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

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

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. Does anyone have 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?

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Answered at MO here. –  Paul McKenney Feb 19 '13 at 16:00

1 Answer 1

I am posting here an answer to the last question I had, since that did not come up at the MO question. (Readers should look there for the answers to the other questions, though.) It is essentially a slight modification of a comment by Farah in his book "Analytic Quotients." (See Example 3.2.3.)

Recall that $\ell^\infty$ is the multiplier algebra of $c_0$. I'll construct a $*$-homomorphism $\phi : \ell^\infty \to \ell^\infty$ which sends $c_0$ into $c_0$ but is not strictly continuous.

Let $\mathcal{U}_n$, $n\in\mathbb{N}$, be a sequence of nonprincipal ultrafilters. Consider the $*$-homomorphism $\phi : \ell^\infty \to \ell^\infty$ defined by

$$\phi(x)(n) = \lim_{k\to\mathcal{U}_n} x(k)$$

Since each $\mathcal{U}_n$ is nonprincipal, it follows that $\phi$ takes $c_0$ into $c_0$; indeed, $\phi$ takes $c_0$ to $0$! But it is obvious that $\phi$ is not strictly continuous; to see this, let $e_k\in\ell^\infty$ be the sequence which begins with $k$-many $1$'s and ends with $0$'s. Then $e_k$ converges in the strict topology to the sequence $e$ with constant value $1$. However, the images $\phi(e_k)$ are all just the constant sequence with value $0$, whereas $\phi(e) = e$.

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Thanks to Martin Sleziak for prompting me to add the answer here. –  Paul McKenney Mar 26 '13 at 14:38

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