Suppose $D$ is the Dirac operator on a closed spin manifold $M$, with spinors $S$. One can take the functional calculus of $D$ with respect to the continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $x\mapsto\frac{x}{\sqrt{x^2+1}}$. Let's denote this operator by


$$f(D):L^2(S)\rightarrow L^2(S),$$


which is now bounded.




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**Part 1 of question:** I've seen it implicitly written that the index of $D$ is unchanged by this modification. Why is this true (perhaps from a functional calculus perspective)?


**Part 2 of question:** Is it possible to see that the index of $D$ is unchanged using the integral expression 


$$f(D)\psi = \frac{2}{\pi}\int\_0^\infty D(D^2+1+\lambda^2)^{-1}\psi\,\,d\lambda?$$


For instance, if $D$ is invertible as an operator $H^1\rightarrow L^2$ (where $H^1$ is the first Sobolev space), with inverse $D^{-1}:L^2\rightarrow H^1$, can we write down an inverse for $f(D):L^2\rightarrow L^2$ using an integral expression that involves $D^{-1}$?




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I should say that although I'm asking this question about Dirac operators in particular, it really applies to more general unbounded operators.


Thanks.



Why is index unchanged after applying functional calculus?
Perhaps the simplest answer is to use the spectral theorem: $L^2(S)$ decomposes as the orthogonal direct sum of $D$-eigenspaces, and $f(D)$ acts on each $\lambda$-eigenspace as multiplication by $f(\lambda)$. Since $f$ vanishes only at $0$, $f(D)$ has the same kernel as $D$. Moreover $f(D^\*) = f(D)^\*$ since $f$ is an odd function, so $f(D)$ has the same cokernel as $D$ as well.


Similar remarks hold for non-compact $M$ as well, with the caveat that you have to carefully specify what you mean by "index" - $D$ is not usually Fredholm in the traditional sense.


