Why is the commutant of a unital $*$-representation a von Neumann algebra?

Question

If $\pi$ is a $*$-representation (i.e. a $*$-homomorphism) of a unital Banach$*$-algebra, then $\pi(\mathscr{A})$, is trivially self-adjoint, but why is $\pi(\mathscr{A})'$ a von Neumann algebra? This is used in the following proof that such a representation is irreducible iff $\pi(\mathscr{A})' = \mathbb{C}\mathbb{1}$: https://planetmath.org/criterionforabanachalgebrarepresentationtobeirreducible

Answer 1

One possible definition of a von-Neumann algebra is that it is the commutant of a $*$-closed set. By that definition, $\pi(\mathscr{A})'$ is von-Neumann.

Another possible definition is that it's a $*$-closed subalgebra of $B(H)$ that is topologically closed in one of several topologies. Say we use the weak topology.

Let $\mathscr{S}$ be a $*$-closed set. Let $\mathscr{M}=\mathscr{S}'$. For $T\in\mathscr{M}$ and $S\in\mathscr{S}$, $S^*$ is also in $\mathscr{S}$, so $TS^*=S^*T$. Adjointing this we get $ST^*=T^*S$. This proves $\mathscr{M}$ is $*$-closed.

If $T_1,T_2\in\mathscr{M}$, $\lambda\in\mathbb{C}$ and $S\in\mathscr{S}$. Since $T_1 S=S T_1$ and $T_2 S=S T_2$ then $\lambda T_1 S=S\lambda T_1$ and $S(T_1+T_2)=(T_1+T_2)S$ and $(T_1 T_2)S=T_1 S T_2=S T_1 T_2$. This proves $\mathscr{M}$ is closed under sum, scalar multiplication and multiplication so it's a $*$-subalgebra.

Finally, topologically, assume $T_\alpha$ is a net in $\mathscr{M}$ that converges weakly to some $T\in B(H)$. Then for all $u,v\in H$ and $S\in\mathscr{S}$, $\langle (T_\alpha S-S T_\alpha) u,v\rangle=0$. Now, $\langle T_\alpha Su, v\rangle\to_\alpha \langle TSu,v\rangle$ because $T_\alpha$ converges weakly to $T$. Similarly, $\langle ST_\alpha u,v\rangle=\langle T_\alpha u,S^*v\rangle\to_\alpha \langle Tu,S^* v\rangle=\langle STu,v\rangle$. It follows that $\langle (TS-ST)u,v\rangle=0$. This is for all $u,v\in H$ so $TS=ST$, which proves that $T\in\mathscr{M}$.

PS: $\mathscr{M}$ is unital because the identity commutes with all operators so it's in $\mathscr{M}$.

The proof for other topologies a von-Neumann algebra is closed in is similar except the topological part which has to be adjusted based on the convergence of nets in that topology.