For a von Neumann algebra, if the elements are all normal, then it is commutative.

Question

Let $\mathcal{A}$ be a von Neumann algebra, if the elements of $\mathcal{A}$ are all normal, then $\mathcal{A}$ is commutative.

I want to know how to prove this, or where I can find the proof. Thank you.

Answer 1

For an operator $T$, let $\Re(T)=(T+T^*)/2$ and $\Im(T)=(T-T^*)/(2i)$. $\Re(T),\Im(T)$ are self-adjoint and $T=\Re(T)+i\Im(T)$.

Verify that $T$ is normal if and only if $\Re(T),\Im(T)$ commute.

Let $\mathcal{A}$ be a von-Neumann algebra.

If all self-adjoints in $\mathcal{A}$ commute then by the real-imaginary decomposition above, all members of $\mathcal{A}$ commute.

Otherwise, there are a couple of self-adjoints $P,Q\in\mathcal{A}$ such that $P$ and $Q$ don't commute. Then $P+iQ$ is not normal.