Isoperimetric inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:38:38Z http://mathoverflow.net/feeds/question/94365 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94365/isoperimetric-inequality Isoperimetric inequality Marc 2012-04-18T06:09:27Z 2012-04-19T19:13:40Z <p>Let $M$ be a complete, non-compact, simply connected Riemannian manifold of dimension $n$ whose sectional curvatures are bounded above by $\kappa&lt;0$. I want to prove that for any open subset $\Omega\subset M$ whose closure in $M$ is compact, the following inequality holds: $$\frac{Vol(\Omega)}{Vol(\partial \Omega)}\leq \frac{1}{(n-1)\sqrt{-\kappa}}$$</p> <p>The constant on the right gives a lower bound for the first Dirichlet eigenvalue of the Laplace operator. If the metric on $M$ is given by $ds^2=g_{ij}dx^idx^j$, then $$\Delta=\frac{1}{\sqrt{\det g}}\sum_{i,j} \frac{\partial}{\partial x^i}\left(\sqrt{\det g} g_{ij} \frac{\partial}{\partial x^j}\right)$$ If $0&lt;\lambda_1&lt;\lambda2&lt;\cdots$ are the Dirichlet eigenvalues of $-\Delta$, by a theorem of Mckean we have an inequality $$\lambda_1(M)\geq \frac{1}{4}(n-1)^2k$$ for a Riemannian manifold satisfying the conditions above.</p> <p>Is there a way to relate the first eigenvalue to the ratio of volumes so as to prove the isoperimetric inequality above or is all this the wrong strategy?</p> <p>Thanks in advance for any insight.</p> http://mathoverflow.net/questions/94365/isoperimetric-inequality/94398#94398 Answer by Benoît Kloeckner for Isoperimetric inequality Benoît Kloeckner 2012-04-18T12:57:22Z 2012-04-19T14:25:38Z <p>I do not know if there is a way to get the isoperimetric inequality from the spectral gap, but both can be proven in almost the same way. The classical references for the linear isoperimetric inequality are S.-T. Yau, "Isoperimetric constants and the first eigenvalue of a compact Riemannian manifold", <em>Ann. Sci. École Norm. Sup.</em> (4) 8 (1975), no. 4, 487–507 and Yurii D. Burago and Victor A. Zalgaller, "Geometric inequalities".</p> <p>I like this proof so let me give it here (this is Burago-Zalgaller presentation). For any unit tangent vector $u$ and positive real $r$, let $s(u,r)$ be the "candle function" defined by <code>$$dy = s(u,r) \,du \,dr$$</code> when $y=\exp_x(ru)$ and $u\in UT_xM$. Up to a normalization, this is simply the jacobian of the exponential map. The curvature hypothesis implies $(\log s(u,r))'\geqslant \sqrt{-\kappa}(n-1)$ where the prime denotes derivative with respect to $r$ (this is a consequence of <em>Günther's inequality</em>).</p> <p>$\Omega$ is contained in the union of all geodesic rays from any fixed point $x_0$ to $\partial \Omega$. Let $U\subset UT_{x_0}M$ be the set of unit vectors generating geodesics that intersect $\Omega$, and for $u\in U$ let $r_u$ be the last intersection time of the geodesic generated by $u$ with $\Omega$. Then<br> $$\mathrm{Vol}(\partial \Omega) \geqslant \int_U s(u,r_u) \,du$$ and $$\mathrm{Vol}(\Omega) \leqslant \int_U \int_0^{r_u} s(u,t) \,dt\,du.$$</p> <p>Now, writing $s(u,r_u)=\int_0^{r_u} s'(u,t) \,dt$ and using Günther's inequality, the desired result comes.</p> <p>I cannot help self-advertising: in fact, the same conclusions (Günther inequality, hence both the linear isoperimetric inequality and the spectral gap of MCKean) hold under a weaker curvature bound (some higher but non-positive sectional curvature can be compensated by enough more negative sectional curvature in other directions). This is explained in an arXiv paper with Greg Kuperberg, "A refinement of Günther's candle inequality" [<a href="http://arxiv.org/abs/1204.3943" rel="nofollow">arXiv:1204.3943</a>].</p> http://mathoverflow.net/questions/94365/isoperimetric-inequality/94558#94558 Answer by Greg Kuperberg for Isoperimetric inequality Greg Kuperberg 2012-04-19T19:13:40Z 2012-04-19T19:13:40Z <p>Benoit gave a good answer in the sense that he revised the question so that it can have a good answer. Instead of trying to derive an isoperimetric inequality from McKean's inequality, the natural thing to do is to prove them both from a logarithmic candle inequality, the property that is discussed in our new paper. But suppose that you still wanted to prove an isoperimetric inequality just from McKean's inequality. Then I don't think that it's possible, because I think that there are manifolds with a large spectral gap, but which have domains with a lot of volume and very little boundary.</p> <p>Suppose that you have a hyperbolic $n$-disk $D$ with radius $\epsilon$ and extreme negative curvature $\kappa \ll 0$. Then you can glue together two copies of $D$ along their boundary, i.e., the doubling construction. This is already a Riemannian manifold with at least a continuous metric; you can also smooth the metric in a tiny neighborhood of the common circle to make it $C^\infty$. If you send $|\kappa|$ to infinity first, before sending $\epsilon$ to 0, then it looks like this surface has a big eigenvalue gap. Since it is a topological sphere, it has a big disk inside with small boundary, namely the complement of a small disk. You can also take a connected sum with this manifold and another manifold, connecting at a small disk.</p> <p>Okay, this type of example is not non-positively curved. So one remaining question is, if $M$ is non-positively curved, then does McKean's inequality imply a linear isoperimetric inequality as stated above? I don't know the answer to that; it could be a good question. But if the intent of the original question was how to prove the existing isoperimetric inequality of $K &lt; \kappa$ manifolds, rather than whether one can prove a new isoperimetric inequality, then I don't see much possibility for McKean's inequality to lead to simplifications.</p> <p>Here is a possibly related fact: There exist <a href="http://mathoverflow.net/questions/56092/amenable-exponential-growth" rel="nofollow">amenable groups with exponential growth</a>. Non-amenability and exponential growth are two other properties that both look something like the properties discussed here, and it turns out that they are not equivalent.</p>