Let $P:C^{\infty}(V) \to C^{\infty}(V)$ be an elliptic self adjoint pseudo differential operator of order $d$. Here $V$ is a vector bundle with a metric over a compact oriented riemannian manifold $M$ and $C^{\infty}(V)$ represents the smooth sections of the vector bundle.
Now since $P$ is elliptic we get that $P:H_d(V) \to L^2(V)$ is a Fredholm operator. Here $H_d(V)$ represents the sobolev space. Now is the index of $P=0$ .
My Try: I checked the link here. He says that the answer is zero but does not give full justification for example like what is the domain being considered and so on. But I checked some examples like for $S^1$ with the operator $\partial$ and the answer was zero.
Edit: I think I have a proof of this fact. Here I do not think we need to worry about the domain issues of self adjointness. Here I will just assume $P$ is symmetric that is $<Pf,g>=<f,Pg>$ here $f,g \in C^{\infty}(V)$.
Proof: Since $<Pf,g>=<f,Pg>$ for $f,g \in C^{\infty}(V)$ and $C^{\infty}(V)$ is dense in $H_d(V)$ this identity also holds for $f,g \in H_d(V)$.
The inner product gives a bilinear non degenerate map $$B: L^2(V) \times L^2(V) \to \mathbb{C}$$ Now we restrict this map to $\textrm{Ker }P \times \textrm{Coker }P$. Here $\textrm{Coker }P$ is thought of as $\textrm{Range }P^{\perp}$(Since the operator is Fredholm and therefore the $\textrm{Range }P$ is closed and whole space breaks as a direct sum of $\textrm{Range }P$ and $\textrm{Range }P^{\perp}$). Now we are done if we show that this map is non degenerate.
Suppose $f \in \textrm{Ker }P$ then there exists a $g \in L^2(V)$ such that $<f,g> \neq 0$. Now we claim that $g \notin \textrm{Range }P^{\perp}$. Suppose not then there exists a function $h \in H_d(V)$ such that $Ph=g$. Now $<Ph,f>=<h,Pf>=0$ since $f$ is in the Kernel. A contradiction. So $g$ is a non trivial element of the Cokernel.
Now we take an element $f \in \textrm{Range }P^{\perp} $ then there exists a $g \in L^2(V)$ such that $<f,g> \neq 0$. Now since $H_d(V)$ is dense in $L^2(V)$ we can assume $g \in H_d(V)$ .
First we observe that $g$ can be chosen to be in $\textrm{Range }P^{\perp}$ because the other component after inner product with $f$ will become $0$. Now $<g,Ph>=0$ for all $h \in H_d(V)$. Therefore $<Pg,h>=0$ for all $h \in H_d(V)$. Therefore $Pg=0$. This proves that the map is non degenerate. Therefore the index is $0$.
So is this proof correct or am I overthinking it?. Thank you.