# Laplacian on Projective plane

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.

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 $$\alpha$$.