Eigenvalues of the laplacian on a compact manifold without boundary Let $ M $ be a compact manifold WITHOUT boundary. It is clear that the first eigenvalue of the Laplace operator $ -\Delta $ is $ \lambda_0=0 $. Now we suppose that M has constant sectional curvature c. My question: is it true that $ \lambda_1= nc $ ($ \lambda_1 $ is the second eigenvalue of the laplacian)?
Thanks
 A: If $c>0$, then $\lambda_1=nc$ implies that $M$ is isometric to the standard round sphere of radius ${1}/\sqrt{c}$. This is known as Lichnerowicz-Obata's theorem. See Theroem 5.1 in Peter Li's "Lecture notes on geometric analysis".
A: The variational principle (e.g., here) says that 
$$\lambda_1(M) = \inf\left\{ \frac{\int_M |\nabla u|^2}{\int_M u^2} : \int_M u=0\right\} \tag1$$ 
If $M$ is a quotient of $S^n$ by group $G$ of isometries, we can pull the functions $u$ back to $S^n$ and conclude that 
$$\lambda_1(M) = \inf\left\{ \frac{\int_{S^n} |\nabla u|^2}{\int_{S^n} u^2} : \int_{S^n} u=0 , \ \forall g\in G \  u\circ g=u \right\} \tag2$$ 
Imposing the symmetry $u\circ g=u$ can increase the infimum. Indeed, consider $\mathbb R P^n = S^n/\{\pm I\}$. On the sphere, $\lambda_1$ is realized by degree $1$ spherical harmonics, i.e., linear functions. But they do not have the antipodal symmetry, and therefore do not descend to  $\mathbb R P^n$. 
More specifically, $\lambda_1(\mathbb R P^n) = \lambda_2(S^n)=2(n+1)$ because the symmetry condition $u(-x)=u(x)$ makes $u$ orthogonal to all degree $1$ spherical harmonics. The first eigenfunction on $\mathbb R P^n$ comes from spherical harmonics of degres $2$.
