Completions of $C(X)$ with respect to the topologies generated by states

I have no intuition in this field so excuse me if this is trivial.

Let $$X$$ be a compact Hausdorff space, and $$C(X)$$ the algebra of continuous functions on $$X$$ with the usual $$\sup$$-norm. This is a $$C^*$$-algebra, and each state $$\omega:C(X)\to{\mathbb C}$$ generates a GNS-representation $$\pi_\omega: C(X)\to B(H_\omega).$$ Let us consider the $$\sigma$$-strong* topology on $$B(H_\omega)$$, i.e. the topology defined in the Bratteli–Robinson book Operator Algebras and Quantum Statistical Mechanics 1 by the seminorms of the form $$A\mapsto \sqrt{\sum_{n=1}^\infty \lVert A\xi_n\rVert^2+\sum_{n=1}^\infty \lVert A^*\xi_n\rVert^2}$$ where $$\xi_n$$ are various sequences in $$H_\omega$$ with the property $$\sum_{n=1}^\infty \lVert\xi_n\rVert^2<\infty.$$ Let $$\tau$$ denote this $$\sigma$$-strong* topology on $$B(H_\omega)$$, and $$B_\tau(H_\omega)$$ denote this space with this topology. So we have a (continuous) mapping $$\pi_\omega: C(X)\to B_\tau(H_\omega).$$ Each state $$\omega:C(X)\to{\mathbb C}$$ generates such a mapping, and together these mappings generate a topology on the linear space $$C(X)$$, i.e. the initial topology on $$C(X)$$ generated by the family of mappings $$\pi_\omega$$. And since $$\tau$$ are locally convex topologies on $$B(H_\omega)$$, and $$\pi_\omega$$ are linear mappings, the corresponding initial topology on $$C(X)$$ is locally convex. My question is the following:

What is the completion of $$C(X)$$ with respect to the initial topology generated by the mappings $$\pi_\omega$$?

I have a suspicion that this must be the space $$U(X)$$ of universally integrable functions (see Leader - On Universally Integrable Functions), but up to now I did not manage to prove it. Is this true?

Actually, if we replace this $$\sigma$$-strong* topology with the other standard operator topologies on the spaces $$B(H_\omega)$$, the corresponding completions of $$C(K)$$ are also interesting for me. I would be grateful if somebody could cast a light on this.

• Completion of a topological space (as opposed to metric space) doesn't seem to be a well defined notion, can you please elaborate on what you mean here. Apr 18, 2022 at 16:43
• @ChristianRemling Christian, all these topologies are locally convex, so $C(X)$ becomes a locally convex space with this topology. I'll correct this. Apr 18, 2022 at 18:39
• Do you think the initial topology above is the same as the initial topology given only by the pure states? An affirmative answer would considerably simplify the problem for $C(X)$. Apr 19, 2022 at 23:34
• @OnurOktay I would think that these are different topologies. Apr 20, 2022 at 8:05
• @OnurOktay but I don't know... And what will the answer be if we rerplace all states by pure states? Apr 20, 2022 at 8:25

1 Answer

Every state on $$C(X)$$ is of the form $$\mu(g) = \int_Xg\ d\mu$$ for some positive measure $$\mu$$, $$\mu(X)=1$$. Thus, $$H_{\mu}=L^2(X,\mu)$$ and $$\pi_{\mu}:C(X)\to B(H_{\mu})$$ is the multiplication operator $$\pi_{\mu}(g)f = gf$$. If $$\sum_n\|\xi_n\|_{L^2_\mu}<\infty$$, then define $$f_n = \sum_{k=0}^n |\xi_k|^2$$ so $$(f_n)$$ is a Cauchy sequence in $$L^1(X,\mu)$$, and so converges to some $$f\in L^1(X,\mu)$$ with $$f\geq 0$$. Conversely, clearly every $$f\in L^1(X,\mu)$$, $$f\geq 0$$ is a limit of a series this form (take $$\xi_0=\sqrt{|f|}$$ and $$\xi_n = 0$$ otherwise). Thus, $$\sum_k\|\pi_\mu(g)\xi_k\|_{L^2_\mu}^2 = \lim_{n\to\infty}\int_X |g|^2f_n\ d\mu = \int_X |g|^2f\ d\mu$$ Consequently, every seminorm above on $$C(X)$$ is of the form $$|g|_{\nu} = \sqrt{\int_X |g|^2\ d\nu} = \|g\|_{L^2_\nu}$$ for some positive bounded measure $$\nu$$ on $$X$$. Hence, a net $$(g_i)_{i\in I}$$ is Cauchy in this topology if and only if converges to some $$G_{\nu}\in L^2(X,\nu)$$ for each $$\nu$$. This induces an element $$G$$ in the space $$\prod_{\nu\in M_b(X)} L^2(X,\nu)$$ with the product topology. The closure of $$C(X)$$ in this space is the completion that you're looking for.

• Onur, this is vague for me. Is it possible that this closure of $C(X)$ in $\prod L^2(X,\nu)$ is exactly the space $U(X)$ of universally integrable functions on $X$ as I conjectured in the question? Or perhaps there is another explicit description. Apr 21, 2022 at 17:49
• Onur, you should remove the limit in $\lim_{n\to\infty}\int_X |g|^2f\ d\mu$. Apr 21, 2022 at 18:06
• @SergeiAkbarov I do not know the answer to your question, but I can say this much: If $(g_n)$ is a bounded sequence in $C(X)$, then $(g_n)$ is Cauchy in the topology that you'd given iff $(g_n)$ converges pointwise. Thus, the space of Baire-1 functions $B_1(X)$ is contained in the completion of $C(X)$ in this topology. Apr 22, 2022 at 19:53