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I would like a source for some Artin-Wedderburn type facts about these algebras which seem to have easy proofs, and are probably written somewhere.

Let $\mathcal{A} \subset M_n(\mathbb{C})$ be an algebra which is closed under taking adjoints. (That is, $X \in \mathcal{A} \Rightarrow X^*\in \mathcal{A}$)

Then, $\mathcal{A}\cong \bigoplus_i M_{n_i}(\mathbb{C})$ for some finite sequence of integers $n_i.$ Furthermore if $\pi:\bigoplus_i M_{n_i}(\mathbb{C}) \rightarrow \mathcal{A}$ is a $*$-isomorphism, then there is a unitary $U$ and integers $m_i$ such that for every $ \bigoplus_i X_i\in\bigoplus_i M_{n_i}(\mathbb{C}),$ we can write the homomorphism via the formula $\pi(\bigoplus_i X_i) = U^* \bigoplus_i (X_i \otimes I_{m_i}) U.$

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These are corollaries of the more general classification of representations of C*-algebras of compact operators; the specific statements you give can be found e.g. in Davidson's "C*-algebras by Example" book, Theorems III.1.1 and III.1.2. They're almost certainly somewhere in Kadison and Ringrose but I don't have those in front of me. –  Mike Jury Jul 24 '13 at 19:48
    
Thanks, Mike! That's exactly what I was looking for! –  J. E. Pascoe Jul 24 '13 at 19:55
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In the opposite direction to @MikeJury's good suggestions: rather than deducing these from results about algebras of compact operators, I think one can find this in Farenick's book Algebras of Linear Transformations (uregina.ca/~farenick/alt_contents.html ) although I don't have a copy at hand to check. The issue is to check, tacitly or explicitly, that such an algebra is semisimple in the ring-theoretic sense, then as you note Artin-Wedderburn gives you the rest –  Yemon Choi Jul 24 '13 at 21:44
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I have Farenick's book, and it definitely constructs the theory of finite-dimensional operator algebras in complete detail; if I recall correctly, he even puts the real case on equal footing with the complex case. –  Branimir Ćaćić Jul 26 '13 at 14:02

2 Answers 2

up vote 6 down vote accepted

$\def\CC{\mathbb C}$Suppose $A$ is a $*$-closed subalgebra of $M_n(\CC)$. $\CC^n$ is a left $M_n(\CC)$-module in a canonical way, and therefore also an $A$-module. Using the fact that $A$ is $*$-closed it is easy to see that if $S\subseteq\CC^n$ is an $A$-submodule, then $S^\perp$ is also an $A$-submodule. This implies that $\CC^n$ is a semisimple $A$-module.

$\CC^n$ is then a semisimple faithful $A$-module: it follows at once that $A$ itself is a semisimple algebra (because the Jacobson radical acts by zero on every semisimple module). The decomposition you want then follows from the usual A-W theorem.

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(My algebras have $1$) –  Mariano Suárez-Alvarez Jul 25 '13 at 7:12
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For the 2nd question, you can start as follows: suppose $A$ is as above, and that $f$, $g:A\to M_n(\CC)$ are two $*$-injections. Then $\CC^n$ is an $A$-module, compatibly with the star, in two ways. Now $A$ has excatly one $n$-dimensional faithful module, so the two modules are isomorphic, and that gives us a matrix $U\in M_n(\CC)$ realizing the iso. &c. (I guess this is a $*$-version of Noether-Skolem's theorem...) –  Mariano Suárez-Alvarez Jul 25 '13 at 7:26
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This is a nice succinct explanation of the fact. I needed to use this to simplify some argument I was doing, and if someone asks why, this kind of logic/perspective will be useful. –  J. E. Pascoe Jul 26 '13 at 3:44

If your main concern is to have a reference to quote, then 1) is Theorem III.1.1 and 2) is essentially Corollary III.1.2 or Lemma III.2.1 in Davidson's book, as mentioned by Mike Jury.

This answer just aims at giving a direct (not using the larger compact case as in Davidson's book) self-contained and elementary operator algebraic approach. Hopefully this is correct...

Your first statement can be seen as the finite-dimensional case of the structure of a type I von Neumann algebra which somehow generalizes Artin-Wedderburn over $\mathbb{C}$. Your second statement is the root of Brattelli diagrams. Both can be proved by elementary manipulations of the projections of the algebra.

Note that if a finite-dimensional $*$-subalgebra $A$ of $B(H)$ does not contain $\mbox{Id}_H$, still the unital $*$-algebra $A\oplus \mathbb{C}\mbox{Id}_H$ is spanned by its projections by functional calculus (or just diagonalization of normal operators), so $A$ is also spanned by its projections (projections of $A\oplus \mathbb{C}\mbox{Id}_H$ are $p\oplus 0$ and $-p\oplus \mbox{Id}_H$ for projections $p$ of $A$). Whence the joint of all the projections of $A$ is a unit for $A$. So a finite-dimensional $*$-subalgebra of $B(H)$ is always a unital von Neumann algebra. Up to reducing $H$, we can assume that this unit is $\mbox{Id}_H$.

1) If $A$ is a finite-dimensional $*$-subalgebra of $B(H)$, then $A$ is unitarily equivalent to $\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$.

In short: a finite-dimensional von Neumann algebra must be of type $\rm{I}$, whence unitarily equivalent to $\sum_{\alpha}^\oplus A_\alpha \otimes B(H_\alpha)$ with $A_\alpha$ Abelian (Theorem I.27 in Takesaki I). In the finite-dimensional case, this gives the assertion.

Here is how the above goes in the finite-dimensional case directly. First, considering the minimal projections of the center $A\cap A'\simeq \mathbb{C}^r$ which decompose $A$ into blocks, we can restrict to the factor case $A\cap A'=\mathbb{C}1$. By compactness of the unit sphere of $A$, there are finitely many Murray-von Neumann equivalence classes of projections since two close enough projections ($\|p-q\|<1$) must be homotopic, whence equivalent. Therefore, there are minimal projections in particular. Since the order on projections is total in a factor, all the minimal projections are equivalent. Finally, take $p_1\oplus \ldots \oplus p_n$ a maximal orthogonal set of minimal projections. By maximality, we must have $p_1\oplus \ldots \oplus p_n=1$. Since these minimal projections must be pairwise equivalent, the partial isometries implementing these equivalences yield a unitary equivalence between $A$ and $M_n(\mathbb{C})\otimes p_1 Ap_1$. And by minimality of $p_1$, $p_1Ap_1\simeq \mathbb{C}$.

Decomposing the range into blocks, your second question follows from the following fact.

2) If $A=\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$, and if $\pi:A\longrightarrow B(H)$ is a $*$-homomorphism, then $\pi$ is unitarily equivalent to an orthogonal direct sum of possibly zero diagonal embeddings $x_j\longmapsto x_j\otimes 1_j$ where $x=(x_1,\ldots,x_r)\in A$.

Without loss of generality, we can assume that $\pi$ is unital, up to reducing $H$.

The key point is that a unital $*$-homomorphism $\pi:A\longrightarrow B(H)$ sends projections to projections (self-adjoint idempotents), and partial isometries to partial isometries.

First, if $A=\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$, then the corresponding orthogonal decomposition of the unit $1=\bigoplus p_j$ is sent to an orthogonal decomposition of the unit of $B(H)$. This allows us to restrict to the case $A=M_n(\mathbb{C})$, and to assume $\pi$ unital again.

Now the matrix units $e_{ij}$ with $1$ in $(ij)$-th position and $0$ elsewhere can be interpreted as partial isometries implementing the Murray-von Neumann equivalence of the projections $e_{ii}$ and $e_{jj}$. This is preserved under the unital $*$-homomorphism $\pi$. It follows that $\pi$ is, up to unitary equivalence, a diagonal embedding.

The case $A=M_2(\mathbb{C})$, with weaker assumptions, is used and explained with somewhat lengthy details in this MSE answer. It works the exact same way for $A=M_n(\mathbb{C})$.

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How do you know at the start that A is a direct sum of matrix algebras? –  Yemon Choi Jul 25 '13 at 6:38
    
@YemonChoi Ok, I've tried to address the first question as well in the same spirit. –  1015 Jul 25 '13 at 18:29
    
This still seems like overkill, given that the original condition makes no reference to norms –  Yemon Choi Jul 25 '13 at 21:46
    
@YemonChoi Note also that the title has "$C^*$-algebras". So I was trying to answer the OP's question from that perspective. Not to give the most elegant proof. –  1015 Jul 25 '13 at 22:17
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This proof is interesting. Luckily, I know enough about the theory vN algebra for the context of this proof to make sense. I understand the feeling that this is somehow the decomposition into factors combined with some manipulatorics. However, I'm certain there are a lot of different ways to approach the problem. I was certainly more interested in the location of known solutions so that I could use this for some simplification of some proof. –  J. E. Pascoe Jul 26 '13 at 4:14

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