A celebrated result in the theory of random matrices is the connection between the extremal eigenvalue of a random matrix sampled from the Gaussian Unitary Ensemble and the Painlevé II equation. In this post I will give a derivation of this that uses the Hankel composition method. This method is given in great generality in the paper of Bothner and was used in our joint work to relate the distribution function of the extremal eigenvalue in the elliptic Ginibre ensemble to an integro-differential equation. By showing how this method works in the case of the GUE it should shed some light on how it works in other cases.

## Background

The Gaussian Unitary Ensemble (GUE) is an ensemble of $$n \times n$$ Hermitian random matrices with probability density

$\frac{1}{Z_{\mathrm{GUE}}} e^{- \frac{1}{2} \mathrm{tr}(H^2)}$

$$Z_{\mathrm{GUE}}$$ is a normalisation constant. We are interested in the distribution of the extremal (rightmost) eigenvalue. A famous result (see Chapter 24 of Mehta’s Random Matrices) shows that the cumulative distribution converges, under an appropriate scaling, to the Fredholm determinant of the Airy kernel. Let $$\lambda_n$$ be the rightmost eigenvalue.

$F(t) \equiv \lim_{n \to \infty} \mathbb{P}\left(\lambda_n \leq \sqrt{2n} + \frac{t}{\sqrt{2}n^\frac{1}{6}}\right) = \det(1 - K)_{L^2(t,\infty)}$

where $$K: L^2(t,\infty) \to L^2(t,\infty)$$ is the operator with kernel

$K(x,y) = \frac{\mathrm{Ai}(x) \mathrm{Ai}^\prime(y) - \mathrm{Ai}^\prime(x)\mathrm{Ai}(y) }{x-y} = \int_0^\infty \mathrm{Ai}(x+s)\mathrm{Ai}(y+s) \, ds$

The motivation for studying this is not simply that the GUE is an easy model to study, but also that this Airy kernel is universal (see Deift’s Orthogonal Polynomials and Random Matrices: A Riemann-Hilbert Approach). That is, suppose we have an ensemble of Hermitian matrices with probability density

$\frac{1}{Z_{V}} e^{- n \mathrm{tr} V(H)}$

for some entire function $$V$$ which grows sufficiently rapidly at $$\pm \infty$$, e.g. a polynomial. For generic $$V$$, the eigenvalues will asymptotically ($$n \to \infty$$) concentrate on disjoint intervals $$[\alpha_1, \beta_1], \dots , [\alpha_m, \beta_m]$$; and the distribution of the extremal eigenvalue at these endpoints $$\alpha_1, \beta_1, \dots, \alpha_m , \beta_m$$ will converge after a suitable rescaling, for “typical” $$V$$, to $$\det(1 - K)_{L^2(t,\infty)}$$. There is a similar universality in the bulk where the “universal” kernel is the sine kernel. Gap probabilities in the sine point process were found to be related to the Painlevé V equation by the group of Jimbo, Miwa, Môri and Sato in 1980 (see here for an accessible introduction to this work). The work of Tracy and Widom on the Airy kernel was strongly inspired by the work of this group.

## The Connection to Painlevé II

Theorem (Tracy and Widom 1993):

$\boxed{F(t) = \exp\left(-\int_t^\infty (s-t)q(s)^2 \, ds\right)}$

where $$q$$ solves Painlevé II

$\boxed{q^{\prime \prime}(t) = t q(t)+ 2 q(t)^3}$

and we have the boundary condition $$q(t) \sim \mathrm{Ai}(t)$$ as $$t \to +\infty$$. $$\triangle$$

We demonstrate the above by showing that

$\frac{d^2}{dt^2} \log F(t) = -q(t)^2$

We then obtain the above formula by integrating twice. To justify this requires showing that $$\log F(t)$$ and $$\frac{d}{dt} \log F(t)$$ tend to zero at $$t = +\infty$$. Showing this requires an asymptotic analysis of the Fredholm determinant $$\det(1 - K)_{L^2(t,\infty)}$$ which is beyond the scope of this post.

The first step is to bring the $$t$$ dependence into the operator. Let

$K_t(x,y) = K(x+t,y+t) = \int_t^\infty \mathrm{Ai}(x+s) \mathrm{Ai}(y+s) \, ds$

Then $$F(t)=\det(1-K)_{L^2(t,\infty)} = \det(1-K_t)_{L^2(\mathbb{R}_+)}$$.

Notation: We let $$\tau_t$$ be the shift operator, so that $$(\tau_t \phi)(x) = \phi(x+t)$$ and $$D$$ be the derivative operator, $$(D\phi)(x) = \phi^\prime(x)$$. We shall be somewhat careless and not specify on what spaces these operators act on. Let us also denote the Airy function $$\mathrm{Ai} = A$$. $$\triangle$$

We see that $$\frac{d}{dt} K_t(x,y) = - A(x+t)A(y+t)$$. Thus

$\frac{d}{dt} K_t = - \tau_t A \otimes \tau_t A$

Remark: I should signpost a point of rigour. We have calculated the derivative with respect to $$t$$ pointwise on the kernel, but in fact what we’d like is for the limit implicit in the derivative to exist in the trace norm, and a complete proof would show this. $$\triangle$$

Then we see by Jacobi’s formula

$\frac{d}{dt} \log F(t) = -\mathrm{tr}_{L^2(\mathbb{R}_+)} \left( (1-K_t)^{-1} \frac{dK_t}{dt} \right) = \mathrm{tr}_{L^2(\mathbb{R}_+)} \left( (1-K_t)^{-1} \tau_t A \otimes \tau_t A \right)$

Remark: Note that $$\mathrm{tr}_{L^2(\mathbb{R}_+)} \left(\psi \otimes \phi\right) = \langle \psi, \phi \rangle_{L^2(\mathbb{R}_+)} = \int_{\mathbb{R}_+} \psi(x) \phi(x) \, dx$$ (we will always take functions to be real valued). Note also that $$K_t$$, and hence $$(1-K_t)^{-1}$$, are symmetric operators with respect to this inner product. $$\triangle$$

Next we use the identity

$\frac{d}{dt} (1-K_t)^{-1} = (1-K_t)^{-1} \frac{dK_t}{dt}(1-K_t)^{-1} = - (1-K_t)^{-1} (\tau_t A \otimes \tau_t A) (1-K_t)^{-1}$

Observe that $$\mathrm{tr}((\alpha \otimes \beta)(\gamma \otimes \delta)) = \mathrm{tr}(\alpha \otimes \delta) \mathrm{tr}(\beta \otimes \gamma)$$.

$\frac{d^2}{dt^2} \log F(t) = 2 \mathrm{tr}_{L^2(\mathbb{R}_+)} \left( (1-K_t)^{-1} D \tau_t A \otimes \tau_t A\right) - \left(\mathrm{tr}_{L^2(\mathbb{R}_+)} \left( (1-K_t)^{-1} \tau_t A \otimes \tau_t A\right)\right)^2$

### A hierarchy of coupled ODEs

Introduce the following notation,

$q_n(t) = ((1-K_t)^{-1} D^n \tau_t A)(0)$ $p_n(t) = \mathrm{tr}_{L^2(\mathbb{R}_+)} \left( (1-K_t)^{-1} D^n \tau_t A \otimes \tau_t A\right)$

In this notation we have $$\frac{d^2}{dt^2} \log F(t) = 2 p_1(t) - p_0(t)^2$$.

We now compute

$\frac{d}{dt} q_n(t) = q_{n+1}(t)- q_0(t)p_n(t)$ $\frac{d}{dt} p_n(t) = p_{n+1}(t)- p_0(t)p_n(t) + \underbrace{\mathrm{tr}_{L^2(\mathbb{R}_+)} \left( (1-K_t)^{-1} D^n \tau_t A \otimes D\tau_t A\right)}_{(\ast)}$

To compute $$(\ast)$$ we integrate by parts

$(\ast) = -q_n(t)(\tau_t A)(0) - \underbrace{\mathrm{tr}_{L^2(\mathbb{R}_+)} \left( D(1-K_t)^{-1} D^n \tau_t A \otimes \tau_t A\right)}_{(\ast \ast)}$

Next we use the identity $$[D,(1-K_t)^{-1}] = (1-K_t)^{-1} [D,K_t] (1-K_t)^{-1}$$. We now give the following important exercise.

Exercise: Let $$\phi \in L^2(\mathbb{R}_+)$$ be a sufficiently nice function (e.g. continuously differentiable and $$\phi^\prime \in L^2(\mathbb{R}_+)$$). Then

$([D,K_t]\phi)(x) = - ((\tau_t A \otimes \tau_t A)\phi)(x) - \phi(0) K_t(x,0)$

This yields the formula

$(\ast \ast) = p_{n+1}(t)- p_0(t)p_n(t)-q_n(t)((1-K_t)^{-1} K_t \tau_t A)(0)$

Using that $$(1-K_t)^{-1} K_t = (1-K_t)^{-1} - 1$$, we obtain a formula for $$\frac{d}{dt} p_n(t)$$. We thus obtain an infinite hierarchy of coupled ODEs, $$n \in \mathbb{N}$$,

$\boxed{\frac{d}{dt} q_n(t) = q_{n+1}(t)- q_0(t)p_n(t) }$ $\boxed{\frac{d}{dt} p_n(t) = -q_{n}(t)q_0(t)}$

Exercise: The quantity $$C = p_0(t)^2 - q_0(t)^2 - 2 p_1(t)$$ is conserved. (There are actually infinitely many such conserved quantities but we only need this one.)

Corollary: It seems reasonable that since the Airy function decreases rapidly at $$+\infty$$ that $$q_n$$ and $$p_n$$ should tend to zero at $$t\to +\infty$$. It therefore follows that $$C=0$$. From this it follows that

$\frac{d^2}{dt^2} \log F(t) = - q_0(t)^2$

Remark: It is “obvious” that since $$K_t$$ is “small” for $$t\to +\infty$$

$q_0(t) \approx (\tau_t A)(0) = \mathrm{Ai}(t)$

This explains the boundary condition. This needs to be rigorously justified but is beyond the scope of this post. $$\triangle$$

### Closing up the system

Everything up until now has been “universal” – in that we haven’t used any properties of $$A$$ – we have only used the Hankel composition structure of $$K$$. In particular, we haven’t used that $$A$$ solves the Airy equation, $$D^2 A = M A$$, where $$M$$ is the operator such that $$(M \phi)(x) = x \phi(x)$$. Such a “non-universal” property allows us to close up the system and obtain an ODE for $$q_0$$. Note that $$(M \phi)(0) = 0$$.

From this we get

$q_2(t) = t q_0(t) + ([(1-K_t)^{-1},M] \tau_t A)(0)$

As before $$[(1-K_t)^{-1},M] = (1-K_t)^{-1} [ K_t, M](1-K_t)^{-1}$$. If we recall our two equivalent formulae for the Airy kernel we see that

$[ K_t, M] = - \tau_t A \otimes D\tau_t A + D\tau_t A \otimes \tau_t A$

This gives

$q_2(t) = t q_0(t) + q_1(t)p_0(t)- q_0(t)p_1(t)$

If we combine this formula with our relation $$p_0(t)^2 - q_0(t)^2 - 2 p_1(t) = 0$$ we find

$\boxed{q_0^{\prime \prime}(t) = tq_0(t) + 2 q_0(t)^3}$

which is Painlevé II.