Let $A$ be a simple $C^*$-algebra. I am trying to prove that $A$ admits a non-zero finite dimensional representation if and only if $A\cong M_n(\mathbb{C})$ for some $n$.

The reverse implication is trivial. For the other one, if $\varphi:A\to B(\mathbb{C}^n)$ is a non-zero finite dimensional representation of $A$, then $\varphi$ is faithful, because $A$ is simple. Since $B(\mathbb{C}^n)\cong M_n(\mathbb{C})$, we have that $A$ is isomorphic to a simple $*$-subalgebra of $M_n(\mathbb{C})$. This is as far as I can go. Any ideas on how to go on?

P.S: I have seen a proof using vN algebras, but the thing is I came across this exercise in a book before the chapter on vN algebras, so I am trying to solve this without vN algebras (or irreducible representations).

Also: I know the classification theorem of finite dimensional $C^*$-algebras, but I can't use this. I want to prove this result in order to classify finite dimensional $C^*$-algebras.

  • $\begingroup$ There's a classification of finite-dimensional $C^*$-algebras as direct sums of matrix algebras. Does this answer your question? $\endgroup$
    – Aweygan
    Mar 31, 2020 at 22:39
  • $\begingroup$ @Aweygan I don't want to use the classification, this is part of my way to the proof of the classification actually. I will edit and specify it. $\endgroup$ Mar 31, 2020 at 22:45
  • $\begingroup$ Where is the proof using vN algebra ? I shall be grateful if you could show me the source of that proof @JustDroppedIn $\endgroup$ Apr 30, 2021 at 18:29
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    $\begingroup$ @Noobmathematician I will demonstrate it as an answer here; I think it exists somewhere in the book of Karen Strung, a draft of which can be found on her site. Anyway, I think it deserves to be on MSE so I'll post it, check the post again later:) $\endgroup$ May 1, 2021 at 11:01
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    $\begingroup$ @Noobmathematician I have added this as an answer; I tried to include as many details as possible. $\endgroup$ May 1, 2021 at 11:28

2 Answers 2


You have to assume $\varphi$ is non-degenerate (i.e., unital). Otherwise you need to restrict the codomain.

Once you have that $\varphi$ is unital, all you need is to consider a unital, simple, C$^*$-subalgebra of $M_n(\mathbb C)$; I will still call it $A$. Consider the center $Z(A)$ of $A$. This is a finite-dimensional, abelian, C$^*$-algebra. Use the Spectral Theorem or Functional Calculus to show that $A$ has a projection; then it has a minimal projection $p$. Now, because $p\in Z(A)$, the subalgebra $Ap$ is an ideal; as $A$ is simple, $p=I$. Thus $Z(A)=\mathbb CI$.

Now use again the Spectral Theorem or Functional Calculus to get a projection $p\in A$; and again since $\dim A<\infty$, there exists a minimal projection $p_1\in A$. If $p_1A(I-p_1)=0$, then for any $a\in A$ we have $$\tag1 p_1a=p_1ap_1+p_1a(I-p_1)=p_1ap_1. $$ If $a=a^*$, taking adjoints in $(1)$ then $p_1a=ap_1$. As selfadjoint elements span the whole algebra, we get that $p_1\in Z(A)$; this would imply $p_1=I$, which is only possible when $n=1$. It follows that $p_1A(I-p_1)\ne0$: that is, there exists $a\in A$ such that $p_1a(I-p_1)\ne0$. Let $vr=p_1a(I-p_1)$ be the polar decomposition.

Note that, as $p_1$ is minimal, the range of $p_1$ agrees with the range of $p_1a(I-p_1)$. Then $v^*v=p_1$. Define $p_2=vv^*$. Note that $v=p_1v(I-p_1)$, so $p_1p_2=v^*vvv^*=0$. Name $v=v_{1}$. Repeat the procedure, now on the algebra $(I-p_1)A(I-p_1)$, and starting with $p_2$, to obtain a minimal projection $p_3\in (I-p_1)A(I-p_1)$ with $p_3p_2=0$ and with a partial isometry $v_{2}$ such that $v_{2}^*v_{2}=p_2$, $v_{2}v_{2}^*=p_3$. As $A$ is finite-dimensional, the process finishes and we end up with pairwise orthogonal minimal projections $p_1,\ldots,p_k$, and partial isometries $v_{s}$, $s=1,\ldots,k-1$, such that $v_{s}^*v_{s}=p_s$, $v_{s}v_s^*=p_{s+1}$. Define $$ E_{rr}=p_r,\ \ E_{1r}=v_{r-1}v_{r-2}\cdots v_1. $$ Then $$ E_{1r}^*E_{1r}=p_1,\ \ \ E_{1r}E_{1r}^*=p_r. $$ Next define $$ E_{r1}=E_{1r}^*,\ \ \ E_{rs}=E_{r1}E_{1s}. $$ It is then easy to check that $$\tag2 E_{rs}E_{vw}=\delta_{sv}\,E_{rw},\ \ \ E_{rs}^*=E_{sr}. $$ It is now straightforward to check that the map $\phi:M_k(\mathbb C)\to A$, given by $[a_{rs}]\longmapsto \sum_{rs}a_{rs}E_{rs}$ is a $*$-isomorphism. Thus $A\simeq M_k(\mathbb C)$.

Note that you cannot expect $k=n$ in general. For instance, you can embed $M_2(\mathbb C)$ as a unital $*$ subalgebra of $M_4(\mathbb C)$ by $$ \begin{bmatrix} a&b\\ c&d\end{bmatrix} \longmapsto \begin{bmatrix} a&0&b&0\\ 0&a&0&b\\ c&0&d&0\\0&c&0&d\end{bmatrix} . $$

  • $\begingroup$ Hello, thank you for a thorough answer (I am still not accepting to see any other solutions). I have two questions though, first, how do you define minimal projections? And, second, do you think that this solution is something that a student can come up with naturally, with no sketch or hints? $\endgroup$ Apr 1, 2020 at 9:56
  • $\begingroup$ A projection is minimal if it has no proper subprojections. In a von Neumann algebra that's equivalent to $pAp=\mathbb Cp$. "Student" is a fairly broad term. Someone who has seen equivalence of projections before (or at least the proof that $M_n(\mathbb C)$ is simple), should not have much trouble with these ideas. $\endgroup$ Apr 1, 2020 at 14:00
  • $\begingroup$ Hello again, you are mentioning the proof that M_n is simple. Are you talking about the *-isomorphism with $B(\mathbb{C}^n)$ and using the fact that all operators on a finite dimensional space are compact (plus that any non-trivial norm-closed ideal contains the compact operators)? Personally this is the only proof I've seen. $\endgroup$ Apr 4, 2020 at 20:33
  • $\begingroup$ That's a lot more machinery than needed. For any ring $R$, the ring $M_n(R)$ is simple if and only if $R$ is simple. There is no need to talk about operators, nor do any analysis. The core of the proof is probably the same, though. $\endgroup$ Apr 4, 2020 at 21:51
  • $\begingroup$ could you please make a reference for a book or notes where I can find the proof of that result? $\endgroup$ Apr 4, 2020 at 21:54

As requested in the comments, I will demonstrate the proof of the result using von Neumann algebras.

So it is obvious that if $A\cong M_n$ then $A$ admits a (faithful) non-degenerate representation on a finite dimensional Hilbert space; just take the isomorphism $M_n(\mathbb{C})\cong\mathbb{B}(\mathbb{C}^n)$.

The converse is the interesting part; so we have a simple $C^*$-algebra $A$ and a non-zero representation $\varphi:A\to\mathbb{B}(H)$ on a finite dimensional Hilbert space $H$. Now say that $\dim(H)=n$, so $H\cong\mathbb{C}^n$ and thus $\mathbb{B}(H)\cong\mathbb{B}(\mathbb{C}^n)\cong M_n(\mathbb{C})$, so we have a $*$-homomorphism $\varphi:A\to M_n(\mathbb{C})$. Since $A$ is simple, the $*$-homomorphism is injective ($\ker(\varphi)$ is an ideal in $A$; $A$ is simple thus $\ker(\varphi)=0$ or $\ker(\varphi)=A$; $\varphi$ is non-zero, so $\ker(\varphi)=0$). This means that $A\cong\varphi(A)\subset M_n(\mathbb{C})$, so we conclude that $A$ is finite dimensional (and $\dim(A)\leq n^2$). Note that it would be "too much to ask" to try to show that $\varphi$ has to be surjective. Martin's answer describes this very nicely at the end.

The trick is the following: Since $A$ is finite-dimensional, we know by the existence of pure states that it admits an irreducible representation say $\psi:A\to\mathbb{B}(K)$. Note that the Hilbert space $K$ has to be finite dimensional: Indeed, since the rep. is irreducible it is cyclic, so we have a cyclic vector $\xi\in K$, i.e. $$K=\overline{\{\psi(a)\xi:a\in A\}}$$ but $A$ is finite dimensional, so $\{\psi(a)\xi:a\in A\}$ is finite dimensional (hence also closed) so $K$ is finite dimensional.

Now by von Neumann's double commutant theorem and the fact that the representation is non-degenerate (since it is irreducible) we have that $\psi(A)''=\overline{\psi(A)}^{SOT}$ (SOT=strong operator topology). But in finite dimensional spaces, the strong operator topology is the same as the norm topology and the image of a $C^*$-algebra through a $*$-homomorphism is always norm-closed, so $\psi(A)''=\psi(A)$. But the representation is irreducible, so $\psi(A)'=\mathbb{C}1_K$, so $\psi(A)''=\mathbb{B}(K)$, so $\psi(A)=\mathbb{B}(K)$.

But if $\dim(K)=k$, then $\mathbb{B}(K)\cong M_k(\mathbb{C})$, so $\psi:A\to M_k(\mathbb{C})$ is an injective (again, because $A$ is simple) and surjective $*$-homomorphism, thus a $*$-isomorphism.

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    $\begingroup$ Thank you for taking time to write this complete proof $\endgroup$ May 1, 2021 at 13:29
  • $\begingroup$ But I was thinking about one thing-- suppose to start with we had $\phi:A\to B(\mathbb C^n)$ an irreducible finite representation on the simple C*-algebra $A$ . Then can we say that $\phi(A)\cong B(\mathbb C^n)$. Right? $\endgroup$ May 1, 2021 at 14:13
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    $\begingroup$ @Noobmathematician yes, of course, this is what is proved in the middle paragraph. The non-trivial part is that $\varphi$ must be surjective, so we employed von Neumann's DCT and some other facts to prove this. $\endgroup$ May 1, 2021 at 14:16

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