Statistical Properties of Eigenvalues of Laplace–Beltrami Operators

Abstract

We study the eigenvalues of a Laplace–Beltrami operator defined on the set of the symmetric polynomials, where the eigenvalues are expressed in terms of partitions of integers. To study the behaviors of these eigenvalues, we assign partitions with the restricted uniform measure, the restricted Jack measure, the uniform measure, or the Plancherel measure. We first obtain a new limit theorem on the restricted uniform measure. Then, by using it together with known results on other three measures, we prove that the global distribution of the eigenvalues is asymptotically a new distribution $\mu$, the Gamma distribution, the Gumbel distribution, and the Tracy–Widom distribution, respectively. The Tracy–Widom distribution is obtained for a special case only due to a technical constraint. An explicit representation of $\mu$ is obtained by a function of independent random variables. Two open problems are also asked.

Appendix

In this section we will prove (1.6), verify (2.77) and derive the density functions of the random variable appearing in Theorem 1 for two cases. They are placed in three subsections.

1.1 Proof of (1.6)

Recall $(2s-1)!!=1\cdot 3\cdots (2s-1)$ for integer $s\ge 1$. Set $(-1)!!=1$. The following is Lemma 2.4 from Jiang [20].

Lemma 3.1

Suppose $p\ge 2$ and $Z_1, \ldots , Z_p$ are i.i.d. random variables with $Z_1 \sim N(0,1).$ Define $U_i=\frac{Z_i^2}{Z_1^2 + \cdots + Z_p^2}$ for $1\le i \le p$. Let $a_1, \ldots , a_p$ be non-negative integers and $a=\sum _{i=1}^pa_i$. Then

$$\begin{aligned} E\left( U_1^{a_1}\cdots U_p^{a_p}\right) = \frac{\prod _{i=1}^p(2a_i-1)!!}{\prod _{i=1}^a(p+2i-2)}. \end{aligned}$$

Proof of (1.6). Recall (1.5). Write $(r-1)s^2=\sum _{i=1}^r x_i^2-r\bar{x}^2$. In our case,

$$\begin{aligned} \bar{x}= & {} \frac{1}{|\mathcal {P}_n(m)|}\sum _{\kappa \in \mathcal {P}_n(m)}\lambda _{\kappa }=E\lambda _{\kappa };\\ s^2= & {} \frac{1}{|\mathcal {P}_n(m)|-1}\sum _{\kappa \in \mathcal {P}_n(m)}\left( \lambda _{\kappa }-\bar{x}\right) ^2 \sim E\left( \lambda _{\kappa }^2\right) - \left( E\lambda _{\kappa }\right) ^2 \end{aligned}$$

as $n\rightarrow \infty$, where E is the expectation about the uniform measure on $\mathcal {P}_n(m)'$. Therefore,

$$\begin{aligned} \frac{\bar{x}}{n^2}=\frac{E\lambda _{\kappa }}{n^2}\ \ \text{ and }\ \ \frac{s^2}{n^4}\sim E\left( \frac{\lambda _{\kappa }}{n^2}\right) ^2 - \left( \frac{E\lambda _{\kappa }}{n^2}\right) ^2. \end{aligned}$$

(3.1)

From Lemma 2.3, we see a trivial bound that $0\le \lambda _{\kappa }/n^2\le 1+\frac{\alpha }{2}m$ for each partition $\kappa =(k_1, \ldots , k_m)\vdash n$ with $k_m\ge 1.$ By Theorem 1, under $\mathcal {P}_n'(m)$,

$$\begin{aligned} \frac{\lambda _{\kappa }}{n^2}\rightarrow \frac{\alpha }{2}\cdot Y\ \ \text{ and }\ \ Y:=\frac{\xi _1^2+\cdots + \xi _m^2}{\left( \xi _1+\cdots + \xi _m\right) ^2} \end{aligned}$$

as $n\rightarrow \infty$, where $\{\xi _i;\, 1\le i \le m\}$ are i.i.d. random variables with density $e^{-x}I(x\ge 0)$. By bounded convergence theorem and (3.1),

$$\begin{aligned} \frac{\bar{x}}{n^2}\rightarrow \frac{\alpha }{2} EY\ \ \text{ and }\ \ \frac{s^2}{n^4} \rightarrow \frac{\alpha ^2}{4} \left[ E(Y^2)-(EY)^2\right] \end{aligned}$$

(3.2)

as $n\rightarrow \infty$. Now we evaluate EY and $E(Y^2)$. Easily,

$$\begin{aligned} EY= & {} m\cdot E\frac{\xi _1^2}{\left( \xi _1+\cdots + \xi _m\right) ^2};\nonumber \\ E(Y^2)= & {} m\cdot E\frac{\xi _1^4}{\left( \xi _1+\cdots + \xi _m\right) ^4} +m(m-1)\cdot E\frac{\xi _1^2\xi _2^2}{\left( \xi _1+\cdots + \xi _m\right) ^4}. \end{aligned}$$

(3.3)

Let $Z_1, \ldots , Z_{2m}$ be i.i.d. random variables with N(0, 1) and $U_i=\frac{Z_i^2}{Z_1^2 + \cdots + Z_{2m}^2}$ for $1\le i \le 2m$. Evidently, $(Z_1^2+Z_2^2)/2$ has density function $e^{-x}I(x\ge 0)$. Then,

$$\begin{aligned} \left( \frac{\xi _i}{\xi _1+\cdots + \xi _m}\right) _{1\le i \le m}\ \ \text{ and }\ \ \ \left( U_{2i-1}+U_{2i}\right) _{1\le i \le m} \end{aligned}$$

have the same distribution. Consequently, by taking $p=2m$ in Lemma 3.1,

$$\begin{aligned} EY= & {} m\cdot E\left( U_1+U_2\right) ^2 \nonumber \\= & {} 2m\left[ E\left( U_1^2\right) +E\left( U_1U_2\right) \right] \nonumber \\= & {} 2m\left[ \frac{3}{4m(m+1)} + \frac{1}{4m(m+1)}\right] =\frac{2}{m+1}. \end{aligned}$$

(3.4)

Similarly,

$$\begin{aligned} E\frac{\xi _1^4}{\left( \xi _1+\cdots + \xi _m\right) ^4}= & {} E\left[ \left( U_1+U_2\right) ^4\right] \\= & {} 2E\left( U_1^4\right) +8 E\left( U_1^3U_2\right) + 6 E\left( U_1^2U_2^2\right) \\= & {} \frac{105}{8}\frac{1}{m(m+1)(m+2)(m+3)} + \frac{15}{2}\frac{1}{m(m+1)(m+2)(m+3)}\\&+ \frac{27}{8}\frac{1}{m(m+1)(m+2)(m+3)}\\= & {} \frac{24}{m(m+1)(m+2)(m+3)} \end{aligned}$$

and

$$\begin{aligned} E\frac{\xi _1^2\xi _2^2}{\left( \xi _1+\cdots + \xi _m\right) ^4}= & {} E\left[ \left( U_1+U_2\right) ^2\left( U_3+U_4\right) ^2\right] \\= & {} 4E\left( U_1^2U_2^2\right) + 8E\left( U_1^2U_2U_3\right) + 4E\left( U_1U_2U_3U_4\right) \\= & {} \frac{9}{4}\frac{1}{m(m+1)(m+2)(m+3)} + \frac{3}{2}\frac{1}{m(m+1)(m+2)(m+3)}\\&+ \frac{1}{4}\frac{1}{m(m+1)(m+2)(m+3)}\\= & {} \frac{4}{m(m+1)(m+2)(m+3)}. \end{aligned}$$

It follows from (3.3) and (3.4) that

$$\begin{aligned}&E(Y^2)=\frac{4m+20}{(m+1)(m+2)(m+3)};\\&E(Y^2)-(EY)^2 = \frac{4m+20}{(m+1)(m+2)(m+3)}-\left( \frac{2}{m+1}\right) ^2\\&\quad = \frac{4m-4}{(m+1)^2(m+2)(m+3)}. \end{aligned}$$

This and (3.2) say that

$$\begin{aligned} \frac{\bar{x}}{n^2}\rightarrow \frac{\alpha }{m+1}\ \ \text{ and } \ \ \frac{s^2}{n^4}\rightarrow \frac{(m-1)\alpha ^2}{(m+1)^2(m+2)(m+3)}. \end{aligned}$$

1.2 Verification of (2.77)

Verification of (2.77)

Trivially, $\Omega (x)$ in (1.12) is an even function and $\Omega (x)' = \frac{2}{\pi } \arcsin \frac{x}{2}$ for $|x| < 2$. Then

$$\begin{aligned} \begin{aligned}&\int _{-2}^2 (\Omega (x) - x)^2 \,\hbox {d}x = \int _{-2}^2 \Omega (x)^2 \,\hbox {d}x + \int _{-2}^2 x^2 \,\hbox {d}x\\&\quad = x \cdot \Omega (x)^2 \Bigr |_{-2}^2 - \int _{-2}^2 x \cdot 2\Omega (x) \cdot \Omega (x)' \,\hbox {d}x + \frac{x^3}{3} \Bigr |_{-2}^2\\&\quad = \frac{64}{3} - \frac{16}{\pi ^2} \int _{0}^2 x \arcsin {\frac{x}{2}} \cdot \left( x \arcsin \frac{x}{2} +\sqrt{4-x^2} \right) \,\hbox {d}x. \end{aligned} \end{aligned}$$

Now, set $x=2\sin \theta$ for $0\le \theta \le \frac{\pi }{2}$, the above integral becomes

$$\begin{aligned}&\int _{0}^{\frac{\pi }{2}} 2\theta \sin \theta \left( 2\theta \sin \theta + 2\cos \theta \right) 2\cos \theta \, \hbox {d}\theta \nonumber \\&\quad = 2\int _{0}^{\frac{\pi }{2}}\left( \theta \sin \theta +\theta \sin (3\theta )+\theta ^2\cos \theta -\theta ^2\cos (3\theta )\right) \,\hbox {d}\theta \end{aligned}$$

(3.5)

by trigonometric identities. It is easy to verify that

$$\begin{aligned}&\theta \sin \theta =\left( \sin \theta -\theta \cos \theta \right) ';\ \ \ \ \theta \sin (3\theta )=\frac{1}{9}\left( \sin (3\theta )-3\theta \cos (3\theta )\right) ';\\&\theta ^2\cos \theta =\left( \theta ^2\sin \theta +2\theta \cos \theta -2\sin \theta \right) ';\\&\theta ^2\cos (3\theta )=\frac{1}{27}\left( 9\theta ^2\sin (3\theta ) +6\theta \cos (3\theta )-2\sin (3\theta )\right) '. \end{aligned}$$

Thus, the term in (3.5) is equal to

$$\begin{aligned} 2\left( 1+\left( -\frac{1}{9}\right) + \left( \frac{\pi ^2}{4}-2\right) -\frac{1}{27}(-\frac{9\pi ^2}{4}+2)\right) =\frac{2}{3}\pi ^2-\frac{64}{27}. \end{aligned}$$

It follows that

$$\begin{aligned} \int _{-2}^2 \left( \Omega (x) - x\right) ^2 \,\hbox {d}x =\frac{64}{3} - \frac{16}{\pi ^2}\left( \frac{2}{3}\pi ^2-\frac{64}{27}\right) =\frac{32}{3}+\frac{1024}{27\pi ^2}. \end{aligned}$$

This completes the verification. $\square$

1.3 Derivation of Density Functions in Theorem 1

In this section, we will derive explicit formulas for the limiting distribution in Theorem 1. For convenience, we rewrite the conclusion as

$$\begin{aligned} \frac{2}{\alpha }\cdot \frac{\lambda _\kappa }{n^2}\rightarrow \nu , \end{aligned}$$

where $\nu$ is different from $\mu$ in Theorem 1 by a factor of $\frac{2}{\alpha }$. We will only evaluate the cases $m=2, 3$. We first state the conclusions and prove them

Case 1. For $m=2$, the support of $\nu$ is $[\frac{1}{2}, 1]$ and the cdf of $\nu$ is

$$\begin{aligned} F(t) =\sqrt{2t-1} \end{aligned}$$

(3.6)

for $t\in [\frac{1}{2}, 1]$. Hence the density function is given by

$$\begin{aligned} f(t)= \frac{1}{\sqrt{2t-1}},\ \ t\in \left[ \frac{1}{2}, 1\right] . \end{aligned}$$

Case 2. For $m=3$, the support of $\nu$ is $[\frac{1}{3}, 1]$, and the cdf of $\nu$ is

$$\begin{aligned} F(t) = {\left\{ \begin{array}{ll} \frac{2}{\sqrt{3}} \pi (t-\frac{1}{3}), &{} \text {if } \frac{1}{3} \le t< \frac{1}{2}; \\ \frac{2}{\sqrt{3}} \left( (t-\frac{1}{3}) (\pi - 3\arccos \frac{1}{\sqrt{6t-2}}) + \frac{\sqrt{6}}{2} \sqrt{t-\frac{1}{2}}\, \right) , &{} \text {if } \frac{1}{2} \le t < 1. \end{array}\right. } \end{aligned}$$

(3.7)

By differentiation, we get the density function

$$\begin{aligned} f(t)={\left\{ \begin{array}{ll} \frac{2}{\sqrt{3}} \pi , &{} \text {if } \frac{1}{3} \le t < \frac{1}{2}; \\ \frac{2}{\sqrt{3}} \left( \pi - 3\arccos \frac{1}{\sqrt{6t-2}} \right) , &{} \text {if } \frac{1}{2} \le t \le 1.\\ \end{array}\right. } \end{aligned}$$

The above are the two density functions claimed below the statement of Theorem 1. Now we prove them.

From a comment below Theorem 1, the limiting law of $\frac{2}{\alpha }\cdot \frac{\lambda _\kappa }{n^2}$ is the same as the distribution of $\sum _{i=1}^m Y_i^2$, where $(Y_1,\ldots ,Y_m)$ has uniform distribution over the set

$$\begin{aligned} \mathcal {H}:=\left\{ (y_1,\ldots ,y_{m}) \in [0,1]^{m};\, \sum _{i=1}^{m}y_i = 1 \right\} . \end{aligned}$$

By (2.2) the volume of $\mathcal {H}$ is $\frac{\sqrt{m}}{(m-1)!}$. Therefore, the cdf of $\sum _{i=1}^m Y_i^2$ is

$$\begin{aligned} F(t)=P\left( \sum _{i=1}^m Y_i^2 \le t\right) = \frac{(m-1)!}{\sqrt{m}} \cdot \text {volume of } \left\{ \sum _{i=1}^m y_i^2 \le t\right\} \cap \mathcal {H},\ \ t\ge 0. \end{aligned}$$

(3.8)

Denote $B_m(t):=\{ \sum _{i=1}^m y_i^2 \le t\} \subset \mathbb {R}^m$. Let V(t) be the volume of $B_m(t) \cap \mathcal {H}$. We start with some facts for any $m\ge 2$.

First, $V(t) = 0$ for $t < \frac{1}{m}$. In fact, if $(y_1, \ldots , y_m)\in B_m(t)\cap \mathcal {H}$, then

$$\begin{aligned} \frac{1}{m}=\frac{(\sum _{i=1}^m y_i)^2}{m} \le \sum _{i=1}^m y_i^2\le t. \end{aligned}$$

Further, for $t>1$, $\mathcal {H}$ is inscribed in $B_m(t)$ and thus $V(t) = \frac{\sqrt{m}}{(m-1)!}$. Now assume $1/m \le t \le 1$.

The proof of (3.6). Assume $m=2$. If $1/2 \le t \le 1$, then $\{(y_1,y_2) \in [0,1]^2: y_1 + y_2 =1 \}\cap \{ y_1^2 + y_2^2 \le t \}$ is a line segment. Easily, the endpoints of the line segment are

$$\begin{aligned} \left( \frac{1+\sqrt{2t-1}}{2}, \frac{1-\sqrt{2t-1}}{2}\right) \quad \text {and} \quad \left( \frac{1-\sqrt{2t-1}}{2}, \frac{1+\sqrt{2t-1}}{2}\right) , \end{aligned}$$

respectively. Thus $V(t) = \sqrt{2(2t-1)}.$ Therefore the conclusion follows directly from (3.8).

The proof of (3.7). We first observe that as t increases from $\frac{1}{3}$ to 1, the intersection $B_{3}(t)\cap \mathcal {H}$ expands and passes through $\mathcal {H}$ as t exceeds some critical value $t_0$; see Fig. 6.

We claim that $t_0 = \frac{1}{2}$. Indeed, the center C of the intersection of $B_{3}(t)$ and the hyperplane $\mathcal {I}:=\{(y_1, y_2, y_3)\in \mathbb {R}^3; y_1+y_2+y_3=1\} \supset \mathcal {H}$ is $C=(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}).$ Thus, the distance from the origin to $\mathcal {I}$ is $d=((\frac{1}{3})^2+(\frac{1}{3})^2+(\frac{1}{3})^2)^{1/2} = \frac{1}{\sqrt{3}}.$ By Pythagorean’s theorem, the radius of the intersection (disc) on $\mathcal {I}$ is

$$\begin{aligned} R(t) = \sqrt{t - d^2} = \sqrt{t - \frac{1}{3}}. \end{aligned}$$

Let $t_0$ be the value such that the intersection $B_3(t)\cap \mathcal {H}$ exactly inscribes $\mathcal {H}$. By symmetry, the intersection point at the (x, y)-plane is $M = (\frac{1}{2},\frac{1}{2}, 0)$; see Fig. 6b. Therefore $|CM| = \sqrt{\frac{1}{6}}.$ Solving $t_0$ from $|CM| = R(t_0),$ we have $t_0 = \frac{1}{2}$.

Fig. 6

Shaded region indicates volume V(t) of intersection as t changes from 1/3 to 1 as $m=3$

When $\frac{1}{3} \le t < \frac{1}{2}$, the intersection locates entirely in $\mathcal {H}$; see Fig. 6a. Then

$$\begin{aligned} V(t) = \pi R(t)^2 = \pi \left( t - \frac{1}{3}\right) . \end{aligned}$$

When $\frac{1}{2} \le t \le 1$, the volume of the intersection part (see Fig. 6c) is given by

$$\begin{aligned} V(t) = \pi R(t)^2 - 3 \cdot V_{\text {cs}}(h(t),R(t)), \end{aligned}$$

where $V_{\text {cs}}(h(t),R(t))$ is the area of circular segment with radius R(t) and height

$$\begin{aligned} h(t)=R(t)- |CM|= \sqrt{t-\frac{1}{3}}-\sqrt{\frac{1}{6}}. \end{aligned}$$

Therefore, it is easy to check

$$\begin{aligned} V(t)= \pi \left( t - \frac{1}{3}\right) - 3\left( t - \frac{1}{3}\right) \arccos \frac{1}{\sqrt{6t-2}} + 3 \sqrt{\frac{1}{6}\left( t-\frac{1}{2}\right) }\,. \end{aligned}$$

This and (3.8) yield the desired conclusion.