# 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

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 $$\phi_2(x)=(x,\lim_{n\to\omega}x(n))$$

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.

• Pro Argerami, I checked the proof,I can only conclude that $\rho(A)K^\perp\subset K^\perp$.How to deduce that $\rho(A)H\subset K$? – math112358 Apr 28 at 8:41
• You are right. I realized that you have no uniqueness when the representation is degenerate. I have edited the answer. – Martin Argerami Apr 28 at 14:38
• No. You can have $\pi|_A=0$; in fact, you can have different such representations. For instance with $c_0$ as above, you can take two different representations of $\ell^\infty$ that are $0$ on $c_0$. – Martin Argerami Apr 28 at 17:22
• Sorry for hijacking a fairly old thread, I just have some question about the proof above. Do we not require that $J$ be a closed ideal? Because otherwise it might be that $J$ is not a $C^*$-subalgebra of $A.$ Would the computation $\lim_{j}\|\pi(ae_j) \|\leq \|a \|$ still work even if $J$ is not closed? I don't think that $\pi$ will even be continuous if we do not force $J$ to be closed. – Kurome Aug 20 at 10:08
• In the C $^*$ world, it is standard that "ideal" means "closed, two-sided". I wouldn't know how to (try to) prove this otherwise. – Martin Argerami Aug 20 at 12:17