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The question may be not appropriate with the title, sine 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?

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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. –  Makoto Yamashita Nov 11 '10 at 4:55
    
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. –  Martin Argerami Nov 11 '10 at 22:48
    
Thanks for your answers. I am still not clear about that... –  Paul Z Nov 12 '10 at 3:27
    
No problem. I wrote most of the details in the answer below. –  Martin Argerami Nov 12 '10 at 5:25
    
Thank you, Martin –  Paul Z Nov 13 '10 at 5:44
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1 Answer

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 \[ \varphi\left(\begin{bmatrix}a&b\\ c&d\end{bmatrix}\right)=\frac{2a+b}3 \] (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 $\xi=0$.

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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. –  Jesse Peterson Apr 3 '11 at 3:59
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