operator-valued semicircular element | Free Probability Theory

Operator-valued semicircular elements are the powerhouse of free probability and random matrix theory. In particular, the intrinsic freeness principle typically replaces random matrices by operator-valued semicircular elements and shows that, in many respects, the former are close to the latter. Thus it is important to understand as much as possible about operator-valued semicirculars.

An operator-valued semicircular element – say over $B=M_m(\mathbb C)$ – is of the form

S=a\otimes 1+\sum_{i=1}^n a_i\otimes s_i\qquad\text{with} \qquad a,a_1,\dots,a_n\in M_m^{sa}(\mathbf C),

where $s_1,\dots,s_n$ are free semicirculars. We denote by ${\rm tr}_m$ the normalized matrix trace on $B$ , and by $\tau$ the trace on the algebra, where the $s_i$ live. All information about such an element $S$ is contained in its mean

({\rm id} \otimes \tau)[S]=a

and its covariance

\eta(b)=:({\rm id}\otimes \tau)[(S-a)b(S-a)]=\sum_{i=1}^n a_i b a_i.

The covariance $\eta$ is a completely positive map from $B$ to $B$ .

We also know how, in principle, to calculate the distribution $\mu_S$ of $S$ . Its scalar-valued Cauchy transform is

g_S(z)=(\tr_m\otimes \tau)[(z1-S)^{-1}].

This can be extracted from the operator-valued Cauchy transform

G_S(b)=({\rm id}\otimes\tau)[(b-S)^{-1}]

simply by

g_S(z)=\tr_m[G_S(z1)].

And $G_S$ is determined by the nice quadratic matrix-valued equation, usually called the Dyson equation,

(b-a)\cdot G_S(b)=1+\eta(G_S(b))\cdot G_S(b).

So everything is determined by mean and covariance. Of course, some quantities are easier, others harder to extract in a meaningful way. For moments of $S$ , for example, one has the operator-valued free version of the Wick/Isserlis formula. The operator norm of $S$ , on the other hand, is not so easily accessible. In principle, this is given by the asymptotics of high moments, but that is not very concrete. So the following celebrated formula of Franz Lehner

\lambda_{\max} \left(a\otimes 1+\sum_{i=1}^n a_i\otimes s_i\right) = \inf_{b\in B, b \gt 0}\lambda_{\max} \left(b+a+\eta(b^{-1})\right)

comes as a nice surprise, as it reduces the upper spectral edge of the infinite-dimensional $S$ to a variational problem for the same quantity on the finite-dimensional matrix algebra $B$ . Here $\lambda_{\max}$ denotes the upper edge of the spectrum; for positive $a$ , Lehner’s original formulation gives the corresponding operator norm.

A few years ago together with Tobias Mai I was trying to understand the determinant (more precisely, the Fuglede–Kadison determinant) of an operator-valued semicircular element, in order to get some info about the accumulation of mass of the distribution of such a semicircular element close to zero. And we succeeded in finding a nice formula in terms of the covariance; more precisely the determinant was essentially given by the capacity of the covariance map. Capacity here is the quantity introduced by Gurvits in connection with operator scaling; it should not be confused with the various notions of channel capacity in quantum information. Unwinding this notion of capacity one can write our result concretely as

\Delta(\sum_{i=1}^n a_i\otimes s_i) = e^{-1/2}\cdot \inf_{b\in B, b>0} \Delta\left(b\cdot \eta(b^{-1})\right)^{1/2}.

$\Delta$ is here the Fuglede–Kadison determinant, which reduces in the finite-dimensional situation to

\Delta(b)=\det (\vert b\vert)^{1/m}\qquad\text{for}\qquad b\in M_m(\mathbb C).

As you see, our result is only for mean $a=0$ . But in this case it has a striking similarity with Lehner’s formula:

\| \sum_{i=1}^n a_i\otimes s_i\| = \inf_{b\in B, b>0} \left\|b+\eta(b^{-1})\right\|,

And of course it raises the question: Can we extend our formula also to the case of general mean. We thought a little bit about this way back then, but did not see how to do this.

Now, in the times of AI hype, it is tempting to come back to this question and ask ChatGPT what it has to say on this. And actually, it has something to say. It proposes the following generalization

\Delta\left(a\otimes 1+\sum_{i=1}^n a_i\otimes s_i\right) = e^{-1/2} \inf_{b \gt 0,\ \mathcal C_b-\eta\succeq 0} \Delta(b)\, \exp\left\{ \frac12 \left\langle a, (\mathcal C_b+\eta)^{-1}(a) \right\rangle_2 \right\},

where

\mathcal C_b(x)=bxb,\qquad\text{and}\qquad\langle x,y\rangle_2 = \operatorname{tr}_m(x^*y)\qquad \text{for} \qquad x,y\in M_m(\mathbb C)

is the Hilbert-Schmidt inner product.

The condition $\mathcal C_b-\eta\succeq 0$ means positivity as an operator on the Hilbert–Schmidt space $M_m(\mathbb C)$ , i.e.

\left\langle x,\mathcal C_b(x)\right\rangle_2 \geq \left\langle x,\eta(x)\right\rangle_2,\qquad\text{or equivalently}\qquad \operatorname{tr}_m(x^*bxb) \geq \operatorname{tr}_m\left(x^*\eta(x)\right), \qquad\text{for all }x\in M_m(\mathbb C).

ChatGPT also produces what it claims is a proof of all this — note that even the reduction of this to my formula with Tobias in the case of $a=0$ is non-trivial — and makes some remarks on the relevance for the Brown measure of an operator-valued circular element. Recall that the Brown measure of an operator $T$ is encoded by the logarithmic potential

z\mapsto \log \Delta (T-z1).

So understanding determinants of shifted operators is exactly what one needs here.

All this looks somehow reasonable, but not very enlightening; and even if I had a Lean certificate for all this (which I don’t), I would still not be too satisfied.

So let’s talk about all this and hope that, together, we can get a better understanding of what is really going on.

Consider a matrix with integer entries, like

$B=\begin{pmatrix} 1&2&1&3\\ 2&2&1&1\\ 1&1&2&3\\ 3&1&3&1 \end{pmatrix}$

How can we convince ourselves that there are no two eigenvalues close to zero; say, you should convince me that two eigenvalues both with absolute value smaller than 0.1 are not possible. Of course, we should not calculate the eigenvalues but decide on this by arguments as soft as possible.

It is easy to get upper bounds for eigenvalues by the operator norm or the normalized trace of the matrix – those are quantities which are usually easy to control.

Not so clear are lower bounds, but actually it’s the determinant which will be of good use here. Namely, since the determinant is the product of the eigenvalues, knowing a lower bound for the absolute value of the determinant, together with the above mentioned easy upper bounds, gives lower bounds for the absolute values of the non-zero eigenvalues. In this example, we can calculate the determinant as -1, estimate the norm against the Frobenius norm of $\sqrt{60}$ and can derive from this that no two eigenvalues of absolute value smaller than 0.1 are possible, because otherwise we would have the following estimate

$\vert \text{det}(B)\vert \leq 0.1\times 0.1\times \sqrt{60}\times\sqrt{60}=.6$

which contradicts the fact that $\vert \text{det}(B)\vert=1$ .

Maybe you don’t like so much the idea of having to calculate the determinant, as this does not feel so soft. Assume however that I told you, as an oracle, that the matrix is invertible, then you can forget about calculating the determinant and just argue that the determinant of a matrix with integer entries must be an integer, too, and since it cannot be zero you can conclude that it must in absolute value at least be 1; and so you can repeat the above estimate.

Those observations are of course not restricted to the above concrete example; actually, I have hopefully convinced you that no invertible matrix with integer entries can have too many eigenvalues close to zero, and the precise estimate depends on the matrix only through its operator norm.

This kind of reasoning can actually be extended to “infinite-dimensional matrices”, which means for us operators in a finite von Neumann algebra. There we are usually interested in the distribution of selfadjoint operators with respect to the trace and in recent work with Johannes Hoffmann and Tobias Mai it has become important to be able to decide that such distributions do not accumulate too much mass in a neighborhood of zero. In finite von Neumann algebras there exists also a nice version of the determinant, named after Fuglede and Kadison, and the main question is whether the above reasoning survives also in such a setting. And indeed, it does for the operators in which we are interested. But this is for reasons which are still nice, but quite a bit more tricky than for ordinary matrices.

The operators we are interested in are operator-valued semicircular elements of the form $S=\sum a_i\otimes s_i$ , where the $s_i$ are free semicircular elements and the $a_i$ are ordinary mxm matrices, with the only restriction that their entries are integers. We assume that such an S is invertible (as an unbounded operator) ; one can show that then its Fuglede-Kadison determinant $\Delta(S)$ is not equal to zero. The main question is whether we have a uniform lower estimate for $\Delta(S)$ away from zero. As there is now no formula for the determinant in terms of the entries of the matrix S, the integer values for the entries of the $a_i$ are of no apparent use. Still we are able to prove the astonishing fact that for all such invertible operator-valued semicircular operators their Fuglede-Kadison determinant $\Delta(S)$ is always $\geq e^{-1/2}$ .

For all this and much more you should have a look at my newest paper Fuglede-Kadison determinants of matrix-valued semicircular elements and capacity estimates with Tobias. If you need more of an appetizer, here is also the abstract of this paper: We calculate the Fuglede-Kadison determinant of arbitrary matrix-valued semicircular operators in terms of the capacity of the corresponding covariance mapping. We also improve a lower bound by Garg, Gurvits, Oliveira, and Widgerson on this capacity, by making it dimension-independent.

If this is still not enough to get you interested, maybe a concrete challenge will do. Here is a conjecture from our work on the limiting behavior of determinants of random matrices.

Conjecture: Let m and n be natural numbers and, for each i=1,…,n, an mxm-matrix $a_i$ with integer entries be given. Denote by $\eta$ the corresponding completely positive map

$\eta:M_m(\mathbb{C})\to M_m(\mathbb{C}), \quad b\mapsto \eta(b):=\sum_{i=1}^n a_iba^*_i$

Assume that $\eta$ is rank non-decreasing. Then consider n independent GUE random matrices $X^{(N)}_1,\dots,X^{(N)}_n$ and let $A_N$ denote the Gaussian block random matrix