User t-' - MathOverflow

Levy-Gromov Isoperimetric Inequality

2012-05-09T14:54:27Z

In his paper "Paul Levy's Isoperimetric Inequality", Gromov gives the following isoperimetric inequality:

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 $$ \frac{ Vol(V_0)}{Vol(V)}= \frac{Vol(B)}{Vol(S^{n+1})}.$$ Then it follows that $$ \frac{Vol( \partial V_0)}{Vol(V)} \geq \frac{Vol(\partial B)}{Vol(S^{n+1})}. $$

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 negative 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.

First Minkowski Formula

2011-01-26T22:16:41Z

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 $\mathcal{M}$.

Many thanks for any suggestions.

Answer by T-' for What is soliton

2012-03-05T19:05:45Z

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.

Partitions of Unity

2012-02-06T19:35:32Z

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 $p \geq 1$ ?

Integration By Parts on Non-compact Manifolds

2012-02-29T10:58:33Z

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?

Compactness of solutions to parabolic equations (parabolic regularity)

2012-01-25T13:51:57Z

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.

For each $s>0$, I have a solution ${u_s}$ of the conjugate heat equation $\Box ^* u_s = 0$ with some 'nice' initial data.

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."

It is proving exceedingly difficult to find good parabolic regularity results written down cleanly somewhere!

Answer by T-' for Is it possible to see Path Spaces as manifold

2011-06-27T10:05:23Z

I give such a manifold structure here: http://mathoverflow.net/questions/65926/the-tangent-bundle-to-an-infinite-dimensional-manifold/65933#65933

Answer by T-' for Proof of Krylov-Bogoliubov Theorem

2011-06-01T16:40:07Z

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 $f$.

Answer by T-' for The tangent bundle to an infinite-dimensional manifold

2011-05-25T08:18:13Z

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.

Answer by T-' for Poisson equation in the plane

2011-04-18T09:11:16Z

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.

Answer by T-' for Geometric bound on the first eigenvalue of Laplace-Beltrami on forms

2011-04-16T09:39:15Z

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: http://arxiv.org/abs/1003.0817. 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.

Compact Hypersurfaces Bounding Compact Domains

2011-02-25T15:54:05Z

The following statement seems to be taken as given in papers I'm reading:

Let $\mathcal{M}^n$ be a compact, embedded hypersurface in $\mathbb{R}^{n+1}$. Then $\mathcal{M}$ is the boundary of some compact domain $\Omega \subset \mathbb{R}^ {n+1}$.

Is this an elementary result? I feel there must be some algebraic topology argument here. Any suggestions?

Hypersurfaces and Elliptic Points

2011-01-19T22:29:26Z

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.

The author then says "As $M$ has one elliptic point, $R$ is a positive constant and the mean curvature is positive somewhere".

I'm lost here - Why does $M$ have an elliptic point? And how does this affect $R$ and the mean curvature?

Thanks for any help.

Answer by T-' for Isometric embedding of 1-manifold

2010-12-24T11:42:34Z

Whilst what you have given is not a Riemannian metric, the general question has a famous answer: http://en.wikipedia.org/wiki/Nash_Embedding_Theorem

Betti Numbers (homology vs cohomology)

2010-12-16T16:12:02Z

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 $H^1(M; \mathbb{R})$.

So my question is essentially: How do we get an isomorphism $H_1(M)\cong H^1(M; \mathbb{R})$?

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.

Homology and submanifolds...

2010-12-02T20:59:45Z

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 $\partial \Omega_1=M=\partial \Omega_2$."

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.

Comment by T-'

2012-05-09T16:03:28Z

Ah sorry, my mistake.

Comment by T-'

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 Poincaré inequality.

Comment by T-'

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...?

Comment by T-'

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.

Comment by T-'

2012-03-01T16:42:19Z

Thanks for this!

Comment by T-'

2012-02-29T19:31:48Z

The question could be rephrased "under what curcumstances do we have the formula \[ \int_{\mathcal{M}} f \Delta g \, dV = \int_{\mathcal{M}} g \Delta f \, dV \]"

Comment by T-'

2012-02-17T19:27:34Z

What type of convergence?

Comment by T-'

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.

Comment by T-'

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.

Comment by T-'

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.

Comment by T-'

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 $s \downarrow 0$.

Comment by T-'

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 $A$.

Comment by T-'

2011-04-16T09:39:54Z

I should have said 'estimates on $\lambda_1$'.

Comment by T-'

2011-02-25T20:03:57Z

Sorry, I should have specified the smoothness in the statement.

Comment by T-'

2010-12-03T23:31:21Z

Thanks, this is great!