Every von Neumann algebra admits nontrivial trace

Question

Let $M$ be a von Neumann. A functional $\tau\colon M_+\to [0,\infty]$ is called trace if it satisfies the following conditions: 1) $\tau(x+y)=\tau(x)+\tau(y), \forall x,y\in M_+$; 2) $\tau(\lambda x)=\lambda \tau(x), \lambda\geq 0, x\in M_+$; 3) $\tau(xx^*)=\tau(x^*x), \forall x\in M$. As wee see from the definition we don't require any topological conditions for trace $\tau$. On the other hand $M\subset B(H)$ for some Hilbert space $H$ and $B(H)$ has it's own standart trace $tr$ which satisfies 1)-3) conditions. If we consider the restriction of $tr$ to $M$, i.e. $\tau = tr|_{M}$ does it follows that $\tau$ is trace on $M$? In other words does it true that every von Neumann algebra admits nontrivial trace?

Answer 1

Yes, of course, and you don't even need to argue as you did. Just define $\tau(x)=\infty$ for all $x$.

Now if you want your tracial weight to be finite on a substantial part of the algebra, that's a different thing. Many von Neumann algebras do not have a semifinite trace: that occurs precisely when they have a nonzero type III component.