Large n Limit for the Product of Two Coupled Random Matrices

Abstract

For a pair of coupled rectangular random matrices we consider the squared singular values of their product, which form a determinantal point process. We show that the limiting mean distribution of these squared singular values is described by the second component of the solution to a vector equilibrium problem. This vector equilibrium problem is defined for three measures with an upper constraint on the first measure and an external field on the second measure. We carry out the steepest descent analysis for a 4 $\times$ 4 matrix-valued Riemann–Hilbert problem, which characterizes the correlation kernel and is related to mixed type multiple orthogonal polynomials associated with the modified Bessel functions. A careful study of the vector equilibrium problem, combined with this asymptotic analysis, ultimately leads to the aforementioned convergence result for the limiting mean distribution, an explicit form of the associated spectral curve, as well as local Sine, Meijer-G and Airy universality results for the squared singular values considered.

Appendix A: Heuristics on the Vector Equilibrium Problem

In this section, we give some heuristic arguments on how to formulate the vector equilibrium problem introduced in Sect. 2.2, which is closely related to the asymptotic analysis of the RH problem for Y.

Recall that the goal of the second transformation $X \rightarrow T$ is to ‘normalize’ the large z asymptotics of X and to prepare for the opening of lenses. We assume that, at this moment, it takes the following form:

$$\begin{aligned} T(z)={\widehat{C}} X(z) {{\,\mathrm{diag}\,}}(e^{n\lambda _1(z)},e^{n\lambda _2(z)},e^{n\lambda _3(z)},e^{n\lambda _4(z)}), \end{aligned}$$

(A.1)

where ${\widehat{C}}$ is a constant matrix and the $\lambda$-functions are of the form

$$\begin{aligned} \begin{aligned} \lambda _1(z)&=\int ^z C^{\mu _1}(s)\,\mathrm {d}s+V_1(z), \\ \lambda _2(z)&=\int ^z C^{\mu _2}(s)\,\mathrm {d}s-\int ^z C^{\mu _1}(s)\,\mathrm {d}s+V_2(z), \\ \lambda _3(z)&=\int ^z C^{\mu _3}(s)\,\mathrm {d}s-\int ^z C^{\mu _2}(s)\,\mathrm {d}s+V_3(z), \\ \lambda _4(z)&=-\int ^z C^{\mu _3}(s)\,\mathrm {d}s+V_4(z). \end{aligned} \end{aligned}$$

(A.2)

In the above formulas, $C^{\mu }(z)$ is the Cauchy transform of a measure $\mu$ given in (2.14), $\mu _1$, $\mu _2$ and $\mu _3$ are three measures satisfying

$$\begin{aligned}&{{\,\mathrm{supp}\,}}\mu _1\subset {\mathbb {R}}_-, \quad {{\,\mathrm{supp}\,}}\mu _2\subset {\mathbb {R}}_+, \quad {{\,\mathrm{supp}\,}}\mu _3\subset {\mathbb {R}}_-, \end{aligned}$$

(A.3)

$$\begin{aligned}&2|\mu _1|=|\mu _2|=2|\mu _3|=1, \end{aligned}$$

(A.4)

and $V_1,V_2,V_3,V_4$ are four functions to be determined.

As $z\rightarrow \infty$, it is readily seen from (A.1) and (5.13) that,

$$\begin{aligned} T(z)= & {} (I_4+\mathcal {O}(z^{-1}))B(z) \\&\times {{\,\mathrm{diag}\,}}\left( z^{\frac{n}{2}}e^{n(\lambda _1-2\alpha z^{\frac{1}{2}})},z^{\frac{n}{2}}e^{n(\lambda _2+2\alpha z^{\frac{1}{2}})},z^{-\frac{n}{2}}e^{n(\lambda _3+2\beta z^{\frac{1}{2}})},z^{-\frac{n}{2}}e^{n(\lambda _4-2\beta z^{\frac{1}{2}})}\right) . \end{aligned}$$

The normalization requirement then invokes us to expect that, as $z\rightarrow \infty$,

$$\begin{aligned} \begin{aligned} \lambda _1(z)-2\alpha z^{\frac{1}{2}}+\frac{1}{2}\log z&= o (1), \qquad \lambda _2(z)+2\alpha z^{\frac{1}{2}}+\frac{1}{2}\log z = o (1), \\ \lambda _3(z)+2\beta z^{\frac{1}{2}}-\frac{1}{2}\log z&= o (1), \qquad \lambda _4(z)-2\beta z^{\frac{1}{2}}-\frac{1}{2}\log z = o (1). \end{aligned} \end{aligned}$$

(A.5)

On the other hand, in view of (3.1), it follows that, as $z\rightarrow \infty$,

$$\begin{aligned} \begin{aligned} \lambda _1(z)&=V_1(z)-\frac{1}{2}\log z + \mathcal {O}(z^{-1}), \qquad \lambda _2(z) =V_2(z)-\frac{1}{2}\log z + \mathcal {O}(z^{-1}), \\ \lambda _3(z)&=V_3(z)+\frac{1}{2}\log z + \mathcal {O}(z^{-1}), \qquad \lambda _4(z) =V_4(z)+\frac{1}{2}\log z + \mathcal {O}(z^{-1}). \end{aligned} \end{aligned}$$

Comparing these asymptotics with (A.5), it is easily seen that we should have

$$\begin{aligned} \begin{aligned} V_1(z)&=2\alpha z^{\frac{1}{2}}, \qquad&V_2(z)=-2\alpha z^{\frac{1}{2}}, \\ V_3(z)&=-2\beta z^{\frac{1}{2}}, \qquad&V_4(z)=2\beta z^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

(A.6)

We next come to the jump condition satisfied by T. Taking into account (A.1), (A.3) and (5.12), it is readily seen that

$$\begin{aligned} T_+(x)=T_-(x)J_{T}(x), \qquad x\in {\mathbb {R}}, \end{aligned}$$

where

$$\begin{aligned} J_T(x)= & {} {{\,\mathrm{diag}\,}}\left( 1,e^{n(\lambda _{2,+}(x)-\lambda _{2,-}(x))},e^{n(\lambda _{3,+}(x)-\lambda _{3,-}(x))},1\right) \\&+ x^{\frac{\nu }{2}}e^{n(\lambda _{3,+}(x)-\lambda _{2,-}(x))}E_{23}, \qquad x\in {\mathbb {R}}_+, \end{aligned}$$

and

$$\begin{aligned} J_T(x)= & {} \Lambda {{\,\mathrm{diag}\,}}\left( e^{n(\lambda _{1,+}(x)-\lambda _{1,-}(x))},e^{n(\lambda _{2,+}(x)-\lambda _{2,-}(x))},e^{n(\lambda _{3,+}(x)-\lambda _{3,-}(x))}, e^{n(\lambda _{4,+}(x)-\lambda _{4,-}(x))}\right) \\&\quad - e^{n(\lambda _{1,+}(x)-\lambda _{2,-}(x))}E_{21}- e^{n(\lambda _{4,+}(x)-\lambda _{3,-}(x))}E_{34}, \qquad x\in {\mathbb {R}}_-, \end{aligned}$$

with

$$\begin{aligned} \Lambda :={{\,\mathrm{diag}\,}}(e^{-\pi i \kappa \sigma _3},e^{\pi i (\nu -\kappa ) \sigma _3}). \end{aligned}$$

We now look at the non-diagonal entries of the jump matrix $J_T$. It is expected that these entries to be constant on the supports of the measures. Taking their real part, we arrive at the following conditions.

• (2, 3)-entry on ${\mathbb {R}}_+$:

  $$\begin{aligned} 2U^{\mu _2}(x)-U^{\mu _1}(x)-U^{\mu _3}(x)+{{\,\mathrm{Re}\,}}\, (V_2(x)-V_3(x))=\ell _2; \end{aligned}$$

• (2, 1)-entry on ${\mathbb {R}}_-$:

  $$\begin{aligned} 2U^{\mu _1}(x)-U^{\mu _2}(x)+{{\,\mathrm{Re}\,}}\, (V_{1,+}(x)-V_{2,-}(x))=\ell _1; \end{aligned}$$

• (3, 4)-entry on ${\mathbb {R}}_-$:

  $$\begin{aligned} 2U^{\mu _3}(x)-U^{\mu _2}(x)+{{\,\mathrm{Re}\,}}\,(V_{3,-}(x)-V_{4,+}(x))=\ell _3, \end{aligned}$$

where $\ell _j$, $j=1,2,3$, is certain constant. From (A.6), we thus find that the potentials $Q_1$, $Q_2$ and $Q_3$ acting on the measures $\mu _1$, $\mu _2$ and $\mu _3$ should be

$$\begin{aligned} Q_1(x)&= {{\,\mathrm{Re}\,}}\,(V_{1,+}(x)-V_{2,-}(x))= 2\alpha (\sqrt{x})_+ + 2\alpha (\sqrt{x})_-=0, \\ Q_2(x)&= {{\,\mathrm{Re}\,}}\,(V_2(x)-V_3(x))= 2(\beta -\alpha )\sqrt{x}, \\ Q_3(x)&= {{\,\mathrm{Re}\,}}\,(V_{3,-}(x)-V_{4,+}(x))=-2\beta (\sqrt{x})_- -2\beta (\sqrt{x})_+ = 0, \end{aligned}$$

as shown in (2.6).

Finally, we explain the upper constraint. The fact that there is an upper constraint for $\mu _1$ but not for $\mu _2,\mu _3$ is connected to the form of the jumps: the equilibrium conditions for $\mu _1$ play a role in a lower triangular block of the jump matrix, whereas for the remaining measures the corresponding equilibrium conditions appear in an upper triangular block. In virtue of the direction of the variational inequalities for the equilibrium problem, we thus expect that associated to $\mu _1$ there should be an upper constraint, but no upper constraint should appear on the remaining measures.

To find the explicit form of the constraint, again some ansatz is needed. We expect that the functions $\lambda _1'$, $\lambda _2'$, $\lambda _3'$ and $\lambda _4'$ should all be solutions to the same algebraic equation (a.k.a. spectral curve). From the sheet structure for the associated Riemann surface, we also expect that $\lambda _1'$ is analytic across the places where $\sigma$ is active, that is, $\lambda _1'$ should be analytic across ${\mathbb {R}}_-{\setminus } {{\,\mathrm{supp}\,}}(\sigma -\mu _1)$. Hence,

$$\begin{aligned} \lambda _{1,+}'(x)-\lambda '_{1,-}(x)=0,\qquad x\in {\mathbb {R}}_- {\setminus } {{\,\mathrm{supp}\,}}(\sigma -\mu _1). \end{aligned}$$

Using the explicit expression for $\lambda _1$ (see (A.2) and (A.6)) and Plemelj’s formula (3.4), we can rewrite the identity above as

$$\begin{aligned}&\frac{1}{2\pi i}\left( C^{\mu _1}_+(x)-C^{\mu _1}_-(x) +V'_{1,+}(x)-V'_{1,-}(x)\right) \\&\quad =\frac{\,\mathrm {d}\mu _1}{\,\mathrm {d}x}(x)+\frac{\alpha }{2\pi i}\left( (x^{-\frac{1}{2}})_+-(x^{-\frac{1}{2}})_+\right) =0,\qquad x\in {\mathbb {R}}_-{\setminus } {{\,\mathrm{supp}\,}}(\sigma -\mu _1). \end{aligned}$$

Taking into account that $\mu _1=\sigma$ on ${\mathbb {R}}_-{\setminus } {{\,\mathrm{supp}\,}}(\sigma -\mu _1)$, the identity above gives us

$$\begin{aligned} \frac{\,\mathrm {d}\sigma }{\,\mathrm {d}x}(x)=\frac{\alpha }{\pi \sqrt{|x|}},\qquad x\in {\mathbb {R}}_-{\setminus } {{\,\mathrm{supp}\,}}(\sigma -\mu _1), \end{aligned}$$

which is (2.7).