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:

  1. Def 2.10, page 19. A Graded $*$-homomoprhism $\phi_D:\mathcal{S} \rightarrow \mathcal{K}(H)$ from spectral theorem.

  2. 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 $Index(D)=Str(P)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.