two commuting $*$ homomorphisms

Question

Suppose there is a nonzero non-trivial $*$ homomorphism $\pi_1:A\rightarrow B(H),\pi_1(A)\neq kId $,Can we construct another nonzero non-trivial $*$ homomorphism $\pi_2:A\rightarrow B(H)$ such that $\pi_1(a)\pi_2(b)=\pi_2(b)\pi_1(a),\forall a,b \in A$?

Answer 1

Not in general. For example, let $A=M_2(\mathbb{C})$, $H=\mathbb{C}^2$ and $\pi_1$ the identity on $M_2(\mathbb{C})$.

Since $M_2(\mathbb{C})$ is simple, any $\ast$-homorphism from $M_2(\mathbb{C})$ to itself is either zero or injective. But every injective linear map from $M_2(\mathbb{C})$ to itself is surjective, and there are of course pairs of matrices in $M_2(\mathbb{C})$ that do not commute. Thus every $\ast$-homorphism $\pi_2\colon M_2(\mathbb{C})\longrightarrow M_2(\mathbb{C})$ that commutes with $\pi_1$ must be zero.