# Computing the Fredholm index in Floer theory

In Salamon's notes on Floer homology, it's claimed that under some non-degenerancy assumptions the operator $$D:= \partial_s+J_0\partial_t+S(s,t): W^{1,p}(\mathbb{R}\times S^1,\mathbb{R}^{2n})\rightarrow L^p(\mathbb{R}\times S^1,\mathbb{R}^{2n})$$ is Fredholm for $$1, and moreover its Fredholm index is given by index $$D=\mu_{CZ}(\Psi^{+})-\mu_{CZ}(\Psi^{-}).$$

Now I was trying to see why this index formula holds up. In the lectures notes the author claims that this formula is true for $$p=2$$, where we are dealing with Hilbert spaces, by using the spectral flow of the family of operators $$J_0\partial_t+S(s,t)$$. However nothing else is said for the case where $$p\neq 2$$. It's clear that in the end this Fredholm index won't depend on the choice of $$p$$, but a priori we would need to prove that this is true so that we get the desired result just by proving the case $$p=2$$. Is this what is happening in this situation ? We can show that $$D$$ has the same Fredholm index for any $$p$$ without using the Conley-Zehnder index ?

Any insight is appreciated, thanks in advance.

The idea is to show that the kernel & cokernel consist of smooth sections, and thus are independent of $$p$$. Since the Fredholm index is the difference between the dimensions of these, it doesn't depend on $$p$$.
• So the fact that the kernel and Cokernel are smooth sections I think will follow from elliptic bootstrapping .By this I mean that every element in those sets will be of class $C$\infty}, I am not sure this is what you mean. But why does the fact that they consist of smooth sections imply that it is independent of $p$? Oct 11, 2021 at 8:51