User hans - MathOverflow

Answer by Hans for Applications of Hodge-de Rham Laplacian on p-forms ($p\neq 0,n$) in physics or engineering

2012-10-10T00:56:05Z

I would suggest you to look at the paper link text

It is written in physics language, so maybe I try to explain:

It is not exactly the Hodge Laplacian which is used, but the Witten Laplacian. The latter
was introduced by Witten in the paper mentioned by Kevin Kordek to study Morse inequalities/complexes on manifolds. It depends on a function $f$ on the manifold (a Morse function in Witten' application) and a small parameter $h>0$ (semiclassical parameter in Witten's application) and is given by

$\Delta_{f,h} = (d_{f,h} + d^*_{f,h})^2$

with $d_{f,h} = h \ e^{-f/h} \ d \ e^{f/h}$ and $d^*_{f,h} = h \ e^{f/h} \ d^* \ e^{-f/h}$

(don't trust me too much for the signs, the point is: the Witten Laplacian is like the Hodge Laplacian, but with the differential distorted by exponential weights. It equals the Hodge laplacian when $f$ is a constant.)

The restriction of the Witten Laplacian to functions ($p=0$) is unitarily equivalent to the generator of a "metastable" stochastic diffusion process (keyword: ground state transformation, it is formula (3) in the paper) moving in the energy landscape given by $f$ (imagien $f$ having several minima). This process follows essentially the (negative) gradient flow of $f$ but a small (with intensity $h$) Brownian noise disturbes the motion and leads to tunneling from one minimum of $f$ to another.

The point of the paper is that metastable properties of this process are much better understood when considering not only the Witten Laplacian (i.e. the generator) for $p=0$ but for every $p=0,dots,n$. For example eigenforms for $p=1$ are related to the metastable transition paths of the process (i.e. the paths along which it tunnels).

There are no boundary conditions here, but of course you can think of similar things by considering a bounded open set of $\mathbb R^n$ and putting boundary conditions.

P.S.: Notation in the mentioned paper: $T$ is my $h$, $E$ is my $f$, $H_{FP}$ is the adjoint of the generator of the diffusion, $H^h$ is the Witten Laplacian (here $h$ stands for hermitian, not to be confused with my $h$). My $p$ is the fermion number. My $d_f,d^*_f$ are denoted by $Q^h, \bar Q^h$.

Answer by Hans for Describe a topic in one sentence.

2012-10-07T18:22:29Z

Probability/Statistical mechanics:

Take a probabilistic model (possibly complicated, involving huge state space, describing a complex system) and rescale it suitably, such that in the limit a simpler "macroscopic" object emerges;

if the latter is still random it's a central limit theorem, if it's deterministic it's a law of large numbers, if you look at fluctuations from the latter it's large deviations; if it is largely independent on the details of the starting probabilsitc model, you have a universality phenomenon (and are happy because when modelling your real system you were forced to add some assumptions just for mathematical comfort); if it changes qualitatively when playing with a parameter of the original model you have a phase transition and want to know the critical values of the parameter.

Answer by Hans for Eliminating 1st order terms in elliptic partial differential equation

2012-03-13T15:06:16Z

Dear Pait,

note that the kind of transformation you are using for constant $a^i$ 's also works if the vector $a=(a^i)_i$ is a gradient, say $a(x)=\nabla g (x)$ for some smooth $g$.

In this case define

$u(x)= v(x)e^{\frac{1}{2}g(x)}$

and you will obtain again an elliptic equation without first order term.

This transformation is sometimes called ground state transformation and it is frequently used to go from reversible diffusion generators to Schroedinger type operators (or the other way around).

Answer by Hans for functions satisfying "one-one iff onto"

2012-02-17T20:49:41Z

An isometry (i.e. distance preserving map) between metric spaces is automatically injective.

Answer by Hans for Reference for estimation gaussian of the heat kernel

2012-02-14T15:21:30Z

A probabilistic proof is given in the book: "Stochastic Analysis on Manifolds" by E. Hsu. See Theorem 5.3.4, which also gives the lower bound.

Answer by Hans for Functions of Pseudodifferential Operators

2012-02-08T23:50:59Z

Another reference is

Dimassi-Sjöstrand: Spectral asymptotics in the semiclassical limit, Cambridge University Press (1999)

See in particular Chapter 8 ("Functional Calculus for pseudodifferentiql operators").

The results are in the semiclassical setting. The result which may interest you is Theroem 8.7 which states that if $P$ is a selfsadjoint h-pseudodiff operator and $f\in C^\infty_0$, then f(P) is again a pseudo with leading symbol $f(p((x,\xi))$ , where $p(x,\xi)$ is the leading symbol of $P$.

The method to prove this is based here on a standard Cauchy formula. (I think in the book of Taylor, cited by Liviu Nicolaescu a different strategy is used).

Answer by Hans for Symmetric Feller processes and Dirichlet Forms

2012-02-08T22:01:53Z

I think that the guess is true under the general assumptions I made, by following Byron Schmuland's reasoning. Let me spell it out the way I understood it.

I denote by $T$ the $L^2(dm)$ semigroup induced by $X$, by $\bar T$ the Feller semigroup generated by the closure of $G$ in $C_0$ and by $\tilde T$ the $L^2(dm)$ semigroup generated by the Friedrichs extension of $G$.

The semigroup $T$ is characterized by

$T=\bar T$ on $L^2(dm)\cap C_0$ (I take this as definition of $T$ as in Fukushima et al.)

So it is enough to show that $\tilde T = \bar T$ on $C_K$ (which is both dense in $L^2(dm)$ and $C_0$).

By Yosida approximation it is enough to show that the corresponding resolvents satisfy for $\lambda>0$

$\tilde R_\lambda = \bar R_\lambda$ on $C_K$

By definition of resolvent $\tilde R_\lambda= \bar R_\lambda$ on $\mathcal F:=(\lambda-G)(\mathcal D)$. The $C_0$-closure of $\mathcal F$ is $C_0$ since $G$ generates, in particular it contains $C_K$. It follows that also the $L^2$ closure of $\mathcal F$ contains $C_K$, so we are done.

Observe that under the assumptions I made in the question, $G$ is automatically essentially selfadjoint, so there is no other selfadjoint extension other than the Friedrichs one. (a criterion for essential selfadjointness is that $(\lambda -G)(\mathcal D)$ is dense in $L^2$ which we have shown above). So the interesting case I was wondering about actually doesn't happen.

Let me know if there is a flaw in what I wrote.

Symmetric Feller processes and Dirichlet Forms

2012-02-01T18:45:28Z

Let $(G, \mathcal D)$ be a densely defined operator on $C_0$ (continuous functions vanishing at infinity on some nice topological space) whose closure $\bar G$ generates a Feller semigroup and let $X$ be a Markov process corresponding to it.

Assume that $\mathcal D\subset C_K$ (continuous functions with compact support), that $G(\mathcal D)\subset C_K$ and that $G$ is symmetric with respect to a Radon measure $m$(Edit: with full support, but not necessarily finite), i.e.

$\int Gf\ g \ dm = \int f \ Gg\ dm$ for every $f,g\in \mathcal D$.

I guess that the Dirichlet form $\mathcal E$ of $X$ (defined as in the book of Fukushima/Oshima/Takeda by using the transition kernel, see (1.4.13) on p.30 in the last edition) is given by the closure of

$\mathcal D\ni f,g \mapsto \int Gf\ g dm$.

In other terms the Friedrichs extension of $G$ in $L^2(dm)$ should be the generator of the $L^2$ semigroup induced by $X$. (Edit: by $L^2$ semigroup induced by $X$ I mean the semigroup corresponding to the Dirichlet form $\mathcal E$ )

Is this true? I didn't find a reference nor a simple argument for showing this.

Or is it possible that a selfadjoint extension other than the Friedrichs one generates the $L^2$ semigroup induced by $X$?

Edit: From the answer of Byron Schmuland it is clear to me that the guess is true if the state space is compact. Observe that in this case $G$ is essentially selfadjoint in $L^2$, so the Friedrichs extension is just the closure of $G$ and there are no other selfadjoint extensions. I'm still confused about the case of noncompact state space. I would also appreciate partial answers which work for some concrete example of $G$ (say elliptic partial differential operators, or discrete operators).

Comment by Hans

2012-10-15T00:26:54Z

I like your answer too, just wanted to clarify some potential ambiguity!

Comment by Hans

2012-10-10T00:05:32Z

Hmmm, isn't it the other way around? He uses supesymmetric quantum mechanics to prove Morse inequalities, as far as I understand! And he does not consider exactly the Hodge Laplacian, but a distorted version of it (now called Witten Laplacian). But I agree that indeed, the Witten Laplacian on the full algebra of forms is used. A bit more precisely: the Witten Laplacian is a Schroedinger type operator for every $p=0,\dots,n$ and he uses semiclassical spectral approximation (through Harmonic oscillators) of these Schroedinger operators to get the Morse inequalities.

Comment by Hans

2012-04-24T00:05:35Z

@Robert Bryant: hmm, this is what I thought at the beginning, but I'm not convinced. It seems to me that the quadratic approximation is not enough to decide. Why should $L^+\geq L$ imply $\phi^+\geq \phi$ near $p$? (We really have $L^+\geq L$ in general and not $L^+ > L$, otherwise it would be clear to me...). Do I miss something?

Comment by Hans

2012-04-22T15:02:50Z

@Robert Bryant. Sorry, I was imprecise: I mean the first option you mentioned.

Comment by Hans

2012-04-22T13:39:59Z

@Robert Bryant. Thanks for this answer. I was wondering if the solution you constructed, uniquely characterized among the smooth solutions by being $0$ in $p$ and nonnegative, could also be uniquely characterized among smooth solutions as the maximal solution. I'm pretty convinced this is true, but I don't see how to prove it. Does it follow somehow directly from your construction or are some other arguments needed? I would be very grateful if you or somebody else can comment on this!

Comment by Hans

2012-03-19T21:28:12Z

@Syang Chen. Many thanks.

Comment by Hans

2012-03-14T19:09:03Z

@Alekk. I'm wondering about your function $h$. What are the typical $h$'s you have in mind? What would be the meaning of $h$ for the Langevin diffusion you mentioned in "Motivations"? Is the $h$ in the paragraph between Example and Motivations the same $h$ of the Question? Could you say a bit more on this?

Comment by Hans

2012-03-04T14:45:43Z

@RadonNikodym. I have two problems with your question: 1) you write "THE self-adjoint extension of..". Is it clear that under your quite general assumptions you have just one selfadjoint extension (i.e. that L is essentially selfadjoint on $C_c^2$) ? 2) Of course any selfadjoint extension has its domain as a core. So what do you ask for exactly? Do you want to know if $C_c^2$ itself is a core? Or do you look for a core having specific properties?

Comment by Hans

2012-02-22T12:47:42Z

No, I would say there are no additional terms. For a precise statement (and proof) of the Feynman-Kac formula with killing and time dependent coefficients you may look at "Brownian Motion and Stochastic calculus" by Karatzas-Shreve (2nd edition), Theorem 7.6 on page 366.

Comment by Hans

2012-02-19T21:17:23Z

@András Bátkai. I agree it is not exactly the square of a group generator. This is why I put the quotation marks when I stated this. But it is not difficult to reduce to the one dimensional case by factorizing as is described in a precise way in the reference I cited (see in particular Example 4.10). Do you agree with this?

Comment by Hans

2012-02-19T18:55:53Z

Besides the method used in the reference given in András Bátkai's answer, there is also another way to prove this fact: one could exploit that the Laplacian is "the square of a group generator". A reference for this is for example "One Parameter semigroups for linear evolution equations" by Engel/Nagel. See in particular Corollary 4.9 and Example 4.10. Also in this case there is no substantial difference in the proof between p=1 and p>1, so I was wondering if Martin knows another type of argument which works only for $p>1$ (I would be interested in this).

Comment by Hans

2012-02-19T18:54:13Z

@András Bátkai. Thanks for pointing out in your answer in which sense the case $p=1$ is different. What do you mean by "the whole space" here above? I just wanted to say that "the Laplace operator with maximal domain on $L^p(\mathbb R^n)$ is the generator of an analytic semigroup on $L^p(\mathbb R^n)$" would be a precise (and true) statement; while, without specification of the domain, it doesn't make much sense.

Comment by Hans

2012-02-18T20:18:33Z

sorry, I mean p>1 of course.

Comment by Hans

2012-02-18T20:12:31Z

Your statement is not very precise, since you don't mention the domain of the Laplacian you are considering. Anyway, I think that it is true also for p=1. And I don't see why the proof should necessarily be different than for p>0. How would you prove it in the latter case?

Comment by Hans

2012-02-18T13:00:07Z

@Yemon Choi: it doesn't really relate to the question. sorry for this disattention. and thanks for warning.