In this note, I will prove two Reilly type formula and show a few applications of it. Of course, these results may not be optimal and the applications I show merely give a taste of it.
Suppose is an orientable Riemannian manifold with smooth boundary . Let be a smooth function on and . Let (resp. ) be the connection on (resp. ) and be the Laplacian on (resp. ). Let be the unit outward normal on and . Let be the second fundamental form and be the mean curvature.
Theorem 1 [Reilly’s formula]
Proof: Let be a local orthonormal frame on such that on , . We denote by etc. Then by Ricci identity, (repeated index are summed from to unless otherwise stated)
We have and for ,
Substitute this into (1), we have
Corollary 2 (Choi-Wang) Let be a compact orientable Riemannian manifold such that its Ricci curvature for some and be an embedded compact orientable minimal hypersurface of , then the first eigenvalue of
Proof: From a result of Lawson [Theorem 2], divides into two components , such that .
Let be a first eigenfunction of , i.e. . Let be the solution of the Dirichlet problem
By Cauchy-Schwarz inequality, , thus by (1), as , we have
Note that the same inequality can be obtained by reversing the choice of (and thus and change sign), so we can assume that . So we have
Since is not constant, we have .
Remark 1 Actually the above proof shows that
It is a standard result that (p.265 Example 2)
Lemma 3 Suppose is a smooth function on and is a smooth function on , then there exists which solves the following Neumann boundary value problem
if and only if
Theorem 4 [A Reilly type formula] Suppose is a vector field on , let be the divergence of w.r.t. . Let and be the tangential component of along and let be the divergence of along . We have
Proof: The proof is very similar to that of Theorem 1. Let be a local orthonormal frame on such that on . We have
- In general, , thus . But in some special case, we can simplify the term . E.g. when is the gradient vector field of some function , then we have , thus .
- Actually, it is not hard to see that
Applying divergence theorem to this equation, we can prove the above result.
Corollary 5 For a Killing vector field (i.e. the Lie derivative ) on a compact orientable Riemannian manifold , we have
Proof: We have the formula . In fact,
From this we have and thus . We also have
In view of Theorem 4, we have
Remark 3 The above result can also be proved by checking that for a Killing vector field ,
By using this Reilly type formula, we recover a result of Bochner:
Corollary 6 (Bochner) If is a compact oriented manifold with , then every field is parallel (i.e. ). If, furthermore, at one point (i.e. is strictly negative definite at that point), then there are no nontrivial Killing vector fields.
Proof: The first statement follows directly from (3). The second statement is true because parallel vector field has constant length.
Let’s return to the application of Theorem 1.
Lemma 7 Let be an embedded compact hypersurface into the Euclidean space. Let be the region enclosed by (by generalized Jordan curve theorem) and be its volume. Let be the position vector on . With the notation as in Theorem 1, we have
Here is the tangential component of (regarding it as a vector at by parallel translation).
Proof: Define on . First of all, for the standard orthonormal frame
Let be an orthonormal frame on such that , then at ,
where we have used . Substitute this into (4),
By divergence theorem,
Corollary 8 (Garay) If is an embedded compact hypersurface which is convex in the sense that and its mean curvature satisfies
where is the radius of a sphere circumscribed around , then is a sphere of radius .
Proof: We can assume the circumscribed sphere is centered at . Then . Thus
On the other hand, by Lemma 7, since . We conclude that is constant and thus . This implies .
Corollary 9 For any hypersuface which is convex in the sense that , then
where is the radius of the smallest sphere which circumscribe . Equality holds if and only if is a sphere of radius .
Proof: We can assume is in the interior of , then by a result of Hadamard, as is convex, we have . So
By Minkowski’s formula (Hon Leung’s theorem), we have
Thus by Lemma 7, we have
If the equality holds, then as . Therefore .
Remark 4 Using Minkowski formula, we can also characterize when is a sphere as follows. We have the two formulas
If , then
However, if is convex, then if is enclosed by , we must have and thus is a sphere (by Aleksandrov’s theorem).
By the above remark, we conclude that
Theorem 10 If is a compact convex hypersurface and
then is a sphere.
Proof: Let , then the gradient of (as a function on ) on is easily seen to be the tangential component of , where are the standard basis of : . By the minimum principle, we have
Thus and so
Let be an orthnormal basis for , then Therefore .
Corollary 12 (Garay) For a convex hypersurface with , is a sphere.
Proof: By Lemma 11, .
If , then
As the equality holds, must be constant and thus is a sphere.