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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!

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

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  • $\begingroup$ 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

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