# Is the extension of $*$ homomorphism unique?

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 $$A$$?

• Interesting. I think a counterexample might lie in the case $A=K(H)$ and $\phi:K(H)\to B(H)$ is the inclusion. It is known that $M(K(H))$ is equal to $B(H)$, so to find a counterexample it would suffice to show that if $\phi:B(H)\to B(H)$ is a $*$-homomorphism which fixes $K(H)$, then $\phi$ must be the identity morphism. – Aweygan Oct 4 '18 at 0:54
• @Aweygan: no such thing exists. Any completely positive map that fixes $K(H)$, fixes $B(H)$; that's (a particular case of) Arveson's Boundary Theorem. It can also be seen via Hamana's notion of rigidity. – Martin Argerami Oct 4 '18 at 1:42
• @Martin Would you like to post this as an answer? – Aweygan Oct 4 '18 at 2:09
• I have actually just posted an actual answer (which is a bit easier since we are dealing with representations and not just cp maps). – Martin Argerami Oct 4 '18 at 2:35

If $$J\subset A$$ is an ideal, then any $$*$$-homomorphism $$\pi:J\to B(H)$$ extends uniquely to $$A$$. This answers the question in the negative, since $$A$$ is an (essential) ideal in $$M(A)$$.
To prove the above, let $$K=\overline{\pi(J)H}$$. Now, given $$a\in A$$, define $$\tilde\pi(a)$$ by (for $$b\in J$$, $$h\in H$$) $$\tilde\pi(a)\pi(b)h=\pi(ab)h,\ \ \ \ \ \ \tilde\pi(a)|_{K^\perp}=0.$$ 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 $$\rho|_A=\pi|_A$$. If $$k\in K^\perp$$, then for all $$b\in J$$, $$h\in H$$, \begin{align} \langle \rho(a)k,\pi(b)h\rangle&=\langle k,\rho(a^*)\pi(b)h\rangle = \langle k,\rho(a^*)\rho(b)h\rangle\\ \ \\ &=\langle k,\rho(a^*b)h\rangle=\langle k,\pi(a^*b)h\rangle=0. \end{align} So $$\rho(A)H\subset \overline{\pi(J)H}$$. And $$\rho(a)\pi(b)h=\rho(a)\rho(b)h=\rho(ab)h=\pi(ab)h=\tilde\pi(a)\pi(b)h.$$ Thus $$\rho(a)$$ and $$\tilde\pi(a)$$ agree on a dense subset of $$H$$ and are thus equal.
• Do you mean $\tilde{\pi}(a)|_{K^\perp}=0$? – mathrookie Oct 4 '18 at 5:31