Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 31 / 1

Let $V_m=\det [x_i^{j-1}{]}_{i,j=1}^m$. It is an alternating polynomial: exchanging two variables exchanges two rows of the Vandermonde matrix and reverses its determinant. Hence $V_m$ vanishes whenever $x_k=x_l$, so each factor $x_l-x_k$ divides it. The distinct linear factors are pairwise coprime polynomials, and their product therefore divides $V_m$. Both have total degree $0+1+\cdots +(m-1)=m(m-1)/2$, soThe coefficient of $x_m^{m-1}$ in the determinant is $V_{m-1}$, by expansion along its last row. The same coefficient in the product is ${\prod }_{k<l<m}(x_l-x_k)$. Thus $C_m=C_{m-1}$, and $C_1=1$. Consequently the Vandermonde determinant identity isThis is a polynomial identity, including repeated coordinates; no division by a possibly zero numerical Vandermonde product is needed.

Write ${\tilde{P}}_j(x)=x^{j-1}+{\sum }_{r=0}^{j-2}{b}_{rj}{x}^r$. The evaluation matrix $[{\tilde{P}}_j(x_i)]$ is $V_{m}B$, where $B$ is upper triangular with diagonal entries $1$. Its determinant is therefore unchanged. This is Vandermonde determinant invariance under monic basis change:For $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, multiply row $i$ by $\sqrt{w(x_i)}$. The requested weighted product becomes the weighted Vandermonde determinant squareThe sign difference between the two orientations of the Vandermonde product disappears on squaring. For $a=0$ interpret the weight at zero by continuity, so $w(0)=1$; for $a>0$, $w(0)=0$.

Choose real monic orthogonal polynomials $p_j$ of degree $j$, $j=0,1,\dots$, for $w(x)=x^a{e}^{-x}$ on $[0,\infty )$, and letExtend each ${\varphi }_j$ by zero to negative $x$. The resulting functions form an orthonormal set in $L^2(\mathbb{R},dx)$. Define the orthogonal polynomial projection kernelThe formula on the right is for nonnegative arguments; the kernel is zero when either argument is negative. With ${\Phi }_{ij}={\varphi }_{j-1}(x_i)$, the kernel evaluation matrix is $\Phi {\Phi }^T$. By the determinant multiplication identity and the preceding monic basis change,Thus initially ${\hat{c}}_m=c_m{\prod }_j{h}_j$.

Use the symmetric labelled joint probability density on the full orthant. Expanding the two copies of $\det \Phi$ into permutations and integrating all variables, orthonormality makes a term vanish unless the two permutations agree. Each of the $m!$ surviving terms integrates to $1$. HenceIf the eigenvalues are instead listed in increasing order on a single ordered chamber, that chamber's density is $m!$ times this symmetric density. We keep the full-orthant convention throughout.

An explicit choice is $p_j(x)=(-1{)}^{j}j!L_j^{(a)}(x)$ in terms of the Generalized Laguerre polynomials. To verify the normalization independently, their Rodrigues' formula givesIntegrating by parts $j$ times proves orthogonality to every lower-degree polynomial. The boundary terms vanish at zero and infinity for $a\ge 0$. Taking the other factor to be $p_j$ and using $p_j^{(j)}=j!$ gives $h_j=j!\Gamma (a+j+1)$, where $\Gamma$ is the Gamma function. The Laguerre projection kernel is thereforefor $x,y\ge 0$.

Write $D_n=\det [K(x_i,x_j){]}_{i,j=1}^n$ and expand it by permutations. The assumptions ensure that the indicated single-variable integrals exist; the finite-rank projection kernel constructed above satisfies them with $r=m$. No symmetry of a general kernel is needed for the following argument.

If a permutation fixes $n$, its product contains the separate factor $K(x_n,x_n)$. Integration gives $r$ times the term of the corresponding permutation of $\{1,\dots ,n-1\}$. These terms contribute $r{D}_{n-1}$.

Otherwise $n$ belongs to a longer permutation cycle. Let $i$ precede $n$ and let $j$ follow it. The only factors containing $x_n$ are $K(x_i,x_n)K(x_n,x_j)$, whose integral is $K(x_i,x_j)$ by the reproducing assumption. Delete $n$ from the cycle to obtain a permutation $\tau$ on $n-1$ letters. Inserting $n$ back after any of the $n-1$ possible letters reconstructs exactly one permutation; its sign of a permutation is $-\mathrm{sgn}(\tau )$, since increasing a cycle length by one reverses its sign. Thus the nonfixed terms contribute $-(n-1)D_{n-1}$.

Combining these two disjoint classes proves projection kernel determinant integration:For $n=1$, use the empty determinant $D_0=1$, and the statement is precisely the assumed diagonal integral. For our kernel, orthonormality directly yieldsso every step of the marginal integration is justified.

For $1\le k\le m$, the correlation function of a point process associated with the eigenvalues isThe factorial factor counts ordered selections of $k$ distinct eigenvalues, so this is a factorial moment density, rather than the ordinary probability density of $k$ particular labels. Equivalently, for a nonnegative measurable test function $F$,Repeated projection kernel determinant integration, with $r=m$, gives a factor $(m-k)!$ on integrating an $m$-by-$m$ kernel determinant down to size $k$. Together with the normalizing $1/m!$ in $f_m$, the factors cancel:In particular $R_1(x)=K_m(x,x)$ and $\int R_{k}d\mathbf{x}=m!/(m-k)!$. Set $R_0=1$ and $R_k=0$ for $k>m$. The eigenvalue configuration is thus a determinantal point process with a rank-$m$ finite-rank projection kernel.