parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/44901 | 11 | On a compact, boundaryless, Riemannian manifold, the dimension of the space of harmonic k-forms is equal to the k-th Betti number. Is this correct (by Hodge theory)? For example, on the surface of a unit sphere $S^2$ and any equivalent topology, the 0-th Betti number is 1, which is consistent with the fact that any harmonic function on a compact, boundless, Riemannian manifold is a constant. More importantly to me, what about the 1st Betti number? I think the only harmonic 1-form on the topological class of $S^2$ is zero --- you can't comb every hair on a sphere towards the same direction --- right?
Basically, it is easy to prove both results (1st betti number = 0 and the only harmonic 1-form is zero) on the ideal case of $S^2$. But if I want to go further, some topological tool has to be involved?
I am a PDE guy, so if there is any PDE-based proof to the above question, it'd be highly appreciated!
| https://mathoverflow.net/users/10576 | Betti number and harmonic forms | Just to add some details to Paul's answer:
It is indeed correct that the $k$-th Betti number of a compact Riemannian manifold $M$ without boundary is equal to the dimension of the space of harmonic $k$-forms. This is the Hodge isomorphism theorem. A good proof of this may be found in Demailly's book (available for free [here](https://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf)) at the beginning of chapter 6.
Basically what happens is this: we use a Riemannian metric to define an inner product on the space $E^k := C^\infty(M, \bigwedge^k T\_M^\*)$ of smooth $k$-forms on $M$. The Riemannian metric also gives us the Laplacian $\Delta$, which turns out to be an elliptic differential operator on $E^k$. Thus we get an orthogonal direct sum decomposition
$$C^\infty(M, \bigwedge^k T\_M^\*) = \mathcal H^k(M) \oplus Im \Delta = \mathcal H^k(M) \oplus Im d \oplus Im d^{\*}$$
where $\mathcal H^k(M) := Ker \Delta$ is the space of harmonic $k$-forms on $M$, and $d^{\*}$ is the formal adjoint of the exterior derivative $d$. The subspace of $d$-closed forms of $E^k$ is exactly $\mathcal H^k(M) \oplus Im d$, and thus we obtain an isomorphism between $\mathcal H^k(M)$ and the $k$-th De Rham cohomology group $H^k(M,\mathbb{R})$, whose dimension is equal to the $k$-th Betti number of $M$.
Morally speaking this isomorphism is interesting because it gives a link between the topological and geometric structures of the manifold. The Betti numbers only depend on the topology of $M$, while the space of harmonic forms is defined by a Riemannian metric. With this theorem we can interpret the $k$-th Betti number as counting the number of linearly independent harmonic $k$-form on the original manifold. In particular, as you say, if a $k$-th Betti number is zero, then the only harmonic $k$-form on $M$ is the zero form.
| 11 | https://mathoverflow.net/users/4054 | 44935 | 28,505 |
https://mathoverflow.net/questions/44930 | 2 | Let $\psi=f+ig=\rho e^{i\theta}$ be a complex function on some open subset of $\mathbb{R}^n$, where $f,g,\rho$ and $\theta$ are real-valued. I happened to find that the identity of differentiation for the polar coordinate expression
\begin{equation}
(1)\qquad\nabla\psi=e^{i\theta}\nabla\rho + i\rho e^{i\theta}\nabla\theta
\end{equation}
is rather formal. In general, $f$ and $g$ has the same regularity as $\psi$: $\psi\in C^1$ iff $f,g\in C^1$; $\psi\in H^1$ iff $f,g\in H^1$, etc. But this is not the case for $\rho$ and $\theta$.
For example, even $f,g\in C^{\infty}$, $\rho=\sqrt{f^2+g^2}$ may not be differentiable at points where $f=g=0$. One can see this by considering the one dimensional example $f(x)=x$, $g(x)=0$, and $\rho=|x|$ is not differentiable at $0$. Assuming $f,g\geq 0$ is a way to remedy this problem. In fact, $f,g\geq 0 $ and $C^1$ implies $\rho\in C^1$. A more natural way is to simply consider weak differentiability. For example if $f,g\in H^1$, I can show $\rho\in H^1$ (of course with weak derivatives in the same form as the usual ordinary ones. But although it seems to be very natural, I found this is a not so easy exercise in real analysis). I proved it by using approximation of $f,g$ by smooth functions. And I failed when using only the definition of weak derivatives. So, if only weak differentiability of $f,g$ is known and there is no approximation theorems as in Sobolev spaces, I don't know whether $\rho$ is also weakly differentiable. But anyway the result in the Sobolev setting may be sufficiently satisfactory.
The big problem comes when considering $\theta$. I have no idea about the regularity of $\theta$ at all. See the second term in the right hand side of identity (1), if $\psi\in H^1$,
we should have $\rho\nabla\theta\in L^2$. But we know $\rho\in L^2$ (in fact $H^1$), so $\nabla\theta\in L^{\infty}$? $\theta\in W^{1,\infty}$? There is also a disturbing
problem that on where $f$ and $g$ are both zero, $\theta$ can in fact be defined arbitrarily. So, a more correct question may be is there a natural choice of $\theta$ so that (1) is true in some sense?
---
Thanks to Denis Serre's references, I think my question may better be divided into two parts.
First, let $f+ig=\rho u$ with $\rho=\sqrt{f^2+g^2}$ and $u$ some complex function with $|u|=1$. Then, what is the best possible regularity of $u$ expected, in terms of $f$ and $g$?
And second, write $u=e^{iv}$, what is the best
regularity of $v$ in terms of $u$?
The references deal with the second question. From their introductions it looks like the natural framework for this question are such spaces as VMO, BMO, etc. But does the first question lead to such answer as $u\in$VMO or $u\in$BMO, under for example the assumption $f,g\in H^1$? I have little knowledge about the spaces BMO and VMO. Is the answer to my last question naturally "yes"?
| https://mathoverflow.net/users/4119 | What is the regularity of the argument of a complex function? | If $f,g$ have weak derivatives, the vanishing of $\rho$ is not the only problem. Even if $\rho$ stays uniformly positive, estimating the regularity of $\theta$ in terms of that of $(f,g)$ is hard and does not always work. Ths has been studied by Bourgain and Brézis (Comm. Pure Appl. Math. 58 (2005) 529–551 ; Publ. Math. Inst. Hautes Etudes Sci. 99 (2004) 1–115), Mironescu (C. R. Math. Acad. Sci. Paris 346 (2008), no. 19-20, 1039–1044) and H.-M. Nguyen (C. R. Math. Acad. Sci. Paris 346 (2008), no. 17-18, 957–962).
| 3 | https://mathoverflow.net/users/8799 | 44940 | 28,508 |
https://mathoverflow.net/questions/44947 | 4 | Let $M,N$ be complex manifolds and $f : M \to N$ be a bijective holomorphic map. Is then $f^{-1}$ also holomorphic?
The open mapping theorem implies that $f^{-1}$ is continuous. In order to apply the inverse function theorem, we need that the differential of $f$ is invertible. This is the case if $M,N$ are open subsets of $\mathbb{C}$. Can we generalize this do higher dimensions? If not, what happens if we assume $dim(M)=dim(N)$?
| https://mathoverflow.net/users/2841 | holomorphy of inverse map | Yes, $f^{-1}$ is holomorphic. In fact, the following result holds, see [Griffiths-Harris, Principles of Algebraic Geometry p. 19].
**Proposition**
If $f \colon U \to V$ is a one-to-one holomorphic map of open sets in $\mathbb{C}^n$, then
$|J\_f| \neq 0$, that is $f^{-1}$ is holomorphic.
The fact that $N$ is smooth is crucial. For instance, if $N \subset \mathbb{C}^2$ is the cuspidal cubic curve of equation $y^2=x^3$ and $f \colon M \to N$ is the normalization map, then $f$ is bijective and holomorphic but it is **not** a biholomorphism, since $f^{-1}$ is not holomorphic at the point $(0,0)$.
| 11 | https://mathoverflow.net/users/7460 | 44952 | 28,514 |
https://mathoverflow.net/questions/44900 | 7 | For Calabi-Yau three-folds we have $\mathcal{mirror \ symmetry}$: a map that associates most Calabi-Yau three-folds $M$ another Calabi-Yau three-fold $W$ such that $ h^{1,1}(M) = h^{2,1}(W)$ and $ h^{1,1}(W) = h^{2,1}(M)$ where $h^{i,j}$ are the Hodge numbers of the Calabi-Yau. In string theory such a duality leads to the conjecture that the type IIA superstring theory compactified on $M$ is equvilalent to the type IIB compactified on $W$.
$ \textbf{Question} :$ Are there extensions of mirror symmetry applied to generalized geometries (in the sense of Hitchin, Cavalcanti, and Gualtieri)? If so, what is the state of the art of this topic/question?
| https://mathoverflow.net/users/6527 | Mirror symmetries for generalized geometries ? | Mirror symmetry is at the most fundamental level an isomorphism of N=(2,2)-supersymmetric conformal field theories attached to different geometric data, which acts on the supersymmetries as a prescribed outer automorphism (switching A- and B-twists). Calabi-Yaus give rise to such SCFTs, hence one can ask for two CYs to be mirror -- though this is not a map in general, rather a correspondence that can then be pinned down more precisely in terms of large volume limits and SYZ fibrations etc. Mirror symmetry also has many rougher manifestations, such as an isomorphism of topological field theories of A- and B-type attached to geometric data, which now no longer need to be CY. In any case the point is that the generalized geometries you mention are precisely appropriate backgrounds to define such (2,2) SCFTs (see eg articles of Kapustin from around 04), so from the beginning they have a mirror symmetry question (look for pairs giving rise to the same SCFT up to this outer automorphism).
| 3 | https://mathoverflow.net/users/582 | 44960 | 28,520 |
https://mathoverflow.net/questions/44713 | 21 | If I have a function and I want to represent it as being the Laplace transform of another, that is, I want to be sure that there is $\hat{f}(s)$ such that my function $f(x)$ can be written as:
$$f(x) = \int ds \hat{f}(s) \exp(-sx)$$
what conditions should I impose over $f(x)$?
In other words, what are the conditions for the Fourier–Mellin–Bromwich integral
$$\hat{f}(s) = \frac{1}{2\pi i} \int\_{\gamma - i\infty}^{\gamma + i\infty} f(x) \exp(sx) dx$$
to exist?
| https://mathoverflow.net/users/757 | When I can safely assume that a function is a Laplace transform of other function? | The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that
$$e^{-ct}\phi(t)\in L^2(0,\infty)\tag{1}\label{1} $$
for some $c\in \mathbb R$. In this case the Laplace transform
$$f(s)=\int\_{0}^{\infty}e^{-st}\phi(t)dt\tag{2}\label{2}$$
can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that
$$\sup\limits\_{\sigma>c}\int\_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\tag{3}\label{3}$$
Now rewriting \eqref{2} as
$$f(\sigma+i\tau)=\int\_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$
we observe that $f$ is just the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended by $0$ to $t\leq 0$) belonging to $L^2(\mathbb R)$ for $\sigma=c$ and to $L^1(\mathbb R)\cap L^2(\mathbb R)$ for $\sigma>c$. Taking the
inverse Fourier transform, we get that
$$ e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int\_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$
and
$$0=\frac{1}{2\pi}\int\_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$
or, equivalently,
$$\lim\limits\_{T\to\infty}\frac{1}{2\pi i}\int\_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\\ \\\ 0, & t\leq 0. \end{cases} $$
One can show also that the Parseval identity
$$\frac{1}{2\pi}\int\_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int\_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt$$
holds, so there is a complete analogy with the standard Fourier transform.
---
**Executive summary.** A function $f$ is the Laplace transform of some function $\phi$ satisfying condition \eqref{1}, if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition \eqref{3} holds. This class of functions is known as the [Hardy space](http://en.wikipedia.org/wiki/Hardy_space) on a (right) half-plane.
| 19 | https://mathoverflow.net/users/5371 | 44988 | 28,532 |
https://mathoverflow.net/questions/44993 | 2 | Hi all. Can you help me with this? I have a square $S$ in euclidean plane with edges $A,B,C,D$ and a closed set $F$ in $S$ such that $F\cap A=F\cap C=\emptyset$, and $F\cap B$ and $F\cap D$ are nonempty. Assume that any curve in $S$ starting on $C$ and ending on $A$ intersects $F$. Does it follow that there exists a curve in $F$ starting on $B$ and ending on $D$?
Intuitively, it seems trivial, but I don't know how to proove it formally. Thanks a lot,
Peter
| https://mathoverflow.net/users/10072 | simple connectedness problem | No, take $S=[-2,2]\times[-2,2]$ and $F$ the closure of the graph of $[-2,2]\setminus\{0\}\ni x\mapsto \sin(1/x)$.
$$\*$$
However, it is true that there is a connected component of $F$ that meets both the (closed) edges $B$ and $D.$ Equivalently, there is a connected component of the set $G:=F\cup B\cup D$ that contains both $B$ and $D.$ Suppose not, by contradiction. Then, the same holds for some open neighborhood $V$ of $G$ in $S$ (this is easily shown recalling that a nested intersection of connected compacta is a connected compact). This amounts to saying that the edges $B$ and $D$ have disjoint open nbds, resp, $U\_B$and $U\_D$ whose union is $V$. The assumption that $S\setminus F$ contains no paths connecting $A$ and $B$ implies that the edges $A$ and $B$ also have disjoint open nbd $U\_A$ resp., $U\_C$, whose union is $S\setminus F.$ In conclusion, we have covered the square with four nbds of the edges, in such a way that nbds of opposite edges do not meet. This leads to a contradiction. Consider a partition of unity subordinate to the covering $\{U\_A, U\_B, U\_C, U\_D\},$ and use it to define a vector field $X$ on $S$ such that on $U\_i\cap U\_j$ the field $F$ is a convex combination of the exterior normal to $U\_i$ and the exterior normal to $U\_j$. This implies that $X$ has topological degree $1$ w.r.to zero, and never vanishes, a contradiction. Otherwise, you may use $X$ to retract the square on its boundary, another contradiction.
| 4 | https://mathoverflow.net/users/6101 | 44994 | 28,536 |
https://mathoverflow.net/questions/44979 | 15 | Let $f:[0,1]\to[0,1]$ be the classical [devil's staircase](http://en.wikipedia.org/wiki/Cantor_function).
Has anybody ever computed (or studied) the fourier coefficient of $f(x)$?
Related question: is the fourier series of $f(x)-x$ normally convergent (with respect to uniform norm)?
| https://mathoverflow.net/users/7979 | Evil Fourier Coefficients | The Fourier transform of the derivative $\mu$ of the Devil staircase is explicitely stated on the wikipedia page of the [Cantor distribution](http://en.wikipedia.org/wiki/Cantor_distribution), in the table at the right,
under the heading "cf" (characteristic function). Its value is
$$ \int\_0^1 e^{itx} d\mu(x) = e^{it/2}\ \ \prod\_{k=1}^\infty \cos(t/3^k)$$
Just multiply by $-1/it$, add $1/it$, and you get the Fourier transform of the Devil staircase.
A word on the proof. The Cantor distribution is the weak limit of the functions obtained by summing the indicator functions of the 2^n intervals generating the Cantor set at the nth step
(after renormalization). The Fourier transform of these sums can be computed explicitely. Then let n goes to infinity.
| 24 | https://mathoverflow.net/users/6129 | 44999 | 28,538 |
https://mathoverflow.net/questions/44876 | 5 | Suppose I have a function $f(x,y)$ from $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ that is convex in both $x$ and $y$. Set
$g(y) = \min\_{x} f(x,y)$
What I would like is for $g(y)$ to be Lipschitz:
$|g(y) - g(y')| \le c \cdot \| y - y' \|$
Unfortunately, $f(x,y)$ may have a very poor Lipschitz constant for general $x$. Are there general conditions on $f$ for which the minima are Lipschitz?
Alternatively, when can we say the minimizer $x^{\ast}(y) = \arg \min\_x f(x,y)$ is Lipschitz in $y$?
I've tried looking in a few convex optimization books for answers, but no luck.
| https://mathoverflow.net/users/1185 | Lipschitz properties of minima/minimizers of convex functions of two variables | I encountered the same problem three years ago and found some relevant literature. Here are a few. See also the refs therein.
Lipschitz Behavior of Solutions to Convex Minimization Problems.
Jean-Pierre Aubin,
Mathematics of Operations Research, Vol. 9, No. 1. (Feb., 1984), pp. 87-111.
Lipschitz continuity of solutions of linear inequalities, programs and complementarity problems.
O. L. MANGASARIAN and T.-H. SHIAU.
SIAM J. CONTROL AND OPTIMIZATION, 25(3), 1987.
Lipschitz Continuity of Solutions of Variational Inequalities with a Parametric
Polyhedral Constraint.
N. D. Yen,
Mathematics of Operations Research, Vol. 20, No. 3. (Aug., 1995), pp. 695-708.
On Lipschitzian Stability of Optimal Solutions of Parametrized Semi-Infinite
Programs.
Alexander Shapiro,
Mathematics of Operations Research, Vol. 19, No. 3. (Aug., 1994), pp. 743-752.
SHARP LIPSCHITZ CONSTANTS FOR BASIC OPTIMAL SOLUTIONS AND BASIC FEASIBLE SOLUTIONS OF LINEAR PROGRAMS.
Wu Li,
SIAM J. CONTROL AND OPTIMIZATION
Vol. 32, No. I, pp. 140-153, January 1994
| 5 | https://mathoverflow.net/users/3736 | 45003 | 28,541 |
https://mathoverflow.net/questions/33401 | 6 | Given an arbitrary Hilbert manifold, can on find a complete Riemannian metric and a Morse function satisfying the Palais-Smale condition?
| https://mathoverflow.net/users/3509 | existence of Morse functions satisfying the Palais-Smale condition | I recently learned that the answer to the question is YES, answered in the ETH preprint "H-cobordism for Hilbert Manifolds" by Dan Burghelea. I found the reference in the article "On the differential topology of Hilbert manifolds" of Eells and Elworthy.
| 8 | https://mathoverflow.net/users/3509 | 45007 | 28,543 |
https://mathoverflow.net/questions/44943 | 9 | For $CAT(\kappa)$ spaces $X$ we have following rigidity result: if equality holds in any of the comparison distances between a triangle $\Delta$ in $X$ and the corresponding comparison triangle $\tilde\Delta$ in the model space $M\_\kappa$ of constant curvature $\kappa$, then the convex hull of $\Delta$ is isometric to the convex hull of $\tilde\Delta$.
In Alexandrov spaces the picture is different. We do not have such a rigidity anymore. A counterexample can be found by considering two copies of a spherical triangle glued by their boundaries.
The rigidity result that we can obtain is following (I think at least, but I do not know any reference): under the assumption of equality in any of the comparison distances we can embed isometrically the convex hull of the comparison triangle $\tilde\Delta$ in $X$, such that two sides coincide with the corresponding two sides of $\Delta$, but it may happen that the third side does not coincide anymore. An example of this can be seen in the counterexample above.
So my question is following. In the example given above the space is singular. Does anybody know an non singular example (i.e. manifold of sectional curvature $\geq \kappa$), where we do not have the same rigidity as in the $CAT(\kappa)$ case?
Remark: We assume for our triangle $\Delta$ that its perimeter is $<2\pi/\kappa$ and each side has length $<\pi/\kappa$.
Edit: Ok, I will be more sprecific with my question:
Is there a Riemannian manifold $M$ with sectional curvature $\geq \kappa$ and a
a triangle $(x,y,z)$ in $M$ and a point $p$ in the side $yz$
such that if $(\tilde x, \tilde y, \tilde z)$ is a comparison triangle in $M^2\_\kappa$, and $\tilde p$ the corresponding point in $\tilde y \tilde z$,
we have the equality $d(x,p)=d(\tilde x,\tilde p)$ but the triangle $(x,y,z)$ (that is, the 1-dimensional object) cannot be filled with a triangle of constant sectional curvature $\kappa$ (that is, the 2-dimensional object). Such an example is easy to construct if we admit singular Alexandrov spaces, but I do not know any manifold examples.
| https://mathoverflow.net/users/23873 | Rigidity of triangle comparison in Alexandrov spaces | The question is not stated precisely.
So I'm free to say anything :)
* If you are interested in "non-uniqueness" then the anser is "NO". In any such triangle $[x y z]$ there are at least two distinct geodesics between directions of $[x y]$ and $[x z]$;
thus the space of directions at $x$ can not be a sphere.
* If you are interested in "the third side is wrong", the answer is "YES". Take $x=(0,0,0)$, $y=(1,0,1)$ and $z=(0,1,-1)$ in $\mathbb R^2\times [-1,1]$.
Then glue $\mathbb R^2\times \{1\}$ to it-self along reflection $(u,v)\mapsto (-u,v)$
and
glue $\mathbb R^2\times \{-1\}$ to it-self along reflection $(u,v)\mapsto (u,-v)$.
You get a singular Alexandrov space with triangle $[x y z]$ where the side $[y z]$ might be wrong for some filling of the hinge at $x$. BUT this triangle admins a flat filling.
It is easy to smooth this spaces near both singular lines to obtain Riemannian manifold with the same property.
* I know one example of a triangle in singular Alexandrov 3-space such that all distances between points on sides are the same as the corresponding distances in the model triangle, but it can not be filled with a flat triangle. This is a bit tricky to construct. I can not make this example to be Riemannian (and I feel that it is impossible).
**P.S.** The answer to the "specific" question is "NO".
Take a 2-dimensional nonnegatively curved manifold $M$ with a pair of points $p,q$ such that there are two minimizing geodesics from $p$ to $q$. One can choose a triangle with vertexes $x=(0,p)$, $y=(1,q)$ and $z=(-1,q)$ in the product $\mathbb R\times M$ which does not admit flat filling.
On the other hand any such triangle satisfies your condition for midpoint of $[y z]$.
| 7 | https://mathoverflow.net/users/1441 | 45014 | 28,547 |
https://mathoverflow.net/questions/44991 | 5 | Is there any known theory for equations like $a\_1 x^{y\_1} + a\_2 x^{y\_2} +..a\_nx^{y\_n}=0$ where the $y\_i$'s are arbitrary irrationals? Can you say anything about the disposition of roots?
| https://mathoverflow.net/users/6495 | Roots of polynomial-like equations with irrational powers | Note that these are simply a special case (generalization? depending on terminology) of [exponential polynomials.](http://en.wikipedia.org/wiki/Exponential_polynomial)
**Or rather, to be more accurate, they can be treated as such.** Let $t=\ln x$, and then your monomials become $a\_i e^{ty\_i}$. The change of variables does introduce some technical issues, obviously (in dimension $n$, you need to do it separately for each of the $2^n$ quadrants, and then possibly iterate on the coordinate zero hyperplanes).
Over the complexes, as was pointed out by Gerry, there is potentially infinitely many roots. Over the reals, the number of roots is still finite, and some of the elementary results like Descartes's rule of signs do translate fairly well to that setting. (The fact that for ordinary polynomials, the derivative is either 0 or has fewer roots is not so crucial in many applications).
In several variables, the theory of [fewnomials](http://en.wikipedia.org/wiki/Pfaffian_function) developed by Khovanskii gives some (pessimistic) upper bounds on the number of real roots (it also gives some restrictions on the patterns of the complex roots).
None of this is overly difficult, but I don't see the point of going into more details without a more specific question.
| 2 | https://mathoverflow.net/users/8212 | 45018 | 28,549 |
https://mathoverflow.net/questions/45020 | 8 | Let $f:[0,1]\to[0,1]$ be a devil's staircase in the [usual sense](http://en.wikipedia.org/wiki/Singular_function). (That is, $f$ is continuous, non-decreasing, $f'=0$ on a set of full Lebesgue measure.) We also require the complement to the set where $f'$ vanishes to have **Hausdorff dimension zero**.
*Question*. Is it true that $f$ is **not** Hölder continuous?
(This looks plausible, since $f$ has `very little room' where it can grow so it has to grow very fast - at least, at some points.)
| https://mathoverflow.net/users/8131 | Non-Hölder continuous devil's staircases | Let $K$ be the bad set.
Assume $f$ is Hölder continuous with exponent $\alpha$.
Since Hausdorff dimension of $K$ is zero,
given $\epsilon>0$ we can cover $K$ by open intervals $\left]a\_i,b\_i\right[$
with length $\ell\_i=b\_i-a\_i$ has such that
$$\sum\_n\ell\_n^\alpha<\epsilon\ \ \ \ \ (\*)$$
and $\ell\_n<\epsilon$ for any $n$.
Set $v\_i=f(b\_i)-f(a\_i)$.
Since $f$ is Hölder continuous,
$$v\_i < C{\cdot}\ell\_i^\alpha.\ \ \ \ \ (\*\*)$$
But clearly
$$\sum v\_i=1$$
which contradicts $( \* )$ and $( \* \* )$.
Did I miss something?
| 9 | https://mathoverflow.net/users/1441 | 45024 | 28,550 |
https://mathoverflow.net/questions/45021 | 7 | Here I am not assuming the factor is represented on a separable Hilbert space. This is quoted on page 370 of Takesaki II, then a bit later on page 381, and I haven't been able to find a proof prior to this point in the book or in Takesaki I.
| https://mathoverflow.net/users/5513 | Why is every factor a tensor product of a $\sigma$-finite factor and a factor of type I? | It is because a von Neumann algebra is $\sigma$-finite if it has a faithful normal state, there is a partion of unity $1 = \sum\_{i\in I} p\_i$ by mutually orthogonal projections equivalent to any given projection $p$ in an infinite factor, and such a decomposition induces the isomorphism $M \sim pMp \bar{\otimes} B(\ell^2I)$.
| 14 | https://mathoverflow.net/users/9942 | 45049 | 28,562 |
https://mathoverflow.net/questions/45043 | 8 | Is there an integer polynomial $ A \in {\mathbb Z} [ X ]$ of degree $d\geq 2$ such that for any integer $n\in {\mathbb Z}$ , $ A(n) $ is a square-free integer?
| https://mathoverflow.net/users/2389 | Integer polynomial (of degree >1) all of whose values are square-free | No. WLOG $A$ is irreducible. Pick a sufficiently large prime $p$ dividing $A(k)$ for some $k$ (there are infinitely many such primes, for example by the argument [here](http://qchu.wordpress.com/2009/09/02/some-remarks-on-the-infinitude-of-primes/)). In particular pick $p$ large enough so that it is relatively prime to the coefficients of $A$ and to $d$, and so that it does not divide the discriminant of $A$. Then the congruence $A(x) \equiv 0 \bmod p^2$ has a solution by Hensel's lemma.
| 13 | https://mathoverflow.net/users/290 | 45053 | 28,564 |
https://mathoverflow.net/questions/45044 | 16 | Let $A:={\mathbb{C}}[x\_0,\ldots,x\_n]=\oplus\_{d=0}^{\infty} {\mathbb{C}}[x\_0,\ldots, x\_n]\_d$ be the graded complex algebra of polynomials in $n+1$ variables, graded by degree.
Suppose $L$ is a line bundle over a projective manifold $X$ such that the ring of plurisections of $L$, i.e. $\oplus\_{d=0}^{\infty} H^0(X,L^d)$ is isomorphic to $A$ as graded complex algebras.
Does that imply that $X$ is necessarily complex projective space ${\mathbb{CP}}^n$ and $L$ is the tautological line bundle?
| https://mathoverflow.net/users/nan | Projective variety with no syzygies but not isomorphic to projective space | The answer is no. Take any projective manifold $X$ mapping $\pi\colon
X\rightarrow\mathbb P^n$ to such that $\pi\_\*\mathcal O\_X=\mathcal O\_{\mathbb
P^n}$ and let $L$ be $\pi^\*\mathcal O(1)$. Then $H^0(X,L^{\otimes
m})=H^0(\mathbb P^n,\mathcal O(m))$ Examples are $X=Y\times\mathbb P^n$ or a
blowing up of a closed smooth subvariety of $\mathbb P^n$.
I do not know if some power of $L$ is necessarily without base points (in which
case $L$ itself is). In any case we can blow up $X$ such that there is a map to
$\mathbb P^n$. The line bundle that gives that mapping may I guess have more sections.
**Addendum**: I was really rambling in the last paragraph. What I meant was that suppose we only know that $\bigoplus\_nH^0(X,L^{\otimes n})$ is a polynomial ring. Does that mean that $L$ is base-point free? It may very well be true but I don't see it at the moment.
| 16 | https://mathoverflow.net/users/4008 | 45056 | 28,566 |
https://mathoverflow.net/questions/45025 | 9 | Let me first pose a trivial question.
>
> Given a Borel probability measure $\mu$ on the real line, is it possible to construct a purely atomic random measure $M$ whose mean is $\mu$?
>
>
>
The answer is obviously yes: take $M = \delta\_{X}$, where $\delta\_x$ is the Diract measure sitting at $x$ and $X$ is a random variable distributed according to $\mu$. In fact take $M$ to be the empirical measure $\frac{1}{n} \sum\_{i=1}^n \delta\_{X\_i}$ where each $X\_i \sim \mu$ suffices.
Now here is my question:
>
> Given a Borel probability measure $\mu$ on the real line and $\delta > 0$, is it possible to construct a purely atomic random measure $M$ whose mean is $\mu$ and $d(M, \mu) < \delta$ almost surely, where $d$ is some metric on the space of probability measures (e.g. the [Wasserstein distance](http://en.wikipedia.org/wiki/Wasserstein_metric), the [Lévy–Prokhorov metric](http://en.wikipedia.org/wiki/L%25C3%25A9vy%25E2%2580%2593Prokhorov_metric) or the Kolmogorov distance, i.e., the sup-distance between distribution functions)?
>
>
>
Of course for an arbitrary distance this is not always possible. For example, if $d$ is the total variation distance and $\mu$ is atomless, then $d(M,\mu) = 2$ a.s. But for a weaker distance, will this be possible? The intuition is the following: consider the empirical measure $\frac{1}{n} \sum\_{i=1}^n \delta\_{X\_i}$ where $X\_i$ are iid generated from $\mu$. Then as $n\to\infty$, it will converge to the mean $\mu$ a.s. under those distances. Therefore within any $\delta$-ball centered at $\mu$, there are lots of atomic measures, i.e., $\mathbb{P} \{d(M,\mu) < \delta\}$ is very close to 1. But can we achieve exactly 1, i.e., can we construct an $M$ supported on those measures which are close to the desired mean?
| https://mathoverflow.net/users/3736 | construction of a random measure with a given mean | Yes -- the idea is to use the random delta-measure described in the first part of your question. However, in order to obtain a good approximation one has to subdivide the real line into small intervals (instead of taking the empirical averages over the whole line).
For simplicity let us consider just the transportation metric $d$. Given a probability measure $\mu$ denote by $\tilde\mu$ the image of $\mu$ under the map $x\mapsto\delta\_x$. Obviously, if the support of $\mu$ is contained in a length $\delta$ interval, then $d(\mu,\lambda)\le \delta$ for $\tilde\mu$-a.e. measure $\lambda$. Now subdivide $\mathbb R$ into intervals $I\_i$ of length $\delta$, denote by $\mu\_i$ the normalized restriction of $\mu$ onto $I\_i$, take independent (actually, independence is not needed here) random measures $\lambda\_i$ with distributions $\tilde\mu\_i$, and put $\lambda= \sum \mu(I\_i)\lambda\_i$. Then $\mathbf E\lambda=\mu$ and almost surely $d(\mu,\lambda)\le\delta$.
| 7 | https://mathoverflow.net/users/8588 | 45057 | 28,567 |
https://mathoverflow.net/questions/45019 | 4 | Let $n\in\mathbb N$ and $X$ be a complete metric space.
>
> Assume that there is $\epsilon>0$ such that
> $$\dim B\_\epsilon(x)\le n$$
> for any $x\in X$.
> Is it true that $\dim X\le n$?
>
>
>
* Here $\dim$ stays for [topological dimension](http://en.wikipedia.org/wiki/Lebesgue_covering_dimension).
* We do not assume that $X$ is separable!!!
| https://mathoverflow.net/users/10330 | Topological dimension, is it local? | From [E]T.7.2.3 p. 484:
1] If a Normal topological space $X$ has a locally finite closed cover $(F\_f)\_{s\in S}$ and $dim F\_s\leq n\ s\in S$ then $dim X \leq n$.
from the subspace theorem ([ED]p.216):
2] For any subspace $M$ of a strongly-hereditarily-normal space (in particular a metric space) $X$ we have $dim M \leq dim X$
***In what follow assume that any open $\epsilon'$-ball ($\epsilon'\leq\epsilon$) B as $dim$-dimention $n$***.
3] If $A\subset X$ contains a open $\epsilon'$-ball and is contained in a open $\epsilon'$-ball ($\epsilon',\epsilon''\leq \epsilon$) then $dim A=n$
PROOF: From [2].
4] There exist a locally finite covering of closed set $(F\_s)\_{s\in S}$ with $dim F\_s\leq n$.
PROOF: COnsider the covering by all open $\epsilon/2$-balls, and let $(B\_s)\_{s\in S}$ a locally finite refinement ($X$ is paracompact), let $F\_s:=Cl(B\_s)$ then $(F\_s)\_{s\in S}$ is a locally finite refinement of the open $\epsilon$-balls covering. Then as in [3] follow that $dim F\_s=n$.
Then from [1] and [4]: $dim(X)\leq n$ and from [2]: $n=dim B\_\epsilon(x) \leq dim X$
[E]: Engelking, General Topology
[ED]: Engelking, Dimention theory
| 4 | https://mathoverflow.net/users/6262 | 45059 | 28,569 |
https://mathoverflow.net/questions/45061 | 2 | Suppose that $n+m$ balls of which $n$ are red and $m$ are blue, are arranged in a linear order, we know there are $(n+m)!$ possible orderings. If all red balls are alike and all blue ball are alike, we know there are $\frac{(n+m)!}{n!m!}$ possible orderings.
For example, 2 red and 3 blue balls:
R1 R2 B1 B2 B3
R2 R1 B2 B3 B1
The above two orderings are equivalent and can be denoted as:
R R B B B
Now here is the problem: what if we further concentrate on the color, and record consecutive balls of the same color with the just ONE color code?
For example the color code for the afore-mentioned example would be:
R B
How many possible color code orderings are there?
| https://mathoverflow.net/users/5115 | linear ordering of color balls | Such a color-code ordering starts with either R or B and continues with strictly alternating R and B. The string can be of any length up to the smaller of $n$ or $m$, meaning it can be twice that smaller value, but that can be followed by one more character if there are enough of the other color. Moreover, every such string is a color-code ordering for some linear ordering of balls. There are a couple of special cases, namely that if either $n$ or $m$ is zero then there is exactly one color-code ordering and there aren't any if both are zero. Also, if neither is zero, we must have at least one instance of each letter.
So:
If $n = m = 0$, the answer is 0.
If exactly one of $n$ and $m$ is zero, the answer is 1.
If $n = m > 0$, the answer is $4n - 2$.
Otherwise, let $p$ be the minimum of $n$ and $m$. The answer is $4p-1$.
| 1 | https://mathoverflow.net/users/10611 | 45066 | 28,573 |
https://mathoverflow.net/questions/44961 | 10 | Is there a group G such that Aut(G) = $C\_3$? What if we replace 3 with a prime number p?
| https://mathoverflow.net/users/10596 | Aut(G) = $C_3$, G = ? | There is no group $G$ (finite or infinite) for which $Aut(G) \cong C\_p$ (the cyclic group of order $p$), if $p > 1$ is an odd number.
Suppose otherwise. The inner automorphism group $Inn(G)$ is a subgroup, also cyclic, and a well-known exercise in group theory is that if $Inn(G) \cong G/Z(G)$ is cyclic, then $G$ is abelian.
An abelian group $G$ has an involution given by inversion. Unless inversion is trivial, we get an element of order 2 in $Aut(G) = C\_p$, contradiction.
If inversion is trivial, then the abelian group $G$ becomes a vector space over $\mathbb{F}\_2$. In that case it is an easy to prove that either $Aut(G)$ is trivial or has an element of order 2; either way we get a contradiction.
**Edit:** After listening to some comments about this at meta, I amended my answer so that it gives less away or leaves a bit more to the imagination, or so I hope.
| 12 | https://mathoverflow.net/users/2926 | 45068 | 28,575 |
https://mathoverflow.net/questions/45008 | 15 | Is it possible to partition any rectangle into congruent isosceles triangles?
| https://mathoverflow.net/users/2884 | Partitioning a Rectangle into Congruent Isosceles Triangles | No. Note that the acute angle of your triangle must divide $\pi/2$ (look at a corner), so there are countably many such triangles (up to similarity), and hence you get only a countable set of possible ratios of sides.
| 24 | https://mathoverflow.net/users/4312 | 45075 | 28,578 |
https://mathoverflow.net/questions/45074 | 1 | Let we have algebraic equation on one variable. Which methods (exept Sturm's theorem and Descartes' rule) exist to find real roots of equation (or real positive)?
| https://mathoverflow.net/users/10613 | real roots of algebraic equation | Well, it depends what you mean by *finding*. Computer algebra systems commonly use isolation methods (approximations) that are based on an improved Uspensky's algorithm (by Rouillier and Zimmerman), but that's based off Descartes's rule of signs.
On the other hand, you can **encode** a real root by specifying the signs of all derivatives of the polynomial at the root. Thom's lemma guarantees that there cannot be more than one root satisfying all the sign conditions. Sturm's theorem helps you find which conditions are met. This does not give you any approximation of the root, but surprisingly enough allows you to compute symbolically with them. More details about this can be found on the book by Basu-Pollack-Roy freely available [here.](http://perso.univ-rennes1.fr/marie-francoise.roy/bpr-ed2-posted1.html)
I realize both of these options are refinements of the things you didn't want to talk about... I don't know of any other methods though (or if I do, they're escaping me right now).
| 3 | https://mathoverflow.net/users/8212 | 45076 | 28,579 |
https://mathoverflow.net/questions/45031 | 7 | I'm going to define an exponential polynomial of degree $k$ as a function $f$ of the form
$f(x) = \sum\_{i=1}^k c\_ie^{\alpha\_ix}$ ($\alpha\_i$s real).
My first question is: is there an algorithm for counting the number of real roots of such an expression, with complexity depending only on the degree $k$?
I strongly suspect that the answer to this question is yes, and that the answer is known (seeing as Tarski's exponential function problem is all but solved), but I can't find it described anywhere.
My second question is: can somebody tell me what this algorithm is? Or give me a hint?
I vaguely remember reading somewhere that there was a known method analogous to the method of Sturm chains for polynomials... But I haven't been able to figure out what it should be, nor have I been able to find where I read that claim. My best guess is that we can get rid of terms of such an expression by first dividing $f$ by an exponential $e^{\alpha x}$ to make a term constant, differentiating, and then multiplying by that same exponential. If we call this operation $D\_{\alpha}$, we get
$D\_{\alpha}f(x) = \sum\_{i=1}^k c\_i(\alpha\_i-\alpha)e^{\alpha\_ix}$.
The nice thing is that $D\_{\alpha}f$ acts analogously to the derivative of $f$, i.e. between any consecutive zeroes of $f$ there is a zero of $D\_{\alpha}f$. The problem is that I can't think of a good analogue to the division algorithm for exponential polynomials (maybe we don't need one?).
**Edit:** When I say that Tarski's exponential function problem is "all but solved," I mean that all that is missing from the full solution is a proof of Schanuel's conjecture. I'm not saying that Schanuel's conjecture is easy, but given this result it seems to me that we should be able to describe some sort of explicit algorithm for deciding problems like this one, although the proof of correctness of such an algorithm might require us to assume Schanuel's conjecture holds.
| https://mathoverflow.net/users/2363 | Sturm chain analogue for exponential polynomials? | Tarski's problem is **not** all but solved, or at least not last I checked. Precisely, the gap between Wilkie's theorem and solving Tarski's problem is *decidability* of exponential systems, for which the best result so far as been that Wilkie and Macintyre showed it was true if you assume that the [Schanuel conjecture](http://en.wikipedia.org/wiki/Schanuel%2527s_conjecture) holds. So if decidability is not known, let's not even get into *counting*!
Of course, this does not mean that we have to be pessimistic about your question, since you restricted yourself to the single variable case. I don't believe there is currently such a result, at least in full generality.
**Added Later:** If we do assume Schanuel's Conjecture, there is a recent algorithm for root isolation (thus I suppose for counting as well). This is the reference I was able to find (I have not read it yet):
*Real root isolation for tame elementary functions*
Adam Strzebonski, Wolfram Research Inc.
In ISSAC '09
<http://dx.doi.org/10.1145/1576702.1576749>
The abstract hits all the important points: it's univariate-only, and relies on Schanuel's conjecture to determine the signs of certain expressions.
| 6 | https://mathoverflow.net/users/8212 | 45077 | 28,580 |
https://mathoverflow.net/questions/45089 | 18 | What is an example of a formal scheme that is not algebraizable?
Recall that, if $X$ is a locally noetherian scheme and $Z$ is a closed subset (of the underlying topological space), then one can form the formal completion of $X$ along $Z$ which is sometimes denoted $X\_{/Z}$. This is a formal scheme whose underlying topological space is $Z$.
What is a formal scheme that is not of this form?
**Update:** Emerton and Francesco Polizzi suggested several examples that arise in the study of deformations of varieties with trivial canonical bundle. It'd be nice to see some more elementary, explicit examples as well.
**Update 2:** In comments, Francesco Polizzi mentioned that further examples can be found in [[Hironaka-Matsumura, "Formal functions and formal embeddings" J. math. soc. Japan 20](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-20/issue-1-2/Formal-functions-and-formal-embeddings/10.2969/jmsj/02010052.full); doi: [10.2969/jmsj/02010052](https://doi.org/10.2969/jmsj/02010052), Theorem 5.3.3] and [[Hartshorne, Ample subvarieties of algebraic varieties](https://link.springer.com/book/10.1007/BFb0067839), [p. 205](https://books.google.sk/books?id=PC58CwAAQBAJ&pg=PA205#v=onepage&q&f=false)].
**This is too long to fit into comments:**
@FP: Thanks! I'm not sure I quite follow the argument for non-algebraizability in the [book](https://books.google.com/books?id=xkcpQo9tBN8C&printsec=frontcover&dq=Sernesi%25252C+Deformations+of+algebraic+schemes&source=bl&ots=7V6WR6vEwJ&sig=k1D4dWjvl0VIhVNMM88W7a4FuYQ&hl=en&ei=Z9XVTOTlEsjDnAfK483OCQ&sa=X&oi=book_result&ct=result#v=onepage&q=Example%202.5.12&f=false). Sernesi states that, if $X \to \text{Spec}(\bar{A})$ is an algebrization, then $X$ would admit a non-trivial line bundle "since $X$ is of finite type over an integral scheme." Furthermore, he states that this line bundle can be chosen to "correspond to a Cartier divisor whose support does not contain $X\_{s}$ [the special fiber] and has nonempty intersection with $X\_{s}$." (note: The notation $X$, $X\_s$ is different in the text.)
It is not clear to me why such a line bundle exists: $\mathbb{A}^n$ is a finite type scheme over an integral scheme that has no non-trivial line bundles.
I understand how this shows that there is no algebraization by a *$\bar{A}$-projective* scheme, but why is there no algebraization by an arbitrary scheme?
I was a little nervous about the argument (Raynaud has an example of a family of Abelian varieties over a nodal curve with non-projective total space), but my concern was needless.
Here is one argument. Let $X\_0/\mathbb{C}$ be an algebraic $K3$-surface. We assume algebraizability and derive a contradiction. The statement about existence of non-algebraic deformations is very strong: In fact, there exists a 1st order deformation $f\_1 \colon X\_1 \to \text{Spec}(\mathbb{C}[t]/(t^2))$ with the property that the restriction of any line bundles $L\_1$ on $X\_1$ to $X\_0$ is numerically trivial. We use this deformation to derive a contradiction.
By definition, there exists a morphism $f\_1 \colon \text{Spec}(\mathbb{C}[t]/(t^{2})) \to \text{Spec}(\mathbb{C}[[x\_1, \dots, x\_{20}]])$ with property that the versal deformation restricts to $X\_1$. Now factor this morphism as $\text{Spec}(\mathbb{C}[t]/(t^{2})) \to \text{Spec}(\mathbb{C}[[t]]) \to \text{Spec}(\mathbb{C}[[x\_1, \dots, x\_{20}]])$ (by lifting the images of $x\_1, \dots, x\_{20}$ under $f\_1^{\*}$).
If $X\_{t} \to \text{Spec}(\mathbb{C}[[t]])$ is the restriction of the versal deformation, then the generic fiber is an algebraic $K3$-surface, hence admits an ample line bundle. The total space $X\_{t}$ of the family is regular, so it is possible to extend this line bundle to a line bundle $L\_{t}$ on $X\_{t}$. But then the restriction of $L\_{t}$ to the special fiber is not numerically trivial (by flatness); however, no such line bundle can even lift to 1st order. Contradiction.
| https://mathoverflow.net/users/5337 | Non-algebraizable Formal Scheme? | I think the following should work.
Let $X$ be a smooth, complex, projective $K3$ surface, and let $\bar{A}$ be the base of the formal semi-universal deformation of $X$. It is well-known that
$\bar{A}=\mathbb{C}[[X\_1, \ldots, X\_{20}]]$.
Let $\mathcal{X} \to \operatorname{Spf}(\bar{A})$ be the corresponding formal scheme. Then $\mathcal{X}$ is not algebraizable. Roughly speaking, the reason is that the general deformation of $X$ is a $K3$ surface which is *not* algebraic.
For a complete proof, see [Sernesi, Deformations of algebraic schemes, Example 2.5.12].
**EDIT.** As it is also remarked in Sernesi's book, this example shows that a smooth, complex, projective variety $X$ need not have an algebraic formally versal deformation, even if the functor $\operatorname{Def}\_X$ is prorepresentable and unobstructed.
| 16 | https://mathoverflow.net/users/7460 | 45090 | 28,589 |
https://mathoverflow.net/questions/45116 | 21 | It is well known that total space of the tautological line bundle $\mathcal{O}(-1)$ over projective space $\mathbb{P}^n$ is closed subvariety of $\mathbb{P}^n\times\mathbb{A}^{n+1}$. My question is how to realize total space of $\mathcal{O}(1)$ over $\mathbb{P}^n$ in such manner, i.e. I need an embedding of $Tot(\mathcal{O}(1))$ in simple variety and defining equations.
Thanks.
| https://mathoverflow.net/users/10626 | Total space of the line bundle $\mathcal{O}(1)$ over $\mathbb{P}^n$ | It is the complement $\mathbb{P}^{n+1} - \{x\}$ of a point in a projective space.
| 34 | https://mathoverflow.net/users/439 | 45121 | 28,604 |
https://mathoverflow.net/questions/45123 | 9 | Let $G$ be a group
generated by $a\_0, a\_1, a\_2$ with relations:
$a\_0 a\_1 a\_0^{-1}=a\_1^4$
$a\_1 a\_2 a\_1^{-1}=a\_2^4$
$a\_2 a\_0 a\_2^{-1}=a\_0^4$
I am wondering if $BS(1,4)=\langle a,b:bab^{-1}=a^4\rangle$ is embedded into G via $a\mapsto a\_1$, $b\mapsto a\_0$
Remark: the group is constructed in analogy to Higman group
| https://mathoverflow.net/users/8699 | Embedding of Baumslag-Solitar group into a certain group | This is a fundamental group of a graph of groups with B-S groups as vertex groups and cyclic groups as edge groups. So yes, each Baumslag-Solitar group naturally embeds in your group.
**Edit.** It is not a graph of groups but a complex of groups: triangle with B-S groups at vertices, cyclic groups as edges and 1 in the triangle. So you need to check the Haefliger condition to prove that it is developable. One can also (better) use Stallings, John R.
Non-positively curved triangles of groups. Group theory from a geometrical viewpoint (Trieste, 1990), 491–503, World Sci. Publ., River Edge, NJ, 1991.
**Edit 2.** No, this method does not work. To compute the Gersten-Stalling angle, one needs to count only the alternating length, not the total length of the word. The alternating length of each relator is not 7 but 4, so each angle is $\pi/2$ and the total angle is $3\pi/2$, too large.
**Edit 3.** In fact the group is finite, so none of the B-S groups embeds. See [this paper](http://www.ma.utexas.edu/users/allcock/research/bs6.pdf) (the link was sent to me by Dani Wise).
| 8 | https://mathoverflow.net/users/nan | 45125 | 28,606 |
https://mathoverflow.net/questions/45102 | 7 | Let $M$ be a factor, and let $\phi:M\to M$ be an irreducible endomorphism
("irreducible" means that the relative commutant of $\phi(M)$ in $M$ is trivial).
Let's also assume that $\phi$ is not invertible.
Is it possible to have $\phi\circ \phi$ conjugate to $\phi$?
In other words, is it possible to have an endomorphism $\phi$, and a unitary $u\in M$, such that $$\phi(\phi(x))=u\phi(x)u^\*,\quad\forall x\in M.$$
If this is possible, I would like to see an example.
---
*Note:* an answer to the above question would also settle [this question](https://mathoverflow.net/questions/41327/subfactor-of-finite-rank-but-infinite-index-is-this-possible).
| https://mathoverflow.net/users/5690 | endomorphism of factor: can it be idempotent up to congugacy? | This is not possible. If it were, then using the notation above, given any $x \in \phi(M)$, we would have $x u^\* = u^\* \phi(x)$, and $\phi(x) u = u x$. Hence, for any $x \in \phi(M)$ we have
$$
x u^\* \phi(u^\*) u^2 = u^\* \phi(x u^\*) u^2
$$
$$
= u^\* \phi(u^\*) \phi \circ \phi (x) u^2 = u^\* \phi(u^\*) u^2 x.
$$ `
Hence $u^\* \phi(u^\*)u^2 \in \phi(M)' \cap M = \mathbb C$ and so $\phi(u^\*) \in \mathbb C \cdot u^\*$. Then, for any $y \in M$ we would have that
$$
\phi \circ \phi (y) = u \phi(y) u^\*
$$
$$
= \phi(u y u^\*).
$$
Since $\phi$ is injective we then have $\phi(y) = u y u^\*$, and hence $\phi$ is invertible.
If you don't require that $\phi(M)$ be irreducible then this is possible.
| 7 | https://mathoverflow.net/users/6460 | 45132 | 28,609 |
https://mathoverflow.net/questions/45050 | 3 | Is there any known bound on sum of independent but not identically
distributed geometric random variables?
I have to show that the tail of the sum drops exponentially (like in
the Chernoff bounds for the sum of iid geom. variables).
Formally, if $X\_i$ ~ Geom($p\_i$), and $X = \sum\_{i=1}^n X\_i$, and it is known that $E[X]=\Theta(n)$,
Is it possible to show that $\Pr(X < 2E[X]) > 1 - \delta ^n$, where $\delta < 1$?
| https://mathoverflow.net/users/10609 | Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v | This isn't true, in general. If you take $p\_0=1/n$ and the other $p\_i=1$ then you get a constant probability for $X>2\mathbb{E}(X)$.
| 2 | https://mathoverflow.net/users/1061 | 45146 | 28,616 |
https://mathoverflow.net/questions/45150 | 16 | There seem to be two conflicting definitions for *p-adic valuation* in the literature.
Firstly, for any non-zero integer n, we have $\nu=\nu\_p(n)$ is the greatest non-negative integer such that $p^\nu$ divides $n$. Secondly, we have $|n|\_p$ which is defined as $1/p^\nu$. [These definitions can be extended to the rationals.]
$\nu$ is defined as the p-adic valuation in Khrennikov, Nilson, *P-adic deterministic and random dynamics* (for example) and $|\cdot|\_p$ is defined as the p-adic valuation in Khrennikov, *P-adic and group valued probabilities*, in Harmonic, wavelet and p-adic analysis (for example).
>
> Question: Is there a preferred definition for p-adic valuation?
>
>
>
| https://mathoverflow.net/users/2264 | What is the p-adic valuation of a number? | I will explain what's going on. We call $\lvert x\rvert\_p$ the $p$-adic absolute value of $x$ and $v\_p(x)$ the $p$-adic valuation of $x$. The distinction that is made by the two terms "absolute value" and "valuation" is completely standard… in English. However, Khrennikov is originally from Russia and in Russian there is one term for both concepts (нормирование = normirovanie, with stress on the second syllable -- I am not making that up, but see comments below this answer about stress on derived words in Russian, like verbs becoming nouns or nouns becoming adjectives). There is a term "absolute value" in Russian, but it is not an abstract concept; it refers only to the usual absolute value on the real or complex numbers (and quaternions?). This is perhaps why Khrennikov is using the term "valuation" incorrectly to refer to an absolute value function.
(I'm giving a course in Moscow this semester and I found this point *frustrating* when I was preparing my initial lectures. In different books I found the same word used for an absolute value and for a valuation and couldn't find the term that exclusively means absolute value. Eventually I determined there isn't one; you just know by context what meaning is intended. Native speakers are welcome to correct me here.)
UPDATE (3 years later): I learned from a student in St. Petersburg that the mathematicians there use separate terms for an absolute value $\lvert\cdot\rvert$ on a field and its corresponding valuation $v$: they call $\lvert x\rvert$ the norm (норма) of $x$ and $v(x)$ the exponent (показатель) of $x$.
UPDATE (11 years later): Consistent with Laurent's answer, the [Russian Wikipedia page for absolute value](https://ru.wikipedia.org/wiki/%D0%90%D0%B1%D1%81%D0%BE%D0%BB%D1%8E%D1%82%D0%BD%D0%B0%D1%8F_%D0%B2%D0%B5%D0%BB%D0%B8%D1%87%D0%B8%D0%BD%D0%B0) (find the English one and then click on Русский) refers to an absolute value as нормирование, a word I mentioned in the first paragraph above, and a valuation as экспоненциальное нормиорование, where the word in front of нормирование is eksponentsialnoe = exponential.
| 48 | https://mathoverflow.net/users/3272 | 45155 | 28,620 |
https://mathoverflow.net/questions/45159 | 25 | I know the definition of symplectic structure, symplectic group, and so on. But what does the word "symplectic" itself mean?
Meta question: I have many other mathematical words whose etymologies are obscure to me. Is it OK for me to ask one question per such word?
| https://mathoverflow.net/users/5420 | What does the word "symplectic" mean? | The term "symplectic group" was suggested in [*The Classical Groups: their invariants and representations*](http://books.google.co.uk/books?id=zmzKSP2xTtYC&printsec=frontcover&dq=classical+groups+weyl&hl=en&ei=FpTWTL6xAcmz4gbSy8yJBw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CDAQ6AEwAA#v=onepage&q&f=false) (1939, p. 165) by Herman Weyl:
>
> The name "complex group" formerly advocated by me in allusion to line complexes, as these are defined by the vanishing of antisymmetric bilinear forms, has become more and more embarrassing through collision with the word "complex" in the connotation of complex number. I therefore propose to replace it by the corresponding Greek adjective "symplectic." Dickson calls the group the "Abelian linear group" in homage to Abel who first studied it.
>
>
>
Take a look at the [Earliest Known Uses of Some of the Words of Mathematics](http://jeff560.tripod.com/s.html) web page.
| 36 | https://mathoverflow.net/users/5371 | 45162 | 28,625 |
https://mathoverflow.net/questions/45149 | 17 | When investigating regular languages, regular expressions are obviously a useful characterisation, not least because they are amenable to nice inductions. On the other hand ambiguity can get in the way of some proofs.
Every regular language is recognized by an unambiguous context-free grammar (take a deterministic automaton which recognises it, and make a production $R \rightarrow tS$ for every edge $R \stackrel{t}{\rightarrow} S$ in the DFA, and $R \rightarrow \epsilon$ for every accepting state $R$).
On the other hand, the natural "grammar" for a regular language is its regular expression. Can *these* be made unambiguous?
To be precise, let's define a parse for a regular expression (this is I think a natural definition, but not one I've seen named before).
* $x$ is an $x$-parse of $x$, if $x$ is a symbol or $x=\varepsilon$
* $(y, 0)$ is an $R\cup R'$-parse of $x$, if $y$ is an $R$-parse of $x$
* Similarly, $(y,1)$ is an $R\cup R'$-parse of $x$, if $y$ is an $R'$-parse of $x$
* $(y\_1, y\_2)$ is an $RR'$-parse for $x\_1x\_2$, if $y\_i$ is an $R$-parse for $x\_i$ for $i=1,2$
* $[]$ is an $R^\*$-parse for $\varepsilon$
* $[y\_1, y\_2, \dots, y\_n]$ is an $R^\*$-parse of $x\_1x\_2\cdots x\_n$, if $y\_i$ is an $R$-parse for $x\_i$ for $1 \le i \le n$
In short, the parses of a string tell us *how* a regular expression matches a string *if* it does.
A regular expression $R$ is *unambiguous* if, for every $x \in L(R)$, there is only one $R$-parse of $x$.
>
> Given a regular expression, is there an unambiguous regular expression which matches the same language?
>
| https://mathoverflow.net/users/10231 | Can regular expressions be made unambiguous? | There's a standard construction of a regular expression from a DFA: define an expression R(i,j,k) for the language of strings that take state i to state j of the DFA while using intermediate states that belong only to the subset of states from state 1 to state k, as follows.
* R(i,j,0) is [xyz...] where x, y, z etc are the symbols that occur as labels of transitions from state i to state j (there can be no intermediate states). If there is no such transition then R(i,j,0)=0 (the expression for the empty language).
* Similarly, R(i,i,0) is e + [xyz...] where e is the expression that represents the empty string.
* For k > 0, R(i,j,k) is R(i,j,k-1) + R(i,k,k-1) R(k,k,k-1)\* R(k,j,k-1). That is, any string that takes you from i to j using intermediate states up to k either goes from i to j without going through k, or can be parsed into a sequence of substrings that go from i to k, k back to itself zero or more times, and then k to j.
* The regular expression for the whole language is then the sum of the expressions R(1,i,n) where i is one of the accepting states.
The first two of these rules obviously give you unambiguous expressions, the third is unambiguous because there's only one way of parsing the string into substrings as described, and the fourth is unambiguous because any given string can only go to a single accepting state. Therefore, the final regular expression constructed in this way is unambiguous.
| 16 | https://mathoverflow.net/users/440 | 45163 | 28,626 |
https://mathoverflow.net/questions/45106 | 9 | Although I think I know the answers to these, I'd just like to collect them all in one place.
What is the quantum PCP theorem, what implications does its proof have for simulation of Hamiltonians and is following Irit Dinur's reproof of the classical version the best/only current mode of attack (and if so why?) What is the sort of math/physics/theoretical CS background needed to approach this problem?
| https://mathoverflow.net/users/10622 | Quantum PCP Theorem | The quantum PCP conjecture (*nobody has proved it, so you can't call it a theorem*) is possibly (*there are a few different ways of generalizing the classical PCP theorem to the quantum regime, and I don't believe any of them deserves the name of **the** quantum PCP conjecture*) given below. Here $k$, $c$, and $d$ are small fixed integer constants, and $\epsilon$ is a constant between 0 and 1.
>
> There is a polynomial time algorithm that does the follows:
> The input is a $k$-local Hamiltonian $H$ with $n$ terms each of total energy 1; the ground state energy of $H$ is either $0$ or greater than $1/n^c$. The output is a new $k$-local Hamiltonian $H'$, with $O(n^{d})$ terms. If the ground-state energy of $H$ is $0$, then $H'$ will also have ground-state energy $0$. If the ground-state energy of $H$ is at least $1/n^c$, then $H'$ will have ground-state energy at least $\epsilon n^{d}$.
>
>
>
If you're using qubits, you should probably take $k=4$, by Bravyi's results on Quantum $k$-SAT. Certainly $k\geq 4$, unless Quantum 3-SAT is QMA-complete. (EDITED June 2013: Quantum 3-SAT is indeed QMA complete. See [this recent paper](http://arxiv.org/abs/1302.0290).)
This would be an incredibly important development in the theory of quantum computing, but I'm not sure whether a proof has any practical implications for simulation of Hamiltonians. What it would show is that it is QMA-complete to tell whether an $n$-term local Hamiltonian has energy $0$ or at least $\epsilon n$.
**Begin(rant)**
Asking whether method $X$ is the best way to attack a big open mathematical conjecture is not a question anybody can answer. If you asked in this forum, for example, what is the best way to prove the Riemann hypothesis, and what mathematics you needed to learn in order to do this, your question would probably be promptly closed.
**End(rant)**
If I had to guess whether a quantum generalization of PCP was even true, I'd probably guess "no." It seems like an incredibly strong statement to me. On the other hand, the classical PCP theorem was also an incredibly strong statement. But just because a miracle happens in the classical regime, should you really be expecting the same miracle in the quantum regime?
| 18 | https://mathoverflow.net/users/2294 | 45167 | 28,629 |
https://mathoverflow.net/questions/45154 | 13 | Let $M$ be a smooth manifold, $\rho(p, q)$ — a differentiable metric on $M$. Can we construct Riemannian metric $g(X,Y)$ on $TM$ that induces $\rho(p, q)$? Under what conditions?
I'm sure this question has been dealt with, I just didn't find it in the quick survey of literature :)
| https://mathoverflow.net/users/44739 | Riemannian metric induced by a metric | Here is a closely related question that may have been what the OP was driving at. Suppose that you ONLY know a Riemannian manifold as a metric space---that is you know the point set and the distance between any two points, but you do not know the metric tensor or even the differentiable structure. Can you nevertheless reconstruct these from the distance function. The answer is that you can. See:
<http://www.ams.org/proc/1957-008-04/S0002-9939-1957-0088000-X/S0002-9939-1957-0088000-X.pdf>
| 18 | https://mathoverflow.net/users/7311 | 45172 | 28,632 |
https://mathoverflow.net/questions/45165 | 9 | Recall that a **standard Borel space** is a measurable space $(X,\mathcal{M})$ (i.e. a set with a $\sigma$-algebra) such that there exists a 1-1 bimeasurable map $\phi$ from $(X,\mathcal{M})$ to $[0,1]$ (the latter equipped with its Borel $\sigma$-algebra $\mathcal{B}\_{[0,1]}$). It is known that any Borel subset of a complete separable metric space is a standard Borel space.
Suppose now that $E$ is a non-Borel subset of $[0,1]$ (or any other complete separable metric space), such as the Vitali set. We can equip $E$ with the $\sigma$-algebra $\mathcal{M}$ induced by its inclusion into $[0,1]$, namely $$\mathcal{M} = \{ B \cap E : B \in \mathcal{B}\_{[0,1]}\}.$$ $\mathcal{M}$ is also the Borel $\sigma$-algebra generated by the subspace topology on $E$, which is also the metric topology on $E$.
Is it possible that $(E,\mathcal{M})$ is a standard Borel space? I would think not. Clearly the inclusion map $E \hookrightarrow [0,1]$ is not bimeasurable (though it is measurable), but it's less clear that no other injection could be bimeasurable.
Wikipedia gives an [example](https://en.wikipedia.org/wiki/Standard_probability_space#A_perforated_interval) using a set $E$ of outer measure 1 and inner measure 0, and shows indirectly that it cannot be standard Borel by equipping it with the probability measure $P(B \cap E)=m(B)$, noting that the inclusion map $X : E \to [0,1]$ is a uniformly distributed random variable, and observing that $X$ does not admit a regular conditional distribution given itself. However, it is not so clear how to extend this to other non-Borel sets (particularly those with outer measure 0), and in any case it would be nice to have a direct proof if possible.
Thanks!
| https://mathoverflow.net/users/4832 | Can a non-Borel set be a standard Borel space? | Strictly speaking, a standard Borel space can also be finite or countable. Keeping in mind this minor point, a subset of $[0,1]$ (endowed with the restriction of the Borel $\sigma$-algebra) is a standard Borel space if and only if it is a Borel subset of $[0,1]$.
This comes from the fact that the image of a Borel subset by an injective Borel map between two standard Borel spaces is a Borel set.
This result is proven for example in Dudley "Real analysis and Probability" or Cohn "measure theory".
Now for the proof. Let $\phi$ be the isomorphism between $([0,1],{\cal B})$ and $(E,{\cal M})$. The space E is a subset of $[0,1]$, so we get a Borel injection from $[0,1]$ to $[0,1]$ whose image is precisely $E$. Hence $E$ is a Borel subset of $[0,1]$.
| 7 | https://mathoverflow.net/users/6129 | 45176 | 28,635 |
https://mathoverflow.net/questions/31696 | 17 | I would like to know if there is a standard technique to check if a singular variety admits a small resolution. What are the main references for these types of questions?
I am mostly interested in threefolds and fourfolds with singularities in codimension 2 or higher.
(By a small resolution, I mean a proper birational transformation $Y\rightarrow X$ such that Y is smooth and the exceptional locus does not contain any divisors.)
| https://mathoverflow.net/users/4046 | Best strategy for small resolutions | I assume that by a *small morphism* you mean a proper birational morphism $f:Y\to X$ such that $f$ does not have an exceptional divisor. If $X$ is smooth in codimension $1$ as in your last sentence, then this is equivalent to that $f$ is an isomorphism in codimension $1$ on both $Y$ and $X$. A small morphism is a *small resolution* if $Y$ is smooth.
There is a nice criterion to check that a variety *does not* admit a small resolution:
Let $f:Y\to X$ be a small morphism. If $Y$ is quasi-projective and $X$ is normal then $X$ is not $\mathbb Q$-factorial (being $\mathbb Q$-factorial means that every Weil divisor is $\mathbb Q$-Cartier, that is, it has a non-zero integer multiple which is Cartier). The proof is very simple: Let $H$ be a Cartier divisor on $Y$ that is not trivial on a curve that gets contracted by $f$, for example an ample divisor on $Y$ will do. Now if $X$ were $\mathbb Q$-factorial, then $m(f\_\*H)$ is a Cartier divisor for some $m\neq 0$. Then by the condition the divisors $mH$ and $f^\*(m(f\_\*H))$ agree. However, the former was chosen to be non-trivial on a curve that is contracted while the latter must be trivial on every such curve as it is a pull-back. Q.E.D.
There is also something one can say for the reverse direction:
If $X$ has klt singularities, then it is possible to construct a morphism $f:Y\to X$ such that $f$ is small and $Y$ is $\mathbb Q$-factorial. I don't know a very easy proof of this. The essence is to take a resolution and then use a well-chosen directed mmp (in the sense of BCHM) to contract all the exceptional divisors.
This implies that the existence of a non-trivial small morphism $f:Y\to X$ where $X$ has klt singularities and $Y$ is quasi-projective is equivalent to $X$ not being $\mathbb Q$-factorial. Whether or not $X$ admits a small resolution is then decided on whether or not there exists a directed mmp so the "$\mathbb Q$-factorial model" obtained by the above method is smooth. It is possible that all choices lead to something $\mathbb Q$-factorial which is still singular and that $X$ does not admit a small resolution after all.
**EDIT:** Removed previous statement about the reverse direction as that was not true as stated. At this time I am not sure how to fix that statement. I replaced it with a different one that I know how to prove, but *the margin is not wide enough to include a proof*, so it is just stated without proof. Sorry. However, the current statement is probably pretty close to what one might hope for.
| 23 | https://mathoverflow.net/users/10076 | 45193 | 28,647 |
https://mathoverflow.net/questions/45212 | 42 | The first time I got in touch with the abstract notion of a sheaf on a topological space $X$, I thought of it as something which assigns to an open set $U$ of $X$ something like the ring of continuous functions $\hom(U,\mathbb{R})$. People said that sections of a sheaf $F$, i.e. elements of $F(?)$, are something which allow 'glueing' like in the example: if two functions $f:U\to\mathbb{R}$ and $g:V\to\mathbb{R}$ coincide on the intersection $U\cap V$ there is an unique function $U\cup V\to\mathbb{R}$ restricting to $f$ and $g$. So a sheaf consists of 'glueable' objects.
A presheaf, say of rings, on a topological space $X$ is a functor $F:Op(X)^{op}\to Rng$ where $Op(X)$ denotes the category of open sets of $X$. One may generalize all this using the terms 'site' and 'topos' but let's consider this easy situation. A sheaf is a presheaf fulfilling an extra condition, so there is an inclusion of categories
$$
Pre(X)\leftarrow Shv(X):i.
$$
Please excuse the awful notation but this inclusion functor admits a left-adjoint, the sheafification functor
$$
f:Pre(X)\leftrightarrow Shv(X):i.
$$
Since I got in touch with schemes, I think of a presheaf or a sheaf as of a space. There is a notation of a 'stalk' $F\_x\in Rng$ at a point $x$ of $X$. I think of a stalk as the point $x$ of the space $F$. The inclusion functor and the sheafification functor both respect the stalks.
For example the sheafification of a constant presheaf is a locally constant sheaf.
My first question is:
>
> How shoud I really think about sheafification?
>
>
>
A presheaf of sets is the canonical co-completion of a category: You take a (small) category $S$ which does not allow glueing (=has not all colimits) and then $Pre(S)$ has all colimits. $S$ is fully and faithfully embedded into $Pre(S)$ with the Yoneda embedding $Y:S\to Pre(S)$. This functor does not respect colimits, so, loosely speaking, the way of glueing is not respected in this transition. Maybe, considering sheaves instead of presheaves is a way of repairing this failure.
My second question is:
>
> With respect to the interpretation above, what really makes the difference between a presheaf and a sheaf and how should I visualize that difference, if I think of a presheaf as if it is a space?
>
>
>
Thank you.
| https://mathoverflow.net/users/2625 | How should one think about sheafification and the difference between a sheaf and a presheaf | There are two ways a presheaf can fail to be a sheaf.
* It has local sections that *should* patch together to give a global section, but don't,
* It has non-zero sections which are locally zero.
When dividing the problems into two classes, it is easy to see what sheafifying does. It adds the missing sections from the first problem, and it throws away the extra sections from the second problem.
The latter kind of sections tend to be easier to notice, but are less common. Usually, when a construction or functor must be sheafified, it has local sections that should patch together but don't.
A simple example of a presheaf with this property is the presheaf $F\_{p=q}$ of continuous functions on the circle $S^1$ which have the same value at two distinct points $p,q\in S^1$. When I restrict to an open neighborhood of $p$ that doesn't have $q$, the condition on their values goes away. Because the same thing is true for open neighborhoods of $q$ which don't contain $p$, the condition on the functions in this presheaf has no effect on sufficiently small open sets. It follows that this presheaf is locally the same as the sheaf of continuous functions. Therefore, for any function on $S^1$ which has different values on $p$ and $q$, I can restrict it to an open cover where each local section is in $F\_{p=q}$, but this function is not in $F\_{p=q}$. This is why $F\_{p=q}$ is not a sheaf.
When we sheafify, we just add in all these sections, to get the full sheaf of continuous functions. This is clear, because any two sheaves which agree locally are the same (though, I mean that the local sections and local restriction maps agree).
This example really does come up in examples. Consider the map $S^1\rightarrow \infty$, where $\infty$ is the topological space which is $S^1$ with $p$ and $q$ identified. If I pull back the sheaf of functions on $\infty$ in the naive way, the resulting presheaf on $S^1$ is $F\_{p=q}$. To get a sheaf, we need to sheafify.
| 97 | https://mathoverflow.net/users/750 | 45218 | 28,663 |
https://mathoverflow.net/questions/45219 | 16 | This question is closely related to [these](https://mathoverflow.net/questions/40309/in-diff-are-the-surjective-submersions-precisely-the-local-section-admitting-map) [two](https://mathoverflow.net/questions/40431/maps-that-admit-local-sections-through-each-point-in-the-domain), but the former doesn't go far enough and the latter didn't attract much attention, and anyway I want to ask the question slightly differently.
Recall that in the category of Topological spaces or in the category of Manifolds, a *submersion* is a (not necessarily surjective!) map $f: X \to Y$ so that for each point $x\in X$, there exists open neighborhood $f(x) \in U \subseteq Y$ and a map $g: U \to X$ splitting $f$, i.e. $f \circ g = \operatorname{id}\_U$. This definition does not generalize well to other categories: it requires at least that "points" know a lot about the objects, and that we know what are "open neighborhoods".
>
> My question is: How much extra "abstract nonsense" structure do I need to put on a category for it to have a good theory of submersions?
>
>
>
On the one hand, the surjective submersions of manifolds are all [regular epimorphisms](http://ncatlab.org/nlab/show/regular+epimorphism) (does this characterize the surjective submersions?), and so I could imagine defining "submersion" to mean a map that factors as a regular epi and a regular mono (I think that the regular monos in manifolds are the open embeddings?). Then it seems that I don't need *any* extra structure, but I have not checked that this conditions characterizes submersions.
On the other hand, (surjective?) submersions form a [Grothendieck pretopology](http://ncatlab.org/nlab/show/Grothendieck+pretopology), and hence determine a Grothendieck topology. Conversely, I would have assumed that a Grothendieck topology (which is extra structure on a category) determines which maps are submersions, although I am sufficiently new to this that I don't have a proposal for such a definition.
| https://mathoverflow.net/users/78 | What abstract nonsense is necessary to say the word "submersion"? | Dear Theo, I think that you're oversimplifying things a bit too much here. The notion of a submersion depends very much on an "admissibility structure" in the sense of Lurie, or a "geometric context" in the sense of Toën-Vezzosi. That is, in addition to a Grothendieck topology, you also need a "geometry" satisfying certain properties to give further structure to your category.
I was confused a while ago about a [similar point](https://mathoverflow.net/questions/18245/simplifying-the-definition-of-a-geometric-context-using-sieves), and after learning more about the subject, I realized that it's not nearly so simple as I had hoped.
For instance, the proper notion of a submersion in the algebro-geometric context is a smooth map, but there is no notion of a smooth map between sheaves on the affine étale site before first discussing what it means for a morphism to be "relatively representable". You may want to check out Toën-Vezzosi's paper "Homotopical algebraic geometry II", where they give an inductive definition of an n-geometric algebraic stack (where stack here means simplicial sheaf) (and this inductive definition holds true for sheaves of sets as well). For a map between schemes to be smooth (resp. a submersion) you need for the map to be "relatively representable" by a "scheme" (resp. a "manifold") (when you restrict to the case of sheaves of sets, $n$ really only varies between $-1$, $0$, and $1$).
This may all sound like gibberish, but if you take a look at Toën-Vezzosi's HAG II chapter 2 and ignore the homotopical stuff, the basic idea should be clear. The moral of the story is that a Grothendieck topology alone cannot characterize the geometry of the sheaves on that site.
| 5 | https://mathoverflow.net/users/1353 | 45235 | 28,675 |
https://mathoverflow.net/questions/45234 | 6 | My question is long on background and motivation, and almost but not quite answered over at the nLab. I'll write up a bunch before asking my question (feel free to skip to the end or look at the title), so that (i) those who know more than me can see exactly what I do and do not understand (ii) those that know less than me might learn something (iii) to clarify my own thoughts.
Background
----------
Let $\mathcal S$ be a reasonably nice category: for example, I want it to at least have either all finite limits, or I want it to have a [good theory of submersions](https://mathoverflow.net/questions/45219/what-abstract-nonsense-is-necessary-to-say-the-word-submersion); and I need some extra conditions (see comments), that I haven't fully thought through, but they should be satisfied in various "geometrical" categories like (your favorite convenient category of) Topological Spaces or Manifolds. A **span in $\mathcal S$** is a diagram $X \leftarrow M \rightarrow Y$, and if I can, I will require that the right map be a submersion. There is a two-category $\operatorname{Span}(\mathcal S)$ constructed in the usual way: 1-morphisms are spans, and 2-morphisms are maps of spans that cover the identity morphisms on the bases.
A **category object in $\mathcal S$** is a span $X = \{X\_0 \leftarrow X\_1 \rightarrow X\_0\}$ and 2-morphisms $$ \{ X\_0 \leftarrow X\_0 \rightarrow X\_0 \} \overset i \to \{X\_0 \leftarrow X\_1 \rightarrow X\_0\}$$ $$ \{X\_0 \leftarrow (X\_1 \underset{X\_0}\times X\_1) \rightarrow X\_0\} \overset m \to \{X\_0 \leftarrow X\_1 \rightarrow X\_0\}$$ in $\operatorname{Span}(\mathcal S)$ making $X$ into an algebra object. (The domain of $m$ is the 1-composition of $X$ with itself in $\operatorname{Span}(\mathcal S)$, and the domain of $i$ is the 1-identity span on $X\_0$.) Writing $\circ$ for the 1-composition in $\operatorname{Span}(\mathcal S)$, the requirement is that the various maps $X\circ X \circ X \to X$ formed from $m,i$ all agree.
A **functor in $\mathcal S$** $\{X\_1 \rightrightarrows X\_0\} \to \{Y\_1 \rightrightarrows Y\_0\}$ is a pair of maps $X\_1 \to Y\_1$ and $X\_0 \to Y\_0$ making some diagrams commute. But there tend not to be enough functors when $\mathcal S$ does not satisfy the axiom of choice. For example, if $\mathcal S$ is the category of manifolds, then certain categories (the groupoids, which I will define in a moment) are supposed to present stacks, but the functor that associates to each manifold $M$ the groupoid of functors $\{M \rightrightarrows M\} \to \{X\_1 \rightrightarrows X\_0\}$ into some groupoid $X$ does not satisfy the right descent axioms.
Instead, the usual thing to do is to define the notion of "right-principal bibundles". Let $X = \{X\_1 \rightrightarrows X\_0\}$ and $Y = \{Y\_1 \rightrightarrows Y\_0\}$ be categories. An **$X,Y$-bibundle** is a span $B = \{X\_0 \leftarrow B\_1 \rightarrow Y\_0\}$ and 2-morphisms $X \circ B \to B$ and $B \circ Y \to B$, such that all the various 2-morphisms $X \circ X \circ B \circ Y \circ Y \to B$ agree (as above, $\circ$ is the 1-composition in $\operatorname{Span}(\mathcal S)$). The "tensor product over $Y$" gives a "composition" of bibundles which is associative up to a canonical associator that satisfies a pentagon.
Given a span $B = \{X\_0 \leftarrow B\_1 \rightarrow Y\_0\}$, there is another span
$$ X\_0 \leftarrow (B\_1 \underset{X\_0 \times Y\_0}\times B\_1) \rightarrow Y\_0 $$
and a "diagonal" map from $B$ to this other span. Let $B$ be an $X,Y$-bibundle. Using the diagonal map, one can build a map
$$ B\_1 \underset{Y\_0}\times Y\_1 \to B\_1 \underset{X\_0}\times B\_1$$
which is actually a map of objects over $X\_0 \times Y\_0 \times Y\_0$. On (generalized) elements, this map is $(b,y) \mapsto (b,by)$. The bibundle $B$ is **right-principal** if this map is an isomorphism. A category $X$ is a **groupoid** if it is right-principal as an $X,X$-bibundle. If $X,Y$ are groupoids, a functor $f: X \to Y$ determines an $X,Y$-bibundle where the middle object is $X\_0 \underset {Y\_0} \times Y\_1$, and it is right-principal.
Then the point is that the 2-category whose objects are groupoids in $\mathcal S$ and whose one-morphisms are right-principal bibundles embeds as a full sub-2-category into the category of "stacks", which I will not define. For a precise version of this story, see for example [C. Blohmann, Stacky Lie groups, 2007](http://arxiv.org/abs/math/0702399).
Question
--------
Any functor of categories in $\mathcal S$ determines a bibundle, but if the categories are not groupoids, then the bibundle is not (usually) right-principal; for example, the identity bibundle is not. I do want the bibundles of groupoids to be "morphisms" of categories, so I don't want to just take functors. On the other hand, as observed in [Op. cit.](http://arxiv.org/abs/math/0702399), if we don't demand some sort of "right-principality" condition, then the 2-category which allows all bibundles as 1-morphisms has neither products nor a terminal object. Hence my question is:
>
> What is the correct notion of "bibundle" for (internal) categories that generalizes "right-principal bibundle of groupoids"?
>
>
>
The answer to my question is almost "[anafunctor](http://ncatlab.org/nlab/show/anafunctor)". An anafunctor seems to have a bit more structure than a bibundle, since it is an object not in $\operatorname{Span}(\mathcal S)$ but in $\operatorname{Span}(\text{categories in }\mathcal S)$. (Conversely, at least when $\mathcal S$ is the category of sets, bibundles wihout any conditions are the same as [profunctors](http://ncatlab.org/nlab/show/profunctor).) If I understood better how to go between anafunctors and bibundles, I would probably be happy.
| https://mathoverflow.net/users/78 | What condition on a "bibundle between categories" generalizes "right-principal bibundle between groupoids"? | Hi Theo. This is something I have been working on lately. In Makkai's original paper on anafunctors he defines a condition on an anafunctor which makes it *saturated*. His motivations are logical, in that he wants the 'image' of a point in the domain category (pullback and pushdown along the span) to be closed under isomorphism. This is satisfied for his examples of (co)limit anafunctors, or an anafunctor arising from a universal property. In the case that he talks about saturated anafunctors between groupoids, this is the same as a right principal bibundle (in Set). When we replace groupoids by categories, then he defines a saturated anafunctor to be an anafunctor such that the underlying span between the cores is saturated. Thus the 'cover' (in Makkai's case, a surjection of sets) is a right principal bibundle for the underlying groupoids.
This is done so that the canonical 2-functor $Cat \to Cat\_{sat.ana}$, where we do *not* assume choice, sends fully faithful, essentially surjective functors to equivalences in the bicategory $Cat\_{ana}$. Actually I'm fudging here, because Makkai defines $Cat\_{sat.ana}$ as a an anabicategory - a category weakly enriched over $Cat\_{ana}$.
Note also that Street at one point defines in his Oberwolfach descent notes a definition of a 'torsor for a category' which is just the same as Makkai's definition. He correctly notes that there are non-invertible maps between such 'torsors', unlike the group(oid) case.
So to cut a long story short, it is possible to define a saturated anafunctor for internal groupoids and hence categories. (email me if you would like some notes on this)
But! This is not the same as a torsor for an internal category, as one could extract or otherwise from various places, e.g. Moerdijk's or Johnstone. See [this answer](https://mathoverflow.net/questions/36156/internal-version-of-a-flat-functor/36182#36182). The two definitions are aiming at very different things. For example, a saturated anafunctor with codomain a topological monoid (as a one-object category) is trivial, but a torsor for the same is not. In general internal saturated anafunctors are about inverting weak equivalences between internal categories (fully faithful and essentially 'surjective': i.e. some map admitting local sections) but internal torsors are about characterising maps between topoi and classifying topoi and stuff (you can tell I know less about the latter).
| 3 | https://mathoverflow.net/users/4177 | 45239 | 28,677 |
https://mathoverflow.net/questions/45147 | 1 | A matrix subspace $S\subset M\_n(C)$ is called "good", if there is two linear independent elements of $S$, says $E\_1,E\_2$ which are simultaneously singular valued decomposable, i.e., $E\_1=UD\_1V$ and $E\_2=UD\_2V$ with $D\_1$, $D\_2$ diagonal and $U,V$ unitary.
Now the question becomes: if all three-dimensional subspace $S\subset M\_3(C)$ is good?
This problem is try to undertand how hard could simultaneously singular valued decomposation be, and how powerful could linear combination of matrix be. $n=3$ is the simplest case.
| https://mathoverflow.net/users/4987 | A 3*3 matrix space problem | wlog one can assume that $||E\_1||=1$.
Let $X$ be the span of the matrix units $E\_{11},E\_{21},E\_{31}$. Then for every $2$ linearly independent operators $E\_1, E\_2$ in this space there exists a unitary matrix $U$ such that $UE\_1=E\_{11}$, $UE\_2=\lambda E\_{21}+E\_{11}$. But $E\_{11}$ and $\lambda E\_{21}+E\_{11}$ are not simultaneously singular valued decomposable.
| 5 | https://mathoverflow.net/users/8699 | 45244 | 28,679 |
https://mathoverflow.net/questions/45240 | 7 | Given a finite non-empty set $N$ of integers, call a subset $M$ of $N$ *good* if $gcd(M)=gcd(N)$. The other subsets are called *bad*.
>
> Does there exist an algorithm which
> computes a good subset of minimal size
> in polynomial time (polynomial in $|N|$)?
>
>
>
Using a greedy strategy, it is easy to find good subsets $M$ which are minimal with respect to inclusion (i.e. every proper subset of $M$ is bad). But it is not difficult to construct examples where such a greedy strategy may fail to find a set of globally minimal size.
Take for example N={6=2\*3, 10=2\*5, 15=3\*5, 1}. Then $gcd(N)=1$, and both {6,10,15} and {1} are good subsets. Both are minimal good subset wrt to inclusion. This can be easily generalized.
So, something more advanced would needed. Obviously, one can test all subsets, but then one gets exponential runtime. Is there a better way? Or can one prove that there isn't? Maybe this is equivalent to efficiently factoring primes?
As it is, I am not even sure whether this problem is in NP...
(Note that this question is about an important special case of an [earlier question of mine](https://mathoverflow.net/questions/44888/finding-globally-minimal-row-subsets-of-an-integer-matrix-which-generate-the-full); I hope it'll attract a few more people by being less technical).
| https://mathoverflow.net/users/8338 | Finding minimal subsets of a finite integer set with gcd equal to the whole set | I claim that the set cover problem (<http://en.wikipedia.org/wiki/Set_cover_problem>) can be reduced to this problem. So this problem is NP-hard.
Given a universe $U$ and a family $S$ that covers $U$, correspond the elements of $U$ to distinct primes and correspond each $A\in{}S$ to the product of the primes that correspond to the elements of $U-A$. Then a subcover of minimum size is equivalent to a good subset of minimum size.
| 10 | https://mathoverflow.net/users/10267 | 45245 | 28,680 |
https://mathoverflow.net/questions/45224 | 5 | Let $S$ be a smooth cubic surface in $\mathbb{P}^3$. I would like to understand that variety $V$ that parametrizes lines $\ell$ such that $\ell \cdot S=3P$ with $P \in S$. At any point $P \in S$, let $\Pi\_P$ be the tangent plane, and let $\Gamma\_P=\Pi\_P \cap S$. Generically, $\Gamma\_P$ is a plane cubic with a node at $P$ and therefore two tangents at $P$. Each of these satisfies $\ell \cdot S=3P$. This leads us to the fact that $V \dashrightarrow S$ is a double cover. I wanted to know if $V$ has a name, and also what can we say about it.
| https://mathoverflow.net/users/4140 | Certain double covers of cubic surfaces | For any surface of degree at least three we can consider more generally a similar construction of the set of lines $\ell$ with $\ell \cdot S \ge 3$. This is called the asymptotic double cover of $S$. See
McCrory, C., Shifrin, T. Cusps of the projective Gauss map, J. Differential Geom. 19 (1984), 257–276.
or my paper:
Surfaces in $P^3$ over finite fields, in Topics in Algebraic and Noncommutative Geometry: Proceedings in Memory of Ruth Michler, C. Melles et al. eds., Contemporary Math. 324 (2003) 219-226.
| 7 | https://mathoverflow.net/users/2290 | 45256 | 28,688 |
https://mathoverflow.net/questions/45087 | 6 | While playing around with certain non-negative matrices, I got stuck at the following question.
Let $A$ be a *strictly* positive-definite $n \times n$ matrix ($n \ge 3$), with ones on the diagonal, and all other entries in the range $[0,1]$. How should I go about proving a tight bound on the sum of the entries of $A^{-1}$. In symbols, how should I go about trying to compute the smallest number $\gamma(n)$ such that $$1^TA^{-1}1 \le \gamma(n).$$
Any pointers to related work, or some possible ways to attack the problem will be very useful. Of course, if you think this question is not well-formulated, please help me improve it!
**Remarks:**
a. Notice that for the special case where $A$ is the identity matrix, we have equality, and $\gamma(n)=n.$
b. If the vector of all ones, i.e., $1$, happens to be an eigenvector of $A$, then also we have an instant answer.
**More background**
The reason I reached the above bounding problem was that I was looking at the matrix
$$\begin{bmatrix} A & 1\\\\
1^T & n\end{bmatrix},$$
and trying to prove that it is positive semidefinite. So, either I could show that $A-11^T/n \succeq 0$, or $1^TA^{-1}1 \le n$. Now, since the bound with $n$ might not always hold, I started searching for a $\gamma(n)$ that holds. Perhaps, I need to further cleanup my question?
| https://mathoverflow.net/users/8430 | Tight bound for sum of entries of the inverse of a nonnegative matrix | When $n\ge3$, there does not exist such a bound. To see this, take $n=3$ and the following matrix
$$A=\begin{pmatrix}
1 & a & 0 \\\\ a & 1 & a \\\\ 0 & a & 1 \end{pmatrix}.$$
If $a< 1/\sqrt2$, it is positive definite with non-negative entries. However
$$1^TA^{-1}1=\frac{3-4a}{1-2a^2}$$
is not bounded as $a\rightarrow1/\sqrt2$.
| 7 | https://mathoverflow.net/users/8799 | 45279 | 28,698 |
https://mathoverflow.net/questions/45266 | 6 | Does anyone know if there is an available (published or unpublished) English translation of Johann Lambert's *Theorie der Parallellinien*? I was able to find it online in German by way of the bibliography of Jeremy Gray's *Worlds out of Nothing* at [this link](http://openlibrary.org/books/OL23347640M/Die_theorie_der_parallellinien_von_Euklid_bis_auf_Gauss), but I have been unable to find it in English. Any help would be appreciated.
| https://mathoverflow.net/users/6793 | English translation of Lambert's Theorie der Parallellinien? | The place to go for everything related to Lambert is
[this one](http://www.kuttaka.org/~JHL/Werke.html).
Asking Maarten Bullynck, the author of this web site, might be
a good idea.
| 6 | https://mathoverflow.net/users/3503 | 45297 | 28,709 |
https://mathoverflow.net/questions/45273 | 29 | Simplicial commutative rings are very easy to describe. They're just commutative monoids in the monoidal category of simplicial abelian groups. However, I just realized that a priori, it's not clear that even some of the simplest facts we can prove for ordinary commutative rings (in particular those that depend integrally on the axiom of choice, or even those that depend on the law of the excluded middle) will hold for simplicial commutative rings. However, we have at least one saving grace. That is, the interesting parts of simplicial commutative algebra come from considering things up to homotopy.
So, for example, as far as it makes sense, can we prove that every simplicial ideal of a simplicial commutative ring is weakly equivalent to one contained in a maximal simplicial ideal? Perhaps a better way to state this would be something like, "every noncontractible simplicial commutative ring admits at least one surjective map to a simplicial commutative ring that's weakly equivalent to a simplicial field", or some variation on where the homotopy equivalence appears. Given that the axiom of choice does not necessarily hold in $sSet$, it doesn't seem reasonable to think that the ordinary theorem will hold.
Is there a version of the Hilbert basis theorem that holds up to isomorphism? How about weak equivalence?
What other well-known theorems will fail, even up to homotopy?
| https://mathoverflow.net/users/1353 | What facts in commutative algebra fail miserably for simplicial commutative rings, even up to homotopy? | Most of the things that stop working are things related to procedures in commutative algebra that don't preserve exactness. The tensor product of simplicial modules always has to be the derived tensor product in order to be meaningful, etc.
One of the sticky points is that "free" is not the same as "polynomial" in higher degrees except in characteristic zero, and this makes building simplicial rings from a generators-and-relations perspective quite difficult.
For example, if R is an ordinary ring viewed as a constant simplicial commutative ring (with homotopy R, concentrated in degree 0), then you can take the free simplicial R-algebra on a class x in degree n; let me call it A. If n=0, the homotopy groups of A are just the polynomial algebra R[x] concentrated in degree zero. If n=1, the homotopy groups of A are an exterior algebra over R on a generator in degree 1. If n=2, the homotopy groups of A are a *divided power algebra* on a over R on a generator in degree 2 (a simplicial commutative ring always has a divided power structure on its higher homotopy groups). If n>3 and 2R = 0, then the answer is always a countable tensor product of exterior algebras over R. If n > 3 and pR = 0, you get algebras whose structure is simply complicated (because it involves iterated Tor starting with a divided power algebra).
Similar remarks apply to trying to construct an ideal generated by a collection of elements in homotopy groups which are anything other than a regular sequence in $\pi\_0$ - you are unlikely to get the "quotient" you expect.
| 20 | https://mathoverflow.net/users/360 | 45304 | 28,715 |
https://mathoverflow.net/questions/45232 | 5 | What are the current best lower bounds for off-diagonal Ramsey numbers $R(k,l)$ with $l$ of order unity and asking for asymptotic behavior for large $k$, such as $R(k,4)$, $R(k,5)$, and so on? (please include any log factors, too!) Other than the more complicated arguments of Kim for $R(k,3)$, are all the other best lower bounds from the Lovasz local lemma?
| https://mathoverflow.net/users/10648 | Best lower bound for off-diagonal Ramsey numbers | The best bounds I know of are due to [Tom Bohman](http://arxiv.org/abs/0806.4375) for $R(k,4)$ and [Bohman and Peter Keevash](http://arxiv.org/abs/0908.0429) for $R(k,5)$ and beyond. Both rely on using the differential equations method to analyze the following process: Start with the empty graph, and at each step add an edge uniformly at random among all edges which do not create a $K\_t$. The bounds they achieve are
$$R(k,t) \geq c\_t \left( \frac{k}{\log k} \right)^{\frac{t+1}{2}} (\log k)^{\frac{1}{t-2}}$$
The final term in this product corresponds to the improvement over the bounds obtained using the Local Lemma. For $t=3$ it matches Kim's bound up to a constant factor.
| 8 | https://mathoverflow.net/users/405 | 45311 | 28,719 |
https://mathoverflow.net/questions/45296 | 0 | $W\_{opt}=\arg \{\max(\pi\_0 F\_{L\_0}(W)-\frac{\pi\_1}{W}\int\_0^W F\_{L\_1}( \alpha )d \alpha )\}$
subject to $\quad \int\_0^W F\_{L\_0} (\alpha)d\alpha <\xi$
We should find analytically the optimal $W >0$ which maximize the first equation subject to the second equation, where $F( \cdot )$ is comulative distribution function (CDF), and $L\_0$ and $L\_1$ are positive random variables. $\xi$, $\pi\_0$, $\pi\_1$ are constant. Also, $0<\pi\_0, \pi\_1<1$ and $\pi\_0 + \pi\_1 =1$. All variables are real. Further, if needed, we can assume that, for example, $L\_0$ and $L\_1$ may have Erlang or exponential distribution.
| https://mathoverflow.net/users/10661 | One-Variable Optimization Problem | Hmm, you could rewrite it this way: (I'm going to assume that $\pi\_0, \pi\_1, \xi'', \mu, \lambda$ are pre-defined constants, and $\alpha,W$ are variables)
$$
\begin{align}
&\max\_{W}\; \pi\_{0} (1 - e^{-\mu W}) - \frac{\pi\_{1}}{W}z\_{1}(W)\\
s.t.\;& \frac{dz\_{0}(\alpha)}{d\alpha} = 1 - e^{-\mu \alpha},\quad z\_{0}(0) = 0\\
& \frac{dz\_{1}(\alpha)}{d\alpha} = 1 - e^{-\lambda \alpha},\quad z\_{1}(0) = 0\\
& z\_{0}(W) + \epsilon \leq \xi''\\
& W \geq 0
\end{align}
$$
where $z\_{0},z\_{1}$ are auxiliary variables and $\epsilon$ is a numerical tolerance value. This then becomes a DAE (differential-algebraic equation) optimization problem, which can be solved numerically. (though given that your decision variable is also the independent variable in the differential equations, some further bilinear transformations may be required. See <http://dx.doi.org/10.1016/j.na.2005.03.066>. Some software packages do this automatically.).
Edit: I just realized, if indeed $\xi''$ is a constant as I have assumed, the inequality constraint can easily be converted into a bound.
$$
\begin{align}
\int\_{0}^{W^U} 1-e^{-\mu \alpha}\,d\alpha = \xi'' - \epsilon\\
\frac{e^{-\mu W^{U}}}{\mu} + W^{U} - \frac{1}{\mu} = \xi'' - \epsilon\\
\end{align}
$$
Solve for $W^{U}$, and replace the above inequality constraints with:
$$ 0 \leq W \leq W^{U}
$$
You may be able to solve this using optimal control methods.
| 0 | https://mathoverflow.net/users/7851 | 45313 | 28,720 |
https://mathoverflow.net/questions/45307 | 7 | Let $f:X\to Y$ be a morphism of varieties over a field $k,$ such that $X(\overline{k})\to Y(\overline{k})$ is bijective. Is $f$ necessarily an affine morphism?
| https://mathoverflow.net/users/370 | affine morphism | No. Here is an example:
Let $g:\mathbb A^2\to Y$ be a morphism that glues two closed points $P$ and $Q$ together and otherwise it is an isomorphism. Now let $X=\mathbb A^2\setminus \{P\}$ and $f$ the restriction of $g$ to $X$.
If you add to the conditions that $f$ is projective, then the statement is true, because then $f$ is finite and hence affine.
**EDIT:**
A minute ago there was another question asking how to define $Y$. It has now disappeared, but perhaps it is still interesting to include some references.
1) See [this](https://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes) MO question and the discussion.
2) See this [paper](http://www.math.utah.edu/~schwede/Papers/SchemeWithoutPoints.pdf) of Karl Schwede. Especially, Theorem 3.4 in general and Corollaries 3.6 and 3.9 in particular.
3) Try to construct it directly. If you get stuck, look at Karl's paper and try to carry out the computation in this special case. It might be a worthy exercise if you have never done anything like this before.
| 9 | https://mathoverflow.net/users/10076 | 45317 | 28,722 |
https://mathoverflow.net/questions/45308 | 2 | I'm not sure if this question is appropriate for mathoverflow, but I can't help but think that other people have wondered about it as well. When anyone first learns about the axiom of choice, the standard example used to convince the listener as to its necessity is the problem of finding a choice function on $\mathscr{P}(\mathbb{R})\backslash\{\emptyset\}$, the powerset of the reals, without the emptyset. I have always wondered: Is the axiom of choice really necessary to construct the function? In other words, is it possible to prove that without the axiom of choice, no such choice function exists. Or, if it is possible to directly construct a choice function on $\mathscr{P}(\mathbb{R})\backslash\{\emptyset\}$, are there any good examples?
| https://mathoverflow.net/users/6856 | Choice Function on the Powerset of the Reals | Quiaochu Yuan and Gerald Edgar have given correct answers based on Solovay's theorem that ZF cannot prove the existence of a Lebesgue non-measurable set. I'd like to add that one doesn't need anything as high-powered as Solovay's model. Cohen's original model for ZF and the negation of AC will do the job. It contains a non-well-orderable set of reals and therefore (by Zermelo's proof) has no choice function on the set of nonempty subsets of the reals.
| 8 | https://mathoverflow.net/users/6794 | 45321 | 28,723 |
https://mathoverflow.net/questions/45324 | 3 | Over rational curve we know that any vector bundle is decomposable to direct sum of line bundles.
In higher dimensions there are examples of indecomposable bundles.
some indecomposable vector bundles have might have proper sub-bundles (all bundles and sub bundles here are in holomorphic category and not topological)
First Question: Over curves, can we have a indecomposable bundle having a proper sub bundle?
For rational curves the answer is negative obviously, what about elliptic curves, and higher genus curves?
Second: same question for Calabi-Yau 3-folds and K3-surfaces? for example for Quintic.
Please provide examples (or give reference) if you know any.
| https://mathoverflow.net/users/5259 | indecomposable vector bundles having proper sub-bundles. | Over a curve any rank $2$ bundle has a rank $1$ subbundle: Choose a subbundle defined over a Zariski dense open set, and then extend it over the missing points by observing that locally the problem of making such an extension is the problem of extending a map into $\mathbb P^1$.
| 7 | https://mathoverflow.net/users/6666 | 45328 | 28,728 |
https://mathoverflow.net/questions/45257 | 13 | First, a rather broad question: has there been any work on what, given a model $M$ of set theory, we can say about those models of set theory $N$ and posets $\mathbb{P}$ such that $\mathbb{P}\in N$ and $M=N[G]$ for some $G$ $\mathbb{P}$-generic over $N$?
Second, a more specific question. Let a poset $\mathbb{P}$ be $detectable$ if there is some sentence $\phi\_\mathbb{P}$ in the language of set theory + a constant denoting $\mathbb{P}$ such that for all $M$ with $\mathbb{P}\in M$, we have $(M, \mathbb{P})\models\phi\iff M=N[G]$ for some model $N$ with $\mathbb{P}\in N$ and $G$ $\mathbb{P}$-generic over $N$. What posets are detectable? [Answered in the comments by Amit Kumar Gupta.]
Finally, an incredibly general question. Let $M$ be a model of $ZFC$, $\mathcal{C}$ a class of posets in $M$. Say $\mathcal{C}$ is $consistent$ if there is some elementary extension $N$ of $M$ such that, for all $\mathbb{P}\in \mathcal{C}$, there is some $N\_\mathbb{P}\models ZFC$ with $\mathbb{P}\in N\_\mathbb{P}$ and some $G$ $\mathbb{P}$-generic over $N\_\mathbb{P}$ such that $N=N\_\mathbb{P}[G]$. What are the consistent classes like? Can we say anything interesting about them?
I'm not sure if these questions are meaningful, or - even assuming they are - if they are interesting. Basically, what I'm interested in is the notion of inverse forcing - similar in an aesthetic sense, at least to me, to inverse Galois theory - and I haven't run into anything along these lines yet.
| https://mathoverflow.net/users/8133 | When can we detect forcing? | Amit Gupta's comment gives the answer to your first question, which I will expand on slightly:
Every poset is detectable, and even more, the class of all posets is uniformly detectable. That is, there is a single formula $\phi(x)$ that takes as input a poset $P$, such that $\phi(P)$ is true in a model $M$ exactly when $M$ is a forcing extension of an inner model $N$ by $P$ (let us call $N$ a ground model of $M$ by $P$). It follows that the same is true for any set of posets (or definable class $C$ of posets) in $M$: we can detect when $M$ is a $C$-forcing extension.
Regarding your second question ("consistent classes"), there are many examples of consistent classes. For example, let $C$ be the class of posets that add a subset to a regular cardinal using the poset as defined in $L$ (that is, $C$ is the class of posets of the form $ Add(\kappa,1)^L $ ). If we start in $V$ and force with the (class) product of $C$, then in the resulting model $V[G]$, for every member $P$ of $C$ there is a ground model of $V[G]$ by $P$. This is because, for a fixed $\kappa$, we can use the commutativity of the product to rewrite the extension as class forcing (adding a subset to each cardinal except $\kappa$), followed by set forcing adding a subset to $\kappa$. This allows us to rewrite $V[G]=V[G^\prime][g]$, and $V[G]$ is an extension of $V[G^\prime]$ by $Add(\kappa,1)^L$ as required.
We can modify this argument in various ways to handle other classes. It will work for any class of posets from $L$ whose product preserves $ZFC$. We can also take our posets from another absolute inner model, such as $L[mu]$, instead of from $L$. Dropping the requirement that the posets be from a designated inner model (for example, taking $C$ to be the class of posets $Add(\kappa,1)$) may still possible but would require a more delicate argument. The issue here is that the definition of the poset $Add(\kappa,1)$ is not absolute, and so the poset defined in $V$ may not exist in every ground model. Nonetheless I suspect that the same model defined in the previous example will work.
There are also some classes that are clearly not consistent, for example the the class $C$ of partial orders that collapse $\kappa$ to $\omega$ (for every $\kappa$). Any model containing generics for all such posets will fail to satisfy $ZFC$, as every cardinal will be collapsed.
For the more general case, one avenue of attack might be to use the Maximality Principle (introduced by Joel Hamkins). This principle says that any statement that can be forced in such a way that further forcing cannot make it false, is already true, and existence of a ground model for a particular poset is just such a statement.
| 12 | https://mathoverflow.net/users/10671 | 45329 | 28,729 |
https://mathoverflow.net/questions/44934 | 21 | Let $H\_k$ be the vector space of degree $k$ homogeneous polynomials in two variables.I'm looking for a reference for the fact that $H^1(SL(2,\mathbb Z);H\_k)=M^0(k+2)\oplus\overline{M^0(k+2)}\oplus E\_{k+2}$, where
1. $M^0(k+2)$ is the space of cuspidal modular forms of weight $k+2$.
2. $\overline{M^0(k+2)}$ is its conjugate.
3. $E\_k$ is $1$-dimensional if $k\geq 4$ is even, and is zero otherwise.
I know how to get the $M^0(k+2)\oplus\overline{M^0(k+2)}$ piece. By the Eichler-Shimura isomorphism, this is the same as the cuspidal cohomology $H^1\_{cusp}(SL(2,\mathbb Z);H\_k)$, spanned by cocycles that vanish on the matrix $T$=[1 1 \ 0 1]. Serge Lang's book on Modular Forms has a detailed explanation of how to get this. I'd really like to know how to get the $E\_{k+2}$ piece, which as far as I understand, corresponds somehow to the Eisenstein series, which is the "extra" modular form of weight $k+2$ that is not cuspidal.
It's not too hard to see that $H^1(SL(2,\mathbb Z);H\_k)=H^1(PSL(2,\mathbb Z);H\_k)$, with $PSL(2,\mathbb Z)$ having the nice presentation $\langle S,T\,|\, (ST)^3=S^2=I\rangle$, so one should be able to represent the $E\_{k+2}$ cocycle by specifying its values on $S,T$.
So, to summarize, I'd be satisfied with any proof that $H^1(SL(2,\mathbb Z);H\_k)=M^0(k+2)\oplus\overline{M^0(k+2)}\oplus E\_{k+2}$, but I'd be happiest with an answer that lets me get my hands on the $E\_{k+2}$ cocycle as explicitly as possible, perhaps even telling me its values on $S$ and $T$. Any comments that would help clarify this are appreciated.
**Edit:** I think I have a better understanding of what the "extra" Eisentstein cocycle is, based on Kevin's comments. It seems that the cuspidal cocycles vanish on $T$, whereas the Eisenstein cocycle vanishes on $S$, although I don't see how to show, for example, that there is only one dimension's worth of cocycles vanishing on $S$, up to coboundaries. (**Edit**: I'm not entirely sure about this.)
**Edit:** Shimura's *Introduction to the Theory of Automorphic Forms* only covers the cuspidal part of the above isomorphism.
| https://mathoverflow.net/users/9417 | Reference request: The first cohomology of SL(2,Z) with coefficients in homogeneous polynomials | The result you're looking for is contained in the following article :
Haberland, Klaus. Perioden von Modulformen einer Variabler and Gruppencohomologie I (German) [Periods of modular forms of one variable and group cohomology I], *Math. Nachr.* **112** (1983), 245-282.
Let $S\_k$ (resp. $M\_k$) be the space of holomorphic cusp forms (resp. holomorphic modular forms) for $\Gamma = SL\_2(\mathbf{Z})$. Let $\Gamma\_{\infty}$ be the stabilizer of $\infty$ in $\Gamma$. Let $V\_k$ be the space of polynomials of degree $\leq k-2$ with complex coefficients. Haberland proves an exact sequence
\begin{equation}
(\*) \qquad 0 \to S\_k \oplus \overline{S\_k} \to H^1(\Gamma,V\_k) \to H^1(\Gamma\_\infty,V\_k) \to 0.
\end{equation}
Let $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in \Gamma\_{\infty}$. There is a natural map $V\_{k+1} \to H^1(\Gamma\_\infty,V\_k)$ sending a polynomial $P$ to the cocycle $c\_P$ determined by $c\_P(T) = P(X+1)-P(X)$. It is easy to check that this map induces an isomorphism $\psi : V\_{k+1}/V\_k \cong H^1(\Gamma\_\infty,V\_k)$, so that the latter space is one-dimensional.
The "Eisenstein cocycle" you're looking for is a natural map $\delta : M\_k \to H^1(\Gamma,V\_k)$ which Haberland constructs the following way (actually I learnt this construction and many other properties of $\delta$ during Zagier's 2002-2003 lectures at the Collège de France).
Let $f \in M\_k$. Let $\widetilde{f}$ be an Eichler integral of $f$, that is any holomorphic function on $\mathcal{H}$ such that
\begin{equation}
\left(\frac{1}{2\pi i} \frac{d}{dz}\right)^{k-1} \widetilde{f}(z) = f(z).
\end{equation}
Note that $\widetilde{f}$ is unique up to adding some element of $V\_k$. Since we integrate $k-1$ times, the function $\widetilde{f}$ should be thought of as a function of "weight" $k-2\cdot (k-1) = 2-k$ (of course this isn't true in the strict sense). Let us make this more precise.
For any $n \in \mathbf{Z}$, let $|\_n$ denote the weight $n$ action of $SL\_2(\mathbf{R})$ on the space of complex-valued functions on $\mathcal{H}$ (so that any $f \in M\_k$ is a fixed vector of the weight $k$ action of $\Gamma$). Note also the weight $2-k$ action gives the usual action of $\Gamma$ on $V\_k$. The crucial fact is that we have
\begin{equation}
\widetilde{f} |\_{2-k} (\gamma-1) \in V\_k \qquad (\gamma \in \Gamma).
\end{equation}
This can be proved using Bol's identity
\begin{equation}
\left(\frac{d}{dz} \right)^{k-1} (F |\_{2-k} g) = \left(\frac{d^{k-1} F}{dz^{k-1}} \right) |\_k g
\end{equation}
which holds for any holomorphic function $F$ on $\mathcal{H}$ and any $g \in SL\_2(\mathbf{R})$.
Since $\gamma \mapsto \widetilde{f} |\_{2-k} (\gamma-1)$ is obviously a coboundary in the space of functions on $\mathcal{H}$, it defines a cocycle in the space $V\_k$. Therefore we get $\delta(f) \in H^1(\Gamma,V\_k)$ and this element doesn't depend on the choice of $\widetilde{f}$. Thus we have constructed $\delta : M\_k \to H^1(\Gamma,V\_k)$.
It is not difficult to check that if $f =\sum\_{n \geq 0} a\_n e^{2i\pi nz}$ then the image of $\delta(f)$ in $H^1(\Gamma\_\infty,V\_k)$ is the image of the polynomial $\frac{a\_0 \cdot (2\pi i)^{k-1}}{(k-1)!} \cdot X^{k-1} \in V\_{k+1}$ under the isomorphism $\psi$ above. In particular $\delta$ is injective, and the exact sequence $(\*)$ gives the isomorphism you want.
Note that there is a distinguished choice of $\widetilde{f}$, namely
\begin{equation}
\widetilde{f} = \sum\_{n \geq 1} \frac{a\_n}{n^{k-1}} e^{2i\pi nz} + \frac{a\_0 \cdot (2\pi i)^{k-1}}{(k-1)!} z^{k-1}.
\end{equation}
Let $c\_f \in Z^1(\Gamma,V\_k)$ be the cocycle associated to this choice of $\widetilde{f}$. Let us compute the value of $c\_f$ on $T$ and $S= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. First as explained above, we have
\begin{equation}
c\_f(T)=\frac{a\_0 \cdot (2\pi i)^{k-1}}{(k-1)!} ((X+1)^{k-1}-X^{k-1}).
\end{equation}
To compute $c\_f(S)$, Haberland uses the natural integral representation of $\widetilde{f}$ in terms of $f-a\_0$, and gets
\begin{equation}
c\_f(S) = \frac{(2\pi i)^{k-1}}{(k-2)!} \int\_0^{\infty} \left(f(z)-\frac{a\_0}{z^k}-a\_0 \right) (z-X)^{k-2} dz
\end{equation}
(there is a similar but more complicated formula for $c\_f(\gamma)$ for any $\gamma \in \Gamma$, see below). Then $c\_f(S)$ can be expressed in terms of the special values of $L(f,s) := \sum\_{n=1}^\infty a\_n/n^s$ at integers $s = 1,\ldots,k-1$. It is then a good exercise to compute $c\_f(S)$ when $f$ is the Eisenstein series $E\_k$, in terms of Bernoulli numbers and of $\zeta(k-1)$ (this is Satz 3 in Haberland's article, Kapitel 1).
Please tell me if something isn't clear in my explanation.
EDIT : I found the following expression for $c\_f(\gamma)$ where $\gamma \in \Gamma$. It is quite complicated (maybe it could be somewhat simplified) :
\begin{equation}
\begin{aligned}
\frac{(k-2)!}{(2\pi i)^{k-1}} c\_f(\gamma) &= \int\_{z\_0}^{\infty} (f(z)-a\_0)(z-X)^{k-2} dz + \int\_{\gamma^{-1} \infty}^{z\_0} \left(f(z) -\frac{a\_0}{(cz+d)^k} \right) (z-X)^{k-2} dz \\
& + \frac{a\_0}{k-1} \left((X-z\_0)^{k-1}-(X-\gamma z\_0)^{k-1} |\_{2-k} \gamma + X^{k-1} |\_{2-k} (\gamma-1) \right)
\end{aligned}
\end{equation}
where $z\_0 \in \mathcal{H}$ is arbitrary and $\gamma= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
| 17 | https://mathoverflow.net/users/6506 | 45337 | 28,735 |
https://mathoverflow.net/questions/45160 | 14 | In *Über die Bestimmung asymptotischer Gesetze in der
Zahlentheorie*, Dirichlet proved his theorem on the asymptotic
behaviour of the divisor function using a Lambert series: let
$d\_n = d(n)$ denote the number of the divisors of $n$; then
Lambert (actually this is due to Euler) observed that
$$ f(z) = \sum\_{n=1}^\infty d\_n z^n
= \sum\_{n=1}^\infty \frac{z^n}{1-z^n} . $$
This series converges for $|z| < 1$, and diverges for $z = 1$.
Setting $z = e^{-t}$ we obtain
$$ g(t) = \sum\_{n=1}^\infty \frac{e^{-nt}}{1-e^{-nt}}
= \sum\_{n=1}^\infty \frac{1}{e^{nt}-1} . $$
Dirichlet writes that "expressing this series by a definite integral
one easily finds" that
$$ g(t) \sim \frac1t \log \frac1t + \frac{\gamma}t $$
as $t \to 0$, where $\gamma$ is Euler's constant.
Dirichlet then claims that the asymptotic behaviour of $g(t)$
would imply that $d\_n$ is, on the average, equal to $\log n + 2 \gamma$,
which in turn implies that
$b\_1 + b\_2 + \ldots + b\_n \approx (n + \frac12) \log n + (2\gamma+1)n. $
He mentions that he has used the integral expressions for $\Gamma(k)$
and its derivative $\Gamma'(k)$ for deriving the first property.
Knopp (Über Lambertsche Reihen, J. Reine Angew. Math. 142) claims
that Dirichlet's proof was "heuristic". I find that hard to believe,
and I am convinced that Dirichlet's sketch can be turned into a valid
proof by someone who knows the tools of the trade.
So here are my questions:
1. How did Dirichlet express "this series by a definite integral" and
derive the asymptotic expression for $g(t)$?
Let me remark that Endres and Steiner ([A new proof of the Voronoi
summation formula](http://www.uni-ulm.de/fileadmin/website_uni_ulm/nawi.inst.260/paper/10/tp10-1.pdf)) use Voronoi summation for proving the sharper estimate
$$ g(t) \sim \frac1t \log \frac1t + \frac{\gamma}t + \frac14 + O(t) $$
as $t \to 0$. But this is not "easily found".
2. How did Dirichlet transform his knowledge about the asymptotic
behaviour of $\sum b\_n e^{-nt}$ as $t \to 0$ into an average
behaviour of $b\_n$? This smells like a Tauberian result, but I'm not
fluent enough in analytic number theory to see how easy this is.
| https://mathoverflow.net/users/3503 | Dirichlet's divisor problem via Lambert series | For part 1 of the question, he would most likely have used the Euler-Maclaurin summation formula
$$
\sum\_{n=1}^{\infty}\frac{1}{e^{nt} - 1} = \int\_{1}^{\infty}\frac{dx}{e^{xt} - 1} +
\frac{1}{2}\frac{1}{e^t - 1} + \int\_{1}^{\infty}S(x)\left(\frac{d}{dx}\frac{1}{e^{xt} - 1}\right)dx
$$
with $S(x)$ the sawtooth function. It is easy to obtain the leading term, because it comes from the first integral
$$
\int\_{1}^{\infty}\frac{dx}{e^{xt} - 1} = \frac{1}{t}\int\_{t}^{\infty}\frac{du}{e^u - 1}
$$
by the change of variable $u = xt$. We have
$$
\int\_{t}^{\infty}\frac{du}{e^u - 1} = \int\_{t}^{1}\frac{du}{e^u - 1} + \int\_{1}^{\infty}\frac{du}{e^u - 1},
$$
and
$$
\frac{1}{e^u - 1} = \frac{1}{u} + \left(\frac{1}{e^u - 1} - \frac{1}{u}\right)
$$
on $0 \leq u \leq 1$, so that
$$
g(t) = \frac{1}{t}\log\left(\frac{1}{t}\right) + O\left(\frac{1}{t}\right).
$$
But to get the second term looks harder, for the integral with the sawtooth function contributes to that term. To go further, one can integrate by parts in that integral, which is the standard approach, or write it as a sum of integrals over the
intervals from $n$ to $n+1$. Also the sawtooth function has a simple Fourier expansion, which may help. I should remark that the integral with the sawtooth function is $O(1/t)$ as one sees when bounding it by passing the absolute value under the integral sign and using $|S(x)| \leq 1/2$. Anyway, I am pretty sure that part 1 is doable with some work.
Part 2 looks trickier. The Lambert series expansion
$$
\sum\_{n=1}^{\infty}(1 + \mu(n))e^{-nt} = \frac{e^{-t}}{1 - e^{-t}} + e^{-t} =
\frac{1}{t} + \frac{1}{2} + O(|t|)
$$
is a little nicer than the one for the divisor function; not only are the coefficients nonnegative, but they are also bounded. Supposing that we have a Tauberian theorem strong enough to yield
$$
\sum\_{n \leq x}(1 + \mu(n)) \sim x,
$$
we would then have proved the Prime Number Theorem from the Lambert series. It seems a little unlikely that Dirichlet had such a strong Tauberian theorem; would he not have proved the Prime Number Theorem if he had? Of course, this argument by analogy is not conclusive, since the two
situations differ by a factor of $\log(x)$.
We shall never know what argument Dirichlet had, and he may have found an approach that
did not use a Tauberian theorem, perhaps exploiting special properties of the divisor function. It is worth noting that Voronoi's first proof of the error term $O(x^{1/3}\log(x))$ for the divisor problem was based on the Euler-Maclaurin summation formula.
| 13 | https://mathoverflow.net/users/8847 | 45341 | 28,738 |
https://mathoverflow.net/questions/45347 | 25 | Hartogs Theorem says every function whose undefined locus is of codim 2 can be extend to the whole domain. I saw people saying this corresponds to the (S2) property of a ring. But I can't see why this is true. Can anybody explain this or give a heuristic argument?
| https://mathoverflow.net/users/1657 | Why does the (S2) property of a ring correspond to the Hartogs phenomenon? | Let $\mathscr F$ be a coherent sheaf on a noetherian scheme $X$ and assume that ${\rm supp}\mathscr F=X$. Let $Z\subset X$ be a subscheme of codimension at least $2$ and $U=X\setminus Z$.
Let $\iota:U\hookrightarrow X$ denote the natural embedding and assume that $\mathcal F\_x$ is $S\_2$ for every $x\in Z$. Now the $S\_2$ assumption implies that
$$
\mathscr H^0\_Z(X,\mathscr F)=
\mathscr H^1\_Z(X,\mathscr F)=0
$$
and the Hartogs type extension is equivalent to
$$
\iota\_\*\iota^\*\mathscr F\simeq \mathscr F.
$$
Finally one has the exact sequence
$$
\mathscr H^0\_Z(X,\mathscr F) \to \mathscr F\to \iota\_\*\iota^\*\mathscr F \to \mathscr H^1\_Z(X,\mathscr F).$$
[See also [this MO answer](https://mathoverflow.net/questions/35736/the-canonical-line-bundle-of-a-normal-variety/46663#46663)]
| 23 | https://mathoverflow.net/users/10076 | 45354 | 28,742 |
https://mathoverflow.net/questions/45352 | 2 | I've run across the statement, "The existence of a strongly inaccessible cardinal implies the consistency of ZFC" in several places (Cohen's Set Theory and the Continuum Hypothesis p. 80, for one). His argument is that the set of all sets of rank less than that large cardinal form a model. It seems to me that since we have access to all of those sets without the large cardinal, we could make a model of ZFC without the large cardinal axiom, and thus prove ZFC consistent. This, of course, is false. So, how does the large cardinal provide a model of ZFC that doesn't exist without the large cardinal?
| https://mathoverflow.net/users/10679 | Large Cardinals Imply a Model of ZFC | A model of ZFC is a *set* $E$ together with a binary relation $R$ satisfying the axioms of ZFC (and we can always **if we can** suppose that $E$ is transitive and that $R=\in|\_{E\times E}$ **then the model is said *standard***).
So we not only need to find a class of sets satisfying the axioms of ZFC, but this class must be itself a set.
If you have a strongly inaccessible cardinal $\kappa$, then there is a *set* $V\_\kappa$ which is a **(standard)** model of ZFC, hence ZFC is consistent.
But without the existence of $\kappa$, even if you could consider the class of all sets of rank “less than $\kappa$”, this would not be a set and the theorem ”there is a model implies the consistency of ZFC” would not be applicable (because there is no model).
| 11 | https://mathoverflow.net/users/10217 | 45358 | 28,745 |
https://mathoverflow.net/questions/45343 | 9 | Let $X$ be a smooth variety and $D \subset X$ be an effective, reduced, irreducible divisor. My question is the following.
1.What is the first order deformation and obstrution for the pair $(X, D)$?
2.In particular, if $D$ is a singular divisor and has a global smoothing such that the line bundle $\mathcal{O}(E)\vert\_E$ also extends to a general smoothing, then under what conditions can we deform the pair so that the induced deformation of the pair gives a smoothing of $D$.
| https://mathoverflow.net/users/10676 | deformation and obstruction of a pair (X, D) | From another point of view, the isomorphism classes of first order deformations of the pair (X,D) are isomorphic to the 1st hypercohomology group of the 2 step complex from the tangent sheaf of X to the normal sheaf of D in X.
The 1st order deformations of just D are given by the 1st hypercohomology group of the 2 step complex from the restriction of the tangent sheaf of X to D, to the same normal sheaf of D in X.
By the long exact sequence induced by the natural map of these complexes, the forgetful map from T^1(X,D) to T^1(D) is an isomorphism whenever the first 2 hypercohomology groups vanish for the 2 step complex from the ideal sheaf of D in X times the tangent sheaf of X, to zero.
This happens for instance if X is a principally polarized abelian variety of dimension 3 or more, and D is the theta divisor, by Kodaira (or abstractly Mumford) vanishing.
ref: Compositio Math 76(1990) pp. 367-398.
| 4 | https://mathoverflow.net/users/9449 | 45365 | 28,750 |
https://mathoverflow.net/questions/45344 | 3 | Hi,
in our lectures we have seen the following:
Let $(M, J)$ be a complex manifold, i.e., $J$ is an integrable almost complex structure. Then a real vector field $X$ is an infinitesimal automorphism of $J$ if and only if its component $X^{1,0}\in \Gamma(T^{1,0}M)$ is holomorphic.
Now let $(M, J)$ be an almost complex manifold and $X \in \chi(M)$ be an infinitesimal automorphism of $J$, i.e., $[X, JY]=J[X, Y]$ for all $Y \in \chi(M)$.
>
> Can we then conclude that the mapping $X^{1,0}f$ with $X^{1,0}:=\frac{1}{2}(X-iJX)$ is pseudoholomorphic for every pseudoholomorphic mapping $f: M \to \mathbb{C}$?
>
>
>
Thank you and best regards,
Differentialgeometer
| https://mathoverflow.net/users/7015 | Pseudoholomorphic vector fields | I can give you a partial answer- a reformulation of your question. Perhaps others will then be able to give a counterexample. (I think it is *not* true in general.) Let's start from the beginning.
A map $h : M \to \mathbb C$ is $J$-holomorphic iff $h\_\* (J Y) = i h\_\* (Y)$ for every vector field $Y$ on $M$. Equivalently, $dh(JY) = i dh(Y)$, or $(JY) (h) = i Y(h)$, or $(Y
+ i JY)h = 0$, or $(Y - iJY)h = 2Yh$ for any $Y$.
The condition that $X$ is an infinitesmal automorphism of the almost complex structure $J$ is that $[X,JY] = J[X,Y]$ for all $Y$. We want to investigate the question of under what conditions on $J$, for such an $X$, is it true that $(X- i JX)f = 2 Xf$ will be $J$-holomorphic for any $J$-holomorphic $f$.
Ignoring the factor of $2$, we see that
\begin{equation\*}
Xf \text{ is $J$-holomorphic} \Leftrightarrow (Y + iJY)(Xf) = 0 \text{ for all $Y$}
\end{equation\*}
\begin{equation\*}
\Leftrightarrow Y(Xf) = -i(JY)(Xf) \text{ for all $Y$}
\end{equation\*}
Recall the definition of the Nijenhuis tensor $N\_J(X,Y) = [JX, JY] - J[JX, Y] - J[X,JY] - [X,Y]$. Since $X$ is an infinitesmal automorphism of $J$, the last two terms cancel. Applying both sides of this equation to a $J$-holomorphic map $f$, we get
\begin{equation\*}
N\_J(X,Y) f = (JX)((JY)f) - (JY)((JX)f) - (J[JX,Y])f = (JX)(iYf) - (JY)(iXf) - i [JX,Y]f
\end{equation\*}
\begin{equation\*}
= i \left( (JX) (Yf) - (JY)(Xf) - (JX)(Yf) + Y((JX)f) \right)
\end{equation\*}
\begin{equation\*}
= i \left( - (JY)(Xf) + i Y(Xf) \right)
\end{equation\*}
which is exactly the quantity that we need to vanish.
Therefore, $Xf$ will also be $J$-holomorphic iff $N\_J(X,Y)f = 0$ for all vector fields $Y$. In particular, this always holds if $J$ is integrable, since then $N\_J$ vanishes identically. I don't know an example where $J$ is not integrable but where such an $X$ and $f$ exist. If there is one (which I would bet there is), then in that case $Xf$ would *not* be $J$-holomorphic.
| 6 | https://mathoverflow.net/users/6871 | 45367 | 28,752 |
https://mathoverflow.net/questions/45331 | 6 | Is there any example of a Riemannian submersion, which is no fibration?
As far as I know, a (any) submersion is locally, but not globally, given by a fibration. The converse holds globally. Nevertheless, I could not think of or find an example.
| https://mathoverflow.net/users/9762 | Examples of Riemannian Submersions | If $E\to B$ is a Riemannian submersion, then so is its restriction onto any open subset, which gives lots of examples of submersions that are not fibrations, e.g. removing a point from $E$ will destroy any fibration structure if there was one.
It is a old theorem of Hermann that any Riemannian submersion $E\to B$ with complete $E$
is a locally trivial fiber bundle (whose structure group need not be a Lie group though). Good sources to learn about Riemannian submersions are Besse's book "Einstein manifolds", chapter 9, and Gromoll-Walschap's recent book "Metric folitations and curvature".
| 11 | https://mathoverflow.net/users/1573 | 45373 | 28,757 |
https://mathoverflow.net/questions/45351 | 15 | Eilenberg and Mac Lane showed that given a group $G$ there exists a pointed topological space $X\_G$ such that $\pi(X\_G,\bullet)\cong G$. It is obviously a way to "invert direction" to the functor $\pi\_1\colon \mathbf{Top}^\bullet\to \mathbf{Grp}$ to a functor $\mathcal K\colon \mathbf{Grp}\to \mathbf{Top}^\bullet$ such that $\pi\_1(\mathcal K(G),\bullet)\cong G$ (almost by definition). This is equivalent to say that there exists a natural transformation (equivalence, in this case) between $\pi\_1\circ\mathcal K$ and $\mathbf{1}\_{\mathbf{Grp}}$, which turns out to resemble some sort of *counity*.
It would be wonderful if I could define an adjunction between the two categories in exam, given by the two functors. Everytime I try to think about some sort of *unity* to this hypotetical adjunction I poorly fail: considering the vast literature in the field of algebraic topology, I believe in only two possible cases. The first, nothing interesting arises from this adjunction. The second, there is no sort of adjunction.
The key point, quite trivial, to answer is: it is well known that an adjunction is uniquely determined by one among *unity* and *counity*, provided the one is universal. But $\boldsymbol\varepsilon\colon \pi\_1\circ\mathcal K\to \mathbf{1}\_{\mathbf{Grp}}$ is an *equivalence*: can I conclude that it is universal?
| https://mathoverflow.net/users/7952 | Does $\pi_1$ have a right adjoint? | In the more general setting the answer is no. Left adjoints preserve colimits and it is not true that $\pi\_{1}(X\vee Y)\cong \pi\_{1}(X)\ast\pi\_{1}(Y)$ for all spaces $X,Y$ (even compact metric spaces). For instance, if $(\mathbb{HE},x)$ is the usual Hawaiian earring, let $X=Y=C\mathbb{HE}=\mathbb{HE}\times I/\mathbb{HE}\times\{1\}$ be the cone on the Hawaiian earring with basepoint the image of $(x,0)$ in the quotient. It is a theorem of Griffiths that $\pi\_{1}(C\mathbb{HE}\vee C\mathbb{HE})$ is uncountable and not the free product of trivial groups.
| 16 | https://mathoverflow.net/users/5801 | 45375 | 28,758 |
https://mathoverflow.net/questions/45376 | 7 | I am interested in pursuing an understanding of K-theory. Primarily, the
$K\_3(\mathbb{Z})$ algebraic K-group over ring of integers of an algebraic number field and its relationship to the $\mathbb{Z}/48$ ring of integers modulo 48.
This is (of course), again, from Terry Gannon's "Moonshine Beyond the Monster"
where he talks about many amazing coincidences with the number 24, the
Riemann Zeta Function $\Sigma\_{n=1}^\infty (1/n)^{-1} = -1/12$, Apery's constant, where
$\Sigma\_{n=1}^\infty (1/n)^2 = \pi^2/6$ (which he states are both synonomous in their relationship to $K\_3(\mathbb{Z})\leftrightarrow \mathbb{Z}/48$....)
A little harder to discern is the (possible) relationship of the Bimonster,
$(M \times M) \rtimes \mathbb{Z}/2 \to M \wr 2$, and the Incidence Graph of the M-13 pseudogroup
with 13 points and lines, (the 13 point, 13 line projective plane, where here,
the coincidence would appear to be the number 26, which is the dimension of Bosonic String Theory (2 + 24 dimensions, the quantum harmonic oscillator on a 2-dimbrane, which relates to -1/12 above per John Baez "My Favorite Number is 24")). It's tempting to see the resemblance of 24 relating to the Monster,
and 48 to the BiMonster, but that seems to obvious. Finally, is there any
relevance in bringing in the M12-Mathieu group here, being so close to the
M13-pseudogroup? I apologize ahead of time if this last paragraph is "shooting
the moon" but hopefully my first two paragraphs are well-stated questions.
| https://mathoverflow.net/users/10350 | Z/48 and Moonshine Beyond the Monster | I am not sure if this helps you since it doesn't contain any speculations on connections with finite simple groups, but the torsion in $K\_{2n+1}(\mathcal{O}\_F)$ is now fairly well understood for arbitrary $n$ and for rings of integers $\mathcal{O}$ of arbitrary number fields, thanks to Suslin, Voevodsky, Rost, et al. The torsion subgroups are essentially certain Tate modules, not only as groups but in fact as Galois modules. In particular, the way 48 appears in the description of $K\_3(\mathbb{Z})$ is that it happens to be twice the biggest $n$, such that the exponent of $(\mathbb{Z}/n\mathbb{Z})^\times$ divides 2, and this observation fits into a bigger picture.
For a sketch of this bigger picture, see e.g. the article by Charles Weibel "Algebraic K-Theory of Rings of Integers in Local and Global Fields" in the Handbook of K-theory. He also gives references for the proofs of the various known results.
Edit: I should also mention that the connection with the Riemann zeta function (or, in the case of number fields, with the Dedekind zeta function of the number field), is - at least conjecturally - not a coincidence either. The special values of Dedekind zeta functions are conjecturally described by sizes of torsion subgroups and volumes of free parts of higher $K$-groups of the rings of integers. This is the Lichtenbaum conjecture. That in turn is also part of a bigger conjectural picture, the Bloch-Kato conjecture on Tamagawa numbers. In short: the connection that you are asking about is a special case of a large conjectural framework of global-local principles, which all take their cue from the analytic class number formula, but are completely out of reach at present.
| 15 | https://mathoverflow.net/users/35416 | 45380 | 28,760 |
https://mathoverflow.net/questions/45378 | 11 | I strongly believe that - given the rules of Conway's [Game of Life](http://en.wikipedia.org/wiki/Conway%27s_Game_of_Life) and an initial configuration - it is not decidable by a Turing Machine whether a given pattern will emerge, let alone as a stable pattern, be it static, moving, and/or rotating.
>
> How can this be proven?
>
>
>
I guess, this kind of uncomputability would go far beyond the "simple" unpredictability of non-linear systems.
| https://mathoverflow.net/users/2672 | Undecidability in Conway's Game of Life | Conway's game of Life [can simulate a universal Turing machine](http://rendell-attic.org/gol/tm.htm) which means that it is indeed undecidable by reduction from the halting problem.
You can program this Turing machine in the game of Life so that it builds some pattern when it halts that doesn't occur while it's still running. Then the pattern will be built if and only if the Turing machine halts.
| 39 | https://mathoverflow.net/users/2294 | 45382 | 28,762 |
https://mathoverflow.net/questions/45252 | 3 | Consider two partially ordered sets $A = \{a< b,a< c\}$, $B=\{x< z,y< z\}$.
Their linear extensions (here we allow equality in linear extensions) for $A, B$ are
$$A\_L=\{A\_1=\{a< b< c\}, A\_2=\{a< b= c\}, A\_3=\{a< c< b\}\}$$
and
$$B\_L = \{ B\_1 = \{x< y< z\}, B\_2=\{y< x< z\}, B\_3 =\{x= y< z\}\}$$
We may define two order isomorphisms
$f\_1: A\_1\to B\_1$ by $f\_1(a)=x, f\_1(b)=y, f\_1(c)=z$ and
$f\_2:A\_3\to B\_2$ by $f\_2(a)=y, f\_2(b)=x, f\_2(c)=z$,
but there is no order isomorphism $f\_3$ from $A\_2$ to $B\_3$ s.t. $t< s\Longleftrightarrow f\_3(t)< f\_3(s)$ and $t=s \Longleftrightarrow f\_3(t)=f\_3(s)$. (And it's easy to see that no other pairing of the $A\_i$ with the $B\_j$ will allow such maps to be chosen.)
There are no order isomorphisms from $A$ to $B$, either.
Is the following conjecture true?
For any partially ordered sets $A,B$, if there exist isomorphisms from $A$'s linear extensions to $B$'s, then there exists an isomorphism from $A$ to $B$.
If not, could you give me a counter example?
| https://mathoverflow.net/users/10410 | Does order isomorphism of linear extensions of two partially ordered sets imply order isomorphism of themselves? | There are two different questions depending on whether the linear extensions are given to us labeled. If we wish to reconstruct a poset $A$ from a labeled list of its linear extensions, then our job is very easy (even if we use "usual" linear extensions with no "=" appearing): for two elements $a, b \in A$, we have $a < b$ in $A$ if and only if $a < b$ in every linear extension of $A$, and so the result follows immediately. (This is the case that Joel David Hamkins addressed a few seconds before I posted :-) .)
If we have unlabeled linear extensions (so really the data that we receive looks like $\{(\* < \* < \*), (\* < \* = \*), (\* < \* < \*)\}$ or the like) then I conjecture that the result is false, with the following potential counter-example.
A Steiner triple system $S$ on a set $X = \{x\_1, \ldots, x\_{13}\}$ of 13 vertices is a collection of 26 blocks $B\_1, \ldots, B\_{26}$, each of which is a set of three elements of $X$, such that every two elements of $X$ occur in exactly one block $B\_i$. To such a system we may associate a poset $P(S)$ whose vertices are $\{x\_1, \ldots, x\_{13}, B\_1, \ldots, B\_{26}\}$ with order relation $\in$. There are two nonisomorphic Steiner triple systems on 13 vertices, yielding two nonisomorphic posets. (Nonisomorphism is easy: the poset conveys exactly the information about which triple contains which elements.) I think that these posets may be a counter-example, for the following (weak) reasons:
Given the list of linear extensions (in your sense) of a poset, it's easy to reconstruct certain basic data about it. In particular, we can find for each $i$ the number of elements whose shortest saturated chain to a minimal (or maximal) element is of length $i$ by looking for the linear extension that is most "bottom-heavy" (or "top-heavy"), in the sense that the initial block of equal elements is as large as possible (which will happen exactly when it consists of all the minimal elements), and the next block of equal elements is as large as possible given that the previous condition is met (which will happen exactly when it consists of all elements that cover a minimal element), and so on. By varying this idea slightly, we can also determine the list of up-degrees of each minimal element in the Hasse diagram (i.e., we know how many elements cover each minimal element), and the list of the number of elements that cover each pair of minimal elements, and so on. This data is not enough to reconstruct the poset on its own, since we don't know how things correspond in the different extensions, and indeed the two posets described above share this data but aren't isomorphic.
Unfortunately, it's clear that knowing all the linear extensions gives us more information than what I've used in the above paragraph. Unfortunately, these posets are too large for me to just slap them into a computer and check whether they really have the same set of extensions. Also, I can't find my copy of Enumerative Combinatorics at the moment to check what is actually known about this sort of thing.
| 2 | https://mathoverflow.net/users/4658 | 45387 | 28,766 |
https://mathoverflow.net/questions/45364 | 0 | Given an algebraic map $f: B^d \to \mathbb{R}$, from the unit ball of dimension $d$ to the real, let $Y = f^{-1}(0)$. Then it is always possible to find a smaller ball $B\_r \subset B^d$ not necessarily centered at $0$, such that for all lines $l$ going through $B\_r$, $|l \cap Y| \le k$, where $k$ is the maximum degree of $f$.
Is the analogue of this for higher codimension still true? By that I mean if $f: B^d \to \mathbb{R}^k$, $k \le d$, and q is a regular point of $f$. Then is it true that we can always find $B\_r$ such that for all affine $n$-planes going through $B\_r$, the number of intersections of the plane and $Y = f^{-1}(q)$ is bounded by the maximum degree of $f$, or perhaps some other polynomial bound depending only on the max degree of $f$?
I feel this should be doable using basic calculus of one variables, eg. mean value theorem. But it seems complicated.
| https://mathoverflow.net/users/4923 | Upper bound on the number of intersections of algebraic manifolds with affine planes | Unless I misunderstood the question, here is a counter-example:
$d=3$, $k=2$, $f:\mathbb R^3\to\mathbb R^2$ is just a linear map, e.g. $f(x,y,z)=(x,y)$. Take $q=(0,0)$, then $Y=f^{-1}(q)$ is a straight line in $\mathbb R^3$.
A generic 2-plane intersects $Y$ at one point. However there are planes containing $Y$ (and hence having infinitely many intersections), moreover these planes cover the entire space. So there is no ball $B\_r$ avoided by such planes.
| 5 | https://mathoverflow.net/users/4354 | 45407 | 28,779 |
https://mathoverflow.net/questions/44844 | 26 | I've stumbled across the family of polynomials
$ f\_p(x) = x^{p-1} + 2 x^{p-2} + \cdots + (p-1) x + p $,
where $p$ is an odd prime. It's not too hard to show that $f\_p(x)$ is irreducible over $\mathbb{Q}$ -- look at the Newton polygon of $f\_p(x+1)$ over $\mathbb{Q}\_p$ and you see that it factors as the product of an irreducible polynomial of degree $p-2$ and a linear. Since $f\_p(x)$ has no real roots (look at the derivative of $f\_p(x) (x-1)^2$) it must be irreducible over $\mathbb{Q}$. It's also not hard to see that the only primes dividing the discriminant are $2, p$ and primes dividing $p+1$. I would expect that the Galois group of a randomish polynomial would be the full symmetric group. Indeed, according to Magma this is true for $f\_p(x)$ for $p=3,5, \dots, 61$ with the exception of $p=7,17$. So my question is -- are these the only exceptions?
Added later: I've had Magma find the Galois group for primes through 101 and it found another exception: $p=97$. So my initial guess was wrong.
Another addition: If one looks at odd $p$ (not just prime) for $p < 100$ there is another exception, 49. Also 241 is not an exception (misread magma's output).
The ideas in the following two papers may be of help:
"On the Galois Groups of the exponential Taylor polynomials" by Robert Coleman, in L'Enseignement Mathematique, v 33 (1987) pp 183-189
and
"On the Galois Group of generalized Laguerre polynomials" by Farshid Hajir, J. Th´eor. Nombres Bordeaux 17 (2005), no. 2, 517–525 (also available on the author's web page).
| https://mathoverflow.net/users/2784 | Galois Groups of a family of polynomials | Let $\alpha$ be a root of a polynomial
$f(x) \in \mathbf{Q}[x]$ of degree $n$, let $K = \mathbf{Q}(\alpha)$,
$L$ be the Galois closure of $K$, and
$G = \mathrm{Gal}(L/\mathbf{Q}) \subset S\_n$.
How does one prove that a permutation group contains $A\_n$?
Following Jordan, the usual method is to show that it
is sufficiently highly transitive. Also following Jordan,
to do this it suffices to construct subgroups of $G$ which
act faithfully and transitively on $n-k$ points and trivially
on the other $k$ points (for $k$ large, $\ge 6$ using CFSG), and
to show that $G$ is primitive. (The standard method for doing this is
to find $l$-cycles for a prime $l$.)
In the context of a Galois group, the most obvious place
to look for "elements" is to consider the decomposition
groups $D$ at places of $\mathbf{Q}$.
If $l$ is unramified in $K$, this corresponds to looking
at a Frobenius element (conjugacy class). In practice
(as far as a computation goes) this is quite useful,
but theoretically it is not so great unless there is a
prime $l$ for which the factorization is particularly clean.
This leaves the places which ramify in $L$.
For example, if $v = \infty$, one is considering
the action of complex conjugation; if there are exactly
two complex roots then $c$ is a $2$-cycle, and from Jordan's
theorem (easy in this case) we see that if $G$ is primitive
then $G$ is $S\_n$.
The proposed method (following Coleman et. al.) for proving that
$G$ contains $A\_n$ is somewhat misguided, I think. The
key point about the polynomial
$\sum\_{k=0}^{n} x^k/k!$ is that the corresponding field is ramified at many primes,
and the decomposition groups at these primes give the
requisite elements. Conversely, the polynomial considered in this
problem corresponds to a field with somewhat limited ramification
- as has been noted, the only primes which ramify divide
$p(p+1)$.
It can be hard to compute Galois groups of random families of polynomials in general. I do not know if this is true in the present case, but given the lack of motivation I won't spend any more time thinking about it than the last hour or two, and instead give some partial results. However, the methods
given here may well apply more generally.
Let $n = p - 1$.
CLAIM: Suppose that $p+1$ is exactly divisible by a prime $l > 3$.
Then $G$ contains $A\_{n}$. (This applies to a set $p$
of relative density one inside the primes.)
STEP I: Factorization of $p$; $G$ is primitive.
Let $f(x) = x^{p-1} + 2 x^{p-2} + \ldots + p$.
Note that
$$(x-1)^2 f(x) = x(x^{p} - 1) - p(x-1) = x^{p+1} - 1 - (p+1)(x-1).$$
We deduce that
$f(x) \equiv x(x-1)^{p-2} \mod p$, and that
$$p = \mathfrak{p} \mathfrak{q}^{p-2}$$
for primes $\mathfrak{p}$ and $\mathfrak{q}$ in
the ring of integers $O\_K$ of $K$ both of norm $p$.
(To show this one needs to check that $[O\_K:\mathbf{Z}[\alpha]]$
is co-prime to $p$ - one can do this by considering the Newton
Polygon of $f(x+1)$.)
Let $D \subset G$ be a decomposition
group at $p$. This corresponds
to choosing a simultaneous embedding of the roots
of $f(x)$ into an algebraic closure of the $p$-adic numbers.
We see that we may write
$f(x) = a(x) b(x)$ as polynomials over the $p$-adic numbers (which
I can't latex at this point for some reason),
where $a(x) \equiv x \mod p$
has degree one and $b(x) \equiv (x-1)^{p-2}$ is
irreducible of degree $p-2$ and
corresponds to a totally ramified extension.
Clearly $D$ acts transitively on the $p-2 = n-1$ roots of $b(x)$ and fixes
the roots of $a(x)$. Since $D \subset G \cap S\_{n-1}$, we
see that $G \cap S\_{n-1}$ is transitive in $S\_{n-1}$ and
so $G$ is $2$-transitive (and hence primitive).
Step II: Factorization of $l$:
Let $l$ be a prime dividing $p+1$. We
assume that $l \ge 5$ and $l$ exactly divides $p+1$.
We see that
$$f(x) \equiv (x-1)^{l-2} \prod\_{i=1}^{k-1} (x-\zeta^i)^{l}$$
where $\zeta$ is a $k$th root of unity and $kl=p+1$.
This suggests that:
$$l = \mathfrak{p}^{l-2} \prod\_{i=1}^{k} \mathfrak{q}^l.$$
This also follows from a Newton polygon argument applied
to $f(x - \zeta^i)$. (Warning, this uses that $l$ exactly divides $p+1$.)
Step III: Some basic facts about local extensions:
Lemma 1. Suppose the ramification degree of $E/\mathbb{Q}\_l$
is $l^m$. Then the ramification degree of the Galois
closure of $E$ is only divisible by primes dividing $l(l^m-1)$.
Proof. Kummer Theory.
Lemma 2. Suppose that $h(x) \in \mathbf{Q}\_l[x]$
is an irreducible polynomial of degree $k$ with $(k,l) = 1$,
such that the
corresponding field $E/\mathbf{Q}\_l$ is totally ramified.
If $F$ is the splitting field of $h(x)$, then
$\mathrm{Gal}(F/\mathbf{Q}\_l) \subset S\_k$ contains a $k$-cycle.
Proof: From a classification of tamely ramified extensions, there
exists an unramified extension $A$ such that $[EA:A] = [E:\mathbf{Q}\_l]$
and $EA/A$ is cyclic and Galois. It follows that
$\mathrm{Gal}(EA/A)$ acts transitively and faithfully on the roots
of $h(x)$, and is thus generated by a $k$-cycle.
Step IV: $G$ contains an $l-2$-cycle.
Consider the decomposition group $D$ at $l$.
The orbits of $D$ correspond to the factorization of $l$ in $O\_K$.
On the factors corresponding to primes of the form
$\mathfrak{q}^p\_i$, the image of $D$ factors through a group
whose inertia has degree divisible only by primes dividing
$l(l-1)$, by Lemma 1. On the other hand, on
the factor corresponding to $\mathfrak{p}^{l-2}$, the image of
inertia contains an $l-2$ cycle, by Lemma 2.
Since $(l(l-1),l-2) = 1$, we see
that $D \subset G$ contains an $l-2$ cycle.
Step V: Jordan's Theorem.
Since $G$ is primitive, and $G$ contains a subgroup
that acts transitively and faithfully on $l-2$ points
(and trivially on all other points), we deduce
(from the standard proof of Jordan's theorem)
that $G$ is $n-(l-2)+1 = n+3-l$ transitive. This is at least $6$
(since $n+2$ is at least $2l$)
and so $G$ contains $A\_n$ (by CFSG).
STEP VI: (for you, dear reader)
Find the analogous argument when $p+1$ is exactly divisible
by $l^k$ for some $k \ge 2$ --- try to construct a cycle
of degree $l^k - 2$, although be careful as it will no
longer be the case (as it was above) that
$[O\_K:\mathbf{Z}[\alpha]]$ was co-prime to $l$.
This still leaves $p-1$ either a power of $2$ or a power
of $2$ times $3$, which might be annoying --- one would
have to think hard about the structure of the decomposition group at $2$ in those cases.
| 12 | https://mathoverflow.net/users/nan | 45412 | 28,782 |
https://mathoverflow.net/questions/45406 | 11 | I read in a paper of Goryunov
(`Functions on space curves',
Journal of The London Mathematical Society, vol. 61 (2000), 807-822;
available on his home page)
that every space curve can be defined as the vanishing
of the N minors of some N by N+1 matrix with entries functions of x, y, z.
How does one prove this?
(I can do it for the rational normal curve, thanks to Harris's book.)
| https://mathoverflow.net/users/2906 | Space Curves as Determinantal Varieties | Let $I\subset R = k[x,y,z]$ be the defining ideal of your curve. Then $R/I$ has dimension one and no embedded components, so has projective dimension $2$ by the Auslander-Buchsbaum formula. Therefore $I$ itself has projective dimension $1$, and so can be fit into a short exact sequence:
$$0 \to F \to G \to I \to 0 $$
with $F,G$ free (a minor point: one needs that projective modules are free here, it is easy if you assume $I \subset (x,y,z) $, since you may as well look at the local ring at the origin). If $N=\text{rank} F$, then $\text{rank} G=N+1$, and the matrix representing the map from $F$ to $G$ is what you want. This is known as the Hilbert-Burch theorem, and details can be found in Chapter 20 of Eisenbud's book "Commutative Algebra with a view..."
| 19 | https://mathoverflow.net/users/2083 | 45413 | 28,783 |
https://mathoverflow.net/questions/45422 | 7 | What is the easiest (preferably without calculations) way to see that the mean value of $\max(x\_1,x\_2,\dots,x\_n)$ on the sphere $\mathbb{S}^{d-1}= \{ (x\_1,\dots,x\_n):\ x\_1^2+\dots+x\_n^2=1 \}$ behaves like $\sqrt{\log(n)/n}$, or at least that is is much more then $1/\sqrt{n}$ for large $n$? The same (and less or more a priori equivalent) question concerns the standard Gaussian measure and expectation of $\infty$-norm w.r.t. it.
The proofs I know (for example, the one which V. Milman attributes to Figiel) use too many integrals.
And by the way, how to put {,} in math here? \ { does not work for me
| https://mathoverflow.net/users/4312 | maximal coordinate on a sphere | For some positive $c$ bounded away from zero, the probability that a standard gaussian variable is larger than $c\sqrt{\log n}$ is $1/n$. It follows that the probability that at least one variable out of $n$ independent standard gaussians is larger than $c\sqrt{\log n}$ is $1-(1-1/n)^n$ which tends to $1-1/e$. From that one gets that the expectation of the maximum of $n$ standard gaussians is at least (basically) $(1-1/e)c\sqrt{\log n}$.
As you said the question about variables on the sphere is equivalent to that.
| 15 | https://mathoverflow.net/users/6921 | 45425 | 28,789 |
https://mathoverflow.net/questions/45427 | 2 | Let X be a Stein manifold and U an open, connected, relatively compact, holomorphically convex subset of X. Is the closure of U in X holomorphically convex?
Also, if X is a Stein *space* with a finite number of singularities, and U is an open, connected, relatively compact, holomorphically convex subset of X containing all the singularities, is it true that the closure of U is holomorphically convex in this case too?
| https://mathoverflow.net/users/3566 | Is the closure of an open holomorphically convex subset of a Stein space holomorphically convex? | The answer is **no**.
In fact, in the first page of the paper
P.J. de Paepe, "Closures of proper analytic polyedra", Compositio mathematica 28 (1974), p. 333-341
there is the example of a relatively compact Stein space $U \subset \mathbb{C}^2$ such that its closure $\bar{U}$ is **not** holomorphically convex.
The example is as follows:
$U:=\{(z, w) \in \mathbb{C}^2 \, | \, |z| < |w|, \frac{1}{2} < |w| <1 \} \cup \{(z, w) \in \mathbb{C}^2 \, | \, |w| < |z|, \frac{1}{2} < |z| <1 \} $.
In fact, the smallest holomorphic set containing $\bar{U}$ is the unit polidisk in $\mathbb{C}^2$, which is different from $\bar{U}$.
**EDIT**. In a previous version of this post I wrote that the answer to the question was positive, since I erroneously assumed that the closure of $U$ was a complex space, see BCnrd's comment below.
| 1 | https://mathoverflow.net/users/7460 | 45430 | 28,790 |
https://mathoverflow.net/questions/45439 | 1 | What are sufficient conditions on $f, g : \mathbb{R}^n \to \mathbb{R}$ such that $\{x : f(x) \geq t \} \cup \{ x : g(x) \geq u\}$ has piecewise-smooth boundary?
Some remarks:
1. I don't mind if the conditions are stronger than necessary. In my application, $f$ and $g$ will be extremely nice functions anyway. My dream is just to have a simple-to-state condition backed up by a citation.
2. Feel free to assume $f$ and $g$ are nonnegative, compactly supported, and $0 < t < \|f\|\_\infty$, $0 < u < \|g\|\_\infty$. I imagine you'll want them to be smooth, too :)
3. To be honest, I'm naive enough that I don't even know sufficient conditions on $f$ such that $\{x : f(x) = t\}$ is a piecewise-smooth $(n-1)$-dimensional hypersurface. Maybe it's enough that $f$ is smooth and $\nabla f$ has only finitely many zeros?
4. Perhaps it depends on the definition of "piecewise"? At first I thought that if $\{x : f(x) \geq t \}$ and $\{ x : g(x) \geq u\}$ both had piecewise-smooth boundary then their union would too. But now that looks to me like it could be wrong. E.g., for $n = 2$ the set of points $\{(x,y) : -1 \leq x \leq 1, y \leq \exp(-1/x^2) \sin(1/x)\}$ has piecewise smooth boundary. If we take its union with $\{(x,y) : -1 \leq x \leq 1, y \leq 0\}$, then the "top part" of the resulting set's boundary is the curve $\max(\exp(-1/x^2) \sin(1/x), 0)$. Is that a piecewise-smooth curve? Seems like you need to break it into a (countably) infinite number of pieces to get all pieces smooth. On the other hand, perhaps one could/should (nonstandardly?) define "piecewise-smooth" to allow for countably many pieces.
5. Why do I even *want* the surface to be "piecewise-smooth"? Well, I want to apply a (higher-dimensional) version of the Cauchy-Crofton formula to it. In the textbooks I've looked at (e.g., Santalo) they usually assume that their surfaces are piecewise-$\mathcal{C}^1$. So really, I only need that, but I'm happy to require piecewise-smoothness if that makes things simpler. What I'd prefer *not* to have to do is to investigate what weaker conditions suffice for these Cauchy-Crofton-type formulas (e.g., is it okay for "piecewise" to allow for countably many pieces?).
Thanks very much, sorry for my naivete!
| https://mathoverflow.net/users/658 | What are sufficient conditions on $f, g : \mathbb{R}^n \to \mathbb{R}$ such that `$\{x : f(x) \geq t \} \cup \{ x : g(x) \geq u\}$` has piecewise-smooth boundary? | If you need a general regularity assumption on $f,g$ but no structural assumption on the shape of the level sets, I think the best one can do is to assume that $f,g$ are real analytic functions. (Of course this requires to drop the assumption of compact support). Then your set is subanalytic and in particular its boundary is made of a finite number of real analytic pieces. Anything less, e.g. $f,g$ in $C^\infty$, may produce a countable number of pieces plus a fat set where the two level sets touch each other (think Cantor). Can't you approximate your problem with real analytic functions?
| 3 | https://mathoverflow.net/users/7294 | 45442 | 28,798 |
https://mathoverflow.net/questions/45429 | 19 | Let $C$ be a category, say with finite products. What can be said about the category $\operatorname{Ab}(C)$ of abelian group objects of $C$? Is it always an abelian category? If not, what assumptions on $C$ have to be made? What happens when $C$ is the category of smooth proper geometrically integral schemes over some locally noetherian scheme $S$?
For example if $C=\mathrm{Set}$, we get of course the abelian category of abelian groups. If $C=\mathrm{Ring}/R$ for some ring $R$, then we get the abelian category $\operatorname{Mod}(R)$ (cf. [nLab](http://ncatlab.org/nlab/show/module#DefWithOCat)). In general, I have already trouble to show that $\operatorname{Hom}(A,B) \times \operatorname{Hom}(B,C) \to \operatorname{Hom}(A,C)$ is linear in the left coordinate if $A$, $B$, $C$ are abelian group objects.
| https://mathoverflow.net/users/2841 | The category of abelian group objects | Is $\mathscr{C}$ is regular/(exact in Barr sense) then for any algeraic theory $T$ the category $T{\operatorname{-Alg}}(\mathscr{C})$ of internal $T$-algebras is regular/(exact), in particular for $\mathscr{C}$ exact and $\operatorname{Ab} ={}$"commutative groups theory" we have $\operatorname{Ab}(\mathscr{C})$ is exact, this is also additive (i.e. abelian, a category is abelian iff is additive and exact):
given $f, g: A\to B$ in $\operatorname{Ab}(\mathscr{C})$ get $f+g$ in either of the following two ways:
1. appling the Yoneda valuation $h^X$, $X\in \mathscr{C}$ to $f, g: A\to B$ (and considering Yoneda's Lemma).
2. $f+g: A \xrightarrow{(f, g)} B\times B \xrightarrow{+}B$
| 12 | https://mathoverflow.net/users/6262 | 45447 | 28,802 |
https://mathoverflow.net/questions/45424 | 9 | While working on a problem in p-adic Hodge theory, and needing to write down a solution to a certain equation involving p-adic power series, I stumbled across a certain sequence of polynomials. Define $h\_j(X)$ for $j \ge 0$ by $h\_0(X) = 1$ and
$$ h\_{j}(X) = \frac{X + 1}{j}\left(- X \frac{\mathrm{d}}{\mathrm{d}\ X} + j\right)h\_{j-1}(X)$$
for $j \ge 1$.
I was interested in these because $h\_j(X)$ is the unique polynomial of degree $j$ such that
$$\left(\frac{t}{e^t - 1}\right)^{j+1} \cdot h\_j(e^t - 1) = 1 + O(t^{j+1}),$$
and in fact it follows from the recurrence that
$$\left(\frac{t}{e^t - 1}\right)^{j+1} \cdot h\_j(e^t - 1) = 1 + (-1)^j \sum\_{n \ge j+1} \binom{n-1}{j} \frac{B\_n t^n}{n!}$$
where $B\_n$ are the usual Bernoulli numbers.
Now, I can't believe that these polynomials $h\_j$ aren't some terribly classical well-studied thing, but they don't match any of the standard sequences of polynomials I could find on the web. Does anyone recognise these?
| https://mathoverflow.net/users/2481 | Is this sequence of polynomials well-known? | The first several are:
$$0! \cdot h\_0(x) = 1$$
$$1! \cdot h\_1(x) = x+1$$
$$2! \cdot h\_2(x) = x^2+3 x+2$$
$$3! \cdot h\_3(x) = x^3+7 x^2+12 x+6$$
$$4! \cdot h\_4(x) = x^4+15 x^3+50 x^2+60 x+24$$
Feeding the sequence $2,3,1,6,12,7,1,24,60$ into the [OEIS](http://www.research.att.com/njas/sequences/) gives the [following page](http://oeis.org/A130850), which contains generating functions, relations, and citations to occurrences of this sequence of polynomials in the literature.
| 16 | https://mathoverflow.net/users/935 | 45457 | 28,806 |
https://mathoverflow.net/questions/34518 | 25 | If $X$ is a compact metric space and $\mu$ is a Borel probability measure on $X$, then the space $C(X)$ of continuous real-valued functions on $X$ is a closed nowhere dense subset of $L^\infty(X,\mu)$, and hence bounded measurable functions are generically discontinuous. Nevertheless, Luzin's theorem says that every measurable function is in fact continuous on a set of arbitrarily large measure. This allows us to gain continuity from measurability at the cost of ignoring a small portion of $X$.
**Question:** *Are there any analogues of Luzin's theorem that allow us to go from continuity to Hölder continuity?*
A direct analogue would be a statement that given a continuous function $f\in C(X)$ and an arbitrary $\epsilon>0$, there exists a set $X\_\epsilon \subset X$ such that $\mu(X\_\epsilon) > 1-\epsilon$ and the restriction of $f$ to $X\_\epsilon$ is Hölder continuous. (For my purposes, it would be all right if the Hölder exponent and coefficient become arbitrarily bad as $\epsilon\to 0$.)
Another possible analogue, and one that I am actually more interested in, would be a statement that given a continuous function $f\in C(X)$ and an arbitrary $\epsilon>0$, there exists a set $X\_\epsilon \subset X$ such that the restriction of $f$ to $X\_\epsilon$ is Hölder continuous (again with arbitrarily bad exponent and coefficient) and instead of an estimate on the *measure* of $X\_\epsilon$, we have
$$
\dim\_H(X\_\epsilon) > \dim\_H(X) - \epsilon,
$$
where $\dim\_H$ is Hausdorff dimension.
*Full motivation*: Ideally I would like to consider the setting where $T\colon X\to X$ is a continuous map, and obtain a similar statement about the restriction of a continuous potential $f\in C(X)$ to a set of large *topological pressure*,
$$
P\_{X\_\epsilon}(f) > P\_X(f) - \epsilon,
$$
such that $f$ restricted to $X\_\epsilon$ has the *Walters property*, which deals with variation on Bowen balls rather than on metric balls. But the purely static version stated above for Hausdorff dimension seems like a good place to start. Does anybody know of any results in this direction? Or counterexamples showing that such a theorem can't be true in full generality?
**Edit:** I've accepted Anonymous's answer, which shows quite nicely that the direct analogue (using measures) fails. However, I remain very interested in the indirect analogue (using dimensions), which seems to still have a chance of holding, so any information in that direction would be welcomed.
| https://mathoverflow.net/users/5701 | Analogues of Luzin's theorem | Unfortunately, no, it is not possible to go from continuity to Hölder continuity in Luzin's theorem. At least, not in the sense of your first statement. We can give counterexamples to the following.
>
> ...given a continuous function $f\in C(X)$ and an arbitrary $\epsilon > 0$, there exists a set $X\_\epsilon\subset X$ such that $\mu(X\_\epsilon) > 1−\epsilon$ and the restriction of $f$ to $X\_\epsilon$ is Hölder continuous.
>
>
>
We can't even make the restriction of f to $X\_\epsilon$ satisfy any given modulus of continuity. Letting $\omega\colon[0,1]\to\mathbb{R}^+$ be continuous and strictly increasing with $\omega(0)=0$, there exist continuous functions $f\colon[0,1]\to\mathbb{R}$ such that $\vert f(x)-f(y)\vert/\omega(\vert x-y\vert)$ is unbounded over $x\not=y$ belonging to any set $S\subseteq[0,1]$ of Lebesgue measure greater than 1/2. Taking, e.g., $\omega(x)=e^{-\vert\log x\vert^{1/2}}=x^{\vert\log x\vert^{-1/2}}$ will show that f is not Hölder continuous on any set of Lebesgue measure greater than 1/2.
The idea is to construct a continuous function $f\colon[0,1]\to\mathbb{R}$ and a sequence of positive real numbers $\epsilon\_n\to0$ such that $\vert f(x+\epsilon\_n)-f(x)\vert/\omega(\epsilon\_n)\ge n$ on a subset of $[0,1-\epsilon\_n]$ of measure at least $1-2\epsilon\_n$.
We can construct this by applying the Baire category theorem to the complete metric space $C$ of continuous functions $[0,1]\to\mathbb{R}$ under the supremum norm. For any $f\in C$ and $K,\epsilon>0$, let $S(f,K,\epsilon)$ denote the set of $x\in[0,1-\epsilon]$ such that $\vert f(x+\epsilon)-f(x)\vert > K\omega(\epsilon)$. Then,
$$
A(K,\epsilon)=\left\{f\in C\colon\mu\left(S(f,K,\delta)\right)>1-2\delta{\rm\ some\ }0 < \delta < \epsilon\right\}
$$
is an open subset of $C$. It is also dense. To see this, first choose a continuously differentiable $f\in C$ and set $g(x)=f(x)+1\_{\{\lfloor x/\delta\rfloor{\rm\ is\ even}\}}K(\omega(\delta)+\sqrt{\delta})$. Choosing $\delta$ small enough, the inequality $\vert g(x+\delta)-g(x)\vert > K\omega(\delta)$ will hold on $[0,1-\delta]$, and there will then be a continuous function $\tilde g$ equal to g outside a set of measure $\delta$ and satisfying $\Vert\tilde g-f\Vert\le K(\omega(\delta)+\sqrt{\delta})$. So, choosing $\delta$ small enough, we have $\tilde g\in A(K,\epsilon)$ and $\tilde g$ as close to $f$ as we like, showing that $A(K,\epsilon)$ is indeed dense in $C$.
The Baire category theorem says that
$$
A=\bigcap\_{n=1}^\infty A(n,1/n)
$$
is nonempty, and any $f\in A$ satisfies the requirements mentioned above.
Now, suppose that $S\subseteq[0,1]$ has measure greater than 1/2. Choosing a random variable X uniformly in $[0,1]$,
$$
\begin{align}
&\mathbb{P}(X,X+\epsilon\_n\in S)\ge\mathbb{P}(X\in S)+\mathbb{P}(X+\epsilon\_n\in S)-1\ge 2\mu(S)-1-\epsilon\_n,\\\\
&\mathbb{P}(X\in S(f,n,\epsilon\_n))\ge 1 - 2\epsilon\_n.
\end{align}
$$
This gives $X,X+\epsilon\_n\in S$ and, simultaneously, $X\in S(f,n,\epsilon\_n)$ with probability at least $2\mu(S)-1-3\epsilon\_n$. For large enough n, this is positive. Therefore, there exist $x,x+\epsilon\_n\in S$ with $\vert f(x+\epsilon\_n)-f(x)\vert/\omega(\epsilon\_n) > n$ for arbitrarily large n.
| 16 | https://mathoverflow.net/users/10698 | 45459 | 28,807 |
https://mathoverflow.net/questions/45460 | 15 | Let $R$ be the ring $R[X,Y]/(X^2+Y^2−1)$. The space of $\mathbb{R}$-rational points of the affine scheme associated to $R$ is the topological circle $S^1$.
An algebraic vector bundle over $R$ is an $R$-algebra $A$ with certain properties or equivalently a finitely generated projective $R$-module $A$.
My first question is: What is the explicit projective $R$-module $A$ corresponding to the topological Moebius bundle over $S^1$? (By 'corresponding' I mean in particular that the $\mathbb{R}$-rational points should induce the topological Moebius bundle and that it has rank one.)
My second question is motivated by the fact that topologically there are only two non-isomorphic rank one bundles over $S^1$: the trivial bundle and the Moebius bundle. What is the analogous algebraic situation? Are there more than two non-isomorphic and rank one projective modules over $R$?
Thank you!
(I apologize that I first asked this question on math.stackexchange, I deleted it there.)
| https://mathoverflow.net/users/2625 | Algebraic analogue of the Moebius bundle over the circle | The internal note "Le ruban de Moebius comme représentation d'un idéal non principal" (Moebius band, as a non-principal ideal), by Daniel Ferrand, contains more or less the material you want. Daniel is retired and doesn't have a website, but, thank you Google!,
the note is downloadable from <http://www.math.unibas.ch/~giordano/Moebius.pdf>
The non-principal ideal you want is the ideal $(1+x,y)$ in the ring $R[x,y]/(x^2+y^2-1)$.
| 16 | https://mathoverflow.net/users/10696 | 45465 | 28,811 |
https://mathoverflow.net/questions/45477 | 44 | Let's call a function f:N→N *half-exponential* if there exist constants 1<c<d such that for all sufficiently large n,
cn < f(f(n)) < dn.
Then my question is this: *can we prove that no half-exponential function can be expressed by composition of the operations +, -, \*, /, exp, and log, together with arbitrary real constants?*
There have been at least two previous MO threads about the fascinating topic of half-exponential functions: see [here](https://mathoverflow.net/questions/12081/does-the-exponential-function-have-a-square-root) and [here](https://mathoverflow.net/questions/4347/ffxexpx-1-and-other-functions-just-in-the-middle-between-linear-and-expo). See also the comments on an [old blog post](http://www.scottaaronson.com/blog/?p=263) of mine. However, unless I'm mistaken, none of these threads answer the question above. (The best I was able to prove was that no half-exponential function can be expressed by *monotone* compositions of the operations +, \*, exp, and log.)
To clarify what I'm asking for: the answers to the previous MO questions already sketched arguments that if we want (for example) f(f(x))=ex, or f(f(x))=ex-1, then f can't even be *analytic*, let alone having a closed form in terms of basic arithmetic operations, exponentials, and logs.
By contrast, I don't care about the precise form of f(f(x)): all that matters for me is that f(f(x)) has an asymptotically exponential growth rate. I want to know: is that hypothesis *already* enough to rule out a closed form for f?
| https://mathoverflow.net/users/2575 | "Closed-form" functions with half-exponential growth | Yes
All such compositions are transseries in the sense here:
G. A. Edgar, "Transseries for Beginners". *Real Analysis Exchange* **35** (2010) 253-310
No transseries (of that type) has this intermediate growth rate. There is an integer "exponentiality" associated with each (large, positive) transseries; for example Exercise 4.10 in:
J. van der Hoeven, *Transseries and Real Differential Algebra* (LNM 1888) (Springer 2006)
A function between $c^x$ and $d^x$ has exponentiality $1$, and the exponentiality of a composition $f(f(x))$ is twice the exponentiality of $f$ itself.
Actually, for this question you could just talk about the Hardy space of functions. These functions also have an integer exponentiality (more commonly called "level" I guess).
| 44 | https://mathoverflow.net/users/454 | 45479 | 28,818 |
https://mathoverflow.net/questions/45474 | 1 | Suppose $X$ and $Y$ are jointly distributed real-valued random variables and for all outcomes $\omega\_1$, $\omega\_2$, we have
$$
X(\omega\_1)\le X(\omega\_2)\quad\Longrightarrow\quad Y(\omega\_1)\le Y(\omega\_2).
$$
**Edit**: As Louigi Addario-Berry's answer below shows, it may be better to consider the following variation:
$$
X(\omega\_1)< X(\omega\_2)\quad\Longrightarrow\quad Y(\omega\_1)\le Y(\omega\_2).
$$
>
> Does this property have a name?
>
>
>
| https://mathoverflow.net/users/4600 | Strongly correlated? Terminology question | This is along the lines of Tom's answer. $X$ induces a partial order on $\Omega$. In fact, it induces a total order on a partition of $\Omega$ into sets $X^{-1}(x)$, $x \in \mathbb{R}$);
simply say $X^{-1}(x) < X^{-1}(y)$ if $x < y$.
By your property, there is then some non-decreasing function $y:\mathbb{R} \to \mathbb{R}$ such that for $x \in \mathbb{R}$, if $\omega\_1, \omega\_2 \in X^{-1}(x)$ then $Y(\omega\_1)=Y(\omega\_2)=y(x)$.
But then we can write $Y(\omega)=y(X(\omega))$. In other words, $Y$ is just a non-decreasing (measurable) function of $X$.
| 0 | https://mathoverflow.net/users/3401 | 45484 | 28,820 |
https://mathoverflow.net/questions/37690 | 16 | **EDIT:** I've tried to alter the question so that its basic nature is clearer, as it's been unclear to a number of people now.
At any prime p, there is a graded polynomial ring $V \cong {\mathbb Z}\_{(p)}[v\_1, v\_2, \ldots]$ carrying two formal group laws. These formal group laws are of the form
$$
F(x,y) = \ell^{-1}(\ell(x) + \ell(y))
$$
for a logarithm $\ell(x) = \sum \ell\_n x^{p^{n+1}} \in ({\mathbb Q} \otimes V)[\![x]\!]$ (where $\ell\_0 = 1$ by convention). *Both* of these formal group laws have the property that they are universal among so-called $p$-typical formal group laws, and see heavy computational use in stable homotopy theory.
These two are based on choices of recursive definition for the logarithm coefficients in terms of the generators $v\_i$ of $V$. The first definition (the Araki generators) satisfies:
$$
p \ell\_n = \sum\_{k=0}^n v\_k^{p^{n-k}} \ell\_k = v\_n + \ell\_1 v\_{n-1}^p + \cdots + \ell\_{n-1} v\_1^{p^{n-1}} + \ell\_n p^{p^n}
$$
The Hazewinkel generators are instead defined by:
$$
p \ell'\_n = \sum\_{k=1}^{n} v\_k^{p^{n-k}} \ell'\_k = v\_n + \ell'\_1 v\_{n-1}^p + \cdots + \ell'\_{n-1} v\_1^{p^{n-1}}
$$
This gives the ring $V$ with two logarithms $\ell$ and $\ell'$, and two distinct universal formal group laws.
My question is: Are these two formal group laws isomorphic? Strictly isomorphic?
**ADDED:** Since the ring is torsion free, any isomorphism between them is of the form $f(x) = (\ell')^{-1} (c \ell(x))$ for a unit $c \in \mathbb{Z}\_{(p)}^\times = V^\times$. They are therefore isomorphic if and only if they are strictly isomorphic. Therefore, the question is equivalent to the following:
**Does the power series $(\ell')^{-1} \circ \ell$ have coefficients in $V \subset V \otimes \mathbb{Q}$?**
(The issue was brought up when thinking about truncated Brown-Peterson spectra ${\rm BP}\langle n\rangle$, whose rings of coefficients are $V/(v\_{n+1},v\_{n+2}, \cdots)$. It then becomes a question as to whether these are equivalent as ring spectra depending on the choice of generators. There are certainly different choices of generators for which they are inequivalent.)
| https://mathoverflow.net/users/360 | Isomorphism between two universal p-typical formal group laws | No, at least when $p=2$ the coefficient of $x^8$ in the relevant power series is not 2-locally integral. I have put a Maple worksheet at <https://strickland1.org/misc/ArHaz.mw> with a PDF version at <https://strickland1.org/misc/ArHaz.pdf>.
| 17 | https://mathoverflow.net/users/10366 | 45486 | 28,822 |
https://mathoverflow.net/questions/45491 | 4 | Assume a dynamical system $\dot{x}=f(x)$, $x \in R^n$ (with $f$ sufficiently smooth -- see below) satisfies the following:
1. The box $B=[-1,1]^n$ is forward invariant: any trajectory that starts in $B$ stays in $B$;
2. The system has a finite number of equilibrium points in $B$;
3. There is a smooth function $V(x) \ge 0$ such that $\dot{V}(x) < 0$ for all points in $B$ except the fixed points. In particular, the only periodic trajectories are the trivial ones corresponding to the fixed points.
I would like to conclude that any trajectory starting in $B$ must approach in the limit one of the fixed points.
>
> Is this actually correct? What hypotheses on $f$ are necessary/sufficient?
>
>
>
This looks like a basic question, so I would be happy with just a reference to a standard text.
| https://mathoverflow.net/users/8460 | Dynamical systems: non-divergence + non-periodicity = convergence? | The function $V$ is constant on the set of limit points of any trajectory, and this set is connected and invariant. It contains the complete forward orbit of every point it contains, so your hypothesis implies that each orbit it contains is a fixed point. Connectedness and the fact that orbits trajectires never leave your compact $B$ then let you conclude.
You can find the needed details in Perko's book *Differential equations and dynamical systems*.
| 5 | https://mathoverflow.net/users/1409 | 45492 | 28,826 |
https://mathoverflow.net/questions/45487 | 6 | Hi all,
I am interested in proofs without using Goedel's completeness theorem.
* Does anyone have a reference to a proof of this theorem that uses Skolem Functions?
* How come Enderton's (Introduction to Logic) has a half a page proof (which looks OK to me) and Boolos (Computability And Logic) has a full chapter of it?
Thanks.
| https://mathoverflow.net/users/10708 | Compactness Theorem for First Order Logic | There's a proof of compactness using Skolem functions in the book "Elements of Mathematical Logic. Model Theory" by Kreisel and Krivine. (I'm assuming here that the English version matches the French, because the latter is the one I checked.) It presupposes the compactness theorem for propositional logic.
The proof runs as follows: Given a finitely satisfiable set $S$ of formulas, Skolemize them to get a set $S'$ of universal formulas that is satisfiable if and only if $S$ is. (Technical detail: If the vocabulary has no constant symbols, adjoin one, so that there are some closed terms for use in the next step.) Let $S''$ be the set of all the formulas you get from those in $S'$ by deleting the universal quantifiers and replacing the variables by closed terms in all possible ways. The formulas in $S''$ are propositional combinations of atomic sentences; by regarding the atomic sentences as propositional variables, we can view $S''$ as a set of formulas of propositional logic. Finite satisfiability of $S$ (in the sense of first-order logic) implies finite satisfiability of $S''$ (in the sense of propositional logic). By propositional compactness, there is a truth assignment satisfying $S''$. That truth assignment amounts to a description of a structure (in the first-order sense) in which every element is the value of some closed term. This structure satisfies $S'$ and therefore $S$.
| 10 | https://mathoverflow.net/users/6794 | 45498 | 28,829 |
https://mathoverflow.net/questions/45466 | 2 | Fix a complete, cocomplete, symmetric monoidal closed category $\mathcal{V}$. I will also assume that there is a forgetful functor $\mathcal{V}\to \mathbf{Set}$ with a left adjoint. By standard results, this adjunction lifts to a 2-adjunction between the 2-category of categories and the 2-category of $\mathcal{V}$-categories; IOW, we can speak of the free $\mathcal{V}$-category on a category.
Now, let $\mathcal{A}$ be a (small) symmetric monoidal closed $\mathcal{V}$-category. Then the $\mathcal{V}$-presheaf category $\mathbf{PShv}(\mathcal{A})$ is symmetric monoidal closed when endowed with the Day convolution structure and the Yoneda embedding is symmetric monoidal closed. Furthermore, any symmetric monoidal $\mathcal{V}$-functor $F:\mathcal{A}\to \mathcal{B}$ with $\mathcal{B}$ cocomplete, lifts uniquely (up to unique isomorphism) to a cocontinuous symmetric monoidal functor $\mathbf{PShv}(\mathcal{A})\to \mathcal{B}$ by taking the left Kan extension $L\_{Y}(F)$ of $F$ along Yoneda $Y$.
I can prove all this to myself without great effort. What has left me stumped is the following question:
Q: assume $\mathcal{B}$ and $F$ closed, is $L\_{Y}(F)$ closed?
Since Yoneda is closed the converse is trivially true. If the answer is negative in general, is there any criteria to have a positive answer? If it helps the answer, my use case is when $\mathcal{A}$ is the free $\mathcal{V}$-category on a Heyting algebra (e.g. the open sets of a topological space) with monoidal structure the intersection. Since in a lattice there is at most one arrow between two objects, Day convolution degenerates into the pointwise tensor product.
Thanks in advance, regards,
G. Rodrigues
| https://mathoverflow.net/users/2562 | When is a left Kan extension closed? | There are various meanings of "closed functor", and the answer will depend on which meaning is adopted.
The meaning that I imagine Buschi Sergio adopted was what might be called "lax closed", where the structural constraint on a lax closed functor $F$ is a map $F(x \Rightarrow y) \to F(x) \Rightarrow F(y)$ satisfying some naturality and coherence conditions. Such a structure comes for free if $F$ is a (lax) monoidal functor, and since the left Kan extension of a lax monoidal functor is again lax monoidal, BS's remark would apply.
But perhaps you meant "strong" (or pseudo) closed, where the structural constraint is an isomorphism. In fact, I'm going to guess that's what you meant, since the other case (just discussed) is more or less obvious.
Since the construction of the internal hom (adjoint to the Day convolution monoidal product) on the presheaf category will inevitable involve limits, and since the left Kan extension does not generally preserve limits, the odds are highly against the left Kan being a closed functor in this strong sense. So the strategy should be to test it against just about the simplest case you can think of, and expect something will go wrong. Indeed, let's take $A$ to be the cartesian closed category $2$ (with two objects $0, 1$ and exactly one non-identity arrow, running in the direction $0 \to 1$). And take $B$ to be $Set$. Take $F: 2 \to Set$ to be the obvious strong closed functor, taking the object $0$ to the empty set and $1$ to the terminal set.
The following assertions are pretty easy to verify:
(1) The Day convolution on $Set^{2^{op}}$ is cartesian product. Thus the internal hom is the usual hom for the cartesian closed structure on the presheaf category.
(2) The left Kan extension $L\_{yon} F$ is easily calculated: $L\_{yon}(F)(X) = X(1)$.
(3) $L\_{yon}(F)(X \Rightarrow Y)$ is accordingly $(X \Rightarrow Y)(1)$, which one may calculate to be the end
$$\int\_{a \in 2} Y(a)^{X(a)}$$
or more explicitly, the pullback of the evident diagram
$$Y(0)^{X(0)} \to Y(0)^{X(1)} \leftarrow Y(1)^{X(1)}$$
(4) But $L\_{yon}(F)(X) \Rightarrow L\_{yon}(F)(Y)$ in $Set$ is just $Y(1)^{X(1)}$.
Since the results of (3) and (4) differ, we see that $L\_{yon}$ is not strong closed.
| 4 | https://mathoverflow.net/users/2926 | 45499 | 28,830 |
https://mathoverflow.net/questions/45494 | 6 | Lets $E\_{\tau}^{\rho}$ be the elliptic curve with complex structure given by $\tau$ in upper half plane and complexified Kahler form $\rho \frac{dz\wedge d\bar{z}}{2}$.( $\rho$ is in upper half plane too)
Then mirror symmetry says that mirror to $E\_{i}^{\rho}$ in A-side is $E^{i}\_{\rho}$ in B-side.(see the paper of Polishchuk and Zaslow)
then what is the mirror for general $E\_{\tau}^{\rho}$ in A-side (i.e. when we change the complex structure on A-side from the one given by $i$ to something else)??
| https://mathoverflow.net/users/5259 | Mirror symmetry for elliptic curves | The mirror of $E^\rho\_\tau$ is $E^\tau\_\rho$, as you may have guessed. The reason this is not discussed in, say, Polishchuk/Zaslow is that the derived category does not depend on the symplectic structure, and the Fukaya category does not depend on the complex structure, so for their purposes the parameter $\tau$ is irrelevant.
Btw, another article on "Mirror symmetry and elliptic curves" that you might find interesting is due to Dijkgraaf's, with title as indicated; it is contained in a 1995 volume on "The moduli space of curves" (see e.g. <http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.8.4194&rep=rep1&type=pdf>)
| 6 | https://mathoverflow.net/users/7437 | 45504 | 28,834 |
https://mathoverflow.net/questions/40736 | 40 | Every (?) algebraic geometer knows that concepts like homotopy groups or singular homology groups are irrelevant for schemes *in their Zariski topology*. Yet, I am curious about the following.
Let's start small. Consider a local ring $A$ with maximal ideal $M$; is the affine scheme $X=Spec(A)$ connected? Sure, because every open subset of $X$ containing $M$ is equal to $X$ itself. Or because the only idempotents of $A$ are $0$ and $1$. But is it path connected? Yes, because if you take any point $P$ in $X$ the following path $\gamma$ joins it to $M$ (reminds you of the hare and the tortoise...):
$ \gamma(t)=P \quad for \quad 0\leq t < 1\quad , \quad \gamma (1)=M $.
The same trick shows that the spectrum of an integral domain is path connected: join the generic point to any prime by a path like above. More generally, in the spectrum of an arbitrary ring $R$ you can join a prime $P$ to any bigger prime $Q$ $(P \subset Q)$ by adapting the formula above:
$ \gamma(t)=P \quad for \quad 0\leq t < 1\quad , \quad \gamma (1)=Q $.
[Continuity at $t=1$ follows from the fact that every neighbourhood of $Q$ contains $P$ and so its inverse image under $\gamma$ is all of $[0,1]$ ]
The question in the title just asks more generally:
**Is a connected scheme path connected ?**
**Edit** (after reading the comments) If an arbitrary topological space is connected and if every point has at least one path connected open neighbourhood, then the space is path connected. But I don't see why the local condition holds in a scheme, affine or not, even after taking into account what I proved about local rings.
| https://mathoverflow.net/users/450 | Is every connected scheme path connected? | There exist connected affine schemes which are not path connected. Let E be a compact connected metric space\* which is not path connected (e.g., the [closed topologist's sine curve](http://en.wikipedia.org/wiki/Topologist%27s_sine_curve)) and consider the following.
>
> $X={\rm Spec}(A)$ where $A$ is the ring of continuous functions $f\colon E\to\mathbb{R}$.
>
>
>
Then X is connected, since any idempotent f satisfies $f(x)\in\{0,1\}$ and, by connectedness of E, $f=0$ or $f=1$. The maximal ideals of A are
$$
\mathcal{m}\_x=\left\{f\in A\colon f(x)=0\right\}
$$
for $x\in E$. There will also non-maximal primes (see [this question](https://mathoverflow.net/questions/35793/prime-ideals-in-c0-1) for example) but, every prime ideal will be contained in one and only one of the maximal ideals\*\*. So, we can define $\pi\colon X\to E$ by $\pi(\mathcal{p})=x$ for prime ideals $\mathcal{p}\subseteq\mathcal{m}\_x$.
In fact, $\pi$ is continuous, using the following argument. For any open ball $B\_r(x)$ in E, choose $f\in A$ to be positive on $B\_r(x)$ and zero elsewhere. Then $D\_f=\left\{\mathcal{p}\in X\colon f\not\in \mathcal{p}\right\}$ is open and $\pi^{-1}(B\_r(x))\subseteq D\_f\subseteq \pi^{-1}(\bar B\_r(x))$.
Writing $B\_r(x)=\cup\_{s < r}B\_s(x)=\cup\_{s < r}\bar B\_s(x)$, this shows that there are open sets $U\_s$ lying between $\pi^{-1}(B\_s(x))$ and $\pi^{-1}(\bar B\_s(x))$. So, $\pi^{-1}(B\_r(x))=\bigcup\_{s < r} U\_s$ is open, and $\pi$ is continuous.
So, $\pi\colon X\to E$ is continuous and onto. If X was path connected then E would be too.
It may be worth noting that ${\rm Specm}(A)$ is also connected but not path connected, being homeomorphic to E.
---
(\*) I assume that E is a metric space in this argument so that the open balls give a basis for the topology, and there are continuous $f\colon E\to\mathbb{R}$ which are nonzero precisely on any given open ball. Actually, it is enough for the topology to be generated by the continuous real-valued functions. So the argument generalizes to any compact Hausdorff space (+ connected and not path connected, of course).
(\*\*) Maybe I should give a proof of the fact that every prime $\mathcal{p}$ is contained in precisely one of the maximal ideals $\mathcal{m}\_x$. Let $V(f)=\{x\in E\colon f(x)=0\}$ be the zero set of f. Then, $V(\mathcal{p})\equiv\bigcap\{V(f)\colon f\in\mathcal{p}\}$ will be non-empty. Otherwise, by compactness, there will be $f\_1,f\_2,\ldots,f\_n\in\mathcal{p}$ with $V(f\_1)\cap V(f\_2)\cap\cdots\cap V(f\_n)=\emptyset$. Then, $f=f\_1^2+f\_2^2+\cdots+f\_n^2\in\mathcal{p}$ would be nonzero everywhere, so a unit, contradicting the condition that $\mathcal{p}$ is a proper ideal. Choosing $x\in V(\mathcal{p})$ gives $\mathcal{p}\subseteq\mathcal{m}\_x$.
On the other hand, we cannot have $\mathcal{p}\subseteq\mathcal{m}\_x\cap\mathcal{m}\_y$ for $x\not=y$. Letting $f,g\in X$ have disjoint supports with $f(x)\not=0, g(y)\not=0$ gives $fg=0\in\mathcal{p}$ and, as $\mathcal{p}$ is prime, $f\in\mathcal{p}\setminus\mathcal{m}\_x$ or $g\in\mathcal{p}\setminus\mathcal{m}\_y$.
| 25 | https://mathoverflow.net/users/10698 | 45507 | 28,836 |
https://mathoverflow.net/questions/45505 | 6 | **What are the odds two uniformly chosen elements of S\_n span the whole group (or just the alternating group)?** Mathematica experements suggest those odds approach 1 - this might have been proven a long time ago. How likely is it to get the alternating group or something much smaller?
Also, **how can you *efficiently* find the size of the subgroup $\langle a,b\rangle$ in S\_n ?** My crude tests consists of randomly multiplying the two permutations and seeing how many different elements you get. Maybe there's a more efficient way to generate all the elements spanned by two permutation.
---
You *can* generate the whole permutation group using a swap (12) and a shift (12...n). I wonder if all two element generating sets are conjugate to this.
| https://mathoverflow.net/users/1358 | What are the odds two permutations in S_n do NOT generate the whole group? | The probability of generation of $A\_n$ or $S\_n$ by two random permutations is $1 - 1/n - O(1/n^2)$. The $1/n$ term comes from both permutations having the same fixed point. This is a classical result of L. Babai: The probability of generating the symmetric group, Journal of Combinatorial Theory, Series A, 1989. Warning: it uses the classification of finite simple groups.
For asymptotics for general simple groups, see [this paper](http://www.springerlink.com/content/r32065tp23lu7847/) by Liebeck and Shalev, and a recent [short survey](http://www.emis.ams.org/journals/DMJDMV/xvol-icm/02/Shalev.MAN.ps.gz) by Shalev.
| 18 | https://mathoverflow.net/users/4040 | 45509 | 28,838 |
https://mathoverflow.net/questions/45449 | 3 | **SymMonCat** is the cartesian 2-category of symmetric monoidal categories, braided monoidal functors, and monoidal natural transformations. The terminal symmetric monoidal category **1** has one object $I$ and $I \otimes I = I$.
A category enriched over a monoidal category $V$ assigns to each pair of objects $X, Y$ an object hom$(X,Y)$ in $V$ and to each object $X$ a morphism $id\_X:I \to \mbox{hom}(X,X)$ in $V$.
When $V = $ **SymMonCat**, the morphism $id\_X:1 \to \mbox{hom}(X,X)$ is a braided monoidal functor; since monoidal functors preserve the monoidal unit and tensor product, it must map the unit $I$ in **1** to the unit $I$ in hom$(X,X)$.
Is there a different way of enriching over **SymMonCat** such that $id\_X$ does not pick out the monoidal unit (other than considering it a subcategory of **Cat**)?
| https://mathoverflow.net/users/756 | Is there a sensible way to enrich over SymMonCat such that id_X is not the monoidal unit? | [Ignore this first part, I'm just leaving it for the context to the comments below.]
It is hard for me to understand why you would want to enrich in symmetric monoidal categories, have an identity, and also want this identity to *not* be the unit of the symmetric monoidal category.
That said, you can always do away with units altogether and consider "enriched categories without identities". Is this what you are after?
---
After Mike's example I am now on board. What you probably want to do is enrich over the symmetric monoidal 2-category of symmetric monoidal categories where the monoidal structure is the "tensor product of symmetric monoidal categories". What is this you ask?
The functor category between two symmetric monoidal categories $Fun^\otimes(B,C)$ is naturally equipped with a symmetric monoidal structure (using pointwise multiplication). The tensor product of symmetric monoidal categories is $(-) \otimes B$ is the (weak) left adjoint to the functor $Fun(B, -)$. Thus $A \otimes B$ is a symmetric monoidal category such that symmetirc monoidal functors from it to C are the same as "bilinear" functors $A \times B \to C$. Now the monoidal unit for this tensor product is the free symmetric monoidal category on one object $\mathbb{F}$ (which is the category of finite sets and permutations).
In this way, if you enrich in (SymCat, $\otimes$) you get a unit being a functor $ \mathbb{F} \to Hom(a,a)$, which is equivalent to just some element, not necessarily the unit object of $Hom(a,a)$.
The prototypical example is the 2-category of symmetric monoidal categories itself.
| 3 | https://mathoverflow.net/users/184 | 45515 | 28,840 |
https://mathoverflow.net/questions/45476 | 4 | Given two smooth projective schemes $X$ and $Y$ over some algebraically closed field $k$, we have $X\times Y$ with the projections $p$ to $X$ and $q$ to $Y$. Furthermore we have a "nice" sheaf of algebras $R$ on $X$, i.e. locally free and of global dimension at most dim($X$), e.g. Azumaya or something similar. Given two $p^{\*}R$-modules $M$ and $N$ on $X\times Y$, which are coherent and torsion free.
Like in the commutative case, i define the i-th relative $\mathcal{E}xt$-sheaf on $Y$ to be: $\mathcal{E}xt^i\_{p^{\*}R,q}(M,N):=(R^i(q\_{\*}\mathcal{H}om\_{p^{\*}R}(M,-))(N)$
Can I expect them to have the same properties as in the commutative case?
For example:
(1) Do we have $\mathcal{E}xt^i\_{p^{\*}R,q}(M,N)=0$ for $i>dim(X)$?
(2) Given $y\in Y$ is there a map $\mathcal{E}xt^i\_{p^{\*}R,q}(M,N)\otimes k(y) \rightarrow Ext\_R^i(M\_y,N\_y)$
(3) If $Ext\_R^i(M\_y,N\_y)=0$ for all $y\in Y$ does this imply $\mathcal{E}xt^i\_{p^{\*}R,q}(M,N)=0$?
(4) Is there a kind of base change theorem for the $\mathcal{E}xt^i\_{p^{\*}R,q}(M,N)$?
Or do I need more conditions for $M$ and $N$ to have the desired properties? I'm especially interested in the case, where $M=p^{\*}P$ for some $R$-module $P$ on $X$.
| https://mathoverflow.net/users/3233 | Base change and relative Ext over noncommutative rings | Questions of this type are discussed in the paper A.Kuznetsov, Hyperplane sections and derived categories, Izvestiya: Mathematics 70:3 (2006) p. 447-547, which is available at
<http://www.mi.ras.ru/~akuznet/publications/HyperplaneSectionsAndDerivedCategories.pdf>
See Appendix D.
| 1 | https://mathoverflow.net/users/4428 | 45522 | 28,846 |
https://mathoverflow.net/questions/45527 | 0 | Let $\mathcal{G} = \mathbb{M}\_n(\mathbb{C})$ be an $n$-by-$n$ matrix algebra over complex numbers. Next let $\mathcal{A} \cong \mathbb{M}\_d(\mathbb{C})$ be a subalgebra of $\mathcal{G}$ and assume $d$ divides $n$. Then is it true that there exists another subalgebra of $\mathcal{G}$ (let's call it $\mathcal{B}$) such that
1) $\mathcal{G} \cong \mathcal{A} \otimes \mathcal{B}$.
2) Every element in $\mathcal{A}$ commutes with every element in $\mathcal{B}$.
3) The intersection of $\mathcal{A}$ and $\mathcal{B}$ are multiples of the identity.
If it is not true, what are the necessary conditions?
| https://mathoverflow.net/users/9003 | Existence of tensor product of subalgebras | Yes, those are true and are consequences of the Double Centralizer Theorem for central simple algebras. See, for example, section 12.7 in Pierce's *Associative Algebras*.
| 4 | https://mathoverflow.net/users/1409 | 45529 | 28,849 |
https://mathoverflow.net/questions/45448 | 23 | The first infinite cardinal, $\aleph\_0$, has many large cardinal properties (or would have many large cardinal properties if not deliberately excluded). For example, if you do not impose uncountability as part of the definition, then $\aleph\_0$ would be the first inaccessible cardinal, the first weakly compact cardinal, the first measureable cardinal, and the first strongly compact cardinal. This is not universally true ($\aleph\_0$ is not a Mahlo cardinal), so I am wondering how widespread of a phenomenon is this. Which large cardinal properties are satisfied by $\aleph\_0$, and which are not?
There is a philosophical position I have seen argued, that the set-theoretic universe should be uniform, in that if something happens at $\aleph\_0$, then it should happen again. I have seen it specifically used to argue for the existence of an inaccessible cardinal, for example. The same argument can be made to work for weakly compact, measurable, and strongly compact cardinals. Are these the only large cardinal notions where it can be made to work? (Trivially, the same argument shows that there's a second inaccessible, a second measurable, etc., but when does the argument lead to more substantial jump?)
EDIT: Amit Kumar Gupta has given a terrific summary of what holds for individual large cardinals. Taking the philosophical argument seriously, this means that there's a kind of break in the large cardinal hierarchy. If you believe this argument for large cardinals, then it will lead you to believe in stuff like Ramsey cardinals, ineffable cardinals, etc. (since measurable cardinals have all those properties), but this argument seems to peter out after a countable number of strongly compact cardinals. This doesn't seem to be of interest in current set-theoretical research, but I still find it pretty interesting.
| https://mathoverflow.net/users/3711 | Aleph 0 as a large cardinal | This probably isn't the kind of thing anyone just knows off hand, so anyone who's going to answer the question is just going to look at a list of large cardinal axioms and their definitions, and try to see which ones are satisfied by $\aleph \_0$ and which aren't. You could've probably done this just as well as I could have, but I decided I'd do it just for the heck of it.
First of all, this doesn't cover all large cardinal axioms. Second, many large cardinal axioms have different formulations which end up being equivalent for uncountable cardinals, or perhaps inaccessible cardinals, but may end up inequivalent in the context of $\aleph \_0$. So even for the large cardinals that I'll look at, I might not look at all possible formulations.
1. **weakly inaccessible** - yes, obviously
2. **inaccessible** - yes, obviously
3. *Mahlo* - no, since the only finite "inaccessibles" are 0, 1, and 2 as noted by Michael Hardy in the comments, and the only stationary subsets of $\omega$ are the cofinite ones
4. weakly compact:
* **in the sense of the Weak Compactness Theorem** - yes, by the Compactness Theorem
* **in the sense of being inaccessible and having the tree property** - yes, by Konig's Lemma
* *in the sense of $\Pi ^1 \_1$-indescribability* - no, it's not even $\Pi ^0 \_2$-indescribable as witnessed by the sentence $\forall x \exists y (x \in y)$
5. *indescribable* - no, since it's not even $\Pi ^0 \_2$-indescribable
6. *Jonsson* - no, the algebra $(\omega, n \mapsto n \dot{-} 1)$ has no proper infinite subalgebra
7. *Ramsey* - no, the function $F : [\omega ]^{< \omega} \to \omega$ defined by $F(x) = 1$ if $|x| \in x$ and $0$ otherwise has no infinite homogeneous set
8. measurable:
* **in the sense of ultrafilters** - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition, i.e. closed under finite intersections
* *in the sense of elementary embeddings* - no, obviously
9. *strong* - no, obviously (taking the elementary embedding definition)
10. *Woodin* - ditto
11. strongly compact:
* **in the sense of the Compactness Theorem** - yes, by the Compactness Theorem
* **in the sense of complete ultrafilters** - yes, as in the case of measurables
* **in the sense of fine measures** - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition
12. supercompact:
* *in the sense of normal measures* - no, if $x \subset \lambda$ is finite and $X$ is the collection of all finite subsets of $\lambda$ which contain $x$, then the function $f : X \to \lambda$ defined by $f(y) = \max (y)$ is regressive, but for any $Y \subset X$, if $f$ is constant on $Y$ with value $\alpha$, then Y avoids the collection of finite subsets of $\lambda$ which contain $\{ \alpha + 1\}$ and hence Y cannot belong to any normal measure on $P\_{\omega }(\lambda)$
* *in the sense of elementary embeddings* - no, obviously
13. *Vopenka* - no, take models of the empty language of different (finite) sizes
14. *huge* - no, obviously (taking the elementary embedding definition)
| 31 | https://mathoverflow.net/users/7521 | 45532 | 28,851 |
https://mathoverflow.net/questions/45549 | 31 | Recently, I was asked to calculate the fundamental group of the space $X= \{a,b,c,d\}$ with open sets generated by $\{ a, c, abc, acd \}$.
Turns out, $\pi\_1(X)\cong \mathbb Z$ and in fact, $X$ is the quotient of $S^1$ (with the northern and southern hemispheres identified). But the result was not so easy to prove and this motivates the questions:
* Is the fundamental group of every connected manifold (other restrictions / generalizations on the manifold are welcome) the fundamental group of a finite space? (Of course, it would not be Hausdorff). (I observe that there are many redundant points on a manifold where homotopy-equivalent loops need not traverse)
* Is calculating $\pi\_1$ of such finite spaces easier than for the given space? (If yes, this gives a method to calculate fundamental groups of many familiar spaces)
Perhaps the fact that -- $\pi\_1$ of any CW complex just depends on its 1-skeleton [edit:2 skeleton]-- may be helpful.
| https://mathoverflow.net/users/2720 | Can the fundamental group of any manifold be realized as the fund grp of a finite space? | In fact, there is the following **theorem**: *Every finite CW complex is weakly homotopy equivalent to a finite topological space, and vice versa.*
For simplicial complexes, this theorem is realized by mapping a complex to its face poset, and using the correspondence between finite posets and finite topological spaces. In the other direction, one maps a poset to its order complex.
In general it is not easy to compute homotopy groups of a finite topological space. I know that there are some techniques in Jonathan Barmak's Ph.D. thesis.
| 67 | https://mathoverflow.net/users/1310 | 45550 | 28,863 |
https://mathoverflow.net/questions/45374 | 1 | [Background:]
Looking at the powerseries for the gamma-function
$ \Gamma(1+x) = 1 + a\_1 x + a\_2 x^2 - a\_3 \* x^3 + ... $
then we can arrive at a decomposition
$ \Gamma(1+x) = r(x) + g(x) $
where g(x) is constructed by the sum of the (taylor-expansions of) geometric series
```
1 1 1 1 1 1
g(x) = --- - --- + ---*--- - ---*--- + ...
1+x 2+x 3+x 2! 4+x 3!
[edit:corrected a sign-error]
```
and from this the powerseries for r(x)
$ r(x) = \Gamma(1+x) - g(x) $
[end background]
That function r(x) begins with
$ r(x) = 1/e + 0.21938 \* x + 0.09784 \*x^2 + \ldots $
The function has then some nice properties. By heuristics and inspection of its powerseries
it seems for instance, that
* apparently it is entire, has no zero except that $ lim\_{x-> \infty} r(-x) = 0 $
* $ r(0) = 1/e $ where $ e = \exp(1) $
* $ r(k) = r(k-1)\*k + 1/e $
Just today I found, that in fact this is the incomplete gamma-
function as defined/implemented in mathematica as "gamma(1+x,1)".
But this may not be of concern here, because I want to understand how to think the other way round:
* Question 1:
Assume we had only the functional relation and the initial value
$ r(x) = r(x-1)\*x +1/e $
$ r(0) = 1/e $
what else would we need to make r(x) unique and arrive at the solution
$ r(x) = \Gamma(1+x)-g(x) $
?
* Question 2:
Is there any way to generalize that construction scheme to
get some function f(x) where the functional equation depends
on a constant parameter c =/= 0
$ f(x) = f(x-1)\*x + c $
For instance let $ c=1/2 $ . What would a -for instance convex - function $ f(x)$ look like?
[update]: Question 2 seems to be easy - at the integer x f(x) is simply a scaling of r(x) by c and e: $ f(x)= r(x)\*c\*e $
so I'd assume the same can be assumed for fractional x.
| https://mathoverflow.net/users/7710 | What are conditions to make f(x) defined by f(x)=f(x-1)*x + 1/e unique(for instance convex)? | I'm answering my own question... :-)
I found that book of Emile Artin on the "Theory of the Gamma-function" and a very nice text of Philip Davis(1959) on "Leonard Euler's Integral" which both dealt very explanative with the uniqueness-problem and the specific topic of convexity.
That answers my question 1) - I'll just have to get some more practice with this.
| 0 | https://mathoverflow.net/users/7710 | 45557 | 28,869 |
https://mathoverflow.net/questions/45560 | 5 | Given a finite group $G$, let $n$ be the smallest integer s.t. $G \subset S\_n$ *à la* Cayley. I guess that if I want to construct the complex irreps (not just the character table) of $G$ then I could take [the irreps of $S\_n$](http://en.wikipedia.org/wiki/Representation_theory_of_the_symmetric_group) and restrict them to $G$. It seems plausible that after decomposition this might yield all the irreps of $G$, but I'm not entirely sure of this. In any event it seems inefficient, even if it works.
>
> So my question is: what is the (is
> there a?) general technique for
> constructing all the inequivalent
> complex irreps of a finite group?
>
>
>
Maybe this is better suited to the underflow site, but since it's come up in actual work (albeit of the documentary sort) I'm posting here.
| https://mathoverflow.net/users/1847 | Constructing inequivalent irreps of finite groups | I think, the article by Vahid Dabbaghian-Abdoly, Journal of Symbolic Computation
Volume 39, Issue 6, June 2005, Pages 671-688, entitled "An algorithm for constructing representations of finite groups", [doi:10.1016/j.jsc.2005.01.002](http://dx.doi.org/10.1016/j.jsc.2005.01.002), and the references in the introduction give you what you are looking for. I think, this is the state of the art to this day. Basically, finding all the irreducible representations can be done in polynomial time.
I am not sure off the top of my head what the complexity of your suggested algorithm will be, but the bottleneck will likely be the fact that $S\_n$ always has an irreducible representation of degree $n$. Generically, decomposing a representation of degree $|G|$ is much more work than you should have to do.
| 9 | https://mathoverflow.net/users/35416 | 45563 | 28,873 |
https://mathoverflow.net/questions/45569 | 3 | Clarification: by "piecewise", I mean a *finite* number of pieces.
I'm sure this must be true, but my search for a citation was in vain (although I did learn the new term "polyconvex").
Thanks!
| https://mathoverflow.net/users/658 | If $K$ and $L$ are compact convex sets with smooth boundary, does their union have piecewise-smooth boundary? | I don't think this is true. Suppose one of the sets is essentially $\{(x,y):y\geq x^2\}$ in the plane (cut off in some smooth way at the top, to make it compact). And suppose the other one is the same except that the parabolic lower boundary has been replaced by the graph of something like $y=x^2+e^{-1/x^2}\sin (1/x)$. In other words add a fierce oscillation but damped so strongly that the region above the curve is still convex (i.e., $d^2y/dx^2$ remains positive). (I haven't done the arithmetic to make sure my $e^{-1/x^2}$ damping is sufficient; if it isn't, then replace it by a more vigorous damping.) The union of the two convex sets will have infinitely many corners, where $\sin (1/x)$ is 0.
| 12 | https://mathoverflow.net/users/6794 | 45572 | 28,878 |
https://mathoverflow.net/questions/45577 | 8 | Hello,
I with to consider the following statement:
If $C$ is a cocomplete category having a dense small full subcategory $D$, then $C$ is complete.
(a full subcategory $D$ is dense in $C$ if every element of $C$ is canonical colimit of elements of $D$...)
I think I know how to prove it (I give proof below), and I want someone to reassure me that this statement is true exactly as stated, as it seems a little bit surprising.
Proof sketch:
Consider the functor $Y : C \to psh(D)$, where $psh(D)$ denotes the category of presheaves on $D$ (Yoneda functor, i.e. $Y(X)(A) = Hom (A, X)$). $D$ being dense in $C$ is equivalent to this functor being fully faithful. Futhermore, we have a functor $F: psh(D) \to C$, namely, the one which extends the inclusion $D \to C$ by cocontinuity (as presheaf categories have the property of being free cocompletions: <http://ncatlab.org/nlab/show/free+cocompletion>). Then one can see that $Y$ is right adjoint to $F$. So this renders $C$ as reflective subcategory of $psh(D)$ ( <http://ncatlab.org/nlab/show/reflective+subcategory>). Now, $psh(D)$ is complete, and so every reflective subcategory of it. hence, $C$ is complete.
Thank you,
Sasha
| https://mathoverflow.net/users/2095 | Any cocomplete category with a dense small full subcategory is complete? | This is a well-known theorem, you can find it for example in [Abstract and concrete categories - the Joy of Cats](http://katmat.math.uni-bremen.de/acc/acc.pdf), Theorem 12.12. The proof there uses that a cocomplete category with a weakly terminal object has a terminal object (the preparation for the Freyd's adjoint functor theorem).
| 5 | https://mathoverflow.net/users/2841 | 45583 | 28,884 |
https://mathoverflow.net/questions/45609 | 0 | Let be $\Omega$ a compact metric space, $\mathcal{B}(\Omega)$ the $\sigma$-algebra of Borelian sets of $\Omega$ and
$\mathcal{M}\_1(\Omega)$ the set of all probabilities defined on $\mathcal{B}(\Omega)$.
Suppose that $\lambda,\mu\in\mathcal{M}\_1(\Omega)$ are extremal points (in the sense of convex combinations) and there is a real number $c$ such that
$$
\lambda(A)\leq c\mu(A)\qquad\text{and}\qquad
\mu(A)\leq c \lambda(A)\qquad
$$
for all $A\in\mathcal{B}(\Omega)$.
Is it true that $\mu=\lambda$ ?
I have proved the above equality in particular cases: 1) $\Omega$ is discrete; 2) $\Omega$ is some subset of $\mathbb{R}$ (the compacity it was not necessary here).
| https://mathoverflow.net/users/2386 | Condition for Uniqueness of Measures | I must be missing the point here. If $\lambda$ gives some Borel set $A$ a measure $p$ strictly between 0 and 1, then $\lambda$ would be a convex combination, $p$ times the conditional probability on $A$ plus $1-p$ times the conditional probability on the complement of $A$. That contradicts the hypothesis that $\lambda$ is an extreme point. So $\lambda$ must take only the values 0 and 1 (it amounts to an ultrafilter in $\mathcal B(\Omega)$). The same goes for $\mu$. Then your inequalities relating $\lambda$ and $\mu$ prevent either of them from taking the value 1 on a set where the other takes the value 0. So they are equal.
| 2 | https://mathoverflow.net/users/6794 | 45611 | 28,894 |
https://mathoverflow.net/questions/45586 | 10 | [This](http://www.oakland.edu/enp/trivia/) site claims that the diameter of the Erdös component of the collaboration graph in 2004 was 23. What is it now? Is it increasing or decreasing with time? Recall that the vertices of the collaboration graph are mathematicians and two vertices are connected if the mathematicians co-author a paper. MathSci allows one to find the collaborative distance between any two mathematicians. So in principle one can find the diameter just by using MathSci. In general (and more seriously) is there a mathematical theory which describes the growth of "real life" networks like the collaborative graph?
**Update 1** A process that I had in mind is something like this. At every step one of the following things can happen.
1. A new vertex $a$ is born with some probability $p$. Usually that vertex is a "student" of some other vertex $x$. We can connect $a$ and $x$ by an edge.
2. A randomly chosen vertex $y$ gets connected with a randomly chosen vertex $z$. But the probability for choosing $z$ is not uniform. Those vertices that are closer to $y$ have more chances getting connected to $y$.
3. A randomly chosen vertex "dies" (with some probability $q$) meaning it does not participate in new edges any longer.
**Update 2.** Many thanks to Balazs and Joseph for their answers. But the first question still remains: what is the diameter of the Erdös component now?
| https://mathoverflow.net/users/nan | The diameter of the Erdös component of the collaboration graph | There is a very large literature on this, written by people doing "network science". One of the names you might want to look up is that of Mark Newman, see for example his papers [The structure of scientific collaboration networks](http://xxx.lanl.gov/abs/cond-mat/0007214) or [Who is the best connected scientist? A study of scientific coauthorship networks](http://xxx.lanl.gov/abs/cond-mat/0011144) as well as [Clustering and preferential attachment in growing networks](http://arxiv.org/abs/cond-mat/0104209). Another important player in this field is Albert-Laszlo Barabasi, whose group had a model (for the web graph) which is somewhat similar to what you suggest, [Emergence of scaling in random networks](http://xxx.lanl.gov/abs/cond-mat/9910332). Some of this work is reviewed nicely in [Statistical Mechanics of Complex Networks](http://xxx.lanl.gov/abs/cond-mat/0106096). The model by Barabasi et al is studied in mathematical terms by Bollobas-Riordan-Spencer-Tusnady's [The degree sequence of a scale-free random graph process](http://onlinelibrary.wiley.com/doi/10.1002/rsa.1009/abstract).
To answer your specific question on the diameter, I would expect it to be quite small ("six degress of separation"); whether it decreases in time must depend on the relative rate of birth of new vertices versus new edges (collaborations). If you take a finite graph and add completely random links, then already a few links lead to a diameter which is logarithmic in the size of the network. I think this remains the case also if you weigh your edge creation mechanism by the distance of the vertices; I once played with a model where the only new edges were created between 2-step neighbours, which still lead to small diameter.
| 10 | https://mathoverflow.net/users/6107 | 45619 | 28,901 |
https://mathoverflow.net/questions/45022 | 4 | A $n\times n$ matrix $A=[a\_{ij}]$ is called "good", if there exists some $k$ and a set of $k\times k$ complex unitaries $U\_i$, $1\leq i\leq n$ , such that $tr(U\_i^{+}U\_j)=ka\_{ij}$, where $U^{+}$ denotes the conjugate transposed matrix of $U$.
Let $S=${$A|A$ is good}, Is there any useful method to check if a given matrix is "good"? Or how to describe $S$ completely?
In quantum information, maximally entangled states are widely concerned, which can be consider as a vector, and really connected to unitaries, the mother problem is:
Describe the set of matrice $A$, where $A=P^{+}P$, with P be a $n\times k$ matrix whose columns are all maximally entangled states.
Is $S$ a closed set? Is it a convex set?
| https://mathoverflow.net/users/4987 | Matrices whose entries are essentially traces of products of unitary matrices. | I think I can give a partial answer.
*(I will use the OP's notation of + for the adjoint)*
1) It is clear that any "good" matrix has every diagonal entry equal to 1.
2) If A is "good", then it is positive. Indeed, being good means that there exist $k\in\mathbb{N}$ and $U\_1,\ldots,U\_n$ unitaries in $M\_k(\mathbb{C})$ with
$$
A=\frac1k\,\mbox{tr}^{(n)}\left(\begin{bmatrix}U\_1\\ U\_2\\
\vdots \\ U\_n\end{bmatrix} \,
\begin{bmatrix}U\_1\\ U\_2\\
\vdots \\ U\_n\end{bmatrix}^+\right),
$$
where $\mbox{tr}^{(n)}$ is the map that replaces each $k\times k$ block by its trace. Since
the trace has abelian range it is completely positive, so $\mbox{tr}^{(n)}$ is positive and thus
$A$ is positive.
3) For $n=2$, the converse holds (i.e. any positive matrix with 1 in the diagonal is "good"). Indeed, let $A=\begin{bmatrix}1&a\\ a&1\end{bmatrix}$, with $|a|\leq1$. Let $k=2$, $U\_1=I\_2$,
$U\_2=\begin{bmatrix}a&\sqrt{1-|a|^2}\\ -\sqrt{1-|a|^2}&a\end{bmatrix}$. Then
$$
A=\frac12\,\mbox{tr}^{(n)}\left(\begin{bmatrix}U\_1\\ U\_2\end{bmatrix} \,
\begin{bmatrix}U\_1\\ U\_2\end{bmatrix}^+\right).
$$
4) So the question remains, is every $n\times n$ positive matrix with diagonal 1, "good"?
For $n\geq3$, I still cannot see too much in this direction. On the one hand, being free to choose $k$ gives an incredible amount of unitaries to play with. On the other, positivity is hard to make precise as a relation among the coefficients, and so I cannot see an obvious way to choose the unitaries in a satisfactory way (somehow the restriction given by positivity should appear in the choice of the unitaries, and I couldn't make sense of it).
5) If the conjecture were true (and it is for $n=2$) this makes the set $S$ closed (a norm-limit of positive matrices with diagonal one is going to be again positive with diagonal 1). Otherwise, from the definition alone, I don't really see it.
| 2 | https://mathoverflow.net/users/3698 | 45630 | 28,906 |
https://mathoverflow.net/questions/45624 | 1 | Really I should first ask this question here on MathOverflow and only then [post it as an open problem](http://garden.irmacs.sfu.ca/?q=op/do_filters_complementive_to_a_given_filter_form_a_complete_lattice) in Open Problem Garden and [propose it as a polymath problem](http://portonmath.wordpress.com/2009/08/30/proposal-conjecture-complementive-filters/). Indeed I did the reverse and now hope that for somebody my "open problem" may be simple.
Let $U$ is a set. A filter (on $U$) $\mathcal{F}$ is by definition a non-empty set of subsets of $U$ such that $A,B\in\mathcal{F} \Leftrightarrow A\cap B\in\mathcal{F}$. Note that unlike some other authors I do not require $\varnothing\notin\mathcal{F}$. I will denote $\mathscr{F}$ the lattice of all filters (on $U$) ordered by set inclusion.
Let $\mathcal{A}\in\mathscr{F}$ is some (fixed) filter. Let $D= \{ \mathcal{X}\in\mathscr{F} | \mathcal{X}\supseteq \mathcal{A} \}$. Obviously $D$ is a bounded lattice.
I will call complementive such filters $\mathcal{C}$ that:
1. $\mathcal{C}\in D$;
2. $\mathcal{C}$ is a complemented element of the lattice $D$.
**Conjecture** The set of complementive filters ordered by inclusion is a complete lattice.
Read also [my current progress on this problem](http://portonmath.wordpress.com/2009/11/02/exposition-complementive-filters/).
| https://mathoverflow.net/users/4086 | Do filters complementive to a given filter form a complete lattice? | The complementive filters ordered by inclusion form a lattice isomorphic to the quotient of the power set of $U$ modulo the filter $\mathcal A$. So, for example, if $U=\omega$ and if $\mathcal A$ is the filter of cofinite sets, then the lattice of complementive filters would not be complete.
To establish the isomorphism, consider a complementive filter $\mathcal C$ and its complement $\mathcal C'$ in $D$. The filter generated by $\mathcal C\cup\mathcal C'$ is the join in $D$, so it must be the top element, the improper filter of all subsets of $U$. So there must be a set $Q\in\mathcal C$ whose complement $U-Q$ is in $\mathcal C'$.
I claim that $\mathcal A\cup\{Q\}$ generates $\mathcal C$. To see this, suppose not, and consider some $Z\in\mathcal C$ that is not in the filter generated by $\mathcal A\cup\{Q\}$. Then $Z\cup(U-Q)$ cannot be in $\mathcal A$ (because, if it were, then its intersection with $Q$ would be in the filter generated by $\mathcal A\cup\{Q\}$, but this intersection is included in $Z$, which is not in that filter). But, since $Z\in\mathcal C$ and $U-Q\in\mathcal C'$, the union $Z\cup(U-Q)$ is in the intersection of these two filters, which is $\mathcal A$ (because they're complements in $D$). This contradiction establishes the claim.
Thus, each complementive $\mathcal C$ in $D$ is generated by the fixed $\mathcal A$ plus one more set $Q$. It is easy to check that the filter generated by $\mathcal A\cup\{Q\}$ and the filter generated by $\mathcal A\cup\{R\}$ are equal if and only if $Q$ and $R$ represent the same element in the quotient Boolean algebra $\mathcal P(U)/\mathcal A$. Better, the filter generated by $\mathcal A\cup\{Q\}$ is included in the one generated by $\mathcal A\cup\{R\}$ if and only if the element of $\mathcal P(U)/\mathcal A$ represented by $R$ is below (in the Boolean algebra) the one represented by $Q$. Thus, the correspondence between $\mathcal C$ and ($Q$ modulo $\mathcal A$) is an order-reversing bijection between the lattice of complementive filters and the Boolean algebra $\mathcal P(U)/\mathcal A$.
Since a Boolean algebra and its dual order are isomorphic, this proves my description of the structure of the lattice of complementive filters.
| 8 | https://mathoverflow.net/users/6794 | 45640 | 28,914 |
https://mathoverflow.net/questions/45626 | 2 | For a finite abelian group $G$ is there an analogue of structure theorem for finitely generated modules like for P.I.D. rings but with $Z[G]$ group ring over integers instead ?
| https://mathoverflow.net/users/10740 | Structure theorem for finitely generated Z[G] modules | Let me give an answer in the opposite direction of the ones already given. It may that be you can't get a precise classification, but nevertheless, you can say something, and it may be enough for the situation you are considering, or at least helpful.
Let's suppose first that your module $M$ is torsion; then it is a product of its $p$-power torsion parts for each prime $p$, so you an suppose it is $p$-power torsion. It is then a module over $\mathbb Z\_p[G]$. You may write $G$ as a product $G\_1\times G\_2,$ where
$G\_1$ is of prime-to-$p$ order and $G\_2$ is of $p$-power order. The group ring $\mathbb F\_p[G\_1]$ is a product of finite fields $k$,
and the ring $\mathbb Z\_p[G\_1]$ is isomorphic to a corresponding product of rings of Witt vectors $W(k)$. So one is left with $p$-power torsion modules over $W(k)[G\_2]$. This may be something of a mess, but one can say something: for example, any Jordan--H\"older factor of such a module is the trivial representation of $G\_2$ over $k$, and so your module is a successive extension of copies of the trivial representation.
Now suppose that you have an $M$ which is not necessarily torsion. Then one can consider
the map $M \to \mathbb Q\otimes\_{\mathbb Z} M$. The kernel is torsion, and so one can get some handle on its structure as above. The image is a $G$-invariant lattice in
the $G$-representation $\mathbb Q \otimes\_{\mathbb Z} M.$ The possibilities for this latter representation (of $\mathbb Q$ over $G$) can be described pretty easily, since $\mathbb Q[G]$ is just a product of fields $K$; it will break up into a product of copies of these various fields $K$. The biggest complication with the image of $M$ is that it typically won't break up as a product.
A useful way to think is to consider Spec $\mathbb Z[G]$, which is an affine scheme mapping finitely to Spec $\mathbb Z$, and then to think of $M$ as quasi-coherent sheaf over Spec $\mathbb Z[G]$. Then tensoring with $\mathbb Q$ corresponds to looking at this sheaf at the generic points; looking at the torsion part of $M$ corresponds to looking at sections supported at closed points; and the possible failure of $M$ to decompose as a product corresponds to the fact that although Spec $\mathbb Z[G]$ will have several irreducible components (if $G$ is non-trivial) --- these correspond to the various factors in the decomposition of $\mathbb Q[G]$ into a product of fields --- it will be connected (the components meet at various closed points, corresponding to the primes dividing the order of $G$).
Number theorists frequently work with rings like $\mathbb Z[G]$, and modules over them.
Depending on what you want to do, it is not necessary to have a complete classification;
a somewhat coarser understanding of the type described above can be quite useful by itself.
| 12 | https://mathoverflow.net/users/2874 | 45642 | 28,915 |
https://mathoverflow.net/questions/45594 | 6 | What does it tell you about two functions if their $L^p$ norms are the same for all $p\in[1,\infty]$? Certainly they could be related by composition with a diffeomorphism with Jacobian of norm 1, or even one could be a "pulled apart version of the other one" in the sense of
$x^2\chi\_{[0,2]}$ vs $x^2 \chi\_{[0,1]} + (x-1)^2 \chi\_{[2,3]}$. To try to ignore the second type of issue, I'll restrict to smooth functions, and ask the following precise question:
Given $f,g\in C^\infty(\mathbb{R})$ such that $\Vert f \Vert\_{L^p(\mathbb{R})} = \Vert g \Vert\_{L^p(\mathbb{R})} <\infty$ for all $p\in [1,\infty]$ is it necessarily true that $f(x) = g(\pm x + C)$ for some constant $C$ and for a choice of either $+$ or $-$ for all $x \in \mathbb{R}$?
As Qiaochu Yuan shows in his answer, smoothness doesn't solve the issue of "pulling apart" at all. Thus, I am interested in the following:
What is the "smallest" (in whatever sense) but still nonempty subset of $S \subset [1,\infty]$ such that there is $f,g\in C^\infty(\mathbb{R})$ such that $S$ is the set of $p$ such that
$\Vert f \Vert\_{L^p} \neq \Vert g \Vert\_{L^p}$?
| https://mathoverflow.net/users/1540 | How small can the set of $p$ such that the $L^p$ norms are different for two fixed functions? | The complement of $S$ in $(1,\infty)$ can't contain an interval or even a sequence converging to a point in $(1,\infty)$. Let $f$ and $g$ be two real functions on ${\mathbb{R}}$ all whose moments exist. Assume $\int|f|^p=\int|g|^p$ for all $p\in S^c$. put $h(z)=\int|f|^z-\int|g|^z$. where $z\in \mathbb{C}$. $h$ is an analytic functions in $\{Real z>1\}$ which we assume is not constantly zero (since $S$ is not empty) so can't vanish on a converging sequence.
| 12 | https://mathoverflow.net/users/6921 | 45649 | 28,918 |
https://mathoverflow.net/questions/45648 | 2 | 1. Given $n$ independent uniformly distributed points on $S^2$, what's the distribution of the distance between two closest points?
2. Consider $n$ iid uniform points on $S^1$, $Y\_1, \ldots, Y\_n$, in counterclockwise order. Now let $I\_1 = Y\_2-Y\_1, \ldots, I\_n = Y\_1 - Y\_n$ be the spacings between consecutive points. Finally order the spacing sequence into $I\_{(1)} < I\_{(2)} < \ldots < I\_{(n)}$. They will also generate a spacing sequence, of size $n-1$, $J\_1 = I\_{(2)} - I\_{(1)}, \ldots, J\_{n-1} = I\_{(n)} - I\_{(n-1)}$. What's the distribution of this last sequence? In particular, what's the mean value of the smallest $J$ and largest $J$?
| https://mathoverflow.net/users/4923 | Two geometric probability questions (one answered, one more to go) | There is an asymptotic formula for the minimal spherical distance when $n$ is large (see e.g. the PhD thesis [*"Random Diameters and Other U-Max-Statistics"*](http://www.imsv.unibe.ch/content/research/publications/theses/2008/e6079/e6261/e6950/Mayer2008_eng.pdf) by M. Mayer, Corollary 3.37):
>
> **Theorem.** Assume that the points $\xi\_1,\xi\_2,...,\xi\_n$ are independent and uniformly distributed on $\mathbb S^{d−1}$. Let $S\_n$ be the smallest central angle formed by point pairs within the sample. Then for $t > 0$
> $$P\{n^{2/(d-1)}S\_n\leq t\}=1-\exp\left(-\frac{\Gamma(\frac{d}{2})}{4\pi^{1/2}\Gamma(\frac{d+1}{2})}t^{d-1} \right)+\mathcal O(n^{-1}).$$
>
>
>
I am not sure if there is a nice explicit formula for finite $n$.
In fact, the knowledge of the exact form of the distribution $P\{S\_n\leq\theta\}$ on $\mathbb S^2$ would lead to a solution of the [Tammes packing problem](http://en.wikipedia.org/wiki/Tammes_problem) (which is only solved for a few values of $n$ to the best of my knowledge).
| 4 | https://mathoverflow.net/users/5371 | 45652 | 28,920 |
https://mathoverflow.net/questions/45653 | 30 | Let's say a **normed division algebra** is a real vector space $A$ equipped with a bilinear product, an element $1$ such that $1a = a = a1$, and a norm obeying $|ab| = |a| |b|$.
There are only four finite-dimensional normed division algebras: the real numbers, the complex numbers, the quaternions and the octonions. This was proved by Hurwitz in 1898:
* Adolf Hurwitz, Über die Composition der quadratischen Formen von beliebig vielen Variabeln, *Nachr. Ges. Wiss. Göttingen* (1898), 309-316.
Are there any infinite-dimensional normed division algebras? If so, how many are there?
| https://mathoverflow.net/users/2893 | Infinite-dimensional normed division algebras | A MathSciNet search reveals a paper by Urbanik and Wright ([Absolute-valued algebras. *Proc. Amer. Math. Soc.* **11** (1960), 861–866](https://doi.org/10.1090/S0002-9939-1960-0120264-6)) where it is proved that an arbitrary real normed algebra (with unit) is in fact a finite-dimensional division algebra, hence is one of the four mentioned in the OP. A key piece of the argument (Theorem 1) is to show that such an algebra $A$ is **algebraic**, in the sense that if $x \in A$, then the subalgebra of $A$ generated by $x$ is finite-dimensional. The authors then invoke a theorem of A. A. Albert stating that a unital algebraic algebra is a finite-dimensional division algebra.
| 27 | https://mathoverflow.net/users/430 | 45663 | 28,929 |
https://mathoverflow.net/questions/45671 | 12 | Consider the Kernel $K\_n$ of the natural group homomorphism from the $n$-th [braid group](https://en.wikipedia.org/wiki/Braid_group) to the symmetric group. Then one can delete the $m$-th braid. This is a well defined homomorphism $d\_m:K\_n\rightarrow K\_{n-1}$. So is there for every $n\in \mathbb{N}$ a braid $1\neq b\in K\_n$ with $d\_m(b)=0$ for all $m$.
This is clearly true for $n=2$, as $K\_1$ is trivial and it is also true for $n=2$ (The "standard" braid does the job). What about higher $n$. Is there a nice construction, that works for every $n$ ?
| https://mathoverflow.net/users/3969 | Borromean braids | Certain elements in the $n-1$st term of the lower central series of the pure braid group should work.
The pure braid group is generated by generators $\beta\_{i,j}$ where the $i$th strand pushes a finger over the intervening strands and hooks with the $j$th strand.
Then, when $n=3$, the commutator $[\beta\_{1,2},\beta\_{1,3}]$ is Brunnian in in your sense. For $n=4$ you can consider $[\beta\_{1,2},[\beta\_{1,3},\beta\_{1,4}]]$, etc. You need to make sure your commutator includes every $\beta\_{1,k}$.
Something like $[\sigma\_1^2,[\sigma\_2^2,\sigma\_3^2]]$ would also work.
The reason this works is that deleting a strand from the braid kills at least one generator involved in the iterated commutator, so that it collapses to $1$. That's why you need to include a generator $\beta\_{i,j}$ that involves each strand.
(This has been edited to remove inaccuracies of previous versions.)
| 14 | https://mathoverflow.net/users/9417 | 45674 | 28,932 |
https://mathoverflow.net/questions/45678 | 5 | In a cryptography book I read that people does not known how to compute the number of points on a Jacobian of a hyperelliptic curve $C$ over a finite field $F\_q$? Is this true? It seems easy to compute it knowing the eigenvalues of the Frobenius action on $H^1(C)$, which could be recovered knowing $\sharp C(F\_{q^l})$ for all $l$ between $1$ and the genus of $C$.
| https://mathoverflow.net/users/2191 | A silly question: is the number of points on a Jacobian (of a curve, over a finite field) known? | Some algorithms working in polynomial time are available, but for high values of the genus the exponent is high and the implementation is difficult. A nice survey is the paper of Gaudry and Harley
"Counting Points on Hyperelliptic Curves over Finite Fields"
Lecture Notes in Computer Science, 2000, Volume 1838/2000, 313-332,
which also contains some explicit computations on Jacobians in the case of genus $2$.
| 8 | https://mathoverflow.net/users/7460 | 45682 | 28,937 |
https://mathoverflow.net/questions/45679 | 1 | Hello. This may not count as a research question, but I guess it's too much for math.stackexchange.
Could we define ZF (Zermelo-Fraenkel Set theory) in classical first-order predicate calculus, then define classical HOLs(Higher order logics) so that ZF can interpret it (via "inhabits" relation (sets)) and would we get that HOLs are interpretable in FOL?
Does that mean that HOLs do not have more expressive power than FOL in principle?
Thank you in advance.
| https://mathoverflow.net/users/6702 | FOL->ZF->HOL (Interpretation) | The interpretation of higher order logic is inherently set-theoretic, since the meaning of the second-order and higher order quantifiers depends on the set-theoretic background in which they are interpreted. Thus, in interpreting and analyzing higher order logic, we should do so in a set-theoretic context. It needn't be ZF or ZFC, but of course ZFC serves as a kind of default set-theoretic background theory for all mathematical enterprises, including the interpretation and analysis of higher order logic.
Meanwhile, there are several ways to set up the semantics for higher order logic. Considering second order logic, one can on the one hand interpret the scope of the second order quantifiers as running through *all* subsets of the model, or one can be explicit and insist that the model come equipped with a second-order part, an explicit family of sets that will be used for the second order quantifiers. The first method is merely an instance of the second, in which the model is equipped with the collection of all subsets. But the second method is somewhat more explicit and flexible, because it allows one to understand how the second order logic semantics are affected by various committments to which sets exist. For example, in second-order analysis, one can imagine including only arithmetically definable sets, or all hyperarithmetic sets, or projective sets, and so on, and these different second order models exhibit different second order natures.
But also, by explicitly listing the sets to be used for the second-order part, the second order logic becomes merely a multi-sorted first order logic---we now have the sets as objects in our structure. In this sense, the difference between higher order logic and first order logic disappears, and one can make the conclusion that any structure in higher order logic can be interpreted in a first order logic structure, simply by inclduing the higher order parts as actual objects.
| 3 | https://mathoverflow.net/users/1946 | 45686 | 28,939 |
https://mathoverflow.net/questions/45660 | 9 | Let $\mu$ be a *finite* Borel measure on $\mathbb{R}^n$. I am interested in how does $\mu(B(x,r))$ behave, where $B(x,r)$ is the open ball of radius $r$ centered at $x$. For instance, as far as I recall, for each $\alpha \in [0,n]$, there exists finite $\mu$ so that $\mu(B(x,r)) \sim r^{\alpha}$ for $\mu$-a.e. $x$, which are called dimensionally-regular measures (with constant local dimension $\alpha$).
Here is my question: does there exist *finite* Borel measure$\mu$ such that that $\mu(B(x,r))$ vanishes superpolynomially fast, say, $\sim e^{-\frac{1}{r}}$, i.e.,
$\mu\left(\left\{x: \liminf\_{r \to 0} r |\log \mu (B(x,r)) | > 0 \right\}\right) > 0$?
| https://mathoverflow.net/users/3736 | local behavior of a finite Borel measure | If I am not wrong, the answer is no (even for $\sigma$-finite measures). We can assume we are working with a finite measure $\mu$ in $[0,1]^d$.
Consider the sets $A\_k =${$ x \ : \ \liminf\_{r\to 0} |\log \mu(B(x,r))| r > 1/k$}. It is enough to show that the sets $A\_k$ have zero measure.
Now, $A\_k = \bigcap\_N \bigcup\_{n>N} B\_n$ where $B\_n =$ {$x \ : \ \mu(B(x,1/n))< e^{-n/k} $}.
To see this, notice that for every $x\in A\_k$, there exists $n\_0$ such that for every $n>n\_0$ we have that $1/n |\log \mu(B(x,1/n))| > 1/k$.
So, it is enough to show that $\sum\_{n>N} \mu(B\_n)\to 0$ as $N\to \infty$.
For every $n>0$ we can divide $[0,1]^d$ in $C n^d$ cubes whose interiors cover $[0,1]^d$ and such that any of them is has diameter smaller than, say, $1/(10n)$.
Now, $\mu(B\_n) \leq \sum\_{i\in I} \mu(Q\_i)$ where the $Q\_i$ are the cubes which intersect $B\_n$. Now, for each $Q\_i$ with $i\in I$ we have that
$$\mu(Q\_i) \leq e^{-n/k}$$
since it is contained in $B(x,1/n)$ of a point $x\in B\_n$.
So, $\mu(B\_n) \leq C n^d e^{-n/k} \to 0$ and this implies that $\sum\_{n>N} \mu(B\_n) \to 0$ as $N\to \infty$.
| 7 | https://mathoverflow.net/users/5753 | 45689 | 28,940 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.