1
$\begingroup$

A von Neumann algebra is defined as a *-algebra which is closed w.r.t. the strong operator topology and a $C^*$-algebra as *-algebra with the property $||A||^2=||A^*A||$.

It was said that every von Neumann algebra is a $C^*$-algebra. I tried to show that every von Neumann algebra fulfils the $C^*$ property. I can not get any further and don't know how to use that the von Neumann algebra is closed. Do you have tips?

Thanks for your help.

$\endgroup$
3
  • $\begingroup$ Are you familiar with the definition of a concrete $C^*$-algebra?\ $\endgroup$
    – Aweygan
    Jul 22, 2022 at 16:56
  • $\begingroup$ $C^*$-algebras can be identitied (by Gelfand-Naimark) theorem with norm closed $*$-subalgebras of $B(H),$ the algebra of all bounded linear operators on a Hilbert space. $\endgroup$
    – Ryszard Szwarc
    Jul 22, 2022 at 18:56
  • $\begingroup$ @Aweygan unfortunately not $\endgroup$
    – Schrödinger's cat
    Jul 22, 2022 at 19:21

1 Answer 1

Reset to default
3
$\begingroup$

A von Neumann algebra $M$ lives in some $B(H)$. The C$^*$ relation $$\|T\|^2=\|T^*T\|$$ holds in $B(H)$. So all you need to check is that $M$ is norm closed; and this comes for free since $M$ is closed in a weaker topology than the norm topology.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .