v-Na generated by

Given two free semicirculars X_1 and X_2 and a projection h in the von-Neumann algebra generated by X_1, how does one show that the von-Neumann algebra generated by {X_1, hX_2(1-h)} is a factor? It is easy to show that the two elements in the generating set are free. But I am unable to see what kind of an object hX_2(1-h) is. It appears in the definition of interpolated free group factor in Radulescu's paper (pre-print 1991) on random matrices, amalgamated free products and subfactors of free group factors of non-integer index.

Let me first point out that $X_1$ and $Y=h X_2 (1-h)$ are not freely independent. This is most easily seen if $h$ has trace 1/2, in which case $Y$ has range and support projections $h$ and $(1-h)$, respectively. But since the support and range projections of $Y$ belong to $W^*(Y)$, it would follow from the assumption that $Y$ and $X$ are free that actually $h$ and $X$ are free. But this is not possible, since they commute.

Now to your question of factoriality. Here is a sketch of the proof. Let us assume for definiteness that $\tau(h) \geq \tau (1-h)$ (otherwise, switch $Y$ and $Y^\ast$). You can then verify that $Y^\ast Y$ and $YY^\ast$ have free Poisson distributions (with different parameters) and that the spectrum of $YY^\ast$ has no atoms. It follows that if you consider the polar decomposition of $Y = V |Y|$, then $V$ is a partial isometry with domain projection $1-h$ and range projection $\leq h$. Using this, you can see that $W^*(X_1,Y)$ is a factor iff $N=h W^*(X_1,Y)h$ is a factor. But $N$ is generated by $hX_1h$ and $Y^\ast Y$; you can prove that these elements are freely independent (in $N$). This either uses a random matrix model (see e.g. Voiculescu's book on free random variables for the proof of the compression formula for free group factors), or can be done directly using operator-valued semicircular systems. Thus $N = W^*(hX_1 H) * W^*(Y^\ast Y)$ which is a free product of two abelian von Neumann algebras, one of which is diffuse and the other not complex numbers. You can then get factoriality (see references in Ueda's paper http://arxiv.org/abs/1011.5017)

• Guess I'm going against the motto of mathoverflow NOT being a discussion platform, but still...I couldn't grasp how the proof is restricted down to a corner of $W^*(X_1, Y)$. I mean it is not true in general that a v-N algebra $M$ is a factor if $hMh$ is so for a projection $h \in M$. Why is it true here? Can we generalize under what properties it will be satisfied? Apr 16 '11 at 22:55
• If $M$ is a finite von Neumann algebra and $h\in M$ is a projection with central support $1$ (i.e., the only central projection in $M$ which majorizes $h$ is $1$), then $M$ is a factor iff $hMh$ is a factor. Proof: if $Z(M)$ is the center of $M$, then $Z\ni x\mapsto hxh \in hZh\subset Z(hMh)$ is injective because of the central support assumption. Thus $M$ not factor implies that $hMh$ not factor. Conversely, if $M$ is a factor, then so is $hMh$ for any projection $h$ in $M$. Apr 26 '11 at 2:42

One way of thinking about the operator $$Y=hX_{2}(1-h)$$ is to work with the random matrix models. More specifically, the operators $X_{1}$ and $X_{2}$ can be thought as the limit as $n\to\infty$ of two independent $n\times n$ Hermitian random matrices where the upper triangular parts are formed by i.i.d. Gaussian random variables of zero mean and variance $1/\sqrt{n}$.

Then $Y$ is the limit of the upper right corner of $X_{2}$. For example, let us assume (for notation simplicity only) that $\tau(h)=1/2$ then by the previous argument you can think of $X_{1}$ and $Y$ as:

$X_{1} = \begin{pmatrix} x & z \\\ z^{*} & y \end{pmatrix}$

$Y = \begin{pmatrix} 0 & w \\\ 0 & 0 \end{pmatrix}$

where $x$ and $y$ are semicircular operators and $z$ and $w$ are circular operators and all of them are free. This will help you to understand the operator $Y$ and deduce all you need (joint moments, factoriality, etc) from the von Neumann algebra generated by $\{X_{1},Y\}$. Note that as Dima is showing you, the elements $X_{1}$ and $Y$ are not free over the the algebra of complex numbers. However, represented as two by two matrices of operators as above they are free over the algebra $M_{2}(\mathbb{C})$.

• Thanks. While trying something smililar to your suggestion I got the freeness statement confused with freeness with respect to amalgamation over $M_2(\mathbb{C})$ and made that wrong statement. Apr 7 '11 at 21:32