Let $\theta:M_3(\mathbb{C}) \to M_7(\mathbb{C})$ be a $*$-homomorphism. Since matrix algebras over a field are simple, we know that $\theta$ must be injective. After reading this I conclude that, up to unitary equivalence,

$$ \theta(A) = \left[\begin{array}{} A & 0_3 & 0 \\ 0_3 & A & 0 \\ 0 & 0 & 0 \\ \end{array}\right] $$

where $0_3$ denotes the $3 \times 3$ zero matrix. My question is the following. We have that $\theta':M_3(\mathbb{C}) \to M_7(\mathbb{C})$ defined by

$$ \theta'(A) = \left[\begin{array}{} A & 0_3 & 0 \\ 0_3 & 0_3 & 0 \\ 0 & 0 & 0 \\ \end{array}\right] $$

is also a $*$-homomorphism. So, how come $\theta(A)$ is unitary equivalent to $\theta'(A)$? In general, why is every possible $*$-homomorphism unitary equivalent to $\theta$? That is, how do you determine the unitary matrix $U \in M_7(\mathbb{C})$ such that $\theta(A)=U^*\theta'(A)U$?

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    $\begingroup$ No, the problem is fine, for unitary equivalence means equality up to composition with an inner automorphism of the codomain, so that in this particular case, $U \in M_7(\mathbb{C})$. $\endgroup$ Mar 17, 2013 at 19:17
  • $\begingroup$ @BranimirĆaćić: Exactly, I was about to point that out (I'll edit my question accordingly). Do you have any insight in this one? I'd really appreciate it since I haven't managed to understand it. $\endgroup$
    – ragrigg
    Mar 17, 2013 at 19:20
  • $\begingroup$ So, there is a problem here, in that Greg Kuperberg's answer actually allows for both of the possibilities you demonstrate. In general, if $n = qm + r$ for $0 \leq r < m$, then up to unitary equivalence, $M_m(\mathbb{C}) \to M_n(\mathbb{C})$ will be given by $A \mapsto (A \otimes I_k) \oplus O_{n-km}$ for $1 \leq k \leq q$, giving rise to both possibilities for $7 = 2 \cdot 3 + 1$. In particular, all these possibilities are manifestly unitarily inequivalent, as Martin points out below. $\endgroup$ Mar 17, 2013 at 19:25
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    $\begingroup$ If $A$ has full rank in $M_3$ then $\theta(A)$ has rank $6$ and $\theta'(A)$ has rank $3$ so $\theta(A)$ and $\theta'(A)$ cannot be unitarily equivalent. Why do you think the two representations should be unitarily equivalent? $\endgroup$
    – Martin
    Mar 17, 2013 at 19:30
  • $\begingroup$ @BranimirĆaćić: Let me now rephrase my question using your last comment. How do you go about proving that you only have those two possibilities? How do you show the existence of a unit matrix (inner automorphism) that will send any other possibility back to either $\theta(A)$ or $\theta'(A)$. I apologize if this is trivial, but I simply don't see it. $\endgroup$
    – ragrigg
    Mar 17, 2013 at 19:44

1 Answer 1


Complete rewrite

Let $\theta : M_m(\mathbb{C}) \to M_n(\mathbb{C})$ be a non-zero $\ast$-homomorphism, let $P = \theta(1_m)$, and let $V = P\mathbb{C}^n$. Then $P$ is the unit of $\theta(M_m(\mathbb{C}))$ as a $C^\ast$-algebra, and in particular, a central orthogonal projection in $\theta(M_m(\mathbb{C}) \subseteq M_n(\mathbb{C})$, so that for any $A \in M_m(\mathbb{C})$, $\theta(A) = P\theta(A)P$, and hence $$ M_m(\mathbb{C}) \to B(V), \quad A \mapsto \theta(A)|_V, $$ is a unital $\ast$-homomorphism. Thus, if $U_0$ is the change of coordinates matrix from the standard orthonormal basis on $\mathbb{C}^n$ to the extension of an orthonormal basis on $V$ to an orthonormal basis on $\mathbb{C}^n$, then $$ U_0 \theta(A) U_0^\ast = \theta_0(A) \oplus 0_{n-s} $$ for a unital $\ast$-homomorphism $\theta_0 : M_m(\mathbb{C}) \to M_s(\mathbb{C})$, where $$ s := \dim V = \operatorname{rank}\theta(1_m). $$

Now, $\theta_0$ defines a unital $\ast$-representation of $M_m(\mathbb{C})$ on $\mathbb{C}^s$. However, $M_m(\mathbb{C})$ is simple with unique (up to unitary equivalence) unital $\ast$-representation given by left matrix multiplication on $\mathbb{C}^m$. Hence, abstractly, there exists a non-zero integer $k$ and a unitary $S : \mathbb{C}^m \to (\mathbb{C}^m)^{\oplus k}$ such that $$ S \theta_0(A) S^\ast = A^{\oplus k}; $$ in particular, $s = km$, so that since $s = \operatorname{rank} \theta(1_m) \leq n$, $1 \leq k \leq q$, where $n = qm+r$ for the quotient $q$ and remainder $r$. In other words, by this unitary isomorphism, there exists an orthonormal basis of $\mathbb{C}^s = \mathbb{C}^{mk}$ such that if $U_1$ is the change of coordinates matrix from the standard orthonormal basis of $\mathbb{C}^{mk}$ to this new basis, then $$ U_1 \theta_0(A) U_1^\ast = A^{\oplus k}. $$

Putting everything together, if $U := (U_1 \oplus 1_{n-km}) U_0$, then $U \in M_n(\mathbb{C})$ is a unitary such that $$ U \theta(A) U^\ast = (A^{\oplus k}) \oplus 0_{n-km}, \; \forall A \in M_m(\mathbb{C}). $$ Note that by construction, $$ k = \frac{\operatorname{rank}\theta(1_m)}{m} $$ is a complete unitary equivalence invariant of $\ast$-homomorphisms $\theta : M_m(\mathbb{C}) \to M_n(\mathbb{C})$. Indeed, this is precisely the label on the single edge of the Bratteli diagram of such a $\ast$-homomorphism between matrix algebras. Explicitly, let $\theta_1, \theta_2 : M_m(\mathbb{C}) \to M_n(\mathbb{C})$ be $\ast$-homomorphisms, and for $i=1,2$, let $$ k_i = \frac{\operatorname{rank}\theta_i(1_m)}{m} $$ and let $U_i \in M_n(\mathbb{C})$ be the unitary such that $$ U_i \theta_i(A) U_i^\ast = A^{\oplus k_i} \oplus 0_{n - k_i m}. $$ If $k_1 = k_2 =: k$, then $$ U_1 \theta_1(A) U_1^\ast = A^{\oplus k} \oplus 0_{n-km} = U_2 \theta_2(A) U_2^\ast, $$ so that $$ \theta_2(A) = U \theta_1(A) U^\ast, \; \forall A \in M_m(\mathbb{C}), \quad U := U_2^\ast U_1 \in U(n), $$ as required. If, instead, $k_1 \neq k_2$, then $\operatorname{rank}\theta_1(1_m) \neq \operatorname{rank}\theta_2(1_m)$, and hence, as Martin pointed out in his comment, $\theta_1(1_m)$ and $\theta_2(1_m)$ cannot possibly be similar, ruling out unitary equivalence of $\theta_1$ and $\theta_2$.

Finally, in the case that you are considering of $\ast$-homomorphisms $\theta : M_3(\mathbb{C}) \to M_7(\mathbb{C})$, since $7 = 2 \cdot 3 + 1$, the unitary equivalence classes of non-trivial such $\ast$-homomorphisms are given by:

  1. the case where $k=1$, so that there exists some unitary $U \in M_7(\mathbb{C})$ such that $$ U \theta(A) U^\ast = \begin{pmatrix} A & 0_{3 \times 4} \\ 0_{4 \times 3} & 0_{4}\end{pmatrix}; $$
  2. the case where $k=2$, so that there exists some unitary $U \in M_7(\mathbb{C})$ such that $$ U \theta(A) U^\ast = \begin{pmatrix} A & 0_3 & 0_{3 \times 1} \\ 0_3 & A & 0_{3 \times 1} \\ 0_{1 \times 3} & 0_{1 \times 3} & 0_1 \end{pmatrix}. $$

In either case, the matrix $U$ is determined, as outlined above, by $P := \theta(1_m)$, and specifically by the unitary $S : P\mathbb{C}^n \to (\mathbb{C}^m)^{\oplus k}$ realising the induced unital $\ast$-representation of $M_m(\mathbb{C})$ on $P\mathbb{C}^n$ as a direct sum of irreducible representations, each of which is necessarily unitarily equivalent to the standard representation on $\mathbb{C}^m$.


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