Function composition on Toeplitz operator.

Question

I am working on Prop 2.3.3 of Higson and Roe Analytic K-theory recently, which says

The map $\alpha : g\mapsto \pi(T_g)$ is an injective *-homomorphism from $C(\mathbb{S}^1)$ to the Calkin algebra $\mathcal{Q}(H^2(\mathbb{S}^1))$.

where $\pi : \mathcal{B}(H)\to\mathcal{Q}(H)$ is the natural projection.

We already know that $\ker \alpha$ is an ideal of $C(\mathbb{S}^1)$, so it must be of the form $\{g\in C(\mathbb{S}^1) : g \ \text{vanishes on some closed}\ X\}$. Then the book says by symmetry $X$ must be rotationally invariant. I am quite confused about the claim.

This is what I have tried: Let $g'=g\circ\exp(\mathrm{i}\varepsilon)$ and it suffices to prove $T_{g'}-T_g$ is a compact operator. But I am stuck dealing with the Toeplitz operator $T_{g'}$ since I do not know how to deal with the function composition of the symbol.

Is there any method to deal with $T_{g'}$? Or some other ways to prove the rotationally invarience?

Any help will be grateful!

Answer 1

Let $\tau:z\mapsto z\exp(\mathrm{i}\varepsilon)$, $g'=g\circ\tau,$ and $U:H^2(\mathbb{S}^1)\to H^2(\mathbb{S}^1),\;f\mapsto f\circ\tau.$ Then, $$T_{g'}=U\circ T_g\circ U^{-1}$$ hence if $T_g$ is compact, so is $T_{g'}.$

Answer 2

In fact, a Toeplitz operator $T_f$ with symbol $f\in L^\infty(\mathbb{S}^1)$ is compact if and only if $f=0$ a.e. .

Perhaps what you should prove is the following:

If $T_g$ is compact, and $g'=g\circ \exp{(i\epsilon)}$ is a rotation of $g$, then $T_{g'}$ is compact.

This is trivial by observing that $$T_{g'}=C_{\epsilon} T_g C_{-\epsilon},$$ where $$(C_{\epsilon}f):=f\circ \exp(i\epsilon)$$ is the composition operator on $H^2(\mathbb{S^1})$.