I am having trouble understanding the following statement:
3.20 Proposition, pg44: Let $D$ be a symmetric, odd graded elliptic operator on a graded vector bundle $S$ over a compact manifold. The element $[\phi_D] \in K(\Bbb C) \cong \Bbb Z$ is the Fredholm index of the operator $D$.
What exactly are used in making sense of this statement:
Def 2.10, page 19. A Graded $*$-homomoprhism $\phi_D:\mathcal{S} \rightarrow \mathcal{K}(H)$ from spectral theorem.
Prop 3.17. There is an isomoprhism $$\Phi:K(A) \rightarrow [\mathcal{S}, A \otimes \mathcal{K}(H)]$$
What the above two means is explain in my other post.
The proof goes as follows:
We have a homotopy of $*$-homomoprhisms $\phi_{s^{-1}D}(f) = f(s^{-1}D)$. At $s=1$ we have $\phi_D$ at $s=0$ we have the homomorphism of $f \mapsto f(0)P$ where $P$ is projection onto the kernel of $D$.
How does one obtain continuity at $s=0$ - why is the resulting map as described?
This corresponds to the integer $\dim(\ker D \cap H_+) - \dim (\ker D \cap H_-)$.
How does one make this computation?
The operator $D$ here is special: If $D$ is elliptic on a compact manifold, it will have compact resolvant by Rellich's lemma: $(1+D^2)^-\frac{1}{2}$ is compact. Therefore the spectrum of $D$ are eigenvalues, they have no limit point except $\lambda=+\infty$. If $f(x)$ is a continuous function of spectrum of $D$, $f(D)$ is well-defined. This will contain something like $\delta_0(x)$ who is 1 when $x=0$ and is 0 elsewhere. This is not continuous on $\mathbb{R}$, but is continuous on the spectrum of $D$!
If $f$ vanishes at the infinity, $f(D)$ is bounded. The continuity of the operators $f(s^{-1}D)$ are considered under the norm topology. Since $$\lim_{s\to 0} f(s^{-1}x)\to \delta_0(x)f(0)$$ As continuous functions over the spectrum of $D$. Also $\delta_0(D)$ is the projection $P$ to the eigenspace for $\lambda=0$. We have the first limit of your question (1).
Also by the result of compact resolvant, the dimension of the eigenspaces are finite dimensional, Let $P_+ ,P_-$ be the projection on $\ker D\cap H_+$, $\ker D\cap H_-$ respectivly. They will be trace-class operators(as operators on $L^2(M)$. We have $$ \dim(\ker D\cap H_+)=Tr(P_+),\quad \dim(\ker D\cap H_-)=Tr(P_-)$$
We define a "supertrace" $Str(P):=Tr(P_+)-Tr(P_-)$.
We have $Index(D)=Str(P)$