User robot - MathOverflow

Tightening Zhang's bound

2013-06-03T11:21:12Z

Inspired by a blogpost by Scott Morrison and ongoing discussion there I decided to create this community wiki to track progress on the original bound of Yitan Zhang.

The original bound was $70,000,000$. The accepted answer should contain latest known improvement.

As of 4.6. 2013 there is a polymath project devoted to improving this bound. The proposal can be found at http://polymathprojects.org/2013/06/04/polymath-proposal-bounded-gaps-between-primes/

Links to various references: http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes

A good place to start is to read notes by Terence Tao and his blog post on the topic.

Answer by robot for Tightening Zhang's bound

2013-06-04T20:18:42Z

Original approach

A set of integers $H$ is called admissible if it avoids at least one residue class modulo $p$ for each prime $p$. In other words $$ \forall p \in \mathcal{P} :\text{cardinality of}\ \{ x \bmod p \ | \ x \in H \} \leq p-1. $$

Let $Q(k_0)$ denote the assertion that for any admissible set $H$ of cardinality $k_0$ there are infinitely many translates $n+H$ that contain at least two primes. The bound on the gap is then $\mathrm{diam}\ H$.

Zhang deduces his bound from the following result:

T1: $Q(3,500,000)$ is true

In Zhang's paper the length $k_0$ is determined by the following inequality (1) that has to hold for some natural number $l_0$

$$ (1+4\varpi) (1-\kappa_2) > \left(1 + \frac{1}{2l_0+1}\right) \left(1 + \frac{2l_0+1}{k_0}\right) (1+\kappa_1), $$ where

$$ \kappa_1 = \delta_1 \left( 1 + \delta_2^2 + k_0 \log\Bigl(1+\frac{1}{4\varpi} \Bigr) \right) \binom{k_0+2l_0}{k_0} $$

$$ \kappa_2 = \delta_1 (1+4\varpi) \left(1 +\delta_2^2 + k_0 \log\Bigl(1+\frac{1}{4\varpi} \Bigr) \right) \binom{k_0+2l_0+1}{k_0-1} $$

$$ \varpi = 1/1168 $$

and

$$ \delta_1 = (1+1/4\varpi)^{-k_0} $$

$$ \delta_2 = \sum_{j=0}^{1/4\varpi} \frac{k_0\log(1+\frac{1}{4\varpi}))^j}{j!}. $$

The admissible set that Zhang uses is $H = \{ p_{k_0+1}, \ldots, p_{2k_0}\}.$

Current record

Terence Tao & Scott Morrison: 4,802,222

Terence Tao established another inequality on $k_0$ that manages to remove most of inefficiency of Zhang estimate. $$ 1+4\varpi > \left(1 + \frac{1}{2l_0+1}\right) \left(1 + \frac{2l_0+1}{k_0}\right) (1+\kappa) $$ where $$ \kappa := \sum_{1 \leq n < 2 + \frac{1}{2\varpi}} \Bigl(1 - \frac{2n \varpi}{1 + 4\varpi}\Bigr)^{k_0/2 + l_0} \prod_{j=1}^{n} \left(1 + 3k_0 \log\Bigl(1+\frac{1}{j}\Bigr)\right). $$ Moreover $l_0$ is allowed to be real number. Scott Morrison then found that for $l_0 = 291.7$ one gets $k_0 = 341,640$ which is the best possible $k_0$ for given $\varpi = 1/1168$.

Paper by Richards suggest to take as admissible set $H_m = \{ \pm 1, \pm p_{m+1}, \ldots, \pm p_{m+k_0/2+1} \}$ for $m$ large enough. This leads to bound $$ 2p_{m+\lceil k_0/2 \rceil + 1} \quad \text{for } k_0 \text{ even} $$ and $$ p_{m+\lfloor{k_0/2}\rfloor-1} + p_{m+\lfloor{(k_0+1)/2}\rfloor-1} \text{ for } k_0 \text{ odd.} $$ For given $k_0=341,640$ program written by Scott Morrison found that $m=5553$ gives the smallest bound of $4,802,222$.

Associated vector bundles of infinite rank and induced connections

2012-02-12T21:04:29Z

Let $\mathbb{V}$ be a representation of a Lie group $G$ and let $P \to M$ be a principal $G$-bundle with a principal connection. If $\mathbb{V}$ is finite-dimensional, then one can associate to this data an associated vector bundle $P\times_G \mathbb{V}$ with linear connection. I thought that basically the same construction should work also when $\mathbb{V}$ is an infinite-dimensional representation, but I haven't found any textbook that would not constrain itself to finite rank. All the textbooks concerning to infinite-dimensional differential geometry that I know of (Michor, Lang, Neeb) doesn't treat associated bundles and induced connections.

Edit:

I now realize that it may not be as straightforward as it seems on a first glance. I want to, in fact, generalize a slightly more complicated construction -- the so called tractor connection induced by a Cartan connection.

Changing the notation a little bit, given a finite-dimensional Lie group $G$ with a closed subgroup $H$, I need to work with an infinite-dimensional vector space $\mathbb{V}$ which is a representation of $\mathfrak{g}$ and also a representation of $H$ (so I can form associated bundles to $H$-principal bundles) with these two representation being compatible. Practically, I am interested mainly in Harish-Chandra modules and their globalizations. I think I am also fine with just a "sort of connection" working on some dense subbundle of the associated bundle and so $L^2$-globalizations are also OK.

I can briefly describe the construction for $\mathbb{V}$ being finite-dimensional representation of $G$ if it is needed.

Answer by robot for Inversion of complex matrix

2013-03-22T15:58:10Z

This is not an answer, but it's too long for comments.

First of all it would be really helpful if you provided some background. Is this a numerical problem or algebraic problem?

Two ideas come to mind.

First, let's say that $A$ is regular, then $$ (A+\lambda \imath B)^{-1} = (A(I + \lambda \imath A^{-1}B)^{-1} = (I + \lambda \imath A^{-1}B)^{-1}A^{-1}. $$

Now if all eigenvalues of $\lambda A^{-1}B$ are smaller than $1$, the converges and you can compute the inverse as $$ (I + \lambda \imath A^{-1}B)^{-1} = \sum_{k=0}^\infty (\lambda\imath)^k (A^{-1}B)^k. $$ If $A^{-1}B$ is nilpotent, then your inverse is just a matrix polynomial in $\lambda$. Otherwise this can be used at least for approximation of the inverse.

Second idea.

If your matrices are not too large you can try to work with analytical inversion which works also for matrices of polynomials. I.e. you can write down the using subdeterminants of $A+ \lambda\imath B$ treating $\lambda$ as a variable.

OK. So the problem seems to be:

Compute (numerically) the sum of elements of the inverse of $A+\lambda\imath B$ for real matrices $A,B$, where $A$ is diagonal, $B$ is symmetric and $\lambda$ is a real parameter. The matrices are of order $1000 \times 1000$.

As was explained in the answer to another problem of OP, the sum of elements of the inverse can be obtained as follows. Let $e$ denote the vector of all ones. Then the sought sum equals to the scalar product of $e$ and the solution $x$ of $(A+\lambda\imath B)x = e$. Without any further information it seems that the best possible course is to rewrite the equation as $$ \Bigl(\frac{-\imath }{\lambda} A + B \Bigr)x = \frac{-\imath }{\lambda}e $$ and google for diagonal updates for iterative methods (e.g. Krylov subspaces method) as Federico Poloni suggested in a comment.

easter problem - egg shapes

2013-03-29T22:39:03Z

Inspired by an exceptionally silly article in today's newspaper I pose the following "egg parametrization problem".

Give an explicit function $ f(x,y,t) : \mathbb{R}^2\times I \to \mathbb{R}$ such that for each $t$ from interval $I$ the solution set of equation $f(x,y,t) = 0$ looks like an egg.

I'm looking for function that provides most of the various egg shapes found in nature.

simple roots of a reflection subgroup

2013-03-22T10:14:48Z

Consider a Hermitian symmetric pair of complex Lie algebras $(\mathfrak{g},\mathfrak{k})$ and split the set of roots into compact roots (i.e. roots of $\mathfrak{k}$) and noncomapt roots $\Delta = \Delta_c \cup \Delta_n$.

Pick a weight $\lambda$ and define the set of $\lambda$-singular roots $\Psi_\lambda$ as the subset of roots orthogonal to $\lambda+\rho$. Now consider a subgroup $W_\lambda$ of the Weyl group of $\mathfrak{g}$ generated by the reflections $s_\beta$ for $\beta \in M_\lambda$, where $M_\lambda$ is the subset of noncompact roots that satisfy the following three conditions

$\beta$ is orthogonal to $\Psi_\lambda$
scalar product of $\beta$ with $\lambda+\rho$ is a natural number
if there is a long root in $\Psi_\lambda$, then $\beta$ is short

It is known (by a result of Dyer) that $W_\lambda$ is in fact Weyl group of a root subsystem $\Delta_\lambda$ of $\Delta$.

The article Resolutions and Hilbert series of the unitary highest weight modules of the exceptional groups provides an algorithm for computation of simple roots of $\Delta_\lambda$ in section 5.2. It uses a partial ordering (see section 4.6) on positive noncompat roots such that $\beta$ covers $\alpha$ iff $\beta = \alpha +\alpha_i$ for some simple root $\alpha_i$. The claim is that the set of simple roots is given by differences of successive elements of $M_\lambda$.

Q1: How to prove this?

Q2: Is this true also for the classical cases?

Q3: Was this order on positive roots already studied?

Note that the weight is not arbitrary but it is such that the irreducible highest weight module $L_\lambda$ is in fact unitarizable.

Answer by robot for Invariant subbundles of tangent bundle of flag variety

2013-03-06T21:08:19Z

This is not an answer. Just a few well known facts.

Each $P$-invariant subset is also invariant with respect to the Levi part $L$ of $P$ and hence it decomposes into irreducibles for $L$.

The representation $\mathfrak{g/p}$ is as a $\mathfrak{p}$-representation isomorphic (via the Killing form) to the nilradical of $\mathfrak{p}$. Now this nilradical is in fact isomorphic to $k$-graded Lie algebra $\bigoplus_{i=1}^k \mathfrak{g}_i$ that is generated (as a Lie algebra) by $\mathfrak{g}_1$. Lie brackets $\mathfrak{g}_i \otimes \mathfrak{g}_j \to \mathfrak{g}_{i+j}$ for $i,j\in {1,\ldots k}$ are $L$-equivariant. (See e.g. section 3.1.2 in Parabolic Geometries I Background and General Theory by Čap and Slovák.)

One always have $P$-invariant subspaces given by filtration components $\mathfrak{g}^j = \bigoplus_{i=j}^k \mathfrak{g}_i$.

Consider a $P$-invariant subspace $V$. If $\emptyset \neq (V\cap \mathfrak{g}_i) \neq \mathfrak{g}_i$, then one can perhaps use the generating property of $\mathfrak{g}_1$ and $L$-equivariance of the Lie bracket $\mathfrak{g}_1\otimes (V\cap \mathfrak{g}_i) \to \mathfrak{g}_{i+1}$ to restrict possible $L$-types occurring in $V$.

But I don't know how the $L$-decompositions of $\mathfrak{g}_i$ look like.

BGG-like resolutions and translations

2013-02-28T20:10:39Z

This question originated from my confusion after I read the following paragraph (page 31, section 4.8) in Resolutions and Hilbert series of the unitary highest weight modules of the exceptional groups.

Suppose $\xi$ and $\nu$ lie in the positive chamber, have integral difference $\xi-\nu$ and equal stabilizers in $W$, i.e., $\{w\in W\, |\, w\xi = \xi\} = \{ w\in W\,|\, w\nu = \nu\}$ . Applying standard techniques of Zuckerman translation we have an equivalence of categories $\Theta : \mathcal{O}_\xi \to \mathcal{O}_\nu$ . Suppose $w\in W^c$ and assume both $w\xi$ and $w\nu$ are $\Delta_c^+$-dominant integral and regular. Then $\Theta N_{w\xi} \simeq N_{w\nu}$ and $\Theta L_{w\xi} \simeq L_{w\nu}$.

Notation here is that $\Delta_c^+$ is the set of positive roots of the Levi part $\mathfrak{l}$ and $W$ and $W^c$ are Weyl groups of $\mathfrak{g}$ and $\mathfrak{l}$ respectively. Symbol $N_\lambda$ denotes the parabolic Verma module induced from a $\Delta_c^+$-dominant integral weight $\lambda$ and $L_\lambda$ is it's simple quotient. Note that this article deals with $|1|$-graded situation only, so the nilradical of the parabolic is abelian.

These isomorphisms are then applied to a BGG-like resolution of unitarizable highest weight modules $L_\lambda$ and from the exactness of the translation functor $\Theta$ the conclusion is stated for invariance of some properties of these resolutions on "unitary strata".

I admit that the fact that translation functor maps parabolic Verma modules to parabolic Verma modules still seems a little bit unclear to me.

From now on I am going to use notation of Humphreys's Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$.

Theorem 9.4 of op.cit. says that the parabolic Verma modules $M_I(\lambda)$ fit into an exact sequence $$ \bigoplus_{\alpha\in I} M(s_\alpha\cdot \lambda) \to M(\lambda) \to M_I(\lambda) \to 0. $$ I guess that using the results on translation functors in section 7.8 coupled with the fact that homomorphisms of Verma modules are essentially unique one arrives at the conclusion $T_\lambda^\nu M_I(\lambda) \simeq M_I(\nu)$.

Q1: How to prove that in this simplest case the translation functor really maps parabolic Verma modules to parabolic Verma modules?

In general I am interested in BGG-like resolutions of simple modules in terms of direct sums of (parabolic) Verma modules. See e.g. Kostant modules in block of category $\mathcal{O}_S$ for examples of such resolutions outside of the classical BGG case.

Sometimes one can use exact functors to create whole families of such resolutions (see e.g. op. cit.). For example some infinitesimal blocks of the ordinary category $\mathcal{O}$ are equivalent to each other via translation functors which allows one to create whole families of resolutions of the same "shape".

Q2a: What are the known results on equivalences between infinitesimal blocks in parabolic category $\mathcal{O}_I$? What are the known results on translation functors in $\mathcal{O}_I$?

Q2b: What functors can be used for such a "translation"? What are known examples?

One example of 2b are Enright-Shelton equivalences constructed from derived translation functors that relate infinitesimal blocks of parabolic categories $\mathcal{O}$ for Lie algebras of different rank.

As was remarked by professor Humphreys in answer to my previous question, some results are contained in Jantzen's Moduln mit einem höchsten Gewicht.

translation functors in parabolic category $\mathcal{O}$

2013-02-13T00:04:25Z

I'm looking for a reference for a treatment of translation functors (as defined e.g. in this [question][1]) in parabolic versions of BGG category $\mathcal{O}$.

I am mainly interested in the following question.

Is it true that for integral weights $\lambda, \mu$ lying in the same facet one gets an equivalence of $\mathcal{O}_\lambda$ and $\mathcal{O}_\mu$ ?

Edit:

To avoid confusion I have created a new question under the title BGG-like resolutions and translations.

Answer by robot for Exceptional Schur-Weyl Duality

2013-02-27T23:28:21Z

There is an article by Jing-Song Huang and Chen-Bo Zhu called Weyl's construction and tensor power decomposition for G2 that builds on an invariant theory for $G_2$ (see the linked review for details). There's quite recent article Algèbres de Jordan et théorie des invariants which deals with these invariants also for other exceptional cases and I would guess there'll be some others.

parabolic subalgebras and Cartan decomposition

2013-01-03T14:40:37Z

Let $\mathfrak{g}$ be a complex simple Lie algebra and $\mathfrak{k}$ its complex subalgebra such that $(\mathfrak{g},\mathfrak{k})$ is a Hermitian symmetric pair; $\mathfrak{g}= \mathfrak{k}\oplus\mathfrak{p}$ is the corresponding Cartan decomposition subject to some Cartan involution $\theta$. Moreover, there is a splitting $\mathfrak{p} = \mathfrak{p}^- \oplus \mathfrak{p}^+$.

Problem: Classify all $\theta$-stable parabolic subgroups $\mathfrak{q}=\mathfrak{l}\oplus\mathfrak{u}$ of $\mathfrak{g}$ such that $\mathfrak{l}\subseteq\mathfrak{k}$ and $\mathfrak{p}^+\subseteq\mathfrak{u}$.

Motivation: In the article Dirac operators and Lie algebra cohomology. Represent. Theory 10 (2006), the authors prove that in such a case there is a Hodge decomposition for $\mathfrak{u}$-homology of a unitarizable $(\mathfrak{g},K)$-module. I am interested for which real parabolic subalgebras of some real form of $\mathfrak{g}$ there is a Hodge decomposition.

Answer by robot for Irreducible representation decomposition of tensor on manifold with metric

2012-10-15T10:16:25Z

There is a "Schur-Weyl theory" for representation of $O(n)$ and $Sp(n)$. The group algebra of the symmetric group is replaced by the Brauer algebra. Basically, you first decompose your tensor product with respect to "the number of traces the vectors contain" and then for each such part you can use the classical $GL$ decomposition.

The relevant representation theory over complex numbers is treated in a book by Wallach and Goodman Symmetry, Representations, and Invariants. In particular look at the appendix F on the linked web page. Alternatively, you can look into the older version of this book which was published under the name Representations and Invariants of the Classical Groups.

You should be careful when dealing with representation of noncompact real groups.

Answer by robot for Jets of Equivariant Vector Bundles

2012-10-14T17:21:09Z

I don't understand your terminology, but I'm gonna try to answer your question anyway. Let $M=G/H$ and let $\mathbb{E}$ be a representation of $H$. By $E$ I denote the associated homogeneous vector bundle $G/H \times_H \mathbb{E}$. The jet space of $E$ is also a homogeneous vector bundle which is induced from the $(\mathfrak{g},H)$-representation $J_{eH}(E)$, i.e. from the representation which is induced on the fiber over identity coset. There is a duality between $J_{eH}(E)$ and $\mathfrak{U(g)\otimes_{U(h)}} \mathbb{E}^*$.

Some details can be found in the appendix of this ESI preprint. For applications to invariant differential operators see Differential opperators on homogeneous spaces by L. Barchini and R. Zierau.

Answer by robot for Curvature of the Cayley projective plane

2012-09-04T12:31:20Z

I'm not quite sure what you are looking for, but explicit computation of the curvature appears in arXiv:math/0702631. The authors define $\mathbb{OP}^2$ via an octonionic atlas, write down explicit isometries with which they prove that it is indeed a homogeneous manifold and then they express the curvature tensor at a point using a coordinate frame. The expressions of the components of $R$ involve only octonionic multiplication and inner product.

Answer by robot for What does the adjective "natural" actually mean?

2012-07-19T23:56:20Z

There is a book called Natural operations in differential geometry, which as far as I know captures what natural means for a differential geometer. Natural bundles are defined basically as functors from the category of manifolds with local diffeomorphisms to the category of fibered manifolds. Natural operator between natural bundles is then a local operator which commutes with local diffeomorphisms.

Answer by robot for Basic software libraries for numerical analysis using modern programming languages?

2012-07-13T11:17:05Z

It is not entirely clear to me what are you looking for and why.

If you want the best performance you are almost surely bound to use compiled language such as Fortran1, C++ or C. Of course, you can always use almost any "higher level language" such as Python, Ruby or whatnot to glue together routines from libraries written in some "low level language" as C, Fortran, etc. Octave, Matlab and Sage come to mind. NumPy is quite good example of this approach, since almost all of its core functions are written in C (i.e. LAPACK).

If, on the other hand, you want to experiment with algorithms themselves, you can implement them in Haskell or (oca)ML or some other "mathematicians friendly" language. Also, succinct syntax and lack of side effects means it's much easier to prove correctness. Moreover there are area specific systems/languages as LiE, Macaulay2, Singular, GAP, ...

Writing basic routines (such as those from Numerical Recipes) in any other language than C, C++ or Fortran means that

you are doing an exercise
you are trying to solve a problem in a language where you can't use functions from libraries written in C/C++/Fortran
you need numerical routines in a big (i.e. not feasible to rewrite in C/C++/Fortran) project where calling an external library causes unwanted overhead (this can be the case for example with Java).

1 Please note that modern Fortran is quite high level language with functional and object-oriented features.

edit:

I am not a Fortran programmer myself, so what follows are information relayed from a friend of mine (physicist and active Fortran programmer). I realize that this is not what you are asking for, but it may be helpful nevertheless.

books:

Metcalf, Reid, Cohen - Modern Fortran Explained
Chapman - Fortran 95/2003 for Scientists & Engineers

When starting, it's best to learn Fortran 95/90 first. For a gentle introduction see http://www.cs.mtu.edu/~shene/COURSES/cs201/NOTES/fortran.html

For an overview of the language see http://en.wikipedia.org/wiki/Fortran_language_features (mostly written by Michael Metcalf; see the book above).

As for OOP, it was possible in Fortran 95 using various hacks, but another revision of the language (Fortran 2003) included OOP features into the language norm itself. For technical account on Fortran 2003 features see ftp://ftp.nag.co.uk/sc22wg5/N1551-N1600/N1579.pdf For tutorial on OOP in Fortran 2003 see http://www.pgroup.com/lit/articles/insider/v3n1a3.htm and http://www.pgroup.com/lit/articles/insider/v3n2a2.htm

Fortran is still evolving, the latest standard being Fortran 2008 which introduced concurrent programming techniques:
http://www.training.prace-ri.eu/uploads/tx_pracetmo/coarrayvideo1.pdf
http://www.training.prace-ri.eu/uploads/tx_pracetmo/coarrayvideo2.pdf
http://www.training.prace-ri.eu/uploads/tx_pracetmo/coarrayvideo3.pdf

I should add that there probably isn't a compiler that supports all the features of the standard (2003).

Software for drawing intervals in Weyl groups

2012-05-31T11:57:30Z

I am looking for an automated way to draw diagrams of intervals in Weyl groups and in their various subsets such as minimal representatives of cosets $W/W_p$ for a parabolic Weyl group $W_p$ or elements of $W_{[\lambda]}$ for some nonintegral weight $\lambda$.

I've looked at sage but their support is very rudimentary. So far the best thing I've found is the package WeylGroups for Macaulay2. It implements minimal representatives and it has output to svg and pgf. However the resulting graph is not very nice imho. Also I would like to be able to label the nodes of the diagram such that the top would be an arbitrary weight and the rest of the nodes would be labelled according to the (affine) Weyl group action on that weight. This is of course doable in Macaulay2 and the output can be improved by hacking the output routines of WeylGroup package and using graphviz. Alas, I have never used M2 before and given the amount of work required I thought that I ask here first. So...

Question: Is there a software that produces nice drawing of intervals in Bruhat order?

Also there are other interesting subsets of Weyl group with a slightly different order which can be of interest. So extendability / hackability is always a plus!

$K$-types of unitarizable highest weight modules

2012-05-17T17:23:34Z

Unitarizable highest weight modules are Harish-Chandra modules which are at the same time simple quotients of (generalized) Verma modules. They were classified in the eighties. The title sums up the question:

Q: What are the $K$-types of unitarizable highest weight modules?

Please note that the group $K$ is also the Levi subgroup of the relevant parabolic subgroup which defines the generalized Verma modules in question. Hence the question can be rephrased as the branching of simple quotients of generalized Verma modules with respect to the Levi subgroup.

I've searched the literature and found only the Hua-Kostant-Schmid formula which gives the result for holomorphic discrete series modules with scalar minimal $K$-type. There are some papers by Toshiyuki Kobayashi on branchings of these beasts, but nothing explicit apart from the already mentioned HKS formula. I suppose there can be some results in the physics literature.

Answer by robot for What is a complex inner product space "really"?

2011-11-03T10:01:21Z

You can view complex inner product space as an object that encompasses a set of real inner product spaces which are not necessarily positive definite. I.e. from $\mathbb{C}^n$ with the standard inner product you can get Euclidean real space as well as Lorentzian by choosing appropriate real form. So if you want to find a real geometric intuition for complex inner product spaces (assuming there is one!) you should probably better start with inner product spaces of indefinite signature.

As an example of use of inner product spaces I suggest the theory of simple Lie algebras. The theory is developed in the complex case and one has a natural (invariant) inner product there - the Killing form. From one complex Lie algebra you can get several real Lie algebras and their properties (i.e. whether they are compact or not) are determined precisely by the signature of the restriction of Killing form.

Answer by robot for Jordan algebra and quaternionic projective space

2012-05-16T09:00:18Z

I am not much educated in this subject so please take my answer with a grain of salt.

The appropriate Jordan algebra is $J_n(\mathbb{H}) := \{ A \in M(n,\mathbb{H})\,|\, \overline{A}^T=A \}$ with multiplication defined as $A\circ B = (AB+BA)/{2}$. The metric is defined by $\mathrm{Tr}(A^2)$. The quaternionic projective space can be then defined as the space of elements of $J_n(\mathbb{H})$ of rank one and trace one (think of these matrices as projectors to one-dimensional subspaces) or alternatively as the (real) projectivization of rank one matrices.

I am little bit more familiar with the case of octonionic projective plane, so let me explain that case. There the relevant algebra is $J_3(\mathbb{O})$ and the plane $\mathbb{OP}^2$ and its metric is defined in the same way^[1] as in the quaternionic case. Now there is a well defined cubic form on $J_3(\mathbb{O})$ which is basically the determinant. The group that preserves this cubic form is $E_6$. In fact one defines some sort of cross product (I think it is called Jordan cross product.) out of the cubic form which then gives the incidence relation of the projective geometry of $\mathbb{OP}^2$. If I remember correctly, the product $A\times B$ is defined by the relation $(A\times B,C) = (A,B,C)$, where on the left hand side the brackets denote the polarization of the quadratic form while on the right hand side the brackets denote the polarization of the cubic form. The group $E_6$ is then the group of collineations of $\mathbb{OP}^2$

The group that preserves the determinant and the quadratic form (the norm) is the group $F_4$. This sheds some light on the fact that $F_4$ is in fact the isometry group of $\mathbb{OP}^2$. It can be proven that $F_4$ is in fact the automorphism group of the Jordan algebra $J_3(\mathbb{O})$!

In the quaternionic case one finds that $E_6$ is replaced by $Sl(n,\mathbb{H})$ and $F_4$ by $Sp(n)$.

The projective plane $\mathbb{OP}^2$ was much studied by Freudenthal and others but I do not know of any reference where the quaternionic case is treated via Jordan algebras.

[1] Of course one needs to be a little careful with the definition of rank because of the nonassociativity of $\mathbb{O}$. Either one uses the fact that any element of $J_3(\mathbb{O})$ can be mapped by the action of $F_4$ to a diagonal matrix and then one defines the rank by the number of nonzero elements in this diagonalization. Or one can experiment a little bit and discover that it is possible to define determinants of 2x2 minors in such a way that their vanishing is equivalent to matrix of being of rank one according to first definition.

Answer by robot for Examples of theorems with proofs that have dramatically improved over time

2012-05-14T22:10:37Z

I think that Gelfand's proof of Wiener's $1/f$ theorem qualifies.

Answer by robot for Type of 26-dimensional representation of different real forms of the complex simple Lie algebra $F_4$

2012-05-09T17:03:44Z

All the forms of $F_4$ can be defined as automorphism groups of some Jordan algebra of three by three matrices with entries in octonions / split octonions / complexified octonions. These algebras are all of dimension 27 over the appropriate field and the subspaces of trace-free matrices are the irreducible 26-dimensional representations of the various forms of $F_4$. The invariant quadratic form is $A\mapsto \mathrm{Tr}(A^2)$. (And the invariant cubic form is $A\mapsto \mathrm{det}(A)$.)

The group $F_4^{-20}$ is according to Yokota (but I guess that one can dig this up also out of the work of Veldkamp or Springer) the automorphism group of the real Jordan algebra $J(1,2,\mathbb{O}) = \{X\in \mathrm{M}(3,\mathbb{O}\otimes_\mathbb{R}\mathbb{C}) \, |\, I_1 \overline{X}^tI_1 = X \}$ where $I_1 = \mathrm{diag}(-1,1,1)$. Now the computation of the signature of $A\mapsto \mathrm{Tr}(A^2)$ is a matter of simple calculation.

The other two real cases $F_4^{-52}$, $F_4^4$ follow similarly since they are the automorphism groups of $J(3,\mathbb{O}) = \{ X\in M(3,\mathbb{O})\,|\, X^t =X \}$ and $J(3,\mathbb{O}) = \{ X\in M(3,\mathbb{O}')\,|\, X^t=X \}$ respectively.

Do infinite products commute with functor of smooth sections?

2012-04-05T06:20:40Z

Similarly to my previous question about direct limits, I have now basically the same question about inverse limits. It seems in fact, that I only need the result for products.

Question: Is there a natural smooth structure on $\prod \mathbb{R}$ such that $\mathcal{C}^\infty(U,\prod \mathbb{R}) = \prod\mathcal{C}^\infty(U,\mathbb{R})$?

Does direct limit commute with functor of smooth sections?

2012-03-12T10:02:46Z

Consider a countable family of finite-rank vector bundles $V_k$ over a finite-dimensional smooth manifold $M$. The direct limit of such a family is still a topological vector bundle even though it may have infinite rank.

The first question is: Is there a natural structure of smooth manifold on the total space of $\varinjlim V_k$?

And the second one: Is it true, that the functor of smooth sections commutes with direct limit? I.e, is it true that $$\mathcal{C}^\infty(M,\varinjlim V_k) = \varinjlim \mathcal{C}^\infty(M,V_k)$$

Or turning things upside down -- is there a choice of smooth structure such that the above equation holds? If so, how does it look like?

topologies on globalizations

2012-03-08T19:39:07Z

I am reading notes by David Vogan on Unitary representations and Complex analysis (pdf / dvi).

The setting is as follows (page 23): Let $X$ be a $(\mathfrak{g},K)$-module and let $X(\mu)$ denote its $\mu$-isotypical component for $\mu\in\hat{K}$. On each $X(\mu)$ there is $K$-invariant scalar product $\|\ \|_\mu$.

Various globalizations of $X$ are given as spaces of sequences $(x(\mu))_{\mu\in\hat{K}}$ with certain conditions on $\| x(\mu)\| _\mu$. These conditions also lead to a canonical topology on the globalizations.

Vogan also mentions that even $X = \bigoplus_{\mu\in\hat{K}} X(\mu)$ itself and $\Pi_{\mu\in\hat{K}} X(\mu) = X^{-K}$ have canonical topologies.

Q1: How are these topologies defined?

Q2: Are the inclusions $X\subseteq X^{\text{glob}} \subseteq X^{-K}$ continuous with dense images?

Q3: Consider Hermitian symmetric case and complexifications. There is a parabolic subgroup of $G^\mathbb{C}$ given by $P=K^\mathbb{C}U$ with $U$ abelian such that the compact symmetric space $G/K$ is isomorphic to $G^\mathbb{C}/P$. Can one extend continuously the action of $K$ on to $P$ for $X$, $X^\text{glob}$ and $X^{-K}$?

Answer by robot for Peter-Weyl theorem as proven in Cartier's Primer

2012-02-24T17:57:43Z

More generally, if $A$ is a bounded $G$-invariant operator acting on a unitary representation $(\rho,\mathcal{H})$ of $G$ and $A^*$ is it's adjoint (i.e. $(Ax,y) = (x,A^*y)$ for all $x,y\in \mathcal{H}$), then we have

$$ (x,A^*\rho(g)y) = (Ax,\rho(g)y) = (\rho(g^{-1})Ax,y) $$

where in the last step we used unitarity of $\rho$. Now use $G$-invariance of $A$ and retrace the steps back

$$ (\rho(g^{-1})Ax,y) = (A\rho(g^{-1})x,y) = (\rho(g^{-1})x,A^*y) = (x,\rho(g)A^*y). $$

QED

Answer by robot for Why are so few operations with arity bigger than 2?

2012-02-05T20:29:30Z

More than two-ary operations pop up in algebraic approach to CSP (constraint satisfaction problem). See e.g. http://www.ams.org/mathscinet-getitem?mr=2470592 or http://www.ams.org/mathscinet-getitem?mr=2137072

ODE with non-continuous right hand side

2012-01-17T13:25:08Z

My brother asked me a question which I didn't know the answer to.

Are there theorems about existence, uniqueness and stability of solutions of ODEs of the followin type

$$ \frac{d^2 y}{dt^2} = f(t,y,\frac{dy}{dt})H(g(y)), $$

where $f$ and $g$ are Lipschitz functions and $H$ is the Heaviside function?

If there are no general theorems that could apply, how does one analyze these problems for given $f,g$?

Answer by robot for Decompose tensor product of type $G_2$ Lie algebras.

2012-01-13T19:22:30Z

I think there is. Actually there is a version of Weyl's theory for $G_2$, i.e. all finite dimensional irreducible representations are given by Schur functors and hence are intimately related to representations of symmetric groups. See article by Jing-Song Huang for details.

$(\mathfrak{g},K)$-modules and parabolic category $\mathcal{O}$

2012-01-05T18:02:54Z

I am trying to get acquainted with various infinite dimensional representations of Lie groups. So a general reference would be appreciated. Right now I am trying to figure out the following question.

Consider a real Lie group $G$ and its maximal compact subgroup $K$. Let $P$ be a parabolic subgroup of the complexification of $G$ such that its Levi factor is a complexification of lie algebra of $K$. What is the relation between category of $(\mathfrak{g},K)$-modules and the parabolic category $\mathcal{O}_P$?

Is there a canonical way to extend the representation of $K$ on the $(\mathfrak{g},K)$-module to a representation of $P$ at least in some special cases (for example in hermitian symmetric setting)?

Of course one can drop the condition on $K$ being the maximal compact subgroup and ask basically the same thing.

Comment by robot

2013-05-23T19:25:52Z

I meant the stuff about hyper-geometric functions. Sorry, I should have been more specific.

Comment by robot

2013-05-23T11:04:08Z

OK. I just wonder whether the second paragraph can be pushed to generalized flag manifolds $G/P$ for $P$ parabolic. Do you know any references here? At least for the Borel case.

Comment by robot

2013-05-22T22:00:36Z

You confused notation in the first paragraph. The bundle should be over $M$ and $H$ should be $B$. Is there any reason why don't you suppose that $G/B$ is a flag variety right from the beginning?

Comment by robot

2013-05-15T15:21:30Z

No need to restrict to complex representations. One just has to assume that the characteristic of the field does not divide the order of $G$.

Comment by robot

2013-04-29T08:00:54Z

bearspace.baylor.edu/Markus_Hunziker/www/… Lemma 2.2

Comment by robot

2013-04-27T16:04:31Z

The efficiency really depends on the construction of his set of vectors, doesn't it?

Comment by robot

2013-04-10T08:53:14Z

In that case you can just renumber indices $i \to i-1$ and write $U_i$, $i\in \mathbb{Z}_n$. Adding some context would be helpful. Are the variables $U_i$ (anti)commutative?

Comment by robot

2013-04-04T12:59:30Z

I am only vaguely familiar with definition of Sobolev spaces on manifolds via patching local definitions. Thus it seems to me, that if you can prove the density locally, you have it also globally.

Comment by robot

2013-04-03T17:39:29Z

@Robert Bryant: I think that users are allowed to edit their posts only if they have high enough reputation. But maybe it applies only to questions and answers can be edited always.

Comment by robot

2013-04-03T17:20:51Z

This could be also a good starting point: en.wikipedia.org/wiki/…

Comment by robot

2013-04-03T17:15:52Z

First hit on google: cs.montana.edu/courses/spring2005/580/papers/…

Comment by robot

2013-03-30T22:47:32Z

Perhaps this could be helpful. mmrc.iss.ac.cn/pub/mm25.pdf/7.pdf

Comment by robot

2013-03-30T22:31:23Z

What is the question here?

Comment by robot

2013-03-22T14:30:54Z

I've emailed Hunziker about a week ago with no reply so far. I guess I should've cc'ed Enright as well.

Comment by robot

2013-03-14T13:40:19Z

I believe that the answer to your question lies in the paper by Hilbert which you can find on the wikipedia page on Hilber matrix.