# What Cayley transformation does to a $*$-homomorphism

We let $$A$$ be a $$C^*$$ algebra. We consider a grading on $$A=C_0(\Bbb R)$$ by even and odd functions whilst a grading on $$M:=M_2(M_\infty(A))$$ by diagonal and off diagonal elements given by grading automorphism, $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} a & -b \\ -c & d \end{pmatrix}$$ Here $$M_\infty(A)$$ denote the infinite matrices over $$A$$.

We consider the cayley transformation of $$\Bbb R \rightarrow S^1, x \mapsto (x+i)(x-i)^{-1}$$.

Suppose we are given a graded $$*$$-homomoprhism $$C_0(\Bbb R) \rightarrow M$$ This induces a $$*$$-homomorphism $$C(S^1) \rightarrow M_+$$where we regard $$M_+ =M_2(M_\infty(A_+))$$, and the same grading on matrix algebras. Prove that the induced homomoprhism sends the generate $$j:S^1 \hookrightarrow \Bbb C$$, of $$C(S^1)$$ to $$u\in M_+$$ such that $$\alpha(u)=u^*$$

Firstly, I don't really see the shape of the elkement $$u$$ gets mapped as in $$M_+$$.

This is my simplification of a claim in Page 43, Proposition 3.17. line 9 into proof. I believe everything I wrote is self contained.

## 1 Answer

I assume with $$M_+$$ and $$A_+$$ you mean a unitisation of $$M, A$$. Specifically the unitisation $$M\oplus \mathbb1\mathbb C$$.

Give $$C(S^1)$$ a graded structure, where $$\alpha(f)[z]=f(\overline{z})$$. Now verify that the map $$\psi:C(S^1)\to C_0(\Bbb R)\oplus \Bbb1\Bbb C$$ is graded. What the map $$\psi$$ does is the following: $$\psi(f) = \left(x\mapsto f(\frac{x+i}{x-i})-f(1), \ f(1)\right).$$ The check that $$\psi$$ preserves the grading then reduces to the following calculation with $$f(1)=0$$:

$$\alpha(\psi(f))[x] = f\left(\frac{-x+i}{-x-i}\right)=f\left(\frac{x-i}{x+i}\right)=f\left(\overline{\frac{x+i}{x-i}}\right)=\psi(\alpha(f))[x].$$

Now $$\alpha(j)=j^{-1}$$ for the map $$j:S^1\to\Bbb C, z\mapsto z$$. Then (denoting the map $$C_0(\Bbb R)\oplus\Bbb1 \Bbb C\to M\oplus\Bbb1\Bbb C$$ with $$\phi$$):

$$\alpha(u)=\alpha(\phi(\psi(j)))=\phi(\psi(\alpha(j)))=\phi(\psi(j^{-1}))=\phi(\psi(j))^{-1}=u^{-1}.$$

• Thank you s.harp! I hope you can also have a look at my post about applying functional calculus to $(T \pm iI )^{-1}$. – CL. Apr 15 at 14:29
• Hi, s.harp, I have also post a follow up question relating to this question, I would be really grateful if you would have the time to comment. – CL. Apr 15 at 14:54