Ultra weakly closed *-subalgebra of B(H)

Question

I'm currently working on a text about von Neumann algebras and the author used without further clarifying that any ultra weakly closed *-subalgebra of $B(H)$ contains a largest projection. Could someone tell me why?

Answer 1

Let $$ V := AH^{\perp} = \{x \in H : x\perp ay \quad\forall a\in A, y\in H\} $$ Let $W$ be the orthogonal complement of $V$, then $W$ is a non-degenerate representation of $A$. It is a consequence of the Double Commutant theorem that $$ \text{id}_W : W\to W $$ is an ultra-strong limit of elements of $A$. Hence, $A$ contains $\text{id}_W$, which can be identified with the orthogonal projection $e\in B(H)$ onto $W$. Furthermore, $$ A = Ae $$ So for any other projection $f\in A = Ae$, we must have $$ f = fe $$ and so $f\leq e$. Hence, $e$ is the largest projection in $A$.

Answer 2

Say $M\subset B(H)$ is your von Neumann algebra. Let $\mathcal P=\{p\in M:\ p \text{ is a projection }\}$. Let $K\subset H$ be the subspace generated by $\bigcup_{p\in\mathcal P}pH$, and write $p_K$ for the orthogonal projection onto $K$.

Then for any $q\in \mathcal P$, $p_Kq=q$. This implies that $q\leq p_K$, and also that $p_kq=qp_K$. For any projection $r\in M'$, $$ rpH=prH\subset pH, $$ so $rK\subset K$. This implies that $p_Krp_K=rp_K$, from where it follows that $rp_K=p_Kr$. So $p_K\in M''=M$.

The above is a sketch of the proof that in a von Neumann algebra arbitrary unions of projections are again in the algebra.