Tag Archives: operator-valued semicircular element

The Story of the Determinant: Why the Determinant Formula is True

There had to be some formula

The logarithmic potential is an integral of the Stieltjes transform. The Stieltjes transform comes from the Dyson equation $G^{-1}+\eta(G)=z-a$ . And the Dyson equation is the Euler–Lagrange equation of a functional, the Dyson action

\mathcal A_z(M)=-\operatorname{tr}_m\log M-\frac12\operatorname{tr}_m\bigl(M\eta(M)\bigr) +\operatorname{tr}_m\bigl((z-a)M\bigr)-1 .

If you differentiate the action at its critical point with respect to $z$ , only the explicit $z$ contributes, and you get $\operatorname{tr}_mG$ . So the action at the Dyson solution is also an integral of the Stieltjes transform, and it is not a surprise that $\log\Delta(S_a-z)=\operatorname{Re}\mathcal A_z(G(z))$ . The shift $a$ plays no role here. It rides along with $z$ .

So the determinant is a critical value. The trouble is that it is the wrong kind of critical value. The action is holomorphic in $M$ , so the Dyson solution is a saddle of its real part, with as many directions going down as going up. A formula with an “ $\inf$ ” in it cannot come from the action directly.

The idea: free the scale

In the centered case the answer is $\det b$ for the matrix $b$ that scales $\eta$ to a doubly stochastic map. This suggested that $G$ should come with a positive “scale” also in general. It does: there is a unique $b\succ0$ with

G=b^{-1/2}\,U\,b^{-1/2},\qquad U\ \text{unitary}.

For scalars this is just $b=1/|G|$ and $U$ is the phase.

The step that was unexpected for me is to forget that $b$ is determined by $G$ and treat it as a new independent variable. Write $G=-u-i\gamma$ and consider

\mathcal K(b,u,\gamma)=\operatorname{tr}_m\log b+\langle a-z,u\rangle -\frac12\langle u,(\mathcal C_b+\eta)u\rangle -\frac12\langle \gamma,(\mathcal C_b-\eta)\gamma\rangle .

When $b$ is the true scale of $-u-i\gamma$ this is the real part of the Dyson action (plus $\frac12$ ), by a direct computation. And the Dyson solution, together with its scale, is still a critical point of $\mathcal K$ in the three variables. So nothing is lost. What is gained is that the saddle has been taken apart: $\mathcal K$ is concave in $u$ and in $\gamma$ , so all the descending directions sit in these two variables, and one can remove them by maximizing.

Two things that come for free

The first is the constraint. I had thought of $\mathcal C_b-\eta\succeq0$ as a condition somebody cleverly imposed. But look at the last term of $\mathcal K$ . Its supremum over $\gamma\succeq0$ is $0$ if $\mathcal C_b-\eta\succeq0$ and $+\infty$ otherwise. The feasible set is simply the set of $b$ for which the problem makes sense. It is the trace that the imaginary part of $G$ leaves behind.

Maximizing over $u$ then gives exactly the function in Roland’s formula.

The second is convexity, and this is the heart of the matter. The function of $b$ that remains is not convex in the usual sense. It is convex along the geodesics $b^{1/2}e^{tX}b^{1/2}$ of the positive cone, and it is so precisely on the feasible set. The scalar semicircle shows what this means. There the function is

F_x(b)=\log b+\frac{x^2}{2(b^2+1)},\qquad b\ge1 .

For positive numbers the geodesics are simply $t\mapsto be^{t}$ , so geodesic convexity is ordinary convexity in the variable $s=\log b$ . In this variable

F_x=s+\frac{x^2}{4}\,\bigl(1-\tanh s\bigr).

The first term is linear. The second is a reflected $\tanh$ : it is convex for $s>0$ and concave for $s<0$ , and its inflection point is at $s=0$ , that is, at $b=1$ , exactly the boundary of the feasible set. So the constraint that came out of the imaginary part is the same as the condition that makes the problem convex. I do not think this is an accident, but I cannot say in one sentence why it must be so.

Once we have convexity, a critical point is a global minimum, and the critical point is the scale of the Dyson solution. This is the whole proof in the upper half-plane. To get to the real axis one adds $\varepsilon\operatorname{tr}_m\gamma$ to $\mathcal K$ , which adds a positive term of order $\varepsilon^2$ to the function of $b$ , and lets $\varepsilon\to0$ . Everything is monotone, and one never needs to know what the minimizers do.

What the picture looks like

In the scalar example the equation $F_x'(b)=0$ is $b+1/b=|x|$ , which is the Dyson equation for $b=1/|G|$ . For $|x|>2$ it has a solution with $b>1$ . For $|x|<2$ it has none, the minimum sits at the wall $b=1$ , and its value is $x^2/4$ , as it should be.

The same holds for matrices, at least at points where the density is nondegenerate. Outside the spectrum the minimizer is in the interior of the feasible set. Inside the spectrum it is on the boundary, and the kernel of $\mathcal C_b-\eta$ at the minimizer contains $\gamma=-\operatorname{Im}G$ . In other words, the spectrum is where the constraint is active, and the density of states is its Lagrange multiplier. This is the form in which I remember the result.

What I do not understand

Is the infimum attained at spectral edges, and at points where the matrix density is nonzero but singular? I do not know. The scales $b_\varepsilon$ might run away, and at points where the density blows up they do. It would be nice to read the type of the singularity from how fast they run.

The question I find most interesting is about the Brown measure. By hermitization the formula applies to any $S=a_0\otimes1+\sum a_i\otimes s_i$ with arbitrary matrices $a_i$ and any complex $z$ . For each fixed $b$ the function under the infimum is a quadratic polynomial in $\operatorname{Re} z$ and $\operatorname{Im}z$ . So the logarithmic potential of the Brown measure is an infimum of quadratic polynomials over a fixed convex set.

Now we can ask whether the minimizer $b_z$ sits in the interior or on the boundary of the feasible set. By the picture above this should detect the spectrum of $S$ : if $S-z$ is invertible, the minimizer is in the interior. Is this enough to detect the support of the Brown measure? That is, is the Brown measure of $S$ supported exactly where the constraint is active? For general operators the spectrum can be strictly larger than the support of the Brown measure. Haagerup and Larsen found this for $R$ -diagonal elements, where it happens when the element has an inverse which is in $L^2$ but is not bounded. Miyagawa has recently shown that a rational function in free semicirculars which is in $L^2$ is bounded, so this mechanism is not available here, and he asks whether there is any gap at all in that setting. Hall and Ho proved that there is none for a self-adjoint element plus a free elliptic one. For matrix semicirculars I do not know the answer. If there is no gap, the variational problem should show it: one would have to prove that the potential cannot be harmonic at a point where the constraint binds.

While this post was waiting to appear, Sheng Yin sent me a different formula for the same determinant, also found with the help of AI. In his formula the infimum is over all positive matrices, with no constraint, and it is attained exactly when the operator is invertible; in Roland’s formula the minimizer sits on the wall inside the spectrum. For self-adjoint coefficients the two are related by an explicit change of variables, which turns the free variable of one into the constrained variable of the other. His manuscript is on Hexagon (hexagon:2610.00180). How the two pictures fit together in general is something we would like to understand.

I should add that I worked on this in long conversations with language models. It felt like working with a co-author, and I cannot separate who contributed what. The responsibility for the correctness is mine.

A Story of the Determinant of Operator-Valued Semicircular Elements

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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.

What do determinants tell us about the eigenvalues of integer-valued matrices? — And what can we say about Fuglede-Kadison determinants of operator-valued semicircular elements?

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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