(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 $\tilde{\phi}:M(A)\to M(B)$.

A good reference is

E.C. Lance, Hilbert C∗-modules, London Mathematical Society Lecture Notes Series 210, Cambridge University Press, Cambridge, 1995.

  • $\begingroup$ 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. $\endgroup$ – Paul McKenney Feb 19 '13 at 15:59
  • 1
    $\begingroup$ Unital in which sense? For $\phi(1)=1$, $A$ has to be unital and then you have $M(A)=A$. $\endgroup$ – Uwe Franz Feb 20 '13 at 9:32
  • $\begingroup$ I think Paul means $\tilde\phi$ unital. $\endgroup$ – David Roberts Jun 20 '18 at 8:04
  • $\begingroup$ If $\phi$ is injective,can we deduce that $\tilde \phi$ is injective? $\endgroup$ – math112358 Nov 16 '18 at 8:51

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