3
$\begingroup$

The question may be not appropriate with the title, since I do not know how to name it. I apologize.

Let $M$ be a finite $II_1$ factor, $\tau$ be the canonical trace. Let $p, q$ be two projections in $M$, if $\tau(p+q)>1$, we know that there exists a nonzero projection $r$, such that $r < p$ and $r < q$ ($r=p\wedge q$ for example).

If we are given an arbitrary state, but not trace, is the statement also true?

$\endgroup$
4
  • 4
    $\begingroup$ Since there is a counter example if you replace the `II$_1$ factor' by $M_2$, there is also a counter example in any II$_1$ factor. $\endgroup$ Nov 11, 2010 at 4:55
  • $\begingroup$ I think that still requires an argument (not a very difficult one, though), because it is not immediately obvious (to me, at least) that two projections with zero intersection in $M_2(\mathbb{C})$ will have zero intersection when viewed in a II$_1$ factor. $\endgroup$ Nov 11, 2010 at 22:48
  • $\begingroup$ Thanks for your answers. I am still not clear about that... $\endgroup$
    – Paul Z
    Nov 12, 2010 at 3:27
  • $\begingroup$ No problem. I wrote most of the details in the answer below. $\endgroup$ Nov 12, 2010 at 5:25

1 Answer 1

7
$\begingroup$

Ok, so here it goes.

First, let us do the $M_2(\mathbb{C})$ case. Let $t\in(0,1)$, and define $$ p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}t&\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix}. $$ Note that $p\wedge q=0$, since their ranges are two distinct lines through the origin. Define a (faithful) state $\varphi$ by $$ \varphi\left(\begin{bmatrix}a&b\\ c&d\end{bmatrix}\right)=\frac{2a+d}3 $$ (any convex combination $ra+sd$ with $r>s$ will do). Now $$ \varphi\left(p+q\right)=\frac{2(1+t)+1-t}3=1+\frac{t}3>1. $$ So that's the counterexample in $M_2(\mathbb{C})$

If now $M$ is a II$_1$ factor, we can use the same idea in the following way: let $p$ be any projection of trace 1/2. Then $p\sim(1-p)$ and there exists a partial isometry $v\in M$ with $v^*v=p$, $vv^*=1-p$. The four operators $p,v^*,v,1-p$ behave exactly as the matrix units $e_{11},e_{12},e_{21},e_{22}.$ So we define $q=tp+\sqrt{t-t^2}(v+v^*)+(1-t)(1-p)$, which is a projection; it is easy to check that $\tau(v)=0$, and that $\tau(q)=1/2$. Let $\varphi$ be the (faithful) state $\varphi(x)=2\tau(2px+(1-p)x)/3$. Then $$ 2p(p+q)+(1-p)(p+q)=2p+2pq+p+q-p-pq=2p+pq+q, $$ and $$ \varphi(p+q)=\frac23\,\tau(2p+pq+q)>\frac23\,\tau(2p+q)=\frac23\,\left(1+\frac12\right)=1. $$ It remains to see that $p\wedge q=0$. Represent $M$ faithfully on a Hilbert space $H$. Suppose that $\xi\in pH\cap qH$. Then $\xi=p\xi=q\xi$. In particular, $(1-p)\xi=0$. Then $v^*\xi=0$, and so $$ \xi=q\xi=tp\xi+\sqrt{t-t^2}v\xi. $$ The last piece of information we need is that $v=(1-p)v$. Then $pv\xi=0$, and $$ \xi=p\xi=pq\xi=tp\xi=t\xi. $$ Since $t\ne1$, this forces $\xi=0$.

$\endgroup$
1
  • 4
    $\begingroup$ It's easy to see from von Neumann's bimommutant theorem that $p \wedge q$ always lives in the von Neumann algebra generated by $p$ and $q$. So $p \wedge q$ does not change if you consider a larger von Neumann algebra. $\endgroup$ Apr 3, 2011 at 3:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.