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}


\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).


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)


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


\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}


\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


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