y$ whose values at $(0,0,0)$ equal $\partial/\partial x$ and $\partial/\partial y$, respectively. Equip this distribution with a left-invariant Euclidean metric. The distribution is completely non-integrable, so we get a Carnot-Caratheodory metric on $H$. Observe that $\phi\_t$ maps $X$ to $e^tX$ and $Y$ to $e^tY$, hence it is a $e^t$-dilatation of the Carnot-Caratheodory metric.
The Carnot-Caratheodory metric is very different from Banach metrics. For example, its Hausdorff dimension equals 4.
| 9 | https://mathoverflow.net/users/4354 | 39969 | 25,595 |
https://mathoverflow.net/questions/39945 | 6 | We can form a braided monoidal category by taking the groupoid coproduct of the Artin braid groups $B\_n$. We can also make the braids into a simiplical set where the ith face operation is removing the i-th strand and the ith degeneracy operation doubling the ith strand. (These operations are not group homomorphisms, so we don't get a simplicial group.)
Are these two notions compatible? For example, can we use the tensor product of the braided monodial category to describe these face and degeneracy operations? or vice versa? is there a 2-category or something that can combine these two notions?
| https://mathoverflow.net/users/nan | Compatibility of braids as a simplicial set and as a braided monoidal category | I think the best way to think about this is in terms of operads.
Braided monoidal categories are representations of an operad $\Pi$ in the category of (small) categories.
The category $\Pi(n)$ has objects parenthesised permutations of $\{1,\ldots,n\}$ like $(4(23))1$. The morphisms $(\sigma) \to (\tau)$ are braids $\beta \in B\_n$ so that $\beta$ maps to $\tau \sigma^{-1}$ under $B\_n\to S\_n$.
The operad structure
$$
\Pi(n) \times \Pi(k\_1) \times \ldots \times \Pi(k\_n) \to \Pi(k\_1+\ldots+k\_n)
$$
is given by replacing the $i$-th sting by the braid on $k\_i$-strings.
Now any representations $\Pi(n) \to \underline{Hom}(\mathcal{C}^n,\mathcal{C}) = End(\mathcal{C})(n)$ induces a braided monoïdal structure with
* $\otimes$ corresponding to the object $(12) \in \Pi(2)$
* the associativity morphism corresponding to the trivial braid $(12)3 \to 1(23)$ in $\Pi(3)$
* the braiding corresponding to the morphism $(12) \to (21)$ in $\Pi(2)$ induced by the generator of $B\_2 = \mathbb{Z}$.
MacLane's coherence theorem tells you that this is an equivalence.
All the operations you're looking at come from this operad structure.
The $\Pi(n)$ have a nice geometric interpretation. They are the fundamental groupoids of the operad of little discs $C\_2(n)$ (or equivalently of $F(\mathbb{C},n)$ the spaces of configurations of $n$ points in the plane) restricted to a suitable collection of basepoints. This generalizes the classical definition $P\_n = \pi\_1(F(\mathbb{C},n),p)$, $B\_n = \pi\_1(F(\mathbb{C},n)/S\_n,\overline{p})$.
I think this makes the whole picture a lot clearer because we get all of these operations as part of the same structure and we get a universal characterisation of
that structure and a geometrical interpretation for it.
This leads to other nice considerations. For example the Grothendieck-Teichmuller group $GT$ defined by Drinfeld is the automorphism group of the (prounipotent completion) operad $\Pi$. This explains why $GT$ is universal for quasi triangular quasi Hopf algebras as their representations form braided monoïdal categories and related to the Galois group of $\mathbb{Q}$ as $GT$ appears as an automorphism group of fundamental groupoïds of the algebraic varieties $F(\mathbb{A}^1\_{\mathbb{Q}},n)$.
| 11 | https://mathoverflow.net/users/1985 | 39981 | 25,602 |
https://mathoverflow.net/questions/39968 | 13 | Given a subfactor $N\to M$ and a conditional expectation $E:M\to N$,
there is a numerical invariant $Ind(E)$ associated to to this situation, called the index of $E$.
The possible values of $Ind(E)$ are
restricted to the set {$4\cdot \cos^2(\pi/n);n=3,4,5,...$} $\cup$ $[4,\infty]$.
The minimal conditional expectation is
the one that minimizes the value of $Ind(E)$.
The *minimal index* of the subfactor is then defined to be the index of its minimal conditional expectation.
Can the minimal index take all values in {$4\cdot \cos^2(\pi/n);n=3,4,5,...$} $\cup$ $[4,\infty]$? In other words, given a real number in the above set, is there a subfactor whose minimal index is that real number?
---
**Remark:** If the factors are of type $II\_1$, there is another preferred conditional expectation: the one
that is compatible with the traces. The corresponding index is called the *Jones index*. This is not the index I care about.
Jones' index agrees with the minimal index in the case of irreducible subfactors,
but not in general.
Jones' index is known to take all the above values. But the subfactors used in the construction are not irreducible (and one can also check that their minimal index is different from their Jones index).
| https://mathoverflow.net/users/5690 | Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]? | There is an irreducible Temperley-Lieb subfactor at every allowed index. For $n\geq 3$, it has index $4\cos^2(\pi/n)$ and principal graph $A\_{n-1}$ (in fact all subfactors of index less than $4$ are irreducible), and for every $r\geq 4$, it has index $r$ and principal graph $A\_\infty$. Doesn't that do the job by your remark?
| 6 | https://mathoverflow.net/users/351 | 39982 | 25,603 |
https://mathoverflow.net/questions/39974 | 3 | This question is about graphical modeling of joint probability functions, Markovian property and Markov random fields.
Suppose we have an undirected graph G where each node represents a random variable and an edge between two nodes says that there is a probabilistic relation in between them. I want to model the joint probability of these variables and to simplify things I assume that the graph/nodes have Markovian property. This should let me write the joint as a factorization over ``local" clique potentials.
In C. Bishop's Pattern Recog. and Machine Learning book, Chapter 8, pp. 386 ([pdf of the chapter](http://research.microsoft.com/en-us/um/people/cmbishop/prml/Bishop-PRML-sample.pdf)), it is said that the joint distribution is written as a product of potential functions over the **maximal cliques** of the graph:
$p(x) = \frac{1}{Z} \prod\_C \psi\_C(x\_C)$ (eq. 8.39)
However, in Stan Li's [book on MRFs](http://www.nlpr.ia.ac.cn/users/szli/MRF_Book/MRF_Book.html), he says this factorization is done over **all possible cliques** of the graph: see equation (1.26) and (1.27) in <http://www.nlpr.ia.ac.cn/users/szli/MRF_Book/Chapter_1/node12.html#SECTION00323000000000000000>.
Stan Li's explanation makes more sense to me. Which one do you think is correct? Or, might they be just different wording of the same fact? Any help would be appreciated.
| https://mathoverflow.net/users/5223 | How to factorize the joint probability of an arbitrary graph whose nodes are random variables? | As I read it, Bishop is asserting that for each maximal clique $C$ we may define the potential as $\psi\_C(x\_C) = \prod\_S U\_S(x\_S)$ where $S$ denotes cliques which are subsets of $C$ (and $U$ is the energy, as in the link from Li). This is how I interpret "[if $C$] is a maximum clique, and we define an arbitrary function over this clique, then including another factor defined over a subset of these variables would be redundant." It is potentially confusing because, as typical in statistics, the symbol $\psi$ is overloaded to allow different functional form for any clique. In any case, Bishop defines each maximal clique potential in terms of a factorization over all subset cliques. Substitution gives back the other definition.
| 2 | https://mathoverflow.net/users/8719 | 39985 | 25,606 |
https://mathoverflow.net/questions/39966 | 7 | What are the minimal conditions on three topological spaces $X,Y$ and $Z$ such that with the compact-open topology the map
$${(X^Y)}^Z \to X^{Y \times Z}$$
given by taking adjoints is a homeomorpism. The map sends $f: Z \to X^Y$ to $g:Y \times Z \to X$ by the relation $g(y,z)=f(z)(y)$.
This result is known for $Z$ Hausdorff and $Y$ locally compact. I'm interested in a proposition of the form the adjoint construction is a homeomorphism of mapping spaces if and only if some statement regarding the spaces $X$, $Y$ and $Z$.
It would also be interesting to see some counterexamples, for example for $Z$ not Hausdorff, etc.
| https://mathoverflow.net/users/9514 | Minimal conditions for the exponential law for compact-open topologies | A very closely related question (and maybe the one you meant to ask?) is: which spaces $Y$ in the category of topological spaces and continuous maps are exponentiable, i.e., for which $Y$ does the functor $- \times Y: Top \to Top$ have a right adjoint? A necessary and sufficient condition is that $Y$ is *core-compact*, as defined at the [nLab](http://ncatlab.org/nlab/show/exponential+law+for+spaces#corecompactness_5). See also the references in that article.
There are various ways of defining core-compactness; perhaps the fastest is that the topology is a continuous lattice. It is a slightly weaker condition than local compactness (if local compactness is defined as meaning that every point has a basis of compact neighborhoods), and coincides with local compactness if $Y$ is Hausdorff.
If $Y$ and $Z$ are core-compact, then for every $X$ one can exhibit a canonical homeomorphism
$$(X^Z)^Y \cong X^{Y \times Z}$$
by abstract nonsense (since any two right adjoints to $- \times (Y \times Z)$, in particular $((-)^Y)^Z$ and $(-)^{Y \times Z}$, are canonically naturally isomorphic).
Your question is also interesting when interpreted for locales. See Johnstone's Stone Spaces, where it is shown that a locale is exponentiable if and only if it is locally compact.
If $Y$ is not core-compact, then it is possible to show that there is no exponential $\mathbf{2}^Y$ where $\mathbf{2}$ is Sierpinski space (two points, one open, one closed). In other words, the functor $\hom\_{Top}(- \times Y, \mathbf{2})$ is not representable. I once went through the detailed argument (in the case where $Y$ is the space of rational numbers, which I think is illustrative) over at the n-Category Café, see [here](http://golem.ph.utexas.edu/category/2007/12/this_weeks_finds_in_mathematic_20.html#c014330) and the ensuing discussion.
| 13 | https://mathoverflow.net/users/2926 | 39987 | 25,607 |
https://mathoverflow.net/questions/39990 | 4 | According to B. Fine, G. Rosenberger, *On restricted Gromov groups*, Comm. Algebra 20 (1992) 2171--2181, Gromov proved the following in his long article introducing word-hyperbolic groups:
>
> Let $x$ and $y$ be elements of a torsion-free word-hyperbolic group. Either the subgroup generated by $x$ and $y$ is cyclic, or there exists $n$ such that the subgroup generated by $x^n$ and $y^n$ is free of rank 2.
>
>
>
As a very special case of this, we get the following corollary
>
> Let $x$ and $y$ be non-commuting elements in the free group on two generators. Then the subgroup generated by $x$ and $y$ contains a copy of the free group on two generators.
>
>
>
I am trying to cite this corollary as efficiently as possible, for background motivation in something I'm writing. Does anyone know of something slightly, erm, more accessible for the non-specialist than Gromov's original article? I don't really know any geometric group theory beyond some of the terminology and Nielsen-Schreier, but the result seems like it shouldn't be too hard to prove directly, modulo some standard results on free groups. Unfortunately, I don't really have space to sketch any proof in what I'm writing.
(So, to clarify, what I'm really hoping for is an answer saying that the result is easily deduced from material in, say, Section Z of Lyndon & Schupp or similar.)
**EDIT/UPDATE:** my thanks to John Stillwell and Ian Agol for pointing out what should have been blindingly obvious, namely that the result is a trivial consequence of Nielsen-Schreier, and for politely *not* pointing out what is just as obvious, that I should think harder before asking questions.
| https://mathoverflow.net/users/763 | Reference request for two-generator subgroups of a free group | By Nielsen-Schreier, the subgroup $F$ of $F\_2$ generated by $x$ and $y$ is free.
Since $x$ and $y$ do not commute, $F$ is not the free group of rank 1, so it must
contain a free group of rank 2
| 7 | https://mathoverflow.net/users/1587 | 39993 | 25,609 |
https://mathoverflow.net/questions/39973 | 12 | Assume that we have two residually finite groups $G$ and $H$. Which properties of $G$ and $H$ could be used to show that their pro-finite (or pro-p) completions are different?
I asked a while ago in the group-pub mailing list whether finite presentability is such a property but Lubotzky pointed out that it is not the case. A finitely presented and an infinitely presented group can have isomorphic pro-finite completions.
| https://mathoverflow.net/users/7307 | Distinguishing pro-finite completions | There's a theorem that two finitely generated residually finite groups have the same profinite completions if and only if they have the same finite quotients. A reference for the statement of this is [Theorem 2 of this paper](http://www.ma.utexas.edu/users/areid/Groth_revised.pdf), but they cite [Ribes and Zalesskii](http://books.google.com/books?id=47ouE_XSJZYC&lpg=PP1&dq=ribes%2520zalesskii&pg=PP1#v=onepage&q&f=false) for the proof.
| 5 | https://mathoverflow.net/users/1345 | 39996 | 25,612 |
https://mathoverflow.net/questions/39986 | 1 | There's a simple relationship between $J\_\nu$, Bessel functions of the first kind, and $I\_\nu$, modified Bessel functions of the first kind, namely $I\_\nu(z) = i^{-\nu} J\_\nu(iz)$. However, there doesn't seem to be any simple relationship between $Y\_\nu$, Bessel functions of the second kind, and $K\_\nu$, modified Bessel functions of the second kind. Is there an identity relating $Y\_\nu$ and $K\_\nu$ and I'm just missing it? I'm interested in arbitrary $\nu$, but a relationship that only holds for integer values would still be welcome.
| https://mathoverflow.net/users/136 | Modified and unmodified Bessel functions of the second kind | $K\_{\nu}(z)$ is conventionally defined as [a linear combination of normal Bessel functions of both kinds of imaginary argument](http://dlmf.nist.gov/10.27.E8), for the simple reason that this particular linear combination of $J\_{\nu}(iz)$ and $Y\_{\nu}(iz)$ is real for real $\nu$ and positive $z$.
| 3 | https://mathoverflow.net/users/7934 | 39997 | 25,613 |
https://mathoverflow.net/questions/40008 | 2 | I am learning perturbation theory and would like to be able to determine where boundary layers are going to occur just by looking at the differential equation.
Let $n\in\mathbb{N}$ and $p\_i(x)$, $0\leq i< n$ some sufficiently well-behaved functions.
Am I able to determine the boundary layers of the following problem just by looking at the $p\_i(x)$ or some other easy to see property?:
$\epsilon \frac{d^n y}{dy^n} + \sum\_{i=0}^{n-1} p\_i(x) \frac{d^i y}{dy^i} =0$
$y(0)=a$ and $y(1)=b$
(I realise that I require more constraints to get a unique solution but I don't think this effects the existence of boundary layers)
Thanks in advance.
| https://mathoverflow.net/users/2011 | Where are the boundary layers? | Of course, you **do** need $n$ boundary conditions, so that your solution is unique. If you have $m$ boundary conditions, with $m < n$, then you have enough choice among the solutions of the differential equation so as to avoid a boundary layer. Here is an example.
$$\epsilon y'''+y''=0,\qquad y(0)=a, y(1)=b.$$
Given $a,b\in{\mathbb R}$, there are a lot of solutions which form a line. All of them have a boundary layer (a scalar times $\exp(-\frac{x}{\epsilon})$), but $\bar y(x)=a+(b-a)x$.
Suppose that the third boundary condition is $y'(0)=c$, then the solution $y\_\epsilon$ tends to some $y$ such that $y''=0$, $y(0)=a$ and $y(1)=b$, which is a Dirichlet problem. There is a boundary layer at $x=0$, where a boundary condition is lost.
If instead the third boundary condition is $y'(1)=d$, then the limit of $y\_\epsilon$ still solves $y''=0$, but with $y(1)=b$ and $y'(1)=d$, which is a Cauchy problem. The boundary layer is still at $x=0$.
In general, consider an equation
$$\epsilon y^{(n}+p\_1(x)y^{(n-1)}+\cdots=0,$$
with $\epsilon>0$ but very small. Assume that $p\_1$ does not vanish, in order that the limit equation be non-singular. Then you can decide whether the boundary layer is at rigt or at left by looking at the sign of $p\_1$. It is at left if $p\_1> 0$, at right if $p\_1< 0$.
| 2 | https://mathoverflow.net/users/8799 | 40026 | 25,629 |
https://mathoverflow.net/questions/39798 | 23 | We have some group word $w$ in $k$ letters. We say a $k$-tuple of group elements $\vec{g} = (g\_1, g\_2, \ldots , g\_k) \in G^k$ satisfies the word $w$ if $w$ gives the identity at $\vec{g}$. More precisely: The word $w$ is an element of $F\_k$, the free group on $k$ letters. The $k$-tuple $\vec{g}$ specifies some homomorphism $\varphi\_{ \vec{g} } : F\_k \longrightarrow G$ by the universal property. In this notation, $\vec{g}$ satisfies $w$ if $\varphi\_{ \vec{g} } (w) = e$.
For a finite group $G$, we are interested in the probability that a random (uniformly chosen) $k$-tuple of group elements satisfies the word $w$. Call this probability $p\_w(G)$.
Now consider some family of finite groups, each injecting into the next:
$$ G\_0 \hookrightarrow G\_1 \hookrightarrow \ldots \hookrightarrow G\_n \hookrightarrow \ldots $$
Is it true that $p\_w(G\_n)$ converges to a limit?
For instance, if the word $w$ is $x\_1 x\_2 x\_1^{-1} x\_2^{-1}$, then an easy group-theoretic argument shows that $p\_w(G\_n)$ decreases monotonically.
For the word $x\_1^2$, the sequence need not be monotonic, but seems to converge anyway.
Does anyone know a proof that the limit exists? Or have a counterexample?
| https://mathoverflow.net/users/9068 | In an inductive family of groups, does the probability that a particular word is satisfied converge? | John, you know this already, and this is far from an answer, but I thought I'd say it here for the benefit of others who may want to think about the problem.
Call a word $w(x\_1,x\_2,\dots,x\_k)$ "groupy" in the variable $x\_i$ if, for fixed values of the other variables, the set of values of $x\_i$ such that $w$ is the identity element is a subgroup of the whole group. Call $w$ "groupy" if it is groupy in all its inputs.
We can show that if $w$ is groupy, and $H \le G$, then $p\_w(H) \ge p\_w(G)$, giving the monotonically decreasing property on $p\_w(G\_n)$.
The word $x$ and the word $e$ are groupy for trivial reasons. Beyond these, the only groupy word I can think of is the commutator word $[x\_1,x\_2]$.
On the other hand, if we restrict ourselves to the variety of abelian groups, all words power words (e.g., $x^2$ or $x^3$) are groupy, hence the monotonically decreasing property holds.
The iterated commutator $[[x\_1,x\_2],x\_3]$ is groupy in $x\_3$ but not (in general) in $x\_1$ or $x\_2$ -- however, it is likely that the groupiness argument can be extended somewhat to cover these kinds of words too.
| 3 | https://mathoverflow.net/users/3040 | 40027 | 25,630 |
https://mathoverflow.net/questions/40018 | 9 | Before stating the questions that I have, which are very specific and probably not so interesting to someone who has never thought about these things, I need to introduce some notation.
Let $p$ be any prime, and let $D$ be the quaternion algebra over $Q$ ramified precisely at $p$ and infinity. Choose a maximal order $R$ inside $D$. For any prime $\ell\neq p$, fix an isomorphism of $D\_\ell:=D\otimes Q\_\ell$ with the algebra $M\_2(Q\_\ell)$ in such a way that
the maximal compact subring $R\_\ell:=R\otimes Z\_\ell$ corresponds to $M\_2(Z\_\ell)$.
If $N$ is an integer $>0$ not divisible by $p$, let $U\_\ell(N)$ be the subgroup of $R\_\ell^\star\simeq GL\_2(Z\_\ell)$ given by those matrices whose bottom row is congruent to $(0$ $1)$ modulo $N$ (equivalently, modulo the highest power of $\ell$ dividing $N$).
The ring $R\_p:=R\otimes Z\_p$ has a unique maximal, two-sided principal ideal $(\pi)$ generated by any uniformizer $\pi$, the residue field $R/(\pi)$ is a finite field with $p^2$ elements. We let $R\_p^\star(1)$ denote the subgroup of $R\_p^\star$ given by the units that are congruent to $1$ modulo $(\pi)$.
Let now $D^\star$ be the multiplicative group of $D$, viewed as an algebraic group over $Q$. For any integer $N>0$ not divisible by $p$, we are going to define an open subgroup $U(1,N)$ of the group $D^\star\_A$ of points of $D^\star$ valued in $A$, the adele ring of $Q$. Namely $U(1,N)$ is the product of all the $U\_\ell(N)$, for $\ell\neq p$; of $R\_p^\star(1)$;
and of the full (connected) component at infinity $D^\star\_\infty$.
Consider the space $S(1,N)$ of complex valued functions on the double coset $D^\star\backslash D^\star\_A/U(1,N)$, which is known to be finite. For any prime $\ell\neq p$ there is an Hecke operator $T\_\ell$ acting on $S(1,N)$ that can be defined in terms of double cosets in the usual way.
(Let $\alpha\_\ell$ be the matrix whose top row is $(\ell$ $0)$ and whose bottom is $(0$ $1)$; decompose $U\_\ell(N)\alpha\_\ell U\_\ell(N)$ as a finite union of left cosets $\gamma\_i U\_\ell(N)$; for $f\in S(1,N)$ define $T\_\ell(f)(x)=\sum f(x\gamma\_i)$).
Let $V$ be the vector space of locally constant, complex valued functions on $D^\star\_A$ that are left invariant by $D^\star$. Observe that $S(1,N)$ can be viewed as a finite dimensional subspace of $V$. Right translation defines an admissible representation of $D^\star\_A$ on $V$ which is known to be completely decomposable into a discrete direct sum of irreducible admissible representations of $D^\*\_A$. If $f\in S(1,N)$, then denote by $V\_f$ the smallest subspace of $V$ that is stable by $D^\star\_A$.
Questions:
1) Let $f\in S(1,N)$, for some $N$. Is it true that the space $V\_f$ is finite dimensional
if and only if $f:D^\star\_A\rightarrow C$ factors through the reduced norm map $Nr:D^\star\_A\rightarrow A^\star$?
2) Does the subspace of $S(1,N)$ given by those functions that factor through the reduced Norm admit an Hecke stable complement?
3) Is the action of $T\_\ell$ on $S(1,N)$, for $\ell\nmid pN$, semisimple?
4) Assuming that 2) holds, and letting $S\_0(1,N)$ be such complement, how do we relate the C-subalgebra $T\_0(N)$ of End($S\_0(1,N)$) generated by the Hecke operators $T\_\ell$, with $\ell\nmid pN$, to a C-algebra of Hecke operators acting on weight 2 cusps forms of a certain level? What I mean is: out of the J-L correspondence, can we read off an isomorphism between $T\_0(N)$ and some Hecke algebra coming from classical modular forms?
Thanks.
[EDIT: In the 2nd and 3rd lines above "Questions:" I should have probably have said "discrete Hilbert direct sum", the "direct sum" being only dense in V]
| https://mathoverflow.net/users/4800 | Finite dimensional automorphic representations of a definite quaternion with prime discriminant and Hecke action | 1) Yes, I think that's true. I guess it follows relatively easy from the statement that an automorphic representation of the algebraic group $D^\times$ is finite-dimensional iff it's 1-dimensional and factoring through the norm. This latter statement probably follows from strong approximation applied to (the adelic points of the algebraic group associated to) the norm 1 elements of $D^\times$, or you can apply Jacquet-Langlands and move to $GL(2)$ and basically use the standard classification of automorphic representations there.
2) Yes, I think it does. There's some analogue of the Petersson inner product here, right? It's just a combinatorial thing and you can use this. The space $S(1,n)$ is not a mysterious thing: multiplicity 1 etc is all true in this setting and the space is a sum of generalised eigenspaces just like classical spaces of cusp forms.
3) Yes. Use either Jacquet-Langlands and classical results, or mimic the standard proof for $GL(2)$ using the pairing on $S(1,N)$.
4) This question has a bit more meat to it. You seem to be asking what the Jacquet-Langlands correspondence is explicitly. I know something this but it's a bit messy. Let me write down some stuff; the answer isn't quite as "clean" as you'd like it to be though, perhaps.
In $S\_0(N)$ you're seeing automorphic representations which have a fixed vector under $R\_p^\*(1)$. These come in two flavours. The first are those which actually have a fixed vector under all of $R\_p^\*$. These contribute one dimension, locally, to $S\_1(N)$ and under J-L correspond to forms of level $Np$ which are new at $p$ with trivial character. The second sort are those without an $R\_p^\*$-fixed vector. These are 2-dimensional (I mean their $\pi\_p$ is 2-dimensional) [Edit: as Tomasso pointed out, this isn't true. There are some more 1-dimensional ones factoring through the local norm map and coming from tamely ramified but not unramified quasicharacters; let's ignore them in what follows.] and all of $\pi\_p$ is fixed by $R\_p^\*(1)$, so here you're getting 2-dimensional spaces in $S\_0(N)$ on which Hecke is acting as scalars (so a super-naive version of multiplicity 1 is failing; what's really going on is that the classical theory of oldforms and newforms works in a completely different way to the behaviour of $S\_0(N)$ at $p$). Applying JL, if memory serves, you see forms which are supercuspidal at $p$; their level will be something like $Np^2$, their character will be something like the norm of the character of $\mathbf{F}\_{p^2}$ associated to the form on $S\_0(N)$. So the answer to your question is "yes, you can write down a classical space of forms that corresponds to $S\_0(N)$" but it's not pleasant to write it down---you have to understand the possibilities of what forms of level $Np^2$ look like, and only choose certain of them. It's much easier to understand from the point of view of representation theory, either of $GL(2,\mathbf{Q}\_p)$ or the local Weil group.
All this is very well-understood by lots of people here, so feel free to correct me or ask more.
| 6 | https://mathoverflow.net/users/1384 | 40038 | 25,637 |
https://mathoverflow.net/questions/40028 | 19 | The "generalized dihedral group" for an abelian group *A* is the semidirect product of *A* and a cyclic group of order two acting via the inverse map on *A*. *A* thus has index two in the whole group and all elements outside *A* have order two. Thus, at least half the elements of any generalized dihedral group have order two.
My question is the converse: if half or more the elements of a finite group *G* have order two, is it necessary that *G* is either an elementary abelian 2-group or a generalized dihedral group? [Note: Actually nontrivial elementary abelian 2-groups are also generalized dihedral, they're an extreme case. Also, note that the direct product of a generalized dihedral group with an elementary abelian 2-group is still generalized dihedral, because the elementary abelian 2-group can be pulled inside the abelian part.]
I have a proof when the order of *G* is twice an odd number *m*. In that case, there is a short an elegant elementary proof -- we consider the set of elements that can be written as a product of two elements of order two and show that this is a subgroup of order *m*, then show that any element of order two acts by the inverse map on it, and hence the subgroup is abelian. It can also be thought of as a toy example of Frobenius' theorem on Frobenius subgroups and complements (though we don't need Frobenius' theorem).
However, I am having some difficulty generalizing this, mainly because there are elements of order two that are inside the abelian group.
Although I have a number of possible proof pathways, I'll refrain from mentioning them for now because the actual proof is likely to be much simpler and I don't want to bias others trying the problem.
| https://mathoverflow.net/users/3040 | Half or more elements order two implies generalized dihedral? | $D\_8\times D\_8$ is such a group (it has 35 involutions) that fails to be either elementary abelian or generalized dihedral.
There is an actual classification of groups with at least half the elements being involutions; it was first done by Miller. A modern reference would be the paper by Wall, ["On Groups Consisting Mostly of Involutions"](https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/abs/on-groups-consisting-mostly-of-involutions/554A7AE2A5C7B537B76CB098A8C6E2E2). A simpler treatment, though with a different goal in mind, is given in Liebeck and MacHale's paper ["Groups with Automorphisms Inverting Most Elements"](https://doi.org/10.1007/BF01142582).
| 25 | https://mathoverflow.net/users/1446 | 40042 | 25,639 |
https://mathoverflow.net/questions/40046 | 0 | Suppose that $X$ is a simply connected metric space, with a non-positively curved metric (for example, Euclidean or hyperbolic space). Let $A,B,C$ be disjoint, convex sets in $X$, and suppose that the shortest path from $A$ to $B$ passes through $C$. Under these hypotheses, it should follow that there does not exist a point in $X$ that is equidistant to $A$, $B$, and $C$.
In the special case where $A,B,C$ are points, this statement amounts to checking inequalities between the sides of a triangle. That is, for any $D \in X$, one of the triangles $ACD$ or $BCD$ -- say, $ACD$ -- will have an obtuse angle at $C$. Then the side $AD$ is longer than $CD$, hence $D$ is not the equidistant point. But I'm stumped about how to show this for more general convex sets.
My hunch is that geometers should have encountered this question before. Does anyone have a reference, an argument, or (gasp) a counterexample?
| https://mathoverflow.net/users/8183 | Equidistant points in negatively curved metric spaces | Hello Dave,
Three disks of equal radius in Euclidean plane with centers on a circle of sufficiently large radius seems to be an easy counter-example.
| 4 | https://mathoverflow.net/users/1811 | 40047 | 25,642 |
https://mathoverflow.net/questions/40077 | 4 | Let $V$ be a vector space over some field $k$ and $T \in \mathrm{GL}(V)$. Then, we can view $T\in \mathrm{GL}(\mathrm{Sym}^k(V))$ where $\mathrm{Sym}^k(V)$ denotes the $k^\mathrm{th}$ symmetric power of $V$ and denote it $T\_k$. Knowing $\det T$, is there a general formula for $\det T\_k$?
| https://mathoverflow.net/users/9035 | Determinant and symmetric power | We have that $\det T\_k$ is a fixed (depending on $n=\dim V$ and $k$ only) power of
$\det T$. To see this, as well as getting the power, one can for instance note
that $\mathrm{SL}(V)$ is the commutator subgroup of $\mathrm{GL}(V)$ (except for extremely small finite fields but we can always increase the size of the field) and hence
if $\det T=1$ then $\det T\_k=1$. We can then write any $T\in\mathrm{GL}(V)$ in
the form $DS$, where $S\in\mathrm{SL}(V)$ and $D$ a diagonal matrix with
diagonal entries $(t,1,1,\ldots,1)$. Then $\det((DS)\_k)=\det(D\_k)\cdot1$ so it
suffices to compute $\det(D\_k)$ but in the standard basis of $\mathrm{Sym}^kV$,
given a basis $e\_1,\ldots,e\_n$ of $V$, $D\_k$ is a diagonal entries and its
determinant is $t^R$, where $R=\sum\_{0\leq i\leq k}is^{n-1}\_{k-i}$. Here
$s^{a}\_{b}=\dim \mathrm{Sym}^bU$ where $\dim U=a$ which equals $\binom{a+b-1}{b}$.
| 7 | https://mathoverflow.net/users/4008 | 40079 | 25,658 |
https://mathoverflow.net/questions/40081 | 3 | I see this term in a paper. There is no abelian variety whatsoever involved. There is an original category which consists of $Z\_p$ modules with some properties, lets denote as $\mathfrak{C}$, then the author says that $\mathfrak{C}\otimes Q\_p$ is its "isogeny category". What is this supposed to mean? Is it some "change of base"? Or are there general notion of "isogeny category"?
Thank you.
| https://mathoverflow.net/users/9539 | what is isogeny category? | As Pete Clark intimates in his comment: If you have a category $\mathfrak C$ in which the Hom sets are $R$-modules, and composition is $R$-bilinear, and if $S$ is an
$R$-algebra, then you can form the category $\mathfrak C \otimes\_R S$ in which you tensor all Hom sets by $S$ over $R$.
If $\mathfrak C$ is the category of abelian varieties, then taking $R = \mathbb Z$ and
$S = \mathbb Q$, this construction leads to the category of abelian varieties "up to isogeny", or "isogeny category" of abelian varieties, hence the name in general.
| 6 | https://mathoverflow.net/users/2874 | 40084 | 25,661 |
https://mathoverflow.net/questions/40089 | 3 | Consider a random infinite binary tree $T(a,b)$, so that $a$ denotes the probability of a left edge branching from any root-connected node,and $b$ denotes the probability of a right edge branching from any root-connected node. We establish an inductive base case, so that $T(0,0)$ contains the root node only.
$T(1,0)$ and $T(0,1)$ trivially has path cardinality $\aleph\_0$ (\*) . $T(1,1)$, the infinite complete binary tree, trivially has Continuum path cardinality.
$T(a,b)$ is finite for $a+b<1$.
$T(a,b)$ has path cardinality $\aleph\_0$ for any $a+b=1$.
Now, is it then true that, $T(a,b)$ has Continuum path cardinality for $a+b>1$?
Intuitively this seems to be the case; Consider for example $T(1,\epsilon )$, for an infinitesimal probability $\epsilon$. On average for every $1/ \epsilon$ left-most nodes there is a right-branch. For each occurrence of a right branching node, we contract the graph, deleting all intermittent nodes which produced no right branch, constructing a new graph node, which branches both left and right. By induction we can then show that $T(1,\epsilon)$ is isomorphic to $T(1,1)$, and therefore has Continuum path cardinality. It seems a similar argument can be used for any $T(a,b)$ with $a+b>1$. Apologies if this is trivial. Does anyone know of any relevant references on path cardinality for random sub-graphs of the infinite complete binary tree?
(\*) path cardinality of $T$ is short for the cardinality of the set of all paths in $T$.
| https://mathoverflow.net/users/1320 | Path cardinality for random $(a+b)$-ary infinite trees |
>
> Yes, $T(a,b)$ has continuum cardinality when $a+b>1$... with positive probability. Of course there is also positive probability that $T(a,b)$ has no infinite paths, if $a<1$ and $b<1$.
>
>
>
$T(1,0)$ has only one infinite path, so ``path'' means finite-or-infinite path.
Background
==========
The relevant result, the Extinction Criterion below, was first stated by Bienaymé in 1845; see Heyde and Seneta (I. J. Bienaymé. Statistical theory anticipated) pp. 116--120 and Lyons and Peres (Probability on Trees and Networks, Cambridge University Press, in preparation. Current version available at <http://mypage.iu.edu/~rdlyons/>), Proposition 5.4. The first published proof of Bienaymé's theorem appears in Cournot (1847, De l'Origine et des Limites de la Correspondance entre l'Algèbre et la Géométrie, Hachette, Paris) pp. 83--86.
Extinction Criterion
====================
Given numbers $p\_{k}\in [0,1]$ with $p\_{1}\ne 1$ and $\sum\_{k\ge 0}p\_{k}=1$, let $Z\_{0}=1$, let $L$ be a random variable with $\mathbb P(L=k)=p\_{k}$, let
$
\{ L^{(n)}\_i \},\quad n,i\ge 1
$
be independent copies of $L$, and let
$$
Z\_{n+1}=\sum\_{i=1}^{Z\_{n}}L\_{i}^{(n+1)}.
$$
Let $q=\mathbb P((\exists n)\, Z\_{n}=0)$. Then $q=1$ iff $\mathbb E(L)=\sum\_{k\ge 0}kp\_{k} \le 1$. Moreover, $q$ is the smallest fixed point of $f(s)=\sum\_{k\ge 0}p\_{k}s^{k}$.
Solution sketch
===============
As stated the Extinction Criterion only shows that with positive probability there is at least one infinite path. Now consider, along this path, the possibility of branching off to create another infinite path. You get infinitely many independent events, so the probability that they all fail to happen is 0. Continuing this way, we get a positive probability of continuum many infinite paths.
| 5 | https://mathoverflow.net/users/4600 | 40097 | 25,669 |
https://mathoverflow.net/questions/40103 | 1 | Suppose I know the following information about a function :
1) Its a polynomial (not an explicit equation, neither the roots nor the degree is known)
2) I have managed to find an algebraic relation between some of the roots (mind you I do not know the roots explicitly, just the form of the algebraic relation is known to me).
Now given this information can one say something about the polynomial itself ?
Now what do I seek for? Well, information on something like the divisors of the degree of the polynomial, or say something about the Galois group of the polynomial may be ....
so you can say am asking an inverse question.
I understand that under these very general condition the problem may not even be well posed. I actually have more information about the polynomial in the particular case I encountered it ... the polynomial is a 0-1 polynomial ...some of the roots lie in the unit circle... etc. etc.
But certainly there would be instances of similar problems (with more information
available about the polynomial/ the nature and number of algebraic relations that are available etc.) which has been dealt with ?
So, I wanted to ask the question in a more general setting. Any variant of this I would say is quite interesting. So you can assume different kind of condition on the roots, coefficient algebraic relation,
I will greatly appreciate if some one can point out where I should be looking. Reference to literature where such a problem has been dealt with would be great.
Regards
Vagabond
| https://mathoverflow.net/users/6766 | A question regarding polynomials whose roots satisfy certain algebraic relation | This is rather vague but regarding relations between roots of a polynomial you may try some of Chris Smyth's papers as a starting point. For instance, this one:
C. J. Smyth, Conjugate algebraic numbers on conics, Acta Arith. 40 (1982), 333–346.
| 4 | https://mathoverflow.net/users/8131 | 40116 | 25,679 |
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