Given two free semicirculars X_1 and X_2 and a projection h in the vonNeumann algebra generated by X_1, how does one show that the vonNeumann algebra generated by {X_1, hX_2(1h)} 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(1h) is. It appears in the definition of interpolated free group factor in Radulescu's paper (preprint 1991) on random matrices, amalgamated free products and subfactors of free group factors of noninteger index.
Let me first point out that $X_1$ and $Y=h X_2 (1h)$ 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 $(1h)$, 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 $\tau(h) \geq \tau (1h)$ (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 $1h$ 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 operatorvalued 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)

$\begingroup$ 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 vN 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? $\endgroup$ Apr 16 '11 at 22:55

2$\begingroup$ 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 $M$. $\endgroup$ Apr 26 '11 at 2:42
One way of thinking about the operator $$ Y=hX_{2}(1h) $$ 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 $M_{2}(\mathbb{C})$.

$\begingroup$ 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. $\endgroup$ Apr 7 '11 at 21:32