4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\alpha$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.