Let $\phi : M_k(\Bbb{C}) \to M_n(\Bbb{C})$ be a homomorphism of $C^*$-algebras.
We know that $\phi$ decomposes as a direct sum of irreducible representations, each of them equivalent to the identity representation (because $M_k$ has no non-trivial invariant subpspaces) with some null part.
So this means that $\phi(a)= u \begin{bmatrix}a&0&0&0&0...&0\\0&a&0&0&0...&0\\0&0&a&0&0...&0\\&&&...&&&\\\\\\0&0&0&0&0...&0\end{bmatrix} u^*$ where $u$ is some unitary.

I would like to understand how we get this unitary.
My approach is: decompose $\phi=\phi_0\oplus \phi_1\oplus \phi_2\oplus...\oplus\phi_j$
Let's try to understand the "null part": $H_0=\{\xi\in \Bbb{C}^n | \phi(a)\xi=0 \forall a\in M_k\}$ $\phi_0:M_k \to B(H_0)$ defined by $\phi_0(a)=\phi(a)|_{H_0}=0?$
Then for $1\leq t\leq j$ $\phi_t:M_k\to B(H_t)$ is irreducible, thus equivalent to the identity representation, i.e. $\exists u_t: \Bbb{C}^k \to H_k$ unitary s.t. $\phi_t(a)=u_tau_t^*$ for all $a\in M_k$.
Actually, we know (from the proof of this result) that $H_1,H_2,...$ are mutually orthogonal, and $H_0 \oplus H_1 \oplus... \oplus H_j = \Bbb{C}^n$, I think that if we denote $\dim H_0 =d$ then $kj+d=n?$
Now, from the $u_t$ and the null part, I want to understand how does $u$ defined?
Thank you for your time!


You have already done all the necessary work: $$ u=\begin{bmatrix}\begin{matrix}u_1&&&\\&\ddots&&\\&&u_j&\\ &&&I_d\end{matrix} \end{bmatrix}. $$

  • $\begingroup$ I thought that would be the u,but wanted to be sure. Thank you! $\endgroup$ – Shirly Geffen May 21 '16 at 6:42
  • $\begingroup$ You are welcome. You did all the work, actually. $\endgroup$ – Martin Argerami May 21 '16 at 6:44

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.