# Is the strict topology on the multiplier algebra of a $C^*$-algebra always finer than the ultrastrong-$*$ topology?

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 $$M(A)=A''$$.

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 $$H$$.

• 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. Dec 22, 2021 at 0:38
• 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. Dec 22, 2021 at 10:00