User hans - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:09:39Z http://mathoverflow.net/feeds/user/21061 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109070/applications-of-hodge-de-rham-laplacian-on-p-forms-p-neq-0-n-in-physics-or-en/109263#109263 Answer by Hans for Applications of Hodge-de Rham Laplacian on p-forms ($p\neq 0,n$) in physics or engineering Hans 2012-10-10T00:56:05Z 2012-10-15T00:28:30Z <p>I would suggest you to look at the paper <a href="http://arxiv.org/pdf/cond-mat/0311273.pdf" rel="nofollow">link text</a> </p> <p>It is written in physics language, so maybe I try to explain:</p> <p>It is not exactly the Hodge Laplacian which is used, but the Witten Laplacian. The latter<br> 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 </p> <p>$\Delta_{f,h} = (d_{f,h} + d^*_{f,h})^2$</p> <p>with $d_{f,h} = h \ e^{-f/h} \ d \ e^{f/h}$ and $d^*_{f,h} = h \ e^{f/h} \ d^* \ e^{-f/h}$</p> <p>(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.)</p> <p>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. </p> <p>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). </p> <p>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> <p>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 $Q^h, \bar Q^h$. </p> http://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence/109087#109087 Answer by Hans for Describe a topic in one sentence. Hans 2012-10-07T18:22:29Z 2012-10-07T18:22:29Z <p>Probability/Statistical mechanics:</p> <p>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;</p> <p>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.</p> http://mathoverflow.net/questions/87072/eliminating-1st-order-terms-in-elliptic-partial-differential-equation/91086#91086 Answer by Hans for Eliminating 1st order terms in elliptic partial differential equation Hans 2012-03-13T15:06:16Z 2012-03-13T15:06:16Z <p>Dear Pait,</p> <p>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 $g$.</p> <p>In this case define </p> <p>$u(x)= v(x)e^{\frac{1}{2}g(x)}$ </p> <p>and you will obtain again an elliptic equation without first order term.</p> <p>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).</p> http://mathoverflow.net/questions/88750/functions-satisfying-one-one-iff-onto/88767#88767 Answer by Hans for functions satisfying "one-one iff onto" Hans 2012-02-17T20:49:41Z 2012-02-17T20:49:41Z <p>An isometry (i.e. distance preserving map) between metric spaces is automatically injective.</p> http://mathoverflow.net/questions/88433/reference-for-estimation-gaussian-of-the-heat-kernel/88436#88436 Answer by Hans for Reference for estimation gaussian of the heat kernel Hans 2012-02-14T15:21:30Z 2012-02-14T15:21:30Z <p>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.</p> http://mathoverflow.net/questions/87937/functions-of-pseudodifferential-operators/87942#87942 Answer by Hans for Functions of Pseudodifferential Operators Hans 2012-02-08T23:50:59Z 2012-02-09T16:12:46Z <p>Another reference is </p> <p>Dimassi-Sjöstrand: Spectral asymptotics in the semiclassical limit, Cambridge University Press (1999)</p> <p>See in particular Chapter 8 ("Functional Calculus for pseudodifferentiql operators").</p> <p>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 $P$.</p> <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).</p> http://mathoverflow.net/questions/87265/symmetric-feller-processes-and-dirichlet-forms/87935#87935 Answer by Hans for Symmetric Feller processes and Dirichlet Forms Hans 2012-02-08T22:01:53Z 2012-02-08T22:01:53Z <p>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. </p> <p>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 $G$.</p> <p>The semigroup $T$ is characterized by</p> <p>$T=\bar T$ on $L^2(dm)\cap C_0$ (I take this as definition of $T$ as in Fukushima et al.)</p> <p>So it is enough to show that $\tilde T = \bar T$ on $C_K$ (which is both dense in $L^2(dm)$ and $C_0$).</p> <p>By Yosida approximation it is enough to show that the corresponding resolvents satisfy for $\lambda>0$</p> <p>$\tilde R_\lambda = \bar R_\lambda$ on $C_K$</p> <p>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. </p> <p>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. </p> <p>Let me know if there is a flaw in what I wrote.</p> http://mathoverflow.net/questions/87265/symmetric-feller-processes-and-dirichlet-forms Symmetric Feller processes and Dirichlet Forms Hans 2012-02-01T18:45:28Z 2012-02-08T22:01:53Z <p>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. </p> <p>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.</p> <p>$\int Gf\ g \ dm = \int f \ Gg\ dm$ for every $f,g\in \mathcal D$.</p> <p>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</p> <p>$\mathcal D\ni f,g \mapsto \int Gf\ g dm$.</p> <p>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 $\mathcal E$ ) </p> <p>Is this true? I didn't find a reference nor a simple argument for showing this. </p> <p>Or is it possible that a selfadjoint extension other than the Friedrichs one generates the $L^2$ semigroup induced by $X$? </p> <p>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). </p> http://mathoverflow.net/questions/109070/applications-of-hodge-de-rham-laplacian-on-p-forms-p-neq-0-n-in-physics-or-en/109073#109073 Comment by Hans Hans 2012-10-15T00:26:54Z 2012-10-15T00:26:54Z I like your answer too, just wanted to clarify some potential ambiguity! http://mathoverflow.net/questions/109070/applications-of-hodge-de-rham-laplacian-on-p-forms-p-neq-0-n-in-physics-or-en/109073#109073 Comment by Hans Hans 2012-10-10T00:05:32Z 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. http://mathoverflow.net/questions/82227/solutions-to-the-eikonal-equation/82284#82284 Comment by Hans Hans 2012-04-24T00:05:35Z 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 $L^+ &gt; L$, otherwise it would be clear to me...). Do I miss something? http://mathoverflow.net/questions/82227/solutions-to-the-eikonal-equation/82284#82284 Comment by Hans Hans 2012-04-22T15:02:50Z 2012-04-22T15:02:50Z @Robert Bryant. Sorry, I was imprecise: I mean the first option you mentioned. http://mathoverflow.net/questions/82227/solutions-to-the-eikonal-equation/82284#82284 Comment by Hans Hans 2012-04-22T13:39:59Z 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! http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59988#59988 Comment by Hans Hans 2012-03-19T21:28:12Z 2012-03-19T21:28:12Z @Syang Chen. Many thanks. http://mathoverflow.net/questions/54557/weighted-poincare-inequality Comment by Hans Hans 2012-03-14T19:09:03Z 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 &quot;Motivations&quot;? Is the $h$ in the paragraph between Example and Motivations the same $h$ of the Question? Could you say a bit more on this? http://mathoverflow.net/questions/85452/core-of-divergence-form-operator-with-unbounded-coefficient Comment by Hans Hans 2012-03-04T14:45:43Z 2012-03-04T14:45:43Z @RadonNikodym. I have two problems with your question: 1) you write &quot;THE self-adjoint extension of..&quot;. 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? http://mathoverflow.net/questions/89179/is-the-feynman-kac-formula-valid-for-a-time-dependent-potential Comment by Hans Hans 2012-02-22T12:47:42Z 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 &quot;Brownian Motion and Stochastic calculus&quot; by Karatzas-Shreve (2nd edition), Theorem 7.6 on page 366. http://mathoverflow.net/questions/88849/analytic-generator Comment by Hans Hans 2012-02-19T21:17:23Z 2012-02-19T21:17:23Z @Andr&#225;s B&#225;tkai. 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? http://mathoverflow.net/questions/88849/analytic-generator Comment by Hans Hans 2012-02-19T18:55:53Z 2012-02-19T18:55:53Z Besides the method used in the reference given in Andr&#225;s B&#225;tkai's answer, there is also another way to prove this fact: one could exploit that the Laplacian is &quot;the square of a group generator&quot;. A reference for this is for example &quot;One Parameter semigroups for linear evolution equations&quot; 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&gt;1, so I was wondering if Martin knows another type of argument which works only for $p&gt;1$ (I would be interested in this). http://mathoverflow.net/questions/88849/analytic-generator Comment by Hans Hans 2012-02-19T18:54:13Z 2012-02-19T18:54:13Z @Andr&#225;s B&#225;tkai. Thanks for pointing out in your answer in which sense the case $p=1$ is different. What do you mean by &quot;the whole space&quot; here above? I just wanted to say that &quot;the Laplace operator with maximal domain on $L^p(\mathbb R^n)$ is the generator of an analytic semigroup on $L^p(\mathbb R^n)$&quot; would be a precise (and true) statement; while, without specification of the domain, it doesn't make much sense. http://mathoverflow.net/questions/88849/analytic-generator Comment by Hans Hans 2012-02-18T20:18:33Z 2012-02-18T20:18:33Z sorry, I mean p&gt;1 of course. http://mathoverflow.net/questions/88849/analytic-generator Comment by Hans Hans 2012-02-18T20:12:31Z 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&gt;0. How would you prove it in the latter case? http://mathoverflow.net/questions/88750/functions-satisfying-one-one-iff-onto/88767#88767 Comment by Hans Hans 2012-02-18T13:00:07Z 2012-02-18T13:00:07Z @Yemon Choi: it doesn't really relate to the question. sorry for this disattention. and thanks for warning.