Under which condition , given a von Neumann algebra $M$ acting on separable Hilbert space $\mathcal{H}$ have uncountable number of faithful normal states?


1 Answer 1


Always, if $M\ne\mathbb C$.

If $\phi$ is a faithful normal state on $M$ and $p\in M$ is a nontrivial projection, let $$\phi_t(x)=t\phi(pxp)+(1-t)\phi((1-p)x(1-p)),\ \ \ \ t\in(0,1).$$ These are all faithful normal states (proof below). And if $\phi_t=\phi_s$, we get $$ t\phi(p)=\phi_t(p)=\phi_s(p)=s\phi(p), $$ so $s=t$.

To see that $\phi_t$ is faithful, if $x=y^*y$ and $\phi_t(x)=0$, then $\phi(pxp)=\phi((1-p)x(1-p))=0$. As $\phi$ is faithful, $pxp=(1-p)x(1-p)=0$. So $$ 0=py^*yp, $$ which implies $yp=0$; similarly, $y(1-p)=0$. Thus $$y=yp+y(1-p)=0,$$ and $$x=y^*y=0.$$

  • $\begingroup$ Given any von Neumann algebra, can we construct a faithful normal state? $\endgroup$ Dec 27, 2020 at 22:42
  • $\begingroup$ No, if it's not separable. That's the whole point of weights. $\endgroup$ Dec 27, 2020 at 23:20
  • $\begingroup$ Would you mind showing me an example of non-separable VNA which has no faithful normal state? $\endgroup$ Dec 28, 2020 at 0:30
  • $\begingroup$ For any uncountable set $X$, the commutative von Neumann algebra $\ell^\infty(X)$ is non-separable, and so it cannot have a normal state. $\endgroup$ Dec 28, 2020 at 3:59

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