Is the extension of $*$ homomorphism unique?

Question

If $\phi:A\to B(H)$ is a $*$ homomorphism, do there exist two different $*$ homomorphisms $\phi_1,\phi_2:M(A)\to B(H)$ which extend $\phi$, where $M(A)$ is the multiplier algebra of $A$?

Answer 1

The extension is unique if $\phi$ is non-degenerate (i.e., if $\phi(A)H$ is dense). Proof below. If $\phi$ is degenerate, the extension is not unique. For instance let $A=c_0(\mathbb N)$, and $\phi:A\to B(\ell^2(\mathbb N))\oplus\mathbb C$, with $\phi(x)=(x,0)$. Here $M(A)=\ell^\infty(\mathbb N)$. Let $\phi_1:M(A)\to B(\ell^2(\mathbb N))\oplus\mathbb C$ be $\phi_1(x)=(x,0)$; for $\phi_2$, fix a free ultrafilter $\omega$, and let $\phi_2(x)=(x,\lim_{n\to\omega}x(n))$

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The proof in the non-degenerate case:

If $J\subset A$ is an ideal, then any non-degenerate $*$-homomorphism $\pi:J\to B(H)$ extends uniquely to $A$. This answers the question in the negative, since $A$ is an (essential) ideal in $M(A)$.

Given $a\in A$, define $\tilde\pi(a)$ by (for $b\in J$, $h\in H$) $$\tag1 \tilde\pi(a)\pi(b)h=\pi(ab)h. $$ This is well-defined: if $\pi(b_1)h_1=\pi(b_2)h_2$, then taking an approximate unit $\{e_j\}$ for $J$, $$ \pi(ab_1)h_1=\lim_j\pi(ae_jb_1)h_1=\lim_j\pi(ae_j)\pi(b_1)h_1 =\lim_j\pi(ae_j)\pi(b_2)h_2=\pi(ab_2)h_2. $$ Also, \begin{align} \|\tilde\pi(a)\pi(b)h\|&=\|\pi(ab)h\|=\lim_j\|\pi(ae_j)\pi(b)h\|\leq\|a\|\,\|\pi(b)h\|. \end{align} This shows that $\tilde\pi(a)$ is bounded on $\pi(J)H$, and so we extend it uniquely to $\overline{\pi(J)H}$. With similar ideas, one proves that $\tilde\pi$ is a $*$-homomorphism.

For the uniqueness, assume that $\rho:M(A)\to B(H)$ is a representation such that $\rho|_A=\pi|_A$. Then $$ \rho(a)\pi(b)h=\rho(a)\rho(b)h=\rho(ab)h=\pi(ab)h=\tilde\pi(a)\pi(b)h. $$ Thus $\rho(a)$ and $\tilde\pi(a)$ agree on a dense subset of $H$ and are thus equal.