On a compact Riemannian manifold $M$ (we assume Dirichlet boundary condition if $\partial M \neq \emptyset$), the Laplace-Beltrami operator $-\Delta$ has a discrete spectrum $0 < \lambda\_1 \leq \lambda\_2 < .....$ going to $\infty$. Consider the eigenspace corresponding to the first eigenvalue $\lambda\_1$. My question is, what is the dimension of this eigenspace?


It seems that the answer should be one, and the otherwise one can locate two nodal domains corresponding to one of the first eigenfunctions, and then $\lambda\_1$ is the first eigenvalue of each of these nodal domains, which should violate domain monotonicity. 


On the other hand, suppose we are given one (even non-sign-changing) eigenfunction $\varphi$ corresponding to $\lambda$. Suppose the manifold $M$ is such that it has a non-trivial isometry $T$. Then $T\varphi (x) := \varphi (Tx)$ is also an eigenfunction, but in general one does not expect that $T\varphi = \varphi$. 


What am I missing here? 



Simplicity of the first Laplace-Beltrami eigenvalue on Riemannian manifolds
Clearly one needs to assume that $M$ is connected (otherwise the first eigenvalue is not necessarily simple). Then indeed the first eigenvalue is simple, the corresponding eigenfunction has a constant sign and if $M$ admits an isometry, so does this eigenfunction. 


The proof is standard variational: the first eigenfunction delivers a minimum to
$\int |\nabla u|^2 \,dx $ under Dirichlet b.c. under assumption $\int u^2\,dx=1$. Then $v=|u|$ delivers the same minimum and thus is also an eigenfunction. But this cannot happen unless $v>0$ in $M\setminus \partial M$. Then $u$ must have a constant sign (we use that $M$ is connected), say $u>0$. 


Then, if $\lambda\_1$ is not simple, there is another eigenfunction $w$ which is orthogonal to $u$, but this is impossible since $w$ also has a constant sign.


