In my recent work on skew-orthogonal polynomials I was interested in deriving a Christoffel-Darboux type formula. To this end Thomas Bothner referred me to a paper of his in collaboration with Marco Bertola which contains a nice derivation of the Christoffel-Darboux formula using only complex analysis arguments (Theorem 2.8). I will present this argument in the simpler context of orthogonal polynomials rather than that of biorthogonal polynomials considered by the authors.

Let us recall the Riemann-Hilbert problem for orthogonal polynomials. Given a measurable function w:R[0,+]w : \mathbb{R} \to [0,+\infty] such that Rw(x)xkdx<+\int_\mathbb{R} w(x) \lvert x \rvert ^k \, \mathrm{d}x < +\infty for all kNk \in \mathbb{N} and where w(x)>0w(x) > 0 on some open set we construct a sequence of monic polynomials Pn(x)=xn+O(xn1)P_n(x) = x^n + \mathcal{O}(x^{n-1}) such that

RPn(x)xkw(x)dx=0\int_\mathbb{R} P_n(x)x^k w(x)\, \mathrm{d}x = 0

for all k=0,,n1k=0, \dots, n-1. We call PnP_n the nnth monic orthogonal polynomial with respect to ww.

We let hn=RPn(x)xnw(x)dxh_n = \int_\mathbb{R} P_n(x)x^n w(x)\, \mathrm{d}x be the (squared) L2(w)L^2(w) norm of PnP_n. We may reformulate this as a Riemann-Hilbert problem.

Riemann-Hilbert problem for orthogonal polynomials (Fokas-Its-Kitaev, 1992):

Find a matrix valued function Xn:CRC2×2X_n : \mathbb{C} \setminus \mathbb{R} \to \mathbb{C}^{2 \times 2} such that

  1. XnX_n is analytic (entry-wise) on CR\mathbb{C} \setminus \mathbb{R}.
  2. XnX_n has continuous non-tangential boundary values up to R\mathbb{R} from above (++) and below (-). We label these Xn±(x)=limϵ0Xn(x±iϵ)X_n^\pm (x) = \lim_{\epsilon \downarrow 0} X_n(x\pm i \epsilon) for xRx \in \mathbb{R}.
  3. These boundary values are related by the jump condition Xn+(x)=Xn(x)(1w(x)01)X_n^+(x) = X_n^-(x) \left( \begin{matrix} 1 & w(x) \\ 0 & 1\end{matrix} \right)
  4. Finally, XnX_n is normalised at infinity by the scaling as zz \to \infty
Xn(z)=(I+O(z1))(zn00zn).X_n(z) = \left( \mathbb{I}+\mathcal{O}(z^{-1}) \right) \left( \begin{matrix} z^n & 0 \\ 0 & z^{-n} \end{matrix} \right).

Proposition: The above RHP has a unique solution given by (for n1n\geq 1)

Xn(z)=(Pn(z)CR(Pnw)(z)2πihn11Pn1(z)2πihn11CR(Pn1w)(z))X_n(z) = \left( \begin{matrix} P_n(z) & C_\mathbb{R} \left( P_n w\right)(z) \\ - 2\pi i h_{n-1}^{-1} P_{n-1}(z) & - 2\pi i h_{n-1}^{-1} C_\mathbb{R} \left( P_{n-1} w\right)(z) \end{matrix} \right)

where

CR(f)(z)=12πiRf(x)xzdxC_\mathbb{R}(f)(z) = \frac{1}{2\pi i} \int_\mathbb{R} \frac{f(x)}{x-z}\, \mathrm{d}x

is the Cauchy transform of the function ff. Furthermore detXn(z)=1\det X_n(z) = 1 identically. If n=0n=0 the solution is X0(z)=(1CR(w)(z)01)X_0(z) = \left( \begin{matrix} 1 & C_\mathbb{R} \left( w\right)(z) \\ 0 & 1 \end{matrix} \right). \triangle

That detXn(z)=1\det X_n(z) = 1 can be seen directly from the RHP since detXn(z)\det X_n(z) has no jump across R\mathbb{R} and has continuous boundary values, and so by Morera’s theorem is entire. detXn(z)1\det X_n(z) \to 1 as zz \to \infty and so by Liouville’s theorem is identically 11. This implies that the RHP can have at most one solution. The reader may then verify that the above is a solution.

Because detXn(z)=1\det X_n(z) = 1 we can introduce a “dual” Riemann-Hilbert problem Xn^(z)=Xn(z)T\widehat{X_n}(z) = X_n(z)^{-\mathsf{T}} (inverse transpose). Xn^\widehat{X_n} solves the following RHP.

Dual Riemann-Hilbert problem for orthogonal polynomials:

Find a matrix valued function Xn^:CRC2×2\widehat{X_n} : \mathbb{C} \setminus \mathbb{R} \to \mathbb{C}^{2 \times 2} such that

  1. Xn^\widehat{X_n} is analytic (entry-wise) on CR\mathbb{C} \setminus \mathbb{R}.
  2. Xn^\widehat{X_n} has continuous non-tangential boundary values up to R\mathbb{R} from above (++) and below (-). We label these Xn^±(x)=limϵ0Xn^(x±iϵ)\widehat{X_n}^\pm (x) = \lim_{\epsilon \downarrow 0} \widehat{X_n}(x\pm i \epsilon) for xRx \in \mathbb{R}.
  3. These boundary values are related by the jump condition Xn^+(x)=Xn^(x)(10w(x)1)\widehat{X_n}^+(x) = \widehat{X_n}^-(x) \left( \begin{matrix} 1 & 0 \\ -w(x) & 1\end{matrix} \right)
  4. Finally, Xn^\widehat{X_n} is normalised at infinity by the scaling as zz \to \infty
Xn^(z)=(I+O(z1))(zn00zn).\widehat{X_n}(z) = \left( \mathbb{I}+\mathcal{O}(z^{-1}) \right) \left( \begin{matrix} z^{-n} & 0 \\ 0 & z^n \end{matrix} \right).

Indeed, we know the unique solution of the dual RHP,

Xn^(z)=(2πihn11CR(Pn1w)(z)2πihn11Pn1(z)CR(Pnw)(z)Pn(z)).\widehat{X_n}(z) = \left( \begin{matrix} -2\pi i h_{n-1}^{-1} C_\mathbb{R} \left( P_{n-1} w\right)(z) & 2\pi i h_{n-1}^{-1} P_{n-1}(z) \\ -C_\mathbb{R} \left( P_n w\right)(z) & P_n(z) \end{matrix} \right).

Let us now derive a pair of recursion relations. We note that Xn+1X_{n+1} satisfies properties 1-3 (of the Fokas-Its-Kitaev RHP), differing only on property 4. Thus if we let Δn(z)=Xn+1(z)Xn(z)1\Delta_n(z) = X_{n+1}(z) X_n(z)^{-1} we see that Δn\Delta_n has no jump across the real axis, has continuous boundary values, and is analytic on CR\mathbb{C}\setminus \mathbb{R}. It is thus entire by Morera’s theorem. Let us expand this at infinity. Let Xn(z)=(I+Anz1+O(z2))(zn00zn)X_n(z) = \left( \mathbb{I}+A_n z^{-1} + \mathcal{O}(z^{-2}) \right)\left( \begin{matrix} z^n & 0 \\ 0 & z^{-n}\end{matrix} \right). Then

Δn(z)=zE1+An+1E1AnE1+O(z1)\Delta_n(z) = z E_1 + A_{n+1}E_1 - A_n E_1 + \mathcal{O}(z^{-1})

where E1=(1000)E_1 = \left( \begin{matrix} 1 & 0 \\ 0 & 0\end{matrix} \right). However since Δn\Delta_n is entire the O(z1)\mathcal{O}(z^{-1}) term is identically zero, so we have

Xn+1(z)=(zE1+An+1E1E1An)Xn(z).X_{n+1}(z) = \left( z E_1 + A_{n+1}E_1 - E_1 A_n \right)X_{n}(z) .

This is the famous “three term recurrence” for orthogonal polynomials, derived by complex analysis arguments. By a similar argument we find

Xn+1^(z)=(zE2An+1TE2+E2AnT)Xn^(z)\widehat{X_{n+1}}(z) = \left( z E_2 - A_{n+1}^\mathsf{T} E_2 + E_2 A_n^\mathsf{T} \right)\widehat{X_{n}}(z)

where E2=(0001)E_2 = \left( \begin{matrix} 0 & 0 \\ 0 & 1\end{matrix} \right). Let us now consider quantity

Yn(z,w):=Xn(z)1Xn(w)=Xn^(z)TXn(w).Y_n(z,w) := X_n(z)^{-1}X_n(w) = \widehat{X_n}(z)^{\mathsf{T}}X_n(w) .

Using our recursion relations for Xn^\widehat{X_n} and XnX_n we may relate YnY_n and Yn+1Y_{n+1} by Yn+1(z)=Xn1(z)Δn(z)1Δn(w)Xn(w)Y_{n+1}(z) = X_n^{-1}(z) \Delta_n(z)^{-1} \Delta_n(w) X_n(w). Δn(z)1Δn(w)\Delta_n(z)^{-1} \Delta_n(w) is a polynomial in two variables, moreover

Δn(z)1Δn(w)=(zw)(An+1)21E21+I\Delta_n(z)^{-1} \Delta_n(w) = (z-w) (A_{n+1})_{21} E_{21} + \mathbb{I}

where E21=(0010)E_{21} = \left( \begin{matrix} 0 & 0 \\ 1 & 0\end{matrix} \right). By our formula for solution XnX_n we find (An+1)21=2πihn1(A_{n+1})_{21} = - 2\pi i h_n^{-1}, and so

Yn+1(z,w)=Yn(z,w)2πihn(zw)Xn1(z)E21Xn(w)Y_{n+1}(z,w) = Y_n(z,w) - \frac{2\pi i}{h_n} (z-w) X_n^{-1}(z) E_{21} X_n(w)

If we now take the (2,1)(2,1) matrix element of both sides we find

Yn+1(z,w)21=Yn(z,w)212πiPn(z)Pn(w)hn(zw).Y_{n+1}(z,w)_{21} = Y_n(z,w)_{21} - 2\pi i \frac{P_n(z)P_n(w)}{h_n} (z-w) .

If we now sum both sides from n=0n=0 to n=N1n=N-1 we find

YN(z,w)21=Y0(z,w)212πi(zw)n=0N1Pn(z)Pn(w)hn.Y_{N}(z,w)_{21} = Y_0(z,w)_{21} - 2\pi i (z-w) \sum_{n=0}^{N-1} \frac{P_n(z)P_n(w)}{h_n} .

Then from our solution Y0(z,w)21=0Y_0(z,w)_{21} = 0 we find

n=0N1Pn(z)Pn(w)hn=12πi(XN(z)1XN(w))21zw\boxed{ \sum_{n=0}^{N-1} \frac{P_n(z)P_n(w)}{h_n} = -\frac{1}{2\pi i} \frac{(X_N(z)^{-1}X_N(w))_{21}}{z-w}}

which is the Christoffel-Darboux formula.

Remark: This way of writing the Christoffel-Darboux formula mirrors nicely with what happens for β=4\beta = 4. Here the relevant quantity that encodes eigenvalue correlation functions is the “pre-kernel,” written as a sum over skew-orthogonal polynomials. Namely,

k=0n1P2k(x)eV(x)ddy(P2k+1(y)eV(y))P2k+1(x)eV(x)ddy(P2k(y)eV(y))2hk\sum_{k=0}^{n-1} \frac{ P_{2k}(x) e^{-V(x)} \frac{\mathrm{d}}{\mathrm{d}y}\left( P_{2k+1}(y) e^{-V(y)} \right) - P_{2k+1}(x) e^{-V(x)} \frac{\mathrm{d}}{\mathrm{d}y}\left( P_{2k}(y) e^{-V(y)} \right)}{2 h_k} =eV(x)V(y)4πi(An(x)1An(y))21xy= - \frac{e^{-V(x)-V(y)}}{4\pi i} \frac{(A_n(x)^{-1}A_n(y))_{21}}{x-y}

where PkP_k is the kkth monic skew-orthogonal polynomial, hkh_k is the skew-norm, and AnA_n is a Riemann-Hilbert problem introduced in my recent paper.