User shu - MathOverflow

locally symmetric space and global symmetric space

2013-05-15T14:34:47Z

Let $Z$ be a compact, connected, orientable (Edit: as Misha point out) and locally Riemannian symmetric space. As a complete, simple connected, locally symmetric space is a global symmetric space. We can write $Z=\Gamma \backslash G/K$. One of such $(G,K)$ is $G=\mathrm{Iso}(\widetilde{Z})$,$K$ is the stablizer of a point in $\widetilde{Z}$.

My question is, if $\widetilde{Z}$ is of non compact type, can we always choose $(G,K,\Gamma)$ such that

$G,K$ is connected.
$\Gamma\subset G$ closed discrete?

If not, under what natural condition, $Z$ is this type.

One of naive choice is to take $G=\mathrm{Iso}^0(\widetilde{Z})$, but it is to always true that $\Gamma\subset \mathrm{Iso}^0(\widetilde{Z})$.

Edit: I pose this question because in papers of H. Moscovici and R.J. Stanton http://link.springer.com/article/10.1007%2FBF01393895 http://link.springer.com/article/10.1007%2FBF01232263 where proved the result for this type of space(maybe I missed something) but state the theorem for locally symmetric space.

distribution of the jumps in a stable Lévy bridge

2013-02-20T20:03:08Z

If $X$ is a spectrally positive (without negative jumps) $\alpha$-stable Lévy process, in other words, its Laplace exponent is given by $\Phi(\lambda)=\lambda^\alpha$ for some $\alpha\in (1,2)$, and let $X^{br}$ denote the associated bridge process (of length $1$, from $0$ to $0$). Is anything known about the distribution of the jumps of $X^{br}$?

Reference for some de Rham cohomologies

2012-06-15T13:58:23Z

I would like find the reference for the following elementray reslut which I would not like to write down the proof in my paper.

If $\mathcal{S}(\mathbf{R})$ is the schwartz space of $\mathbf{R}$. Let $$\Omega^\cdot_\pm(R)=e^{\pm x^2}\mathcal{S}(\mathbf{R},\Lambda^\cdot\mathbf{R}^*).$$ So, we have $$H^\cdot(\Omega^\cdot_+(R),d)=H^\cdot_{dR}(\mathbf{R}), \ \ H^\cdot(\Omega^\cdot_-(R),d)=H^\cdot_{dR,c}(\mathbf{R}).$$

I would like also the kunneth formula in this context.

Topologic or geometric mean of the structure constants of a semi simple lie algebra

2012-05-22T15:31:41Z

Let $G$ be a semi simple Lie group (or real reductive), $\mathfrak{g}$ its lie algebra and $B$ its killing form. We can defined the 3-form $k$ by $$k(X,Y,Z)=B([X,Y],Z).$$ with $X,Y,Z\in \mathfrak{g}$. In fact, $k$ is nothing but the structure constants.

It is easy to prove that $k$ is a closed forme on $G$. For example, let $\nabla$ be the connection on $TG$ defined by $\nabla_XY=0$. Its torsion is $T(X,Y)=-[X,Y]$. Then $$d=e^i\wedge\nabla_{e_i}+i_{T}.$$ where $e_i$ is a base of $\mathfrak{g}$ and $e^i$ is the dual base. It is easy to verifier $e^i\wedge\nabla_{e_i}k=0$ and $i_{T}k=0$. So $dk=0$.

Can someone give some explanations of the 3 closed form $k$? For example,

What is the topology mean of its cohomology class $[k]\in H^3(G)$.
How about the case $[k]=0$ or $[k]\neq0$?
If $G$ is a compact semi simple Lie group, can we say more about the form $k$?

Is there some explication for the transformation of the eigenvalues in Selberg trace formula

2011-09-13T11:15:22Z

Selberg trace formula http://en.wikipedia.org/wiki/Selberg_trace_formula says $$\sum_nh(r_n)=...$$ where $1/4+r_n^2$ is the eigenvalue of the Laplacian.

My question is "what is the geometric exlication for the transformation $r_n\to r_n^2+1/4$?"

Or it just makes the formula beautiful.

Reference for estimation gaussian of the heat kernel

2012-02-14T14:23:02Z

Let $(M,g^{TM})$ a Riemannian manifold of dimension $n$ and $\Delta$ the Laplace–Beltrami operator. I would like to find a reference (analytic or probabilistic) for the following classic result.

If $p_t(x,y)$ is the kernel of the semigroup $e^{t\Delta}$, then there exist $C,c>0$, such that $$p_{t}(x,y)\leq \frac{C}{t^{n/2}}e^{-cd(x,y)^2/t}.$$

Path integral and harmonic oscillator

2012-01-26T16:56:41Z

Maybe this is not a research level question. I post it because I heard that the path integral can be rigorous by Brownian motion. But my knowledge of probability is so limited.

If $$L=\frac{1}{2}(-\frac{d^2}{dx^2}+x^2),$$ we know that $Sp(L)={1/2,3/2,5/2,...}$. So we get $$\mathrm{Tr}[e^{-L}]=\frac{1}{2\sinh1/2}.$$

I would like to recover it by following "method".

If $E_x$ denote the expectation of the Brownian motion $x_.$ start from x. By Feymann-Kac formula, we have $$e^{-L}f(x)=E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds}f(x_1)].$$

If $p(x,y)$ denote the kernel of $e^{-L}$, we get $$p(x,y)=E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds};x_1=y]\frac{e^{-\frac{1}{2}|x-y|^2}}{\sqrt{2\pi}}$$ where $E_x[...;x_1=y]$ is the conditional expectation.

So we get $$\mathrm{Tr}e^{-L}=\int_{x\in \mathbb{R}}\frac{dx}{\sqrt{2\pi}}E_x[e^{-\frac{1}{2}\int_0^1x_s^2ds};x_1=x]$$

All the thing is rigorous until now. But in some physics book, it follows that the right side is $$\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\dot{x}_s^2+x^2_sds}\mathcal{D}x =\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\langle-\Delta +1x,x\rangle} \mathcal{D}x$$ where $$\mathcal{D}x=\mathrm{det}^{1/2}(-\Delta)\frac{dx}{(2\pi)^{\infty/2}}.$$ As the finite cas, $$\int_{periodic\ path} e^{-\frac{1}{2}\int_{0}^1\langle-\Delta +1x,x\rangle} \mathcal{D}x=\frac{\mathrm{det}^{1/2}(-\Delta )}{\mathrm{det}^{1/2}(-\Delta +1)}=\mathrm{det}^{-1/2}(1+(-\Delta)^{-1}).$$ As we know, $Sp(-\Delta)={4\pi^2k^2,k\in \mathbb{Z}}$. We have $$\mathrm{det}^{1/2}(1+(-\Delta)^{-1})=\Pi_{k=1}^{\infty}(1+\frac{1}{4k^2\pi^2})=2\sinh{1/2}$$

It also follows the right answer.

So my question is "How to make it rigorous?"

First, it will need a gaussian measure on the periodic path. But I can not find a natural one.

Edit: Thanks to Alexander Chervov, he give a interesting measure by Fourier Analysis. It is a right one in some sense. But it is not even clear for me that its support is the contious path. And with this measure how can we get the final answer rigorousment.

Edit2 and Answer Thanks to A.J. Tolland and Glimm & Jaffe's book. I just complete the answer to my question.

Let $P$ is the measure of the brownian bridge.

Consider the operator forme $C^{-\infty}(S^1)$ to $C^{\infty}(S^1)$, $$-\Delta+1, $$ There is a unique gaussian mesure $Q$ on $C^{-\infty}(S^1)$(in fact its support is $C^0(S^1)$) whose matrix of covariance is $$(-\Delta+1)^{-1}$$

By the uniqueness, we have $$\frac{dx}{\sqrt{2\pi}}e^{-\frac{1}{2}\int_{0}^1(x_s+x)^2ds}dP=\frac{\mathrm{det}^{1/2}(-\Delta)}{\mathrm{det}^{1/2}(-\Delta+1)}dQ$$

After integral, we have get the resultat.

Remark, the existance and the uniqueness of the gaussian measure is the big theorem in the Appendix A.4 of Glimm & Jaffe's book.

Answer by shu for Metric Connections on a Lie Group

2012-01-27T12:15:00Z

If I understand correctly, the answer is Yes.

the +/-/0 connections can be defined by, if $X,Y$ is the left invariant vector $$\nabla_{X}Y=a[X,Y]$$ where $a=1,-1,0$.

The connection is metric for left invariant metric iff $$0=\langle\nabla_{X}Y,Z\rangle+\langle Y,\nabla_{X}Z\rangle.$$

This is trival for the bi-invariant metric.

Vanishing of $\hat{A}$ genus and positive scalar curvature

2011-09-28T11:43:30Z

Classicly, for a spin Riemannian manifold $M$, the $\hat{A}(M)$ genus will be $0$, if the scalar curvature is positive.

The proof is to use the Lichnerowicz formula. we have the index of the Dirac operator will be $0$, i.e., $$Ind(D_+)=0.$$ On the other hand, by the index theorem of Atiyah and Singer, we have $$Ind(D_+)=\hat{A}(M).$$ So we get $$\hat{A}(M)=0.$$

My question is "Can we have a different method to proof this result? without using the Lichnerowicz formula or without using the index theorem" Maybe a formal proof or explanation.

Comment by shu

2013-05-17T14:28:42Z

Yes, I do agree.

Comment by shu

2013-05-15T16:00:12Z

Thank you for the answer. But if we suppose that $Z$ is orientable or even more is spin. Can this be true?

Comment by shu

2013-05-07T10:47:43Z

As you said the the heat kernel proof of the index theorem had already been achieved. But not for the familly index theorem... Quillen's formalism gives a strategy for the heat kernel proof of the familly index theorem. This strategy is achieved by Bismut later via Bismut supperconnection.

Comment by shu

2012-12-13T11:46:35Z

Yes, there exist such a proof. You can find it in Bismut's paper "The Witten complex and the degenerate Morse inequalities." projecteuclid.org/… But I would like to say he also use the Morse complex but more closed to the heat kernel method. The essential is the same as Witten's proof...

Comment by shu

2012-05-29T18:43:58Z

Thanks for the answer, I like so much Cartan's book. And planetmath.org/encyclopedia/… tells the whole story.

Comment by shu

2012-03-03T20:22:08Z

By the symbole argument and self adjointness, $D^2-\nabla^*\nabla$ is a 0 order operator on your manifold. The thing rest is to understand the curvature term. It is rather subtile. If you compare the diffusion of tow side, the thing is more or less like passing the Ito calculs to Stratonovich calculs. The later is coordinate free.

Comment by shu

2012-03-03T20:08:41Z

You can proof it by lifting in principal bundle which does not need a local coordinate system. But as Paul Siegel pointed out, there ...may not be a conceptual argument.

Comment by shu

2012-02-16T10:17:40Z

Thank you, mfolz. It is a very interesting paper.

Comment by shu

2012-02-15T07:48:06Z

Thank you!I find it also in Davies's book.

Comment by shu

2012-01-30T21:00:41Z

@Alexander Chervov, your answer is always wonderful!

Comment by shu

2012-01-30T20:55:50Z

@Tolland, Thanks for your answer. After reading Glimm & Jaffe's book, I understand the whole histoire.

Comment by shu

2012-01-26T19:57:35Z

@Alexander Chervov, indeed, this measure works formally. I want to make it rigourous. It is not even clear for me that its support is the contious path.

Comment by shu

2011-11-05T21:02:13Z

@Otis:For the spin^c and spin , the reference is the lovely book write by Morgan. press.princeton.edu/titles/5866.html.

Comment by shu

2011-11-04T20:02:23Z

The manifold $X$ is spin iff the loop space $LX$ is orientable. But I don't think this will help. If you look at $spin^c$, almost complexe manifold $M$ is $spin^c$. All of the 4 dimention manifold is $spin^c$. So if more over, $det^(1/2) TM$ exist, $M$ is spin.

Comment by shu

2011-10-04T15:53:05Z

As Johannes Ebert said "The first accident is that the Euler class has a direct geometric-topological meaning, which allows to relate it to another topological invariant." But i want to say the $\hat{A}$ genus has also some formal "geometric-topological meaning". Like the Euler number which is 'the number' of zero of a generic section of the tangent bundle, the $\hat{A}$ genus is the 'number' of zero of some section of the tangent bundle $TLM$ of loop space $LM$. It's the idea of Witten's formal proof of the index theorm. Unfortunately his point view does not appear the scalar curvature.