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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 $A$?

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    $\begingroup$ 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. $\endgroup$ – Aweygan Oct 4 '18 at 0:54
  • $\begingroup$ @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. $\endgroup$ – Martin Argerami Oct 4 '18 at 1:42
  • $\begingroup$ @Martin Would you like to post this as an answer? $\endgroup$ – Aweygan Oct 4 '18 at 2:09
  • $\begingroup$ I have actually just posted an actual answer (which is a bit easier since we are dealing with representations and not just cp maps). $\endgroup$ – Martin Argerami Oct 4 '18 at 2:35
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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.

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  • $\begingroup$ I saw your question, but I don't know the answer. $\endgroup$ – Martin Argerami Oct 4 '18 at 11:49
  • $\begingroup$ Pro Argerami, I checked the proof,I can only conclude that $\rho(A)K^\perp\subset K^\perp$.How to deduce that $\rho(A)H\subset K$? $\endgroup$ – math112358 Apr 28 at 8:41
  • $\begingroup$ You are right. I realized that you have no uniqueness when the representation is degenerate. I have edited the answer. $\endgroup$ – Martin Argerami Apr 28 at 14:38
  • $\begingroup$ I have another doubt.Suppose $A$ is non-unital and we have a unital $*$homomorphism $\pi:M(A)\to B(H)$,then we can obtain a representation $\pi_A:A\to B(H)$ by restricting $\pi$ to A.Can we conclude that there is only one extension $\pi$ of M(A) which extends $\pi_A$? $\endgroup$ – math112358 Apr 28 at 16:34
  • $\begingroup$ 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 $c_0$. $\endgroup$ – Martin Argerami Apr 28 at 17:22

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