# Finite dimensional C*-algebra

Let $A$ be a simple and finite dimensional C*-algebra. We first note that $aAb\neq 0$ for every nonzero $a,b\in A$. Let $B$ be a maximal abelian self-adjoint subalgebra of $A$. Being finite dimensional, the spectrum of $B$ is finite, say $\{x_1,...,x_n\}$. For each $i$, $1\leq i\leq n$, let $e_i$ denote the projection of $B$ corresponding to the characteristic function of the one point set $\{x_i\}$. we have then the $\{e_i\}$ are orthogonal and $\sum_{i=1}^n e_i = 1$, and $B$ is isomorphic to $\Bbb C e_1 \oplus ... \oplus\Bbb C e_n$. It follows then that $e_iAe_i, 1\leq i\leq n$, commutes with every $e_j$, $1\leq j\leq n$. Since $B$ is maximal abelian, $e_iAe_i$ is contained in $B$ because the $\{e_j\}$ generate $B$, which means that $e_iAe_i = \Bbb C e_i$.

I could not see $e_iAe_i$ is abelian and it's containd in $B$. Please give me a hint. Thanks in advance.

## 1 Answer

Hint: Since $B$ is isomorphic to $\Bbb C e_1 \oplus \cdots \oplus\Bbb C e_n$ and $B$ is commutative all projections $e_j$ has to be of rank $1$. Hence $e_j A e_j$ is onedimensional.

• Supposed $dim(e_iAe_i) >1$, so there is orthogonal projections $p_1,p_2 \in e_iAe_i$(because $e_iAe_i$ is a von Neumann algebra). Also for $j\neq i$, $e_j\perp p_1$ and $e_j\perp p_2$. Put $C = C^*(p_1,P_2,e_2,...,e_n)$, which is abelian. Clearly $dim(C) >dim(B)$ and it's a contradiction. Is my process correct? – niki Feb 26 '15 at 8:13
• Yes, it seems to be correct. – Janko Bracic Feb 26 '15 at 9:08
• Thanks for your attention. – niki Feb 26 '15 at 9:55