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 $U^*AU$?
Best Answer
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.