Equilibrium Measures for a Class of Potentials with Discrete Rotational Symmetries

Abstract

The asymptotic analysis of the eigenvalue distribution of $N\times N$ random normal matrix models in the large $N$ limit naturally leads to a logarithmic energy problem with external potential in the complex plane. In the present paper, we consider this variational problem for the class of matrix models whose associated external potential is of the special form $|z|^{2n}+tz^d+\bar{t}\bar{z}^d$, where $n$ and $d$ are positive integers satisfying $d\le 2n$. By exploiting the discrete rotational invariance of such potentials, a simple symmetry reduction procedure is used to calculate the equilibrium measure for all admissible values of $n,d$, and $t$. It is shown that, for fixed $n$ and $d$, there is a critical value $|t|=t_\mathrm{cr}$ such that the support of the equilibrium measure is simply connected for $|t|<t_\mathrm{cr}$ and has $d$ connected components for $|t|>t_\mathrm{cr}$.

Appendices

Appendix A: Univalency

1.1 Pre-critical Case

Definition 2

A map $f:\{|u|>1\}\rightarrow \mathbb {C}$ is starlike if it is univalent and the compact complement of its image is star-shaped w.r.t. the origin.

Theorem 2

[in [16]] Let $f:\{|u|>1\}\rightarrow \mathbb {C}$, then $f$ is starlike if and only if

$$\begin{aligned} \mathfrak {R}\left( u\frac{f'(u)}{f(u)}\right) >0, \qquad |u|>1. \end{aligned}$$

(28)

In order to prove that our pre-critical map

$$\begin{aligned} f(u)=ru\left( 1-\frac{\alpha }{u}\right) ^{\frac{d}{n}} \end{aligned}$$

is univalent, it suffices to show that (28) holds, i.e.,

$$\begin{aligned} \mathfrak {R}\left( 1+\frac{d}{n}\frac{\alpha }{u-\alpha }\right) >0. \end{aligned}$$

Thus we just need to check that

$$\begin{aligned} \mathfrak {R}\left( \frac{\alpha }{u-\alpha }\right) >-\frac{1}{2} \quad \text{ for }\ |u|>1. \end{aligned}$$

It is easy to see that $\frac{\alpha }{u-\alpha }$ maps the exterior of the circle of radius $|\alpha |$ into the half-plane $\mathfrak {R}(z)>-\frac{1}{2}$. This proves that our conformal map is univalent in $|u|>1$.

1.2 Post-critical Case

Let us consider the map

$$\begin{aligned} f(u)=-\frac{r}{\bar{\alpha }}\frac{1-\bar{\alpha }u}{u-\alpha }u\left( 1-\frac{\alpha }{u}\right) ^{\frac{d}{n}}. \end{aligned}$$

We can rescale $u\rightarrow \frac{|\alpha |}{\bar{\alpha }}u$ so that

$$\begin{aligned} f(u)=-\frac{r}{\bar{\alpha }}\frac{1-|\alpha |u}{u-|\alpha |}u\left( 1-\frac{|\alpha |}{u}\right) ^{\frac{d}{n}}=-\frac{r}{\bar{\alpha }}(1-|\alpha |u)\left( 1-\frac{|\alpha |}{u}\right) ^{\frac{d}{n}-1}. \end{aligned}$$

We need the following lemma:

Lemma 2

(in [16]) Let $f$ be a map analytic in $\{|u|\ge 1\}$ and injective on $\{|u|=1\}$; then $f$ is univalent in $\{|u|\ge 1\}$.

So we just need to prove that the image of unit circle has no self–intersections, i.e., $f(u)=f(v)$ if and only if $u=v$. We first look for the points $u=\text{ e }^{i \mu }$ and $v=\text{ e }^{i \nu }$ such that $|f(u)|=|f(v)|$:

$$\begin{aligned} \left| \frac{1-|\alpha |u}{u-|\alpha |}\right| |u|\left| 1-\frac{|\alpha |}{u}\right| ^{\frac{d}{n}}&= \left| \frac{1-|\alpha |v}{v-|\alpha |}\right| |v|\left| 1-\frac{|\alpha |}{v}\right| ^{\frac{d}{n}},\\ \left| 1-\frac{|\alpha |}{u}\right| ^{\frac{d}{n}}&=\left| 1-\frac{|\alpha |}{v}\right| ^{\frac{d}{n}},\\ \left| 1-|\alpha |\text{ e }^{-i \mu }\right| ^{\frac{d}{n}}&=\left| 1-|\alpha |\text{ e }^{-i \nu }\right| ^{\frac{d}{n}},\\ \left| 1-|\alpha |\text{ e }^{-i \mu }\right| ^2&=\left| 1-|\alpha |\text{ e }^{-i \nu }\right| ^2,\\ 2-2|\alpha |\cos (\mu )&=2-2|\alpha |\cos (\nu ),\\ \cos (\mu )&=\cos (\nu ). \end{aligned}$$

Thus $|f(\text{ e }^{i \mu })|=|f(\text{ e }^{i \nu })|$ if and only if $\nu =\pm \mu$.

Now we just need to show that $f(\text{ e }^{i \mu })\ne f(\text{ e }^{-i \mu })$ for $\mu \ne k\pi$.

$$\begin{aligned} -\frac{r}{\bar{\alpha }}(1-|\alpha |\text{ e }^{i \mu })\left( 1-|\alpha |\text{ e }^{-i \mu }\right) ^{\frac{d}{n}-1}&=-\frac{r}{\bar{\alpha }}(1-|\alpha |\text{ e }^{-i \mu })\left( 1-|\alpha |\text{ e }^{i \mu }\right) ^{\frac{d}{n}-1},\\ (1-|\alpha |\text{ e }^{i \mu })\left( 1-|\alpha |\text{ e }^{-i \mu }\right) ^{\frac{d}{n}-1}&=(1-|\alpha |\text{ e }^{-i \mu })\left( 1-|\alpha |\text{ e }^{i \mu }\right) ^{\frac{d}{n}-1}. \end{aligned}$$

The left-hand side and the right-hand side are complex conjugates, so this equation has solution if and only if the image of $\text{ e }^{i \mu }$ through the map

$$\begin{aligned} (1-|\alpha |\text{ e }^{i \mu })\left( 1-|\alpha |\text{ e }^{-i \mu }\right) ^{\frac{d}{n}-1} \end{aligned}$$

(29)

is real.

Let us change the variable $y:=1-|\alpha |\text{ e }^{-i \mu }$: since $|\alpha |<1$, $y$ is contained in the unit disk centered at $1$, which is contained in the half-plane $\mathfrak {R}(y)>0$. In terms of $y$, (29) assumes the simple form

$$\begin{aligned} \frac{\bar{y}}{y}y^{\frac{d}{n}}. \end{aligned}$$

(30)

The map $w=y^{\frac{d}{n}}$ sends the half-plane $\mathfrak {R}(y)>0$ into the set $\left\{ w\in \mathbb {C}|-\frac{d}{2n}\pi <\arg (w)\right.$ $\left. <\frac{d}{2n}\pi \right\}$, which, since $0<d<2n$, is contained in $\mathbb {C}\setminus \mathbb {R}_{-}$.

Written in its polar form, $y=\rho \text{ e }^{i\lambda }$, (30) becomes

$$\begin{aligned} \rho ^{\frac{d}{n}}\text{ e }^{-i\frac{2n-d}{n}\lambda }. \end{aligned}$$

Thus, since $-\frac{\pi }{2}<\lambda <\frac{\pi }{2}$, the previous number is real if and only if $\lambda =0$, i.e., $y\in \mathbb {R}$. This implies that $1-|\alpha |\text{ e }^{-i\mu }\in \mathbb {R}$, but this can happen if and only if $\mu =k\pi$.

Therefore Lemma 2 gives that the post-critical map $f(u)$ is univalent in $|u|>1$.

Appendix B: Analysis of the Equation for the Conformal Radius

Let us consider equation (22),

$$\begin{aligned} r^{\frac{4n}{d}-2}-\frac{T}{n}r^{\frac{2n}{d}-2}+\frac{n-d}{n}\frac{d^2}{n^2}|t|^2=0 \end{aligned}$$

for the conformal radius $r$ as a function of $t$. We need to show that (22) has a unique positive solution $r=r_0$ such that

$$\begin{aligned} |\alpha |=\frac{d}{n}|t|r_0^{1-\frac{2n}{d}} <1, \end{aligned}$$

(31)

or equivalently,

$$\begin{aligned} |\alpha |^{2}=\frac{n}{n-d}\left( \frac{T}{n}r_{0}^{-\frac{2n}{d}}-1\right) <1. \end{aligned}$$

(32)

Solving the critical equation $|\alpha |=1$ and using (22), we can obtain the critical values $t_\mathrm{cr}$ and $r_\mathrm{cr}:=r(t_\mathrm{cr})$ given by

$$\begin{aligned} t_\mathrm{cr}&=\frac{n}{d}\left( \frac{T}{2n-d}\right) ^{1-\frac{d}{2n}},\\ r_\mathrm{cr}&=\left( \frac{T}{2n-d}\right) ^{\frac{d}{2n}}. \end{aligned}$$

Clearly, the formulae above make sense only for $d\ne n$ and $d\ne 2n$. However, these cases are trivial, so we can restrict ourselves to the study of the cases $0< d< n$ and $n<d<2n$.

Proposition 3

Assume that $|t|<t_\mathrm{cr}$.

• For $0< d<n$, Eq. (22) has two positive solutions $r_{\pm }(|t|)$, with

  $$\begin{aligned} 0\le r_{-}(|t|)<r_{+}(|t|)\le \left( \frac{T}{n}\right) ^{\frac{d}{2n}} \quad \text{ and }\quad r_{-}(0)=0,\ r_{+}(0)= \left( \frac{T}{n}\right) ^{\frac{d}{2n}}. \end{aligned}$$

  With the choice $r=r_{+}(|t|)$, the inequality (31) is satisfied, whereas the other solution $r=r_{-}(|t|)$ is not compatible with (31).

• For $n< d < 2n$, Eq. (22) has a unique positive solution $r_{0}(|t|)$ that is compatible with the inequality (31).

Proof

Let $0< d < n$. Consider the function

$$\begin{aligned} y(r)= r^{\frac{4n}{d}-2}-\frac{T}{n}r^{\frac{2n}{d}-2} \end{aligned}$$

on the non-negative real axis. The only roots of $y(r)=0$ are

$$\begin{aligned} r=0 \quad \text{ and } \quad r= \left( \frac{T}{n}\right) ^{\frac{d}{2n}}, \end{aligned}$$

and $y(r)$ has a unique minimum at

$$\begin{aligned} r_\mathrm{min}=\left( \frac{T}{n}\frac{n-d}{2n-d}\right) ^{\frac{d}{2n}}, \end{aligned}$$

with

$$\begin{aligned} y(r_\mathrm{min})= -\frac{T}{2n-d}\left( \frac{T}{n}\frac{n-d}{2n-d}\right) ^{\frac{n-d}{n}}. \end{aligned}$$

Now it is easy to see that there exist precisely two solutions

$$\begin{aligned} 0 < r_{-} < r_\mathrm{min} < r_{+} < \left( \frac{T}{n}\right) ^{\frac{d}{2n}} \end{aligned}$$

for $0<|t|<t_\mathrm{max}$, where

$$\begin{aligned} t_\mathrm{max}=\frac{n^2}{d}\left( \frac{T}{n}\frac{n-d}{2n-d}\right) ^{\frac{n-d}{2n}}\sqrt{\frac{T}{n(n-d)(2n-d)}}. \end{aligned}$$

Since

$$\begin{aligned} \frac{t_\mathrm{max}}{t_\mathrm{cr}}=\left( \frac{n}{n-d}\right) ^{\frac{d}{2n}}<1, \end{aligned}$$

this condition is always satisfied for $0<|t|<t_\mathrm{cr}$. Moreover,

$$\begin{aligned} \frac{r_\mathrm{cr}}{r_\mathrm{min}} =\left( \frac{n}{n-d}\right) ^{\frac{d}{2n}}>1. \end{aligned}$$

Therefore with the choice $r=r_{+}$, we have

$$\begin{aligned} |\alpha |^{2}=\frac{n}{n-d}\left( \frac{T}{n}r_{+}^{-\frac{2n}{d}}-1\right) <\frac{n}{n-d}\left( \frac{T}{n}r_\mathrm{cr}^{-\frac{2n}{d}}-1\right) =\frac{n}{n-d}\left( \frac{2n-d}{n}-1\right) {=}1. \end{aligned}$$

In order to prove that $r_{-}$ does not satisfy (32), we write

$$\begin{aligned} |\alpha |^{2}= & {} \frac{n}{n-d}\left( \frac{T}{n}r_{-}^{-\frac{2n}{d}}-1\right) >\frac{n}{n-d}\left( \frac{T}{n}r_\mathrm{min}^{-\frac{2n}{d}}-1\right) \\= & {} \frac{n}{n-d}\left( \frac{2n-d}{n-d}-1\right) =\left( \frac{n}{n-d}\right) ^2>1. \end{aligned}$$

Therefore the only positive solution compatible with (31) is $r=r_{+}(t)$.

For $n< d< 2n$, the function $y(r)$ is strictly increasing with

$$\begin{aligned} y(r)\rightarrow -\infty , \quad r\rightarrow 0_{+}, \quad \text{ and } \quad y(r) \rightarrow \infty , \quad r \rightarrow \infty , \end{aligned}$$

and therefore (22) has a unique solution for every value of $|t|$. Since the unique root of $y(r)=0$ is given by

$$\begin{aligned} r=\left( \frac{T}{n}\right) ^{\frac{d}{2n}}, \end{aligned}$$

for $0<|t|<t_\mathrm{cr}$, we have

$$\begin{aligned} \left( \frac{T}{n}\right) ^{\frac{d}{2n}}<r_{0}(|t|)<r_\mathrm{cr} \end{aligned}$$

and hence

$$\begin{aligned} |\alpha |^{2}{=}\frac{n}{d-n}\left( 1-\frac{T}{n}r_{0}^{-\frac{2n}{d}}\right) <\frac{n}{d-n}\left( 1-\frac{T}{n}r_\mathrm{cr}^{-\frac{2n}{d}}\right) =\frac{n}{d-n}\left( 1-\frac{2n-d}{n}\right) =1, \end{aligned}$$

as required. $\square$