# When is a $*$-homomorphism between multiplier algebras strictly continuous?

(This question was posted on MSE here but didn't get any answers.)

The strict topology on the multiplier algebra M(A) of a C*-algebra A is that generated by the seminorms

$$x\mapsto \|ax\|\quad x\mapsto \|xa\| \qquad (a\in A, x\in M(A))$$

Whereas a ∗-homomorphism $\phi : M(A)\to M(B)$ between two multiplier algebras is necessarily norm-continuous, if I understand things correctly it will not always be continuous with respect to the strict topologies on either side. Where is there a good reference for this?

On the other hand an easily-proven theorem states that $\phi$ is strictly continuous if the image of $\phi$ contains B. This is not necessary, however; take $\phi : \mathcal{B}(\ell^2)\to\mathcal{B}(\ell^2)$ to be the map $x\mapsto sxs^*$ where $s$ is the unilateral shift. This is strictly continuous even though its image doesn't contain $\mathcal{K}(\ell^2)$. Are there other conditions which guarantee $\phi$ to be strictly continuous?

I'm particularly interested in the case where $\phi$ maps A into B, and both are nonunital. Is this enough to show that $\phi$ is strictly continuous?

Non-degenerate *-homorphism from $A$ (or $M(A)$) to $M(B)$ are strict (where non-degenerate means that $\phi(A)B$ is total in $B$). An important property of strict maps $\phi:A\to M(B)$ is that they possess a unique strict extension $\tilde{\phi}:M(A)\to M(B)$.
• Thanks, Uwe; upon looking at Lance's book, it seems that nondegeneracy is also a necessary condition for $\phi$ to be strict (assuming $\phi$ is unital). I'm still unsure of the answer for the third question, though. – Paul McKenney Feb 19 '13 at 15:59
• Unital in which sense? For $\phi(1)=1$, $A$ has to be unital and then you have $M(A)=A$. – Uwe Franz Feb 20 '13 at 9:32
• I think Paul means $\tilde\phi$ unital. – David Roberts Jun 20 '18 at 8:04
• If $\phi$ is injective,can we deduce that $\tilde \phi$ is injective? – mathrookie Nov 16 '18 at 8:51