parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/45688 | 9 | Hi,
It is well known that the fundamental group of a topological group is abelian, and that every group is the fundamental group of some topological space.
My question is: Does every abelian group arise as the fundamental group of some topological group or are there other restrictions?
| https://mathoverflow.net/users/10217 | Abelian groups as fundamental groups of topological groups | Yes, if $G$ is an abelian group, its classifying space $BG$ is an *abelian* topological group whose $\pi \_1$ is $G$. You can find details in John Baez's wonderful post, hearteningly called "Classifying Spaces Made Easy"
<http://math.ucr.edu/home/baez/calgary/BG.html>
or in the answer by Chris Schommer-Priess to the following question on this site:
[Classifying Space of a Group Extension](https://mathoverflow.net/questions/12144/classifying-space-of-a-group-extension)
| 16 | https://mathoverflow.net/users/450 | 45690 | 28,941 |
https://mathoverflow.net/questions/45687 | 3 | Hi,
Is it true that for every locally compact separable metric space $E$ there exists a sequence $(K\_n)\_{n\in\mathbb{N}}$ of compact subsets of $E$ such that $K\_n\subset\stackrel{\circ}{K\_{n+1}}$ and $\cup K\_n = E$ ?
I’m almost sure this is false but I can’t find a counterexample.
Thank you.
| https://mathoverflow.net/users/10217 | Locally compact separable metric spaces | I think the following argument works under your hypothesis:
Consider $\mathcal{B}=\{B\_n\}$ a countable basis of the topology of $E$ such that $\overline{B\_n}$ is compact for any $n$ (this exists since $E$ is a separable metric space, thus, it has a countable basis and then a basis like this is constructed using local compactness).
Now, start with $K\_0= \overline{B\_0}$. Now, given $K\_n$, define $K\_{n+1}$ by the union of $\overline{B\_{n+1}}$ with the closure of a finite subcovering of $K\_n$ so, it is compact (being finite union of compact sets) and $K\_n \subset int(K\_{n+1})$.
We get that $\bigcup K\_n =E$ since it contains $\bigcup B\_n=E$.
| 4 | https://mathoverflow.net/users/5753 | 45691 | 28,942 |
https://mathoverflow.net/questions/45683 | 1 | Hi, I am new to queueing theory. I am interested in a question that I feel should be fairly basic, yet I haven’t really found a clear solution to it. Hopefully somebody here can help me.
We have a single server system, with an infinite queue, and with slotted time. At the beginning of every slot, a number of jobs arrive in the queue. The number of jobs $X$ is a random variable over the non-negative integers, with expectation $\mu$. After these jobs arrive, the server processes some jobs, which leave the queue. The number of jobs the server can process is a Bernoulli random variable $C$. That is, $C = 1$ with some probability $p$, and $0$ otherwise. To state what is probably obvious, if $C = 1$, the queue size is reduced by $1$ (if the queue was non-empty), and the queue remains unchanged if $C= 0$ or if the queue was empty. Both $C$ and $X$ are iid across time.
I want to understand the conditions under which this system is stable. By stable, I mean $\sup\_{n \geq 1} E(Q(n)) < M$ for some finite $M$, where $Q(n)$ is the size of the queue at the beginning of time slot $n$, and $E(Q(n))$ is the expectation of $Q(n)$. I am not necessarily interested in a explicit value of $M$, just knowledge that it is finite is fine. I am hoping that the condition would be $\mu < p$ or something like that.
I realize that probably some sort of assumption on the distribution on $X$ is needed, which is fine. Assumptions like finite variance, strong law of large numbers, or even large deviation inequalities are OK with me.
**Edit:** Additionally, I am interested in what would happen if $E(C)$ was not a fixed $p$ but $p(t)$ (ie, a function of time). Here $p(t)$ itself is a random variable where $E(p(t)) = p$ for all $t$, and $p(t)$ converges to $p$ almost surely. This question appears to be related to "time dependent Markov chains". However, the references for time dependent Markov chains that I could find do not consider $p(t)$ to be a random variable it self (such as <http://portal.acm.org/citation.cfm?id=990738.990783>). Asmussen’s book talks about time dependent properties of Markov chains, but that appears to be quite different.
| https://mathoverflow.net/users/5873 | Stability of discrete queue (new twist) | You are asking the queue $Q\to (Q+Y)^+$ to be ergodic, where $Y$ is your $X-C$, and the stationary distribution of this queue to be integrable. Ergodicity requires that $E(Y)<0$, i.e. $\mu < p$. Integrability holds as soon as $Y$ (or, equivalently, $X$) is square integrable.
Amongst many other places, you might want to check example I.5.7 of *Applied Probability and Queues* by Søren Asmussen. (Are you sure this is not HW?)
| 2 | https://mathoverflow.net/users/4661 | 45701 | 28,947 |
https://mathoverflow.net/questions/45700 | 9 | The de-Rham cohomology ring of U(n) is the exterior algebra generated by the odd-dimensional classes x\_1, x\_3, ..., x\_(2n-1). Moreover, on a Lie group every cohomology class is represented by a unique invariant form (both left and right). I ask two questions:
1) if we represent U(n) with matrices U = [z\_ij], what is an explicit expression of a generator as an invariant form of U(n), in terms of the differentials dz\_ij, for each odd degree between 1 and 2n-1?
2) in the 1-dimensional space of the invariant forms of a fixed degree (multiples of x\_i), which are the two (opposite to each other) which represent the real image of a generator of the integral cohomology?
| https://mathoverflow.net/users/10758 | Cohomology of the unitary group | Here's an answer to the first question.
Let $\theta = U^{-1} dU$ be the left-invariant Maurer-Cartan one-form: it is a matrix of one-forms. Then
$$\omega\_{2n+1} = \mathrm{Tr} \theta^{2n+1}$$
is the desired bi-invariant form. Here $\theta^{2n+1}$ stands both for the wedge and matrix product of $\theta$ with itself $2n+1$ times.
I'm not sure I understand the second question: what "two" representatives are you talking about? I think the canonical representative will be certain multiple of $\omega\_{2n+1}$. I'll try to fish out a reference later on.
| 11 | https://mathoverflow.net/users/394 | 45702 | 28,948 |
https://mathoverflow.net/questions/45094 | 3 | Let $u^i\_j$, $i,j = 1, . . . N$, and det$\_q^{-1}$ be the standard generators of the quantum group $U\_q(N,C)$, and define the matrices $U$ and $U^{\ast}$ by setting $U\_{ij} := u^i\_j$ and $U^{\ast}\_{ij}:=(u^j\_i)^{\ast}$. It is "well known" that $U^{\ast}U=UU^{\ast}=1$. How does one prove this?
Moreover, how does this imply that $u^i\_j(u^r\_s)^{\ast} = (u^r\_s)^{\ast}u^i\_j$, for $r\neq i,s\neq j$?
| https://mathoverflow.net/users/2612 | Prove that $U^*U=UU^*=1$ for $U_q(N,C)$ | I assume that $U^\ast\_{ij}=S(u\_j^i)$, where $S$ is the antipode. (See for example the book by Klimyk & Schmüdgen, Section 9.2.4.) If so, then $UU^\ast=1=U^\ast U$ follows from the antipode axiom
$\mu\circ(\mathrm{Id}\otimes S)\circ \Delta =\varepsilon =\mu\circ( S\otimes\mathrm{Id})\circ \Delta$
after applying both sides to an arbitrary generator $u\_j^i$.
To prove $u\_j^iS(u\_s^r)=S(u\_s^r)u\_j^i$ for $r\neq i, s\neq j$, one can multiply both sides by the quantum determinant and use that it is central to get the equivalent statement $u\_j^iM\_s^r=M\_s^r u\_j^i$, where $M\_s^r$ are quantum minors of size $(N-1)\times(N-1)$.
Now one can observe that the subalgebra generated by $u\_b^a$ with $a\neq r, b\neq s$ is isomorphic to $M\_q(N-1,C)$, the quantized algebra of regular functions on all $(N-1)\times(N-1)$ complex matrices. $u\_j^i$ and $M\_s^r$ belong to this subalgebra and moreover $M\_s^r$ is the quantum determinant in $M\_q(N-1,C)$, hence it commutes with $u\_j^i$.
| 2 | https://mathoverflow.net/users/10756 | 45707 | 28,951 |
https://mathoverflow.net/questions/45697 | 5 | I would like to classify the integers $m \geq 2$ for which the four quadratic polynomials
$3k^2$, $3k^2+2k$, $3k^2+3k+1$, and $3k^2+5k+2$ together represent all integers modulo $m$. That is, every integer modulo $m$ should be in the range of at least one of these polynomials (where all operations are carried out modulo $m$). Computer evidence suggests that this holds if and only if $m$ is one of the following:
$7, 10, 19, 2^j, 3^j, 5^j, 11^j, 13^j, 41^j, 2\cdot3^j, 5\cdot3^j$, where $j \geq 1$.
Does someone see how to prove this? Thank you.
| https://mathoverflow.net/users/10757 | Four polynomials representing all integers modulo m | For a prime $p>2$, fix a nonsquare $c$. If you find $y$ such that $y/3$ is a non-square (i.e. $y/3=cx^2, x\ne0$) and $y/3 - 1/9 = cz^2, z\ne 0$, then $y$ is not represented by the first two polynomials and I can't be bothered completing the square to write the conditions for the other two. Bottom line is, you find such a $y$ if you can find a point on a curve over the finite field
$\mathbb{F}\_p$. By Weil, this will happen as soon as $p$ is large enough. So your $m$ can only have prime factors from a finite set. Should be downhill from here.
| 7 | https://mathoverflow.net/users/2290 | 45709 | 28,952 |
https://mathoverflow.net/questions/45708 | 3 | Apparently there is a notion of for example a $G$-connection on a discrete set. I've understood that this is a standard tool in for example lattice gauge theory. I'm looking for references to learn more about this (i.e. discrete 1-forms, connections, etc.) in the discrete setting.
More specifically, suppose I have a set $V$ of vertices and a set $E$ of directed edges. Let $G$ be a finite (perhaps non-abelian) group. What is the right notion of $G$-connection on $(V,E)$? Is there some classification of connections? In particular, I suppose it might matter if the directed graph can be drawn on a torus or on a sphere (or the plane).
Any references to literature where the basic notions are described is greatly appreciated.
| https://mathoverflow.net/users/9545 | Discrete G-connections | I'd begin by looking at Oeckl's "Discrete gauge theory":
[http://www.amazon.com/Discrete-Gauge-Theory-Lattices-Tqft/dp/1860945791](http://rads.stackoverflow.com/amzn/click/1860945791)
For a finite group $G$, the notion of a $G$-connection is easy to define; it is usually done when you have not just a graph but a 2-complex and add a condition that the monodromy of connection around the boundary of each 2-cell is trivial. If you apply it to a cell decomposition of a closed surface, you get the space of gauge equivalence classes of $G$-bundles on the surface, so teh result does not depend on the choice of cell decomposition.
This is a special (and well-known) example of Turaev-Viro theory, which in this case is also known as Chern-Simons theory with a finite gauge group (see paper of Freed and Quinn in CMP vol. 156).
| 6 | https://mathoverflow.net/users/10745 | 45712 | 28,954 |
https://mathoverflow.net/questions/45670 | 6 | In group cohomology, for $H$ a finite-index subgroup of $G$ and $M$ a $G$-module, there is a transfer (or corestriction) map $Cor : H^\* (H;M) \to H^\*(G;M)$.
In homotopy theory, there is a transfer map for finite covering spaces $\bar{X} \to X$, and it exists for all coefficient systems on $X$. It is given on homology (say) by sending a small simplex in $X$ to its finitely-many lifts to $\bar{X}$. The group transfer is obtained from the topological one by the finite covering space $BH \to BG$.
There is a fancier transfer, due to Becker and Gottlieb, for any map $E \to B$ whose homotopy fibre is stably equivalent to a finite complex. Can this be extended to give a transfer map for coefficient systems on $B$?
| https://mathoverflow.net/users/318 | Transfer homomorphisms with coefficients | I believe the answer is yes. The idea should be to generalize the construction of the group-theoretic transfer. I'll describe how I think this ought to go. I'm pretty sure what I'm describing is a known construction, but I haven't been able to dig up a reference which describes it.
Let $f:X\to Y$ be a map of spaces (CW-complexes), and assume (for simplicity) that both are path connected. Then there is an inclusion of topological groups $\phi: G\to H$, such that $B\phi: BG\to BH$ is equivalent to $f:X\to Y$.
A *local system* on $X$, in the most general possible context, is a spectrum $M$ equipped with a $G$-action. (I'm not doing anything fancy here; in particular, I'm not doing equivariant stable homotopy theory. Just spectra with a $G$-action; a $G$ map $f:M\to N$ of such things is a weak equivalence if it is a weak equivalence of the underlying spectra.)
Let $S\_G$ be the category of $G$-spectra. It'll be important to note that this is a closed monoidal category: the smash product $X\wedge Y$ of two objects in $S\_G$ is defined in $S\_G$ (smash the spectra, and use the diagonal $G$-action), as well as a function object $\mathrm{Hom}(X,Y)$ (take the spectrum function object, with diagonal $G$-action).
The Becker Gottlieb transfer $f\_\!$ of $f$ associated to an $M\in S\_H$ should be a map $M\_{hH}\to M\_{hG}$ (where "$hG$" is "homotopy orbits"). So, if $M=S^0$ with trivial $H$-action, you get a map $S^0\_{hH}\to S^0\_{hG}$, which is a map $\Sigma^\infty Y\_+\to \Sigma^\infty X\_+$. You should interpret the map $\pi\_\* f\_\!$ as being a map
$$
H\_\*(Y,M) \to H\_\*(X,f^\*M),
$$
where these are "homology with coefficients in the local systems $M$ and $f^\*M$". There should be a similar way to get a cohomolgy transfer.
Let $F=\Sigma^\infty(H/G)\_+$, as a spectrum with $H$-action. Note that the space $H/G$ is just the fiber of $f:X\to Y$.
Here is a sequence of maps in $S\_H$:
$$
M\to \mathrm{Hom}(F,F\wedge M) \leftarrow \mathrm{Hom}(F,M)\wedge F \to \mathrm{Hom}(F,M)\wedge F\wedge F \to M\wedge F \approx M\wedge\_G H.
$$
The first is smashing with the identity map of $F$; the second comes from smashing a map $F\to M$ with a map $S^0\to F$; the third comes from the diagonal map $H/G\to H/G\times H/G$; the fourth is "evaluation".
If $F$ has the homotopy type of a *finite* CW-spectra (i.e., if $H/G$ is finitely dominated), then the backwards map in this sequence is a weak equivalence. In this case, we get a map $M\to M\wedge\_G H$ of $H$-spectra, and taking homotopy orbits gives $M\_{hH}\to M\_{hG}$.
If $Y$ is contractible, this amounts to a map
$$
S^0 \to \mathrm{Hom}(F,F) \leftarrow \mathrm{Hom}(F, S^0)\wedge F \to \mathrm{Hom}(F,
S^0)\wedge F\wedge F \to S^0\wedge F,
$$
which is to say, a map $S^0\to \Sigma^\infty X\_+$, and in cohomology this will send $1\in H^0(X)$ to $\chi(X)\in H^0(\mathrm{point})$. So apparently we recover the Becker-Gottlieb transfer.
If the spaces aren't connected, you can still do all this, but you need to allow $H$ and $G$ to be groupoids.
(I learned this way of thinking from some paper of John Klein about the "dualizing spectrum" of a toplogical group; I can't find one where he addresses the BG transfer this way.)
| 7 | https://mathoverflow.net/users/437 | 45724 | 28,961 |
https://mathoverflow.net/questions/41334 | 6 | This is a fairly minor, technical question, but I'll toss it out in case someone has a good idea on it.
Suppose $(X,<\_X)$ and $(Y,<\_Y)$ are well-founded orderings (not necessarily linearly ordered, though I don't think it matters). Consider the ordering ${<}$ on $X\times Y$ given by $(x',y') < (x,y)$ if $x'\leq x$ and $y'\leq y$, and either $x' < x$ or $y' < y$. Note that this is not the lexicographic ordering; indeed, it's symmetric.
Obviously $X\times Y$ is well-founded. Suppose I want to prove this carefully (by which I really mean "in the formal theory $ID\_1$"); more precisely, let's take $X$ to be a set with two properties:
$$Cl\_X:\forall x(\forall x'<\_X x. x'\in X)\rightarrow x\in X$$
and
$$Ind\_X: \forall Z[\forall x(\forall x'<\_X x. x'\in Z)\rightarrow x\in Z)\rightarrow X\subseteq Z]$$
and similarly for $Y$. (These just characterize that $X$ is its own well-founded part.) I want to prove that for all $(x,y)\in X\times Y$, $(x,y)$ are in the well-founded part of $X\times Y$ under ${<}$; call the well-founded part of $X\times Y$ $Acc(X\times Y)$.
I know one way to prove this: for each $x\in X$, define $Y\_x=\{y\in Y\mid (x,y)\in Acc(X\times Y)\}$. Let $X'$ be the set of $x\in X$ such that $Y\subseteq Y\_x$. Then it would be good enough to show that $X'$ satisfies the closure property, so I can apply $Ind\_X$. To do this, in turn, I show that, if $Y\subseteq Y\_{x'}$ for all $x'<\_X x$ then $Y\_x$ satisfies the closure property, so I can apply $Ind\_Y$.
Of course, that means I know I second way: I could swap $X$ and $Y$ in the above proof. Moreover, when one works through the details, it's clear that I'm really proving that the lexicographic ordering is well-founded, and using the fact that ${<}$ is a subrelation of the lexicographic ordering.
Which brings me to my question: is there a proof that $Acc(X\times Y)=X\times Y$ which is *symmetric*?
| https://mathoverflow.net/users/8991 | Symmetric Proof that Product is Well-Founded | I've found a symmetric proof.
First, every well-founded relation admits an ordinal rank function, an assignment of points to ordinals that respects the relation. For example, in your case $\alpha\_x=\sup\{\ \alpha\_u+1\mid u\mathrel{\lt\_X} x\ \}$ is the canonical rank function for $X$ and $\beta\_y=\sup\{\ \beta\_w+1\mid w\mathrel{\lt\_Y} y\ \}$ is the canonical rank function for $Y$.
Second, the key idea is to use the symmetric version of ordinal addition, called [natural sum](http://en.wikipedia.org/wiki/Cantor_normal_form#Natural_operations), an associative and commutative addition operation on ordinals. Specifically, the natural sum $\alpha\mathop{\sharp}\beta$ of two ordinals is the supremum of the order types arising in any linear completion of the disjoint sum partial order $\alpha\sqcup\beta$. Alternatively, if you express $\alpha$ and $\beta$ in Cantor normal form, then $\alpha\mathop{\sharp}\beta$ is the ordinal obtained by mixing the Cantor normal forms together and putting the terms in the correct order. The natural sum is defined in a completely symmetric way, and this is why it is commutative.
In your product order $X\times Y$, let us associate to the point $(x,y)$ the ordinal $\alpha\_x\mathop{\sharp}\beta\_y$. This ordinal assignment is completely symmetric, since we would assign the same ordinal to $(y,x)$ in $Y\times X$, as the natural sum is commutative. The point now is that this ordinal assignment serves as a rank function in $X\times Y$, since if $(x',y')\lt (x,y)$, then we know that $\alpha\_{x'}\leq \alpha\_x$ and $\beta\_{y'}\leq \beta\_y$, and at least one of these is strict. It follows that $\alpha\_{x'}\mathop{\sharp}\beta\_{y'}\lt \alpha\_x\mathop{\sharp}\beta\_y$, essentially because $(\alpha+1)\mathop{\sharp}\beta=(\alpha\mathop{\sharp}\beta)+1$, and so this really does rank the product relation, and so it is well-founded.
| 3 | https://mathoverflow.net/users/1946 | 45725 | 28,962 |
https://mathoverflow.net/questions/45680 | 3 | How can I prove that the Cartier dual of αp is again αp (using the Yoneda lemma)? It should be something like $\alpha\_p(R) \to (\alpha\_p(R) \to \mu\_p(R)),x \mapsto (y \mapsto exp\_{p−1}(x+y)$, where $exp\_{p−1}$ is the truncated exponential sequence. My problem is that this isn't a homomorphism.
| https://mathoverflow.net/users/nan | Cartier dual of \alpha_p | It is probably a bad idea to try to compute the Cartier dual but better to let
Cartier do that for you... If $G$ is a flat commutative finite group scheme with
affine algebra, the commutative and cocommutative $A$ which is the flat over the
base ring $R$. Then the Cartier dual is the spectrum of the dual Hopf algebra
$A^\ast$ of $A$. The proof of this is simple enough; an $R$-algebra homomorphism
$A\rightarrow R$ corresponds to a $\varphi\in A^\ast$ of multiplicative type,
$\Delta^\ast(\varphi)=\varphi\otimes\varphi$, which in turn corresponds to a Hopf
algebra map $R[t,t^{-1}]\rightarrow A^\ast$. As this can be done for all
$R$-algebras we get an isomorphism of functors.
Doing this for $\alpha\_p$ which has $A=R[x]/(x^p)$ we get that $A^\ast$ has a
basis dual to $x^i$ of the form $1/i!\partial^i/\partial x^i$. Unravelling the
definitions one gets the formula $s\mapsto(t\mapsto \exp\_{p-1}(st))$.
| 7 | https://mathoverflow.net/users/4008 | 45727 | 28,964 |
https://mathoverflow.net/questions/45738 | 1 | Given two smooth projective surfaces $X$ and $Y$ over some algebraically closed field.
Given a torsion free coherent sheaf $M$ on $X$. One has the projections $\pi\_X$ and $\pi\_Y$ from the product $X\times Y$. Then we have $\pi\_X^{\\*}M$ on $X\times Y$.
Question: Is $\pi\_X^{\\*}M$ flat over $Y$?
One has to show that for each $z\in X\times Y$ the $O\_{X\times Y,z}$-module $\pi\_X^{\\*}M\_{z}$ is a flat $O\_{Y,\pi\_Y(z)}$-module. Now $\pi\_X^{\\*}M\_z=M\_{\pi\_X(z)} \otimes\_{O\_{X,\pi\_X(z)}} O\_{X\times Y,z}$. I don't see why this should be flat over $O\_{Y,\pi\_Y(z)}$. I just know that $O\_{X\times Y}$ is flat over $O\_X$ and $O\_Y$, but that doesn't seem to help me.
Background: I'm reading www.imsc.res.in/library/pdf/shaves.pdf, page 144, Theorem 6.1.8. When they want to compute the relative $Ext$-sheaves explicitely, they choose a certain complex of sheaves for whose existence they cite an article by Banica, Putinar and Schumacher. Now i looked that article up, and the complex is constructed using that both sheaves are flat over the base scheme. In this proof the sheaf $\mathcal{E}$ is a quasi universal family, so it is flat by definition, but what about the pullback sheaf?
| https://mathoverflow.net/users/3233 | Are pullbacks from a factor of a product scheme flat over the other factor? | Certainly it is flat, and you don't even have to assume that $M$ is torsion free. Indeed, the question is local, so you can assume that $X$ and $Y$ are affine, so the question is: given $k$-algebras $A$ and $B$ and an $A$-module $M$ check that $M\otimes\_A(A\otimes\_k B)$ is flat over $B$. But this is easy --- $M\otimes\_A(A\otimes\_k B) \cong M\otimes\_k B$, so it is a free $B$-module.
| 3 | https://mathoverflow.net/users/4428 | 45741 | 28,968 |
https://mathoverflow.net/questions/45613 | 3 | Can Anyone prove the following conjecture?
Consider $k$ rational function vectors $V\_1(x\_1,\cdots,x\_n),\cdots,V\_k(x\_1,\cdots,x\_n)$. They are called \textbf{linearly dependent} if there exists rational functions $\alpha\_1(x\_1,\cdots,x\_n),\cdots,\alpha\_k(x\_1,\cdots,x\_n)$ which are not identically zero such that
\begin{align}
\alpha\_1(x\_1,\cdots,x\_n)V\_1(x\_1,\cdots,x\_n)+\cdots+\alpha\_k(x\_1,\cdots,x\_n)V\_1(x\_1,\cdots,x\_n)=0
\end{align}
This defines the rank of a matrix.
Conjecture:
(Rank-1 Decomposition Conjecture)
Let a $l \times m$ polynomial matrix
\begin{align}
A(x\_1,\cdots,x\_n)=
\left(
\begin{array}{ccc}
a\_{11}(x\_1,\cdots,x\_n) & \cdots & a\_{1m}(x\_1,\cdots,x\_n) \newline
\vdots & \ddots & \vdots \newline
a\_{l1}(x\_1,\cdots,x\_n) & \cdots & a\_{lm}(x\_1,\cdots,x\_n)
\end{array}
\right)
\end{align}
is rank $k$ where $a\_{ij}$ are linear functions whose coefficients are taken from $\mathbb{C}$. Then, there exists $l \times m$ polynomial matrices $A^{(1)}(x\_1,\cdots,x\_n),\cdots,A^{(k)}(x\_1,\cdots,x\_n)$ that satisfies the following three properties:
(i) $A(x\_1,\cdots,x\_n)=A^{(1)}(x\_1,\cdots,x\_n)+ \cdots + A^{(k)}(x\_1,\cdots,x\_n)$.
(ii) The rank of $A^{(i)}(x\_1,\cdots,x\_n)$ is $1$.
(iii) The elements of $A^{(i)}(x\_1,\cdots,x\_n)$ are linear functions on $x\_1,\cdots,x\_n$ with coefficients taken from $\mathbb{C}$.
Examples :
(1)
$\left(
\begin{array}{ccc}
x\_1 & x\_2 & x\_3 \newline
x\_4 & x\_5 & x\_6
\end{array}
\right)$
=
$\left(\begin{array}{ccc}
x\_1 & x\_2 & x\_3 \newline
0 & 0 & 0
\end{array}\right)$
+
$\left(\begin{array}{ccc}
0 & 0 & 0 \newline
x\_4 & x\_5 & x\_6
\end{array}\right)
$
(2)
$\left(
\begin{array}{cccc}
x\_1 & x\_2 & x\_5 & x\_5 \newline
x\_1 & x\_2 & x\_6 & x\_6 \newline
x\_1 & x\_2 & x\_7 & x\_7 \newline
x\_8 & x\_9 & x\_{10} & x\_{11}
\end{array}\right)
$
$=
\left(
\begin{array}{cccc}
x\_1 & x\_2 & 0 & 0 \newline
x\_1 & x\_2 & 0 & 0 \newline
x\_1 & x\_2 & 0 & 0 \newline
0 & 0 & 0 & 0 \\
\end{array}
\right)
+
\left(
\begin{array}{cccc}
0 & 0 & x\_5 & x\_5 \newline
0 & 0 & x\_6 & x\_6 \newline
0 & 0 & x\_7 & x\_7 \newline
0 & 0 & 0 & 0
\end{array}
\right)
+
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \newline
0 & 0 & 0 & 0 \newline
0 & 0 & 0 & 0 \newline
x\_8 & x\_9 & x\_{10} & x\_{11}
\end{array}
\right)
$
(3)
$\left(
\begin{array}{cc}
x\_1 & x\_4 \newline
x\_2 & x\_5 \newline
x\_3 & x\_6
\end{array}\right)$
$=
\left(
\begin{array}{cc}
x\_1 & 0 \newline
x\_2 & 0 \newline
x\_3 & 0
\end{array}
\right)
+
\left(
\begin{array}{cc}
0 & x\_4 \newline
0 & x\_5 \newline
0 & x\_6
\end{array}\right)
$
(4)
$
\left(
\begin{array}{ccc}
x\_1 & 2 x\_1 & 3 x\_1 \newline
2x\_2 & 4 x\_2 & 6 x\_2 \newline
x\_3 & 2 x\_3 & 4 x\_3
\end{array}
\right)$
$=
\left(
\begin{array}{ccc}
x\_1 & 2 x\_1 & 3 x\_1 \newline
2x\_2 & 4 x\_2 & 6 x\_2 \newline
x\_3 & 2 x\_3 & 3 x\_3
\end{array}
\right)
+
\left(
\begin{array}{ccc}
0 & 0 & 0 \newline
0 & 0 & 0 \newline
0 & 0 & x\_3
\end{array}
\right)
$
| https://mathoverflow.net/users/10735 | Rank-1 decomposition conjecture for matrix with linear function elements | The answer is **NO**. Take $l=m=3$ and the generic skew-symmetric matrix
$$A:=\begin{pmatrix} 0 & x & y \\\\ -x & 0 & z \\\\ -y & -z & 0 \end{pmatrix}.$$
The rank is $k=2$, yet it cannot be written $A=A^1+A^2$ where both $A^j$ would be linear in the indeterminates $x,y,z$ and rank-one.
>
> Sketch of the proof. Such matrices $A^j$ must be of the form $CR$ where one of the column $C$ or the row $R$ is constant and the other one is linear (Hint: use Gauss' Lemma). If the decomposition holds, we have $A=C^1R^1+C^2R^2$ with three possible cases. Either both rows $R^j$ are constant, or both column $C^j$, or one row and one column. In the first case, the range of $A(x,y,z)$ would be contained in the fixed plane spanned by $R^1$ and $R^2$, which is false. The second case is treated as well. There remains the case where $C^1$ and $R^2$ are constant, whereas $C^2$ and $R^1$ are linear. But then the fixed vector $C^1$ belongs to the range of $A(x,y,z)$ for every $(x,y,z)$ such that $R^1$ is not parallel to $R^2$, therefore for almost every $(x,y,z)$. Again, this is absurd.
>
>
>
| 2 | https://mathoverflow.net/users/8799 | 45743 | 28,969 |
https://mathoverflow.net/questions/45432 | 2 | I am looking for a comprehensible material covering Plancherel formula for $SL(n,\mathbb{R})$ and $SL(n,\mathbb{C})$. Of course, I wouldn't mind reading an explanation for general semisimple Lie groups.
(I am reading Varadarajan's Introduction to harmonic analysis on semisimple Lie groups and I'm a bit lost there.)
| https://mathoverflow.net/users/6818 | Plancherel formula for special linear group | Knapp's book *Representation theory of semisimple groups : an overview based on examples* has a chapter on various versions of the Plancherel theorem. It is also done (very carefully, with all the details, I suppose) in volume 2 of Wallach's *Real Reductive Groups*.
| 3 | https://mathoverflow.net/users/9962 | 45744 | 28,970 |
https://mathoverflow.net/questions/45735 | 6 | To a (projective smooth) algebraic surface $S$ over an algebraically closed field and a divisor $D$ of $S$, we can associate $n= \dim |D|$ and $g$, the genus of a generic member of $|D|$. I would like to fix $g$ and vary $S,D$ so as to make $n$ as large as possible. Is $n$ unbounded and, if not, what is the optimal bound? I am mainly interested in the case of positive characteristic but an answer over the complex numbers would be welcome too.
| https://mathoverflow.net/users/2290 | Maximal dimension of linear system of curves of fixed genus on a surface | $n$ is unbounded, as it is shown by the following example.
EDIT: my example did not work, as pointed out by quim in the comments.
The example he suggests however works: take $S$ the blowup of $P^2$ at a point $x$ and $D$ the strict transform of a curve of degree $d$ with a singular point of multiplicity $d-1$ at $x$. The general $D$ is smooth and $|D|$ has dimension $2d$.\
Of course, a similar construction can be used to construct examples with $g>0$.
On the other hand, if the Kodaira dimension of $S$ is $\ge 0$ and $D$ is irreducible, then $n=\dim|D|$ is bounded by $g$ by the following argument.
Up to blowing up $S$ we may assume that the general $D$ is smooth.
Let $m>0$ be such that $mK\_S\ge 0$. If $n>0$, then $D$ is not in the fixed part of $|mK\_S|$, hence $K\_SD\ge 0$. Hence by the adjunction formula $D|\_D$ is a divisor of $D$ of degree $\le 2g-2$ and therefore it satisfies $2\dim|D|\_D|\le \deg D$ (this is Clifford's theorem if $D$ is special and it is trivially true otherwise).
So we have $\dim |D|\le \dim|D|\_D|+1\le D^2/2+1\le (D^2+K\_SD)/2+1=g$.
I, hence $K\_S|\_D$ is effective and $D|\_D$ is special. Then Clifford's theorem and the adjunction formula give $2(n-1)\le D^2\le D^2+K\_SD=2g-2$.
| 11 | https://mathoverflow.net/users/10610 | 45749 | 28,973 |
https://mathoverflow.net/questions/20771 | 12 | **Background:**
Let $G$ be a profinite group. If $M$ is a discrete $G$-module, then $M=\varinjlim\_U M^U$, where the direct limit is taken with respect to inclusions over all open normal subgroups of $G$, and one naturally has $H^n(G,M)\simeq\varinjlim H^n(G/U,M^U)$, where the cohomology groups on the right can be regarded as the usual abstract cohomology groups of the finite groups $G/U$ (this is sometimes, as in Serre's Local Fields, taken as the definition of $H^n(G,M)$).
More generally if one has a projective system of profinite groups $(G\_i,\varphi\_{ij})$ and a direct system of abelian groups $(M\_i,\psi\_{ij})$ such that $M\_i$ is a discrete $G\_i$-module and the pair $(\varphi\_{ij},\psi\_{ij})$ is compatible in the sense of group cohomology for all $i,j$, then $\varinjlim M\_i$ is canonically a discrete $\varprojlim G\_i$-module, the groups $H^n(G\_i,M\_i)$ form a direct system, and one has $H^n(\varprojlim G\_i,\varinjlim M\_i)\simeq\varinjlim H^n(G\_i,M\_i)$. The statement and straightforward proof of this more general result can be found, for instance, in Shatz' book on profinite groups.
**Question:**
In general, I'm wondering if there are, under appropriate hypotheses, any similar formulae for projective limits of discrete $G$-modules. Now, given a projective system of discrete $G$-modules $(M\_i,\psi\_{ij})$, it isn't even obvious to me that the limit will again be a discrete $G$-module, and at any rate, while each $M\_i$ is discrete, the limit (in its natural topology) will be discrete if and only if it is finite. So, for the sake of specificity, I'll give a particular situation in which I'm interested. If $R$ is a complete, Noetherian local ring with maximal ideal $\mathfrak{m}$ and finite residue field and $M$ is a finite, free $R$-module as well as a discrete $G$-module such that the $G$-action is $R$-linear, then the canonical isomorphism of $R$-modules $M\simeq\varprojlim M/\mathfrak{m}^iM$ is also a $G$-module isomorphism (each $M/\mathfrak{m}^iM$ is a discrete $G$-module with action induced from that of $M$). Moreover, in this case, one can see that the limit is a discrete $G$-module (because it is isomorphic to one as an abstract $G$-module!). There is a natural homomorphism $C^n(G,M)\rightarrow\varprojlim C^n(G,M/\mathfrak{m}^iM)$ where the projective limit is taken with respect to the maps induced by the projections $M/\mathfrak{m}^jM\rightarrow M/\mathfrak{m}^iM$, and this induces similar map on cohomology. I initially thought the map at the level of cochains was trivially surjective, just because of the universal property of projective limits. However, given a ``coherent sequence" of cochains $f\_i:G\rightarrow M/\mathfrak{m}^iM$, the property gives me a map $f:G\rightarrow M$ that is continuous when $M$ is regarded in its natural profinite topology, which is, as I noted above, most likely coarser than the discrete topology, so this might not be a cochain. So, what I'd really like to know is whether or not the map on cohomology is an isomorphism.
**Why I Care:** The reason I'd like to know that the map described above is an isomorphism is to apply it to the particular case of $G=\hat{\mathbb{Z}}$. It is well known (and can be found, for instance, in Serre's Local Fields) that $H^2(\hat{\mathbb{Z}},A)=0$ for $A$ a torsion abelian group. In particular the higher cohomology of a finite $\hat{\mathbb{Z}}$-module vanishes, and I'd like to be able to conclude that the same is true for my $M$ above, being a projective limit of finite abelian groups.
Thanks!
* Keenan
| https://mathoverflow.net/users/4351 | In what sense (if any) does the cohomology of profinite groups commute with projective limits? | Hi Keenan,
You're right that the projective limit of discrete $G$-modules is not necessarily discrete. To take the cohomology of such "topological $G$-modules" you can use continuous cochain cohomology and this continuous cochain cohomology commutes with inverse limits under certain conditions. See section 7 of chapter II of Cohomology of Number Fields by Neukirch, Schmidt & Wingberg.
| 3 | https://mathoverflow.net/users/10766 | 45751 | 28,975 |
https://mathoverflow.net/questions/45748 | 0 | Here are some direct questions at the interface of algebraic and differential geometry:
(1) Is there an easy characterisation of those affine algebraic varieties which are Kahler?
(2) Is there an easy characterisation of those affine algebraic varieties which are symmetric spaces?
(3) Is there an easy characterisation of those affine algebraic varieties which are both? (From the first comment below, it seems that we can rephrase this question as: which affine algebraic varieties are symmetric?)
(4) What happens if I then also require compactness?
| https://mathoverflow.net/users/1867 | What are the Compact Symmetric Kahler Algebraic Varieties? | If we ignore the trivial case of the affine line, then irreducible symmetric spaces come in pairs compact - non-compact. The compact ones are naturally projective varieties, while the non-compact ones are affine varieties. Thus question (4) is problematic, unless you mean "locally symmetric" or a more general notion of symmetric space than I understand here (i.e. "globally symmetric Riemannian symmetric space"). As far as non-compact symmetric spaces are concerned, they are Kähler if and only if they are biholomorphic to a bounded symmetric domain. Equivalently, there exists a compact quotient with non-trivial H^2 or, equivalently, the point stabilizer of the automorphism group has infinite center... I could give many more characterizations, but I do not quite see what you are after, so maybe you can provide more detailed information?
| 1 | https://mathoverflow.net/users/9927 | 45753 | 28,977 |
https://mathoverflow.net/questions/45730 | 1 | Suppose $G$ is a semigroup (i.e., closed under matrix multiplication) of invertible $2\times 2$ real matrices. Suppose also that $G$ is transitive i.e., for any two non-zero vectors $u$ and $v$ there exists a matrix in $G$ that maps $u$ to $v$. Are there any non-trivial examples of such a $G$?
| https://mathoverflow.net/users/3960 | Transitive Semigroups of $2\times 2$ matrices | Here is a complete answer:
*Every semigroup* $S$ *of invertible* $2\times 2$\*-matrices which is transitive on\* $\mathbb R^2$ *is either conjugate to* $SO\_2(\mathbb R) \times \mathbb R^+$ *or* $SO\_2(\mathbb R) \times \mathbb R$ *or it is a product of* $SL\_2(\mathbb R)$ *and a multiplicative subgroup of* $\mathbb R$\*.\*
**Proof:** Let S be such a semigroup. Then the intersection $S\_0$ with $SL\_2(\mathbb R)$ is a subsemigroup of $SL\_2(\mathbb R)$. By a theorem of Hilgert and Hofmann (see their beautiful paper on "Old and new on $SL\_2$") there are only three choices for $S\_0$: Either $S\_0$ is all of $SL\_2(\mathbb R)$, a circle group or contained in a conjugate of the elements of $SL\_2(\mathbb R)$ with only positive entries. If $S\_0$ is a circle group, then $S$ will be conjugate to $SO\_2(\mathbb R) \times \mathbb R^+$ or $SO\_2(\mathbb R) \times \mathbb R$. If $S\_0$ happens to be all of $SL\_2(\mathbb R)$, then we have $SL\_2(\mathbb R)\subset S \subset GL\_2(\mathbb R)$, so $S$ is a product of $SL\_2(\mathbb R)$ and a multiplicative subgroup of $\mathbb R$. In the third case, we may assume that $S\_0$ is actually contained in the semigroup described above. Then $S\_0$ maps every vector with two positive entries to a vector with two posiitve entries, hence $S$ maps the upper right quadrant to a subset of itself and the lower left quadrant. In particular, $S$ cannot be transitive.
| 3 | https://mathoverflow.net/users/9927 | 45757 | 28,978 |
https://mathoverflow.net/questions/45731 | 8 | The concept of a subobject classifier is of course standard and ubiquitous. But is there any nontrivial example of an unrestricted slice classifier?
Specifically, what I mean by this is, is there any non-preorder category with pullbacks with a morphism m into an object X such that ALL other morphisms can be taken as a pullback of m along some morphism into X? And, if so, is it even possible to have furthermore that parallel morphisms from any object Y into X are equal just in case the pullbacks of m along them are isomorphic as objects of the slice category over Y?
Naturally, if we demand further structure on the category (e.g., local cartesian closure), this becomes impossible by Cantor type arguments in its internal logic, but if we only demand pullbacks, can it be done?
| https://mathoverflow.net/users/3902 | Is it possible for a nontrivial category to have a slice classifier? | It looks like such categories may be rather easy to construct. The following example should give the general idea: take the category of sets $V\_\alpha$ of cardinality less than or equal to $\alpha$, for some infinite cardinal $\alpha$. The morphism classifier will be the set $C$ of cardinals up to and including $\alpha$, and the universal morphism should be $S \to C$ where the fiber over a cardinal $\beta$ is a set of cardinality $\beta$.
Given any function $f: Y \to X$, the classifying morphism $\chi\_f: X \to C$ takes $x$ to the cardinal number of $f^{-1}(x)$.
Hopefully I haven't made any dumb mistakes...
| 7 | https://mathoverflow.net/users/2926 | 45768 | 28,982 |
https://mathoverflow.net/questions/45715 | 10 | Let $I\subseteq{\mathbb C}[X\_1,\dotsc,X\_n]$ be an ideal, and
let $V\subseteq{\mathbb C}^n$ be the corresponding algebraic set
($V$ consists of those $x$ at which all $f\in I$ vanish).
Is it true that then there exists an integer $N$ such that:
if a function $f\in{\mathbb C}[X\_1,\dotsc,X\_n]$
and all its partial derivatives up to order $N$
vanish on $V$,
then $f\in I$?
MOTIVATION. I believe this would imply an affirmative answer to my question about commutative algebras and representations of the category of finite sets:
[Commutative algebras and Gamma-modules](https://mathoverflow.net/questions/41440/commutative-algebras-and-gamma-modules)
| https://mathoverflow.net/users/9878 | Strong Nullstellensatz | The answer is yes. Let $I = q\_1 \cap q\_2 \cap \cdots \cap q\_k$ be the primary decomposition of $I$. Let $p\_i$ be the radical $\sqrt{q\_i}$ and let $N\_i$ be large enough that $q\_i \supseteq p\_i^{N\_i}$. I claim that we can take $N= \max(N\_i)$. We'll abbreviate $\mathbb{C}[x\_1, \ldots, x\_n]$ to $A$.
Let $f$ be a polynomial as in the statement of the question. In order to show that $f \in I$, it is enough to show that $f \in q\_i$ for every $i$. We focus on one $q$ to pay attention to, so we can drop the subscript $i$ and just talk about $p$, $q$ and $N$. Let $W$ be the variety of $p$.
Now, $A$ is regular, so $A\_p$ is regular. In other words, if $p$ has codimension $d$, there are transverse smooth hypersurfaces $y\_1$, $y\_2$, ..., $y\_d$ which generate the maximal ideal of $A\_p$. So there is some $u \not \in p$ such that $u^{-1} p = \langle y\_1, \ldots, y\_d \rangle$ in the localization $A[u^{-1}]$.
We claim that $f$, as an element of $A[u^{-1}]$ is in $u^{-1} p^N$. Proof: Write
$$f=\sum\_{i\_1+\cdots+i\_d < N} b\_{i\_1 \ldots i\_d} y\_1^{i\_1} \cdots y\_d^{i\_d} + r$$
with $r \in u^{-1} p^N$ and with all the $b$'s either $0$ or not in $p$. We want to show the $b$'s are zero. If not, let $(i\_1, \ldots, i\_d)$ be minimal with the corresponding $b$ nonzero. Since $b$ is not in $p$, it is not zero on $W$. Choose $w$ in $W$ where neither $b$ nor $u$ vanishes.
Since the $y\_i$'s are transverse hypersurfaces, we can take $\sum i\_j$ derivatives to obtain a polynomial with leading term $\mbox{nonzero stuff} \cdot \prod (i\_j)! \cdot b$ at $w$. Since we are in characteristic zero, this doesn't vanish at $w$. (In finite characteristic, the factorials might be zero.) But $w \in W \subseteq V$, so this derivative is supposed to vanish at $w$. This contradiction shows that $f \in u^{-1} p^N$.
Since $q \supseteq p^N$, we have $f \in u^{-1} q$ so $u^k f \in q$ for some $k$. But $q$ is primary and $u \not \in p = \sqrt{q}$. So this shows $f \in q$, as desired.
| 10 | https://mathoverflow.net/users/297 | 45777 | 28,985 |
https://mathoverflow.net/questions/45782 | 6 | I'm looking for the name of a certain n-category definition. (Someone explained it to me a couple of years ago. I remember the definition, but not the name. Without the name it's difficult to search for a citation. I want the citation in order to explain something we're *not* doing in a paper.)
For background, consider the Moore loop space $\Omega\_r$ of loops of length $r$ (that is, parameterized by the interval $[0,r]$). We have a strictly associative composition $\Omega\_r\times \Omega\_s\to \Omega\_{r+s}$. The main idea of an "xxxx" n-category is to imitate this idea in higher dimensions. The $k$-morphisms are parameterised by $k$-dimensional rectangles with sides of lengths $r\_1,\ldots,r\_k$. Gluing rectangles together gives $k$ different strictly associative ways to compose $k$-morphisms.
Question: What is "xxxx" above?
Bonus question: What's the best (or any) citation for this idea?
---
EDIT: It turns out the definition I was trying to remember is unpublished work of Ulrike Tillmann. But the version from Ronnie Brown linked to in David Roberts' answer is pretty similar (for my purposes, at least).
| https://mathoverflow.net/users/284 | What's the name of this flavor of n-category? | Ronnie Brown has a related idea, contained in this article:
>
> Moore hyperrectangles on a space form a strict cubical omega-category
>
> [arXiv](http://arxiv.org/abs/0909.2212)
>
>
>
discussed briefly [here at the nLab](http://ncatlab.org/nlab/show/Moore+path+category).
If you are instead thinking of a globular $n$-category, the closest I know of is a [Trimble n-category](http://ncatlab.org/nlab/show/Trimble+n-category), but that doesn't use Moore paths, but paths of length 1 and the $A\_\infty$-co-category structure on $[0,1]$.
| 7 | https://mathoverflow.net/users/4177 | 45783 | 28,987 |
https://mathoverflow.net/questions/45784 | 76 | Suppose $f\_n$ is a sequence of real valued functions on $[0,1]$ which converges pointwise to zero.
1. Is there an uncountable subset $A$ of $[0,1]$ so that $f\_n$ converges uniformly on $A$?
2. Is there a subset $A$ of $[0,1]$ of cardinality the continuum so that $f\_n$ converges uniformly on $A$?
Background: Egoroff's theorem implies that the answer to (2) is yes if all $f\_n$ are Lebesgue measurable. It is not hard to show that the answer to (1) is yes if you change "uncountable" to "infinite".
Motivation: I thought about this question while teaching real analysis this term but could not solve it even after looking at some books, googling, and asking some colleagues who are much smarter than I, so I assigned it as a problem (well, an extra credit problem) to my class. Unfortunately, no one gave me a solution.
ADDED 11-12-10: Thanks for all the great answers. I accepted Jonas' answer since it was the first one.
| https://mathoverflow.net/users/2554 | Does pointwise convergence imply uniform convergence on a large subset? | I did some Googling and came up with something that looks relevant, [Theorem 10](https://books.google.com/books?id=WwmvxtDlz9UC&lpg=PA124&ots=lcdSy9gacd&dq=point%2520set%2520theorem%2520morgan&pg=PA88#v=onepage&q&f=false) quoted below from Morgan's *Point set theory*. It cites works of Sierpiński from the late 1930s, but I can't tell what works are cited because the preview won't let me see that page in the references.
>
> The existence of a linear set having the power of the continuum that is concentrated on a denumerable set is equivalent to the existence of a pointwise convergent sequence of functions of a real variable that does not converge uniformly on any uncountable set.
>
>
>
| 21 | https://mathoverflow.net/users/1119 | 45786 | 28,989 |
https://mathoverflow.net/questions/45770 | 4 | I encountered this problem in my research and it is turning out to be a surprisingly difficult one(for me, at least).
Suppose we have a univariate nonlinear function $f(x)$ where $x \in [L,U]$. Our goal is to approximate this nonlinear function with $n$ piecewise-continuous linear functions $g\_{i}(x)$ within the given domain. We assume that $n$ is a pre-specified number. We define each line segment as follows:
$$
g\_{i}(x) = \frac{f(a\_{i}) - f(a\_{i-1})}{a\_{i} - a\_{i-1}} (x - a\_{i-1}) + f(a\_{i-1})\text{ for }a\_{i-1} \leq x \leq a\_{i}
$$
where $a\_{i}$ are knot points in $[L,U]$ and $i = 1,\ldots,n$. The first and the last knot points are fixed at the boundaries, that is, $a\_{0} = L, a\_{n} = U$. Also, the knot points are ordered and unique: $ a\_{i} > a\_{i-1}$ for $i=1,\ldots,n$.
I want to find the optimal placements for the knot points $a\_{1},\ldots,a\_{n-1}$, such that the overall squared-approximation error $e$ is minimized. We can pose the objective as follows:
$$
\min\_{a\_{1},\ldots,a\_{n-1}} \left\{ e = \int\_{L}^{U} [f(x) - g\_{i}(x)]^2 dx \right\}
$$
This picture illustrates the problem:
[Piecewise Linear functions http://dl.dropbox.com/u/6809582/linearfunctions.png](http://dl.dropbox.com/u/6809582/linearfunctions.png)
The final optimization problem looks like the following (after a simple reformulation into a optimal-control-like form):
$$
\begin{align\*}
&\min\_{a\_{1},\ldots,a\_{n-1}} e(U)\\
s.t.\quad & \frac{de(x)}{dx} = [f(x) - g\_{i}(x)]^2, \quad e(L) = 0\\
&g\_{i}(x) = \frac{f(a\_{i}) - f(a\_{i-1})}{a\_{i} - a\_{i-1}} (x - a\_{i-1}) + f(a\_{i-1})\text{ for }a\_{i-1} \leq x \leq a\_{i}\\
& a\_{0} = L, a\_{n} = U\\
& a\_{i} \geq a\_{i-1} + \epsilon,\quad i=1,\ldots,n
\end{align\*}
$$
This optimization problem is extremely difficult to solve numerically, owing to its nonsmoothness and nonconvexity.
Question: How do I solve this problem to global optimality? Can anyone provide any attacks (even partial ones)? Any simplifying properties?
| https://mathoverflow.net/users/7851 | Optimal knot placement for fitting piecewise-continuous linear functions to a nonlinear function | In the limit of fine subdivision, the local goodness of fit depends on only two things: the absolute value of the local second derivative, and the local density of knots. For a segment of constant second derivative $a$ over an interval of length $c$, the integrated squared error over the interval comes out to $a^2c^5/120$. Given a small approximation interval (small enough that the second derivative does not vary by much over its length) whose squared error contribution is $E$, replacing that interval by two approximation intervals of half the width reduces the total squared error by approximately $E- 2(E/32)=(15/16)E$. The best segment to subdivide (according to this approximation) is thus whichever one contributes the most to the error. It follows that in the limit of many knots the optimum will have an equal contribution to the error coming from each segment. This is true, in the limit, when the local density of knots at $f(x)$ is proportional to $f^{\prime\prime}(x)^{2/5}$. A very good approximation to your optimal assignment problem can thus be obtained by making a plot, from $L$ to $U$, of $\int\_L^x f^{\prime\prime}(s)^{2/5}\ ds$, dividing it into equally spaced horizontal strips, and placing your knots at the horizontal values where the integral plot crosses from one horizontal strip to the next.
Incidentally, if you keep the same number of straight-line approximations but allow them to cross the function (and back) in their interiors rather than at their endpoints, you can improve the integrated squared error by (in the limit) a factor of 6.
| 5 | https://mathoverflow.net/users/7936 | 45793 | 28,994 |
https://mathoverflow.net/questions/45776 | 1 | Given a toric ideal, say $J$, in a polynomial ring $k[x\_1,...,x\_n]$ we can find a finite
generating set for $J$. Is it possible, perhaps under additional assumptions on the structure of $J$, to give a finite minimal generating set for $J$ such that every subset of generators also generates a toric ideal.
If not, are there any known counterexamples in the general case?
If yes, could you provide a reference? Does it generalize to lattice ideals?
Motivation: For particularly chosen, generating sets of toric ideals there exist subsets that also generate a toric ideal. For e.g. the 2xn determinantal ideal is toric. A generating set is given by the 2x2 minors of the defining matrix, say $M$. Then the ideal generated by those minors which correspond to a subset of the columns of the $M$ is also toric.
| https://mathoverflow.net/users/10775 | Do subsets of generators of a toric ideal generate a toric ideal? | I believe this is a counterexample. Toric ideals are prime by definition (assuming that I am remembering correctly). Then I think that the ideal of the twisted cubic $C\subseteq \mathbb P^3$ ought to do it.
$J=\langle xz-y^2, xw-yz,wy-z^2\rangle$.
Any two of the three generators will intersect in the union of $C$ and a line. For instance, the vanishing of $I=\langle xz-y^2, xw-yz\rangle$ is $C\cup L$ where $L=V(x,y)$. So $I$ can't be prime.
| 2 | https://mathoverflow.net/users/4 | 45797 | 28,998 |
https://mathoverflow.net/questions/45787 | 10 | An element $F\in \mathbb{C}[[x,y]]$ defines a germ of plane curve.
We assume $F(0,0)=0$.
The multiplicity $mult$ of the germ is defined to be a minimal number $i$
such that $F\in m^i$ where $m=(x,y)$ is the maximal ideal in $\mathbb{C}[[x,y]]$.
Other standard invariants of the germ are Milnor number:
$$
\mu=\dim \mathbb{C}[[x,y]]/(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y})
$$
and delta invariant:
$$ \delta=\frac{\mu+r-1}{2}$$
where $r$ is number of branches of curve $F=0$ at $(0,0)$.
Question: What is the minimum of $\delta$ and $\mu$ among the germs
of given multiplicity?
| https://mathoverflow.net/users/10578 | Minimum of Milnor number for the curve singularities of fixed multiplicity | As Roy remarked in his answer, the delta invariant of the germ of plane curve singularity $f(x,y)=0$ at $p$ is equal to
$\delta(f) = \sum \frac{m\_q(m\_q-1)}{2}$,
where the sum is extended over all the points $q$ which are "infinitely near" to $p$ and $m\_q$ denotes the multiplicity at $q$.
Then $\delta(f)$ is minimal among germs of a given multiplicity when there are no infinitely near points, in other words when the first blow-up of the germ is smooth. Of course there can be many analitically distinct germs satisfying this property: for instance, both the node and the ordinary cusp do the job among double points.
Now, since the Milnor number is equal to
$\mu(f)=2 \delta(f)-r(f)+1$,
where $r(f)$ is the number of branches, it follows that $\mu(f)$ is minimal among plane singularities of given multiplicity $n$ when $\delta(f)$ is minimal and the number of branches is maximal, in other words when $f=0$ is the "ordinary" $n$-ple point.
As an example, for the ordinary double point (node) $y^2=x^3+x^2$ we have
$(\delta, \mu)=(1,1)$,
whereas the ordinary cusp $y^2=x^3$ satisfies
$(\delta, \mu)=(1,2)$.
Summing up, the ordinary $n$-ple point is the only germ of plane curve singularity which minimizes *both* $\delta$ and $\mu$, and the corresponding values are
$(\delta, \mu)= (\frac{n(n-1)}{2}, (n-1)^2)$.
| 7 | https://mathoverflow.net/users/7460 | 45803 | 29,001 |
https://mathoverflow.net/questions/45802 | 53 | I believe this is the right place to ask this, so I was wondering if anyone could give me advice on research at the undergraduate level.
I was recently accepted into the [McNair Scholars program](http://www.unh.edu/mcnair/). It is a preparatory program for students who want to go on to graduate school. I am expected to submit a research topic proposal in the middle of the spring semester and study it during the summer with a mentor.
Since I am currently in the B.S. Mathematics program and I want to get my Masters later. I figured that while my topic can be in any area, it should be in math since it is my main interest as well.
I am a junior at the moment and taking: One-Dimensional Real Analysis, Intro to Numerical Methods, and Abstract Algebra. I frequently search MathWorld and Wikipedia for topics that interest me, although I don't consider myself a brilliant student or particularly strong. I have begun speaking with professors about their research also.
I have not met any other students doing undergraduate math research and my current feeling is that many or all the problems in math are far beyond my ability to research them. This may seem a little defeatist but it seems mathematics is progressively becoming more specialized. I know that there are many areas emerging in Applied mathematics but they seem to be using much higher mathematics as well.
My current interest is Abstract Algebra and Game Theory and I have been considering if there are possibilities to apply the former to the latter.
So my questions are:
1) Are my beliefs about the possibilities of undergraduate research unfounded?
2) Where can I find online math journals?
3) How can I go about finding what has been explored in areas of interest. Should I search through Wikipedia and MathWorld bibliographies and or look in the library for research?
Thanks I hope someone can help to clarify and guide me.
| https://mathoverflow.net/users/nan | Undergraduate math research | Since you are a student who's already interested in going on to graduate school and is specifically asking about finding a topic to study at your undergraduate level program at McNair, please *disregard the negative nattering nabobs* whose answers and comments suggest that undergraduates have no place or business in trying to perform research, whether it's research as defined for all scientists or the "research experience" that is put together for undergraduates and for advanced high-school students. Undergraduates can definitely perform research, or even benefit from going through a structured and well-administered "research experience".
I agree with Peter Shor about finding a mentor, or multiple mentors, as soon as possible. There's no reason you have to be limited to getting advice from just one professor or teacher.
I agree with Ben Webster, specifically about speaking with professors in order to get a reasonable idea about the level of work that would be needed for you to perform useful research at an undergraduate level. A few other suggestions come to mind:
* if you are at an institution that offers Masters and Ph.D. level degrees in mathematics, then your institution's library should have multiple **research journals in hard-copy**. I have found that it is much easier to go to the stacks in the library and browse through one or two year's worth of Tables of Contents and Abstracts in one journal in an afternoon or evening. This will familiarize you with the types of research papers being published currently, and make you aware of what "quanta" of research is enough to be a single research article.
* make sure to attend Seminars, Colloquia, and (if your school's graduate students have one) any graduate research seminar courses that you can find time for. This will allow you to become more familiar with various subtopics within the topics of your interests, and to see what the current areas of interest are for local and visiting faculty members.
* Colloquia are great as they often start by including a brief history of the topic by an expert in that field.
* Seminars are great because they allow students to see the social aspect of math, including the give-and-take and the critical comments and requests for more detail and explanation, even by tenured faculty who don't follow a speaker's thought processes.
* Graduate student seminar presentations are great because a student observes how graduate students can falter during presentations, how they are quizzed/coached/criticized/mentored/assisted by faculty during their presentations.
* I'll admit that I'm not sure attending dissertation defenses would be of any serious benefit to the undergraduate student, other than observing the interaction level (animosity level?) between faculty and graduate students.
* absolutely make sure to schedule some time to meet with mathematics professors who specialize in the fields of your interest, and communicate your desire to do research while you are an undergraduate, and communicate your desire to go on to graduate studies in mathematics.
* look on the internet and search for undergraduate opportunities for research in mathematics. I guarantee you will find quite a number of web sites that can give you more information. MIT has an undergraduate research opportunity program that many of their students take advantage of. Your institution may have professors who can speak with you and give you advice.
Also, make sure to speak with more than one professor, and **do not take any single person's advice as being the final word.** Mathematicians are human beings too, and subject to the foibles and inclinations and disinclinations that all human beings have. If you run into disgruntled and critical individuals, do not let that dissuade you from going on into mathematics or decrease your desires. If you run into overly optimistic individuals who praise you too much and are too eager to take you on to do "scut work" computer programming, thank them for their time and let them know you'll come back to speak with them after you've spoken with other professors and weighed your options. Don't turn anyone down immediately. Always be polite in speaking with professors and teachers. Ask them how they chose their topics for their degrees, and you'll learn a lot.
| 45 | https://mathoverflow.net/users/8676 | 45810 | 29,008 |
https://mathoverflow.net/questions/45812 | 8 | I'm on my last semester of a math B.Sc. and about to start studying for a math M.Sc in the same institute.
It now seems like a good time to start thinking of a PhD.
I'm interested in both algebra and algorithms. So I read a little in some computational algebra books (comp. group theory to be precise) and it looks like a great choice for me.
My questions:
1. Is this a "hot topic" in mathematics? Are there many serious people researching it?
2. Where are the major places in the world to research comp. algebra and to get a PhD?
Thank you!
| https://mathoverflow.net/users/nan | Computational algebra: where? | The first piece of advice I would give you is not to prematurely limit yourself to a narrow area, unless your M.Sc only takes a year. If it takes at least two years, then you will still learn plenty of exciting mathematics before you have to make a reasonably definitive choice. If you are thinking of doing a PhD in the states, then even the PhD itself is so long, that being at a generally strong department might be more beneficial than to have an isolated area of expertise around you.
Having said that, if you are interested in computation mathematics of an algebraic nature (computational group theory, number theory, geometry, more general algebra), then a natural choice would be some place where one of the big computer algebra packages is being developed. The sites I list below also provide a good list of contributors, which will help you in choosing a shortlist of possible advisors. Some examples include
1. [MAGMA](http://magma.maths.usyd.edu.au/magma/htmlhelp/ackn.htm) in Sydney
2. [GAP](https://www.gap-system.org/) in various places; this one is particularly focused on group theory
3. [PARI/GP](http://pari.math.u-bordeaux.fr/) in Bordeaux
4. [SAGE](https://www.sagemath.org/) in Washington, whose site also provides [a nice map](https://www.sagemath.org/development-map.html) of where the developers are based. Thanks to Suvrit for fixing this unforgivable omission.
If I can think of more, I will add them here, but these should give you a good place to start. You can also browse through some of the dedicated journals, like the [Journal of symbolic computation](https://www.journals.elsevier.com/journal-of-symbolic-computation) or the [LMS Journal of computation](https://www.lms.ac.uk/publications/jcm) and see who the prolific contributors are. Reading some of those articles might also help you decide whether this is what you want to do.
As for your question whether this is a "hot topic", that's harder to answer. It certainly is in the places mentioned above and in various others, where there are strong computational algebra groups, but I know of some places where this area of mathematics is frowned upon by the more theoretical community. But I suppose that that's probably true of almost any area of mathematics and you shouldn't let that deter you.
| 16 | https://mathoverflow.net/users/35416 | 45816 | 29,011 |
https://mathoverflow.net/questions/45844 | 23 | The standard proof of the Hahn-Banach theorem makes use of Zorn's lemma. I hear that, however, Hahn-Banach is strictly weaker than Choice. A quick search leads to many sources stating that Hahn-Banach can be proven using the ultrafilter theorem, but I cannot seem to find an actual proof. So...
1. What is the ultrafilter theorem; and
2. How does the said theorem imply the Hahn-Banach theorem?
Any easily-accessible reference would be quite enough; thanks in advance!
| https://mathoverflow.net/users/8452 | Hahn-Banach without Choice | The ultrafilter theorem is the statement that any filter on a set can be extended to an ultrafilter. It is perhaps more common to see it sated as the (Boolean) Prime ideal theorem: Every Boolean algebra admits a prime ideal.
The Hahn-Banach theorem is actually equivalent to the statement that every Boolean algebra admits a real-valued measure, but this is not entirely straightforward (see Luxemburg, "Reduced powers of the real number system and equivalents of the Hahn-Banach extension theorem", Intern. Symp. on the applications of model theory, (1969) 123-127).
For a discussion of Hahn-Banach vs. Choice and some additional remarks and references, see Jech "The axiom of choice", North-Holland, 1973.
You may also be interested in the references I include in [this answer](https://mathoverflow.net/questions/42215/does-constructing-non-measurable-sets-require-the-axiom-of-choice/42226#42226).
| 22 | https://mathoverflow.net/users/6085 | 45846 | 29,031 |
https://mathoverflow.net/questions/45863 | 13 | A topologist came to me with this question, but everything I think should work doesn't.
How many triangulations are there of a polyhedron with n vertices?
By a "triangulation" of a polyhedron P we mean a decomposition of P into 3-simplices whose interiors are disjoint, whose vertices are vertices of P, and whose union is P. Since this obviously depends on the polyhedron, let's say that P is the convex hull of n points on the curve (t, t^2, t^3). (I think this is general, but a proof of that would be nice too.) In particular, this means that all of the faces are triangles, since no four vertices are coplanar.
Since triangulations of a polygon are counted by the Catalan numbers, a reasonable first guess is that these are counted by the generalized Catalan numbers $C\_{n,k} = \frac{1}{(k-1)(n+1)} {kn \choose n}$, which count k-ary trees (among other things). But just at n=5 we run into trouble: there are 2 (not 3) such triangulations, and *they don't even contain a fixed number of pieces*: one of them triangulates P into two tetrahedra, and one breaks it into three.
This seems obvious enough that someone would have asked it before, but I'm not finding anything. Of course, answers to the obvious generalization (triangulations of k-polytopes whose vertices lie on (t, t^2, ..., t^k)) are welcome as well.
| https://mathoverflow.net/users/1060 | Triangulations of polyhedra | You should read Section 6.1 in the excellent monograph "Triangulations" by De Loera, Rambau and Santos (here is a [slightly dated version](http://www.math.ucdavis.edu/~deloera/BOOK/OLDVERSIONS/april2010.pdf)). It deals with triangulations of cyclic polytopes - exactly the subject of your question. Not only it answers your question, it is also the state of art for the rest of the subject.
| 21 | https://mathoverflow.net/users/4040 | 45865 | 29,045 |
https://mathoverflow.net/questions/45704 | 2 | Suppose we have a PEL type $(H,\phi ,\*;T,O,V)$ where H is a rational nonsplit quaternion algebra, $\phi$is an embedding of Q-algebra $\phi : H-->M(2,R)$, and \* is a positive anti involution of H; O is the maximal order of H , and V level structure . Associate these datum a shimura curve parametric fake elliptic curves in a standar way.
My question is :what is the group for the shimura datum ,is it the group H\*, the invertible element of H ? And since this family parametirc abelian two folds , what is the map from this group to $\mathrm{GSP}(4,\mathbb{Q})$?
| https://mathoverflow.net/users/3945 | Shimura datum of family of fake elliptic curves | Actually, fake elliptic curves are discussed in Chapter 9 of Lang's Introduction to algebraic and abelian functions.
In order to describe the map to $GSP(4,Q)$, recall that there is the standard antiinvolution $x\to x'$ on the quaternion $Q$-algebra $H$ such that both $tr(x)=x+x'$ and $Norm(x)=xx'=x'x$ are rational numbers for all $x\in H$. Every positive antiinvolution of $H$ is of the form $x \to \gamma^{-1}x'\gamma$ where $\gamma$ is a fixed nonzero (invertible) element of $H$ such that $\gamma^2$ is a negative rational number and therefore $\gamma'=-\gamma$. This gives rise to the alternating nondegenerate $Q$-bilinear form
$$E: H \times H \to Q, x,y \mapsto tr(\gamma x y').$$
Now let us consider the following faithful action of the multiplicative group $G$ of $H$ on the $Q$-vector space $V=H$:
$$u(x)=x u'$$ for $x \in V=H$ and $u \in G$. Then
$$E(u(x),u(y))=tr(\gamma x u'u y')=(u'u)tr(\gamma x y'),$$
which means that
$$E(u(x),u(y))=Norm(u) E(x,y).$$
This gives us the embedding $G \to GSP(V,E)\cong GSP(4.Q)$.
The same construction over arbitrary commutative $Q$-algebras $R$ gives us the desired embedding of the corresponding $Q$-algebraic groups.
| 11 | https://mathoverflow.net/users/9658 | 45872 | 29,049 |
https://mathoverflow.net/questions/26821 | 64 | Last year a paper on the arXiv (Akhmedov) claimed that Thompson's group $F$ is not amenable, while another paper, published in the journal "Infinite dimensional analysis, quantum probability, and related topics" (vol. 12, p173-191) by Shavgulidze claimed the exact opposite, that $F$ is amenable. Although the question of which, if either, was a valid proof seemed to be being asked by people, I cannot seem to find a conclusion anywhere and the discussion of late seems to have died down considerably. From what I can gather, Shavgulidze's paper seems unfixable, while the validity of Akhmedov's paper is undecided (although it may of course be decided now).
So, does anyone know if either of these papers is valid?
| https://mathoverflow.net/users/6503 | Is Thompson's Group F amenable? | While I did not participate in most of the checking of Shavgulidze's
argument, I can offer the following partial account of the situation. I
am told the paper was correct except for a lemma (or sequence of them)
claiming that a sequence of auxiliary measures had certain properties.
These were Borel measures on the $n$-simplex (one for each $n$). I
believe it was shown that the original proposed auxiliary sequence of
measures did *not* have one of the two properties. Shavgulidze
proposed other sequences of measures. The most recent attempt that I am
aware of (which was presented during his 2010 trip to the US mentioned
by Mark Sapir in the above comment) involved the direct construction of
Folner sets for the action of $F$ on the finite subsets of dyadic
rationals (see the next paragraphs). The details were somewhat sparse
and the definitions involved many unspecified numerical parameters, but
it appeared to be the case that these sets could not be Folner in the
necessary sense (see below for a clarification of "necessary sense").
This is because they would likely
both contradict the iterated exponential lower bound on the Folner
function which I have demonstrated and because they appear to violate
the qualitative properties which I have demonstrated that Folner sets of
trees must have (see the pre-print on my webpage; the qualitative
condition appears in lemma 5.7, noting that marginal implies measure
0 with respect to any invariant measure).
Meanwhile I was able to provide a direct elementary proof that the
existence of such a sequence having these properties implied the amenability
of $F$. In fact the proof gives an explicit procedure for constructing
(weighted) Folner sets from the sequence of measures satisfying the hypotheses
mentioned above. A note containing the details was circulated to a few people around the
time of Shavgulidze's visit to Vanderbilt.
While I am reluctant to speak for anyone else (including the author), it
appears to me that after the dust had settled (which took a considerable amount of time),
the problem with the proof seems to have at least some of its roots
in the following observation (which I now include for the sake of prosperity). $F$
acts on the finite subsets of the dyadic rations (let's call this set
$\mathcal{D}$) by taking the set-wise image (here I am utilizing the
piecewise linear function model of $F$). Now let $\mathcal{T}$ denote
the finite subset of $[0,1]$ which contain $0$ and $1$ and are such that
any consecutive pair is of the form $p/2^q,(p+1)/2^q$ (for natural numbers $p,q$). $F$ only acts
*partially* on $\mathcal{T}$: the action $T \cdot f$ is defined if $f'$
is defined on the complement of $T$ in $[0,1]$ (there may be other cases
when $T \cdot f$ is in $\mathcal{T}$, but let's restrict the domain of
the action as above). The full action of $F$ on $\mathcal{D}$ *is*
amenable. The point here is that the action of the standard generators
on the sets $\{0,1-2^{-n},1\}$ is the same for large enough $n$
and thus we can build
Folner sets as in a $\mathbb{Z}$ action. The amenability of partial
action of $F$ on $\mathcal{T}$ is, on the other hand, equivalent to the
amenability of $F$ (this is well known, but see the preprint above to see
this spelled out in the present jargon).
Now here is the catch, if we also require that the invariant
measure/Folner sets for the action of $F$ on $\mathcal{D}$ to
concentrate on sets of mesh less than $1/16$, then one again arrives at
an equivalent formulation of the amenability of $F$. The author was
aware of the need for the mesh condition, but (in the most recent example)
arranged it only in a
modification after the fact (which interferes with invariance).
Incidentally the hypotheses on the sequence of measures mentioned above
are a condition requiring that the measures concentrate on sets of
arbitrarily small mesh as $n$ tends to infinity and a condition which
is an analog of translation invariance.
I apologize if this borders on ``too much information.''
[Added 1/28/2011]
Shavgulidze's 1/14/2011 posting to the ArXiv is essentially a more detailed version of what he was saying in notes, seminars, and private communication in January 2010 during his visit to the US mentioned in Mark Sapir's post above. The present note is still sufficiently vague and full of sufficiently many errors (many typographical in nature) that it is hard (or easy, if you like) to say explicitly which line of the proof is incorrect. It is possible, however to point to places where crucial details are missing and where there are certainly going to be errors (specifically the problems will be on page 11, if not elsewhere as well). The comments from my answer above still apply equally well to the present version. It appears that the present version (or any perturbation of it) still would violate the lower bound on the growth of the Folner function which I have established. The present version still totally ignores that the combinatorial statements on page 11 themselves readily imply the amenability of F, without the involvement of any analytical concepts.
[Added 2/3/2011]
Details on what is incorrect with Shavgulidze's proof of the amenablity of $F$ can be found [here](http://arxiv.org/abs/1102.0747).
[Added 10/3/2012]
Well, well, well: now *I'm* in the position of having announced a proof that $F$ is amenable only to have an error be found. The error was finally found by Azer Akhemedov after being overlooked for roughly 4 weeks by myself and 9 or more people who had checked the proof and found no problems. The basic strategy of the proof still may be valid: it began by considering an extension of the free binary system $(\mathbb{T},\*)$ on one generator to the finitely additive probability measures on this system:
$$\mu \* \nu (E) = \int \int \chi\_E(s \* t) d \nu (t) d \mu (s).$$
It was shown (correctly) that any idempotent measure is $F$-invariant (there is a natural way of identifying $\mathbb{T}$ with the positive elements of $F$).
The difficulty came in constructing the idempotent measure.
A version of the Kakutani Fixed Point Theorem was used to construct approximations $K\_{\mathcal{B},k,n}$ to the set of idempotent measures.
The error occurs in attempting to intersect these compact families of measures.
In the proof, it was claimed that the parameter $k$ could be stablized along the an ultrafilter
(Lemma 4.13 in the most recent version), allowing one to take a directed intersection of nonempty compact sets.
This lemma is likely false and at least is not proved as claimed.
One may still be able to argue that a relevant intersection of these approximations is nonempty and hence that there is an idempotent.
This seems to require new ideas though.
| 54 | https://mathoverflow.net/users/10774 | 45891 | 29,064 |
https://mathoverflow.net/questions/45881 | 1 | I guess this is a well known fact/definition for many people. It is mentioned in many places that if $\Gamma$ is a lattice of a vector space(vector bundle/affine bundle) $V$, then there is a dual lattice $\check{\Gamma}$ in $V^\* $ and the torus $V/\Gamma$ has a dual torus $V^\*/\check{\Gamma}$. What does this mean? When is it meaningful? I want to know the answer because I hope that we can get an easy topological description of ("toy model" of) mirror manifold (I have to admit that I am not sure if I really understand what's "mirror manifold"). For example, if $V$ is a vector bundle and $\Gamma$ is a lattice in $V$ with some extra structure or information, how can we find a torus bundle dual to $V/\Gamma$? What kind of extra structure or information we need? If $V$ is described by local chart and transitive groups {$U\_\alpha,\mu\_{\alpha \beta}$}, can we find a dual set {$U^\* \_\alpha,\mu^\* \_{\alpha \beta}$} to describe $V^\* $? Is it unique or canonical?
The other question is about the canonical complex structure of tangent bundle. Of course, it relate to the above question. Again, like the symplectic structure on cotangent bundle, I guess it is well known, but I really can't find any reference. One source I found is a slide of Mark Gross (<http://math.mit.edu/~auroux/frg/mit08-notes/M.%20Gross%20-%20Slides%20-%20From%20affine%20manifolds%20to%20complex%20manifolds.pdf>). I am not sure if this is the standard one we understand. The complex structure looks much less natural then the canonical symplectic structure.
| https://mathoverflow.net/users/10799 | What's dual torus and mirror manifold? | The usual answer is that the dual lattice is
$\check{\Gamma}=\{f\in V^\* | f(\gamma)\in \mathbb{Z}\ \forall \gamma \in \Gamma\}$. It is defined for any lattice $\Gamma\subset V$ - no extra information needed.
| 4 | https://mathoverflow.net/users/10745 | 45895 | 29,065 |
https://mathoverflow.net/questions/44420 | 2 | I am reading some aspects of Mirror Symmetry and in mirror symmetry the $N=2$ SCFT on a Calabi Yau Manifold can be divided into two sectors each of which is a topological sigma model, A-Model and B-Model. After some research through some literature about the topological models, it seems that the topological models are constructed only on supersymmetric theory.
Are there any non -Supersymmetric topological sigma models?
Are there some topological models where the target space is not a Calabi-Yau manifold (or in general a Kahler manifold)?
| https://mathoverflow.net/users/9534 | Are there non-supersymmetric and/or non-Calabi-Yau topological sigma models? | I believe that A-model does not require a Calabi-Yau target space. In fact, A-model is well-defined on any almost complex manifold, which was Witten's original construction (Comm. Math. Phys. Volume 118, Number 3 (1988), 411-449). On the other hand, B-model can only be defined on a Calabi-Yau manifold, which follows from anomaly cancelation.
In general, topological field theories have many different types (not necessarily supersymmetric). As an example, Chern-Simons theory is topological. Try <http://en.wikipedia.org/wiki/Topological_quantum_field_theory> for some general discussion.
| 2 | https://mathoverflow.net/users/7035 | 45897 | 29,067 |
https://mathoverflow.net/questions/45880 | 1 | ### Trying to draw the Amoeba
With Mathematica, it's possible to graph $e^{-k x} + e^{-k y} = 1,e^{-k x} - e^{-k y} = 1$ and $e^{-k x} + e^{-k y} = -1$ to get the amoeba of **1 + x + y** when **k = 1**. Then by plotting when $k \to \infty$, these graphs converge to the tropical polynomial
<http://www.freeimagehosting.net/uploads/7a0cc0b6f3.jpg>
However, I meant to draw **Min(1,x,y)** which I fixed by shifting my coordinates **x,y** by the vector **(1,1)**. This is a bit ad-hoc and I am probably not understanding how constant functions "tropicalize".
### Main question
I would like to "fatten" **Min[x, y, 1, x + y + 1]** into its amoeba, so I thought the right curve should be **1 + x + y + xy**. However, my "neck" is disappearing in the scaling limit. *How should I scale the coefficients correctly to get my amoeba?*
Following Zeb's suggestion (but a few second before he posted it) I came up with this imge
<http://www.freeimagehosting.net/uploads/b612de3bf9.gif>
However, this "dequantization" procedure doesn't always produce the whole tropical curve. Here's the curve I drew to "requantize" **Min[1, x , y , x+ y + 1, -2 + 2x , 2y]**. A line has to be missing b/c of the zero tension condition, as in [Tropical Mathematics](http://arxiv.org/abs/math/0408099) by David Speyer and Bernd Sturmfels.
<http://www.freeimagehosting.net/uploads/50219e2142.gif>
Here is the code. You have to draw 4 different versions of the curve to get all the absolute values. Maybe this should somehow involve complex phases as well.
```
q[x_] := E^( x)
f[a_, b_, c_] := c + a + b + c a b + (1/c^2) a^2 + b^2;
{x0, y0} = { -1, -3};
L = 5;
k = 8;
ContourPlot[
{ f[q[k x], q[k y], q[k ]] == 0, f[-q[k x], q[k y], q[k ]] == 0,
f[q[k x], -q[k y], q[k ]] == 0, f[-q[k x], -q[k y], q[k ]] == 0},
{x, x0, x0 + L}, {y, y0, y0 + L}
]
```
Ideally, I want to take any tropical curve and fatten it into its amoeba. Tropical conics and cubics seem the best starting point.
---
In anticipation of comments, by "amoeba" here I think I mean the boundary of the 2D region which is usually called "amoeba".
| https://mathoverflow.net/users/1358 | How to Tropicalize a Polynomial in Two Variables? | For the first amoeba you mentioned, I think your equations should be $e^{-kx}\pm e^{−ky}=\pm e^{-k}$, not $e^{-kx}\pm e^{−ky}=\pm e^{0}$.
For the main question, I think you should be using an equation like $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} = 0$... so really you want a curve like $c+x+y+cxy$, where when you rescale $x$ and $y$ by raising them both to a power, you also rescale the coefficient $c$ by raising it to the same power.
Edit: Ok, for the second problem you are having, I think this is coming up because plugging in different signs of $x$ and $y$ doesn't give you all the different sign possibilities for your detropicalized polynomial.
So, if you want to get the tropical curve $Min(1,x,y,1+x+y,2x-2,2y)$, you want to use all of the equations $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} \pm e^{-k(2x-2)} \pm e^{-k(2y)} = 0$.
In fact, I think you don't need to use all $32$ of those equations, you just need to use enough of them that every pair of terms have opposite signs in one of your equations, such as the following three:
$e^{-k}+ e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} - e^{-k(2y)} = 0$
$e^{-k}- e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} + e^{-k(2x-2)} - e^{-k(2y)} = 0$
$e^{-k}+ e^{-kx}- e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} + e^{-k(2y)} = 0$
In the limit this will give you the right amoeba, but I'm cheating a bit, because really we should be doing fancy stuff with logarithms of complex numbers.
| 3 | https://mathoverflow.net/users/2363 | 45899 | 29,069 |
https://mathoverflow.net/questions/45879 | 7 | Let $f\colon \mathbb R^1\to \mathbb R^3$ be a continuous and injective map. Is $\mathbb R^3\setminus f(\mathbb R^1)$ a path-connected space?
| https://mathoverflow.net/users/4298 | Space-discriminating injective curve | Yes. In fact, $\mathbb R^3$ could be any 3-manifold and $f(\mathbb R^1)$ any countable union of embedded segments.
**Lemma 1**. Let $U$ be an open ball in $\mathbb R^3$ and $f:I\to\mathbb R^3$ an embedding. Then $U\setminus f(I)$ is path-connected.
*Proof.* See [Hatcher](http://www.math.cornell.edu/~hatcher/AT/ATpage.html), Proposition 2B.1 on page 169. This proposition proves (among other things) that $S^3$ minus any embedded segment is path-connected. Extract the proof of this particular statement and repeat it word-by-word with $U$ in place of $S^3$. (The proof uses only the following facts about $U$: $U$ minus any closed sub-segment of $f(I)$ is open; $U$ minus any point of $\mathbb R^3$ is simply connected.)
**Lemma 2.** Let $f:I\to\mathbb R^3$ be an embedding and $s:I\to\mathbb R^3$ a path with endpoints outside $f(I)$. Then for any $\varepsilon>0$ there exists a path $s\_\varepsilon:I\to\mathbb R^3$ which is $\varepsilon$-close to $s$ (in $C^0$), connects the same endpoints, and avoids $f(I)$.
*Proof.* Divide the domain of $s$ into intervals so small that the diameters of their images are less than $\varepsilon/10$. Let $p\_0,p\_1,\dots,p\_n\in\mathbb R^3$ be the images of the division points ($p\_0$ and $p\_n$ are the endpoints of $s$). For each $p\_i$, let $q\_i$ be a point outside $f(I)$ such that $|p\_i-q\_i|<\varepsilon/10$ (for $i=0$ and $i=n$, choose $q\_i=p\_i$). Such a point $q\_i$ exists since $f$ is injective and hence cannot cover a set with nonempty interior. Let $U\_i$ be the ball of radius $\varepsilon/3$ centered at $p\_i$. By Lemma 1, $U\_i\setminus f(I)$ is path-connected. Therefore we can connect $q\_i$ to $q\_{i+1}$ by a path contained in $U\_i$ and avoiding $f(I)$. The (suitably parametrized) product of these paths is the desired $s\_\varepsilon$, q.e.d.
Now return to the original problem. The set $f(\mathbb R)$ is a countable union of sets $J\_k$, $k\in\mathbb N$, where each $J\_k$ is an injective image of a segment. We want to prove that any two points $p,q\in\mathbb R^3$ can be connected by a path avoiding $\bigcup J\_k$.
Let $s\_1$ be a path from $p$ to $q$ avoiding $J\_1$ (such a path exists by Lemma 1). Let $\varepsilon\_1$ be the minimum distance from $s\_1$ to $J\_1$, divided by 10. By Lemma 2, there exists a path $s\_2$ which is $\varepsilon\_1$-close to $s\_1$ and avoids $J\_1$. Let $\varepsilon\_2$ be the minimum of $\varepsilon\_1$ and the minimum distance from $s\_1$ to $J\_1$, divided by 10. Then there is a path $s\_3$ which is $\varepsilon\_2$-close to $s\_2$ and avoids $J\_2$, and so on. The resulting sequence $s\_1,s\_2,\dots$ converges (since $C^0$ is complete) to some path $s$ which avoids every set $J\_k$.
*Remark*. The closure of $f(\mathbb R)$ can separate the space, for example, $f(\mathbb R)$ can be a dense subset of the torus.
| 14 | https://mathoverflow.net/users/4354 | 45902 | 29,071 |
https://mathoverflow.net/questions/45923 | 6 | While skimming the book *Concrete Mathematics*, (edit: first edition) I came across the following problem, which is listed there as a Research Problem: (Chapter 5, Exercise 96)
>
> Is ${2n \choose n}$ divisible by the square of a prime for all $n > 4$.
>
>
>
This problem looked to me much simpler than a divisibility problem that I found on MO [(look here)](https://mathoverflow.net/questions/11335/a-binomial-sum-is-divisible-by-p2), but then again, I guess in number theory, the simpler the problems looks, the harder it usually is!
The nice form of this problem has made me very curious to find out more about it. But because I do not have more than a fleeting acquaintance with number theory, I don't know what search keywords would be useful to gain more information about this problem.
Thus, could somebody please tell me more about this problem and its current status?
| https://mathoverflow.net/users/8430 | Divisibility of a binomial coefficient by $p^2$ -- current status | This is/was known as the Erdős square-free conjecture, and seems to now be solved. See the bottom of [this page](http://en.wikipedia.org/wiki/Square-free_integer#Erd.C5.91s_Squarefree_Conjecture).
| 13 | https://mathoverflow.net/users/8103 | 45929 | 29,080 |
https://mathoverflow.net/questions/45928 | 27 | Inspired by a recent Math.SE question entitled [Where do we need the axiom of choice in Riemannian geometry?](https://math.stackexchange.com/questions/10102/where-do-we-need-the-axiom-of-choice-in-riemannian-geometry), I was thinking of the [Arzelà--Ascoli theorem](http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem). Let's state a very simple version:
>
> **Theorem.** Let $\{f\_n : [a,b] \to [0,1]\}$ be an equicontinuous sequence of functions. Then a subsequence $\{f\_{n(i)}\}$ converges uniformly on $[a,b]$.
>
>
>
The proofs I have seen operate as follows: Take a countable dense subset $E$ of $[a,b]$. Use a "diagonalization argument" to find a subsequence converging pointwise on $E$. Use equicontinuity to conclude that this subsequence actually converges uniformly on $[a,b]$.
The "diagonalization" step goes like this: Enumerate $E$ as $x\_1, x\_2, \dots$. $\{f\_n(x\_1)\}$ is a sequence in $[0,1]$, hence has a convergent subsequence $\{f\_{n\_1(i)}(x\_1)\}$. $\{f\_{n\_1(i)}(x\_2)\}$ now has a convergent subsequence $\{f\_{n\_2(i)}(x\_2)\}$, and so on. Then $\{f\_{n\_i(i)}\}$ converges at all points of $E$.
Of course, to do this, at each step $k$ we had to choose one of the (possibly uncountably many) convergent subsequences of $\{f\_{n\_{k-1}(i)}(x\_k)\}$, so some sort of choice is needed here (I guess dependent choice is enough? I am not a set theorist (IANAST)). Indeed, we have proved that $[0,1]^E$ is sequentially compact (it is metrizable so it is also compact).
On the other hand, we have not used (equi)continuity in this step, so perhaps there is a clever way to make use of it to avoid needing a choice axiom.
So the question is this:
>
> Can the Arzelà--Ascoli theorem be proved in ZF? If not, is it equivalent to DC or some similar choice axiom?
>
>
>
| https://mathoverflow.net/users/4832 | Does Arzelà-Ascoli require choice? | There is a canonical way of checking the literature for most questions of this kind. Since they come up with some frequency, I think having the reference here may be useful.
>
> First, look at "Consequences of
> the Axiom of Choice" by Paul Howard
> and Jean E. Rubin, Mathematical
> Surveys and Monographs, vol 59, AMS,
> (1998).
>
>
> If the question is not there, but has
> been studied, there is a fair chance
> that it is in the database of the book
> that is maintained online,
> <http://consequences.emich.edu/conseq.htm>
>
>
>
Typing "Ascola" on the last entry at the page just linked, tells me this is form 94 Q. Note the statement they provide is usually called the classical Ascoli theorem:
>
> For any set $F$ of continuous functions
> from ${\mathbb R}$ to ${\mathbb R}$, the following conditions
> are equivalent:
>
> 1. Each sequence in $F$ has a subsequence that converges
> continuously to some continuous
> function (not necessarily in $F$ ).
>
> 2. (a) For each $x \in{\mathbb R}$ the set $F (x) =\{f (x) \mid f \in F \}$ is bounded, and
>
> (b) $F$ is equicontinuous.
>
>
>
To see the other equivalent forms of entry 94, type "94" on the line immediately above.
From there we learn: Form 94 is "Every denumerable family of non-empty sets of reals has a choice function."
There are some other equivalent forms that may be of interest. For example:
* (94 E) Every second countable topological space is Lindelöf.
* (94 G) Every subset of ${\mathbb R}$ is separable.
* (94 R) Weak Determinacy. *If* $A$ is a subset of ${\mathbb N}^{\mathbb N}$ with the property that
$\forall a \in A\forall x \in{\mathbb N}^{\mathbb N}($ if $x(n) = a(n)$ for $n = 0$ and $n$ odd, then $x\in A)$, *then* in the game [$G(A)$](http://en.wikipedia.org/wiki/Determinacy#Games) one of the two players has a winning strategy.
* (94 X) Every countable family of dense subsets of ${\mathbb R}$ has a choice function.
Proofs and references are provided by the website and the book. A reference that comes up with some frequency in form 94 is Rhineghost, Y. T. "The naturals are Lindelöf iff Ascoli holds". Categorical perspectives (Kent, OH, 1998), 191–196, Trends Math., Birkhäuser Boston, Boston, MA (2001).
| 45 | https://mathoverflow.net/users/6085 | 45931 | 29,082 |
https://mathoverflow.net/questions/45932 | 10 | André and Quillen both gave constructions of the relative cotangent complex for commutative rings, so pretty immediately that gives us that we understand the cotangent complex for affine schemes. Illusie generalized the cotangent complex construction from "rings over A" for a ring A to "rings over $\mathcal{O}\_X$" for a base ring object of an arbitrary Grothendieck topos. At least for ordinary schemes, it doesn't seem too hard to believe that we could glue together relative cotangent complexes along affine opens, but for things like algebraic spaces and/or formal schemes it seems conceivable to me that it might be substantially harder to glue the local modules together while preserving their simplicial structure.
What difficulties with globalizing the local definition of the cotangent complex lead to the topos-theoretic approach used by Illusie? (This is not a history question. I'm just wondering what the motivation is for the greater generality, since I'm currently reading André's book, which only covers the "classical" case of a commutative $A$-algebra for set-theoretic commutative ring $A$.)
| https://mathoverflow.net/users/1353 | Is Illusie's generalization of the cotangent complex to arbitrary ringed toposes necessary in algebraic geometry? | As BCnrd points out, gluing cotangent complexes is a nontrivial thing. You might still ask whether it is really necessary for Illusie to work in the generality of a ringed topos. Would using a ringed space suffice? For standard deformation problems (deformation of a morphism or deformation of a scheme) working on the underlying ringed space would be enough. For more "interesting" deformation problems, like deformation of a morphism $X \rightarrow Y$, where $X$, $Y$, and the morphism are all allowed to vary, one needs something more sophisticated. Illusie constructs a ringed topos that encodes all of these data and then applies the machinery for ringed topoi that he has already developed.
| 11 | https://mathoverflow.net/users/32 | 45941 | 29,087 |
https://mathoverflow.net/questions/45905 | 13 | If $K$ is a number field, is it always possible to find a finite extension $L/K$ such that $L$ is the composite of fields $L\_1,\ldots, L\_n$, with the property that at most one prime ramifies in $L\_i/\mathbb{Q}$? Equivalently, is the composite of all extensions ramified at $\leq 1$ place all of $\overline{\mathbb{Q}}$?
The first question has an affirmative answer when $K$ is abelian, but for the general case, the equivalent second question sounds too strong to be true. Any ideas?
| https://mathoverflow.net/users/5513 | Expressing a number field as a composite of extensions ramified at one place | I believe the answer is `No', and Franz Lemmermeyer's example $K=Q(2^{1/3})$ and his strategy of the proof do the trick.
Suppose this particular $K$ is contained in the compositum $F$ of $L\_i$, with every $L\_i$ ramified at only one prime. Assume each $L\_i$ is Galois over $Q$ (otherwise replace it by its Galois closure) and that no two $L\_i$ ramify at the same prime $p$ (otherwise replace this pair by their compositum). The $L\_i$ are then linearly disjoint over $Q$, since their pairwise intersections would have to be unramified at all primes. So $G=Gal(F/Q)$ is the direct product of $Gal(L\_i/Q)$'s.
Now the group $G$ has a 2-dimensional irreducible representation $\rho$, the one that factors through the Galois closure of $K/Q$ (an $S\_3$-extension of $Q$). As $G$ is the direct product of groups, we can write $\rho=\rho\_1\otimes...\otimes\rho\_n$ uniquely, with $\rho\_i$ irreducible representations of $Gal(L\_i/Q)$. Moreover, $\rho$ is self-dual, so all the $\rho\_i$ are self-dual as well. Of these $\rho\_i$ one must be 2-dimensional and the others are 1-dimensional. Because 1-dimensional self-dual characters have order 2, this shows that at all primes $p$ except at most one (the one corresponding to the 2-dimensional $\rho\_i$) inertia $I\_p$ acts on $\rho$ through a quotient of order 2. But $I\_2$ and $I\_3$ act through quotients of order 3 and 6 respectively, contradiction!
| 15 | https://mathoverflow.net/users/3132 | 45949 | 29,093 |
https://mathoverflow.net/questions/45950 | 39 | Occasionally I find myself in a situation where a naive, non-rigorous computation leads me to a divergent sum, like $\sum\_{n=1}^\infty n$. In times like these, a standard approach is to guess the right answer by assuming that secretly my non-rigorous manipulations were really manipulating the Riemann zeta function $\zeta(s) = \sum\_{n=1}^\infty n^{-s}$ and its cousins. Then it's reasonable to guess that the "correct" answer is, for example, $\sum\_{n=1}^\infty n = \zeta(-1) = -\frac1{12}$. Thus the zeta function and its cousins are a valuable tool for other non-number-theoretic problem solving: it's always easier to rigorously prove that your guess is correct (or discover, in trying to prove it, that it's wrong) than it is to rigorously derive an answer from scratch.
I recently found myself wishing I could do something similar for the sum of the *quantum* integers. Recall that at quantum parameter $q = e^{i\hbar}$, *quantum $n$* is the complex number $$[n]\_q = \frac{q^n - q^{-n}}{q - q^{-1}} = q^{n-1} + q^{n-3} + \dots + q^{3-n} + q^{1-n}.$$ The point is that $[n]\_1 = n$.
>
> **Question:** Are there established methods to sum the divergent series $\sum\_{n=1}^\infty [n]\_q $ and its cousins? For example, is there some well-behaved function $\zeta\_q(s)$ for which the series is naturally the $s=-1$ value?
>
>
>
Note that when $n$ is a root of unity, the series truncates, and it would be nice (but maybe too much too hope for) if the regularized series agreed with the truncated series at these values.
I should mention also that I consider the following answer tempting but inaccurate, as it definitely doesn't work at roots of unity, which I do care about:
$$ \sum\_{n=1}^\infty [n]\_q = \frac1{q-q^{-1}} \sum\_{n=1}^\infty (q^n - q^{-n}) = \frac1{q-q^{-1}} \left( \sum\_{n=1}^\infty q^n - \sum\_{n=1}^\infty q^{-n}\right) = $$
$$ = \frac1{q-q^{-1}} \left( \frac{q}{1-q} - \frac{q^{-1}}{1-q^{-1}}\right) = \frac{q+1}{(q-q^{-1})(q-1)}$$
| https://mathoverflow.net/users/78 | Is there a "quantum" Riemann zeta function? | The paper by Cherednik [On q-analogues of Riemann's zeta function](https://doi.org/10.1007/s00029-001-8095-6 "Sel. math., New ser. 7, 447–491 (2001). zbMATH review at https://zbmath.org/?q=an:1001.11033") ([arXiv:math/9804099](https://doi.org/10.48550/arXiv.math/9804099)) gives precisely the definition you're after:
$$
\zeta\_q(s)=\sum\limits\_{n=1}^\infty q^{sn}/[n]\_q^s
$$
His paper also contains a brief discussion of the properties of this $q$-zeta function.
On the other hand, the term *quantum zeta function* appears to have a somewhat different meaning, see e.g. the paper [On the quantum zeta function](https://doi.org/10.1088/0305-4470/29/21/014 "J. Phys. A, Math. Gen. 29, No. 21, 6795–6816 (1996). zbMATH review at https://zbmath.org/?q=an:0905.58040") by R.E. Crandall.
| 42 | https://mathoverflow.net/users/2149 | 45958 | 29,098 |
https://mathoverflow.net/questions/45936 | 3 | Hi there,
Assuming X and Y are modal formulae and diamond X is satisfiable and diamond Y is satisfiable, how would one show that they X AND Y is satisfiable?
I don't think it requires much effort?
I think you need to choose one world and one model where X AND Y is true and that would mean it is satisfiable?
So assuming I'm going about it correctly, any ideas what model and world I should select to show this X AND Y is satisfiable?
Any advice would be great,
Thank you.
P.S. NO appropriate tags for this type of most, maybe someone should create a modal logic one (I can't as I'm a new user)
| https://mathoverflow.net/users/10814 | Modal logic - satisfiability | Your question as originally written (which Henry correctly diagnosed as problematic in two ways) does not match the more reasonable aim reflected in your comments to Henry's answer. Specifically, your comments make it sound like you want to show that the satisfiability of both $\diamond X$ and $\diamond Y$ implies the satisfiability of $\diamond X \wedge \diamond Y$, **not** the satisfiability of $X\wedge Y$ as your original phrasing states.
If your notion of satisfiability of a formula $Z$ is simply that there is some Kripke model $\mathcal{M}$ (with no restrictions on its accessibility relation) and some world $w$ in it such that $\mathcal{M},w\models Z$, then this weaker form of the question isn't too difficult to answer.
Let $\mathcal{M},w\models\diamond X$ and $\mathcal{N},v\models\diamond Y$. In particular, there is a world $u$ in $\mathcal{N}$ which is accessible from $v$ and satisfies $Y$. Now just form a new model $\mathcal{P}$ whose set of worlds is the union of those of $\mathcal{M},\mathcal{N}$, and whose accessibility relation is the union of those of $\mathcal{M},\mathcal{N}$, plus we set $u$ to be accessible from $w$. Then $\mathcal{P},w\models\diamond X \wedge \diamond Y$.
Henry's point about underspecification is still pertinent. I'm not sure I've gotten at what you want, and if you were to be limited to special kinds of Kripke frames, for instance, then the argument would need to say a bit more (ensuring we end up with an appropriate $\mathcal{P}$). I hope this is helpful.
| 6 | https://mathoverflow.net/users/4137 | 45959 | 29,099 |
https://mathoverflow.net/questions/45871 | 7 | There seem to be intimate connections between the different definitions of von Neumann module. The two that I'm aware of are Hilbert von Neumann modules and correspondences (in the sense of Connes). I was wondering if there is any significant interplay between the two - for instance if any of the ideas around Jones' submodule theory (which uses *correspondences* of von Neumann algebras) have been extended or transfered to Hilbert von Neumann modules?
Some observations: Suppose we have a left Hilbert $N$-module (so inner-product $N$-linear in the first variable) with a faithful state $\varphi$ on $N$. Then define a complex-valued inner product on $E$ via
$$\langle x,y\rangle\_\varphi:=\varphi(\langle x,y\rangle).$$
If we complete then we get a vector space $\widetilde{E}$ and a von Neumann algebra $N$ acting on it - so a left von Neumann module in a sense similar to that of correspondences.
Questions:
(i) Given a faithful representation of $N$ on a Hilbert space $H$ can we always find a state $\varphi$ and left Hilbert $N$-module $E$ so that the representation of $N$ on $\widetilde{E}$ is equivalent to the one on $H$?
(ii) Can this idea be taken over to the case of correspondences and Hilbert bimodules (for instance in the case of $II\_1$ factors)?
(ii) What results (if any) can be pulled up to the Hilbert bimodule level?
I'm aware that question (iii) is vague, but any input or references would be appreciated!
| https://mathoverflow.net/users/10779 | Subfactor theory and Hilbert von Neumann Algebras | Answers: (i) Yes, if we replace states by weights (not every
von Neumann algebra admits a faithful state);
(ii) Yes (for all von Neumann algebras); (iii) All of them.
Suppose M is an arbitrary von Neumann algebra and p≥0 is a real number.
Then we define a right L\_p(M)-module as a right M-module equipped
with an inner product with values in L\_{2p}(M), satisfying the same algebraic
properties as for Hilbert W\*-modules together with the appropriate completeness
condition (we require completeness in the measurable topology,
which coincides with the σ-weak topology for p=0 and with the norm topology for p>0).
Here L\_p(M)=L^{1/p} denotes the L\_p-space of M, in particular,
L\_0(M)=L^∞(M)=M, L\_1(M)=L^1(M)=M\_\* (the predual), L\_{1/2}(M)=L^2(M)=the Hilbert
space of half-densities on M. (The subscript notation is much more natural than the superscript notation
because L\_p-spaces form a graded algebra, p being the grading.)
A morphism of right L\_p(M)-modules is defined as a morphism of algebraic right M-modules
that is continuous in the measurable topology.
It turns out that right L\_p(M)-modules form a W\*-category.
We observe that the category of representations of M on Hilbert spaces is equivalent
to the category of right L\_{1/2}(M)-modules.
If we have a right L\_{1/2}(M)-module X with an inner product x,y↦(x,y)∈L\_1(M),
then x,y↦tr(x,y)∈C is a complex-valued inner product on X,
which turns X into a Hilbert space together with an action of M.
Vice versa, if X is a Hilbert space equipped with an action of M,
then x,y→(w∈M↦(x,yw)∈C)∈L\_1(M) is the corresponding L\_1(M)-valued inner product.
Suppose 0≤p≤q are real numbers. We define a functor from the category
of right L\_p(M)-modules to the category of right L\_q(M)-modules
by sending a right L\_p(M)-module X to X⊗L\_{q-p}(M).
Here ⊗ denotes the algebraic tensor product, without any kind of completion.
Although it is non-obvious, in the end this tensor product turns out to be complete.
Likewise, we define a functor from the category of right L\_q(M)-modules
to the category of right L\_p(M)-modules by sending a right L\_q(M)-module Y
to Hom\_M(L\_{q-p}(M),Y).
Here Hom\_M denotes the space of algebraic homomorphisms preserving the right action of M,
without any kind of continuity property.
Again it is a non-obvious fact that this space is actually a right L\_p(M)-module.
One can prove that the two functors defined above form an adjoint unitary equivalence
of the W\*-categories of right L\_p(M) and L\_q(M) modules.
In particular, the category of Hilbert W\*-modules over M and the category
of representations of M on Hilbert spaces are equivalent.
The result above extends to bimodules.
An M-L\_p(N)-bimodule is a right L\_p(N)-module X together
with a morphism of von Neumann algebras A→End\_N(X). (The algebra
of endomorphisms of any object in a W\*-category is a von Neumann algebra.)
Since the above equivalence is an equivalence of W\*-categories,
we can immediately extend it to an equivalence of categories of M-L\_p(N) and M-L\_q(N) bimodules.
In particular, the category of Hilbert W\*-bimodules from M to N is equivalent to the category
of Connes' correspondences from M to N.
Moreover, one can observe that the bicategory of von Neumann algebras, Connes' correspondences, which compose via Connes' fusion, and their intertwiners
is equivalent to the bicategory of von Neumann algebras, Hilbert W\*-bimodules,
which compose via the completed tensor product, and their intertwiners.
This result is also valid for arbitrary p.
References:
* The equivalence in the last paragraph of the answer was apparently first proven by Baillet, Denizeau, and Havet in their 1988 paper Indice d'une espérance conditionnelle.
* L\_p(M) modules were defined by Junge and Sherman in their 2005 paper Noncommutative L^p modules.
* I am not aware of any paper that proves the above equivalences for arbitrary p, but I will include a proof of these statements in my thesis.
| 7 | https://mathoverflow.net/users/402 | 45964 | 29,102 |
https://mathoverflow.net/questions/45966 | 2 | Hello,
i was reading your article about non metrizability of \*R.
i was able to prove that the interval open topology is not metrizable by proving that the intersection of decreasing hyper-intervals contains an interval. But i do not get how you used this argument for any metric, since we do not know how our nested balls look like.
Plus how do you prove that \*R is not connected?
thanks a lot
| https://mathoverflow.net/users/10820 | Metrization of hyperreals | I am not sure whom you are addressing in your question, but
some of your remarks relate to issues brought up at [this
MO
question](https://mathoverflow.net/questions/10870/which-topological-spaces-admit-a-nonstandard-metric). If not, could you let us know to which post you were referring?
It is quite common to consider hyperreal structures
${}^\*\mathbb{R}$ that exhibit uncountable cofinality, which
means that every countable subset of the hyperreals is
bounded. For example, this is true in all the hyperreal
models obtained by the ultrapower method, and it is a
consequence of the $\omega\_1$-saturation property that is often
insisted upon. (But it is possible though rarely done to construct hyperreal
models with the full transfer principle, but mere countable cofinality.)
In any hyperreal model with uncountable cofinality, no
point has a countable local basis in the order topology,
since for any countable collection of neighborhoods, one
can find an infinitesimal ball around the point contained
in them all. Thus, such a space cannot be metrizable, as
metric spaces always have a countable local basis at every
point, a countable collection of open neighborhoods of the
point, such that every open neighborhood of the point
contains one of them.
In a hyperreal model ${}^\*\mathbb{R}$ with *countable* cofinality (rare but possible), however, then the order topology is metrizable. To see this, fix an increasing unbounded $\omega$-sequence of nonstandard reals $r\_n$, which will eventually include many infinitely large hyperreals, and assume $r\_{n+1}\geq 2r\_n$. The balls of radius $\frac{1}{r\_n}$ form a countable local basis around any point. Define a metric $d$ on ${}^\*\mathbb{R}$ by $d(x,y)=\frac{1}{2^n}$, if $n$ is first such that $\frac{1}{r\_n}\leq |y-x|$. It generates the same topology since the ball of radius $\frac{1}{2^n}$ corresponds basically to the interval of radius $\frac{1}{r\_n}$ in ${}^\*\mathbb{R}$.
Finally, the hyperreals are never connected, since we may
partition the space into the infinitesimals and the
non-infinitesimals, and these sets are both disjoint and
open. In the case that there are increasingly tiny levels of infinitesimality, a consequence of saturation, then the same argument shows that the space is totally disconnected, since for any two points I may consider the lower-level infinitesimals around $p$ and the complement of this set, which are both open.
| 9 | https://mathoverflow.net/users/1946 | 45972 | 29,109 |
https://mathoverflow.net/questions/45954 | 4 | As far as I understood, there are two way to assign weights to a mixed Hodge complex (a way that does not depend on shifts, and another one that does). There is a clever way to relate these to ways using Deligne's decalage. Does there exist a conceptual explanation of these matters (including decalage)?
| https://mathoverflow.net/users/2191 | Do Deligne's decalage and two filtrations for mixed Hodge complexes have a conceptual explanation? | Unfortunately, I don't have a good conceptual explanation for any of this.
So here's a somewhat more technical explanation of how *decalage* is used in Hodge theory.
The actual business of constructing mixed Hodge structures tends to be quite involved.
In Deligne's approach, the Hodge and weight filtrations $F$, $W$ are
constructed directly on certain complexes $K,\ldots$ subject to a bunch of
axioms that can be verified in natural situations. This datum is a so called mixed Hodge complex.
One seemingly technical, but very important, consequence of these conditions
is that:
>
> **Theorem** The spectral sequences associated to $F$ and $W$ degenerate at $E\_1$ and $E\_2$ respectively.
>
>
>
The $E\_1$ degeneration is a very natural condition, it means that
$F$ is strictly compatible with differentials. It can viewed as an abstract generalization of the Hodge decomposition. (Note that this doesn't do away with harmonic theory, since
one needs it to verify the above axioms in the first place.)
However, the $E\_2$ degeneration condition for $W$ is a bit harder to work with.
Here decalage becomes very convenient, in that it converts $E\_2$ degeneration to $E\_1$
degeneration for a new filtration $Dec(W)$. This makes certain additional arguments much
more manageable. In fact, this trick of passing to $Dec(W)$ is needed in the proof of the above theorem.
**Added** Perhaps another instructive illustration of these ideas can be found in
Beilinson's paper *Notes on absolute Hodge cohomology*. In the paper, he introduces
a variant of a mixed Hodge complex called a $\tilde p$-Hodge complex. Localizing the
category of these with respect to quasi-isomorphism results in a triangulated
category $D^b\_{\tilde H^p}$. He proves (or more accurately sketches a proof) that this is equivalent to the bounded derived category of polarizable mixed Hodge structures $D^b(H^p)$.
The functor
$$D^b(H^p)\to D^b\_{\tilde H^p}$$
is easy to write down, but the inverse
involves decalage. I mention this because
it seems related to your original question about the two filtrations.
This has got to be one least understandable answers that I've ever written here, but
when you have complexes with several filtrations the story *is* going to be technical. The miracle is that it isn't worse than it is.
| 6 | https://mathoverflow.net/users/4144 | 45977 | 29,114 |
https://mathoverflow.net/questions/45953 | 18 | I'm curious about the following:
Is every real $n$-manifold isomorphic to a quotient of $\mathbb{R}^n$?
Thanks.
EDIT: As Tilman points out, the manifold should be connected. Also, yes, I'm thinking about topological quotients. Specifically, is there a surjective map $\mathbb{R}^n\to M$ such that $M$ has the quotient topology?
EDIT': I guess an interesting addendum to the question is "when is it true?"
| https://mathoverflow.net/users/2857 | Is every real n-manifold isomorphic to a quotient of $\mathbb{R}^n$? | Hahn–Mazurkiewicz Theorem: Suppose $X$ is a nonempty Hausdorff topological space.
Then the following are equivalent:
1. there is a surjection $[0,1]\to X$,
2. $X$ is compact, connected, locally connected and second-countable.
It follows that a Hausdorff space satisfying the conditions of (2) is a quotient
of $I = [0,1]$.
Cor: Every connected compact manifold is a quotient of $I$.
Since $I$ is a quotient of $\mathbb{R}^n$, we have your answer.
Cor: Every compact connected $m$-manifold is a quotient of $\mathbb{R}^n$ for any $n\geq 1$.
| 33 | https://mathoverflow.net/users/3634 | 45980 | 29,116 |
https://mathoverflow.net/questions/45756 | 10 | Considering the success of a [previous question](https://mathoverflow.net/questions/29118/need-help-proving-that-sum-limits-j0k-1-1j1k-j2k-2-binom2k1) involving Eulerian numbers, I thought I might throw this question into the mix. It comes from some localization computations in GW theory, but in this form is purely combinatorial.
Eulerian numbers of the second kind are defined by the recursion relation
$$\left\langle\left\langle n\atop m\right\rangle\right\rangle = (m+1)\left\langle\left\langle n-1\atop m\right\rangle\right\rangle+(2n-m-1)\left\langle\left\langle n-1\atop m-1\right\rangle\right\rangle$$
with the initial conditions $\left\langle\left\langle n\atop 0\right\rangle\right\rangle=1$ and $\left\langle\left\langle n\atop m\right\rangle\right\rangle$ = 0 for $m\geq n$.
For references, see:
<http://en.wikipedia.org/wiki/Eulerian_number> and <http://oeis.org/classic/A008517>.
The following three statements are known:
1. $\sum\_{m=0}^{n}
(-1)^mm!(2n-m-2)!\left\langle\left\langle
n\atop
m\right\rangle\right\rangle=0$ for
all $n$;
2. $\sum\_{m=0}^{n}
(-1)^m(m)m!(2n-m-2)!\left\langle\left\langle
n\atop
m\right\rangle\right\rangle=0$ for
odd $n$;
3. $\sum\_{m=0}^{n}
(-1)^m(m+1)!(2n-m)!\left\langle\left\langle
n\atop
m\right\rangle\right\rangle=0$ for
even $n$;
(2 and 3 are equivalent).
Question: Show that the expression in the second statement is non-zero for even $n$, i.e. show
$\sum\_{m=0}^{n}
(-1)^m(m)m!(2n-m-2)!\left\langle\left\langle
n\atop
m\right\rangle\right\rangle\neq0$ for
even $n$.
Certainly we've been checking this on a computer for modest values of $n$.
| https://mathoverflow.net/users/6240 | Expressions involving Eulerian numbers of the second kind: trying to show $\sum_{m=0}^{n} (-1)^m(m)m!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle\neq0$ for even $n$. | Following Mike Spivey's comment above I will consider $B(n)$ in (3). It turns out that your conjecture is true, because for odd $n$ the sum $B(n)$ is a certain weighted $L^2$ norm of the Eulerian polynomial of the second kind of order $\frac{n+1}{2},$ with the sign of $(-1)^{\frac{n-1}{2}}\, . $
The product of the two factorials in $B(n)$ may be expressed in terms of the Eulerian Beta integral
$$(m+1)!(2n−m)!=(2n+2)!\int\_0^1 t^{m+1 }(1-t)^{2n-m}dt,$$
so that dividing it by $(2n+2)!
$ we have
$$ \frac {B(n)}{(2n+2)!}= \int\_0^1 \sum\_{m\ge0}(-1)^m\left\langle\!\!\left\langle n\atop m\right\rangle\!\!\right\rangle t^{m+1 }(1-t)^{2n-m}dt = $$
$$= \int\_0^1 t(1-t)^{2n}\sum\_{m\ge0} \left\langle\!\! \left\langle n\atop m\right\rangle\!\! \right\rangle \Big(\frac{t}{t-1}\Big)^{ m} dt= \int\_0^1 t(1-t)^{2n} E\_n \Big(\frac{t}{t-1}\Big) dt. $$
Changing variable with $x:=\frac{t}{t-1}$ this becomes:
$$ \int\_{-\infty}^0 x(x-1)^{-2n-3} E\_n(x) dx, $$
where $E\_n$ denotes the Eulerian polynomial of the second kind
$$E\_n(x):=\sum\_{m\ge0} \left\langle\!\!\left\langle n\atop m\right\rangle\!\! \right\rangle x^m,$$
and satisfies the recursive relation (corresponding to the relation for the coefficient that you gave in your question):
$$(x-1)^{-2n-2}E\_{n+1}(x)=\left( -x(x-1)^{-2n-1}E\_n(x) \right)^{\prime}.$$
By the above formula it is now easy to show, integrating by parts repeatedly, that $B(n)=0$ for even $n$ while for odd $n=2p-1$
$$B(2p-1)=(-1) ^ { p + 1 } (4p)! \int\_0^{+\infty} E\_p (-x)^2 (x+1)^{-4p-1} x dx . $$
(To check this, it is convenient to introduce the sequence of rational functions $U\_ n(x):= (x-1)^{-2n}E\_n(x)$ that satisfy the recurrence $U\_ {n+1}= \big (\frac{x}{1-x} U\_ n \big) ^ {\prime} $ with initial condition $U\_0=1.$ Hence for all $n+m>0$ we have $\int\_{-\infty}^0 U\_{n+1}U\_{m}\frac{x}{1-x}dx=-\int\_{-\infty}^0 U\_ {n}U\_ {m+1}\frac{x}{1-x}dx\, .$ The integral found above for $B(n) / (2n+2)!\, $ was $-\int\_{-\infty}^0 U\_{n}U\_{1}\frac{x}{1-x}dx$ ).
| 12 | https://mathoverflow.net/users/6101 | 45985 | 29,117 |
https://mathoverflow.net/questions/45534 | 9 | Recall that an $R$-algebra $R\to S$ is called formally smooth (resp. formally unramified resp. formally étale) if given any lifting problem of the form
$$\begin{matrix}
R&\to &T\\
\downarrow&{}^?\nearrow&\downarrow\\
S&\to&T/J\end{matrix}$$
where $J$ is a square-zero nilpotent ideal of $T$, there exists at least one (resp. at most one, resp. exactly one) lift $S\to T$ making the diagram commute.
In Quillen's paper "Homology of Commutative Rings", he cites the following construction (Not sure who came up with it, but my copy of the paper has "Hopf?" written next to the relevant proposition):
Given an $R$-algebra $S$ and an $S$-module $M$, we let $S\oplus M$ be the commutative ring on the underlying $S$-module $S\oplus M$ with the multiplication given by the composite
$$(S\oplus M)\otimes\_R (S\oplus M)\cong (S\otimes\_R S) \oplus (S\otimes\_R M) \oplus (M\otimes\_R S) \oplus (M\otimes\_R M)\to S\times M\cong S\oplus M$$
Where:
$S\otimes\_R S\to S$ is the multiplication of $S$
$S\otimes\_R M\to M$ is the left action of $S$ on $M$
$M\otimes\_R S\to M$ is the right action of $S$ on $M$
$M\otimes\_R M\to M$ is the zero map $M\otimes\_R M\to 0\to M$
This ring is commutative, unital, and is equipped with a canonical $R$-linear projection $\phi:S\oplus M\to S$ with kernel $S$-linearly isomorphic to $M$. Also notice that $ker(\phi)$ is square-zero.
Interestingly, given any $R$-algebra $A$ extending $S$, i.e. $A\in (R\operatorname{-Alg}\downarrow S)$, it is easily verified that $\operatorname{Hom}\_{(R\operatorname{-Alg}\downarrow S)}(A,S\oplus M)\cong \operatorname{Der}\_R(A,Res\_{A\to S}(M))$, that is, lifts $A\to S\oplus M$ are in bijective correspondence with derivations $A\to Res\_{A\to S}(M)$ where $Res\_{A\to S}$ is the restriction of scalars by the map $A\to S$.
In particular, it seems, at least to my untrained eye, that there's a connection between this construction and the definition of formally smooth (resp. formally unramified resp. formally étale) algebras. The lifts we see, at least when $T$ is of the form $U\oplus M$ for some $U$-module $M$ correspond exactly to $R$-derivations.
Quillen then proves that the exact objects corresponding to the modules of the form $S\oplus M$ are the abelian group objects of the category $(R\operatorname{-Alg}\downarrow S)$. In particular (back to our original notation), this does not necessarily hit all square-zero extensions of $T/J$, since it misses those square-zero extensions that don't admit a section $T/J\to T$. Since the module of relative Kähler differentials carries all of the data of these derivations, this appears to explain exactly why Kähler differentials are only sufficient to characterize formally unramified algebras.
So far, I follow.
So here's the question: Why, morally, do we need to look at the $S$-modules over the cofibrant replacement of $S$ (we can give a functorial cofibrant replacement by looking at a simplicial resolution of $S$ by defining a simplicial ring consisting of free algebras over $A$ in every degree (c.f. André 1974)) (with all of the homotopical baggage we need to characterize the model structure) to capture the lifting data from the rest of the square-zero extensions that we would need to characterize formal smoothness (resp. formal étaleness)?
That is, similar to how every trivial extension is of the form $S\oplus M$ for an $S$-module M, can we give a homotopical version of the trivial extension such that the nilpotent extensions of $S$ are exactly those that that are trivial over some cofibrant replacement $QS$?
| https://mathoverflow.net/users/1353 | Formally smooth morphisms, the cotangent complex, André-Quillen cohomology, and representability of nilpotent extensions as trivial extensions over a cofibrant replacement | Your question is:
>
> Why, morally, do we need to look at the S-modules over the cofibrant replacement
> of S to capture the lifting data from the rest of the square-zero extensions
> that we would need to characterize formal smoothness (resp. formal étaleness)?
>
>
>
Here's how I think about it, though I don't know if it addresses what's bothering you about this.
You'd like to understand surjective maps of rings $f:T\to T'$ whose kernel $M$ (an $S$-module) is square-zero, but where $f$ might not admit a section.
There's a universal example of such an extension (in $R$-algebras over $S$); it lives in *simplicial* rings, i.e., in $(s(R\operatorname{-Alg})\downarrow S)$. Namely, form the simplicial $S$-module $BM$ by shifting (so $\pi\_1BM=M$, $\pi\_iBM=0$ for $i\neq1$), then consider the simplicial ring $S\oplus BM$, formed with trivial multiplication on the $BM$ part.
Now let $g: S\to S\oplus BM$ be the evident inclusion map into the first factor. The "kernel" of this map of simplicial rings is the homotopy fiber of $g$ (calculated in simplicial $S$-modules), and this homotopy fiber is equivalent to $M$ (concentrated in dimension $0$).
So $g$ has "kernel" $M$.
The claim is that isomorphism classes of square-zero extensions (possibly not split) over $S$ is the same as homotopy classes of maps $S\to S\oplus BM$ in $(s(R\operatorname{-Alg})\downarrow S)$. Calculating homotopy classes requires using a cofibrant model for the domain $S$. (In fact, there are no nontrivial maps $S\to S\oplus BM$ at all.)
Given such a map $f:QS\to S\oplus BM$ from a cofibrant replacement, pull back $g$ along it to get an extension $g'\colon E\to QS$. The simplcial ring $E$ has homotopy concentrated in degree $0$, and in fact there's an exact sequence $0\to M\to \pi\_0 E\to \pi\_0 QS=S$.
(Note that $(s(R\operatorname{-Alg})\downarrow QS)$ and $(s(R\operatorname{-Alg})\downarrow S)$ are Quillen equivalent model categories, so in fact giving an "extension" over $QS$ is the same as giving one for $S$.)
This story is morally just like the one you can tell for extension of $R$-modules: to compute extensions of $M$ by $N$, note that there's a universal extension by $N$ in the derived category of $R$-modules, of the form $N\to C\to N[1]$ (where $H\_\*C=0$). So extensions over $M$ are $\mathrm{Hom}\_{D(R)}(M, N[1])$, and you can compute this group by replacing $M$ with a projective resolution (i.e., cofibrant replacement).
| 7 | https://mathoverflow.net/users/437 | 45987 | 29,119 |
https://mathoverflow.net/questions/46011 | 16 | Let $K$ be a compact metric space, and $(E,d\_E)$ a complete separable metric space.
Define $C:=C(K,E)$ to be the continuous functions from $K$ to $E$ equipped with
the metric $d(f,g)=\sup\_{k\in K}\ d\_E (f(k),g(k))$. Is the space $C$ separable?
The result is true when $E$ is the real line; this is Corollary 11.2.5 in Dudley's
book *Real Analysis and Probability*.
The result is also true when $K=[0,1]$ (if I'm not being too careless) by
considering $C$ as a subspace of the Skorohod space $D\_E[0,1]$, which is
complete and separable by Theorem 5.6 in Ethier and Kurtz's book *Markov Processes:
Characterization and Convergence*.
For general $K$, it is not so obvious how to find an explicit countable
dense set in $C$, but I suspect one could modify Ethier and Kurtz's approach and get a proof.
But surely this result is known, and stated in some book? I've searched through
my library without success.
---
**Update:** This result is also **Theorem 2.4.3** of S. M. Srivastava's book *A Course on Borel Sets*. His proof is the same as Kechris's. I have also found an alternative, but false, published proof using the "fact" that $C(K,E)$ is $\sigma$-compact. Beware!
| https://mathoverflow.net/users/nan | Is the space of continuous functions from a compact metric space into a Polish space Polish? | Yes, it appears e.g. as Theorem 4.19 in Chapter I of Kechris' *Classical Descriptive Set Theory*. (The relevant page is visible in Google Books if it's not in your library.)
| 13 | https://mathoverflow.net/users/4137 | 46015 | 29,136 |
https://mathoverflow.net/questions/46024 | 2 | It is known that solving systems of linear equations is reducible to SVD in a straightforward way; if you want to solve $\mathbf{Ax}=\mathbf{b}$, then you can perform SVD on $\mathbf{A}$ and minimize $||\mathbf{UDVx}-\mathbf{b}||$.
However, is there a reverse reduction that is also very efficient? That is, if you can solve linear equations, you can solve SVD?
EDIT: Because of Denis's comment/answer below, it looks like there isn't a reduction *in general*. But I'm interested in these problems over $\mathbb{C}$; so, the new question is: If we can solve linear equations over $\mathbb{C}$ exactly or approximately, can we perform an "approximate" SVD (for some suitable notion of "approximate")?
The answer still seems to be in the negative, but I defer to people who actually know something about this.
| https://mathoverflow.net/users/5534 | What is the relationship between singular value decomposition and solving linear systems? | I don't think so. Solving linear equations is an algebraic problem, where the scalar field is arbitrary: $\mathbb R$ or $\mathbb C$, but also $\mathbb Q$, $\mathbb F\_{p^n}$, $k(X)$, $\mathbb Q\_p$, $\mathbb Q(\alpha)$ ($\alpha$ algebraic). In many cases, there is no analogue of SVD at all. This is why Gauss elimination remains meaningful.
When the scalar field is $\mathbb R$ or $\mathbb C$, SVD is one aspect of resolution. But it has the flaw that it cannot be done exactly in finitely many operation. Thus you have to choose between (costly) exact methods ($LU$ factorization, which is reminiscent to Gauss) or iterative approximations (relaxation, SSOR, conjugate gradient, SVD, ...) Notice that the conjugate gradient is theoretically an exact method, but in practice it is used in an iterative way.
You may have a look to my book *Matrices : Theory and Applications*, Grad. Text in Math. 216, Springer-Verlag. The second edition is released now.
| 6 | https://mathoverflow.net/users/8799 | 46026 | 29,143 |
https://mathoverflow.net/questions/46019 | 20 | Suppose $F: C\to D$ is an additive functor between abelian categories and that
$$0\to X\xrightarrow f Y\xrightarrow g Z\to 0$$
is and exact sequence in $C$. Does it follow that $F(X)\xrightarrow{F(f)} F(Y)\xrightarrow{F(g)} F(Z)$ is exact in $D$? In other words, is $\ker(F(g))=\mathrm{im}(F(f))$?
**Remark 1:** If the answer is no, a counterexample must use a *non-split* short exact sequence. This is because additive functors send split exact sequences to split exact sequences. A splitting is a pair $s:Y\to X$ and $r:Z\to Y$ so that $id\_Y=f\circ s+r\circ g$, $id\_X=s\circ f$, and $id\_Z=g\circ r$. An additive functor preserves these properties, so $F(s)$ and $F(r)$ will split the sequence in $D$.
**Remark 2:** You probably know you know lots of left exact and right exact additive functors, but you also know lots of *exact in the middle* additive functors. $H^i$ and $H\_i$ for any (co)homology theory are neither left or right exact, but they are exact in the middle by the long exact sequence in (co)homology.
| https://mathoverflow.net/users/1 | Is there an additive functor between abelian categories which isn't exact in the middle? | Consider the abelian category of morphisms of vector spaces, i.e., the objects are linear maps $f:U\to V$, and the morphisms are commutative squares. Let the functor $Im$ assign to a morphism $f$ its image $Im(f)$. Consider the short exact sequence of morphisms $(0\to V)\to (U\to V)\to (U\to 0)$. The functor $Im$ transforms it to the sequence $0\to Im(f)\to 0$, i.e. $Im$ is not exact in the middle.
On the other hand, notice that $Im$ is epimorphic and monomorphic, i.e., transforms epimorphisms to epimorphisms and monomorphisms to monomorphisms.
| 28 | https://mathoverflow.net/users/2106 | 46035 | 29,148 |
https://mathoverflow.net/questions/45716 | 7 | Consider the classical Schwartz space $\mathcal{S}(\mathbb{R})$ together with the Fourier transform $\mathcal{F} : \mathcal{S}(\mathbb{R}) \rightarrow \mathcal{S}( \mathbb{R})$.
Consider the subspace $V$ of the even, smooth functions on the interval $[-1,1]$.
Can you construct a (bounded) operator $D:\mathcal{S}(\mathbb{R}) \rightarrow \mathcal{S}(\mathbb{R}) $ such that
$$ D \mathcal{F} v = 0, \quad Dv=v \qquad\forall v \in V ?$$
Observe that by Paley-Wiener, the intersection $\mathcal{F}V \cap V =0$ is trivial.
What is the associated Schwartz kernel?
| https://mathoverflow.net/users/10400 | A Question concerning the Fourier Transform of $\mathbb{R}$ | You don't need to do things the rough way; there is enough freedom for the smooth approach.
Take any even $C\_0^\infty$ descent $\Phi$ from $[-1,1]$ and define $Pf=\Phi f$ and $Qf=\mathcal F^{-1}(\Phi\mathcal F f)$. Now take the standard $D=I-(I-PQ)^{-1}(1-P)$. This works in $L^2$ for the same reason as it does with orthogonal projections: $PQ$ is a contraction. The good news is that $PQ$ maps $L^2$ to $S$ continuously and $(I-PQ)^{-1}=I+PQ(I-PQ)^{-1}$, so the resulting operator is bounded from $S$ to $S$ as well.
The kernel can be "found" by expanding $D$ into power series that converges geometrically but, since this construction involves an arbitrary smooth cutoff, to write an explicit formula seems quite hopeless.
| 5 | https://mathoverflow.net/users/1131 | 46045 | 29,156 |
https://mathoverflow.net/questions/46044 | 0 | Hi -- is there any analogue or adjustment of, say, Schwartz Bayesian (or other) information criterion that would be applicable to model selection in ridge regression with a given ridge parameter $\eta$, i.e. $\hat{\beta}\_\eta = (X'X+\eta I)^{-1}X'Y$?
Thanks!!
| https://mathoverflow.net/users/10837 | Information criteria for ridge regression | The ridge estimator corresponds to the posterior mean under a Normal linear regression model with a conjugate Normal-inverse-gamma prior on the regression coefficients: $\beta \mid \sigma^2, \lambda \sim \mbox{N}(0, \lambda^{-1}\sigma^2 \mbox{I})$ and $\sigma^2 \sim \mbox{IG}(a,b)$ for known hyperparameters $a$ and $b$. One may additionally put a prior distribution over $\lambda$. If you consider a discrete number of possible values for $\lambda$ then one may compute posterior probabilities for each of these values or compute Bayes factors to compare different values.
As BIC and AIC and other such "information criterions" can be viewed as approximations to Bayes factors, this may answer your question. Usually, as you probably know, one simply checks prediction error for the different values via cross-validation (at least in prediction contexts) and selects lambda that way.
| 1 | https://mathoverflow.net/users/8719 | 46053 | 29,160 |
https://mathoverflow.net/questions/46046 | 4 | Let $k$ be a positive integer. Is it true that any finite group $H$ of cardinal $4k+2$ whose center contains an element $h$ of order $2$ is isomorphic to the direct product $H=(\mathbb{Z}/2\mathbb{Z})\times G$, where $G=H/\{1,h\}$?
An equivalent statement would be: Let $G$ be a finite group of odd cardinal. Is it true that the second cohomology group $H^2(G,\mathbb{Z}/2\mathbb{Z})$ with respect to the trivial action of $G$ on $\mathbb{Z}/2\mathbb{Z}$, vanishes?
| https://mathoverflow.net/users/10675 | A question about finite groups. | There is a reasonably simple argument that any group $H$ of twice odd order has a normal subgroup $N$ of index 2. Given that, if you know also that $H$ has a central subgroup $Z$ of order 2, then it is straightforward to show that $H \cong N \times Z$.
The argument goes like this. By Cayley's Theorem, there is an isomorphism $\phi$ that maps $G$ to a permutation group of degree $|G|$ in which all non-identity elements of $G$ act without fixed points. So the elements of order 2 in $G$ are mapped by $\phi$ onto odd permutations, and hence the inverse image $N$ in $G$ of the intersection of the image of $\phi$ with the alternating group has index 2 in $G$.
| 11 | https://mathoverflow.net/users/35840 | 46056 | 29,162 |
https://mathoverflow.net/questions/45831 | 33 | This is an excellent question asked by one of my students. I imagine the answer is "no", but it doesn't strike me as easy.
Recall the set up of the [stable marriage problem](http://en.wikipedia.org/wiki/Stable_marriage_problem). We have $n$ men and $n$ women. Each man has sorted the women into some order, according to how much he likes them, and each women has likewise ranked the men. We want to pair off the men with the women so that there do NOT exist any pairs $(m,w)$ and $(m', w')$ where
* $m$ prefers $w'$ to $w$ and
* $w'$ prefers $m$ to $m'$.
It is a theorem of Gale and Shapley that such an assignment is always possible.
Here is a potential way you could try to find a stable matching. Choose some function $f: \{ 1,2,\ldots, n \}^2 \to \mathbb{R}$. Take the complete bipartite graph $K\_{n,n}$, with white vertices labeled by men and black vertices by women, and weight the edge $(m,w)$ by $f(i,j)$ if $w$ is $m$'s $i$-th preference, and $m$ is $w$'s $j$-th preference. Then find a perfect matching of minimal weight, using standard algorithms for the [assignment problem](http://en.wikipedia.org/wiki/Assignment_problem).
>
> Is there any function $f$ such that this method works for all preference lists?
>
>
>
| https://mathoverflow.net/users/297 | Can assignment solve stable marriage? | David's question is sufficiently more precise than the question of Donald Knuth referred to in the answer below. I believe it can be answered in the negative for $n \geq 3$, as follows.
Let $\{m\_1,m\_2,\ldots,m\_n\}$ and $\{w\_1,w\_2,\ldots,w\_n\}$ be the vertices of the parts of our graph $K\_{n,n}$. Consider a choice of preferences where for $i\geq 3$ the man $m\_i$ and the woman $w\_i$ are each other's top choice. Then any stable marriage must match them to each other.
We consider now the preferences of $m\_1,m\_2,w\_1$ and $w\_2$. I will write $m\_1 : w\_1 \to 1, w\_2 \to 3$ to denote that $w\_1$ is $m\_1$-th first choice and $w\_2$ is his third, etc.
* $m\_1 : w\_1 \to 1, w\_2 \to 2,$
$m\_2 : w\_1 \to 2, w\_2 \to 3,$
$w\_1 : m\_1 \to 2, m\_2 \to 3,$
$w\_2 : m\_1 \to 1, m\_2 \to 2.$
Here $m\_1w\_1, m\_2w\_2$ is a stable matching and therefore if the function $f$ is as desired then we must have
$f(1,2)+f(3,2) < f(2,1)+f(2,3),$
where on the left we have the weight of the stable matching and on the right is the weight of the matching $m\_1w\_2, m\_2w\_1$.
But the left and right side of the inequality above are symmetric and so by switching the $m$'s and $w$'s we can design another set of preferences implying
$f(2,1)+f(2,3) < f(1,2)+f(3,2).$
It follows that the function $f$ as specified in the question can not exist.
| 19 | https://mathoverflow.net/users/8733 | 46059 | 29,164 |
https://mathoverflow.net/questions/46068 | 104 | Let $n$ be a large natural number, and let $z\_1, \ldots, z\_{10}$ be (say) ten $n^{th}$ roots of unity: $z\_1^n = \ldots = z\_{10}^n = 1$. Suppose that the sum $S = z\_1+\ldots+z\_{10}$ is non-zero. How small can $|S|$ be?
$S$ is an algebraic integer in the cyclotomic field of order $n$, so the product of all its Galois conjugates has to be a non-zero rational integer. Using the utterly crude estimate that the magnitude of a non-zero rational integer is at least one, this gives an exponential lower bound on $S$. On the other hand, standard probabilistic heuristics suggest that there should be a polynomial lower bound, such as $n^{-100}$, for $|S|$. (Certainly a volume packing argument shows that one can make $S$ as small as, say, $O(n^{-5/2})$, though it is unclear to me whether this should be close to the true bound.) Is such a bound known? Presumably one needs some algebraic number theoretic methods to attack this problem, but the only techniques I know of go through Galois theory and thus give exponentially poor bounds.
Of course, there is nothing special about the number $10$ here; one can phrase the question for any other fixed sum of roots, though the question degenerates when there are four or fewer roots to sum.
| https://mathoverflow.net/users/766 | How small can a sum of a few roots of unity be? | In this paper they talk about this problem for 5 instead of 10 roots.
<http://www.jstor.org/stable/2323469>
EDIT: In view of Todd Trimble's comment, here's a summary of what's in the paper.
Let $f(k,N)$ be the least absolute value of a nonzero sum of $k$ (not necessarily distinct) $N$-th roots of unity. Then
$f(2,N)$ is asymptotic to $cN^{-1}$, where $c$ is $2\pi$ for even $N$, $\pi$ for odd $N$,
$f(3,N)$ is asymptotic to $cN^{-1}$, where $c$ is $2\pi\sqrt3$ for $N$ divisible by 3, $2\pi\sqrt3/3$ otherwise,
$f(4,N)$ is asymptotic to $cN^{-2}$, where $c$ is $4\pi^2$ for even $N$, $\pi^2$ for $N$ odd,
$f(k,N)>k^{-N}$ for all $k,N$,
$f(2s,N)<c\_sN^{-s}$ for $N$ even and $s\le10$,
$f(k,N)<c\_kN^{-[\sqrt{k-6}]-1}$ for $N$ even and $k>5$, and
If $N$ is twice a prime, and $k<N/2$, then there exists $k'<2k$ such that $f(k',N)\le2k2^{k/2}\sqrt{k!}N^{-k/2}$.
The only result in the paper for 5 roots of unity is (the trivial) $f(5,N)>5^{-N}$, but it is suggested that maybe $f(5,N)>cN^{-d}$ for some $d$, $2\le d\le3$, and some $c>0$.
| 61 | https://mathoverflow.net/users/10769 | 46069 | 29,169 |
https://mathoverflow.net/questions/43267 | 13 | I am reading the Witten's topological twisting for $N = 2$ Superconformal Field Theory(SCFT) <http://arxiv.org/abs/hep-th/9112056>
In this paper Witten constructed 2 TQFTs i.e. A-model and B-model from an $N=2$ SCFT with a Kahler target manifold. My queries are the following :
1. When we do the twist, we're actually changing the interpretation of the fermionic fields e.g. we're taking the field $\psi\_{+}^{i}$ to be a section of $\Phi^{\star}(T^{0,1}X)$ instead of a section of $K^{\frac{1}{2}}$ tensor $\Phi^{\star}(T^{0,1}X)$.Mathematically, this seems OK. But on the physics side, are we changing any physics.
2. Are the topological A and B models still supersymmetric. Does the twisting preserves any supersymmetric.
3. And is this type of twisting always possible, for any SCFT.
I am sorry for the question is not very clear, feel free to modify.
| https://mathoverflow.net/users/9534 | Witten's topological twisting | First, a historical note: the twisting procedure for $N=2$ SCFTs is due to [Eguchi and Yang](http://inspirebeta.net/record/296815); although a twisting of sorts had already appeared in Witten's Topological Quantum Field Theory paper of 1988.
Let me give quick answers to your questions:
1. Twisting *per se* does not change the physics: it merely allows one to identify a particular subsector of the theory. The topological theory is obtained by restriction to that subsector.
2. The topological A and B models are particular subsectors of the sigma model quantum field theory. Supersymmetries will generally not preserve those subsectors, hence the topological theories are no longer supersymmetric.
3. Any $N=2$ SCFT can be twisted in this way, but more generally one can twist also other conformal field theories. For example, one can twist string theories or also theories based on a Kazama algebra, such as the $G/G$ gauged Wess-Zumino-Witten model. More generally still, it used to be the case that all known two-dimensional topological conformal field theories are "cohomologically equivalent" to a twisted $N=2$ SCFT. (My information is probably out of date, since I last looked at this topic 15 years ago!)
| 12 | https://mathoverflow.net/users/394 | 46071 | 29,171 |
https://mathoverflow.net/questions/46028 | 3 | A category is called **well-powered** if the partial order of subobjects of any object is a set. Is there a German translation which is well-established in the literature? I don't want to use the English term in a German text.
| https://mathoverflow.net/users/2841 | German translation for "well-powered category" | Pareigis and Schubert both use "lokal klein" in their textbooks.
Of course there is a clash with the other meaning of "locally small" (i.e. that
Hom(a,b) is always a set), but this is usually satisfied by their definitions
of category.
| 3 | https://mathoverflow.net/users/10840 | 46072 | 29,172 |
https://mathoverflow.net/questions/46079 | 4 | So I just started learning about quasicategories... Alright, that's an understatement: I just listened to Julie Bergner talk about quasicategories, and then started reading Moritz Groth's short course about them. I did a quick google search for this question, but there's a high probability that I just wasn't using the right terminology, or that I'm being silly. In any case, here is my question:
Suppose you have yourself a quasicategory (or $\infty$-category, or weak kan complex, whatever you call it). Is it always possible to embed this into a finitely bicomplete, pointed quasicategory? (I purposefully didn't ask for a way to make it a stable $\infty$-category, though that may be equally possible). An existence statement would be nice, a construction or reference to an instruction would be even nicer.
A bit more down to earth: Can this even be done with small categories? If you have a small category, can you always add objects to it in some way to make it finitely bicomplete? I suppose if I wanna sound fancy then I might want to be asking for an adjoint to the inclusion functor from complete categories into categories... or something like that.
There's a good reason I'd like to know all of this... but it's Level 1 Classified by the NSA.
(In other words... I'd like to think about it a bit more to make sure I don't embarrass myself playing with machinery I don't understand yet.)
NB: Easy on the higher categorical language... I'm good with categories, and I'm good with simplicial sets. If you need to use something bigger, then please give a reference, or a gentle description. Thanks!
| https://mathoverflow.net/users/6936 | Do quasi-categories have a `completion'? | Maybe this is not exactly what you are looking for, but it might be a place to start.
There is an analog of the Yoneda embedding for quasi-categories which takes the form $$j : \mathcal C \rightarrow Map(\mathcal C^{op}, \mathcal S)$$ where here $\mathcal S$ is the quasicategory of spaces (that is, the coherent nerve of the category of Kan complexes.) This map is an embedding which can be regarded as the free completion of $\mathcal C$ with respect to colimits, just as in ordinary category theory, and moreover preserves all limits which exist in $\mathcal C$. This is of course quite a bit more than just finitely bicomplete, but Lurie's book on Higher Topos Theory outlines a method for adjoining any class of colimits, and presumably this would work for finite colimits as well if that's what you want.
| 5 | https://mathoverflow.net/users/4466 | 46085 | 29,181 |
https://mathoverflow.net/questions/46094 | 7 | Let $H$ be a separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$.
Is the Hilbert space $L^2(H,\gamma)$ separable?
| https://mathoverflow.net/users/238 | If $H$ is a separable Hilbert space, is $L^2(H)$ separable? | By Example 7.14.13 in Volume 2 of Bogachev's *Measure Theory*, every Radon measure on $H$ is separable, so that $L^2(H,\gamma)$ is also separable. It is not necessary that $H$ is a Hilbert space, just that every compact subset of $H$ be metrizable.
| 14 | https://mathoverflow.net/users/nan | 46097 | 29,191 |
https://mathoverflow.net/questions/46022 | 11 | I just found an English translation of Serre's FAC at Richard Borcherds' Algebraic Geometry course web page. I really want to read it sometime. I am beginner in Algebraic Geometry, just started learning Scheme theory (Hartshorne Ch II). My question is :
Shall I read coherent sheaves and the cohomology from the translation of Serre's FAC or from the Hartshorne, this is going to be my first encounter with Coherent sheaves. Do you recommend FAC as an introduction to Coherent Sheaves.
My prime interest at present is the derived categories of coherent sheaves on projective varieties, specially the work of Tom Bridgeland and the Mirror Symmetry.
| https://mathoverflow.net/users/9534 | Serre's FAC versus Hartshorne as an introduction to sheaves in algebraic geometry | As always, the source you use may be related to what your goals are. To give some perspective, recall there are several ways to define sheaf cohomology, and Serre and Hartshorne feature different methods. Serre used Cech cohomology, and there the important long exact sequence property does not always hold. He was able to prove however that it does hold for "coherent" sheaves. One big advantage of Cech theory is its easier computability in specific cases, such as on projective space. Hartshorne presents first Grothendieck's theory of derived functor cohomology, but then proves it agrees with Cech cohomology before using that theory to compute the cohomology of coherent sheaves on projective space. But if you want to learn the derived functor theory you must choose Hartshorne over Serre.
The distinction made above between schemes and varieties is also relevant. Serre teaches Cech cohomology on varieties,and Hartshorne presents derived functor cohomology on schemes. If you are only interested in varieties, or prefer learning cohomology in the easier setting of varieties, then you may prefer Serre's FAC. Another good source is the book Algebraic Varieties by George Kempf, where the derived functor theory is presented on varieties and used for basic computations, including coherent cohomology of projective space and even the full Riemann Roch theorem, before being linked with Cech theory. So if you want to learn to make computations with the abstract derived functor theory you might prefer George's treatment, although some details are missing there, and some misprints exist.
Finally there are slight differences in Serre's and Hartshorne's results which can be relevant in some settings. E.g. in Beauville's book on surfaces, he uses Cech theory to relate rank two vector bundles on curves with ruled surfaces. To prove that all ruled surfaces arise from vector bundles he then uses Serre's result that Cech H^2 vanishes on a curve with coefficients in any sheaf coherent or not. (He also gives a second argument.) But this vanishing theorem for Cech cohomology does not follow from Hartshorne's treatment, since he proves vanishing for derived functor cohomology but does not relate derived functor and Cech cohomology on non coherent sheaves above degree one.
There is a sentence in Hartshorne, at the end of chapter III, section 2, page 212 in the 1977 edition, which says that Serre proved vanishing for "coherent sheaves on algebraic curves and projective algebraic varieties", whereas the correct statement would be that he proved it "for curves, and for coherent sheaves on projective algebraic varieties". Since Robin is very careful, one wonders whether some well intentioned copy editor did not change this sentence's meaning unwittingly to make it flow better.
| 18 | https://mathoverflow.net/users/9449 | 46099 | 29,192 |
https://mathoverflow.net/questions/46087 | 28 | The category of simplicial sets has a standard model structure, where the weak equivalences are those maps whose geometric realization is a weak homotopy equivalence, the cofibrations are monomorphisms, and the fibrations are Kan fibrations.
Simplicial sets are combinatorial objects, so morally their model structure should not be dependent on topological spaces. Are there any approaches to this model structure which do not use the geometric realization functor, and do not use topological spaces?
| https://mathoverflow.net/users/1709 | Model structure on Simplicial Sets without using topological spaces | Quillen's original proof (in *Homotopical Algebra*, LNM 43, Springer, 1967) is purely combinatorial (i.e. does not use topological spaces): he uses the theory of minimal Kan fibrations, the fact that the latter are fiber bundles, as well as the fact that the classifying space of a simplicial group is a Kan complex. This proof has been rewritten several times in the literature: at the end of
S.I. Gelfand and Yu. I. Manin, *Methods of Homological Algebra*, Springer, 1996
as well as in
A. Joyal and M. Tierney [An introduction to simplicial homotopy theory](http://hopf.math.purdue.edu/cgi-bin/generate?/Joyal-Tierney/JT-chap-01)
(I like Joyal and Tierney's reformulation a lot). However, Quillen wrote in his seminal Lecture Notes that he knew another proof of the existence of the model structure on simplicial sets, using Kan's $Ex^\infty$ functor (but does not give any more hints).
A proof (in fact two variants of it) using Kan's $Ex^\infty$ functor is given in my Astérisque 308: the fun part is not that much about the existence of model structure, but to prove that the fibrations are precisely the Kan fibrations (and also to prove all the good properties of $Ex^\infty$ without using topological spaces); for two different proofs of this fact using $Ex^\infty$, see Prop. 2.1.41 as well as Scholium 2.3.21 for an alternative). For the rest, everything was already in the book of Gabriel and Zisman, for instance.
Finally, I would even add that, in Quillen's original paper, the model structure on topological spaces in obtained by transfer from the model structure on simplicial sets. And that is indeed a rather natural way to proceed.
| 30 | https://mathoverflow.net/users/1017 | 46112 | 29,200 |
https://mathoverflow.net/questions/46061 | 5 | I'm searching for a translation for the term "intertwiner" in German.
| https://mathoverflow.net/users/10400 | What is the German translation for intertwiner? | I (native speaker) learned the term "Vertauschungsoperator" in my undergraduate courses. Unfortunately I cannot cite any reference right now, except the fact that the lecturer of the courses, Prof H.S. Holdgruen, is very sensible in his use of the german language. Therefore I estimate the probability very high that this is a common term in german texts on representation theory.
| 11 | https://mathoverflow.net/users/43085 | 46118 | 29,203 |
https://mathoverflow.net/questions/46104 | 1 | Consider an entire function $f : \mathbb{C} \rightarrow \mathbb{C}$! We search the function
$$ g: (a,b) \rightarrow \mathbb{C},$$ which solves the following equation locally: $g'(t)=f(g(t))$ and $g(0)=f(x\_0)$.
I can compute the inverse $G$ of $g$, if $f(x\_0) \neq 0$, i.e.
$$ G(y) = \int\limits\_{f(x\_0)}^y \frac{d s}{f(s)}.$$
I also known how to compute the Taylor expansion recursively, whose radius of convergence is positive (see below). Also we can give suitable approximations of the solution in terms of Picard iterations. I am not interested in such a solution!
Is there an alternative to this integral expression?
| https://mathoverflow.net/users/10400 | A simple ordinary differential equation | It is hard to guess what you are looking for. Take the apparently simpler case where $f$ is a polynomial, say of degree $d$. If $d = 1$ you have an explicit solution in terms of the exponential function (because your $G$ is logarithmic). If $d = 2$ the solution can be written in terms of trigonometric functions. If $d = 3$ you need elliptic functions to express the solution explicitly. As soon as $d$ is greater than $3$, I don't know of any standard naming for the functions you get or any interesting theory of these functions.
| 2 | https://mathoverflow.net/users/7311 | 46119 | 29,204 |
https://mathoverflow.net/questions/46106 | 9 | Let $\mathfrak{M}\_g$ denote the moduli stack of Riemann surfaces of genus $g$, it is a smooth complex analytic stack, and is the analytic stack underlying $\mathsf{M}\_g$, the moduli stack of complex algebraic curves of genus $g$. Alternatively, it is the quotient stack $\mathcal{T}\_g // \Gamma\_g$ of the mapping class group acting on Teichmuller space (by biholomorphisms).
There are two things we might mean by the Picard group of $\mathfrak{M}\_g$, using either holomorphic line bundles or algebraic line bundles. The first is $Pic\_{hol} := H^1(\mathfrak{M}\_g;\mathcal{O}^\times\_{\mathfrak{M}\_g})$ and the second is $Pic\_{alg} := H^1(\mathsf{M}\_g;\mathcal{O}^\times\_{\mathsf{M}\_g})$. As $\mathfrak{M}\_g$ has a finite cover by the quasiprojective variety $\mathfrak{M}\_g[3]$ of curves with a level 3 structure, I believe the latter group can also be defined to be the subgroup of $Pic\_{hol}$ of those holomorphic line bundles which become algebraic on $\mathfrak{M}\_g[3]$. If not, this defines yet another notion of Picard group.
It seems to me that it is $Pic\_{alg}$ which is usually discussed, for example in Mumford's paper on Picard groups of moduli problems. Is $Pic\_{hol}$ discussed explicitly anywhere? On the other hand, presumably these groups agree (certainly $Pic\_{alg} = \mathbb{Z}$ is a summand of $Pic\_{hol}$). Is this explicitly stated and proved anywhere?
Any other observations about this distinction are welcome too.
EDIT: As a related question, suppose that one only knew about the analytic object $\mathfrak{M}\_g$. Can one directly show that $c\_1 : Pic\_{hol} \to H^2(\mathfrak{M}\_g;\mathbb{Z})$ is injective? (Granted $H^1(\mathfrak{M}\_g;\mathbb{Q})=0$, say.)
| https://mathoverflow.net/users/318 | Picard group of $\mathfrak{M}_g$ | This will be a bit sketchy, but hopefully you can fill in the necessary steps.
If you see a problem, let me know.
To avoid getting distracted with stacks, let me use a level $n\ge 3$ structure, and
write $M=M\_{g}[n]$. Let $j:M\hookrightarrow \bar M$ be the Satake (not Deligne-Mumford) compactification.
This is the normalization of the closure of the image of $M$ in the Satake compactification
of $A\_{g}[n]$. The point is that $codim (\bar M-M)\ge 2$. So given a holomorphic line
bundle $L$ on $M$, $j\_\*L$ would be a coherent analytic sheaf. Therefore by GAGA
$j\_\*L$ is coherent algebraic, so that $L$ is an algebraic line bundle. This
implies surjectivity of map $Pic\_{alg}\to Pic\_{hol}$ in your notation.
This should finish it, since as you said,
you already know injectivity of the above map.
| 9 | https://mathoverflow.net/users/4144 | 46128 | 29,209 |
https://mathoverflow.net/questions/46124 | 3 | Let $H$ be an infinite-dimensional, separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$.
Let $H^\*$ denote the space of continuous linear functionals on $H$. By the Riesz representation theorem, $$H^\* = \{ \langle k, \cdot \rangle : k \in H \}.$$ By the assumption that $\gamma$ has covariance operator $I$, for all $k \in H$, $$\int\_H |\langle k, h \rangle|^2 \, \mathrm{d}\gamma(h) = \langle k, k \rangle < \infty,$$ so $H^\*$ is contained in the [separable](https://mathoverflow.net/questions/46094/if-h-is-a-separable-hilbert-space-is-l2h-separable) Hilbert space $L^2(H)$.
**My question:** Is $H^\*$ dense in $L^2(H)$?
| https://mathoverflow.net/users/238 | If $H$ is a separable Hilbert space, is its dual dense in $L^2(H)$? | (Rewritten to give an answer more useful to future visitors.)
First of all, as noted in comments, there is no (countably additive) Gaussian measure on $H$ with covariance operator the identity.
However, if we take $\gamma$ to be some other Gaussian measure, the answer is no, $H^\*$ is not dense in $L^2(H, \gamma)$. One way to see this is that every $f \in H^\*$, considered as a random variable on $(H, \gamma)$, has a centered Gaussian distribution (in which we include the "degenerate" Gaussian distribution which is the constant 0). In particular $f$ has mean zero, and the mean-zero random variables are a proper closed subspace of $L^2(\gamma)$. (So for instance, the nonzero constants are not in the closure of $H^\*$.)
Indeed, $L^2$ limit of centered Gaussian random variables is again centered Gaussian (this actually holds if you replace "$L^2$" by "in distribution"). Thus every random variable in the $L^2$ closure of $H^\*$ is centered Gaussian, and hence non-Gaussian $L^2$ functions on $H$ are not in the closure of $H^\*$ either.
It is true that if you allow *polynomials* $F(x) = p(f\_1(x), \dots, f\_n(x))$ where $f\_1, \dots, f\_n \in H^\*$, then all such functions $F$ are in $L^2(H,\gamma)$, and they form a dense subspace. Indeed, if you let $p$ range over all Hermite polynomials of degree $n$, then the closed span of all corresponding $F$ gives you the space $\mathcal{H}\_n$ which is the $n$th Wiener chaos, and we have an orthogonal decomposition $L^2(H, \gamma) = \bigoplus \mathcal{H}\_n$. These spaces are also the eigenspaces of the Ornstein-Uhlenbeck operator $N$ (aka number operator).
Back to the "measure" with covariance operator $I$: we can consider such a "measure" as a finitely additive measure on the cylinder sets of $H$. I suppose it might be possible to study an $L^2$ space with respect to this finitely additive measure. I don't know much about such spaces, but I would guess that a similar argument would show that $H^\*$ is still not dense.
| 7 | https://mathoverflow.net/users/4832 | 46135 | 29,213 |
https://mathoverflow.net/questions/46133 | 12 | This question was asked on NMBRTHRY by Kurt Foster:
If $p$ is a prime number and $\mathbb{F}\_p$ the field of $p$ elements, the zeroes of the Artin-Schreier polynomial
$x^p - x - 1 \in \mathbb{F}\_p[x]$
obviously have multiplicative order dividing $1 + p + p^2 + \dots + p^{p-1} = (p^p - 1)/(p-1)$ (express the norm as the product of the compositional powers of the Frobenius map)
Once upon a time, long long ago, I read that it had been conjectured (by Shafarevich IIRC) that this is the *exact* multiplicative order for every prime $p$. Can anyone supply a reference?
| https://mathoverflow.net/users/2784 | Multiplicative order of zeros of the Artin-Schreier Polynomial | I've never seen it ascribed to Shafarevich, but it is an old question. As a question, it equivalent to determining the period mod p of the sequence of Bell numbers discussed, e.g. in:
Levine, Jack; Dalton, R. E.
Minimum periods, modulo p, of first-order Bell exponential integers.
Math. Comp. 16 1962 416–423.
But they refer to even older papers. Any conjecture is wishful thinking since we can't get past $p=29$ or so with current technology. (Edit: As Kevin points out in the comments, I am seriously underestimating current technology, so this comment applies only to last century.)
Incidentally, I proved that the order is at least $2^{2.54p}$ in JTNB 16 (2004) 233-239.
| 10 | https://mathoverflow.net/users/2290 | 46137 | 29,214 |
https://mathoverflow.net/questions/46136 | 3 | An object $M$ of an abelian category is called of finite type iff for every directed set of subobjects $M\_i$ of $M$ whose sum is $M$ there exists some $i$ with $M = M\_i$. Is the direct sum $M \oplus N$ of two objects $M,N$ of finite type again of finite type?
So let $P\_i \subseteq M \oplus N$ be a directed set of subobjects whose sum is $M \oplus N$. If $M\_i$ denotes the projection of $P\_i$ to $M$ and $N\_i$ the one to $N$, then it is easy to see that there is some $i$ with $M\_i = M$ and $N = N\_i$. But does not show yet $P\_i = M \oplus N$!
More general: If $0 \to M' \to M \to M'' \to 0$ is an exact sequence, where $M',M''$ are of finite type, does this imply that $M$ is of finite type?
If there are counterexamples: Is it at least true in a Grothendieck-category? What are other reasonable definitions for "finitely generated" which generalize the ones for modules over rings or quasi-coherent modules over nice schemes?
| https://mathoverflow.net/users/2841 | $M \oplus N$ is of finite type if $M,N$ are of finite type? | At least if we have a Grothendieck category everything seems OK: Suppose
$0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ is exact with $M'$ and
$M''$ of finite type. Assume $\{M\_i\}$ is a directed collection of subobjects of
$M$ such that $\sum\_i M\_i=M$. We then have $M'=M'\bigcap\sum\_i M\_i=\sum\_i
M'\bigcap M\_i$ and hence $M'=M'\bigcap M\_{i\_0}$ for some $i\_0$. Throwing away
all indices which are not $\geq i\_0$ we may assume $M'\subseteq M\_i$ for all
$i$. We then get that $M''=\sum\_i M\_i/M'$ and hence $M''=M\_{i\_1}/M'$ for some
$i\_1$
**Addendum**: Stealing some ideas from Sándor's reply we can get the
statement without extra axioms. Note that finite generation is formulated in
terms of $\sum\_iM\_i=M$ which is the same as $\mathrm{lim}M\_i\to M$ being surjective
(as the sum is image of the limit). Now, with notations as before we put $M''\_i$
to be the image of $M\_i$ in $M''$. As $\mathrm{lim}M\_i\to M$ is surjective we get
that so is $\mathrm{lim}M\_i''\to M''$ and hence $M''\_i=M''$ for some $i$ and after
throwing away we can assume this is always true. This means that we get an exact
sequence $0\to M'\_i\to M'\to M/M\_i\to0$ and as $\mathrm{lim}M/M\_i=0$ (by right exactness of
directed colimits) we get that $\mathrm{lim}M'\_i\to M'$ is surjective (again by
right exactness) and hence that $M'\_i=M'$ for some $i$ but then $M\_i=M$ as $M''\_i=M''$,
which means that $M=M\_{i}$.
| 3 | https://mathoverflow.net/users/4008 | 46148 | 29,217 |
https://mathoverflow.net/questions/46155 | 3 | Let $X$ be an n point finite set equipped with a metric $d$. An isometry is a map $\varphi$ $:X \mapsto X$ satisfying that $d(\varphi(x),\varphi(y))=d(x,y)$ for any $x,y \in X$. Identify $X$ with the set {$1,2,...,n $}. We can say every isometry is an element of $S\_n$(the permutation group of n elements) and the isometry group of $X$ (denote it by $Iso(X,d)$) is a subgroup of $S\_n$. The question is the following: if there is no isometry moving $i$ to $j$,what can we say about $Iso(X,d)$ ?
Are there any books about this literature? Thanks.
| https://mathoverflow.net/users/7360 | A question about the isometry group of a finite metric space | As secretman says, the condition that there exist $i$ and $j$ in $X$ such that no isometry carries $i$ to $j$ is precisely to say that the action of the isometry group on $X$ is not *transitive*. If that's really your question, that seems to be all that can be said, and I doubt there any books on the subject.
One might simply ask: what can be said about isometry groups of finite metric spaces? This is a rather broad question, but nevertheless it seems interesting. Here is what I was able to come up with via a short amount of thought and googling:
1. Every finite group $G$ occurs up to abstract group isomorphism as the full isometry group of a finite metric space. Indeed, in [this 1976 paper](http://alpha.math.uga.edu/%7Epete/Asimov76.pdf), D. Asimov proved that if $G$ is finite of cardinality $k$, there exists a finite subset $X\_G$ of Euclidean $k-1$-space (with the induced metric), of cardinality $k^2-k$, such that the full isometry group of $X\_G$ is isomorphic to $G$.
2. On the other hand, it is not true that every permutation group is the isometry group of a finite space: i.e., if $X = \{1,\ldots,n\}$ and $G$ is subgroup of $S\_n$, then there need not be a metric $\rho$ on $X$ such that $G = \operatorname{Aut}(X,\rho)$. For a simple example, take $n = 3$ and let $G$ be the subgroup of order $3$, i.e., $G$ is generated by $\sigma = (123)$. Since $\sigma$ is an isometry,
$\rho(1,2) = \rho(\sigma(1),\sigma(2)) = \rho(2,3) = \rho(\sigma(2),\sigma(3)) = \rho(1,3)$.
Thus all pairwise distances between the three elements of $X$ are equal and the isometry group is the full $S\_3$.
More generally, let $\rho$ be a metric on $X$ and $G \subset \operatorname{Sym}(X)$ the
isometry group. Let $X^{(2)}$ be the set of (unordered!) two-element subsets of $X$: there is a
natural action of $\operatorname{Sym}(X)$ -- and hence also of $G$ -- on $X^{(2)}$. Now the observation here is that if $G$ acts transitively on $X^{(2)}$ -- e.g. if $G$ is *doubly transitive* as a permutation group on $X$, but this condition is weaker -- then that means that all pairs of points in $X$ have the same distance, and therefore $G = \operatorname{Sym}(X)$.
~~It strikes me that one could take this construction a step further: we may define a graph $G$ with vertex set $X^{(2)}$ by decreeing two vertices $\{x\_1,x\_2\}$, $\{y\_1,y\_2\}$ to be adjacent iff $\rho(x\_1,x\_2) = \rho(y\_1,y\_2)$. Then $G$ is precisely the group of graph automorphisms of $G$. This is sort of a "combinatorialization" of the problem, although I don't know whether it's actually useful for anything.~~ This is false, as Aleksander Horawa has pointed out to me. I'm not sure what I was thinking when I wrote this. As he points out, it is at least true that when $\# X \geq 3$, the induced homomorphism from the isometry group of $X$ to the automorphism of the graph is injective (even this fails when $\# X = 2$).
| 4 | https://mathoverflow.net/users/1149 | 46169 | 29,226 |
https://mathoverflow.net/questions/46156 | 22 | Let $F\subset\mathbb{Q}^2$ a closed set. Does there exists some closed and connected set $G\subset\mathbb{R}^2$ such that $F=G\cap\mathbb{Q}^2$?
For example if $F=\{a,b\}$, you can take $G$ the reunion of two lines of different irrational slopes passing through $a$ and $b$. This is a connected set and the intersection with $\mathbb{Q}^2$ is $\{a,b\}$ because the slopes are irrationnals.
But I don’t know how to prove it in general (and I don’t know if it’s true). When there are many connected components this is not clear how to connect them without adding new rational points.
| https://mathoverflow.net/users/10217 | Is every closed set of Q² the intersection of some connected closed set of R² with Q² | Enumerate all rational points outside your set. Then cover these points by open balls by induction as follows: the next ball is centered at the first rational point not covered so far, its radius is so small that is does not intersect $F$ and the previous balls and is chosen so that the boundary of the ball does not contain rational points. Then the complement of the union of these balls is path-connected: to connect two points, draw a segment between them and go around every ball intersected by this segment.
Note that this works for any countable set, not just $\mathbb Q^2$.
| 45 | https://mathoverflow.net/users/4354 | 46174 | 29,229 |
https://mathoverflow.net/questions/46116 | 18 | I have been reading "The Geometry of Schemes" by Eisenbud and Harris and have a question about Exercise III-43. There, one should show that there is a bijection between the sets
$\{(n+1)\mbox{-tuples of elements of }A\mbox{ that generate the unit ideal }\}$
and
$\{ \mbox{maps} \mbox{ Spec} A \to \mathbb{P}^n\_A$ such that the composite $\mbox{Spec} A \to \mathbb{P}^n\_A\to \mbox{Spec}A=id\} $,
i.e. $A$-valued points of $\mathbb{P}^n\_A$.
Now, of course, $(n+1)$-tuples of elements of $A$ give $A$-valued points, but if $A$ is not a ring such that every invertible $A$-module is free of rank one, I don't see why the converse should work:
Let us take, e.g. a number field $K$ such that $A=\mathcal{O}\_K$ is not a PID. Then, up to multiplication by a unit, an $A$-valued point corresponds to an invertible $A$-module $P$ and an epimorphism $A^{n+1}\to P$ by the characterization of morphisms from $\mbox{Spec}A$ to $\mathbb{P}^n\_{\mathbb{Z}}$ (Corollary III/42 in Eisenbud+Harris).
Starting with an $(n+1)$-tuple generating $A$, I clearly get an epimorphism $A^{n+1}\to A$ and $A$ is a projective $A$-module, so I get an $A$-valued point.
However, if I start with an $A$-valued point corresponding to an epimorphism $A^{n+1}\to P$ and the invertible module $P$ is not free, how can I choose an $(n+1)$-tuple of points of $A$ which generate the unit ideal? Moreover, don't these $A$-valued points give "additional" points, which do not come from $(n+1)$-tuples of elements of $A$?
Most books just consider the case when $A$ is a field, there everything works just fine.
Thanks!
| https://mathoverflow.net/users/10844 | A-valued points of projective space | Yes, you (and BCnrd) are absolutely correct and the quoted statement is wrong.
Over any scheme $S$, the $S$-points of $\mathbb P^n$ are the surjections $\mathcal O\_S^{\oplus n+1} \to F$ with invertible $\mathcal O\_S$-module $F$. More generally, the $S$-points of the grassmannian $Gr(m,k)$ are the surjections $\mathcal O\_S^{\oplus m}\to F$ with $F$ locally free of rank $k$.
Note: no Noetherian assumptions on $S$ are necessary. This is the first step for Grothendieck's construction of Hilbert schemes, without Noetherian assumption.
So, for a ring $A$, the $A$-points of $\mathbb P^n$ are the surjections $A^{n+1}\to P$ with $P$ locally free (equivalently, projective) $A$-modules of rank 1.
| 12 | https://mathoverflow.net/users/1784 | 46175 | 29,230 |
https://mathoverflow.net/questions/46176 | 10 | I'm wondering if finite unramified morphism between reduced schemes decomposes as closed immersions and etale morphisms. Suppose I have a morphism between reduced schemes which is finite, surjective and unramified, is it necessarily etale? I think this is certainly true if both source and target are curves, but I'm not sure about higher dimensional examples. Thanks
EDIT: to avoid trivial example let's assume the source and target are connected. What I'm wondering is precisely when one can deduce flatness from these conditions.
| https://mathoverflow.net/users/1657 | What are unramified morphisms like? | Finite, surjective, and unramified does not imply etale. E.g. suppose that $Y$ is a proper closed subscheme of $X$, and we consider the map $X \coprod Y \to X$ defined as the disjoint union of the identity on $X$, and the given closed immersion $Y \to X$
on $Y$.
Then this map is finite, unramified, and surjective, but not etale. (See Sandor's answer for the missing condition,
which is flatness!)
Added: A more interesting example is given by letting $X$ be a nodal cubic, letting
$\tilde{X}$ be the normalization, and considering the natural map $\tilde{X} \to X.$
This map is not flat and certainly not etale, but it is unramified. (Formally, each branch
through the node maps by a closed immersion into the nodal curve.)
| 15 | https://mathoverflow.net/users/2874 | 46179 | 29,232 |
https://mathoverflow.net/questions/46181 | 3 | The following is a corollary of the Briançon-Skoda theorem:
If $R$ is a regular Noetherian ring of Krull dimension $d$ and $f\_1,f\_2,...,f\_{d+1}\in R$. Then, $f\_1^df\_2^d...f\_{d+1}^d \in (f\_1^{d+1},f\_2^{d+1},...,f\_{d+1}^{d+1})R$
Now, given $R=k[[x\_1,...,x\_d]]$ where $k$ is a field, then $R$ is a regular local ring, so the corollary applies. I was wondering if there is direct proof of the above corollary for the case of the power series ring. I remember reading that there is a direct proof for the case when $f\_1,...,f\_{d+1}$ are polynomials. Can anyone point to a reference for this.
| https://mathoverflow.net/users/10775 | About a corollary of the Briançon-Skoda theorem | It depends on what do you mean by "direct". There is no elementary proof, as far as I known, even for polynomials when $d=2$, see page 8 of this [note](http://www.math.lsa.umich.edu/%7Ehochster/balt.ps). When $d=1$, the statement is an easy exercise: one can write $f\_1= da, f\_2=db$ with $(a,b) = R$ since $R$ is an UFD.
There are a few proofs of this very interesting theorem, analytic (the original one, over complex numbers), using duality theory ([Lipman-Sathaye](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-28/issue-2/Jacobian-ideals-and-a-theorem-of-Brianc-con-Skoda/10.1307/mmj/1029002510.full), for all regular rings) and reduction to characteristic $p$ + tight closure (Hochster-Huneke, when $R$ contains a field, see the note in the first paragraph for some exposition of this method) but I am not sure any of them can be called direct. Any new proof will be very exciting!
| 3 | https://mathoverflow.net/users/2083 | 46190 | 29,238 |
https://mathoverflow.net/questions/46196 | 3 | I just recently learned the Ham Sandwich Theorem in my algebraic topology class. If we take the measure to be the counting measure and let $n=2$, then the theorem tells us that given a set of black and white points in the plane, we can draw a line that'll divide the plane so that there is an equal number of black and white points on either side of the line. But this theorem is existential only. Is there an algorithm for actually computing the line for this discrete case? If so, what's the complexity?
| https://mathoverflow.net/users/8162 | Algorithm for Ham Sandwich with Points | The paper by
Lo, Matoušek, and Steiger entitled *"Algorithms for Ham-Sandwich Cuts"* gives an $O(n)$ algorithm, where $n$ is the number of points. That's the best you can do, since you need to consider all such points.
| 3 | https://mathoverflow.net/users/491 | 46200 | 29,242 |
https://mathoverflow.net/questions/46183 | 0 | Suppose X is a pure dimensional projective complex scheme, reducible and non-reduced but without embedded components of lower dimension. Let $X=\cup X\_i$ be the decomposition such that $X\_i$ is set-theoretically irreducible and $\dim(X\_i\cap X\_j)<\dim(X\_i)$.
Let $F\_X$ be a torsion free sheaf of $\mathcal{O}\_X$-modules. If $F$ is generically locally free on each $X\_i$ then one can define its multi-rank. What to do if $F|\_{X\_i}$ is not generically locally free?
For example let $X:=(x^p=0)\subset\mathbf P^2$ be the multiple line. Let $F\_j$ be the $\mathcal{O}\_X$ module generated by $x^j$. For $0 < j < p$ it is not locally free at any point.
My guess: let $pt\in X\_i$ be generic enough closed point. Let $p\_i$ be the generic multiplicity of $X\_i$. At the point $pt\in X$ consider the intersection of $X\_i$ with the generic plane of the complementary dimension. This gives the sub-scheme $Y\subset X $ supported on $pt$. Take the fraction $\frac{length(F\otimes\mathcal{O}\_Y)}{p\_i}$ as the rank on $X\_i$.
In the previous example this gives the rank of $F\_j$: $\frac{p-j}{p}$.
Is this the commonly accepted definition? References?
| https://mathoverflow.net/users/2900 | How the multi-rank of a torsion free sheaf on a non-reduced scheme is defined? | I don't know about commonly accepted, but here is one definition of rank (not multi-rank) that appears in theory of [stable sheaves](http://www.numdam.org/item?id=PMIHES_1994__79__47_0). Fix an ample line bundle $L$ and consider the Hilbert polynomial $P\_{L}(F, t)$, defined by
$$
P\_{L}(F, n) = \chi(F \otimes L^{\otimes n}).
$$
The leading term of $P\_{F, L}(t)$ is of the form $ \frac{r}{d!} t^d$, where $d$ is the dimension of the support. The number $r$ is called the ``rank" (with respect to $L$).
You can try to define multi-rank by doing this component-by-component.
**Added:** Here is an example. Consider the projective plane $\mathbb{P}^2\_{\mathbb{C}}$ with coordinates $X, Y, Z$. A closed subscheme $C$ is defined by $X^p=0$, and the hyperplane class defines a very ample line bundle $\mathcal{L}$ on $C$. What are $\mathcal{L}$-ranks of some sheaves on $C$?
The Hilbert polynomial of the structure sheaf is:
$$
P\_{\mathcal{L}}(\mathcal{O}\_{C}, T) = p T + 1 - \frac{(p-1)(p-2)}{2}.
$$
In particular, the $\mathcal{L}$-rank is not $p$ and not $1$! More generally, you can check that every line bundle on $X$ has $\mathcal{L}$-rank $p$. This suggests that the $\mathcal{L}$-rank takes into account both the degree of the polarization and some information about the generators of $F$.
Now let's consider a sheaf that is generically non-zero, but also generically non-free. If $i \colon \mathbb{P}^1 \to C$ is the inclusion of the reduced subscheme, then
$$I := i\_{\*}(\mathcal{O}\_{\mathbb{P}^1})$$
has this property. An application of the projection formula shows that the Hilbert polynomial of this function is:
$$
P\_{\mathcal{L}}(I, T) = T + 1.
$$
In particular, the $\mathcal{L}$-rank is $1$!. The ratio of the $\mathcal{L}$-rank and the degree of $\mathcal{L}$ is equal to $1/p$. This looks closer to the number you are expecting.
| 2 | https://mathoverflow.net/users/5337 | 46201 | 29,243 |
https://mathoverflow.net/questions/46149 | 10 | Following along a similar line to the question asked here: [Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V?](https://mathoverflow.net/questions/38082/is-there-an-explicit-bound-on-the-number-of-tetrahedra-needed-to-triangulate-a-hy)
Let $K$ be a (hyperbolic) knot in $S^3$. Let $n$ be the minimal number of crossings of any diagram of $K$ and let $M = S^3 \backslash K$ be its complement. By Moise’s theorem the 3-manifold $M$ can be triangulated by tetrahedra. But is there any known bound on the number of tetrahedra needed to triangulate $M$ as a function of $n$? I am particularly interested in any known lower bounds, but upper bounds would also be interesting.
| https://mathoverflow.net/users/3121 | Lower bound on number of tetrahedra needed to triangulate a knot complement | **Revision**
Let $t(K)$ be the minimal number of tetrahedra needed to triangulate a knot complement. Let $c(K)$ be the minimal crossing number of a knot.
As Ryan Budney points out in the comments, $t(K)\leq C c(K)$ for some constant $C$. One may show that this lower bound is optimal, indirectly using a result of Lackenby.
For alternating knots, one obtains a linear lower bound on the hyperbolic volume of the knot complement, and therefore on the number of tetrahedra needed to triangulate the complement (as noted in Thurston's answer to the other question), in terms of the *twist number* of the alternating knot, by [work of Lackenby](http://www.ams.org/mathscinet-getitem?mr=2018964). One may find hyperbolic alternating knots $K\_i$ where the twist number is equal to the crossing number $c(K\_i)$, and thus $ c(K\_i)\leq c\_1 Vol(S^3-K\_i) \leq c\_2 t(K\_i)$ for some constants $c\_1, c\_2$. This shows that the linear lower bound is optimal.
For the other direction,
of course, there is some sort of upper bound, since there are only finitely many knot complements with a triangulation by $\leq n$ tetrahedra, so just take the one with the maximal crossing number! We may attempt to make this relation more explicit, using a result of [Simon King](http://www.ams.org/mathscinet-getitem?mr=1983086). I think I can show that $c(K)\leq e^{p(t(K))}$, for some polynomial $p(n)$.
Inverting this inequality gives a lower bound on $t(K)$ in terms of $c(K)$.
King shows that if one has a triangulation $\tau$ of $S^3$ with $m$ tetrahedra, and a knot $K$ in the 1-skeleton of $\tau$, then the crossing number $c(K)$ is bounded by $C^{m^2}$ for some $C$.
Given a triangulation of the knot complement, one wants to estimate the crossing number. To apply King's result, one must extend the triangulation of $S^3-K$ to a triangulation of $S^3$. The difficulty is that the meridian of the knot may be a very complicated curve in the triangulation of the torus boundary of the knot complement. Work of [Jaco and Sedgwick](http://www.ams.org/mathscinet-getitem?mr=1958532) allow one to (in principle) estimate the combinatorial length of the meridian in the triangulation of the boundary torus. Their work also allows one to extend the triangulated knot complement along a solid torus to get a triangulation of $S^3$ with the knot in the 1-skeleton. I think one could obtain from their work an estimate on the number of tetrahedra of $\tau$ which is polynomial in the number of tetrahedra $t(K)$. Together with Simon King's result, this should give a bound $c(K)\leq e^{p(t(K))}$ ($p(n)$ a polynomial) on the crossing number in terms of the number of tetrahedra. I expect the answer to be exponential though.
Exponential lower bounds are realized, for example, by the torus knots. The crossing number is bounded linearly below by the genus. One may find sequences of torus knot complements triangulated by $n$ tetrahedra, but with genus growing exponential in $n$, and therefore crossing number growing exponentially. Estimates of the number of tetrahedra needed to triangulate Seifert fibered spaces were given by Martelli and Petronio, in *[Complexity of geometric three-manifolds](https://arxiv.org/abs/math/0303249)*. The point is that you can get a triangulation of a $(p,q)$ torus knot with the number of tetrahedra growing like the continued fraction expansion of $p/q$, which can be like $log(|p|+|q|)$, but the genus is $(p-1)(q-1)/2$.
| 13 | https://mathoverflow.net/users/1345 | 46204 | 29,245 |
https://mathoverflow.net/questions/46143 | 10 | We approach the problem of finding a metric of constant curvature on a surface (i.e. a $C^\infty$ 2-manifold). Specifically, what we want to do is, given a surface $M$ and a metric $g\_0$, show that there exists a new metric $g$ of the form $g=e^{2u}g\_0$ for some $u\in C^\infty (M)$ such that $g$ has constant curvature. It is well known that if we let $K\_0$ and $K$ be the curvatures associated to $g\_0$ and $g$ respectively, then they must satisfy the curvature equation
\begin{equation}
K=(K\_0-\triangle u)e^{-2u}
\end{equation}
or equivalently,
\begin{equation}
\triangle u-K\_0+Ke^{2u}=0
\end{equation}
where $\triangle$ stands for the Laplace-Beltrami operator associated to the original metric $g\_0$. Thus, solving our problem amounts to solving the curvature equation (i.e. finding $u$) for different values of $K$.
We have essentialy three cases (thanks to the Gauss-Bonnet theorem), which are $K=0$, $K=-1$, and $K=1$. The first two cases are fairly straightforward (simple solutions can be found in an article by Melvin S. Berger 1971, or the book on PDE's by Taylor), but the case when $K=1$ is more challenging. I have read solutions which use the Riemann-Roch theorem on one hand or Ricci flow on the other.
The question is: is there any other way to find a solution to the curvature equation (in the case $K=1$)? The reason I want to know this is because I am writing my thesis and I would like to give the simplest proof of this result to make it more accessible. Thanks!
| https://mathoverflow.net/users/10850 | Finding constant curvature metrics on surfaces for the case of positive Euler characteristic | If you like the case $K=-1$ better, one way to do this is to choose 3 points $\{x,y,z\} \subset S^2$, and use the uniformization theorem to find a complete conformally equivalent metric on $P=S^2-\{x,y,z\}$ with constant curvature $K=-1$. There is a unique such metric on $P$, which is conformally equivalent to $\mathbb{CP}^1-\{0,1,\infty\}$. Then fill in the punctures to get a conformally equivalent metric on $S^2$. One way to understand why uniformization of $S^2$ is a bit harder than the other cases is that there are Mobius transformations of $S^2$ which are conformal transformations but not isometries, so there is not a unique metric with $K=1$. By choosing three points, though, one gets rid of these conformal symmetries.
**Addendum:** Incidentally, the first proof that the Ricci flow on $S^2$ converges to the round metric was due to Bennett Chow, using a type of entropy defined specially on $S^2$ (Chow finished off a case not resolved by work of Hamilton). I think Perelman's work gives a new proof in the case of $S^2$, which I'm sure he realized, but I'm not sure has been properly disseminated. The idea is that if a singularity forms in finite time for Ricci flow on $S^2$, then one may take a rescaled limit to get a $\kappa$-non-collapsed positive curvature ancient solution (Perelman proof of this works in arbitrary dimensions, so is not special to $S^2$). In two dimensions, the only such solutions are solitons, which are either Hamilton's cigar or $S^2$ with the round metric (plus some non-orientable examples). But the cigar is collapsed, so the only possibility is $S^2$, which implies that the metric converges at the singular time to the round metric on $S^2$.
| 8 | https://mathoverflow.net/users/1345 | 46205 | 29,246 |
https://mathoverflow.net/questions/46138 | 16 | Every book which treats dual spaces of normend spaces states that $(c\_0)' = \ell^1$ and $(\ell^1)' = \ell^\infty$ and some also describe $(\ell^\infty)'$.
However, is anything known about higher order duals in general? Does taking the normed dual of a given normed space $X$ stabilize? In other words: Defining $X^{(1)} = X'$ and
$X^{(n)} = (X^{(n-1)})'$. Is there always an integer $n$ such that $X^{(n+1)} = X^{(n)}$?
| https://mathoverflow.net/users/9652 | Does "taking the dual space" stabilize? | As noted previously, we can restrict our attention to Banach spaces.
A Banach space $X$ is reflexive if the canonical embedding into its double-dual is onto.
Now let $X$ be any Banach space. We define $X^{(\alpha)}$ for all ordinals $\alpha$ as follows:
$X^{(0)}=X$, $X^{(\alpha+1)}=(X^{(\alpha)})^\prime$, and for limit ordinals $\delta$ let $X^{(\delta)}$ be the direct limit of the $X^{(\alpha)}$, $\alpha<\delta$, $\alpha$ is even.
For a given space $\alpha$, we can ask whether $X^{(\alpha)}$ will ever become reflexive.
The key fact is that a closed subspace of a reflexive space is again reflexive.
(It was pointed out before that $X$ is reflexive iff its dual is. But for the following discussion we have to think about subspaces of reflexive spaces.)
This implies that if $X$ is not reflexive, then the sequence $X^{(\alpha)}$, $\alpha$ ordinal, will never stabilize, i.e., no $X^{(\alpha)}$ will be reflexive, like in the $\ell^\infty$ example above (which can be iterated through all the ordinals).
I find this fact rather striking. It was pointed out to me by either Dirk Werner or Ehrhard Behrends. I don't remember exactly who.
Now remarkably, this needs the axiom of choice (which comes from the use of the Hahn-Banach Theorem in the proof). I have convinced myself a while ago that if you don't have a nontrivial atomless finitely additive probability measure on $\mathbb N$ (which can happen if AC fails),
then the sequence for $X=c\_0$ is $c\_0$, $\ell^1$, $\ell^\infty$, $\ell^1$ and so on.
I.e., $\ell^1$ and $\ell^\infty$ are both reflexive in this situation and duals of each other.
This is because without the nontrivial measure, every functional on $\ell^\infty$ that vanishes on $c\_0$ vanishes on all of $\ell^\infty$.
If I remember correctly, there is some remark about this in Solovay's paper on the model of set theory in which all sets of reals are measurable. In this model there is no nontrivial
finite, finitely additive, atomless measure on the integers.
| 16 | https://mathoverflow.net/users/7743 | 46210 | 29,250 |
https://mathoverflow.net/questions/46207 | 7 | Let $X$ be a smooth projective variety defined over $\mathbb{C}$. In the context of the minimal model program it is often important to understand the geometry of the maps defined by the complete linear systems associated to powers of the canonical class $nK\_{X}$. I would like to understand some examples of how intricate these maps can be for small $n$.
For simplicity let's restrict to the case of algebraic surfaces. We assume that $X$ is of general type and minimal. A concrete question that I have in mind is the following. One expects that for large $n$ the complete linear systems become basepoint free, but how large is large? What explicit examples are known where the systems $nK\_{X}$ all have a basepoint for $n \leq N$ but become free when $n>N$ ? Is there some absolute bound on $N$, and if so for what surface is the bound saturated?
| https://mathoverflow.net/users/5124 | Basepoints in the canonical system of algebraic surfaces | In the general case when $n=\dim X$ is arbitrary
Kollár proves an analogue for minimal varieties of general type, although the bound is much worse than in the surface case.
In the situation at hand it implies that $\vert ((n+3)!)K\_X\vert$ is basepoint-free (actually one can write $(n+2)(n+2)!$ instead of $(n+3)!$, but this is shorter).
See:
Kollár, János [Effective base point freeness](https://link.springer.com/article/10.1007/BF01445123). Math. Ann. 296 (1993), no. 4, 595–605.
A closely related problem is *Fujita's conjecture*, which states that if $L$ is an ample divisor on $X$, then $|K\_X+mL|$ is basepoint-free if $m\geq \dim X+1$ and very ample if $m\geq \dim X +2$.
There are many related results, but the conjecture is still open.
Angehrn and Siu prove that if $L$ is ample, then $|mL+K\_X|$ is base point free if $m\geq \frac 12(n^2+n+2)$.
Angehrn, Urban; Siu, Yum Tong
Effective freeness and point separation for adjoint bundles.
Invent. Math. 122 (1995), no. 2, 291–308.
32J25 (14C20 32L10)
Helmke also has some related results:
Helmke, Stefan . The base point free theorem and the Fujita conjecture.
School on Vanishing Theorems and Effective Results in Algebraic
Geometry (Trieste, 2000),
215--248, ICTP Lect. Notes, 6, Abdus Salam Int. Cent. Theoret. Phys., Trieste, 2001.
Helmke, Stefan . On global generation of adjoint linear systems.
Math. Ann. 313 (1999), no. 4, 635--652.
Helmke, Stefan . On Fujita's conjecture.
Duke Math. J. 88 (1997), no. 2, 201--216.
| 8 | https://mathoverflow.net/users/10076 | 46213 | 29,251 |
https://mathoverflow.net/questions/41386 | 4 | Can anybody give a definition of the equalizer completion of a cartesian category?
Is the method to get more or less as the regular and exact completions in the way that are given in: <http://ncatlab.org/nlab/show/regular+and+exact+completions>?
How is in that case the behaviour of the forgetful functor FL-->FP (where *FL* are categories with finite limits and *FP* categories with finite products)?
| https://mathoverflow.net/users/3338 | Equalizer completion | There are general results about how to freely add limits or colimits to a category. They are formally dual, but people normally state the colimit variety because they involve the category of presheaves.
To freely add some class of colimits to a given category $C$, you form the closure of the representables in the presheaf category $[C^{op},\mathsf{Set}]$ under the given type of colimit.
If you want to do this in such a way that certain existing colimits are preserved, then you form the closure not in the presheaf category, but in the full subcategory of those presheaves which send the existing colimits in C into limits in Set.
So to freely complete a category $C$ with finite products to a category with finite limits, you should look at the Yoneda embedding of $C$ into the opposite $FP(C,Set)^{op}$ of the category $FP(C,\mathsf{Set})$ of finite-product-preserving functors from $C$ to $\mathsf{Set}$. Now take the closure of the representables under finite limits. That is, finite limits in $FP(C,\mathsf{Set})^{op}$, or finite colimits in $FP(C,\mathsf{Set})$. This then gives the value at $C$ of a left biadjoint to the forgetful
2-functor from the 2-category LEX of categories with finite limits to the 2-category FP of categories with finite products.
There is an explicit, syntactic description, due to Andy Pitts. It can be found, for example, in the paper
>
> M. Bunge and A. Carboni, The symmetric topos, Journal of Pure and Applied Algebra, 1995.
>
>
>
An object is a pair of maps $f,f':X \to X'$ equipped with a common retraction $r$. A morphism
from such a pair to $g,g':Y \to Y',s$ consists of an equivalence class, under a suitably defined equivalence relation, of pairs $(a,a')$ where $a:X \to Y$, $a':X'\to Y'$, and the evident diagrams commute.
| 12 | https://mathoverflow.net/users/10862 | 46214 | 29,252 |
https://mathoverflow.net/questions/46185 | 2 | The nodes of a surface are special cases of more general singularities. For example, the Cayley cubic has four nodes.
The full set of singularities of a surface can be characterized by finding all points where the partial derivatives are all zero. However, not all singularities are nodes. Some are cusps or other kinds of singularities. For example, the Cayley cubic has five singularities, four of which occur at the nodes of the internal elliptope and the fifth which occurs at zero and is apparently not a node.
Is there a simple way to check which singularities are surface nodes? Or, more interestingly, is there a way to compute the full set of nodes of a surface directly?
| https://mathoverflow.net/users/10859 | How can I compute the full set of nodes of a surface? | A node (as in Cayley's surface) is a double point with nondegenerate tangent cone.
To check whether a given point on a surface in A^3 is a node in this sense, change coordinates so that it is the origin and write the equation as 0=F\_2+F\_3+... with F\_i homogeneous of degree i. The point is a node iff F\_2 is irreducible.
I am assuming the base field is algebraically closed.
| 4 | https://mathoverflow.net/users/1939 | 46216 | 29,254 |
https://mathoverflow.net/questions/46217 | 3 | A colleague of mine recently asked me if this set family had a name (see definition of *this* below) . I didn't know the answer, so I thought I would consult the MO oracle.
Let $\mathcal{S}:=\{ S\_1, \dots, S\_k \}$ be a family of subsets of $[n]$. Consider the family $\mathcal{F}\_{\mathcal{S}}$ formed by taking all sets of the form
$
S\_1' \cap \dots \cap S\_k'
$
where each $S\_i'$ is either $S\_i$ or the complement of $S\_i$. Note that we are forced to intersect exactly $k$ such sets.
>
> Do set families arising in this way have a well-established name?
>
>
>
| https://mathoverflow.net/users/2233 | Does this set family have a name? | I am not sure how standard this is, but It makes sense to call this family the [atoms](http://en.wikipedia.org/wiki/Atom_(order_theory)) of the corresponding lattice of sets obtained from $\mathcal S$. They can be illustrated as the different regions in a Venn diagram.
| 4 | https://mathoverflow.net/users/2384 | 46218 | 29,255 |
https://mathoverflow.net/questions/46224 | 2 | Is it possible to find an example of an $\mathbb{R}$-Cartier divisor $D$ on an irreducible variety $X$ that is non-trivial, nef, effective and numerically rigid?
By "numerically rigid" I mean that if $E$ is another $\mathbb{R}$-Cartier effective divisor such that $E$ is numerically equivalent to $D$ then $D=E$.
For curves this clearly cannot be the case, since an effective non-trivial divisor is always ample.
| https://mathoverflow.net/users/7276 | Numerically rigid nef divisor | Take a minimal surface $S$ of general type with $p\_g=1$, $q=0$ and zero torsion.
Then $S$ contains a unique effective canonical curve $K$, which is nef and numerically rigid.
In fact, since $q=0$ and there is no torsion, we have $\textrm{Pic}^0(S)=0$, the Neron - Severi group $\textrm{NS}(S)$ coincides with the Picard group $\textrm{Pic}(S)$ and any two numerically equivalent divisors on $S$ are linearly equivalent.
Examples of these surfaces, with $K^2=2$, are given in the paper of Debarre and Catanese
"Surfaces with $K^2=2$, $p\_g=1$, $q=0$",
J. reine angew. Math. 395 (1989), 1-55.
| 3 | https://mathoverflow.net/users/7460 | 46225 | 29,257 |
https://mathoverflow.net/questions/46220 | 2 | Is there a classification of Sasaki or Sasaki-Einstein manifolds?
What are important examples?
| https://mathoverflow.net/users/7015 | Classification of Sasaki manifolds | The book [Sasakian Geometry](http://books.google.com/books?id=ERYZAQAAIAAJ) by Charles Boyer and the late Krzysztof Galicki contains a wealth of information on this topic. As far as I know there is no classification, but there are of course tons of examples.
| 4 | https://mathoverflow.net/users/394 | 46226 | 29,258 |
https://mathoverflow.net/questions/46202 | 3 | Today in my introductory algebraic geometry class we defined the so-called Rees algebra associated with an ideal $I$ of a ring $R$ (with strong conditions on $R$, if you like: I don't mind restricting to finitely generated reduced algebras $R$ over an algebraically closed field $k$). If we want to think of (maximal) Proj applied to the Rees algebra as Spec $R$ blown up along the vanishing set of $I$, what happens when $I$ is not radical?
In particular, if $R = k[X\_1,\cdots,X\_n]$ and $I = (X\_1,\cdots,X\_n)$ we get $\mathbb{A}^n$ blown up at the origin. Have people tried to interpret geometrically the Rees algebra of $I^m$ for $m > 1$, or even $I^2$? I asked my professor, and (in an unusual turn of events) he did not have an answer available off the top of his head.
| https://mathoverflow.net/users/3544 | Rees algebra for non-radical ideals | The blowup of $Spec R$ along $I$ and $I^m$ give isomorphic results. This is Hartshorne exercise II.7.11.a. In general, any birational projective morphism is realized as the blowing up of the target along some ideal sheaf, which in general will be quite complicated (e.g. not radical, but not a power of a radical ideal either).
| 5 | https://mathoverflow.net/users/397 | 46233 | 29,263 |
https://mathoverflow.net/questions/41756 | 4 | I'm playing with MATLAB's svd function to compute the svd of
```
[ 1 4 7 10
2 5 8 11
3 6 9 12 ]
```
When I type [U1, ~, ~] = svd(X), I get
```
U1 =
-0.5045 0.7608 0.4082
-0.5745 0.0571 -0.8165
-0.6445 -0.6465 0.4082
```
But when I compute the svd of the transpose of X with [~, ~, U2] = svd(X'), I get
```
U2 =
0.5045 0.7608 0.4082
0.5745 0.0571 -0.8165
0.6445 -0.6465 0.4082
```
The first singular vectors seem to be pointing to the opposite directions but the others are the same. I know that svd is not unique and the solution is correct because the first component of V1 and V2 are pointing opposite directions as well. But, I would expect MATLAB to return the same answers. I thought to add a postprocessing and check the singular vector pairs and turn their directions to make them consistent under transpose but I couldn't find a mathematically reasonable way of doing it.
Do you know why MATLAB (or LAPACK as it says MATLAB uses LAPACK) computes this way? Do you have any suggestions how to make it consistent? Thanks.
**Update:** the reason I ask for it is that I wanted to see what happens if I apply higher order singular value decomposition (HOSVD) to 2 dimensional matrices. In theory, it holds. Using the same mathematical notation, SVD is formulated as follows:
Every $\mathbf{X} \in \mathbb{C}^{I\_1 \times I\_2}$ can be written as the product
.$$\mathbf{X} = \mathbf{U}^{(1)} \cdot \mathbf{S} \cdot \mathbf{U}^{(2)^H} = \mathbf{S} \times\_1 \mathbf{U}^{(1)} \times\_2 \mathbf{U}^{(2)}$$
in which
$\mathbf{U}^{(1)} = \left[ \begin{array}{cccc} U\_{1}^{(1)} & U\_{2}^{(1)} & \dots & U\_{I\_1}^{(1)} \end{array} \right] \in \mathbb{C}^{I\_1\times I\_1}$ is unitary.
$\mathbf{U}^{(2)} = \left[ \begin{array}{cccc} U\_{1}^{(2)} & U\_{2}^{(2)} & \dots & U\_{I\_I}^{(2)} \end{array} \right] \in \mathbb{C}^{I\_2\times I\_2}$ is unitary.
$\mathbf{S} \in \mathbb{C}^{I\_1\times I\_2}$ has the the following properties:
* *pseudodiagonality:* $\newcommand{\diag}{\mathop{\mathrm{diag}}} \mathbf{S} = \diag\left(\ \sigma\_1, \sigma\_2, \dots, \sigma\_{\min(I\_1,I\_2)} \right)$
* *ordering:* $\sigma\_1 \geq \sigma\_2 \geq \ldots \geq \sigma\_{\min(I\_1, I\_2)} \geq 0$
So, to compute it I found the right and singular vectors as if they were computed independently because I want to extend it to multidimensions later. To compute them independently, I calculated the eigenvalues of $\mathbf{XX'}$ and $\mathbf{X'X}$ as follows:
```
[U{1}, ~] = eig(X*X');
U{1} = fliplr(U{1});
[U{2}, ~] = eig(X'*X);
U{2} = fliplr(U{2});
```
but they seem to be as I explained above. Actually, it doesn't effect what I need but I would like to make the singular values positive. This way, they might become negative since the right and left singular vectors are not as I expected.
**Update 2:** Alternatively, I used the code with svd() function as below and it gives similar results. With the 'econ' option, only min(nDim1,nDim2) singular vectors are returned and I set the remaining as 0 vector. The economy option (i.e. computing the non-zero vectors makes it practical when the higher order matrices [n-way arrays] are turned into matrices and svd is applied to them.)
```
[U{1}, ~, ~] = svd(X, 'econ');
U{1} = [U{1} zeros(size(U{1},1), size(U{1},1)-size(U{1},2))];
[U{2}, ~, ~] = svd(X', 'econ');
U{2} = [U{2} zeros(size(U{2},1), size(U{2},1)-size(U{2},2))];
```
| https://mathoverflow.net/users/5287 | Making MATLAB svd robust to transpose operation | You may be interested in the following article which addresses this issue:
R. Bro, E. Acar and T. G. Kolda. Resolving the sign ambiguity in the singular value decomposition. Journal of Chemometrics 22(2):135-140, February 2008. (doi:10.1002/cem.1122)
The MATLAB code can be downloaded here:
<http://www.mathworks.com/matlabcentral/fileexchange/22118-sign-correction-in-svd-and-pca>
Best wishes,
Tammy Kolda
| 7 | https://mathoverflow.net/users/10872 | 46257 | 29,274 |
https://mathoverflow.net/questions/46247 | 7 | It is well-known that the universal cover $\tilde G$ of a connected Lie group $G$ has a Lie group structure such that the covering projection $\tilde G\to G$ is a Lie group morphism. Of course $\tilde G$ might not be linear even though $G$ is, but this is not the point here.
My question is: assume that $G$ is a not necessarily connected Lie group. Does there exist a Lie group $\tilde G$ and an onto Lie group morphism $\tilde G\to G$ whose restriction to the identity component of $\tilde G$ is the universal cover of the identity component of $G$?
I assume that the answer is "no" in general, but I could not find any counter-example.
---
@Jim: Of course, the terminology "universal cover" would have been inappropriate even though such a cover existed (which as you and André pointed out, is not the case).
I came to this question from some other direction. Namely, the $PSL\_2(R)$ action on $RP^1$ lifts to a $\widetilde{PSL\_2(R)}$ action on the universal cover of $RP^1$, and this action extends to an action of a two-sheeted cover of $PGL\_2(R)$. It is tempting to denote this cover by $\widetilde{PGL\_2(R)}$, and I wondered whether such a construction was standard.
| https://mathoverflow.net/users/10675 | Is there a good definition of the universal cover for non-connected Lie groups? | The group $Pin\_-(2)$ is an example of what you're looking for.
It can be described explicitly as a subgroup of the group of unit quaternions:
$Pin\_-(2)=$ { $a+bi| a^2+b^2=1$ } $\cup$ { $cj+dk| c^2+d^2=1$ } $\subset \mathbb H^\times$.
Its main interesting properties are:
- The conjugation action of $\pi\_0$ on its Lie algebra is non-trivial.
- All the elements of the non-identity component have a non-trivial square.
There is no Lie group that is diffeomorphic to $\mathbb Z/2\times \mathbb R$
and that shares those properties.
| 12 | https://mathoverflow.net/users/5690 | 46260 | 29,275 |
https://mathoverflow.net/questions/46258 | 14 | [Belyi's theorem](http://en.wikipedia.org/w/index.php?title=Belyi%27s_theorem) states that the following properties of a nonsingular projective algebraic curve $X$ are equivalent:
1) $X$ is defined over $\overline{\mathbb{Q}};$
2) There exists a meromorphic function $\phi: X\to\mathbb{P}^1\mathbb{C} $ ramified at most at $0,1,$ and $\infty$;
3) $X$ is isomorphic to $\Gamma \backslash \mathbb{H}$ (compactified at cusps) for a finite index subgroup $[\mathrm{PSL}\_2(\mathbb{Z}), \Gamma]<\infty.$
The remaining question is:
$$\boxed{\text{
Is there a way to treat singularities in this or a similar framework? }}$$
The following of my original questions have been answered:
Can this be generalized to arbitrary projective nonsingular varities of higher dimensions? (I discussed this with one professor here in Goettingen. That seems to be ongoing research. Please see also comment of David Roberts.)
What compactification do they mean here? $\Gamma \backslash \mathbb{H} \cup \mathbb{Q}$! (see the answer of Robin Chapman)
What is a nice reference for the proof of Belyi's theorem? (see answer of YBL and Koeck + the comments of Emerton)
How does the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ enter the picture? (see comment of Ariyan and answer of YBL)
Where can I find nice examples where these computations have been done explicitly?
(see answer of Andy Putnam and JSE)
| https://mathoverflow.net/users/10400 | Generalizations of Belyi's theorem | The compactification is the usual one coming up in
the theory of modular forms, with the cusps being
orbits of $\Gamma$ on $\mathbb{Q}\cup\{\infty\}$.
As for the proof, I like
[this paper](http://uk.arxiv.org/abs/math/0108222) by Bernhard Koeck.
| 9 | https://mathoverflow.net/users/4213 | 46261 | 29,276 |
https://mathoverflow.net/questions/46180 | 10 | If $X\_1, \ldots , X\_k$ are i.i.d normal random variables with mean $0$ and variance $1$, then is there a way to sample $Y\_1, \ldots , Y\_m$ for $m=\omega(k)$ such that each of the $Y\_i$'s is a normal random variable with mean $0$ and variance $1$ and they are pairwise independent?
| https://mathoverflow.net/users/10858 | How to sample pairwise independent gaussians | Here is the answer I promised in my last comment.
Instead of considering ${\rm N}(0,1)$ variables, we may consider uniform$[0,1)$ variables.
Indeed, if $Z\_i$ are i.i.d. ${\rm N}(0,1)$ variables, then, with $\Phi(\cdot)$ denoting the ${\rm N}(0,1)$ distribution function, $U\_i := \Phi (Z\_i)$ are i.i.d. uniform$[0,1)$ variables. In turn, if $\tilde U\_i$ are pairwise independent uniform$[0,1)$ variables, then $\tilde Z\_i := \Phi^{-1} (\tilde U\_i)$ are pairwise independent ${\rm N}(0,1)$ variables.
The rest of this answer is based on the recent paper "Recycling physical random numbers",
available at [1](http://projecteuclid.org/euclid.ejs/1262617417) or [2](http://www-stat.stanford.edu/~owen/reports/recycle.pdf). Henceforth, we use the same letters as in that paper. Suppose that $U\_1,\ldots,U\_n$ are independent uniform$[0,1)$ variables. Fix $2 \leq m \leq n$, and define $N\_m = {n \choose m}$. Now let $X\_i$, for $i = 1,\ldots,N\_m$, comprise all $N\_m$ distinct sums of the form $U\_{r\_1 } \oplus U\_{r\_2 } \oplus \cdots \oplus U\_{r\_m }$, for $1 \le r\_1 < r\_2 < \cdots < r\_m \le n$. Here $U\_{r\_1 } \oplus U\_{r\_2 } \oplus \cdots \oplus U\_{r\_m }$ is the sum modulo $1$ of the $U\_{r\_i}$, given explicitly by
$$
U\_{r\_1 } \oplus U\_{r\_2 } \oplus \cdots \oplus U\_{r\_m } = U\_{r\_1 } + U\_{r\_2 } + \cdots + U\_{r\_m } - \left\lfloor {U\_{r\_1 } + U\_{r\_2 } + \cdots + U\_{r\_m } } \right\rfloor ,
$$
where $\left\lfloor \cdot \right\rfloor$ is the floor function. Then, the $X\_i$ are pairwise independent uniform$[0,1)$ variables. In particular, by letting $m=2$, we can efficiently construct $n(n-1)/2$ pairwise independent uniform variables from $n$ independent ones.
Finally, for general purposes it might be worth stating the following simple fact (Proposition 2 in the aforementioned paper). For $N \geq 2$, let $Y\_1,\ldots,Y\_N$ be pairwise independent random variables with common mean $\mu$ and common variance $\sigma^2 < \infty$. Define $\bar Y = \frac{1}{N}\sum\nolimits\_{i = 1}^N {Y\_i }$ and $s^2 = \frac{1}{{N - 1}}\sum\nolimits\_{i = 1}^N {(Y\_i - \bar Y)^2 }$. Then, ${\rm E}(\bar Y) = \mu$, ${\rm Var}(\bar Y) = \sigma^2/N$, and ${\rm E}(s^2) = \sigma^2$. Combined with the previous paragraph, a straightforward implication is that for a square-integrable function $f$ defined on $[0,1)$, we can approximate the integral $\mu = \int\_{[0,1)} {f(x)\,{\rm d}x}$ using a modest number $n$ of independent random inputs. Indeed, note that $n$ independent random inputs can be used to get unbiased Monte Carlo estimates for $\mu$ with the same variance as with $N\_m = {n \choose m}$ independent random inputs.
| 9 | https://mathoverflow.net/users/10227 | 46264 | 29,278 |
https://mathoverflow.net/questions/46102 | 8 | I am searching for a conformal mapping from the upper halfplane onto a hyperbolic polygon, i.e. the sides of the polygon have to be geodesics.
The classical Schwarz Christoffel theorem does the job for euclidean polygons (see e.g. <http://en.wikipedia.org/wiki/Schwarz-Christoffel_mapping>).
Does anybody know of a similar constructions in hyperbolic geometry?
Does anybody know of similiar constructions for any other domains?
Any idea will be very wellcomed! I am far from being an expert in conformal mappings and do only know some isolated examples!
| https://mathoverflow.net/users/10400 | Conformal Mappings for hyperbolic polygon | See Harmer and Martin's work on [Conformal Mappings from the Upper Half Plane to Fundamental Domains on the Hyperbolic Plane](http://www.math.auckland.ac.nz/Research/Reports/view.php?id=499).
Some of the ideas developed by [Christopher Bishop](http://www.math.sunysb.edu/~bishop/) in the context of computational geometry may also be of interest.
See his [talks](http://www.math.sunysb.edu/~bishop/lectures/lec.html) and [papers](http://www.math.sunysb.edu/~bishop/papers/papers.html) on conformal maps.
| 6 | https://mathoverflow.net/users/5372 | 46269 | 29,283 |
https://mathoverflow.net/questions/46262 | 1 | Some time ago, I asked about inite interpolation by
a nondecreasing polynomial here at [Finite interpolation by a nondecreasing polynomial](https://mathoverflow.net/questions/16673/finite-interpolation-by-a-nondecreasing-polynomial). This turned out to be an already solved problem; it also turned out that the degree of the solution could not be bounded in terms of the number of interpolation points alone.
My new question is: if we are willing to replace polynomials by another
vector space V of indefinitely differentiable functions, then we can we achieve something better than with polynomials, in the sense that V is finite-dimensional?
Formally, fix $x\_1 \lt x\_2 \lt \ldots \lt x\_n$ and let $y\_1 \leq y\_2\leq \ldots \leq y\_n$ vary. We consider the system $(S)$ made of the $n$ interpolation constraints $f(x\_i)=y\_i$ for $i$ between $1$ and $n$. Is there a finite-dimensional subspace $V$ of ${\cal C}^{\infty}([x\_1,x\_n],{\mathbb R})$ such that for any $y\_1 \leq y\_2\leq \ldots \leq y\_n$, there is a solution to $(S)$ which is nondecreasing and also in $V$?
| https://mathoverflow.net/users/2389 | Finite interpolation by nondecreasing indefinitely differentiable functions in a finite-dimensional space | This is to expand Qiaochu Yuan's comment. For $1\le i < n$ let $b\_i:=(x\_{i+1}+x\_i)/2$ be the mid-point of the $i$-th interval, and let $0 < \epsilon \leq \min\_{1\le i < n} (x\_{i+1}+x\_i)/2\, .$ Start with the linear space $V\_0$ of all continuous functions on $\mathbb{R}$ that are affine on each component interval of $\mathbb{R}\setminus \{b \_i\, : \, 1\leq i < n \}$ and take $V$ to be the image of $V\_0$ via the (linear) convolution operator $u\mapsto u\*\phi$, with a symmetric kernel $\phi\ge0$, $C^\infty\_c(\, ]-\epsilon,+\epsilon[\, )\, $ and $\int\_\mathbb{R} \phi = 1 .$ The convolution with $\phi$ preserve monotonicity and the values at the nodes $x\_i$, so that the interpolation function in $V$ is just the mollification of the interpolation function on $V\_0.$
| 2 | https://mathoverflow.net/users/6101 | 46272 | 29,285 |
https://mathoverflow.net/questions/46249 | 9 | Thinking about the question [Four polynomials representing all integers modulo m](https://mathoverflow.net/questions/45697/four-polynomials-representing-all-integers-modulo-m/45709#45709) lead me to the following complementary question:
If $S$ is a set of positive integers, say that a positive integer $m$ is *covered* by $S$ if every congruence class $\bmod m$ has a representative in $S$. Denote by $C(S)$ the set of positive integers covered by $S$. If $x>0$ let $S(x) = \{ k \in S : k \le x \}$ and the *lower density* of $S$, $\ell(S) := \lim \inf\_{x \rightarrow \infty} |S(x)|/x$. My question: is there a non-trivial lower bound on $\ell(C(S))$ in terms of $\ell(S)$? That is, is there a continuous function $f : [0,1] \rightarrow [0,1]$, not identically 0, such that $\ell(C(S)) \ge f(\ell(S))$.
The set in the question I referred to has density 0, so my question wouldn't apply to it. However, I wondered if there were a simple argument in the case of positive lower density. This has the smell of the kind of question that Erdos would ask, so I wouldn't be surprised to see it there.
| https://mathoverflow.net/users/2784 | Density of congruence classes covered by a set | Denote by $P$ the set of prime powers not covered by $S$ (for each prime $p$, take only the smallest non-covered its power). If $\sum\_{x\in P} 1/x=+\infty$, then $\prod\_{x\in P} (1-1/x)$ is 0, so the product over some finite subset is arbitrarily small. But this product is a density of numbers without forbidden remainders modulo respective prime powers. So, $S$ has density 0. A contradiction. Hence $a=\prod\_{x\in P} (1-1/x)$ is positive and $\ell(S)\leq a$. But then complement of $C(S)$ is the set of numbers divisible by at least one element of $P$. Density of such numbers equals $1-a$ (this is rather technical, but standard). So, we get that $\ell(C(S))\geq \ell(S)$.
| 3 | https://mathoverflow.net/users/4312 | 46277 | 29,287 |
https://mathoverflow.net/questions/46252 | 134 | Without prethought, I mentioned in class once that the reason the symbol $\partial$
is used to represent the boundary operator in topology is
that its behavior is akin to a derivative.
But after reflection and some research,
I find little support for my unpremeditated claim.
Just sticking to the topological boundary (as opposed to the boundary
of a manifold or of a simplicial chain),
$\partial^3 S = \partial^2 S$
for any set $S$.
So there seems to be no possible analogy to
Taylor series. Nor can I see an analogy with the
fundamental theorem of calculus.
The only tenuous sense in which I can see the boundary
as a derivative is that $\partial S$ is a transition between $S$
and the "background" complement $\overline{S}$.
I've looked for the origin of the use of the symbol $\partial$ in topology without luck.
I have only found [references](http://en.wikipedia.org/wiki/%E2%88%82)
for its use in calculus.
I've searched through *History of Topology* (Ioan Mackenzie James)
online without success (but this may be my poor searching).
Just visually scanning the 1935 *Topologie* von Alexandroff und Hopf, I do not
see $\partial$ employed.
I have two questions:
>
> **Q1**. Is there a sense in which the boundary operator $\partial$ is
> analogous to a derivative?
>
>
> **Q2**. What is the historical origin for the use of the symbol
> $\partial$ in topology?
>
>
>
Thanks!
**Addendum** (*2010*). Although **Q2** has not been addressed
[was subsequently addressed by @FrancoisZiegler], it seems appropriate to accept one
among the wealth of insightful responses to **Q1**. Thanks to all!
| https://mathoverflow.net/users/6094 | Is the boundary $\partial S$ analogous to a derivative? | The surface area $|\partial S|$ of a (bounded, smooth) body $S$ is the derivative of the volume $|S\_r|$ of the $r$-neighbourhoods $S\_r$ of $S$ at $r=0$:
$$ |\partial S| = \frac{d}{dr} |S\_r| |\_{r=0}.$$
Thus, for instance, the boundary $\partial D\_r$ of the disk $D\_r$ of radius $r$ has circumference $\frac{d}{dr} (\pi r^2) = 2\pi r$.
More generally, one intuitively has the Newton quotient-like formula
$$ \partial S = \lim\_{h \to 0^+} \frac{S\_h \backslash S}{h};$$
the right-hand side does not really make formal sense, but certainly one can view $S\_h \backslash S$ as a $[0,h]$-bundle over $\partial S$ for $h$ sufficiently small (in particular, smaller than the radius of curvature of $S$).
In a similar spirit, one informally has the "chain rule"
$$ {\mathcal L}\_X S "=" (X \cdot n) \partial S $$
for the "Lie derivative" of $S$ along a vector field $X$, where $n$ is the outward normal. (There may also be a divergence term, depending on whether one is viewing $S$ as a set, a measure, or a volume form.) Again, this does not really make formal sense, although Stokes' theorem already captures most of the above intuition rigorously (and, as noted in the comments, Stokes' theorem is probably the clearest way to link boundaries and derivatives together).
EDIT: A more rigorous way to link boundaries with derivatives proceeds via the theory of distributions. The weak derivative $\nabla 1\_S$ of the indicator function of a smooth body $S$ is equal to $-n d\sigma$, where $n$ is the outward normal and $d\sigma$ is the surface measure on $\partial S$. (This is really just a fancy way of restating Stokes' theorem, after one unpacks all the definitions.) This can be used, for instance, to link the Sobolev inequality with the isoperimetric inequality.
In a similar spirit, $\frac{1\_{S\_h} - 1\_S}{h}$ converges in the sense of distributions as $h \to 0$ to surface measure $d\sigma$ on $\partial S$, thus providing a rigorous version of the intuitive difference formula given previously.
| 137 | https://mathoverflow.net/users/766 | 46285 | 29,288 |
https://mathoverflow.net/questions/46115 | 27 | Let $$G=\mathbb{Z}/p\_1^{e\_1}\times\cdots\times\mathbb{Z}/p\_n^{e\_n}$$ be any finite abelian group.
What are $G$'s subgroups? I can get many subgroups by grouping the factors and multiplying them by constants, for example: If $$G=\mathbb{Z}/3\times \mathbb{Z}/9\times \mathbb{Z}/4\times \mathbb{Z}/8,$$ then I can take $$H=3(\mathbb{Z}/3\times \mathbb{Z}/9)\times 2(\mathbb{Z}/4) \times \mathbb{Z}/8.$$ Do I get all subgroups that way? (I'm interested in all subgroups, not just up-to-isomorphism).
Which are the subgroups $H$ in $G$ for which $G/H$ is primary cyclic?
| https://mathoverflow.net/users/nan | Subgroups of a finite abelian group | These three answers were originally comments. I am answering the part of the question which was deleted:
**Is there anything else (interesting) to say about the collection of subgroups of an [finite] abelian group.**
1. This paper: Ganjuškin, A. G. Enumeration of subgroups of a finite abelian group (theory). Computations in algebra and combinatorial analysis, pp. 148–164, Akad. Nauk Ukrain. SSR, Inst. Kibernet., Kiev, 1978 gives an algorithm for enumerating all subgroups of a finite Abelian group.
**Update.** The answer by Amritanshu Prasad, comments by Derek Holt and Robin Chapman here provide much more (very interesting) information about enumerating subgroups.
1. On the other hand, the elementary theory of pairs $(A,H)$ where $A$ is a finite Abelian group and $H$ is its subgroup is undecidable (see Taĭclin, M. A. Elementary theories of lattices of subgroups, Algebra i Logika 9 1970 473–483 and references there). Hence there cannot be a nice description of subgroups of finite Abelian groups (say, one cannot represent a pair $(A,H)$ as a direct product of pairs of sizes bounded in terms of the period of $A$).
2. This is a better reference than Taiclin. Slobodskoĭ, A. M.; Fridman, È. I.: Theories of abelian groups with predicates that distinguish subgroups. Algebra i Logika 14 (1975), no. 5, 572–575.
**Update** In fact, in the paper, Sapir, M. V. Varieties with a finite number of subquasivarieties. Sibirsk. Mat. Zh. 22 (1981), no. 6, 168–187, I proved that for every prime $p$ one cannot find finitely many pairs $(A\_i,H\_i)$ such that every pair $(A,H)$ where $A$ is Abelian group of period $p^6$ (it is not true for $p^5$) is a direct product of copies of $(A\_i,H\_i)$.
| 4 | https://mathoverflow.net/users/nan | 46300 | 29,296 |
https://mathoverflow.net/questions/45892 | 9 | I am not sure if this is appropriate for MO. If not, I shall be happy to take it to SE.
For a local ring $(R,m)$, given any proper ideal $I$, the (Krull) dimension (from here on dimension means Krull dimension) of the associated graded ring of $R$ with respect to $I$, $gr\_I(R)=\oplus\_{n\geq 0}\frac{I^n}{I^{n+1}}$ is equal to the dimension of $R$ itself. The only proof I know for this involves writing the associated graded ring as a quotient of the extended Rees ring $R[It,t^{-1}]$ and using dimension formulas for the latter. I was wondering if anyone was aware of a proof that does not route via the extended Rees ring. Any references would be appreciated. I googled, but could not stumble upon anything useful.
| https://mathoverflow.net/users/10775 | (Krull) dimension of any associated graded ring of a ring R equals the dimension of R | Though I heartily agree with Victor Protsak's comment, I will add some references. These might be useful for you, at least if you haven't seen them before. The references add a restriction, however, by assuming that $I$ is an ideal of finite co-length.
Then Corollary 12.5 of Eisenbud's Commutative Algebra uses the theory of Hilbert-Samuel polynomials to prove that $\text{dim}(R)=\text{dim}\text{ gr}\_I(R)$.
Alternately, you might also be interested in Corollary 10.12 of the same book. This second corollary assumes that $I=\mathfrak m$, but the proof makes use of "Going down for flat extensions", which has a somewhat different flavor than the Rees ring approach.
| 4 | https://mathoverflow.net/users/4 | 46307 | 29,301 |
https://mathoverflow.net/questions/45668 | 3 | We consider *finite* algebras for a given signature, in the sense of universal algebra (for example, they might be groups, rings, or lattices). An algebra $A$ is *irreducible* when $A \cong B \times C$ implies that $B$ or $C$ is the one-point algebra.
Is it the case that a $\Sigma$-algebra can be expressed as a cartesian product of irreducible algebras in an essentially unique way, i.e., unique up to permutation of factors? I suspect that this is either a theorem with somebody's name attached to it, or there is a counterexample in groups.
| https://mathoverflow.net/users/1176 | Is the decomposition of an algebra into irreducible components essentially unique? | Let $A, B$ be the algebras with two elements $0,1$ under addition mod $2$ and unary operation $x'=x$ in $A$ and $x'=1-x$ in $B$. Then $A \times B \cong B \times B$, though $A$ and $B$ are not isomorphic.
B. Jonsson
Construct a 12-element commutative semigroup which does not have the unique factorization property.
R. McKinsey
(exercises on p. 170 of Birkhoff, *Lattice Theory* (3rd edition))
| 3 | https://mathoverflow.net/users/454 | 46310 | 29,304 |
https://mathoverflow.net/questions/46312 | 4 | Let $A$ be a closed set and let $B$ be a connected set such that $A \subset B$.
Does there always exist a closed connected subset $C$ of $B$ that contains $A$?
What if $B$ is path connected, is there always a path-connected $C$? A connected $C$?
| https://mathoverflow.net/users/10876 | Closed connected subset of a connected set | The answer to your first question is no. Let $A$ be a sequence on the $x$ axis converging to the origin in the real plane $\mathbb{R}^2$, which is the background topological space for these examples. Let $B$ be a cone over the sequence, connecting each point in the sequence to the point $(0,1)$, together with the origin, but not connecting the origin to $(0,1)$. Then $B$ is connected, but not closed, and there is no connected closed set $C$ contained in $B$ and containing $A$, since any connected set $C$ contained in $B$ and containing $A$ would have to contain all the lines and thus be all of $B$, which is not closed in the plane.
The example can be extended to an answer for your path-connected question as well. We simply add a roundabout path from the origin to $(0,1)$. That is, $A$ is the sequence $(\frac{1}{n},0)$ plus $(0,0)$ in the real plane. The set $B$ contains lines from every $(\frac{1}{n},0)$ to $(0,1)$, and $B$ also includes a circular curve connecting $(0,0)$ to $(1,0)$ through the left half-plane. So $A$ is closed, $B$ is path-connected, but any closed path-connected set $C$ with $A\subset C\subset B$ will have to contain all those lines and hence $C=B$, but this is not closed in the plane.
Similarly, for your last question, any closed connected $C$ with $A\subset C\subset B$ for the same example will have to contain all the lines from the points on the sequence to $(0,1)$, and hence not be closed (since it will miss the line joining the origin to $(0,1)$, which is not in $B$.
| 5 | https://mathoverflow.net/users/1946 | 46313 | 29,305 |
https://mathoverflow.net/questions/46305 | 11 | For a non-Kahler complex manifold $M$, we still have the decomposition of differential forms into differential forms of type $(p,q)$ and we can write $d=\partial+\bar\partial$ and we can define cohomology classes $H^{p,q}\_{\bar\partial}(M)$.
In general is there any relation between $h^r(M)$ and $h^{p,q}(M)$?
| https://mathoverflow.net/users/5259 | Non-Kahler Complex manifolds | For any compact complex manifold there is a spectral sequence with $E\_1$ term $H^{p,q}(M)$ which converges to $H^{p+q}(M)$. If $M$ were Kahler, then this spectral sequence would degenerate at the $E\_2$ page, giving the familiar Hodge decomposition on cohomology.
In general, there is still a filtration on the cohomology, and the associated graded pieces will be sub-quotients of the $H^{p,q}(M)$.
So there is such a relationship between Hodge numbers and Betti numbers, but being able to write it down depends on being able to calculate the differentials in the spectral sequence.
Voisin's book, Hodge Theory and Complex Algebraic Geometry has more information. There is also an interesting exercise in which you can calculate the precise relationship in the case when $M$ is a complex surface (not necessarily Kahler).
| 13 | https://mathoverflow.net/users/7762 | 46316 | 29,306 |
https://mathoverflow.net/questions/46321 | 6 | Suppose we have two closed-form expressions with $k$ unknowns which are hard to test for equality but easy to evaluate numerically over $\mathbb{R}^k$. One could then approach the problem of equality testing by checking equality numerically at several points. The interesting questions are then -- for which kinds of expressions can you do it, how to pick sampling points and how many points are needed.
Google Scholar gives 0 hits for "numeric equality testing"
Has this kind of problem been studied before? What are the right keywords to search for?
| https://mathoverflow.net/users/7655 | Numeric equality testing? | Maple has a procedure **testeq** which is a "random polynomial-time equivalence tester". It works in this way. The 1986 paper [New results for random determination of equivalence of expressions](http://portal.acm.org/citation.cfm?id=32465&dl=ACM&coll=DL&CFID=111332393&CFTOKEN=61560774) by Gaston H. Gonnet might be a starting point for checking that out.
| 7 | https://mathoverflow.net/users/8008 | 46327 | 29,312 |
https://mathoverflow.net/questions/46325 | 8 | I'm interested in looking at the details of Gauss' method of determining the sign of the Gauss sum in his "Summatio Quarumdam Serierum Singularium", and I was wondering if anyone knew if there was an English or French translation available at all? I've tried searching on the internet and haven't found anything. Also, are there any modern presentations of the Gauss' method that you would recommend?
Thanks for your help.
| https://mathoverflow.net/users/5431 | English or French translation of Gauss' "Summatio Quarumdam Serierum Singularium" | ["*The determination of Gauss sums*"](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-5/issue-2/The-determination-of-Gauss-sums/bams/1183548292.full) by Berndt and Evans (Bull. Amer. Math. Soc., Vol. 5, Number 2 (1981), 107-129.) contains an exposition of the original proof due to Gauss. It also includes a short historical outline of various classical and modern proofs due to Dirichlet, Cauchy and others.
| 9 | https://mathoverflow.net/users/5371 | 46329 | 29,314 |
https://mathoverflow.net/questions/46253 | 1 | If I would ask for $\phi'(x) = f( \phi(x))$ and $\phi(0)=f(0)$, I would get that the inverse of $\phi$ is forced to be of the form:
$$\phi^{-1}(z) = \int\_{0}^z \frac{1}{f(x)} d x.$$
Now it is natural to ask whether there is something similiar for the problem $\phi''(x) =f( \phi(x),\phi'(x))$?
Or a little bit mor restrictive:
Given a (homogenous) polynomial $f\_j(X\_0, \dots X\_n) \in \mathbb{C}[X\_0, \dots, X\_n]$, where $j=1, \dots, m$, such that the variety $V(f\_1, \dots, f\_m)$ is a non singular variety over $\mathbb{C}$.
Q1: Is it known in general, if there exists a function $\phi: \Omega \rightarrow \mathbb{C}$, such that
$$f\_j( \phi(s), \phi'(s), \dots, \phi^{(n)}(s))=0$$
or perhaps the easier problem with isolated variables
$$\left( \phi^{(n+1)}(s) \right)^{\mathrm{deg} (f\_j)}= f\_j( \phi(s), \phi'(s), \dots, \phi^{(n)}(s)).$$
for all $s \in \Omega$? Here $\Omega$ is any open subset of $\mathbb{C}$. Are there global meromorphic solutions? What is the algebraic invariant, which counts the linear independent solutions here?
Q2: (see Denis Serre's answer): What is known in the case $m=1$? Or what if require all $f\_j$'s to have the same degree?
Q3: Is there a general theory/algorithm, how to compute the solutions in special cases?
Q4: Are there some easy examples, which give an intuition, what we can expect to be true and what not?
I make this community wiki, since I do not know wheter my question is naive. I have little intuition about this stuff. Please comment if you do some changes. As a motivation, the Weierstrass $\wp $ function does the job for elliptic curves.
| https://mathoverflow.net/users/10400 | Solving nonlinear ODE's | The answer is **NO** in general because your differential system is overdetermined: it has several equations and only one unknown function. Take the polynomials $\omega^2 X\_0-X\_2$ and $X\_2^2-X\_1^2-X\_0^2$ with $\omega^4\ne1$. The solutions of the first equation are $ae^{\omega s}+be^{-\omega s}$, but none of them satisfy the second equation, but $\phi\equiv0$.
A more interesting question should be which varieties $V$ yield non-trivial solutions. The answer must be of geometrical nature. Overdetermined linear systems have been studied for a long time (dating back to Goursat and E. Cartan), but the situation for nonlinear ones might be widely open.
| 5 | https://mathoverflow.net/users/8799 | 46334 | 29,319 |
https://mathoverflow.net/questions/46314 | 6 | My question is related to several notions of hyperbolicity, applied to Kähler manifolds (projective, in general). Kähler hyperbolicity was introduced in [this paper of Gromov's](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-33/issue-1/K%C3%A4hler-hyperbolicity-and-L_2-Hodge-theory/10.4310/jdg/1214446039.full). He calls a Kähler manifold *Kähler hyperbolic* if the lift to the universal cover of the symplectic form(=imaginary part of hermitian metric) is the differential of a bounded $1$-form. A $1$-form is bounded if its norm is pointwise absolutely bounded (the norm is induced by the pull-back of the metric to the tangent space).
As he notes in the introduction, this notion implies Kobayashi hyperbolicity. My question is regarding the converse, namely to find an example of a Kobayashi hyperbolic manifold that isn't Kähler hyperbolic.
---
My guess is that an example of an algebraic variety with the properties described below exists (so it would be nice if an algebraic geometer could say a few things here).
For a Kähler hyperbolic manifold, Gromov proves that its Euler characteristic is $(-1)^n$, where $n$ is its *complex* dimension. He also shows a Kähler hyperbolic manifold has quasi-ample canonical bundle (namely, it has Kodaira dimension $n$). Another observation is that a Kähler hyperbolic manifold cannot have amenable fundamental group. So any algebraic variety which fails one of the above tests but still is Kobayashi hyperbolic would fit the bill.
For somewhat different reasons, I would be happier if this would be a projective variety, which also doesn't have any $(2,0)$ cohomology. As a side question, just to check my understanding, is it true that for a variety with no $(2,0)$ classes in cohomology, any homology class that comes from a map of a surface (pushing forward the fundamental class) can in fact be realized by some algebraic curve in the variety?
| https://mathoverflow.net/users/10857 | Hyperbolicity for algebraic varieties and relation to curves on them | Take a very general hypersurface $j \colon X \hookrightarrow \mathbb{P}^n$ of sufficiently high degree, $3 \leq n \leq 4$. Then $X$ is Kobayashi hyperbolic (Kobayashi actually conjectured that this is true for all $n$ and Siu recently outlined a strategy for proving this, see Diverietti's comment below).
On the other hand, Lefschetz hyperplane theorem says that the natural map
$\pi\_1(X) \stackrel{j\_\*}{\longrightarrow} \pi\_1(\mathbb{P}^n)$
is an isomorphism. Then $X$ is simply connected, in particular its fundamental group is amenable and so $X$ cannot be Kahler hyperbolic.
If $n=4$, the projective variety $X$ also satisfies your second request. If fact, again by Lefschetz theorem, see for instance [Dimca, Singularities and topology of hypersurfaces, Theorem 2.6 p. 151], for all $n \geq 4$ we also have an isomorphism
$H^2(\mathbb{P}^n) \stackrel{j^\*}{\longrightarrow} H^2(X)$.
Since $H^2(\mathbb{P}^n)=\mathbb{C}$, Hodge decomposition yields
$H^{2,0}(X)=H^{0,2}(X)=0, \quad H^{1,1}(X)=\mathbb{C}$.
If Kobayashi's conjecture were true, then the hypersurfaces $X$ would provide examples in all dimensions.
| 8 | https://mathoverflow.net/users/7460 | 46339 | 29,324 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.