# Geometric conditions for isoperimetric, Sobolev, Poincar\'e inequalities on a riemannian manifold

By a theorem of Lichnerowicz, on a riemannian manifold $M^{(m)}$ with positive Ricci curvature, the reciprocal of Sobolev constant(ie. the first eigenvalue of laplacian) can be bounded from below by the quantity $inf |Ric|$. It has been generalised to bounding the Sobolev constant via $K=||Ric||_{L^p}$ by S.Gallot, wherein $p > \frac{m}{2}$. I assume that this holds even if the metric degenerates to a semi-positive metric on a subset of $M$ while keeping $K$ bounded. Theorem of Lichnerowicz does not, obviously, hold for degenerate metrics. My question is whether other geometric -sufficient- conditions are known that can be used to bound the Sobolev constant on manifolds with -possibly- degenerate metrics.

Edit: The Poincar\'e constant I consider is $C$ in the following: $||u - \bar{u}||_{2}$ $\leq C ||\nabla u||_{2}$.

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Your first statement is not correct! Think about a family of flat tori $S^1\times S^1$ with product metrics where you make the length of one of the two $S^1$ factors (or both!) go to zero. The Sobolev constant in this case blows up. By results of Croke, Li-Yau and others, you can bound the Sobolev and Poincare' constants uniformly if you have a uniform lower bound for the Ricci curvature, a lower bound for the total volume, and a diameter upper bound. – YangMills Jun 24 '12 at 0:33
Do you have a precise definition of "degenerate metric"? – YangMills Jun 27 '12 at 19:36
By a degenerate metric, I mean a metric that might be semi-definite at some points. (Let's assume that this set is a no wehere dense subset). What probably is important is the `rate' of degeneration. One concrete example might be the following: let $g$ be a positive definite metric and consider $u g$, obtained by a conformal change, wherein $u$ is merely non-negative. Under what conditions on $u$, or geometric properties of the metric $u g$ can we guarantee a bound on Poincar\'e constant? By the way, I think the definition of Sobolev constant in your example is the reciprocal of mine. – S.A.A Jun 27 '12 at 20:44
You haven't stated Gallot's theorem properly. – Deane Yang Sep 26 '12 at 0:42
@Deane Yang: Do you mean the fact that it requires the $L^p$ bound of the negative part of the Ricci curvature? – S.A.A Sep 26 '12 at 0:47

Thank you Dr. Rajala. One further question: do you know of an analogue of ricci bound from below in terms of bounding some integral of the quantity $-inf_{|x|=1} Ric(x,x)$? – S.A.A Jun 27 '12 at 2:56