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https://mathoverflow.net/questions/33436 | 2 | Let $M$ be a compact manifold, and let $M\_1,\ldots, M\_k$ (k>2) be embedded submanifolds. Suppose that $p\in\cap\_{i=1}^k M\_k$ and that for any subset $S$ of $\{1,\ldots, k\}$ and any $j\notin S$ that $\cap\_{i\in S}M\_i$ intersects $M\_j$ transversally at $p$.
I believe that in this case the fact that $\cap\_{i=1}^k M\_k$ is nonempty is stable (still true after homotoping each $M\_i$ a little bit). Does anyone have a reference for this fact?
| https://mathoverflow.net/users/5399 | Normal intersections of submanifolds | The matter being local, we can restrict to a nbd $U$ of $p$ and think that $M\_i$ is the zero set of some local submersion $g\_i:U\to\mathbb{R}^{n\_i}$. If I'm not wrong your transversality assumption then translates into the surjectivity of the differential at $p$ of the map $g:=(g\_1,\dots,g\_k):U\to\mathbb{R}^m$ (here $m:={n\_1+\dots+n\_k}$). So $0\in\mathbb{R}^m$ is a regular value for $g$, which implies your thesis. Note that the compactness assumption on $M$ plays no role.
| 3 | https://mathoverflow.net/users/6101 | 33446 | 21,677 |
https://mathoverflow.net/questions/32843 | 4 | Hello, this is my first post here. I hope that it is not too vague; I will be as precise as I can, but I have more of a meta-problem so please forgive me if this is inappropriate. My question is essentially of the form "what mathematical tools are most appropriate for investigating another problem?" Of course, if it turns out that you can answer the underlying problem directly that would be great too :-)
Background and definitions
--------------------------
Given a Hermitian operator $H$, there exists a unitary operator $U(t)=\exp(-iHt)$ for any real $t$.
Let the set $\lbrace v\_i\rbrace\_{i=1}^N$ be a basis for the Hilbert space on which $H$ and $U$ act.
Let $V\_n=\text{span}(\lbrace v\_i\rbrace\_{i=1}^n)$ be the span of a given subset of the full basis.
The question
------------
Let $a\in V\_n$. In general, $U(t)\cdot a\notin V\_n$, but it may be the case that for some $t\_0$, $U(t\_0)\cdot a\in V\_n$. I am looking for the more specific case in which this is true for all $a\in V\_n$, for a single $t\_0$. So,
>
> For a given $H$, what properties might determine whether or not there exists a $t\_0$ such that $U(t\_0)$ maps $V\_n$ onto itself? What tools would best tackle this problem?
>
>
>
It seems to me that this is equivalent to asking whether there is a $t\_0$ for which $U$ is block-diagonalizable such that one of the blocks acts only on $V\_n$. I also think that my question may be related to orbits in a Lie algebra, but I am not certain how to rigorously state it in that language.
Additional information
----------------------
Thanks to Helge, I have realized that I should have included a bit more motivation and am doing so here.
There are at least two ways in which the problem stated above could be answered trivially:
1. Helge's solution of rational eigenvalues.
2. $H$ could already be block-diagonal, so that the states of $V\_n$ are never mapped out of it at any $t$.
I am interested in determining those $H$ that behave as specified for some $t\_0$, but which do mix the states of $V\_n$ with the rest of the Hilbert space for $t\in(0,t\_0)$, and which result in $U(t\_0)$ acting non-trivially on $V\_n$.
| https://mathoverflow.net/users/7814 | Under what conditions will a unitary matrix fix a subspace which does not diagonalize the generating Hamiltonian? | OK, I think I have a comprehensive answer. Since this is physics-inspired, I'll use bra-ket notation.
First, in my comment above, you can see that if $U\_{t\_0}$ fixes a subspace $V\_n \subset V$, then $U\_{t\_0}$ is block diagonal:
>
> If we write $U$ in the basis $\{|v\_1\rangle,...,|v\_N\rangle\}$ and see how it acts on the basis of $V\_n$, we must have the lower left block of $U$ be zero. Then, since the columns of unitary matrices form an orthonormal basis of the whole space, the columns of the upper left block of $U$ must be an orthonormal basis of $V\_n$ . So, the columns of the right side of $U$ must span ${V\_n}^{\perp}$, so that the upper right hand block is zero.
>
>
>
Next, let $W:=U\_{t\_0}|\_{V\_n}$ be the upper block of $U\_{t\_0}$ which acts on $V\_n$:
$\quad U\_{t\_0} = \\left[ \begin{matrix} W & 0 \\\ 0 & W' \end{matrix} \right]$
$W$ is unitary $n \times n$ matrix, and therefore can be diagonalized. Without loss of generality, assume the orthonormal vectors $\{|v\_1\rangle,...,|v\_n\rangle\}$ diagonalize $W$ with respective eigenvalues $\{\exp(i w\_i t\_0)\}$. Also, let $\{|h\_1\rangle,...,|h\_N\rangle\}$ be the orthonormal basis (in general, completely different than the $|v\_i\rangle$) which diagonalizes $H$ with eigenvalues $h\_j$.
Then, for all $i$ with $1 \le i \le n$,
$\quad \exp(i w\_i t\_0) = \langle v\_i| U\_{t\_0}|v\_i\rangle = \langle v\_i|\exp(i H t\_0) |v\_i\rangle = \sum\_{j=1}^N \exp (i h\_j t\_0) |\langle v\_i | h\_j \rangle|^2 $.
After multiplying both sides by $\exp(-i w\_i t\_0) $ we get
$\quad 1 = \sum\_{j=1}^N \exp [i (h\_j-w\_i) t\_0] |\langle v\_i | h\_j \rangle|^2$.
Now, since $\{|h\_j\rangle\}$ is an orthonormal basis, the values $|\langle v\_i | h\_j \rangle|^2$ sum to unity (for fixed, normalized $|v\_i\rangle$). That means *all* the exponentials on the rhs for which $|\langle v\_i | h\_j \rangle|^2 \neq 0$ must be equal to unity. So,
$\quad (h\_j-w\_i) \frac{t\_0}{2 \pi} \in \mathbb{Z}$
and
$\quad (h\_j-h\_{j'}) \frac{t\_0}{2 \pi} \in \mathbb{Z} \quad \forall j,j'$ such that $\langle v\_i | h\_j \rangle $ and $ \langle v\_i | h\_{j'} \rangle \neq 0$.
In other words, for all $|h\_j\rangle$ that have non-vanishing inner product with $|v\_i\rangle$, the corresponding $h\_j$ are separated by integer multiples of $\frac{2 \pi}{t\_0}$.
It's easy to show that this is also a sufficient condition. Given $H$ with eigenvalues $h\_j$ and any subset $\mathbf{H}\_{t\_0} \subset \{h\_j\}$ such that for all $h\_j,h\_{j'} \in \mathbf{H}\_{t\_0}$,
$\quad (h\_j-h\_{j'}) \frac{t\_0}{2 \pi} \in \mathbb{Z}$
then *any* vector $|v\rangle$ in the span of eigenvectors corresponding to $\mathbf{H}\_{t\_0}$ will be an eigenvector of $U\_{t\_0} = \exp(i H t\_0)$ with eigenvalue $\exp(i h\_j t\_0)$ (which is the same for all $h\_j \in \mathbf{H}\_{t\_0}$).
(Everything below is old)
Example (Old)
-------------
Still not an answer, but here's a simple but non-trivial example:
Let $U\_t$ be the family of unitaries acting on $C^3$ which spatially rotate $R^3$ about the vector $(1,1,0)$ with period $T=2\pi$. Let $t\_0 = T/2=\pi$. Then in the standard basis,
$\quad H = \frac{i}{\sqrt{2}}\left[ \begin{matrix} 0 & 0 & 1 \\\ 0 & 0 & -1 \\\ -1 & 1 & 0 \end{matrix} \right] \quad ,$
$\quad U\_{t\_0/2} = \frac{1}{2}\left[ \begin{matrix} 1 & 1 & -\sqrt{2} \\\ 1 & 1 & \sqrt{2} \\\ \sqrt{2} & -\sqrt{2} & 0 \end{matrix} \right]\quad , \quad U\_{t\_0} = \left[ \begin{matrix} 0 & 1 & 0 \\\ 1 & 0 & 0 \\\ 0 & 0 & 1 \end{matrix} \right] \quad , \quad U\_{T} = I,$
so $U\_{t\_0}V\_2 = V\_2$, where $V\_2 = \mathrm{span}\{(1,0,0),(0,1,0)\}$. The eigenbasis is $\{(1,-1,-i\sqrt{2}),(1,-1,i\sqrt{2}),(1,1,0)\}$ with respective eigenvalues $\{1,-1,0\}$ for $H$.
Note that $H$ and $U\_{t\_0 /2}$ are not block diagonal in the standard basis, but $U\_{t\_0}$ is. In particular, no subset of the eigenvectors of $H$ form an orthonormal basis of $V\_2$.
Commutation (Old)
-----------------
I think the confusion over when you can infer that $H$ is block diagonal comes from the non-uniqueness of matrix roots. Positive matrices have unique positive roots, of course, but unitary matricies do not have unique unitary roots. For example, the $2 \times 2$ identity matrix has itself and
$\quad \left[ \begin{matrix} 0 & 1 \\\ 1 & 0 \end{matrix} \right]$
as square roots, but only the latter mixes the first and second rows. Likewise, I suspect that all non-trivial cases of unitaries fixing subspaces (where non-trivial is defined by Michael Underwood above) will have $t\_0$ exactly such that non-degenerate eigenvectors of $H$ are *degenerate* for $U\_{t\_0}$.
| 1 | https://mathoverflow.net/users/5789 | 33449 | 21,680 |
https://mathoverflow.net/questions/33443 | 2 | On p.8 of <http://www.msri.org/publications/books/Book39/files/marker.pdf>, the author writes $\Gamma(\bar{d})$, when $\Gamma$ is, first of all, a set of formulas (not a single one), and it is a formula which has variables, not constants. This doesn't make sense. And what does he mean by $T+\Gamma(\bar{d})$? This would have to mean that we are working in a different language. And why does it imply facts about $\psi\_i(\bar{v})$ when we have $\bar{d}$, i.e. a set of constants, rather than variables, when $\bar{v}$ is a set of variables, not a set of constants?
Similarly, in the proof of the "CLAIM," in the sentence that begins "If $\Sigma$ is inconsistent...", how can we go from $\psi\_1(\bar{d})$ to $\psi\_1(\bar{v})$? One takes constants, the other takes variables.
| https://mathoverflow.net/users/1355 | Confusion about model theory notes | The notation $\Gamma(\bar d)$ means that for every formula $\psi(\bar v)\in\Gamma(\bar v)$ we substitute $\bar d$ for $\bar v$. The set $T+\Gamma(\bar d)$ is just the union $T\cup\Gamma(\bar d)$.
The interchanging of variables $\bar v$ and constants $\bar d$ in the proof comes from the fact that the constants $\bar d$ are new constants that do not appear in $T$. Thus, for example, if $T\models \phi(\bar d)$, then also $T\models\forall \bar v (\phi(\bar v))$.
| 3 | https://mathoverflow.net/users/400 | 33452 | 21,683 |
https://mathoverflow.net/questions/33432 | 2 | Extending my [earlier question about linear transformations](https://mathoverflow.net/questions/33303/linear-transformation-takes-a-polygon-to-another-one), what's the easiest way to test if there exists a projective linear transformation that takes one polygon to another (in $\mathbb{R}\mathbb{P}^2$)?
| https://mathoverflow.net/users/2503 | Projective transformation between polygons. | The answer Tim Gowers and Will Jagy gave you in response to your earlier question can be extended straightforwardly. A homogeneous linear transformation on R^3 has 3x3 - 1 = 4x2 degrees of freedom, so there is a unique projective-linear transformation between any given pair of non-degenerate quadrilaterals. Thus do with quadrilaterals what was done in the previous question with triangles.
| 2 | https://mathoverflow.net/users/2036 | 33459 | 21,688 |
https://mathoverflow.net/questions/33478 | 63 | The coefficients of lowest and next-highest degree of a linear operator's characteristic polynomial are its determinant and trace. These have well-known geometric interpretations. But what about its intermediate coefficients?
For a linear operator $f : V \to V$, we have the beautiful formula
$$\chi(f) = det(f - t) = \sum\_{i=0}^n (-1)^i\ tr(\wedge^{n-i}(f))\ t^i,$$
where $\wedge^{p}(f)$ is the map induced by $f$ on grade $p$ of $V$'s exterior algebra.
While this formula is rarely mentioned (at least I haven't seen it in any of the standard textbooks), it is not too surprising if you have a good grasp of exterior algebra. It presents $\chi(f)$ as a generating function for the exterior traces of $f$.
My question is whether these traces have a simple geometric interpretation on par with $tr$ and $det$.
| https://mathoverflow.net/users/2036 | Geometric interpretation of characteristic polynomial | A rather simple response is to differentiate the characteristic polynomial and use your interpretation of the determinant.
$$det(I-tf) = {t^n}det(\frac{1}{t}I-f) = (-t)^ndet(f-\frac{1}{t}I)= {(-t)^n}\chi(f)(1/t)$$
So if we let $\chi(f)(t) = \Sigma\_{i=0}^n a\_it^i$, then ${(-t)^n}\chi(f)(1/t) = (-1)^n\Sigma\_{i=0}^n a\_it^{n-i}$
But $I-tf$ is the path through the identity matrix, and $Det(A)$ measures volume distortion of the linear transformation $A$.
$$det(I-tf)^{(k)}(t=0) = (-1)^nk!a\_{n-k}$$
and a change of variables ($t\longmapsto -t$) gives (and superscript $(k)$ indicates $k$-th derivative)
$$det(I+tf)^{(k)}(t=0) = (-1)^{n+k}k!a\_{n-k}$$
So the coefficients of the characteristic polynomial are measuring the various derivatives of the volume distortion, as you perturb the identity transformation in the direction of $f$.
$$a\_k = \frac{det(I+tf)^{(n-k)}(t=0)}{(n-k)!}$$
| 75 | https://mathoverflow.net/users/1465 | 33482 | 21,701 |
https://mathoverflow.net/questions/33470 | 12 | In general, the tensor product of two local rings is not local. For example, $\mathbb{C} \otimes\_{\mathbb{R}} \mathbb{C}\ $ is not a local ring.
Let $\mathbb{F}\_{p}$ denote the finite field with $p$ elements. Let $A,B$ be two complete local noetherian $\mathbb{Z}\_p$-algebras with residue field $\mathbb{F}\_p$. Let $m\_A, m\_B$ denote the maximal ideals of $A,B$, respectively.
Question:
Is it true that $A \otimes\_{\mathbb{Z}\_p} B\ $ is a local ring?
Clearly, the ideal $m\_A \otimes B + A \otimes m\_B$ is a maximal ideal of $A \otimes\_{\mathbb{Z}\_p} B\ $ with the residue field $F\_p$. Is it the only maximal ideal of $A \otimes\_{\mathbb{Z}\_p} B\ $?
| https://mathoverflow.net/users/1816 | Is tensor product of local algebras local? | Let $A=\mathbb F\_p[[t]],B=\mathbb F\_p[[u]]$. Then, $1\otimes1-t\otimes u$ is neither in $\mathfrak m\_A\otimes B+A\otimes\mathfrak m\_B$ nor a unit, so it is contained in some other maximal ideal of $A\otimes B$. (Proof that $1\otimes1-t\otimes u$ is not a unit: An element of $\mathbb F\_p[[t]][[u]]$ coming from $A\otimes B$ has the property that its coefficients with respect to $u$ span a finite-dimensional subspace of $\mathbb F\_p[[t]]$, but this fails for the coefficients $t^k$ of $(1-tu)^{-1}=\sum t^ku^k$.)
| 7 | https://mathoverflow.net/users/2035 | 33488 | 21,706 |
https://mathoverflow.net/questions/32322 | 7 | Consider the Koch curve $G \subseteq \mathbb{R}^2$. Clearly $G$ is the invariant set (IS) of the iterated function system (IFS) $\lbrace \phi\_1, \phi\_2, \phi\_3, \phi\_4 \rbrace$. Where (not wanting to jump between $\mathbb{R}^2$ and $\mathbb{C}$ but doing so for ease):
$\phi\_1(x) = \frac{1}{3} x$,
$\phi\_2(x) = \frac{1}{3} (x \exp(\frac{i \pi}{3}) + 1)$,
$\phi\_3(x) = \frac{1}{3} (x \exp(-\frac{i \pi}{3}) + 1 + \exp(\frac{i \pi}{3}))$,
$\phi\_4(x) = \frac{1}{3} (x + 2)$
However can we do better? i.e. can we find an IFS consisting of fewer contractions such that its IS is $G$?
In this case, yes. The IFS $\lbrace \psi\_1, \psi\_2 \rbrace$ also has $G$ as its IS where:
$\psi\_1(x) = \frac{1}{\sqrt{3}} x \exp(-\frac{5 i \pi}{6}) + \frac{1}{3} (1 + \exp(\frac{i \pi}{3}))$,
$\psi\_2(x) = \frac{1}{\sqrt{3}} x \exp(\frac{5 i \pi}{6}) + 1$
And as we know that an IFS consisting of a single contraction has a single point as its IS, we know that this is the best that we can do.
But what about in general?
>
> If $G \subseteq \mathbb{R}^n$ is the IS of the IFS $\lbrace \phi\_1, \phi\_2, \ldots, \phi\_m \rbrace$ when can we tell if there exists an IFS with $G$ as its IS and consisting of strictly less than $m$ contractions?
>
>
>
As a specific example: how about the Sierpinski gasket / carpet? Can we do better that the obvious 3 / 8 construction IFS?
| https://mathoverflow.net/users/3121 | Minimum number of contractions needed to obtain a particular invariant set | An interesting question. Of course there is some ambiguity in the formulation "when can we tell".
Certainly in explicit examples, one may be able to apply ad-hoc methods. For example, things are easier for the Sierpinski gasket and carpet, since these have identifiable features in terms of their complementary regions.
For example, if we wish to write the Sierpinski gasket as a union of smaller copies, it should be fairly easy to see that each complementary region must be mapped to another complementary region. But this means that each contraction must correspond to one of the "smaller triangles" that appear in the usual gasket contraction, and we need at least three of these to make up the whole gasket.
The same type of argument should work for the Sierpinski carpet.
EDIT: Let me provide a few additional arguments to illustrate what I mean in the case of the Sierpinski gasket and carpet.
Lemma
-----
Any equilateral triangle contained in the Sierpinski gasket is the triangle surrounding one of the "children" in the usual construction.
Proof
-----
An exercise for the reader. (Hint: Note that the only way for a straight line in the gasket to begin in one of the standard triangles but not end there is to pass through the two points by which it is connected to the rest of the gasket.)
Corollary
---------
If $A$ is an affine similarity that maps the Sierpinski gasket into itself, then $A$ can be written as a finite composition of the three maps from the "standard" IFS that generates the gasket.
---
A similar argument works for the carpet. Here it is not enough to consider the outer square, but if we add the first generation squares to it, the same works. To state this, let A be the union of the boundaries of nine squares that are joined together to form a larger square; e.g.
$$A=\{(x,y)\in[0,3]^2: x\in\{0,1,2,3\} \text{ or } y\in\{0,1,2,3\}\}$$.
Let us call any image of A under an affine similarity a "3-by-3 grid".
Lemma
-----
The outer boundary of any nine-by-nine grid contained in the Sierpinski gasquet is the boundary of one of the squares occuring in the usual construction.
Again, I will leave the proof as an exercise. The claim that the usual system is optimal then follows immediately once more.
| 1 | https://mathoverflow.net/users/3651 | 33492 | 21,709 |
https://mathoverflow.net/questions/33489 | 12 | Let $X$ be a locally ringed space (or a scheme) and $M,N$ two $\mathcal{O}\_X$-modules such that $M \otimes N \cong \mathcal{O}\_X$. Does it follow that $M$ is invertible in the usual sense, namely that $M$ is locally free of rank $1$?
It is true if $M$ is locally of finite type (which is, of course, also necessary).
Proof: Let $x \in X$. Then $M\_x \otimes N\_x \cong \mathcal{O}\_{X,x}$. Now tensor with the residue field of $\mathcal{O}\_{X,x}$ and use linear algebra to conclude that $M\_x / \mathfrak{m}\_x M\_x$ is $1$-dimensional. Since $M\_x$ is of finite type over $\mathcal{O}\_{X,x}$, Nakayama shows that $M\_x$ is generated by just one element. Since $M$ is of finite type in a neighborhood of $x$, it follows that the generator at $x$ is also a generator in a neighborhood of $x$. Also $N$ has one generator, and their tensor product is a generator of $M \otimes N \cong \mathcal{O}\_X$, which must be free. Thus also the generators of $M$ and $N$ are free.
But I don't know what happens in the general case. Here are some intermediate questions:
* Does it follow that $M$ is flat?
* Is the resulting morphism $M \to Hom(N,\mathcal{O}\_X)$ an isomorphism?
* Is the claim true for $X$ a point, i.e. a local ring?
* Is the claim true if $X$ is an affine scheme and $M,N$ are quasi-coherent? (Thus in the question, replace $\mathcal{O}\_X$ by a usual ring.)
| https://mathoverflow.net/users/2841 | Justification of the term "invertible sheaf" | Any $M'\to M$ that induces an epi after tensoring with $N$ must also induce an epi after tensoring with $N\otimes M$ and therefore must be an epi. And locally such a finitely generated $M'$ must exist.
| 6 | https://mathoverflow.net/users/6666 | 33497 | 21,713 |
https://mathoverflow.net/questions/32444 | 2 | Ie, is there a way to probe it for regions of depth that involves a function, the domain of which is the Mandelbrot set itself, or a part of that set?
| https://mathoverflow.net/users/7739 | Is there a way to find regions of depth in the Mandelbrot set other than simply poking around? | It is a little bit difficult to answer the question as posed, because there is a question as to what you mean by "depth".
One of the previous answers mentions Misiurewicz points - parameters where the critical orbit is pre-periodic. Examples are the "branch points" and "tips" in the Mandelbrot set. However, these are not likely to be good candidates for what you are looking for. Indeed, rescalings of the Mandelbrot set near such a parameter will converge, with the same limit as corresponding rescalings in the dynamical plane, by a theorem of Tan Lei. This does not seem like what you are looking for.
If you pick a "Feigenbaum point" (an infinitely renormalizable parameter of bounded type, such as the famous Feigenbaum value which is the limit of the period-2 cascade of bifurcations), then Milnor's hairiness conjecture, proved by Lyubich, states that rescalings of the Mandelbrot set converge to the entire complex plane. So there is certainly a lot of thickness near such a point, although again this may not be what you are looking for. It may also prove computationally intensive to produce accurate pictures near such points, because the usual algorithms will end up doing the maximum number of iterations for almost all points in the picture.
In this case you are left with either non-renormalizable parameters (those that do not belong to any small copies of the Mandelbrot set) or infinitely renormalizable parameters with interesting behaviour.
An example of the formal is given by the so-called "Fibonacci parameter"; see "Parameter scaling for the Fibonacci point" by Wenstrom for information (and a picture) about the parameter space structure near this point. Rodrigo Perez investigated the structure of parameter space near such pieces in much more detail.
To imagine creating a zoom sequence for infinitely renormalizable parameters, imagine the following: Begin with a Misiurewicz parameter c\_0. Then pick a little Mandelbrot copy M\_1 near c\_0, and choose a Misiurewicz point c\_1 in this little copy. Pick a little Mandelbrot copy M\_2 very close to c\_1, and again a Misiurewicz point c\_2 in there, and so on. Zooming in on the limit parameter is essentially the same as zooming in to the Misiurewicz parameter c\_0, then changing track and zooming into the little Mandelbrot copy M\_1, centering on c\_1, and so on. Of course instead of using Misiurewicz points, you could at various types use any other types of points that are on the boundary of the Mandelbrot set (boundary points of interior components, Feigenbaum parameters, etc).
| 8 | https://mathoverflow.net/users/3651 | 33499 | 21,714 |
https://mathoverflow.net/questions/32442 | 4 | Using Matlab, how to generate a net of 3^10 points that are evenly located (or distributed) on the 8-dimensional unit sphere?
Thanks for any helpful answers!
| https://mathoverflow.net/users/7738 | How to generate a net on a 8-dimensional sphere | If it's really important for the points to be evenly distributed, and you don't mind doing a lot of calculation to get them that way, you can start with a randomly distributed set and then iterate over the entire set repeatedly, allowing each point in turn to make whatever small adjustment improves your chosen definition of uniformity, and repeat this until the set of points converges. If you're even pickier than that, and not satisfied by just a locally optimal arrangement, the canonical next thing to try is simulated annealing.
For picking points at random, I agree with Peter Shor that taking the time to implement a one-to-one volume-preserving map from a product of intervals to a high-dimensional sphere would be much more wasteful (of time; you would learn a lot) than throwing away 98% of your random numbers. It's an interesting question, though, whether systematically chosen points in a product of intervals can be well-distributed under one of these volume-preserving (but distance-destroying) maps. The first interesting case of such a map is the axial projection from the curved surface of a cylinder of height 2 and radius 1 to the surface of the unit sphere it contains: projecting straight to the axis, one direction gets stretched out in exact counterbalance to the compression of the other direction. Call the coordinates of the cylinder surface *z* ∈ [$-1$, $1$] and *θ* ∈ [$0$, $2\pi$]. Choosing an ordinary regular rectangular grid in *z* and *θ* does terrible things to the projection. On the other hand, for any $N$, setting *zi* = $(-N+2 i - 1)$/$N$ and *θi* = $2\pi (\phi i$ mod $1$), where $\phi$ is the golden mean, actually gives a very nice distribution of points after projection. It's possible that in any dimension there is such a lattice in the cube that projects nicely, for any *N*, to the sphere.
| 4 | https://mathoverflow.net/users/7936 | 33501 | 21,716 |
https://mathoverflow.net/questions/33477 | 6 | I just took a look at the nlab entry: [Nikolai Durov](http://ncatlab.org/nlab/show/Nikolai+Durov). It seems that Skoda never mentioned that what Durov introduced was a special case of generalized scheme theory. I did not read his [dissertation](http://arxiv.org/abs/0704.2030) carefully or completely. I wonder whether his "generalized scheme" is a special case of noncommutative scheme in the sense of A. Rosenberg.
I will give a talk on noncommutative schemes in a few days. Now I am collecting interesting examples of noncommutative schemes (quasi schemes). Examples that I already know are:
commutative schemes; D-modules; quantum D-modules; almost schemes by Gabber; general Grothendieck category (abelian category); Artin-Zhang noncommutative projective schemes; quantum flag variety in the sense of Rosenberg and Lunts; holonomic D-modules.
Thanks!
| https://mathoverflow.net/users/1851 | Did Durov's work give an example of noncommutative schemes? | No, there is no need for the Rosenberg [noncommutative scheme](http://ncatlab.org/nlab/show/noncommutative+scheme) that the categories be abelian in general; whatever being said in his 1998 paper he does not mean so. He defines a relative scheme over a base category given by exactness properties of direct and inverse image functors describing covers used for gluing and the affinity property. In general one has to be careful, with formulating correctly the exactness properties for nonabelian context.
Of course, to justify this one needs to say that the Durov's schemes are determined by the categories of quasicoherent modules. I do not know if the reconstruction theorem a la Gabriel-Rosenberg holds for the generalizes schemes of Durov. Durov glues schems along categorical localizations which he calls for some reason "pseudolocalizations". They just have the correct exactness properties. On the other hand, for commutative monads, Durov has two nice versions of prime spectra; now one should compare those with a version of Rosenberg's spectrum for nonabelian context. Now one version if the spectrum for right exact categories of Rosenberg. What is a right exact structure for the case of categories like the ones in Durov's work ? Well Durov has spent some time to develop a theory of [vectoids](http://ncatlab.org/nlab/show/vectoid) which generalize topoi, but also the categories of modules over finitary monads in Set and, more generally, the categories of quasicoherent sheaves of $\mathcal{O}$-modules over generalized schemes. Durov wrote a draft text in Russian about vectoids (which I have seen but is not released yet), and a video of a talk at Steklov. Is there a canonical right exact structure on a vectoid for which the spectrum of the right exact category in the sense of Rosenberg gives a sensible reconstruction theorem, it would be very interesting to investigate.
Remark: I believe that that the quantum flag variety of Lunts-Rosenberg is isomorphic in the case of $SL\_n$ to the flag variety studied in my thesis from the dual point of view and with explicit Ore localizations. I never had time to check and publish all the details.
| 6 | https://mathoverflow.net/users/35833 | 33514 | 21,724 |
https://mathoverflow.net/questions/33513 | 4 | Let $A$ be a local ring with nilpotent maximal ideal $\mathfrak{m}$ (i.e., some power of $\mathfrak{m}$ vanishes), and $M$ an $A$-module (not necessarily finitely generated). Let $\bar{S}\subset M/\mathfrak{m}M$ be a set of generators and $S$ a set of representatives of $\bar{S}$ in $M$. Then is it true that $S$ is a set of generators of $M$? This is a common form of Nakayama's lemma with the assumption of finite generation of $M$ replacing the nilpotence of $\mathfrak{m}$. A passage in Matsumura's book "Commutative Ring Theory" (see Theorem 7.10) seems to imply this result, and I can't figure out why.
| https://mathoverflow.net/users/5292 | Non-finite version of Nakayama's lemma? | Dear Kwan,
Let $N$ be the submodule of $M$ generated by $S$. Then by assumption
$M = N +\mathfrak m M.$ Iterating this, we find that
$$M = N + \mathfrak m (N + \mathfrak m M) = N + \mathfrak m^2 M = \cdots
= N + \mathfrak m^n M$$
for any $n > 0.$ If we take $n$ large enough then $\mathfrak m^n = 0$ (by hypothesis).
Thus $M = N,$ as desired.
P.S. I've found this to be quite a useful fact!
| 19 | https://mathoverflow.net/users/2874 | 33517 | 21,726 |
https://mathoverflow.net/questions/31279 | 10 | I've decided to start a wiki to do collaborative mathematics. However I don't have access or control over a server. So I need a wiki farm. I've tried out pbworks and wikidot, but their latex support is not as straightfoward as say wordpress.
Do you have a suggestion of which wiki farm to use?
| https://mathoverflow.net/users/nan | Suggestions for wiki farm with good latex support | If you're looking for a wiki that can handle LaTeX-style equations, then you should take a look at instiki. Not only does it display mathematics properly, it can also export pages to LaTeX.
<http://www.instiki.org/show/HomePage>
| 4 | https://mathoverflow.net/users/45 | 33521 | 21,730 |
https://mathoverflow.net/questions/30493 | 2 | Let $P$ be a (finitely generated) pro-$p$ group, and let $E$ be an infinite elementary abelian normal subgroup. Does $E$ necessarily contain a non-trivial finite normal subgroup of $P$? We can think of $E$ as consisting of sequences of elements of $C\_p$, with open subgroups $O\_X$, where $X$ is a finite subset of the indexing set and $O\_X$ consists of the sequences that are zero on $X$. However, I can't think of a way of making $P$ act on these sequences that doesn't leave some finite subgroup invariant. Acting on the indexing set is no good because the orbits would have to be finite, and you'd have a finite normal subgroup consisting of sequences that are zero outside some given orbit.
| https://mathoverflow.net/users/4053 | Elementary abelian normal subgroups of a pro-p group | After talking to Charles Leedham-Green, I now have an example that answers the question (I think). See <http://mathoverflow.net:80/questions/33533/name-this-pro-p-group>. More interesting examples would still be nice though, particularly if they do not have $C\_p \wr C\_p$ as an image.
| 1 | https://mathoverflow.net/users/4053 | 33536 | 21,740 |
https://mathoverflow.net/questions/33540 | 4 | Does a module (over a commutative ring) always possess a minimal generating set? When the module is not finitely generated, the typical Zorn's lemma type argument doesn't seem to work. More precisely, if $(S\_{\alpha})\_{\alpha}$ is a chain of generating subsets of a module $M$, $M=\cap \_ {\alpha}(S \_ {\alpha})\supset (\cap\_{\alpha}S\_{\alpha})$ is not in general (I think) equality (the parentheses in the last equation indicate submodule generated by).
| https://mathoverflow.net/users/5292 | Existence of a minimal generating set of a module | $\mathbb{Q}$, as a $\mathbb{Z}$-module, has no minimal generating set.
By the way, in the paper "A characterization of left perfect rings" by Yiqiang Zhou, it is proven that a ring $R$ is left perfect if and only if every generating set of some $R$-module contains a minimal generating set.
| 9 | https://mathoverflow.net/users/2841 | 33542 | 21,742 |
https://mathoverflow.net/questions/33468 | 10 | I seem to remember reading somewhere that ZF+AD proves that $\omega\_1$ and $\omega\_2$ are measurable cardinals.
Is that right?
If so, can someone [point me to or give here] a [sketch or proof] of these results?
| https://mathoverflow.net/users/nan | Measurable cardinals under Axiom of Determinacy | An alternative proof for $\omega\_1$ goes via the set $D$ of Turing degrees (also known as degrees of unsolvability). $D$ is upward directed by the ordering $\leq\_T$ of Turing reducibility, so the cones $C\_d=\{p\in D:d\leq\_Tp\}$ generate a filter on $D$. The axiom of determinacy implies that this filter is an ultrafilter, i.e., that every subset $X$ of $D$ either includes a cone or is disjoint from a cone. (Proof: Define a game where the two players alternately choose natural numbers, producing an infinite sequence $x$; player 1 wins iff the Turing degree of $x$ is in $X$. Any strategy for either player, being a function from finite sequences of natural numbers to natural numbers, can be coded as a sequence $s$ of natural numbers. Any Turing degree $p$ in the cone $C\_s$ [pedantically, the subscript should be the degree of $s$] is the degree of some play in which the player who has strategy $s$ uses it, because the other player can play any sequence of degree $p$ and then the overall play will also have degree $p$. So, if $s$ is a winning strategy for player 1 (resp. player 2), then all (resp. none) of the degrees in $C\_s$ must be in $X$.) So AD gives an ultrafilter on $D$, called the cone ultrafilter. It's non-principal (because there is no highest Turing degree) and countably complete (because AD implies that all ultrafilters are countably complete --- this follows from the fact that there are, under AD, no non-principal ultrafilters on $\omega$).
To get from $D$ to $\omega\_1$, use the function $f$ assigning to each Turing degree $d$ the ordinal usually called $\omega^{CK}(d)$, the first ordinal that is not the order-type of an ordering of $\omega$ of degree $\leq\_Td$. The image of the cone ultrafilter under this map $f$ is again a countably complete ultrafilter, now on $\omega\_1$. It is non-principal because $f$ takes arbitrarily large countable ordinals as values on any cone.
The ultrafilter obtained in this way is actually the same as the club filter mentioned in other answers, but the proof via Turing cones needs somewhat less recursion theory than Solovay's proof (which used Kleene's boundedness theorem for $\Sigma^1\_1$ sets).
One can also prove the measurability of $\omega\_2$ by projecting the cone filter from $D$ to $\omega\_2$ via a suitable map. (I don't remember right now whether the suitable map sends a degree $d$ to the successor cardinal in $L[d]$ of (genuine) $\omega\_1$ or to the next Silver-indiscernible over $d$ after (genuine) $\omega\_1$; maybe both of them work.)
| 15 | https://mathoverflow.net/users/6794 | 33546 | 21,743 |
https://mathoverflow.net/questions/33541 | 2 | Suppose we are working in an "arrows-only" definition of a category such as given in
Mac Lane's "Categories of the Working Mathematician" (1998) p.279 or on
[nlab](http://ncatlab.org/nlab/show/single-sorted+definition+of+a+category). How can we formulate the definition of a product in such a category? I do not see how to do this without implicitly referring to objects (identity arrows).
| https://mathoverflow.net/users/7779 | What would be an "arrows-only" defintion of a product in a category? | Using the notation of nlab, the following is a fibered product: if $x, y$ are arrows with $t(x) = t(y)$, then their fiber product is the pair of arrows $u, v$ with $t(u) = s(x)$, $t(v) = s(y)$, and $s(u) = s(v)$ such that for any pair of arrows $a, b$ with $s(a) = s(b)$, $t(a) = t(u) \, (= s(x))$ and $t(b) = t(v) \, (= s(y))$ and such that $x \circ a = y \circ b$, there is a unique arrow $c$ having $s(c) = s(a) = s(b)$, $t(c) = s(u) = s(v)$, and $a = u \circ c$, $b = v \circ c$.
To define a plain product, suppose the category has a final object (that is, of course, that there exists an arrow $f$ such that for any arrow $x$ there exists a unique arrow $x'$ with $s(x') = s(x)$ and $t(x') = f$) and replace $x$ and $y$ by $x'$ and $y'$ in the above. If it doesn't have one, of course you can just add one.
Can't get any farther away from identity arrows than that; you need to be able to specify sources and targets to define composition.
| 3 | https://mathoverflow.net/users/6545 | 33547 | 21,744 |
https://mathoverflow.net/questions/33539 | 12 | Suppose that $G$ is a finitely presented group and $H$ is a finitely generated normal subgroup such that $G/H$ is infinite cyclic. Is it true that $H$ is finitely presented?
| https://mathoverflow.net/users/7307 | Finitely generated subgroups with infinite cyclic quotient | No. [Ollivier & Wise's version](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=ollivier&s5=wise&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq) of the Rips Construction gives, for any finitely presented group $Q$, a finitely presented group $G$ of cohomological dimension 2 and a surjection $G\to Q$ such that the kernel $K$ satisfies:
1. $K$ is finitely generated; and
2. $K$ has Kazhdan's property T, in particular $K$ has at most one end.
Now it follows from Theorem 5.3 of [a paper of Bieri](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=bieri%252C%2520r%2a&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=39&mx-pid=390078) that $K$ is only finitely presented if $Q$ is finite.
**Note:** In my original answer, I only mentioned [the unadulterated Rips Construction](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=rips%252C%2520e%2a&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=18&mx-pid=642423). Using Ollivier and Wise's version is overkill, but it makes the application of Bieri's theorem cleaner.
I should also mention another, famous and beautiful (though I suppose less general) counterexample. In its simplest cases this example is more elementary.
Given a flag complex $L$, [Bestvina & Brady](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=bestvina&s5=brady&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq) consider the corresponding right-angled Artin group $A\_L$ and the kernel $K\_L$ of the map $A\_L\to\mathbb{Z}$ that sends each generator to $1$. They prove:
1. $K\_L$ is finitely generated if and only if $L$ is connected; and
2. $K\_L$ is finitely presented if and only if $L$ is simply connected.
So just take $L$ to be your favourite connected, non-simply connected flag complex to construct a counterexample. The square graph with four vertices and four edges is a good choice for $L$, in which case $A\_L$ is just the direct product of two copies of the free group on two generators. In this simple case, it's easy to see that $K\_L$ is finitely generated; one should be able to prove (though I haven't tried) that $K\_L$ is not finitely presented by messing around with some spectral sequences...
| 15 | https://mathoverflow.net/users/1463 | 33549 | 21,746 |
https://mathoverflow.net/questions/31007 | 22 | What are the units in $R[X,X^{-1}]$, where $R$ is a commutative ring with $1$? I know that the question for polynomial rings is a standard textbook exercise. However, I couldn't find a reference for Laurent polynomials, since most people only seem to consider coefficients in an integral domain.
| https://mathoverflow.net/users/7422 | What are the units in the ring of Laurent polynomials? | You can find a more general result in the [paper [1]](http://dx.doi.org/10.1007/s10440-008-9370-8), which determines the units and nilpotents in arbitrary group rings $\rm R[G]$ where $\rm G$ is a unique-product group - which includes ordered groups. As the author remarks, his note was prompted by an earlier [paper [2]](http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf) which explicitly treats the Laurent case.
[1](http://dx.doi.org/10.1007/s10440-008-9370-8) Erhard Neher. Invertible and Nilpotent Elements in the Group Algebra of a Unique Product Group
Acta Appl Math (2009) 108: 135-139
<http://dx.doi.org/10.1007/s10440-008-9370-8>
<http://homepage.uibk.ac.at/~c70202/jordan/archive/note/note.pdf>
[2](http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf) Ottmar Loos. Remarks on Holger P. Petersson's "Idempotent 2-by-2 matrices"
<http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf>
| 7 | https://mathoverflow.net/users/6716 | 33550 | 21,747 |
https://mathoverflow.net/questions/33504 | 5 | I got to thinking about this problem while sifting through the [math puzzles for dinner](https://mathoverflow.net/questions/29323?sort=votes&page=1#sort-top) thread. There's a fun puzzle by rgrig which asks the guests to prove that when they came to dinner two of them shook hands the same amount of times. The solution is to make the handshake graph, and apply the pigeonhole principle on the vertex degrees. The key observation is that if there are $n$ guests, then the degrees 0 and $n-1$ cannot both occur.
Yaakov Baruch made the astute comment that the result is false if the forgetful mathematicians shake hands twice with each other. However, note that mathematicians are probably not so forgetful to shake hands *three* times with someone. This leads us to the following questions:
**Question 1.** Is it true that for large $n$, every loopless multigraph on $n$ vertices with at most two parallel edges between any two vertices has two vertices of the same degree? This is false for $n=3,4,5,6,7$.
**Question 2.** Is there a nice characterization of such degree sequences?
I'll finish with what is known. If we allow loops, then the problem is trivial, because all sequences whose sum is even are degree sequences. To see this, put a matching on the odd degree vertices, and then add loops. For simple graphs, there are many nice characterizations. For example, the Erdos-Gallai theorem says that a decreasing sequence $(d\_1, \dots, d\_n)$ is a degree sequence of a simple graph if and only if $\sum\_i d\_i$ is even and for all $k \in [n]$
\[
\sum\_{i=1}^k d\_i \leq k(k-1)+ \sum\_{i=k+1}^n \min (k, d\_i).
\]
This sort of answers Question 2, since a degree sequence of a multigraph with multiplicity at most 2 is the sum of two degree sequences of simple graphs. However, this is a rather convoluted characterization, and it is unclear how to answer Question 1 from it.
I'll end by mentioning that if we do not bound the multiplicity of edges, then there is a nice characterization of degree sequences of multigraphs by Hakimi (1962).
| https://mathoverflow.net/users/2233 | Degree sequences of multigraphs with bounded multiplicity | The answer to Question 1 is no. Here's a construction. First we construct a simple graph $G\_n$ on $n$ vertices for each $n\geqslant3$ which has every degree from $1$ to $n-1$ inclusive occurring, with the degree $n-1$ occurring only once. You do this recursively: to construct $G\_n$, just take the complement of $G\_{n-1}$ and add a vertex joined to all the others.
Now to construct the desired multigraph: label the vertices of $G\_n$ as $v\_1,\dots,v\_n$ in such a way that the degrees of these vertices are increasing; then these degrees are $1,2,\dots,i,i,i+1,\dots,n-1$ for some $i < n-1$. Now just duplicate the edge from $v\_j$ to $v\_n$ for $j=i+1,\dots,n-1$, and you have your multigraph with distinct degrees $1,2,\dots,n-1,2n-i-2$.
| 3 | https://mathoverflow.net/users/6771 | 33562 | 21,755 |
https://mathoverflow.net/questions/33545 | 3 | I have a question about surgery.
Let $G= \mathbb{Z}\_m \times \mathbb{Z}$ and $M$ be a oriented 3-manifold with G-action. i.e. There exists a map $f\colon M/G \to BG$, where $BG$ is classifying space.($BG=K(G,1)=K(\mathbb{Z}\_m \times \mathbb{Z},1)).$ Therefore, we can regard $(M/G,f)\in \Omega^{SO}\_3(K(G,1))$, where $\Omega^{SO}\_3(X)$ is (3-dimensional oriented) bordism group of $X$. By using Leray-Serre spectral sequence, we can easily see that $\Omega^{SO}\_3(K(G,1))=H\_3(G)=\mathbb{Z}\_m$ (By using Kunneth fomula) Therefore, $\exists$ r>0 such that $r(M/G,f)$ is nullbordant over $G$. (we can choose $r =m$) That is, there exists a 4-manifold $V$ with G-action such that $\partial V = M$ and $\partial (V/G) = M/G$ and induced map $g\colon V/G\to BG$ is an extension of $f\colon M/G \to BG$.
$\underline{Question}$ : Can I do a surgery on $V/G$ to make $\pi\_1(V/G)=G$ without losing $G$ action on $V$?
| https://mathoverflow.net/users/7776 | Equivariant Surgery problem | Yes. You need to extend the map to $BG$ over the surgery cobordism, which is possible.
First let me add that all your $G$-actions seem to be free and that the boundary of $V$ would consist of $r$ copies of $M$.
Now for the surgeries:
first take connected sum of V/G with 2 copies of $S^1\times S^2$'s (surgeries on embeddings $S^0\times D^3$) and extend the map to BG as $S^1\times S^2\to S^1$ and then to generators of $G=\pi\_1(BG)$ to make $V'/G\to BG$ 1-connected. Then find $S^1$'s in $V'/G$ generating the kernel of $\pi\_1$, thicken to embeddings of $S^1\times D^2$ (they have trivial normal bundle) and do surgery on them, extending the map to $BG$ by nullhomotopies of their images in $BG$.
| 2 | https://mathoverflow.net/users/1090 | 33565 | 21,757 |
https://mathoverflow.net/questions/33563 | 14 | I recently read the model-theoretic proof of the Nullstellensatz using quantifier elimination (see www.msri.org/publications/books/Book39/files/marker.pdf). I'm convinced that the Nullstellensatz is true, i.e. by showing that $\exists y\_1 \cdots \exists y\_n (\bigwedge\_{i} (f\_i(y) = 0)$ is equivalent to a quantifier free formula, hence if it is true in an algebraically closed extension, it is true in the original algebraically closed field. However, I don't have any intuition for how this formal sequence of steps leads to the truth of the Nullstellensatz. In particular, what I don't see is what the quantifier-free formulas are that these are equivalent to. For example, what is a quantifier-free formula in $a,b,c,d,e,f,g,h,i,j,k,l$ which is equivalent to the system $ax^2+bxy+cy^2+dx+ey+f=gx^2+hxy+iy^2+jx+ky+l=0$ having a non-trivial solution? And generalizations?
Can one simplify the model-theoretic proof by showing more directly that the existence of solutions to systems of equations (or non-equations) in algebraically closed fields are equivalent to polynomial conditions on the coefficients? I.e. apply essentially the same argument but restrict to the case of algebraically closed fields and avoid general results like Godel's Completeness Theorem to make the argument more clear. My expectation is that this might involve proving an algorithm for determining the quantifier-free formula from the formula with quantifiers.
| https://mathoverflow.net/users/1355 | Intuition for Model Theoretic Proof of the Nullstellensatz | Your example is not an illuminating one, because you used two equations in two variables. You would therefore expect there, generically, to be a solution for most values of $(a,b,\ldots, l)$. I'll get back to your example, but let's start with the more informative case of two equations in one variable. Then I'll use this as an excuse to talk about properness.
---
Starting very small, when do $ax+b=0$ and $cx+d=0$ have a common root? Answer: We must have $ad-bc =0$. In addition, we must either have $a$ and $c \neq 0$, or else $a=b=0$, or else $c=d=0$. So, there is a polynomial $ad-bc$ which basically cuts out the location where the two polynomials have a common root, but we then have to remove some smaller strata by inequalities.
More generally, when do $f\_m x^m + f\_{m-1} x^{m-1} + \cdots + f\_0=0$ and $g\_n x^n + g\_{n-1} x^{n-1} + \cdots + g\_0=0$ have a common root? The polynomial which plays the role of $ad-bc$ is the determinant
$$R(f,g) := \det \left( \begin{matrix}
f\_m & f\_{m-1} & \cdots & f\_0 & 0 & \cdots & 0 \\
0 & f\_m & f\_{m-1} & \cdots & f\_0 & \cdots & 0 \\
\ddots \\
0 & \cdots & 0 & f\_m & f\_{m-1} & \cdots & f\_0 \\
g\_n & g\_{n-1} & \cdots & g\_0 & 0 & \cdots & 0 \\
0 & g\_n & g\_{n-1} & \cdots & g\_0 & \cdots & 0 \\
\ddots \\
0 & \cdots & 0 & g\_m & g\_{m-1} & \cdots & g\_0
\end{matrix} \right)$$
There are $n$ rows in the first block and $m$ in the second block. The determinant $R(f,g)$ vanishes if and only if there are polynomials $a$ and $b$, of degree $n-1$ and $m-1$, so that $af+bg=0$. It is easy to see that this is true if and only if either (1) $f$ and $g$ have a common root or (2) $f\_m = g\_n =0$. The little nuisance terms come up in ruling out case (2).
The polynomial $R$ is called the resultant. If you wnat to learn more about this, including generalizations to more polynomials in more variables, the terms to search for are "resultant" and "elimination theory". Standard references are *Using Algebraic Geometry* by Cox, Little and O'Shea and *Discriminants, resultants, and multidimensional determinants* by Gelfand, Kapranov and Zelevinsky.
---
The awkward thing about your chosen example is that the analogue of the resultant is $0$, and you only have correction terms. I don't really feel like working it out.
---
The theory looks much prettier if we use homogeneous polynomials. For example, $ax+by$ and $cx+dy$ have a common nonzero root if and only if $ad-bc=0$, with no special cases. Two homogeneous quadratics in three variables *always* have a common root.
Here is the big result about homogeneous polynomials: Let $S$ be the ring $\mathbb{C}[t\_1, \ldots, t\_k, x\_0, x\_1, \ldots, x\_n]$. We think of the $t$'s as parameters and the $x$'s as the homogeneous coordinates. Let $F\_1(x, t)$, ..., $F\_N(x, t)$ be any collection of polynomials, homogeneous in the $x$'s. Then there are polynomials $G\_1(t)$, ..., $G\_M(t)$ such that, given a point $(s\_1, \ldots, s\_k) \in \mathbb{C}^k$, there is a nonzero common root of the $F\_i(s, x)$ if and only if $G\_1(s) = \cdots = G\_M(s)=0$.
This theorem is better phrased in the language of the Zariski topology. I don't know how much algebraic geometry you've taken. A subset of $\mathbb{C}^k$ is called Zariski closed if it is the zero locus of a set of polynomials. So the statement here is the following:
>
> If we have a Zariski closed subset of
> $\mathbb{CP}^n \times \mathbb{C}^k$,
> then its projection to $\mathbb{C}^k$
> is Zariski closed.
>
>
>
In other words,
>
> The mapping $\mathbb{CP}^n \times \mathbb{C}^k \to \mathbb{C}^k$ is closed.
>
>
>
The corresponding statement is NOT true for $\mathbb{C}^n$. Consider the subset of $\mathbb{C} \times \mathbb{C}$ cut out by $xy=1$. This hyperbola is closed. But the projection onto one of the coordinates is $\{ x \neq 0 \}$, which is not closed.
Now, in fact, a stronger statement is true:
>
> For any variety $B$, the mapping $\mathbb{CP}^N \times B \to B$ is closed.
>
>
>
We express this by saying that $\mathbb{CP}^N$ is universally closed.
If you get far enough in algebraic geometry, you'll run across the theorem that projective space is proper. The definition of proper is universally closed, plus some other conditions which we can ignore for now. So this theorem is the highly abstract way of making the statement about $F$'s and $G$'s above. Hopefully, I have given you something to look forward to as you continue in algebraic geometry!
| 19 | https://mathoverflow.net/users/297 | 33570 | 21,760 |
https://mathoverflow.net/questions/33571 | 3 | I'm studying category theory for the first time in a very succint book for computer scientists (I'm not actually a computer scientist, I'm a physicist, but my interest in cat theory is related to purely functional programming languages). But, as the book is very succint, may be it lacks some information so I have a question.
Supose I have a category with a finite number of objects {a1, a2, ..., an}. Supose also that, if there is a morphism f : ai -> aj connecting two objects, then it is unique. Does this category always correspond to some partial order in the set {a1, a2, ..., an}?
I convinced myself drawing some diagrams that this could be the case (couldn't find a counter example), but I'm not sure.
---
edit:
I see your point. I don't have antisymmetry garanteed.
Suppose I further restrict things: I'm not going to allow a morphism (b -> a) it there's already a (a -> b). Then I fix this, right? It'll be a partial order with no equalities, right?
| https://mathoverflow.net/users/757 | Finite categories and partial orders | Almost, this has nothing to do with finiteness: any category where the
homsets have at most one element each is a preorder. Define
$a\le b$ if there is an arrow from $a$ to $b$. Then $\le$ is reflexive
and transitive, by the category axioms. But it may fail to be
antisymmetric: one may have $a\le b\le a\ne b$. But a preorder
induces a partial order on the equivalence classes with respect to
the relation $a\sim b$ if $a\le b\le a$: in the category these
are isomorphism classes of elements.
| 3 | https://mathoverflow.net/users/4213 | 33573 | 21,762 |
https://mathoverflow.net/questions/33522 | 36 | The following statement is well-known:
Let $A$ be a commutative Noetherian ring and $M$ a finitely generated $A$-module. Then $M$ is flat if and only if $M\_{\mathfrak{p}}$ is a free $A\_{\mathfrak{p}}$-module for all $\mathfrak{p}$.
My question is: do we need the assumption that $A$ is Noetherian? I have a proof (from Matsumura) that doesn't require that assumption, but the fact that other references (e.g. Atiyah, Wikipedia) are including this assumption makes me rather uneasy.
| https://mathoverflow.net/users/5292 | Flatness and local freeness | By request, my earlier comments are being upgraded to an answer, as follows. For finitely generated modules over any local ring $A$, flat implies free (i.e., Theorem 7.10 of Matsumura's CRT book is correct: that's what proofs are for). So the answer to the question asked is "no". The CRT book uses the "equational criterion for flatness", which isn't in Atiyah-MacDonald (and so is why the noetherian hypothesis was imposed there). This criterion is in the Wikipedia entry for "flat module", but Wikipedia has many entries on flatness so it's not a surprise that this criterion under "flat module" would not be appropriately invoked in whatever Wikipedia entry was seen by the OP.
An awe-inspiring globalization by Raynaud-Gruson (in their overall awesome paper, really with authors in that order) is given without noetherian hypotheses: if $A$ has finitely many associated primes (e.g., any noetherian ring, or any domain whatsoever) and if $M$ is a finitely generated flat $A$-module then it's finitely presented (so Zariski-locally free!). See 3.4.6 (part I) of Raynaud-Gruson (set $X=S$ there). By 3.4.7(iii) of R-G, the finiteness condition on the set of associated primes cannot be removed, as any absolutely flat ring that isn't a finite product of fields provides a counterexample. (An explicit counterexample is provided by the link at the end of Daniel Litt's answer, namely a finitely generated flat module that is not finitely presented, over everyone's favorite crazy ring $\prod\_{n=0}^{\infty} \mathbf{F}\_2$.)
| 35 | https://mathoverflow.net/users/3927 | 33574 | 21,763 |
https://mathoverflow.net/questions/33486 | 12 | In an earlier thread I had asked whether or not one can find a smooth 4-dimensional submanifold of $S^4$ whose fundamental group has an unsolvable word problem. The answer is yes, and the reference is in that thread.
see: [Word problem for fundamental group of submanifolds of the 4-sphere](https://mathoverflow.net/questions/32787/word-problem-for-fundamental-group-of-submanifolds-of-the-4-sphere)
I'm curious if one can go a step further. Can one find a smooth $4$-dimensional submanifold $M$ of $S^4$ such that both $\pi\_1 M$ and $\pi\_1 (S^4 \setminus M)$ have unsolvable word problems?
My motivation for this question has to do with the decidability of a classical 3-manifold theory problem. Given a smooth $3$-manifold, determine whether or not it admits a smooth embedding into $S^4$.
It's possible to algorithmically generate all smooth embeddings of $3$-manifolds in $S^4$ using normal surface theory for triangulations of the $4$-sphere ((a) start with a given triangulation of $S^4$, (b) enumerate vertex-normal $3$-manifolds, (c) barycentrically subdivide, goto (b)). This is an unfortunately horribly memory-intensive algorithm, but it's exponential run-time so it could be a lot worse.
Whether or not the embedding problem is algorithmically decidable boils down to whether or not there is a computable function $\beta : \mathbb N \to \mathbb N$ such that if a $3$-manifold $M$ has a triangulation with $n$ tetrahedra, you want to know that you only need to barycentrically subdivide a triangulation of $S^4$ at most $\beta(n)$ times to find $M$ as a vertex-normal 3-manifold in the triangulated $4$-sphere.
Another way to look at this is that perhaps some $3$-manifolds have to embed in $S^4$ in extremely "twisted" ways. If $M$ embeds in $S^4$ such that both $4$-manifolds that it separates $S^4$ into have unsolvable word problem, that qualifies for "really twisted" in my opinion.
This also gets to the issue of invariants. If you want to say that a $3$-manifold does not embed in $S^4$, perhaps invariants sensitive to the fundamental group should be important. But any such invariant would have to be relatable to $\pi\_1$ of the two $4$-manifolds that $M$ separates $S^4$ into. So issues of computability also come up here. At present I know of no obstruction to a $3$-manifold embedding in $S^4$ that sees really "deeply" into $\pi\_1 M$ but it seems like such obstructions are a realistic possibility. Alexander modules are known to play a role as an obstruction. Twisted Alexander modules should play a role as well, for example, although the literature on this is currently contained in the (closely related) knot-concordance world.
Presumably this is an open problem (my apologies) but I'm curious if there are any insights out there.
An offshoot of this question would be whether or not every $3$-manifold that admits an embedding in $S^4$ has *some* embedding in $S^4$ where the two $4$-manifolds it splits $S^4$ into have solvable fundamental group word problems.
| https://mathoverflow.net/users/1465 | 4-manifolds in the 4-sphere such that it, *and* its complement have unsolvable word problem | The answer to the question in the beginning should be YES and it follows from the answer to your previous question. We just need to use the fact that if $G$ has unsolvable word problem then $G\*F$ too, where $F$ is a group and $G\*F$ is the free product.
To construct the example take the solution to the previous question, namely a $4$-manifold $M^4 $embeddable in $S^4$ with unsolvable word problem. Now take an open ball $B^4$ in $M^4$ and cut from it a small copy of $M^4$, that we call $N^4$. Finally dig a wormhole that connects $N^4$ with $S^4\setminus M^4$. This divides the sphere into two connected 4-manifolds each of which is homotopic to a connected sum of two manifolds, one of which has same fundamental group as $M^4$.
| 6 | https://mathoverflow.net/users/943 | 33577 | 21,766 |
https://mathoverflow.net/questions/33510 | 11 | I am writing an article on Fermat's work in number theory and feel uncomfortable everytime I have to write "Fermat's Little Theorem": it's clumsy and belittles the fundamental character of Fermat's result. "Fermat's Theorem" is too ambiguous, and I don't really like acronyms such as Flt or Flit. Has anyone ever seen a better name for this result (or a new suggestion)?
| https://mathoverflow.net/users/3503 | Christening Fermat's Little Theorem | I think you shouldn't change the name. It's universally known as Fermat's Little Theorem, and especially if you're writing a survey or historical article, you're not in a place to try to revolutionize established mathematical nomenclature. There are many instances of unfortunate terminology in mathematics, but in my opinion, once they are in general use, they become part of the lore and the culture. I would make exceptions only in a few cases, such as:
a) it's on the level of adjectives such as "good" and "admissible",
b) it's crediting the wrong person (Cayley numbers, Burnside's lemma), or
c) it's very recent, with the inventor implicitly begging to attach his name to it
And if your life work is going to become known as "Lemmermeyer's dirty trick", well, take it with humor.
| 8 | https://mathoverflow.net/users/4183 | 33578 | 21,767 |
https://mathoverflow.net/questions/31249 | 15 | I define a 1-isomorphism between two groups as a bijection that restricts to an isomorphism on every cyclic subgroup on either side. There are plenty of examples of 1-isomorphisms that are not isomorphisms. For instance, the exponential map from the additive group of strictly upper triangular matrices to the multiplicative group of unipotent upper triangular matrices is a 1-isomorphism. Many generalizations of this, such as the Baer and Lazard correspondences, also involve 1-isomorphisms between a group and the additive group of a Lie algebra/Lie ring.
Consider the following function *F* associated to a finite group *G*. For divisors $d\_1$, $d\_2$ of *G*, define $F\_G(d\_1,d\_2)$ as the number of elements of *G* that have order equal to $d\_1$ and that can be expressed in the form $x^{d\_2}$ for some $x \in G$.
Question: Suppose *G* and *H* are finite groups of the same order such that $F\_G = F\_H$. Does there necessarily exist a 1-isomorphism between *G* and *H*?
Note that the converse is obviously true: if there exists a 1-isomorphism between *G* and *H*, then $F\_G = F\_H$.
Incidentally, *just* knowing the orders of elements does not determine the group up to 1-isomorphism. There are many counterexamples of order 16, with two non-abelian groups (one being the direct product of the quaternion group and the cyclic group of order two, and the other a semidirect product of cyclic groups of order four) having the same statistics on orders of elements as $\mathbb{Z}\_4 \times \mathbb{Z}\_4$, but neither being 1-isomorphic to it because they don't have the same number of squares.
Similarly, just knowing how many elements are there of the form $x^d$ for each divisor *d* of the order is not sufficient to determine the group up to 1-isomorphism. Again, there are counterexamples of order 16.
| https://mathoverflow.net/users/3040 | Order information enough to guarantee 1-isomorphism? | Here is a counterexample of order $32$.
$G$ and $H$ will each have $3$ elements of order $2$ and $28$ elements of order $4$. In both cases all three elements of order $2$ will have square roots. That insures that $F\_G=F\_H$. But in $G$ one of them will have $4$ square roots while the others each have $12$, and in $H$ one of them will have $20$ square roots while the others each have $4$. That rules out a $1$-isomorphism.
Let $Q$ be a quaternion group of order $8$ and let $C\subset Q$ be a (cyclic) subgroup of order $4$. Inside $Q\times Q$ there are three subgroups of index $2$ that contain $C\times C$. Let $G$ be $Q\times C$ and let $H$ be the one that is neither $Q\times C$ nor $C\times Q$.
| 16 | https://mathoverflow.net/users/6666 | 33583 | 21,769 |
https://mathoverflow.net/questions/33543 | 9 | Let $M$ be a filtered module over a filtered algebra $A$, and suppose $gr(M)$ is flat over $gr(A)$, where $gr$ means the associated graded module and algebra, respectively.
What can one say in general about the flatness of $M$ over $A$, or with relevant assumptions (for instance in the above, we should assume both filtrations are complete to avoid dumb counterexamples)? Are there good references for this sort of question? I have played with the various definitions of flatness trying to find an obvious relationship, but I find flatness proofs confusing.
The particular examples I have in mind are comparing $U(\mathfrak{g})$-modules to $S(\mathfrak{g})$-modules, and $D(X)$-modules to $O(T^\*X)$-modules for affine varieties $X$, if it helps. I suspect the answer doesn't depend on any of the details though.
| https://mathoverflow.net/users/1040 | Associated graded and flatness | Let me suppose, as in your examples, that we have a base field $k$.
It is well known that to check that a right $A$-module $M$ is flat it is enough to show that whenever $I\leq\_\ell A$ is a left ideal, the map $M\otimes\_AI\to M\otimes\_A A$ induced by the inclusion $I\to A$ is injective. This condition can be rewritten: $M$ is flat iff for each left ideal $I\leq\_\ell A$ we have $\mathrm{Tor}^A\_1(M,A/I)=0$.
So now suppose $A$ and $M$ are (exhaustively, separatedly, increasingly from zero) filtered in such a way that $\mathrm{gr}M$ is a flat $\mathrm{gr}A$-module.
Pick a left ideal $I\leq\_\ell A$; notice that the filtration on $A$ induces a filtration on the quotient $A/I$. We can compute $\mathrm{Tor}^A\_\bullet(M,A/I)$ as the homology of the homologically graded complex $$\cdots\to M\otimes\_kA^{\otimes\_kp}\otimes\_kA/I\to M\otimes\_kA^{\otimes\_k(p-1)}\otimes\_kA/I\to\cdots$$ with certain differentials whose formula does not fit in this margin, coming from the bar resolution. Now the filtrations on $M$, on $A$ and on $A/I$ all collaborate to provide a filtration of our complex. We've gotten ourselves a positively homologicaly graded with a canonically bounded below, increasing, exhaustive and separated filtration. The corresponding spectral sequence then converges, and its limit is $\mathrm{Tor}^A\_\bullet(M,A/I)$. Its $E^0$ term is the complex
$$\cdots\to\mathrm{gr}M\otimes\_k\mathrm{gr}A^{\otimes\_kp}\otimes\_k\mathrm{gr}(A/I)\to \mathrm{gr}M\otimes\_k\mathrm{gr}A^{\otimes\_k(p-1)}\otimes\_k\mathrm{gr}(A/I)\to\cdots$$
with, again, the bar differential, and its homology, which is the $E^1$ page of the spectral sequence, is then precisely $\mathrm{Tor}^{\mathrm{gr}A}\_\bullet(\mathrm{gr}M,\mathrm{gr}(A/I))$. Since we are assuming that $\mathrm{gr}M$ is $\mathrm{gr}A$-flat, this last $\mathrm{Tor}$ vanishes in positive degrees, so the limit of the spectral sequence also vanishes in positive degrees. In particular, $\mathrm{Tor}^A\_1(M,A/I)=0$.
**NB:** As Victor observed above in a comment, Bjork's *Rings of differential operators* proves in its Proposition 3.12 that $\mathrm{w.dim}\_AM\leq\mathrm{w.dim}\_{\mathrm{gr}A}\mathrm{gr}M$ (here $\mathrm{w.dim}$ is the *flat dimension*) from which it follows at once that $M$ is flat as soon as $\mathrm{gr}M$ is; the argument given is essentialy the same one as mine. I am very suprised about not having found this result in McConnell and Robson's!
| 8 | https://mathoverflow.net/users/1409 | 33584 | 21,770 |
https://mathoverflow.net/questions/32911 | 10 | An important invariant of a knot in $S^3$ is its *Alexander polynomial*, related also to *Reidemeister torsion*. Is there something like that for knotted surfaces in $S^4$? If not, what are the difficulties?
| https://mathoverflow.net/users/5196 | Alexander polynomial or Reidemeister torsion for knotted surfaces? | Yes for $S^2$, and more generally, depending on what you are after.
Given any torsion module $M$ over the PID $Q[t,t^{-1}]$, the order of $M$ is well defined in
$Q[t, t^{-1}]$ up to units, in the usual way. Moreover $M\otimes Q(t)=0$. These two facts are at the heart of why the Alexander polynomial is related to Reidemeister torsion. The way it works is that if $C$ is a f.g. chain complex over $Q[t, t^{-1}]$ its homology is torsion iff $C\otimes Q(t)$ is acyclic. Then the two notions of torsion are related by
$$\prod\_k order(H\_k(C))^{(-1)^k}=\tau(C)$$
If $S^n\subset S^{n+2}$ and $X$ its complement, then $X$ has a $Z$ cover $\tilde{X}$ and hence an exact sequence of chain complexes $0\to C(\tilde{X})\to C(\tilde{X})\to C(X)\to 0.$ The corresponding long exact sequence shows that $H\_k(\tilde{X})$ is torsion for all $k$, since $H\_k(X)=H\_k(S^1)$.
Setting $\Delta\_k$ to be the order of $H\_k(\tilde{X})$ gives the $k$-th Alexander polynomial. You get Reidemeister torsion equals the multiplicative Euler characteristic:
$$\tau= \prod \Delta\_k^{(-1)^k}$$
In the case of $S^1\subset S^3$ Poincare duality relates $\Delta\_2$ and $\Delta\_1$, and $\Delta\_0$ is independent of the knot, so you recover Alexander poly "=" Reidemeister torsion. But in general $\tau$ combines all the $\Delta\_k$.
All this (including how to pick the basis of the acyclic complex) is explained in Milnor's article "infinite cyclic covers."
As far as what happens more generally, all this extends, but you have to be careful when the homology is not torsion, which can happen for surfaces in $S^4$. Then you have to view Reidemeister torsion as a function of the homology , etc. Milnor's other Torsion articles are worthwhile reading for these topics.
| 7 | https://mathoverflow.net/users/3874 | 33587 | 21,772 |
https://mathoverflow.net/questions/32511 | 9 | Edited question:
Are there any other non-trivial \*-homomorphisms between matrix algebras apart from the unitary homomorphisms?
Original question:
Does there exist a surjective (but not bijective) \*-homomorphism between matrix algebras over the complex numbers? If so, are there any nice examples?
(I had not realized the matrix algebras were simple but since they are, the answer to the original question is indeed obvious)
| https://mathoverflow.net/users/6985 | *-homomorphisms between matrix algebras | The algebra $M\_n(\mathbb{C})$ of $n \times n$ complex matrices is Morita equivalent to $\mathbb{C}$. Which implies: Every unital representation of $M\_n(\mathbb{C})$ is isomorphic to a direct sum of copies of the defining representation. Thus your homomorphism $\rho:M\_n \to M\_k$ exists precisely when $k$ is a multiple of $n$, and after a change of basis in the target, $\rho(A)$ is just copies of $A$ on the diagonal. This is so for all homomorphisms, whether or not they are \*-homomorphisms, but the answer for \*-homomorphisms is the same. The only difference is that instead of a general change of basis, everything is equivalent up to a unitary change of basis.
The nonunital homomorphisms are not much more general. Up to a change of basis, you can pad a unital homomorphism with extra rows and columns that are all 0.
There is a similar result for a direct sum of matrix algebras. It is summarized in the concept of a "Bratteli diagram" to describe a homomorphism between two direct sums of matrix algebras. The homomorphism can be thought of as a bin packing -- packing items in bins --- with allowed repetition of the items. The Bratteli diagram shows how many copies of each item (matrix summand of the domain) goes into each bin (matrix summand of the target). In the unital case, the bins have to be filled exactly.
| 16 | https://mathoverflow.net/users/1450 | 33592 | 21,774 |
https://mathoverflow.net/questions/33591 | 0 | If there is an arrow $h: (A\times B) \to (A\times B)$, then, necessarily $h = \left\lt f,g\right\gt$ for some $f: (A\times B) \to A$ and $g (A\times B) \to B$?
The book I'm studying defines a product $A\times B$ to be an object provided with two projecting arrows $\pi\_A : A\times B \to A$ and $\pi\_B: A\times B \to B$ such that, for any other object C and arrows $f: C \to A$ and $g : C \to B$ there's exactly one arrow $\left\lt f,g\right\gt: C \to (A\times B)$ such that:
$\pi\_A \cdot \left\lt f,g\right\gt = f$
$\pi\_B \cdot \left\lt f,g\right\gt = g$
It says nothing about $\left\lt f,g\right\gt$ being the only arrow from $C$ to $A\times B$, but I got that impression later when he asked to prove that all products are isomorphic: I could only prove assuming that any arrow from $A \times B$ to itself was of the form $\left\lt f,g\right\gt$ for some f and g from $A\times B$ to A and B, respectively.
This suggests me that $\left\lt f,g\right\gt$ was meant to be the only possible arrow from C to $A \times B$.
Is this right or did I just misunderstood the book?
| https://mathoverflow.net/users/757 | Products of objects in categories | The reason you're having trouble proving the uniqueness is that you need to use the projections. There is not a unique map from $C$ to $A \times B$. But there is a unique map *given* specified maps from $C$ to $A$ and $B$ (i.e. such that the composition of the map from $C$ to the product with the projection gives the map from $C$ to $A$ or $B$). If $C\_1$ and $C\_2$ are two products, we can use the universal property to construct maps between them. The compositions are then the identity. Why? It is not true that there is only one map from $C\_1$ to itself. BUT, there is only one map from $C\_1$ to itself which, when composed with $\pi\_A$ and $\pi\_B$, gives $\pi\_A$ and $\pi\_B$. If you construct the maps correctly, this will be the case, and you will be able to prove the isomorphism between $C\_1$ and $C\_2$.
| 3 | https://mathoverflow.net/users/1355 | 33593 | 21,775 |
https://mathoverflow.net/questions/33581 | 11 | After reading Cohen and Voronov's notes on string topology, one can find the following construction: Suppose we have a topological space $X$ with continuous action of $S^1$. This means we have a map $\rho: S^1 \times X \to X$. If we choose a fundamental class $[S^1]$ of the circle, we can form the operator $\Delta: H\_\ast(X) \to H\_{\ast+1}(X)$ by setting $\Delta(a) = \rho\_\ast([S^1] \times a)$. For dimensional reasons, $\Delta$ squares to zero and therefore turns $H\_\ast(X)$ into a cochain complex. Let's denote the cohomology of this complex by $H\_\ast^\Delta(X)$. Because in this cohomology we have as representatives homology classes in $X$ which are annihilated by the action of $S^1$ "in a homological sense", one would think this is relevant to the $S^1$-equivariant homology of $X$.
However, a few simple examples show that $\Delta$-cohomology and equivariant homology are certainly not equal. For example, take the point with trivial action, then $H^\Delta\_\ast(pt)$ is $\mathbb{Z}$ in degree 0 and zero in all other degrees, but $H^{S^1}\_\ast(pt) = \mathbb{Z}[a\_2]$ where $|a\_2| = 2$. Another example is $LS^1$.
So my question is: is there a different (hopefully geometric) description of $\Delta$-cohomology? Is it related to $S^1$-equivariant homology in any way?
If this is not possible in the general case, is it at least possible for $X$ of the form $LM$, with $M$ a manifold? My main motivation for considering $\Delta$-cohomology is string topology, where $\Delta$ is also known as the BV-operator.
| https://mathoverflow.net/users/798 | What does this naive attempt at $S^1$-equivariant homology describe? | By $G$-equivariant homology of $X$ you mean the homology of the Borel construction or homotopy orbit space $EG\times\_GX$. (I only mention this because the same phrase can refer to other kinds of homology for $G$-spaces, satisfying a weaker kind of homotopy axiom. This "Borel homology" gives you an isomorphism for every $G$-equivariant map $X\to Y$ that is nonequivariantly a homotopy equivalence.)
Yes, the operator that you call $\Delta$ is related to $S^1$-homology. There is a first-quadrant spectral sequence for the $G$-homology of $X$, coming from the fact that $EG\times \_GX$ is a bundle over $BG$ with fiber $X$. The $E^2$ is $H\_i(BG;H\_j(X)$. It is a spectral sequence of modules for the ring $H^{-\*}BG$. When $G=S^1$, the groups are $E^2\_{2k,j}=H\_j(X)$ if $k\ge 0$ and otherwise $0$, and the differential $d^2\_{2k,j}\to E^2\_{2k-2,j+1}$ is independent of $k$. This first differential $d^2$ is the operator $\Delta$. The $2$-periodicity continues in $E^r$ until the differentials start crossing the vertical axis. So basically your $\Delta$-homology is $E^3=E^4$ and then there is an operator $d^4$ going from $ker(\Delta)/im(\Delta)$ to $ker(\Delta)/im(\Delta)$ and raising $j$ by $3$; and so on.
This is related to cyclic homology.
| 14 | https://mathoverflow.net/users/6666 | 33595 | 21,777 |
https://mathoverflow.net/questions/33602 | 17 | In my answer to [this question](https://math.stackexchange.com/questions/640/why-are-differentiable-complex-functions-infinitely-differentiable/820#820) on MU, I suggested that the OP think about the difference between real-differentiable and complex-differentiable functions by using a sort of finitary analogue. One way to make this precise is the following. The finitary analogue of a real-differentiable function is just a function $f : \mathbb{Z} \to \mathbb{Z}$, whose finitary derivative is the first difference $f(z + 1) - f(z)$. The first difference can be arbitrarily nasty; in particular it does not have to grow in any kind of smooth way, which is a reasonable analogue of what happens for arbitrary real-differentiable functions.
The finitary analogue of (the real part of) a complex-differentiable function, on the other hand, is a discrete harmonic function $f : \mathbb{Z}^2 \to \mathbb{Z}$ e.g. one which satisfies
$$\frac{f(x+1, y) + f(x-1, y) + f(x, y+1) + f(x, y-1)}{4} = f(x, y).$$
Such functions are clearly more constrained than arbitrary functions $\mathbb{Z} \to \mathbb{Z}$. Is there a reasonable finitary statement about the properties of such functions which is analogous to the statement that harmonic functions are smooth?
(The best case would be if one could deduce the infinitary statement from the finitary one, but maybe this is difficult. Also, I can't think of good tags for this question, so feel free to retag.)
| https://mathoverflow.net/users/290 | What is a reasonable finitary analogue of the statement that harmonic functions are smooth? | There are quite a few properties shared by discrete and continuous harmonic functions. They also generalize in various forms to graphs. I don't know if the analogy is so close that there is a unique concept of smoothness for the discrete case, but see the last item for one point of contact.
1. Poisson formula: $f(p)$ is a weighted average of the values of $f$ on a "curve" of points surrounding $p$ (that is, a minimal set of points disconnecting it from infinity). As in the continuous case the weighting is the distribution of hitting probabilities, $h\_p(x,y)$ = probability that a random walk started at $p$ first hits the curve at $(x,y)$.
2. Maximum principle. Because values in the interior are weighted averages of values on the boundary.
3. Growth constraints. A bounded discrete-harmonic function is constant. More generally there is a discrete analogue of the theorem that if a harmonic $f(x,y)$ has polynomial-bounded growth, $|f(x,y)| < C(|x|+|y|+1)^n$ then $f(x,y)$ is a (harmonic) polynomial in $x$ and $y$, of degree at most $n$.
4. Harnack inequalities. It is these that are a form of smoothness. They say that if $a$ and $b$ are nearby points far away from the boundary curve used in the Poisson formula, $f(a) - f(b)$ is small (relative to $|f|$ on the boundary). There are many expressions of this principle, see for example papers of Fan Chung and her collaborators ( www.math.ucsd.edu/~fan ) for inequalities in the case of graphs with a vertex-transitive symmetry group. By phrasing the theory in terms of eigenfunctions of the Laplacian one can give the discrete and continuous Harnack inequalities a very parallel form. But in the discrete case they also seem to play a more basic role of showing that smoothness is present in some form, which in the continuous case is automatic because $f$ is assumed to have at least two derivatives.
| 15 | https://mathoverflow.net/users/6579 | 33610 | 21,787 |
https://mathoverflow.net/questions/33597 | 16 | I'm trying to refresh myself on quantum algorithms and have been skimming Childs and van Dam's [2008 RMP paper](http://arxiv.org/abs/0812.0380v1) among other things. From my preliminary surfing it looks like the known quantum algorithms still essentially fall into (in that their nontrivial quantum aspect is governed by) four major classes, viz.
1. Linear
algebra
2. Quantum Fourier transform and hidden
subgroup problems
3. Quantum search
4. Quantum simulation/annealing
and I'm wondering: does this classification clearly miss anything?
(NB. Because I'm hoping for answers that discuss *why* a given algorithm does not fall into one of these classes, I'm not making this CW.)
| https://mathoverflow.net/users/1847 | Are there any known quantum algorithms that clearly fall outside a few narrow classes? | Does the [Farhi-Goldstone-Gutman game tree evaluation algorithm](http://arxiv.org/abs/quant-ph/0702144) and the extensions of it fall into one of these classes? You might put it in quantum simulation/annealing because of the technique used, but I think that would be a mistake. You might also put it in quantum search because it doesn't give a super-polynomial speed-up, but it has a very different flavor than all the other quantum search algorithms.
| 15 | https://mathoverflow.net/users/2294 | 33612 | 21,789 |
https://mathoverflow.net/questions/33469 | 3 | The following integral came up in one of my applications:
$\int\_{-1}^1P\_n(x)T\_j(x)T\_k(x)\mathrm{d}x$
where $P\_n(x)$ is a Legendre polynomial, $T\_k(x)$ is a Chebyshev polynomial, and $j$, $k$, and $n$ are nonnegative integers.
I want to ask if there might be a closed-form representation for this integral. I have a feeling it will involve gamma functions and Pochhammer symbols, but I seem to be unable to figure out how to proceed.
Alternatively, since I am aware that Legendre polynomials can be expressed as a linear combination of Chebyshev polynomials, it might be easier to instead simplify the integral
$\int\_{-1}^1T\_n(x)T\_j(x)T\_k(x)\mathrm{d}x$
or in trigonometric form
$\int\_{0}^{\pi}\cos(n\theta)\cos(j\theta)\cos(k\theta)\sin(\theta)\mathrm{d}\theta$
but I do not know of any closed form for this either.
I have already tried looking in Abramowitz and Stegun, the DLMF, Gradshteyn and Ryzhik, and the Wolfram functions site to no avail.
(edit:
I had neglected to exploit the identity
$T\_j(x)T\_k(x)=\frac1{2}\left(T\_{j+k}(x)+T\_{j-k}(x)\right)$
when I first formulated my question. I now amend my question to asking for a closed form for
$\int\_{-1}^1P\_n(x)T\_j(x)\mathrm{d}x$
of which the only fact I know about it is that it is 0 if $j<n$ by virtue of the orthogonality of the Legendre polynomial.)
| https://mathoverflow.net/users/7934 | Closed form for an orthogonal polynomial integral? | It turns out the identities I needed for resolving
$\int\_{-1}^1P\_n(x)T\_j(x)\mathrm{d}x$
into a closed form was well-hidden in Abramowitz and Stegun and Gradshteyn and Ryzhik.
As I had mentioned in the edit to my original question, the integral is 0 if $j<n$ by virtue of the orthogonality property of the Legendre polynomials.
I now considered the following integral:
$\int\_{-1}^1P\_n(x)T\_{n+k}(x)\mathrm{d}x\quad k\geq0$
To dispose of an elementary case first, I noted that $P\_n(x)T\_{n+k}(x)$ is an odd function iff $k$ is odd and even iff $k$ is even; the integral is therefore 0 for odd $k$.
The even $k$ case I had solved by making use of two identities: this [series representation](http://functions.wolfram.com/Polynomials/ChebyshevT/06/01/02/01/0009/) for $T\_{n}(x)$ (also in Abramowitz and Stegun as [22.3.6](http://people.math.sfu.ca/%7Ecbm/aands/page_775.htm)), and an integral I derived from a more general form in Gradshteyn and Ryzhik:
$\int\_0^1x^{n+2\rho}P\_n(x)\mathrm{d}x=\frac{\left(2\rho+1\right)\_n}{2^{n+1}\left(\rho+\frac1{2}\right)\_{n+1}}$
where $\left(a\right)\_n$ is the Pochhammer symbol. (The identity actually listed in G&R was an integral for a Gegenbauer (ultraspherical) polynomial, of which the Legendre polynomial is a special case.)
I only needed to retain terms in the series greater than or equal to $n$, again due to orthogonality of the Legendre polynomial. Applying the integral formula to each term (with an additional factor of 2 because the integrand is even), and feeding the resulting sum to *Mathematica* netted the following closed form:
$\int\_{-1}^1P\_n(x)T\_{n+2k}(x)\mathrm{d}x=-\frac1{4}\frac{\left(n+2k\right)\Gamma\left(n+k\right)\Gamma\left(k-\frac1{2}\right)}{\Gamma\left(k+1\right)\Gamma\left(n+k+\frac{3}{2}\right)}$
(The original result returned by *Mathematica* 5.2 had nasty cosecant factors, which I disposed of using the reflection formula for the gamma function).
This can then be applied to the original integral with the three polynomials by exploiting the product-sum identity for the Chebyshev polynomial.
| 5 | https://mathoverflow.net/users/7934 | 33613 | 21,790 |
https://mathoverflow.net/questions/33614 | 23 | Let $D$ be an elliptic operator on $\mathbb{R}^n$ with real analytic coefficients. Must its solutions also be real analytic? If not, are there any helpful supplementary assumptions? Standard Sobolev methods seem useless here, and I can't find any mention of this question in my PDE books.
I began thinking about this because I overheard someone using elliptic regularity to explain why holomorphic functions are smooth. Aside from the fact that I find that explanation to be in poor mathematical taste (I regard the beautiful regularity properties of holmorphic functions as fundamentally topological phenomena), it occurred to me that standard elliptic theory falls short of exhibiting a holomorphic function as the limit of its Taylor series. So I'm left wondering if this is an actual limitation of elliptic regularity which could vindicate and entrench my topological bias.
In the unfortunate event of an affirmative answer to my question, I would be greatly interested in geometric applications (if any).
| https://mathoverflow.net/users/4362 | Does elliptic regularity guarantee analytic solutions? | While probably not the fastest approach I think that Hörmander: The analysis of linear partial differential equations, IX:thm 9.5.1 seems to give a (positive) answer to your question. It is overkill in the sense that it gives you a microlocal statement telling you that for $Pu=f$, $u$ is analytic in the same directions as $f$ is.
| 7 | https://mathoverflow.net/users/4008 | 33615 | 21,791 |
https://mathoverflow.net/questions/28241 | 8 | Can the lattice stick number of a knot be bounded
by the stick number of the knot?
The stick number
$S(K)$ of a knot $K$ is the fewest number of segments
needed to realize it by a simple 3D polygon.
The lattice stick number $S\_L(K)$ is the fewest segments in a realization in the cubic
lattice, with all segments parallel to a coordinate axis.
For example, the stick number of the trefoil knot $K=3\_1$
is 6, and its lattice stick number is 12
(the latter a result of Huh and Oh from 2005).
My question is whether it is possible to bound $S\_L(K)$
by $m S(K)$, where $m$ is some multiplier factor.
Ideally $m$ would be a constant, but perhaps it is more realistic
to expect it to depend
on the complexity of the knot (e.g., on its crossing number
$cr(K)$).
What I have in mind is replacing each stick in a stick realization
by a bounded lattice path.
**Addendum**. Tracy Hall's clever example below indicates that it is unlikely that $m$ could be a constant.
| https://mathoverflow.net/users/6094 | Lattice Stick Number vs. Stick Number of Knot | I wouldn't be surprised by something like a quadratic bound, or possibly something reasonable in terms of another complexity measure for the knot, but I see no hope for making $m$ constant. Consider the following construction: given $m$, choose some large number like $N=(10m)^6$ of points uniformly at random in the unit sphere, and connect them sequentially in a cycle with straight line segments to define a knot $K$. By construction $K$ has stick number no more than $N$, but each stick has a long narrow tunnel that it must traverse in a very precise direction, which is difficult to do with only $m$ lattice sticks. Of course any one tunnel can be made shorter and wider with an affine transformation (or any small collection of tunnels, with a piecewise affine transformation) but I am convinced (without attempting a rigorous proof) that with probability approaching $1$ a knot so constructed has a lattice stick number much higher than $mS\_L(K)$.
| 8 | https://mathoverflow.net/users/7936 | 33616 | 21,792 |
https://mathoverflow.net/questions/33622 | 7 | The Poincare-Hopf theorem tell us that the sum of the indices of a vector field at isolated zeros on a compact, oriented manifold is the same as the Euler characteristic of the manifold. But how to construct a vector fiedls with isolated zeros?
| https://mathoverflow.net/users/3896 | How to construct a vector fields with isolated zeros? | Your question isn't very well defined. A manifold on its own is not an object where constructions come by easily. But there is a generic way to construct vector fields with isolated zeros. *Any* vector field can be approximated by one with isolated zeros. This is a consequence of Sard's theorem. So start off with the zero vector field and choose any small random perturbation of that, and there you go.
If you want a more constructive answer you'll have to assume a more constructive situation. Like say if your manifold is triangulated, or has a handle decomposition, or a morse function.
Chapman describes the Morse situation so I'll give the triangulation situation.
The vector field has these properties:
There is a critical point at the barycentre of every cell in the triangulation. The vertices are repellors. The barycentres of the top-dimensional simplices are the attractors. A 1-simplex is a (1,n-1)-index critical point -- meaning there's two orbits approaching (along the 1-simplex) and an n-2-dimensional family of reverse orbits attracting. Etc. A j-simplex barycentre has a j-1-dimensional family of attracting orbits, and an n-j-1-dimensional family of reverse orbits attracting.
That isn't quite explicit as one needs an explicit smoothing of the triangulation to put this all together. But it gives you the idea.
| 4 | https://mathoverflow.net/users/1465 | 33627 | 21,798 |
https://mathoverflow.net/questions/33476 | 6 | Sometimes, given an object A in an Abelian category, the Yoneda product on Ext(A, A) is graded-commutative, for example in cases where it coincides with the cup-product in singular cohomology. Are there any nice theorems about when the Yoneda product is graded-commutative in general? Thanks in advance.
| https://mathoverflow.net/users/7935 | When is the Yoneda product graded commutative? | I move this to a more proper answer to discuss some subtle points of the
question. The Eckman-Hilton argument (or more concrete calculations) shows, as
Chris points out, that $\mathrm{Ext}(A,A)$ is commutative when $A$ is the unit
for a monoidal category. The subtleties appear when we consider for instance the
ring $R=k[x]/(x^2)$ for $k$ a field and $A=k$. Then $A$ has a uniform resolution
$\dots\xrightarrow{x}R\xrightarrow{x}R\xrightarrow{x}R\to k\to 0$ giving
$\mathrm{Ext}^i(A,A)=k$ for all $i$. Using the definition of the Yoneda product
in terms of maps of resolutions we get that $\mathrm{Ext}(A,A)$ is the
polynomial ring on $\mathrm{Ext}^1(A,A)$. This is graded commutative only when
the characteristic is $2$ (and then it is not graded commutative in the strict
sense of the square of odd degree elements being zero). However, it is exactly
in characteristic $2$ that $R$ is the affine algebra of a finite group scheme
(with $x\mapsto x\otimes1+1\otimes x$ as coproduct) with $k$ the unit for the
associated monoidal structure on the category of $R$-modules. Hence we have a
monoidal reason for the $\mathrm{Ext}$-algebra being graded commutative in
characteristic. On the other hand we have a uniform description of the
$\mathrm{Ext}$-algebra in all characteristics which just happens to fulfil the
definition of being graded commutative in characteristic $2$.
| 6 | https://mathoverflow.net/users/4008 | 33633 | 21,801 |
https://mathoverflow.net/questions/33625 | 11 | [Qiaochu's question](https://mathoverflow.net/questions/33602/what-is-a-reasonable-finitary-analogue-of-the-statement-that-harmonic-functions-a) on a discrete analogue of harmonic function theory reminded me of some thoughts I had a long time ago about the relationship between cubical cohomology and de Rham cohomology.
The main reason de Rham cohomology is often used as a first introduction to algebraic topology is due to the few technical prerequisites. You can set up the differential graded ring structure with very little work. One of the important results is the Stokes theorem,
$$\int\_C d\omega = \int\_{\partial C} \omega,$$
where $\omega$ is a $p$-form and $C$ is a $(p+1)$-chain. Conceptually, it says the differential operator $d$ is adjoint to the boundary operator $\partial$ with respect to the non-degenerate bilinear pairing $\int$.
It is sometimes remarked that you could turn around and use this result to justify defining $d$ as the adjoint of $\partial$. Indeed, that is how simplicial cohomology is usually defined in terms of simplicial homology. But it seems clear that the better analogy is with cubical cohomology:
Establishing homotopy invariance and the cup product for singular simplicial cohomology is a bit technical. In contrast, it's easy with singular cubical cohomology: $I^m\ I^n$ = $I^{m+n}$, so cubical prisms and products are manifestly cubical, no subdivision needed. Demonstrating these properties for de Rham comology can be done with similar ease. The advantages and disadvantages of cubes over simplices have been discussed here [before](https://mathoverflow.net/questions/3656/cubical-vs-simplicial-singular-homology).
The analogy goes further. If you write out products and exterior derivatives of differential forms in coordinates and compare them with the corresponding formulas in cubical cohomology in a lattice-like neighborhood, they are exactly analogous in the same way that freshman calculus and the calculus of finite differences are analogous. If you have trouble wrapping your head around the geometrical interpretation of concepts in de Rham cohomology like products and differentials and especially more advanced concepts in Hodge theory, the analogy with cubical cohomology is very enlightening (or at least it was for me).
My question is whether anyone has worked out this relationship in detail. Maybe in a context where infinitesimals have a concrete existence as in algebraic geometry and synthetic differential geometry.
| https://mathoverflow.net/users/2036 | Cubical cohomology and de Rham cohomology | M. Carmen Minguez in [this article](http://archive.numdam.org/article/CTGDC_1988__29_1_59_0.pdf) constructs a homomorphism between de Rham and Cubical Singular Cohomology without showing that is an isomorphism. This is done in the context of Synthetic Differential Geometry.
In general Synthetic Differential geometers seem to be quite aware of the cubical setup, probably because differential forms naturally get a cubical structure if defined via infinitesimals. This is very visible in chapter IV, section 1 of Moerdijk/Reyes' [book](http://books.google.com/books?id=OSzvAAAAMAAJ&q=moerdijk+reyes+smooth+infinitesimal+analysis&dq=moerdijk+reyes+smooth+infinitesimal+analysis&hl=de&ei=HOtPTIzOIsGPOIq2iKcB&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCsQ6AEwAA) on synthetic differential geometry. In remark 1.8 (p. 145) they mention a bridge to cubical (co)homology - then, however, they choose to establish the isomorphism between de Rham and *simplicial* singular cohomology.
| 7 | https://mathoverflow.net/users/733 | 33635 | 21,803 |
https://mathoverflow.net/questions/33638 | 5 | I refer to "Sheaves in Geometry and Logic", by S. MacLane.
Let **C** be a category. Dealing with a *subobject* of an object $D \in \text{Ob}\_{\mathbf C}$, one defines an equivalence relation between morphisms towards *D*:
>
> Two monomorphisms $f:A\to D$, $g:B\to D$ with a common codomain *D* are called *equivalent* if there exists an isomorphism $h\colon A\to B$ such that *gh*= *f*.
> A *sbobject* of *D* is an equivalence class of monos towards *D*. The collection Sub**C**(*D*) of subobject of *D* carries a natural partial order [...].
> Then Sub**C**(*D*) is **the set** of all subobjects of *D* in the category **C**.
>
>
>
I can't figure out *why* Sub**C**(*D*) is a set, rather than a proper class! Indeed, we are considering something like an qeuivalence relation on
$\displaystyle \coprod\_{A\in \text{Ob}} \text{Hom}\_{\bf C}(A,D)$
which is not a set, as soon as **C** isn't small.
So, how can I avoid the problem?
| https://mathoverflow.net/users/7952 | Is the subobject functor really a presheaf? | For a general category the subobjects do indeed not have to form a set.
In the context of MacLane/Moerdijk you only look at toposes and there one has a natural isomorphism $Sub\_{\mathbf{C}}(D) \cong Hom\_{\mathbf{C}}(D,\Omega)$, where $\Omega$ is the subobject classifier.
So it follows from the axioms of a topos, *(edit, thanks Mike:) if it is locally small*, that $Sub\_{\mathbf{C}}(D)$ is a set. When you prove that the basic examples, sheaves, finite sets, products of those, etc. are toposes you exhibit an object $\Omega$ and establish the above bijection. Before that point the left hand side could a priori be a proper class but the right hand side is a set, since you know that your category is locally small, and your bijection then shows that subobjects form a set.
Knowing this you can conclude that objects in full subcategories *(edit, thanks again:) whose embedding preserves monos (e.g. if they are reflective)* of toposes also have a set of subobjects, e.g. in all locally presentable categories...
| 10 | https://mathoverflow.net/users/733 | 33643 | 21,809 |
https://mathoverflow.net/questions/32999 | 7 | This problem arises while studying the complexity of algorithms and I am quite unfamiliar with the subject.
Consider the set F of injective functions from {1..N} to {1..M}
we can define an association scheme on F x F by
(f,f') and (g,g') are in the same class if there is a permutation $\pi\in S\_M$ and a permutation $\tau \in S\_N$ such that $g = \pi \circ f \circ \tau$ and $g' = \pi \circ f' \circ \tau$.
I checked that this really defines an association scheme. In a way it is an "ordered" version of the Johnson scheme. It seems to me that it is a natural extension of the Johnson scheme, but I did not find any reference about it.
>
> Q1: Has this association scheme ever been studied? What is its name?
>
>
> Q2: Can this scheme be obtained by a combination (tensor product? suprema?) of the Johnson scheme and another quantity?
>
>
>
More precisely, I am interested in the "Bose-Mesner Algebra" point of view on this scheme. It is known that all the matrices in the algebra defined by this association scheme diagonalize in the same basis.
>
> Q3: How can we construct/characterize these eigenspaces?
>
>
>
--
Some background on Association Schemes.
An association scheme is a set of symmetric boolean matrices $A\_1, \dots , A\_S$ such that
1) $\sum\_{i=1}^s A\_i =J$ the all-one-matrix
2) $A\_1 = I$ the identity matrix
3) $\forall i,j \; A\_iA\_j \in {\rm span} ( A\_i )$
The matrices $A\_i$ can be seen as adjency matrix for some graph (but I don't think it might help here)
The span{$A\_i$} defines an algebra called the Bose-Mesner Algebra. Condition (3) implies that all matrices commute so they diagonalize in the same basis.
--
In the case I'm considering here, the dimension of the $A\_i$ is ${M \choose N}N!\times {M \choose N}N!$. The $A\_i$ are not explicitly defined but we know that $[A\_i]\_{fg}=[A\_i]\_{f'g'}$ if there is a permutation $\pi\in S\_M$ and a permutation $\tau \in S\_N$ such that $g = \pi \circ f \circ \tau$ and $g' = \pi \circ f' \circ \tau$.
--
About the Johnson scheme: The $A\_i$ have size ${M \choose N}$. The rows and the columns of the matrices are labeled by subsets of size $N$ of {$1,\dots,M$}. (in my case, the labels are injective functions, ie. ordered sets of subsets of size $N$ of {$1,\dots,M$}.
$[A\_i]\_{ab}=[A\_i]\_{a'b'}$ if there is a permutation $\pi\in S\_M$ such that $\pi(a) = a'$ and $\pi(b) = b'$. (where $\pi(a)$ denotes the subset of {$1,\dots,M$} obtained by applying the permutation $\pi$ to the elements of the sets $a$.
| https://mathoverflow.net/users/6673 | Association scheme on injective functions | I don't think this scheme has a particular name, and am not aware of any study of it. Its Bose-Mesner algebra is the commutant of a multiplicity-free representation of the wreath product of $S\_m$ by $S\_n$. To get the eigenspaces you need to find the decomposition of the representation into irreducibles.
The most useful reference I know is "Harmonic Analysis on Finite Groups: Representation Theory, Gelfand Pairs and Markov Chains" by Ceccherini-Silberstein, Scarabotti and Tolli.
for what it's worth, I think there's a chance that getting the decomposition is actually
doable, but it would not be quick :-(
| 5 | https://mathoverflow.net/users/1266 | 33651 | 21,815 |
https://mathoverflow.net/questions/33676 | 3 | Does the existence of exotic smooth structure in $\mathbb{R}^4$ imply the existence of an atlas which has a $C^0$ mapping to the Cartesian atlas, but not a $C^k$ mapping (for some finite $k$)? Does the nonexistence of exotic smooth structure in $\mathbb{R}^n$, $n\neq 4$ imply that all atlases therein have smooth mappings to the Cartesian atlas?
| https://mathoverflow.net/users/7956 | exotic smooth structure clarification | Regarding your 1st question, perhaps you meant to ask something else? Any atlas can be composed with a non-smooth homeomorphism to produce an atlas that isn't smooth in the standard sense. For example, $\mathbb R \to \mathbb R$ defined by $t \longmapsto t^{1/3}$ is an atlas on $\mathbb R$ but it's not $C^1$. This answers your 2nd question in the negative.
Alternatively, some exotic smooth $\mathbb R^4$'s are diffeomorphic to open subsets of the standard $\mathbb R^4$, so even for exotic smooth $\mathbb R^4$'s you could potentially have only a one-map atlas, which is smooth in the standard sense.
You might like to read this article: <http://en.wikipedia.org/wiki/Exotic_R4>
| 7 | https://mathoverflow.net/users/1465 | 33677 | 21,827 |
https://mathoverflow.net/questions/25484 | 7 | Given two modules $M$ and $N$ there is a nice scheme parametrizing extensions
$0 \rightarrow M \rightarrow E \rightarrow N \rightarrow 0$
namely $\operatorname{Ext}^1(N,M)$ or, leaving out the trivial extension, the projective space $P(\operatorname{Ext}^1(N,M))$.
There are (at least) two natural generalizations
1. $n$-step extensions
$M \rightarrow E\_1 \rightarrow E\_2 \rightarrow \dots \rightarrow E\_n \rightarrow N$
between $N$ and $M$.
2. Filtered modules: Parametrize modules $E$ which admit a filtration
$0 \subset F\_1 \subset F\_2 \dots F\_n=E$
with fixed graded objects $E\_i=F\_i/F\_{i-1}$.
I suppose in the first case one can use the group $\operatorname{Ext}^n(M,N)$, although I never
saw a construction of a universal family. Is there a good reference?
In the second case, I do not have a clue. So my main question is:
**Is there a nice moduli space of filtered objects?**
| https://mathoverflow.net/users/5714 | Moduli of extensions of modules | in the situation you have in mind (sheaves on an algebraic variety),
such spaces are not too difficult to construct as Artin stacks. If you omit the condition that the i-th filtration quotient is isomorphic to a given one, then such a universal Artin stack is e.g. constructed in Bridgeland's introduction to Hall-algebras (arXiv:1002.4372, he calls them $\mathcal M^{(n)}$), but of course also in earlier articles by Joyce. Basically it follows from the existence of relative quot schemes.
These universal extension stacks have evaluation morphisms to $\mathcal M$, the stack of all sheaves, sending the filtration to its i-th quotients, so you can take a base change via the map from $\operatorname{Spec} k \to \mathcal M \times \dots \times \mathcal M$ given by your set of objects $E\_i$, and the fiber product will be the Artin stack you are looking for.
If you want a scheme instead of an Artin stack - then I would ask back "why?" :) Nevertheless, it would be useful to understand this fiber product better when $n > 2$.
| 4 | https://mathoverflow.net/users/7437 | 33678 | 21,828 |
https://mathoverflow.net/questions/33533 | 3 | Let $K$ be the set of open-closed subsets of $\mathbb{Z}\_p$. Let $M$ be the set of functions from $K$ to $C\_p$ that are additive under disjoint unions. Then $M$ can be regarded as an elementary abelian pro-$p$ group: multiplication is pointwise in $C\_p$, and a base of open subgroups is given by $\{ U\_n \}$, where $U\_n$ consists of those functions which map cosets of $p^n \mathbb{Z}\_p$ to $0$. Moreover, $\mathbb{Z}\_p$ has a translation action on $K$, and hence a continuous action on $M$ in which each of the $U\_n$ is normalised. We use this action to construct a group $G = M \rtimes \mathbb{Z}\_p$. This is realisable as an inverse limit of the groups $C\_p \wr C\_{p^n}$ (regular wreath product); in particular, it is a $2$-generator pro-$p$ group.
Now for the question: Is the group $G$ above isomorphic to a group arising from some standard construction, and if so does it have a nice name? I would be surprised if nobody has used this group before as an example of something or other. Also, there seems to be quite a general construction behind this.
Edit: the group described is a subgroup of the pro-$p$ group $C\_p \wr\_K \mathbb{Z}\_p$. The unusual part is the extra condition that the functions should be additive (motivation: I wanted $G$ to have relatively few normal subgroups). Are there alternative conditions one could impose that would result in an isomorphic group? In particular, I'm relying on the fact that $C\_p$ is abelian, which is bad for generalisations.
| https://mathoverflow.net/users/4053 | Name this pro-$p$ group | I would call this group the pro-$p$ completion of $C\_p \wr \mathbb{Z}$. Alternatively, it is the group given by the pro-$p$ presentation <$ a, b| a^p, [a, a^b]>$, along similiar lines.
It looks a lot like $C\_p\wr \mathbb{Z}\_p$, except we have the product of as many copies of $C\_p$ as open subgroups of $\mathbb{Z}\_p$ rather than elements. This does seem a natural, interesting, general construction, and unless I am mistaken is a generalistion of the pro-$p$ wreath product. (This particular example is nice especially by virtue of having a balanced pro-$p$ presentation).
| 1 | https://mathoverflow.net/users/4100 | 33682 | 21,831 |
https://mathoverflow.net/questions/33672 | -2 | I'm trying to solve the following least squares problem:
$\underset{x}{\text{min}} ||Ax - \tilde{b}||\_2$
where $Ax = b$ and $\tilde{b} = b + w$
### Question:
How do I determine which probability distribution fits $w$ best?
Also, $A \in \mathbb{R}^{n\times n}$ is a large, and sparse Toeplitz matrix, $\tilde{b} \in \mathbb{R}^{n \times 1}$
$w$ is not your typical measurement noise and is probably not Gaussian. Also, I have access to $b$ (training data) and therefore, $w$.
Also, in a related question, I'm using LSQR to determine $x$ --- since I have my training data, is there any way to do cross-validation to determine how to best solve for $x$?
For example, I tried Tikhonov regularization to find the best $\lambda$ but I'm looking for a better method.
| https://mathoverflow.net/users/1899 | Determine noise distribution | There's no way to answer in general what distribution data has. It has whatever distribution it has, but what you really want to know is whether a distribution which someone has identified adequately fits your data. So you have to propose a specific distribution first, then test for goodness of fit. Searching on "goodness of fit" will give you lots of resources, but a good general-purposed goodness of fit test is Kolmogorov-Smirnov, often abbreviated K-S.
You could look at a histogram of your data to start your search. Is it symmetric or skewed? Long tails or short tails? Unimodal or bimodal? If you post an image, someone may be able to suggest some distributions to test.
Your question may be more appropriate for the [Statistical Analysis](https://stats.stackexchange.com/) Stack Exchange site.
| 2 | https://mathoverflow.net/users/136 | 33694 | 21,838 |
https://mathoverflow.net/questions/33629 | 13 | Here are two well known facts, which put together leave me confused.
First, it's well known that intuitionistic logic is the logic of constructive mathematics. From every intutionistic proof, you can extract an algorithm which will compute the witness to that theorem (i.e., the famous "existence principle" of intutionistic logic).
Second, it's also well-known that topologies give rise to models of intuitionistic mathematics. We can equate propositions with open sets, and then the fact that a topology is a Heyting algebra gets us off to the races. (Caveat: this is an oversimplification.)
In fact, you can get pretty far with the dictionary "computability is continuity" -- which is precisely what puzzles me!
Intuitionistically, we can construct the real numbers pretty much as usual, eg via Cauchy sequences, which can be viewed as functions $\mathbb{N} \to \mathbb{Q}$ subject to the usual conditions. Now recall that classically, the only continuous functions $f : \mathbb{R} \to \mathbb{Z}$ are the constant functions. So according to this dictionary, we should not expect to give an intuitionistically valid function which computes the integer part of a real number.
So far, everything makes sense. For example, the Cauchy sequence $0, 0.9, 0.99, 0.999, \ldots$ has $1$ as its limit, but given a black-box $f : \mathbb{N \to \mathbb Q}$, we can't tell that it's this sequence without looking at every $f(n)$ -- which is obviously uncomputable.
However, suppose that we have in our hand an actual computer program of type $\mathbb{N} \to \mathbb{Q}$ (say, in Haskell), which we happen to know represents a Cauchy sequence. We can actually run this program, and use some prefix of the Cauchy sequence to print something close to the integer part of the real to the computer screen. This operation obviously isn't a *function* on reals, since it can give different results for different representations of the same real. For example, our program might print "1" for the Cauchy sequence $1, 1, 1, \ldots$, and "0" for the Cauchy sequence "$0.9, 0.99, 0.999, \ldots$". But equally obviously this a perfectly sensible *computer program* to write.
So that's my question: what *is* this operation, and how can we axiomatize its logical behavior? (What kind of crazy gadget doesn't respect equality!?)
| https://mathoverflow.net/users/1610 | What happens when we print the digits of a real number? | [**Edit:** Following Carl Mummert's comments, I have improved the part about transformation of Cauchy sequences to digit expansions. Thanks to Carl for interesting observations.]
Your question is best divided into two subquestions:
1. Does every real number have a digit expansion?
2. What is the nature of the "process" of going from a real number to its digit expansion,
assuming it has one.
For the purposes of this discussion let us say that the real numbers are defined a la Cauchy, i.e., $\mathbb{R}$ is the metric completion of $\mathbb{Q}$ equipped with the Euclidean metric. Let me also assume countable choice, so that $\mathbb{R}$ can be shown to exist. In particular, we can take $\mathbb{R}$ to be the set of Cauchy sequences quotiented by a suitable equivalence relation. (I am talking informal intuitionistic mathematics here and do not want to tell you what formal system I am in, but if you force me, I can tell you about various possibilities.)
We need to be careful about what is understood by "digit expansion" of a real number. Digit expansions are special kind of a Cauchy sequences: they are monotone and have a fixed rate of convergence. We have the following observations, none of which is hard to establish (assuming countable choice):
* for every Cauchy sequence there exists one with a fixed rate of convergence (say the
$n$-th term is within $2^{-n}$ of its limit), but a function mapping from the former
to the later cannot be shown to exist.
* every Cauchy sequence with a fixed rate of convergence (let me call such *fast* Cauchy sequences) has a monotone fast Cauchy sequence, and there is a function mapping the former to the later, as pointed out by Carl in the comments.
* a decimal digit expansion is an infinite sequence, i.e., a member of $D = \lbrace 0, 1, \ldots 9\rbrace^\mathbb{N}$. We cannot prove constructively that every monotone fast Cauchy sequence has a corresponding digit expansion (again, by a continuity argument).
* if we try to weasle out of the problem by taking only those reals which have a decimal digit expansion, then basic operations such as addition and multiplication cannot be shown to exist because they would be discontinuous.
* as is well known, if we allow *negative* digits so that a digit expansion is a member of $\lbrace -9, -8, \ldots, 8, 9\rbrace^\mathbb{N}$ then every Cauchy sequence has a corresponding digit expansion, but now every real has very many digit expansions.
So I am going to assume that by "digit expansion" you mean one which can contain negative digits (otherwise your premise that every real has an expansion cannot be established). As you notice, there is no continuous map from reals to (negative) digit expansions. Nevertheless, one can write a program which takes a real number (represented by a rational Cauchy sequence with an explicit modulus of convergence, or some other equivalent form, such as a sequence of ensted intervals with rational endpoints) and produces *one* of its digit expansions. Such programs can be understood in several ways:
1. As realizers for the statement $$\forall x \in \mathbb{R} . \exists d \in D . \;\text{$d$ is digit expansion of $x$}.$$ This is my favorite. When we reason about real numbers in analysis, we rarely talk about digit expansions, so it does not matter if the computation of a digit expansion appears only implicitly as a realizer for a $\forall\exists$ statement.
2. We can formalize a distinction between an *operation* and a *function*. The former
can be *intensional*, i.e., it does not have to repsect the equality of its domain,
while the later must be *extensional* so that it maps equals to equals. This introduces
certain subtleties that only logicians can appreciate. In this setting, the program
is an operation which is not a function.
3. Such programs can be seen as realizers witnessing the totality of a *multi-valued*
function that maps a real to the set of its digit expansions. Such a map is easily seen to exist. This is more manageable
than option 2 and is popular in Type Two Effectivity.
In practice, you cannot hope to implement reals in such a way that each real receives only one representation. Even having finitely many representations is out of the question – some reals will necessarily have infinitely many representatives, or something will go wrong (arithmetical operations won't be computable, or the structure won't be closed under limits of (computable) Cauchy sequences, or the strict linear order won't be semidecidable). Classical decimal digit expansions are *not* computable, even though every single computable real has a decimal digit expansion. This shows the importance of thinking about *spaces* as whole entities, rather than each of its points in isolation. Or to be even more blunt: spaces are *not* determined by their points!
| 10 | https://mathoverflow.net/users/1176 | 33702 | 21,844 |
https://mathoverflow.net/questions/33687 | 7 | I'm writing some software to automatically compute things like Alexander modules, Alexander ideals, Milnor signatures and Farber-Levine pairings for 1 and 2-knot complements. So among other things I'd like to have a quick way of comparing ideals in the Laurent polynomial ring $\mathbb Z[t^{\pm 1}]$.
So my question:
Is there an efficient way of determining whether or not two ideals in $\mathbb Z[t^{\pm 1}]$ are equal? What's the most pleasant procedure you can think of?
My ideals are given by a finite number of generators.
The knot theory literature tends to stop at "$\mathbb Z[t^{\pm 1}]$ *isn't a PID*" but I'm hopeful there's some reasonable tools out there. Understanding ideals in $\mathbb Z[t^{\pm 1}]$ is essentially the same thing as understanding cyclic group actions on finitely generated abelian groups so I could imagine answers might be more complicated than I like. But I hope to be pleasantly surprised by some sort of "division algorithm + details" type algorithm.
For example, is there a better way to go about this problem than:
Given an ideal $I \subset \mathbb Z[t^{\pm 1}]$, consider the finitely-generated abelian group $G = \mathbb Z[t^{\pm 1}]/I$ together with the action of $\mathbb Z$ on $G$ given by multiplication by $t$. We want to determine $G$ as a $\mathbb Z$-module. So presumably you would do this on the $\mathbb Z$-torsion subgroup of $G$ (call it $\tau G$), then work out the conjugacy class of the action of $\mathbb Z$ on $G/\tau G$ (which amounts to the conjugacy problem in $GL\_n(\mathbb Z)$), then there would be an extension problem to deal with.
I'm hoping there's a simpler way to deal with this problem -- simpler in the sense of implementation.
| https://mathoverflow.net/users/1465 | Ideals in the ring of single-variable Laurent polynomials with integer coefficients | It is fairly straightforward to adapt standard Gröbner basis techniques to such algebras, e.g. see the [paper [1]](https://doi.org/10.1007/s002000050108). See also the [paper [0]](https://doi.org/10.1007/11870814_12) which applies such algorithms to the problem at hand.
[[0](https://doi.org/10.1007/11870814_12)] Jesus Gago-Vargas; Isabel Hartillo-Hermoso; Jose Marya Ucha-Enryquez
Algorithmic Invariants for Alexander Modules. LNCS 4194, 149-154
<https://link.springer.com/chapter/10.1007/11870814_12>
Abstract. Let G be a group given by generators and relations. It is possible
to compute a presentation matrix of a module over a ring through
Fox's differential calculus. We show how to use Gröbner bases as an algorithmic
tool to compare the chains of elementary ideals defined by the
matrix. We apply this technique to classical examples of groups and to
compute the elementary ideals of Alexander matrix of knots up to 11
crossings with the same Alexander polynomial.
[[1](https://doi.org/10.1007/s002000050108)] Franz Pauer, Andreas Unterkircher.
Gröbner Bases for Ideals in Laurent Polynomial Rings and their Application to Systems of Difference Equations.
AAECC 9, 271-291 (1999)
<https://link.springer.com/article/10.1007/s002000050108>
*Abstract.* We develop a basic theory of Gröbner bases for ideals in the algebra
of Laurent polynomials (and, more generally, in its monomial subalgebras). For
this we have to generalize the notion of term order. The theory is applied to
systems of linear partial difference equations (with constant coefficients) on
${\mathbb Z}^n$. Furthermore, we present a method to compute the intersection of an ideal
in the algebra of Laurent polynomials with the subalgebra of all polynomials.
| 5 | https://mathoverflow.net/users/6716 | 33707 | 21,846 |
https://mathoverflow.net/questions/33713 | 5 | Let $T: X \to X$ be an Anosov diffeomorphism. Suppose $f: X \to \mathbb{R}$ is Holder continuous (say with exponent $\alpha$). The question arises as to when $f$ can be written as $g \circ T - g $ for some $\alpha$-Hölder $g: X \to \mathbb{R}$. It is easily checked that a necessary condition is that the periodic data vanishes, i.e. the sum of $f$ over every periodic orbit is zero. It is also a (nontrivial) theorem of Livsic that the converse is true, namely vanishing periodic data implies that this cohomological equation is solvable.
I'm interested in the following variant of the cohomological equation. Let $A: \mathbb{R}^k \to \mathbb{R}^k$ be an isomorphism. Suppose $f: X \to \mathbb{R}^k$ is $\alpha$-Hölder. When is the equation $f = A( g\circ T) - g$ solvable in $g$?
There seem to be sufficient conditions known for the case of compact groups. Are there conditions known in the noncompact but abelian case?
| https://mathoverflow.net/users/344 | Existence conditions for twisted cohomological equations? | Part I:
The answer is yes under additional conditions:
1. Periodic data conditions are satisfied. That is, for any periodic point $p$
$$
\sum\_{x\in O(p)}f(x)=0.
$$
2. Exponent $\alpha$ is sufficiently close to 1.
3. Transformation $A$ is dominated by $T$. That is, the map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.
Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that there exists an $\alpha$-Holder solution $g$. The result is more general: the target group is any Lie group with a bi-invariant metric and the equation is the cohomological equation for two cocycles rather than just coboundary equation.
Part II:
Notice however that if $A\neq Id$ then the periodic conditions may be no longer necessary. Let's restrict to the case when $k=1$ then our equation takes form
$$
f=\lambda g\circ T-g,
$$
where $\lambda<1$. Direct computation shows that
$$
g=-\sum\_{i\ge 0} \lambda^if\circ T^i
$$
is a solution. It is also clear that $g$ is Holder continuous. Moreover, in this case uniqueness is clear as well since the above formula for $g$ is obtained recurrently from
$$
g=-f+\lambda g\circ T.
$$
It seems that this generalizes rather straightforwardly to the case when $A$ is hyperbolic. And I think it's worthwhile to see if anything interesting happens in the case then $A$ has some eigenvalues on the unit circle and some off.
| 7 | https://mathoverflow.net/users/2029 | 33725 | 21,855 |
https://mathoverflow.net/questions/33706 | 4 | This question is related to a previous question of mine:
[Determinacy interchanging the roles of both players](https://mathoverflow.net/questions/32966/determinacy-interchanging-the-roles-of-both-players)
Given any set A of sequences of natural numbers, every strategy (no matter for which player) is either winning (W), or losing (L), or neither(N) for A.
So depending on A, the set of all strategies available for any of both players can be, a priori, of any of the seven kinds: W (all winning), L (all losing), N (all neither), WL (some winning and the rest losing), WN (some winning and the rest neither), LN (some losing and the rest neither) and WLN (some winning, some losing, and the rest neither).
This makes a total of 49 situations (now taking into account both players). Of course, not all of them are possible because we have the following restrictions:
a. If one player has a winning (losing) strategy, the other one cannot have a winning (losing) strategy.
b. If one player has only winning (losing) strategies, the other one only has losing (winning) strategies.
I don't know of any other restrictions (not derivable from them, for example it follows that if one player is WL for A, the other one can only be N).
This leaves us with the following possible situations:
(I'd draw a table here, but unfortunately I don't know how to edit it; I tried html without sucess)
1. W for I and L for II
2. L for I and W for II
3. N for I and N for II
4. N for I and WL for II
5. N for I and WN for II
6. N for I and LN for II
7. N for I and WLN for II
8. WL for I and N for II
9. WN for I and N for II
10. WN for I and LN for II
11. LN for I and N for II
12. LN for I and WN for II
13. WLN for I and N for II
My question is, could one find examples (prove the existence of subsets of sequences of naturals) for each situation only assuming ZFC?
Some are obvious, like the empty set for 2 or the set of all sequences for 1, or like the set of all sequences with a 1 in the odd positions for 8, but others may be not.
| https://mathoverflow.net/users/6466 | Subsets of sequences of natural numbers vs. strategies under ZFC | 4 is not possible, the others that you list are.
By observing that a strategy is winning for player X in the game $G(A)$ if and only if it is losing for X in $G(A^c)$ we can reduce the total number of games necessary to construct by noting some equivalences. For example, if there is a game of the form 5, then by taking the complement we get a game of the form 6. This also tells us that games for 3,5,6,9,11 cannot be constructed in ZF alone since they (are equivalent to) games where neither player has a winning strategy. The others (not 4) can be realized by simple games that can be described in ZF. I'll describe these simple games below. Those that require some choice I'll just indicate it; the diagonalization is no harder than the one described at [Determinacy interchanging the roles of both players](https://mathoverflow.net/questions/32966/determinacy-interchanging-the-roles-of-both-players)
$1$. (and equivalently 2). As you said, taking $A=\omega^\omega$ works.
$3$. Can be done in ZFC.
$4$. Not possible. Suppose $G(A)$ were such a game. Let $\tau\_0$ be a winning strategy for II, and $\tau\_1$ a losing strategy. Consider the strategy $\tau$ where when I plays 0, II follows $\tau\_0$, and otherwise II follows $\tau\_1$. Then $\tau$ falls under the neither category.
$5$. (and equivalently 6). Can be done in ZFC.
$7$. Take $A$ to be $\{x\in\omega^\omega:x(1)=0\}$ (an example of a neither strategy for II: respond to 0 with 0, respond to 1 with 1).
$8$. Take $A=\{x\in\omega^\omega:x(0)=0\}$.
$9$. (and equivalently 11). This can be done in ZFC (in fact you can check a game in which I has a winning strategy. and the complementary game is not determined is necessarily of this form, so existence here follows from the earlier question).
$10$. (and equivalently 12). Take $A=\{x\in\omega^\omega:x(0)\not=0\}\cup\{x\in\omega^\omega:x(0)=0,x(1)=0\}$. (IIs losing strategy is to always play 0)
$13$. Take $A=\{x\in\omega^\omega:x(0)=0\}\cup\{x\in\omega^\omega:x(0)=1,x(1)=0\}$.
| 7 | https://mathoverflow.net/users/2436 | 33728 | 21,858 |
https://mathoverflow.net/questions/33717 | 5 | What is a good reference for a geometrical viewpoint on the calculus of variations for physics, using differential forms etc. to derive Yang-Mills equations and other topics of the standard model?
Thanks for ideas.
| https://mathoverflow.net/users/7626 | literature on geometrical viewpoint on calculus of variations for physics | I think that the book
* David Bleecker: *Gauge Theory and Variational Principles*, Addison-Wesley, 1981
contains exactly what you are looking for.
| 4 | https://mathoverflow.net/users/3473 | 33729 | 21,859 |
https://mathoverflow.net/questions/33724 | 7 | Background
----------
When constructing the exterior algebra of a (finite-dimensional, complex) vector space $V$, there are two equivalent pictures. The first is the quotient picture. First you define the tensor algebra $T(V)$, define $\mathcal{J}$ to be the 2-sided ideal generated by elements of the form $x\otimes y + y \otimes x$, and then define the exterior algebra to be the quotient $\Lambda(V) = T(V)/\mathcal{J}$.
The other viewpoint is via embedding the exterior algebra in the tensor algebra. This is done as follows. There is an action $\rho$ of the symmetric group $S\_n$ on $V^{\otimes n}$ for each $n$, given by
$$\rho\_\pi (v\_1 \otimes \dots \otimes v\_n) = v\_{\pi(1)} \otimes \dots \otimes v\_{\pi(n)}.$$
Then the map
$$A\_n = \frac{1}{n!} \sum\_{\pi \in S\_n}sgn(\pi) \rho\_\pi$$
is idempotent, i.e. satisfies $A\_n^2 = A\_n$. If we define
$$ A = \bigoplus\_{n=0}^\infty A\_n$$
on $T(V)$, then it turns out that the kernel of $A$ is equal to $\mathcal{J}$, so that $\Lambda(V) \simeq \mathrm{im}(A)$.
Why I care
----------
I'm trying to understand the quantum analogue of this, where $V$ is the fundamental representation/vector representation of $U\_q(\mathfrak{sl}\_N)$. The problem is that there is no longer an action of the symmetric group on $V^{\otimes n}$; instead there is an action of the braid group. Both the quotient picture and the embedding picture have analogues in the quantum setting. In particular, denote by $\sigma$ the braiding on $V \otimes V$, and let $\sigma\_i$ be the automorphism of $V^{\otimes n}$ given by $\sigma$ acting in the $i$ and $i+1$ spots.
One constructs the $q$-antisymmetrizer as follows. Define $\tau\_i$ to be the adjacent transpositions $(i, i+1)$ in $S\_n$. For a permutation $\pi$, write
$\pi= \tau\_{i\_1} \dots \tau\_{i\_k},$
where $k$ is the minimal number of adjacent transpositions needed. Then define
$$\sigma\_\pi = \sigma\_{i\_1} \dots \sigma\_{i\_k}.$$
It is a theorem that this is well-defined, i.e. that any two minimal decompositions of $\pi$ as products of adjacent transpositions can be transformed into one another using only the braid relations. Anyway, once this is known, you can define the $q$-analogue of the antisymmetrizer map as
$$A\_n = \frac{q^{\binom{n}{2}}}{[n]!} \sum\_{\pi \in S\_n} sgn(\pi) q^{-\ell(\pi)} \sigma\_\pi,$$
where $[n]$ is the $q$-number and $\ell(\pi)$ is the length of a minimal decomposition of $\pi$. The significance of the $\binom{n}{2}$ is that it is the length of the longest word in $S\_n$. As far as I can tell, this was first defined by Jimbo in his 1986 paper "A $q$-analogue of $U(\mathfrak{gl}\_{N+1})$, Hecke Algebra, and the Yang-Baxter Equation". He states that $A\_n^2 = A\_n$, and refers to Gyoja's paper "A $q$-Analogue of Young Symmetrizer." Gyoja's paper certainly does the trick, but I find that something is lost in the abstraction.
I've verified that $A\_n^2 = A\_n$ for $n = 3$, which already involves 36 terms, and it wasn't that enlightening. The problem is that the braid generators $\sigma\_i$ aren't idempotent, they instead satisfy $\sigma\_i^2 = 1 + (q - q^{-1})\sigma\_i$.
The question
------------
Does anybody know a nice way of seeing that the $q$-antisymmetrizer is idempotent, or a nice presentation of it somewhere?
| https://mathoverflow.net/users/703 | Idempotency of the q-antisymmetrizer | This element has many expressions. It is characterised up to a scalar multiple by the property that $\sigma\_iA\_n=-q^{-1}A\_n$ for $i=1,2,\ldots ,n-1$. It also satisfies
$A\_n\sigma\_i=-q^{-1}A\_n$ for $i=1,2,\ldots ,n-1$. In particular it is central.
Using this property you can calculate $A\_n^2$. Note that
$\sigma\_\pi A\_n=(-q)^{-\ell(\pi)}= sgn(\pi)q^{-\ell(\pi)}$. So
$$A\_n.\sum\_\pi sgn(\pi)q^{-\ell(\pi)}\sigma\_\pi=A\_n.\sum\_\pi q^{-2\ell(\pi)}$$
You fix the scalar factor by the condition $A\_n^2=A\_n$. Alternatively in the one dimensional representation $\sigma\_i \mapsto -q^{-1}$ you require $A\_n \mapsto 1$.
| 7 | https://mathoverflow.net/users/3992 | 33734 | 21,863 |
https://mathoverflow.net/questions/33568 | 0 | There is only one differentiable structure permitted in R^2, meaning, I think, that all atlases in R^2 are diffeomorphic to the Cartesian atlas. But, doesn't the polar coordinate system represent an atlas that is not truly diffeomorphic to the Cartesian atlas, due to the coordinate singularity it has at its origin?
| https://mathoverflow.net/users/7956 | differentiable structure and coordinates in R^2 | I think Matt Noonan's comment technically answered my question. The polar coordinate chart is not a valid atlas on $\mathbb{R}^2$ because the angular coordinate is discontinuous across the non-negative x-axis.
| -1 | https://mathoverflow.net/users/7956 | 33735 | 21,864 |
https://mathoverflow.net/questions/33567 | 8 | I'm trying to find the minimal (monic) polynomial $M(x)$ (over the rationals) for an algebraic number. I know the degree of the polynomial (call it $d$) and I have $d+1$ data points of the form $(x\_i, |M(x\_i)|)$. The $x\_i$ are all rational numbers, so the $|\cdot|$ is just regular absolute value.
If it wasn't for the absolute value sign, it'd be a straight-forward polynomial fit. However, to only way I've been able to solve for $M(x)$ is fitting a polynomial for each of the $2^d$ choices of plus/minus on the second coordinate, and then finding one that's monic.
Luckily, so far this has always produced one and only one such polynomial. Anybody know a more feasible/elegant way to do this?
(For more motivation than you actually care for, see <http://course1.winona.edu/eerrthum/Papers/MAANCS081018.pdf> where on slide 8 (page 34) I mention the brute force method.The last slide contains an example calculation.)
(Feel free to re-tag as appropriate.)
| https://mathoverflow.net/users/3400 | Monic polynomial from absolute value information | Here's a solution using lattice reduction:
1) Find degree $d$ polynomials $p\_i(x)$ such that $p\_i(x\_j) = |M(x\_i)| \delta\_{i j}$.
2) Let $c\_i$ be the coefficient of $x^d$ in $p\_i(x)$, and $c$ the $d+1$ long column vector whose coordinates are $c\_i$.
3) Find a matrix $U \in SL\_{d+1}(\mathbb{Z})$ such that $U c = e$, where $e$ is the $d+1$ long column vector with a 1 in the first coordinate and zeros elsewhere. Added later:
not quite. Let $D$ be a common denominator of all the elements of $c$, and form a $d+1 \times d+1$ matrix, $A$, whose first column is $cD$ and the lower $d \times d$ block is the identity matrix (with the rest of the top row 0). Find the Hermite normal form: $U \in SL\_{d+1}(\mathbb{Z})$, $U A = H$. The first column of $H$ will be zeros below the first entry, which will be a positive integer $r$. In order for there to be a solution it is necessary that $r | D$. Form a new matrix $U'$ by multiplying the top row of $U$ by $D/r$.
4) The answer (see below) is a vector in the $\mathbb{Z}$-lattice generated by the bottom $d$ rows of $U'$ which is close to the top row.
Namely, form the matrix $W$ by adjoining a $d+1 \times d+1$ identity matrix to the right of $c$. Since only the coefficient of $x^d$ matters in the answer, we can see that an answer will be given by some integer linear combination of the rows of $W$ which has $\pm 1$ in the last $d$ coordinates. The squared Euclidean length of that vector will be $d+1$, which is quite short. There are a number of algorithms for finding a closest vector (in theory for general lattices it's a hard problem, but in practice in a lattice like this it's not too hard). For a nice account of how to do it look in <http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.81.8089> starting around page 14.
The idea here is to prepend a column to $U'$ which is a unit vector with 1 in the first position and 0's everywhere else. Now multiply the rest of the matrix (all columns but the first) by a large scaling factor, $s$. This will make sure that the first row will show up in the linear combination forming the shortest basis vector.
Lattice reduction will supply a short vector in the lattice which we know that our answer is. We then read off the coefficients of the $p\_i$ in the last $d+1$ coordinates.
I've programmed this, and tested it on random polynomials of degree 20, and it successfully finds the $\pm 1$ combination leading to a monic polynomial.
In Tony Scholl's example of three different polynomials having the same data, the lattice generated has a lot of short vectors, so in that case one needs to enumerate short vectors to pick out the answers.
| 2 | https://mathoverflow.net/users/2784 | 33737 | 21,865 |
https://mathoverflow.net/questions/33720 | 1 | Dear all;
Let $\Sigma$={a,b,c,d}, and $\delta$ be a function that returns a string $S$ of infinite length over $\Sigma$, where each character $s \in S$ has been chosen uniformly at random. My questions are the following:
1. What would be an intuitive notation
for $\delta$?
2. Is $\delta$, as I've
defined it above equivalent to a
random permutation with repetition
over the kleene closure of $\Sigma$?
Thank you,
| https://mathoverflow.net/users/7995 | Permutation with repetitions of a vocabulary | For your first question, I would simply let $X$ be a random infinite string from the alphabet $\Sigma$. I don't see anything wrong with calling it $\delta$, either. I would treat it as a random variable, though. I don't really see the problem with treating it as a nondeterministic function either, but I'm not a computer scientist.
To answer your second question, let $\Sigma$ be any finite alphabet. Then $\Sigma^\star$, the Kleene closure, is the set consisting of the empty string, together with all finite strings consisting of elements of $\Sigma$. it is countably infinite, then: to list the elements, I put an ordering on $\Sigma$, and list the empty string, then the length 1 strings, then the length 2 strings, and so on, in lexicographic order. $\Sigma^\star$ contains no infinite strings.
If each string is chosen with probability $p$, then $\sum\_{S \in \Sigma^\star} = 0$ (if $p$ is 0) or is undefined (if $p$ > 0). In either case, we do not have a probability measure -- there is no uniform distribution on *any* countably infinite set, so you cannot hope to sample uniformly from $\Sigma^\star$ unless $\Sigma = \emptyset$.
The reason this argument doesn't "break" the uniform distribution on $[0,1]$, say, is because we cannot sum over an uncountable index set.
In any case, there is no uniform distribution on $\Sigma^\star$, and your random variable is sampling from a different space.
| 1 | https://mathoverflow.net/users/35336 | 33738 | 21,866 |
https://mathoverflow.net/questions/33736 | 2 | $(A,\mathfrak{m})$ a Noetherian local ring, $M\neq 0$ a finitely generated $A$-module. As I understand, $\mbox{Ext }^{j}(A/\mathfrak{m}, M) = 0$ for $j<\mbox{depth }(M)$ and for $j>\mbox{inj. dim }(M)$, while $\mbox{Ext }^{j}(A/\mathfrak{m}, M) \neq 0$ for $j=\mbox{depth }(M)$ and $j=\mbox{inj. dim }(M)$. And I cannot help but wonder if $\mbox{Ext }^{j}(A/\mathfrak{m}, M) \neq 0$ for every $j$ between $\mbox{depth }(M)$ and $\mbox{inj. dim }(M)$ ?
| https://mathoverflow.net/users/5292 | Homological dimensions of module | Yes. See Fossum, Foxby, Griffith, and Reiten, ["Minimal injective resolutions with applications to dualizing modules and Gorenstein modules"](http://archive.numdam.org/article/PMIHES_1975__45__193_0.pdf) (Theorem 1.1) and also Roberts, "[Two applications of dualizing complexes over local rings](http://archive.numdam.org/article/ASENS_1976_4_9_1_103_0.pdf)". An earlier paper by Foxby, "[On the mu\_i in a minimal injective resolution](http://www.mscand.dk/article.php?id=2032)" settles several special cases, including when $A$ or $M$ is CM, $\mathrm{depth} M \geq \mathrm{depth} A$, or $M$ has finite injective dimension.
| 4 | https://mathoverflow.net/users/460 | 33743 | 21,869 |
https://mathoverflow.net/questions/33774 | 26 | Let $k$ be a field (I'm mainly interested in the case where $k$ is a number field, however results for other fields would be interesting), and $X$ a smooth projective variety over $k$.
By a zero cycle on $X$ over $k$ I mean a formal sum of finitely many (geometric) points on $X$, which is fixed under the action of the absolute Galois group of $k$. We can define the degree of a zero cycle to be the sum of the multiplicities of the points.
Now, if $X$ contains a $k$-rational point then it is clear that $X$ contains a zero cycle of degree one over $k$.
What is known in general about the converse? That is, which classes of varieties are known to satisfy the property that the existence of a zero cycle of degree one over $k$ implies the existence of a $k$-rational point? For example what about rational varieties and abelian varieties?
As motivation I shall briefly mention that the case of curves is easy. Since here zero cycles are the same as divisors we can use Riemann-Roch to show that the converse result holds if the genus of the curve is zero or one, and there are plently of counter-examples for curves of higher genus. However in higher dimensions this kind of cohomological argument seems to fail as we don't (to my knowledge) have such tools available to us.
| https://mathoverflow.net/users/5101 | Existence of zero cycles of degree one vs existence of rational points | There has been a lot of work on this problem, although nothing like a general answer is known. By way of abbreviation, the **index** of a nonsingular projective variety is the least positive degree of a $k$-rational zero cycle, so you are asking about the relationship between index one and having a $k$-rational point.
First, you ask whether rational varieties and abelian varieties with index one must have a rational point. Here you probably mean $k$-forms of such things: i.e., *geometrically* rational varieties and *torsors* under abelian varieties. (Both rational varieties and abelian varieties have rational points, the latter by definition, the former e.g. by the theorem of Lang-Nishimura which says that having rational points is a birational invariant of a nonsingular projective variety.) I can answer this:
1. A torsor under an abelian variety has index one iff it has a rational point. This follows from the cohomological interpretation of torsors as elements of $H^1(k,A)$.
2. A geometrically rational surface of index one need not have a rational point: this is a theorem of Colliot-Thelene and Coray. (A reference appears in the link below.)
On to the general question. A very nice recent paper which proves a big result of this type and gives useful bibliographic information about other results is Parimala's 2005 paper on homogeneous varieties:
[Link](https://projecteuclid.org/journals/asian-journal-of-mathematics/volume-9/issue-2/Homogeneous-varieties---zero-cycles-of-degree-one-versus/ajm/1144070587.full)
Finally, there are some fields $k$ for which every geometrically irreducible projective variety has index one -- most notably finite fields. In this case any variety without a rational point over such a field gives a counterexample to "index one implies rational point". For instance, for any finite field $\mathbb{F}\_q$ and all sufficiently large $g$, one can easily write down a hyperelliptic cuve over $\mathbb{F}\_q$ of genus $g$ without rational points. [N.B.: What I had written before was too strong: if instead you fix $g$ and let $q$ be sufficiently large, then by the Weil bounds you must have a rational point.] There are also K3 surfaces over finite fields without rational points, and so forth.
Some further discussion of fields over which every (geometrically irreducible) variety has index one occurs in the appendix of a recent paper of mine:
[http://alpha.math.uga.edu/~pete/trans.pdf](http://alpha.math.uga.edu/%7Epete/trans.pdf)
There are many more results than the ones I've mentioned so far. If you have further questions, please don't hesitate to ask!
| 29 | https://mathoverflow.net/users/1149 | 33778 | 21,884 |
https://mathoverflow.net/questions/33710 | 3 | Let $K$ be a non-Archimedean local field, i.e., complete with respect to a non-trivial, non-archimedean discrete absolute value, with finite residue field $k$ of characteristic $p\neq 0$. Also let $K\_s$ be a fixed separable closure of $K$, and $K\_{un}$ (resp. $K\_t$) the maximal unramified (resp. tamely ramified) extensions of $K$ inside $K\_s$. Finally let $I\_K=Gal(K\_s/K\_{un})$ be the inertia group of $K$ and $P\_K=Gal(K\_s/K\_t)$. My basic question is whether or not the following statement is true:
**For every positive integer $e$ prime to $p$, there exists a unique open normal subgroup of $I\_K$ of index $e$.**
I don't recall ever seeing this explicitly stated, but I think it's plausible for the following reason. An open normal subgroup of $I\_K$ of index $e$ (with $e$ as above) is of the form $Gal(K\_s/F)$ with $F/K\_{un}$ Galois of degree $e$. Such an extension is necessarily totally tamely ramified (totally ramified because the residue field of $K\_{un}$ is algebraically closed and tame because $e$ is prime to $p$). An example of such an extension is $K\_{un}(\pi^{1/e})$, where $\pi$ is a uniformizer for $K$, which is Galois of degree $e$ since $K\_{un}$ contains $\mu\_e$ and $X^e-\pi$ is Eisenstein (over the integers of $K\_{un}$). In fact, $K\_t$ is the union of such extensions over integers prime to $p$.
If I knew (as in complete case) that every TTR extension of $K\_{un}$ of degree $e$ had this form, it would imply that $K\_{un}(\pi^{1/e})$ is necessarily the unique extension of $K\_{un}$ of degree $e$ (since the unit group of $K\_{un}$ is $e$-divisible by Hensel's lemma), which gives the statement I'm after (unless I've done something wrong).
My guess is that maybe the assertion relating TTR extensions and $e$-th roots of uniformizers really only requires a valuation ring where Hensel's lemma is valid (I guess these are called Henselian), but I've also never seen this asserted before, so I'm sort of skeptical.
| https://mathoverflow.net/users/4351 | What are the open normal subgroups of the inertia group of a local field? | As all the responses indicate, the answer to my question is "yes." The most direct route seems to be the one suggested by KConrad. Explicitly, if $F/K\_{un}$ is Galois of degree $e$ (inside $K\_s$), then the ring of integers of $F$ is a DVR, and if $\Pi$ is a uniformizer for $O\_F$, then because $F/K\_{un}$ is totally ramified, $F=K\_{un}(\Pi)$ and the minimal polynomial $f$ for $\Pi$ over $K\_{un}$ is Eisenstein (of degree $e$). Taking $K^\prime$ to be the *finite* (necessarily unramified) extension of $K$ obtained by adjoining the coefficients of $f$, $K^\prime(\Pi)/K^\prime$ is TTR of degree $e$. It is totally ramified because $f$ is still Eisenstein when viewed in $O\_{K^\prime}$ (since $K^\prime$ is unramified over $K$). Thus by the result alluded to in my question, $K^\prime(\Pi)=K^\prime((\pi^\prime)^{1/e})$ for *some* uniformizer $\pi^\prime$ in $K^\prime$, and as a result, $F=K\_{un}(\Pi)=K\_{un}((\pi^\prime)^{1/e})$. The last extension is equal to $K\_{un}(\pi^{1/e})$ since both $\pi$ and $\pi^\prime$ are uniformizers in $O\_{K\_{un}}$ and $O\_{K\_{un}}^\times$ is $e$-divisible.
| 3 | https://mathoverflow.net/users/4351 | 33782 | 21,887 |
https://mathoverflow.net/questions/33784 | 1 | As I understand, if $0\rightarrow A\rightarrow X\rightarrow B\rightarrow 0$ is a short exact sequence of abelian groups, $\mbox{Ext }\_{\mathbb{Z}}^{1}(B,A)$ gives all the isomorphism classes of what can come in as $X$. But when I consider $0\rightarrow \mathbb{Z}\rightarrow X\rightarrow \mathbb{Z}/(3)\rightarrow 0 $,$\ \ \ $ $\mbox{Ext } \_{\mathbb{Z}}(\mathbb{Z}/(3),\mathbb{Z})=\mathbb{Z}/(3)\ \ $ but all I can think of for $X$ are only two, $\mathbb{Z}$ and $\mathbb{Z}\oplus\mathbb{Z}/(3)$. Am I missing something or am I not understanding the result of extension problem correctly?
| https://mathoverflow.net/users/5292 | Extension problem | $\mathrm{Ext} = \mathrm{Ext}^1$ does not classify the middle terms up to isomorphism. It classifies the short exact sequences up to equivalence, where the equivalence relation is generated by commutative diagrams having equalities on each end. It's possible for two inequivalent short exact sequences to have isomorphic middle terms. See Eisenbud's book, Exercises A3.26, A3.27, and especially A3.29 for more.
As for your particular example, consider the pullback of the exact sequence $0 \to \mathbb{Z} \rightarrow \mathbb{Z} \to \mathbb{Z}/(3) \to 0$ by the map given by multiplication by 2 on $\mathbb{Z}/(3)$. The middle term is isomorphic to $\mathbb{Z}$, but the short exact sequence is not equivalent to the original.
| 6 | https://mathoverflow.net/users/460 | 33789 | 21,890 |
https://mathoverflow.net/questions/33145 | 3 | Suppose we have a [Markov Random Field](http://en.wikipedia.org/wiki/Markov_random_field) P(X1,...,Xn) on graph G. Suppose we know P(Xi,Xj) for every edge (i,j). Can we recover P(X1,...,Xn)?
If G is a tree, then there's a formula for joint (product of edge marginals divided by product of node marginals). Is there a nice formula that works for some non-tree graphs?
Edit: this is essentially equivalent to the following problem - given an exponential family, how do you write the joint in terms of mean parameters? There's a closed form solution when sufficient statistics are 2 variable functions defined on (Xi,Xj) pairs where (i,j) are edges in some tree graph, is there a closed form solution for other graphs?
Motivation: given an approximate marginalization method, can you fit parameters of a distribution by maximizing joint likelihood of the data under the model "implied" by this marginalization method?
| https://mathoverflow.net/users/7655 | Recovering joint distribution from marginals | No, a counter-example can be constructed as follows:
Let G be the complete graph on 3 vertices, where each vertex is a binary random variable. Let the joint distribution for each pair of vertices be independent Bernoulli with probability 1/2.
There are multiple joint distributions which satisfy such edge marginals. For example:
* Independent Bernoulli for each vertex, with probability 1/2 (complete mutual independence)
* Each variable could be the XOR (sum modulo 2) of the other two variables (i.e. functionally dependent)
* The probability of each outcome could be 3/16 if 3 or 1 vertices are 1's, and 1/16 if 2 or 0 are 1's.
So what do you require to be able to uniquely determine the joint distribution? If the graph is decomposable (aka chordal or triangulated), you require the joint distribution for each clique (maximal complete subset) of the graph. Then the joint density is then:
$p(X) = \frac{\prod\_{\text{cliques }C} p(X\_C)}{\prod\_{\text{separators }S} p(X\_S)}$
(see [Dawid and Lauritzen (1993)](http://projecteuclid.org/euclid.aos/1176349260), Lemma 2.5)
If the graph is not decomposable, then the problem is a bit trickier: the only result I know of is [Lauritzen (1996)](http://books.google.co.uk/books?id=mGQWkx4guhAC&printsec=frontcover), Lemma 3.14. Basically, given the clique marginal distributions, uniqueness is determined when the sample space is finite, and each clique marginal density is the limit of a sequence of positive densities. I suspect this result could be made stronger in some way, but I am not aware of any efforts to do so.
| 2 | https://mathoverflow.net/users/8019 | 33794 | 21,892 |
https://mathoverflow.net/questions/33798 | 7 | Let $E\_k$ be the normalized Eisenstein series of weight k and let p be an odd prime. Then
$$
E\_{p^m(p-1)} = 1 \mod p^{m+1},
$$
and so the p-adic limit $\lim E\_{p^m(p-1)} = 1$ is a p-adic modular form of weight 0. (It is even overconvergent.)
>
> **Question:** Suppose f is a p-adic modular form whose q-expansion is a polynomial. Is the q-expansion of f a constant?
>
>
>
| https://mathoverflow.net/users/2 | Can a the q-expansion of a p-adic modular form be a non-constant polynomial? | It is. I want to argue the following way: if the polynomial is non-constant then after scaling it has integral coefficients and so the reduction of the p-adic form mod p^n will be a *classical* form whose q-expansion is a non-constant polynomial. But I think Katz proved in his Antwerp paper that the only modular forms which are polynomials in q are the constants, which is a contradiction. One needs to dot some i's and cross some t's here, but I am optimistic.
| 13 | https://mathoverflow.net/users/1384 | 33800 | 21,895 |
https://mathoverflow.net/questions/33805 | 5 | The fact that the atlas using $\phi: x \mapsto x^{1/3}$ on $\mathbb{R}$ is diffeomorphic to the trivial atlas using $\psi: x \mapsto x$ on $\mathbb{R}$ highlights my ignorance of diffeomorphisms and atlases. Apologies in advance for clustering several questions, but I'm not sure how to disentangle them.
First of all, which is the more relevant/correct statement here: that the atlases are diffeomorphic, that the manifolds are diffeomorphic ($M \mapsto N$), that the manifold-atlas pairs are diffeomorphic ($(M,\psi) \mapsto (N,\phi)$), or that the manifolds and their differential structures are diffeomorphic ($(M,\mathcal{A}) \mapsto (N,\mathcal{B})$)?
Is the relevant diffeomorphism patently obvious here given that the domain of $x \mapsto x^{1/3}$ is the same as that of $x \mapsto x$?
If we had two homeomorphic manifolds with atlases that were not diffeomorphic, would it be obvious?
| https://mathoverflow.net/users/7956 | Why is the x->x^1/3 atlas on R diffeomorphic with the x->x atlas on R? |
>
> The fact that the atlas using $x \mapsto x^{1/3}$ on $\mathbb{R}$ is diffeomorphic to the trivial atlas using $x \mapsto x$ on $\mathbb{R}$ highlights my ignorance of diffeomorphisms and atlases.
>
>
>
Other differential topologists should weigh in on this to confirm or deny, but as a differential topologist I have never come across the notion of a diffeomorphism between two **atlases**, or even two smooth structures. Moreover, what you have here is not two atlases but two charts. These may seem like picky points, but if you find yourself getting confused then one good technique to learn is to sharpen your definitions. By that, I mean be a bit more careful about distinguishing between things that although often used synonymously are actually distinct.
So you have two charts, $x \mapsto x^{1/3}$ and $x \mapsto x$. As both of these have image $\mathbb{R}$, they each define an atlas: $\{x \mapsto x^{1/3}\}$ and $\{x \mapsto x\}$. Each of these atlases then defines a smooth structure on $\mathbb{R}$. Each of these smooth structures defines a smooth manifold with underlying topological space $\mathbb{R}$. Although each of these constructions follows in a unique way from the previous step, technically each is a different thing.
Back to the confusion about diffeomorphisms. We talk of two atlases being **equivalent** if they generate the same smooth structure, or if they define the same smooth manifold. In concrete terms, we can test this by looking to see if the **identity** map is smooth in both directions (using the atlases to test smoothness).
But I can have inequivalent atlases that nonetheless define diffeomorphic manifolds. This is because the condition of being *equivalent* is stronger than that of defining diffeomorphic manifolds. Equivalence rests on the smoothness of the identity map (in both directions), the manifolds being diffeomorphic rests on the smoothness of *some* map (and its inverse). So although the two atlases given are inequivalent, they define diffeomorphic manifolds because I'm free to take the map $x \mapsto x^{1/3}$ and its inverse to define the diffeomorphism.
It's a good exercise to help with sorting out the definitions to check that you really understand why these two manifold structures on $\mathbb{R}$ are diffeomorphic. That is, write down the map and write out its compositions with the transition maps and **see** that it works.
| 21 | https://mathoverflow.net/users/45 | 33818 | 21,904 |
https://mathoverflow.net/questions/33816 | 1 | I have a Banach space $B$ and a continuous linear transformation $F:B \rightarrow B\times B$. One of the induced transformations $F(1):B \rightarrow B$ and $F(2):B \rightarrow B$ into the factors of the product has closed range. Must F have closed range? I have the max norm on the product, i.e., $\|F(x) \| = max\{\|F(1)(x)\|, \|F(2)(x)\|\}$ for $x$ in $B$. I was hoping to use the minimum moduli of the
$F(i)$ to provide an affirmative answer.
| https://mathoverflow.net/users/8027 | Closed range for a continuous linear transformation | No. Consider $F(x)=(f(x),0)$ where $f$ does not have closed range.
| 2 | https://mathoverflow.net/users/2554 | 33824 | 21,908 |
https://mathoverflow.net/questions/33788 | 11 | This question is inspired by [Emerton's question](https://mathoverflow.net/questions/28776/does-the-ideal-class-of-the-different-of-a-number-field-have-a-canonical-square) whether the ideal class of the different has a canonical square root.
Consider the diagram (of elements; the groups these lie in are the ideal class group of $L$ in the top row and that of $K$ in the bottom row)
$$ \matrix{
? & \to & [diff(L/K)] \cr
\downarrow & & \downarrow \cr
[St(L/K)] & \to & [disc(L/K)]} $$
where the horizontal maps are squaring and the vertical maps are taking norms from $L$ down to $K$. Here $diff(L/K)$ is the different of an extension of number fields, and $disc(L/K)$ its discriminant. The element $[St(L/K)]$ of the class group of the base field $K$ is the Steinitz class (see KConrad's comment to Emerton's question).
My question is whether there exists an element in the class group $Cl(L)$ that is at home in the left upper corner of this diagram. The most simple question would be whether the Steinitz class is always a norm; and if it is, whether it is the norm of a class whose square is the different class. The last question would be whether this element "?" is unique up to elements that lie in the intersection of the kernel of the norm and that of squaring.
| https://mathoverflow.net/users/3503 | Does the ideal class of the different have a functorial square root? | Taking $L$ to be the Hilbert class field of $K$, such a construction
would imply that the Steinitz class of $L/K$ is always trivial.
Yet this is false - take $K = \mathbb{Q}(\sqrt{-15})$ for example.
(EDIT): If $L = K(\sqrt{\alpha})$ is a tamely ramified extension of $K$, then
the Steinitz class is represented by an ideal $I$ such that $I^2 (\alpha) = \Delta\_{L/K}$.
In particular, if $L/K$ is unramified, so $(\alpha) = \mathfrak{n}^2$, then the Steinitz class is trivial if and only if $\mathfrak{n}$ is principal, i.e., if and only if we may take
$\alpha$ to be a unit in $K$. Clearly this is not the case for $L/K$ above.
In fact, I just found a source that works out this example in explicit detail - see Theorems 2.2 and 3.1 of the following:
<http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/notfree.pdf>
| 12 | https://mathoverflow.net/users/nan | 33827 | 21,911 |
https://mathoverflow.net/questions/29347 | 3 | Given a finite set of statements known to be true, I need to derive all the "non-redundant" statements in disjunctive form using only literals that can be derived from this set of statements, i e all statements on the form (a ∨ b ∨ c ∨ ...) where a, b ... are literals. By non-redundant I mean that I do not wish to include a statement if one of the literals could be removed and it would still always be true (given the initial set of statements).
Another way of looking at this would be that I want a statements in conjunctive normal form, but that includes all non-redundant statements as the inner groups, not just the minimal set equivalent to the initial set of statements.
A naive algorithm for this would be to test all assignments of truth values to literals, keeping only the ones that 1) are true given the initial set of statements 2) are non-redundant (tested by assigning false to each of the literals in the statement, checking if it still remains slow). However, the number of literals I'll be working with will be so large that this naive approach is not feasible.
**Edit:**
What I'm looking for is an algorithm that will efficiently produce the set of statements as described above, such as the strategies that can be used to convert to a conjunctive normal form.
| https://mathoverflow.net/users/7049 | Deriving the complete set of "non-redundant" true statements in disjunctive form in propositional logic | I believe your problem is: Given a boolean function $\phi$, find the set of its prime clauses (aka prime implicates). This is equivalent to finding all the prime implicants of $\lnot\phi$.
You will find in the paper [A Knowledge Compilation Map](http://scholar.google.co.uk/scholar?cluster=4020517321592725796) many references to papers that discuss variants of this problem. For example, the paper [Algorithms for Selective Enumeration of Prime Implicants](http://scholar.google.co.uk/scholar?cluster=3018518818059820501) focuses on the case when $\phi$ is given in DNF or CNF (and also has many good references). Another interesting paper is [Implicit and Incremental Computation of Primes and Essential Primes of Boolean Functions](http://scholar.google.co.uk/scholar?cluster=9792089307967946883) that represents the resulting set as a BDD, which may be much smaller than a simple list. (Though a ZBDD, invented later, might work even better.)
In short, there are lots of algorithms, none works fast all the time, but some might be fast enough in practice.
(Note: Knuth uses the term conjunctive prime form for the conjunction of prime clauses. This appears in fascicle 0 of volume 4 of [TAoCP](http://www-cs-faculty.stanford.edu/~uno/taocp.html) from 2005-2009. But the name didn't yet caught on.)
| 2 | https://mathoverflow.net/users/840 | 33835 | 21,914 |
https://mathoverflow.net/questions/33817 | 32 | It is an open problem to prove that $\pi$ and $e$ are algebraically independent over $\mathbb{Q}$.
* What are some of the important results leading toward proving this?
* What are the most promising theories and approaches for this problem?
| https://mathoverflow.net/users/4361 | Work on independence of pi and e | People in model theory are currently studying the complex numbers with exponentiation.
Z'ilber has an axiomatisation of an exponential field (field with exponential function) that looks like the complex numbers with exp. but satisfies Schanuel's conjecture.
He proved that there is exactly one such field of the size of $\mathbb C$.
I would find it odd if Z'ilber's field turned out to be different from the complex numbers.
By results of Wilkie, the reals with exponentiation are well understood, and the complex numbers with exponentiation is in some way the next step up. The model theoretic frame work
(o-minimality) that works for the reals with exp. fails for the complex numbers, but there might be a similar theory that works for the complex field with exponentiation.
| 35 | https://mathoverflow.net/users/7743 | 33837 | 21,916 |
https://mathoverflow.net/questions/33833 | 8 | I've been reading the book of Hilton, Mislin, and Roitberg on Localization of Nilpotent Groups and Spaces. In Section II.2 they define a principal refinement at stage $n$ of a Postnikov system $$\cdots \to X\_n\overset{p\_n}{\to} X\_{n-1}\to \cdots$$
to be a factorization of $p\_n$ into a finite sequence of fibrations
$$X\_n=Y\_c \overset{q\_c}{\to}\cdots\to Y\_1\overset{q\_1}{\to}Y\_0=X\_{n-1}$$
whose fibers are Eilenberg-MacLane spaces $K(G\_i,n)$.
But isn't the point of a Postnikov system that $p\_n$ is already a fibration whose fibers are Eilenberg-MacLane spaces? So I don't understand why the condition of the definition isn't satisfied trivially at every stage. Perhaps there's some subtlety involving the condition also given that each $q\_i$ be induced by a map $g\_i: Y\_{i-1}\to K(G\_i, n+1)$, but aren't all fibrations with fiber $K(G\_i, n)$ induced this way since $K(G\_i, n+1)$ is the base of a path-space fibration with fiber $K(G\_i, n)$?
| https://mathoverflow.net/users/6646 | What is a principal refinement of a Postnikov system? | The key idea is that not all fibrations $E \to B$ with fibre an Eilenberg-MacLane space $K(\pi,n)$ can be constructed by pulling the principal path fibration $K(\pi,n) \to PK(\pi,n+1) \to K(\pi,n+1)$ along a classifying map $B \to K(\pi,n+1)$. If you can construct the fibration in this way then the classifying map is the Postnikov $k$-invariant. Clearly this at least requires that the group $\pi$ is abelian.
Now, it is a nice little exercise to check that existence of a principal refinement of the Postnikov tower is equivalent to $\pi\_1$ being nilpotent and acting nilpotently on all of the higher homotopy groups.
(Recall that a group $G$ acts nilpotently on a group $H$ if $H$ has a finite sequence of $G$-invariant subgroups $H \supset H\_1 \supset H\_2 \supset \cdots H\_k = 1$ such that $H\_i/H\_{i+1}$ is abelian and the action of $G$ on it is trivial.)
The general idea of Hilton, Mislin, and Roitberg is that is it obvious how to localise abelian groups, and nilpotent groups are those which can be assembled from abelian groups one layer at a time. So we can localise nilpotent groups by working one layer at a time, and then we can localise nilpotent spaces by working up the refined Postnikov tower one stage at a time.
| 10 | https://mathoverflow.net/users/4910 | 33845 | 21,919 |
https://mathoverflow.net/questions/33842 | 23 | The Suzuki and Ree groups are usually treated at the level of points. For example, if $F$ is a perfect field of characteristic $3$, then the Chevalley group $G\_2(F)$ has an unusual automorphism of order $2$, which switches long root subgroups with short root subgroups. The fixed points of this automorphism, form a subgroup of $G\_2(F)$, which I think is called a Ree group.
A similar construction is possible, when $F$ is a perfect field of characteristic $2$, using Chevalley groups of type $B$, $C$, and $F$, leading to Suzuki groups. I apologize if my naming is not quite on-target. I'm not sure which groups are attributable to Suzuki, to Ree, to Tits, etc..
Unfortunately (for me), most treatments of these Suzuki-Ree groups use abstract group theory (generators and relations). Is there a treatment of these groups, as *algebraic groups* over a base field? Or am I being dumb and these are not obtainable as $F$-points of algebraic groups.
I'm trying to wrap my head around the following two ideas: first, that there might be algebraic groups obtained as fixed points of an algebraic automorphism that swaps long and short root spaces. Second, that the outer automorphism group of a simple simply-connected split group like $G\_2$ is trivial (automorphisms of Dynkin diagrams mean automorphisms that preserve root lengths).
So I guess that these Suzuki-Ree groups are inner forms... so there must be some unusual Cayley algebra popping up in characteristic 3 to explain an unusual form of $G\_2$. Or maybe these groups don't arise from algebraic groups at all.
Can someone identify and settle my confusion?
Lastly, can someone identify precisely which fields of characteristic $3$ or $2$ are required for the constructions of Suzuki-Ree groups to work?
| https://mathoverflow.net/users/3545 | Suzuki and Ree groups, from the algebraic group standpoint | It is not really a question of inner forms. What happens is that the
*algebraic group* $G\_2$ has an extra endomorphism $\varphi$ whose square
is the Frobenius map (over the appropriate finite field). Just as for any
algebraic group over a finite field $F$ its rational points over $F$ are the
fixed points of the Frobenius endomorphism the Suziki groups are, by definition,
the fixed points of $\varphi$. Again, just as the Frobenius, on points over the
algebraic closure of $F$ $\varphi$ is an automorphism of the abstract group.
However, that is misleading, the essential points is that it is an endomorphism
(which definitely is not an automorphism) of the algebraic group. Most of the
properties of points over $F$ of a semi-simple algebraic group $G$ defined over
$F$ follows from the algebro-geometric theory of $G$ and the properties of the
Frobenius endomorphism. Similarly, most of the properties of Suziki groups
follows from the algebro-geometric theory of $G\_2$ together with the properties
of $\varphi$. As $\varphi$ is very similar to the Frobenius endomorphism this
works almost the same way as if $\varphi$ were indeed a Frobenius endomorphism.
**Addendum**: As one simple example of the similarity of $\varphi$ to a Frobenius consider the problem of computing the order of the Suzuki groups. As the square of $\varphi$ is the Frobenius, the action of it on the tangent space at any fixed point is nilpotent. This implies that such a fixed point appear with multiplicity one in the Lefschetz fixed point formula and the order of its group of fixed points is thus equal to the Lefschetz trace on (étale) cohomology of the algebraic group $G\_2$. That cohomology can be canonically expressed in terms the action of the Weyl group on the character group of the maximal torus (see for instance example in SGA 4 1/2) and how $\varphi$ acts on that character group is essentially part of the definition of $\varphi$.
| 17 | https://mathoverflow.net/users/4008 | 33847 | 21,921 |
https://mathoverflow.net/questions/32986 | 27 | This may be more of a recreational mathematics question than a research question, but I have wondered about it for a while. I hope it is not inappropriate for MO.
Consider the standard assumptions for ruler and compass constructions: We have an infinitely large sheet of paper, which we associate with the complex plane, that is initially blank aside from the points 0 and 1 being marked. In addition we have an infinite ruler and a compass that can be stretched to an arbitrary length.
Let us define a *move* to be one of the two actions normally associated with a ruler and compass:
1. Use the ruler to draw the line defined by any two distinct points already marked on the paper.
2. Stretch the compass from any one marked point to another and draw the resulting circle.
Assume that all intersection points among lines and circles drawn by these operations are automatically marked on the paper.
Now define $D(n)$ to be the maximum distance between any two marked points that can be constructed in this way with $n$ moves.
Questions:
1. Is anyone aware of results about the function $D(n)$ or something equivalent?
2. It is not difficult to prove $D(n) > 2^{2^{cn}}$ for some positive constant $c$ for sufficiently large $n$. Can one do better? If so, can one prove an upper bound on $D(n)$?
| https://mathoverflow.net/users/7641 | How fast are a ruler and compass? | *Edit on July 31: Now the upper bound is tight (up to replacing n by O(n)). The improvement over the older version is in the argument after Lemma 1. Now we consider the Weil absolute logarithmic height of an algebraic number instead of the length.*
Here is a proof that D(n) < 22cn for some positive constant c>0 for sufficiently large n. In other words, the answer to the “can one do better” part of the question 2 is negative.
As Terry Tao pointed out, this problem can be rephrased in the algebraic form. We have z0=0 and z1=1, and any zn (n≥2) can be obtained from earlier numbers by addition, subtraction, multiplication, division or square root. The claim follows if |zn| < 22O(n).
**Lemma 1**: The degree of zn over ℚ is at most 2n.
Proof: Let Fn=ℚ(z0, …, zn) be the minimum field containing ℚ∪{z0, …, zn}. Then F0=ℚ, and Fn is either equal to Fn−1 or an extension of Fn−1 obtained by adjoining a square root. Since adjoining a square root of a non-square element gives an extension of degree 2, the extension degree [Fn:Fn−1] is either 1 or 2. By the degree formula, it holds that [Fn:ℚ] = [Fn:F0] = [Fn:Fn−1][Fn−1:Fn−2]…[F1:F0] ≤ 2n. Therefore, the degree of every element in Fn over ℚ is also at most 2n. (end of proof of Lemma 1)
There is a function called the *Weil absolute logarithmic height* h(α) defined on algebraic numbers α which takes nonnegative real values. See Section 3.2 of [Wal00] for its definition and the proof of the following properties:
1. If α is an algebraic number of degree d, then |α| ≤ exp(dh(α)).
2. If p and q are integers which are relatively prime, then h(p/q) = ln max{|p|,|q|}.
3. If α and β are algebraic numbers, then h(α+β) ≤ h(α) + h(β) + ln 2.
4. If α and β are algebraic numbers, then h(αβ) ≤ h(α) + h(β).
5. If α is an algebraic number and n is an integer, then h(αn) = |n|h(α). In particular, h(√α)=h(α)/2.
By using the properties 2–5 and the mathematical induction, we can prove that h(zn) ≤ 2n ln 2. By combining the property 1 and Lemma 1, we obtain that |zn| ≤ 222n.
References
[Wal00] Michel Waldschmidt: *Diophantine Approximation on Linear Algebraic Groups: Transcendence Properties of the Exponential Function in Several Variables*, Springer, 2000.
| 17 | https://mathoverflow.net/users/7982 | 33857 | 21,928 |
https://mathoverflow.net/questions/33854 | 6 | How does one solve the diophantine equation $x^2 + y^2 = z^2 + w^2 $? Can solutions be parameterized in three variables analogously to the Pythagorean triples case?
| https://mathoverflow.net/users/1358 | integer solutions to quadratic forms | Here is the standard geometric argument: after extracting common factors, you are asking for rational points on the quadric $Q\colon x^2-w^2=z^2-y^2$ in $\mathbb{P}^3$, which is isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ (as Matt Young's comment explains). Projection from a point on $Q$ gives a birational map $p:Q\to \mathbb{P}^2$ hence a rational parameterisation of $Q$. Explicitly, projecting from $P\_0=(0:1:0:1)$ to the $(w=0)$-plane gives the parameterisation
$$
p : (a:b:c)\in\mathbb{P}^2 \mapsto (2ab:c^2-a^2+b^2:2bc:c^2-a^2-b^2) \in Q
$$
of all the points except for the two lines on $Q$ passing through $P\_0$ - i.e. the intersection of $Q$ with the tangent plane at $P\_0$, which are $z=w$, $x=\pm y$. Notice also that the points $\{b=0,a\ne\pm c\}$ all go to $P\_0$, for the same reason.
The advantage of this method is that there is nothing particularly special about the quadratic form $x^2+y^2-z^2-w^2$ here; the same argument parameterises the rational zeros of any nonsingular quadratic form in 3 or more variables, as soon as it has one non-trivial zero.
| 15 | https://mathoverflow.net/users/5480 | 33864 | 21,932 |
https://mathoverflow.net/questions/33861 | 10 | Looking at the chart of cardinals in Kanamori's book, one realizes that all large cardinals are implied by stronger ones and imply weaker ones. For instance measurable implies Jonsson which implies zero sharp which implies weakly compact which implies Mahlo which implies inaccessible. So it seems as if all these large cardinal assumptions are linearly ordered by consistency strength. Is there a some assumption above ZFC that is not implied by and does not imply any of the linearly ordered large cardinals?
| https://mathoverflow.net/users/3859 | Large cardinals | By the well-known Levy-Solovay theorem, large cardinal properties are preserved under "small" forcing. Therefore CH is an assumption above ZFC which is not settled by large cardinal axioms.
| 10 | https://mathoverflow.net/users/3532 | 33870 | 21,935 |
https://mathoverflow.net/questions/32801 | 8 | A secret sharing scheme such as [Shamir's secret sharing](http://en.wikipedia.org/wiki/Shamir%27s_Secret_Sharing) allow to perform addition and multiplication for secret values so far as there is at least 3 participants. Addition of two secret values is done locally at each party by adding the corresponding local shares, so it is possible to do addition even with only 2 parties. In the case of multiplication, a degree reduction step is obligatory, because multiplication increases the degree of the shares and this makes it impossible for 2 parties to perform multiplications without the aid of a third one.
It is possible to construct a multiplicative sharing scheme that works for two parties, but it wouldn't be additive.
>
> Is any one aware of a secret sharing
> scheme for two parties (without the
> need for a third party) that is BOTH
> additive and multiplicative, or is it
> impossible ?
>
>
>
So far I know that it is impossible to construct scalar product protocol with unconditional security for two parties. But I don't suppose that it necessarily prevents the construction of an algebraic (additive+multiplicative) secret sharing scheme.
Update: I am aware of homomorphic encryption and the existence of algebraic homomorphic encryption schemes. However homomorphic encryption is not unconditionally secure unlike secret sharing which is.
| https://mathoverflow.net/users/7685 | Is there a two-party multiplicative and additive secret sharing scheme ? | Unconditionnaly secure 2-party computation does not exist (unfortunately). This is derived from the impossibility of Oblivious Transfer. Also note that unconditionnaly secure OT is also impossible if the 2 parties are quantum.
| 1 | https://mathoverflow.net/users/6673 | 33884 | 21,942 |
https://mathoverflow.net/questions/33883 | 5 | Suppose $E$ is a vector bundle with structure group $G$ and let $P = F(E)$ be the frame bundle. Let $\mathfrak{g}\_P$ denote the associated bundle to the adjoint representation of $G$ on its Lie algebra (i.e. $\mathfrak{g}\_P = P \times\_G \mathfrak{g}$ where $\mathfrak{g}$ is the Lie algebra of $G$). Given a section of $\mathfrak{g}\_P$, I should be able to "exponentiate" it pointwise to get a gauge transformation. How is this defined? I couldn't come up with anything well-defined.
For some context, I was reading Chapter 2 of Donaldson and Kronheimer's "The Geometry of 4-Manifolds," and they mention this pointwise exponential in passing on p. 33. I'm guessing they assume the reader is familiar with it from a more elementary text, but I looked in a few other books and couldn't find it.
| https://mathoverflow.net/users/8041 | How do you exponentiate a section of the adjoint bundle to get a gauge transformation? | I'm not sure what it is that you tried, but the exponential map should work. First of all, let $U\_i$ be a trivialising cover for $P$ and its associated bundles. A section through $\mathfrak{g}\_P$ is given by functions $\omega\_i: U\_i \to \mathfrak{g}$ which, on overlaps, transform according to
$$\omega\_i(p) = \operatorname{Ad}\_{g\_{ij}(p)} \omega\_j(p) \qquad \forall p \in U\_i \cap U\_j,$$
where I use $\operatorname{Ad}$ to mean the adjoint representation of $G$ on its Lie algebra $\mathfrak{g}$.
Now simply compose $\omega\_i$ with the exponential map $\exp: \mathfrak{g} \to G$, resulting in functions $\exp\omega\_i : U\_i \to G$ which, on overlaps, transform according to
$$\exp\omega\_i(p) = g\_{ij}(p) \exp\omega\_j(p) g\_{ij}(p)^{-1} \qquad \forall p \in U\_i \cap U\_j.$$
But this is just a section through the associated fibre bundle usually denoted $\operatorname{Ad} P$, and that is the same thing as a gauge transformation.
| 7 | https://mathoverflow.net/users/394 | 33887 | 21,944 |
https://mathoverflow.net/questions/33877 | 10 | Let A be an algebra (or dg algebra). Where can I find a proof of HH\_\*(A) = HH\_\*(Mod\_A) and HH^\*(A) = HH^\*(Mod\_A)? (And does this hold for any A?) Here Mod\_A is, e.g., the category of left A-modules.
One reason why this is interesting/important/useful is because many categories which arise "in nature" are of the form Mod\_A. For example, there is a theorem of Bondal and van den Bergh which states that derived categories of a large class of varieties (I forget their exact hypotheses) are equivalent to Mod\_A for some A. Dyckerhoff also proved that categories of matrix factorizations are of this form. By mirror symmetry, Fukaya-type categories should be of this form as well...
Anyway, so to compute HH of such a category, it suffices to find this A and then compute HH(A). I think that it generally(?) should be easier to compute HH of an algebra than HH of a category. (Of course finding this A can be a very nontrivial task.)
| https://mathoverflow.net/users/83 | Hochschild (co)homology of A and of Mod_A | Basically this follows from the fact that the derived category of bimodules over two algebras is equivalent to the (suitably defined) functor category between the derived category of modules of each algebra. Say, Toen's paper on derived Morita equivalence. Then, the identity functor is given by the algebra itself interpreted as a bimodule, so the Hochschild cohomology is $\mathrm{Ext}^i\_{A-A}(A,A)$. You can compute this using the bar resolution and a quick calculation gives you the usual definition of Hochschild cohomology.
| 8 | https://mathoverflow.net/users/947 | 33889 | 21,945 |
https://mathoverflow.net/questions/33893 | 4 | There is a result on the dimension bound for ${M\_{g,n}}/S\_n$, (the moduli space for Riemann surfaces of genus $g$ with $n$ marked points) that is
$H\_{i}({M\_{g,n}}/S\_n)=0$, for $i\ge 6g-7+2n$ except $(g,n)=(0,3),(0,2),(0,1),(0,0),(1,1)$. This result (see Costello: Gromov-Witten potential associated to a TCFT) can be derived from the virtual cohomology dimension of the mapping class group (see J. L. Harer, The virtual cohomological dimension of the mapping class group of an orientable surface, Invent. Math). I am wondering if there is such a theorem for surfaces with boundary. For example, is there any similar result for mapping class groups of orientable surfaces with boundary(and marked points if necessary)? Then can we get a result similar to the above dimension bound for moduli spaces of Riemann surfaces with boundary and marked points? I just want to know if such a result already exists.
Can we use reduced homology so that we don 't need exclude those low dimensional cases? I mean if it is natural?Thanks!
| https://mathoverflow.net/users/2391 | Homology dimension of the mapping class group of a surface with boundary | There is a fibration sequence
$$\mathbb{S}(\Sigma\_g) \to \mathcal{M}\_{g}^1 \to \mathcal{M}\_g$$
where $\mathcal{M}\_g$ is the moduli space of Riemann surfaces, $\mathcal{M}\_{g}^1$ is the moduli space of Riemann surfaces with a single boundary, and $\mathbb{S}(\Sigma\_g)$ is the sphere bundle associated to the tangent bundle of a surface of genus $g$.
As the homology of $\mathcal{M}\_g$ vanishes in degrees at least $6g-7$, and $\mathbb{S}(\Sigma\_g)$ is a 3-manifold, the Serre spectral sequence implies that the homology of $\mathcal{M}\_{g}^1$ vanishes in degrees at least $6g-4$.
Iterating, the homology of $\mathcal{M}\_{g, n}^b$ vanishes in degrees at least $6g-7+2n+3b$.
| 2 | https://mathoverflow.net/users/318 | 33899 | 21,950 |
https://mathoverflow.net/questions/33896 | 25 | I was always bothered by the definition of the cross product given in e.g. a calculus course because it's never made clear how one would go about defining the cross product in a coordinate-free manner. I now know, not one, but *two* ways of doing this, and I can't quite see how they're related:
* The cross product is the Lie bracket in the Lie algebra of $\text{SO}(3)$.
* The cross product is the Hodge star map $\Lambda^2(V) \to V$ where $V$ is an oriented $3$-dimensional real inner product space.
Okay, so there's one obvious relation here: $V$ has automorphism group $\text{SO}(3)$. But for some reason I can't figure out where to go from here. A good starting point would be to exhibit a canonical isomorphism between an oriented $3$-dimensional inner product space $V$ and the Lie algebra of $\text{Aut}(V)$. Maybe this is obvious. In any case, I would appreciate some clarification.
| https://mathoverflow.net/users/290 | How are these two ways of thinking about the cross product related? | To expand on Victork Protsak's comment, if $V$ is an $n$-dimensional real vector space with inner-product, the inner-product gives an isomorphism $V\to V^\*$ and hence $V\otimes V \to \mathrm{End}(V)$. Under this isomorphism, $\Lambda^2(V)$ is identified with skew-adjoint endomorphisms of $V$, which is precisely the Lie algebra $\mathfrak{so}(V)$.
In the case $\dim V =3,$ the Hodge star gives an isomorphism $\Lambda^2(V) \to V$ and so in total we see that $V$ is canonically isomorphic to $\mathfrak{so}(V)$. A more direct way to see this isomorphism is to send the vector $v \in V$ to the generator of the right-handed rotation about the axis in the direction of $v$ with speed $|v|$.
The use of the phrase "right-handed" makes it clear that in order to identify $V$ and $\mathfrak{so}(V)$ we have used an orientation on $V$; indeed, you need that for the Hodge star. What is interesting is that if you reverse the orientation on $V$, the map to $\mathfrak{so}(V)$ changes sign. This means that what ever orientation you chose on $V$, the push-forward to $\mathfrak{so}(V)$ is the same. Conclusion: $\mathfrak{so}(3)$ is *naturally oriented*. This is analogous to the natural orientation on $\mathbb{C}$. A more prosaic way to describe the orientation is to pick two independent elements $x,y \in \mathfrak{so}(3)$ and then use $[x,y]$ to complete them to an oriented basis. (Of course, you then need to check that this doesn't depend on your choice of $x,y$.)
| 22 | https://mathoverflow.net/users/380 | 33903 | 21,953 |
https://mathoverflow.net/questions/33888 | 6 | For finite sets $A$ and $B$, it is clear that $A \subseteq B$ and $|A| \geq |B|$ implies $A = B$. While an obvious fact, it can sometimes be a nice shortcut in proofs.
Analogously, if $V$ and $W$ are finite-dimensional vector spaces such that $V \subseteq W$ and $dim\ V \geq dim\ W$ then $V = W$. This is an especially useful tool when $V$ is defined parametrically and $W$ is defined implicitly. Then you can easily prove $V \subseteq W$ by plugging the parametric expression for $V$ into the equations for $W$. Counting the dimensions can take more work, but you sometimes get lucky.
For a while I've been wondering how this extends to algebraic varieties.
Here's an attempted application to proving the spectral theorem. Fix the dimension $n$; all matrices will be $n \times n$. The theorem says that for every symmetric matrix $S$ there exists an orthogonal matrix $Q$ and a diagonal matrix $D$ such that
$$Q^T\ D\ Q = S.$$
Let $A$ and $B$ respectively denote the matrices of the form on the left-hand and right-hand side. $A$ is defined parametrically by a function $f$ from $D$ and $Q$, and $B$ is defined implicitly by the symmetry condition. We want to prove $A = B$. It is clear that $A \subseteq B$:
$$(Q^T\ D\ Q)^T = Q^T\ D^T\ (Q^T)^T = Q^T\ D\ Q,$$
so $Q^T\ D\ Q$ is symmetric.
The domain of $f$ has dimension $dim\ D + dim\ Q$ where $dim\ D = n$ and $dim\ Q = (n-1) + \cdots + 1$, while $S$'s space has dimension $n + (n-1) + \cdots + 1$, so the dimensions seem to match.
But what about $f$'s degree of injectivity? It isn't perfectly injective: if $D$ and $D'$ equal the identity matrix then $Q^T\ D\ Q = Q'^T\ D'\ Q'$ for any independent combination of $Q$ and $Q'$.
My question is thus twofold:
>
> What is the right generalization of the theorem for vector spaces to algebraic varieties?
>
>
>
and
>
> Can the attempted proof of the spectral theorem be salvaged with a genericity argument?
>
>
>
I'm happy with the extant proofs of the spectral theorem, so this is more curiosity than anything else.
| https://mathoverflow.net/users/2036 | Proving equality of varieties by dimension counting | I guess this answers at least part of the question, so I will post it as an answer. As in my comment above: the correct generalization is that if $X \subset Y$ is a closed subvariety, if $X$ and $Y$ are both irreducible, and if $\dim X = \dim Y$, then $X = Y$. At this level of generality this is an application of the noetherian property of finite-dimensional varieties over a field together with the definition of irreducible. From this you can deduce that even if $X$ is *locally closed* in $Y$, it must be dense (since $X$ has the same dimension as its closure). An even more general form of this theorem is that if you have merely a dominant map $X \to Y$ (that is, its image is dense) then the minimal dimension of a fiber is $\dim X - \dim Y$ and this occurs generically ($X$ and $Y$ should still be irreducible).
This is often combined with another theorem: if $X$ is irreducible and $X \to Y$ is surjective, then so is $Y$; conversely, if $Y$ is irreducible and the fibers of that map are irreducible of the same dimension, then so is $X$. Now you can perform the following trick beloved of algebraic geometers (e.g. anything in Harris' "Algebraic Geometry: A First Course", under the aegis of "incidence correspondences"; this theorem is 11.14): to compute the dimension of $Y$, cover it by $X$ and then have $X$ cover some other, irreducible $Z$ whose dimension is easy to compute. Then apply both the irreducibility and dimension theorems twice. Unfortunately, unless they are all projective varieties (which they often are in applications of this method) this is more of a way of computing dimension than of showing that maps are surjective (however, see for example the appendix to section 4 of Mumford's "Abelian Varieties" for an impressive example of this).
To show that $X \to Y$ is surjective, one general method that comes to mind is to show that it is dominant and has some kind of "closure" property. Either it could be actually closed, or perhaps you have a group acting on both $X$ and $Y$, transitively on $Y$, for which the map is equivariant. IANAAG (I am not an algebraic geometer) so this is probably hopelessly naive as far as general methods go.
| 8 | https://mathoverflow.net/users/6545 | 33908 | 21,957 |
https://mathoverflow.net/questions/15204 | 13 | $\bf Definition.$ We define the space bounded communication in the following way. A and B are
supernatural beings capable of computing anything but
they only have a limited amount of memory and that is shared. The
minimum size of this common memory that they can use to evaluate
a given function $f$ for which both of them possesses one half of the input, resp. $x$ and $y$,
shall be denoted by $S(f)$. At the
beginning it is filled with zeros. Then in each step one of the
players can put there an arbitrary message depending only on the
previous message and his input. They are finished when both of
them knows the value of $f(x,y)$. We can also imagine this as two
people communicating who have no memory at all (however, they can
remember their own input) and are allowed to send each other a
rewritable disk. The question is how big the disk has to be if
both of them wants to know the value of $f(x,y)$.
Define the identity function as $I(x,y): \{0,1\}^n\times \{0,1\}^n\rightarrow \{0,1\}$ with $I=1$ if and only if $x=y$.
$\bf Question.$ How much is $S(I)$?
$\bf Remarks.$ I know it is between $\log n$ and $\log n - \log \log n$, but which? Is it possible to solve it in $\log n-\omega(1)$ space? Anyone heard of any related things?
$\bf Example/Easy Claim.$ $S(I) \le \log(n) + O(1)$.
$\bf Proof.$ We present a construction. A sends her bits one after the other along with their ordinal number and a leading 1, meaning that it is up to B to speak. B replies to each message with his bit with the same ordinal number and a leading 0. This requires $2 + \log n$ space. If in a step his bit differs from her, they know that the answer is 0, the algorithm is over. If they finish sending all their bits, the answer is 1. Therefore, $S(I) \leq \log n + O(1)$.
| https://mathoverflow.net/users/955 | Space Bounded Communication Complexity of Identity | Here's an argument that I believe shows that $\log n - \omega(1)$ is impossible. (This argument came out of a discussion I had with Steve Fenner.)
Let Alice's input be $x\in\{0,1\}^n$ and let Bob's input be $y\in\{0,1\}^n$. Assume the shared memory stores states in $\{0,1\}^m$, and its initial state is $0^m$. We are interested in lower-bounding $m$ for any protocol that computes EQ$(x,y)$.
A given protocol is defined by two collections of functions $\{f\_x\}$ and $\{g\_y\}$, representing the functions applied to the shared memory by Alice on each input $x$ and Bob on each input $y$, respectively, along with some answering criterion. To be more specific, let us assume that if the shared memory ever contains the string $1^{m-1}b$ then the output of the protocol is $b$ (for each $b\in\{0,1\}$). For convenience, let us also assume that $f\_x(1^{m-1}b) = 1^{m-1}b\:$ and $g\_y(1^{m-1}b) = 1^{m-1}b\:$ for every $x,y\in\{0,1\}^n$ and $b\in\{0,1\}$. In other words, Alice and Bob don't change the shared memory once they know the answer.
Now, for each $x,y\in\{0,1\}^n$, consider what happens when Alice and Bob run the protocol on the input $(x,y)$. Define $A\_{x,y}\subseteq\{0,1\}^m$ to be the set of all states of the shared memory that Alice receives at any point in the protocol, and likewise define $B\_{x,y}\subseteq\{0,1\}^m$ to be the states of the shared memory that Bob receives. Also define
$$
S\_{x,y} = \{0w\,:\,w\in A\_{x,y}\} \cup \{1w\,:\,w\in B\_{x,y}\}.$$
We will assume Alice goes first, so $0^m \in A\_{x,y}$ for all $x,y$. Let us also make the following observations:
1. By the definition of $A\_{x,y}$ and $B\_{x,y}$, it holds that $f\_x(A\_{x,y}) \subseteq B\_{x,y}$ and $g\_y(B\_{x,y}) \subseteq A\_{x,y}$ for all $x,y$.
2. For every $x,y$ with $x\not=y$ it holds that $1^{m-1}0\in A\_{x,y} \cup B\_{x,y}$, because Alice and Bob output 0 when their strings disagree.
3. For every $x$ it holds that $1^{m-1}0\not\in A\_{x,x} \cup B\_{x,x}$, because Alice and Bob do not output 0 when their strings agree.
Now let us prove that $S\_{x,x}\not=S\_{y,y}$ whenever $x\not=y$. To do this, let us assume toward contradiction that $x\not=y$ but $S\_{x,x} = S\_{y,y}$ (i.e., $A\_{x,x} = A\_{y,y}$ and $B\_{x,x} = B\_{y,y}$), and consider the behavior of the protocol on the input $(x,y)$.
Let $w\_t$ be the contents of the shared memory after $t$ turns have passed in the protocol, so we have $w\_0 = 0^m$, $w\_1 = f\_x(w\_0)$, $w\_2 = g\_y(w\_1)$, and so on. It holds that
$$
\begin{array}{c}
w\_0 = 0^m \in A\_{x,x}\\
w\_1 = f\_x(w\_0) \in f\_x(A\_{x,x}) \subseteq B\_{x,x} = B\_{y,y},\\
w\_2 = g\_y(w\_1) \in g\_y(B\_{y,y}) \subseteq A\_{y,y} = A\_{x,x},
\end{array}
$$
and in general
$$
\begin{array}{c}
w\_{2t + 1} = f\_x(w\_{2t}) \in f\_x(A\_{x,x}) \subseteq B\_{x,x} = B\_{y,y}\\
w\_{2t+2} = g\_y(w\_{2t+1}) \in g\_y(B\_{y,y}) \subseteq A\_{y,y} = A\_{x,x}
\end{array}
$$
for each $t\in\mathbb{N}$. It follows that $A\_{x,y} \subseteq A\_{x,x}$ and $B\_{x,y} \subseteq B\_{y,y}$.
But now we have our contradiction, assuming the protocol is correct: given that $A\_{x,y}\subseteq A\_{x,x}$ and $B\_{x,y} \subseteq B\_{y,y}$, it follows that $1^{m-1}0 \in A\_{x,x}\cup B\_{y,y}$, so Alice and Bob output the incorrect answer 0 on either $(x,x)$ or $(y,y)$.
Each $S\_{x,x}$ is a subset of $\{0,1\}^{m+1}$, so there are at most $2^{2^{m+1}}$ choices for $S\_{x,x}$. Given that the sets $S\_{x,x}$ must be distinct for distinct choices of $x$, it follows that $2^{2^{m+1}} \geq 2^n$, so $m \geq \log n - 1$.
| 8 | https://mathoverflow.net/users/7641 | 33912 | 21,959 |
https://mathoverflow.net/questions/33910 | 5 | There is classical description of cohomology ring of projective bundle. Is there an analog in quantum cohomology?
| https://mathoverflow.net/users/8051 | Quantum cohomology of projective bundles | There is a quantum Leray-Hirsch Theorem, do to Maulik and Pandharipande [here](http://arxiv.org/abs/math/0412503). It relates the Gromov-Witten theory of the projective line bundle to that of the base.
| 4 | https://mathoverflow.net/users/622 | 33926 | 21,968 |
https://mathoverflow.net/questions/33879 | 7 | Take multi-sorted first-order logic with equality, complex scalars, 1xn vectors, nx1 vectors, nxn matrices, addition and multiplication for each pair of sorts they make sense for, and hermitian transpose (which is conjugation on scalars). Is it decidable what sentences are [true for all n]? (there are 4 sorts, what sentences are true simultaneously for all n)
(For each particular n, it is decidable by interpreting in a real ordered field.)
What if we also add real scalars and ≤ for them?
| https://mathoverflow.net/users/nan | Decidability of matrix algebra | The second problem (where real scalar variables and the comparison relation are also allowed) is equivalent to the first problem. Here is a standard argument showing this:
* A complex scalar variable z can be restricted to real values by requiring $z=\bar{z}$.
* The comparison x≤y can be replaced by $\exists z.\ x+z\bar{z}=y$, where z is a fresh complex scalar variable.
Back to the original question, the following paper may be related (or even answer your question) but I do not have enough knowledge to understand the content completely. Mihai Putinar: [Undecidability in a Free \*-Algebra](https://scholar.google.com/scholar?hl=en&q=Putinar+Undecidability+Free+Algebra), preprint, April 2007, <https://www.ima.umn.edu/sites/default/files/2165.pdf> ([Wayback Machine](https://web.archive.org/web/20220709100927/https://www.ima.umn.edu/sites/default/files/2165.pdf)).
| 2 | https://mathoverflow.net/users/7982 | 33927 | 21,969 |
https://mathoverflow.net/questions/33923 | 4 | Let $(\Omega, \mathcal{F}, \mathbb{P}, \mathcal{F}\_t)$ be a given
probability space with usual conditions, on which $W$ is a standard
Brownian motion. For $x \ge 0$, consider
$$X(t) = x + \int\_0^t \sigma (X(s)) dW(s)$$
Assume $\sigma \in C^{0,1/2}\_{loc}$, $\sigma(0) = 0$, $\sigma>0$ on $(0,\infty)$.
By [Karatzas and Shereve 98], there exists a unique strong solution with
absorbing state at zero. Denote the running maximum by $X^\*(T) =
\sup\_{s\in [0,T]} X(s)$.
Question: For a fixed $T$, is this possible to show that
$\mathbb{P} ( X^\*(T) \ge \beta) = o(\beta^{-1})$ as $\beta \to \infty$?
I am trying to use time-changed Brownian motion, i.e. $X(t) = x +
B([X]\_t)$, where $B$ is BM, and $[X]$ is quadratic variation. There is
also density function available for running maximum $B^\* (T)$, i.e.
$\mathbb{P}(B^\*(T) \ge \beta) = 2 - 2 \Phi(\beta/\sqrt{T}) =
o(\beta^{-1})$, where $\Phi(\cdot)$ is c.d.f of standard normal
distribution. But, I could not succeed using those facts to prove it.
Thank you for your time.
| https://mathoverflow.net/users/5656 | Distribution of running maximum of a local martingale | No. It is true that $\mathbb{P}(X^\*\_T>\beta)=O(\beta^{-1})$, but you don't have a`little-o' bound. In fact it fails, and $\beta\,\mathbb{P}(X^\*\_T>\beta)$ converges to a strictly positive value, precisely when $X$ fails to be a martingale.
If $S$ is the first time at which $X$ hits $\beta>x$ then continuity gives
$$
X\_{S\wedge T} = \beta 1\_{\{X^\*\_T>\beta\}}+1\_{\{X^\*\_T\le\beta\}}X\_T
$$
Take expectations, and use $\mathbb{E}[X\_{S\wedge T}]=x$, which follows from the fact that the first term is a local martingale stopped at time $S$, so is bounded (and hence a proper martingale).
$$
x=\beta\,\mathbb{P}(X^\*\_T>\beta)+\mathbb{E}[1\_{\{X^\*\_T\le\beta\}}X\_T].
$$
The final expectation converges to $\mathbb{E}[X\_T]$ as $\beta$ goes to infinity, by monotone convergence. This gives
$$
\lim\_{\beta\to\infty}\beta\,\mathbb{P}(X^\*\_T>\beta)=x-\mathbb{E}[X\_T].
$$
Now, it is a well known result that if $X$ is a nonnegative local martingale and $X\_0$ is integrable then it is a supermartingale, so $\mathbb{E}[X\_T]\le\mathbb{E}[X\_0]$, and equality holds precisely when it is a martingale over the range $[0,T]$. So, in our case, $\mathbb{P}(X^\*\_T>\beta)=o(\beta^{-1})$ exactly when $\mathbb{E}[X\_T]=x$ and $X$ is a martingale over the range $[0,T]$.
An example when solutions to your SDE fails to be a martingale is $\sigma(x)=x^2$, $dX=X^2\,dW$. The solution to this SDE can be written as $X=1/\Vert B\Vert$ for a 3-dimensional Brownian motion $B$ started from the point $(x^{-1},0,0)$. You can calculate $\mathbb{E}[X\_t]$ and determine that it is decreasing in $t$, so $X$ is not a martingale - just a local martingale. This example appears in Roger's & Williams book [Diffusions, Markov Processes and Martingales](http://books.google.co.uk/books?id=bDQy-zoHWfcC&lpg=PP1&dq=rogers%2520and%2520williams&pg=PP1#v=onepage&q&f=false) as an example of a local martingale which is not a proper martingale.
| 4 | https://mathoverflow.net/users/1004 | 33938 | 21,974 |
https://mathoverflow.net/questions/33924 | 9 | Background: I am trying to work out some Ext calculations for finite flat group schemes over a ring where p is nilpotent. I know how to do these calculations for finite group schemes over a finite field $k$ using the anti-equivalence with Dieudonne modules. I also know how to do this for finite flat group schemes over the ring of Witt vectors $W(k)$ by using the anti-equivalence with finite Honda systems.
Question: Is there a similar anti-equivalence that classifies finite flat group schemes over a ring where $p$ is nilpotent?
(Although this is probably no simpler, I would be satisfied with knowing the answer when $p^2 = 0$.)
| https://mathoverflow.net/users/8055 | How does one classify finite flat group schemes over a ring where p is nilpotent? | Since the case of interest is $W\_2(k)$ with perfect $k$ of characteristic $p > 2$, the answer is given by Ioan Berbec's 2009 paper "Group schemes over artinian rings and applications. In that paper (esp. section 3) he defines an essentially surjective additive functor to a certain semi-linear algebra category and proves that it is full (i.e., surjective on Hom's) and that the isomorphism property can be read off on either side. The same method of proof for the isomorphism property (which amounts to passing to a computation on the special fiber, where classical Dieudonne theory applies) works just as well for closed immersions and quotient maps. Thus, it is a simple exercise to deduce that the induced homomorphism between Ext-groups is an isomorphism. So that answers the question in the cases of interest. For more general bases things will be tougher; already for $W\_n(k)$ with $n > 2$ I don't know a proved result which is suitable for doing hands-on Ext computations (but maybe Berbec's result can be generalized a bit, possibly imposing a "truncated Barsotti-Tate" condition).
| 10 | https://mathoverflow.net/users/3927 | 33941 | 21,977 |
https://mathoverflow.net/questions/23943 | 59 | NEW CONJECTURE: There is no general upper bound.
Wadim Zudilin suggested that I make this a separate question. This follows
[representability of consecutive integers by a binary quadratic form](https://mathoverflow.net/questions/23690/)
where most of the people who gave answers are worn out after arguing over indefinite forms and inhomogeneous polynomials. Some real effort went into this, perhaps it will not be seen as a duplicate question.
So the question is, can a positive definite integral binary quadratic form
$$ f(x,y) = a x^2 + b x y + c y^2 $$
represent 13 consecutive numbers?
My record so far is 8: the form $$6x^2+5xy+14y^2 $$
represents the 8 consecutive numbers from 716,234 to 716,241. Here we have discriminant $ \Delta = -311,$ and 2,3,5,7 are all residues $\pmod {311}.$ I do not think it remotely coincidental that $$6x^2+xy+13 y^2 $$ represents the 7 consecutive numbers from 716,235 to 716,241.
I have a number of observations. There is a congruence obstacle $\pmod 8$ unless, with
$ f(x,y) = a x^2 + b x y + c y^2 $ and $\Delta = b^2 - 4 a c,$ we have $\Delta \equiv 1 \pmod 8,$ or
$ | \Delta | \equiv 7 \pmod 8.$ If a prime $p | \Delta,$ then the form is restricted to either all quadratic residues or all nonresidues $ \pmod p$ among numbers not divisible by $p.$
In what could be a red herring, I have been emphasizing $\Delta = -p$ where $p \equiv 7 \pmod 8$ is prime, and where there is a very long string of consecutive quadratic residues $\pmod p.$ Note that this means only a single genus with the same $\Delta = -p,$ and any form is restricted to residues. I did not anticipate that long strings of represented numbers would not start at 1 or any predictable place and would be fairly large. As target numbers grow, the probability of not being represented by any form of the discriminant grows ( if prime $q \parallel n$ with $(-p| q) = -1$), but as the number of prime factors $r$ with $(-p| r) = 1$ grows so does the probability that many forms represent the number if any do. Finally, on the influence of taking another $\Delta$ with even more consecutive residues, the trouble seems to be that the class number grows as well. So everywhere there are trade-offs.
EDIT, Monday 10 May. I had an idea that the large values represented by any individual form ought to be isolated. That was naive. Legendre showed that for a prime $q \equiv 7 \pmod 8$ there exists a solution to $u^2 - q v^2 = 2,$ and therefore infinitely many solutions. This means that the form $x^2 + q y^2$ represents the triple of consecutive
numbers $q v^2, 1 + q v^2, u^2$ and then represents $4 + q v^2$ after perhaps skipping $3 + q v^2$.
Taking $q = 8 k - 1,$ the form $ x^2 + x y + 2 k y^2$ has no restrictions $\pmod 8,$ while an explicit formula shows that it represents every number represented by $x^2 + q y^2.$ Put together, if $8k-1 = q$ is prime, then $ x^2 + x y + 2 k y^2$ represents infinitely many triples. If, in addition, $ ( 3 | q) = 1,$ it seems plausible to expect infinitely many quintuples. It should be admitted that the recipe given seems not to be a particularly good way to jump from length 3 to length 5, although strings of length 5 beginning with some $q t^2$ appear plentiful.
EDIT, Tuesday 11 May. I have found a string of 9, the form is $6 x^2 + x y + 13 y^2$ and the numbers start at $1786879113 = 3 \cdot 173 \cdot 193 \cdot 17839$ and end with
$1786879121$ which is prime. As to checking, I have a separate program that shows me the particular $x,y$ for representing a target number by a positive binary form. Then I checked those pairs using my programmable calculator, which has exact arithmetic up to $10^{10}.$
EDIT, Saturday 15 May. I have found a string of 10, the form is $9 x^2 + 5 x y + 14 y^2$ and the numbers start at $866988565 = 5 \cdot 23 \cdot 7539031$ and end with
$866988574 = 2 \cdot 433494287.$
EDIT, Thursday 17 June. Wadim Zudilin has been running one of my programs on a fast computer. We finally have a string of 11, the form being $ 3 x^2 + x y + 26 y^2$ of discriminant $-311.$ The integrally represented numbers start at 897105813710 and end at 897105813720. Note that the maximum possible for this discriminant is 11. So we now have this conjecture: For discriminants $\Delta$ with absolute values in this sequence
<http://www.oeis.org/A000229>
some form represents a set of $N$ consecutive integers, where $N$ is the first quadratic nonresidue. As a result, we conjecture that there is no upper bound on the number of consecutive integers that can be represented by a positive quadratic form.
| https://mathoverflow.net/users/3324 | Can a positive binary quadratic form represent 14 consecutive numbers? | I just wanted to remark that if $p$ is a prime such that
$\ell$ splits in $F = \mathbb{Q}(\sqrt{-p})$ for all $\ell \le N$,
then one may prove the existence of $N$ consecutive integers which
are norms of integers in $\mathcal{O}\_F$, providing one is willing
to *assume* a standard hypothesis about prime numbers, namely,
Schinzel's Hypothesis H.
First, note the following:
Lemma 1: If $C$ is an abelian group of odd order, then there exists a finite
(ordered) set $S = \{c\_i\}$ of elements of $C$ such that every element in $C$ can be
written in the form
$\displaystyle{\sum \epsilon\_i \cdot c\_i}$
where $\epsilon\_i = \pm 1$.
Proof: If $C = A \oplus B$, take $S\_C = S\_A \cup S\_B$. If $C
= \mathbb{Z}/n \mathbb{Z}$ then take $S = \{1,1,1,\ldots,1\}$ with
$|S| = 2n$.
Let $C$ be the class group of $F$. It has odd order, because
$2$ splits in $F$ and thus $\Delta\_F = -p$.
Let $S$ be a set as in the lemma. Let $A$ denote an ordered
set of distinct primes $\{p\_i\}$ which split in $\mathcal{O}\_F$ such
that one can write $p\_i = \mathfrak{p}\_i \mathfrak{p}'\_i$ with
$[\mathfrak{p}\_i] = c\_i \in C$, where $c\_i$ denotes a set of elements
whose existence was shown in Lemma 1.
Lemma 2: If $n$ is the norm of some ideal $\mathfrak{n} \in \mathcal{O}\_F$,
and $n$ is not divisible by any prime $p\_i$ in $A$, then
$$n \cdot \prod\_{A} p\_i$$
is the norm of an algebraic integer in $\mathcal{O}\_F$.
Proof: We may choose
$\epsilon\_i = \pm 1$ such that
$\displaystyle{[\mathfrak{n}] + \sum \epsilon\_i \cdot c\_i = 0 \in C}$.
By assumption, $[\mathfrak{p}\_i] = c\_i \in C$ and thus
$[\mathfrak{p}'\_i] = -c\_i \in C$. Hence the ideal
$$\mathfrak{n}
\prod\_{\epsilon\_i = 1}
{ \mathfrak{p}}
\prod\_{\epsilon\_i = -1}
\mathfrak{p}'$$
is principal, and has the desired norm.
By the Chebotarev density theorem (applied to the Hilbert class field
of $F$), there exists a set $A$ of primes as above which avoids any fixed finite set of
primes. In particular, we may
find $N$ such sets which are pairwise distinct and which contain
no primes $\le N$. Denote these sets by $A\_1, \ldots, A\_N$.
By the Chinese remainder theorem, the set of integers $m$ such that
$$m \equiv 0 \mod p \cdot (N!)^2$$
$$m + j \equiv 0 \mod \prod\_{p\_i \in A\_j} p\_i, \qquad 1 \le j \le N$$
is of the form $m = d M + k$ where $0 \le k < M$, $d$ is arbitrary, and $M$
is the product of the moduli.
Lemma 3: Assuming Schinzel's Hypothesis H, there exists infinitely many integers
$d$ such that
$$ P\_{dj}:= \frac{dM + k + j}{j \cdot \prod\_{p\_i \in A\_j} p\_i}$$
are simultaneouly prime for all $j = 1,\ldots,N$.
Proof: By construction, all these numbers are coprime to $M$ (easy check).
Hence, as $d$ varies, the greatest common divisor of the product of these
numbers is $1$, so Schinzel's Hypothesis H applies.
Let $\chi$ denote the quadratic character of $F$. Note that
$dM + k + j = j \mod p$, and so
$\chi(dM + k + j) = \chi(j) = 1$ (as all primes less than $N$ split in $F$).
Moreover, $\chi(p\_i) = 1$ for all
primes $p\_i$ in $A\_j$
by construction. Hence
$\chi(P\_{dj}) = 1$. In particular, if $P\_{dj}$ is prime, then
$P\_{dj}$ and $j \cdot P\_{dj}$ are norms of (not necessarily principal) ideals in
the ring of integers of $F$. By Lemma 2, this implies that
$$dM + k + j = j \cdot P\_{dj} \prod\_{p\_i \in A\_j} p\_i$$
is the norm of some element of $\mathcal{O}\_F$ for all $j = 1,\ldots, N$.
---
One reason to think that current sieving technology will not be sufficient
to answer this problem is the
following: when Sieving produces a non-trivial lower bound, it usually
produces a pretty good lower bound. However, there are no good (lower) bounds known for the following problem: count the number of integers $n$
such that $n$, $n+1$, and $n+2$ are all sums of two squares. Even for the
problem of estimating the number of $n$ such that $n$ and $n+1$ are both sums of squares is tricky - Hooley implies that the natural sieve does not give lower bounds
(for reasons analogous to the parity problem). Instead,
he relates the problem to sums of the form
$\displaystyle{\sum\_{n < x} a\_n a\_{n+1}}$ where
$\sum a\_n q^n = \theta^2$ is a modular form. In particular, he
implicitly uses
automorphic methods which won't work
with three or more terms.
| 22 | https://mathoverflow.net/users/nan | 33961 | 21,992 |
https://mathoverflow.net/questions/33962 | 5 | Is the countability of the set of irrational algebraic numbers somehow reflected in a characteristic property of their decimal expansions?
| https://mathoverflow.net/users/8064 | Question on the decimal expansion of algebraic numbers | The answer is not known, and it is a conjecture of Borel that the answer is no. See [Words and Transcendence](http://arxiv.org/abs/0908.4034) by M. Waldschmidt for references, in particular your previous claim that every digit in irrational algebraic numbers occurs infinitely often is an open problem. We don't know if there are irrational algebraic numbers in the Cantor set.
| 5 | https://mathoverflow.net/users/2384 | 33967 | 21,995 |
https://mathoverflow.net/questions/33945 | 17 | Let $\mathcal{O}(\mathbb{C})$ be the ring of entire functions, that is, those functions $f : \mathbb{C} \to \mathbb{C}$ which are holomorphic for all $z \in \mathbb{C}.$ For each $z\_0 \in \mathbb{C}$.
Are there any other maximal ideals in $\mathcal{O}(\mathbb{C})$ besides these obvious ones?
If anyone can give a concise description of $\text{Spec }\mathcal{O}(\mathbb{C})$, that would be extremely helpful. I'm trying to understand wether or not knowing the closed subset $V(f)$ of $\text{Spec }\mathcal{O}(\mathbb{C})$ of ideals containing $f$ gives you more information about $f$ than simply knowing the vanishing set of $f$ in the classical sense.
| https://mathoverflow.net/users/4872 | What is the spectrum of the ring of entire functions? | Here's a more analytic description of exactly what knowing $V(f)$ tells you. Let us say $f$ ~ $g$ if their vanishing sets are the same, and moreover there exist positive constants $c,C$ such that $c\cdot ord\_g(z)< ord\_f(z)< C\cdot ord\_g(z)$ as $z$ ranges over the vanishing set. Then knowing $V(f)$ is the same is as knowing the equivalence class of $f$. Indeed,
$V(f) = V(g)$ $\iff$ there exists $n$ such that $f^n\in (g)$ and $g^n\in (f)$ $\iff$ $f$ ~ $g$.
| 12 | https://mathoverflow.net/users/5513 | 33969 | 21,997 |
https://mathoverflow.net/questions/33963 | 3 | Let $C\_1, \dots, C\_n$ be a family of disjoint simple curves in a surface $\Sigma$. If $C$ is any simple curve in $\Sigma$, it turns out that we can map $C$ to a curve $C'$ (via a homeomorphism of $\Sigma$) such that $C'$ only intersects each $C\_i$ at most twice. I believe I have a proof of this result that works for all surfaces, but I'm pretty sure this is classical stuff.
Indeed, in John Stillwell's book *Classical Topology and Combinatorial Group Theory* he mentions that the above result was proven by Lickorish (1962) for orientable surfaces. In the case of orientable surfaces, the homeomorphism can in fact be achieved via Dehn twists and isotopies. Unfortunately, the Lickorish proof doesn't work for non-orientable surfaces.
**Question:** Can someone please provide a reference of the above result for non-orientable surfaces?
| https://mathoverflow.net/users/2233 | Removing intersections of curves in surfaces | There are two questions here.
1) The fact about Dehn twists and isotopies is really a consequence of the fact that the [mapping class group](http://en.wikipedia.org/wiki/Mapping_class_group) of an orientable surface is generated by Dehn twists. For non-orientable surfaces, this is not true -- you also need the so-called "crosscap slides". For a discussion of this, see Lickorish's paper "Homeomorphisms of non-orientable two-manifolds" and Chillingworth's paper "A finite set of generators for the homeotopy group of a non-orientable surface".
2) For the fact about simple closed curves, there is a very general trick using the classification of surfaces that has become known as the "change of coordinates principle" for a surface. It can prove almost any statement of this type. For a nice discussion of it, see Section 1.3 of Farb and Margalit's book "A primer on mapping class groups", which is available
[here](http://www.math.utah.edu/~margalit/primer/).
| 3 | https://mathoverflow.net/users/317 | 33970 | 21,998 |
https://mathoverflow.net/questions/33942 | 12 | I believe this question is due to Erdős and Graham, and I think it is still open: does the base 3 expansion of $2^n$ avoid the digit 2 for infinitely many $n$?
If we concatenate the digits of $2^i$, $i \geq 0$, we produce the number $0.110100100010000...$. This number is not simply normal in base 2, so it is not normal. Is it simply normal in base 3? I think even that result would not imply that for sufficiently large $n$, 2 doesn't appear in the base 3 expansion of $2^n$.
The number 20 here is not special:
$2^{20} = 1222021101011\_3, \;\;\;\; 2^{21} = 10221112202022\_3, \;\;\;
2^{22} = 21220002111121\_3$
Statistically, we seem to be flipping a fair 3-sided coin, and statistical analysis for larger $n$ bears this out (in the past, I did a p-test on the digits, but don't have the data available here). If we actually produced these digits by flipping this 3-sided coin, for fixed $n$ we would have probability about
$$(2/3)^{n\ln2/\ln3}$$
of having no 2s in the base-3 digit expansion.
What is the state of the art for this problem? Is there a good number-theoretic reason why this problem should be very difficult (e.g. an analogy with other supposed-hard problems)? Are there related problems that have been solved?
| https://mathoverflow.net/users/35336 | Do the base 3 digits of $2^n$ avoid the digit 2 infinitely often -- what is the status of this problem? | As of a few months ago, the status of the problem was: still unsolved. See the slides [Jeff Lagarias](http://www.math.lsa.umich.edu/~lagarias/) put up from a talk he gave in September 2009: <http://www.math.lsa.umich.edu/~lagarias/TALK-SLIDES/ternary-fields-2009sep.pdf>
An older reference is
<http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.65.6976&rep=rep1&type=pdf> (Brian Hayes, Third Base, American Scientist) which says the problem was still open in late 2001; also that Ilan Vardi searched up to $2^{6973568802}$ without finding any 2-less powers of 2 (other than $2^2$ and $2^8$).
| 6 | https://mathoverflow.net/users/3684 | 33980 | 22,003 |
https://mathoverflow.net/questions/33964 | 4 | Let $\Delta \subset \mathbb{R}^n$ be an $n$-dimensional integral polytope, let $f$ be a Laurent polynomial in $n$-variables with coefficients in an extension of the integers and Newton polytope $\Delta$. We can view $f$ as both a Laurent polynomial over $\mathbb{C}$ and $\mathbb{F} \_ q$ for some prime power $q$. Suppose that $f$ is $\Delta$-regular (for a definition, see <http://www.math.uci.edu/~dwan/gottingen.pdf> pg. 16) over both $\mathbb{C}$ and $\overline{\mathbb{F}} \_ q$. Let $X\_\mathbb{C}\subset(\mathbb{C}^\*)^n$ be the affine hypersurface defined by the vanishing of $f$ in the torus (viewing $f$ as a polynomial over $\mathbb{C}$) and $X\_{\overline{\mathbb{F}}\_q} \subset (\overline{\mathbb{F}} \_ q ^ \*)^n$ be the analog over $\overline{\mathbb{F}} \_ q$.
My question is:
Is it true that $$\dim (H^i(X\_\mathbb{C})) = \dim H^i(X\_{\overline{\mathbb{F}}\_q},\mathbb{Q} \_ \ell),$$ where the cohomology on the left hand side is the standard topological cohomology? If this is true, I would greatly appreciate a reference if you have one.
Thanks!
| https://mathoverflow.net/users/8044 | Relationship between topological cohomology and $\ell$-adic cohomology | The way to study the topology of the situation was introduced by Khovanski in
"Newton polyhedra, and toroidal varieties" Funkcional. Anal. i Priložen. 11
(1977), no. 4, 56--64, 96. His result (if I have interpreted it correctly) is
that $X$ may be compactified as a hypersurface in a projective toric variety to
a smooth variety with normal crossings such that each stratum is of the same
form as $X$. As far as I can see this construction works uniformly so that we
would get the same construction over a (suitable) mixed characteristic discrete
valuation ring. Then desired isomorphism then follows from the smooth and proper
base change theorem. (I have some vague recollection that this comment is also to be
found somewhere in SGA but I am not going to do any wading looking for it...)
*Addendum*: Let me first note that the right setup to even formulate the
question is a scheme $S$ with functions on it giving the coefficients of $f$.
The latter polynomial should then be non-degenerate in the sense that all its
fibres over (geometric) points of $S$ should be non-degenerate. The statement is
then that if $\pi\colon X\to S$ is the scheme of zeroes of $f$ in the constant torus
over $S$, then $R^i\pi\_\ast\mathbb Z\_\ell$ is locally constant commuting with base
change where $\ell$ is invertible in $\mathcal O\_S$ (together with comparison
theorem of $\ell$-adic cohomology and classical for a complex point of $S$). As
this statement is only dealing with the $\ell$-adic sheaves $R^i\pi\_\ast\mathbb Z\_{\ell}$
we may use the definition of $\ell$-adic sheaf introduced by Jouanolou in SGA V.
Kohvanski's method then should give a compactification of $X$ by a smooth
$S$-scheme with complement of relative normal crossings. The two theorems used,
proper base change and vanishing of vanishing cycles, then follows directly from
the case of finite coefficients (by Jouanolou's very definition).
| 6 | https://mathoverflow.net/users/4008 | 33981 | 22,004 |
https://mathoverflow.net/questions/25030 | 10 | Suppose that we have a 2d-regular graph whose edges are colored such that the edges of each color form a cycle of length 2d. (So if the graph has 2n vertices, then there are n colors.) Is it true that there always is a perfect matching containing one edge of each color?
Remarks. For d=2 there is a simple proof by Zoltan Kiraly who also invented the above formulation of the problem. I even do not know the answer for d=3.
| https://mathoverflow.net/users/955 | Can we select a rainbow matching if each degree is 6 and each colorclass is a C_6? | This is a little embarrassing, but it turned out that not even a (non-rainbow) matching is guaranteed to exist. The problem was solved on [this](http://www.renyi.hu/~emlektab/) workshop by a number of people, presented by Tamas Terpai. They raised the same question for bipartite graphs, for which a matching must always exist.
| 2 | https://mathoverflow.net/users/955 | 33992 | 22,008 |
https://mathoverflow.net/questions/32889 | 24 | [K] refers to Kontsevich's paper "Deformation quantization of Poisson manifolds, I".
Background
----------
Let $X$ be a smooth affine variety (over $\mathbb{C}$ or maybe a field of characteristic zero) or resp. a smooth (compact?) real manifold. Let $A = \Gamma(X; \mathcal{O}\_X)$ or resp. $C^\infty(X)$.
Denote the dg *Lie algebra* of polyvector fields on $X$ (with Schouten-Nijenhuis bracket and zero differential) by $T$. Denote the dg *Lie algebra* of the shifted Hochschild cochain complex of $A$ (with Gerstenhaber bracket and Hochschild differential) by $D$.
Then the Hochschild-Konstant-Rosenberg theorem states that there is a quasi-isomorphism of dg *vector spaces* from $T$ to $D$. However, the HKR map is *not* a map of dg *Lie algebras*. It is *not* a map of dg *algebras*, either (where the multiplication on $T$ is given by the wedge product and the multiplication on $D$ is given by the cup product of Hochschild cochains).
I believe "Kontsevich formality" refers to the statement that, while the HKR map is not a quasi-isomorphism --- or even a morphism --- of dg *Lie algebras*, there is an $L\_\infty$ quasi-isomorphism $U$ from $T$ to $D$, and therefore $D$ is in fact formal as a dg *Lie algebra*.
The first "Taylor coefficient" of the $L\_\infty$ morphism $U$ is precisely the HKR map (see section 4.6.2 of [K]).
Moreover, this quasi-isomorphism $U$ is compatible with the dg *algebra* structures on $T$ and $D$ (see section 8.2 of [K]), and it yields a "corrected HKR map" which is a dg algebra quasi-isomorphism. The "correction" comes from the square root of the $\hat{A}$ class of $X$. See [this previous MO question](https://mathoverflow.net/questions/14861/is-there-a-refinement-of-the-hochschild-kostant-rosenberg-theorem-for-cohomology/).
Questions
---------
(0) Are all of my statements above correct?
(1) In what way is the $L\_\infty$ morphism $U$ compatible with the dg *algebra* structures? I don't understand what this means.
(2) When $X$ is a smooth (compact?) real manifold, I think that all of the statements above are proved in [K]. When $X$ is a smooth affine variety, I think that the statements should all still be true. Where can I find proofs?
(3) Moreover, the last section of [K] suggests that the statements are all still true when $X$ is a smooth *possibly non-affine* variety. For a general smooth variety, though, instead of taking the Hochschild cochain complex of $A = \Gamma(X;\mathcal{O}\_X)$, presumably we should take the Hochschild cochain complex of the (dg?) derived category of $X$. Is this correct? If so, where can I find proofs?
In the second-to-last sentence of [K], Kontsevich seems to claim that the statements for varieties are corollaries of the statements for real manifolds, but I don't see how this can possibly be true. In the last sentence of the paper, he says that he will prove these statements "in the next paper", but I'm not sure which paper "the next paper" is, nor am I even sure that it exists, since "Deformation quantization of Poisson manifolds, II" doesn't exist.
P.S. I am not sure how to tag this question. Feel free to tag it as you wish.
| https://mathoverflow.net/users/83 | A few questions about Kontsevich formality | To (1): Daniel is right, there is a map of homotopy Gerstenhaber algebras between the two algebras. However the full story is quite complicated and to show that the hochschild cochains form a homotopy Gerstenhaber algebra is hard, it's known as the Deligne conjecture. I don't know the details of the proof.
Recall that a Poisson algebra is a commutative algebra with a Lie bracket and these two products satisfy a Leibniz identity. A Gerstenhaber algebra is a bit like a Poisson algebra, except the Lie bracket is of degree 1 not 0. The bracket satisfies a graded Leibniz identity wrt to the commutative algebra structure.
The formality morphism as homotopy Gerstenhaber algebras restricts to a formality morphism as homotopy Lie algebras and to a formality morphism as homotopy commutative algebras.
In my view the simplest proof of the formality of the Hochschild cochains of a nice enough algebra as a homotopy Gerstenhaber algebra is contained in
<http://arxiv.org/abs/math.KT/0605141>
| 6 | https://mathoverflow.net/users/109 | 34002 | 22,015 |
https://mathoverflow.net/questions/33990 | 3 | can you give me a good paper (in the sense of a simple introduction) about Sasaki-Einstein manifolds?
Thank you and best regards
Florian M.
| https://mathoverflow.net/users/7015 | Paper about Sasaki-Einstein manifolds | Well, now there is a great *textbook* for Sasaki-Einstein geometry, by Boyer-Galicki: Sasakian Geometry.
[Here is a link to the book at Oxford University Press](https://global.oup.com/academic/product/sasakian-geometry-9780198564959), DOI: [10.1093/acprof:oso/9780198564959.001.0001](https://doi.org/10.1093/acprof:oso/9780198564959.001.0001)
I would definitely start there.
| 4 | https://mathoverflow.net/users/6871 | 34006 | 22,019 |
https://mathoverflow.net/questions/34016 | 2 | Let $k$ be a field. What are the $k$-rational points of the affine $k$-scheme $\mathrm{Spec}(k[[t]])$, where $k[[t]]$ is the power series ring over $k$ (equivalently, what are the $k$-algebra morphisms $k[[t]] \rightarrow k$?)
I'm only sure about one point, namely the map $t \mapsto 0$. Do I have to assume some sort of completeness of $k$ to get more points?
Is there a nice presentation of $k[[t]]$, i.e. a quotient of some polynomial ring that is isomorphic to $k[[t]]$?
| https://mathoverflow.net/users/8070 | What are the k-rational points of k[[t]]? | $k[[t]]$ is a local ring with maximal ideal $(t)$ and the kernel of every $k$-homomorphism $k[[t]] \to k$ is a maximal ideal, thus *the* maximal ideal. Thus it factors as $k[[t]] \to k[[t]]/(t) = k \to k$ and $t \mapsto 0$ is the unique $k$-rational point.
| 5 | https://mathoverflow.net/users/2841 | 34018 | 22,026 |
https://mathoverflow.net/questions/33995 | 17 | In the first pages of SGA4 I read
>
> [...] *Cependant le seul univers connu est l'ensemble des symboles du type* {Ø,{Ø},{Ø,{Ø}}, ... } *etc. (tous les éléments de cet univers sont des ensembles finis et cet univers est dénombrable). En particulier, on ne connaît pas d'univers qui contienne un élément de cardinal infini.* [...]
>
>
>
(the sole known universe is like {Ø,{Ø},{Ø,{Ø}}, ... }, and we don't know any universe with a infinite cardinal).
*Mais, c'est vrai?* I wonder if during all these years somebody discovered a universe "bigger" than that exhibited by Grothendieck.
| https://mathoverflow.net/users/7952 | Is {Ø,{Ø},{Ø,{Ø}}, ... } the only known universe? | The universe that Grothendieck intends to suggest by his notation is known in set theory as HF, the class of hereditarily finite sets, the sets that are finite and have all elements finite and elements-of-elements, and so on (the transitive closure should be finite). The set HF is the same as $V\_\omega$ in the Levy hiearchy, and can be built by starting with the emptyset and iteratively computing the power set, collecting everything together that is produced at any finite stage. This is the smallest nonempty transitive set that is closed under power set. It satisfies all the Grothendieck universe axioms, except that it doesn't have any infinite elements, since none appear at any finite stage of this consrtruction.
There is an interesting presentation of this universe by a simple relation on the natural numbers. Namely, define $n\ E\ m$ if the $n^{\rm th}$ bit in the binary expansion of $m$ is $1$. The structure $\langle\mathbb{N},E\rangle$ is isomorphic to $\langle HF,{\in}\rangle$ by the map $\pi(n)=\{\pi(m)\,|\,m\,E\,n\}$, which set-theorists will recognize as the Mostowski collapse of $E$.
Since HF doesn't have any infinite elements, it is a rather impoverished universe for many applications of that concept. And so we naturally seek larger universes. But the difficulty is that we cannot prove they exist. The difficulty is not one of "discovery," but rather just that we can prove that the hypothesis of the existence of a univese containing infinite sets is too strong for us to prove from our usual axioms. The reason is, as has been remarked in some of the other answers and comments, all other Grothendieck universes have the form $H\_\kappa$, the hereditarily size less than $\kappa$ sets, for an inaccessible cardinal $\kappa$. So this is just like HF, which is $H\_\omega$, but on a higher level, and in this sense, these higher universes are not so mysterious. They are intensely studied in set theory, a part of the research effort in large cardinals.
In [this MO answer](https://mathoverflow.net/questions/24552/what-interesting-nontrivial-results-in-algebraic-geometry-require-the-existence-o/28913#28913), I mention a number of weaker universe concepts that we can prove exist, and which I believe serve most of the uses of the universe concept in category theory, if one wanted to care more about such set theoretic issues.
| 21 | https://mathoverflow.net/users/1946 | 34024 | 22,031 |
https://mathoverflow.net/questions/34007 | 81 | Nowadays, the usual way to extend a measure on an algebra of sets to a measure on a $\sigma$-algebra, the Caratheodory approach, is by using the outer measure $m^\* $ and then taking the family of all sets $A$ satisfying $m^\* (S)=m^\* (S\cap A)+m^\* (S\cap A^c)$ for every set $S$ to be the family of measurable sets. It can then be shown that this family forms a $\sigma$-algebra and $m^\*$ restricted to this family is a complete measure. The approach is elegant, short, uses only elementary methods and is quite powerful. It is also, almost universally, seen as completely unintuitive (just google "Caratheodory unintuitve" ).
Given that the problem of extending measures is fundamental to all of measure theory, I would like to know if anyone can provide a perspective that renders the Caratheodory approach natural and intuitive.
I'm familiar with the fact that there is a topological approach to the extension problem (see [here](http://terrytao.wordpress.com/2009/01/03/254a-notes-0a-an-alternate-approach-to-the-caratheodory-extension-theorem/) or [link text](http://arxiv.org/abs/0712.2270)) for the $\sigma$-finite case due to M.H. Stone (Maharam has actually shown how to extend it to the [general case](http://purl.pt/3098/1/)), but it doesn't give much of an insight into why the Caratheodoy approach works and that is what I`m interested in here.
| https://mathoverflow.net/users/35357 | Demystifying the Caratheodory Approach to Measurability | Here is an argument that may give some intuition:
Assume that $m^{\*}$ is an outer measure on $X$, and let us assume furthermore that this outer measure is finite:
$m^\* (X) < \infty$
Define an "inner measure" $m\_\*$ on $X$ by
$m\_\* (E) = m^\* (X) - m^\* (E^c) $
If $m^\*$ was, say, induced from a countably additive measure defined on some algebra of sets in $X$ (like Lebesgue measure is built using the algebra of finite disjoint unions of intervals of the form $(a,b]$), then a subset of $X$ will be measurable in the sense of Caratheodory if and only if its outer measure and inner measure agree.
From this viewpoint, the construction of the measure (as well as the $\sigma$-algebra of measurable sets) is just a generalization of the natural construction of the Riemann integral on $\mathbb{R}^n$ - you try to approximate the area of a bounded set $E$ from the outside by using finitely many rectangles, and similarly from the inside, and the set is "measurable in the sense of Riemann" (or "Jordan measurable") if the best outer approximation of its area agrees with the best inner approximation of its area.
The point here (which often isn't emphasized when Riemann integration is taught for the first time) is that the concept of "inner area" is redundant and can be defined in terms of the outer area just as I did above (you take some rectangle containing the set and consider the outer measure of the complement of the set with respect to this rectangle).
Of course, Caratheodory's construction doesn't require $m^\*$ to be finite, but I still think that this gives some decent intuition for the general case (unless you think that the construction of the Riemann integral itself is not intuitive :) ).
| 52 | https://mathoverflow.net/users/7392 | 34029 | 22,035 |
https://mathoverflow.net/questions/34010 | 43 | In an answer to the popular question on common false beliefs in mathematics
[Examples of common false beliefs in mathematics](https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23568#23568)
I mentioned that many people conflate the two different kinds of formal Laurent series field in two variables. Let $K$ be any field. Then we have the **joint Laurent series field**, the field of fractions
$K((x,y)) = \operatorname{Frac}(K[[x,y]])$
and also the **iterated Laurent series field**
$K((x))((y)) = \left(K((x))\right)((y))$.
These fields are not the same, roughly because when we write out an arbitrary element of the iterated field as a formal Laurent series in $y$, for each (say non-negative) $n$, the coefficient of $y^n$ is allowed to be an arbitrary formal Laurent series in $x$. In particular, as $n$ varies, arbitrarily large negative powers of $x$ may appear.
However, this is rather far from a convincing argument. Indeed, I gave the following explicit (and fallacious!) example: $\sum\_{n=0}^{\infty} x^{-n} y^n$. But in a comment to my answer, user AS points out that this element is equal (in the iterated field, say) to $\frac{1}{1-\frac{y}{x}}$ and therefore it must lie in the fraction field of $K[[x,y]]$. Evidently the fallacy here is that the fraction field of $K[[x,y]]$ is the field of all formal Laurent series which are finite-tailed in both $x$ and $y$. But as this example shows, the latter isn't even a field, unlike the one-variable case.
[At least when $K = \mathbb{C}$, by less explicit means one can see that these two fields are very different: e.g., the joint field is Hilbertian so has nonabelian Galois group, whereas the iterated field has Galois group $\widehat{\mathbb{Z}}^2$.]
AS offered to write down, with proof, an explicit element of $K((x))((y)) \setminus K((x,y))$, so I decided to post a question asking for such a guy. Of course, there is more than one such element -- or better put, more than one type of construction of such elements -- so I would be interested to see multiple answers to:
>
> Please exhibit, with proof, an explicit element of $K((x))((y)) \setminus K((x,y))$.
>
>
>
| https://mathoverflow.net/users/1149 | Explicit elements of $K((x))((y)) \setminus K((x,y))$ | Suppose $R$ is a domain with field of fractions $F$. Let $f\in F[[y]]$ and suppose that $f\in Frac(R[[y]])$. Then $f=h/g$ with $g,h\in R[[y]]]$ and we may assume that $g=b\_0+b\_1y+\dots$ with $b\_0\ne0$. Therefore
$$
b\_0g^{-1}=(1+(b\_1/b\_0)y+\dots)^{-1}\in R[[y/b\_0]]
$$
and so $f\in b\_0^{-1}R[[y/b\_0]]$.
So when $R=k[[x]]$, any $f=\sum y^n/x^{r(n)}$ with $r(n)/n \to \infty$ will work.
An analytic paraphrase of this argument (which is relevant to Makhalan's comment) is that (regarding $F=k((x))$ as a valued field) any element of $Frac(k[[x,y]])$ is a ratio of power series converging on the open unit disc in $F$. So it is enough to write down a power series in $y$ with zero radius of convergence.
| 15 | https://mathoverflow.net/users/5480 | 34033 | 22,036 |
https://mathoverflow.net/questions/34030 | 3 | Let $f \colon X \to S$ be a proper morphism form a seperated algebraic space $X$ to an affine noetherian scheme $S$.
Given a coherent sheaf $F$ on $X$, we know from Knutson's book, that the finiteness theorem holds: The $R^qf\_\*F$ are finite $\mathcal O\_S$-modules. For complete $S$, we also have the theorem of formal functions.
Now, let us assume that $F$ is flat over $S$. If $X$ were a scheme, we would know that the function on $S$, given by the dimension of the cohomology of $F$ restricted to fibers, is upper semicontinous. Is the same also true for $X$ an algebraic space?
Moreover, if we knew that for some $i \ge 0$ the function $s \mapsto h^i(X\_s, F\_s)$ is constant, than for $X$ a scheme, it would follow that the base change map $R^i f\_\* (F) \otimes k(s) \to H^i(X\_s, F\_s)$ is an isomorphisms. What about algebraic spaces?
many thanks in advance.
| https://mathoverflow.net/users/5273 | Semicontinuity and cohomological flatness for algebraic spaces | This is definitely true and there ought to (but may not) exist a reference. One
way to prove it is just to check that the usual proof for schemes extends. If we
follow Hartshorne for instance the crucial part is Proposition III:12.2 which
can be proven using an étale affine cover instead of an open affine one.
| 7 | https://mathoverflow.net/users/4008 | 34035 | 22,038 |
https://mathoverflow.net/questions/34041 | 1 | Hi--
Where can I find a proof of this theorem: For each $r \in \mathbb{Z}\_{+}$,
there exists a complex entire function $f(z)$ such that $f(r) \neq 0$ but
$f(r+1)=f(r+2)=\cdots =0$,
i.e. $f(z) \in I\_{r+1}$ but $f(z) \neq I\_{r}$, where
$I\_{r}= \{ f \in R \ | f(r)=f(r+1)= \cdots =0\}$
where $R$ is the ring of complex entire functions
| https://mathoverflow.net/users/1483 | Weierstrass Theorem | Why don't you start with a function with zeros at the integers,
for instance $\sin\pi z$, and then somehow eliminate the zero at $r$?
| 5 | https://mathoverflow.net/users/4213 | 34043 | 22,043 |
https://mathoverflow.net/questions/34044 | 6 | I have seen that if $G$ is a finite group and $H$ is a proper subgroup of $G$ with finite index then $ G \neq \bigcup\limits\_{g \in G} gHg^{-1}$. Does this remain true for the infinite case also?
| https://mathoverflow.net/users/1483 | Group cannot be the union of conjugates | Not in general. Every matrix in $\text{GL}\_2(\mathbf C)$ is conjugate to an invertible upper triangular matrix (use eigenvectors), and the invertible upper triangular matrices are a proper subgroup.
| 23 | https://mathoverflow.net/users/3272 | 34046 | 22,045 |
https://mathoverflow.net/questions/33936 | 16 | Recall the notion of *Lie algebroid* ([n Lab](http://ncatlab.org/nlab/show/Lie+algebroid), [Wikipedia](http://en.wikipedia.org/wiki/Lie_algebroid)). One motivation for studying Lie algebroids is that they are infinitesimal versions of Lie groupoids, and Lie groupoids present stacks. In particular, Lie groupoids are the objects of a 2-category whose 1-morphisms are (left-principal) biactions of groupoids, and whose 2-morphisms are smooth biequivariant maps thereof. A "smooth stack" is then a Lie groupoid up to equivalence in this category.
I would like to understand the infinitesimal version of this category. I presume that this would require understanding what are:
* smooth left- and right-actions of Lie algebroids,
* when a smooth left action by one Lie algebroid on a space and a smooth right action by another Lie algebroid on the same space commute,
* when a Lie algebroid action is principal.
Hence the question in the title. In particular, I'd like some intuition for whether two Lie algebroids are equivalent. As a consistency check, there should be a functor from the 2-category of groupoids to the 2-category of algebroids that on objects takes a groupoid to its tangent algebroid.
Bonus points: there is a wonderful description of the *1-category* whose objects are Lie algebroids as the full subcategory of the category of dg manifolds (morphisms are graded smooth maps that relate the homological vector fields) on the dg manifolds generated in degrees $0$ and $1$. Is there a similar "dg" description of the *2-category*?
| https://mathoverflow.net/users/78 | What is the 2-category whose 0-objects are Lie algebroids? | Here is one answer, but also sort of a no-go observation.
We are interested in creating a bicategory whose objects are Lie algebroids. The 1-morphisms and 2-morphisms are something yet to be determined. We have a couple simple requirements that we are going to demand:
1. We want a functor from the bicategory of Lie groupoids and bibundles to this bicategory.
2. This functor should take a Lie groupoid to its underlying Lie algebroid.
3. The naive notion of morphism of Lie algebroid gives rise to a 1-morphism in this bicateogry, and isomorphism become equivalences.
Now I will describe a bicategory which minimally satisfies these conditions, but I don't think that this bicategory of Lie algebroids is going to be very interesting. The reason is that the above requirements force you to identify too many things.
For example take any space X and any cover U of the space. We can form two groupoids out of this which are Morita equivalent. These are X, viewed as a groupoid with just identities and the Cech groupoid $U \times\_X U \rightrightarrows U$. These are equivalent groupoids in the bibundle bicategory and so must be sent to equivalent Lie algebroids. (In a certain sense the bibundle bicategory is what you get when you take Lie groupoids, functors, etc. and force these two types of groupoids to be equivalent. More on this below.)
However their Lie algebroids are very simple and appear very different. They are just the trivial Lie algebroids over X and U, respectively. In particular this second one has no information about the fiber products $$U \times\_X U.$$ This means that the tangent bundle of X and the tangent bundle of U must be equivalent objects in this hypothetical bicategory of Lie algebroids, whenever U covers X.
Following through with similar examples allows us to see that any time we pull a Lie algebroid back by a cover of its base we get an "equivalent" Lie algebroid. In particular this means that every Lie algebroid will be equivalent to one on a trivial bundle.
Maybe I am wrong and this is still an interesting bicategory, but it feels like we're loosing too much information. In any event this suggests what your hypothetical bicategory of Lie algebroids actually looks like.
Let's set up some terminology first. Suppose I have a Lie algebroid with base space X. Suppose further that I have a surjective submersion of the base $U \to X$. Then I can pull-back the Lie algebroid on X to one on U. We will call the morphism of Lie algebroids from U to X a \*weak equivalence".
So the bicategory you are after should have object Lie algebroids and the morphisms should be spans of Lie algebroids $$X \stackrel{\sim}{\leftarrow} U \rightarrow Y$$ where $U \to X$ is a weak-equivalence.
The 2-morphisms are not just naive morphisms of spans. They are more complicated. This is why your [previous MO question](https://mathoverflow.net/questions/32528/do-all-equivalences-in-the-2-category-of-spans-come-from-isomorphisms) doesn't apply. What we are doing is inverting the weak equivalences to make them, well, equivalences. We want to do it in such a way that we still have a bicategory and that it has the obvious universal property (functors out of it are the same as weak equivalence inverting functors out of Lie algebroids). This is sometimes called the derived localization.
Fortunately there is some systematic machinery to accomplish this, devoloped by Dorette Pronk. Some of it is [described at the nlab](http://ncatlab.org/nlab/show/bicategory+of+fractions), but the full story is in her paper "Etendues and stacks as bicategories of fractions". Among other things, this paper contains a description of the bicategory of Lie groupoids and bibundles as an example of this sort of derived localization, and so that is a good example to compare with.
From this description it is also clear that the derived localization of Lie algebroids along the weak equivalences is going to be the universal (initial) thing which satisfies the three properties outlined above.
The question remains though: is this an interesting bicategory? I don't have an answer for this. Perhaps you have a use for it?
| 7 | https://mathoverflow.net/users/184 | 34048 | 22,047 |
https://mathoverflow.net/questions/34055 | 7 | Can anyone suggest me an ingenious proof of the transcendence of $\pi$. I have seen Lindemann's proof but it appears intricate.
| https://mathoverflow.net/users/1483 | Transcendence of PI | There is a very nice book, "Irrational Numbers" by Ivan Niven. Available in paperback from the M.A.A. Evidently he gives a proof in the M.A.A.'s American Mathematical Monthly, volume 46 (1939) pages 469-471. His comment in the notes for chapter 9 of the book has "Proofs of the transcendence of $e$ and $\pi$ are not so difficult as the proof of the more general Theorem 9.1" And his 9.1 is indeed Hermite-Lindemann-Weierstrass. $$ $$ See also
[Proof that pi is transcendental that doesn't use the infinitude of primes](https://mathoverflow.net/questions/21367/proof-that-pi-is-transcendental-that-doesnt-use-the-infinitude-of-primes)
which had a specific emphasis.
| 11 | https://mathoverflow.net/users/3324 | 34056 | 22,050 |
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