Does there exist a non-commutative $C^*$-algebra which does not contain a completely isometric copy of $M_n$ for any $n>1$?

Question

It is clear that if a $C^*$-algebra contains a homomorphic copy of $M_n$ for some $n>1$, then it can not be commutative. I guess that having a copy of $M_n$ completely isometrically must be equivalent to having it via a $*$-homomorphism. Any comments on this would also be appreciated.

Essentially my question is, whether the non-commutativity of a $C^*$-algebra is due the presence of some $M_n$'s.

Answer 1

No. Every non-commutative C*-algebra contains a completely isometric copy of $M_2.$ Let $A$ be a non-commutative C*-algebra.

Suppose first that $A$ is a Type I C*-algebra. Since $A$ is not commutative it admits an irreducible representation $\pi$ that is not 1-dimensional. Since $A$ is Type I, $\pi(A)$ contains the compact operators (see for example, Theorem 6.8.7 in Pedersen's "C*-algebras and their automorphism groups").

Therefore $\pi(A)$ contains a (C*-algebraic) copy of $M_2$ (because $\pi$ is not 1-dimensional). Since $\pi(A)$ is a nuclear C*-algebra, by the Choi-Effros lifting theorem there is a ucp lift $\phi:\pi(A)\rightarrow A$ of $\pi$, i.e. $\pi\phi$ is the identity on $\pi(A).$ Hence $\phi$ is a complete isometry. Since there is a copy of $M_2$ inside of $\pi(A)$ we obtain a completely isometric copy of $M_2$ inside of $A.$

Suppose now that $A$ is not Type I. Then a result of Blackadar ("Nonnuclear Subalgebras of C*-algebras" JOT 1985) shows there is a subalgebra $B\subseteq A$ such that $B$ quotients onto the Cuntz algebra $\mathcal{O}_2.$ It is easy to see that $\mathcal{O}_2$ contains a C*-algebraic copy of $M_2$ (indeed span$\{ s_is_j^*:1\leq i,j\leq 2 \}\cong M_2$ where $s_i$ are standard generators of $\mathcal{O}_2$). Again since $\mathcal{O}_2$ is nuclear we obtain a completely isometric lift from $\mathcal{O}_2$ into $B\subseteq A$ by the Choi-Effros lifting theorem, and hence a completely isometric copy of $M_2$ inside of $A.$

Answer 2

• In finite dimension: every finite-dimensional C$^*$-algebra is a direct sum of full matrix algebras, so any non-commutative C$^*$-algebra of finite dimension contains a copy of $M_n(\mathbb C)$ for some $n$.

• In infinite dimension: there are plenty C$^*$-algebras which are projectionless, while $M_n(\mathbb C)$ obviously has projections. One notable example is $C_r^*(\mathbb F_2)$, the reduced C$^*$-algebra of the free group on two generators. Having no projections, it doesn't contain a copy of $M_n(\mathbb C)$ for any $n>1$.