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 $A$?

  • 1
    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 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 at 1:42
  • @Martin Would you like to post this as an answer? – Aweygan Oct 4 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 at 2:35
up vote 3 down vote accepted

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.

  • Thanks for your time and patience,Pro Argerami – mathrookie Oct 4 at 5:26
  • Do you mean $\tilde{\pi}(a)|_{K^\perp}=0$? – mathrookie Oct 4 at 5:31
  • I saw your question, but I don't know the answer. – Martin Argerami Oct 4 at 11:49

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