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https://mathoverflow.net/questions/40796 | 22 | Here is my precise question. Let $M, \omega$ be a symplectic manifold and let $H: M \to \mathbb{R}$ be any smooth function. The symplectic form gives rise to an isomorphism between the tangent bundle and cotangent bundle of $M$, and in this way we can associate to the 1-form $dH$ a vector field $X\_H$ which is characterized by the property that $\omega(X\_H, Y) = Y(H)$ for any vector field $Y$. The one parameter group of diffeomorphisms associated to $X\_H$ is the "Hamiltonian flow" associated to $H$.
An interesting special case of this construction is furnished by Riemannian geometry. For any manifold $M$, there is a canonical symplectic structure on $T^\*M$ (regarded as a manifold in its own right) defined as follows. Given a tangent vector $X \in T(T^\*M)$ sitting over a covector $p \in T^\*M$, define $\eta\_p(X) = p(d\pi\_p(X))$ where $\pi: T^\*M \to M$ is the natural bundle projection. Then $\eta$ is a 1-form on $T^\*M$, and one checks that $d\eta$ is a symplectic form. If $M$ is equipped with a Riemannian metric $g$ then the metric yields an isomorphism between $TM$ and $T^\*M$, and the construction of the previous paragraph produces a Hamiltonian flow associated to any smooth function on $TM$. If we consider the smooth function $H: TM \to \mathbb{R}$ given by $H(V) = g(V, V)$, then it is a fact that the resulting Hamiltonial flow $F\_t$ is precisely the geodesic flow for $M$. In other words, given a tangent vector $W \in T\_p M$, $F\_t(W)$ is the velocity vector at time $t$ of the unique geodesic $\gamma$ with $\gamma(0) = p$, $\gamma'(0) = W$.
So I am wondering if there are interesting invariants - dynamical, geometric, topological, or otherwise - which help to determine whether or not a given Hamiltonian system is secretly the geodesic flow on some Riemannian manifold. This is kind of a screwy question from a geometric point of view, because it essentially asks if given a smooth function $H: M \to \mathbb{R}$ on a symplectic manifold there is a submanifold $N$ of $M$ such that there is a diffeomorphism $M \to TN$ which carries $H$ to a positive definite quadratic form on each fiber. But dynamically it boils down to a fairly natural question: how can one characterize geodesic flows among all Hamiltonian dynamical systems?
If this question has any sort of reasonable answer, I can think of half a dozen follow-up questions. Is there a natural notion of equivalence up to which $N$ is unique? To what extent does $H$ constrain the geometry and topology of $N$? If a Lie group acts on the pair $M, H$ can we choose $N$ which is invariant under the group action? For example, one idea along these lines that comes to mind immediately is the assertion that if the Hamiltonian flow for $H$ is not ergodic relative to a prescribed smooth invariant measure then $N$ cannot have nonpositive curvature. If you have an answer to this question and you can elaborate on the relationship between $H$ and the geometry of $N$, please do so.
| https://mathoverflow.net/users/4362 | When is the time evolution of a Hamiltonian system described by the geodesic flow on a Riemannian manifold? | My reading of the question is this: we're given $H\in C^\infty(M)$ with $M$ symplectic, and we want to know whether there's a submanifold $L\subset M$, a Riemannian metric $g$ on $L$, and a symplectomorphism $T^\ast L \cong M$ under which $H$ pulls back to the norm-square function. And we want to know if $(L,g)$ is unique.
Uniqueness is easy: we recover $L$ as $H^{-1}(0)$, and $g$ as the Hessian form of $H$ on the vertical tangent bundle (determined by the symplectomorphism) along $L$.
Basic necessary conditions:
(1) $L:=H^{-1}(0)$ is a Lagrangian submanifold of $M$.
(2) $L$ is a non-degenerate critical manifold of $H$ of normal Morse index 0.
These conditions imply that a neighbourhood of $L$ embeds symplectically in $T^\ast L$, and also (by the Morse-Bott lemma) that $H$ is quadratic in suitable coordinates near $L$. These two sets of coordinates needn't be compatible, so let's replace (2) by something much stronger (but still intrinsic):
(3) There's a complete, conformally symplectic vector field $X$ (i.e., $\mathcal{L}\_X\omega=\omega$), whose zero-set is exactly $L$, along which $H$ increases quadratically (i.e., $dH(X)=2H$).
I claim that (1) and (3) are sufficient. With these data, you can locate a point $x\in M$ in $T^\ast L$. Flow $X$ backwards in time starting at $x$ to obtain the projection to $L$; pay attention to the direction of approach to $L$ to get a tangent ray, and use the metric (i.e., the Hessian of $H$ on the fibres of projection to $L$) to convert it to a cotangent ray. Pick out a cotangent vector in this ray by examining $H(x)$. If I'm not mistaken, this will single out a symplectomorphism with the desired properties.
| 10 | https://mathoverflow.net/users/2356 | 40871 | 26,126 |
https://mathoverflow.net/questions/40857 | 8 | If we have a Serre fibration $p: E \rightarrow B$ with fiber of homotopy type $S^{k-1}$, then we can create a fibration with contractible fiber by first taking the mapping cylinder $M\_p$ of $p$ to get a map $M\_p \rightarrow B$ (not necessarily a fibration) with contractible "fiber" and then applying the "space of paths" construction to get a fibration $M\_p \times\_B B^I \rightarrow B$. My question is, is this last part of the construction necessary, or is the mapping cylinder $M\_p$ already a Serre fibration?
I tried lifting a homotopy $f\_t: X \times I \rightarrow B$ with starting point $\tilde f\_0: X \rightarrow M\_p$ by cutting $X$ into the closed preimage $C$ of $B \subset M\_p$ and the open preimage $U$ of $E \times [0,1) \subset M\_p$. On $C \times I$ we set $\tilde f\_t(x) = f\_t(x) \in B \subset M\_p$. On $U \times I$ we lift $f\_t|U: U \times I \rightarrow B$ to $g\_t: U \times I \rightarrow E$ and then set $\tilde f\_t(x) = (g\_t(x),(1-t)(t$ coordinate of $\tilde f\_0(x)) + t(1))$. This defines a continuous lift on $C$ and on $U$ separately. If the continuous lift on $U$ extends to the closure of $U$ then we're done. The map $U \rightarrow E$ could be nasty though near the boundary of $U$. Perhaps a better approach is to first construct a map from $X \times I$ that is only "close to" a lift, then use obstruction theory (I'm not an expert on this) to show that it is homotopic to some lift.
| https://mathoverflow.net/users/1874 | Is the mapping cylinder of a Serre fibration also a Serre fibration? | Waldhausen, Jahren and myself proved a
fiber gluing lemma for Serre fibrations,
in the context of simplicial sets, that
may be useful. In Propositions 2.7.10
and 2.7.12 of
"Spaces of PL manifolds and categories
of simple maps"
<http://folk.uio.no/rognes/papers/plmf.pdf>
we prove that given:
* a diagram of
simplicial sets $Z\_1 \leftarrowtail
Z\_0 \to Z\_2$, where one map is a
cofibration,
* a sufficiently
nice base simplicial set $B$ (a
simplicial complex will do), and
* compatible maps $Z\_i \to B$ that become
Serre fibrations upon geometric
realization,
then the pushout map
$Z\_1 \cup\_{Z\_0} Z\_2 \to B$ becomes
a Serre fibration upon geometric
realization.
Mapping cylinders are a special
case of pushouts. If $p \colon E \to B$
becomes a Serre fibration upon
realization, then so do
the obvious map $E \times \Delta^1
\to B$ and the identity map
$B \to B$. The pushout map is
your map $M\_p \to B$, and our
conclusion is that its realization
is a Serre fibration.
Our proof depends on working with
simplicial sets. The technical
condition on $B$ is that each
nondegenerate simplex is embedded.
* John
| 9 | https://mathoverflow.net/users/9684 | 40873 | 26,128 |
https://mathoverflow.net/questions/39567 | 6 | Suppose we have an odd composite $N$ and want to find numbers $a\_1,\ldots,a\_k$ such that each $a\_i^2$, reduced mod $N$, is $b$-smooth. Of course we can use the quadratic sieve algorithm (minus the matrix step) to find such $a\_i$. With a factorization oracle, they could be found directly by factoring small squares—the quadratic sieve without the "sieve" part. (I assume there's no faster way in this case than generating [presumably sequentially, to avoid actual divisions] and testing?)
1. If instead of a general factoring oracle we had access to an oracle for the factors of N, could we improve on the speed of the quadratic sieve?
2. Harder: If we had only a partial factorization of $N$ (say, of size $N^{2/5}$), could we improve on the speed?
Algorithms, heuristics, and reductions to known hard problems would be welcome. You may assume that $k$ and $b$ are reasonable: there are $\gg k$ solutions.
| https://mathoverflow.net/users/6043 | Speeding the quadratic sieve with an oracle | If we know the factorization of $N$ then we can take square roots of small numbers that are quadratic residues mod all primes dividing $N$. Knowing a partial factorization of size $M\sim N^\alpha$, square a number close to $N^{1/2−\alpha/2}$ mod $N/M$, find a root mod $M$, then use crt to find the root mod $N$. We have a found a square that is of size $N^{1/2−\alpha/2}$ instead of $N^{1/2}$.
| 2 | https://mathoverflow.net/users/2024 | 40881 | 26,133 |
https://mathoverflow.net/questions/40882 | 1 | We know that Morrey's inequality says $W^{1,p} \subset C^{0,\gamma}$ for $\gamma = 1 - n/p$ where $n$ is the dimension. However, in 1D, following the proof of Evans "Partial Differential Equations" (first edition, pp. 267), we have for any function $f$ in $W^{1,p}(0,1)$, and $x, y \in (0,1)$
$u(x) - u(y) = \int\_0^1 \frac{d}{dt}u(tx + (1-t)y) dt = \int\_y^x Du(s) (x-y) ds$.
Take absolute value on both sides and use the basic inequalities, we have
$|u(x)-u(y)| \le |Du|\_{L^p(0,1)} |x-y|$.
We obtain something more than Morrey's inequality would indicate.
Is there anything wrong?
| https://mathoverflow.net/users/9754 | In 1D, is a $W^{1,p}$ function always Lipschitz, for $p\ge1$? | This does not look right. You have $u(x)-u(y) = \int\_y^x Du(s)(x-y)ds$ but the correct expression is clearly $\int\_y^x Du(s)ds$. I think a change of variables went wrong somewhere.
A counterexample: take $u(x) = x^{3/4}$ on $(0,1)$, with $p=2$. Then $Du(x) = \frac{3}{4}x^{-1/4} \in L^2(0,1)$, so $u \in W^{1,2}(0,1)$, but $u$ is not Lipschitz.
| 5 | https://mathoverflow.net/users/4832 | 40884 | 26,135 |
https://mathoverflow.net/questions/40828 | 2 | Let $X$ be a nonsingular complex projective variety. Suppose $X$ is embedded as a nonsingular real subvariety of complex projective space ${\mathbb{CP}}^n$.
When can we embed ${\mathbb{CP}}^n$ in some larger complex projective space ${\mathbb{CP}}^N$ such that the image of $X$ is now a nonsingular *complex* subvariety of this larger ${\mathbb{CP}}^n$?
| https://mathoverflow.net/users/nan | Obstruction for real subvariety to be embedded as complex subvariety | I am assuming from your comment that you are demanding that the embedding $X \to \mathbb{CP}^N$ preserves the complex structure that $X$ originally had (although it is still not completely clear). In this case, the real embedding $X \to \mathbb{CP}^n$ has to be a complex embedding. Otherwise, the actions of multiplication by $i$ on the tangent spaces of $X$ and $\mathbb{CP}^N$ will be incompatible.
For example, you can take $X = \mathbb{CP}^1$, with complex conjugation as the real embedding into $\mathbb{CP}^1$. This is a real-algebraic isomorphism, but orientation-reversing. Then, you will never have a compatible way to embed both varieties as complex subvarieties in $\mathbb{CP}^N$ by complex algebraic maps.
| 1 | https://mathoverflow.net/users/121 | 40909 | 26,150 |
https://mathoverflow.net/questions/40893 | 3 | By Tychonoff Theorem $\prod\_{\mathbb R} [0,1]$ is compact and since $\mathbb R=2^{\omega}$, if for $\alpha \in 2^{\omega}$, $x\_n(\alpha)=\alpha(n)$ then if we consider a subsequence $x\_{n\_0}, x\_{n\_1}, x\_{n\_2},...$ where $ n\_0 < n\_1< n\_2<... $ and where for $\alpha \in 2^{\omega}, \alpha(n\_0)=1, \alpha(n\_1)=0, \alpha(n\_2)=1,...$, then $(\alpha(n\_0), \alpha(n\_1), \alpha(n\_2),...)=(0,1,0,...)$ does not converge. This is of course an example of a compact space which not sequentially compact.
But is it possible to do it without the axiom of choice? I would like to show an explicit example of a sequence with an explicit limit point that does not have a convergent subsequence in $\prod\_{\mathbb R} [0,1]$. Thx.
| https://mathoverflow.net/users/3859 | A sequence with no convergent subsequence without choice | I think this question is fairly clear now, in its edited form. Assuming AC,
$[0,1]^{\mathbb R}$ is not sequentially compact, witnessed by the sequence specified in the question. What happens without AC? Is there a sequence with a limit point that has no convergent subsequence? I think this is an interesting question.
I cannot give a complete answer to your question right now, but I hope the following clarifies things.
For simplicity, I take the exponent to be $2^\omega$ instead of $\mathbb R$ (the two sets have the same size even without choice). Take the sequence $(x\_n)$ in $[0,1]^{2^\omega}$
defined by $x\_n(y)=y(n)$ for $n\in\omega$ and $y\in 2^\omega$.
Now, if $z\in[0,1]^{2^\omega}$ is a limit point of the sequence, then
the collection of all sets of the form $\{n\in\omega:x\_n\in U\}$, $U$ an open neighborhood of $z$, generates a free ultrafilter on $\omega$.
The non-existence of a free ultrafilter on $\omega$ is consistent with ZF.
Hence, the sequence $(x\_n)$ has no convergent subsequence (we don't need AC for this)
but it might also not have limit point.
I don't know whether you can construct without AC a sequence in $[0,1]^{2^\omega}$
that has no convergent subsequence but a limit point. My guess would be no, and an argument
might go like this: Let $(x\_n)$ be a sequence without convergent subsequence.
No projection of the sequence to a countable set of coordinates can have this property.
Figure out a way to show (without choice) that the sequence behaves essentially as the
particular sequence that we have defined above, i.e., from a limit point we can extract a nontrivial ultrafilter.
---
Just one more remark on the Tychonov theorem: Tychonov for Hausdorff spaces is not equivalent to full AC. You need to consider spaces that are not Hausdorff to get an equivalent of AC. The argument above shows that from the compactness of $[0,1]^{2^\omega}$ one gets a free ultrafilter on $\omega$, whose existence cannot be proved without AC.
Hence we cannot prove the compactness of this particular product of spaces without AC
(same with $[0,1]$ replaced by $\{0,1\}$).
| 4 | https://mathoverflow.net/users/7743 | 40915 | 26,155 |
https://mathoverflow.net/questions/40807 | 0 | For a surface $z(x,y)$, let $p = z\_x$ and $q = z\_y$. I have a pair of coupled semilinear PDEs in p and q.
PDE1: $p\_x - a p\_y = b q$.
PDE2: $q\_x - a q\_y = -b p$.
Note that $a$ and $b$ are functions of $(x,y)$. Is there a general way to solve such coupled systems of semilinear PDEs? (We may or may not assume any relationship between $p$ and $q$. Since they come from a surface, we know $p\_y = q\_x$, but I'll be happy if the coupled system can be solved even without that integrability constraint.)
Thanks!
| https://mathoverflow.net/users/9728 | Coupled semilinear PDEs | First, disregard the constraint $p\_y=q\_x$. Consider the integral curves of the vector field $V:=(1,-a)$, namely the solutions of the ODE
$$\frac{dy}{dx}=-a(x,y).$$
They are parametrized by $x$. Your equations are ODEs along these curves,
$$\dot p=bq,\qquad\dot q=-bp.$$
Set $q+ip=\rho e^{i\theta}$. Then the ODEs become $\dot\rho=0$, $\dot\theta=b$. You see that you need an initial data over a curve transversal to $V$, for instance along a curve $x={\rm cst}$.
If we now take in account the constraint, the system becomes *over-determined*. Such systems usually do not have a solution, unless the data (here $a,b$ and the initial data) are very special. To see how it begins, let us set $h:=p-aq$. Then we have $h\_x=(b-a\_x)q$ and $h\_y=-bp-a\_yq$. By Schwarz, we obtain a new equation
$$(a\_xq-bq)\_y=(a\_yq+bp)\_x.$$
This is a fourth differential equation. With the three others, you may solve
$$(p\_x,p\_y,q\_x,q\_y)=F(p,q,x,y),$$
unless a $4\times4$ determinant vanishes. And the tale does not end there. You can continue and find other first-order PDEs, which yield incompatible in just one more step. Again, unless you are very lucky.
**edit after a few hours.** Once you have the first derivatives in terms of $p$ and $q$, you may apply Schwarz, $(p\_x)\_y=(p\_y)\_x$ and $(q\_x)\_y=(q\_y)\_x$. This gives $\partial\_yF\_1(p,q,x,y)=\partial\_xF\_2(p,q,x,y)$ and $\partial\_yF\_3(p,q,x,y)=\partial\_xF\_4(p,q,x,y)$. Eliminating first derivatives, there remain two equations $p=P(x,y)$ and $q=Q(x,y)$. Remark that you may not any more impose initial data, because you obtain explicit form of $p,q$ without really solving a differential equation. You have only use elimination and Schwarz identity. Finally, it happens in general that $P$ and $Q$ do not solve at all the differential equation.
In theory, you could explicit a necessary and sufficient condition in terms of $a$ and $b$ in order that your overdetermined system admit a solution. A very classical and simple situation is the system $u\_x=f$, $u\_y=g$, where the condition for having a solution is $f\_y=g\_x$.
| 3 | https://mathoverflow.net/users/8799 | 40916 | 26,156 |
https://mathoverflow.net/questions/40895 | 7 | The question is about the specific case of reflection theorems (copied straight from Franz Lemmermeyer's "Class Groups of Dihedral Extensions"):
>
> Let $k^+ = \mathbb{Q}(\sqrt{m})$ with $m\in \mathbb{N}$, and put $k^- = \mathbb{Q}(\sqrt{-3m})$; then the 3-ranks $r\_3^+$ and $r\_3^-$ of $Cl(k^+)$ and $Cl(k^-)$ satisfy the inequalities $r\_3^+ \le r\_3^- \le r\_3^+ + 1$.
>
>
>
The proofs I have seen either use p-adic arguments or galois actions.
### Question
Is there an explicit surjective map from $Cl(k^-)[3]$ to $Cl(k^+)[3]$ that might, as the theorem suggests, have kernel of size 3?
At the least, an algorithm for such a map?
| https://mathoverflow.net/users/2024 | Explicit map for Scholz reflection principle | One way to think the reflection principle, which is similar to what you are proposing in your question, is a relation between the index 3 subgroups of $Cl(k^{+})$, which I'll call $I\_{3}(m)$, and the subgroups of $Cl(k^{-})$ order 3 which I'll call $S\_{3}(-3m)$. It is not difficult to see that $$|S\_{3}(-3m)|=\frac{3^{r\_{3}^{-}}-1}{2}$$ and that $$|I\_{3}(m)|=\frac{3^{r\_{3}^{+}}-1}{2},$$ hence any injective map $$ \Phi\_{m}: I\_{3}(m) \rightarrow S\_{3}(-3m)$$ would yield to $r\_{3}^{+}\leq r\_{3}^{-}$. It is a result of Hasse that the set $I\_{3}(m)$ is in bijection with the isomorphism classes of cubic fields of discriminant $m$ (notice that here I'm assuming that $m$ is fundamental, i.e., $m=disc(k^{+})$) hence what we are looking is for a map $\Phi$ that takes a cubic field $K$ and produces a subgroup of $Cl(k^{-})$ of order $3$. In other words given a cubic field $K$ of discriminant $m$ we need to associate a primitive, binary quadratic form of discriminant $-3m$ with the extra condition that the form has order 3 under Gauss composition. To shorten the exposition I'll assume $(3,m)=1$ however all that I'm saying can be worked out in full generality. One natural way to define $\Phi\_{m}$ is as follows: Let $O\_{K}^{0}$ be the set of integral elements in $K$ with zero trace. Let $q\_{K}(x):=Tr(x^{2})/2$. Then, one can show that $q\_{K}(x)$ is a primitive, binary quadratic form of discriminant $-3d$. Moreover, as an element of the class group $q\_{K}^{2}$ has order $3$. It is possible to show that the map $\Phi\_{m}$ sending $K$ to the group generated by $q\_{K}^{2}$ is injective, so the result follows.
All the above results should be appearing at some point soon in ANT, but I can email you a copy of the article if you are curious of the details.
**Added:** In response to Alex comment I should say that the other inequality can be also derived with the same ideas I explained above. Now, you start with $I\_{3}(-3m)$ and you notice that $$|I\_{3}(-3m)|=\frac{3^{r\_{3}^{-}}-1}{2}$$ Moreover, $S\_{3}(-3(-3m))=S\_{3}(m)$ hence by using the trace you get a map $$ \Phi\_{-3m}: I\_{3}(-3m) \rightarrow S\_{3}(m).$$ The diference here is that this map is not injective, but it can be shown that roughly the map is 3-to-1 hence the other inequality. So summarizing Scholz reflection principle is a relation between index 3 subgroups in one class group and subgroups of order 3 in the other, and one way to make this relation explicit is via the trace form.
One place to see where the difference in the behavior $\Phi\_{m}$ and $\Phi\_{-3m}$ is, as professor Lemmermeyer already pointed out, Bhargava's first paper on Higher composition laws, more specifically Corollary 15. Another place to look at this is J. W. Hoffman and J. Morales, *Arithmetic of binary cubic
forms*, Enseign. Math. (2) **46**, 2000, 61-94.
| 8 | https://mathoverflow.net/users/2089 | 40917 | 26,157 |
https://mathoverflow.net/questions/40922 | 0 | I want to maximize $||x-y||$ with $x$ and $y$ in $C$ where $C$ is the intersection of some discs. We assume the intersection is nonempty, and closed. I am thinking, how to formulate it as a semidefinite programmimg problem? Does anyone know how?
| https://mathoverflow.net/users/9557 | max length or size of a convex set | I hope that the word "disc" indicates that we work in 2-dimensional Euclidean space.
Then an intersection of discs would be something like a polygon, just that the sides are not straight lines but arcs of a circle. It should be possible to save the intersection in some data structure. And then one can compute the maximum distance from each arc to each other arc and take the maximum value.
To compute the maximum distance between to given Arcs $A\_1$ and $A\_2$ let $x\_i\in A\_i$ be two points such that $d(x\_1,x\_2)=\sup\{d(x,y)|x\in A\_1,y\in A\_2\}$. (Existence follows from compactness). If you take a unit speed parametrization $\gamma\_1,\gamma\_2$ of the arcs, and times $t\_i$ with $\gamma\_i(t\_i)=x\_i$. Assuming, that $x\_1$ is not the endpoint of the arc $A\_1$, then the function $\mathbb{R}\rightarrow \mathbb{R} \qquad t\mapsto d(\gamma\_1(t),x\_2)$ obtains a maximum at $t$, so its differential must be $0$ at $t\_1$. But this is just the scalar product of $\gamma'\_1(t\_1)$ and the gradient of $d(-,x\_2)$ at $\gamma(t)$, which can be seen as the direction of the line connecting $x\_1$ and $x\_2$.
This should make it possible to compute the distance by looking at finitely many cases. (Both endpoints, one inner point, two inner points).
Hope I understood the question right.
| 1 | https://mathoverflow.net/users/3969 | 40925 | 26,159 |
https://mathoverflow.net/questions/40920 | 99 | The title of the question is also the title of a talk by Vladimir Voevodsky, available [here](http://video.ias.edu/voevodsky-80th).
Had this kind of opinion been expressed before?
**EDIT.** Thanks to all answerers, commentators, voters, and viewers! --- Here are three more links:
[Question arising from Voevodsky's talk on inconsistency](https://mathoverflow.net/questions/41217) by [John Stillwell](https://mathoverflow.net/users/1587/john-stillwell),
[Nelson's program to show inconsistency of ZF](https://mathoverflow.net/questions/36693), by [Andreas Thom](https://mathoverflow.net/users/8176),
[Pierre Colmez, La logique c’est pas logique !](http://images.math.cnrs.fr/La-logique-c-est-pas-logique.html)
**EDIT.** [Here](http://www.cs.nyu.edu/pipermail/fom/2011-May/thread.html "link") the link to the FOM list discussing these themes.
| https://mathoverflow.net/users/461 | What if Current Foundations of Mathematics are Inconsistent? | The talk in question was given as part of a celebration (this past weekend) of the 80th anniversary of the founding of the Institute for Advanced Study in Princeton. As you might guess there were quite a few very well-known mathematicians and physicists in the audience. (To name just a few, Jack Milnor, Jean Bourgain, Robert Langlands, Frank Wilczek, and Freeman Dyson, all of whom also spoke during the weekend.) The talk was a gem, and what did come as a surprise, at least to me, was that towards the end of his talk Voevodsky let on that he hoped that someone did find an inconsistency---and that by that time there was no audible gasp from the audience. There was of course a very lively discussion after the talk, and nobody seemed willing to say they felt that the "Current Foundations" (whatever they are) are definitely consistent. Of course Voevodsky was NOT saying that he felt that the body of theorems making up the "classic mathematics" that we normally deal with might be inconsistent, that is quite a different matter. What we should keep in mind is that a hundred years ago an earlier generation of mathematicians were quite surprised by not one but several "antinomies", like Russell's Paradox, The Burali-Forti Paradox, etc., (and that was followed by the greatest century in the history of Mathematics). As to the question "Had this kind of opinion been expressed before?", yes of course it has, but perhaps not so forcefully or in such a high-level forum. One person who has been expressing such ideas in recent years is my old friend Ed Nelson, who was also in the audience. (You can see his ideas in a recent paper: <http://www.math.princeton.edu/~nelson/papers/warn.pdf>). I spoke with him after the talk and he seemed pleased that it was now becoming acceptable to discuss the matter seriously.
| 81 | https://mathoverflow.net/users/7311 | 40927 | 26,160 |
https://mathoverflow.net/questions/40918 | 3 | I heard it said that the cohomology rings of some Lie groups and Grassmannians can be read from the Dynkin graph. Can someone give me any reference?
| https://mathoverflow.net/users/8152 | the relation between cohomology and Dynkin graphs of lie groups | Your question involves a complex (semi)simple Lie group, I guess, and its Dynkin diagram. The topology of Lie groups and their homogeneous spaces $G/P$ (such as Grassmannians) is an old and rich subject (E. Cartan, Stiefel, Samelson, Bott, Kostant, Chevalley, Borel, ...) which can be approached from a number of viewpoints such as deRham cohomology and Morse theory. Most basic is the reduction of the problem to determination of the topology of a compact real form. Cohomology rings turn out to be intimately related to the invariant theory of Weyl groups and to the alcove geometry of associated affine Weyl groups (Stiefel diagram). Some of this is encoded in the Dynkin diagram, but working out for example the coinvariant algebra of the Weyl group takes further work.
I'm not a specialist in this area, but am aware of numerous textbook treatments of compact Lie groups: structure, representations, cohomology. Research papers and surveys by some of those mentioned above are often useful. A couple of older examples:
MR0064056 (16,219b),
Armand Borel,
Sur l’homologie et la cohomologie des groupes de Lie compacts connexes. (French)
Amer. J. Math. 76 (1954), 273–342.
MR0072426 (17,282b),
Armand Borel,
Topology of Lie groups and characteristic classes.
Bull. Amer. Math. Soc. 61 (1955), 397–432.
One of Bott's surveys *The geometry and representation theory of compact Lie groups* (pp. 65–90) can be found in the volume:
MR568880 (81j:22001),
Representation theory of Lie groups.
Proceedings of the SRC/LMS Research Symposium held in Oxford, June 28–July 15, 1977.
London Mathematical Society Lecture Note Series, 34.
Cambridge University Press, Cambridge-New York, 1979.
Newer treatments exist, of course, but the most important results are fairly old
and hard to improve on. The answer to your question depends a lot on what you already know and what sources are accessible, such as the lecture notes by Hiller explaining Borel's approach to cohomology:
MR649068 (83h:14045)
Howard Hiller,
Geometry of Coxeter groups.
Research Notes in Mathematics, 54.
Pitman (Advanced Publishing Program), Boston, Mass.-London, 1982.
| 6 | https://mathoverflow.net/users/4231 | 40929 | 26,162 |
https://mathoverflow.net/questions/40934 | 1 | Hello,
this may be a trivial question, but I am not very familiar with the topic.
Let (M,g) be a Riemannian Manifold. (In fact, we don't need the metric here.)
What exactly does it take for two k-submanifolds $S$ and $S'$ to lie in the same homology class? And why does that imply $~ \int\_S ~ \omega = \int\_{S'} ~ \omega ~$, where $\omega$ is a closed differential form on M.
This must be somehow a consequence of Stokes' Theorem. I would be pleased with an answer for Dummies. :)
$~$
Greetings,
Henry
| https://mathoverflow.net/users/9762 | When do submanifolds lie in the same homology class? | By triangulating each manifold $S$ and $S'$, you can regard them as sums of singular simplices. (A singular simplex is a continuous map $f\colon \Delta^n\to M$.) So the very simple answer is that $S$ and $S'$ are the same homology class if and only if they are the boundary of some sum of singular simplices. This can happen if, for example, there is a map of an oriented manifold of one higher dimension into $M$ whose boundary is the disjoint union $S\cup -S'$. If all the singular simplices are smooth maps, then Stokes theorem indeed tells you that the integral along the boundary of each singular simplex is zero, which then implies that the integral $\int\_{S\cup-S'}\omega=0$, or $\int\_S\omega=\int\_{S'}\omega$. Someone else can probably comment on why you can pass from the topological to the smooth category.
| 5 | https://mathoverflow.net/users/9417 | 40937 | 26,164 |
https://mathoverflow.net/questions/40944 | 5 | Let $G\_0$ be a finitely generated group, and suppose there are groups $G\_i$ and $K\_i$ as in the following short exact sequences
$1\to K\_i\to G\_{i+1}\to G\_i\to 1$
with $K\_i$ free and nonabelian (you may assume finitely generated), and $G\_i$ commutative transitive. (If $a$ is nontrivial and $b$ and $c$ both commute with $a$, then $b$ and $c$ commute.) Does it follow that $\mathrm{rank}(G\_i)\to\infty$ as $i\to\infty$? Are there examples of extensions of this sort where the rank doesn't increase?
| https://mathoverflow.net/users/2225 | Ranks of iterated extensions of a group by free groups. | I assume that you consider the infinite cyclic group to be free. Then take the free nilpotent group $G\_n$ of class $c\gg 1$ with 2 generators. It has an infinite cyclic central subgroup $K\_n$, the factor-group $G\_{n-1}=G\_n/K\_n$ is nilpotent and torsion-free, it has an infinite cyclic subgroup $K\_{n-1}$, and so on. Every group $G\_i$ is 2-generated, the chain can be arbitrary long ($n$ depends on $c$).
Edit 1: Since you do not want to consider $\mathbb Z$ free enough, here is another example. Take the Baumslag-Solitar group $BS(2,3)=\langle a,t | ta^2t^{-1}=a^3\rangle$. It is non-Hopfian, and has a free normal subgroup $K$ such that $BS(2,3)/K$ is isomorphic to $BS(2,3)$. So all $G\_i$ are isomorphic to $BS(2,3)$, all $K\_i$ are infinitely generated free groups. Is that what you want?
Edit 2: Since you have another condition now, "commutative transitive", then here is another example. Take a non-elementary torsion-free hyperbolic group $G$. It has a free normal subgroup $N$ such that $G/N$ is still non-elementary torsion-free and hyperbolic (that is proved by Olshanskii, and also by several others, including Delzant). You can continue as long as you wish. All $G\_i$ will be hyperbolic and torsion-free (hence commutative-transitive), all $K\_i$ will be infinitely generated free groups.
Just to anticipate a future change in the formulation of the question: if you really insist that $K\_i$ are finitely generated, the question becomes harder and I am not sure the answer is still the same.
Edit 3: Since you want to have an infinite sequence, here is what to do. For every hyperbolic group $G$ with 2 generators, there exists another 2-generated hyperbolic group $G'$ and a free normal subgroup $N \le G'$ such that $G'/N=G$. This can be done using Rips' construction. In the original Rips' construction (and in all modifications) $N$ was not free, because he wanted $N$ to be finitely generated. But if you do not want $N$ to be finitely generated, it is easy to modify Rips' construction to make $N$ free. Using this you can construct your sequence $G\_0=G, G\_1=G', G\_2=G\_1', \ldots$.
Edit 4:In fact Rips' construction does not quite work because the number of generators increases. Certainly if $G$ is free of rank 2, $G'$ cannot be of rank 2. But here is another construction. Take a (torsion-free) lacunary hyperbolic group given by an infinite presentation satisfying a small cancellation condition (see <https://arxiv.org/abs/math/0701365>). Let $r\_1,r\_2,...$ be the presentation of $G$. Then $G$ is commutative transitive (it is easily deduced from the fact that $G$ is an inductive limit of hyperbolic groups and surjective homomorphisms). Now the group $G'$ given by the same presentation but without $r\_1$ is again lacunary hyperbolic, $G$ is a factor-group of $G'$ over the normal subgroup $N$ generated by $r\_1$. It is possible to prove that $N$ is free. Indeed, if some product of conjugates of $r\_1$ is equal to 1 in $G$, consider the corresponding van Kampen diagram. The boundary of that diagram has parts labeled by $r\_1$ and parts labeled by the conjugators. By Greendlinger lemma, if the diagram has cells, it must have a cell with more than, say, $90\%$ of its boundary common with the boundary of the diagram (take the small cancelation condition $C'(1/300)$). Then more than a half of that part of the boundary must be inside a conjugator, the conjugator can be shortened, and a shorter product of conjugates of $r\_1$ is equal to 1 in $G'$. Since $G'$ is lacunary hyperbolic again and satisfies the same small cancellation condition as $G$, we can repeat the construction. Since the presentation is infinite, the process will continue indefinitely.
Edit 5: A more clean way to prove that $N$ is free in Edit 4 is the following. suppose that some product of conjugates of $r\_1$ is equal to 1 in $G'$. Consider the corresponding van Kampen diagram $\Delta$. Its boundary label is equal to 1 in the group given by 1 relator $r\_1$. Consider the diagram $\Psi$ corresponding to that equality. Now identify the boundaries of $\Psi$ and $\Delta$. We get a diagram over the presentation of $G$ on a sphere: $\Delta$ occupies the northern hemisphere, $\Psi$ occupies the southern hemisphere, and the product of conjugates of $r\_1$ labels the equator. Reduce that diagram. Since we can assume that $\Psi$ is reduced, and the $r\_1$-cells are only in the south, the $r\_1$-cells won't cancel. Hence we shall have a reduced non-empty diagram over the presentation of $G$ on a sphere which is impossible because of the Greendlinger lemma (the boundary of the spherical diagram is empty).
| 11 | https://mathoverflow.net/users/nan | 40946 | 26,169 |
https://mathoverflow.net/questions/23981 | 1 | This question is related to [On a positivity of a matrix with trace entries.](https://mathoverflow.net/questions/16983/on-a-positivity-of-a-matrix-with-trace-entries)
Let $A\_1, \cdots, A\_m$ be strictly contractive $n\times n$ complex matrices .Is it true that
$$\left(\begin{array}{cccc}Tr\{(I-A\_1^\*A\_1)^{-1}\}&Tr\{(I-A\_1^\*A\_2)^{-1}\}&\cdots &Tr\{(I-A\_1^\*A\_m)^{-1}\}\\Tr\{(I-A\_2^\*A\_1)^{-1}\}&Tr\{(I-A\_2^\*A\_2)^{-1}\}&\cdots &Tr\{(I-A\_2^\*A\_m)^{-1}\}\\ \cdots&\cdots&\cdots&\cdots\\Tr\{(I-A\_m^\*A\_1)^{-1}\}&Tr\{(I-A\_m^\*A\_2)^{-1}\}&\cdots &Tr\{(I-A\_m^\*A\_m)^{-1}\}
\end{array}\right)$$
is positive semidefinite.
| https://mathoverflow.net/users/3818 | A matrix with trace entries. | I guess, in the meanwhile you might have already proved that this matrix is **not** positive-semidefinite. I ran a *brute force* experiment, using $2 \times 2$ symmetric, real matrices, which shows that the above conjecture is not true.
I tried different values of $m$, and indeed, the smaller the $m$, the lower the (empirical) probability for a set of random (e.g., uniform), symmetric, real matrices to yield a counterexample. Here is an explicit example with $m=5$, where each $\|A\_i\|<1$:
$$A\_1= \begin{pmatrix}
0.68 &0.21\\\\ 0.21 &0.84
\end{pmatrix}$$
$$A\_2= \begin{pmatrix}
0.58 &0.31\\\\
0.31 &0.74
\end{pmatrix}
$$
$$A\_3=\begin{pmatrix}
0.20 &0.56\\\\
0.56 &0.58
\end{pmatrix}$$
$$A\_4=\begin{pmatrix}
0.31 &0.39\\\\
0.39 &0.75
\end{pmatrix}$$
$$A\_5=\begin{pmatrix}
0.42 &0.34\\\\
0.34 &0.77
\end{pmatrix}$$
The corresponding matrix $M$ with entries $m\_{ij}=\text{trace}((I-A\_iA\_j)^{-1})$, has the following eigenvalues:
(127.8507, 7.4835, **-0.3282**, 0.3286, 0.9082)
| 4 | https://mathoverflow.net/users/8430 | 40956 | 26,177 |
https://mathoverflow.net/questions/40960 | 9 | Let $X$ be an integral scheme of finite type over a field. Then there is a surjective finite map $\tilde{X} \to X$ from the normalization $\tilde{X}$ of $X$.
Is this going to be bijective?
In the simplest non-normal case, namely the spectrum of $k[x^2, x^3] = k[t,u]/(t^3 -u^2)$, the map is bijective, because the curve is geometrically just a cubic curve (the set of all $(v^2, v^3)$ in affine 2-space, geometrically). I can't find it asserted anywhere that the map is necessarily bijective, though.
| https://mathoverflow.net/users/344 | Is the normalization map bijective? | Your example is not really the simplest case, in that a cusp is not the simplest possible curve singularity. Rather, the simplest curve singularity is a node, e.g. as in
$y^2 = x^3 - x^2$. There are then two branches passing through the singularity, and
the normalization map separates them. (Exercise: The normalization is again given by a map
$t \mapsto x(t),y(t)$; find this map!)
This is the typical phenomenon with normalization: it separates the different branches passing through a singularity. The situation of having just a single branch (and hence having a bijective normalization map) is somewhat unusual (and has a special adjective to describe it: *unibranch*).
| 20 | https://mathoverflow.net/users/2874 | 40963 | 26,181 |
https://mathoverflow.net/questions/40945 | 78 | This question is inspired by [an answer of Tim Porter](https://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/40010#40010).
Ronnie Brown pioneered a framework for homotopy theory in which one may consider multiple basepoints. These ideas are accessibly presented in his book Topology and Groupoids. The idea of the fundamental groupoid, put forward as a multi-basepoint alternative to the fundamental group, is the highlight of the theory. The headline result seems to be that the van-Kampen Theorem looks more natural in the groupoid context.
I don't know whether I find this headline result compelling- the extra baggage of groupoids and pushouts makes me question whether the payoff is worth the effort, all the more so because I am a geometric topology person, rather than a homotopy theorist.
>
> Do you have examples in geometric topology (3-manifolds, 4-manifolds, tangles, braids, knots and links...) where the concept of the fundamental groupoid has been useful, in the sense that it has led to new theorems or to substantially simplified treatment of known topics?
>
One place that I can imagine (but, for lack of evidence, only imagine) that fundamental groupoids might be useful (at least to simplify exposition) is in knot theory, where we're constantly switching between (at least) three different "natural" choices of basepoint- on the knot itself, on the boundary of a tubular neighbourhood, and in the knot complement. This change-of-basepoint adds a nasty bit of technical complexity which I have struggled with when writing papers. A recent proof (Proposition 8 of [my paper with Kricker](http://www.msp.warwick.ac.uk/agt/2009/09-03/p049.xhtml)) which would have been a few lines if we hadn't had to worry about basepoints, became 3 pages. In another direction, what about fundamental groupoids of braids?
Have the ideas of fundamental groupoids been explored in geometric topological contexts? Conversely, if not, then why not?
| https://mathoverflow.net/users/2051 | Compelling evidence that two basepoints are better than one | Here is an interesting example where groupoids are useful. The mapping class group $\Gamma\_{g,n}$ is the group of isotopy classes of orientation preserving diffeomorphisms of a surface of genus $g$ with $n$ distinct marked points (labelled 1 through n). The classifying space $B\Gamma\_{g,n}$ is rational homology equivalent to the (coarse) moduli space $\mathcal{M}\_{g,n}$ of complex curves of genus $g$ with $n$ marked points (and if you are willing to talk about the moduli orbifold or stack, then it is actually a homotopy equivalence)
The symmetric group $\Sigma\_n$ acts on $\mathcal{M}\_{g,n}$ by permuting the labels of the marked points.
**Question:** How do we describe the corresponding action of the symmetric group on the classifying space $B\Gamma\_{g,n}$?
It is possible to see $\Sigma\_n$ as acting by outer automorphisms on the mapping class group. I suppose that one could probably build an action on the classifying space directly from this, but here is a much nicer way to handle the problem.
The group $\Gamma\_{g,n}$ can be identified with the orbifold fundamental group of the moduli space. Let's replace it with a fundamental groupoid. Fix a surface $S$ with $n$ distinguished points, and take the groupoid where objects are labellings of the distinguished points by 1 through n, and morphisms are isotopy classes of diffeomorphisms that respect the labellings (i.e., sending the point labelled $i$ in the first labelling to the point labelled $i$ in the second labelling).
Clearly this groupoid is equivalent to the original mapping class group, so its classifying space is homotopy equivalent. But now we have an honest action of the symmetric group by permuting the labels on the distinguished points of $S$.
| 36 | https://mathoverflow.net/users/4910 | 40965 | 26,183 |
https://mathoverflow.net/questions/40887 | 9 | This question - so far as I know - has no broader mathematical significance, but it occurred to me a while ago and I haven't been able to make any headway.
Any knot diagram $D$ splits the plane into a finite number of pieces. For example, in a standard diagram for a trefoil (e.g., <http://en.wikipedia.org/wiki/File:Trefoil_knot_left.svg>), the plane is split into five pieces: one lying "outside the diagram," three "lobes" of the knot, and a central region around which the three "lobes" are arrayed. The outer region has three crossings on its boundary, as does the inner region; each of the lobes only have two crossings each on their boundaries.
A diagram for a more complicated knot may split the plane into many more regions, and one of these regions may have many crossings on its boundary. My question is: Does there exist some $n$ such that for any knot $K$, there is a diagram $D$ of $K$ such that any region created by $D$ has at most $n$ crossings on its boundary?
If the answer is no, a (very soft) question that then arises is: given a knot $K$, how hard is it to find the least $n$ ($=n\_K$) such that for some diagram $D$ of $K$, every region created by $D$ has at most $n$ crossings on its boundary? For example, for any knot $K$, we can easily see that $n\_K>2$.
| https://mathoverflow.net/users/8133 | Polygons arising from knot diagrams | Colin Adams, Reiko Shinjo and Kokoro Tanaka have a paper (<http://arxiv.org/abs/0812.2558>) that shows that for any knot you can find a diagram which has only regions with 2, 4 and 5 sides.
| 10 | https://mathoverflow.net/users/9717 | 40966 | 26,184 |
https://mathoverflow.net/questions/40938 | 13 | Well, the title is self-explaining, I guess - I am referring to the complete Segal space model structure of Theorem 7.2 in Rezk's article "[A model for the homotopy theory of homotopy theories](http://www.math.uiuc.edu/~rezk/rezk-ho-models-final-changes.pdf)".
Has anyone tried to find out at all, whether this model category is right proper? And, if yes, then produced a reference?
Thanks!
P.S.: I know that I can get a Quillen equivalent right proper model catgory by [the techniques of Thomas Nikolaus](http://arxiv.org/abs/1003.1342), but in that setting it is hard to tell what a general cofibration looks like, which lets me shy away from that step...
| https://mathoverflow.net/users/733 | Is the model category of Complete Segal Spaces right proper? | The model structure for complete Segal spaces is not right proper. To see this, one can first prove that the model structure for quasi-categories is not right proper: for instance, the map $\delta^1\_2:\Delta\_1\to\Delta\_2$ is a fibration between fibrant objects in the Joyal model category (because it is the nerve of a fibration of the canonical model structure on the category of small categories), but its pullback along the inner horn $\Lambda^1\_2\to\Delta\_2$ is the boundary $\partial\Delta\_1$. Now, given a quasi-category $X$, there is a canonical complete Segal space $N(X)$ associated to it (denoted like this because this is an homotopic version of the classical nerve): the space of $n$-simplices of $N(X)$ is the Kan complex $Map(\Delta\_n,X)$, where $Map$ means the mapping space for the Joyal model category structure. A nice explicit model for $Map(\Delta\_n,X)$ is just $k(\underline{Hom}(\Delta\_n,X))$, where $\underline{Hom}$ is the internal Hom in simplicial sets, and where $k(A)$ denotes the maximal Kan complex contained in the quasi-category $A$. For this explicit model of $N(X)$, if $X=\Delta\_m$, then we get that $N(X)$ is just the classical nerve of the poset corresponding to $\Delta\_m$ (because there are no other isomorphisms than the identity in $\Delta\_m$). In other words, $\Delta\_m$ is a complete Segal space already. Therefore, the counter-example for right properness given above for quasi-categories gives a counter-example for complete Segal spaces. For the same reason, the model structure for Segal categories is not right proper.
Edit: Just to avoid any further hope: I considered implicitely that we worked with the model structures for which the cofibrations are the monomorphisms. However, the notion of (right) properness only depends on the class of weak equivalences, so that there just no hope to get right properness for these homotopy theories of $(\infty,1)$-categories, even if we change the notion of fibration.
| 19 | https://mathoverflow.net/users/1017 | 40967 | 26,185 |
https://mathoverflow.net/questions/40930 | 11 | Suppose $\pi$ and $\rho$ are cuspidal automorphic representations on $GL(n)$ and $GL(m)$ respectively. Then the L-function $L(s,\pi \times \rho)$ has a pole iff and $m=n$ and $\pi$ is isomorphic to the contragradient of $\rho$ by some twist. Does anyone know some reference containing the proof of this fact?
I checked Rankin-Selberg convolution paper by Jacquet-P.S-Shalika. It mentioned this result and said the proof would appear somewhere.
Many thanks.
| https://mathoverflow.net/users/1832 | reference help needed on a fact about poles of L-functions | It seems to be in the Cogdell-PS paper "Remarks on Rankin-Selberg Convolutions" in the Shalika volume, though I haven't read through it myself.
| 5 | https://mathoverflow.net/users/6518 | 40980 | 26,193 |
https://mathoverflow.net/questions/40962 | 6 | Suppose that I have $n$ unknown variables $x\_1,\ldots,x\_n$. I wish to compute their sum:
$$Sum(x) = \sum\_{i=1}^nx\_i$$
However, the only access to these variables is through products: that is, for any subset $S \subset [n]$ I may compute:
$$P(S) = \prod\_{i \in S}x\_i$$
That is, I wish to find some number of subsets $S\_1,\ldots,S\_k$, compute $P(S\_1),\ldots,P(S\_k)$, and then apply some postprocessing $f$ to find the sum of the variables:
$$f(P(S\_1),\ldots,P(S\_k)) = Sum(x)$$
My question is: How large must $k$ be? Clearly, $k = n$ suffices, since with $k$ subsets I may uniquely identify each $x\_i$ and then sum the values myself. Is it possible to do with $k < n$? With $k = O(1)$?
| https://mathoverflow.net/users/9769 | How many products specify a sum? | Here is an extreme case: I will tell you that either every variable is zero or possibly a single one of them is 1. So your task is to decide if the sum is 0 or it is 1. Any product of more than one term gives no information at all. To rule out all zero you need to check each variable.
| 30 | https://mathoverflow.net/users/8008 | 40986 | 26,199 |
https://mathoverflow.net/questions/36988 | 2 | I am not familiar with much semigroup theory, but this question came up in my research and I've been unable to find much on it.
| https://mathoverflow.net/users/8434 | what conditions can one place on a finitely generated periodic semigroup that will ensure the semigroup is finite? | This question has been studied extensively more than 20 years ago. Most results are about semigroups satisfying identities. the first such result is by Morse and Hadlund: there exists an infinite 3-generated (infinite 2-generated) semigroup satisfying $x^2=0$ (resp. $x^3=0$). The latest results are in Sapir, M. V. Problems of Burnside type and the finite basis property in varieties of semigroups. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 51 (1987), no. 2, 319--340, 447; translation in Math. USSR-Izv. 30 (1988), no. 2, 295--314. Here is a statement from that paper: All finitely generated periodic semigroups with identity $u=v$ in $n$ variables are finite if and only if all periodic finitely geneated groups with that identity are finite and the Zimin word $Z\_{n+1}$ is not an isoterm for $u=v$ or $v=u$. Here Zimin words are defined by $Z\_1=x\_1, ..., Z\_{n+1}=Z\_nx\_{n+1}Z\_n$. A word $Z$ is an isoterm for $u=v$ if for every substitiution $\phi$, if $\phi(u)$ is a subword of $Z$, then $\phi(u)=\phi(v)$ (by a substitution I mean a map assigning words to variables). If a periodic semigroup has some algebraic structure properties, then its finiteness can be deduced sometimes from Shevrin's results (see, for example, Shevrin, L. N.
Certain finiteness conditions in the theory of semigroups. (Russian)
Izv. Akad. Nauk SSSR Ser. Mat. 29 1965 553--566 or Shevrin, L. N.
A general theorem on semigroups with certain finiteness conditions. (Russian)
Mat. Zametki 15 (1974), 925--935. ). For example, if the semigroup is a union of groups (i.e. every element $x$ satisfies $x=x^p$ for some $p=p(x)$), then the semigroup is finite if and only if all subgroups are locally finite. There are many other results but I do not have time to list them here.
| 5 | https://mathoverflow.net/users/nan | 40990 | 26,201 |
https://mathoverflow.net/questions/40997 | 16 | Given a large number $q$ (say, a prime) and a number $a$ between 2 and $q-1$ what is the fastest algorithm known for computing the inverse of $a$ in the group of residue classes modulo $q$?
| https://mathoverflow.net/users/2344 | Fast computation of multiplicative inverse modulo q | The fast Euclidean algorithm runs in time $O(M(n)\log n)$, where $n=\log q,$ and $M(n)$ is the time to multiply two $n$-bit integers. This yields a bit-complexity of
$$
O(n\log^2 n\log\log n)
$$
for the time to compute the inverse of an integer modulo $q$, using the standard Schonhage-Strassen algorithm for multiplying integers (a slightly better asymptotic result can be obtained using Furer's multiplication algorithm).
To understand the basic idea behind the fast Euclidean algorithm, recall that the standard Euclidean algorithm computes the (extended) gcd of two integers $r\_0 > r\_1$ by successively computing $m\_i=\lfloor r\_{i-1}/r\_i\rfloor$ and setting
$$
r\_{i+1} = r\_{i-1} - m\_ir\_i,
$$
until it obtains $r\_{k+1} = 0$, at which point $r\_k = \gcd (r\_0, r\_1)$. This can be expressed in matrix form as
$$
R\_1 = \begin{bmatrix}
r\_0\newline
r\_1
\end{bmatrix};\qquad
R\_{i+1} = \begin{bmatrix}
r\_i\newline
r\_{i+1}\end{bmatrix} = M\_iR\_i;\qquad
M\_i=\begin{bmatrix}
0&1\newline
1&-m\_i
\end{bmatrix},
$$
and if we compute the matrix $S\_k=M\_kM\_{k-1}\ldots M\_1$, we have $R\_{k+1}=S\_kR\_1$ which expresses the entry $r\_k=\gcd(r\_0,r\_1)$ as a linear combination of $r\_0$ and $r\_1$. Assuming this gcd is 1, we can then read off the inverse of $r\_1$ modulo $r\_0$ (and vice versa) from the top row of $S\_k$.
As described above, this involves $O(k)$ arithmetic operations on integers of size $O(n)$, and one can show that $k=O(n)$, leading to a running time that is roughly quadratic in $n$. The fast Euclidean algorithm achieves a quasi-linear running time by only computing $O(\log n)$ of the matrices $S\_i$. Roughly speaking (and ignoring many important details), this is done by directly computing $S\_{k/2}$ using what is known as a "half-gcd" algorithm, computing $R\_{k/2+1}=S\_{k/2}R\_1$, and then proceeding recursively. The half-gcd algorithm, in turn, works by recursively calling itself. The depth of the recursion is $O(\log n)$ and this yields an $O(M(n)\log n)$ complexity bound.
This algorithm also works over polynomial rings and is often described in this setting. Further details can be found in the (incomplete) list of references below:
Chapter 11 of von zur Gathen and Gerhard, "Modern Computer Algebra," Cambridge University Press, 2003.
Chapter 2 of Yap, "Fundamental Problems of Algorithmic Algebra," Oxford University Press, 2000.
N. Moller, "[On Schonhage's algorithm and subquadratic integer GCD computation](http://www.lysator.liu.se/~nisse/archive/S0025-5718-07-02017-0.pdf)," Mathematics of Comutation 77(261), pp. 589-607 (2008).
Stehle and Zimmerman, "[A binary recursive GCD algorithm](http://perso.ens-lyon.fr/damien.stehle/downloads/recbinary.pdf)," ANTS-VI, LCNS 3076, pp. 411-425, 2004.
| 30 | https://mathoverflow.net/users/4757 | 41004 | 26,210 |
https://mathoverflow.net/questions/40989 | 2 | I am reading "Fundamentals of Diophantine Geometry" by Serge Lang.
Let V be a (absolute) variety, W be a simple subvariety of V. Then we know that the local ring of W is a discrete valuation ring, hence induces a discrete valuation v. But why is v well-behaved (in the sense of Lang's book)?
Can anyone help me find the answer or recommend me a book proving this assertion? Thanks in advance!
| https://mathoverflow.net/users/4621 | Why is an absolute value generated by a simple subvariety of a variety V well-behaved? | According to Lang the valuation $v$ of the field $K$ is well-behaved, if for every finite extension $E/K$ the equation
$
[E:K] =\sum\limits\_{w|v} [E\_w:K\_v]
$
holds, where the summation runs over all extension $w$ of $v$ to $E$, and $K\_v$, $E\_w$ are the completions of the fields $K$, $E$ with respect to the valuations $v$, $w$ respectively.
For a discrete valuation $v$ the completion $K\_v$ is equal to the field of fractions of the $M\_v$-adical completion $\widehat{O\_v}$ of the local ring $O\_v$, where $M\_v$ is the maximal ideal of $O\_v$.
The discrete valuation $v$ we are discussing here by assumption is a localization of an integral, finitely generated $k$-algebra($k$ the field over which the variety $V$ lives). Such an algebra has a finite normalisation in every finite extension of their field of fractions. This property is inherited by localisations of the algebra, thus $O\_v$ has this property too: the normalization $O\_v(E)$ of $O\_v$ in a finite extension $E$ of the field of fractions $K$ is a finitely generated, torsion-free $O\_v$-module. It is well-known that such modules over a factorial ring are free - and $O\_v$ is factorial. The rank of $O\_v(E)$ msut be $[E:K]$ - just localise at $0$.
There is a bijection between the valuation rings $O\_w$ of the extensions $w$ of $v$ to $E$ and the localisations of $O\_v(E)$ at maximal ideals $M$.
The product of all these maximal ideals is some ideal $I$. The $I$-adical completion $\widehat{O\_v(E)}$ of $O(E)$ satisfies:
$
\widehat{O\_v(E)} = \prod\limits\_{w|v}\widehat{O\_w}
$
(Matsumura, Thm. 8.15).
Since a power of $I$ lies in the ideal $M\_vO\_v(E)$ and a power of $M\_vO\_v(E)$ lies in $I$, the completions of $O\_v(E)$ with respect to these two ideals coincide.
The completion of $O\_v(E)$ with respect to $M\_vO\_v(E)$ on the other hand equals the tensor product $
O(E)\otimes\_{O\_v}\widehat{O\_v} $. Since the extension $\widehat{O\_v}/O\_v$ is faithfully flat this tensor product is a free $\widehat{O\_v}$-module of rank $[E:K]$.
Altogether we see now:
$
\prod\limits\_{w|v}\widehat{O\_w}
$
is a free $\widehat{O\_v}$-module of rank $[E:K]$, from which we get
$
\prod\limits\_{w|v}E\_w
$
is a free $K\_v$-module of rank $[E:K]$.
Hagen
| 2 | https://mathoverflow.net/users/3556 | 41007 | 26,211 |
https://mathoverflow.net/questions/41006 | 4 | Let $G$ be a transitive permutation group on a set of size $n$, and suppose $Z(G)=1$ (for instance $G$ is a direct power of a non-abelian simple group). What can we say about the centraliser $K$ of $G$ in $Sym(n)$? I'm interested firstly if there are any restrictions on $K$ independent of degree, and secondly on what role the degree plays.
| https://mathoverflow.net/users/4053 | Centralisers of transitive permutation groups | Of course, there is the classical result that $C\_{Sym(n)}(G)$ is a semi-regular subgroup of $Sym(n)$ of cardinality $|Fix(G\_{0})|$, where $G\_{0}$ is the stabilizer of a point and $Fix(G\_{0})$ is the set of points fixed by $G\_{0}$.
| 6 | https://mathoverflow.net/users/4706 | 41012 | 26,214 |
https://mathoverflow.net/questions/40939 | 8 | Let $X$ be a regular integral noetherian scheme of dimension 2 and let $D$ be a simple normal crossings divisor in $X$.
EDIT: Let $U = X-D$.
Consider a finite etale morphism $V\longrightarrow U$ with $V$ connected. Let $\pi:Y\longrightarrow X$ be the normalization of $X$ in the function field of $V$. So $Y$ is a $2$-dimensional normal noetherian scheme and $\pi$ is finite.
One can show that $\pi$ is surjective, flat and that $Y$ is CM.
Q: Is $\pi$ a local complete intersection?
| https://mathoverflow.net/users/4333 | Is this finite surjective flat morphism of 2 dimensional schemes a local complete intersection | Let $k$ be a field of characteristic prime to $n$ and $\zeta\in k$ a primitive $n$-th root of unity. Let $a\in \mathbb{Z}$ and let $Y$ be the quotient of $\mathop{\rm{Spec}}k[u,v]$ by the automorphism $(u,v)\mapsto (\zeta u,\zeta^a v)$.
Then $Y$ is a tame cyclic cover of $X=\mathop{\rm{Spec}}k[u^n,v^n]$ etale away from $D=\{uv=0\}$. Typically $Y$ is not an lci. For example, if $(n,a)=(3,1)$ then
$$Y=\mathop{\rm{Spec}} k[u^3,v^3,uv^2,u^2v]=\mathop{\rm{Spec}}k[x,y,z,t]/(xy-zt,xz-t^2,yt-z^2).$$
If you allow wild ramification I suspect you can produce an example in which $D$ is even a smooth divisor.
| 7 | https://mathoverflow.net/users/5480 | 41014 | 26,216 |
https://mathoverflow.net/questions/41024 | 7 | Given any group $G$, is there an amenable group $A(G)$ together with a morphism $G\rightarrow A(G)$, such that every other morphism $G\rightarrow A'$ to another amenable group $G'$ uniquely factorizes through $A'$?
That is the question. My approach would be to consider the set of normal subgroups with amenable quotient $S:=\{N\unlhd G|G/N $ is amenable $\}$. Then $\bigcap S$ is a normal subgroup of $G$. But I don't know, whether $G/\bigcap S$ is amenable. It embeds into the group $\prod\_{H\in S}G/H$. It is not clear, that a infinite product of amenable groups is amenable again. But maybe one can embed $G/\bigcap S$ in a smaller group.
| https://mathoverflow.net/users/3969 | Is there an amenabilization of groups ? | Infinite cartesian product of amenable groups is not necessarily amenable. For example, the free group is a subgroup of the infinite (cartesian) product of finite groups. In general, there are finitely generated residually finite just infinite groups (for example, lattices in semi-simple Lie groups of higher ranks, say, $SL\_3({\mathbb Z})$, by the Margulis normal subgroup theorem). These groups do not have amenabelization in your sense.
Update: As Andreas pointed out in a comment below, any residually finite non-amenable group does not have an amenabelization. Indeed, if $G$ is this group. $A$ is the amenabelization, then every hom. $G\to S$ from $G$ to a finite group should factor through $A$. Since the homomorphisms of $G$ onto finite groups separate all elements, the natural homomorphism $G\to A$ is an embedding, but $A$ is amenable while $G$ is not, a contradiction.
| 12 | https://mathoverflow.net/users/nan | 41029 | 26,223 |
https://mathoverflow.net/questions/41005 | 7 | Given a compact Riemannian manifold (with a fixed metric) and a Morse function on it (also fixed). Is there a bound (depending on the metric and the Morse function) on the length of the Morse trajectories? (You can assume the Morse-Smale condition if helpful.)
EDIT (In response to Dick's answer): The Morse function and the metric are fixed. I am just looking for something like $\int\_{-\infty}^{\infty}\| \nabla f(\phi\_t(p))\|dt\leq C$ and $C=C(f,g)$ where $f$ is the Morse function in question, $g$ stands for the metric and $\phi$ denotes the flow of $-\nabla f$. Note here that the constant is independent of the starting point $p$, as it is easy to see that such a constant additionally depending on $p$ exists (you use the hyperbolicity of $\nabla f$ to deduce exponential convergence towards a critical point). Furthermore it is also easy to see that the above integral is bounded if you include a $2$ in the exponent of the norm (a.k.a. $L^2$), as $\| \nabla f(\phi\_t(p))\|^2=-\frac{d}{dt}f(\phi\_t(p)).$
EDIT2 (In response to Bill's answer): I changed the statement to make it abundantly clear that the bound may depend on the metric and the Morse function. Maybe I was somewhat unclear in my formulation - sorry for that.
| https://mathoverflow.net/users/3509 | Is there a bound on the length of the longest Morse trajectory? | As explained by Bill Thurston, on a given Riemann manifold $(M,g)$ you may really have Morse functions with gradient lines of any length. However, as now you are asking for a bound in terms of the Morse function $f$, the following argument shows why lengths of all Morse trajectories (MT) are bounded, and how to get a quantitative bound provided you are able to evaluate simpler geometric quantities.
Thanks to the hyperbolic structure of the flow near critical points, any critical point $x$ has a nbd $U\_x$ such that any flow line crosses $U\_x$ in an arc interval, and the length of the arc interval is bounded above by a constant $c=c(f,g).$ In order to do this quantitatively you may use the Morse lemma. Also, if the $U\_x$ are taken small enough, and assuming the flow is Morse-Smale, any flow line or MT can possibly meet them only in strictly decreasing order of the relative Morse indices: in particular, it can meet at most $\dim(M)+1$ of them. In conclusion, the contribute to the length of any flow line or MT inside the set $U:=\cup \_ {x\in\operatorname{crit}(f)} U \_ x$ is bounded by $(\dim(M)+1)c.$
Now consider an arc of gradient line $\gamma$ in $M\setminus U$ of length say $L$.You can (re)parametrize it with respect to arc-length on the interval $[0, L].$ It solves the ODE $\gamma'(s)=-\frac{\nabla f(\gamma(s))}{\|\nabla f(\gamma(s))\|},$ so this time the derivative of $f$ along $\gamma$ wrto $s$ is $\frac{d}{ds}f(\gamma(s))=-\|\nabla f(\gamma(s))\|$ and integrating you get
$f(\gamma(0))-f(\gamma(L))= \int\_0^L \|\nabla f(\gamma(s))\|ds \geq L\\ \min\_ {M\setminus U} \|\nabla f\| $.
In conclusion, for any gradient line or MT $\Gamma$, summing over the components of $\Gamma\setminus U$ and $\Gamma\cap U$ one gets this bound on the length
$$\mathrm{length}(\Gamma)\leq (\dim(M)+1)c + \frac{\max\_M(f)-\min\_M(f)}{ \min\_ {M\setminus U} \|\nabla f\| }$$
(BTW, note that, as a consequence, you can reparametrize on $[0,1]$ all Morse trajectories so as to be Lipschitz of constant $C(f,g)$, and that by Ascoli-Arzelà these are therefore a compact set in the uniform distance).
[*edit*] A more realistic variant, from a quantitative point of view. If the flow is not asssumed to be Morse-Smale one can simply bound the first term (the contribute to the length inside $U$) with a bound $N$ on the number of critical points, times the constant $c$. This way the choice of the nbd's $U\_x$ is only subjected to local properties (on the contrary, telling how small they have to be in order to ensure the mentioned monotonicity of the Morse indices, requires informations on global properties of the flow).
| 15 | https://mathoverflow.net/users/6101 | 41038 | 26,229 |
https://mathoverflow.net/questions/41033 | 4 | Let $\mathfrak{g}$ = $\mathfrak{gl}\_{\infty}$.
To each positive integer $k$ one can associate the level $k$ Fock space $\mathcal{F}\_{k}$.
For a dominant weight $\lambda$ of level $k$, one can define an action of $\mathfrak{g}$ on $\mathcal{F}\_{k}$
so that it has a highest weight vector of weight $\lambda$ which generates the corresponding irreducible $\mathfrak{g}$-module. Let us denote $\mathcal{F}\_{k}$ with this action as
$\mathcal{F}\_{k}(\lambda)$.
My question is about realizing these spaces and actions. In the lectures of Kac and Raina "Highest-weight representations of infinite dimensional Lie algbebras", there is a construction of $\mathcal{F}\_{1}(\lambda)$ (here $\lambda$ is a fundamental weight) as a subspace of the semi-infinite wege space. In this realization the action of $\mathfrak{g}$ is just the natural action on wedge products. Is there an analogous realization of
$\mathcal{F}\_{k}(\lambda)$
for higher $k$? What about for
$\mathfrak{g}$ = $\hat{\mathfrak{sl}}\_{p}$?
| https://mathoverflow.net/users/316 | Realizing higher level Fock spaces | For $\mathfrak{gl}\_{\infty}$ the answer is easy, you realize the Fock space as a direct limit of polynomial representations of finite $\mathfrak{gl}\_n$ modules. You can read about the construction [here](http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=842325&vfpref=html&r=3&mx-pid=2473335). I worked on the $\hat{sl}\_p$ case a few years ago, but got stuck. If you could do it, it would be very nice.
| 2 | https://mathoverflow.net/users/4366 | 41053 | 26,237 |
https://mathoverflow.net/questions/41042 | 6 | For simplicity fix a base field $k$ of characteristic zero, and consider smooth affine algebraic $k$-groups. (It is understood that unipotent groups in positive characteristic are more complicated, as one might have interesting non-smooth ones.)
Question 1: forms of unipotent groups
If $k$ is algebraically closed, then it is clear that every connected unipotent $k$-group is a successive extension of $\mathbb{G}\_\mathrm{a}$'s. Then what about the case $k$ not algebraically closed? Is there non-trivial $K$-forms of, say, the upper trangular unipotent $k$-group of $GL\_n$, and of the unipotent radicals of Levi $k$-subgroups of $GL\_n$, etc?
If $K$ is a finite Galois extension of $k$ of Galois group $\Gamma$, then a $K$-form of the split $k$-torus is the same as a $\Gamma$-module structure on the group of characters $\mathbb{Z}^d$. Is there analogous results for $K$-forms of unipotent $k$-groups?
For example, with $K$ a finite extension field of $k$. the scalar restriction $$Res\_{K/k}\mathbb{G}\_{\mathrm{m}}$$
is not split as $k$-torus, but
$$Res\_{K/k}\mathbb{G}\_\mathrm{a}$$
splits into a direct sum of $\mathbb{G}\_\mathrm{a}$, becasue $K$ is a finite vector space over $k$ viewed additively. It is from this example that I want to know if there are interesting examples of forms of unipotent groups.
Question 2: representations of unipotent groups
If one has the one dimensional unipotent group $U=\mathbb{G}\_\mathrm{a}$, then an algebraic representation of $U$ on $V$ a finite dimensional $k$-vector space is the same as a unipotent operator on $V$. One can then extend this description naturally to obtain the Tannakian category of finite dimensional algebraic representations of $U$.
And what about general unipotent $k$-group $U$? By the theorem of Lie-Engel, we know that such a representation of $U$ on $V$ is upper-triangular: it stabilizes a full flag of $V$, and acts trivially on the successive quotients (because of unipotence). Is there more precise information one can find about these representations so as the determine the Tannakian category of representations of $U$?
Again, let $K$ be a finite Galois extension of $k$ with Galois group $\Gamma$, and $U$, $W$ two connected unipotent $k$-groups that are isomorphic over $K$. Then how can one distinghuish the representations of the two groups by some "action" of $\Gamma$ one the representations spaces, in the spirit one finds in representations of the $k$-torus $$Res\_{K/k}\mathbb{G}\_\mathrm{m}$$
thanks!
| https://mathoverflow.net/users/9776 | unipotent groups, their forms and representations | As Torsten points out, unipotent groups correspond naturally to nilpotent Lie algebras in characteristic 0. This is dealt with nicely on the scheme level, for example, in IV.2.4 of Demazure-Gabriel *Groupes algebriques*. They also treat in Chapter IV some questions about prime characteristic, which get quite tricky outside the commutative case. Both of your questions are more conveniently studied in the Lie algebra framework, I think, where standard Lie algebra methods for discussing forms in are available and where there is quite a bit of literature on structure, representations, and (in small dimensions) classification in characteristic 0. See for example Jacobson's 1962 book *Lie Algebras*.
Representation theory is potentially very complicated for nilpotent Lie algebras (say over the complex or real field), even in the finite dimensional situation: unlike the semisimple case, there is no nice general structure based on highest weights, etc. Dixmier and others have studied infinite dimensional representations extensively in connection with Lie groups. Classification of nilpotent Lie algebras is just about impossible in general, but up to dimension 7 or so there are lists. Anyway, there is a lot of literature out there. (Tori are on the other hand also studied a lot over fields of interest in number theory. They have at least the advantage of being commmutative.)
[ADDED] An older seminar write-up might be worth consulting, especially in prime characteristic, along with the relatively sparse literature published since then and best searched through MathSciNet: *Unipotent Algebraic Groups* by T. Kambayashi, M. Miyanishi, M. Takeuchi, Springer Lecture Notes in Math. 414 (1974). But as their treatment suggests, the main research challenges have occurred in prime characteristic. Over finite fields, there has been quite a bit of recent activity in studying the characters of finite unipotent groups related to the unipotent radical of a Borel subgroup.
| 4 | https://mathoverflow.net/users/4231 | 41071 | 26,245 |
https://mathoverflow.net/questions/41043 | 23 | Let me take this question again from the top.
I would like to know what a special parahoric subgroup is.
I think this is a "real" question, though not an especially good one -- it indicates my complete lack of expertise in this area, but it is certainly a question of interest to research mathematicians (who else?).
The reason that I want to know -- and that I would prefer a short, relatively simple answer rather than an invitation to read the original paper of Bruhat and Tits -- is that I have been asked to write a MathReview for a paper using this concept. Now, perhaps I shouldn't have agreed to review this paper, and I didn't, exactly, but it was sent to me about three months ago so it seems a little late to object. Anyway, the introduction is clear, so I think my final product will be a reasonable specimen of the form. It's just that, for my own benefit, when I write sentences like
"The stabilizer of a self-dual periodic lattice chain is a parahoric subgroup (or, in some cases, contains a parahoric subgroup of index 2, which one recovers by intersecting with the kernel of the Kottwitz homomorphism). The author restricts to subsets $I$ for which the parahoric so obtained is special in the sense of Bruhat-Tits theory."
it would be nice if I understood a little better what that meant.
I am looking ideally for a quick answer that increases my knowledge at least a little bit together with a place to read up on it when I get the time / inclination to deepen my knowledge.
| https://mathoverflow.net/users/1149 | What is a special parahoric subgroup? | I'd recommend first that you and your friend spend more time with Tits :), "Reductive groups over local fields", from the Corvallis volume (free online, last time I checked). Undoubtably there are other references, like papers of Prasad-Raghunathan mentioned by Greg Kuperberg, and any paper that treats Bruhat-Tits theory for unitary groups. I'll try to provide a background/basic treatment here.
You're certainly used to Bruhat-Tits theory for $SL\_2$, or at least for $PGL\_2$, over $Q\_p$, where one encounters the $(p+1)$-regular tree. As you know, $PGL\_2(Q\_p)$ acts on this tree, and the stabilizer of a point is a maximal compact subgroup that is conjugate to $PGL\_2(Z\_p)$, and the stabilizer of an edge is an Iwahori subgroup. This I assume is a familiar picture.
To understand the "special point" subtleties, you should first think about a group like $G=SU\_3$ -- the $Q\_p$-points of a quasisplit form of $SL\_3$, associated to an *unramified* quadratic field extension $K/Q\_p$ with $p$ odd. Let $S$ be a maximal $Q\_p$-split torus in $G$. Then $S$ has rank one, though the group $G$ has absolute rank two. It follows that the Bruhat-Tits building for $SU\_3$ is again a tree, though not as simple as the $SL\_2$ case. In fact, in this unramified situation, the building can be seen as the fixed points of the building of $G$ over $K$ (which is the building of $SL\_3$), under the Galois involution.
Now, one must think about the *relative* roots of $G$ with respect to $S$ -- i.e., decompose the Lie algebra of $G$ with respect to the adjoint action of $S$. There are four eigenspaces with nontrivial eigenvalue -- these are the root spaces for the relative roots which I'll call $\pm \alpha$ and $\pm 2 \alpha$. The root spaces ${\mathfrak g}\_{\pm \alpha}$ are two-dimensional.
Now, let $A$ be the apartment of the building associated to $S$ -- $A$ is a principal homogeneous space for the one-dimensional real vector space $X\_\bullet(S) \otimes\_Z R$. After choosing a good base point (a hyperspecial base point, using the fact that $K/Q\_p$ is unramified), $A$ may be identified with $X\_\bullet(S) \otimes\_Z R$ and the affine roots are functions of the form $\pm \alpha + k$ and $\pm 2 \alpha + k$, where $k$ can be any integer.
Let $h$ be a generator of the rank 1 $Z$-module $X\_\bullet(S)$, so that $\alpha^\vee = 2 h$, and $A = R \cdot h$. The affine roots are given by:
$$[\pm \alpha + k](r h) = r + k, [\pm 2 \alpha + k](r h) = 2r + k.$$
The vanishing hyperplanes of these affine roots are the points:
$$r h : r \in \frac{1}{2} Z.$$
These are the vertices of the building, contained in the apartment $A$.
Now consider a vertex $nh$, where $n$ is an integer. The affine roots $\pm (\alpha - n)$ and $\pm (2 \alpha - 2n)$ vanish at the vertex $n$. The *gradients* of these affine roots are the roots $\pm \alpha$ and $\pm 2 \alpha$. These are *all* of the roots in the original (relative) root system. That's why these vertices are hyperspecial vertices.
On the other hand, consider a vertex $(n + \frac{1}{2}) h$, where $n$ is an integer. The affine roots vanishing at this vertex are $\pm (2 \alpha - 2n - 1)$. The gradients of these affine roots are the roots $\pm 2 \alpha$. These are *not* all of the original roots, but all original roots are *proportional* to these roots. You can see how this phenomenon requires the setting of a non-reduced root system to happen. These "half-integral" vertices are special points, since the original root system does not occur in the system of gradients, but it does up to proportionality. At these special (but not hyperspecial) points, the Bruhat-Tits group scheme underlying the parahoric has special fibre with reductive quotient isomorphic to $PGL\_2$ (I think... or is it $SL\_2$) over the residue field. At the hyperspecial points, the group would be a quasisplit $SU\_3$ over the residue field.
If it's not clear from above, a special point in the building occurs where the set of gradients of affine roots vanishing at that point is equal, modulo proportionality, to the set of relative roots. That's the general definition.
Hope this helps - see Tits for more.
| 20 | https://mathoverflow.net/users/3545 | 41073 | 26,246 |
https://mathoverflow.net/questions/41039 | 5 | Let $B(x)$ be infinitely differentiable with respect to $x$.
Drop the use of parentheses on $B$ to delimit the argument $x$
and use them instead to hold the order of the derivative with respect to $x$.
i.e. $B(0) = B$,
$B(1) = dB/dx$, etc.
Let parentheses on $x$ hold the order of the derivative of $x$ with respect to $t$.
So
\begin{align\*}
x\_1 & {}= B(x) = B \\
x\_2 & {}= BB\_1 \\
x\_3 & {}= BB\_1^2 + B^2 B\_2.
\end{align\*}
Is there a "nice" formula for the integer coefficient of an arbitrary monomial
$B(0)^u(0) \cdot B(1)^u(1) \cdot\dotsb\cdot B(n-1)^u(n-1)$ in $x(n)$?
The first few terms are:
\begin{align\*}
x(1) & {}= B, \\
x(2) & {}= B\cdot B(1), \\
x(3) & {}= B\cdot B(1)^2 + B^2\cdot B(2), \\
x(4) & {}= B\cdot B(1)^3 + 4\cdot B^2\cdot B(1)\cdot B(2) + B^3\cdot B(3).
\end{align\*}
One of my (many) approaches involved defining $A = 1/B$ so that
$A(x)dx/dt = 1$. Then integrating and solving with the Lagrange Inversion Formula yields
$x(n)$ = sum over all sequences $S=(s(2),s(3),\dotsc)$ of nonnegative integers such that
$\sum (i-1)\cdot s(i)$ equals $n-1$ of
$$(-1)^{T(S)} \cdot (T(S)+n-1)!A^{(-n-T(S))}\cdot \prod\_{i=2}^n \frac1{s(i)!}\left(\frac1{i!}\cdot\frac{d^{(i-1)}A}{dx^{(i-1)}}\right)^{s(i)} $$
where $T(S) = \sum\_{i=2}^n i\cdot s(i)$.
I know I can simply make the upper limits in these products be infinity, because all but finitely many of the terms in the products are 1, because all but finitely many of the
$s(i)$ in any sequence $S$ are zero.
But, for future computational purposes, I want to drag around the finite limits to remind myself when it comes time to implement on a computer.
So then I tried substituting the Faa da Bruno formula for
$$
A(n) = d^A/dx^n = \text{sum of $(-1)^k\cdot k!\cdot B^{-1-k}\cdot\sum\_{V=(v(1),v(2),\dotsc)}\prod \left(\frac{B(i)}{i!}\right)^{v(i)}\frac1{v(i)!}$}
$$
into the equation above and expanding and collecting all similar monomials in the $B$s.
But, I cannot visualize a simple formula for the way all the terms combine.
---
So now I tried computing the terms of this sequence directly.
Homogeneity immediately tells you that any monomial product $B(i)^u(i)$ from $i = 0$ to $n-1$
appearing with nonzero coefficient in $x(n)$ satisfies $u(0) = 1 + \sum\_{i = 2}^{n - 1} (i-1)\cdot u(i)$ and $u(1) = n-1 - \sum\_{i = 2}^n i\cdot u(i)$.
I solved for $u(0)$ and $u(1)$ in terms of the "slack" (?) variables $u(2), \dotsc, u(n-1)$
because the terms whose coefficients I CAN compute are most easily expressed in this form.
So, all I've got so far is
\begin{align\*}
x(n) ={} & B\cdot B(1)^{n-1} + (2^{n-1} - n)\cdot B^2\cdot (B(1))^{n-3}\cdot B(2) +{} \\
& {}(1/4)(3^n - 3 - 2^{n+1}\cdot(n-1) + 2(n-1)^2)\cdot B^3\cdot(B(1))^{n-5}\cdot(B(2))^2 \\
& {}+ (1/4)\cdot(3^{n-1} - 2^{n+1} + 2\cdot n+1)\cdot B^3\cdot (B(1))^{n-4}\cdot B(3) \\
& {}+ \cdots? +{} \\
& {}+ \left(\frac{n^2-3\cdot n+4}2\right)\cdot B^{n-2}\cdot B(1)\cdot B(n-2) + B^{n-1}\cdot B(n-1).
\end{align\*}
I could keep going, deriving longer and longer formulae for more of the terms,
and then HOPE that I can guess the general pattern for all of them.
$B(i)^{u(i)}$ means $B(i)$ raised to the $u(i)$-th power.
| https://mathoverflow.net/users/8789 | Formula for n-th iteration of dx/dt=B(x) | These expansions can be described in terms of [rooted trees](https://en.wikipedia.org/wiki/Tree_(graph_theory)). The first few coefficients are easy to derive by hand, but rooted trees provide you a way of generating the coefficients to arbitrary order, see [here](http://en.wikipedia.org/wiki/Butcher_group). You start with the trivial tree, and at each stage of the derivation you add another level to the tree according to specific rules. The coefficients that occur in the tree have various number-theoretic properties.
This is a very interesting part of combinatorics, with applications in numerical analysis and quantum field theory (the only two fields that I understand). As for numerical analysis, the design of RK methods of arbitrary order didn't go anywhere precisely because of the calculatory issues that you experienced. It was actually a pretty big achievement by John Butcher in the 1960s that he was able to describe these prolongations of ODEs in a compact way, and hence provide a description of RK methods of arbitrary order.
| 11 | https://mathoverflow.net/users/3909 | 41082 | 26,253 |
https://mathoverflow.net/questions/41054 | 2 | Firstly, I have to say that I don't understand Padé approximation well.
But I discovered that, it is more precise than Taylor series.
I have to create approximation for these functions: Log(x) and Tanh. And I have to create iterative algorithms (I must compute result with variable precision).
So my questions are:
Is Padé approximation usable (and more efficient than simple Taylor series) for this task?
If yes, is there any paper or something about this?
If no, is there any better way to approximate these functions?
| https://mathoverflow.net/users/9787 | Padé approximation - usability in iterative algorithms | A qualitative reason for using rational approximations (e.g. Padé) instead of polynomial ones (e.g. Taylor) is that rational approximations can exhibit behavior (e.g. poles and asymptotes) that polynomials are hard-pressed to emulate; they thus tend to be slightly more accurate (there are always exceptions, though).
What you can do that is equivalent to using a Padé approximant (SFAICT, there's no simple method for generating the coefficients of the numerator and denominator polynomials for the two functions you have, except by solving the appropriate Toeplitz system) is to use continued fraction expansions, which [$\ln(1+x)$](http://functions.wolfram.com/ElementaryFunctions/Log/10/) and [$\tanh(x)$](http://functions.wolfram.com/ElementaryFunctions/Tanh/10/) have by their virtue of being expressible as hypergeometric functions.
Of course, for proper use, you have to perform appropriate argument reductions (e.g. for $\tanh$, compute $x^{\star}=\frac{x}{2^n}$ where $n$ is an appropriate integer such that $x^{\star}$ is "small" enough, evaluate the continued fraction at $x^{\star}$, and use the double-argument formula for $\tanh$ to undo your previous transformation).
For evaluating continued fractions, a (reasonably) robust way of going about it is to use the "modified Lentz method" due to Lentz, Thompson, and Barnett; the algorithm's details are in [*Numerical Recipes*](http://rads.stackoverflow.com/amzn/click/0521880688) or Gil/Segura/Temme's [*Numerical Evaluation of Special Functions*](http://rads.stackoverflow.com/amzn/click/0898716349).
| 4 | https://mathoverflow.net/users/7934 | 41085 | 26,255 |
https://mathoverflow.net/questions/41027 | 11 | Given any space $X$ of Lebesgue dimension at most $n$. Suppose a group $G$ acts on $X$ continuously. Can the dimension of the quotient $G\backslash X$ exceed the dimension of $X$?
I know examples, where quotient maps increase the dimension. But I don't know an example, where this quotient is given by dividing out a group action.
| https://mathoverflow.net/users/3969 | Can dividing out a group action can increase the Lebesgue dimension ? | There are examples of dimension-raising orbit maps arising from actions of p-adic groups. Quoting from the MathSciNet review of the following paper: Raymond, Frank; Williams, R. F. Examples of $p$-adic transformation groups. Ann. of Math. (2) 78 1963 92--106, review by P. Conner: "The authors of this paper show that if $A\_p$, $p$ prime, is the compact 0-dimensional $p$-adic group, then for any integer $n\geq 2$ there is a compact $n$-dimensional metric space $X$ together with an action of $A\_p$ upon $X$ as a group of transformations so that the dimension of the quotient space, $X/A\_p$, is $n+2$."
It is unknown whether a $p$-adic group can act effectively on an $n$-manifold. It is known that if such an action exists, then the orbit space necessarily has higher dimension. The conjecture that no such action exists is known as the Hilbert-Smith Conjecture.
| 9 | https://mathoverflow.net/users/1822 | 41089 | 26,258 |
https://mathoverflow.net/questions/41103 | 9 | I have the following situation: let $m,n$ be integers such that $m|n$ and let $\zeta\_m$, $\zeta\_n$ denote primitive $m$ and $n$th roots of unity. Then we have the inclusion of fields
$$\mathbb{Q}\subset \mathbb{Q}(\zeta\_m) \subset \mathbb{Q}(\zeta\_n)$$
Now suppose we also have primes (where $(p,n)=1$)
$$(p)\subset \mathbb{Z}$$
and then
$$\mathfrak{p}\subset \mathbb{Q}(\zeta\_m)$$
lying over $(p)$ and
$$\mathfrak{P}\subset \mathbb{Q}(\zeta\_n)$$
lying over $\mathfrak{p}$.
I have a congruence in $\mathbb{Q}(\zeta\_n)$ of the form $a\equiv b \pmod{\mathfrak{P}}$, where $a,b$ are actually elements of $\mathbb{Q}(\zeta\_m)$.
What can I say about the congruence properties of $a,b$ in $\mathbb{Q}(\zeta\_m)$? More importantly, if I take the trace or the norm down to $\mathbb{Q}$, can I say anything about their congruence properties there? Ideally I'd like a congruence of *something* in the integers.
Thanks!
Edit: Are there any assumptions that you can make that might give congruences mod a prime power?
| https://mathoverflow.net/users/9769 | Congruences mod primes in Galois extensions | Sure. $a\equiv b\pmod{\mathfrak{P}}$ just means $a-b\in\mathfrak{P}$. Taking norms to any subfield $K$ of $\mathbb{Q}(\zeta\_n)$ (e.g., $\mathbb{Q}$ or $\mathbb{Q}(\zeta\_m)$) gives you $N\_{\mathbb{Q}(\zeta\_n)/K}(a-b)\in N\_{\mathbb{Q(\zeta\_n)}/K}\mathfrak{P}.$
For $K=\mathbb{Q}$, the latter norm is just $p^f$ where $f$ is the order of $p\pmod{n}$.
For $K=\mathbb{Q}(\zeta\_m)$, the former norm is $(a-b)^{\phi(n)/\phi(m)}$ and the latter is $\mathfrak{p}^{f'}$, where $f'$ is the easily-calculated relative residue degree.
This doesn't give you an explicit congruence between $a$ and $b$, but given Gerry's answer, that might have been too much to ask for anyway. On the other hand, if $\phi(n)/\phi(m)$ is small or (as in Alex's answer) if $p$ has few factors in $\mathbb{Q}(\zeta\_m)$, you get something at least slightly non-stupid out.
| 5 | https://mathoverflow.net/users/35575 | 41105 | 26,268 |
https://mathoverflow.net/questions/41032 | 5 | Given a once-punctured surface $F$ and an orientation preserving self homeomorphism $\phi$, let $M\_\phi$ be the bundle over $S^1$ with fiber $F$ and monodromy $\phi$.
In Sakuma's survey article [The topology, geometry and algebra of unknotting tunnels](http://ams.org/mathscinet/search/publdoc.html?extend=1&pg1=IID&s1=203701&vfpref=html&r=16&mx-pid=1628753) (and in [this paper](http://ams.org/mathscinet/search/publdoc.html?extend=1&pg1=IID&s1=203701&vfpref=html&r=17&mx-pid=1664979)), Johannson and Kobayashi are credited for proving that any unknotting tunnel for $M\_\phi$ is isotopic to a tunnel $\alpha$ which lies on a fiber $F$ such that $\alpha\cap\phi(\alpha)=\emptyset$. This leads to a classification of unknotting tunnels in once-punctured torus bundles.
The references are talks. Does anyone know if this is written down anywhere?
| https://mathoverflow.net/users/4325 | Unknotting tunnels in surface bundles | [Scharlemann and Thompson](http://www.ams.org/mathscinet-getitem?mr=1990938) proved that one can isotope/slide a tunnel
to be disjoint from a minimal genus Seifert surface (which was what I had in mind
in my comment above). Their argument does not depend on it being a knot complement.
I also found a [citation for a paper of Sakuma](http://www.ams.org/mathscinet-getitem?mr=1483404) which claims to prove the
result for punctured torus bundles, so it might have the result you're looking
for (although I haven't looked at the article yet).
| 1 | https://mathoverflow.net/users/1345 | 41107 | 26,270 |
https://mathoverflow.net/questions/41117 | 2 | I have been reading about Riemann Zeta function $\zeta(s)$ and have been thinking about it for some time. I did some calculations, and reached a conclusion where $\Re(\rho) \le \log\_2(3) - 1$ as $\Im(\rho) \to \infty$ where $\rho$'s are the roots of Riemann Zeta function in the critical strip. Anyways, I know its not the place to discuss claimed proofs and similar stuff, but just to give a background of where I am coming from. So straight to the question.
>
> Is there any similar result regarding upper
> bound ($< 1$) for the real part of the zeros zeta
> function as their imaginary
> parts tend to infinity?
>
>
>
Thanks
| https://mathoverflow.net/users/2865 | Upper bound for real part of Riemann Zeta function zeros | There is no known non-trivial (less than 1) bound for real parts of Zeta zeros (I guess, it is even called "weak Riemann conjecture" to find such a bound). So, your result is very-very interesting, maybe the most interesting result in mathematics for many years.
| 9 | https://mathoverflow.net/users/4312 | 41120 | 26,279 |
https://mathoverflow.net/questions/41138 | 2 | Let $A$ be a complete regular local noetherian ring of dimension $d>1$ and $B$ an $A$-algebra, finite and free as $A$-module. Assume moreover that there exists an open subset $U$ of $\textrm{Spec}\ A$ of primes of height $d-1$ such that there is exactly one prime in $\textrm{Spec}\ B$ over a prime of $U$. Is it then true that $\textrm{Spec}\ B$ is an irreducible topological space?
Under various choices of ancillary hypotheses, results like this one follow from Zariski's connectedness principle; see e.g [A useful form of principle of connectedness](https://mathoverflow.net/questions/9060/a-useful-form-of-principle-of-connectedness).
| https://mathoverflow.net/users/2284 | A weaker form of Zariski's connectedness principle | Yes. Assume not. Since $B$ is finite flat over $A$, each irreducible component of $\mathrm{Spec}(B)$ is pure $d$-dimensional, finite and *surjective* over $\mathrm{Spec}(A)$. If $Y\neq Z$ are two of them, then the image of $Y\cap Z$ in $\mathrm{Spec}(A)$ is a proper closed subset, hence does not contain $U$. Every point of $U$ has a preimage in $X$ and one in $Y$.
| 3 | https://mathoverflow.net/users/7666 | 41140 | 26,287 |
https://mathoverflow.net/questions/41141 | 61 | I want to cite a paper which is on arxiv.org but is not published or reviewed anywhere, and no publication or review seems to be in the pipeline. Would citing this arxiv.org paper be bad? Should I wait for a paper to be peer reviewed before I cite it?
Added:
I don't actually know whether a 'real' publication is in the pipeline. The alternative to citing the paper would probably be to ignore it; I have a way to extend the results in the paper if the paper's results are true, but I don't have the skill or time to verify that the arxiv.org paper is correct.
| https://mathoverflow.net/users/9501 | Should I not cite an arxiv.org paper which otherwise seems to be unpublished? | [It is] Not really [bad to cite an arXiv paper]\*. If the paper on arXiv provides the result you want, you are free to cite it. Before the arXiv, citing "private communication" or "pre-print" is not unheard of. On the other hand, since it hasn't been peer reviewed, you probably should double check and make sure you understand and believe the paper before you cite it (if you use one of its results crucially) (not that you shouldn't do the same for peer-reviewed papers, just that one may want to be extra careful with referring to pre-prints).
Note that there are two reasons for citations. The first is to give credit where credit is due: you do not want to look like you are appropriating someone else's result (or in some cases, inadvertently slighting somebody by sin of omission). The second is to provide references for assertions made without proof in your paper. Obviously if you are citing for the former reason, a paper is arXiv is really no different from a paper in a published journal. If the author's right, you covered your bases. If he was wrong, then better for you, perhaps. It is with the latter case you need to be more careful. If the paper has been on arXiv for a long time and not appeared in any journals (definition of "long time" of course vary from field to field), you may want to be a bit cautious in deciding whether the foundation to your house is sound.
Also, how do you know "no publication or review seems to be in the pipeline"? I know several people (myself included) who would only include the journal ref on arXiv after it has been accepted for publication. Perhaps you should double check with the original author whether it has been submitted, and if not, why not?
\* As Joel pointed out in his comments to the original question, and Emerton in his comments to this answer, there is some ambiguity as to which question I was answering.
| 70 | https://mathoverflow.net/users/3948 | 41146 | 26,290 |
https://mathoverflow.net/questions/41136 | 1 | Given $S$ a $K3$ surface and $M$ the moduli space of simple sheaves of rank 2 and fixed Chern classes on $S$, under which conditions does a universal family on $S\times M$ exist?
| https://mathoverflow.net/users/33841 | Universal family over the moduli space of simple sheaves on a K3 surface | Let us introduce the Mukai pairing on the cohomology $H^\*(X,\mathbb{Z})=H^0\oplus H^2\oplus H^4$ of a K3 surface X, as
$<(r,l,s).(r',l',s')>=l.l'-rs'-r's$
where $l.l'$ is the intersection pairing on $H^2$.
The Mukai vector of a sheaf $E$ is defined to be
$v(E)=ch(E).\sqrt{Td(X)}=(rk(E),c\_1(E),rk(E)+c\_1^2(E)/2-c\_2(E))$.
If the Mukai-vector $v \in H^\*(X,\mathbb{Z})$ is indivisible (i.e. there is a $v'$ with $< v.v' >=1$ for the Mukai-pairing) then there is a polarization $h$ such that the Moduli space $M\_h(v)$ is fine.
This criterion is due to Mukai: "On the Moduli space of Bundles on K3 surfaces, I" (Tata Inst. Fund. Res. Stud. Math., 11, Tata Inst. Fund. Res., Bombay, 1987).
You can find it also in the famous book by Huybrechts and Lehn.
| 2 | https://mathoverflow.net/users/5714 | 41147 | 26,291 |
https://mathoverflow.net/questions/41137 | 2 | Hello,
Is there any software for calculating Green polynomials (of type A)? Or, at least, where can I find tables of Green polynomials? Also, I would be interested in some formulas for Green polynomials in simple cases.
| https://mathoverflow.net/users/6772 | Green polynomials | I am not really an expert but it looks like the Morita paper [Decomposition of Green polynomials of type A and Springer modules for hooks and rectangles](http://dx.doi.org/10.1016/j.aim.2006.07.007) could be of some help here. In particular, at page 483 one finds an explicit formula for these polynomials for one of the simplest special cases. This paper also gives an explicit formula for the Green polynomials for an arbitrary hook (see Proposition 8).
| 3 | https://mathoverflow.net/users/2149 | 41148 | 26,292 |
https://mathoverflow.net/questions/41128 | 5 | Given a smooth function $f:M\rightarrow \mathbb R$ on a manifold, its local homology at a critical point $x$ is the group
$$ C\_\star(x) := H\_\star ( M\_{ < c} \cup \{ x \} , M\_{ < c} ) ,$$
where $H\_\star$ denotes singular homology (with any coefficient group), $c=f(x)$, and $M\_{ < c}$ is the space of those points $x\in M$ such that $f(x) < c$.
If $x$ is a non-degenerate critical point, then $C\_\star(x)$ is completely determined by the Morse index of $f$ at $x$: the group $C\_j(x)$ is equal to the coefficient group of the homology for $j=\mathrm{ind} (x)$, and is trivial for other values of $j$.
If $x$ is degenerate, the knowledge of $\mathrm{ind}(x)$ and $\mathrm{nul}(x)$ (this latter being the nullity of $f$ at $x$) is not enough to determine $C\_\star (x)$. It is easy to build examples of functions on $\mathbb R^2$ having a critical point $x$ with local homology $C\_1(x)= G\oplus ...\oplus G$ ($k$ times, where $k>1$ and $G$ is the coefficient group) and $C\_j(x)=0$ for $j\neq 1$. For instance, consider the function $f:\mathbb R^2\rightarrow\mathbb R$ given by
$$ f(x,y)=(y-2x^2)(y-x^2)(y+x^2)(y+2x^2). $$
Here, the origin is a critical point whose local homology (say, with $\mathbb Z\_2$ coefficients) should be $C\_1(0)=\mathbb Z\_2\oplus\mathbb Z\_2\oplus \mathbb Z\_2$ and $C\_j(0)=0$ for $j\neq 1$.
Does anybody know examples of functions having critical points whose local homology is nonzero in more then one degree?
If the answer to the previous question is yes (as I would expect), is it true that given $(n\_1,d\_1), ..., (n\_r,d\_r)$ there exists a function $f:M\rightarrow\mathbb R$ with a critical point $x$ whose local homology is given by $C\_{d\_j}(x)=G^{ \oplus n\_j }$ and $C\_d(x)=0$ for $d\neq d\_1,...,d\_r$?
| https://mathoverflow.net/users/9736 | Local homology of degenerate critical points | $(x^2+y^2)z^2-c(x^2+y^2+z^2)^2$ for small positive $c$.
More generally $f-cr^{2d}$ where $f\ge 0$ is a homogeneous polynomial function of degree $2d$ in $n$ variables. The local homology at the origin should be essentially the homology of the set of points in $S^{n-1}$ where $f=0$.
| 5 | https://mathoverflow.net/users/6666 | 41156 | 26,298 |
https://mathoverflow.net/questions/41158 | 19 | Let $f:X\to Y$ be a morphism of complex analytic spaces. Assume $f$ is flat (or, more generally, that there is a coherent sheaf on $X$ with support $X$ which is $f$-flat). Is $f$ an open map?
The rigid-analytic analogue is true (via Raynaud's formal models): see Corollary 7.2 in S. Bosch, Pure Appl. Math. Q., 5(4) :1435–1467, 2009. I don't know about the Berkovich side.
In the algebraic case it's also true (that's what led me to the question, see <http://arxiv.org/abs/1010.0341>). Specifically, if $K$ is an algebraically closed field with an absolute value, and $f:X\to Y$ is a universally open morphism of $K$-schemes of finite type, the the induced map on $K$-points is open (for the strong topology).
Note that in the complex analytic case, I don't know any reasonable substitute for "universally open". If I believe in the analogy, the result ($f$ is open) should be true assuming for instance that $Y$ is locally irreducible and $f$ is "equidimensional" in some sense (e.g. surjective, $X$ irreducible and the fiber dimension is constant). In this setting the case of fiber dimension 0 is known.
| https://mathoverflow.net/users/7666 | Are flat morphisms of analytic spaces open? | The answer is **yes**.
In fact, there is the following result, see
Banica-Stanasila, *Algebraic methods in the global theory of Complex Spaces*, Theorem 2.12 p. 180.
>
> **Theorem.**
> Let $f \colon X \to Y$ be a morphism of complex spaces and let $\mathscr{F}$ be a coherent analytic sheaf on $X$, which is flat with respect to $f$. Then the restriction of $f$ to supp($\mathscr{F}$) is an open map.
>
>
>
In particular, every flat morphism is open.
| 19 | https://mathoverflow.net/users/7460 | 41163 | 26,303 |
https://mathoverflow.net/questions/41122 | -1 | The Frobenius, or Hilbert-Schmidt, norm of an $n$ by $n$ matrix $A$ is defined as $\|A\|\_2 = \sqrt{\sum\_{i,j=1}^n |A\_{ij}|^2}$. The absolute value of $A$ is the unique positive matrix $|A|$ satisfying $|A|^2 = A^\* A$. Are there any known relations between $\| |A| \|\_2$ and $\|A\|\_2$?
| https://mathoverflow.net/users/9807 | Absolute values and Frobenius norm | It holds that
|||A|||\_2 =||A||\_2
Proof: Let the singular value decomposition of A be given by $A=U\Sigma V^H$. Since the HS-norm is invariant under unitary transformations, $||A||\_2= ||\Sigma||\_2$ holds. As for $|A|$, we obtain $|A|= |A^HA|=|V\Sigma U^HU\Sigma V^H|=|V\Sigma^2 V^H|= V\Sigma V^H$. Again, unitary invariance yields $|| |A| ||\_2=||\Sigma ||\_2=||A||\_2$, qed.
| 0 | https://mathoverflow.net/users/9810 | 41169 | 26,308 |
https://mathoverflow.net/questions/41167 | 10 | Suppose we are working in a category of schemes over a scheme $S$. The scheme $S$ itself is geometrically a ``point''. Let $G$ be a group scheme that acts on a scheme $X$. The quotient stack $[X/G]$ looks in a way as a factor of $X$ by a free action of $G$, i.e. if the action is indeed free and there is a scheme $Y$ such that $X$ is a principal bundle over $Y$, isomorphic as a $G$-scheme to $X$, then $Y$ is isomorphic to $[X/G]$.
In particular, if we take $X=S$, we obtain a classifying stack. There is a surjective map $S \to [S/G]$ and every principal $G$-bundle can be obtained as a pullback of it. This strongly resembles the construction of the classifying space in topology, where we have a space $EG$ with a homotopy type of a point and a cover $EG \to BG$ obtained as a factor by a free action of $G$ on $EG$.
Here is my question (which might sound a bit naive): how this similarity to algebraic topology can be explained? I am mostly curious about how it happens that the construction of the classifying stack is not related to the notion of the homotopy type, whereas the construction of the classifying space relies on the fact that EG has the homotopy type of a point.
| https://mathoverflow.net/users/2234 | Classifying stacks and homotopy type of a point | This is all best explaied by working with groupoids in schemes. By this, I mean, groupoid objects in the category of schemes over $S$. A groupoid object $\mathcal G$ consists of two schemes $\mathcal G\_0$ (the objects) and $\mathcal G\_1$ (the morphisms) and a whole bunch of maps between them:
• a map $s:\mathcal G\_1\to \mathcal G\_0$ that sends an arrow to its source.
• a map $t:\mathcal G\_1\to \mathcal G\_0$ that sends an arrow to its target.
• a map $e:\mathcal G\_0\to \mathcal G\_1$ that sends an object to its identity arrow.
• a map $i:\mathcal G\_1\to \mathcal G\_1$ that sends an arrow to its inverse.
• a map $m:\mathcal G\_1\times\_{\mathcal G\_0}\mathcal G\_1\to \mathcal G\_1$ that composes arrows.
subject to even more axioms.
From now on, I will simply say "groupoid" instead of groupoid object in schemes.
---
Let $G$ be a group over $S$.
Then, there is a groupoid called $EG$, whose objects are $G$, and whose arrows are $G\times\_S G$. The groupoid $EG$ is equivalent to $S$, viewed as a groupoid with only identity morphisms. The group $G$ acts freely on $EG$ (this is all happening in the category of schemes over $S$), and the quotient $EG/G$ is $BG$. Here, $BG$ is the groupoid with objects $S$ and morphisms $G$.
If $X$ is a scheme over $S$, then $[X/G]$ is the groupoid whose objects are $X$, and whose morphisms are $X\times \_S G$.
This groupoid can also be described as the quotient of the (free) diagonal action of $G$ on the groupoid $X\times EG$.
---
You're asking:
"why is the construction of the classifying stack not related to the notion of the homotopy type, whereas the construction of the classifying space relies on the fact that $EG$ has the homotopy type of a point?".
There are two ways of answering your question:
- One is to make the algebraic-geometric story look a little bit more like what people do in topology. That's what I did above. In particular, the fact that the groupoid $EG$ is equivalent to $S$ is the analog of the fact that $EG$ is contractible in topology.
- The other is to make the topological story a little bit like what people do in algebraic geometry. Namely, instead of defining $BG$ as $EG/G$, define it as the space that represents the functor $X\mapsto$ {iso-classes of $G$-bundles over $X$} where I'm now working in the category of topological spaces and homotopy classes of maps.
---
---
I should maybe add:
There is a stack (over the category of $S$-schemes) associated to any groupoid object in $S$-schemes. The converse is not true for general stacks. But it is true, essentially by definition, for Artin stacks. If you restrict yourself to Deligne-Mumford stacks, then you can also assume that the groupoid is particularly nice, namely, that the maps $s$, $t$, $e$, $i$ and $m$ are étale.
| 17 | https://mathoverflow.net/users/5690 | 41172 | 26,310 |
https://mathoverflow.net/questions/41177 | 20 | I am stuck on solving an apparently simple ODE. I have checked numerous texts, references, software packages and colleagues before posting this...
$$y(t)^n+a(t)\frac{dy(t)}{dt}=ba(t)$$
If the RHS had a $y$ term it would simply be Bernoulli's equation. Does the $n$ term prevent a solution?
| https://mathoverflow.net/users/3754 | A nonlinear first order ordinary differential equation: $y(t)^n+a(t)\frac{dy(t)}{dt}=ba(t)$ | First, rewrite your equation as
$$
\frac{dy(t)}{dt}=b+f(t)(y(t))^n, \qquad \qquad \qquad \qquad \qquad (\*)
$$
where $f(t)=-1/a(t)$.
This is a special case of the so-called [Chini equation](http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Chini)
(Equation 1.55 in the Kamke's book mentioned below)
$$
\frac{dy(t)}{dt}=f(t)(y(t))^n+g(t)y(t)+h(t)
$$
which generalizes the Riccati and the Abel equations and is in general not solvable by quadratures
but some of its special cases are, see e.g. the book (in German)
E. KAMKE, [Differentialgleichungen: Lösungen und Lösungsmethoden, Band I: Gewöhnliche Differentialgleichungen](http://openlibrary.org/books/OL15487458M/Differentialgleichungen_Lo%25CC%2588sungsmethoden_und_Lo%25CC%2588sungen), Leipzig, 1951,
and [this list](http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc1.htm) and references therein.
It is known that if the Chini invariant
$$
C=f(t)^{-n-1}h(t)^{-2n+1}(f(t)
dh(t)/dt-h(t)df(t)/dt+n f(t)g(t)h(t))^n n^{-n}
$$
is independent of $t$, there is a straightforward recipe (described in the Kamke's book)
for solving the equation. However, in the case under study (both for general $a(t)$ and for $a(t)$
linear in $t$ as suggested by the original poster in the comment to this reply) this invariant for (\*) does depend on $t$ (unless I messed up the computations :)), so the recipe in question does not apply.
The only case when $C$ is independent of $t$ occurs (again modulo possible errors in computations :)) if $((1/f(t))df(t)/dt)^n=\alpha f(t)$, i.e., when $f=(-\alpha(t+\beta)/n)^n$, where $\alpha$ and $\beta$ are arbitrary constants.
Now let us turn to the particular cases with small values of $n$.
For $n=1$ you have a linear inhomogeneous ODE which is easily solved.
For $n=2$ you get a special case of the so-called [general Riccati equation](http://eqworld.ipmnet.ru/en/solutions/ode/ode0123.pdf)
$$
\frac{dy(t)}{dt}=b+f(t)(y(t))^2,
$$
solving which is equivalent to solving a second-order linear ODE.
Indeed, upon introducing a new independent variable $\tau(t)=\int f(t) dt$
you end up with the "standard" Riccati equation
$$
\frac{dy(\tau)}{d\tau}=h(\tau)+(y(\tau))^2,
$$
where the function $h$ is defined so that $h(\tau(t))=b/ f(t)$,
and putting $y(\tau)=-(1/z(\tau))dz(\tau)/d\tau$ yields
$$
d^2z(\tau)/d\tau^2+h(\tau)z(\tau)=0.
$$
However, this linear equation in general is not necessarily [solvable by quadratures](http://eom.springer.de/I/i051760.htm).
For $n=3$ (\*) is a special case of the Abel differential equation of the first kind,
see e.g. [here](http://eom.springer.de/A/a010110.htm) and references therein for details.
| 31 | https://mathoverflow.net/users/2149 | 41181 | 26,315 |
https://mathoverflow.net/questions/41183 | 11 | Is it true that any manifold homotopy equivalent to a k-dimensional CW-complex admits a proper Morse function with critical points all of index <= k? I believe this is not true, so I would like to see a counterexample.
| https://mathoverflow.net/users/9800 | index of morse functions and homotopical dimension | Take a contractible $3$-manifold which is not homeomorphic to $\mathbb R^3$ -- like the Whitehead manifold.
If such a Morse function existed on the Whitehead manifold, it would be a Morse function with only one critical point, the minimum, and therefore the Whitehead manifold would be an open 3-ball. The proof of this has two steps: (1) $f$ can have at most one critical point, WLOG a minimum by homotopy-type considerations, and it can't have less than one critical point by the "edit" below. Step (2) if it has one critical point let it be $f(p)=0$, the minimum. By the Morse Lemma, $f^{-1}[0,\epsilon]$ is diffeomorphic to a closed 3-ball. The flow the the gradient of $f$ gives a diffeomorphism between $f^{-1}[\epsilon,\infty)$ and $S^2 \times [\epsilon, \infty)$. Pasting these two diffeomorphisms together gives you a diffeomorphism between the manifold and $\mathbb R^3$.
edit: Well, I guess there's the special case to consider that the Morse function could have no critical points but then you could deal with this by the argument that the Whitehead manifold isn't a product of a surface and $\mathbb R$, which by the classification of surfaces amounts to saying that the Whitehead manifold isn't homeo/diffeomorphic to $\mathbb R^3$.
Welcome to MO, Victor!
| 10 | https://mathoverflow.net/users/1465 | 41184 | 26,316 |
https://mathoverflow.net/questions/41180 | 4 | What's the easiest (by which I mean uses the least fancy machinery) proof of the direct summand conjecture in dimension 2?
Recall that the direct summand conjecture says that:
**Conjecture** (Hochster): If $R$ is a regular ring and $S$ is a module finite integral extension, then $R \to S$ splits as a map of $R$-modules.
It is trivial in characteristic zero (via the trace map) and not that hard in characteristic $p > 0$ using Frobenius-type methods. In mixed characteristic it is known up to dimension 3.
| https://mathoverflow.net/users/3521 | Reference request, direct summand conjecture in dimension 2 | You may assume that $R,S$ are complete and $S$ is a domain. Now take the integral closure $T$ of $S$, which is $S$-finite. Since we are in dimension $2$, $T$ is maximal Cohen-Macaulay module over $R$, so $T$ is $R$-free. Thus the composition map $R\to T$ splits (as it takes $1$ to $1$) whence the map $R\to S$ splits.
The moral of this is that existence of small Cohen-Macaulay modules implies a lot of things, and you can get that for free in dimension $2$ via integral closure.
| 6 | https://mathoverflow.net/users/2083 | 41185 | 26,317 |
https://mathoverflow.net/questions/41150 | 31 | This question is prompted by a recent MO question on explicit computations of Weyl group invariants for certain exceptional simple Lie algebras:
[37602](https://mathoverflow.net/questions/37602/polynomial-invariants-of-the-exceptional-weyl-groups). Like some others who started graduate study in the 1960s with almost no physics background but with an interest in abstract mathematics, I was drawn to algebraic Lie theory for mainly esthetic reasons. I also had no background in differential geometry or Lie groups. So when I bought a copy of Jacobson's newly-published book on Lie algebras at a bookstore in Ithaca I had no appreciation of the historical connections of the subject.
The eminent Dutch physicist H.B.G. Casimir was apparently the first to introduce an explicit second degree invariant (unique up to scalars) in the center of $U(\mathfrak{g})$, now called the *Casimir element* or *Casimir invariant*.
Roughly speaking, this involves fixing a basis of $\mathfrak{g}$ along with its dual basis under the Killing form, then adding the respective products. Sometimes it is convenient to recast the answer in terms of PBW monomials for the given ordered basis.
On the mathematical side, Chevalley and Harish-Chandra determined the full center of $U(\mathfrak{g})$: it is a polynomial algebra in $\ell$(= rank of $\mathfrak{g}$) variables. Generators can be taken to be homogeneous of uniquely determined degrees. Moreover, the center is isomorphic in a natural way (but requiring a subtle $\rho$-twist) to the algebra $U(\mathfrak{h})^W$ of invariants of the Weyl group relative to a fixed Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$. The earlier MO question involved this algebra and its (non-unique) homogeneous generators of degrees $2 = d\_1, d\_2, \ldots, d\_\ell$. Key papers were those by Harish-Chandra (Trans. AMS 1951) and Chevalley (Amer. J. Math. 1955), the latter generalizing the fundamental theorem on elementary symmetric polynomials for $W=S\_n$.
My question then is:
>
> What role did Casimir's work play in this mathematical development?
>
>
>
A related matter is the practice of referring to homogeneous generators of the center of the enveloping algebra as "Casimir operators": how far was Casimir himself involved in this direction beyond his degree 2 invariant?
ADDED: The short article referenced by mathphysicist is illuminating and may be the
best published indication of Casimir's influence on subsequent representation theory. I was at first hoping to find a more definite paper trail, but this may not exist and would probably reach back before Math Reviews. What struck me most in browsing through the first volume of Harish-Chandra's collected papers (1944-54) was the abrupt transition around 1948 from his work in physics like *Motion of an electron* to mathematics like *Faithful representations of Lie algebras* and of course his foundational 1951 paper I mentioned. Nowhere along the way do I see any direct citation of Casimir's papers, though the 1950 paper *Lie algebras and the Tannaka duality theorem* does quote the "Casimir operator" in rank 1 as well known and uses it as a stepping-stone to the general case. Since Harish-Chandra studied physics with Dirac in his early years, it was probably he who imported Casimir's idea into representation theory. But Chevalley was at the time also a major influence on Harish-Chandra's thinking, so it's all hard to document. (They both taught at Columbia for some time.)
| https://mathoverflow.net/users/4231 | What was Casimir's precise role in describing the center of the universal enveloping algebra of a semisimple Lie algebra? | At the first glance it appears that he more or less just gave the first nontrivial example(s) of what was later called the Casimir operators.
His [obituary](http://dx.doi.org/10.1088/0143-0807/22/4/320) says:
*On 1 May 1931 he wrote a letter from The Hague to the famous Gottingen mathematician
Hermann Weyl and announced: ‘While studying the quantum-mechanical properties of the
asymmetic rotator I arrived at some ‘results’ (?) concerning the representation of continuous
groups.’ He then sketched his findings on the matrix elements of the irreducible representations
for the three-dimensional rotation group, and a possible extension for semi-simple groups in
general, where he introduced what was later called the ‘Casimir operator’. This operator turned
out to be a multiple of the unit-operator and may be used to characterize in an elegant way
the irreducible representations of a given continuous group. To Casimir’s question, ‘Whether
the case is worth considering?’, Weyl answered definitely ‘Yes’. Hence the Leiden doctoral
candidate published his mathematical results in a paper, communicated by Ehrenfest to the
meeting of 27 June 1931 of the Amsterdam Academy [7], and he also included them as Chapter
IV of his dissertation, which he defended on 2 November 1931 at the University of Leiden [8].*
| 16 | https://mathoverflow.net/users/2149 | 41193 | 26,322 |
https://mathoverflow.net/questions/41195 | 1 | More accurately, let $\displaystyle A=\sum\_{i=0}^{\infty}A\_i$ be a finitely generated graded algebra over say $\mathbb{Q}$ but $\dim A\_i=\infty$ for each $i.$ Is it possible?
| https://mathoverflow.net/users/9645 | Need an example of finitely generated graded algebra such that each its graded subspace has infinite dimension. | $\mathbb Q[x,y]$ with $x$ in degree 0, and $y$ in degree 1.
If you want your generators to be in positive degrees, then what you're asking for is impossible.
| 7 | https://mathoverflow.net/users/5690 | 41196 | 26,323 |
https://mathoverflow.net/questions/41118 | 12 | Does the usual development of category theory (within Goedel-Bernays set theory, for example) require the axiom of replacement? I would have asserted that this was obviously true, but it seems to be common wisdom that the axiom of replacement is an exotic axiom not used outside of axiomatic set theory. Also, sadly the overlap between things that I have thought were obviously true and were in fact false is larger than I would like to admit...
The axiom of replacement basically says that if a class is the same size as a set, then it *is* a set. This allows us to identify classes that are sets as being *small* and those that are proper classes as *large*. Without replacement, you could have countable classes that not sets.
I don't see how to construct most limits and colimits in familiar categories such as *Set* or *Top* without the use of replacement. Already, for an infinite set $X$, I don't see how you construct
$$
\coprod\_{i=0}^\infty P^i X,
$$
where $P^i$ is the $i$-th power set of $X$ (i.e. $P^0 X = X$ and $P^i X = P (P^{i-1}X$)).
Using replacement, it's easy to construct. You form the set
$$
\bigcup\_{i=0}^\infty P^i X,
$$
and then the coproduct is isomorphic to the set of pairs $(i, x)$, where $x \in P^i X$. Am I completely misunderstanding the issue here, and this coproduct can be proven to exist without replacement?
I suppose you could cook up a definition of diagram so that in the absence of replacement, you cannot even form the diagram for this particular coproduct, but that would be sort-of unsatisfying.
| https://mathoverflow.net/users/3711 | Axiom of Replacement in Category Theory | There’s one issue underlying a lot of the discrepancies between people’s answers, I think:
>
> How are we defining “$f$ is a function $s \to V$”, where $s$ is a set and $V$ is a (possibly proper) class?
>
>
>
(hence also, how we define subsequent things like “a small-category-indexed diagram of sets”) There are at least two main options here:
1. $f$ is a class of pairs, such that…
2. $f$ is a set of pairs, such that…
At least in most traditional presentations, I think it’s defined as the latter, but some people here also seem to be using the former. The answer to this question depends on which we take.
If we take the “a function is just a class” definition, then as suggested in the original question, and as stated in François’ answer, we definitely have some big problems without replacement: **Set** is no longer complete and co-complete, etc. (nor are the various important categories we construct from it); we can’t easily form categories of presheaves; and so on. Under this approach, we certainly get crippling problems in the absence of replacement.
On the other hand, if we take the “a function must be a set” definition, we get some different problems (as pointed out in Carl Mummert’s comments), but it’s not so clear whether they’re big problems or not. We now *can* form limits of set-indexed families of sets; presheaf categories work how we’d hope; and so on. The problem now is that we can’t form all the set-indexed families we might expect: for instance, we if we’ve got some construction $F$ acting on a class (precisely: if $F$ is a *function-class*), we can’t generally form the set-indexed family $\langle F^n(X)\ |\ n \in \mathbb{N} \rangle$.
This is why we still can’t form something like $\bigcup\_n \mathcal{P}^n(X)$, or $\aleph\_\omega$. On the other hand, such examples don’t seem to come up (much, or at all?) outside set theory and logics themselves! Most mathematical constructions that do seem to be of this form — e.g. free monoids $F(X) = \sum\_n X^n$, and so on — can in fact be done without replacement, one way or another.
Now… I seem to remember having been shown an example that was definitely “core maths” where replacement was needed; but I can’t now remember it! So if we take this approach, then we certainly still lose something; but now it’s less clear quite how much we really needed what we lost.
(This approach is very close to the question “What maths can be developed over an arbitrary elementary topos?”.)
| 9 | https://mathoverflow.net/users/2273 | 41200 | 26,326 |
https://mathoverflow.net/questions/41202 | 3 | I'm looking for information about how representations of $S\_n$ decompose under restriction.
I know about the branching rule: That is, in characteristic 0, irreducible modules $L(\lambda)$ for $S\_n$ decompose into a direct sum of irreducible modules of $S\_{n-1}$ and there is a nice combinatorial rule for determining the multiplicities of the $S\_{n-1}$-modules in the decomposition.
Are there similar results for the embedding of $S\_k \times S\_{n-k}$ in $S\_n$? That is, is there a way of computing the multiplicity of the irreducible $S\_k \times S\_{n-k}$-module $L(\nu) \otimes L(\mu)$ in $L(\lambda) \downarrow^{S\_n}\_{S\_k \times S\_{n-k}}$?
| https://mathoverflow.net/users/9821 | Decompositions for Symmetric Groups | To elaborate on Alexander Woo's comment, the multiplicity of $L(\nu) \otimes L(\mu)$ in $L(\lambda)$ restricted is the Littlewood-Richardson coefficient $c^\lambda\_{\nu, \mu}$. See <http://en.wikipedia.org/wiki/Littlewood-Richardson_coefficient> for the statement. For a proof (i.e., why this is related to symmetric functions or representations of the general linear group), you can see Section 7.18 of Stanley's *Enumerative Combinatorics vol. 2*.
| 4 | https://mathoverflow.net/users/321 | 41209 | 26,332 |
https://mathoverflow.net/questions/41217 | 21 | This question arises from the talk by Voevodsky mentioned in
[this recent MO question](https://mathoverflow.net/questions/40920/what-if-current-foundations-of-mathematics-are-inconsistent). On one of his slides, Voevodsky says that
>
> a general formula even with one free variable describes a subset of
> natural numbers for which one can prove, using an argument similar to the
> one which is used in Goedel's proof, that there is not a single number n
> which can be shown to belong to this subset or not to belong to it.
>
>
>
And in his spoken commentary he adds that there is a formula defining
>
> a subset about which you can prove that it is impossible to say
> anything about this subset, whatsoever.
>
>
>
I interpret this as the claim that there is an arithmetically definable
set $S$ for which there is no theorem of Peano arithmetic of the form
$n\in S$ or $n\not\in S$. Perhaps I am misinterpreting, but can anyone
supply (informally) the definition of such a set?
| https://mathoverflow.net/users/1587 | Question arising from Voevodsky's talk on inconsistency | Let $S$ be a first order definable Martin-Löf random set such as Chaitin's $\Omega$. If Peano Arithmetic, or ZFC, or any other theory with a computable set of axioms, proves infinitely many facts of the form $n\in S$ or $n\not\in S$ then it follows that $S$ is not immune or not co-immune and hence not ML-random after all.
(The set of theorems of our theory of the form $n\in S$ (or $n\not\in S$) is computably enumerable and infinite, hence has an infinite computable subset. Being immune means having no infinite computable subset.)
So only finitely many such facts can be proved. Now using an effective bijection between $\mathbb N$ and $\mathbb N\times \mathbb N$, decompose $S$ into infinitely many "columns", $S=S\_0\oplus S\_1\oplus\cdots$. Then one of these columns has the required property.
The theory should be strong enough to deal effectively with breaking a definable set up into columns and associating values in a column with values in the original set, but this is certainly doable in PA or ZFC.
| 21 | https://mathoverflow.net/users/4600 | 41220 | 26,337 |
https://mathoverflow.net/questions/41127 | 7 | Suppose $p\_n$ is $n$-th prime, $g\_n=p\_{n+1}-p\_n$ is the corresponding [prime gap](http://en.wikipedia.org/wiki/Prime_gap). What is the highest number $C$, such that $p\_N>C$ can be proven for $N=\min\{n\mid g\_n\geq 1.609\cdot 10^{18}\}$.
**Motivation:** I've read about [Goldbach's weak conjecture](http://en.wikipedia.org/wiki/Goldbach%27s_weak_conjecture). The number $C$ above is obvious lower bound for the first odd number, which does not admit a representation as a sum of three primes, which follows from check of Goldbach's conjecture up to $1.609\cdot 10^{18}$, which is done already by computers. I just want to know, how big is it.
| https://mathoverflow.net/users/8134 | What is the best known estimate for the place of the prime gap with length 1.609*10^18? | The title and body ask different questions, so I'll address both.
A reasonable estimate for the first prime gap of length $L$ is $e^{\sqrt L}$, so no prime gap this large would be expected below $\exp(1.268\cdot10^9)$, a 550,886,759-digit number.
As for a lower bound, Dusart 2010 [1] shows that for $x\ge396,738$, there is a prime between x and $x\left(1+\frac{1}{25\log^2x}\right)$, so 113353896002617492536754 (about $1.13\cdot10^{23}$) is a lower bound.
Under the Riemann hypothesis ([2]), the bound can be improved to 15373988432858515871940264945439 (about $1.5\cdot10^{31}$).
[1] <http://arxiv.org/abs/1002.0442>
[2] Lowell Schoenfeld, 'Sharper Bounds for the Chebyshev Functions $\theta(x)$ and $\psi(x)$. II'. Mathematics of Computation, Vol 30, No 134 (Apr 1976), pp. 337-360.
| 8 | https://mathoverflow.net/users/6043 | 41222 | 26,338 |
https://mathoverflow.net/questions/41173 | 3 | This is a followup to [an earlier question](https://mathoverflow.net/questions/33597/) on a taxonomy for quantum algorithms in which I ultimately concluded in a comment that all known nontrivial quantum algorithm speedups (in [Jordan's quantum zoo](http://www.its.caltech.edu/~sjordan/zoo.html)) could be regarded as arising from four basic classes of quantum subroutines: 1) quantum Fourier transform; 2) amplitude amplification [incl. Grover]; 3) quantum simulation/annealing; 4) quantum walks.
Now I wonder if it would be more fair to say *three* classes...
Today I read in more than one place specifically that [amplitude amplification](http://en.wikipedia.org/wiki/Amplitude_amplification) can be regarded as a [quantum walk](http://en.wikipedia.org/wiki/Quantum_walk) algorithm, but have not been able to find a definitive demonstration or reference (I have glanced at [Santha's paper](http://arxiv.org/abs/0808.0059) and similar things; while I may have missed something, this smells like it would be a MO-hard reference request) other than the overly general result that [quantum walks are universal for quantum computation](http://prl.aps.org/abstract/PRL/v102/i18/e180501).
The nearest statement I can feel comfortable with (especially having not gone over any details yet for myself) having some obvious meaning is that Grover search can be regarded as a quantum walk algorithm, but it's not clear to me why this would entail that amplitude amplification would be realizable as a quantum walk algorithm. Perhaps I'm missing some "amplitude amplification is Grover-hard" result...?
>
> Anyway, how is amplitude
> amplification explicitly realizable as a quantum
> walk algorithm?
>
>
>
---
As a more philosophical followup, if quantum walks are universal for quantum computation, does it really make sense to talk about them as subroutines?
[This makes me think that I might be justified in going so far as to classify all known quantum algorithm speedups as relying in principle on either a) the quantum Fourier transform or b) an arbitrary **BQP**-complete computational primitive (e.g., simulation or walk). In this setting the taxonomic question would be appropriately recast about what sorts of problems are better suited to solution via a given **BQP**-complete primitive, and what architectures are better suited to implementing a given **BQP**-complete primitive.]
| https://mathoverflow.net/users/1847 | Amplitude amplification as a quantum walk algorithm | To understand how amplitude amplification naturally fits in as a particularly simple quantum walk, I'd recommend understanding properly how Grover search works as a quantum walk. Try reading [Andrew Childs' lecture notes](http://www.math.uwaterloo.ca/~amchilds/teaching/w08/co781.html), particularly lectures 14 and 15. He explains how to view Grover search as a discrete-time and continuous-time quantum walk. I think after understanding that, the case of amplitude amplification should be clear.
To answer your philosophical question, consider the classical case. Linear programming is P-complete, and thus every problem in P can be solved by linear programming. But you wouldn't apply linear programming to every problem you see. Some problems have natural LP formulations, and some don't. So you can't just say "Linear programming is P-complete, so I'll solve everything using linear programming."
Similarly, quantum walks are BQP-complete, but not all quantum algorithms have a natural quantum walk formulation. Calling any BQP-complete problem a computational primitive is a bit like calling the Circuit value problem and the Horn-satisfiability problem primitives for designing polynomial time algorithms. While linear programming is considered a classical algorithmic technique, it's not because of its P-completeness. It's because of the wide variety of problems that can be cast as LPs and solved in polynomial time. Similarly, quantum walks are a quantum algorithmic technique because many problems can be solved naturally using quantum walks, but not all BQP-complete problems can be considered to be algorithmic techniques.
| 5 | https://mathoverflow.net/users/8075 | 41233 | 26,347 |
https://mathoverflow.net/questions/41229 | 1 | A *weakly initial set* in a category C is a set of objects I of C such that every object a of C has at least one arrow from an object contained in I.
The question is then, does Fields have a weakly initial set? This is equivalent to the collection of prime fields being a set.
The converse is, is there a (fairly natural) example of a category without a weakly initial set? Aside from obvious things like the discrete category on the objects of a large category.
| https://mathoverflow.net/users/4177 | Weakly initial sets - examples and nonexamples | Regarding the second question: I'm not sure what should count as "natural", but couldn't you just work with examples where the solution set condition in an adjoint functor theorem fails? The solution set *is* a weakly initial set in a comma category.
For example, there is no left adjoint to the underlying-set functor $U$ from complete Boolean algebras to sets, and in particular no free complete Boolean algebra on a countably infinite set. But the category of complete Boolean algebras is small-complete and $U$ preserves all small limits. So it's the solution set condition that fails, and therefore the comma category
$$\mathbb{N} \downarrow U$$
has no weakly initial set.
Edit: After reading David's request for really simple, I offer instead $Ord^{op}$, where $Ord$ is the class of ordinals ordered by inclusion. I acknowledge the influence of Laurent's answer.
| 1 | https://mathoverflow.net/users/2926 | 41235 | 26,348 |
https://mathoverflow.net/questions/41194 | 18 | I have heard through the academic rumor mill (my advisor heard from so-and-so about a result they heard from big-name who saw it in some journal, etc.) of the following theorem:
**Theorem:** Almost all [strongly regular graphs](http://en.wikipedia.org/wiki/Strongly_regular_graph) have trivial automorphism group.
This contrasts that most known families of strongly regular graphs have high symmetry, due to their constructions using algebraic objects.
Does anyone know the reference for this theorem? Also, what is the measure used to describe "almost all"?
| https://mathoverflow.net/users/4167 | Are "almost all" strongly regular graphs rigid? | The article *Random strongly regular graphs?* by Peter Cameron <http://www.maths.qmul.ac.uk/~pjc/preprints/randsrg.pdf> provides some information about what is known and why someone might make that claim.
First an example: There are 11,084,874,829 strongly regular graphs with parameters SRG(57,24,11,9) which arise from a Steiner triple system with 19 points (and 57 blocks); Of these 11,084,710,071 are rigid. (There might be other SRG(57,24,11,9))
MR2059752 (2005b:05035)
Kaski, P; Östergård, P
*The Steiner triple systems of order 19.*
Math. Comp. 73 (2004), no. 248, 2075--2092
<http://www.ams.org/journals/mcom/2004-73-248/S0025-5718-04-01626-6/S0025-5718-04-01626-6.pdf>
Cameron explains that the SRG with smallest eigenvalue -m are of 4 types:
1) a complete multipartite graph with km blocks of size m (so $v=km$)
2) Produced from $m-2$ mutually orthogonal $k\times k$ Latin squares (so $v=k^2$, nodes connected if they are in the same row or column, or have the same symbol in one of the squares)
3) The vertices are the blocks of a Steiner system with blocks of size m (so $v={\binom{k}{2}}/{\binom{m}{2}}$.
4) A finite list of exceptions $\mathcal{L}(m)$.
Type 1 has a huge automorphism group, but there are not very many of them.
Type 2: For $m=3$, there are on order of $n^{n^2/6}$ latin squares of order $n$, most with trivial automorphism group.
Type 3: For $m=3$ one has Steiner triple systems as above, there are on order $n^{n^2}$ and most are rigid.
Much less is known about sets of $m$ mutually orthogonal latin squares and about Steiner systems with block size $m$.
There are also graphs whose lower two eigenvalues are irrational conjugates (in some ring).
Any graph with $n$ vertices is an induced subgraph of a SRG with at most $4n^2$ vertices. On the other hand, every finite group is the automorphism group of a SRG (if I recall correctly). So the feeling is that there are lots of SRG with lots of freedom to construct them and most are rigid.
The notion of switching **is** useful. In a STS a Pasch configuration is a set of 6 points and 4 triples abc ade fbe fcd. This would correspond to a 4-clique in the corresponding graph. Switching these to abe acd fbc fde would still leave a 4 clique in the graph but would shift around the connections to the rest of the graph. There can be more elaborate switches too (I think). With enough room one can probably destroy all automorphisms this way. Of the rigid STS(19) above, 2538 don't have any Pasch configurations but over 1,000,000,000 have 14 (similarly for 15 and 16).
| 11 | https://mathoverflow.net/users/8008 | 41240 | 26,352 |
https://mathoverflow.net/questions/41174 | 32 | I am trying to find the precise statement of the correspondence between stable Higgs bundles on a Riemann surface $\Sigma$, (irreducible) solutions to Hitchin's self-duality equations on $\Sigma$, and (irreducible) representations of the fundamental group of $\Sigma$. I am finding it a bit difficult to find a reference containing the precise statement. Mainly I'd like to know the statement for the case of stable $GL(n,\mathbb{C})$ Higgs bundles. But if anyone knows the statement for more general Higgs bundles that would be nice too.
Just at the level of say sets and not moduli spaces, I think the statement is that the following 3 things are the same, if I am reading Hitchin's original paper correctly:
* stable $GL(n,\mathbb{C})$ Higgs bundles modulo equivalence,
* irreducible $U(n)$ (or is it $SU(n)$?) solutions of the Hitchin equations modulo equivalence,
* irreducible $SL(n,\mathbb{C})$ (or is it $GL(n,\mathbb{C})$? $PSL$? $PGL$?) representations of $\pi\_1$ modulo equivalence.
Is this correct? Is there a reference?
Hitchin's original paper (titled "Self duality equations on a Riemann surface") does some confusing maneuvers; for example he considers solutions of the self-duality equations for $SO(3)$ rather than for $U(2)$ or $SU(2)$, which would seem more natural to me. Moreover, for instance, he doesn't look at all stable Higgs bundles, but only a certain subset of them - but I think this is just for the purpose of getting a *smooth* moduli space. And finally, Hitchin looks at $PSL(2,\mathbb{C})$ representations of $\pi\_1$ rather than $SL(2,\mathbb{C})$ representations or $GL(2,\mathbb{C})$ representations, which confuses me as well...
Thanks in advance for any help!!
**EDIT:** Please note that I am only interested in the case of a Riemann surface. Here it appears that *degree zero* stable Higgs bundles correspond to $GL(n,\mathbb{C})$ representations. But the question remains: are stable Higgs bundles of *arbitrary degree* related to representations? If so, which representations, and how are they related? Moreover, I think that general stable Higgs bundles should correspond to solutions of the self-duality equations -- but what's the correct group to take? ("Gauge group"? Is that the correct terminology?) I think it's $U(n)$ but I am not sure.
For example, in Hitchin's paper, he considers the case of rank 2 stable Higgs bundles of *odd degree* and fixed determinant line bundle, with trace-zero Higgs field (see Theorem 5.7 and Theorem 5.8). As for the self-duality equations, he uses the group $SU(2)/\pm 1$. We get a smooth moduli space. In the discussion following Theorem 9.19, it is shown that this moduli space is a *covering* of the space of $PSL(2,\mathbb{C})$ representations. It seems that this should generalize...
| https://mathoverflow.net/users/83 | What is the precise statement of the correspondence between stable Higgs bundles on a Riemann surface, solutions to Hitchin's self-duality equations on the Riemann surface, and representations of the fundamental group of the Riemann surface? | See Ó. Garcia-Prada's appendix to the third edition of R.O. Wells' book Differential Analysis on Complex Manifolds. It addresses most of what you are asking quite explicitly, in the context of Riemann surfaces, and has references to the original papers. Also the book of Lübke and Teleman on the Kobayashi-Hitchin correspondence might be helpful.
| 8 | https://mathoverflow.net/users/9471 | 41241 | 26,353 |
https://mathoverflow.net/questions/41214 | 59 | Mathematics has undergone some rather nice developments recently with the adoption of new techologies, things like on-line journals, the arXiv, this website, etc. I imagine there must be many further developments that could be quite useful.
What I'm thinking of is a website where anyone can contribute formal proofs of theorems. In particular there would be many proofs of the same theorem provided the proof is different -- like a constructive proof of Brouwer's fixed point theorem, and non-constructive proof, etc.
The idea would be to build up a large web of formal proofs, one building on another so that one could eventually do searches through this space of formal proofs to find out what the most efficient proofs are, in the sense of how many ASCII characters it would take to write-up the proof using Zermelo-Frankel set theory. One hope would be to have a big, active database of verified formal proofs. Another would be to have a webpage where you could hope to discover whether or not there are simpler proofs of theorems you know, that you may have not been be aware of.
Being a web-page there would be certain useful efficiencies -- the webpage could "compile" your proof and check to see it's valid. Being a wiki would make it relatively easy for people to contribute and build on an existing infrastructure. And you'd be free to use pre-existing proofs (provided they've been verified as valid) in any subsequent proofs. One could readily check what axioms a proof needs -- for example to what extent a proof needs the axiom of choice, and so on.
Is there any efforts towards such a development? Such a tool would hopefully function like the publishing arm of some sort of modern internet-era Bourbaki.
| https://mathoverflow.net/users/1465 | Has anyone thought about creating a formal proof wiki with verifier? | There are lots of sites for formal proofs, but no wiki that I am aware of.
Typical examples are:
archive of formal proofs at <https://www.isa-afp.org/>
Mizar <http://mizar.org/>
Lots of proofs are contained in the distributions of various interactive theorem provers: Isabelle, Hol, hol light, Coq, acl2 etc etc
As stated in another post, there is no agreement on foundations (formulas, axioms and rules of inference). A typical split is between classical (Hol et al.) and non-classical (Coq et al.) systems, but the differences are typically much more subtle than that. As a result all these systems are effectively unable to reuse proofs from other systems. Occasionally someone writes a translator from one system to another, but the problem here is that the translation typically does not produce a readable proof in the target system; a readable proof is necessary if the translated proof is to be maintainable. If you fix on ZFC+(maybe some other axioms), then Mizar probably has the most extensive library.
Every few years, someone proposes a big database of formal proofs, but these projects invariably die for various reasons related to the issues above. An example is the QED project:
<http://en.wikipedia.org/wiki/QED_manifesto>
My personal view is that constructing formal proofs, and maintaining them, is currently too difficult. Having said that, in the long run this is clearly an idea whose time will come.
| 34 | https://mathoverflow.net/users/9831 | 41242 | 26,354 |
https://mathoverflow.net/questions/41212 | 16 | I would be interested to learn if the following generalization of the classical [Looman-Menchoff theorem](https://en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem) is true.
---
Assume that the function $f=u+iv$, defined on a domain $D\subset\mathbb{C}$, is such that
1. $u\_x$, $u\_y$, $v\_x$, $v\_y$ exist almost everywhere in $D$.
2. $u$, $v$ satisfy the Cauchy–Riemann equations almost everywhere in $D$.
3. $f=f(x,y)$ is separately continuous (in $x$ and $y$) in $D$.
4. $f$ is locally integrable.
>
> **Question:** Does it follow that $f$ is analytic *everywhere* in $D$?
>
>
>
---
**Remark 1.** Condition 3 is essential (take $f=1/z$).
**Remark 2.** G. Sindalovskiĭ proved analyticity of $f$ under conditions 2-4 when the partial derivatives exist everywhere in $D$, *except on a countable union of closed sets of finite linear Hausdorff measure* ([link](https://iopscience.iop.org/article/10.1070/SM1987v056n02ABEH003041 "DOI: 10.1070/SM1987v056n02ABEH003041")).
| https://mathoverflow.net/users/5371 | The Cauchy–Riemann equations and analyticity | No.
Let $c$ be the [Cantor function](https://en.wikipedia.org/wiki/Cantor_function) on $[0,1]$, so that $c$ is continuous, $c' = 0$ a.e., but $c$ is not constant. Then take $u(x+iy) = v(x+iy)=c(x)c(y)$. We have $u\_x=u\_y=v\_x=v\_y=0$ a.e. so the Cauchy–Riemann equations are trivially satisfied, and $f(z)=u(z)+iv(z)$ is bounded and continuous on the unit square, but certainly not analytic.
Almost everywhere differentiability is almost never the right condition for solutions to a PDE. A better condition would be to have $u,v$ in some Sobolev space.
| 27 | https://mathoverflow.net/users/4832 | 41268 | 26,366 |
https://mathoverflow.net/questions/41254 | 1 | Suppose $\mathcal{A}$ is a quasi coherent sheaf of algebras over a group scheme $\mathcal{G}$. Suppose it is generated by global section. Then , what can we say about the external tensor product $\mathcal{A}\boxtimes\mathcal{A}$? Will this sheaf also be generated by the tensor product of the global section with itself? Or is it bigger than this?
| https://mathoverflow.net/users/9492 | External tensor product of sheaves | What do you mean by $A\boxtimes A$? If this is the sheaf $p\_1^\*A\otimes p\_2^\*A$ on $G\times G$ then certainly it is globally generated by the tensor square of global sections. It follows easily from right-exactness of the pullback and of the tensor product --- let $V$ be the space of global sections, then $V\otimes O\_G \to A$ is surjective, hence $V\otimes O\_{G\times G} \to p\_i^\*A$ is surjective, hence $V\otimes V\otimes O\_{G\times G} \to A\boxtimes A$ is surjective.
| 4 | https://mathoverflow.net/users/4428 | 41269 | 26,367 |
https://mathoverflow.net/questions/41264 | 4 | Given a diagram $X\_1 \rightarrow X\_{12} \leftarrow X\_2$ of spaces (though I think the question applies more generally), there are two cosimplicial resolutions which I've seen used to compute the homotopy pullback. The first one, which I'll call $\mathcal A$ is given by $$\mathcal A^p = X\_1 \times X\_{12}^{\times p} \times X\_2$$ and the coface maps are like in a bar (cobar?) construction. This is the one you see, for example, in the construction of the Eilenberg-Moore spectral sequence.
For the other one, let's let $\mathscr P$ denote the pullback category, and $F$ our functor. Then the standard cosimplicial replacement, which I'll write $\mathcal B$, is given by $$\mathcal B^p = \prod\_{\substack{x\_0 \rightarrow \cdots \rightarrow x\_p \\ \in \mathscr P}} F(x\_p)$$
It seems that I always encounter one or the other, but not both, which leads me to my question: how do I know they give the same answer? The first one computes the homotopy pullback basically by definition, whereas the second one is a bit mysterious to me. Is there a simple way to compare these two cosimplicial spaces? They do not appear to be level-wise equivalent. Can we say anything about the relationship of their $\mathrm{Tot}$ towers?
| https://mathoverflow.net/users/4466 | Equivalence of cosimplicial models for homotopy pullbacks | The first one has codegeneracies as well as cofaces, yes? And when you say that it computes the htpy pullback basically by definition I suppose you mean that its Tot is homeomorphic to the space whose points are triples consisting of a point in $X\_1$ a point in $X\_2$ and a path connecting their images in $X\_{12}$? That's how I think of it.
And if you think about Tot of the second one the same way, you should get the space in which a point is a $5$-tuple, one point in each of $X\_1$, $X\_{12}$, $X\_2$ and two paths in $X\_{12}$ connecting the middle point with the images of the other two. Which is homeomorphic to the same old thing.
EDIT In reply to Eric's comment: The way I think about it is that holim is the derived functor of lim, therefore well-defined (up to canonical weak eq in some strong sense) when it exists; and then the question is how do you make it? In a simplicial model cat you can always make it by Tot of Bousfield-Kan cosimplicial replacement, I guess, though in general you have to make things fibrant first. For particular indexing categories, there are sometimes even simpler models. For pullbacks, one model is replace the two maps in the diagram by fibrations and then make lim (= pullback). This passes some general test for being a derived functor, so no need for direct comparison with the B-K model. Likewise the cobar construction model should pass that test.
| 5 | https://mathoverflow.net/users/6666 | 41272 | 26,370 |
https://mathoverflow.net/questions/41262 | 10 | What can be said about the structure of maximal ideals of $R=k[\{x\_i\}\_{i \in I}]$, or geometric properties of $\text{Spm } k[\{x\_i\}\_{i \in I}]$? Here $k$ is an arbitrary field and $I$ is an *infinite* set. Kernels of evaluation homomorphisms yield an injective map
$\overline{k}^I / Aut(\overline{k}/k) \to \text{Spm } k[\{x\_i\}\_{i \in I}]$.
The image consists of those maximal ideals whose residue field is algebraic over $k$. If $I$ is finite, every residue field is algebraic (Noether Normalization). However, if $I$ is infinite and $|I| \geq |k|$, for example $k(t)$ is a residue field which is not algebraic. What happens if $|k| > |I|$? Is there a description in the general case?
| https://mathoverflow.net/users/2841 | maximal ideals of $k[x_1,x_2,...]$ | If $|k| > |I|$ then the usual cheap proof of Nullstellensatz still works: let $K$ be a residue field. Then $\dim\_k K \le \dim\_kR = |I|$, but if $t\in K$ is transcendental over $k$, the elements $1/(t-a)$ for $a\in k$ are $k$-linearly independent. So $K/k$ is algebraic.
| 16 | https://mathoverflow.net/users/5480 | 41275 | 26,372 |
https://mathoverflow.net/questions/41279 | 1 | Let $D$ be a set, $\mathbb{N\_0}$ the set of natural numbers including zero. Let $P$ be the set of all functions from $D$ to $\mathbb{N\_0}$, i.e. $P = \lbrace m \mid m: D \rightarrow \mathbb{N\_0} \rbrace$.
Let $f, g \in P$. Then we define that $f \leq g$ holds iff $\forall d\in D f(d) \leq g(d)$. So $\leq$ is a partial order on $P$.
Now, $(P,\leq)$ is a lattice but it is not a complete lattice because some subsets of $P$ don't have a supremum in $P$.
So let us allow the functions to map to an infinity, $\infty$, too, i.e. $\forall n\in \mathbb{N\_0} n < \infty$. Let $P' = \lbrace m \mid m: D \rightarrow \mathbb{N\_0} \cup \infty \rbrace$.
Is $(P',\leq)$ a complete lattice? Resp., I am quite sure it is and I would like to know if there is some reference I could cite or if it is so trivial that I can simply state it.
| https://mathoverflow.net/users/9711 | A complete lattice of functions | Once one adds infinity, it is easy to check that $(P', \leq)$ is indeed a complete lattice. For any subset $A$ of $P'$, it is easy to see that $\vee A$ is just the pointwise supremum of members of $A$. In the case that $A$ is empty, we have that $\vee A$ is the zero function. So, $(P', \leq)$ is a complete join semi-lattice, and hence a complete lattice.
| 2 | https://mathoverflow.net/users/2233 | 41281 | 26,376 |
https://mathoverflow.net/questions/41234 | 18 | Fix some $n \geq 3$. It's hopeless to classify the torsion elements in $\text{GL}\_n(\mathbb{Z})$, but I have a couple of less ambitious questions. It's well-known that for any odd prime $p$, the map $\phi\_p : \text{GL}\_n(\mathbb{Z}) \rightarrow \text{GL}\_n(\mathbb{Z}/p)$ is ``injective on the torsion''. In other words, if $F$ is a finite subgroup of $\text{GL}\_n(\mathbb{Z})$, then $\phi\_p|\_F$ is injective.
Fixing an odd prime $p$, I have two questions about the map $\phi\_p$.
1. Are there any interesting restrictions on elements $y \in \text{GL}\_n(\mathbb{Z}/p)$ such that there exists some finite-order $x \in \text{GL}\_n(\mathbb{Z})$ with $\phi\_p(x) = y$?
2. Do there exist any non-conjugate finite order elements $x,x' \in \text{GL}\_n(\mathbb{Z})$ such that $\phi\_p(x) = \phi\_p(x')$? The hypothesis that $x$ and $x'$ are non-conjugate is to rule out the silly example where $x' = g x g^{-1}$ with $g \in \text{ker}(\phi\_p)$.
| https://mathoverflow.net/users/317 | Torsion in GL_n(Z) | Here is a loose collection of partial answers, most of which address a simpler question than that Andy actually asked. I am particularly focusing on Andy's question in the comments "can we classify all liftable elements of ${\rm GL}\_n(\mathbb{Z}/p\mathbb{Z})$ of order $k$ for some fixed $k$?". Please take all this with a grain of salt, because I am improvising as I go and it's already late at my end.
The first remark is that I don't think that restricting the order of the element makes the problem any easier. That's because, suppose that you have an element $X$ of order $k$ that lifts from ${\rm GL}\_n(\mathbb{Z}/p\mathbb{Z})$ to ${\rm GL}\_n(\mathbb{Z})$. Then, for any invertible $A\in {\rm GL}\_n(\mathbb{Z}/p\mathbb{Z})$, to find out whether $AXA^{-1}$ lifts, you will almost certainly have to determine whether $A$ lifts. But $A$ can be of arbitrary order.
To fix this, I will instead consider the question: for a given $k$, what modular representations of the cyclic group of order $k$ can be lifted to integral ones. This also bypasses another issue: namely that the order of the lift could potentially be higher than that of the original element.
Edit: please ignore that last sentence. By injection of torsion, this can't happen, as pointed out by vytas.
One fact that I will implicitly use all the time is that if two integral representations give rise to the same rational representation, then their reductions have the same semi-simplifications.
**Case 1:** Suppose that $k$ is co-prime to $p$ and that the order of $p$ in $(\mathbb{Z}/k\mathbb{Z})^\times$ is $\varphi(k)$, in other words that the extension of $\mathbb{F}\_p$ obtained by adjoining the $k$-th roots of unity is as big as possible. This ensures that it's "no easier" for a representation of $C\_k$ to be defined over $\mathbb{F}\_p$, than over $\mathbb{Q}$ (equivalently than over $\mathbb{Z}$). Then, by construction, there is a bijection between indecomposable mod $p$ representations and indecomposable integral representations of $C\_k$, thus any representation lifts. But as I remarked at the beginning, that probably doesn't mean that any element of order $k$ lifts. (Or does it? Can someone fix this issue?)
**Case 2:** Let's just assume that $k$ is co-prime to $p$. The previous case should give a hint for this more general one: there is a bijection between representations of $C\_k$ over a sufficiently large extension of $\mathbb{F}\_p$ and representations over a sufficiently large extension of $\mathbb{Q}$. For example if $k$ is prime, then the irreducible rational representations are the trivial and a $p-1$-dimensional. Therefore, a representation can be lifted if and only if the numbers of copies of all the non-trivial representations of $C\_k$ over $\mathbb{F}\_p(\zeta\_k)$ in this representation are equal. Note that talking about liftable representations, rather than matrices allows me to ignore the very subtle issue of non-isomorphic integral representations sitting in the rational one - they all give the same reduction modulo $p$. If $k$ is composite, then a similar analysis can be carried out.
**Case 3:** Suppose that $k=p$ or $k=p^2$. There is one indecomposable modular representation of dimension $n$ for each $n$ between 1 and $p$. Their semi-simplifications are all sums of trivial representations. Moreover, if I remember correctly, the reduction of any integral representation sitting inside the irreducible rational one of dimension $p-1$ remains indecomposable over $\mathbb{F}\_p$. Thus, a representation is liftable, if and only if it is a direct sum of indecomposable representations of dimensions 1 and $p-1$. I believe that with some more work, the case $k=p^2$ can be worked out in a similar vein.
**Case 4:** $k$ is divisible by $p^3$. Here, all hell breaks loose, because now, there are infinitely many isomorphism classes of indecomposable integral representations and they can have arbitrarily large ranks. No one knows how to classify them, so it's hopeless to try listing their reductions. However, since you are fixing the dimension $n$, there is some good news:
**Jordan-Zassenhaus Theorem:** there exist only finitely many isomorphism classes of integral representations giving rise to any given rational representation.
If I remember correctly, the proof constructs these for you (it breaks up into two parts: first showing that there are only finitely many isomorphism classes in each genus (i.e. integral representations that are locally isomorphic everywhere) and then showing that there are finitely many genera in any given rational representation). So for any given $n$ and $k$, you can follow through the proof to get a classification of integral representations and therefore of their reductions. As far as I know, there is no uniform description of the answer, you just have to compute it in any given case.
I hope, I didn't write completely off topic, seeing as I didn't really address the actual question.
For most of the facts about integral representations together with proper references, see Reiner's [excellent survey article](http://projecteuclid.org/euclid.bams/1183531476). The modular stuff can be found e.g. in Benson's book, or, in those cases, where I messed up, not at all.
| 11 | https://mathoverflow.net/users/35416 | 41290 | 26,378 |
https://mathoverflow.net/questions/41273 | 4 | Suppose we have $n$ normal variable $X\_1,X\_2,\dots,X\_n$, with corresponding mean $\mu\_1,\dots,\mu\_n$ and sd $\sigma\_1,\dots,\sigma\_n$. What is the probability of $X\_1 < X\_2 < \dots < X\_n$, i.e. $P(X\_1 < X\_2 < \dots < X\_n)$.
Numerical method is also okay. Actually I have thought about how to do the multi-variable integration but I cannot get it done. When $n=2$, it is easy to see that $X\_1-X\_2$ is also a normal variable so $P(X\_1-X\_2<0)$ is quite easy to calculate. However, I think when $n=3$, because of $P(X\_1-X\_2)$ and $P(X\_2-X\_3)$ are not independent, the result is not easy to calculate. Currently I am thinking about whether we can use some method to approximately calculate the value the that probability. I need to come up with an algorithm to compute that.
Thank you!
| https://mathoverflow.net/users/9836 | How to calculate the probability of N normal variable being in increasing order? | We can use multivariate normal distribution to solve this problem. Denote the difference between consecutive elements with a (n-1)-dimensional random vector $Y=(X\_1-X\_2,X\_2-X\_3,\dots, X\_{n-1}-X\_n)$, what we need to find is $Y<(0,0,...,0)$.
Random vector $Y$ has a multivariate normal distribution.
The mean vector $\mu = (E(X\_1-X\_2), \dots, E(X\_{n-1}-X\_n)) = (\mu\_1 - \mu\_2, \dots, \mu\_{n-1} - \mu\_n)$.
The covariance matrix $\Sigma = [Cov[X\_i-X\_{i+1}, X\_j-X\_{j+1}]], i=1,\dots, n-1, j=1,\dots, n-1$.
For each entry of the matrix,
$Cov\_{i,i} = Var(X\_i-X\_{i+1}) = \sigma\_i^2 + \sigma\_{i+1}^2$
$Cov\_{i,i+1} = Cov\_{i+1,i} = E((X\_i-X\_{i+1})(X\_{i+1}-X\_{i+2})) - E(X\_i-X\_{i+1})E(X\_{i+1}-X\_{i+2})$
$ = E(X\_iX\_{i+1} + X\_{i+1}X\_{i+2}-X\_{i+1}^2-X\_iX\_{i+2})-(\mu\_i-\mu\_{i+1})(\mu\_{i+1}-\mu\_{i+2})$
Since $X\_i, X\_{i+1}, X\_{i+2}$ are independent, we have the above expression
$= E(X\_i)E(X\_{i+1})+E(X\_{i+1})E(X\_{i+2})-E(X\_{i+1}^2)-E(X\_i)E(X\_{i+2}) - (\mu\_i-\mu\_{i+1})(\mu\_{i+1}-\mu\_{i+2})$
Apart from $E(X\_{i+1}^2)$, other terms are easy to calculate now. For $E(X\_{i+1}^2)$, it can be calculate using chi-square distribution of order one. The result turns out to be $\mu\_{i+1}^2+\sigma\_{i+1}^2$. Plug in the above equation, we have $Cov\_{i,i+1}=-\sigma\_{i+1}^2$.
Other entries in the covariance matrix is zero.
Finally, use numerical method to calculate the probability. One implementation of the cdf of multivariate normal distribution can be found in <http://www.math.wsu.edu/faculty/genz/software/software.html>, the MVNDST function.
| 5 | https://mathoverflow.net/users/9836 | 41293 | 26,379 |
https://mathoverflow.net/questions/41283 | 3 | Given a pizza, represented by the unit disk $D\_1(0,0)=\{(x,y)\in\mathbb{R}^2\mid \|(x,y)\|\leqslant 1\}$, and given $N$ slices of $r$-pepperoni, represented by disks $D\_r(a\_i,b\_i)=\{(x,y)\in\mathbb{R}^2\mid \|(x,y)-(a\_i,b\_i)\|\leqslant r\}$, for $i\in[1..N]$.
Assume that the $N$ slices are randomly distributed on the pizza and that:
1. for each $i\in[1..N]$, $D\_r(a\_i,b\_i)\subseteq D\_1(0,0)$; and
2. for each $i,j\in[1..N]$, if $i\neq j$ then $D\_r(a\_i,b\_i)\cap D\_r(a\_j,b\_j)=\emptyset$.
Suppose that the pizza is shared by $2$ people. Is it always possible to slice the pizza along a single straight line such that the two parts have the same area (same amount of pizza) and the same total area of pepperoni?
What about generalizations to $K$ people and cuts along line segments?
| https://mathoverflow.net/users/9839 | Is always possible to slice a pizza in a fair way | Intermediate Value Theorem (for two people).
We draw a $x$-axis through the centre of the disk. For an $\theta \in R$ we draw an axix which makes an angle of $\theta$-degrees with the x-axis (note that the axix for $\theta+\pi$ gives exactly the opposite direction).
Lets look now at the two half disks defined by this $\theta$-axis.
Let $f(\theta)$ denote the quantity of peperoni on the left half disk minus he quantity of peperoni on the rigth half disk (since the axis is oriented, left-rigth make sense, one looks in the direction of the axis).
$f$ is continuous in $\theta$, and since $f(\theta)+f(\theta+\pi)=0$, either $f \equiv 0$ or
$f$ takes both positive and negative values. Now the IVT completes the proof.
| 2 | https://mathoverflow.net/users/9498 | 41295 | 26,380 |
https://mathoverflow.net/questions/41257 | 4 | For given continuous real functions $f$ and $g$ defined on $[-1,1]$, let's define
$$
D(f,g) = \sup\_{x \in [-1,1]} \left|{\frac{f(x)-g(x)}{f(x)}}\right|
$$
(in this context, let's take $0/0$ to be $0$ and $x/0 = \infty$ for every $x \not= 0$).
I am looking for a theory of approximation by polynomials with respect to this error that parallels the existing theory of approximation by polynomials with respect to the $L\_\infty$-norm.
For example: is it true that for every real $f$ continuous on $[-1,1]$ and every degree $n$ there exists a best approximating polynomial $p$ of degree $n$ in the sense that $D(f,p)$ is minimal among all polynomials of degree $n$? In such a case, can we find it or characterize it?
I would like to know if this has been studied before and where.
**Edit**: From the answers I got below, perhaps I should add that, in the application I have in mind, $f(x)$ may or may not be $0$ exactly at one place: at $x = 0$. This rules our infinitely many zeros in $[-1,1]$, and in case $f(0) = 0$, the best approximator $p(x)$ should be such that $p(0) = 0$ (my convention to take $0/0 = 0$ and $x/0 = \infty$ for $x \not= 0$ comes from this).
| https://mathoverflow.net/users/9355 | Relative error approximation by polynomials | According to Johan's answer, the problem is not well posed if the function $f$ has zeroes in $[-1,1]$ (if the number of zeroes is finite, perhaps something could be done). In the following I assume that $f(x)\ne0$ for all $x\in[-1,1]$. Let $n$ be a positive integer and let $\mathbb{P}(n)$ be the set of all real polynomials, and denote by $\|\cdot\|$ the $L^\infty$ norm on $[-1,1]$. Then you ask for the existence of $P\in\mathbb{P}(n)$ such that
$$\|1-P/f\|=\inf\_{p\in\mathbb{P}(n)}\|1-p/f\|.$$
This can be brought under the theory of approximation in the $L^\infty$ norm. Let $\phi\_k(x)=x^k/f(x)$, $1\le k\le n$. Each $\phi\_k$ is a continuous function, they are independent and generate an $n$-dimensional subspace of the space of continuous functions on $[-1,1]$ that we denote by $V$, which is nothing but the space of all functions of the form $P/f$ with $P\in\mathbb{P}(n)$. Moreover, each $\phi\in V$ has at most $n$ zeros in $[-1,1]$, so that it satisfies what is called *the Haar condition*. The original problem is now recast as: find the best approximation in $V$ of the constant function $1$.
The general theory of approximation shows that there is a unique best approximation, and that it can be characterized as follows: $\phi\in V$ is the best approximation in $V$ of the constant function $1$ if and only if there exist $x\_1 < x\_2 < \dots < x\_{n+2}$ in $[-1,1]$ such that
1. $|e(x\_k)|=\|e\|$, $1\le k\le n+2$,
2. $e(x\_k)=-e(x\_{k+1})$, $1\le k\le n+1$,
where $e(x)=1-\phi(x)$ is the error of the approximation. This is Tchebyshev's alternance theorem.
Finally, Remes' algorithm can be used to construct the best approximation.
**Edit** in response to your comment
If $f$ has a finite number of zeroes in $[-1,1]$ and can be written as $f=q\cdot h$ where $q$ is a polynomial and $h$ is a continuous function such that $h(x)\ne0$ for all $x\in[-1,1]$, then you may consider the space $V$ of all functions of the form $q\cdot p/f=p/h$ whith $p\in\mathbb{P}(n)$. Then
$$\frac{|f(x)-q(x)p(x)|}{|f(x)|} =|1-\frac{p(x)}{h(x)}|.$$
If $\phi$ is the best approximation to $1$ in $V$, then $q\cdot \phi$ will be a best approximation to $f$ of degree $n+\hbox{degree}(q)$ in the relative error sense.
| 5 | https://mathoverflow.net/users/1168 | 41298 | 26,382 |
https://mathoverflow.net/questions/41211 | 26 | Let $X$ be a finite-dimensional Banach space whose isometry group acts transitively on the set of lines (or, equivalently, on the unit sphere: for every two unit-norm vectors $x,y\in X$ there exist a linear isometry from $X$ to itself that sends $x$ to $y$). Then $X$ is a Euclidean space (i.e., the norm comes from a scalar product).
I can prove this along the following lines: the linear isometry group is compact, hence it admits an invariant probability measure, hence (by an averaging argument) there exists a Euclidean structure preserved by this group, and then the transitivity implies that the Banach norm is proportional to that Euclidean norm.
But this looks too complicated for such a natural and seemingly simple fact. Is there a more elementary proof? I mean something reasonably short and accessible to undegraduates (so that I could use it in a course that I am teaching).
**Added.** As Greg Kuperberg pointed out, there are many other ways to associate a canonical Euclidean structure to a norm, e.g. using the John ellipsoid or the inertia ellipsoid. This is much better, but is there something more "direct", avoiding any auxiliary ellipsoid/scalar product construction?
For example, here is a proof that I consider "more elementary", under the stronger assumption that the isometry group is transitive on two-dimensional flags (that is, pairs of the form (line,plane containing this line)): prove this in dimension 2 by any means, this implies that the norm is Euclidean on every 2-dimensional subspace, then it satisfies the parallelogram identity, hence it is Euclidean.
Looking at this, I realize that perhaps my internal criterion for being "elementary" is independence of the dimension. So, let me try to transform the question into a real mathematical one:
* Does the fact hold true in infinite dimensions (say, for separable Banach spaces)?
| https://mathoverflow.net/users/4354 | Easy proof of the fact that isotropic spaces are Euclidean | It is a famous problem (due to Banach and Mazur) whether a separable infinite dimensional Banach space which has a transitive isometry group must be isometrically isomorphic to a Hilbert space. Of course, if every two dimensional subspace has a transitive isometry group, then the space is a Hilbert space since then the norm satisfies the parallelogram identity. For counterexamples in the non separable setting, consider the $\ell\_p$ sum of uncountably many copies of $L\_p(0,1)$ with $p$ not $2$.
For a recent paper related to the Banach-Mazur rotation problem, which contains some other references related to the problem, see
<http://arxiv.org/abs/math/0110202>.
| 17 | https://mathoverflow.net/users/2554 | 41302 | 26,384 |
https://mathoverflow.net/questions/41285 | 10 | By the Riemann-Hilbert correspondence, there is an equivalence between
(1)
$\mathcal{D}\operatorname{-mod}(X)$
, the (derived) category of holonomic D-modules on a complex variety X, and
(2)
$D^b\_c(X)$
, the (derived) category of constructible sheaves on X.
There is a "naive" t-structure we can put on both categories. In
$\mathcal{D}\operatorname{-mod}(X)$
, we can look at a t-structure whose heart
$\mathcal{D}\operatorname{-mod}^\heartsuit$
is a complex (of D-modules) concentrated in degree 0. In
$D^b\_c(X)$
, we can look at the naive t-structure whose heart
$D^{b \heartsuit}\_c$
is a complex (of constructible sheaves) concentrated in degree 0.
It's known that if we transfer the naive t-structure on
$\mathcal{D}\operatorname{-mod}(X)$
to
$D^b\_c(X)$
(using the equivalence above),
$\mathcal{D}\operatorname{-mod}^\heartsuit$
is identified with "perverse sheaves" on X.
My question is:
>
> If we map
> $D^{b\heartsuit}\_c$
> to the category of D-modules using the Riemann-Hilbert correspondence, what subcategory of
> $\mathcal{D}\operatorname{-mod}$
> do we get? Does this have a well-known name?
>
>
> More generally, is there some geometric/nice description of what the naive t-structure on
> $D^b\_c$
> becomes on
> $\mathcal{D}{\operatorname{-mod}}$
> ?
>
>
>
| https://mathoverflow.net/users/3593 | How does one interpret the naive t-structure on constructible sheaves as a t-structure on D-modules? | The t-structure is described in
[this paper of Kashiwara](http://arxiv.org/abs/math/0302086).
It looks essentially like Donu and t3suji suggest in their comments: defined by conditions that look like middle-perversity support conditions.
| 13 | https://mathoverflow.net/users/2628 | 41304 | 26,385 |
https://mathoverflow.net/questions/41310 | 13 | The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice its not possible.
Formally: Let's say that $q\_{i}$ is the probability that we throw $i$ on the first die and $p\_{i}$ is the same for the second die. $p\_{i},q\_{i} \in [0,1]$ for all $i \in 1\ldots 6$. The question is that with these constraints are there $q\_{i}$s and $p\_{i}$s that satisfy the following equations:
$ q\_{1} \cdot p\_{1} = \frac{1}{11}$
$ q\_{1} \cdot p\_{2} + q\_{2} \cdot p\_{1} = \frac{1}{11}$
$ q\_{1} \cdot p\_{3} + q\_{2} \cdot p\_{2} + q\_{3} \cdot p\_{1} = \frac{1}{11}$
$ q\_{1} \cdot p\_{4} + q\_{2} \cdot p\_{3} + q\_{3} \cdot p\_{2} + q\_{4} \cdot p\_{1} = \frac{1}{11}$
$ q\_{1} \cdot p\_{5} + q\_{2} \cdot p\_{4} + q\_{3} \cdot p\_{3} + q\_{4} \cdot p\_{2} + q\_{5} \cdot p\_{1} = \frac{1}{11}$
$ q\_{1} \cdot p\_{6} + q\_{2} \cdot p\_{5} + q\_{3} \cdot p\_{4} + q\_{4} \cdot p\_{3} + q\_{5} \cdot p\_{2} + q\_{6} \cdot p\_{1} = \frac{1}{11}$
$ q\_{2} \cdot p\_{6} + q\_{3} \cdot p\_{5} + q\_{4} \cdot p\_{4} + q\_{5} \cdot p\_{3} + q\_{6} \cdot p\_{2} = \frac{1}{11}$
$ q\_{3} \cdot p\_{6} + q\_{4} \cdot p\_{5} + q\_{5} \cdot p\_{4} + q\_{6} \cdot p\_{3} = \frac{1}{11}$
$ q\_{4} \cdot p\_{6} + q\_{5} \cdot p\_{5} + q\_{6} \cdot p\_{4} = \frac{1}{11}$
$ q\_{5} \cdot p\_{6} + q\_{6} \cdot p\_{5} = \frac{1}{11}$
$ q\_{6} \cdot p\_{6} = \frac{1}{11}$
I don't really now how to start with this. Any suggestions are welcome.
| https://mathoverflow.net/users/9834 | Any sum of 2 dice with equal probability | You can write this as a single polynomial equation
$$p(x)q(x)=\frac1{11}(x^2+x^3+\cdots+x^{12})$$
where $p(x)=p\_1x+p\_2x^2+\cdots+p\_6x^6$ and similarly for $q(x)$.
So this reduces to the question of factorizing $(x^2+\cdots+x^{12})/11$
where the factors satisfy some extra conditions (coefficients positive,
$p(1)=1$ etc.).
This is a standard method (generating functions).
| 12 | https://mathoverflow.net/users/4213 | 41311 | 26,388 |
https://mathoverflow.net/questions/41296 | 17 | In his 1951 report [Sur la théorie du corps de classes](https://projecteuclid.org/euclid.jmsj/1261734944), Weil writes that
>
> *La recherche d'une interprétation de* $C\_k$ *si* $k$ *est un corps de
> nombres*, analogue en quelque manière
> à l'interprétation par un groupe de
> Galois quand $k$ est un corps de
> fonctions, *me semble constituer l'un
> des problèmes fondamentaux de la
> théorie des
> nombres* à l'heure actuelle; il se
> peut qu'une telle interprétation
> renferme la clef de l'hypothèse de
> Riemann ….
>
>
>
As requested by [@PeteL.Clark](https://mathoverflow.net/questions/41296/lun-des-probl%C3%A8mes-fondamentaux-de-la-th%C3%A9orie-des-nombres#comment97379_41296), a translation (by [@TonyScholl](https://mathoverflow.net/questions/41296/lun-des-probl%C3%A8mes-fondamentaux-de-la-th%C3%A9orie-des-nombres#comment97392_41296)):
>
> The search for an interpretation for $C\_k$, where $k$ is a number field—in some way analogous to its interpretation by a Galois group when $k$ is a function field—seems to me to be one of the fundamental problems of number theory today; perhaps such an interpretation contains the key to the Riemann hypothesis ….
>
>
>
Here, $C\_k$ is of course the idèle class group of the number field $k$.
I've heard that some people working in noncommutative geometry have thought about this problem.
**Question.** What progress has since been made towards such an interpretation?
| https://mathoverflow.net/users/2821 | L'un des problèmes fondamentaux de la théorie des nombres | Interpreting "progress" in a different (perhaps more controversial!) way than in Tony Scholl's answer, one could also mention that Langlands (followed by Kottwitz and perhaps others) has introduced a hypothetical group $L\_k$ which should be even bigger than $W\_k$, normally referred to as the Langlands group, which should bear the same relationship to arbitrary automorphic forms as $C\_k$ does to Grossencharacters.
It might help to remark that the algebraic Grossencharacters correspond geometrically to abelian varieties with CM, and so the algebraic envelope of $C\_k$ (the associated pro-algebraic torus through which all algebraic Grossencharacters factor) is the Tannakian group of the category of motives over $k$ generated by the Tate motive together with the motives of all abelian varieties over $k$ which have CM defined over $k$.
The algebraic envelope of $W\_k$ (which is now a non-commutative reductive pro-algebraic group) is the Tannakian group of the category of motives over $k$ generated by motives which are *potentially* CM, i.e. which become CM motives (i.e. belong to the category considered in the
preceding paragraph, i.e. are classified by an algebraic Grossencharacter on $C\_l$) over some extension $l$ of $k$. This category contains all Artin motives, for example.
The algebraic envelope of $L\_k$ should be the Tannakian group of the category of all motives over $k$.
So the problem of constructing $L\_k$ can be thought of, from this point of view, as the problem of enlarging the category of motives so that one can make sense of motives with "non-integral Hodge grading" (i.e. has $h^{p,q}$ for $p$ and $q$ complex numbers
that are not necessarily integral); $L\_k$ would then be (some version of) the Tannakian group of this category.
Going back to $C\_k$, from this optic one would like to generalize the notion of CM abelian variety to include objects with non-integral Hodge gradings, which would give rise to non-algebraic Grossencharacters in the way that usual CM abelian varieties correspond to algebraic Grossencharacters.
And of course, for $W\_k$, once wants to generalize potentially CM abelian varieties in the same way.
| 13 | https://mathoverflow.net/users/2874 | 41318 | 26,392 |
https://mathoverflow.net/questions/41314 | 7 | Let $X$ be a separable Banach space, $M \subset X$ a linear subspace. Must $M$ be a Borel set in $X$?
I believe the answer is "no," since I have seen authors who are careful to talk about "Borel subspaces". But I have not been able to find a counterexample.
If the answer is indeed "no", does every infinite-dimensional separable Banach space contain a non-Borel dense linear subspace?
| https://mathoverflow.net/users/4832 | Non-Borel subspace of Banach space | This is an answer, but not the "right" answer".
Presumably you mean that $X$ is infinite dimensional and hence has Hamel dimension the continuum $c$. For every subset of a given Hamel basis you get the linear subspace spanned by the subset, and these subspaces are different for different subsets of the basis. Thus $X$ has $2^c$ linear subspaces but only $c$ Borel sets (since $X$ is separable).
EDIT: I just noticed the second question. Consider subsets of the basis that all contain one fixed countable subset whose span is dense.
| 11 | https://mathoverflow.net/users/2554 | 41319 | 26,393 |
https://mathoverflow.net/questions/41289 | 8 | This question consists of two parts. I will try to be as short and clear as possible.
Let $S$ be a Dedekind scheme of characteristic zero. The main examples are $\mathbf{P}^1\_k$, with $k$ a field of characteristic 0 and $\textrm{Spec} \ \mathbf{Z}$.
A fibered surface is a projective flat morphism $X\longrightarrow S$ with $X$ an integral scheme of dimension 2.
Suppose $S=\mathbf{P}^1\_k$ with $k$ a field of characteristic 0. Then $X$ is an algebraic surface over a field of characteristic zero. The theory of rational singularities for $X$ is then explained in Chapter 5 of Kollar and Mori: Birational geometry of algebraic varieties.
Now, I would like to know if there are analogous results for when $S=\textrm{Spec} \ \mathbf{Z}$ as in *loc. cit.*. Let me make this more precise.
Assume that $X$ is normal. (This will suffice for my applications.)
Call a resolution of singularities $\rho:Y\longrightarrow X$ rational if it satisfies
$$R^i\rho\_\ast \mathcal{O}\_Y =0 \ \textrm{for} \ i>0.$$
We say that $X$ has rational singularities if all resolutions of $X$ are rational.
To have a good theory, we should probably show that $R^i \rho\_\ast \omega\_{Y/S} = 0$ for $i>0$.
**Question 1:** Is it true that $R^i \rho\_\ast \omega\_{Y/S} = 0$ for $i>0$ for any resolution of singularities?
Compare my last question with Theorem 5.10 of Kollar-Mori.
**Question 2:** Is it true that the following properties
A. $X$ has a rational resolution
B. $X$ has rational singularities
C. For every resolution of singularities $\rho:Y\longrightarrow X$, we have that $\rho\_\ast \omega\_{Y/S} = \omega\_{X/S}$.
are equivalent?
I guess one could try to see if some of the arguments given in Kollar-Mori carry over to this setting. For example, it probably suffices to show that a resolution is rational if and only if it satisfies C.
Here is a possible application: Assume $X$ to be regular from now on.
Let $S$ be $\mathbf{P}^1\_{\mathbf{C}}$. Let $\pi:Y\longrightarrow X$ be a finite surjective morphism with branch locus a normal crossings divisor. Then, it is easy to see that the singularities of $Y$ are quotient. (Consider the analytic topology or the etale topology and use Abhyankar's lemma). It is well-known that quotient singularities are rational.
I guess Abhyankar's lemma is still valid and therefore I was hoping to show that $Y$ in case $S= \textrm{Spec} \mathbf{Z}$ still has rational singularities. (The notion of quotient singularities seems only to work over $\mathbf{C}$...)
| https://mathoverflow.net/users/4333 | Rational singularities for fibered surfaces | Ariyan,
EDIT: This contains some substantial edits and added references.
Lipman has defined the following notion (EDIT: twice):
**Definition** (Lipman ; Section 9 of "*Rational singularities with applications to algebraic surfaces and factorization*"): If $X$ is 2-dimensional and normal, $X$ has two *pseudo-rational singularities* if for every proper birational map $\pi : W \to X$ there exists a proper birational normal $Y$ over $W$ where, $R^1 \pi\_\* \mathcal{O}\_Y = 0$
**Definition** (Lipman-Teissier ; Section 2 of "*Pseudo-rational local rings and a theorem of Briancon-Skoda about integral closures of ideals*"): $X$ has *pseudo-rational singularities* if $X$ is CM (Cohen-Macaulay) and if for every proper birational map $\pi : Y \to X$ with $Y$ normal, $\pi\_\* \omega\_Y = \omega\_X$.
If these are the same in dimension 2, this seems pretty close to what you want in dimension 2.
EDIT2: These are the same in dimension 2, I was in Purdue and asked Lipman about question 1, which holds, and certainly implies this.
He also points out that regular schemes are pseudo-rational. In particular, this implies that if $\pi\_\* \omega\_Y = \omega\_X$ for *one* resolution of singularities, it also holds for *every* resolution of singularities (in fact, for every proper birational map with normal domain).
In dimension 2, he also studies relations between this condition and the local-finiteness of the divisor class group.
On the other hand, I'm pretty sure this is different from the definition of rational singularities you gave above at least in higher dimensions (with the appropriate $R^i$ vanishing instead of just $R^1$).
With regards to your specific questions:
**Question #1:** That vanishing, called Grauert-Riemenschneider vanishing, is known to fail for $\dim X > 2$ outside of equal characteristic zero. I believe the answer should hold in the two-dimensional case, certainly it should assuming that Lipman's various definitions of pseudo-rational singularities are consistent.
EDIT: This holds in dimension 2, see Theorem 2.4 in Lipman's "Desingularization of two-dimensional schemes".
In any dimension, that vanishing has recently been proven in equal characteristic $p > 0$ over a smooth variety (or a variety with tame quotient singularities), see arXiv:0911.3599.
**Question #2:** In higher dimensions, I'm pretty confident that the answer is no. In the 2-dimensional case, probably this is done by Lipman? In view of question #1, in order to find such a counter example in higher dimensions, one should look at various cones probably over 3 or 4-dimension schemes with negative Kodaira dimension (probably Fano's) but which fail Kodaira vanishing.
I have some thoughts on some other definitions of rational singularities which might be better in mixed characteristic, but I'm not sure I want to post them on MathOverflow right now. If you email me, I'd be willing discuss it a bit.
Quotient singularities can behave a little different outside of characteristic zero as well (see various papers of Mel Hochster from the 70s for instance). This can also lead one to look at questions like the Direct Summand Conjecture.
| 7 | https://mathoverflow.net/users/3521 | 41321 | 26,394 |
https://mathoverflow.net/questions/41253 | 61 | I am teaching a course leading up to Tate's thesis and I told the students last week, when defining ideles, that the first topology that was put on the ideles was not so good (e.g., it was not Hausdorff; it's basically the profinite topology on the ideles, so archimedean components don't get separated well). You can find this mentioned on the second page of the memorial article *Claude Chevalley (1909–1984)* by Dieudonné and Tits in Bulletin AMS **17** (1987) (doi:[10.1090/S0273-0979-1987-15509-1](https://doi.org/10.1090/S0273-0979-1987-15509-1)), where they also say that Chevalley's introduction of the ideles was "a definite improvement on earlier similar ideas of Prüfer and von Neumann, who had only embedded $K$ [the number field] into the product over the *finite* places" (emphasis theirs). [Edit: Scholl's answer says in a little more detail what Prüfer and von Neumann were doing, with references.]
I have two questions:
1) Can anyone point to a specific article where Prüfer or von Neumann used a product over just the finite places, or at least indicate whether they were able to do anything with it?
2) Who introduced the restricted product topology on the ideles? (In Chevalley's 1940 paper deriving global class field theory using the ideles and not using complex analysis, Chevalley uses a different topology, as I mentioned above.) I would've guessed it was Weil, but BCnrd told me that he heard it was due to von Neumann. Any answer with some kind of evidence for it is appreciated.
Edit: For those wondering why the usual notation for the ideles is $J\_K$ and not $I\_K$, the use of $J\_K$ goes right back to Chevalley's papers introducing ideles. (One may imagine $I\_K$ could have been taken already for something related to ideals, but in any event it's worth noting the use of "$J$" wasn't some later development in the subject.)
| https://mathoverflow.net/users/3272 | who fixed the topology on ideles? | I know nothing about work of ``idelic nature'' by Von Neumann or Pruefer. Already in the 1930's
Weil understood that Chevalley was wrong to ignore the connected component, because Weil understood already then that Hecke's characters were the characters of the idele class group for the right topology on that. I don't know of any place before his paper dedicated to Takagi where he defined the ideles explicitly as a topological group, but he must have understood the situation way before that
When I wrote my thesis I used what seemed to me to be the obvious topology without going into the history of the matter.
| 161 | https://mathoverflow.net/users/9848 | 41332 | 26,398 |
https://mathoverflow.net/questions/41266 | 4 | An [Enriques surface](http://en.wikipedia.org/wiki/Enriques_surface) (in characteristic zero) is an algebraic surface which is the quotient of a K3 surface by a fixed-point-free involution. Such a surface has a rank 10 lattice of divisors.
>
>
> >
> > (1) What are the ample and effective cones of an Enriques surface?
> >
> >
> >
>
>
>
In particular,
>
>
> >
> > (2) Is it the case that there are ample divisors with arbitrarily large numbers of global sections which do not split nontrivially as the sum of two effective divisors?
> >
> >
> >
>
>
>
**Edit:** As per Damiano's comment,
>
>
> >
> > (3) Is there *any* surface for which the answer to (2) is yes?
> >
> >
> >
>
>
>
| https://mathoverflow.net/users/4707 | What is known about the ample and effective cones of an Enriques surface? | The answer to question 2 is negative thanks to Kamawata-Morrison conjecture about nef cones of Calabi-Yau varieties that is proven for surfaces (a recent reference is here <http://arxiv.org/abs/1008.3825>), and to the comment of Damiano below.
Indeed, it is known that on a minimal two-dimesnional surface of Kodaira dimesnion 0
the fundamental domain of the action of automorphism of the surface on the nef cone
is rational polyhedral (see the above reference). So it is sufficient to consider only ample classes that belong to one rational polyhedral fundamental domain. Notice that every nef integral class (apart from $K$) on each Enriques surface is effective (see the proof of Damiano). Now notice that since the fundamental domain
is rational polyhedral, the semi-group of integer vectors in it is finitely generated, so only finitely many integer vectors can not be presented as a sum of two others.
In other words, only finite number of ample divisors in a given fundamental domain are not linearly equivalent to a sum of several divisors. Since by definition all domains are the same, the statement is proven.
| 2 | https://mathoverflow.net/users/943 | 41340 | 26,403 |
https://mathoverflow.net/questions/41288 | 4 | Let $G$ be a finite soluble group, let $x$ be a non-identity element of $G$, let $v$ be an $n$-tuple in $G$, and let $w$ be a word in $n+2$ variables. Does there exist $(G,x,v,w)$ satisfying the following equations?
$w(1,1,v)=w(x,1,v)=w(1,x,v)=1; w(x,x,v)=x.$
As for why the soluble condition is there: let $w(a,b,c) = [a,c^{-1}bc]$, and suppose we have chosen $x,v \in G$ such that $[x,v^{-1}xv]=x$. This seems quite a plausible equation in a simple group (at first glance, anyway) but obviously has no non-trivial solutions in a soluble group, because the left-hand side is further down the derived series than the right-hand side. But perhaps a more complicated word does not have this limitation.
| https://mathoverflow.net/users/4053 | Can a soluble group compute OR? | After thinking about it some more, I realise it can't be done. Suppose $x$ is contained in a normal subgroup $M$ of $G$, but not in $M'=[M,M]$. Then modulo $M'$ we have
$w(x,1,v) = w(1,1,v)m\_1; w(1,x,v) = w(1,1,v)m\_2; w(x,x,v) = w(1,1,v)m\_1m\_2$
where $m\_1,m\_2 \in M$ (just 'pull through' all occurrences of the first two variables up to conjugacy, eg $xd$ can be rewritten $dx^d$). The first set of equations force $m\_1 = m\_2 = 1$ (all modulo $M'$), so $w(x,x,v)$ is in $M'$, in particular $w(x,x,v)$ does not equal $x$.
| 3 | https://mathoverflow.net/users/4053 | 41341 | 26,404 |
https://mathoverflow.net/questions/41343 | 8 | Let $X$ be a real smooth manifold, and $M$ a locally-finitely-generated sheaf of $\mathcal C^\infty(X)$-modules. (If $X$ is not compact, I will also insist that there be a global bound on the number of generators I might need in different regions; maybe this is part of the usual meaning of the words "locally finitely generated".)
I would like to find finite-dimensional vector bundles $E\_1,\dots,E\_n$ over $X$ and maps of $\mathcal C^\infty$-modules
$$ 0 \to \Gamma(E\_n) \to \dots \to \Gamma(E\_1) \to M \to 0$$
so that the sequence is exact. Can I always do this? And is there an explicit bound on the number of vector bundles needed, e.g. $n = \dim X$ or $n = \dim X+1$?
| https://mathoverflow.net/users/78 | Is there a "Hilbert syzygy theorem" for smooth manifolds? Or: does every finitely generated $C^\infty$ module have a finite-length resolution in vector bundles? | No. Let $X$ be $\mathbb R$. In the ring $C^{\infty}(X)$ let $I$ be the ideal of all functions vanishing to infinite order at $0$. The module $C^{\infty}(X)/I$ does not have a finite resolution by finitely generated projective modules.
Edit:
Still no if you want the finitely generated module to be contained in a finitely generated projective module. For the same $X$ pick a function $f$ such $f(x)$ vanishes precisely when $x<0$. let $J$ be the ideal generated by $f$. The module $J$ does not have a finite resolution by finitely generated projective modules.
For both of these examples, the method I have in mind is this: If a module $M$ has a finite projective resolution $P\_\bullet$ then for every point in $p\in X$ the alternating sum of the $k\_p$ vector space dimension of $Tor\_n(M,k\_p)$ is independent of $p$ because it's the alternating sum of the rank of $P\_n$. I believe that in the first example this Euler number comes out to be $1$ if $p$ is the origin and otherwise $0$, and in the second it's $1$ if $p> 0$ and $0$ if $p< 0$.
| 13 | https://mathoverflow.net/users/6666 | 41344 | 26,406 |
https://mathoverflow.net/questions/41349 | 6 | If X and Y are non-isomorphic objects, then "[is / is not] isomorphic to [ X / Y ]" are invariants that distinguish X and Y. You can also do things like take an object Z that is not isomorphic to Y, and then "is isomorphic to X or isomorphic to Z" is another invariant that distinguishes X and Y. Similarly, if W is isomorphic to X, then "is isomorphic to W" works too.
Are there any objects X and Y known to be non-isomorphic, but all known distinguishing invariants
(a) use the concept or definition of isomorphism
or
(b) use the concept or definition of being isomorphic to some object Z, where Z is isomorphic to X or Y
| https://mathoverflow.net/users/nan | non-isomorphic objects with no known nontrivial distinguishing invariants | Hereditarily indecomposable Banach spaces are strange objects that fail to be isomorphic to any of their proper subspaces. However, in a certain sense those subspaces are not "interestingly different" from the spaces themselves. So there is no hope of finding an invariant to distinguish between the space and a subspace. The way the proof actually works is that one proves that every operator from the space to itself can be approximated, in a certain sense, by a multiple of the identity, and hence either has no hope of being an isomorphism or is Fredholm with index zero. But an isomorphism to a proper subspace would have to be Fredholm with positive (or infinite) index.
| 15 | https://mathoverflow.net/users/1459 | 41355 | 26,409 |
https://mathoverflow.net/questions/41353 | 9 | This must be common knowledge.
Where exactly in the development of homological algebra does one need the axiom that makes ~~additive~~pre-abelian and abelian categories different? (I mean this statement: for every morphism $u: X \to Y$, the canonical morphism $\bar{u}: \mathrm{Coim}\ u \to \mathrm{Im}\ u$ is an isomorphism.)
My gut feeling is that it should be necessary for the Snake lemma, but I couldn't find a step in the proof that would use it.
| https://mathoverflow.net/users/2234 | abelian categories vs. additive categories | A good example of the situation you are thinking about is the category of filtered modules (over some ring). As Yemon Choi notes in a comment, Banach spaces give another example (and in general, filtrations behave very similarly to topologies, and are a closely related notion); but (at least for me) it is a bit easier to write down filtered modules concretely.
Suppose that you have two exact sequences in the category of filtered modules (exact in the strong sense that the filtrations on the outer two terms are induced by the filtration on the middle term, which is to say that these morphisms satisfy the condition that coimage = image) and a morphism between them; the snake lemma will then give a long sequence which is exact as a sequence of morphisms of modules, but which is no longer exact in the strong sense that its morphisms need no longer be strict: i.e. they need not satisfy the condition that
coimage = image.
A trivial example is given by considering any morphism
$$( 0 \to A \to B \to C \to 0) \longrightarrow (0 \to A' \to B'\to C' \to 0)$$
of exact sequences of non-trivial $R$-modules, and then declaring $A$, $B$, and $C$ to be filtered by having $F^0A = A$ and $F^1A = 0$, and similarly with $B$ and $C$,
while equipping $A'$, $B'$, and $C'$ with filtrations such that $F^1A' = A'$ and similarly with $B'$ and $C'$.
The boundary map in the snake lemma from $ker(C \to C')$ to $coker(A\to A')$ now will not be strict (because $F^1$ on the kernel is $0$, while $F^1$ on the cokernel is everything).
One can nevertheless work out homological algebra in this context (e.g. one can form the "filtered derived category", which is the derived category of filtered modules) but one has to take care with the details, because of the phenomenon described above. See Laumon's paper "Sur la categorie derivees des $\mathcal D$-modules filtres" (Lecture Notes in Math 1016) for a careful development of the theory.
| 14 | https://mathoverflow.net/users/2874 | 41381 | 26,422 |
https://mathoverflow.net/questions/41369 | 4 | I've heard it said that the reason why the homology groups of a space are a *computable* invariant is because they are a *stable invariant* in the sense that they are stable under suspension.
I'm familiar with the standard computation which shows that for reduced homology ${\tilde H\_n}(S X)\simeq {\tilde H\_{n-1}}(X)$ for all $n$, where $S X$ denotes the suspension of a topological space $X$; just let $A$ be $S X$ with the top vertex removed and $B$ be $S X$ with the bottom vertex removed, and apply Mayer-Vietoris, noting that $A$ and $B$ are contractible, and that $A \cap B$ is homotopy equivalent to $X$.
Is this computation the sense in which my first sentence should be understood? That it enables us to compute the homology of certain spaces (such as spheres) which are built out of each other by suspension, by some kind of inductive argument? But other than the obvious example of spheres, are there many other examples of spaces whose homology can be computed by taking suspensions in this way? Are there other reasons why this "stability" of homology groups under suspension makes it much more computable than other invariants?
| https://mathoverflow.net/users/1148 | Homology is computable because it is stable under suspension | Stability alone surely does not give you much. It gives you just the values on the spheres if you have the value on $S^0$. But many interesting spaces (eg all smooth manifolds) are CW-complexes and can be built out of spheres. And if you have a way to handle cofiber sequence $X\to Y \to CX\cup\_X Y$ with your theory, you have some hope to compute something. In essence, you have then (togehter with homotopy invariance) exactly the notion of a reduced homology theory. And homology theories are comparatively computable.
But as said already in the comments above, that might be deceptive. Because
1) you need to know the value on $S^0$. This is, of course, the case in singular homology, but in a general theory stable under suspension it is far from true. Eg the stable homotopy groups of spheres are unknown in general (but far better known than the unstable ones).
2) not every space has a CW-structure and even if it has one, you need not know it. For example, you can write down smooth manifolds by giving it as a zero-set of suitable polynomial equations. It might be quite hard to compute its homology then.
3) even if you have a CW-structure and can compute singular homology of the space, you may not be able to compute other homology theories of it (even if you know their coefficients), although these are stable under suspension, of course, as well; you need to go to the Atiyah-Hirzebruch spectral sequence, which need not collapse.
| 9 | https://mathoverflow.net/users/2039 | 41382 | 26,423 |
https://mathoverflow.net/questions/40801 | 15 | Is the following fact true?
>
> Let $v\_1,\ldots, v\_k \in \mathbb{R}^2$, $\|v\_i\|\leq 1$, be vectors that add up to zero. Does there exist a permutation $\sigma\in S\_k$ and vectors $w\_1,\ldots, w\_k \in \mathbb{R}^2$, $\|w\_i\|\leq 1$, such that $v\_{\sigma(i)}=w\_i-w\_{i-1}$? (here, I assume $w\_0=w\_k$)
>
>
>
**Edit. Related (known) facts:**
1. The same fact in $\mathbb{R}^1$ is true. (can be easily proven by choosing $w\_0=0$ and $\sigma(i)$ such that $\|w\_i\|\leq 1$ for $w\_i:=w\_{i-1}+v\_{\sigma(i)}$)
2. For each $\epsilon>0$ there exists a family of vectors $v\_i$, such that $\max\_i\|w\_i\|>1-\epsilon$. See my comment below for the proof.
| https://mathoverflow.net/users/8134 | Representation of vectors in $\mathbb{R}^2$ via differences of small vectors. | I have a counterexample in $\mathbb R^2$.
Here's how it goes.
Pick two numbers $n$ and $N$, with $N>>n>>1$.
The collection {$v\_i$} consists of:
* $N(n-1)$ times the vector $(\frac{n-2}{n},\frac{1}{N(2n-3)})$
* $N(n-2)$ times the vector $(-\frac{n-1}{n},\frac{1}{N(2n-3)})$
* The vector $(0,-1)$ once.
The smallest ball into which those vectors can be fit back-to-back has diameter $\sqrt{5}-\varepsilon$.
| 8 | https://mathoverflow.net/users/5690 | 41384 | 26,424 |
https://mathoverflow.net/questions/41351 | 3 | Everything below is defined over $\mathbb C$.
Let $T$ be a smooth affine variety, $\pi : \mathscr X \to T$ be a smooth family of smooth
projective varieties, and $\mathcal F$ be a locally free coherent sheaf on $\mathscr X$.
As usual write $\mathscr X\_t$ for the fiber of $\pi$ over $t$ and $\mathcal F\_t$ for the
restriction of $\mathcal F$ at it.
Suppose $h^0(\mathscr X\_t, \mathcal F\_t) >0 $ for every $t \in T$ and fix $t\_0 \in T$.
Can I conclude the existence of a non-zero section of $\mathcal F\_{t\_0}$ which extends
to a section of $\mathcal F$ on a neighborhood of $\mathscr X\_{t\_0}$?
**Clarification**: By a smooth family of smooth projective varieties, I mean that $\mathscr X$ is smooth, $\pi$ is a submersion, and the fibers are smooth and connected. In particular, the family is topologically trivial. Also, I do not need a section at an affine neighborhood of $\mathscr X\_{t\_0}$: an analytic neighborhood would suffice.
| https://mathoverflow.net/users/605 | Existence of non-zero sections | No, this is not true. Here is the simplest example. Take $T = \mathbb A^2$, $X = \mathbb A^2 \times \mathbb P^1$. The morphims $\pi \colon X \to T$ is the projection. We will denote by $\cal O(d)$ the pullback of $\cal O\_{\mathbb P^1}(d)$ to $X$. Let $x$ and $y$ be the coordinates on $\mathbb A^2$, and $u$ and $v$ the homogeneous coordinates on $\mathbb P^1$. Set $R = \mathbb C[x,y]$.
Let $F$ be the generic extension of ${\cal O}(1)$ by ${\cal O}(-2)$. We have an isomorphism
$Ext^1({\cal O}(1), {\cal O}(-2)) \simeq R^2$;
if $a$, $b$ denotes a basis of $Ext^1({\cal O}(1), {\cal O}(-2))$ as an $R$-module, then $F$ corresponds to the element $xa+yb$. By restricting to the fiber over $t$, we get the split exension for $t = (0,0)$, and a non-split extension for $t \neq (0,0)$. It is easy to see that $h^0(F\_t)$ is 2 for $t = (0,0)$ and 1 otherwise. We get an exact sequence
$$
0 \longrightarrow H^0(F) \longrightarrow H^0({\cal O}(1)) \longrightarrow H^1({\cal O}(-2))
$$
that is
$$
0 \longrightarrow H^0(F) \longrightarrow R^2 \longrightarrow R;
$$
the image of the boundary map $R^2 \longrightarrow R$ is the maximal ideal $(x,y)$. Hence, the composite $H^0(F) \to H^0({\cal O}(1)) \to H^0({\cal O}(1)\_{t\_0})$ is 0. Since $H^0(F\_{t\_0})$ maps isomorphically onto $H^0({\cal O}(1)\_{t\_0})$, we conclude that the restriction map $H^0(F) \to H^0(F\_{t\_0})$ is 0. This gives the counterexample.
This can be usefully viewed using the result of Grothendieck saying that, if $Y = \mathop{\rm Spec}R$, there is a finite $R$-module $Q$ such that for every finite $R$-module $M$ there is a canonical isomorphism of $R$-modules $H^0(X, F\otimes\_R M) \simeq Hom\_R(Q,M)$, which is functorial in $M$. In the counterexample above, $Q$ is the maximal ideal $(x,y)$. When $Y$ is 1-dimensional, we can use the structure theorem for modules over Dedekind rings, and conclude that $Q$ is the direct sum of a projective module and a torsion module. If $h^0(F\_t) \neq 0$ for all $t$, then the projective summand must be non-zero, so the statement does hold.
| 7 | https://mathoverflow.net/users/4790 | 41392 | 26,427 |
https://mathoverflow.net/questions/41388 | 5 | Let $\Lambda$ denote the ring of symmetric functions in variables $x\_1,x\_2,\dots$ and with coefficients in $\mathbf{Q}$. Then $\Lambda$ is freely generated as an $\mathbf{Q}$-algebra by $p\_1,p\_2,\dots$, where $p\_n$ denotes the $n$-th power sum function $x\_1^n+x\_2^n+\cdots$. Let $\Delta^+$ and $\Delta^{\times}$ denote the $\mathbf{Q}$-algebra maps $\Lambda\to\Lambda\otimes\_{\mathbf{Q}}\Lambda$ determined by $\Delta^+(p\_n)=1\otimes p\_n + p\_n\otimes 1$ and $\Delta^{\times}(p\_n)=p\_n\otimes p\_n$ for all $n\geq 1$. Let $\mathbf{Q}\_+$ denote the sub-semiring $\{a\in\mathbf{Q}| a\geq 0\}$ of $\mathbf{Q}$.
Consider the following properties on subsets $S$ of $\Lambda$:
1. $S$ is a $\mathbf{Q}$-linear basis of $\Lambda$.
2. All finite sums and products of elements of $S$ are contained in the $\mathbf{Q}\_+$-linear span of $S$. (That is, the span is a sub-$\mathbf{Q}\_+$-algebra.)
3. The subsets $\Delta^+(S)$ and $\Delta^{\times}(S)$ of $\Lambda\otimes\_{\mathbf{Q}}\Lambda$ are contained in the $\mathbf{Q}\_+$-linear span of $S\otimes S = \{s\otimes s' | s,s'\in S\}$.
4. For all $s,s'\in S$, the composition $s\circ s'$ is contained the $\mathbf{Q}\_+$-linear span of $S$, where $\circ$ denotes plethysm.
These properties are satisfied if $S$ is the set of Schur functions or the set of monomial symmetric functions, for example. But the Schur functions give a smaller example in the sense that
they're contained in the $\mathbf{Q}\_+$-linear span of the monomial symmetric functions.
I have many imprecise questions about subsets $S$ satisfying these properties, but in the interest of fair play, I'll ask a yes/no one:
>
> Are the Schur functions the smallest example? That is, if $S$ satisfies the properties above, does its $\mathbf{Q}\_+$-linear span contain all the Schur functions?
>
>
>
(Apologies if this is standard, but I don't know much about Schur functions. I didn't find anything about it in Macdonald's book, and rather than emailing random experts, it's more fun to ask it here.)
| https://mathoverflow.net/users/1114 | Are the Schur functions the minimal basis of the ring of symmetric functions with the following properties? | There is another simple set of symmetric functions fulfilling your properties 1 to 4: the set of products of power sums,
$$
p\_1, p\_2, p\_1^2, p\_3, p\_1 p\_2, ...
$$
The convex cone they generate is not contained in the cone generated by the Schur functions. For instance, $p\_2=s\_2-s\_{1,1}$.
EDIT: and the Schur basis is not contained in the convex cone generated by the power sums: $s\_{1,1}=\frac{p\_1^2-p\_2}{2}$.
| 4 | https://mathoverflow.net/users/6768 | 41401 | 26,432 |
https://mathoverflow.net/questions/41397 | 2 | When I work out the James construction for a discrete pointed space, it appears that the
induced map $\pi\_0 (J(X)) \to \pi\_0( \Omega\Sigma X)$ is the inclusion of the free monoid on $\pi\_0(X)$ into the free group on $\pi\_0(X)$, so $J(X) \to \Omega\Sigma X$
is not a homotopy equivalence; and it seems clear that the same phenomenon holds
for other disconnected spaces (which is why I made the distinction between $X$ and $\pi\_0(X)$).
I also think that $J(X) \to \Omega \Sigma X$ will induce isomorphisms on $\pi\_n$ for $n \geq 1$, which would mean that the restriction to the basepoint components, which I'll denote
$J\_0(X) \to \Omega\_0 \Sigma X$, should be a (weak) equivalence. But I don't recall having seen this claim anywhere, so: is this written up somewhere?
| https://mathoverflow.net/users/3634 | James Construction for Disconnected Spaces | No. Let $X$ be the disjoint union of a point and a circle, and compare the two different $J(X)$ that you get according to what point you call the base point. Now think about $\pi\_1(\Omega \Sigma X)=\pi\_2(S^1\vee S^2)$. Same as $\pi\_2$ of the universal covering space ...
| 5 | https://mathoverflow.net/users/6666 | 41402 | 26,433 |
https://mathoverflow.net/questions/41287 | 10 | Let $G$ be the mapping class group of a closed surface $S\_{g}$. Bestvina-Bromberg-Fujuwara <https://arxiv.org/abs/1006.1939> recently constructed a finite index subgroup $B$ of $G$ such that for every essential closed simple curve $\gamma$ on $S\_{g}$ and every $h\in B$ either $h\gamma=\gamma$ or $h\gamma$ intersects $\gamma$ (everything is modulo isotopy of $S\_{g}$). The construction is not difficult: $B$ is just the subgroup of $G$ fixing $\pi\_1(S\_{g})/N$ for some characteristic subgroup $N$ of the $\pi\_1$ of finite index. The $N$ can be found by intersecting kernels of the first homology mod $6$ with kernels of the first homology mod 2 of all index 2 subgroups of the $\pi\_1$. Question: can one find another finite index subgroup $B'$ of $G$ with the same property but with a smaller index. Most probably $B'$ cannot be above Torelli subgroup, but can it be "not far from Torelli". I am interested in $B'$ that does not act non-trivially on a simplicial tree. The motivation is here: <https://arxiv.org/abs/1005.5056>.
**Update:** I read somewhere (I do not remember where now) that if $g$ is an element of the Torelli subgroup, $\gamma$ is a simple closed curve, then at least two of the three curves $\gamma, g\gamma, g^2\gamma$ must intersect. From this I deduced that two curves are not enough and that a Bestvina-Bromberg-Fujiwara subgroup cannot contain the Torelli subgroup. Now the question is whether it may contain the Johnson kernel. For this one needs first to answer the following question: if $g$ is the Dehn twist about a separating curve, $\gamma$ is a simple closed curve, is it true that $g\gamma$ intersects $\gamma$ or $g\gamma=\gamma$?
| https://mathoverflow.net/users/nan | A finite index subgroup of the Mapping Class Group | Regarding the new question, you're correct that $f\in \mathcal{I}$ doesn't imply $f(\gamma)$ meets $\gamma$: for example, if $f$ is a product of disjoint bounding pairs and $\gamma$ is a (nonseparating) curve meeting each curve-defining-$f$ in at most 1 point, then $f\in\mathcal{I}$ but $f(\gamma)$ is disjoint from $\gamma$.
For the rest of the question, the following theorems are taken from Farb-Leininger-Margalit, "The lower central series and pseudo-Anosov dilatations", American Jour. Math, 130 (2008), no. 3, 799--827. ([PDF](http://www.math.uchicago.edu/~farb/papers/dil.pdf))
Lemma 2.2: if $f\in \mathcal{I}$, $\gamma$ is separating, and $f(\gamma)\neq\gamma$, then $i\big(f(\gamma),\gamma\big)\geq 4$.
Lemma 2.3: if $f\in \mathcal{I}$ and $\gamma$ is nonseparating, and $f(\gamma)\neq\gamma$ then $i\big(f(\gamma),\gamma\big)\geq 2$ or $i\big(f^2(\gamma),\gamma\big)\geq 2$.
Proposition 3.2: if $f\in\mathcal{K}$ and $f(\gamma)\neq \gamma$, then $i\big(f(\gamma),\gamma\big)\geq 4$.
In particular 3.2 answers your question about a separating twist. The proof of the proposition is as follows: $\mathcal{K}$ contains no nonseparating bounding pairs (use the Johnson homomorphism). By Lemma 2.2 we can assume $\gamma$ is nonseparating. But if $f(\gamma)$ is disjoint frmo $\gamma$, then $[f,T\_{\gamma}]=f T\_{\gamma}f^{-1} T\_{\gamma}^{-1}=T\_{f(\gamma)}T\_{\gamma}^{-1}$ is a bounding pair, which lies in $\mathcal{K}$ if $f$ does, yielding a contradiction.
---
I do appreciate the nice calming green background, but I'm not sure we're done. If $f\in\mathcal{K}$, and $g\in \mathcal{BBF}$ (the group defined by Bestvina-Bromberg-Fujiwara), how do you know that $fg$ has the desired property?
Here is a sketch of a possible construction of a BBF-subgroup which properly contains the Johnson kernel. Let $G$ be the subgroup generated by the Johnson kernel $\mathcal{K}$ together with ($\mathcal{BBF}\cap\mathcal{I}$). Since $G$ is contained in $\mathcal{I}$, separating curves are taken care of by Lemma 2.2 above. If $\gamma$ is a nonseparating curve such that $f(\gamma)$ is disjoint from $\gamma$, then $[f,T\_\gamma]$ is a nonseparating bounding pair; $[f,T\_\gamma]$ lies in $G$ if $f$ does, since $G$ is normal.
Now the idea is to show that $G$ contains no nonseparating bounding pairs. Certainly $\mathcal{BBF}$ doesn't, but we need to show that this is still true after throwing in $\mathcal{K}$. Since $G$ contains $\mathcal{K}=\ker \big(\tau\colon\mathcal{I}\to \bigwedge^3 H/H\big)$, this is the same as showing that $\tau(\mathcal{BBF}\cap \mathcal{I})$ does not contain the image under $\tau$ of a genus $k$ bounding pair. We know the image of such a BP, and you should be able to write down a condition on $\tau(f)$ for $f\in\mathcal{BBF}$ coming from the triviality of the action on the homology of the double cover. I have the feeling that this condition will rule out such a bounding pair, which would conclude the proof. I don't know how hard this approach would be if you wanted to actually carry it out, though.
| 5 | https://mathoverflow.net/users/250 | 41408 | 26,437 |
https://mathoverflow.net/questions/41373 | 5 | Hi
I really need a proof for the following statement by Baumgartner:
There exists a stationary subset of $[\omega\_2]^{\omega}$ of size $\aleph\_2$.
This is Exercise 38.15. in Jechs Book (2003) and you can find a hint there which goes like this: For each $\alpha < \omega\_2$, let $f\_{\alpha} : \alpha \to \omega\_1$ be one to one. If $\alpha < \omega\_2$ and $\xi < \omega\_1$ set $X\_{\alpha, \xi} =$ { $\beta < \alpha : f\_{\alpha} (\beta) < \xi$ }. Then $S:=$ { $X\_{\alpha, \xi} : \alpha < \omega\_2, \xi < \omega\_1$} is our desired stationary subset.
But so far my attempts to proof this didn't work, because the sequence of the $f\_{\alpha}$s doesn't have any nice regularity properties.
Thank you.
| https://mathoverflow.net/users/4753 | Why is this set stationary? | Given $F:[\omega\_2]^{<\omega}\to[\omega\_2]^{\aleph\_0}$ as above, we first claim the existence of an ordinal $\omega\_1\leq\alpha<\omega$ that is closed under $F$, i.e., $s\in [\alpha]^{<\omega}$ implies $F(s)\subseteq\alpha$. For this, let $\alpha$ be the limit of the sequence $\omega\_1=\alpha\_0<\alpha\_1<\cdots$ where $\alpha\_{n+1}$ is sufficiently large that $F(s)\subseteq \alpha\_{n+1}$ for $s\in [\alpha\_{n}]^{<\omega}$.
Given $\alpha$ as above, construct similarly the ordinal $\omega\leq\xi<\omega\_1$ so that $f^{-1}\_\alpha[\xi]$, that is, $\{\beta<\alpha:f\_\alpha(\beta)<\xi\}$, is closed under $F$. This can be done similarly: let $\xi$ be the limit of the sequence $\omega=\xi\_0<\xi\_1<\cdots$ where $\xi\_{n+1}$ is chosen so that if $s$ is a finite subset of $\{\beta<\alpha:f\_\alpha(\beta)<\xi\_n\}$, then $F(s)$ (which is a subset of $\alpha$) is a subset of $\{\beta<\alpha:f\_\alpha(\beta)<\xi\_{n+1})\}$. Now $X\_{\alpha,\xi}$ is closed under $F$.
| 10 | https://mathoverflow.net/users/6647 | 41422 | 26,448 |
https://mathoverflow.net/questions/41429 | 2 | Given two smooth elliptic curves $C\_1$ and $C\_2$ over $\mathbb{C}$. Assume they are not isogenous. I'm interested in the structure of $Pic(A)$ and $Pic^{0}(A)$ for $A:=C\_1 \times C\_2$.
Reading Birkenhake/Lange - Complex Abelian Varieties, i think this has to do with correspondences of curves. Since an elliptic curve is its own Jacobian and the two curves are not isogenous, we have $Hom(C\_1,C\_2)=0$. So the space of correspondences $Corr(C\_1,C\_2)$ is trivial, i.e. every line bundle $L$ on $A$ is of the form $L=q^{\\*}M\otimes p^{\\*}N$, where q and p are the projections on the factors and $M$ and $N$ are line bundles on the factors. This implies $Pic(A)=Pic(C\_1)\times Pic(C\_2)$.
Does this impliy $Pic^{0}(A)=Pic^{0}(C\_1)\times Pic^{0}(C\_2)$? That is, is the Picard variety of $A$ isomorphic to $A$ in this case?
| https://mathoverflow.net/users/3233 | Line bundles on special abelian surfaces | Yes: in fact $Pic^0(C\_1\times C\_2)=Pic^0(C\_1)\times Pic^0(C\_2)$ for any pair of curves. The fact that $C\_1$ and $C\_2$ are not isogenous in your case only affects the Neron-Severi group $Pic/Pic^0$ of $C\_1\times C\_2$, exactly for the reasons you describe.
| 2 | https://mathoverflow.net/users/5480 | 41431 | 26,451 |
https://mathoverflow.net/questions/41389 | 20 | For my finals, I am digging through the book by Varadarajan [An introduction to harmonic analysis on semisimple Lie groups](http://books.google.com/books?id=seycdRenNfoC&lpg=PP1&dq=varadarajan%2520harmonic%2520analysis%2520on%2520semisimple%2520lie%2520groups&pg=PP1#v=onepage&q&f=false%20%22Varadarajan%20-%20An%20introduction%20to%20harmonic%20analysis%20on%20semisimple%20Lie%20groups%22). I find it a rather hard read and I feel it's a bit outdated now. Any recommendation of a more modern (and/or) introductory treatment to the topics covered by this book would be greatly appreciated.
I am more interested in the representation theory, although I find the connection to harmonic analysis intriguing. Having learned finite-dimensional representation theory I wanted to move on to the infinite-dimensional one. From harmonic analysis I know only the classical Pontryagin duality, which I thought is enough to get me started, but I've found Varadarajan's approach based on examples difficult to follow.
| https://mathoverflow.net/users/6818 | Harmonic analysis on semisimple groups - modern treatment | Speaking as a nonexpert, I'd emphasize that the subject as a whole is deep and difficult. Even leaving aside the recent developments for $p$-adic groups, the representation theory of semisimple Lie groups has been studied for generations in the spirit of harmonic analysis. So there is a lot of literature and a fair number of books (not all still in print). Having heard many of Harish-Chandra's lectures years ago, I know that the subject requires enormous dedication and plenty of background knowledge including classical special cases. Some books are certainly more accessible for self-study than others, but a lot depends on what you already know and what you think you want to learn.
Access to MathSciNet is helpful for tracking books and other literature, as well as some insightful reviews. Without attempting my own assessment, here are the most likely books to be aware of besides the corrected paperback reprint of Varadarajan's 1989 Cambridge book (I have the original but not the corrected printing, so don't know how many changes were made):
MR2426516 (2009f:22009),
Faraut, Jacques (F-PARIS6-IMJ),
Analysis on Lie groups.
An introduction.
Cambridge Studies in Advanced Mathematics, 110.
Cambridge University Press, Cambridge, 2008.
MR1151617 (93f:22009),
Howe, Roger (1-YALE); Tan, Eng-Chye (SGP-SING),
Nonabelian harmonic analysis.
Applications of SL(2,R).
Universitext.
Springer-Verlag, New York, 1992.
MR0498996 (58 #16978),
Wallach, Nolan R.,
Harmonic analysis on homogeneous spaces.
Pure and Applied Mathematics, No. 19.
Marcel Dekker, Inc., New York, 1973.
This old book by Wallach as well as another by him on Lie group representations are presumably out of print. In any case, textbooks at an introductory level which emphasize both Lie group representations and harmonic analysis (often in the direction of symmetric spaces) are relatively few and far between. That probably reflects the practical fact that graduate courses aren't often attempted and are inevitably rather advanced. On the other hand, there are some modern graduate-level texts on compact Lie groups and related harmonic analysis as well as books on Lie groups and their representations with less coverage of harmonic analysis and symmetric spaces.
| 7 | https://mathoverflow.net/users/4231 | 41434 | 26,454 |
https://mathoverflow.net/questions/41391 | 4 | Let $I =\langle f\_1,\cdots,f\_m\rangle \subset K[x\_1,\cdots,x\_n]$be an ideal,
where $f\_k\in K[x\_1,\cdots,x\_n].$
$K[e\_1,\cdots,e\_n]$ the polynomial algebra generated by the elementary symmetric polynomials
$e\_1,\cdots,e\_n\in K[x\_1,\cdots,x\_n].$
Is there any method(algorithm) to compute the K-algebra $I \cap K[e\_1,\cdots,e\_n]$ of the intersection of $I$ and $K[e\_1,\cdots,e\_n]? $
Since $K[e\_1,\cdots,e\_n]$ is not an ideal of $ K[x\_1,\cdots,x\_n],$ it fails to compute elimination ideal. Anyway, $I \cap K[e\_1,\cdots,e\_n]$ is an ideal of the ring
$K[e\_1,\cdots,e\_n].$ What I want to do is to give the generating sets of this ideal by
polynomials in $K[e\_1,\cdots,e\_n].$
| https://mathoverflow.net/users/8152 | Finding generators of subalgebra of polynomial algebra $K[x_1,\cdots,x_n]$ that are invariant under the action of symmetric group | UPDATE: Sorry! As pointed out in a comment, my previous answer was incorrect. So I've edited my answer. The following simpler algorithm seems to me that it should work (at least assuming I'm understanding the question\dots).
From the geometric perspective, the inclusion map $k[e\_1, \dots, e\_n]\subseteq k[x\_1, \dots, x\_n]$ corresponds to the quotient $\phi: \mathbb A^n\to \mathbb A^n/S\_n$. The ideal $J:= I\cap k[e\_1, \dots, e\_n]$ is the contraction of the ideal $I$. Hence, geometrically, it seems to me that computing $J$ is equivalent to computing the ~~Zariski~~ scheme-theoretic closure of $\text{Spec}(k[x\_1, \dots, x\_n]/I)$ under the map $\phi$.
So the ~~Zariski closure~~ Kernel of a Ring Map algorithm on page 84 of Greuel and Pfister's "A Singular Introduction to Commutative Algebra" would seem to be applicable. To apply the algorithm, you define a ring $R:=k[x\_1,\dots,x\_n, t\_1, \dots, t\_n]$ and then define an ideal $N\subseteq R$ by
$
N:=I+\langle t\_1-e\_1(\mathbf{x}), \dots, t\_n-e\_n(\mathbf{x}) \rangle
$
where $e\_i(\mathbf{x})$ is the $i$'th symmetric polynomial. Compute the elimination ideal $N\cap k[t\_1, \dots, t\_n]$, and say that it equals $\langle p\_1(\mathbf{t}), \dots, p\_r(\mathbf{t})\rangle$. Then $I\cap k[e\_1, \dots, e\_n]$ will equal the ideal $\langle p\_1(\mathbf{e}), \dots, p\_r(\mathbf{e})\rangle$.
I hope this helps.
| 5 | https://mathoverflow.net/users/4 | 41436 | 26,455 |
https://mathoverflow.net/questions/41259 | 3 | Let $\mathcal{P}(\mathbb{N})$ be the set of all probability mass functions on $\mathbb{N}=\{1,2,\dots \}$. Let $E$ be a closed(with respect to pointwise convergence, or equivalently the total variation metric) subset of $\mathcal{P}(\mathbb{N})$ and $Q\notin E$. Let $0<\beta<1$.
Now $\displaystyle \sum\_{x\in N} P(x)^{\beta}Q(x)^{1-\beta}\le 1$ for any $P$ and $Q$ by Holder's inequality. Let us suppose that $0< s:=\displaystyle \sup\_{P\in E}\sum P(x)^{\beta}Q(x)^{1-\beta}$.
Let $\{P\_n\}$ be a sequence in $E$ such that $\sum P\_n(x)^{\beta}Q(x)^{1-\beta}\to s$
Now my goal is to examine whether $\{P\_n\}$ have a convergent subsequence converging to a true probability distribution. By the diagonal argument I can always extract a convergent subsequence, but the limit need not be a probability distribution; it could be a defective probability distribution. But in this problem can one somehow argue that we can extract a convergent subsequence converging to a probability distribution?
I am also thinking of showing that $P\_n$ are tight, because in the examples which I have, mass cannot escape to infinity.
I hope, if we show tightness, we can extract a convergent subsequence converging to a true probability distribution.
| https://mathoverflow.net/users/7699 | Tightness of probabilty distributions | I think this conjecture is false, that is, there does not necessarily exist a subsequence that converges to a true probability distribution. Consider the following situation:
Let $Q=(1,0,0,0,...)$, i.e. the probability distribution with all mass at $x=1$.
Define the distribution $R\_n$, for $n=2,3,...$, as $$R\_n(1)=\frac{1}{2} - \frac{1}{n}$$
$$R\_n(n)=\frac{1}{2} + \frac{1}{n}$$
and $R\_n(x)=0$ for all other values.
Let $E$ be the set of all $R\_n$, for $n\geq 2$. Note that $E$ is closed (because any distribution not in $E$ can be separated from it by a sufficiently small $\epsilon$ ball) and that $s=(1/2)^\beta>0$.
Next, we argue that any convergent distributions $P\_n$ contained in $E$ can be viewed as a subsequence of the $R\_n$. For any $R\_n$, $$\sum\_{x=1}^\infty R\_n^\beta (x) Q^{1-\beta}(x)=\left(\frac{1}{2}-\frac{1}{n}\right)^\beta$$
As $n$ increases, this value increases monotonically to $s$. Therefore, if we have any set of points $P\_n$ in $E$ such that $$\lim\_{n\rightarrow \infty} \sum P\_n^\beta (x) Q^{1-\beta}(x) =s$$
then $P\_n$ has a convergent subsequence that is a subsequence of the $R\_n$.
Finally, if we consider any subsequence of $R\_n$, it does not converge to a probability distribution since half of its probability wafts off to infinity. (A formal proof is straightforward.)
| 2 | https://mathoverflow.net/users/8938 | 41438 | 26,456 |
https://mathoverflow.net/questions/41428 | 5 | If I have a Deligne-Mumford stack $\Pi : X \to (\mathrm{Sch}/k)$ for some field $k$, can it be reconstructed from $\Pi^{-1}(C) \subset X$ for some "small" subcategory $C \subset (\mathrm{Sch}/k)$? For example, let's say we assume the fppf topology (to be concrete), then does the restriction to the fully faithful subcategory $T$ with objects
$\mathrm{Ob}(T) := $ { $I\_n := \mathrm{Spec~} k[\epsilon]/(\epsilon^n) \mid n \in \mathbf{N}$ }
completely determine the stack? I think I've read that the restriction to $(\mathrm{Aff}/k)$ is enough, but since the fibers over the $I\_n$ determine the $n^{th}$-order formal neighborhoods ($\mathrm{HOM}(I\_n, X) \cong X(I\_n)$), I wonder if the restriction to $T$ is enough?
| https://mathoverflow.net/users/5935 | Can Deligne-Mumford stacks be characterized by their restriction to a small subcategory? | Restricting to Aff is certainly enough, but Aff isn't small (there are e.g., polynomial algebras on arbitrary sets). If your DM stack is *finitely presented* over $k$ (which is probably good to include in the definition, to avoid these issues), then it is determined by it's restriction to finitely-presented affines (which is essentially small).
Without some finiteness hypothesis, no set of finitely-presented algebras can suffice (even for affine schemes, nevermind DM stacks). (And I suppose no small category of test objects can suffice: Take Spec of a field generated by a set of cardinality larger than that of global sections of any of your test schemes.)
The set you give is insufficient even for smooth varieties over an alg. closed field: you will have a morphism whenever you have an (arbitrary) map on $k$-points.
| 7 | https://mathoverflow.net/users/1921 | 41443 | 26,459 |
https://mathoverflow.net/questions/41390 | 10 | Let $f:X \rightarrow Y$ be a morphism between two smooth projective varieties $X,Y$ which are defined over an algebraically closed field $k$. I am looking for some criteria which guaranties the projectivity of $f$.
For instance if $f$ is finite it is projective. Here we don't need the projectivity of the varieties $X,Y$.
Is the morphism $f$ projective if
Question1: The fibers of $f$ are finite?
Question 2: $f$ is one-to-one?
Question 3: $f$ is onto?
Does the assumption $k=\mathbb{C}$ make the questions easier?
| https://mathoverflow.net/users/5286 | Morphism between projective varieties | Here is an attempt to prove Angelo's comment (it seems too simple to use a reference for it):
$X,Y$ defined over $S$. If they are both proper over $S$, then so is $f$ by Hartshorne, II.4.8(e). In particular $f$ is separated and universally closed.
If $X$ is projective over $S$ then for some $n$ there exists $\iota: X\to \mathbb P^n\_{S}$, a universally closed separated immersion.
The morphism $\nu:X\to\mathbb P^n\_S\times\_S Y=:\mathbb P^n\_Y$ defined by $x \mapsto (x,f(x))\in \mathbb P^n\_S\times\_SY$ is the composition of the base extension of $f$ by the projection $\pi:\mathbb P^n\_Y\to Y$; $f\_{\pi}$, the embedding $\iota$ base extended by the identity of $X$; $X\times\_SX\to \mathbb P^n\_S\times\_S X$ and the diagonal morphism of $X$; $\Delta\_X: X\to X\times\_S X$. I.e., $\nu=f\_{\pi}\circ (\iota\times\_S{\rm id}\_X)\circ \Delta\_X: X\to \mathbb P^n\_S\times\_S Y$. Actually, this might be a better definition than the one with "coordinates". Since $f$ is separated and universally closed and $\iota$ is universally closed, it follows that $\nu$ is closed. It is obviously an embedding. Now $f=\pi\circ\nu$ and hence it is projective.
Well, may be it was not that simple, and Angelo might tell me that it is wrong....
| 9 | https://mathoverflow.net/users/10076 | 41446 | 26,461 |
https://mathoverflow.net/questions/41329 | 0 | (This might look like just a post to you and you might think I shouldn't have submitted it as a question here but in reality it is some questions put together, so I hope you don't close it)
I only started studying about this topic two days ago and with my math not being really strong (yet) I am struggling to survive. But I am quite interested so I will, somehow!
I decided to begin by drawing the picture of the different categories of problems and this is what I have so far:
Computational Problems are divided in four categories: Decision, Search, Counting and Optimization
Then, we have another set of categories: NP, NPC, NPH and P
P is the class of problems that are solved in polynomial time by a deterministic Turing machine.
NP is the class of problems that are solved in polynomial time by a non-deterministic Turing machine but their solution can be verified by a deterministic Turing machine in polynomial time.
Including some other categories as well, a representation image is featured on the [Important Complexity Classes Wikipedia entry](http://en.wikipedia.org/wiki/Computational_complexity_theory#Important_complexity_classes).
**Question 1:** I understand that Turing Machines are only used to help us examine the the capabilities of computers but do non-deterministic machines "exist"?
**Question 2:** The image above is from a point of view with non-deterministic Turing machines. If they DO NOT exist, the representation is false, as P is included in NP because NTMs can solve both P and NP problems in polynomial time. Am I right or am I missing something?
**Question 3:** Is there a reason we call them P and NP (which at first confuses because it looks like **P** olynomial and **N** ot **P** olynomial instead of DP (Deterministic Polynomial) and NDP (Non-Deterministic Polynomial)?
Then, NPC is the class of NP problems that all other NP problems can be reduced to, in polynomial time.
And NPH is the class of problems (not essentially in NP) that all NP problems can be reduced to, in polynomial time.
**Question 4:** Since not all NPH problems are in NP, why is it part of their name?
**Question 5:** There should exist more than just one categories like NPC, no? To me, which means "to a person with no great math background", NPC is a set of some NP problems that look identical so we could find more NP problems that look identical and have more categories like NPC.
| https://mathoverflow.net/users/3550 | A few questions about Computational Problems Complexity Classification | 1) It depends what you mean by "exist", but probably the answer is no: we can't build an actual, physical, non-deterministic machine. (Of course, we can't literally build a Turing machine either, since the definition of a Turing machine generally requires infinite amounts of memory; but even if you're willing to fake that by counting a computer as an "existing" Turing machine, we still can't make non-deterministic ones.) Non-deterministic machines are essentially a thought experiment.
2) I don't quite understand this question, but I'll try to restate the relevant facts. A problem is in NP if it is solvable by a non-deterministic Turing machine in polynomial time. So any problem in P is also a problem in NP: since the problem is in P, it can be solved by a deterministic Turing machine in polynomial time. But every deterministic Turing machine IS a non-deterministic machine; it just ignores the additional capabilities. So the problem is also in NP.
3) The short answer is history. P may have been named before people were even considering non-deterministic machines (I'm not sure about the timing). DP would be a bit odd because it suggests that the only relevant question is determinism; but really, the point is that P is the class solved by "ordinary" machines, and other machines are defined from that. If you were thinking about P versus, say, programs solvable by quantum machines in polynomial time, it would be weird to emphasize that P was a deterministic class, because the relevant question would be quantum or not, rather than deterministic or not.
It is a bit confusing for people new to the area, but one gets used to it quickly. The potential confusion of NP with "not polynomial" doesn't last long, because "not polynomial" isn't a natural category to talk about.
4) I've never seen the NP hard problems referred to as "NPH", and google seems to think that you're the first person to use "NPH" to refer to a complexity class. But the term NP appears in the name of the "NP hard" problems because the problems are defined relative to NP. In general, if C is a class, it makes sense to talk about the problems hard for C (relative to some appropriate kind of reduction), and the NP hard problems are a special case of that.
5) The NP complete problems are more than just some class of similar NP problems; they're the "maximal" class. But yes, there could be others, and indeed, if P!=NP, it's a theorem that there are others. But as far as I know, there aren't any "natural" classes: no one has found a groups of problems that all reduce to each other, and are candidates for being not P but NP complete.
| 5 | https://mathoverflow.net/users/8991 | 41452 | 26,464 |
https://mathoverflow.net/questions/41447 | 6 | Definitions
===========
I believe set theorists have studied all of the following three notions in the context of forcing extensions of a model of ZFC, $M$ (hopefully the terminology is the standard one).
1. A function $f:\mathbb N\rightarrow\mathbb N$ is *eventually different* if for each function $g:\mathbb N\rightarrow\mathbb N$, $g\in M$, the set $\{n: f(n)=g(n)\}$ is finite.
2. A real $r\in [0,1]$ is a *Solovay random real* if for each measure-zero subset $S$ of $\mathbb R$ with $S\in M$, we have $r\not\in S$.
3. A function $f:\mathbb N\rightarrow\mathbb N$ is *dominating* if for each function $g:\mathbb N\rightarrow\mathbb N$ in the ground model $M$, the set $\{n: f(n)\le g(n)\}$ is finite.
Motivation
==========
An eventually different function that is not too fast-growing is reminiscent of a random real. Can we always use it to construct a random real? The analogous problem in computability theory was quite difficult but has been solved by Kumabe and Lewis (J. LMS, 2009).
Question
========
>
> **I.** Is it possible to add an eventually different function to $M$ while adding neither a Solovay real nor a dominating function?
>
>
>
>
>
EDIT: Now stating the question in the strongest possible form, which is the one Andrés Caicedo answers below.
| https://mathoverflow.net/users/4600 | An eventually different function adding no Solovay real nor dominating function? | Hi Bjørn, and congratulations to you and Bonnie!
The answer to I is yes. In fact, there is a standard way of doing this, with the "eventually different forcing ${\mathbb E}$". This notion does not add random or dominating reals, and adds an eventually different function.
Conditions have the form $(s, A)$ where $s\in\omega^{<\omega}$ and $A\in[\omega^\omega]^{<\omega}$, with $(s, A)\le(s',A')$ iff $s\supseteq s'$, $A\supseteq A'$, and for all $f\in A'$ and $j\in[|s'|,|s|)$, we have $s(j)\ne f(j)$. (For me, $p\le q$ means that $p$ is stronger.)
This is a nice forcing: It is ccc, in fact, $\sigma$-centered, since any two conditions with the same first coordinate are compatible. But no $\sigma$-centered forcing adds random reals.
That ${\mathbb E}$ does not add dominating reals is a tad more work. But you can find a written proof in section 7.4.B of "Set Theory: On the structure of the real line", by Tomek Bartoszy´nski and Haim Judah. Let me know if you do not have access to a copy.
| 6 | https://mathoverflow.net/users/6085 | 41461 | 26,470 |
https://mathoverflow.net/questions/41286 | 8 | Suppose we are given a finite collection of finite binary strings $\mathcal{S}$, of various lengths. Our task is to express any binary sequence $x\in 2^\mathbb{N}$ as juxtaposition of strings taken from $\mathcal{S}:$
$$x=\sigma\_1\sigma\_2\sigma\_3\dots$$
For any such sequence of "bricks", $\sigma\in\mathcal{S}^\mathbb{N},$ we consider
$$L^\*(\sigma):=\limsup\_{n\to\infty}\frac{1}{n}\sum\_{k=1}^n \mathrm{length}(\sigma\_k)$$
as a kind of parameter of quality of the factorization: we appreciate a factorization if the mean length of the composing pieces is frequently high.
*Question:*
>
> Given the set $\mathcal{S},$ how to
> compute the best $L^\*(\sigma)$ which
> is always attainable, whatever is
> $x\in 2^\mathbb{N}?$ That is, the quantity (depending on $\mathcal{S}$ only)
>
>
> $$\lambda(\mathcal{S}):=\inf\_{x\in2^\mathbb{N}} \max \{L^\*(\sigma) : \sigma\in\mathcal{S}^\mathbb{N}, x=\sigma\_1 \sigma\_2 \sigma\_3\dots \}.$$
>
>
> Also, are there special assumptions on the collection $\mathcal{S}$ that may simplify the analysis?
>
>
>
*Example.* Let $\mathcal{S}:=\{ 0,\ 1,\ 00,\ 01,\ 11 \}.$ Then, any binary sequence $x$ can be broken into a sequence of strings in $\mathcal{S}$, with average length larger than or equal to $3/2$.
*Remark.* For my purposes, we can assume that the collection $\mathcal{S}$ always enjoys the property of being "stable for extraction of sub-strings", that is, if $\sigma=\epsilon\_1 \epsilon\_2 \dots \epsilon\_n\in\mathcal{S}$, then also $\epsilon\_p \epsilon\_{p+1} \dots \epsilon\_q\in\mathcal{S},$ for any $1\leq p < q\leq n.$ If I am not wrong, this allows quite a simple inductive procedure for a canonical optimal factorization $\sigma$ of a binary sequence $x$: having chosen $\sigma\_1,\dots,\sigma\_k,$ take $\sigma\_{k+1}$ as the longest admissible element of $\mathcal{S}$: by the above property of $\mathcal{S}$ any other factorization $\tau$ of $x$ can be easily compared with $\sigma$, showing $L^ \*(\sigma) \geq L^ \*(\tau)$. So in this case I'd expect there is some hope of being able of computing the quantity $\lambda(\mathcal{S}).$
| https://mathoverflow.net/users/6101 | Breaking efficiently a binary sequence into given strings | Not an answer, but perhaps of some use:
It would help if S, the set of blocks, were finite. In any case, if S contains all words of length k, you know that for sufficiently long words L will have k as a lower bound.
(Computing k for a given S closed under consecutive substrings is feasible, but may not be
polynomial time in the length of a representation of S.) If S contains almost all words of length k, then the lower bound for L will still be k on those sigma which avoid the missing words, otherwise the lower bound for L will approach
the lower bound for L of those missing words which occur more frequently as you read more of sigma. It may then make sense to compute L for short words not in S, or infinite repetitions of such short words, and use this in aiding the computation of L of sigma.
If S has infinitely many finite blocks (and the alphabet is still finite), then there
is the potential for L to assume values of infinity. If you don't care about that or
if you are focussing on the cases where L will be finite, then you have some mathematics
which may touch upon descriptive set theory, and you may encounter a large cardinal or
other surprising axiom to consider. S being closed under consecutive substrings will
in many cases produce a larger k (k as in the above paragraph), and will have to be
pretty special otherwise to avoid any finite substring. There may be other tools in mathematical logic (e.g. filters on finite subsets of the language) that might tell
you what properties you want S to have.
Another poster referenced Huffman encoding. Outside of symbolic dynamics (and
related fields of automata theory and semigroup theory), that and similar string-processing
algorithms is a first choice of where to look for previous work. If you are considering extra-world (surreal? non-real?) applications using infinite S might you then turn to mathematical logic.
Gerhard "Ask Me About System Design" Paseman, 2010.10.07
| 3 | https://mathoverflow.net/users/3568 | 41463 | 26,472 |
https://mathoverflow.net/questions/41484 | 6 | Let G be a semisimple Lie group and let k be some number field. Let A be a finite index subgroup of some group B which is discrete and Zariski dense in G. Suppose A is in G(k). Then can we expect that B is in G(k') where k' is a finite Galois extension of k? When can we expect k'=k?
| https://mathoverflow.net/users/9891 | Finite index subgroup of discrete subgroup | I'll prove this when $G=SL(n,\mathbb{C})$. So assume $B\leq SL(n,\mathbb{C})$ is discrete, and $A\leq B$ is a finite-index subgroup. Let $A'\lhd A$ such that $A'\lhd B$ is the core, which is also finite index (by considering the kernel of the action on cosets of $A$ in $B$). Suppose that $A\subset SL(n,k)$ where $k$ is a number field, and assume that $k$ is the smallest field containing the entries of $A$. Since $B$ is Zariski dense in $SL(n)$, $A'$ is also Zariski dense. Take a maximal collection $a\_1,\ldots, a\_m\in A'$ of linearly independent elements. Since $A$ is finite index in $B$, the field generated over $\mathbb{Q}$ by the entries of $B$ will be spanned over $k$ by the entries of finitely many coset representatives of $A$ in $B$, and therefore will be a finitely generated extension field $k'$ of $k$. Consider the variety $X$ of representations $\rho:B\to SL(n,\mathbb{C})$ such that $\rho\_{| A}=Id$. This is a variety again, because it is determined by finitely many matrices which are coset representatives of $A$ in $B$. Considering the fact that $\rho(b)^{-1} a\_i \rho(b)=\rho(b^{-1}a\_i b)=b^{-1}a\_i b\in A'$, we see that $b\rho(b)^{-1}\in Z(a\_i)$. But since the $a\_i$ are linearly independent, we see that $b \rho(b)^{-1}\in Z(A')=Z(SL(n,k))$ since $A'$ is Zariski dense. Thus, we see that there are only finitely many possibilities for $\rho(b)$ for each $b$, and therefore the variety $X$ is finite. This in turn implies that $k'$ is a finite extension of $k$, since a transcendental extension would give rise to a variety $X$ of dimension $\geq 1$, since you could vary the transcendental generators at will.
Addendum: One may actually show that $k'$ is a Galois extension of $k$. Since $Z(SL(n,\mathbb{C}))=\{ \zeta I: \zeta^n=1\}$, we see that $\rho(b)=\zeta b$, for some $\zeta \in k', \zeta^n=1$. Thus, for different Galois embeddings $\sigma:k'\to \mathbb{C}$ which restrict to the identity on $k$, the entries of $\sigma(b)$ will change by multiplication by a cyclotomic number. So we conclude $b\_{ij}^n \in k$ for each matrix entry $b\_{ij}$ of $b$. This implies that $k'$ is a Galois extension of $k$ obtained by adding finitely many $n$-th roots to $k$.
| 7 | https://mathoverflow.net/users/1345 | 41485 | 26,484 |
https://mathoverflow.net/questions/41427 | 4 | In which sense is it possible to solve $\Delta u=0$, $\partial\_\nu u=\phi$, for $\int\phi=0$ on a closed domain, say a ball $B^3\subset\mathbb R^3$?
For example would a $\phi\in L^p(\partial B^3)$, $1< p<2$ make sense?
In other words, is $W^{1,p}$ really the right trace space, or else, which is?
Where can I find this kind of results?
Thanks!
| https://mathoverflow.net/users/5628 | Trace space and Neumann boundary condition | You can solve the problem with even less regularity than in Rekalo's answer. If $u\in W^{1,p}(\Omega)$, it does not have a normal trace in general. But if you assume in addition that $\Delta u\in L^p(\Omega)$, then the normal trace is well-defined and belongs to $W^{-1/p',p}(\partial\Omega)$, where $p'$ is the conjugate exponent. This space is of negative order, thus is not contained in any $L^q$. It is defined as the dual space of $W^{1/p',p'}(\partial\Omega)$. Since the latter contains the function ${\bf 1}$, it makes sense to say that the integral of $\phi$ is zero: just write $\langle\phi,{\bf 1}\rangle=0$. Under this condition, the Neumann boundary value problem admits a solution, unique up to an additive constant.
| 6 | https://mathoverflow.net/users/8799 | 41487 | 26,486 |
https://mathoverflow.net/questions/41464 | 1 | The short form of my question is: Can we find two formulae (in the multiplicative fragment of linear logic (MLL), that is, without additives or exponentials) A and B such that (1st) A is provable and B is provable, (2nd) A ⅋ B (A par B) is provable?
The context of the question is this: If we can find two such formulae, we can easily show that the multiplicative fragment of linear logic is not truth functional.
By truth-functionality, we should have:
1. ϕ(A) = ϕ(B) ==> ϕ(~A) = ϕ(~B)
2. ϕ(A1) = ϕ(A2) AND ϕ(B1) = ϕ(B2) ==> ϕ(A1 ⊗ B1) = ϕ(A2 ⊗ B2)
3. ϕ(A1) = ϕ(A2) AND ϕ(B1) = ϕ(B2) ==> ϕ(A1 ⅋ B1) = ϕ(A2 ⅋ B2)
Let us assume, for reductio, that there exists a function ϕ, from the set of linear formulae to the elements of an arbitrary set E, that satisfies (1)-(3), and which is such that ϕ(A) = e if and only if A is provable in linear logic.
(Remark: We know that it is possible, in fact, to find a truth-functional value-assignment that gives every provable formula the same value---i.e. that assigns "true" to a formula whenever it is provable. This won't concern us here. What we want to show is that any such function always assigns truth to too many formulae, that "truth", on any assignment, may be construed as a necessary condition for provability, but by no means a sufficient condition for provability.)
What we want to do now is find formulae {A1, A2, B1, B2} which are each, on their own, provable, and so which will be such that:
ϕ(A1) = ϕ(A2) = e
ϕ(B1) = ϕ(B2) = e
But which are such that (A1 ⅋ B1) is not provable, while (A2 ⅋ B2) is provable. If we find such formulae, then we will have shown that our assumption leads to a contradiction: by (I), ϕ(A1 ⅋ B1) ≠ ϕ(A2 ⅋ B2), but by (II), ϕ(A1 ⅋ B1) = ϕ(A2 ⅋ B2).
This is where I hit the wall. Finding multiplicative A1, B1 is simple enough: Let them each be the constant **1**. **1** ⅋ **1** is easily shown to be *unprovable* (which by itself tells us a lot about the peculiarities of linear logic---**1** ⅋ **1** *looks* like the linear counterpart of True v True). If we allow ourselves to dip into the additives, we can let A2 = **1** and let B2 = (**1**⊕⊥), and the resulting A2 ⅋ B2 will be *provable*. This is enough to show that LL, as a whole, is not truth-functional, because it leads to the destructive dilemma:
Either ϕ(**1** ⅋ **1**) = ϕ(1 ⅋ (**1** ⊕ ⊥)) = e, in which case ϕ(x) = e is not equivalent to “x is provable”; since ϕ and e are arbitrary, this shows that no truth-functional value assignment captures linear provability.
Or else ϕ(**1** ⅋ **1**) ≠ ϕ(**1** ⅋ (**1** ⊕ ⊥)), which contradicts truth-functionality requirement (3). To the extent that it violates (3), ϕ is not truth-functional.
This is all well and good, but can we get a similar result for the multiplicative fragment, that is, can we find two *multiplicative*, MLL-provable formulae whose multiplicative disjunction is also provable?
(PS: I'm using Girard's notation, not Troelstra's, so **1** and ⊥ should be read as the multiplicative constants, the neutrals for tensor and par, respectively.)
| https://mathoverflow.net/users/9881 | Can the Multiplicative Fragment of Linear Logic be shown to be non-truth-functional? | I don't have a complete proof, but I'm rather skeptical of the existence of such formulas. Certainly in the unit-free fragment of MLL it's hopeless, by applying a proof net criterion. For example, using the Danos-Regnier criterion for validity of proof nets, if you have a sequent $\to A \wp B$ (which is equivalent to $\to A, B$; I'm using $\wp$ to denote par) and if $A$ and $B$ are provable, then for any way of setting the $\wp$-switches on subformulas, the resulting graphs for $A$ and $B$ are acyclic and connected, so the graph for $\to A, B$ would be disconnected, and hence the sequent cannot be provable.
Generally speaking, the presence of units (or neutrals as you call them) make provability of sequents much harder to decide. About all I can say at the moment is that if you have proofs of $A$ and $B$, then any proof of $A \wp B$ cannot make any use of the proofs of $A$ and $B$ whatsoever, for essentially the same reason as given above.
| 2 | https://mathoverflow.net/users/2926 | 41498 | 26,491 |
https://mathoverflow.net/questions/41041 | 4 | Let $I,J$ be homogeneous ideals in the algebra of polynomials in $n$ variables over the complex numbers.
Let $V(I)$ and $V(J)$ be the **affine** algebraic varieties that are determined by $I$ and $J$ (**not** the projective varieties). Suppose that $V(I)$ and $V(J)$ are isomorphic as algebraic varieties. By this I mean that there are polynomial maps $f$ and $g$ from $\mathbb{C}^n$ to itself, such that $f$ restricted to $V(I)$ is a bijection onto $V(J)$, and such that $g$ restricted to $V(J)$ is its inverse.
The question is this: does it follow that there exists a linear map on $\mathbb{C}^n$ that maps $V(I)$ onto $V(J)$?
Thanks to discussions with colleagues (thank you David and Mike), I am quite convinced that if we assume that the origin is the only singular point in $V(I)$ then the answer is yes. Is this true in general?
I think this question is equivalent to the following (see my partial answer below): **Is it true that whenever there is an isomorphism between $V(I)$ and $V(J)$, there is also isomorphism that fixes $0$?**
| https://mathoverflow.net/users/1193 | If two "homogeneous" algebraic varieties are isomorphic, are they necessarily related by a linear map? | Trying to generalize Torsten's answer: It seems that if the cones are isomorphic then the isomorphism can indeed be chosen to preserve the origin.
For a projective variety $V$ let's denote the affine cone by $C(V)$. Torsten says that if $V$ is not a (projective) cone then in $C(V)$ the "cone point" $0$ is the unique point of maximum multiplicity.
$V$ is a projective cone if it is the join of a point $\mathbb P^0$ in the ambient projective space with a variety $W$ in a hyperplane (a hyperplane not containing that point). In this case $C(V)$ is the product of $C(\mathbb P^0)=\mathbb A^1$ and $C(W)$. In the general case $V$ is the join of a linear $\mathbb P^{d-1}$ with some $W$ which is not itself a projective cone, and then $C(V)=C(\mathbb P^{d-1})\times C(W)=\mathbb A^d\times C(W)$. Surely Torsten's statement generalizes to say that the points of maximal multiplicity in $C(V)$ are now those in $\mathbb A^d\times 0$.
So, given $V\_1$ and $V\_2$ such that $C(V\_1)$ and $C(V\_2)$ are isomorphic, the two numbers $d\_1$ and $d\_2$ must be equal, and if the isomorphism does not carry $0$ to $0$ then it can be adjusted to do so using translations in $\mathbb A^d$.
EDIT in response to Orr's comment: Here's what I mean, in your notation. Let $I$ be a homogeneous ideal in $k[x\_1,\dots ,x\_n]$ corresponding to some "homogeneous" affine variety $V(I)\subset\mathbb A^n$ and let $P(I)\subset \mathbb P^{n-1}$ be the variety that the same ideal defines. (I believe $V(I)$ is called the affine cone on $P(I)$.)
It might happen that after some linear change of coordinates $I$ becomes an ideal generated by a homogeneous ideal $I\_1$ in $k[x\_1,\dots ,x\_p]$ and a homogeneous ideal $I\_2$ in $k[x\_{p+1},\dots ,x\_n]$. If so, then $V(I)$ is the product of $V(I\_1)\subset \mathbb A^p$ and $V(I\_2)\subset \mathbb A^{n-p}$, and I believe that $P(I)$ would be called the join of the projective varieties $P(I\_1)\subset \mathbb P^{p-1}$ and $P(I\_2)\subset \mathbb P^{n-p-1}$. In particular if $p=n-1$ and $I\_2=0$ then $V(I)=V(I\_1)\times \mathbb A^1$ and $P(I)$ is called the projective cone on $P(I\_1)$.
Torsten is arguing that if $P(I)$ is not a cone, i.e. if there is no linear change of variable such $I$ is generated by polynomials not involving the last coordinate, then the origin is intrinsically characterized as the unique point in $V(I)$ of maximal multiplicity. I am saying that one can treat the general case in the same way, as follows: Suppose that $P(I)$ is a cone, or a cone on a cone, or ... as far as you can go. That is, make a linear change of variables so that $I$ is generated by polynomials in the first $p$ coordinates with $p$ as small as possible. Thus $V(I)$ is the product of some $V(I\_1)$ with $\mathbb A^{n-p}$ and $P(I)$ is the join of the corresponding $P(I\_1)$ with $\mathbb P^{n-p-1}$. Now in $V(I)=V(I\_1)\times \mathbb A^{n-p}$ the points of $0\times \mathbb A^{n-p}$ are the points of maximum multiplicity, and furthermore any one of them is carried to $0=(0,0)$ by some automorphism of $V(I)$ since $\mathbb A^{n-p}$ has an automorphism group that acts transitively.
The idea of iterated singular locus is not quite so successful. In most cases if the projective variety $S(P(I))$ is $P(J)$ then the homogeneous affine variety $S(V(I))$ will be $V(J)$. In the extreme case when $P(I)$ is smooth, so that $S(P(I))$ is empty, $S(V(I))$ will be $0$, with the exception that if $P(I)$ is a projective space (linearly embedded in $\mathbb P^{n-1}$) then $V(I)$ will be an affine space (linearly embedded and containing the origin) whose singular locus is empty rather than $0$. Thus in the sequence $V(I)$, $S(V(I))$, ... the last nonempty thing will be an affine space, possibly $0$ or possibly bigger. But it will not always be the same thing as before (the maximal affine space such that $V(I)$ is in a linear fashion the product of it with something). For example, if $n=3$ and $P(I)$ is a projective curve which has exactly one singular point but which is not simply the union of lines through that point, then the singular locus of the homogeneous surface $V(I)\subset A^3$ will be a line through the origin but there will be no automorphism of $V(I)$ moving $0$ to another point in that line (or to anything other than $0$).
| 1 | https://mathoverflow.net/users/6666 | 41509 | 26,496 |
https://mathoverflow.net/questions/41497 | 1 | Let $A$ be a $n \times n$ matrix all of whose entries has modulus 1.
Suppose the matrix $A$ is singular.
We will assume without loss of generality that all the entries in the first row and the first column of the matrix are 1.
Observe when $n=2$ the matrix $A$ can be then singular if and only if $a\_{2,2}=1$ as well.
A slightly less trivial observation is that the same thing happens when $n=3$,
that is the matrix $A$ is singular if and only if two of the rows or columns
are identical.
\begin{equation}
\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & \alpha\_{2,2} & \alpha\_{2,3} \\
1 & \alpha\_{3,2} & \alpha\_{3,3} \\
\end{array}\right| = 0
\end{equation}
So the matrix $A$ is singular iff $(\alpha\_{2,2}-1)(\alpha\_{3,3}-1)=(\alpha\_{2,3}-1)(\alpha\_{3,2}-1)$.
Let us assume without loss of generality that $\alpha\_{2,2} \neq 1$ and $\alpha\_{3,2} \neq 1$.
Consider the circle $C\_1(t)= (\alpha\_{2,2}-1) (e^{2 \pi i t}-1) $ and $C\_2(t)=(\alpha\_{2,3}-1) (e^{2 \pi i t}-1), t\in [0,1]$.
Since, the two circles either are identical and in that case $\alpha\_{i,2}=\alpha\_{i,3}$ that is the second and third columns are identical, or else as two distinct circles can intersect in at most two points we get similarly two of the rows or columns are identical.
Now, probably it is too much to expect the same result for all $n$.
>
> But my requirement is only for $n=4$,
> is it true that a similar result holds
> for $n=4$ ?
>
>
> Edit: I forgot to mention that I am interested in the case when the matrix is singular > > and none of its sub matrices are singular. (thanks @ Gerry Myerson for pointing it out)
>
>
>
Thankyou,
| https://mathoverflow.net/users/6766 | structure of singular matrices whose entries have modulus one | Your first calculation for $3\times3$ matrices applies in full generality: If your matrix writes blockwise $[1 \quad e^T ; e \quad M]$, with $e^T=(1,\ldots,1)$, and if $M$ is non-singular (implied by your assumption), then the property that $\det A=0$ is equivalent to $e^T\hat M e=\det M$, that is $e^TM^{-1}e=1$, or to $\det(J-M)=0$, with $J=ee^T$ the matrix with $1$s everywhere. Just use the Sherman-Morrison formula $\det(B+xy^T)=(\det B)(1+y^tB^{-1}x)$.
**Edit** after 3 hours. Take the $3\times3$ matrix $N:=[1 \quad i \quad -1;i \quad -1 \quad1;-1 \quad1 \quad-1]$. We have $N^{-1}e=(-i-1,-2i,-i)^T$. Let $D$ be a diagonal matrix with unit entries on the diagonal, so that $M:=ND^{-1}$ is still an admissible matrix. Then
$$e^TM^{-1}e=e^TDN^{-1}e=(-i-1)z\_1-2iz\_2-iz\_3.$$
**Claim**: There exist unit numbers $z\_j$ such that the right-hand side equals $1$. There exists actually a lot of them Consequence: the matrix $A$ is singular. Yet it does not have two equal rows or columns. *Proof* of the claim: we may search for unit numbers $y\_j$ such that $\sqrt2 y\_1+2y\_2+y\_3=1$. Taking $y\_3=y\_1$, we just have $(1+\sqrt2)y+2y'$, which covers a corona $(\sqrt2-1)\le |z|\le 3+\sqrt2$. In particular the equation $(1+\sqrt2)y+2y'=1$ has a solution. If instead we choose $y\_3=e^{i\epsilon}y\_1$ with a small enough $\epsilon$, the number $\sqrt2 y\_1+y\_3$ covers a circle of raidus $\rho$ close to $\sqrt2-1$, and the corona obtained by adding $2y\_2$ still contain $1$.
| 2 | https://mathoverflow.net/users/8799 | 41518 | 26,502 |
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