CW-complex, Hausdorff space, second-countable space, sober space
connected space, locally connected space, contractible space, locally contractible space
An operator topology is an abbreviation of a topology on a space of (continuous linear) operators between topological vector spaces over a fixed field $k$ of reals or complexes (possibly also p-adics, skewfield of quaternions etc.). In other words the hom-sets in the category of topological vector spaces as objects and continuous linear operators as morphisms are equipped with an operator topology.
There are many widely used topologies, some with standard names. Let $L(V,W) = Hom_{TVS}(V,W)$ be the set of continuous linear operators.
weak operator topology on $L(V,W)$ is given by the basis of open neighborhoods of zero given by sets of the form $U(x,f) = \{A\in L(V,W) : |f(A(x)) \lt 1 \}$ where $x\in V$ and $f\in W^* = Hom_{TVS}(W,k)$. A sequence $(A_n)$ converges to $A$ in weak operator topology iff the sequence $(A_n(x))$ converges to $A(x)$ in the weak topology on $W$. We write $A_n\stackrel{w}\longrightarrow A$ or $w-lim A_n = A$.
strong operator topology: the basis of neighborhoods of zero is given by sets $N(x,U) = \{A\in L(V,W) \,|\, A v \in U\}$, where $v\in V$ and $U$ is a neighborhood of zero in $W$. For convergence of sequences, we write $A_n\stackrel{s}\longrightarrow A$ or $s-lim A_n= A$.
uniform operator topology: here we assume that $V,W$ are normed spaces with norms $p_V$, $p_W$. Then $L(V,W)$ has a uniform operator topology induced by the norm given by the formula
The reason that in the definition of a unitary representation, the strong operator topology on $\mathcal{U}(\mathcal{H})$ is used and not the norm topology, is that only few homomorphisms turn out to be continuous in the norm topology.
Example: let $G$ be a compact Lie group and $L^2(G)$ be the Hilbert space of square integrable measurable functions with respect to its Haar measure. The right regular representation of $G$ on $L^2(G)$ is defined as
and this will generally not be continuous in the norm topology, but is always continuous in the strong topology.