Laplacian on Projective plane

Question

The Laplacian of unit sphere with standard metric is well studied. Now I am wondering what is spectrum of Laplacian on Projective space. Since projection map is locally isometry, I guess they have same Laplacian operator, so do they have same Laplacian spectrum? This also sounds weird to me. Thanks for your help.

Answer 1

Let $u$ be $\Delta u=\lambda u$ on the projective space $RP^n$, pull $u$ back to $S^n$ we get an eigenfunction $u^*$ with the same eigenvalue. One can find $\alpha>0$ with $\alpha(\alpha-n-1)=-\lambda$, so that $p(x)=r^\alpha u^*(\theta)$ is harmonic on ${\mathbb R}^{n+1}-\{0\}$, where $(r, \theta)$ is the polar coordinate on ${\mathbb R}^{n+1}$. One can check this using separation of variables. $p$ must be a homogeneous polynomial of degree $\alpha$ (first by removable singularity theorem we see $p$ is actually harmonic on the whole ${\mathbb R}^{n+1}$, then one can use Gilbarg-Trudinger Theorem 2.10 to show $p$ is a polynomial) which turns out to be an integer. Now by the construction of $u^*$ we see $p$ is even, i.e. $p(-x)=p(x)$. Thus $\alpha$ is an even integer.

So the spectrum of $RP^n$ is a part of the spectrum of $S^n$, missing those $\lambda$ that correspond to odd $\alpha$.