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?
 A: 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.
A: 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 $B(H)$?
