when is an element of $M_n(M)$ $\ast$-free from $M_n(\mathbb{C})$ for a $\ast$-non-commutative probability space $M$.

Question

From "Lectures on the combinatorics of free probability" by Nica and Speicher we have a necessary sufficient criteria for an element of $M_n(M)$ being free from $M_n(\mathbb{C})$ for a non-commutative probability space (NCPS) $M$. Do we have a similar result for such an element being $\ast$-free from $M_n(\mathbb{C})$ for a $\ast$-non-commutative probability space $M$? I can't think of even a good necessary condition.

The result that I mentioned above is: Let $x_{i,j} \in$ an NCPS $(M,\phi)$, $i,j=1, \cdots, n$. TFAE:

(1) The matrix $(x_{i,j}$ is free from $M_n(\mathbb{C})$ in $(M_n(M),Tr)$ where $M_n(M) = M_n(\mathbb{C}) \otimes M$ and $Tr=tr \otimes \phi$.

(2) Free cumulants of $\{x_{i,j}\}$ in $(M,\phi)$ are such that only cyclic cumulants $\kappa_m(x_{i(1),i(2)},x_{i(2),i(3)}, \cdots, x_{i(m),i(1)})$ are possibly different from 0, the value depending only on $m$, not the tuple $(i(1), \cdots, i(m))$.

I am also defining a $\ast$-non-commutative probability space $M$ and $\ast$-freeness below:

Def 1- A non-commutative probability space (NCPS) is a couple $(M,\phi)$ where $M$ is a unital algebra and $\phi$ is a unital linear functional on it. $(M, \phi)$ is called a $\ast$-non-commutative probability space if $M$ has an involution $\ast$ on it and $\phi$ is positive w.r.t $\ast$. My interest lies in the case where $M$ is a $II_1$ factor and $\phi$ is tracial state. But the result we have is for general NCPS.

Def 2- A family $(A_i)$ of unital $\ast$-subelgebras of $M$ is called free if $\phi(a_1 \cdots a_n)=0$ whenever $n \ge 1, a_j \in A_{i(j)}, \phi(a_j)=0$ and $i(j) \ne i(j+1)$ for all $j=1, \cdots, n$. $(a_i)$, a family of elements of $M$ are called $\ast$-free if the family $(alg(1,a_i,a_i^*))$ is free in $M$.

Answer 1

Yes, one can get a similar characterization, only you will need to talk about $\ast$-cumulants instead of cumulants (in other words, you will need to allow both $x_{ij}$'s and $x_{ji}^\ast$'s in the formula involving the cyclic cumulants above, and require that all other cumulants vanish).

Let me explain the source behind the formula of Nica-Speicher. Associated to $x\in M_n(M)$ there are two $R$-transforms: the scalar-valued one (i.e., Voiculescu's $R$-transform computed for the element $x$ viewed as belonging to the non-commutative probability space $(M_n(M), \frac{1}{n} Tr\circ \tau)$ and the $M_n(\mathbb{C})$-valued $R$-transform (i.e., the $R$-transform of $x$ viewed as belonging to the non-commutative probability space $(M_N(M), E=1\otimes \tau : M_n(M)\to M_n(\mathbb{C})$. Then a way to characterize freeness of $x$ is to say that the matrix-valued $R$-transform is the composition of the scalar-valued $R$-transform with the trace $\tau$ (so that the matrix-valued $R$-transform "factorizes" through the trace). For details, see section 3 of our paper with Nica and Speicher (http://arxiv.org/pdf/math.OA/0201001). When translated combinatorially, this corresponds to the characterization you mentioned. A similar characterization is available (in the paper above) for the case that you compare $D$-freeness and $B$-freeness where $D\subset B$ is some subalgebra (the discussion above being for $\mathbb{C}=D\subset B=M_{n\times n}(\mathbb{C})$.

Now the exact same characterization is available for the case of $\ast$-probability spaces (or, if you like, $n$-tuples of operators; indeed, to consider the $\ast$-operation all you need to do is to view $(x,x^\ast)$ as a pair). Another (equivalent) trick is to encode the $\ast$-distribution of $x$ as the $M_{2\times 2}(\mathbb{C})$ valued distribution of the matrix $$\left(\begin{matrix} 0 & x \cr x^\ast & 0 \end{matrix}\right)$$ and then write (in terms of factorization of $R$-transforms) that this matrix is free with amalgamation over $D=M_{2\times 2}\otimes 1$ from $B=M_{2\times 2}(\mathbb{C})\otimes M_{n\times n}(\mathbb{C})$.