Free probability: A unitary group heuristic for the relationship between additive free convolution and free compression

Question

From one perspective, free probability is the study of how the eigenvalues of large random matrices interact under the basic matrix operations. The free probability operations of free additive convolution, free multiplicative convolution and free compression describe respectively the addition, multiplication, and taking minors of large random matrices. I have a question about an observed relationship between free additive convolution and free compression.

Free additive convolution. Let $N$ be large, and let $A = \mathrm{diag}(a_1,\ldots,a_N)$ and $B = \mathrm{diag}(b_1,\ldots,b_N)$ be diagonal matrices with real eigenvalues chosen in such a way that their empirical spectra approximate probability measures $\mu$ and $\nu$ in the sense that $$\frac{1}{N} \sum_{i=1}^N \delta_{a_i} \approx \mu$$ and $$\frac{1}{N} \sum_{i=1}^N \delta_{b_i} \approx \nu.$$

Let $U$ be Haar distributed on the unitary group. According to a result in free probability, the empirical spectrum of the random matrix $A+U^*BU$ approximates a probability measure $\mu \boxplus \nu$ on the real line. This measure is known as the additive free convolution of $\mu$ and $\nu$.

Free compression. On the other hand, with $A$ as above, set $j = \tau N$. Then the empirical spectrum of the $j \times j$ principal minor of $U^*AU$ approximates a probability measure $[\mu]_\tau$ on the real line. Call this measure the $\tau$-compression of $\mu$.

The relationship between free additive convolution and free compression. Let $k \geq 1$ be an integer. Set $\tau = 1/k$. Then it has been observed, for instance in this paper

Shlyakhtenko, Dimitri; Tao, Terence, Fractional free convolution powers, ZBL07642254. with a version freely available on arXiv here

that up to a rescaling, the $k$-fold additive free convolution $\mu^{\boxplus k}$ and the $\tau$-compression of $\mu$ are the same measure. (More precisely, one needs to pushforward the $\tau$-compression under the map $x \mapsto kx$ to get the same measure.)

This fact can be proved using formal calculations in free probability; see e.g. Section 2.5.4 on Free probability in Tao's book on Random Matrix Theory.

Question. Without appealing to any free probability, but just properties of the unitary group, is there a heuristic way to see that free compression and free additive convolution are the same up to rescaling? I have in mind a possible geometric heuristic, associating the Haar unitary matrix $U$ as a uniformly chosen orthonormal frame of $\mathbb{C}^N$.

Honing in on the case $k=2$: Is there a heuristic way of guessing that for large $N$, the empirical spectrum of $A+U^*AU$ is, up to scaling, approximately the same as the top-left corner of $N/2 \times N/2$ of $U^*AU$?

Answer 1

I have a suggestion that seems to bring the two sides to be compared much closer together.

Take a block-diagonal matrix $B$ with $k$ blocks each a copy of $A$. This has the same eigenvalue measure as $A$. So the fact to be explained is that for $U$ a random $Nk\times Nk$ unitary matrix, the eigenvalue measure of the top $N \times N$ block of $U B U^*$ is similar to the eigenvalue measure of the sum of $k$ different unitarily-conjugated copies of $A$.

If we write $U$ as consisting of $k^2$ different $N \times N$ blocks, $U^{ij}$, then the top $N \times N$ block of $U BU^*$ is $\sum_{i=1}^k U^{1i} A (U^{1i})^*$.

So the claim is that $k$ adjacent $N \times N$ blocks of a unitary matrix behave, for this purpose, like $k$ different $N \times N$ unitary matrices. I'm not sure what the right heuristic for this is. I want to say that the entries of unitary matrices behave approximately like i.i.d Gaussians and restricting to a block preserves this property, but this isn't literally true and can't be the whole story.

Specifically to check that two bounded measures on $\mathbb R$ are approximately equal, it suffices to check that their moments are approximately equal. The relevant moments here are $$ \operatorname{tr} ( \sum_{i=1}^k U^{1i} A (U^{1i})^* )^m$$ and $$ \operatorname{tr} ( \sum_{i=1}^k V^i A (V^i)^* )^m$$ where $V^i$ are independent Haar-random unitary matrices.

One can expand these powers-of-sums out and cancel certain terms, but maybe that is getting too close to the free probability calculation.

Answer 2

One can get a certain way towards this goal via a sort of "dimensional analysis". This isn't a completely satisfying heuristic argument - in particular, it only partially specifies what compression must look like - but it does at least (a) use the matrix-based definition of free convolution and free compression, and (b) avoids extensive computations.

From the matrix-based definition of free convolution we see that the operation $\boxplus$ is both commutative and associative, and also torsion-free. This is consistent with the existence of a large number of (semigroup) homomorphisms from probability measures equipped with $\boxplus$ to the real line equipped with addition. And indeed we have the free cumulants $\kappa_1(\mu), \kappa_2(\mu), \dots$ with $$ \kappa_n(\mu \boxplus \nu) = \kappa_n(\mu) + \kappa_n(\nu).$$ (These are secretly encoding the $R$-transform, but we will not need to explicitly work with the $R$-transform here.)

Now one has to take the following facts on faith:

1. The free cumulants $\kappa_n(\mu)$ uniquely determine the measure.
2. We have one free cumulant for each homogeneity: if $\lambda_* \mu$ denotes the pushforward of $\mu$ by the dilation map $x \mapsto \lambda x$, then $\kappa_n(\lambda_* \mu) = \lambda^n \kappa_n(\mu)$.

Now we turn to compression. From the matrix-based definition we see that compression is an endomorphism on probability measures with free convolution:

$$ (\mu \boxplus \nu)^{\boxplus k} = \mu^{\boxplus k} \boxplus \nu^{\boxplus k}.$$

Thus (by Fact 1), compression $\mu \mapsto \mu^{\boxplus k}$ is (morally at least) a linear map from the tuple $(\kappa_n(\mu))_{n=1}^\infty$ of free cumulants of $\mu$ to the tuple $(\kappa_n(\mu^{\boxplus k}))_{n=1}^\infty$. On the other hand, it is clear from the matrix-based definition that compression also preserves homogeneity: $$ ( \lambda_* \mu )^{\boxplus k} = \lambda_* (\mu^{\boxplus k}).$$ Thus this linear map must completely decouple in the cumulants (which describe the "isotypic components" of the homgeneity symmetry, by Fact 2), and we must have a proportionality relationship of the form $$\kappa_n(\mu^{\boxplus k}) = C_{n,k} \kappa_n(\mu)$$ for some constants $C_{n,k}$. The matrix-based definition also gives the semigroup law $$ (\mu^{\boxplus k})^{\boxplus l} = \mu^{\boxplus kl}$$ so this suggests (heuristically at least) that the relationship is of power type in $k$: $$\kappa_n(\mu^{\boxplus k}) = k^{C_n} \kappa_n(\mu).$$ The one thing I don't have a slick explanation for is why $C_n=1$. For $n=2$ one can see this just by plugging in the semicircular law which corresponds to GUE, using the matrix-based fact that the compression of GUE is just a rescaled version of GUE. Unfortunately for higher $n$ the cumulants of the semicircular law vanish. Perhaps there is another test distribution with non-vanishing cumulants for which the compression can be easily computed. One could in principle work things out using an asymptotic analysis as $k \to \infty$ but this level of calculation begins to reach the same level as the usual free probability computations.

Answer 3

I am not sure whether this counts as heuristics, as it goes even deeper into free probability results, but it might give some high-level kind of idea why this result should be true. The main fact about a unitarily-invariant matrix $X=U^*AU$ is that asymptotically the free cumulants $\kappa_n(X)$ of its eigenvalue distribution are given by scaled limits of classical cumulants $c_n$ of the entries $x_{ij}$ of the matrix, with a cycle structure in the indices, namely $$\kappa_n(X)=\lim_{N\to\infty} N^{n-1} c_n(x_{i_1i_2},x_{i_2i_3},\dots,x_{i_ni_1}).$$ The actual value of the $i_k$ does not matter, because asymptotically they all have the same leading order, so we might just take $c_n(x_{12},x_{23},\dots,x_{n1})$. This fact is surely not obvious and buried in the literature; it can be found, e.g., in Theorem 2.6 of this paper. Taking this for granted then the statement about the free compression is obvious. Take as new matrix $Y$ the upper $\tau N\times \tau N$-corner of our $N\times N$-matrix $X$. Then $Y$ is still unitarily-invariant, so we get its free cumulants as the limits of classical cumulants in its entries, but for the entries we can take the same cycle as for $X$. So we have \begin{align*}\kappa_n(Y)&=\lim_{N\to\infty} (\tau N)^{n-1} c_n(x_{i_1i_2},x_{i_2i_3},\dots,x_{i_ni_1})\\&= \frac 1\tau \lim_{N\to\infty} N^{n-1} c_n(\tau x_{i_1i_2}, \tau x_{i_2i_3},\dots,\tau x_{i_ni_1})= \frac 1\tau \kappa_n(\tau X).\end{align*} By realizing the free convolution as the sum of independent unitarily-invariant random matrices, one sees that one has the same result also in this case (by taking into account that mixed classical cumulants in independent variables vanish).