Radial symmetry of the first eigenfunction

Question

Let $M$ be a simply connected space form (i.e. $\mathbb R^n$, sphere, or hyperbolic space) and $B$ be a ball in $M$. Let $\phi$ be the first Laplacian eigenfunction on $B$, with respect to the Dirichlet boundary condition $\phi=0$ on $\partial B$.

In Cheng's remarkable paper "Eigenvalue comparison theorems and its geometric applications", it is remarked that (p. 290) "since all simply connected space forms are two-point homogeneous, $\phi$ is a radial function". I think two-point homogeneity means that there always exists an isometry mapping one pair of points to any other pair of points which have the same distance apart (is it?). I don't understand how this condition is used.

I think the reason why $\phi$ is radial is that we can apply the spherical mean of it to obtain $\widetilde \phi$, which is also an eigenfunction as rotation preserves the metric. If $\widetilde \phi\ne \phi$, then $h=\phi-\widetilde \phi$ is a first eigenfunction that changes sign, which is a contradiction. But I don't see where the "two-point homogeneous condition" is used. (Is there an alternate proof?)

If my argument is correct it should also apply to balls centered at 0 of the warped product space $dr^2+f(r)^2 d\theta^2$. Is this well-known? If so, is there a quick reference?

Answer 1

For me it is a consequence of the Alexandrov moving plane method. In the euclidean case you have the following result due to Gidas, Ni and Nirenberg, see Theorem 2.34 of the book of Han and Lin, Elliptic Partial Differential equations,

Theorem: Suppose $u\in C^2(B)\cap C(\overline{B})$ is a positive solution of $$\Delta u =f(u) $$ with $$u=0 \hbox{ on } \partial B .$$ If $f$ is Lip then $u$ is radially symettic.

The proof rely on the moving plane method which consists in applying the maximum principle to some "reflected" solution. Hence, Cheng say that as soon as you can build such a re flexion between any pair of points the result is still true. Of course it works only for the first eigenvalue since, you need the positivity.

Answer 2

The fact that the first eigenfunction is radial follows from the uniqueness (up to scale) of the first eigenfunction, that the eigenvalue equation is invariant under isometries fixing $B$ and (1) the fact that the isotropy group at the orgin of the ball is transitive. Fact (1) is shown below.

Now if $\Delta_g u = \lambda_1 u$ with $\|u\|_{L^2(B)}=1$ and $\psi:B\to B$ is an isometry then $\tilde u (y) := u(\psi(y))$ satisfies the same equation and its $L^2$-norm is the same. Since the eigenspace is one-dimensional this means $u=\tilde u$. Because the isotropy group at the center $x$ of $B$ acts transitive on each $\partial B_s(x)\subset B=B_r(x)$ it holds $u(y)=u(z)$ for all $y,z\in B$ with $d_g(y,x)=d_g(z,x)$. This shows $u$ must be radial.

The whole argument does not require linearity of the equation only a form of invariance under scaling is needed (i.e. if $u$ is a solution so is $\alpha u$). In such a case it suffices to show that the Rayleigh quotient is invariant under isometries and that the space of solution is one-dimensional. An example where this argument applies is the first eigenfunction of the p-Laplace equation $\operatorname{div}(|\nabla|^{p-2}\nabla u) = \lambda_1 u $.

However, if the equation is linear then via averaging one can show that for each eigenvalue there is a radial solution.

(1) The fact that $B=B_r(x)$ is a ball in a two-point homogeneous space $(M,g)$ means that for all $y,z \in B$ with $d_g(y,x)=d_g(z,x)$ there is an isometry $\psi$ of $M$ with $\psi(x)=x$ and $\psi(y)=\psi(z)$. Since $x$ is fixed and $\psi$ is distance preserving it holds $\psi(B)=B$, i.e. $\psi$ fixes the ball.