# compact-open topology on $B(H)$

In topology, it is common to use the compact-open topology on the set of continuous maps between two given topological spaces.

Let now $H$ be a Hilbert space and $B(H)$ the set of continuous linear maps from $H$ to itself. In functional analysis, there are many topologies that people like to use on $B(H)$. Does any one of them agree with the compact-open topology? If yes, which one?

• Why not mention your attempts to answer this? In fact, there are two common topologies for $H$, giving 4 different compact-open topologies for $B(H)$. – Gerald Edgar Jan 12 '14 at 22:59
• My attempts did not go very far, mostly because I do not understand very well the (norm-)compact subsets of $H$. By the way, I was thinking of the norm topology on $H$. The question might also have some merit for the weak topology (and the norm-weak and weak-norm combinations). – André Henriques Jan 12 '14 at 23:44
• In any Banach space, the every norm compact set is contained in the closed convex hull of a sequence that converges in norm to zero. – Bill Johnson Jan 13 '14 at 0:06
• For any Banach space $X$, the topology on $B(X)$ of uniform convergence on compact subsets of $X$ (as analysts like to describe the topology) comes up in studying the approximation property (AP). A Banach space has the AP iff the identity operator is in the closure of the finite rank operators in this topology. – Bill Johnson Jan 13 '14 at 0:11
• If $X$ is reflexive with the AP, then the identity operator is in the closure of the finite rank operators that have norm at most one. Of course, on bounded subsets of $B(X)$, the strong topology-the topology of pointwise convergence-agrees with the topology of uniform convergence on compact sets. – Bill Johnson Jan 13 '14 at 0:14

It's easy to see that the compact-open topology agrees with the strong operator topology on norm-bounded subsets of $B(H)$. Bill Johnson mentioned this in a comment. I think this shows that of all the "usual" topologies on $B(H)$ the only candidates for agreeing with the compact-open topology are the strong and the ultrastrong toplogies.

However, I believe compact-open is strictly stronger than ultrastrong (which is itself strictly stronger than strong). To see this, fix an orthonormal basis $(e_n)$ of $H$ and consider the compact set $K = \{0\} \cup\{n^{-1/2}e_n: n = 1, 2, \ldots\} \subset H$. Then the set $U$ of all operators in $B(H)$ which take $K$ into the open unit ball of $H$ is open for the compact-open topology, but it is easily seen to not be strongly open --- given any operator $A$ in $U$ and any finite list of vectors $v_1, \ldots, v_k \in H$ you can easily find $B \in B(H)$ such that $Bv_i = Av_i$ for $1 \leq i \leq k$ but $\|Be_n\| > n^{1/2}$ for some sufficiently large $n$. (For large $n$ the vector $e_n$ is almost orthogonal to ${\rm span}(v_1, \ldots, v_k)$, so we have freedom in defining $Be_n$.)

It seems to me that a similar argument shows that the set $U$ is not even ultrastrongly open. Given any finitely many positive operators $C_i$ in the predual of $B(H)$, since their eigenvalues are square-summable, as $n$ goes to infinity we're going to have ${\rm max}_i \langle C_i e_n,e_n\rangle = o(n^{-1/2})$, so that again you can find $B$ which approximates the behavior of $A$ when tested against each $C_i$ but has $\|Be_n\| > n^{1/2}$ for some large $n$. That's just a sketch but I think the idea is sound.

• Are you saying that the compact-open topology is the topology of uniform convergence on sets of the form $a(H_1)$ for $a$ a compact operator, whereas the ultrastrong topology is the topology of uniform convergence on sets of the form $a(H_1)$ for $a$ a Hilbert-Schmidt operator (and that those two topologies disagree)? Here, $H_1$ is the unit ball of $H$. – André Henriques Jan 13 '14 at 14:07
• @André, that sounds plausible, let me think about it. – Nik Weaver Jan 13 '14 at 14:20
• Yes, that is true, Andre--every compact subset of $H$ is contain in the image of the unit ball of $H$ under a compact operator on $H$. More generally, any compact operator into $H$ factors through a compact operator on $H$. – Bill Johnson Jan 13 '14 at 15:13

As an introductory remark, there has been a considerable amount of work done on topologies on spaces of operators in Hilbert spaces, in particular with regard to their relevance to the theory of von Neumann algebras. In my opinion there are two essential criteria which one should apply: the topology should be complete and its dual should be a natural space of operators. The compact open topology fails the second one---its dual is the space of finite rank operators. (Incidentally, the most immediate natural candidate, the norm topology, also passes the first one but fails the second---in this case, the dusl is too large). Further candidates (which have already been mentioned)---the weak, strong, ultraweak and ultrastrong topologes---also fail this test. A suitable family of natural topologies which pass both tests with flying colours (whereby the dual spaces are the space of nuclear operators) was introduced 40 years ago in the Comptes Rendues paper "Topologies dans l'espace des operateurs sur les espaces de Hilbert" (1973). These are the finest locally convex topologies which agree with the topology of compact convergence on the unit ball. (There are four, since we can consider convergence with resepct to the weak or strong topologies and also their symmetric versions, i.e., ones for which the operation of taking adjoints is continuous). Some of their basic properties can be found in the above article.

This shows that the answer to your question is pretty well always no and goes on to answer the (implicit?) continuation---what is the "correct" topology on $B(H)$?

• The concept of Smith space (which is new to me) seems to coincide with the special case of a Saks space where the unit ball is compact. These are known to functional analysts as Waelbroeck spaces but Smith would appear to have priority. I will look into the article cited in your link. – 7891user Jan 13 '14 at 8:53
• Interesting. I always thought that the ultraweak topology is the "correct" topology on $B(H)$. Why do you say that it fails the test of having a good dual? Its dual is the space of trace class operator. Isn't that a natural space of operators? – André Henriques Jan 13 '14 at 14:01
• Do I understand correctly that you proposal for a topology better than the ultraweak topology is the "finest locally convex topology which agrees with the ultraweak topology on the unit ball of $B(H)$". Also, another issue with the ultraweak topology (and many other topologies) is that the multiplication $B(H)\times B(H)\to B(H)$ is not jointly continuous -- I don't know if that's fixable... – André Henriques Jan 13 '14 at 15:26
• @André. Multiplication is indeed not continuous for the topology you mention. This difficulty is intrinsic. It is also one of the reasons for considering several related, but not identical, topologies. Thus for the corresponding topology for the ultrastrong topology, multiplication is at least continuous on the unit ball. These two topologies, while distinct, have many properties in common (same duality, same bounded sets, same concepts of analyticity or measurability for operator-valued functions) and one gets by juggling between the two. – 7891user Jan 13 '14 at 22:01
• The finest topology for which multiplication is continuous would be the discrete topology, not very helpful. Anyway the difficulty with the continuity is intrinsic and has to be dealt with accordingly. There is no easy fix of this type available---well,that's my opinion for what it's worth. – 7891user Jan 14 '14 at 20:58