User t-' - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:55:16Z http://mathoverflow.net/feeds/user/11266 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96456/levy-gromov-isoperimetric-inequality Levy-Gromov Isoperimetric Inequality T-' 2012-05-09T14:54:27Z 2012-05-09T22:25:51Z <p>In his paper "Paul Levy's Isoperimetric Inequality", Gromov gives the following isoperimetric inequality:</p> <blockquote> <p>Let $V$ be a closed $(n+1)$-dimensional Riemannian Manifold with $\mathrm{Ric}(V) \geq n \space (= \mathrm{Ric}(S^{n+1}))$. Let $V_0 \subset V$ be a domain with smooth boundary and let $B$ be a round ball in $S^{n+1}$ such that <code>$$\frac{ Vol(V_0)}{Vol(V)}= \frac{Vol(B)}{Vol(S^{n+1})}.$$</code> Then it follows that $$\frac{Vol( \partial V_0)}{Vol(V)} \geq \frac{Vol(\partial B)}{Vol(S^{n+1})}.$$</p> </blockquote> <p>Now my question: in a (slightly earlier) article 'Isoperimetric Inequalities In Riemannian Manifolds', Gromov states that the above inequality will still be true even if $V$ only admits a <strong>negative</strong> lower bound on its Ricci curvature. Does anyone have a reference for a proof of this, or is the statement obvious? It just seems to me that the hypothesis compares the curvature of $V$ to that of $S^{n+1}$, so allowing $\mathrm{Ric}(V)$ to be negative will obscure this. </p> http://mathoverflow.net/questions/53412/first-minkowski-formula First Minkowski Formula T-' 2011-01-26T22:16:41Z 2012-04-14T04:24:38Z <p>Does anyone know of a modern proof of the First Minkowski Formula for a compact embedded hypersurface $\psi \colon \mathcal{M}^n \hookrightarrow \mathbb{R}^{n+1}$ ? The integral formula is $$\int_{\mathcal{M}} H \langle \psi , \nu \rangle \mathrm{d}A +A = 0$$ where $A$ is the area of $\mathcal{M}$, $H$ is the mean curvature of $\mathcal{M}$ and $\nu$ is the outward normal field on $\mathcal{M}$.</p> <p>Many thanks for any suggestions.</p> http://mathoverflow.net/questions/90296/what-is-soliton/90303#90303 Answer by T-' for What is soliton T-' 2012-03-05T19:05:45Z 2012-03-05T19:05:45Z <p>A soliton (at least in my field) is a 'self-similar solution' to a PDE. For instance a solution $(g_t)$ to the Ricci flow equation $$\frac{\partial g }{ \partial t} = - 2 \mathrm{Ric}(g(t))$$ is a Ricci soliton if it takes the form $g(t)= \alpha (t) \phi_t^* (g(0))$ where the $\alpha(t)$'s are scalars and the $\phi_t$'s are diffeomorphisms, i.e. the metric at time $t$ differs from the initial metric by the action of diffeomorphisms and/or dilation. </p> http://mathoverflow.net/questions/87708/partitions-of-unity Partitions of Unity T-' 2012-02-06T19:35:32Z 2012-03-01T09:51:44Z <p>Fix a metric $g$ on a smooth, closed manifold $\mathcal{M}$. Take a finite subcover of the manifold from its atlas. Is it true that any smooth partition of unity subordinate to this cover has uniformly bounded derivatives in $L^p (\mathcal{M})$ for each $p \geq 1$ ? </p> http://mathoverflow.net/questions/89855/integration-by-parts-on-non-compact-manifolds Integration By Parts on Non-compact Manifolds T-' 2012-02-29T10:58:33Z 2012-02-29T13:42:15Z <p>This is undoubtedly a very easy question, but perhaps there are some subtleties. Under what circumstances can we integrate by parts over a non-compact Riemannian manifold? I am aware that having bounded curvature is sufficient (is there a reference for this?), but can this be weakened?</p> http://mathoverflow.net/questions/86629/compactness-of-solutions-to-parabolic-equations-parabolic-regularity Compactness of solutions to parabolic equations (parabolic regularity) T-' 2012-01-25T13:51:57Z 2012-01-25T16:05:02Z <p>I am working on a Ricci flow $(\mathcal{M} , g(t))$, for which the conjugate heat operator is $\Box ^* := - \partial _t - \Delta + R$, where $R$ is the scalar curvature.</p> <p>For each $s>0$, I have a solution ${u_s}$ of the conjugate heat equation $\Box ^* u_s = 0$ with some 'nice' initial data.</p> <p>My question is: does anyone have good sources for results like "If we have such-and-such an estimate on the ${u_s}$, then we get convergence ${u_s} \to u$ smoothly on compact subsets, where $u$ is the fundamental solution of the conjugate heat equation."</p> <p>It is proving exceedingly difficult to find good parabolic regularity results written down cleanly somewhere! </p> http://mathoverflow.net/questions/68906/is-it-possible-to-see-path-spaces-as-manifold/68916#68916 Answer by T-' for Is it possible to see Path Spaces as manifold T-' 2011-06-27T10:05:23Z 2011-06-27T10:05:23Z <p>I give such a manifold structure here: <a href="http://mathoverflow.net/questions/65926/the-tangent-bundle-to-an-infinite-dimensional-manifold/65933#65933" rel="nofollow">http://mathoverflow.net/questions/65926/the-tangent-bundle-to-an-infinite-dimensional-manifold/65933#65933</a> </p> http://mathoverflow.net/questions/66669/proof-of-krylov-bogoliubov-theorem/66671#66671 Answer by T-' for Proof of Krylov-Bogoliubov Theorem T-' 2011-06-01T16:40:07Z 2011-06-01T16:40:07Z <p>First, fix $x \in X$ and let $\mu_1 := \delta_x$ be the Dirac measure supported at $x$. Then define a sequence of probability measures $\mu_n$ such that for any $f \in C^0 (X)$, $$\int_X f(y) \mathrm{d} \mu_n (y) = \frac{1}{n} \sum_{k=0}^{n-1} \int_X f \circ T^k (y) \mathrm{d} \mu_1 (y).$$ Apply the Banach-Alaouglu Theorem to deduce there exists a subsequence $\mu_{n_j}$ which converges in the weak-$\star$ topology. It is then very easy to prove that this limit measure is in fact T-invariant, using the formulation that $\mu$ is T-invariant if and only if $$\int_X f \circ T \mathrm{d} \mu = \int_X f \mathrm{d}\mu$$ for all continuous $f$. </p> http://mathoverflow.net/questions/65926/the-tangent-bundle-to-an-infinite-dimensional-manifold/65933#65933 Answer by T-' for The tangent bundle to an infinite-dimensional manifold T-' 2011-05-25T08:18:13Z 2011-05-25T08:18:13Z <p>Perhaps I am misunderstanding your question, but to put a Banach manifold structure on mapping spaces between finite dimensional manifolds is not that hard. Start by choosing $f \in C^\infty$. Using that the exponential map is a local diffeomorphism, we can find a neighbourhood of the zero section in $C^0 \Gamma f^* (TB)$ (i.e. continuous sections of the pullback of the tangent bundle), called $V_0^f$, say, such that the map $\phi_0^f \colon V_0^f \to C^0 (A,B)$ given by $\phi_0^f (\mathcal{S})(a):= \mathrm{exp}_{f(a)}(\pi^* (f) \mathcal{S}(a))$ is locally invertible. Then choose these inverse maps to be your charts around $f$. By restricting the domains of these charts appropriately, you get a Banach manifold structure on any space $C^r (A,B)$ and indeed a Hilbert manifold structure on $H^s (A,B)$ which is rather nice.</p> http://mathoverflow.net/questions/62092/poisson-equation-in-the-plane/62102#62102 Answer by T-' for Poisson equation in the plane T-' 2011-04-18T09:11:16Z 2011-04-18T09:11:16Z <p>The usual way to approach this would be to first search for radial solutions, i.e. look for functions $v$ such that $v(|(x,y)|)=u(x,y)$ and $v$ solves the equation. This should reduce the problem to an ODE in this special case, from which it is easy to find a solution. </p> http://mathoverflow.net/questions/61901/geometric-bound-on-the-first-eigenvalue-of-laplace-beltrami-on-forms/61904#61904 Answer by T-' for Geometric bound on the first eigenvalue of Laplace-Beltrami on forms T-' 2011-04-16T09:39:15Z 2011-04-16T09:39:15Z <p>Closely connected to what you are talking about is Reilly's Formula (it deals with manifolds with boundary, but estimates on $\Delta$ on manifolds without boundary are possible using it). Here is a paper extending this to forms: <a href="http://arxiv.org/abs/1003.0817" rel="nofollow">http://arxiv.org/abs/1003.0817</a>. I'm sure there are bound to be interesting estimates in there, or allusions as to where else to look. Beware that the curvature terms you have to control aren't as pretty as just Ricci curvature, unfortunately.</p> http://mathoverflow.net/questions/56642/compact-hypersurfaces-bounding-compact-domains Compact Hypersurfaces Bounding Compact Domains T-' 2011-02-25T15:54:05Z 2011-02-25T22:38:02Z <p>The following statement seems to be taken as given in papers I'm reading:</p> <blockquote> <p>Let $\mathcal{M}^n$ be a compact, embedded hypersurface in $\mathbb{R}^{n+1}$. Then $\mathcal{M}$ is the boundary of some compact domain $\Omega \subset \mathbb{R}^ {n+1}$.</p> </blockquote> <p>Is this an elementary result? I feel there must be some algebraic topology argument here. Any suggestions?</p> http://mathoverflow.net/questions/52554/hypersurfaces-and-elliptic-points Hypersurfaces and Elliptic Points T-' 2011-01-19T22:29:26Z 2011-01-20T03:33:08Z <p>I'm reading a paper, in which we have $M^n$ an n-dimensional compact hypersurface embedded in $\mathbb{R}^{n+1}$. We take the scalar cuvature $R$ to be the elementary symmetric polynomial of degree 2 in the principal curvatures of $M$. We know that $R$ is constant.</p> <p>The author then says "As $M$ has one elliptic point, $R$ is a positive constant and the mean curvature is positive somewhere".</p> <p>I'm lost here - Why does $M$ have an elliptic point? And how does this affect $R$ and the mean curvature?</p> <p>Thanks for any help.</p> http://mathoverflow.net/questions/50280/isometric-embedding-of-1-manifold/50285#50285 Answer by T-' for Isometric embedding of 1-manifold T-' 2010-12-24T11:42:34Z 2010-12-24T11:42:34Z <p>Whilst what you have given is not a Riemannian metric, the general question has a famous answer: <a href="http://en.wikipedia.org/wiki/Nash_Embedding_Theorem" rel="nofollow">http://en.wikipedia.org/wiki/Nash_Embedding_Theorem</a></p> http://mathoverflow.net/questions/49659/betti-numbers-homology-vs-cohomology Betti Numbers (homology vs cohomology) T-' 2010-12-16T16:12:02Z 2010-12-16T16:16:05Z <p>I'm somewhat confused about the definitions of Betti numbers for Riemannian manifolds. Working with the first Betti number as an example, I have usually taken the definition to be the rank of the homology group $H_1(M)$, where $M$ is the manifold in question. I'm also aware that through a Hodge-theoretic argument, we have that the first Betti number is equal to the dimension of the space of harmonic 1-forms on $M$, and that in fact this space is isomorphic to $H^1(M; \mathbb{R})$.</p> <p>So my question is essentially: How do we get an isomorphism $H_1(M)\cong H^1(M; \mathbb{R})$?</p> <p>I know that through Poincaré duality we have the isomorphism $H_1(M) \cong H^{n-1}(M; \mathbb{Z})$ but I can't see how this helps.</p> http://mathoverflow.net/questions/48093/homology-and-submanifolds Homology and submanifolds... T-' 2010-12-02T20:59:45Z 2010-12-02T21:14:32Z <p>I'm reading a paper which includes the following line, and I can't find a reference anywhere to the result the authors mention: "Let M be a compact orientable embedded minimal hypersurface of a compact orientable Riemannian manifold N. Suppose we know that the first Betti number is zero. Then using that M,N are both orientable and chasing through exact sequences of homology groups, it is easy to see that M divides N into two components $\Omega_1$ and $\Omega_2$ such that $\partial \Omega_1=M=\partial \Omega_2$."</p> <p>It would be great if someone could help me with this - I can't imagine the argument is too complicated but I can't see where to go. Thanks.</p> http://mathoverflow.net/questions/96456/levy-gromov-isoperimetric-inequality Comment by T-' T-' 2012-05-09T16:03:28Z 2012-05-09T16:03:28Z Ah sorry, my mistake. http://mathoverflow.net/questions/91434/inequality-for-the-solution-to-the-heat-equation Comment by T-' T-' 2012-03-18T14:02:24Z 2012-03-18T14:02:24Z There should indeed be a minus sign in the exponential. I suppose this is a little interesting since it involves the optimal constant in the Poincar&#233; inequality. http://mathoverflow.net/questions/91434/inequality-for-the-solution-to-the-heat-equation Comment by T-' T-' 2012-03-17T00:11:06Z 2012-03-17T00:11:06Z Hint: Doesn't the inequality you are trying to prove look very much like an inequality solutions to an ODE might solve...? http://mathoverflow.net/questions/86629/compactness-of-solutions-to-parabolic-equations-parabolic-regularity Comment by T-' T-' 2012-03-01T16:48:48Z 2012-03-01T16:48:48Z I also forgot about this. A great reference is 'Linear and Quasi-Linear Equations of Parabolic Type' by Ladyzhenskaya, Solonnikov and Ural'ceva. It isn't the nicest to read, but the results are very general. It turns out that to answer my question, $L^1$ estimates on the $u_s$ are sufficient, since by a duality argument these can be improved to $L^p$ estimates for some $p$ slightly greater than 1, and then parabolic Calderon-Zygmund-type results will do the rest, setting up an Arzela-Ascoli-type argument. But we get $L^1$ bounds for free by the fact the functions solve the equation. http://mathoverflow.net/questions/87708/partitions-of-unity/89946#89946 Comment by T-' T-' 2012-03-01T16:42:19Z 2012-03-01T16:42:19Z Thanks for this! http://mathoverflow.net/questions/89855/integration-by-parts-on-non-compact-manifolds Comment by T-' T-' 2012-02-29T19:31:48Z 2012-02-29T19:31:48Z The question could be rephrased &quot;under what curcumstances do we have the formula $\int_{\mathcal{M}} f \Delta g \, dV = \int_{\mathcal{M}} g \Delta f \, dV$&quot; http://mathoverflow.net/questions/88752/sequence-of-smooth-functions-converging-to-sgnx Comment by T-' T-' 2012-02-17T19:27:34Z 2012-02-17T19:27:34Z What type of convergence? http://mathoverflow.net/questions/87708/partitions-of-unity Comment by T-' T-' 2012-02-06T23:31:11Z 2012-02-06T23:31:11Z I already have - I only asked about the more general case out of interest. I expect a counter-example would not be too hard to construct. http://mathoverflow.net/questions/87708/partitions-of-unity Comment by T-' T-' 2012-02-06T21:14:26Z 2012-02-06T21:14:26Z Thanks for your input, very interesting. In practice, all I really need are for $\nabla \phi_j$, $\Delta \phi _j$ to be bounded in each $L^p$ for each of my finite number of $j$'s, which is surely true. http://mathoverflow.net/questions/87708/partitions-of-unity Comment by T-' T-' 2012-02-06T20:19:58Z 2012-02-06T20:19:58Z I mean that $|| \nabla ^k \phi _j ||_{ L^p (\mathcal{M})} \leq K$ for each $k \geq 0$, where $K$ can depend on $j$ but is independent of $k$. Here, $\phi _j$ is one function in the partition of unity. In terms of trying it for myself, I have given it some thought but was hoping I was missing something trivial. http://mathoverflow.net/questions/86629/compactness-of-solutions-to-parabolic-equations-parabolic-regularity Comment by T-' T-' 2012-01-25T23:03:30Z 2012-01-25T23:03:30Z To be more explicit, the initial data for the $u_s$ are 'generalised Gaussians' on the manifold, i.e. the Euclidean heat kernel where the $|x-y|$ part is replaced with $d_g( \cdot , x_0)$ for some fixed $x_0 \in \mathcal{M}$. I am taking the limit as $s \downarrow 0$. http://mathoverflow.net/questions/65926/the-tangent-bundle-to-an-infinite-dimensional-manifold/65933#65933 Comment by T-' T-' 2011-05-25T08:39:46Z 2011-05-25T08:39:46Z Oh, and then it is not hard to show that we can identify the tangent spaces by $T_f C^r (A,B)= \{v \colon A \to TB | v(a) \in T_{f(a)}B\}$. So in fact $T_{id}C^r(A,A)$ is precisely the space of $C^r$ vector fields on $A$. http://mathoverflow.net/questions/61901/geometric-bound-on-the-first-eigenvalue-of-laplace-beltrami-on-forms/61904#61904 Comment by T-' T-' 2011-04-16T09:39:54Z 2011-04-16T09:39:54Z I should have said 'estimates on $\lambda_1$'. http://mathoverflow.net/questions/56642/compact-hypersurfaces-bounding-compact-domains/56656#56656 Comment by T-' T-' 2011-02-25T20:03:57Z 2011-02-25T20:03:57Z Sorry, I should have specified the smoothness in the statement. http://mathoverflow.net/questions/48093/homology-and-submanifolds/48095#48095 Comment by T-' T-' 2010-12-03T23:31:21Z 2010-12-03T23:31:21Z Thanks, this is great!