# Is finite dimensional representation of $C^*$ algebra surjective?

Suppose $A$ is a $C^*$ algebra,$B$ is a finite dimensional $C^*$ algebra,$\phi:A\rightarrow B$ is a nonzero $*$ homomorphism.

1.Can we deduce that $\phi$ is surjective

2.Does there exist a nonzero $*$ homomorphism $\psi:A\rightarrow C$ ,where $C$ is a $C^*$ subalgebra of $B$ containg the unit of $B$?

• Can $A$ be finite dimensional with dimension smaller than that of $B$? because then the answer to 1. is simply "no". – uniquesolution Aug 20 '18 at 16:42
• Look at the inclusion into the first summand $\mathbb C \hookrightarrow \mathbb C \oplus \mathbb C$. – user42761 Aug 20 '18 at 18:25
• If $A$ is inifinte dimensional ,what about the conclusion? – math112358 Aug 21 '18 at 3:51

1. We can deduce no such thing. Consider the map $\varphi:C([0,1])\to M_2(\mathbb C)$ given by $$\varphi(f)=\begin{pmatrix}f(0)&0\\0&f(1)\end{pmatrix}.$$ This is a $*$-homomorphism, from an infinite-dimensional $C^*$-algebra into a finite-dimensional $C^*$-algebra which is not surjective.
2. In general the answer is no. Consider the map $\psi:M_2(\mathbb C)\oplus M_2(\mathbb C)\to M_4(\mathbb C)$ given by $\psi(a,b)=\operatorname{diag}(a,b)$, and consider the unital $*$-subalgebra $\mathbb C^4\subset M_4(\mathbb C)$. If $\theta:M_2(\mathbb C)\oplus M_2(\mathbb C)\to \mathbb C^4$ is a non-zero $*$-homomorphism, then $\dim\ker\theta=0$ or $4$ (since $M_2(\mathbb C)$ is simple). Then $\dim(\operatorname{Im}(\theta))=4$, so $\theta$ is surjective. But $\mathbb C^4$ is abelian, and no quotient of $M_2(\mathbb C)\oplus M_2(\mathbb C)$ is abelian, a contradiction.
If you want an infinite-dimensional counterexample to 2, consider the same thing as above, but replace $M_2(\mathbb C)\oplus M_2(\mathbb C)$ by $M_2(\mathbb C)\oplus M_2(\mathbb C)\oplus A$, where $A$ is an infinite-dimensional simple $C^*$-algebra (such as $K(H)$ for a separable infinite-dimesional Hilbert space $H$).
• $M_2(\mathbb C)$ is simple,the only two ideals of $M_2(\mathbb C)$ are 0 and $M_2(\mathbb C)$ ,how to conclude that $\dim\ker\theta=0$ or 4? – math112358 Aug 21 '18 at 15:37
• The kernel of $\theta$ is an ideal of $M_2(\mathbb C)\oplus M_2(\mathbb C)$. Since $M_2(\mathbb C)$ is simple, the only ideals of $M_2(\mathbb C)\oplus M_2(\mathbb C)$ are $0$, $0\oplus M_2(\mathbb C)$, and $M_2(\mathbb C)\oplus 0$. – Aweygan Aug 21 '18 at 15:42