## Cheeger’s isoperimetric constant and the first eigenvalue of manifolds with a negative curvature upper bound

[Update: a simplified proof (9-10-2011), Rauch comparison proved (12-10-2011) ]

For a compact Riemannian manifold ${M}$ with boundary ${\partial M}$, if the metric is ${ds^2= g_{ij}dx^i dx^j}$, we define the Laplace operator

$\displaystyle \Delta = \frac{1}{\sqrt{\det g} } \sum _{i,j}\frac{\partial}{\partial x^i}( \sqrt {\det g }\;g^{ij} \frac{\partial }{\partial x^j}) \ \ \ \ \ (1)$

on ${C^\infty(M)}$ where ${(g^{ij})= (g_{ij})^{-1}}$ and ${\det g= \det{(g_{ij})}}$. We define ${W^{1,2}_0(M)}$ by the completion of ${C^\infty_0(M)}$ with the norm

$\displaystyle |\phi|^2=\int_M (\phi^2+|\nabla \phi|^2)$

and ${\Delta}$ can be extended as a self-adjoint non-positive definite operator from $W^{1,2}_0(M)$ to $L^2(M)$. Let ${0<\lambda_1<\lambda_2\leq \lambda_3\leq\cdots}$ be the (Dirichlet) eigenvalues of ${-\Delta }$, i.e. the eigenfunctions ${\phi_i}$ satisfy

$\displaystyle -\Delta \phi_i= \lambda_i\phi_i,\quad \phi_i |_{\partial M}=0.$

[Physically, the eigenvalues can be thought of as the fundamental modes (all possible frequencies) if you imagine $M$ as a drum. ] We are interested to find out the first eigenvalue ${\lambda_1}$, or at least estimate it (lower bound is often more important) in terms of the geometric conditions. First we give a definition.

 Definition 1 (Cheeger) For a compact Riemannian manifold ${M}$ with boundary ${\partial M\neq \emptyset}$, define the isoperimetric constant $\displaystyle h(M):= \inf\left\{\frac{\text{Area}(\partial \Omega)}{\text{Vol}(\Omega)}\mid \Omega\Subset M\right\}$

An important result is

 Theorem 2 (Cheeger) For a compact Riemannian manifold ${M}$ with boundary, $\displaystyle \lambda_1\geq \frac{1}{4} h(M)^2.$

Proof: For an eigenfunction ${f}$ with respect to ${\lambda_1}$, integrating $-f\Delta f=\lambda_1 f^2$ and applying divergence theorem, as $f|_{\partial M}=0$,

$\displaystyle \int_M |\nabla f|^2=\lambda_1 \int _M f^2.$

Now we claim that for any smooth function ${\phi}$ with ${\phi|_{\partial M}=0}$,

$\displaystyle \int_M |\nabla \phi|\geq h(M) \int_M|\phi|.$

If this is true, then taking ${\phi=f^2}$, we have

$\displaystyle \begin{array}{rcl} h(M) \int_M f^2 \leq \int_M |\nabla (f^2)|= 2 \int_M |f||\nabla f|\leq 2 (\int_M f^2)^{\frac{1}{2}}(\int_M |\nabla f|^2)^{\frac{1}{2}}, \end{array}$

thus

$\displaystyle \lambda_1 \int_M f^2=\int_M |\nabla f|^2 \geq \frac{h(M)^2}{4}\int_M f^2,$

which is what we want. It remains to prove our claim. We have to use the co-area formula: ${\int_M |\nabla \phi|= \int_{-\infty}^\infty \text{Area}(\phi=t)\,dt}$. Using this we have

$\displaystyle \begin{array}{rcl} \int_M |\nabla \phi|&=& \int_{-\infty}^\infty \text{Area}(\phi=t)dt\\ &=& \int_{-\infty}^\infty \frac{\text{Area}(\phi=t)}{\text{Vol}(\phi\geq t)}\cdot\text{Vol}(\phi\geq t) dt\\ &\geq& \inf_t \frac{\text{Area}(\phi=t)}{\text{Vol}(\phi\geq t)}\int_{-\infty}^\infty \text{Vol}(\phi\geq t) dt\\ &=& \inf_t \frac{\text{Area}(\phi=t)}{\text{Vol}(\phi\geq t)}\int_M |\phi|\\ &\geq& h(M)\int_M |\phi| \end{array}$

where we have used the layer-cake representation formula in the last equality and the definition of ${h(M)}$ is the last inequality. $\Box$

Let us apply the above theorem to estimate the first eigenvalue for a Riemannian manifold which have a negative upper bound for its sectional curvature.

 Definition 3 For a complete noncompact simply-connected Riemannian manifold ${M}$ we define $\displaystyle \lambda_1(M):= \lim _{r\rightarrow \infty } \lambda _1(B(x_0, r))$ where ${x_0\in M}$. Since the RHS of the above is non-increasing and is positive, the limit exists.

We now state a theorem of McKean.

 Theorem 4 If ${M}$ is a complete non-compact simply-connected Riemannian manifold $M^n$ with sectional curvature ${ K \leq -k < 0}$, then $\displaystyle \lambda_1(M)\geq \frac{(n-1)^2 k}{4}.$

 Remark 1 This estimate is sharp, in the sense that ${\lambda_1}$ for the hyperbolic space ${\mathbb{H}^n}$ (of curvature ${-1}$) is exactly ${\frac{(n-1)^2}{4}}$.

Since I don’t quite understand McKean’s original proof myself, I will give a modified proof here.
Proof: As ${M}$ is simply connected with negative curvature, by Cartan-Hadamard theorem, the distance function ${r(x)=d(x_0,x)}$ is differentiable on ${M\setminus \{x_0\}}$. So for ${\Omega \Subset M}$,

$\displaystyle \text{Area} (\partial \Omega) =\int_{\partial \Omega}1\geq \int_{\partial \Omega} \frac{\partial r}{\partial \eta}= \int_\Omega \Delta r$

where ${\eta}$ is the outer normal of ${\partial \Omega}$. We will show that

$\displaystyle \Delta r\geq (n-1)\sqrt{k}. \ \ \ \ \ (2)$

If this is true, then the result would follow immediately by Cheeger’s theorem.

 Lemma 5 ${\nabla ^2 r}$ is the second fundamental form ${h}$ of the geodesic sphere ${S(x_0,r)}$ (at the point ${(r,v)}$) centered at ${x_0}$ with radius ${r}$. In particular, ${\Delta r= H}$ is the mean curvature of ${S(x_0,r)}$.

Proof: For tangent vector fields ${X,Y}$ to ${S(x_0,r)}$, as ${\partial _r}$ is the unit outward normal,

$\displaystyle \nabla ^2r(X, Y)= X(Y(r))-(\nabla_X Y)(r)=-\langle \nabla _X Y,\partial_r\rangle=h(X,Y).$

$\Box$

We now show that the eigenvalues of ${h}$ are ${\geq \sqrt k}$, then by Lemma 5, (2) will follow immediately.To see this, we need the following

 Proposition 6 (Rauch’s comparison theorem) With the same assumptions as in Theorem 4, for any tangent vector ${X}$ to ${S(x_0,r)}$ at ${(r,v)}$, $\displaystyle h(X, X)\geq \sqrt{k} \coth (\sqrt k r)g(X,X).$

It then follows that all the eigenvalues of ${h}$ are bounded from below by ${\sqrt k}$.

Proof: We can assume the tangent vector ${|X|=1}$ at ${(r_0,v)}$. Let ${Y}$ be the Jacobi field on ${\exp_{x_0}(rv)}$ with ${Y(0)=0}$ and ${Y(r_0)=X}$. As $[Y,\partial_r]=0$, we have

$\displaystyle h_{(r,v)}(Y,Y)=\langle\nabla _Y \partial _r, Y\rangle=\langle \nabla_r Y,Y\rangle=\frac{1}{2}\partial_r (|Y|^2)= |Y||Y|'. \ \ \ \ \ (3)$

i.e.

$\displaystyle |Y|'= \langle Y',Y\rangle |Y|^{-1}.$

So

$\displaystyle \begin{array}{rcl} |Y|''&=& (\langle Y'',Y\rangle+|Y'|^2)|Y|^{-1}-|Y|^{-3}\langle Y',Y\rangle\\ &=& (|Y'|^2|Y|^2- \langle Y',Y\rangle^2- |Y|^2\langle R(Y,\partial_r)\partial_r,Y\rangle)|Y|^{-3}\\ &\geq& - |Y|^{-1}\langle R(Y,\partial_r)\partial_r,Y\rangle\\ &\geq& k|Y|. \end{array}$

Thus

$\displaystyle |Y|''-k|Y|\geq 0.$

Consider ${s_k(r)= \frac{|Y|'(0)}{\sqrt k}\sinh(\sqrt kr)}$, then we have ${s_k(0)=0=|Y|(0), s_k'(0)=|Y|'(0)}$ and

$\displaystyle s_k''-k s_k=0.$

Then we have

$\displaystyle (s_k |Y|'- s_k'|Y|)'\geq 0.$

As ${s_k |Y|'- s_k'|Y|=0}$ at ${r=0}$, we have

$\displaystyle s_k(r) |Y|'(r)\geq s_k'(r)|Y|(r), \quad i.e. \;|Y|'(r)\geq \sqrt k\coth (\sqrt kr)\geq \sqrt k.$

But then (3) implies ${ h(X, X)\geq \sqrt k}$. Our claim is proved. $\Box$

$\Box$