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.


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 $\phi$):


  • $\begingroup$ Thank you s.harp! I hope you can also have a look at my post about applying functional calculus to $(T \pm iI )^{-1}$. $\endgroup$ – CL. Apr 15 at 14:29
  • $\begingroup$ 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. $\endgroup$ – CL. Apr 15 at 14:54

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