Why doesn't the argument of circular law convergence of Ginibre spectrum give the same result for GUE?

Question

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.

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 !