It appears I am profoundly confused in the following nice argument of Ginibre and Mehta and beautifully presented in Djalil Chafai's blog http://blog.djalil.chafai.net/2010/11/02/aspects-of-the-complex-ginibre-ensemble/ In particular look at theorem 3 and 4.

To summarize the argument, we know that the joint eigenvalue density is given by $$\displaystyle p(z_1, \ldots, z_n) = C_n \exp(-\sum_{i=1}^n |z_i|^2) \prod_{i<j} |z_i - z_j|^2$$ for some constant $C_n = \frac{\pi^{-n^2}}{\prod_{k=1}^n k!}$. Now we can write the Vandermonde factor as $$\displaystyle \prod_{i<j}|z_i - z_j|^2 = \det(z_j^{k-1})_{j,k=1}^n \det(\bar{z}_j^{k-1})_{j,k=1}^n. $$

Also we can distribute the factor $\frac{1}{\prod_{k=1}^n k!}$ into the rows of the two determinants above and transpose the second determinant matrix to finally get

$$\displaystyle p(z_1,\ldots, z_n) = \pi^{-n^2} \exp(-\sum_{i=1}^n |z_i|^2) \det( \sum_{k=0}^{n-1} \frac{(z_i \bar{z}_j)^k}{k!})_{i,j=1}^n.$$

Using the fact that this is in determinantal form (one can always distribute the exponential factor into the determinantal using multilinearity), one can integrate out the variables $z_n,z_{n-1}, \ldots, z_2$ one at a time to arrive at the single point marginal density

$$\displaystyle p_1^{(n)}(z) = C_n \exp(-|z|^2) \sum_{k=0}^{n-1} \frac{(z_i \bar{z}_i)^k}{k!}.$$

From this the circular law is almost immediate, as the last factor converges to $\exp(|z|^2)$ for $z < N$.

My question is, why doesn't this argument work for GUE? I.e., instead of the complicated Hermite polynomial computation, why not just use the above simple basis $z^k$ without worrying about orthogonality? Also it would be ridiculous if it does work for GUE, as it would give the analogue of circular law in one dimension, namely segment law, instead of semicircle law.

I suspect the determinantal contraction is not the same in the two cases, but I can't think of a reason why it would fail, without having to examine the whole proof for a few days.

  • $\begingroup$ @John Why not ask the man? I just mentioned your question to Djalil. $\endgroup$
    – Did
    Mar 9, 2011 at 17:52
  • $\begingroup$ The LaTeX seems to have suffered a rendering incongruity, can someone with the appropriate powers fix it :-) ? $\endgroup$
    – Suvrit
    Mar 9, 2011 at 18:05
  • $\begingroup$ Actually I did ask him. But I was too impatient and wanted a quick answer. Sorry for the spam. $\endgroup$
    – John Jiang
    Mar 9, 2011 at 18:15

1 Answer 1


Dear John, to my knowledge, for the GUE, we have basically the same story with a determinantal kernel and nice orthogonal polynomials, which are the Hermite polynomials. Nothing goes wrong (see e.g. the article in EJP by Ledoux). The polynomials of the complex Ginibre ensemble are just particularly simple. So simple that they allow a direct argument without any extra analysis about their equilibrium measure ! This is the holomorphic power of the complex Ginibre Ensemble, free of any annoying symmetries. From this point of view, the complex Ginibre ensemble is the simplest Gaussian matrix ensemble ever, simpler than the GUE !

  • $\begingroup$ Thanks for pointing out that the $z_j^k$s are in fact the orthogonal polynomials for the Gaussian measure in two dimensions (please correct me if I misunderstood, but I will take a close look at Ledoux's paper). I guess I am missing some link in how orthogonal polynomial plays a role getting the 1-point marginals from the $n$-point density function (I understand how it facilitates the use of Christoffer Darboux formula). $\endgroup$
    – John Jiang
    Mar 9, 2011 at 22:50

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