Let me be precise about what I mean in the title. Let $A$ be a $C^*$-algebra, which we identify with its image of its universal representation $(\pi, H)$, so the second dual of $A$ is canonically identified with the von Neumann algebra $A''$ generated by $\pi(A) = A$. Denote $\widetilde{A}$ the minimal unitalization of $A$ in $A''$, and $M(A)$ the maximal one, i.e. $M(A) = \{x \in A'' \mid xA \cup Ax \subset A\}$, which is of course a copy of the multiplier algebra of $A$. It is known (see e.g. Pedersen's book, Theorem 3.12.9) that $M(A)_{\mathrm{sa}} = (\widetilde{A}_{\mathrm{sa}})_m \cap (\widetilde{A}_{\mathrm{sa}})^m$, where $(\widetilde{A}_{\mathrm{sa}})_m$ denotes the set of strong limits in $A''$ of bounded decreasing nets in $\widetilde{A}_{\mathrm{sa}}$, and $(\widetilde{A}_{\mathrm{sa}})^m$ the set of strong limits of bounded increasing nets in $\widetilde{A}_{\mathrm{sa}}$. Of course $M(A)_{\mathrm{sa}} \subset A''_{\mathrm{sa}}$ and the inclusion is strict in general. There are a variety of topologies on $M(A)$. Besides the ones that inherited from $B(H)$ as a subspace, one also often considers the strict topology on $M(A)$, which is the locally convex topology defined by the family of semi-norms $\|(\cdot)a\|$ and $\|a(\cdot)\|$ with $a$ running through $A$. It is also known that $M(A)$ is in fact the strict completion of $A$ as a locally convex space.

Now von Neumann's double commutation theorem says that the ultra-strong-$*$ closure of $A$ in $B(H)$ is $A''$. On $M(A)$, denote the ultra-strong-$*$ topology that is inherited from $H$ by $\sigma^*$, and the strict topology by $\beta$. If $\sigma^* = \beta$, then $M(A)$ is ultra-strongly-$*$ complete since it is strictly complete, in particular $M(A)$ is ultra-strongly-$*$ closed in $B(H)$, forcing $M(A) = A''$, but we know this is not always the case, although we do have $M(A) \subset A''$ by the above discussion, i.e. the $\beta$-closure of $A$ is always contained in the ultra-strong-$*$ closure of $A$. Since finer topology gives smaller closure, my question is, is it true that we always have $\beta$ finer than $\sigma^*$?

Here's a special case to get started. Note that if $A = K(H)$, then $A''=B(H)$. In this case, it is fairly easy to see that the strict topology on $M(A)$ is finer than the strong-$*$ topology on $M(A)$. Indeed, take any rank-one projection $p_\xi$ onto $\mathbb{C}\xi$ with $\xi$ being an arbitrary unit vector of $H$, one has, for any $x \in B(H) = M(K(H)) = K(H)''$, that $\|xp_\xi\| = \|xp_\xi x^*\|^{1/2} = \|x\xi\|$ and $\|x^*p_\xi\| = \|x^* p_\xi x\|^{1/2} = \|x^*\xi\|$. As $p_\xi \in K(H)$, and $\|(\cdot)\xi\|$ together with $\|(\cdot)^*\xi\|$ form a generating family of semi-norms of the strong-$*$ topology, we've shown that indeed the strict topology on $M(K(H)) = B(H)$ is finer than the strong-$*$ one. It is known that the strict topology, the strong-$*$ topology, the ultra-strong-$*$ topology and the Arens-Mackey topology on $B(H)$ all agree on bounded parts. So it is reasonable to compare the strict topology to other topologies that are finer than the strong-$*$ one but still agree with it on bounded parts. Two such topologies are already mentioned, the ultra-strong-$*$ topology and the Arens-Mackey topology. We know the dual of $B(H)$ equipped with the strict topology is exactly the predual $B(H)_*$ of $B(H)$, so the Arens-Mackey topology is finer than the strict one. The question remains about the comparison between the ultra-strong-$*$ topology and the strict topology, even in this special case where we do have $M(A)=A''$.


1 Answer 1


The answer is "yes".

Use the standard technique: if necessary, replace $H$ by $H\otimes\ell_2$ to ensure that for each $\omega\in B(H)_*$ (the predual of $B(H)$, the trace-class operators on $H$) there is $\xi\in H$ so that $\omega(y) = (y(\xi)|\xi)$ for each $y\in A''$. As $A$ acts non-degenerately on $H$ (because $A''$ must be unital) and as replacing $H$ by $H\otimes\ell_2$ does not change this, we may apply the Cohen–Hewitt factorization theorem. This allows us to find $a\in A, \xi'\in H$ with $a(\xi') = \xi$.

Now let $(x_i)$ be a net in $M(A)$ which converges strictly (that is, for the $\beta$ topology) to $x\in M(A)\subseteq A''$. This means that $x_i a \rightarrow xa$ and $a^*x_i\rightarrow a^*x$ in norm, so also $x_i^*a\rightarrow x^*a$, in norm. In particular, $$ \lim_i \omega((x_i-x)^*(x_i-x)) = \lim_i \|(x_i-x)\xi\|^2 = \lim_i \|(x_i-x)a(\xi')\|^2 = 0, $$ and similarly $\omega((x_i-x)(x_i-x)^*) \rightarrow 0$. As $\omega$ was arbitrary, this shows that $x_i\rightarrow x$ in the $\sigma$-strong$^*$-topology, that is, for the $\sigma^*$ topology.

In fact, we can avoid the "standard technique". Restriction of functionals defines a quotient map $B(H)_* \rightarrow M_*$ from the trace-class operators to the (unique) predual of our von Neumann algebra $M$. Then the ultra-strong$^*$-topology is given by the seminorms $$ M\ni x \mapsto ( \omega(x^*x) + \omega(xx^*) )^{1/2} $$ as $\omega$ varies through the positive part of $M_*$. In this way, we see that there is no dependence on $H$. (I guess there is an extra argument needed here: that a positive member of $M_*$ lifts to a positive trace-class operator).

In our case, $M=A''\cong A^{**}$ the bidual, as the representation of $A$ is universal. So $M_* \cong A^*$ the dual of $A$. For $\mu\in A^*$ and $a\in A$ define $a\cdot\mu\in A^*$ to be the functional $A\ni b\mapsto \mu(ba)$. There are various ways to show that the linear span of $\{ a\cdot\mu : a\in A, \mu\in A^* \}$ is norm-dense in $A^*$. (For example, polar decomposition of functionals and the GNS construction.) We can hence directly apply Cohen--Hewitt to $A^*$ to show that for each $\mu\in A^*$ there is $a\in A, \mu'\in A^*$ with $\mu = a\cdot\mu'$. Similarly we can regard $A^*$ as a right $A$-module, and then the same argument shows that we can find $a'\in A,\mu''\in A^*$ with $\mu' = \mu''\cdot a'$, so that $\mu = a\cdot\mu''\cdot a'$ (this being a bimodule, so the order of left/right actions doesn't matter).

Now the same argument works: if $(x_i)$ in $M(A)\subseteq A^{**}$ converges strictly to $x\in M(A)$ then $\|(x_i-x)a\| \rightarrow 0$ and $\|a'(x_i-x)^*\| = \|(x_i-x)a'^*\| \rightarrow 0$, and so $$ \mu((x_i-x)^*(x_i-x)) = \mu''(a'(x_i-x)^*(x_i-x)a) \rightarrow 0. $$ Thus we have converted the argument to not depend on $H$.

  • $\begingroup$ I see. Nice argument! To summarize just to make sure that I understand, the key point is the Cohen-Hewitt factorization theorem. Before we do the trick of replacing $H$ by $H \otimes \ell^2$, we can already prove that $\beta$ is finer than the strong-$*$ topology using the same argument, and the $\sigma^*$ case follows by this trick. $\endgroup$ Dec 22, 2021 at 0:38
  • $\begingroup$ Yes, that's right. In fact, I think we can make the argument a bit slicker: I'll add an extra bit to my answer in a second. $\endgroup$ Dec 22, 2021 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.