# Relationship between R-transform and free convolution of random matrices?

I've been using the R-transform to calculate the free convolution of the eigenvalue spectra of two random matrices and I am trying to understand how it works, and in particular how it relates to noncrossing partitions.

As far as I understand it, the R-transform consists of the following steps:

1. Given a probability distribution function $f(t)$ over some domain $D$ (which I usually take to be $\mathbb R$), find its Cauchy transform $$g(s) = \int_D \frac {f(t)} {t-s} dt$$
2. Calculate the functional inverse $g^{-1}(w)$ and subtract $\frac 1 w$ to obtain the R-transform $$r(w) = g^{-1}(w) - \frac 1 w$$

and that the free convolution of two pdfs $f_1 \boxplus f_2$ consists of:

1. Adding the two R-transformed functions $$r_s (w) = r_1(w) + r_2(w)$$
2. Adding $\frac 1 w$ to the sum, then computing the functional inverse of $$g_s^{-1}(w) = r_s(w)+\frac 1 w$$
3. Computing the inverse Cauchy transform using the Plemelj relation $$f_s(t) = \frac 1 \pi \Im g_s (s)$$

I am trying to understand the mechanics of R-transform (why does it work?) and relating it to calculating the free convolution using random matrices, i.e. that if we have matrices $A$ and $B$ with eigenvalue spectra $f_A$ and $f_B$, that by taking a random orthogonal matrix $Q$ of Haar measure you can form the sum $$A + Q B Q^T$$ that has spectrum $f_A \boxplus f_B$ in the large $N$ limit.

I can see that expanding the resolvent in the Cauchy transform produces a formal power series in $s$, but I don't really see how the coefficients come out to free cumulants. What is $\frac 1 w$ doing? What is the relationship with noncrossing partitions? My intuition is that they must play a role in counting the number of terms with the same value in this expansion, and that the noncrossing must arise from where the $Q$ and $Q^T$s show up in the series (effecting a change of basis), but I don't quite see it yet.

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Your claim that A+QBQ^T has spectrum f_A boxplus f_B is only true in the large N limit (isn't this a result of Pastur and someone else?) –  Yemon Choi Sep 24 '11 at 23:56
In my personal opinion, as a non-expert in random matrices, your question is a bit too unfocused than is ideal for a MathOverflow question, and has too many "moving parts". Maybe you could split this into two more focused questions, one of which might be "why does the R-transform linearize free convolution, and how do the noncrossing partitions come up?" –  Yemon Choi Sep 25 '11 at 0:00
Yemon: Yes, thank you for pointing out the limit I left out. Will update. Perhaps you were thinking of the Marchenko-Pastur distribution? As far as I understand things, the two pieces are actually related, and I am missing some pieces in the middle to figure out how. The linearization allowed by free convolution has something to do with the vanishing of certain moments that go into the computation of free cumulants when the two distributions are freely independent, just like for the usual classical cumulants for independent distibutions. –  Jiahao Chen Sep 25 '11 at 16:35