We consider the Riemannian structure on the sphere $\mathbb{S}^n$ seen as a submanifold of $\mathbb{R}^{n+1}$ and the Laplace-Beltrami operator defined on $C^\infty(\mathbb{S}^n)$ by the equation
$\Delta f= -\operatorname{div}\operatorname{grad} f = -\frac{1}{\sqrt{g}}\frac{\partial}{\partial u^i}\left(\sqrt{g}g^{ij}\frac{\partial f}{\partial u^j}\right).$
We regard $C^{\infty}(\mathbb{S}^n)$ as a dense subspace of the Hilbert space $L^2(\mathbb{S}^n)$.
Question Is it true that $\Delta$ has compact resolvents, meaning that there exists $\lambda \in \mathbb{R}$ such that the closure of $\Delta-\lambda$ is invertible and its inverse operator is compact?
I think that we can easily work out the special case $n=1$: in this case the equation $\Delta u-\lambda u = v$ reduces to the standard Sturm-Liouville problem
$\begin{cases} -\frac{d^2}{dt^2}u-\lambda u = v & t\in (-\pi, \pi) \\ {}\\ u(-\pi)=u(\pi) \\ u'(-\pi)=u'(\pi)\end{cases}$
which admits Green's function for, say, $\lambda=-1$ (actually any $\lambda \notin \{0, 1, 4, 9 \ldots\}$ will do).
So the inverse of $-d^2/dt^2+1$ is an integral operator and in particular it is compact. I suspect that, similarly, the operator $\Delta_{\mathbb{S}^n}+1$ admits Green's function in any dimension $n$, but I am unable to prove (or disprove) this.
Thank you for reading.