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https://mathoverflow.net/questions/35910 | 4 | A k-coloring or k-labeling of the vertices of a single-component undirected graph G with $n$ vertices can be a proper coloring or not. If it is not a proper coloring, such that each vertex has neighbors on its edges which are of a different color or label, then for each possible labeling, it possible to count the number of vertices which have a proper coloring locally as an integer $x \in \{0,n\}$, and those which do not $y = n-x$. Let us call the vertices which are locally properly colored "stable", and those which have at least one neighbor with the same color label "unstable." The set of all possible labelings, of which there are $k^n$ can be partitioned into $n+1$ sets by the metric of how many vertices are unstable for each coloring.
What is the size of each partition from 0-unstable to n-unstable? *Is there a particular name for this type of partitioning?* Obviously, if we are setting $k=2$, then the size of the partition of 0-unstable is 2 if the graph G is bipartite and allows for a proper-2-coloring; if $k=3$ and the graph is 3-colorable, then the size of the partition of 0-unstable is 6; etc.
Of course, this partitioning can be calculated by brute force methods in exponential time ($k^n$) by enumerating over all possible labelings with $k$ labels on a graph with $n$ vertices.
For what classes of graphs is the problem tractable? For example, as described below in one answer, for star graphs $S\_m$ with $m$ leaves and $n=m+1$ vertices, the distribution is almost the binomial distribution, with $2$ as the size of the 0-unstable partition, zero as the size of the 1-unstable partition, and $2 {m \choose j-1}$ as the size of the $j$-unstable partition where $1\lt j \le n$.
ABmd
| https://mathoverflow.net/users/8567 | Tractably Partitioning the possible vertex k-colorings of a graph by local stability and instability. | If you limit it to specific classes of graphs, say for example star graphs, you can come up with some answers. For a star graph $S\_m$, with a vertex at the center and $m$ vertices connected to the center, yielding a graph $G$ with $n=m+1$ vertices and $m$ edges, it can be calculated that for $k=2$
If the center vertex is labeled black, then the only "0-unstable" coloring is where all of the leaves are white. If any of the leaves are also black, say $j$ of the $m$ leaves are black while the center is also black, then the center and those $j$ black leaves are unstable, leaving $m-j$ leaves as stable nodes. There are $m \choose j$ = ($m$ choose $j$) ways to color $j$ of the $m$ leaves as black.
The same is true with the color labels reversed if the center is labeled white.
Thus for $k=2$, for two-color labeling of a star-graph $S\_m$ with $m$ leaves and $n=m+1$ vertices, the sizes of the partitions of all of the possible two-colorings are as follows:
|{0 unstable}| = 2
|{1 unstable}| = 0
|{r unstable}| = $2 \times$ ${m}\choose{r-1}$ for $ 2 \le r \le n$, with $r \in Z $
The size of the 1-unstable partition is always zero for this family of graphs. The size of the 1-unstable partition is always zero for any graph and for any $k$ because instablity occurs over an edge linking two vertices with the same color label, thus always creating two unstable vertices if there are any unstable vertices at all.
The sum of all of these partitions sizes is $2^n$, thus all of the possible $2^n$ colorings of the $n=m+1$ vertex star graph $S\_m$ have been accounted for. A similar calculation can be made for star graphs for $k>2$.
Apurva
| 3 | https://mathoverflow.net/users/8636 | 36115 | 23,269 |
https://mathoverflow.net/questions/36088 | 1 | What are the functions $f$ so that a set $\{a \cdot f(x+b) : a \in \mathbb{R}, b \in \mathbb{R}\}$ is a finite dimensional linear vector space ?
Is there a complete characterization of such functions?
$e^{c x}$ (where $c$ is some constant) is a good example of a base of one dimensional space.
$sin (c x), cos(c x)$ seems to be two dimensional example.
| https://mathoverflow.net/users/8631 | Linear space of translatable functions. | The question, as stated, is about the *set* of multiples of translates, but from the example quoted, $\sin x,$ I suspect that OP really meant the *span*.
**Theorem** Let $f$ be a continuous complex-valued function on $\mathbb{R}.$ Then the following conditions are equivalent:
1. The translates $\{f(x+b) : b\in\mathbb{R}\}$ span a finite-dimensional vector space;
2. $f$ satisfies a homogeneous constant coefficient linear differential equation;
3. $f$ is a finite linear combination of functions $f\_{k,\lambda}(x)=x^k e^{\lambda x}.$
*Proof.* If $f$ is assumed infinitely differentiable then all derivatives of $f$ belong to the $\mathbb{R}$-span of translates of $f.$ Thus condition 1 implies that $f$ and its derivatives of order up to $n$ are linearly dependent over $\mathbb{R},$ which is condition 2. The smoothness assumption may be removed by using the Fourier or Laplace transform.
The equivalence of conditions 2 and 3 is a basic fact of ODEs. Finally, a direct computation shows that $f\_{k,\lambda}(x)$ spans the $(k+1)$-dimensional vector space $\{P(x)e^{\lambda x}: P\text{ is a polynomial of degree} \leq k\}$, so condition 3 implies condition 1. $\square$
---
Condition 1 – 3 have the following representation-theoretic interpretation. The additive group of $\mathbb{R}$ acts on itself by the right multiplication. This gives rise to a linear representation of $\mathbb{R}$ on the functions on $\mathbb{R}$ via translations called the *right regular representation*, and condition 1 states that $f$ belongs to a finite-dimensional subrepresentation $V$. Finite-dimensionality of $V$ implies that $V$ contains an irreducible subrepresentation $W$, which must be one-dimensional (Schur's lemma), hence $W$ is spanned by a character of $\mathbb{R}.$ All *continuous* characters are the exponential functions $e^{\lambda x}$ for various $\lambda\in\mathbb{C}$; however, using a Hamel basis of $\mathbb{R},$ it is easy to see that there are uncountably many others.
Condition 2 is the Lie algebra analogue of condition 1: viewing $\mathbb{R}$ as a one-dimensional Lie group, the content of Lie's theorem is that its finite-dimensional (continuous) representations correspond (by differentiation and exponentiation) to f.d. representations of the abelian one-dimensional Lie algebra, i.e. to a single linear transformation on $V.$ The span $V\_{n,\lambda}$ of the functions $f\_{k,\lambda}$ with $0\leq k\leq n-1$ from condition 3 is an $n$-dimensional indecomposable representation of $\mathbb{R},$ whose infinitesimal version is a Jordan block of order $n$ with $\lambda$ on the diagonal. Moreover, any subrepresentation isomorphic to $V\_{n,\lambda},$ i.e. corresponding to the same Jordan block, must be $V\_{n,\lambda}$ itself.
| 6 | https://mathoverflow.net/users/5740 | 36119 | 23,272 |
https://mathoverflow.net/questions/36091 | 8 | If we consider the action of the $U\_p$ operator on overconvergent $p$-adic modular forms, then we can get some information about the field over which the eigenforms are defined by looking at the slopes. For instance, my paper in Math Research Letters ([MR2106238](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=RVCN&pg6=ALLF&pg7=ALLF&pg8=ET&review_format=html&s4=kilford&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=9&mx-pid=2106238)) proves that the slopes of $U\_2$ acting on 2-adic overconvergent modular forms of level 4 with primitive Dirichlet character are distinct, so the field of definition has to be $\mathbf{Q}\_2$. However, there are cases when the slopes fail to be distinct; for instance, in Emerton's thesis it is proved that the lowest slopes of T\_2 acting on level 1 forms of weight congruent to 14 modulo 16 are 6 and 6.
For classical modular forms of level 1, we have Maeda's Conjecture which says that the field of definition is essentially as large as it can be; the Hecke polynomial is irreducible with Galois group $S\_n$ where $n$ is the dimension. However, there is no reason that this should be true for overconvergent modular forms, and in fact it isn't. Discussions with Robert Coleman led me to the concrete example of 2-adic overconvergent modular forms of tame level 1 and weight 142, where there are two eigenforms of slope 6 which are both defined over the ground field $\mathbf{Q}\_2$.
The question is, what should one expect here? Can one tell any more about the field of definition from the slopes than the absolute minimum?
| https://mathoverflow.net/users/4555 | Fields of definition for p-adic overconvergent modular eigenforms | Professor Buzzard raises the a question of whether *every* normalized eigenform of level $1$ is defined over a quadratic extension of $\mathbb{Q}\_2$.
(This is Question 4.3 of <http://www2.imperial.ac.uk/~buzzard/maths/research/papers/conjs.pdf>)
In contrast, the multiplicity of the valuation of the set of $2$-adic slopes at level $1$ can be arbitrarily large,
as can be observed as follows:
Consider the space
$S\_k:=S^{new}\_k(\Gamma\_0(2))$ of newforms of weight $k$. Every newform has slope $(k-2)/2$. Thus, by
work of Coleman, the number of slopes of valuation $(k-2)/2$ at level $1$ and weight $k + 2^n$ for sufficiently large $n$ will be at least $\mathrm{dim}(S\_k)$, which is unbounded as $k$ increases.
EDIT: The point of the last example is that the answer to the original question is "not much", i.e., there can be many forms of the same slope, but all the forms are defined over a small (or even trivial) extension of $\mathbb{Q}\_2$.
| 2 | https://mathoverflow.net/users/nan | 36120 | 23,273 |
https://mathoverflow.net/questions/36126 | 26 | Next semester I will be teaching an introductory course on geometric group theory and there is a basic question that I do not know the answer to. Let $G$ be a finitely generated group with finite symmetric generating set $S$ and let $\Gamma$ be the corresponding Cayley graph. For each $n \geq 1$, let $B\_{n}$ be the closed ball of radius $n$ in $\Gamma$ about the unit element $e$ and let
$b\_{n} = |B\_{n}|$. Then it is known that $\lim b\_{n}^{1/n}$ always exists. (For example, see de la Harpe's book.) My question is whether $\lim b\_{n+1}/b\_{n}$ always exists?
| https://mathoverflow.net/users/4706 | On the size of balls in Cayley graphs | In the article
[R. Grigorchuk and P. De La Harpe, On problems related to growth, entropy, and spectrum in group theory, Journal of Dynamical and Control Systems, Volume 3, Number 1, 51-89](http://www.springerlink.com/content/475575746034r344/)
on the lower part of page 58 the authors mention the manuscript
A. Machi, Growth functions and growth matrices for finitely generated groups. Unpublished manuscript, Univ. di Roma La Sapienza, 1986.
and explain an example due to Machi. Machi showed that the convergence of $b\_{n+1}/b\_n$ can fail for one generating set of ${\mathbb Z}\_2 \star {\mathbb Z}\_3$ and hold for another. In particular, the limit does not always exist. The two generating sets are $\lbrace s,t\rbrace$ and $\lbrace s,st\rbrace$, where $s$ and $t$ are the natural generators with $s^2=t^3=e$.
| 31 | https://mathoverflow.net/users/8176 | 36132 | 23,279 |
https://mathoverflow.net/questions/29256 | 6 | I have a vertex set $V$ and a collection of disjoint arc sets $E\_1, \ldots, E\_t$ such that $$G\_i = (V, E\_i),\quad\forall i = 1, \ldots t,$$ are directed acyclic graphs (DAGs) and $$G = (V, E\_1 \cup \ldots \cup E\_t)$$ is a tournament. We note that the individual DAGs may be disconnected and that $G$ may not be acyclic. However, suppose there exists a bipartition of the arc set indices $\alpha \cup \beta$ such that $$G' = (V, E\_\alpha\cup E\_\beta^T)$$ is an **acyclic** tournament where $$E\_\alpha = E\_{\alpha\_1} \cup \ldots \cup E\_{\alpha\_p}$$ and $$E\_\beta = E\_{\beta\_1} \cup \ldots \cup E\_{\beta\_q}$$ and $E^T$ is the transpose of $E$ (all the arcs are reversed).
Does anybody know of any results relating to the above? In particular, does anybody know of a method of determining a bipartition $\alpha \cup \beta$, given that at least one exists, other that enumerating all possible bipartitions and checking if the resulting $G'$ is acyclic?
| https://mathoverflow.net/users/685 | Combining DAGs into an acyclic tournament | The problem you pose, of finding a bipartition if one exists, is of polynomially equivalent difficulty to the decision problem of determining whether a bipartition exists. The decision problem in turn is NP-complete, by reduction from 3-SAT (and the fact that a solution is easily checked.)
Given an instance of 3-SAT with $n$ clauses, we construct a family of DAGs on $4n$ vertices. All edges in the complement of $n$ disjoint $4$-cycles will be singleton DAGs. One "universal" DAG consists of a single edge in each $4$-cycle, and establishes a potential (forbidden) orientation on each $4$-cycle. Then for every variable in the 3-SAT instance we define a DAG consisting of an edge in each of the $4$-cycles corresponding to the clauses in which that variable appears, with the direction depending on whether the variable appears negated in the clause, in such a way that the forbidden orientation imposed by the universal DAG is achieved in a given $4$-cycle if and only if no literal in the corresponding clause is true, where a variable is considered true when its DAG lies on the same side of the bipartition as the universal DAG and is considered false otherwise. Then an acyclic bipartition of the DAGs exists if and only if the instance of 3-SAT has a satisfying assignment.
| 2 | https://mathoverflow.net/users/7936 | 36137 | 23,283 |
https://mathoverflow.net/questions/36084 | 0 | Hello to all,
While sprucing up my knowledge of group (co)homology,I stumbled onto the following question: The first step you usually take to compute various (co)homologies is to construct the infamous "bar resolution" which resolves $\mathbb{Z}$ by free $\mathbb{Z}[G]$-modules (I'll assume everyone knows which one I mean).
Now, in the case of the (co)homology of cyclic groups, one creates a 2-periodic resolution by splicing together certain exact sequences involving the norm element of $\mathbb{Z}[G]$. I was wondering if it was possible to distill this 2-periodic resolution somehow out of the standard bar-resolution above in some natural way ?
In the case of $\mathbb{Z}\_2$, this is quite trivial, but the higher cases are a mystery to me!
Thank you and merry Fields day
| https://mathoverflow.net/users/4863 | (co)homology of cyclic groups |
>
> Louis asked: "I was wondering if it was possible to distill this 2-periodic resolution somehow out of the standard bar-resolution above in some natural way ?"
>
>
>
By a result of Benson and Carlson [Complexity and Multiple Complexes. Math. Zeit. 195(1987), 221-238, Theorem 4.4], for finite groups there is a general procedure that produces a resolution that is the tensor product of r periodic complexes where r is the rank of the group.
Given a projective resolution and a set of r cocycles that represents a homogeneous system of parameters of the integral cohomology ring, the construction of the periodic complexes is explicit and quite simple. If a cocycle has degree d than the corresponding periodic complex is d-periodic.
Now consider a finite cyclic group. Then r = 1. If you figure out a cocycle of the bar resolution that generates the second integral cohomology group and apply the construction of Benson-Carlson to this cocycle then you'll end up with the usual 2-periodic resolution.
| 4 | https://mathoverflow.net/users/8644 | 36144 | 23,286 |
https://mathoverflow.net/questions/36154 | 1 | I have an intuition that Average of twin prime pairs is always Abundant number except for 4 and 6.
For example:
12 < 1+2+3+4+6=16
18 < 1+2+3+6+9=21
...
But I can't prove this. Could you give me any good idea?
---
2010-08-22
I think that Any prime is a factor of average of twin prime pair.
Do you agree with me?
| https://mathoverflow.net/users/8140 | Average of twin prime pairs is Abundant number, except for 4 and 6. Any prime is a factor of average of twin prime pair | Prove the sum is always a multiple of 6, then prove that multiples of 6 are abundant.
| 9 | https://mathoverflow.net/users/3684 | 36155 | 23,291 |
https://mathoverflow.net/questions/36169 | 6 | Hi all,
I would like to propose the following problem:
Given two representations $\rho$ and $\tau$ of a group $G$ over complex number, we would like to know if there exists an automorphism $\phi$, such that $\rho\circ\phi$ and $\tau$ are equivalent.
Is there any mathematical results concerning this problem? It seems that to understand the action of automorphisms on the set of irreducible representations is crucial.
Thank you!
Youming
| https://mathoverflow.net/users/8012 | to test equivalence of representations under automorphism | I assume that $\phi$ is an automorphism of $G.$ Note that if $\phi$ is inner then trivially $\rho$ and $\rho\circ\phi$ are equivalent, thus the answer depends only on the image of $\phi$ in the outer automorphism group $Out(G).$
If $G$ is a finite group (or, more generally, compact group) and representations are finite-dimensional, so that they are determined up to isomorphism by their characters, then this problem admits a complete theoretical solution using the character theory. The automorphism group $Aut(G)$ acts on the set $\{C\_i\}$ of the conjugacy classes of $G$, this action factors through the action of $Out(G),$ and
$$\chi\_{\rho\circ\phi}(C)=\chi\_{\rho}(\phi(C)),\qquad (\*)$$
where $\chi\_\rho$ is the character of $\rho$ and $C$ is any conjugacy class. Since representations are determined by their characters,
$$\rho\circ\phi\simeq\sigma \iff \chi\_{\rho\circ\phi}=\chi\_\sigma,$$
which can be tested using $(\*).$ Of course, applying this method in practice requires good knowledge of the character table of $G$ and the outer automorphism group $Out(G).$
| 13 | https://mathoverflow.net/users/5740 | 36170 | 23,296 |
https://mathoverflow.net/questions/36175 | 14 | In the elementary group theory we know that for the symmetric groups $S\_n$, except $S\_6$, we have $Aut(S\_n) \cong S\_n$. Then the following question is natural:
What is the necessary and sufficient condition for $G$ such that $Aut(G) \cong G$?
| https://mathoverflow.net/users/3926 | Is there any criteria for whether the automorphism group of G is homomorphic to G itself? | This answer is essentially a series of remarks, but ones which I hope will be helpful to you.
(1) There are two ways to interpret the condition that $G$ be isomorphic to its automorphism group: canonically and non-canonically.
a) Say that $G$ is [complete](http://en.wikipedia.org/wiki/Complete_group) if every automorphism of $G$ is inner (i.e., conjugation by some element of $G$) and $G$ has trivial center. In this case, there is a canonical isomorphism $G \stackrel{\sim}{\rightarrow} \operatorname{Aut}(G)$.
The linked wikipedia article gives some interesting information about complete groups. As above, by definition having trivial center is a necessary condition; all nonabelian simple groups satisfy this. On the other hand, an interesting sufficient condition is that for any nonabelian simple group $G$, *its* automorphism group $\operatorname{Aut}(G)$ is complete, i.e., we have canonically $\operatorname{Aut}(G) = \operatorname{Aut}(\operatorname{Aut}(G))$.
b) It is possible for a group to have nontrivial center and outer automorphisms and for these two defects to "cancel each other out" and make $G$ noncanonically isomorphic to $\operatorname{Aut}(G)$. This happens for instance with the dihedral group of order $8$.
2) It seems extremely unlikely to me that there is a reasonable necessary and sufficient condition for a general finite group to be isomorphic to its automorphism group in either of the two sense above.
But a lot of specific examples are certainly known: see for instance
<http://en.wikipedia.org/wiki/List_of_finite_simple_groups>
in which the order of the outer automorphism group of each of the finite simple groups is given. So, for instance, exactly $14$ of the $26$ sporadic simple groups have trivial outer automorphism group, hence satisfy $G \cong \operatorname{Aut}(G)$.
I wouldn't be surprised if the outer automorphism groups of all finite groups of Lie type were known (they are not all known to me, but I'm no expert in these matters).
| 14 | https://mathoverflow.net/users/1149 | 36177 | 23,299 |
https://mathoverflow.net/questions/36128 | 9 | Let $q$ and $q'$ be complex numbers with $0<|q|,|q'|<1$, and let $m$ and $n$ be positive integers.
Suppose that $q^m={q'}^n$. Then the map
$$
f:\mathbb{C}^\times/q^{\mathbb{Z}} \to \mathbb{C}^\times/{q'}^{\mathbb{Z}}\qquad \text{defined by}\qquad u\mapsto u^m
$$
gives an isogeny of (analytic) elliptic curves over $\mathbb{C}$.
The Tate curve $\mathrm{Tate}(q)$ is an (algebraic) elliptic curve over the Laurent series ring $\mathbb{Z}((q))$ which can be used to give a uniformization of the curve $\mathbb{C}^\times/q^\mathbb{Z}$ by means of certain well known explicit formulae.
My question is:
>
> Does there exist an isogeny $\mathrm{Tate}(q)\to \mathrm{Tate}(q')$ of elliptic curves defined over $\mathbb{Z}((q,q'))/(q^m-{q'}^n)$ which "lifts" the map $f$ above, and if so, how do you prove it exists?
>
>
>
It should suffice to construct such an isogeny for $(m,n)=(m,1)$, and use dual isogenies and composition to get the general case.
(I'm being vague about "lifts", because one has to worry about convergence somewhere. Probably you want to say that the isogeny is defined over some subring of $\mathbb{Z}((q,q'))/(q^m-{q'}^n)$ of power series which are analytically convergent near $q=0$, or something like that.)
I presume (though I probably can't prove) that the existence of the analytic isogenies means that such a map of schemes is defined over $\mathbb{C}((q,q'))/(q^m-{q'}^n)$, so that this is just a question about integrality.
This is very closely related to exercise 5.10 in Silverman, *Advanced Topics in the Arithmetic of Elliptic Curves*. There, he apparently asks us to show that for a $p$-adic field $K$, if $q,q'\in K$, $0<|q|,|q'|<1$, and $q^m={q'}^n$, then the function $\overline{K}^\times/q^\mathbb{Z}\to \overline{K}^\times/{q'}^{\mathbb{Z}}$ defined by $u\mapsto u^m$ lifts to an isogeny $E\_q\to E\_{q'}$ of elliptic curves over $K$, where $E\_q$ and $E\_{q'}$ are defined by the Tate curve equations. (An answer to my question solves his exercise, right?)
Unfortunately, I have no idea how to do Silverman's exercise either (he marks it as difficult). Any hints?
| https://mathoverflow.net/users/437 | Isogenies between Tate curves | No matter how you define Tate(q), it should have the following properties:
(a) for any $n$ it contains a subgroup $M\_n$ canonically isomorphic to $\mu\_n$ (which corresponds tho $\mu\_n\subset\mathbb{C}^\times$ in the complex model),
(b) the (co)tangent space along the unit section is canonically trivialized (by $d\log u$ in the complex model).
Let me first treat the case $n=1$, as Charles suggests. The sought-for isogeny Tate($q$)$\to$Tate($q^m$) is characterized by two conditions:
(a') its kernel is $M\_m$ (i.e. it induces an isomorphism Tate($q$)$/M\_m\to$Tate($q^m$)),
(b') it induces multiplication by $m$ on the tangent space, modulo the identification (b).
Consider the scheme $X\to S:=\mathrm{Spec}\:\mathbb{Z}((q))$ parametrizing isomorphisms Tate($q$)$/M\_m\to$Tate($q^m$). This is an unramified $S$-scheme, and in fact it is finite because Tate($q$) has no complex multiplication in any fiber over $S$ (I guess this has to be checked). Since it has a section over $\mathrm{Spec}\:\mathbb{C}((q))$ it is dominant, hence surjective over $S$. Since $S$ is normal it suffices to find a section at the generic point. But by flat descent, condition (b') guarantees that the above section over $\mathrm{Spec}\:\mathbb{C}((q))$ descends to the fraction field of $\mathbb{Z}((q))$. QED.
Remark: I am a bit uncomfortable about the "no CM" stuff, but we can probably avoid it by noting that $X\to S$ satisfies the valuative criterion of properness, even when it is not of finite type. This (together with unramifiedness) is enough to imply that a section at the generic point extends over a normal base.
For arbitrary $n$, observe that we have just constructed $\alpha\_m:\text{Tate}(q)\to \text{Tate}(q^m)$ with kernel killed by $m$, so multiplication by $m$ factors as $\beta\_m\circ\alpha\_m$ for some $\beta\_m:\text{Tate}(q^m)\to \text{Tate}(q)$. You can now treat the general case by taking the composition
$$\text{Tate}(q)\to \text{Tate}(q^m)=\text{Tate}(q'^n)\to \text{Tate}(q')$$
where the two maps are $\alpha\_m$ and $\beta\_n$ respectively.
| 7 | https://mathoverflow.net/users/7666 | 36179 | 23,300 |
https://mathoverflow.net/questions/36178 | 13 | Warning: This is a very stupid question regarding a basic misunderstanding that I have. I realize that the question is very elementary, but I guess asking stupid questions is better than remaining ignorant.
To be explicit, consider the $\mathrm{SU}(2)$ Chern-Simons action on some very nice $3$-manifold $M$, i.e. the number
$S(A) = \frac{k}{4\pi}\int\_M \mathrm{tr} \left(A\wedge\mathrm{d}A + \frac{2}{3}A\wedge A\wedge A\right)$,
where $A$ is an $\mathfrak{su}(2)$-valued $1$-form on $M$.
What I simply cannot wrap my head around, and what is obviously a very silly basic question, is: What trace is this?! As I understand it, there is a certain abuse of notation in $\wedge$ on vector bundle-valued forms (namely, with $E$ the bundle, the wedge of two $E$-valued forms is an $E\otimes E$-valued one), but in the case of the Chern-Simons action [this answer](https://mathoverflow.net/questions/31905/some-basic-questions-about-chern-simons-theory/31930#31930) suggests that the $E\otimes E=\mathfrak{g}\otimes\mathfrak{g}\rightarrow\mathfrak{g}$ is supposed to be the Lie bracket on $\mathfrak{g}$. Anyway, that leads me to think that the trace is the trace on $\mathfrak{g}=\mathfrak{su}(2)$, which of course vanishes everywhere.
What am I misunderstanding here?
| https://mathoverflow.net/users/8653 | What is the trace in the Chern-Simons action? | The trace is simply a (properly normalised) ad-invariant inner product on the Lie algebra; that is, a nondegenerate symmetric bilinear form $\langle-,-\rangle$ which obeys the "associativity" condition
$$\langle [x,y],z \rangle = \langle x, [y,z] \rangle$$
for every $x,y,z$ in $\mathfrak{g}$.
Lie algebras admitting such inner products are said to be *metric*. The normalisation of the inner product is such that $k$ is an integer. This only makes sense for indecomposable metric Lie algebras; that is, those which are not isomorphic to the direct product of perpendicular proper ideals.
The notation "tr" stems from the fact that if $\rho: \mathfrak{g} \to \operatorname{End}(V)$ is a faithful finite-dimensional representation, then
$$\langle x, y\rangle := c \operatorname{tr}\rho(x)\rho(y)$$
works for a suitable nonzero $c$. (For a simple Lie algebra, just take $\rho$ to be the adjoint representation.)
For the explicit case of $\mathfrak{g}$ the Lie algebra of SU(2) you can take $\rho$ to be the fundamental representation and $c= -\frac12$, I believe.
---
**Edit**
Notice that $\operatorname{tr}(A \wedge dA)$ is really $\langle A \stackrel{\wedge}{,} dA \rangle$, where $\langle -\stackrel{\wedge}{,}-\rangle$ means that we are both taking the wedge product of the forms and the inner product on the Lie algebra. Similarly,
$$\operatorname{tr}(A \wedge A \wedge A) = \frac12 \langle [A\stackrel{\wedge}{,}A] \stackrel{\wedge}{,} A \rangle,$$
with a similar notational caveat about $[A\stackrel{\wedge}{,}A]$.
---
**Further addition**
In response to Anirbit's comment, I would say that there is, in general, no trace on vector-valued differential forms; although if the forms take values in endomorphisms, then of course there is: simply compose with the trace of endomorphisms to obtain a map
$$\Omega^\bullet(M;\operatorname{End}(V)) \to \Omega^\bullet(M).$$
| 19 | https://mathoverflow.net/users/394 | 36180 | 23,301 |
https://mathoverflow.net/questions/36156 | 2 | I'm working out of *Sheaves in geometry and logic*, for reference.
There is a characterisation of flat functors $A:C \to Set$ as those such that the Grothendieck construction $\int\_C A$ is a filtering category. There are more general versions of this result, in which $Set$ is replaced by a more general topos. One should also be able to characterise those discrete opfibrations that arise from flat functors (up to iso/equiv?). How about if we replace $C$ by an internal category, in a topos $E$ say? Then functors out of $C$ are replaced by discrete opfibrations over $C$ in $E$.
My question is this:
>
> What sort of thing should be considered as the analogue of a flat functor in the internal setting?
>
>
>
| https://mathoverflow.net/users/4177 | internal version of a flat functor? | A flat internal presheaf/discrete fibration $F \to C$ is simply one whose total category F is filtered in the internal sense (see *Topos Theory* 2.51 (filteredness), 4.31 (flatness); *Elephant* B.2.6.2, B.3.2.3). The definition just reexpresses filteredness by requiring that certain maps (e.g. $F\_0 \to 1$) are regular epis (presumably you can replace these with covers for your favourite Grothendieck topology).
I *think* that if C = BG is a group then these are exactly the G-torsors in E. I'm not sure whether this extends to flat = locally representable for C a general internal category, though.
| 5 | https://mathoverflow.net/users/4262 | 36182 | 23,303 |
https://mathoverflow.net/questions/36190 | 1 | I have the following question about short-time existence and uniqueness
results for non-linear schrodinger equations (NLSE) where the non-linearity involves
a loss of derivatives (in my case, it is a non-local non-linearity involving
a loss of two derivatives.) It seems that most current techniques allow
some small number of derivatives in the non-linearity, except for a series of papers by Poppenberg, which use Nash-Moser. (Unfortunately he seems to have left math.)
Question: are there any short-time existence and uniqueness results for NLSE with loss
of several derivatives for a space of initial conditions smaller than $C^\infty$? I have in mind for example a space introduced by Floer which is a dense subspace of $C^\infty$
that carries the structure of a Banach space whose norm is a combination of all C^k norms.
One could imagine doing something similar by combining all Sobolev norms.
I would be quite happy with a short-time existence and uniqueness result for initial
conditions in some dense subspace of C^\infty.
p.s. I am posting in a second a related question about Floer's Banach space with a symplectic geometry tag.
| https://mathoverflow.net/users/6223 | Short-time Existence/Uniqueness for Non-linear Schrodinger with Loss of Several Derivatives | Just a few thoughts. The answer to your question depends on a number of key factors. To focus, let us consider a nonlinear term like $F(D\_x^k u)$ and let us work in one space dimension.
1) How large is $k$? If $k\le2$ you can linearize the equation and work in Sobolev spaces. Of course you need some structural assumption on the nonlinearity (otherwise you may take e.g. $F(u\_{xx})=\pm 2i u\_{xx}+...$, and create all sorts of difficulties). There are some classical works by Kenig, Ponce, Vega on this (see "The Cauchy problem for the quasilinear S.e." around 2002 I think) which more or less give the complete picture from the classical point of view i.e. without trying to push below critical Sobolev etc. So if this is the case what are you exactly looking for? if you prefer to work in smaller spaces, what you need is a 'regularity' result, i.e., if the data are in some smaller space, this additional regularity propagates and the solution stays in the same space for some time. There are some results of this type, in classes of analytical or Gevrey functions; but see below.
2) If $k\ge3$, then the same remark as in (1) applies, you need some strong structural assumptions on the nonlinearity. Indeed, now the poor $u\_{xx}$ is no longer the leading term and the character of the (linearized) equation is entirely determined by the Taylor coefficient of $D^ku$ in the expansion of $F$. So then a careful case-by-case discussion is necessary. Unless...
3) unless, and we come maybe closer to your question, you decide to work in MUCH smaller spaces than $C^\infty$. Gevrey classes are roughly speaking classes of smooth functions such that the derivatives of order $j$ grow at most like $j!^s$. For $s=1$ you get analytic functions. For $ 1 < s < \infty $ you get larger classes $G^s$, with quite nice properties (to mention just one, you have compactly supported functions in these classes). For $ 0 < s < 1 $ the classes $G^s$ are rather small, strictly contained in the space of analytic functions. The only reason why $G^s$ for $s<1$ are useful is that you can prove a sort of very general Cauchy-Kowalewski theorem, local existence in Gevrey classes, for any evolution equation $u\_t=F(D^k\_xu)$, provided $s<1/k$. No structure is required on $F$, only smoothness. Contrary to the appearance this is a weak result, e.g. you can solve locally both $u\_t=\Delta u$ and $u\_t=-\Delta u$ in $G^{1/2}$ (and globally in $G^s$ for $s<1/2$ ), so you are essentially trivializing the equation and forgetting all of its structure. But if this is what you need I can give you pointers.
EDIT (since this does not fit in the comments):
I am a bit rusty on these topics, now I start to remember more details.
1) $s<1$. There is a problem with working in $G^s$ for $s<1$, and it is that this space is unstable for product of functions. So also in this case you need very special nonlinearities to work with. But the linear theory is straightforward and you can find an account maybe [here](http://onlinelibrary.wiley.com/doi/10.1002/mana.19911500104/abstract). Also Beals in some old paper developed semigroup theory in Gevrey classes. BTW, if you find a way to handle products this might be quite interesting.
2) $s>1$. The Mizohata school and other japanese mathematicians worked on NLS in Gevrey classes quite a lot, see e.g. [this paper](http://ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/ojm33_04_05.pdf)
3) The Nash-Moser approach might be useful provided you can prove that the linearized of your operator is solvable in every Sobolev class, with a fixed loss of derivatives. If you want to try this route, the best introduction to the theory I know of is in Hamilton's 1992 (?) Bull. AMS paper. It's very long but extremely readable, give it a try.
| 5 | https://mathoverflow.net/users/7294 | 36196 | 23,308 |
https://mathoverflow.net/questions/36183 | 13 | This is slightly an open-ended invitation to discuss references and reasons for excitement about the linear and non-linear sigma model.
I gauge from some other interactions that it has considerable interest even in mathematics (geometry?) apart from QFT.
I would like if people can type in what are the canonical explanatory/expository references about the linear and non-linear sigma model from both the QFT and mathematical points of view. It would be illuminating if experts can shed light on why is this model exciting again from both the QFT and mathematics point of view.
{As an example of the kind of expository reference I am looking for I would like to mention [this one](http://arxiv.org/abs/hep-ph/9802419) as an example for another interesting model in QFT}
| https://mathoverflow.net/users/2678 | Linear/Non-linear sigma model | I don't know anything about the QFT side, so I'll refrain from saying things about it.
For the mathematics, one of the reasons that there aren't that many expository/introductory references for it maybe because the development of the (non-linear) theory is rather incomplete. (The linear theory is sort-of trivial: it boils down to decoupled linear wave equations.) The simplest version of the non-linear sigma model is the harmonic map/wave map systems (the former is Riemannian/elliptic, the latter is Lorentzian/hyperbolic).
Perhaps I should say a few words here to establish notation. Here sigma model generally means a Lagrangian theory of maps for $\phi: M\to N$, where $M$, endowed with a pseudo-Riemannian metric $g$, is called the source manifold, and $N$ the target. The Lagrangian density is given by $\mathcal{L} = L dvol\_g$, where in index notation $L = g^{ij}k\_{AB}\partial\_i\phi^A\partial\_j\phi^B$ where $k\_{AB}$ is some symmetric tensor depending, possibly, on the map $\phi$ and its first jet.
Then the linear sigma model can be interpreted as when $N$ is some finite dimensional vector space and $k$ an inner product on $N$.
For the harmonic/wave map systems, $N$ is endowed with a Riemannian metric $h$, and $k\_{AB}$ is set to be equal to the metric $h$. So we can also view $L$ to be the $g$-trace of the pull-back metric $\phi^\*h$.
A lot of words have been written about harmonic maps. For an introduction, Jost's book *Riemannian Geometry and Geometric Analysis* has a good section on it. The notes of Helein *Harmonic Maps, Conservation Laws, and Moving Frames* is also quite nice. Schoen and Yau's *Lectures on Harmonic Maps*, as well as Eells and Lemaire's book *Selected Topics in Harmonics Maps*, are both very good.
One instance where the Riemannian harmonic maps have come into play is the study of stationary axisymmetric solutions to Einstein's equations in general relativity. I refer you to the works of [Gilbert Weinstein](http://www.math.uab.edu/weinstei/index.shtml) or to Luc Nguyen's PhD Thesis at Rutgers University.
For the Lorentzian version, the question is much more open. In the general case, local well-posedness follows from general theory (the system of PDEs forms a semilinear hyperbolic system of equations). As far as I know, all further work (blow-up or global existence for various target manifolds) have been done only with the source manifold being the Minkowski space. A reasonably complete set of references can be found at the [Dispersive Wiki](http://tosio.math.utoronto.ca/wiki/index.php/Wave_maps). Some of the notable news recently are the blow-up results of Rodnianski-Sterbenz and Raphael-Rodnianski, and the global wellposedness results for negatively curved targets due to Tao and Krieger-Schlag (you can find all these on the arXiv).
Now, the harmonic/wave map systems can be described as the simplest of a family of nonlinear sigma models for maps between pseudo-Riemannian manifolds. Assume now $(M,g)$ and $(N,h)$ are the source and target manifolds, and $\phi: M\to N$ some map. We shall write $D^\phi$ for the (1,1)-tensor field given by $g^{-1}\circ \phi^\*h$. $D^\phi$ induces at every point a linear map from $T\_pM$ to itself. Note that the harmonic/wave-map Lagrangian is characterized by $L = \mathop{tr} D^\phi$, or the first invariant $\lambda\_1$ of the matrix $D^\phi$. Naturally one asks whether Lagrangian field theories with the Lagrangian being (linear combinations of) other invariants $\lambda\_k$ are interesting.
There are two special cases which I know that are actively studied. The case where $L = \lambda\_1 + \lambda\_2$ is known as the Skyrme model (there are, again, a Riemannian and a Lorentzian version depending on the signature in the source manifold). You can read a lot more about it in Manton and Sutcliffe's book *Topological Solitons* (for the traditional case where the target manifold is $SU(2)$ with the bi-invariant metric and the source manifold is either Minkowski space or $\mathbb{R}^3$). This model originally arose in nucleon physics. One interesting fact is that the harmonic map system from $\mathbb{R}^3\to \mathbb{S}^3$ does not admit finite energy solutions; but it looks like the Riemannian Skyrme model might (it is still an open problem). You may want to look up the works of Lev Kapitanski if you are interested in theoretical work in this direction. For the Lorentzian version not much is known (it is one of the things I am working on; a pre-print from Jared Speck and me should be available after the job application season).
The other special case which has been studied is the case where $L = (\det D^\phi)^p$ ... roughly speaking. (In the case where $M$ and $N$ are both Riemannian and have the same dimensions, this is correct; in the case where $M$ is Lorentzian and has one more dimension than $N$, the determinant should be thought of as being restricted to space-like slices [else it vanishes identically].) This is the case of non-linear elasticity and fluid dynamics (though this is not how most books in elasticity and fluids formulate their theory, the various formulations are roughly equivalent). I'm not sure who the active participants are for this model, but I vaguely remember Lars Andersson having some interest in it. (For a bit more about the mathematical set-up of this model, a good reference is, if I recall correctly, Demetrios Christodoulou's 1998 AIHP paper "On the geometry and dynamics of crystalline continua" as well as his book *Action Principle and Partial Differential Equations*.)
| 8 | https://mathoverflow.net/users/3948 | 36198 | 23,310 |
https://mathoverflow.net/questions/36140 | -1 | I've seen some definition of the relative entropy between two states of a C\*algebra. However this definitions work only for finite dimensional C\*algebras and I don't know if there is a correspondent notion for only one state.
| https://mathoverflow.net/users/8642 | Is there a general notion of entropy for the states of a C*algebra? | There are several different definitions of entropy associated with operator algebras. It would be good if you knew which one you were referring to. Do you have a reference? I expect you're talking about the generalization of Kullback-Leibler relative entropy to matrix algebras.
There's a perfectly good definition of entropy for density matrices (positive self-adjoint trace 1 matrices); this is called [von Neumann entropy](http://en.wikipedia.org/wiki/Von_Neumann_entropy), and although I haven't thought about it I think you should be able to extend it to a definition of entropy for self-adjoint trace 1 operators for II$\_1$ factors in von Neumann algebras in the same way that you can extend Shannon entropy to continuous distributions by using differential entropy. In fact, I'd be surprised if that hasn't already been done.
| 1 | https://mathoverflow.net/users/2294 | 36200 | 23,311 |
https://mathoverflow.net/questions/33675 | 2 | The basis for the deterministic polynomial-time algorithm for primality of Agrawal, Kayal and Saxena is (the degree one version of) the following generalization of Fermat's theorem.
---
Theorem
-------
Suppose that P is a polynomial with integer coefficients, and that p is a prime number. Then
$(P(X))^p\equiv P(X^p)\ (\mod p)$.
---
Surely this result was known previously, but I have not been able to find a reference in the literature on the AKS algorithm (which means that the authors also did not know of a reference). Does anyone here know of one?
Furthermore, there is a converse to the lemma in the AKS paper:
---
Lemma
-----
If n is a composite number, then $(X+a)^n\not \equiv X^n+a\ (\mod n)$ whenever a is coprime to n.
---
Again, it is easy to generalize this statement. For example, if P is a polynomial which has at least two nonzero coefficients and such that all nonzero coefficients are coprime to n, then $P(X)^n\not\equiv P(X^n)\ (\mod n)$ for composite n.
On the other hand, clearly some conditions are necessary; for example $(3X+4)^6\equiv 3X^6+4\ (\mod 6)$.
Is there a best possible statement? And, again, is there a reference?
| https://mathoverflow.net/users/3651 | Fermat for polynomials, as used in the AKS (Agrawal-Kayal-Saxena) algorithm | Your first theorem occurs as an easily proved statement on p. 287 of Schönemann's article *Grundzüge einer allgemeinen Theorie der höhern Congruenzen, deren Modul eine reelle Primzahl ist*, J. Reine Angew. Math. 31 (1846), 269--325. Schönemann was one of the first mathematicians (not counting Gauss, who eliminated the corresponding Section 8 from his Disquisitiones at the last minute; see G. Frei's article "The Unpublished Section Eight: On the Way to Function Fields over a Finite Field" in *The shaping of arithmetic after C.F. Gauss's Disquisitiones Arithmeticae*) who studied the arithmetic of polynomials modulo primes. It might very well occur somewhere in Galois's papers, but it surely was considered to be essentially trivial by all of them.
This lemma also has a habit of showing up in various proofs of the irreducibility of the cyclotomic equation.
| 5 | https://mathoverflow.net/users/3503 | 36202 | 23,313 |
https://mathoverflow.net/questions/36199 | 7 | What kinds of Yoneda-like situations induce an embedding that preserves the tensor product for some arbitrary monoidal category?
The cases where the monoidal product is given by a limit or colimit give this immediately for the usual Yoneda embedding, but this breaks down for "real" monoidal categories like $(Vect, \otimes)$.
Are there $V$-enriched cases where the generalised embedding
$$ Y : C \to V^{C^{op}} $$
does preserve the tensor product for interesting monoidal categories $C$?
| https://mathoverflow.net/users/800 | In what cases does a Yoneda-like embedding preserve monoidal structure? | Day showed that, for suitable V, any monoidal structure on a (V-)functor category $[C^{\mathrm{op}}, V]$ is essentially determined by its restriction to the representables as
$$ F \otimes G = \int^{A,B} F A \otimes G B \otimes P(A,B,-) $$
where $P(A,B,-) = C(-, A) \otimes C(-, B)$ is a profunctor $C \otimes C \otimes C^{\mathrm{op}} \to V$. P (together with a unit and the usual structural isos) is said to endow C with a *promonoidal* structure.
If C is already a monoidal V-category, then there is a canonical promonoidal structure on it given by
$$ C(-, A) \otimes C(-, B) = C(-, A \otimes B) $$
In that case, the Yoneda embedding is strong monoidal by definition. In fact it is the unit for the monoidal cocompletion of C.
| 5 | https://mathoverflow.net/users/4262 | 36203 | 23,314 |
https://mathoverflow.net/questions/36208 | 1 | Let $A$ be a commutative euclidean ring, (probably) with 2 a unit in $A$. I'm trying to compute Witt and Grothendieck-Witt rings and since $A$ is a PID any f.g.p. module over it is free so I only need think about forms on $A^n$.
Question: Is every (non-degenerate) quadratic form over $A$ diagonalizable?
A form $q$ is diagonalizable we can perform a base change on $A^n$ such that the matrix for $q$ becomes diagonal. Edit: Non-degenerate here means that any matrix associated to the form is invertible.
From Milnor-Husemoller's book I know this is true if $A$ is local. If I can show that any non-degenerate quadratic form on $A^n$ represents some unit ($q(x) = u$ a unit for some $x \in A^n$) then the statement holds by cor. I.3.3 in Mil-Hus.
| https://mathoverflow.net/users/1123 | Diagonalization of quadratic forms over euclidean rings | Quadratic forms over $\mathbb{Z}$ don't diagonalize in general.
Even positive definite rank two forms like $3x^2+2xy+5y^2$ can't be diagonalized.
Inverting $2$ won't help things.
| 3 | https://mathoverflow.net/users/4213 | 36210 | 23,318 |
https://mathoverflow.net/questions/36050 | 9 | This is a follow up question to my answer here [How do you define the Euler Characteristic of a scheme?](https://mathoverflow.net/questions/35156/how-do-you-define-the-euler-characteristic-of-a-scheme/36038#36038)
A real analytic space is a ringed space locally isomorphic to $(X,O/I)$ where $X$ is the zero locus of some number of real analytic functions $f\_1,\ldots, f\_k$ on an open set $U$ of $\mathbf{R}^n$, $O$ is the sheaf of germs of real analytic functions on $U$ and $I$ is the ideal sheaf generated by $f\_1,\ldots, f\_k$ (see e.g. <http://eom.springer.de/a/a012430.htm>) I would like to ask if it is true that each real analytic space with a countable base can be embedded as a closed analytic subset of some Euclidean space.
The motivation behind this comes from the triangulation theorem for complex algebraic varieties: the only proof of that that I know of (Hironaka's 1974 notes) is based on triangulating analytic subvarieties of Euclidean spaces. So to apply this one must embed a complex algebraic variety as a real subvariety of a Euclidean space. This is easy for projective varieties and is probably possible in general, but I don't know a reference for the general case. (I'm mainly interested in the complex algebraic case, but I don't see why it should be any easier that embedding arbitrary real analytic spaces; however if it is easier, I'd be interested to know.)
A related question: is it possible to prove the triangulation theorem (for complex algebraic varieties or in general) without using embeddings in Euclidean spaces?
| https://mathoverflow.net/users/2349 | Embeddings and triangulations of real analytic varieties | If you just want a proper 1-1 real analytic map whose image is a real analytic variety then the result is theorem 2 page 593 of a paper of Tognoli and Tomassini in Ann.Scuola.Norm.Pisa
(3) vol 21 yr 1967 pages 575-598. This means there is no control over the differential of the
map.I am assuming that the real analytic space has finite dimension.If the dimension of the
Zariski tangent spaces of a connected reduced real analytic space is bounded then it can be
real analytically embedded in euclidean space, see paper by Aquistapace Broglia and Tognoli
Ann.Scuola.Norm.Pisa (4) vol 6 yr 1979 p 415-426.
| 5 | https://mathoverflow.net/users/4696 | 36213 | 23,319 |
https://mathoverflow.net/questions/36222 | 32 | I just discovered that something I've been working with has the structure of an operad. So I'm wondering what natural basic questions does one ask about operads? For example, if I knew I had the structure of a group, I'd ask if it is abelian or has torsion, etc. So what are these questions for operads?
| https://mathoverflow.net/users/8667 | What are natural questions to ask about an operad? | There's a lot of things you could ask.
* Operads act on things, that's their point. What things does your operad act on? Presumably this is how you found your operad. Moreover, once you know it acts on something you can ask if that action is maximal, whether or not your operad fits into a bigger operad that also acts on the thing in question, etc. Similarly, you can ask what does that operad tell you about the thing its acting on.
* Operads can have sub-operads, do you have any interesting ones? That would lead to other related questions, like is your operad an extension of other operads? Take a look at the Markl and Stasheff operad book to get a sense for some of the operads out in the literature, and what they're good for.
* There are things like totalizations and bar constructions for operads. What might that look like for your operad?
* Operads induce other operads, for example, the homology of a topological operad is another operad. Does your operad have any related operads that are known or otherwise interesting?
* (Edit, idea from Jeff's cyclicity suggestion) Operads sometimes fit into even larger higher-algebraic structures. Jeff mentions cyclic operads, but there are also PROPs, for example. You might want to consider that maybe you're dealing with something that's "more than" an operad.
That'd be a start.
| 19 | https://mathoverflow.net/users/1465 | 36223 | 23,322 |
https://mathoverflow.net/questions/36225 | 1 | Suppose *f* is a real-valued function of one variable, and suppose *f* is of differentiability class C1. My question is, if $\Gamma$ is the graph of *f*, then must $\dim\_H(\Gamma)=1$? If anyone knows of a published proof of the answer, I'd appreciate a reference greatly.
| https://mathoverflow.net/users/8665 | Do all graphs of C1 functions have Hausdorff dimension 1? | Let $f \colon I \to \mathbb{R}$. Since $f$ is $C^1$, the graph $\Gamma\_f$ is locally bilipschitz to $I$, via the projection. It follows that Hausdorff dimension is the same as that of $I$ (being defined in terms of the metric space structure only), so it is $1$.
Disclaimer: I haven't seen these topics for quite a while, so I may have said something stupid.
| 2 | https://mathoverflow.net/users/828 | 36233 | 23,325 |
https://mathoverflow.net/questions/36218 | 9 | It is well known that the automorphisms of a group $G$ form a group under composition, and that the group of inner automorphisms $\phi (x)=gxg^{-1}$ forms a normal subgroup of $\mbox{Aut}(G)$. Thus, $\mbox{Aut}(G)$ is simple if and only if either $\mbox{Inn}(G)=\mbox{Aut}(G)$ or $\mbox{Inn}(G)$ is trivial. In the second case, since $G/Z(G)=\mbox{Inn}(G)$, $G$ must be abelian. My question is, when does $\mbox{Inn}(G)=\mbox{Aut}(G)$? Or, as it is unlikely that the general case is not fully understood, are there nice classes of groups for which there are a nice set of criteria for $\mbox{Inn}(G)=\mbox{Aut}(G)$.
| https://mathoverflow.net/users/6856 | Criteria for Aut(G) to be simple | Here is an approximation of an answer to "For what finite groups is Aut(G) simple?"
As Daniel Miller mentioned, Inn(G) is a normal subgroup of Aut(G), so for Aut(G) to be simple either Inn(G) = 1, in which case G is abelian, or Inn(G) = Aut(G) is simple. The former case should be somewhat easy to handle assuming G is finite. In the latter case, we have that G/Z(G) is simple. If G is also perfect, then G is called **quasi-simple**. Of course, G need not be perfect as G ≅ A5 × 2 shows. However, I believe this is the only obstruction, so ignoring a possible cyclic direct factor of order 2, G/Z(G) is simple, and G is quasi-simple. The finite quasi-simple groups and their automorphism groups are classified, but the classification is a bit long. For a fixed simple group, X = G/Z(G), there are only finitely many isomorphism classes of quasi-simple groups D such that D/Z(D) = X. In fact there is a unique largest one called the Schur cover, that I'll call D. If Z(D) is cyclic, then in fact Aut(G) = Aut(X) = Aut(D) does not pay any attention to the center. So all we need to do is find all X with Aut(X) = X [and each one works], and all X with Z(D) non-cyclic [and check which ones work].
Having done most, but not all, of that, I thought it might help to record the basic result:
If G = H×T where T=1 if H is abelian and T is cyclic of order dividing 2 otherwise, and where H is on the following list, then Aut(G) is simple:
* cyclic of order 3, 4, or 6
* elementary abelian of order 2n for n ≥ 3
* M11, 2.Sz(8), J1, 2.Sp(6,2), M23, M24, Ru, 2.Ru, Co3, Co2, Ly, Th, Fi23, Co1, 2.Co1, J4, B, 2.B, E7(2), M
* Ω(2n+1,2) for n ≥ 3
* Sp(2n,2) for n ≥ 3
* E8(p) for any prime p
* F4(p) for any prime p
* G2(p) for any prime p ≥ 5
Additionally if Aut(G) is simple, then G = H×T as above, except possibly H/Z(H) is on the following list:
* L3(4), U4(3), U6(2), 2E6(2)
* Ω+(4n,q) for certain q
These are groups with non-cyclic multiplier other than Sz(8) [definitely an example] and Ω+(8,2) [not an example]. The Ω+(4n,q) case should be mostly easy, as there are too many automorphisms to kill. The others would be easy in an ideal world, but as far as I know our computational knowledge of these groups is limited and/or flawed. Of course, I also need to check the abelian case carefully, but I think 3,4,6 and 2^n are the only abelian examples.
It would make another good answer: **For what torsion abelian groups G is Aut(G) simple?** This would handle the abelian groups here, as well as some of the original poster's interest, without delving into the nastier aspects of abelian groups.
| 21 | https://mathoverflow.net/users/3710 | 36239 | 23,331 |
https://mathoverflow.net/questions/33125 | 2 | Consider a fourth order linear (biharmonic) PDE in two variables of the form
$\nabla^4u + c\nabla^2u-\lambda u = F(x,y)$; $(x,y) \in \Lambda$
To have uniqueness, we must specify two equations per point on $\partial \Lambda$. Now consider the limit where $c\rightarrow \infty$. The solution, $u$, is approximated by $\tilde u$, given by
$c\nabla^2\tilde u-\lambda \tilde u = F(x,y)$; $(x,y) \in \Lambda$
for which we must only specify one equation on the boundary.
The question is: what boundary conditions do we pick for $\tilde u$?
If I use a "clamped boundary" $u=u'=0$, the solution is approximated by using the boundary condition $\tilde u = 0$. If I use a "free boundary" $u''=u'''=0$, the appropriate approximation is $\tilde u'=0$.
| https://mathoverflow.net/users/5312 | What happens to the boundary conditions as a PDE is approximated by a lesser order PDE? | The question seems to contain a misprint, as Harald Hanche-Olsen pointed out above. I am assuming that the real situation is as follows: $ \epsilon\nabla^4u + c\nabla^2u-\lambda u=F(x,y)$, where $ (x,y)\in\Gamma $, with a pair of boundary conditions, say $ u=u'=0$, where $ (x,y)\in\delta\Gamma $.
When $\epsilon$ is "small" it is natural to assume that solutions of this PDE should be "close" in some sense to solutions of the reduced PDE $ c\nabla^2u'-\lambda u'=F(x,y) $.
This is a typical singular perturbation problem (see Van Dyke's "Perturbation methods in fluid mechanics" or any other text on asymptotic expansions that contains a discussion of singular perturbation problems). It is clear that since the order of the original equation has been reduced, one can now only satisfy one boundary condition, not two of the original problem. This basically indicates the presence of a boundary layer. This boundary layer is a solution of another asymptotic limit of the problem, the one for which $\epsilon\nabla^4u$ can no longer be omitted (see the comment by Willie Wong). The problem can generally be non-trivial in 2D; let me just show the idea how this type of problems is solved in 1D.
Consider for $x\ge 0$
$$ \epsilon^2u\_{,xxxx}-c^2u\_{,xx}-\lambda u=F(x) \quad\mbox{subject to}\quad u(0)=u\_{,x}(0)=0 $$
(I made some parameter changes to simplify the forthcoming algebra and changed the sign of $c^2$ as it actually makes more sense to pre-tension beam anyway). The "regular" solution $u\_r$, as you pointed out, satisfies $c^2u\_{r,xx}+\lambda u\_r=-F(x)$. The boundary layer solution is typically extracted by "zooming" into the vicinity of the boundary by using a coordinate transformation $\xi=x/\epsilon^{\alpha}$ (so that $\partial/\partial x=\epsilon^{-\alpha}\partial/\partial\xi$). How do we find $\alpha$? Intuitively, we want to keep the previously vanishing term $\epsilon^2u\_{,xxxx}$ in the equation and the non-trivial answer should balance it with at least one other term of the original equation. If $\alpha=1/2$ then first term is balanced by the third at $O(1)$, but the second term becomes $O(\epsilon^{-1})$. Hence, one needs to take $\alpha=1$ to balance the first two terms with the resulting equation
$$ \epsilon^{-2}u\_{,\xi\xi\xi\xi}-c^2\epsilon^{-2}u\_{,\xi\xi}-\lambda u=F(x). $$
Clearly, when $\epsilon$ is "small", the solution of this equation is approximated by the solution of boundary layer equation $u\_{bl,\xi\xi}-c^2u\_{bl}=0$. What happens next depends on the sign of $c^2$. I shall take $c^2$ to be strictly positive. Then the boundary layer solution is a rapidly decaying exponent $u\_{bl}=U\_{bl}\exp(-c\xi)=U\_{bl}\exp(-cx/\epsilon)$.
Let us now assume that the general solution of our problem is given by sum $u=u\_r+u\_{bl}$ (this is so-called Vishik-Lyusternik method). This means that when you were posing boundary condition $u\_{,x}(0)=0$, you effectively posed
$$ u\_{r,x}(0)+u\_{bl,x}(0)=0 $$
Since $u\_{bl}=U\_{bl}\exp(-cx/\epsilon)$, then $u\_{bl,x}=-cU\_{bl}\exp(-cx/\epsilon)/\epsilon=-cu\_{bl}/\epsilon$. Therefore, we can conclude that
$$ u\_{r,x}(0)+u\_{bl,x}(0)=u\_{r,x}(0)-cu\_{bl}(0)/\epsilon=0 \quad\mbox{so that}\quad u\_{bl}(0)=\epsilon u\_{r,x}(0)/c. $$
If we now turn our attention to the original first boundary condition, the same line of argument gives
$$ u(0)=u\_r(0)+u\_{bl}(0)=u\_r(0)+\epsilon u\_{r,x}(0)/c=0. $$
The conclusion then is: instead of solving the original fourth-order ODE with clamped boundary conditions, for sufficiently small $\epsilon$ one can simply solve the following second order boundary value problem
$$ c^2u\_{r,xx}+\lambda u\_r=-F(x) \quad\mbox{subject to the "effective" condition}\quad u\_r(0)+\epsilon u\_{r,x}(0)/c=0. $$
The resulting solution error will be of the order $O(\epsilon^2)$ (except in the close vicinity of the boundary where we made error $O(\epsilon)$). Your own suggested answer $u\_r(0)=0$ is only accurate to within $O(\epsilon)$ everywhere. My explanation is quite loose here, but I hope it should give you the idea of how to tackle this kind of problems.
| 3 | https://mathoverflow.net/users/8670 | 36246 | 23,335 |
https://mathoverflow.net/questions/36251 | 10 | I'm looking into a secret sharing scheme that has a secret permutation $\theta$ which has the cycle structure (n/2)+(n/2) (i.e. two (n/2)-cycles).
The permutation $\theta$ is decomposed into two permutations $\alpha$ and $\beta$, where $\alpha$ is generated uniformly at random. So with knowledge of both $\alpha$ and $\beta$, we can find $\theta$, while with knowledge of $\alpha$ xor $\beta$, we cannot find $\theta$ (although, we could guess).
At this point, I want to make public $\beta\alpha(L)$ (L is actually a Latin square, but this is not too relevant for the question I want to ask). It is possible that an attacker could find $\beta\alpha$ from $\beta\alpha(L)$. However, I worry that knowledge of $\beta\alpha$ might give information about $\theta$.
>
> If I know $\theta=\alpha\beta$, and I'm given the permutation $\beta\alpha$, what can I say about $\theta$? (without a priori knowledge of $\alpha$, $\beta$ or $\theta$)
>
>
>
| https://mathoverflow.net/users/2264 | What can I say about the permutation $\alpha\beta$ if I know the permutation $\beta\alpha$? | $\theta$ could be any permutation of the form $\alpha (\beta \alpha) \alpha^{-1}$; in other words, it could be any permutation conjugate to $\beta \alpha$, so knowing $\beta \alpha$ tells you only the cycle type of $\theta$, no more and no less. Since you already specified the cycle type this means an attacker gains no information (assuming $\alpha$ and $\beta$ really are chosen randomly).
| 25 | https://mathoverflow.net/users/290 | 36252 | 23,338 |
https://mathoverflow.net/questions/31116 | 3 | Hello,
Could you name a couple of books or downloadable lecture notes that discuss spectral graph theory and its connection to spectral problems in hyperbolic Riemann surfaces ? You could also mention some papers if you know.
Thank you !
| https://mathoverflow.net/users/6953 | Books that discuss spectral graph theory and its connection to eigenvalue problems in hyperbolic geometry | Lubotzky and Zuk's book [on property ($\tau$)](http://www.ma.huji.ac.il/~alexlub/) discusses expander graphs and the minimal eigenvalue of the Laplacian on covers of Riemann surfaces. See for example Prop. 2.9 in the book. There's also his book D[iscrete groups, expanding graphs and invariant measures](http://books.google.com/books?id=aNURlzNuotEC&lpg=PA1&dq=lubotzky%2520expanders&pg=PP1#v=onepage&q=lubotzky%2520expanders&f=false), but it's not available online. I don't know of relations between eigenvalues of graphs and eigenvalues of surfaces deeper into the spectrum, since the correspondence is only coarse. The papers of Robert Brooks and his collaborators are quite readable.
| 4 | https://mathoverflow.net/users/1345 | 36253 | 23,339 |
https://mathoverflow.net/questions/35396 | 5 | Let $F\_n$ where $n \ge 3$ be a free group and let $(\mathcal A\_n(k))$ where $k \ge 1$ be the
kernel of the homomorphism $Aut(F\_n) \to Aut(F\_n/\gamma\_{k+1}(F))$
determined by the natural homomorphism $F\_n \to F\_n/\gamma\_{k+1}(F).$
($(\mathcal A\_n(k) : k \ge 1)$ is called the Johnson filtration of $Aut(F\_n);$
$\gamma\_k(G)$ denotes the $k$-th terms of the lower central series of a group $G,$
$\gamma\_1(G)$ being equal to $G$).
I do not know an example of a group homomorphism
$f : Aut(F\_n) \to G$ which
takes ***all*** terms of the Johnson filtration
$(\mathcal A\_n(k))$ to the same ***nontrivial***
subgroup:
$$
1 \ne f(\mathcal A\_n(1))=f(\mathcal A\_n(2)) = \ldots = f(\mathcal A\_n(k)) = \ldots
$$
I would be very grateful for such an example, or for an argument
that homomorphisms like that do not exist.
| https://mathoverflow.net/users/7983 | Can all terms of the Johnson filtration be hom-mapped onto the same nontrival group? | The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A\_q$, a *simple* one (see his comment above). Let $K=\mathcal A(1).$ Then
$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad (m\quad times) \qquad (\*)$
Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F\_n) \to A\_q$ which doesn't vanish at $\alpha$.
Then
$$
A\_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K).
$$
It follows that
$$
\Delta( [K,K,\ldots,K])=A\_q
$$
and by $(\*)$ $\Delta( \mathcal A(m))=A\_q $ for every $m \ge 1.$
| 5 | https://mathoverflow.net/users/7983 | 36258 | 23,342 |
https://mathoverflow.net/questions/36263 | -3 | There are three boxes. B1, B2, B3 The probability of selecting them is 0.2, 0.2 , 0.6 respectively.
B1 contains 3 red balls and 7 green balls.
B2 contains 5 red balls and 5 green balls.
B3 contains 2 red balls and 8 green balls.
If we select a box and then a ball from the box what is the probability that the ball is of red color.
If we select the a ball and it turns out to be of green color what is the probability that it comes from B3 ?
| https://mathoverflow.net/users/8246 | Porbability of selecting balls from boxes | Well, this looks more like someone trying to get their homework done, but for the first part:
$ p = 0.2 \* \frac{3}{3+7} + 0.2 \* \frac{5}{5+5} + 0.6 \* \frac{2}{2+8}$
$ p = 0.06 + 0.10 + 0.12 $
$ p = 0.28 $
Showed the work for you too.
So if the probability that a chosen ball is red is 28%, then the probability that a chosen ball is green is 72%.
So what is {probability that chosen ball came from B3 | chosen ball is green}? Look up conditional probability, look up bayesian, etc.
$ p\_g = 0.2 \* \frac{7}{3+7} + 0.2 \* \frac{5}{5+5} + 0.6 \* \frac{8}{2+8}$
$ p\_g = 0.14 + 0.10 + 0.48 $
$ p\_g = 0.72 $
{ $p\_g3$ | green ball} = (picked from box 3 and green) / (picked green)
= 0.48 / 0.72
= 2 /3
Please do your homework yourself.
Showed the work for you too.
| 0 | https://mathoverflow.net/users/8676 | 36265 | 23,345 |
https://mathoverflow.net/questions/36276 | 14 | Perhaps this question will not be considered appropriate for MO - so be it. But hear me out before you dismiss it as completely elementary.
As the question suggests, I would like to know when $\sin(p\pi/q)$ can be expressed in radicals (in the way that $\sin(\pi/4) = \sqrt{2}/2$ and $\sin(\pi/3) = \sqrt{3}/2$ can). Let $\alpha = \sin(x)$, and consider the field extension $\mathbb{Q}[\alpha]$. Using $(\cos(x) + i\sin(x))^k = \cos(kx) + i \sin(kx)$ together with the binomial formula and the Pythagorean identity relating sine and cosine, we can see that $\sin(kx)$ lies in a solvable extension of $\mathbb{Q}[\alpha]$. Thus $\sin(p\pi/q )$ is expressible in radicals if $\sin(\pi/q)$ is.
To handle $\sin(\pi/q)$, we start by using the same trick (which most people also learn in an elementary trig class). Write $-1 = (\cos(\pi/q) + i\sin(\pi/q))^q$, use the binomial theorem to expand, compare imaginary parts, and express the right-hand-side in terms of sine using the Pythagorean identity. This gives an explicit equation for any $q$ one of whose solutions is $\sin(\pi/q)$. This equation is not a polynomial in $\sin(\pi/q)$ since it involves terms of the form $\sqrt{1 - \sin^2(\pi/q)}$, but it is enough to prove that $\sin(\pi/q)$ is algebraic.
So I am curious about the number theoretic properties of this equation. What can be said about the Galois group of its "splitting field" over $\mathbb{Q}$? Can we at least determine when it is solvable? Note that if the prime factors of $q$ are $p\_1, \ldots p\_k$ and we can express each $\sin(\pi/p\_j)$ in radicals, then the same is true for $\sin(\pi/q)$. So it suffices to consider the case where $q$ is prime. That's about all the progress I have made.
| https://mathoverflow.net/users/4362 | When is sin(r \pi) expressible in radicals for r rational? | As $\cos x=\pm\sqrt{1-\sin^2 x}$ and $e^{ix}=\cos x +i\sin x$, and
$\sin x=(e^{ix}-1/e^{ix})/2i$ then $\sin x$ is in a radical extension of $\mathbb{Q}$
iff $e^{ix}$ is. For rational $r$ with denominator $d$, $e^{2\pi i r}$
is a primitive $d$-th roots of unity. The extension of $\mathbb{Q}$
generated by a root of unity is a [cyclotomic field](http://en.wikipedia.org/wiki/Cyclotomic_field).
Every cyclotomic field is an abelian extension of $\mathbb{Q}$.
(By the Kronecker-Weber theorem any abelian extension is contained
in a cyclotomic field.) The Galois group of the $n$-th cyclotomic
field is isomorphic to the multplicative group
$(\mathbb{Z}/n\mathbb{Z})^\*$
So we can obtain any $\sin r\pi$ for $r$ rational in terms of radicals,
both in a trivial way if you allow an $n$-th root of unity as $1^{1/n}$,
and also in a stricter sense, if you insist that you ascend through
a chain of fields by adjoining at each stage a root of $x^m-a$
where this polynomial is irreducible over the previous field.
However if you insist on this more exacting definition, you will
need radicals of non-real numbers unless $n$ is a product of
distinct Fermat primes. All this is well-known.
| 29 | https://mathoverflow.net/users/4213 | 36277 | 23,352 |
https://mathoverflow.net/questions/36278 | 6 | Let $K$ be a local field of characteristic zero, $k$ its residue field, $R$ its ring of integers and $p$ the characteristic of the residue field $k$. Let $G$ be the Galois group of $K$, $I\subset G$ the inertia group and $P$ the maximal pro-$p$ subgroup of $I$. Let $I\_t:=I/P$.
Let $A\_0$ be an abelian scheme over $R$ with generic fibre $A$. Then $A[p]$ is an $I$-module.
Let $V$ be a Jordan-Hölder quotient of the $I$-module $A[p]$.
I am interested in the representation $I\to Aut(V)$.
Question (\*): Is it true that $P$ acts trivially on $V$?
(I have seen that there are results
of Raynaud and Serre on the "action of $I\_t$ on $V$". I want to study these things, but I am already stuck
with Question (\*) at the moment, i.e. with the question whether $I\_t$ acts at all.)
Maybe someone can help?
| https://mathoverflow.net/users/8680 | Abelian varieties over local fields | $V$ is an irreducible $\mathbb{F}\_p$-representation of $I$. As $P$ is a pro-$p$ group, $V^P\ne 0$, and $P$ is normal in $I$ so $V^P$ is stable under $I$. Therefore $V^P=V$. You might like to look at Tate's article on finite flat group schemes in "Modular Forms and Fermat's Last Theorem" for more information.
| 7 | https://mathoverflow.net/users/5480 | 36280 | 23,353 |
https://mathoverflow.net/questions/36281 | 6 | Since I'm dealing with the distinction between sequential continuous and continuous maps at the moment I came to ask myself once again what can be said about spaces where these two notions agree ([sequential spaces](http://en.wikipedia.org/wiki/Sequential_space)). Of course we all know that metric spaces and more generally first-countable spaces are sequential and in the literatur it seems that often metrizability or first-countability is only assumed in order to not need to distinguish between sequential continuity and continuity.
I'm mostly interested in spaces that arise naturally in functional analysis, i.e. subspaces of topological vector spaces. A well known theorem says that a hausdorff topological vector space is metrizable iff it is first-countable. I tried to find out what could be said about sequential t.v.s. Are sequential t.v.s. metrizable too? Are there any reasonable t.v.s. that are sequential but not metrizable?
| https://mathoverflow.net/users/3041 | Sequential topological vector spaces | The space of tempered distributions is sequential (for its usual strong topology). See, e.g., [Dudley](http://www.ams.org/journals/proc/1971-027-03/S0002-9939-1971-0270145-X/S0002-9939-1971-0270145-X.pdf), and the references therein.
| 12 | https://mathoverflow.net/users/2508 | 36300 | 23,362 |
https://mathoverflow.net/questions/36309 | 0 | What is the number of lines that pass through the center of an n-dimensional tic-tac-toe grid?
| https://mathoverflow.net/users/8682 | tic-tac-toe n-dimensional | The squares tic tac toe grid can be represented with $(x^1,x^2,...,x^n)$ where each can be 0, 1 or 2.
A line passing through the center can be identified Given any coordinate except $(0,0,...,0)$.
There are $3^n-1$ possible coordinates to start a line from.
For each line through center, there will be two endpoints. This should help you get your answer.
| 2 | https://mathoverflow.net/users/8602 | 36310 | 23,369 |
https://mathoverflow.net/questions/36105 | 9 | I recall hearing about a result, or maybe a cluster of results, in some area of complexity theory, probably algebraic, to the effect that there are known, specific, short formulas whose minimal derivation is known to be exceedingly long. Or perhaps it is a specific function that requires an exceedingly deep curcuit. The "philosophy" seemed to be that such examples would threaten to make the older asymptotic question obsolete: "Who cares about asymptotics if the constants are huge." Can anyone, given these hints, describe ( and direct me to) the result I overheard?
| https://mathoverflow.net/users/1643 | nonasymptotic complexity results | Perhaps you were told about
>
> Larry J. Stockmeyer, Albert R. Meyer: Cosmological lower bound on the circuit complexity of a small problem in logic. J. ACM 49(6): 753-784 (2002)
>
>
>
From the abstract: *"An exponential lower bound on the circuit complexity of deciding the weak monadic second-order theory of one successor (WS1S) is proved. Circuits are built from binary operations, or 2-input gates, which compute arbitrary Boolean functions. In particular, to decide the truth of logical formulas of length at most 610 in this second-order language requires a circuit containing at least $10^{125}$ gates. So even if each gate were the size of a proton, the circuit would not fit in the known universe.*
This demonstrates a specific function for which all inputs of size 610 cannot be determined within the physical universe. The proof is essentially a diagonalization argument, but is very carefully executed to take all constants into account.
(Note this is actually based on an asymptotic result, but it took some work to get a concrete lower bound out of it.)
| 4 | https://mathoverflow.net/users/2618 | 36313 | 23,371 |
https://mathoverflow.net/questions/36286 | 5 | Is every holomorphic vector bundle a direct summand of a trivial vector bundle on submanifolds of C^n? What about projective varities? I believe Swan's theorem says something about the first question. But I wanted to make sure.
| https://mathoverflow.net/users/3709 | Holomorphic vector bundles and Swan's theorem | The statement for Stein manifolds follows indeed from the analogue of the Serre-Swan theorem for Stein manifolds, which was proven first in 1967 in "Zur Theorie der Steinschen Algebren un Moduln" by O. Forster. The situation is a bit more complicated than the affine scheme or manifold case, but the final result relevant for the purposes of the question is the same
The category of locally free sheaves of finite rank is the same as the category of finitely generated projective modules over the global sections $\Gamma(O\_X)$. Then one notes that a f.g. projective module is always a direct summand of a finite free module.
| 8 | https://mathoverflow.net/users/6986 | 36315 | 23,372 |
https://mathoverflow.net/questions/36230 | 3 | Let $F\colon C\to D$ be a functor. Given a functor $\delta\colon D\to Sets$, one can compose with $F$ to get $\delta\circ F\colon C\to Sets$. This process is functorial in the category of $D$-sets, and we denote it $$F^\ast\colon D-set\to C-set.$$ The functor $F^\ast$ has both a left and a right adjoint, denoted $F\_!$ and $F\_\ast$ respectively.
It turns out that the two left adjoints in this picture, $F^\ast$ and $F\_!$ have nice descriptions in terms of the Grothendieck construction. $F^\ast$ is given by pullback: given a functor $\delta\colon C\to Sets$, take its Grothendieck construction $\int\delta\to D$ and form the product with $C\to D$ over $D$; this is $F^\ast\delta$. The "left pushforward" functor $F\_!$ is given by a factorization system on $Cat$; given $\gamma\colon C\to Sets$, we have a composition $$\int\gamma\to C\to D$$ which we can factor as an initial functor followed by a discrete op-fibration. The discrete op-fibration is the Grothendieck construction applied to $F\_!\gamma$.
Does the "right pushforward" functor $F\_\ast$ have any kind of description in terms of Grothendieck constructions? If not, is there a nice reason?
| https://mathoverflow.net/users/2811 | Does F_* have a description in terms of the Grothendieck construction? | I guess it depends what you would accept as a "description in terms of the Grothendieck construction."
For each $d \in D$, we have the projection $\pi\_d : d/F \rightarrow C$. Given some $\gamma : C \rightarrow Set$, I can form the composite $$ \gamma \circ \pi\_d : d/F \rightarrow C \rightarrow Set$$
This determines a discrete opfibration $\int \gamma \circ \pi\_d$ over $d/F$. Let $\Gamma\_{d/F}$ denote the *set* of sections of the projection $\int \gamma \circ \pi\_d \rightarrow d/F$. Then I believe you will find that $F\_\* \gamma$ is the Grothendieck construction on the functor $d \mapsto \Gamma\_{d/F}$.
Of course, this is not really that interesting, since it is really nothing more than the observation that the limit of a functor $F : C \rightarrow Set$ can be calculated as the set of sections of the natural projection $\int F \rightarrow C$, together with definition of the right Kan extension.
On the other hand, it suggests (to me at least) that there is unlikely to be a more "global" description in terms of the Grothendieck construction since the object we are trying to describe is like a "union of a collection of maps," and these to operations tend not to commute (maps from a union is the *product* of the maps on each component). Probably you knew all this, but maybe somebody will find it useful . . .
| 2 | https://mathoverflow.net/users/4466 | 36317 | 23,374 |
https://mathoverflow.net/questions/36312 | 4 | Consider a Seifert fiber space. Is it always possible to find a finite cover that is a circle bundle and the preimage of any fiber is a finite union of circles?
| https://mathoverflow.net/users/3375 | Getting rid of exceptional fibers by passing to finite covers? | If the Seifert fiber space is compact, then this is true, as long as the base orbifold is "good", which means that it has a finite-sheeted manifold cover, which is a compact surface. This induces a cover of the Seifert fiber space which is a circle bundle over the surface. If the base orbifold is bad, then no such covering will exist. This can happen for a Seifert fibering of $S^3$ over a football orbifold with distinct orders of torsion points, or over a teardrop orbifold.
If the Seifert fiber space is non-compact, then there may be infinitely many exceptional fibers, and the base orbifold might have torsion of arbitrarily large order, so there is no hope of finding a finite-index cover which is a circle bundle.
See the draft of [Thurston's book](http://math.berkeley.edu/~ianagol/276.S10/bookdraft.pdf) for more information on orbifolds and Seifert fibered spaces. Exercise 5.7.10 is on the Seifert fibering of $S^3$ over bad orbifolds.
| 6 | https://mathoverflow.net/users/1345 | 36323 | 23,378 |
https://mathoverflow.net/questions/36326 | 33 | I'm finishing up a PhD in math and am thinking about options outside of academia. So far, I've really only focused on pure mathematics, but I have a year left of grad school. Suppose I am interested in looking for a job in software engineering, finance, or some other quantitative field. What should I be doing in the next year? What classes should I take?
| https://mathoverflow.net/users/8694 | How to transition from pure math PhD to nonacademic career? | This is the perspective of someone who went from a math PhD to the media industry and then to software/tech (and enjoyed it immensely). I can't speak for finance.
Don't spend time on classes; instead, figure out the skills you want to acquire, and learn them yourself. One important reason to sidestep classes is that from now on you will have to learn things all the time without the benefit of formal instruction! So part of what you're going to be teaching yourself is how to learn quickly outside the classroom. Also, while it's always helpful to read theoretical material, you'll probably want to spend the majority of your time learning by doing.
A good approach is to actually do the job you want. Software engineering is a very welcoming world and there are a lot of avenues. Follow and then join an open-source project. Put up your own web site that does something interesting. Help out a nonprofit. Here's a more offbeat thought: almost every newspaper and magazine in the US is currently (a) in a financial crisis; and (b) desperate for technology help. It's quite possible that, even with few qualifications, you can get an unpaid internship at a name-brand place that will net you good recommendations and excellent experience. Note that this will be helpful even if you ultimately want to work in an industry that's not desperate and in a financial crisis :-)
Finally, keep in mind that a big benefit of the "do the job you want" approach is that you'll discover whether you actually like the job before you commit full-time. If you don't like it, choose something else. And keep this attitude long after you've graduated: one of the wonderful things about the non-academic world is the total freedom to change your career when you want.
| 23 | https://mathoverflow.net/users/1227 | 36327 | 23,381 |
https://mathoverflow.net/questions/36321 | 8 | Let $X$ be a (irreducible) variety (over $\mathbb{C}$ if necessary, smooth orbifold if necessary), and $U\subset X$ a nonempty open subset, and let $G$ be a finite group with an algebraic action on $U$.
When does the action extend to an action of $G$ on $X$? Are there any conditions that are reasonable to verify, as I'd prefer to avoid having to construct the action on $X\setminus U$ (I know the action exists for other reasons, but the method I'm using gives easily that I have an action on a nonempty open, rather than the whole thing)
| https://mathoverflow.net/users/622 | Extending group actions on varieties | Well, not always, e.g. take $X=\mathbb P^2$ with homogeneous co-ordinates $x,y,z$, $U$ the locus defined by $xyz\ne 0$ and $\sigma$ the involution of $U$ defined by $(x,y,z)\mapsto (yz,xz,xy)$ (the standard quadratic Cremona transformation of $X$). Then $\sigma$ is of order $2$ and does not extend to $X$. On the other hand some sufficient conditions, in characteristic zero, are: there are no rational curves in $X\setminus U$; $X$ has ample canonical class and is smooth (this can be weakened to having canonical singularities).
[Update: Charles Siegel asked for references, and I have none to hand, although this is all well known.]
For the Cremona example, recall that every automorphism of $\mathbb P^2$ is linear, and $\sigma$ clearly isn't. More geometrically, $\sigma$ blows up the vertices of the triangle $xyz =0$ and collapses its sides.
No rational curves in $X\setminus U$: fix $g\in G$ and think of it as a rational map $g:X\to X$. By Hironaka, there is a minimal composite $p:Z\to X$ of blow-ups such that $g\circ p$ is regular. If $X\setminus U$ has no rational curves, then the rational curves in the exceptional divisor of the last blow-up are contracted, so the last blow-up was redundant, contradicting minimality.
Ample canonical class $\omega\_X$: then $X=Proj(R(X,\omega\_X))$, with $R(X)= \oplus\_{n\ge 0}H^0(X,\omega\_X^{\otimes n})$, the canonical ring. This is a birational invariant of $X$, so $G$ acting on $U$ acts on $R(X)$, so on $X$.
| 6 | https://mathoverflow.net/users/8698 | 36337 | 23,388 |
https://mathoverflow.net/questions/36306 | 20 | I am confused about the "bigger picture" when one goes from classical modular forms on $SL\_2(\mathbb{Z})$ and its subgroups to automorphic forms (possibly non-holomorphic).
For classical modular forms we have cusp forms and Eisenstein series. There is the Peterson product defined when one of the forms is a cusp form (since cusp forms vanish at the cusps) and Eisenstein series are the orthogonal complement of cusp forms. For cusp forms there is Atkin-Lehner theory but not (as far as I can understand) for Eisenstein series.
Now from reading various sources, in order to generalise this to nonholomorphic forms, the correct thing is to look at $L^2(\Gamma\backslash SL\_2(\mathbb{R}))$. I have figured out how to make a modular form into a function on $SL\_2(\mathbb{R})$, but unless it is a cusp form it doesn't look like it is in $L^2$ (the inner product is just the generalization of Peterson, isn't it?)
On the other hand there are series which look like Eisenstein series like
$$
E\_\phi(z) = \sum\_{\gamma\in\Gamma\_\infty\backslash\Gamma} \phi(\mathrm{Im}(\gamma(z)))
$$
where $\phi$ is a compactly supported $C^\infty$-function on the positive reals. These obviously vanish at the cusps (and are in $L^2$) but they are (apparently) not cusp forms.
So my specific questions are: what is the right way to generalize modular forms so that holomorphic Eisenstein series still stay in the picture? Where do these $E\_\phi$ series fit in? And what happens to Hecke theory in this setting?
I would like to also understand the role of weight 2 "almost-holomorphic" Eisenstein series in a more conceptual way.
(I don't know much functional analysis, so Gelbart's book is proving rather difficult for me. Do I have to understand all the functional analysis to get a feeling for what is going on?)
| https://mathoverflow.net/users/8688 | Cusp forms and L^2 | This is a long answer because the question asks quite a lot of things. I agree that Gelbart's book, although inspirational, is hard for someone without a strong analytic background. The Boulder and Corvallis proceedings are full of articles which are worth studying if you want to get an understanding of automorphic theory. It is hard to summarise but here's a (necessarily oversimplified) sketch of the "big picture" you may be looking for.
First, two important things to get straight:
1) "Cuspidal" is not the same as "vanishing at the cusps". Sticking to the upper half-place for the moment, a function is cuspidal if at each cusp the 0-th Fourier coefficient vanishes. For a holomorphic function the Fourier coefficients are constant, so cuspidal and vanishing at cusps are the same thing. But for those functions $E\_\phi$ the 0-th Fourier coefficient at $\infty$ is some function $c\_0(y)$ vanishing in a neighbourhood of infinity, but certainly non-zero in general. In fact, $f\in L^2(\Gamma\backslash\mathfrak{H})$ is cuspidal iff it is orthogonal to all the $E\_\phi$ (compute the Petersson product to see this), so $\{E\_\phi\}$ generates a complement to the cusp forms in $L^2$. Arithmetically, though, they aren't interesting.
2) Automorphic forms and $L^2(\Gamma\backslash G)$ are different beasts entirely. An automorphic form is a smooth function on $\Gamma\backslash G$ satisfying various properties (moderate growth, K-finite, and killed by an ideal of finite codimension in the centre of the universal enveloping algebra). It does not have to be in $L^2$. For example, holomorphic Eisenstein series are automorphic forms which are not in $L^2$.
Hecke theory: for $SL\_2(\mathbb{Z})$ you can just mimic the classical definitions in this more general setting, but for groups other than $GL\_2/\mathbb{Q}$ you really need to look at the adelic setting. Here an automorphic form (for a reductive group $G/\mathbb{Q}$) is a smooth function on $G(\mathbb{Q})\backslash G(\mathbb{A})$, satisfying a bunch of properties (see Borel and Jacquet's article in Corvallis for a precise definition, indeed for just about everything here). The finite adelic group $G(\mathbb{A}\_f)$ acts by right translation on the space of automorphic forms, and this is the right generalisation of Hecke operators. The Lie group $G(\mathbb{R})$ does *not* act on the space of automorphic forms (K-finiteness is not preserved) but there is a suitable algebra of invariant differential operators (the Hecke algebra of $G(\mathbb{R})$) which does.
Cuspidal automorphic forms - in the adelic setting - are those $F$ for which the function
$$
F\_N(g) = \int\_{N(\mathbb{Q})\backslash N(\mathbb{A})} F(hg)\ dh
$$
vanishes for suitable unipotent subgroups $N$. For $GL\_2$ these are just the unipotent radicals of Borel subgroups (defined over $\mathbb{Q}$), and since these are all conjugate it's enough to verify the vanishing for the upper triangular unipotent subgroup.
A classical cusp newform $f$ of weight $k$ then gives rise to a cuspidal automorphic form $F$ on $GL\_2/\mathbb{Q}$. There are two different things one can now do with $F$:
1) The translates of $F$ under $GL\_2(\mathbf{A}\_f)$ generate an irreducible representation $V\_f$, which encodes the action of the Hecke operators (and more). Elements of this representation are none other than oldforms associated to $f$. More precisely, $V\_f$ is an infinite tensor product of representations $V\_p$ of $GL\_2(\mathbb{Q}\_p)$. If $p$ doesn't divide the level of $f$, then $V\_p$ tells you the Hecke $T\_p$ and $R\_p$ eigenvalues. At bad primes, it contains much more delicate information than classical Atkin-Lehner theory does (one reason to use the adelic approach even for $GL\_2$).
2) The translates of $F$ under the Hecke algebra at infinity, on the other hand, generate an irreducible representation $V\_\infty$ of a particular type (discrete series with parameter given by $k$), inside which $F$ is characterised as the lowest weight vector (this is the group-theoretic interpretation of the holomorphy of $f$).
The space of *all* translates of $F$ (i.e. by the finite adelic group and the Hecke algebra at infinity) is just $V\_\infty\otimes V\_f$. It is an example of an automorphic representation. (In general, an automorphic representation is any irreducible subquotient of the spaces of automorphic forms under these actions.)
The map $F\mapsto F\_N$ allows one to describe the quotient (automorphic forms)/(cusp forms) by induced representations from parabolic subgroups. Although explicit, this quotient is quite a complicated representation - in particular, it is very far from being semisimple, even for $GL\_2$.
So, if you are looking at this from a number-theorist's perspective, why care about $L^2$ ? The reason is the trace formula, which needs to be formulated in the setting of Hilbert spaces. There is essentially no difference between an $L^2$-form which is cuspidal and an automorphic cusp form, so the trace formula, appropriately wielded, can tell you a lot about cusp forms. It is a powerful and indispensable tool. But to apply it you need also to look at the rest of the $L^2$, in which lives the continuous spectrum, accounted for by real-analytic Eisenstein series at $Re(s)=1/2$. Here there is a lot of analysis but you can usually find a friendly expert to help you out.
PS: Some people like to define a cuspidal automorphic representation as an irreducible subspace of $L^2\_{\mathrm{cusp}}$. This has some advantages: (a) it's concise, and (b) at infinity one is working with genuine (unitary) representations of the real Lie group, rather than $(\mathfrak{g},K)$-modules (equivalently, representations of the real Hecke algebra). But it only gives the correct answer for cuspidal representations.
| 29 | https://mathoverflow.net/users/5480 | 36343 | 23,392 |
https://mathoverflow.net/questions/36345 | 34 | I'm wrapping up a summer project that involved a computation in Morava $E$-theory. As background knowledge I had to look into how the Johnson-Wilson theories $E(n)$ and Morava $K$-theories were constructed. This was manageable since I'd been part-way down that road already and there's lots of support in, for example, the form of Hopkins' course notes. Then, in early May I spent some time digging around for a construction of Morava $E$-theory, which led me to some conclusions:
* The words "Morava $E$-theory" don't determine what object you're talking about; there's a whole bunch of slightly different Morava $E$-theories.
* People frequently conflate Morava $E$-theory with (completed) Johnson-Wilson theory. One source even claimed (without citation) that one was a finite free module over the other, and that I therefore shouldn't worry about the difference.
In the end, I didn't need to really know much about $E\_n$ beyond a couple formal properties to work out the broad strokes of my computation, so I let the whole thing slide and pretended there existed a spectrum that did what I'd hoped.
However, I'm now getting to a point where understanding what I'm actually doing would be valuable. In decreasing order of importance, can someone provide a reference that...
* ... constructs a family of Morava $E$-theories. Any family would be a start! I am particularly interested in one with a coefficient ring of the form $\mathbb{Z}\_p[\![v\_1, \ldots, v\_{n-1}]\!][v\_n^{\pm 1}]$.
* ... illustrates that $\mathrm{spf}\,(E\_n)^\* \mathbb{C}P^\infty$ is the universal deformation of $\mathrm{spf}\,K(n)^\* \mathbb{C}P^\infty$ to formal groups over formal spectra of complete, local rings with residue field $\mathbb{F}\_p[v\_n^{\pm 1}]$. The remark above about the comparison between Johnson-Wilson theory and Morava $E$-theory made me particularly uncomfortable in this respect; it's not clear to me that the formal group associated to Johnson-Wilson theory should be thought of as the universal deformation of the Honda formal group. Clearing that up would be nice too.
* ... also shows that the reduction of the universal deformation to the "mod $p$" case exists as a spectrum, and the reduction map exists as a map of spectra. That is, there is a complex-oriented, structured ring spectrum $E\_n/p$ with coefficient ring $\mathbb{F}\_p[\![v\_1, \ldots, v\_{n-1}]\!][v\_n^{\pm 1}]$ whose associated formal group is the universal deformation of the Honda formal group to complete, local rings of characteristic $p$.
* ... also shows that the reduction of the universal deformation modulo the $n$th power of its maximal ideal exists as a spectrum, and the reduction map exists as a map of spectra.
* ... demonstrates this fact about $E\_n$ being a finite free module over $E(n)$, at least for an appropriate interpretation of the symbol "$E\_n$".
It may not be the case that points 3 and 4 are even true, but I'm hopeful. Still, surely this is all catalogued somewhere!
| https://mathoverflow.net/users/1094 | Construction of Morava E-theory | So far as what Morava E-theory should be: Morava E-theory *always* implicitly comes with a choice of a perfect residue field of positive characteristic and a formal group law of finite height over this field. Sometimes people take a very specific formal group law, but there is no reason to be restrictive. The underlying homotopy type of spectrum may not change in a very interesting way, but the multiplicative structure *does* (just as many rings that are different in interesting ways may have the same underlying abelian groups). This actually becomes a more serious issue with things like $BP\langle n\rangle$ and the Johnson-Wilson theories because there are multiple possible inequivalent orientations that look basically the same when you write down the rings of homotopy groups.
Your points in order:
* Constructing a family of Morava E-theories. (The coefficient ring you list is implicitly the coefficient ring of a completed Johnson-Wilson theory, and that's only if I interpret your power series brackets as "invert vn and then complete" due to the grading issue.) Johnson-Wilson theories and completed Johnson-Wilson theories, with the names you've given to the generators, all satisfy Landweber's criterion and so you can produce them the easy way as spectra via the Landweber exact functor theorem. Morava E-theories are slightly more difficult; they're still Landweber exact, but to prove that you need to know that the universal deformation ring of a formal group law of finite height has a particular structure relative to the coordinates of its p-series. However, if you want to construct any of these as genuine commutative ring spectra then you've got to use the Goerss-Hopkins-Miller theorem on the Morava E-theories, use it with its functoriality properties for the completed Johnson-Wilson theories, and you're out of luck for the uncompleted Johnson-Wilson theories except in a handful of very specific cases (some of which may require you to be slightly flexible about what "Johnson-Wilson" means). For references for the Hopkins-Miller theorem in the associative case, there are of course Charles Rezk's Notes on the Hopkins-Miller theorem, although you have a far better source on hand. The Goerss-Hopkins paper [Moduli problems for structured ring spectra](http://www.math.northwestern.edu/~pgoerss/spectra/obstruct.pdf) covers the obstruction theory for making these objects commutative.
* So far as illustration goes, the method I was proposing in the previous paragraph *started* with the universal deformation ring and produced the E-theory as its manifestation, meaning that by the time you get here you've already finished this step. Note that you have to be a little careful about the grading issue. The coefficient ring of a Morava E-theory in its full, graded glory is the universal deformation ring of a formal group law equipped with a choice of generator of the relative cotangent space (a trivial torsor for the multiplicative group over the universal deformation of the formal group law).
* The mod-p reduction exists as a spectrum, but not as a *commutative* ring spectrum. Any commutative ring spectrum with $p=0$ in its homotopy groups is a module over the Eilenberg-Mac Lane spectrum ${\rm H}\mathbb{F}\_p$ (this is because this Eilenberg-Mac Lane spectrum is actually the free algebra over the little 2-disks operad with $p=0$!) and these residue fields don't qualify. You can show that these residue objects exist using obstruction theory; there are a number of references but let me specifically plug Vigleik Angeltveit's [Topological Hochschild homology and cohomology of A-infinity ring spectra](http://www.msp.warwick.ac.uk/gt/2008/12-02/p022.xhtml).
* Modulo the n'th power of the maximal ideal is a little trickier and I would have to go check to see if the obstruction theory necessarily worked out. If you're willing to instead kill the n'th powers of the generators of the maximal ideal then the obstruction theory from the previous point still works, because you are killing a regular sequence of generators. Again, this only produces associative objects rather than commutative ones.
* The fact that $E\_n$ is a finite free module over the completed $E(n)$ is a formal consequence of the computation of the homotopy groups of both involved. The left-hand object has homotopy groups
$$
\mathbb{Z}\_{p^n}[\![u\_1,\ldots,u\_{n-1}]\!][u^{\pm 1}]
$$
with $|u\_i| = 0, |u| = 2$ and the right-hand object has homotopy groups as the subring
$$
\mathbb{Z}[v\_1,\cdots,v\_n,v\_n^{-1}]^\wedge
$$
where $v\_i = u^{p^i - 1} u\_i$ for $1 \leq i < n$, $v\_n = u^{p^n - 1}$, and the completion is taken with respect to the intersection of the maximal ideal with this subring. Once you have figured out what this entails, you find that the homotopy groups are a finite free module and as a consequence the spectrum itself is a finite free module.
| 31 | https://mathoverflow.net/users/360 | 36349 | 23,394 |
https://mathoverflow.net/questions/36299 | 11 | Was providing an alternative proof of the **PNT** one of the main impulses that led to the *discovery* of the Tauberian theorem of Wiener and Ikehara or the other way around?
In any case, do you know who was the **first** individual to realize that, in order to prove the **PNT**, all one needs to have is the non-vanishing of the Riemann zeta function on the line $\sigma =1$ plus a Tauberian theorem of the Wiener-Ikehara persuasion?
I thank you in advance for your insightful replies.
**References**
[**1**] P. T. Bateman & H. G. Diamond. *A hundred years of prime numbers*. Amer. Math. Monthly **103** (1996), no. 9, pp. 729-741.
| https://mathoverflow.net/users/1593 | The Wiener-Ikehara approach to the PNT | This is an elaboration on my comment below John's answer. Its goal is to give a very brief summary of the history underlying the question; I hope that it is more or less correct.
I think that it is fair to say that from the beginning it was understood that non-vanishing on the line $\Re(s) = 1$ was the main requirement for proving the prime number theorem.
However, in Hadamard and de la Vallee Poussin's approaches, this non-vanishing was fed into an explicit formula (which gives a Fourier-type expansion of the prime counting function, or some variant), which was in turn obtained from the $\zeta$-function by various Mellin transform games. In particular, if I understand correctly, establishing the explicit formula involves pushing the
contour of integration to the left of $s = 1$ (since it involves a sum over the zeroes
of the zeta function), and so doesn't just involve analysis in the
region $\Re(s) \geq 1$.
As noted in anon's answer, Landau formulated an approach to PNT via a Tauberian theorem involving just analysis on the region $\Re(s) \geq 1,$ but as well as the crucial condition
of there being no zeroes on the line $\Re(s) = 1$, there was a growth condition.
In his paper, Wiener discusses earlier Tauberian approaches, including Landau's, and then
refers to Ikehara's theorem (proved in Ikehara's thesis, I believe, under Wiener's supervision) as being the "true theorem" (I'm fairly confident that I'm remembering his language correctly here), i.e. the one with the correct condition, those conditions being simply that there are no zeroes on the line $\Re(s) = 1$.
| 5 | https://mathoverflow.net/users/2874 | 36354 | 23,396 |
https://mathoverflow.net/questions/36314 | 2 | I've been hearing about using the sign of the real part of eigenvalues of the multivariate derivative of the change vector field of an ODE at an equilibrium point to determine whether that equilibrium is stable or unstable. Unfortunately, the source is not very good at things like getting the statement of theorems correct, so I was wondering:
Is there a statement and proof of such a theorem I could access online?
I'm particularly interested in if anything can be said about the non-diagonalizable case where the eigenvalues have zero real part.
| https://mathoverflow.net/users/nan | Linearization at equilibrium points | I got curious about "the non-diagonalizable case" with purely imaginary eigenvalues
and came up with the system
$$ y' = A y $$ where
$$
A \; = \; \left( \begin{array}{rrrr}
0 & -1 & 1 & 0 \\\
1 & 0 & 0 & 1 \\\
0 & 0 & 0 & -1 \\\
0 & 0 & 1 & 0
\end{array}
\right) .
$$
Taking the matrix of eigenvectors and "generalized eigenvectors"
(see <http://en.wikipedia.org/wiki/Generalized_eigenvector> )
$$
B \; = \; \left( \begin{array}{rrrr}
1 & i & 0 & 0 \\\
i & 1 & 0 & 0 \\\
0 & 0 & 1 & i \\\
0 & 0 & i & 1
\end{array}
\right) .
$$
we get a permuted Jordan form in the shape I prefer for this case,
$$
B^{-1} A B \; = \; \left( \begin{array}{rrrr}
-i & 0 & 1 & 0 \\\
0 & i & 0 & 1 \\\
0 & 0 & -i & 0 \\\
0 & 0 & 0 & i
\end{array}
\right) .
$$
This is something I made up years ago and forgot, with an original real matrix and a repeated characteristic value $\alpha$ with non-diagonal Jordan block, we get the same block for the complex conjugate $\bar{\alpha}.$ For both the original real matrix $A$ and
$B^{-1} A B $ we can rearrange everything into convenient 2 by 2 blocks, in particular the off-diagonal 1's become little 2 by 2 identity matrices. Try it for a six by six example where there is a 3 by 3 block for $\alpha$ with two off-diagonal ones, then a 3 by 3 block for $\bar{\alpha}$ with two off-diagonal ones. Permute the diagonal elements to
$ \alpha, \; \bar{\alpha}, \; \alpha, \; \bar{\alpha}, \; \alpha, \; \bar{\alpha} $ and see what happens. By the way, these "forms" probably have standard names.
After a bunch of work I found that the "fundamental matrix" for the system is
$$
e^{A t} \; = \; \left( \begin{array}{rrrr}
\cos t & - \sin t & t \cos t & - t \sin t \\\
\sin t & \cos t & t \sin t & t \cos t \\\
0 & 0 & \cos t & - \sin t \\\
0 & 0 & \sin t & \cos t
\end{array}
\right) .
$$
which means that any solution of the system is $ y = e^{A t} y\_0.$
So, if the third and fourth entries in $y\_0$ are $0,$ the orbit is a circle. If not, the orbit leaves the origin as $t$ increases. The trick from my first answer, making a not quite linear system, can force the periodic orbits to switch to attracting.
| 1 | https://mathoverflow.net/users/3324 | 36355 | 23,397 |
https://mathoverflow.net/questions/36358 | 13 | I wonder if anybody can help me with this problem.
I'm trying to compute the Mertens function for large $n$. The most obvious algorithm is just to compute all primes up to $\sqrt{n}$ and then to sieve. That takes at least an order of $n\log n$ operations, and really even more.
The most recent article that I could find that discusses methods to compute the function directly is dated 1994, and it proposes to do exactly that.
Are there any known algorithms that let you compute Mertens faster than by sieving? I know that $\pi(n)$ can be computed in $O(n^{2/3})$, I looked into that algorithm but it does not seem to be easily adaptable to my task.
Alternatively, I could use an algorithm to compute $M(n+dn)-M(n)$ for $dn\ll n$ (say $dn\sim \sqrt{n}$ ) in $O(\sqrt{n})$ time or less.
| https://mathoverflow.net/users/8701 | Computing the Mertens function | [This](https://projecteuclid.org/journals/experimental-mathematics/volume-5/issue-4/Computing-the-summation-of-the-M%C3%B6bius-function/em/1047565447.full) article presents an algorithm to compute Mertens function in $O(x^{2/3}(\log \log x)^{1/3})$ time and $O(x^{1/3}(\log \log x)^{2/3})$ space, I wonder if it is the same one you are referring to. On the other hand people sometimes make use of certain recursions such as the results in [this paper](https://www.unirioja.es/cu/jvarona/downloads/Benito-Varona-TOMATJ-Mertens.pdf) to compute things about the Mertens function. [This](https://web.archive.org/web/20170808045325/http://www.emis.ams.org/journals/EM/expmath/volumes/13/13.4/Kotnik.pdf) paper seems to claim that these algorithms haven't been improved upon.
| 11 | https://mathoverflow.net/users/2384 | 36360 | 23,399 |
https://mathoverflow.net/questions/36350 | 5 | My attempts to search via Google seem to be failing, so I thought of asking here.
All the derivatives of the function
$r\_n(z):=\frac{J\_n(z)}{J\_{n-1}(z)}$
where $J\_n(z)$ is the Bessel function of the first kind are expressible in terms of $r\_n(z)$, for instance $\frac{\mathrm{d}}{\mathrm{d}z}r\_n(z)=r\_n(z)^2-\frac{2n-1}{z}r\_n(z)+1$ . I've been trying to derive a (linear?) differential equation (hopefully just second-order) that might be satisfied by $r\_n(z)$, but my manipulative ability does not seem to be up to snuff.
Probably my problem can be resolved in two ways:
1. Are there any papers where generating functions/differential equations of ratios of Bessel functions have been studied? ; or
2. How can I derive a differential equation for $r\_n(z)$ with the knowledge that all higher derivatives are expressible in terms of $r\_n(z)$?
(On the other hand, the *difference* equation for $r\_n(z)$ (and thus its continued fraction representation) is easily derived, so no problem for me there.)
I will be interested in any input. Thanks!
**EDIT:**
Per Pietro's request, I now tip my hand and reveal my reason for interest: I saw [this paper](http://dx.doi.org/10.1016/S0010-4655%2898%2900193-3) many years ago on a neat method for computing the first few roots of the Bessel function of the first kind. Some time later, I came across J.F. Traub's "Iterative Methods for the Solution of Nonlinear Equations", where he shows the [construction of iteration functions involving derivatives](http://books.google.com/books?id=se3YdgFgz4YC&pg=PA78) and can be constructed to have quadratic, cubic... convergence. (Newton's method is but the first member of this family). I also came across this [short note](http://www.ams.org/journals/mcom/1958-12-061/S0025-5718-1958-0099752-6/S0025-5718-1958-0099752-6.pdf) by D.J. Hofsommer on how one might profitably exploit the methods derived by Traub if the function of interest satisfies a simple differential equation (Essentially, one just constructs the Newton correction $u=\frac{f(z)}{f^{\prime}(z)}$, and the high-order iteration functions are merely a series in powers of $u$). That got me wondering on how one might recursively generate iteration functions with increasing order of convergence for the case of finding the roots of the Bessel function. (On another note, I was able to successfully use the ideas of Traub and Hofsommer for the generation of Gaussian quadrature rules, e.g. Legendre, Lobatto, Radau, and was hoping things might be just as successful for Bessel function root-finding).
| https://mathoverflow.net/users/7934 | Differential equation for a ratio of consecutive Bessel functions | No, you will not find a linear ordinary differential equation (with polynomial coefficients) for $r\_n$. This is because $J\_{n-1}$ has infinitely many zeroes which are not cancelled out by the zeroes of $J\_n$, so that $r\_n$ has infinitely many poles. Holonomic functions, aka solutions of LODEs with polynomial coefficients, can only have a finite number of singularities.
The transformation between LODEs and Riccati equations involves transformations between an equation for a quantity $y(z)$ [your $r\_n$] and an LODE for $-u'/u$. So what you'll get is indeed an LODE for a single Bessel $J$ function!
| 8 | https://mathoverflow.net/users/3993 | 36374 | 23,407 |
https://mathoverflow.net/questions/36127 | 13 | Let $(A,\mathfrak{m})$ be a local Artinian ring with
finite residue field, which I'm happy to assume is $\mathbf{F}\_3$.
(In particular, $A$ has finitely many elements.)
I would like to do some computations of the following kind, as $I$ ranges over
all of the ideals of $A$.
(0) A way to enumerate all the ideals of $A$.
(1) For an ideal $I$ of $A$, compute the length of $I/I^2$.
(2) For an ideal $I$ of $A$, compute the ideal $J = \mathrm{Ann}(I)$.
(3) For an ideal $I$ of $A$, decide if $I$ is principal. (By computing the length of
$I/\mathfrak{m} I$ or otherwise.)
The ring $A$ itself will be given explicitly as a quotient of a power series
ring over $W(\mathbf{F}\_3) = \mathbf{Z}\_3$. For example, $A$ might be
given as $\mathbf{Z}\_3[[x]]/(27,9x,x^3)$ or $\mathbf{Z}\_3[[x]]/(9,x^2)$.
My question: What is the computer algebra package that is best suited to carry
out these computations? (I would like something that can be semi-automated for various possible $A$.) I would be interested in even a very simple one like $\mathbf{Z}\_3[[x]]/(9,x^2)$
EDIT 2: There seems to be a consensus in the comments that this problem is significantly more manageable if $A$ is actually an algebra over its residue field. For example, in MAGMA, it is only possible to create ideals and quotient rings in univariate polynomial rings over fields. Other computer algebra packages have similar issues when the coefficient ring is not a field, although SINGULAR (for example) has some functionality with polynomials in several variables. As it happens, the problem I was interested in studying is still of interest for such fields.
| https://mathoverflow.net/users/nan | Computational Question about finite local rings: | This is fleshed out from comments of Sam Lichtenstein; Some (or all) of what I have written is surely not the most elegant programming (at best), feel free to improve. One can do the following with MACAULAY2 (if you copy and paste this into the MACAULAY2 prompt it should work:)
`R = GF(2)[x,y]/(x^3,y^3);
m = ideal(x,y);
ModSquare = ideal -> length(ideal/(ideal*ideal));
Generators = ideal -> length(ideal/(ideal*m));
I := ideal(x^2 + y^2);
J := ann(I);
[Generators(I),Generators(J),ModSquare(I),ModSquare(J)]`
The function ModSquare applied to an ideal $I$ computes the length of $I/I^2$, and
the function Generators computes the minimal number of generators of $I$. Similarly, ann computes the annihilator of $I$. From this we may compute that $I$ is principal, $J$ has two generators, and both $I/I^2$ and $J/J^2$ have length $4$. These functions only work for homogenous ideals. Following James Parson's suggestion, I also considered MAGMA (again with the restriction to affine algebras), and the following works for arbitrary ideals:
`A:=AffineAlgebra`<`GF(2),x,y|x^3,y^3`>`; x:=A.1; y:=A.2; AssignNames(~A,["x","y"]);
m:=ideal`<`A|x,y`>`;
Minus := func`<`ideal | Dimension(quo`<`A|ideal`>`)`>`;
Generators := func`<`ideal | Minus(ideal*m) - Minus(ideal)`>`;
ModSquare := func`<`ideal | Minus(ideal*ideal) - Minus(ideal)`>`;
ann := func`<`ideal | Annihilator(ideal)`>`;
I:=ideal<A|x^2+y^2>;
J:=ann(I);
[Generators(I),Generators(J),ModSquare(I),ModSquare(J)];`
| 5 | https://mathoverflow.net/users/nan | 36375 | 23,408 |
https://mathoverflow.net/questions/36362 | 20 | Let me admit right at the outset that I have a very superficial outsider's knowledge of homotopy theory. Nevertheless, I was trying to gain some understanding of Hopkins' ICM lecture ['algebraic topology and modular forms.'](http://arxiv.org/abs/math/0212397)
In section 6, he mentions two constructions. To a map
$$\phi: MSpin\rightarrow KO$$
of $E\_{\infty}$ ring spectra, he associates a characteristic power series $$K\_{\phi}(x)\in \mathbb{Q}[[x]].$$ Similarly, to an $E\_{\infty}$-map
$$\psi: MO\langle 8\rangle \rightarrow tmf,$$
he associates a power series $$K\_{\psi}(x)\in MF\_{\mathbb{Q}}[[x]],$$ where $tmf$ is the topological modular form spectrum and $MF\_{\mathbb{Q}}=MF\otimes \_{\mathbb Z}\mathbb Q$ is the ring of modular forms with rational coefficients.
I wonder if someone could give a brief outline of how these associations are carried out. I presume it is something elementary having to do with the homotopy groups of $MSpin$ and $MO\langle 8\rangle$, but I don't quite have the resources right now to track these down.
As usual with questions of this sort, I'm sure my level of ignorance is incongruous with the words I am employing already, but thank you in advance for any tolerant answers or references.
Added:
Maybe I should summarize the point of this question for fellow number-theorists who are too busy to look into the paper. In the notation above, one associates to $\phi$ a characteristic sequence
$$b(\phi)=(b\_2, b\_4, b\_6,\ldots)$$
via the formula
$$\log(K\_{\phi}(x))=-2\sum\_{n>0} b\_n\frac{x^n}{n!}.$$
Incredibly, this procedure sets up a bijection:
homotopy classes of $E\_{\infty}$ maps from $MSpin$ to $KO$ $\leftrightarrow$ the set of sequences of rational numbers $(b\_i)$ as above that satisfy
(1) $b\_n\equiv B\_{n}/n \ \ \mod \mathbb{Z}$, where the $B\_n$ are the Bernouilli numbers;
(2) for each odd prime $p$ and $p$-adic unit $c$,
$$m\equiv n \ \mod p^k(p-1) \Rightarrow (1-c^n)(1-p^{n-1})b\_n \equiv (1-c^m)(1-p^{m-1})b\_m \ \mod p^{k+1};$$
(3) for each $2$-adic unit $c$,
$$m\equiv n \ \mod 2^k \Rightarrow (1-c^n)(1-2^{n-1})b\_n \equiv (1-c^m)(1-2^{m-1})b\_m \ \mod 2^{k+2}.$$
In the case of the homotopy classes of maps from $MO\langle 8\rangle$ to $tmf$, one gets similar congruences involving Eisenstein series instead of their constant terms. Incidentally, perhaps these congruences imply the ones above?
| https://mathoverflow.net/users/1826 | Characteristic power series for maps of E_{\infty} ring spectra | In short, the series $K\_\phi$ is the "Hirzebruch characteristic series" which arises in the construction/calculation of genera, and in Hirzebruch-Riemann-Roch. The first few chapters of *Manifolds and modular forms* by Hirzebruch et al. describe the classical version of this pretty well.
If I have a one dimensional formal group law $F$ over a ring $A$, then over $A\_{\mathbb{Q}}$ there is an isomorphism $\mathrm{exp}\_F: G\_a\to F$ with the additive formal group. Let $K(x)=x/\mathrm{exp}\_F(x)$.
Now suppose $R$ is a "complex orientable cohomology theory", which means we are given a suitable isomorphism of rings $R^\*(CP^\infty)\approx \pi\_\*R[[x]]$. Such a theory has an associated formal group law $F$ (induced by the map $CP^\infty\times CP^\infty\to CP^\infty$ which classifies tensor product of line bundles), and thus there is an assocated series $K(x)=x/\mathrm{exp}\_F(x)$ in $\pi\_\*R\_\mathbb{Q}[[x]]$.
It turns out that a map of ring spectra $\phi:MU\to R$ corresponds exactly to giving a complex orientation of $R$. By Thom, elements of $\pi\_\*MU$ correspond to cobordism classes of stably-almost-complex manifolds, and there is a standard calculus due to Hirzebruch of calculating the effect of the map $\pi\_\*MU \to \pi\_\*R\_{\mathbb{Q}}$ using $K(x)$, which I might as well call $K\_\phi(x)$, since it depends on $\phi$. The formula (if I remember correctly), is that, if $[M]\in \pi\_\*MU$ is the class corresponding to a manifold of dimension $2n$, then
$$
\phi(M) = \langle K\_\phi(x\_1)\dots K\_\phi(x\_n), [M] \rangle,
$$
where the $x\_i$ are the "chern roots" of the tangent bundle of $M$, and $[M]\in H\_{2n}M$ is the fundamental class.
There is a "universal example" of a $K\_\phi$, corresponding to the identity map $\phi\colon MU\to MU$. It turns out that $\pi\_\*MU\_{\mathbb{Q}}$ is a polynomial ring on the coefficients of $K\_\phi$, so that $K\_\phi(x)=\sum a\_{i-1}x^i$ (with $a\_0=1$) and $\pi\_\*MU\_{\mathbb{Q}}=\mathbb{Q}[a\_1,a\_2,\dots]$. (I'll need this later.)
In his talk, Mike isn't talking about complex orientations, but rather orientations with respect to $MSpin$ or $MO\langle 8\rangle$ (instead of $MO\langle 8\rangle$, we call it $MString$ these days, for some reason).
There is a map of ring spectrum $MU\to MSO$, induced by the apparent homomorphisms $U(n)\to SO(2n)$ of Lie groups. There is also a map $MSpin\to MSO$, induced by the double cover of lie groups. Although $MSpin\neq MSO$, we have that $\pi\_\*MSpin\_{\mathbb{Q}}\to \pi\_\*MSO\_{\mathbb{Q}}$ is an isomorphsism. Thus, a map $\phi\colon MSpin\to R$ induces
$$ \pi\_\*MU\_{\mathbb{Q}}\to \pi\_\*MSO\_{\mathbb{Q}}\approx \pi\_\*MSpin\_{\mathbb{Q}}\to \pi\_\*R\_{\mathbb{Q}},$$
and we can get $K\_\phi(x)$ from this.
The $MO\langle 8\rangle$ case is a little trickier.
There is a map $MU\langle 6\rangle \to MO\langle 8\rangle$, so a ring spectrum map $\phi\colon MO\langle 8\rangle\to R$ gives rise to a map
$$ MU\langle 6\rangle\_{\mathbb{Q}} \to R\_{\mathbb{Q}}.$$
On the other hand, the effect of the map $MU\langle 6\rangle\to MU$, on homotopy groups tensored with $\mathbb{Q}$, is
$$ \mathbb{Q}[a\_3,a\_4,\dots]\to \mathbb{Q}[a\_1,a\_2,a\_3,\dots].$$
So a map $\phi\colon MO\langle 8\rangle\to R$ gives us elements $\phi(a\_i)\in \pi\_{2i}R$ for $i\geq3$, which we can use as the coefficients of a series $K\_\phi(x)\in \pi\_\*R\_{\mathbb{Q}}$.
I must point out: there is actually an error in the statement of (2) and (3) given in Mike's talk. What he writes down are the "Kummer congruences"; but what one really needs to require are the "generalized Kummer congruences", which are basically the collection of all possible $p$-adic congruences involving Bernoulli numbers, not just the ones listed in (2) and (3). This comes from the theory of the "Mazur measure": the generalized Kummer congruences imply that the sequence $b\_n(1-p^{n-1})(1-c^n)$ can be interpolated to a function $f$, so that $f(n)$ for $n$ an integer is the moment of a measure on $\mathbb{Z}\_p^\times$. With (2) and (3) replaced by "interpolates to the moments of a measure on $\mathbb{Z}\_p^\times$", the result is correct.
Finally: There is a writeup of this at <http://www.math.uiuc.edu/~mando/papers/koandtmf.pdf>, which may or may not be of any use to you!
| 16 | https://mathoverflow.net/users/437 | 36381 | 23,412 |
https://mathoverflow.net/questions/36379 | 31 | Just a basic point-set topology question: clearly we can detect differences in topologies using convergent sequences, but is there an example of two distinct topologies on the same set which have the same convergent sequences?
| https://mathoverflow.net/users/7005 | Is a topology determined by its convergent sequences? | In a metric (or metrizable) space, the topology is entirely determined by convergence of sequences. This does not hold in an arbitrary topological space, and Mariano has given the canonical counterexample. This is the beginning of more penetrating theories of convergence given by [nets](http://en.wikipedia.org/wiki/Net_(topology)) and/or [filters](http://en.wikipedia.org/wiki/Filter_(mathematics)). For information on this, see e.g.
[http://alpha.math.uga.edu/~pete/convergence.pdf](http://alpha.math.uga.edu/%7Epete/convergence.pdf)
In particular, Section 2 is devoted to the topic of sequences in topological spaces and gives some information on when sequences are "topologically sufficient".
In particular a topology *is* determined by specifying which nets converge to which points. This came up as a [previous MO question](https://mathoverflow.net/questions/19285/how-do-you-axiomatize-topology-via-nets). It is not covered in the notes above, but is well treated in Kelley's *General Topology*.
| 33 | https://mathoverflow.net/users/1149 | 36382 | 23,413 |
https://mathoverflow.net/questions/36319 | 2 | Often in theorems of pcf theory one has the assumption that the length of sequences of functions has to respect some bound so things can be well-defined. For instance, let $a=[\aleph\_2,...,\aleph\_n,...:n<\omega]$ be a set of regular cardinals, say you have a sequence $f\_\beta$ in $\prod a$ of length at most $|a|^+$. Then $sup\_\beta f\_\beta \in \prod a$ since $|a|^+ < min(a)$. But why is this true? If you have for example an $\omega\_2$ sequence of functions $f:\kappa \rightarrow \kappa$ such that $f(\kappa)\in \kappa$, $\kappa$ some $\aleph\_n$, $n$ not 0 and not 1,then why is $f\_\beta$ for $\beta=\omega\_2$ outside of the product, as far as we know, we don't know if $2^{\aleph\_0}= \aleph\_2 $ since $a$ is a countable set of regular cardinals (say the set of $\aleph\_n$'s)? Thanks
| https://mathoverflow.net/users/3859 | An assumption in pcf theory | Let $a$ be a set of regular cardinals. An element of $\prod a$ is a function $f:a \to \sup a$ such that $f(\kappa) < \kappa$ for every $\kappa \in a$. Suppose we are given a family $f\_i$, $i \in I$, of elements of $\prod a$. In order to ensure that $\sup\_{i \in I} f\_i \in \prod a$ we need to make sure that $\sup\_{i \in I} f\_i(\kappa) < \kappa$ for every $\kappa \in a$. A sufficient condition for this is that $|I| < \min a$. Indeed, since $|I| < \mathrm{cf}(\kappa) = \kappa$, we then have $\sup\_{i \in I} f\_i(\kappa) < \kappa$ for every $\kappa \in a$. Thus the assumption $|a| < \min a$ ensures that the supremum of any sequence of elements of $\prod a$ with length less than $|a|^+$ has a supremum in $\prod a$.
| 3 | https://mathoverflow.net/users/2000 | 36383 | 23,414 |
https://mathoverflow.net/questions/36296 | 6 | Let $M$ be a compact subset of $\mathbb R^2$ with smooth boundary, and let $g$ be a Riemannian metric on $M$. If $g'$ is another Riemannian metric which is "close" to $g$, then they should have almost identical curvature profiles. I would like to prove a concrete estimate on the total difference of their curvatures in terms of the distance of $g'$ to $g$. Before I state the question precisely, I need to introduce some notation.
Write $\operatorname{Sym}$ for the space of symmetric $2\times 2$ real matrices, and let $\operatorname{SPD} \subseteq \operatorname{Sym}$ be those matrices which are also positive-definite. Consider the function space $\Omega = C^2(M, \operatorname{SPD})$. Denote partial derivatives of $g\_{ij} \in \Omega$ by additional subscripts following a comma, so that $\tfrac{\partial}{\partial x^k} g\_{ij} = g\_{ij,k}$, et cetera. Endow the space $\Omega$ with the norm $$\|g\| = \sup\_{x \in M} \max\_{i,j,k,l} \left\{|g\_{ij}(x)|, |g\_{ij,k}(x)|, |g\_{ij,kl}(x)| \right\},$$ so that it has the structure of an open cone within the Banach space $C^2(M, \operatorname{Sym})$.
Each $g \in \Omega$ defines a Riemannian structure on $M$ via the inner product $\langle v, g(x) v' \rangle$ for $v, v' \in T\_x M$. Let $K(g,x)$ be the scalar curvature of the metric $g$ at the point $x \in M$.
**What I want to prove:** For each $g \in \Omega$, there exist constants $C$ and $\epsilon$ so that if $g' \in \Omega$ with $\|g - g'\| < \epsilon$, then $$\sup\_{x \in M} \left| K(g,x) - K(g',x) \right| \le C\|g- g'\|.$$
My current approach to this is quite clunky, and involves calculating everything directly from the Christoffel symbols of the metrics. Is there a better, more geometric approach to this than brute force calculations?
I'm sure this type of lemma is well known to geometric analysts. Is a proof of a similar result written down somewhere?
| https://mathoverflow.net/users/238 | How does curvature change under perturbations of a Riemannian metric? | This is a straightforward consequence of the fact that $K(x)$ is a continuous function of $g(x)$, $\partial g(x)$, and $\partial^2(g)$.
| 5 | https://mathoverflow.net/users/613 | 36385 | 23,416 |
https://mathoverflow.net/questions/6444 | 14 | Let $R\_n$ be a simple random walk with $R\_0 = 0$, and let $T$ be the smallest index such that $k\sqrt{T} < |R\_T|$ for some positive $k$.
What is an expression for the probability distribution of $T$?
| https://mathoverflow.net/users/2003 | How long for a simple random walk to exceed $\sqrt{T}$? | For a Brownian motion, [Novikov](http://siamdl.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=TPRBAU000016000003000449000001&idtype=cvips&gifs=Yes) finds an explicit expression for any real moments (positive and negative) of the random variable $(\tau(a,b,c)+c)$, where
$$
\tau(a,b,c) = \inf(t \geq 0, W(t) \leq -a +b(t+c)^{1/2})
$$
with $a \geq 0$, $c \geq 0$, and $bc^{1/2} < a$. [Shepp](https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-38/issue-6/A-First-Passage-Problem-for-the-Wiener-Process/10.1214/aoms/1177698626.full) provides similar results but with W(t) replaced by |W(t)| in the definition, and the range of permissible $a,b,c$ restricted accordingly. Shepp also cites papers by Blackwell and Freedman (1964), Chow, Robbins, and Teicher (1965), and Chow and Teicher (1965), which look like they prove similar but weaker results when the Brownian motion is replaced by a random walk with finite variance. I don't have time to read those references at the moment but I figure these papers should lead you to your answer.
| 7 | https://mathoverflow.net/users/3401 | 36413 | 23,432 |
https://mathoverflow.net/questions/36405 | 19 | The Prime Number Theorem was originally proved using methods in complex analysis. Erdos and Selberg gave an elementary proof of the Prime Number Theorem. Here, "elementary" means no use of complex function theory.
Is it possible that any theorem in number theory can be proved without use of the complex numbers?
On the one hand, it seems a lot of the theorems using in analytic number theory are about the distributions of primes. Since the Prime Number Theorem has an elementary proof, this might suggest that elementary proofs exist in other cases.
On the other hand, the distribution of primes is intimately related to the zeros of the Riemann Zeta function. Perhaps the proofs of other statements in analytic number theory require more direct references to the Riemann Zeta function.
This topic is more of a fascination for me, as I am not a number theorist. I would be interested if there are other examples of elementary proofs of theorems originally proved with complex analytic methods.
| https://mathoverflow.net/users/8358 | Complex and Elementary Proofs in Number Theory | Yes, there is a theorem to this effect by Takeuti given in his book "Two applications of logic to mathematics". He shows roughly that complex analysis can be developed in a conservative extension of Peano arithmetic.
| 17 | https://mathoverflow.net/users/51 | 36414 | 23,433 |
https://mathoverflow.net/questions/35560 | 4 | This question is related to the previous discussion [here](https://mathoverflow.net/questions/16393/finding-a-cycle-of-fixed-length).
Due to the [result](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.42.3377) of Noga Alon et al., there is an $O((2k)^kn)$ algorithm for deciding whether a planar graph $G$ contains a fixed subgraph $H$ of size $k$, and the time complexity is reduced to $O(2^kn)$ if the graph $H$ is of bounded treewidth. Take $k = O(\log n)$ yields a polynomial time algorithm for the latter case, say the $k$-path problem mentioned by Ryan Williams in [this paper](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V0F-4TX797W-1&_user=7761201&_coverDate=02%252F28%252F2009&_rdoc=1&_fmt=high&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1430356488&_rerunOrigin=google&_acct=C000051951&_version=1&_urlVersion=0&_userid=7761201&md5=903daeb9157174e1d0630f4b232f46e0).
There is an open problem in the [result](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.42.3377):
>
> If we want to solve $k$-path problem in a planar graph
> with slightly larger $k$, say $k = O(\log^2 n)$, is there a polynomial
> time solution at this point? If so, what is the best time complexity at present?
>
>
>
| https://mathoverflow.net/users/4248 | Finding a subgraph with slightly large size in planar graphs | I remember thinking about this a while ago, and stopped because it seemed unlikely that $log^2 n$ paths can be found in polynomial time. This was my argument, if I remember correctly.
The best known algorithm for Hamiltonian path is $O^\*(2^n)$. I think improving this to sub-exponential time, like $O(2^{o(n)})$ would violate the exponential time hypothesis (ETH).
Now if we had a polynomial time algorithm for finding a path of length k, where $k=O(\log^2 n)$, then we could solve Hamiltonian path on a graph with k vertices in time polynomial in n. Since $k=O(\log^2 n)$, polynomial time in n translates to time $O(2^{\sqrt{k}})$ in terms of k. This is a sub-exponential time algorithm for Hamiltonian path, which violates the ETH.
If this reasoning is correct, then it's quite unlikely that $\log^2 n$ length paths can be found in polynomial time on general graphs. As for planar graphs, I think ETH gives a lower bound of $\Omega(2^{\sqrt{n}})$, so maybe $\log^2 n$ graphs can still be found, but not any larger, like $\log^{2.1} n$ paths.
| 4 | https://mathoverflow.net/users/8075 | 36415 | 23,434 |
https://mathoverflow.net/questions/36403 | 12 | Let $X$ be an infinite dimensional separable Hilbert Space with norm $||\cdot||$ and let $\mu$ be a Gaussian measure on $X$ such that $\mu(X) = 1$. What do we know about $\mu(B(0,1))$, where $B(0,1)$ is the unit ball w.r.t the norm?
This seems to me like a fundamental question but I cannot seem to find anything. Any information/references would be most appreciated.
EDIT: A related question which is of interest to me: Do there exist asymptotically tight bounds to
$\int\_{||u||> K}||u||^2 \mu(du)$?
| https://mathoverflow.net/users/4047 | What is known about the Gaussian measure of the unit ball in a Hilbert Space? | You can't talk about "the" Gaussian measure on an infinite-dimensional Hilbert space, for the same reason that you can't talk about a uniform probability distribution over all integers. It doesn't exist; see Richard's answer. However, there are a lot of non-uniform Gaussian measures on infinite dimensional Hilbert spaces.
Consider the measure on $\mathbb{R}^\infty$ where the $j$th coordinate is a Gaussian with mean 0 and variance $\sigma\_j^2$, where $\sum\_{j=1}^{\infty} \sigma\_j^2 < \infty$ (and different coordinates are independent). This is almost surely bounded in the $\ell\_2$ metric, and any projection onto a finite-dimensional space has a Gaussian distribution. The squared length of a vector drawn from this measure is a sum of squares of Gaussians, and so follows some kind of generalized $\chi$-square distribution. If I knew more about generalized $\chi$-square distributions, I might be able to tell you what the measure of the unit ball was.
This kind of Gaussian distribution is very important in quantum optics. In fact, in quantum optics, a thermal state is Gaussian, so "the" Gaussian measure actually makes some sense.
| 10 | https://mathoverflow.net/users/2294 | 36417 | 23,436 |
https://mathoverflow.net/questions/36396 | 13 | Suppose that $(A,\mathfrak{m})$ is a local Artinian ring.
If $A$ is Gorenstein, then $A$ admits a dualizing functor
on finite length modules defined by $D(M):= Hom\_A(M,A)$ which preserves
lengths. If $M$ is a finite length $A$ module, let $\mathrm{length}(M)$
denote the length of $M$.
If $I$ is an ideal of $A$ and $J = \mathrm{Ann}(I)$, and
$A$ is Gorenstein, it is not hard to show that
$I = \mathrm{Ann}(J)$.
Question: Suppose that $A$ is a local Gorenstein Artinian ring, $I$ is an ideal,
and $J = \mathrm{Ann}(I)$. If $I$ is principal, is it the
case that
$$\mathrm{length}(J/J^2) \ge^{?} \mathrm{length}(I/I^2)$$
If so, can one characterize when equality holds?
Remarks:
1. If $J$ is also principal, then (reversing $I$ and $J$) one
predicts there should be an equality. One can show that this is the case.
However, equality sometimes occurs without $J$ being principal, as can be seen from
the example:
$A = \mathbf{F}\_2[x,y]/(x^3,y^3)$, $I = (x^2+y^2)$, $J = (x^2 + y^2,xy)$, with
$\mathrm{length}(I/I^2) = \mathrm{length}(J/J^2) = 4$.
2. My reasons for believing that the answer to the question is "yes" are somewhat obscure, and I would not be entirely surprised if it turns out to be false.
3. I do not know a counterexample to the claim that $\mathrm{length}(J/J^2) \ge
\mathrm{length}(I/I^2)$ even *without* the Gorenstein hypothesis. I would
be interested in seeing such a counterexample, if one exists (although I'm more interested in the Gorenstein case).
EDIT: As usual with bounty questions, this has been "answered" although not yet completely solved --- further comments still welcome.
| https://mathoverflow.net/users/nan | Length of I/I^2 versus Ann(I)/Ann(I)^2 in Artinian rings. | UPDATE 10/04/10:
Here are some partial results in the graded case. Let $R=\oplus\_0^s R\_i$ be a graded Gorenstein algebra over $k=R\_0$ ($s$ is the socle degree). $R$ is said to have *strong Lefschetz property* (SLP) if for a general linear form $l$ in $R$, the multiplication map $\times l^a: R\_i \to R\_{i+a}$ has maximal rank for all $i\geq 0, a \geq 1$ (so it is either injective or surjective). Such $l$ is called a Lefschetz element.
By a result of Stanley, monomial complete intersections of characteristic $0$, e.g. $R=k[x\_1\cdots,x\_d]/(x\_1^{a\_1},\cdots, x\_d^{a\_d})$ have SLP (apply the Hard Lefschetz Theorem!). Many more classes of rings with SLP are known, and I think it is conjectured to hold for all complete intersections, at least in char. $0$ (the keywords are Weak Lefschetz and Strong Lefschetz Property, the literature is quite big, see for example [this paper](http://atlas.mat.ub.es/personals/miro/MMR3.pdf)).
Now, suppose $R$ is Gorenstein with SLP, and $x$ be a Lefschetz element. I claim that such an element would satisfy your inequality. Let $L$ be the annihilator of $x^2$. By Fact 1 we have that $\text{length} (x/x^2) = \text{length}(L) - \text{length}(J)$.
Let $h\_i = \text{dim}\_k R\_i$. Then since $R$ is Gorenstein with SLP, the sequence $h\_0,\cdots,h\_s$ is unimodal and symmetric. Now since the maps $\times x$ and $\times x^2$ are always injective or surjective, it is not hard to compute that $\text{length}(L)=h\_n+h\_{n+1}$ and $\text{length}(J)=h\_n$ with $n=\lfloor s/2\rfloor$. So $\text{length} (x/x^2) =h\_{n+1}$.
If $s$ is odd, then $h\_n=h\_{n+1}$ and $J$ must live in degrees at least $n+1$, thus $J^2=0$ and equality actually holds.
If $s$ is even, then $\text{length} (x/x^2)=h\_{n+1}$. Now if $h\_n= h\_{n+1}$, then as above $J^2=0$ and equality holds. If $h\_n > h\_{n+1}$, then $J$ lives only in degree $2n=s$, but $h\_s=1$ (remember $R$ is Gorenstein) so $\text{length} (J^2)$ is at most $1$. So $\text{length}(J/J^2) \geq h\_n-1\geq h\_{n+1}$.
I think one can push this argument to show the inequality for $x=l^a$ for any Lefschetz element $l$ and $a>0$. Because the SLP is an open condition, this would imply the inequality for general forms of degree $a$.
I also have some examples of equality in positive characteristic with pretty interesting patterns, but I need more time to think about them.
End of UPDATE
(Some history: Yesterday when I saw this question I posted a simple solution, which I immediately realized is wrong. After a few email exchanges with FC we were both convinced that my first attempt would not work.)
I have thought about the question a bit today but could not quite prove it. Since I may not have time to work on it more in the next few days, I will put some of my thoughts here in case they help anyone.
Let $I=(x)$ and $J$ be the annihilator of $(x)$. Also, I will use $R$ instead of $A$.
>
> Fact 1: $(x) \cong R/J\cong D(R/J)$
>
>
>
Proof: For the first isomorphism, just look at the map $R\to R$ by multiplying with $x$. Now
$D(R/J) = \text{Hom}(R/J,R)$ is isomorphic to the annihilator of $J$, which is $(x)$ again.
>
> Fact 2: $x/x^2 \cong R/(J+x)$.
>
>
>
Proof: $x/x^2 = (x)\otimes R/(x) =R/J\otimes R/(x)$
>
> Fact 3: $R/J$ is itself Gorenstein.
>
>
>
Proof: The canonical module of $R/J$ is $D(R/J)$ which is isomorphic to $R/J$ by Fact 1.
>
> Fact 4: Let $(S,m,k)$ be a Gorenstein, artinian local ring. An $S$-module map $S\to M$ is injective iff the image of the generator of the socle of $S$ (which is $\text{Hom}(k,S)$ and is 1-dimensional as $S$ is Gorenstein) is non-zero.
>
>
>
Proof: If the kernel is some non-zero ideal $K$, then some element in $K$ would have $m$ as the annihilator. But then such element has to be inside the socle of $S$, which is a 1-dim vector space.
>
> Fact 5: For any ideal $L$, $L/L^2 = \text{Tor}\_1(R/L,R/L)$
>
>
>
Proof: tensor $0\to L \to R\to R/L$ with $R/L$.
Here are a couple of approaches I tried with some comments:
**Random thoughts A:**
By Fact 2 we need to prove $$\text{length}(J) -\text{length}(J^2)\geq \text{length}(R)- \text{length}(J+x) $$
Rearranging, one needs to prove $$\text{length}(R/J) \leq \text{length}((J+x)/J^2)$$
Let $S=R/J$ and $M=(J+x)/J^2$. One obvious thing to try is to show that $S$ (which is Gorenstein by Fact 3) can embed in $M$. Let $s\in R$ be a lift of the socle generator of $S$. By Fact 4, for a counter example one would need
$$(J+x)s \subseteq J^2 $$
This rules out many potential counter examples because of degree reasons, or if $J^2$ is too small. Note that $xs$ represents the socle generator of $R$.
**Random thoughts B:** Here is a formulation that only involves $x$. By Fact 5 we need to prove $$\text{length}(\text{Tor}\_1(R/(x),R/(x)) \leq \text{length}(\text{Tor}\_1(R/J,R/J)$$
Which, by Fact 1 is really:
$$\text{length}(\text{Tor}\_1(R/(x),R/(x)) \leq \text{length}(\text{Tor}\_3(R/(x),R/(x))$$
This equivalent statement actually makes me a little doubtful. It might be true, and sort of make sense, because the free resolution of $R/(x)$ will typically gets bigger and bigger, but many similar homological statements about Gorenstein rings turn out to be false (although it is often not easy to cook up examples).
So I would say that the statements is likely to be true for small (in terms of lengths or degrees) rings, because of Thoughts A, but might be false in general. Of course, I would be very happy to be wrong, and may be I was missing something really simple.
PS: thanks for asking a nice question in commutative algebra (-:
| 9 | https://mathoverflow.net/users/2083 | 36419 | 23,437 |
https://mathoverflow.net/questions/36420 | 12 | Let a problem instance be given as $(\phi(x\_1,x\_2,\dots, x\_J),M)$ where $\phi$ is a diophantine equation, $J\leq 9$, and $M$ is a natural number. The decision problem is whether or not a given instance has a solution in natural numbers such that $\sum\_{j=1}^J x\_j \leq M$. With no upper bound M, the problem is undecidable (if I have the literature correct). With the bound, what is the computational complexity? If the equation does have such a solution, then the solution itself serves as a polytime certificate, putting it in NP. What else can be said about the complexity of this problem?
| https://mathoverflow.net/users/8719 | Is the solution bounded Diophantine problem NP-complete? | A particular quadratic Diophantine equation is NP-complete.
$R(a,b,c) \Leftrightarrow \exists X \exists Y :aX^2 + bY - c = 0$
is NP-complete. ($a$, $b$, and $c$ are given in their binary representations. $a$, $b$, $c$, $X$, and $Y$ are positive integers).
Note that there are trivial bounds on the sizes of $X$ and $Y$ in terms of $a$, $b$, and $c$.
Kenneth L. Manders, Leonard M. Adleman: NP-Complete Decision Problems for Quadratic Polynomials. STOC 1976: 23-29
| 20 | https://mathoverflow.net/users/8723 | 36424 | 23,441 |
https://mathoverflow.net/questions/36410 | 0 | How to prove $U \otimes Ind W = Ind(Res(U) \otimes W)$? where U is a representation of G and W is a rep of H, a subgroup of G. $Ind(W)$ is the induced rep and $Res(U)$ is the restrict rep.
I got the answer, by both approaches: groups algebra and constructing isomorphic map. Thanks for all the very helpful comments and answer.
| https://mathoverflow.net/users/5737 | How to prove $U \otimes Ind W = Ind(Res(U) \otimes W)$ | This is really a comment which got too long.
Personally, I always find this one rather confusing. If you think in terms of modules over group rings, we want to show that $U \otimes (\mathbb{C}[G]\otimes\_{\mathbb{C}[H]} W) \cong \mathbb{C}[G] \otimes\_{\mathbb{C}[H]} (U\otimes W)$. The $G$-equivariant isomorphism is not given by sending $u\otimes (x\otimes w)$ to $x \otimes (u\otimes w)$. There's a lot wrong with this formula, but that's not the point.
The point is that to get the right formula, one really needs to remember exactly how the universal property of induction works. I don't have Fulton and Harris in front of me to see what they say, but Serre's book has a good discussion of induction which will lead one right to the answer.
Also, unless I'm confused, this really seems to depend on the structure of $\mathbb{C}[G]$ as both a ring and as a $\mathbb{C}[H]$-module. One needs to know that it's a free $\mathbb{C}[H]$-module, and that it has a decomposition as a $C[H]$-module into summands isomorphic to $C[H]$ that are permuted by the units of the ring $C[G]$. One could ask, for morphisms of rings $C\to R\to S$, when it's true that for an R-module M and an S-module N we have the formula $N \otimes\_C (S\otimes\_R M) \cong S \otimes\_R (N \otimes\_C M)$ (as S-modules). (Above I wrote $\otimes$ instead of $\otimes\_{\mathbb{C}}$; now C is the ground ring). I don't know how to prove this without assuming S has the sort of structure mentioned above (free as an R-module, etc.).
| 4 | https://mathoverflow.net/users/4042 | 36428 | 23,445 |
https://mathoverflow.net/questions/36392 | 5 | This is a direct consequence of my previous question: [Extending group actions on varieties](https://mathoverflow.net/questions/36321/extending-group-actions-on-varieties)
In his answer, inkspot said that group actions can be extended if the variety has ample canonical class and is smooth, but mentions that canonical singularities can be allowed. Now, my situation doesn't involve a smooth variety, but instead I have an orbifold.
However, I know that for surfaces, the canonical singularities are the duVal singularities, and are all orbifold points (they're all $\mathbb{C}^2$ modulo a finite subgroup of $\mathrm{Sl}\_2$.)
Now, I've not studied general singular points of surfaces, so I could be wrong already with surfaces, but are orbifold singularities canonical?
| https://mathoverflow.net/users/622 | Are orbifold singularities canonical? | For quotient singularities there is the so-called Reid-Tai criterion to check whether the singularity is canonical or not.
Suppose $G$ is a finite subgroup of $GL\_n(\mathbb{C})$ without quasi-reflections. Let $m=|G|$ and fix a primitive $m$-th root of unity $\zeta$. Let $g\in G$ and let $0\leq a\_i < m$ be such that $\zeta^{a\_1},\dots,\zeta^{a\_n}$ are the eigenvalues of $g$. Then the Reid-Tai sum of $g$ is defined as $\Sigma(g):=1/m(\sum a\_i)$.
The Reid-Tai criterion states that $\mathbb{C}^n/G$ has a canonical singularity at 0 if and only if $\Sigma(g)\geq 1$ for all $g\in G, g\neq id$.
(Taking a quotient by quasi-reflections does not yield a singularity, the Reid-Tai sum depends on the choice of $\zeta$, the criterion does not.)
For more on this see e.g. M. Reid's Young person's guide to canonical singularities (Bowdoin 1985 proceedings).
| 11 | https://mathoverflow.net/users/8621 | 36430 | 23,447 |
https://mathoverflow.net/questions/35567 | 16 | If I have committed to a number x by revealing g^x mod p, can I prove that 0 < x mod (p-1) < (p-1)/2, i.e. that x is positive, without leaking any more information about x?
My bounty is ending in 4 days and I am unsatisfied with the current answers so I would like to provide more context and also expand the question for the limited time remaining. Consider the following situation:
Paul and Quentin are wealthy and competitive with each other and they frequently settle their account with great variance: one week Paul is ahead by a million dollars, the next week Quentin is ahead by a billion, the next week Paul is ahead by only a thousand. Paul and Quentin have a wealthy rival Raul, whom they shun, but all 3 persons patronize the same accountant Verne. Verne is honest and discreet and frugal and he will never make a payment to a client on credit, but he will pay an owed amount to a client on demand. Raul can profit from information about Paul's account, indirectly costing Paul, and everyone knows this. How can Verne manage his accounts without having to buy insurance against Paul's legal accusation of a conflict of interest?
| https://mathoverflow.net/users/2003 | Zero-knowledge proof of positivity | This answer combines my three comments to the question and expands them a little.
Following [BM84], let’s call the integers $g^x \bmod p$ for $0 < (x \bmod (p−1)) < (p−1)/2$ *principal square roots*. We call the problem of deciding, given $p$, $g$ and $y$, whether an integer $y$ is a principal square root or not the *principal square root problem*.
For the original question, the answer is positive assuming one-way functions. This is because if one-way functions exist, every problem in NP has a computational zero-knowledge interactive proof system [GMW91]. Note that the principal square root problem is clearly in NP.
As the questioner pointed out, this construction has a drawback that it requires a reduction from the principal square root problem to the $3$-colorability problem, which involves Cook’s reduction and blows up the instance size (polynomially). In addition, this construction requires the assumption that one-way functions exist.
I do not know a direct way to construct a zero-knowledge interative proof system for the principal square root problem. However, [GK93] shows an interesting result related to the question: the principal square root problem *under a promise that $(x \bmod (p−1)/2)$ is not too close to $(p−1)/2$* has a perfect zero-knowledge interactive proof system. The construction is direct and does not use any cryptographic assumptions.
#### References
[BM84] Manuel Blum and Silvio Micali. How to generate cryptographically strong sequences of pseudorandom bits. *SIAM Journal on Computing*, 13(4):850–864, Nov. 1984. [DOI 10.1137/0213053](https://doi.org/10.1137/0213053). [Zbl 0547.68046](https://zbmath.org/?q=an:0547.68046)
[GK93]
Oded Goldreich and Eyal Kushilevitz. A perfect zero-knowledge proof system for a problem equivalent to the discrete logarithm. *Journal of Cryptology*, 6(2):97–116, June 1993. [DOI 10.1007/BF02620137](https://doi.org/10.1007/BF02620137). [Zbl 0783.68039](https://zbmath.org/?q=an:0783.68039)
[GMW91]
Oded Goldreich, Silvio Micali and Avi Wigderson. Proofs that yield nothing but their validity or all languages in NP have zero-knowledge proof systems. *Journal of the ACM*, 38(3):690–728, July 1991. [DOI 10.1145/116825.116852](https://doi.org/10.1145/116825.116852). [Zbl 0799.68101](https://zbmath.org/?q=an:0799.68101)
| 11 | https://mathoverflow.net/users/7982 | 36433 | 23,449 |
https://mathoverflow.net/questions/36418 | 2 | My specific situation is that I have a non-spacelike continuous future directed curve $\gamma:[0,a)\to M$ in a Lorentzian manifold. The curve must necessarily satisfy a local Lipschitz condition and therefore the derivative of the curve exists almost everywhere. In these circumstances, as I understand it, one can show (using Carathéodory's existence theorem?) that given $v\in T\_{\gamma(0)}M$ there exists an absolutely continuous vector field $V$ along $\gamma$ satisfying a local Lipschitz condition with $V(0) = v$ to the equation $\nabla\_{\gamma'}V=0$.
I am specifically interested in the details of the proof of this and other ODE existence and uniqueness theorems in order to gain deeper insight into the relation of the differentiability conditions to conditions on the solutions (existence, uniqueness, differentiability, etc...).
Thus I am looking for a reference that treats such theorems in full generality with regards to the assumptions used. I'm not so concerned about the generality of the domain of the equations, $\mathbb{R}^n$ is ok for me. Though a reference that covered function spaces would also be interesting. I'm interested in a reference that provide full (or near to full) proofs for each result. Think "Real Analysis" by Haaser and Sullivan or "Introduction To Commutative Algebra" by Atiyah and MacDonald.
| https://mathoverflow.net/users/5993 | Reference for existence and uniqueness of differential equations for low differentiability? | I do not know any recent book on ODE's satisfying yours criterium. The classical books below do a fairly good job though. The first being more comprehensive and the second more elementary(no measure theory needed).
1. Coddington, Earl A.; Levinson, Norman.
Theory of ordinary differential equations.
2. Petrovski, I. G.
Ordinary differential equations.
| 2 | https://mathoverflow.net/users/605 | 36436 | 23,451 |
https://mathoverflow.net/questions/36388 | 8 | Complex moduli space (or Teichmuller space) of a Quintic Calabi-Yau 3-fold is a
101-dimensional complex orbifold. Does it have a toric structure?
| https://mathoverflow.net/users/5259 | Is the complex moduli of Quintic Calabi-Yau toric? | The complex moduli space does not admit a toric strucutre, since the orbifold fundamental group of a toric orbifold must be abelian. Indeed, $\pi\_1(\mathbb C^\*)^n$ surjects on the orbifold fundamental group. Also, the orbifold stabisier of each point on a toric orbifold
is a finite abelian group. At the same time the stabiliser of the quintic $\sum\_i z^5=0$
is a non-comutative group. Also I am sure that the orbifold fundamental group of the moduli space of quintics contains free (non-abelian) subgroups, but I don't know how to prove it.
Also it should be true that the Tiechmuller space is not algebraic. It least this happen in lower dimensions for cubics in $\mathbb CP^2$ and for quartics in $\mathbb CP^3$.
In the first case the Theichmuiller space is a disk, and in the second it is
a hermitian domain of type IV. Moduli spaces of polarised K3 are discussed here
for example, here:
<http://people.bath.ac.uk/masgks/Papers/k3moduli.pdf>
| 7 | https://mathoverflow.net/users/943 | 36438 | 23,452 |
https://mathoverflow.net/questions/36445 | 8 | Given a coherent $\mathcal{D}\_X$-Module $M$, one can assign to it its characteristic variety
$$ch(M)\subseteq T^\*X$$
or one could look at its support
$$supp(M)\subseteq X$$
as an $\mathcal {O}\_X$ module. Is there a relation between these two spaces?
Edit: Is it true, that the characteristic variety of a holonomic $\mathcal{D}\_X$-Module is the conormal space of its support?
| https://mathoverflow.net/users/2837 | Relation between characteristic variety and support of D-Module | I'm not an expert either but, at least for holonomic D-modules, the relation you're looking for should be
$$
supp(M) = ch(M) \cap T^\*\_X X
$$
where $T^\*\_XX$ is the zero section of the cotangent bundle identified with $X$.
You can check it in the basic examples: $X = \mathbb{C}$, $M$ the trivial bundle or the $\delta$-module at $x\in X$. The corresponding $ch(M)$ is $T^\*\_XX$ or $T^\*\_xX$. The general case should be an easy exercise in linear algebra.
| 8 | https://mathoverflow.net/users/1985 | 36456 | 23,460 |
https://mathoverflow.net/questions/36457 | 4 | My question: is the set of potential games closed under convex combinations?
An n player game with action set $A = A\_1 \times \ldots \times A\_n$ and payoff functions $u\_i$ is called an exact potential game if there exists a potential function $\Phi$ such that:
$$\forall\_{a\in A} \forall\_{a\_{i},b\_{i}\in A\_{i}} \Phi(b\_{i},a\_{-i})-\Phi(a\_{i},a\_{-i}) = u\_{i}(b\_{i},a\_{-i})-u\_{i}(a\_{i},a\_{-i})$$
A game is a general (ordinal) potential game if there exists a potential function $\Phi$ such that:
$$\forall\_{a\in A} \forall\_{a\_{i},b\_{i}\in A\_{i}} sgn(\Phi(b\_{i},a\_{-i})-\Phi(a\_{i},a\_{-i})) = sgn(u\_{i}(b\_{i},a\_{-i})-u\_{i}(a\_{i},a\_{-i}))$$
Potential games are interesting because they always have pure strategy Nash equilibria: in particular, a sequence of best-responses must eventually converge to one.
Say that we have two games on the same action set, with utility functions $u\_i$ and $u'\_i$ respectively, for each player $i$. For any $0 \leq p \leq 1$, there is a convex combination of these two games, again on the same action set, where the utility function for each player $i$ is now $u^p\_i(\cdot) = (1-p)u\_i(\cdot) + pu'\_i(\cdot)$.
Clearly, the convex combination of two exact potential games is also an exact potential game: just take the same convex combination of the two potential functions.
But is it possible to have two (general) potential games such that their convex combination is not a potential game?
| https://mathoverflow.net/users/3027 | Is the convex combination of two potential games a potential game? | No, the set of ordinal potential games is not convex. I will think about an example, but it follows from a little fact that I've yet to get around to publishing:
Theorem: Any convex set of games strictly containing the exact potential games contains a game with no pure Nash equilibrium.
In particular, since every ordinal potential game has a pure Nash equilibrium, convexity of the set thereof would contradict this theorem.
EDIT: Bimatrix games A and B below are ordinal potential games, but the average of these two games is the zero-sum game "matching pennies" which obviously has no pure Nash equilibrium and so is not an ordinal potential game.
```
Game A: Potential A:
(4,-1) (0,1) 3 4
(-2,1) (-2,-1) 2 1
Game B: Potential B:
(-2,-1) (-2,1) 1 2
(0,1) (4,-1) 4 3
Matching pennies: (No potential)
(1,-1) (-1,1)
(-1,1) (1,-1)
```
| 6 | https://mathoverflow.net/users/5963 | 36465 | 23,467 |
https://mathoverflow.net/questions/36443 | 3 | Let $K$ be a field (of course, of positive characteristic, unless you want a trivial question). Let $G$ be a finite group, and $V$ and $W$ be two completely reducible (finite-dimensional) representations of $G$ over $K$. Is the (interior) tensor product $V\otimes\_K W$ a completely reducible representation of $G$ ?
I know that this holds for exterior tensor products:
Let $K$ be a field. Let $G$ and $H$ be two finite groups, and $V$ and $W$ be two completely reducible (finite-dimensional) representations of $G$ and $H$, respectively, over $K$. Then, the tensor product $V\otimes\_K W$ is a completely reducible representation of $G\times H$.
(Proof: Combine Curtis/Reiner "Methods of Representation Theory I" Theorems 7.10 and 10.38 (i).)
Note that my above question is equivalent to the Jacobson radical of the group ring $KG$ being a coideal (the coalgebra structure on $KG$ is the canonical one, of course: $\Delta g=g\otimes g$). It may be total nonsense but unfortunately I don't have any nontrivial examples of modular representations to check with.
| https://mathoverflow.net/users/2530 | Innocent question on tensor products of modular representations | Jim has already given the correct answer "no".
Here is a hopefully instructive example. Let $k$ be an alg. closed field of pos. char $p$, and let $G = SL\_2(k)$. Write $V = k^2$ for the "natural" 2-dimensional representation of $G$ say with basis $e\_1,e\_2$. Let $W = S^pV$ be the $p$-th symmetric power of $V$. Then $W$ contains a 2 dimensional submodule $A$ spanned by the $p$-th powers $e\_1^p$ and $e\_2^p$; the module $A$ is isom. to the "first Frobenius twist" of $V$.
It is an exercise to check that there is no $G$-stable complement to $A$ in $W$; i.e. the SES
$$0 \to A \to W \to W/A \to 0$$
is not split.
Thus
$W$ is not completely reducible. Evidently there is a surjective mapping
$V^{\otimes p} \to W$, thus also the $p$-th tensor power $V^{\otimes p}$ is not completely
reducible.
But $V$ is a simple (hence completely reducible) $G$-module; thus tensor powers of a completely reducible module are not in general completely reducible. In fact,
the $(p-1)$-th tensor power $V^{\otimes p-1}$ *is* completely reducible;
arguing as before, one sees that $V \otimes (V^{\otimes p-1})$ is not completely
reducible; thus in general the tensor product of two completely reducible modules
is not completely reducible.
I gave some further remarks about semisimplicity of tensor products in an answer to
[this question.](https://mathoverflow.net/questions/22707/if-a-tensor-product-of-modules-is-semi-simple-are-the-tensor-factors-semi-simple/31477#31477)
| 7 | https://mathoverflow.net/users/4653 | 36473 | 23,471 |
https://mathoverflow.net/questions/36444 | 42 | Can anyone give me a plain-and-simple
definition of an E-infinity algebra without using
the words "operad," "ring spectrum," or
"stable homotopy"?
Sorry, but I honestly couldn't find it using
all on-line resources at my disposal.
Thanks!
| https://mathoverflow.net/users/1186 | Definition of an E-infinity algebra | In characteristic 0, Kadeishvili has a notion of $C\_{\infty}$ algebra which models rational homotopy theory. See the last paragraph of the introduction of his paper arXiv:0811.1655. His point of view is to simply consider $A\_{\infty}$ algebras whose operations satisfy a certain property with respect to shuffle maps. So your computer doesn't have to remember any new operations, just check that the old ones are right.
In characteristic $p$, things are probably hopeless.
Added Remark: I just want to make clear that this does not give a "trivial proof" that a commutative dga is formal as a commutative dga if the underlying dga is formal in the "non-commutative" sense. The reason is that when you transfer from cochains from cohomology, you are restricted in the kind of morphisms allowed if you are interested in the commutative theory. So, just as in the answers to this [question](https://mathoverflow.net/questions/9146/noncommutative-rational-homotopy-type), there is some work to be done if you want results like that (to be completely honest, there is not yet a proof that I completely understand, so declare myself agnostic).
| 14 | https://mathoverflow.net/users/6948 | 36478 | 23,475 |
https://mathoverflow.net/questions/36479 | 6 | Let f:X→Y be proper birational morphism between two quasi-projective varieties over an algebraically closed field $k$. I am particularly interested in the case where the characteristic of $k$ is positive; Y is singular, X is smooth and both X and Y are not projective.
Let D be a closed subset of Y, let E=f-1(D) and assume that f|X\E:X\E→Y\D is an isomorphism.
Based on results in characteristic 0 (and on results on rigid cohomology) I expect that there is a long exact sequence of étale cohomology groups
Hi(Y,ℚl)→
Hi(X,ℚl)⊕
Hi(D,ℚl)→
Hi(E,ℚl)→
Hi+1(Y,ℚl).
I hoped this exact sequence is well-known and would appear in a standard text, but I had trouble identifying such a text. I can think of a prove using Cox's étale version of tubular neighbourhoods for E in X and D in Y and you might be able to compare them using the Mayer-Vietoris sequences in étale cohomology, but such a proof seems quite involved (you need to define the image of an étale tubular nhd under a proper morphism (which seems non-trivially to me) and check that for certain exact sequences taking direct or projective limits turns out to be an exact functor.)
Before working out the details I would like to ask whether anyone knows a reference for the above sequence or knows a simpler/nicer proof.
| https://mathoverflow.net/users/8621 | Exact sequence in étale cohomology related with proper birational morphism | Let $V := X\setminus E$ and $U := Y\setminus D$ and $j\colon U \rightarrow Y$ and
$k\colon V \rightarrow X$ the inclusions. We have exact sequences
$\cdots\rightarrow H^\ast(X,k\_!\mathbb Z\_\ell)\rightarrow H^\ast(E,\mathbb
Z\_\ell)\rightarrow H^\ast(X,\mathbb Z\_\ell)\rightarrow\cdots$ and a similar one for
$Y$, $D$ and $j$. Furthermore, $f$ induces a map of these long exact
sequences. A standard diagram chasing (also used for instance to the
Mayer-Vietoris sequence from excision) can be used to show that to get the
desired exact sequence it is enough to show that $f$ induces an isomorphism
$H^\ast(Y,j\_!\mathbb Z\_\ell)\rightarrow H^\ast(X,k\_!\mathbb Z\_\ell)$. We have that
$H^\ast(X,k\_!\mathbb Z\_\ell)=H^\ast(Y,Rf\_\ast k\_!\mathbb Z\_\ell)$ but $f$ is proper so
that $Rf\_\ast k\_!\mathbb Z\_\ell=f\_!k\_!\mathbb Z\_\ell$. However,
$f\_!k\_!=(fk)\_!=(jf')\_!=j\_!f'\_!$, where $f'=f\_{|V}$, and as $f'$ is an isomorphism
we have $f'\_!\mathbb Z\_\ell=\mathbb Z\_\ell$.
| 8 | https://mathoverflow.net/users/4008 | 36482 | 23,476 |
https://mathoverflow.net/questions/36481 | 1 | I've got what might be a couple of very basic questions on the fundamental representations of locally isomorphic semi-simple Lie groups and their relationship to representations of the corresponding Lie algebra. I know (from reading Cornwell, 'Group Theory in Physics') that every representation of a semi-simple Lie algebra exponentiates to give a representation of the unique, simply connected semi-simple Lie group associated with that algebra. (So for example, every representation of the algebra su(2) exponentiates to give a representation of the group SU(2).) I also know that, at least in some cases (such as with the group SO(3), which also has su(2) as its algebra), it is not the case that every representation of the algebra exponentiates to give a representation of a given non-simply-connected group associated with that algebra. The first non-trivial representation of SO(3), for example, is not obtainable by exponentiating the fundamental (2-dimensional) representation of su(2) (with highest weight 1/2), but is rather obtained by exponentiating the 3-dimensional representation of the algebra (with highest weight 1).
So I have two questions on the same theme as the above.
1. Is it always the case that the fundamental representation(s) of the various non-simply-connected semi-simple Lie groups associated with a given semi-simple Lie algebra are not obtainable by exponentiating the fundamental representation(s) of the algebra?
2. Do the fundamental representation(s) of each of the different non-simply-connected groups correspond to different representations of the algebra? (That is, will two groups that are locally but not globally isomorphic always have different fundamental representations?)
Any knowledge anyone can bring to bear on this would be really appreciated. (I'm pretty sure the second in particular is trivial, but I'm not sure as at the level I'm at in physics we pretty much always work with representations of the algebras and I'm not familiar with how they relate to the representations of the groups.) Thanks a lot.
| https://mathoverflow.net/users/7317 | Reps of groups and reps of algebras | Let us consider for simplicity the complex semi-simple case. As you mention, every irreducible representation $\rho$ of a semi-simple finite dimensional Lie algebra $g$ integrates to a representation of the corresponding simply connected Lie group $G$. But even if we suppose $\ker\rho=0$, the representation of $G$ may well have a kernel, a finite central subgroup $H$ of $G$. So we obtain representations of $G/H'$ where $H'$ is a finite central subgroup contained in $H$, but not of other groups with Lie algebra $g$.
So one could ask: given a Lie group $G'$ with Lie algebra $g$, when does a representation $\rho$ of $g$ with highest weight $\Lambda$ integrates to a representation of $G'$? The answer: if and only if $\Lambda$ belongs to the character lattice of a maximal torus $T$ of $G$. (Recall that the character lattice of $T$ is the discrete additive subgroup of the dual of the Lie algebra of $T$ spanned by the differentials of the homomorphisms $T\to\mathbf{C}^{\ast}$.)
One can also ask a similar question: given a representation $\rho$ of $g$ with $\ker\rho =0$ and highest weight $\Lambda$, how to compute the kernel of the resulting representation $R$ of the simply connected group $G$? The answer is as follows. The dual of the Lie algebra $t$ of $T$ contains the weight lattice $P$ and the root lattice $Q$. Recall that the weight lattice is the lattice spanned by the fundamental weights and the root lattice is spanned by the roots i.e. the weights of the adjoint representation. We have $Q\subset P$. Consider the dual lattices $Q^\vee\subset P^\vee$; these are formed by all elements $a$ of the Lie algebra of $T$ such that $l(a)\in\mathbf{Z}$ for all $l\in P$, resp. all $l\in Q$.
Inside $P^\vee$ there is a sublattice formed by all $a\in t$ such that $\Lambda(a)\in\mathbf{Z}$; it contains $Q^\vee$ and the quotient by $Q^\vee$ is naturally identified with the kernel of $R$ (by exponentiating). There is a similar result when $\rho$ is reducible -- in that case we just consider the sublattice formed by all elements of the tangent space such that all weights that take integral values on them.
Example: when $G=SL\_2(\mathbf{C})$, we can identify $Q$ with the sublattice of $\mathbf{R}$ spanned by 1 and $R$ will be spanned by $\frac{1}{2}$. Then the representations with integral weight will have kernel $\mathbf{Z}/2$ and the representations with half-integral weight will have trivial kernel.
Hopefully one of these two questions also covers the questions of the posting.
| 2 | https://mathoverflow.net/users/2349 | 36490 | 23,481 |
https://mathoverflow.net/questions/36502 | 3 | In a round-robin tournament with $n$ teams, each team plays every other team exactly once. Thus, there are $n(n-1)/2$ total games played. How many different standings can result? By a "standing" I mean the ordered sequence $(W\_1, \ldots, W\_n)$ where $W\_i$ is the number of wins by the $i$th player. Assume that no game ends in a tie.
I wrote a program to calculate this sequence. For $n$ up to 13, it agrees with the number of forests on $n$ labeled nodes, which is Sequence [A001858](https://oeis.org/A001858) in the [OEIS](https://oeis.org/). But I can't see the correspondence between tournament standings and forests with labeled nodes. Can anyone explain this?
| https://mathoverflow.net/users/4758 | Round-Robin Tournaments and Forests | The bijection between score vectors and forests on labeled nodes is due to Kleitman and Winston. ([This](https://doi.org/10.1007/BF02579176 "Kleitman, D.J., Winston, K.J. Forests and score vectors. Combinatorica 1, 49–54 (1981). zbMATH review at https://zbmath.org/?q=an:0491.05028") paper)
A small clarification, your question about the cardinalities being equal was answered by Stanley and Zaslavsky, see Stanley's paper "[Decompositions of rational convex polytopes](https://doi.org/10.1016/S0167-5060(08)70717-9 "Stanley, R.P. Decompositions of rational convex polytopes. Ann. Discrete Math. 6, 333–342 (1980). zbMATH review at https://zbmath.org/?q=an:0812.52012")", but the proof was not bijective.
| 9 | https://mathoverflow.net/users/2384 | 36503 | 23,488 |
https://mathoverflow.net/questions/36348 | 7 | Let $(a\_{mn})\_{m,n\in\mathbb{N}}$ and $(b\_m)$ be sequences of complex numbers.We say that $(a\_{mn})$ and $(b\_m)$ constitute *an infinite system of linear equations in infinitely many variables* if we seek a sequence $(x\_n)$ of complex numbers such that $\forall m\in\mathbb{N}:$ $\sum\_{n=1}^{\infty}a\_{mn}x\_n=b\_m$. Note that in general the order of summation matters.
I am sort of a undergraduate student with focus on number theory and have some background in functional analysis (2 semesters functional analysis, 1 semester non-linear functional analysis, 1 semester operator algebras, 2 semesters PDEs), so I am sort of a becoming number-theorist with bias for functional analysis :-) That is also why I am fascinated by the above defined object as a sort of natural extension of a practical problem from linear algebra.
We have never dealt with this type of objects and I wasn´t able to find much on google that I could start something with, maybe partly because I have searched in the wrong way. That is why I have a request if you could recommend some **introductory** literature *focused* on such infinite systems of linear equations in infinitely many unknowns over $\mathbb{C}$.
Thanks in advance!
| https://mathoverflow.net/users/1849 | infinitely many linear equations in infinitely many variables | The systems of this kind are fairly common in applications. For example, they naturally appear when solving boundary value problems for linear partial differential equations using the method of separation of variables.
Predictably, the problem is not meaningful for *any* sequences {$a\_{nm}$}, {$b\_m$}, but only for sufficiently well-behaved ones. If, for example, you were to consider systems of the form
$$ x\_n+\sum\_{m=1}^{\infty}a\_{nm}x\_m=b\_n,\quad\mbox{such that}\quad \sum\_n\sum\_m a\_{nm}^2<\infty \quad\mbox{ and }\quad \sum\_nb\_n^2<\infty, $$
then this system possesses a unique solution in the Hilbert space $l\_2$ such that $\sum\_n x\_n^2<\infty$ (assuming that the problem is not singular, i.e. that $\det(I+A)\ne0$). These requirements are too restrictive for some applications, hence there is a body of literature concerned with various kinds of *regularity* conditions involving {$a\_{nm}$} and {$b\_m$}, weaker than above, which ensure the well-posedness of the problem and enable numerical solution of such systems (which is usually done by truncation; see the appropriate accuracy estimates in F. Ursell (1996) "Infinite systems of equations: the effect of truncation", *Quarterly Journal of Mechanics and Applied Mathematics*, 49(2), 217--233).
One good old book that discusses these systems in some detail was written by By L. V. Kantorovich and V. I. Krylov and is called "Approximate methods of higher analysis" (New York: Interscience Publishers, 1958).
| 10 | https://mathoverflow.net/users/8670 | 36504 | 23,489 |
https://mathoverflow.net/questions/18938 | 7 | Every partially ordered set gives a triangulation of (the geometric realisation of) its order complex. (The n-simplices of the order complex are the chains $x\_0\leq x\_1\leq\cdots\leq x\_n$.) However, there are triangulations of topological spaces that do not arise this way.
Is there a name for triangulations having this special property of "coming from a poset?"
EDIT: Apparently, the following formulation of my question is cleaner: *what conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset?*
| https://mathoverflow.net/users/1291 | Triangulations coming from a poset. Or: What conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset? | Here are necessary and sufficient conditions for an abstract, finite simplicial complex $\mathcal{S}$ to be the order complex of some partially ordered set.
(i) $\mathcal{S}$ has no missing faces of cardinality $\geq 3$; and
(ii) The graph given by the edges (=$1$-dimensional simplices) of $\mathcal{S}$ is a comparability.
[Definitions.
(a) A *missing face* of $\mathcal{S}$ is a subset $M$ of its vertices (=$0$-dimensional simplices) such that
$M \not \in \mathcal{S}$, but all proper subsets $P\subseteq M$ satisfy $P\in \mathcal{S}$.
(b) A graph (=undirected graph with no loops nor multiple edges) is a *comparability* if its
edges can be transitively oriented, meaning that whenever edges $\{p, r\_1\}, \{r\_1, r\_2\},\ldots, \{r\_{u−1}, r\_u\}, \{r\_u, q\}$ are oriented as $(p, r\_1), (r\_1, r\_2),\ldots, (r\_{u−1}, r\_u), (r\_u, q)$, then there exists an edge $\{p, q\}$ oriented as $(p, q)$.]
This characterisation appears with a sketch of proof $-$ which is not hard, anyway $-$ in
>
> M. M. Bayer, *Barycentric subdivisions*. Pacific J. Math. 135 (1988), no. 1, pp. 1-16.
>
>
>
As Bayer points out, the result was first observed in
>
> R. Stanley, *Balanced Cohen-Macaulay complexes*, Trans. Amer. Math. Soc, 249 (1979), pp. 139-157.
>
>
>
@Rasmus and @Gwyn: The characterisation might perhaps disappoint you if you were expecting something more topological. However, it's easy to prove that no topological characterisation of order complexes is possible, and therefore a combinatorial condition such as the one on comparabilities *must* be used. For this, first check that the barycentric subdivision of *any* simplicial complex indeed is an order complex. Next observe that barycentric subdivision of a simplicial complex does not change the homeomorphism type of the underlying polyhedron of the complex. Finally, conclude that for any topological space $T$ that is homeomorphic to a compact polyhedron, there is an order complex whose underlying polyhedron is homeomorphic to $T$.
I hope this helps.
| 13 | https://mathoverflow.net/users/8735 | 36505 | 23,490 |
https://mathoverflow.net/questions/36466 | 16 | Lets $X$ be a complex manifold (algebraic variety), $N$ an integer, and consider the sheaf $F$ defined by:
$F(U)$ ={ holomorphic maps $f: U\rightarrow GL(N,\mathbb{C})$ } with multiplicative structure.
Can we define $H^i(X,F)$ ? Note that for $N=1$, this would be just $H^i(X,O\_X^\*)$.
(Please give reference for your claims)
| https://mathoverflow.net/users/5259 | Do we have non-abelian sheaf cohomology? | The quick reply is: not really for $i \gt 2$, and not in the way you perhaps expect for $i=2$, see below.
---
**EDIT (Feb 2017):** Debremaeker's PhD thesis [0] has now been translated into English and placed on the arXiv: [Cohomology with values in a sheaf of crossed groups over a site](https://arxiv.org/abs/1702.02128), arXiv:1702.02128
---
The comment on Charles' answer about 'teaching you never to ask that question again' is partly true, partly not. The lesson to learn from Giraud is that really one does not use groups for coefficients of higher cohomology. For a start, Giraud's $H^2(X,G)$ is not functorial with respect to group homomorphisms $G\to H$! One also does not get the exact sequences that one expects (this is due to the lack of functoriality). But this is not a problem with his definition of the cohomology set, but a problem with what category you believe the coefficients lie in. This is because the coefficient object of Giraud's cohomology is actually the crossed module $AUT(G) = (G \to Aut(G))$, and the assignment $G \mapsto AUT(G)$ is not functorial. (Aside: Giraud contains lots of other important things on stacks and gerbes and sites and so on, so the book is not a waste of time by any means)
But little-known work by Debremaeker[1-3] from the 1970s fixed this up and showed that really the Giraud cohomology was functorial with respect to morphisms of crossed modules. This has been recently extended by Aldrovandi and Noohi [4] by showing that it is functorial with respect to weak maps of crossed modules aka butterflies/papillion.
It was realised by John E. Roberts (no relation) and Ross Street that the most general nonabelian cohomology has as coefficient objects higher categories. In fact, we now know that the coefficients of $n^{th}$ degree cohomology is an $n$-category (usually an $n$-groupoid, though), even when we are talking about usual abelian cohomology.
Everything I've talked about is just for groups etc in Set, but it can all be done internal to a topos, i.e. for sheaves of groups, and more generally a Barr-exact category (and probably weaker, but Barr-exact means that the monadic description of cohomology therein due to Duskin (probably going back to Beck) works fine).
---
[0] R. Debremaeker, Cohomologie met waarden in een gekruiste groepenschoof op een situs, PhD thesis, 1976 (Katholieke Universiteit te Leuven). English translation: Cohomology with values in a sheaf of crossed groups over a site, arXiv:[1702.02128](https://arxiv.org/abs/1702.02128)
[1] R. Debremaeker, Cohomologie a valeurs dans un faisceau de groupes croises sur un site. I, Acad. Roy. Belg. Bull. Cl. Sci. (5), 63, (1977), 758 -- 764.
[2] R. Debremaeker, Cohomologie a valeurs dans un faisceau de groupes croises sur un site. II, Acad. Roy. Belg. Bull. Cl. Sci. (5), 63, (1977), 765 -- 772.
[3] R. Debremaeker, Non abelian cohomology, Bull. Soc. Math. Belg., 29, (1977), 57 -- 72.
[4] E. Aldrovandi and B. Noohi, Butterflies I: Morphisms of 2-group stacks, Advances in Mathematics, 221, (2009), 687 -- 773.
| 42 | https://mathoverflow.net/users/4177 | 36508 | 23,493 |
https://mathoverflow.net/questions/36528 | 3 | In the plane, the exterior angle of a vertex is $\pi -$ the standard ("interior") angle, which may be negative in some cases. The following is true for non-weird polygons:
>
> The sum of the exterior angles at each vertex is a full turn ($2\pi$ radians).
>
>
>
I am informally calling polygons with self-crossing edges or holes as "weird" -- please do let me know what the standard terminology is. I have seen an extension to 3-dimensional polytopes of this form:
>
> The sum of the exterior angles of a polytope is $4\pi$ radians.
>
>
>
In this case, the exterior angle of a vertex is $2\pi -$ (the sum of the face angles at that vertex). I have not seen a proof, but I think it is true for non-weird polytopes, and a modified version is probably true for polytopes with nonzero genus.
My question is: is there a general n-dimensional version of these properties?
| https://mathoverflow.net/users/2498 | An exterior angle theorem for n-dimensional polytopes? | What you call exterior angles are the curvatures at the vertices, and the result you are referring to is the combinatorial Gauss-Bonnet theorem. It says that in an angled 2-complex $K$ the sum of the curvatures at the vertices plus the sum of the curvatures at the polygonal faces is $2\pi \chi(K)$ (Euler characteristic). When your complex is piecewise Euclidean then the curvature at each face is zero implying your result in the case of polyhedra. For polyhedra, another way to see this is to perform barycentric subdivision and sum the angles of all faces to reduce your identity to $V+F-E=2$. A possible reference is [this](http://arxiv.org/abs/0910.5688) paper (theorem 4.8).
| 1 | https://mathoverflow.net/users/2384 | 36534 | 23,507 |
https://mathoverflow.net/questions/36511 | 4 | Let $M$is a closed oriented 2n-dimensional smooth manifold, $E$ is a 2n-dimensional oriented real vector bundle on $M$, with inner product on each fibers.
Let $\tau=(\sqrt{-1})^{n}c(e\_{1})c(e\_{2})...c(e\_{2n})$, use it we can define a $\mathbb{Z}\_2$-grading on $\wedge(E^{\*})\otimes\mathbb{C}=\wedge\_{+}(E^{\*})\otimes\mathbb{C}\oplus\wedge\_{-}(E^{\*})\otimes\mathbb{C}$, the Clifford action is defined by $c(v)\alpha=\varepsilon(v)\alpha-\iota(v)\alpha$.
I read a formula about Pfaffian like this:
$$Str[exp(-R^{\wedge(E^{\*})\otimes\mathbb{C}})]=2^{n}(\sqrt{-1})^{-n}{\rm Pf}(-R^{E})$$
here $R^{E}$ and $R^{\wedge(E^{\*})\otimes\mathbb{C}}$ are curvatures,
$E^{\*}$ is the dual to $E$.
How to proof this formula? or any matrial about this? If anyone can tell me something I will be very thanks.
| https://mathoverflow.net/users/3896 | A question about a formula of Pfaffian | The supertrace can be evaluated either by Berezin Gaussian integration, or equivalently by
summation over a Clifford algebra. Here is a description of the second method.
Let $\omega$ be a skew-symmetric 2n by 2n matrix. Let $\{{\ e\_1, e\_2, . . . . e\_{2n} \}}$ be a real
2n-dimensional Clifford algebra. Then:
$exp(\Sigma\_{k,l=1}^{2n} \omega\_{kl} e\_k e\_l) = \Sigma\_{|K| even} Pf(\omega\_K) \hat{e}\_K$
where: $K$ is a subset of $\{{ 1, 2, . . . . 2n \}}$ and $\hat{e}\_K$ is the corresponding wedge product of the Clifford generators and $\omega\_K$ is the skew-symmeterized submatrix containg only the rows and columns in $K$.
Now, the supertrace selects the coefficient of the top form, giving you the required formula.
A proof of this result can be found for example in the following [book](http://books.google.com/books?id=2yJIwWbh1isC&pg=PA204&lpg=PA204&dq=Pfaffian+%252B+%2522Berezin+integral%2522&source=bl&ots=ex-Shnj--R&sig=jRCLVTeXucsrtVVYi-rZDc1PCMI&hl=en&ei=ssBzTMqXMM_uOZuy2OUI&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q=Pfaffian%2520%252B%2520%2522Berezin%2520integral%2522&f=false) by: José Gracia Bondía, Joseph C. Várilly, Héctor Figueroa.
| 2 | https://mathoverflow.net/users/1059 | 36551 | 23,515 |
https://mathoverflow.net/questions/36552 | 0 | Disclaimer: This is not a homework problem. I stumbled on this puzzle on internet and I also have the answer. However I am not able to figure out whats the method to be used to arrive at the answer.
The puzzle is as below:
The product of the ages of David's children is the square of the sum of their ages. David has less than eight children. None of his children have the same age. None of his children is more than 14 years old. All of his children is at least two years old. How many children does David have, and what are their ages?
The answer happens to be 2,4,6,12.
Please suggest ways to solve this problem systematically.
| https://mathoverflow.net/users/8246 | How to tackle this puzzle? | Well, we know that the sum is at most $14+13+12+11+10+9+8+7=84$, so the product is at most $7056$.
If there are $7$ or more children, then the product is at least $8!>7056$, so there are at most $6$ children.
Furthermore, if there are $6$ children, the sum is at most $84-8-7=69$, so the product is at most $69^2=4761$, but the product is at least $7!=5040$, so there cannot be $6$ children.
Let $S$ denote the sum and $P$ the product. By the AM-GM inequality, we have $\frac{S}{n} \ge \sqrt[n]{P} = \sqrt[n]{S^2}$, so $\frac{S^n}{n^n} \ge S^2$, or $S^{n-2} \ge n^n$, where $n$ is the number of children. This means that $n \ge 3$, since $n^n \ge 1$, which would contradict $n=2$.
I'm not sure there's much else you can do without getting into some messy casework. To rule out $3$ and $5$, you can divide into cases like "Suppose at least two children are older than 11," etc, and use similar arguments regarding sums and products as above. To then find the result given that $n=4$, you'll need to use some divisibility arguments and, yes, a little bit of casework.
| 2 | https://mathoverflow.net/users/1355 | 36558 | 23,518 |
https://mathoverflow.net/questions/36550 | 1 | Hey I'm trying to understand what kind of boundary data I can pose for the wave equation. Let's work in one dimension for now. It appears I should be able to pose any Neumann, Dirichlet or Robin boundary conditions.
I've heard that you need your boundary data to be 'consistent with the Cauchy data'.
I would like to understand this better.
If I think about the infinite string problem so solve
$u\_{tt} - u\_{xx} = 0$ on $[0,\infty)$ and I consider a characteristic $x+t = x\_0$ I must have $(u\_t(x\_0 - t, t) - u\_x(x\_0-t,t))$ is constant (I'm thinking of a point (x=0,t) on the boundary x=0 and a point from my the x-axis (x\_0,0) with a characteristic connecting them). This tells me that I can only choose one of $u(x=0,t)$ or $u\_x(x=0,t)$ but not both since the must agree on the bounary.
Is this what it means for "cauchy data to be consistent with boundary data"? Is there an analagous statement for the heat equation?
| https://mathoverflow.net/users/8755 | Can I pose any bounary data for the wave equation on $[0,\infty)$ for given Cauchy data? | I am just going to answer here the problem actually posed in the title.
The one dimensional wave equation can be re-written as $\partial\_u\partial\_v \phi = 0$, where $u$ and $v$ are the null variables $x + t$ and $x - t$. The initial data then is prescribed on $u+v \geq 0, u-v = 0$, while the boundary is $u+v = 0, u-v \geq 0$.
So we see that the wave equation implies that the function $\psi(x,t) = \partial\_u \phi(x,t)$ solve a transport equation with negative velocity (cf. my second comment above). Thus if your boundary condition is given such that $\partial\_u \phi(x,t)$ is well-determined along the boundary by just the data given there, you will reach an inconsistency. This is one of the ways of seeing why you cannot prescribe simultaneously Dirichlet and Neumann conditions at the same time.
(On the other hand, to make the IBVP well-posed, you *need* to specify $\partial\_v \phi(x,t)$ along the boundary, since it satisfies a transport equation with positive velocity. That just one of Dirichlet or Neumann conditions suffice follows from the fact that $\partial\_v \phi(0,t)$ can be solved from $\partial\_u \phi(0,t)$ [transported from Cauchy data] and the boundary data.)
| 2 | https://mathoverflow.net/users/3948 | 36561 | 23,520 |
https://mathoverflow.net/questions/36269 | 1 | Given a smooth projective surface $X$, let $A$ be a sheaf of maximal orders in a division ring.
Let us for simplicity assume $A$ ramifies in one curve $C$ with ramification index $e$. Let $A^\*$ be the dual sheaf.
How can I see that the determinant map is a map from $A^\*$ to $O\_X((1-\frac{1}{e})C)$?
And how to understand the invertible sheaf $O\_X((1-\frac{1}{e})C)$? How to handle the rational coefficients?
Since $A$ only ramifies in C, we have that $A$ is Azumaya on $U:=X\backslash C$. So on
$U$ we have $A^\*\cong A$. There the determinant map induces a map $A^{\\*} \rightarrow O\_U$ since $A$ is just a matrix algbera etale locally. So I see that we have a map $A^\* \rightarrow O\_X(rC)$. But how to find $r=1-\frac{1}{e}$? How can i determine $A^\*$ on $C$?
The question arose reading Theorem 7.1.4. on page 157 of this script: <http://www.math.lsa.umich.edu/courses/711/ordersms-num.pdf>
| https://mathoverflow.net/users/3233 | Q-Divisor and Determinant Map on a Maximal Order | The determinant should map to ${\cal A}^\* \rightarrow {\cal O}\_X(e(1-1/e)$. You can see this along $C$ in codimension one since if you \'etale localize at the generic point of $C$ then the structure Theorem for maximal orders says that ${\cal A}$ localizes to something Morita
equivalent to
R tR ... tR
R R ... tR
... ... ...
R ... R tR
R ... ... R
with equal size blocks. Where $R$ is the strictly henselian d.v.r at the generic point of $C$. Now if you if you dualize and count $t^{-1}$, you can have at most a pole of order $e-1$
in the determinant.
| 1 | https://mathoverflow.net/users/8762 | 36567 | 23,523 |
https://mathoverflow.net/questions/36574 | 0 | Given an unweighted graph $G = (V, E)$, let the cut function on this graph be defined to be:
$C:2^V \rightarrow \mathbb{Z}$ such that:
$$C(S) = |\{(u,v) \in E : u \in S \wedge v \not\in S\}|$$
Suppose you have the ability to query $C$, but otherwise have no knowledge of the edge set $E$. Is it possible to reconstruct $G$ by making only polynomial (in $|V|$) many queries to the cut function?
| https://mathoverflow.net/users/3027 | Reconstructing a graph given access to its cut function | Asking individual vertices, you figure out the valence of each vertex with $n$ questions. Asking for pairs of vertices, you can then decide if each pair has an edge between them or not: they have an edge if and only if the "degree" on the pair is two less than the sum of the "degrees". Thus you can reconstruct the graph with $n+\binom{n}{2}$ questions.
| 5 | https://mathoverflow.net/users/8761 | 36578 | 23,530 |
https://mathoverflow.net/questions/36549 | 2 | Usually one shows the density of the functions $\sin(kx)$ in $L^2([0,1])$ using the Fourier transform. This in fact comes from the Stone-Weierstrass theorem however and then uses the density of continuous functions in $L^2([0,1])$.
However, the Stone–Weirstrass theorem can be used to show, for example, that the functions $e^{ikx}$ are dense in $C([0,1])$ and hence dense in $L^1([0,1])$ as well. So we obtain (not-necessarily-unique) coefficients $c\_k$ such that $f\_k(x) =c\_ke^{ikx}$ converge to any given $f \in L^1([0,1])$. How should I think about these coefficients? How do they relate to the Fourier series of $f$ (with basis $e^{ikx})$?
| https://mathoverflow.net/users/8755 | Coefficients from Stone–Weierstrass versus Fourier transform | Just a comment if you choose coefficients $c\_{k,n}$ such that
$$
\lim\_{n\to\infty} \left(\sum\_{k=-n}^{n} c\_{k,n} e^{2\pi i n x}\right) \to f (x)
$$
in some sense, e.g. $L^1$, then these are not unique. It is even known that the obvious choice $c\_{k,n} = \hat{f}(k) = \int e^{-2\pi i n x} f(x) dx$ is not the best. It's much better to choose
$$
c\_{k,n} = \left(1 - \frac{|k|}{n}\right) \hat{f}(k).
$$
Then one Cesaro sums the Fourier series, and this is known to converge.
As pointed out by Zen Harper below, I should mention that with the choice $c\_{k,n} = \hat{f}(k)$ for $-n \leq k \leq n$, the Fourier series of a $L^1$ function must not converge. In fact it can diverge almost-everywhere.
Having said these things, the obvious advantage of this is, that everything is explicit and does not rely on any abstract hocus pocus.
I realized one more thing: Consider the case $f \in L^2$. Then the choice $c\_{k,n} = \hat{f}(k)$ for $-n \leq k \leq n$ is optimal. This follows from easy Hilbert space theory!
| 7 | https://mathoverflow.net/users/3983 | 36585 | 23,535 |
https://mathoverflow.net/questions/36583 | 3 | Given an unweighted graph $G = (V, E)$, let the cut function on this graph be defined to be:
$C:2^V \rightarrow \mathbb{Z}$ such that:
$$C\_G(S) = |\{(u,v) \in E : u \in S \wedge v \not\in S\}|$$
For any two vertices $i,j \in V$, let the $(i,j)$ min-cut in a graph $G$ be:
$$\alpha\_{i,j}(G) = \min\_{S \subset V : i \in S, j \not \in S}C\_G(S)$$
Now, suppose we have two unweighted graphs on the same vertex set, $G = (V,E)$ and $H = (V,E')$ such that they are identical with respect to all $(i,j)$ min-cuts:
$$\forall i,j \in V, \alpha\_{i,j}(G) = \alpha\_{i,j}(H)$$
How much can $H$ and $G$ differ with respect to their cuts? That is, how large can the following quantity be:
$$\Delta(H,G) = \max\_{S \subset V} |C\_G(S) - C\_H(S)|$$
Note that if the graphs are allowed to be weighted (or to be multigraphs), then for any $G$, there is a tree $T$ that agrees with $G$ on all min-cuts (A Gomory-Hu tree). But I am interested in the case of unweighted graphs...
| https://mathoverflow.net/users/3027 | To what degree do min-cuts specify the cut function of a graph? | If G and H are graphs with n vertices, Δ(G,H) can be Θ(n2).
Here is an example which shows this. Let n=4k+1 be a prime. Define two graphs G=(V,E) and H=(V,E′) by V={0,1,2,…,4k}, E = {{i,j} | 1 ≤ ((j−i) mod n) ≤ k}, E′ = {{i,j} | k+1 ≤ ((j−i) mod n) ≤ 2k}. Note that G and H are both unions of k edge-disjoint Hamiltonian circuits, which implies that αi,j(G)=αi,j(H)=2k for any distinct i and j. Let S={0,1,2,…,2k−1}. Then Δ(G,H) ≥ CH(S)−CG(S) = k(3k+1)−k(k+1) = 2k2 = Θ(n2).
There is no reason to believe that this is the maximum of Δ(G,H) for given n, but it is obviously optimal up to a constant factor.
| 6 | https://mathoverflow.net/users/7982 | 36589 | 23,537 |
https://mathoverflow.net/questions/36546 | 3 | I am interested in the properties of the following subclass of split graphs:
The class consists of all split graphs $G=(C\cup I)$ where $C$ is a clique and $I$ an independent set, and *every* pair of vertices in $I$ have at least one common neighbor in $C$.
Does this class of graphs have a special name? Has this class and its properties been studied? If so, what would be some good references for this?
| https://mathoverflow.net/users/1667 | Have this subclass of split graphs been studied before? | Split graphs of diameter 2?
| 7 | https://mathoverflow.net/users/7170 | 36591 | 23,539 |
https://mathoverflow.net/questions/36588 | 0 | While reading some papers translated from the Russian literature, I've noticed that a delta symbol can be used to denote a FDTD stencil that discretizes a PDE. For example, in [1], a fourth order mixed partial derivative term is denoted by
$
2\frac{{\partial ^4 u}}{{\partial ^2 x\partial ^2 y}} = \Delta \_{xy}^4 u^{k + 1} \_{i + 1,j + 1} + \Delta \_{xy}^4 u^k \_{i - 1,j - 1}
$
where an example is given of
$\Delta \_{xy}^4 u\_{i + 1,j + 1} = \Delta \_x^2 u\_{i + 1,j + 2} - 2\Delta \_x^2 u\_{i + 1,j + 1} + \Delta \_x^2 u\_{i + 1,j}$
Notice that this example given in the paper does not have the $\{ k,k + 1\} $ superscipts.
Clearly ${i,j}$ are spatial indices and $k$ is the timestep. But what is being implied by the use of the delta symbol? I suspect that this is a differential, but I have never seen a differential with $u\_{i,j}$ and $i,j$ indices. The author does not define the symbol in his paper, so I think that it should be implicitly understood. I am also unsure as to whether such a notation has also been used by other authors.
How would I write out $\Delta \_{xy}^4 u\_{i + 1,j + 1}$ and $\Delta \_{xy}^4 u\_{i - 1,j - 1}$ using a 5-point stencil or 7-point stencil? Are there any other papers which use similar notation?
[1] V. Saul'yev, “A difference method for solving parabolic equations of any order,” Computational Mathematics and Mathematical Physics, vol. 36(12), 1996, pp. 1697-1700.
| https://mathoverflow.net/users/7875 | Delta notation used for describing numerical stencil | If I saw that symbol alone, I would guess that $\Delta^2\_x u\_{i,j}:=u\_{i+1,j}-2u\_{i,j}+u\_{i-1,j}$. But with that definition the stencil $\Delta^4\_{xy}u\_{i+1,j+1}$ looks very odd.
| 1 | https://mathoverflow.net/users/1898 | 36598 | 23,544 |
https://mathoverflow.net/questions/36592 | 5 | To be slightly more precise: let $M\subset B(H)$ be a finite von Neumann algebra equipped with a faithful normal trace $\tau$, and let $L^0(M,\tau)$ be the completion of $M$ in the measure topology; this is an algebra, whose elements can be identified with those densely-defined and closed operators on $H$ that are affiliated with $M$. (See e.g. E. Nelson, *Notes on noncommutative integration*, JFA 1974). Let $e$ be an idempotent in $L^0(M,\tau)$, not necessarily self-adjoint; then it is not hard to show that $R=\{ \xi\in H : e\xi=\xi\}$ is a closed subspace of $H$.
**Question:** is the orthogonal projection onto $R$ affiliated with $M$?
I suspect the answer is yes (and would like it to be, for some calculations I'm doing at the moment) but am having difficulties nailing the argument down. Given that this should, if true, be a pretty basic bit of operator algebra theory, and standard knowledge, I'd be grateful if someone could point me to a reference. (I currently have somewhat limited library access, but this might well be covered in Kadison & Ringrose for instance.)
---
**Edit/update:** both Martin Argerami and Jonas Meyer have given straightforward proofs of the desired result, and a quick check in Kadison & Ringrose vol. 1 has not turned up any explicit statement (probably because the result turns out to be so basic). Since I can't accept both their answers, I'm accepting Martin's on grounds of personal preference.
| https://mathoverflow.net/users/763 | range projection of an unbounded idempotent affiliated to a finite von Neumann algebra | I couldn't find it in Kadison & Ringrose. But what about this: let $\xi\in R$ and let $u$ be a unitary in $M'$. Since $e$ is affiliated with $M$, $ue=eu$. So $u\xi=ue\xi=eu\xi\in R$. This shows that $uR\subset R$ for any unitary $u$, and so $uR=R$ for any unitary $u$ in $M'$. This in turn is equivalent to $ur=ru$, where $r$ is the orthogonal projection onto $R$. As $u$ was any unitary in $M'$, we conclude that $r\in M$.
| 5 | https://mathoverflow.net/users/3698 | 36605 | 23,548 |
https://mathoverflow.net/questions/36532 | 12 | Suppose you have a labeled tree $T$ on vertices $V=\lbrace 1,\ldots,n\rbrace$ that is drawn uniformly at random from the set of all $n^{n-2}$ such trees. I am seeking an $f$ satisfying the following desiderata:
D1. $f(T)$ is a (random) tree on the vertex set $V'=\lbrace 1,\ldots,n+1\rbrace$;
D2. the distribution of $f(T)$ is uniform over the set of all $(n+1)^{n-1}$ labeled trees on $V'$;
D3. $f$ is a simple graph-theoretic recipe.
This $f$ can be a random function. That is, it can flip coins in deciding what to do with $T$: for instance, removing an edge uniformly at random.
One obvious way to satisfy D1 and D2 is:
1. convert $T$ to its Prüfer sequence $s$;
2. independently, with probability $1/(n+1)$, change each entry of $s$ to $n+1$ (otherwise leave it fixed);
3. then add a random number to the end of $s$, drawn uniformly from $V'$;
4. let $f(T)$ be the tree corresponding to the new sequence.
But what this procedure "really does" to $T$ seems not so easy to describe in graph-theoretic terms. I am looking for a recipe that satisfies D1-3 by manipulating the graph "directly" (i.e., adding and removing edges) without opaque steps like (1) and (4) in the above procedure.
| https://mathoverflow.net/users/7967 | Is there a simple inductive procedure for generating labeled trees uniformly at random, without direct recourse to Prüfer sequences? | This is an interesting question. For any fixed positive integer $d \geq 2$, write $T\_d^{\infty}$
for the complete infinite rooted $d$-ary tree (by this I mean every node has exactly $d$ children). [Luczak and Winkler](http://onlinelibrary.wiley.com/doi/10.1002/rsa.20011/abstract) proved the existence of a procedure which will generate a sequence $(T\_{n,d})\_{n \geq 0}$ such that for all $n \geq 1$,
(a) the distribution of T\_{n,d} is uniformly random over all $n$-node subtrees of $T^{\infty}\_d$ that contain the root of $T^{\infty}\_d$; and
(b) $T\_{n,d}$ is a subtree of $T\_{n+1,d}$.
It is not hard to show that (a) implies that for all $n$, $T\_{n,d}$ is distributed as a Galton-Watson tree with offspring distribution $\mathrm{Bin}(d,1/d)$, conditioned to have total size $n$.
Since $\mathrm{Bin}(d,1/d)$ tends to a $\mathrm{Poisson}(1)$ distribution as $d$ becomes large,
this means that as $d \to \infty$, the distribution of $T\_{n,d}$ tends to that of a Galton-Watson tree with offspring distribution $\mathrm{Poisson}(1)$ conditioned to have $n$ nodes (let me write $\mathrm{PGW}\_n(1)$ for this distribution). The latter distribution is the same as that of a uniformly random labelled rooted tree on labels $1,\ldots,n$. (At least, the latter is true once we label the Galton-Watson tree uniformly at random, or alternately remove the labels of the labelled rooted tree.)
As noted by [Lyons et al.](http://arxiv.org/abs/0711.1893) (Theorem 2.1), all this implies in particular that one can define a similar sequence $(T\_n)\_{n \geq 1}$ such that for all $n$, $T\_n$ is a subtree of $T\_{n+1}$ and $T\_n$ has distribution $\mathrm{PGW}\_n(1)$.
However, the construction in the Luczak-Winkler paper uses flows, and it is not 100% obvious how it "passes through to the $d=\infty$ limit." (I say this with the caveat that I didn't make any serious attempt at figuring this out.) As a consequence, while it is known that there exists a generation procedure of the type you are looking for, I am not aware of an explicit description of the actual rule for where the leaf should be attached to $T\_n$ to create $T\_{n+1}$. I asked Peter Winkler about this at a conference last year and he also didn't know (though I don't know whether he had thought about this specific question in depth, either).
| 2 | https://mathoverflow.net/users/3401 | 36606 | 23,549 |
https://mathoverflow.net/questions/35226 | 8 | Let $f:X\to Y$ be a birational morphism, $X, Y$ projective, $X$ smooth (threefold if this helps). Let $Exc(f)\subseteq X$ be the exceptional locus of $f$ and let $E\subseteq Exc(f)$ be an irreducible divisor. Is it true that for any curve $C\subseteq E$ contracted by $f$ one has $C\cdot E<0$? I can see this is true if $C$ is not contained in any other divisor sitting in $Exc(f)$, but what if it is?
| https://mathoverflow.net/users/5419 | Negativity of contraction | Dear Carlos, the statement is false in general. For example let $Y$ be $\mathbb{C}^3$, let $f\_1 : X\_1 \rightarrow Y$ be the blowup of a point on $Y$, and $f\_2 : X \rightarrow X\_1$ the blowup of a point on the exceptional divisor of $f\_1$. Let $f : X \rightarrow Y$ be the composition. The exceptional locus of $f$ has two components: a copy $F$ of $\mathbb{P}^2$ (the exceptional divisor of $f\_2$) and a copy $E$ of the blowup of $\mathbb{P}^2$ in one point (the strict transform of the exceptional divisor of $f\_1$). The intersection $C = E \cap F$ is a copy of $\mathbb{P}^1$, it is a line on $F$ and the $f\_2$-exceptional curve on $E$. Now $E \cdot C$ equals $C^2$ computed on $F$ (because $C=E \cap F$),
so $E \cdot C = +1$.
| 5 | https://mathoverflow.net/users/8770 | 36612 | 23,553 |
https://mathoverflow.net/questions/36613 | 6 | Suppose that $K$ is a number field, and (writing $G\_K=\mathrm{Gal}(\overline{K}/K)$), suppose that $\phi:G\_K\to \overline{\mathbb{Q}}$ is a finite order character of $G\_K$. I believe that the obstruction to taking a square root of $\phi$ (that is, the obstruction to finding some finite order $\chi:G\_K\to \overline{\mathbb{Q}}$ with $\chi^2=\phi$) can be identified with a 2-torsion element of the Brauer group of $K$, and I believe that the proof just involves taking a long exact sequence from a cunningly-chosen short exact sequence; but I cannot now reconstruct it. Can anyone remind me?
| https://mathoverflow.net/users/3513 | obstruction to taking the square root of a Galois character | With luck, this will be blunder-free (and if it's not, please tell me!):
Consider the short exact sequence
$$1 \to \mu\_2 \to \overline{\mathbb Q}^{\times} \to \overline{\mathbb Q}^{\times}
\to 1,$$
with the third arrow being squaring, and with trivial $G\_k$-action. Passing to cohomology,
the sequence of $H^0$s is just this same sequence, and the sequence of higher cohomology
becomes
$$0 \to Hom(G\_K,\mu\_2) \to Hom(G\_K,\overline{\mathbb Q}^{\times}) \to
Hom(G\_K,\overline{\mathbb Q}^{\times}) \to H^2(G\_K,\mu\_2),$$
with the last arrow being the obstruction you asked about.
Added: See Brian Conrad's comment below for a cleaner point of view, showing
that $H^2(G\_K,\mu\_2)$ is the precise obstruction space.
| 12 | https://mathoverflow.net/users/2874 | 36618 | 23,555 |
https://mathoverflow.net/questions/36607 | 0 | Suppose X is compact and totally disconnected space, and that phi a homeomorphism of X.
We say a subset Z of X is phi-invariant if phi(Z) = Z. A phi-invariant set is minimal if it is closed, phi-invariant, nonempty and the smallest of all such sets. We say (X,phi) is minimal if X itself is a minimal set.
An orbit of x in X is the set {phi^n(x) : n an integer}
A system (X,phi) is minimal iff every orbit is dense.
Given (X,phi) as above, and any point y in X. The system is "essentially minimal" if one of the following equivalent conditions hold:
1) For all x in X, y in { phi^n(x) : n >= 0, n an integer }.
2) For all x in X, y in { phi^n(x) : n < 0, n an integer }.
3) X contains a unique minimal set Y, and y in Y.
If a system is minimal, then condition 3 is satisfied (setting Y := X), and is hence essentially minimal.
Does essential minimality imply minimality?
| https://mathoverflow.net/users/8769 | Does essentially minimal imply minimal? | As far as I understand the author tried (with missing closure in conditions (1) and (2), as it was already pointed out) to reproduce the standard definition of an essentially minimal dynamical system as one which has a unique minimal subset, see, for instance, Definition 1.2 from
MR1194074 (94f:46096)
Herman, Richard H.(1-MD); Putnam, Ian F.(3-DLHS); Skau, Christian F.(N-TRND)
Ordered Bratteli diagrams, dimension groups and topological dynamics.
Internat. J. Math. 3 (1992), no. 6, 827--864.
There is a lot of examples of essentially minimal systems which are not minimal, see the Introduction to
MR1944409 (2003k:37020)
Matui, Hiroki(J-CHIBES-MI)
Topological orbit equivalence of locally compact Cantor minimal systems. (English summary)
Ergodic Theory Dynam. Systems 22 (2002), no. 6, 1871--1903.
The simplest is just to take the shift on the integers and to extend it to the one-point compactification.
| 4 | https://mathoverflow.net/users/8588 | 36620 | 23,557 |
https://mathoverflow.net/questions/36611 | 5 | It is known that a valuation domain is a principal ideal ring if and only if its prime ideals are principal. Is it a principal ideal ring when its unique maximal ideal is principal?
| https://mathoverflow.net/users/5775 | Is a valuation domain PID when its maximal ideal is principal? | I assume that by a valuation domain you mean an integral domain $R$ with fraction field $K$ such that: for all $x \in K^{\times}$, at least one of $x,x^{-1}$ lies in $R$.
In this case, I believe the answer is **no**. Let $R$ be any valuation domain whose value group $K^{\times}/R^{\times}$ is isomorphic, as a totally ordered abelian group, to $\mathbb{Z} \times \mathbb{Z}$ with the lexicographic ordering. (It is known that every totally ordered abelian group is the value group of some valuation domain, e.g. by a certain generalized formal power series construction due to Neumann.) In this case, the maximal ideal consists of all elements whose valuation is strictly greater than $(0,0)$, but the valuation of any such element is at least $(0,1)$ and therefore any element of valuation $(0,1)$ gives a generator of the maximal ideal.
For some information on valuation rings, see e.g. Section 17 of
[http://alpha.math.uga.edu/~pete/integral.pdf](http://alpha.math.uga.edu/%7Epete/integral.pdf)
| 9 | https://mathoverflow.net/users/1149 | 36636 | 23,570 |
https://mathoverflow.net/questions/36639 | 2 | Suppose I have a singular projective variety defined by some homogeneous equations in complex projective space. Is the resolution of singularities effective? That is, do I actually know which smooth centers to blow up?
| https://mathoverflow.net/users/nan | Is resolution of singularities effective? | Yes, in the sense that resolution of singularities is implemented in the computer algebra package Singular. See the manual of Singular for references. (There might be other/better references.) However, if I remember correctly the centers are not unique.
| 5 | https://mathoverflow.net/users/8621 | 36640 | 23,572 |
https://mathoverflow.net/questions/36568 | 6 | To do Algebraic K-theory, we need a technical condition that a ring $R$ satisfies $R^m=R^n$ if and only if $m=n$. I know some counterexamples for a ring $R$ satisfies $R=R^2$.
Are there any some example that $R\neq R^3$ but $R^2 = R^4$ or something like that?
(c.f. if $R^2=R^4$, then we need that $R^3=R^5=\ldots =R^{2n+1}$ for any $n>1$)
| https://mathoverflow.net/users/7776 | Subtle counterexample to $m\neq n$ but $R^m=R^n$ for some ring $R$ ? | Rings that satisfy the condition
$R^n \cong R^m \iff n=m$ are said to have *invariant basis number* or the *invariant basis property*. P. M. Cohn has constructed examples of rings which fail to have this property, even giving examples of (non commutative) integral domains for which e.g. $R^3\cong R$ but $R^2\neq R$.
Suppose $R^n\cong R^m$ for some $n,m$. Then there must exist integers $h,k$ such that
$$
R^m\cong R^{m'} \iff m=m'\ \mbox{or}\ m,m'\ge h \ \mbox{and} \ m\equiv m'\ (\mod k)
$$
A ring satisfying this condition is of type $(h,k)$. In [P. M. Cohn, Some remarks on the invariant basis property, Topology 5 (1966), 215-228] a fairly simple construction is given for rings of type $(h,k)$ for any $h,k$.
There are earlier constructions of rings of type $(h,k)$ by Leavitt [W. G. Leavitt:
Modules without invariant basis number, Proc. Am. Math. Sot. 8 (1957), 322-328], [W. G. Leavitt:
The module type of a ring, Trans. Am. Math. Sot. 103 (1962), 113-130], but they are far more complicated.
Certainly we can do algebraic K-theory over rings without the invariant basis property; we just need to be a little more careful. For example we won't necessarily have $K\_0(R)\cong {\bf Z} \oplus \tilde{K}\_0(R)$.
| 4 | https://mathoverflow.net/users/6701 | 36644 | 23,574 |
https://mathoverflow.net/questions/36625 | 2 | Say $L\mathbb{C}^\times$ is the loop group of smooth maps $S^1 \to \mathbb{C}^\times$. There is a submonoid $L\_{poly}\mathbb{C}^\times$ of loops that look like $w\_0 + w\_1z +w\_2z^2 + \cdots + w\_nz^n$ where $z = e^{i\theta}$ (as Andrew notes below this is not a group because its not closed under taking inverses). Equivalently $L\_{poly}\mathbb{C}^\times$ as a set is just polynomials $p(z) \in \mathbb{C}[z]$ such that $p(z) \ne 0$ for $|z| = 1$.
If we mod out by scaling and rotation then the set of polynomial loops describe a subset $X$ of $\mathbb{P}(\oplus\_{n \in \mathbb{N}} \mathbb{C})$ (by identifying a loop with its vector of coefficients). I want to look at $X$ from an algebro-geometric point of view, but I have no intuition about how bad or nice $X$ may be; i.e. can it be a variety?
The way I've been thinking about it is that $X = \cup\_{n \in \mathbb{N}} X\_n$ where $X\_n \subset \mathbb{P}^n$ are the loops of degree at most $n$. I think $X\_1$ is the image under the projection $\mathbb{C}^2 - 0 \to \mathbb{P}^1$ of the set {$(w\_0,w\_1):|w\_0|\ne |w\_1|$}. So it seems describable as the complement of a hypersurface in $\mathbb{R}^4$ but probably its not a complex variety.
But already trying to figure out what $X\_2$ is seems difficult. Also I feel I don't have any `sophisticated' way of thinking about this stuff meaning my attempts to describe $X\_2$ seems to always degenerate to just fumbling around with planar geometry.
Some specific questions regarding this setup:
0) What is the dimension of $X\_n$?
1) Which if any of the $X\_n$ or $X$ are a variety over $\mathbb{C}$ or $\mathbb{R}$?
2) If $X$ or $X\_n$ are not varieties can you find any positive dimensional varieties contained in them?
3) Can you suggest any tools that might be useful for answering any of the previous questions?
Of course if any of this seems to easy you are welcome to replace $\mathbb{C}^\times$ with $GL(n,\mathbb{C})$, polynomials with rational functions or with power series convergent in an annulus containing $|z| = 1$.
An extra thought: So $L\_{poly}\mathbb{C}^\times$ is not a group, but it seems you do get a group if you look at convergent series in non positive powers of $z$; i.e. loops that look like $\sum\_{j \in \mathbb{N}} c\_j z^{-j}$. I wonder if there's anything interesting you can say about the $c\_j$.
| https://mathoverflow.net/users/7 | parameterizing polynomial loops in $\mathbb{C}^\times$ | This is mostly a series of comments, but guided by the questions you asked.
First of all, I will only talk about $X\_n$, interpreting it as the space of non-zero complex polynomials $p$ of degree at most $n$ such that no root of $p$ lies on the unit circle, taken up to non-zero scaling. We may as well think of the polynomials as homogeneous of degree exactly $n$ in two variables, so that each point of $X\_n$ defines a subset of $n$ points of the complex projective line $\mathbb{CP}^1$ (the set of roots of the polynomial) that is disjoint from the unit circle. Therefore, $X\_n$ is certainly a non-empty open subset in the standard topology of the projective space of homogeneous polynomials of degree $n$, and therefore **$X\_n$ has (real) dimension $2n$**. Observe that the unit circle disconnects $\mathbb{CP}^1$, and that the points of $X\_n$ are likewise distributed into (at least) $n+1$ connected components, corresponding to how the $n$ points in $\mathbb{CP}^1$ are distributed with respect to the two halves obtained by removing the unit circle (recall that $\mathbb{CP}^1$ is topologically a two-dimensional sphere and that the unit circle can be identified with the equator of the sphere, so that, among the $n$ points we are talking about, there are some that are in one hemisphere and some that are in the other). On the other hand, a non-empty Zariski open subset of an irreducible algebraic variety is connected. Thus, as a subset of the space of homogenous polynomials of degree $n$, **the space $X\_n$ is certainly not a complex algebraic subvariety**.
But, we may decide to analyze further the space $X\_n$, zooming in on the locus $X\_n^k$ where, for a fixed integer $k$, there are $k$ points on the half containing the origin and $n-k$ points on other half. Clearly, within each hemisphere, the points are free to roam around! Thus $X\_n^k$ is homeomorphic (and in fact diffeomorphic with the natural choice of differentiable structure) to the space of ordered pairs $(p\_1,p\_2)$ where $p\_1$ is a monic polynomial of degree $k$ and $p\_2$ is a monic polynomial of degree $n-k$: expand each hemisphere to a whole complex plane and "code" the $k$ points on one half by the unique monic polynomial having them as a root (and do the same to the other half). Thus each space $X\_n^k$ is connected and in fact diffeomorphic to $\mathbb{C}^k \times \mathbb{C}^{n-k}$. As a complex manifold you can also say that $X\_n^k$ is the product of the symmetric product of $k$ copies of the unit disk with the symmetric product of $n-k$ copies of the unit disk. Thus, again using a structure induced from the ambient space of homogeneous polynomials, any complex algebraic subvariety of $X\_n$ would be a complex algebraic subvariety of a symmetric product of unit disks: **I think that this means that the only complex algebraic subvarieties of $X\_n$ are the points**.
Finally, let me make a small stab at getting your hand on $X$, by mentioning one description of the "glueing" of $X\_n$ inside $X\_{n+1}$. From the point of view of non-homogeneous polynomials, this corresponds to simply realizing that a polynomial of degree at most $n$ is also a polynomial of degree at most $n+1$. From the point of view of their homogenizations, the inclusion corresponds to replacing $z^i$ by $x^iy^{n+1-i}$ instead of $x^iy^{n-i}$. Effectively, we are adding the point at infinity as one of our roots (namely the extra root $y=0$). Thus in terms of the description above, we observe that the two hemispheres are not "identical": one of them has a point that is special, namely the point at infinity. The points of $X\_{n+1}$ that come from points of $X\_n$ are the points that correspond to $(n+1)$-tuples one of whose elements is the point at infinity.
| 5 | https://mathoverflow.net/users/4344 | 36651 | 23,581 |
https://mathoverflow.net/questions/36658 | 4 | Hi,
let $E$ be a vector bundle over a smooth projective variety $X$ and $\pi:\mathbb{P}(E)\rightarrow X$ its projectivization, $T\_{\pi}:=ker(\pi\_{ \* })$ where $\pi\_{\*}:T\_{\mathbb{P}(E)}\rightarrow \pi^{\*}T\_X$, let $\mathcal{O}\_E(-1)\hookrightarrow\pi^{\*}E$ be the "tautological" bundle over $\mathbb{P}(E)$. The following it is not a proof but a first reasoning to understand things, if i restrict to a point $x\in X$ i have the usual Euler sequence
$0 \rightarrow \mathcal{O}\_{E\_x}(-1)\rightarrow ({\pi^{\*}}E)\_x \rightarrow (T\_{\pi})\_{x}\otimes \mathcal{O}\_{E\_x}(-1)\rightarrow 0 $
so my thought is that the generalized euler sequence becomes
$0 \rightarrow \mathcal{O}\_{E}(-1)\rightarrow \pi^{\*}E\rightarrow T\_{\pi}\otimes \mathcal{O}\_{E}(-1)\rightarrow 0 $
is this the right path or i'm wrong?
thank you in advance.
| https://mathoverflow.net/users/4971 | Which is the correct generalization of Euler sequence to the projectivization of a vector bundle? | This sequence is indeed exact. Once you check that the maps are globally defined, exactness can be checked fiberwise (remember that we are dealing with a sequence of *vector bundles*, not an arbitrary sequence of sheaves), and in this case it follows by the usual Euler sequence for the projective space.
| 5 | https://mathoverflow.net/users/828 | 36661 | 23,586 |
https://mathoverflow.net/questions/36494 | 3 | Maybe this is too technical and elementary, but I cannot make up my mind, nor find a reference.
The situation is the following: let $X$ be a double cochain (right half-plane) complex of abelian groups and let
$$
(\mathbf{Tot}^{\prod} X)^n = \prod\_{p+q = n}X^{pq}
$$
denote its total-product complex. The first filtration on $X$,
$$
{}\_I F^s(X) =
\begin{cases}
X^{pq} & \ \text{if} \ p \geq s , \\\
0 & \ \text{otherwise}
\end{cases}
$$
gives you the filtration on $\mathbf{Tot}^\prod X$:
$$
(F^s \mathbf{Tot}^\prod X)^n = \prod\_{p+q=n , p\geq s} X^{pq}
$$
and you have, as with any filtered differential complex $(A, F, d)$, an induced filtration in cohomology:
$$
F^pHA = \mathbf{im} (HF^pA \longrightarrow HA) \ .
$$
My question is the following: is the filtration induced by ${}\_I F$ on $H(\mathbf{Tot}^\prod X)$ Hausdorff? That is,
$$
\bigcap\_p F^pH(\mathbf{Tot}^\prod X ) = 0 \ ?
$$
I couldn't find an answer in the literature. Weibel's book says that in this situation the spectral sequence arising from the first filtration is "convergent". Unfortunately, for Weibel this only means that you have an isomorphism $E\_0 HA = E\_\infty A$. Cartan-Eilenberg "Homological Algebra" doesn't work with the total-product complex, but with the total-sum one:
$$
(\mathbf{Tot}^{\bigoplus} X)^n = \bigoplus\_{p+q = n}X^{pq}
$$
For this one, I think, the answer is "yes": if I had some cohomology class
$$
[x] \in \bigcap\_p F^pH^n(\mathbf{Tot}^\bigoplus X ) \quad \Longleftrightarrow \quad [x] \in F^pH^n(\mathbf{Tot}^\bigoplus X ) \ \text{for all} \ p
$$
then I could find representatives for $[x]$ like
$$
(0, \stackrel{p-1}{\dots}, 0, x^{p,n-q}, x^{p+1, n-q-1}, \dots )
$$
for all $p \geq 0$. Since there is only a finite number of $x^{pq} \neq 0$ for each element of $(\mathbf{Tot}^\bigoplus X)^n$, in a finite number of steps, I can be sure that I can find a representative for $[x]$ which is zero, so $[x]=0$.
But this reasoning doesn't work with $\mathbf{Tot}^\prod X$: you can only state with certainty that, for every $p$ there is some $x\_p \in F^p$ and $b\_p$ such that
$$
x - x\_p = db\_p \quad \Longleftrightarrow \quad x- db\_p \in F^p \ .
$$
These equations have a nice interpretation: if you topologize $\mathbf{Tot}^\prod X$ taking as basic open sets $x + F^p$ for all $x \in \mathbf{Tot}^\prod X$ and $p$, they read:
$$
(db\_p) \longrightarrow x \ .
$$
That is, $x$ is a limit of coboundaries. But, unless the filtration is finite, this doesn't imply that $x$ itself *is* a coboundary, does it?
So I could ask my question this way: in this situation, is the set of coboundaries closed?
| https://mathoverflow.net/users/1246 | Is the first filtration Hausdorff? | No, this filtration is not necessarily Hausdorff. The problem is connected to nonexactneess of the inverse limit functor. Here is a family of examples to illustrate the basic issue.
Suppose $\cdots \to A\_2 \to A\_1 \to A\_0$ is an inverse system of abelian groups. Define a complex by
$$
X^{p,-p} = X^{p+1,-p} = A\_p
$$
for all natural numbers $p$, with $X^{p,q} = 0$ otherwise. The horizontal differentials $X^{p,-p} \to X^{p+1,-p}$ are the identity maps $A\_p \to A\_p$, and the vertical differentials $X^{p,-p} \to X^{p,1-p}$ are the maps $A\_{p} \to A\_{p-1}$ from the inverse system for $p > 0$.
The total complex has only two nonzero groups in degrees 0 and 1:
$$
\prod A\_p \to \prod A\_p
$$
and the indicated map is the map Milnor introduced whose kernel is $\lim A\_p$ and whose cokernel is $\lim^1 A\_p$. (Possibly up to sign, depending on what your conventions for a double complex are.) Therefore
$$
H^0 Tot^{\Pi} X = \lim A\_p,\ H^1 Tot^\Pi X = {\lim}^1 A\_p
$$
However, the spectral sequence associated to this complex has the amusing behavior that the "diagonal" entries $E\_r^{p,-p}$ busily spawn differentials that erase everything on the "superdiagonal" $E\_r^{p+1,-p}$. Therefore, the entirety of $H^1(Tot^\Pi X)$ is in the intersection of the images of the filtrations.
(As a mostly nonmathematical aside, $\lim^1$ is the bane of many an argument.)
If you're interested in convergence properties, I'd suggest Boardman's thesis, mentioned in [this answer](https://mathoverflow.net/questions/32762/convergence-of-right-half-plane-spectral-sequence-bounded-on-the-right/32769#32769) of Tilman's.
| 2 | https://mathoverflow.net/users/360 | 36662 | 23,587 |
https://mathoverflow.net/questions/36642 | 15 | My question is about $m \times n$ binary matrices (aka $\{0,1\}$-matrices), whose rows all sum to the same value, and whose columns all sum to the same value (but these two values may be different).
The first question is simply: is there a standard name for such matrices? They correspond to the biadjacency matrices of so-called "biregular bipartite graphs", but this terminology doesn't appear to be commonly used.
Second, are there any "interesting" constructions of families of such matrices, in particular that are connected to other combinatorial objects?
Two simple examples of constructions of these matrices are the $\binom{n}{k} \times n$ matrix whose rows consist of every $n$-bit string with Hamming weight $k$; and the $2^n \times 2^n$ [Sylvester-Hadamard](https://en.wikipedia.org/wiki/Hadamard_matrix#Sylvester.27s_construction) matrices with the first row and column removed.
I did find a [paper](https://doi.org/10.1016/0024-3795(80)90105-6) by Brualdi titled "Matrices of Zeros and Ones with Fixed Row and Column Sum Vectors", but this seems to be more concerned with the question of existence of these matrices, rather than constructing them.
| https://mathoverflow.net/users/7767 | Binary matrices with constant row and column sums | To answer your question about interesting combinatorial objects: Your Sylvester-Hadamard matrix example generalizes in at least two ways.
1. The incidence matrix of any balanced incomplete [block design](https://en.wikipedia.org/wiki/Block_design) or, more generally, $t$-design has constant row and column sums. Specifically, a BIBD$(v,b,r,k,\lambda)$ represented as a $v\times b$ incidence matrix has row sums $r$ and column sums $k$. Normalizing an $n\times n$ Hadamard matrix so that the first row and column consist entirely of 1s, and then removing this row and column while replacing $-1$s with 0s gives a design with parameters $v=b=n-1$, $r=k=n/2-1$, $\lambda=n/4-1$. So among Hadamard matrices, Sylvester-Hadamard matrices are not special in this regard. Finite projective planes are also designs, with parameters $v=b=q^2+q+1$, $r=k=q+1$, $\lambda=1$, where $q$ is the order of the plane.
2. Sylvester-Hadamard matrices have the additional property that if the first row and column are removed, and the remaining rows and columns are suitably permuted, one obtains a circulant matrix. (Most Hadamard matrices do not have this property, but Paley-Hadamard matrices constructed using quadratic residues in $\mathbb{F}\_p$, $p\equiv3\pmod{4}$ also do. More generally, Hadamard matrices constructed from [difference sets](https://www.dmgordon.org/diffset/) do.) But any $n\times n$ circulant matrix whatsoever will have constant row and column sums (with row sums equal to column sums). (This is not combinatorially so interesting in general.)
**Addendum:** In answer to your first question, design theorists call an incidence structure whose incidence matrix has constant row and column sums a **tactical configuration**. This is a $t$-design with $t=1$. The balanced incomplete block designs in the first part of my answer are $t$-designs with $t=2$, but any $t$-design is also a $(t-1)$-design. The condition that any two points be incident with exactly $\lambda$ blocks gives BIBDs a lot of additional interesting structure that tactical configurations do not typically have. 3-, 4-, and 5-designs are more interesting still. See the [PlanetMath page](https://planetmath.org/incidencestructure) for more information.
**Added 12 July 2022:** I just updated the link above to the La Jolla Difference Set Repository, which is now the La Jolla Combinatorics Repository. Dan Gordon's slides [here](http://www.cargo.wlu.ca/ArasuFest/ArasuFest_talks/Dan_Gordon.pdf) have examples showing how to query the database.
| 8 | https://mathoverflow.net/users/484 | 36668 | 23,591 |
https://mathoverflow.net/questions/36665 | 7 | I have a triangle $ABC$ with side lengths $a,b,c$ (edges $BC, CA, AB$ respectively).
I have a point $p$ with barycentric coordinates $u:v:w$.
These are normalised: $u+v+w=1$.
$1:0:0$ corresponds to point $A$, $0:1:0$ is $B$ etc.
Is there a simple expression for the distance $d$ of the point $p$ from $A$ ?
(My initial naive guess based on $d(1:0:0)=0, d(0:1:0)=b, d(0:0:1)=c$ was that $d$ was linear $d(u,v,w)=v\*b+w\*c$ but this is clearly wrong as in the case of an equilateral triangle $a=b=c=1$ it returns $d=2/3$ for the centroid ($u:v:w = 1/3:1/3:1/3$), when the correct answer should be $1/\sqrt 3$ (the radius of the circumscribed circle)).
| https://mathoverflow.net/users/3519 | Distance of a barycentric coordinate from a triangle vertex | There isn't really a simple formula, but you can use vector methods.
Let $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ be position vectors
of the vertices. A point $P$ with normalized barycentric coordinates
$(u,v,w)$ has position vector $\mathbf{p}=u\mathbf{a}+v\mathbf{b}+w\mathbf{c}$.
Therefore $\mathbf{p}-\mathbf{a}=v(\mathbf{b}-\mathbf{a})+w(\mathbf{c}-\mathbf{a})$.
This leads to
$$|AP|^2=v^2|AB|^2+w^2|AC|^2+2vw|AB||AC|\cos\alpha$$
where $\alpha$ is the angle at $A$. Of course one can express
$|AB||AC|\cos\alpha$ in terms of the three side-lengths of the triangle
using the cosine rule.
This shows that neither $|AP|$ nor $|AP|^2$ is a linear function
of the barycentric coordinates (actually this is geometrically evident too).
But there is a simpler formula for the distance of $P$ to a given side
of the triangle.
| 4 | https://mathoverflow.net/users/4213 | 36669 | 23,592 |
https://mathoverflow.net/questions/36674 | 1 | I'm looking for a proof in the literature of the following fact:
Let $A\_t$ be a $C^1$-function of one argument $t \in (a,b)$ taking values in the self-adjoint $N \times N$ matrices. Suppose that for every $t$ the spectrum of $A\_t$ is simple. Denote by $\lambda\_t$ a continuous parametrization of an eigenvalue of $A\_t$. Then there exists a $C^1$ function $\psi\_t$ taking values in $\mathbb{C}^N$ such that
$$
A\_t \psi\_t = \lambda\_t \psi\_t
$$
and $\psi\_t$ is normalized $\sum\_{j=1}^{N} |\psi\_t(n)|^2 = 1$.
It is not hard to give a proof, but it is somewhat awkward to write down, that's why I am looking for a reference of this fact, I couldn't find one in Reed--Simon 4 and Kato. Here is a short sketch of my idea how to proof it: $\psi\_t$ lives in $S^N$, whose tangent space at the point $\psi\_t$ is $T\_{\psi\_t} S^N \cong \{\psi\_t\}^{\perp}$. Next, we have that $A\_t - \lambda\_t|\_{\psi\_t ^{\perp}}$ is invertible (here the simplicity of the spectrum is used). Using this we can define $\psi\_t$ as the solution of the ODE
$$
\dot{\psi\_t} = (A\_t - \lambda\_t|\_{\psi\_t ^{\perp}})^{-1} (\dot{A}\_t - \dot{\lambda}\_t) \psi\_t.
$$
Now, various computations show that everything is well-defined and indeed give the desired solution.
| https://mathoverflow.net/users/3983 | Parametrizing eigenvectors | I do not know a reference, but here is an easy argument. Consider the space $\overline{M}$ of pairs $(A,\lambda)$ where $A$ is a (self-adjoint) $N \times N$ matrix and $\lambda$ is a root of the characteristic polynomial of $A$. Over the open subset of $\overline{M}$ lying above the operators with distinct eigenvalues, the projection onto the first factor is clearly a proper smooth map, with finite fibers; in other words, the space $\overline{M}$ is a covering space over this open set. Since everything in sight is differentiable, the result that you want follows from the lifting of paths in the base. You need to choose an eigenvalue for the lifting to be unique, but you seem to know which one to choose!
Btw, you only need the eigenvalue $\lambda$ to be simple for the same argument to work. In this case you can use the Implicit Function Theorem to lift the path, since the assumption on the simplicity of the root imply that the differential at the point is surjective.
| 3 | https://mathoverflow.net/users/4344 | 36677 | 23,597 |
https://mathoverflow.net/questions/36638 | 13 | Let $f(x\_1,\ldots, x\_n)$ be a real polynomial in several variables. Is there an effective algorithm to test whether $f$ is positive (or nonnegative) on the whole of ${\mathbb{R}}^n$?
| https://mathoverflow.net/users/nan | Effective algorithm to test positivity | If by effective you mean "is this computable", then yes, the computational versions of Tarski-Seidenberg such as cylindrical algebraic decomposition give you a finite algorithm. (I suppose this is assuming your polynomial has rational coefficients, or at least algebraic coefficients each given by a polynomial they satisfy along with an interval isolating them from other roots. I would guess the problem is not computable if your coefficients are given by Turing machines which compute [successively better approximations to] the reals in question, but I'm guessing that's not the problem you're asking about.)
If by effective you mean "can this be done in polynomial time", the answer is probably not; the problem is NP-hard. In particular, a matrix $A$ is defined to be **copositive** if $x^T A x\geq 0$ for all elementwise nonnegative column vectors $x$. That is to say, $A$ is copositive if and only if $\left[\begin{smallmatrix}x\_1^2 & \cdots & x\_n^2\end{smallmatrix}\right]A\left[\begin{smallmatrix}x\_1^2 & \cdots & x\_n^2\end{smallmatrix}\right]^T\geq 0$ for all real $x$, so checking copositivity is a particular example of the kind of problem you have mentioned. Murty and Kabadi's 1987 paper shows that checking if an integer matrix is *not* copositive is NP-complete. In particular this means that checking nonnegativity is NP-hard even in the degree four case.
If by effective you mean "are there polynomial-time methods that work well in practice", the answer is yes. In particular one can check sufficient conditions like whether $f$ is a sum of squares of polynomials (and a hierarchy of tighter conditions) in polynomial time using semidefinite programming, and often this allows one to solve such problems. Such methods often enable one to compute the global minimum of a polynomial and so can often even give a "no" answer despite the fact that they are a priori just sufficient conditions for nonnegativity. If you are interested in such techniques I would recommend the papers by Parrilo, Lasserre, Nie, etc. and in particular Parrilo's course on MIT OpenCourseWare.
| 14 | https://mathoverflow.net/users/5963 | 36685 | 23,602 |
https://mathoverflow.net/questions/10029 | 14 | Let $k$ be a number field and $F$ a $1$-variable function field over $k$ (a finitely generated extension of $k$, of transcendence degree $1$, in which $k$ is algebraically closed). If $F$ becomes the rational function field over every completion $k\_v$ of $k$, then $F$ is the rational function field over $k$. This is a restatement of the local-to-global principle for the existence of rational points on $k$-conics.
Now suppose that $F$ is a $2$-variable function field over $k$, and that $F$ becomes the rational functional field over every completion $k\_v$ of $k$. Does it follow that $F$ is the rational function field over $k$ ? (In other words, if a $k$-surface is birational of ${\mathbb P}\_2$ over every $k\_v$, is it $k$-birational to ${\mathbb P}\_2$ ?)
I don't think the answer is known, but perhaps someone can put together what is known about the group of $\bar k$-automorphisms of the rational function field ${\bar k}(x,y)$ (the Cremona group) to decide one way or the other.
| https://mathoverflow.net/users/2821 | A local-to-global principle for being a rational surface | It seems to me that there are irrational surfaces over $\mathbb Q$ that are $\mathbb Q\_v$-rational for all $v$. (I couldn't find them in the literature, but didn't look very hard. Almost certainly they are to be found there, in papers by either Iskovskikh or Colliot-Thelene.)
Take the affine surface $S$ given by $y^2+byz+cz^2=f(x)$, where $f$ is an irreducible cubic and $b^2-4c$ equals the discriminant $D(f)$ of $f$, up to a square in $\mathbb Q^\*$, and $D(f)$ is not a square. According to Beauville-Colliot--Thelene--Sansuc--Swinnerton-Dyer $S$ is not $\mathbb Q$-rational, but is stably rational. (Irrationality is Iskovskikh I think, in fact.) Via projection to the $x$-line a projective model $V$ of $S$ is a conic bundle over $\mathbb P^1$ with $4$ singular fibers (one is at infinity). There is an embedding of $V$ into a weighted projective space $\mathbb P(2,2,1,1)$; the defining equation is $Y^2+bYZ+cZ^2=F(X,T)T$, where $F$ is the homogeneous version of $f$. By construction the Galois action on the $8$ lines that comprise the singular fibers is via the symmetric group $S\_3$: the two lines in the fiber at infinity are conjugated, and the other six are permuted transitively.
Claim: Assume that $D(f)$ is square-free and prime to $6$. Then $S$ is $\mathbb Q\_v$-rational for all $v$.
Proof: Suppose that the decomposition group $G\_v$ at $v$ is cyclic. Whatever its order ($1,2$ or $3$) there are at least $2$ disjoint lines among the $8$ that are $G\_v$-conjugate, so they can be blown down to give a conic bundle over $\mathbb P^1$ with at most $2$ singular fibres and a $\mathbb Q\_v$-point; it is well known that such a surface is $\mathbb Q\_v$-rational.
Now suppose that $G\_v= S\_3$. Then $v$ is non-archimedean and $V$ has bad reduction there. In fact, exactly two of the singular fibers are equal modulo $v$; it follows that $G\_v=S\_3$ is impossible, and we are done.
E.g., $f=x^3+x+1$, of discriminant $-31$, $c=8$, $b=1$.
(This doesn't use stable rationality, but rather the fact that these surfaces, although irrational, are very close to being rational, in the sense that the action of $Gal\_{\mathbb Q}$ on the lines is as small as possible subject to the surface being irrational, and the action of the decomposition groups is even smaller.)
| 5 | https://mathoverflow.net/users/8726 | 36687 | 23,604 |
https://mathoverflow.net/questions/35092 | 2 | A [Median graph](http://en.wikipedia.org/wiki/Median_graph) is graph with the property, that for each three vertices $x,y,z$ there is a unique vertex $m(x,y,z)$ lying on shortest paths from $x$ to $y$, from $y$ to $z$ and from $z$ to $x$. Examples are trees, the Cayley graph of $\mathbb{Z}^n$ (with the standart generating set) and cross products of other median graphs.
Suppose, that $x$ and $x'$ are connected by an edge. Is it true, that $m(x,y,z)$ and $m(x',y,z)$ are also connected by an edge ?
EDIT: OK forgot about the case, that $m(x,y,z)=m(x',y,z)$. So I should better ask, whether $d(x,x')\le 1$, so that they are either connected by an edge or equal.
| https://mathoverflow.net/users/3969 | Stability of medians in Median graphs | Here is an attempt to prove that the answer is **yes**.
**Claim 1**: Median graphs are bipartite.
This surely appears in the literature and is easy to verify. (Consider for a contradiction the shortest odd cycle and a median of 3 vertices on it: a pair of adjacent ones and a third one "opposite" of this pair.)
**Claim 2**: If $z \neq m(x,y,z)$ then there exists a vertex $z'$ adjacent to $z$ such that $d(x,z')=d(x,z)-1$ and $d(y,z')=d(y,z)-1$. Further, for each such vertex $z'$ we have $m(x,y,z')=m(x,y,z)$.
Let $m=m(x,y,z)$ and let $P(z,m)$ be as in Tony's comment. Then the neighbor of $z$ on $P(z,m)$ satisfies the claim. The second part of the claim holds as one can extend to $z$ the shortest paths between $z'$ and $x$ and $y$.
**Main argument**: By induction on $d(x,y)+d(x,z)+d(y,z)$. By Claim 1 $d(x,y)=d(x',y)\pm 1$ and $d(x,z)=d(x',z)\pm 1$. If the signs in both of these identities are the same then $m(x,y,z) = m(x',y,z)$ by Claim 2. Thus, wlog, $d(x,y)=d(x',y)+1$ and $d(x,z)=d(x',z)-1$.
If $z \neq m(x,y,z)$ then let $z'$ be as in Claim 2. We have $m(x,y,z')=m(x,y,z)$. As $d(x',z') \leq d(x,z')+1 = d(x',z)-1$, by the second part of the claim we have $m(x',y,z')=m(x',y,z)$. We can now replace $z$ by $z'$ and apply induction hypothesis.
We assume therefore that $z = m(x,y,z)$. Symmetrically, $y=m(x',y,z)$. We have
$(d(x,z) + d(z,y)) + (d(x',y)+d(y,z))=d(x,y)+d(x',z) \leq (d(x',y)+1)+(d(x,z)+1)$.
Thus $d(y,z) \leq 1$, as desired.
| 4 | https://mathoverflow.net/users/8733 | 36702 | 23,611 |
https://mathoverflow.net/questions/36691 | 11 | A configuration of queens on an 8 by 8 chessboard (or n by n if you like) is a *queen domination* if every square on the board lies in the same row, column, or diagonal as at least one of the queens. The Queens Domination Problem is to find the minimum number of queens necessary for a queen domination. A *solution* to the Queens Domination Problem is a queen domination using the minimum number of queens.
For the 8 by 8 chessboard, brute force has shown that 5 queens is the minimum number. More discussion about that [here](https://mathoverflow.net/questions/30330/is-there-a-good-argument-for-why-you-cant-place-4-queens-which-cover-a-chessboar).
The adjacency relation on the set of solutions to the Queens Domination Problem is defined as follows: we say solutions are *adjacent* if they differ by only the placement of one queen. For example: C3, E4, D5, B6, F4 is adjacent to C3, E4, D5, B6, F2.
>
> How many equivalance classes are there in the equivalance relation generated by adjacency?
>
>
>
In other words, starting with one solution, can we reach any other solution by moving one queen at a time, such that the result of each move is itself a solution?
| https://mathoverflow.net/users/1231 | Is the space of solutions to the Queens Domination Problem connected? | Revised and correct:
589 equivalence classes.
<http://www.math.ucsd.edu/~etressle/classes.txt>
It seems that the pairs (#vertices,#components) are (1,388) (2,100) (3,40) (4,34) (5,20) (18,4) (20,2) (3804,1) – damiano 15 hours ago
```
1: Class I.D. 2, 4, 5, 6, 7,..., 589.
2: Class I.D. 22, 29, 35, 47, 48,..., 579.
3: Class I.D. 28, 33, 38, 49, 58,..., 574.
4: Class I.D. 13, 90, 95, 126, 141,..., 576.
5: Class I.D. 32, 37, 62, 89, 97,..., 395.
18: Class I.D. 17, 31, 185, 303.
20: Class I.D. 3, 75.
3804: Class I.D. 1.
```
| 11 | https://mathoverflow.net/users/35336 | 36703 | 23,612 |
https://mathoverflow.net/questions/36708 | 12 | In singular (co)homology, if $\alpha\in C^\*(X)$ and $x\in C\_\*(X)$, then the cap product $\alpha \cap x$ is generally defined by the following process:
1. Apply to $x$ the diagonal map $C\_\*(X)\to C\_\*(X\times X)$ followed by some choice of Alexander-Whitney chain equivalence $ C\_\*(X\times X)\to C\_\*(X)\otimes C\_\*(X)$ to obtain an element $\sum y\_i\otimes z\_i$.
2. Apply $\alpha$ to $\sum y\_i\otimes z\_i$ by the slant product, or, in other words and roughly speaking, apply $\alpha$ to ``half of the factors''.
3. Depending on your favorite conventions (and what you're trying to accomplish), there may also be a sign (which might also be part of your definition of the slant product - but let's ignore this).
The reason I'm being so cagey with wording in part 2 is directly related to my question: In almost all major textbook sources I have consulted, step 2 is performed by forming $\sum y\_i \alpha(z\_i)$, which strikes me as somewhat unnatural, forcing the $\alpha$, which starts off on the left to jump all the way over the $y\_i$ terms to get to the $z\_i$ terms on the right. Is there a good mathematical reason for this convention? Why not define the cap product to be $\sum \alpha(y\_i) z\_i$?
The one major exception to this convention seems to be Hatcher. He does form $\sum \alpha(y\_i) z\_i$, but he also writes cap products as $x\cap \alpha$, so his cochain also has to jump, but it jumps over the $z$s instead!
(For the record, I'm not asking this question out of idle pickiness. Jim McClure and I have been doing a lot of work with cap products recently, and we're trying to be consistent amongst various conventions for various issues, but preferably with good reasons thrown in!)
| https://mathoverflow.net/users/6646 | Conventions for definitions of the cap product | This is just an expanded version of Tyler's comment, I think.
Let's use a, b, c for cochains, x, y, z for chains, [a,x] for the value of a cochain on a chain. I'll be lazy and write $ab$ for $a\cup b$ and $ax$ for $a\cap x$. Let $[a\otimes b,y\otimes z]=(-1)^{|b||y|}[a,y][b,z]$.
I like to define $bx$ in such a way that $[a,bx]$ is $[ab,x]$. Then associativity and unit of cup makes cap into a module structure: $(bc)x=b(cx)$ because $[a,(bc)x]=[a(bc),x]=[(ab)c,x]=[ab,cx]=[a,b(cx)]$, and $1x=x$ follows from $[a,1x]=[a1,x]=[a,x]$.
If you have a product $a\otimes b\mapsto ab$ of cochains and a coproduct of chains defined in such a way that $[ab,x]=[a\otimes b,y\_i\otimes z\_i]$ where the coproduct of $x$ is $\Sigma y\_i\otimes z\_i$, then that means I have to define $bx$ to be $\Sigma (-1)^{|y\_i||b|}[b,z\_i]y\_i$, so as to get
$[a,bx]=[ab,x]=[a\otimes b,\Sigma y\_i\otimes z\_i]=\Sigma (-1)^{|y\_i||b|}[a,y\_i][b,z\_i]=[a,\Sigma (-1)^{|y\_i||b|}[b,z\_i]y\_i]$.
The sign is a bit unexpected, as is the jump you mention, but it's worth it for the neat formulas in the third paragraph above.
It's not a bad idea to define $xa$ as well. I'd do it by first declaring $[x,a]$ to be $(-1)^p[a,x]$ when $a$ is a $p$-cochain and $x$ is a $p$-chain, then defining $xa$ in such a way that $[xa,b]=[x,ab]$. This insures $x(ab)=(xa)b$ and $x1=x$. The chain-level formula is no better and no worse than the formula for $ax$.
Of course, $ax$ and $xa$ end up differing only by a sign when you get to homology, but the sign is hard to remember; in working it out you have to use the commutativity law for the cup product.
| 8 | https://mathoverflow.net/users/6666 | 36718 | 23,623 |
https://mathoverflow.net/questions/36714 | 3 | Is there a standard notation for a graph (on a given set of vertices) without any edges?
| https://mathoverflow.net/users/7732 | Notation for a graph without any edges? | There are many ways to define a *graph*, but a pretty standard one is a pair $(V,E)$ where $V$ is a finite set of points and $E \subset \binom{V}{2}$. So, what you are looking for is $(V, \emptyset)$; which would be pretty widely understood.
| 6 | https://mathoverflow.net/users/2233 | 36732 | 23,633 |
https://mathoverflow.net/questions/36729 | 3 | Consider the category of finite graphs with graph homomorphisms as morphisms.
>
> Are there interesting graph properties
> that can be defined in categorical
> language? Can for example
> connectedness be defined in
> categorical language?
>
>
>
| https://mathoverflow.net/users/2672 | Graph properties, categorically defined | Disjoint union is the coproduct in the category of finite graphs, so connected graphs are precisely the noninitial objects in this category that can not be expressed as a coproduct of two nonempty subobjects. See the entry on [connected object](http://ncatlab.org/nlab/show/connected+object) in the nlab.
If you want to read more about arguments that intertwine classical results in graphs theory with category theory I suggest the book "Graphs and homomorphisms" by J. Nesetril.
| 5 | https://mathoverflow.net/users/2384 | 36740 | 23,635 |
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