## Definition

A bounded operator $T$ on a separable Hilbert space $\mathcal{H}$ is called *compact* if the set
$$
T(\mathcal{H}_1) = \{Tv \colon v\in\mathcal{H},||v||\leq 1\}
$$
is a relatively compact subset of $\mathcal{H}$. Bounded operators are continuous, and hence map compact sets to compact sets, thus the composition of two compact operators is another compact operator. Sums of compact subsets of $\mathcal{H}$ are aain compact, and linear combinations of compact operators are compact, so the compact operaotrs on $\mathcal{H}$ form a subalgebra $\mathcal{H}(\mathcal{H})$ of $B(\mathcal{H})$.

It turns out that $\mathcal{K}(\mathcal{H})$ is also closed under taking adjoints and closed in the norm topology.

## Automorphism group

Endow $B(\mathcal{H})$ with the strong operator topology, namely the topology in which a net $\{T_i\}$ converges to $T$ iff $||T_i v - Tv|| \to 0$ for all $v\in\mathcal{H}$.

Given a $C^\ast$-algebra $A$, consider the group $\operatorname{Aut}(A)$ of automorphisms. In order to make this group into a topology group, it needs to be endowed with the *point-norm topology*, the topology of pointwise convergence of functions on $A$, *i.e.*, $f_i \to f$ iff $||f_i(a)-f(a)||\to 0$ for all $a\in A$.

For $A = \mathcal{K}(\mathcal{H})$, there is a short exact sequence of topological groups. Let $i(z) = z\operatorname{Id}$ and $\operatorname{Ad}_U T = UTU^\ast$, then $$ 1\to U(1) \xrightarrow{i} U(\mathcal{H}) \xrightarrow{\operatorname{Ad}} \operatorname{Aut}(\mathcal{K}(\mathcal{H})) \to 1 $$ where $U(\mathcal{H})$ is endowed with the strong operator topology and $\operatorname{Aut}(\mathcal{K}(\mathcal{H}))$ with the point-norm topology.

The automorphism group $\operatorname{Aut}(\mathcal{K}(\mathcal{H}))$ is called the projective unitary group (of a Hilbert space) $PU(\mathcal{H})$.