research35 min studyLesson 857 in Mathematics

Orthogonal polynomials and determinantal kernels

Build a finite-rank projection kernel and distinguish factorial correlation functions from probability densities.

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What you’ll understand

  • Construct a Lebesgue-reference kernel from weighted orthogonal polynomials.
  • Use determinant correlation formulas with the correct counting normalization.
  • Calculate a small Hermite kernel and its two-point repulsion explicitly.

The squared Vandermonde in a unitary ensemble can be rewritten as a product of two determinants. Orthogonalizing the polynomial columns then converts a many-variable normalization problem into a finite-rank projection problem. The resulting kernel describes expected eigenvalue counts and their correlations. Its formula depends on the reference measure, so the weight must be placed consistently either inside the kernel or inside the integration measure.

Absorb the weight into orthonormal functions

Consider a unitary polynomial ensemble whose unordered joint density is proportional to the squared Vandermonde times ∏jw(xj)\prod_j w(x_j). Assume the positive weight has all moments required through the polynomial degrees being used, and the polynomial inner product is nondegenerate. Choose real orthonormal polynomials pjp_j in the measure w(x)dxw(x)dx. Multiplying each polynomial by the square root of the weight gives functions ϕj\phi_j orthonormal in ordinary Lebesgue measure. Their finite sum defines the kernel in the display.

This kernel is the integral kernel of the orthogonal projection onto the span of those first N functions. Orthonormality proves the projection identity by integrating and collapsing the double sum. Its diagonal integrates to N, the rank of the projection. One can instead keep the weight outside and use the polynomial sum as a kernel relative to weighted measure, but mixing that expression with Lebesgue integration loses the weight and changes every correlation formula. Here all correlation kernels use Lebesgue reference measure.

ϕj(x)=pj(x)w(x),KN(x,y)=∑j=0N−1ϕj(x)ϕj(y),∫KN(x,u)KN(u,y) du=KN(x,y).\phi_j(x)=p_j(x)\sqrt{w(x)},\qquad K_N(x,y)=\sum_{j=0}^{N-1}\phi_j(x)\phi_j(y),\qquad \int K_N(x,u)K_N(u,y)\,du=K_N(x,y).
The projection identity uses ordinary Lebesgue integration because the square-root weights have already been absorbed into the functions.↔ Scroll wide equations horizontally.

Determinants count distinct eigenvalue tuples

Replacing monomials in the Vandermonde by orthogonal polynomials changes only a constant. Expanding the two determinants and integrating with orthonormality yields the normalized unordered density as the determinant of the full kernel matrix divided by N factorial. Integrating out coordinates gives factorial correlation functions: they weight ordered tuples of distinct particles. In particular the first correlation is expected number density, not the probability density of one uniformly selected eigenvalue. Divide it by N to obtain that latter normalized density.

For two particles the determinant subtracts a squared cross-kernel term from the product of diagonal intensities. At coincidence the two rows agree and the determinant vanishes, expressing repulsion. Integrating the first correlation over an interval gives the expected count there. Projection also gives count variance as this expectation minus the double integral of the squared kernel over that interval, hence a variance no larger than the expected count. These scalar determinantal formulas describe the beta-two ensemble; GOE correlations require a different, Pfaffian structure.

pN(x1,…,xN)=1N!det⁡[KN(xi,xj)]i,j=1N,ρk=det⁡[KN(xi,xj)]i,j=1k,ρ2(x,y)=KN(x,x)KN(y,y)−∣KN(x,y)∣2.p_N(x_1,\ldots,x_N)=\frac1{N!}\det[K_N(x_i,x_j)]_{i,j=1}^N,\qquad \rho_k=\det[K_N(x_i,x_j)]_{i,j=1}^k,\qquad \rho_2(x,y)=K_N(x,x)K_N(y,y)-|K_N(x,y)|^2.
The correlation rho_k integrates to N factorial divided by (N minus k) factorial, rather than to one.↔ Scroll wide equations horizontally.

Hermite functions connect exact formulas to limits

For the Gaussian weight w(x)=e−x2/2w(x)=e^{-x^2/2}, the polynomial system is the probabilists’ Hermite system with its appropriate normalization. These coordinates are unscaled Gaussian eigenvalues: to recover the normalized GUE convention of the previous lesson, divide each coordinate by the square root of N. A kernel transforms under this rescaling with a Jacobian factor. Merely replacing its arguments while leaving its amplitude unchanged would produce the wrong expected count.

Christoffel–Darboux identities compress a long polynomial sum into an expression involving two adjacent degrees. This becomes a practical starting point for asymptotic analysis: different spatial rescalings of the Hermite functions lead to bulk and edge kernels. The limiting formulas require asymptotic estimates, not just finite-dimensional orthogonality. The finite projection calculation already explains exact normalization and repulsion, while locating a sine or Airy limit needs additional information about how the polynomial degree and spatial coordinate grow together.

A two-function Hermite projection

Use w(x)=e−x2/2w(x)=e^{-x^2/2} and N equal to two to compute the kernel and verify its total diagonal mass. These are unscaled coordinates.

  1. The first two orthonormal functions are ϕ0(x)=(2π)−1/4e−x2/4\phi_0(x)=(2\pi)^{-1/4}e^{-x^2/4} and ϕ1(x)=xϕ0(x)\phi_1(x)=x\phi_0(x). Their norms are one and their inner product vanishes by Gaussian symmetry.

  2. Summing their products gives K2(x,y)=(2π)−1/2e−(x2+y2)/4(1+xy)K_2(x,y)=(2\pi)^{-1/2}e^{-(x^2+y^2)/4}(1+xy). In particular K2(0,0)=1/2πK_2(0,0)=1/\sqrt{2\pi}.

  3. The diagonal is the standard normal density times 1+x21+x^2. Its integral is 1+1=21+1=2. Thus K2(x,x)/2K_2(x,x)/2, rather than the diagonal itself, is the density of a uniformly selected eigenvalue.

The result

The kernel is a rank-two Lebesgue projection with diagonal mass two; the selected-eigenvalue density has mass one.

Explore the model

Interactive lab

Interactive lab · vectors

Orthogonality makes a projection sum collapse

Pause/Reset, set angle 0 degrees and magnitude 1: projection onto the horizontal unit direction is 1. For a fresh comparison Pause/Reset, set both angle 90 degrees and magnitude 1: that projection is 0. Compare the unit vector with the first direction and then its orthogonal direction.

Orthogonality makes a projection sum collapsePlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOrthogonality removes cross terms in projection calculations. The two-dimensional vectors illustrate that algebra; the lab does not evaluate weighted Hermite functions or a determinantal point process.

Check your understanding

For the Lebesgue-reference projection kernel of an N-particle unitary ensemble, what does the integral of K_N(x,x) over an interval represent?

Try it yourself

For the worked kernel, calculate ρ2(0,a)\rho_2(0,a), evaluate it at a equal to one, and obtain the unordered joint density at that pair.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the two-by-two determinant. The off-diagonal squared term cancels the constant part of the diagonal product.

Compare with the solution

The diagonal product is (2π)−1e−a2/2(1+a2)(2\pi)^{-1}e^{-a^2/2}(1+a^2), while ∣K2(0,a)∣2=(2π)−1e−a2/2|K_2(0,a)|^2=(2\pi)^{-1}e^{-a^2/2}. Therefore ρ2(0,a)=a2e−a2/2/(2π)\rho_2(0,a)=a^2e^{-a^2/2}/(2\pi), which vanishes at coincidence. At a equal to one it is e−1/2/(2π)≈0.0965324e^{-1/2}/(2\pi)\approx0.0965324. The unordered joint density is half the factorial correlation, hence e−1/2/(4π)≈0.0482662e^{-1/2}/(4\pi)\approx0.0482662 at that pair. This factor two is the same ordering multiplicity found by direct Gaussian normalization.

Take it with you

  • Specify whether the kernel uses Lebesgue or weighted reference measure.
  • Projection rank gives the integral of the diagonal and distinguishes counts from probabilities.
  • Beta-two repulsion is a determinant cancellation, with finite kernels providing a route to asymptotic statistics.

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