Irreducible representation Suppose $H$ is a separable Hilbert space, and $K$ is a Hilbert- schmidt space on $H$. We know $K$ is a Hilbert space. 
Consider representation $\pi : B(H) \to B(K)$ such that $\pi(a)x:= ax$. Proveing 
 $\pi$ is an irreducible representation.
For this purpose, I tried to show $\overline{\pi(B(H))}^{sot} = B(K)$, but I was not successful. 
Please just give me a hint. Thanks in advance.
 A: If you have a representation (of groups or operator algebras), then the following property is equivalent to irreducibility:
Let $\pi : G \to GL(V)$ be a representation, if the commutant $\pi(G)' := \{A \in GL(V): A\pi(g)=\pi(g)A\ \ \forall g \in G \}$ consists only of maps proportional to the identity, then the representation is irreducible.
You said you only want a hint and not the full proof, so I'll be a bit vague now and rewrite the reply when you give me the go-ahead.
So what you have to show is that if for an $f \in B(K)$ you have $f(Ax) = Af(x)$ $\forall A \in B(H), \forall x \in K$ it follows that $f$ is proportional to the identity. You might note that span of projections onto 1 dimensional subspaces of $H$ are a dense subset of $K$. So if you can show that the restriction of $f$ onto this subset must be proportional to identity, $f$ itself must be proportional to identity.
A: This can  be verified just by definition of irreducible representation.
Supoose $L\neq 0$ is a invariant closed subspace of $K$ for $\pi(B(H))$, i.e., $$ax\in L,\forall a\in B(H)\mbox{ and } x\in L.$$ It is equivalent to  $L$ is a non-zero left ideal of $B(H)$. Thus the space $F(H)$ of all finite-rank operators on $H$ is contained in $L$.  Note that $F(H)$ is dense in $K$. 
