# Simple C*-algebras with finite representations are matrix algebras

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.

• There's a classification of finite-dimensional $C^*$-algebras as direct sums of matrix algebras. Does this answer your question? Mar 31, 2020 at 22:39
• @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. Mar 31, 2020 at 22:45
• Where is the proof using vN algebra ? I shall be grateful if you could show me the source of that proof @JustDroppedIn Apr 30, 2021 at 18:29
• @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:) May 1, 2021 at 11:01
• @Noobmathematician I have added this as an answer; I tried to include as many details as possible. May 1, 2021 at 11:28

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 $$\begin{bmatrix} a&b\\ c&d\end{bmatrix} \longmapsto \begin{bmatrix} a&0&b&0\\ 0&a&0&b\\ c&0&d&0\\0&c&0&d\end{bmatrix} .$$

• 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? Apr 1, 2020 at 9:56
• 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. Apr 1, 2020 at 14:00
• 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. Apr 4, 2020 at 20:33
• 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. Apr 4, 2020 at 21:51
• could you please make a reference for a book or notes where I can find the proof of that result? 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.

• Thank you for taking time to write this complete proof May 1, 2021 at 13:29
• 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? May 1, 2021 at 14:13
• @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. May 1, 2021 at 14:16