Let $-\Delta: W^{2,2} \subset L^2(\mathbb{S}^2) \rightarrow L^2(\mathbb{S}^2)$. Then it is "easy" to show that $-\Delta $ is self-adjoint. Now, I am looking for closed operators $T$ and $T^*$ of order $1$ and $-1$ respectively such that we have $-\Delta = T^*T$. Notice, that the canonical square root by the functional calculus does not work, as this gives me again a self-adjoint operator of order 0.

Let $M$ be closed Riemannian manifold (e.g, $M={\mathbb S}^2$). Let $\{\varphi_k\}_{k\in\mathbb N_0}$ be an orthonormal basis for $L^2=L^2(M)$ consisting of eigenfunctions of $-\Delta$, where $-\,\Delta \varphi_k =\lambda_k^2\varphi_k$, $0=\lambda_0<\lambda_1\leq \lambda_2\leq \dots$ Then the operator $T\colon H^1\subset L^2 \to L^2$ defined by $$ T\varphi_k = \begin{cases} 0, & k=0, \\ \lambda_k\varphi_{k-1}, & k\geq1, \end{cases} $$ has the desired properties. Notice that $T^\ast \varphi_{k-1}=\lambda_k\varphi_k$ for $k\geq1$.

anyselfadjoint operator is $0$. $\endgroup$ – Liviu Nicolaescu Jan 20 '15 at 20:492more comments