# Functional Calculus and Fredholm index

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.

• The index of any selfadjoint operator is $0$. – Liviu Nicolaescu Jan 20 '15 at 20:49
• ah, you are right, now I see it. Do you know if there is a way to decompose the Laplacian on this sobolev space into operators that have an index 1? I have edited my question, cause this is essentially what I wanted to find out, just had the wrong idea. – Ethan Co Jan 20 '15 at 21:07
• @LiviuNicolaescu I just played around with a decomposition $T^*T= -\Delta$ and I think I just showed that then in any case, we would have $ind(T)=0$. Now this is extremely strange, as this would mean that for ANY self-adjoint operator, the Witten index would vanish. – Ethan Co Jan 20 '15 at 21:39
• I don't know what you meant by witten index. – Liviu Nicolaescu Jan 20 '15 at 22:24
• @LiviuNicolaescu actually, I still mean Fredholm index ( was talking physicists' language) – Ethan Co Jan 20 '15 at 22:44

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 $k\geq1$.