Eigenvector distribution in the critical regime of BBP transition

Abstract

In this paper, we study the random matrix model of Gaussian Unitary Ensemble (GUE) with fixed-rank (aka spiked) external source. We will focus on the critical regime of the Baik–Ben Arous–Péché (BBP) phase transition and establish the distribution of the eigenvectors associated with the leading eigenvalues. The distribution is given in terms of a determinantal point process with extended Airy kernel. Our result can be regarded as an eigenvector counterpart of the BBP eigenvalue phase transition [6]. The derivation of the distribution makes use of the recently re-discovered eigenvector–eigenvalue identity, together with the determinantal point process representation of the GUE minor process with external source.

A Proof of results in Sect. 2

Proof of Lemma 6

We only give a sketch of the proof of Lemma 6, because the case that $j_1 = j_2$, and all $\alpha _j = \sqrt{N}$ (equivalent to $a_{k - j + 1} = 0$) for $j = 1, \dotsc , k$ has already been proved in [46] which follows closely the method used in [6]. Our proof is an adaption of that in [46]. The main difference between our lemma and the results in [6] and [46] is that we require $\epsilon$ to be large enough while in [6] and [46], $\epsilon$ is only required to be positive, because they essentially assume all $a_j = 0$. Our assumption on $\epsilon$ implies that the operators $e^{\epsilon (x - y)} K^{k_1, k_2}_{{{\,\mathrm{Airy}\,}}, {\mathbf {a}}}(x, y)$ and $e^{\epsilon (x - y)} K^{j_1, j_2}_{N, {{\,\mathrm{scaled}\,}}}(x, y)$ are both trace class.

Proof

Below in this proof we are going to use notation in [46] that is quite different from the notation used elsewhere in our paper.

We define, analogous to [46, Formula (16)],

$$\begin{aligned} K'_{N, j_1, j_2}(x, y) = N^{-1/6} N^{(j_1 - j_2)/6} e^{N^{1/3}(y {-} x)} {\widetilde{K}}^{j_1, j_2}_{{{\,\mathrm{GUE}\,}}, \varvec{\alpha }}(2\sqrt{N} {+} N^{-1/6} y, 2\sqrt{N} {+} N^{-1/6} x).\nonumber \\ \end{aligned}$$

(A.1)

We only need to consider the convergence of $e^{\epsilon (y - x)} K'_{N, j_1, j_2}(x, y)$ to $e^{\epsilon (y - x)} {\widetilde{K}}^{k_1, k_2}_{{{\,\mathrm{Airy}\,}}, {\mathbf {a}}}(y, x)$ (pointwise and in trace norm). Analogous to [46, Formulas (18) and (19)], we denote $F(z) = z^2/2 - 2z + \log z$ (see [46, Formula (17)]), ${\tilde{w}}_c = 1 + \epsilon N^{-1/3}$ (see [46, Formula (16)]), and define

$$\begin{aligned} H_{N, j_2}(x) = {}&\frac{N^{\frac{1}{3}}}{2\pi } \int _{\varGamma } \left[ z^k \prod ^k_{i = j_2 + 1} \frac{1}{z - \frac{\alpha _i}{\sqrt{N}}} \right] \exp (-N F(z)) \exp (N^{\frac{1}{3}} x(z - {\tilde{w}}_c)) \mathrm {d}z, \end{aligned}$$

(A.2)

$$\begin{aligned} J_{N, j_1}(y) {=} {}&\frac{N^{\frac{1}{3}}}{2\pi } \int _{\gamma } \left[ w^{-k} \prod ^k_{i {=} j_1 {+} 1} \left( w {-} \frac{\alpha _i}{\sqrt{N}} \right) \right] \exp (N F(w)) \exp ({-}N^{\frac{1}{3}} y(w {-} {\tilde{w}}_c)) \mathrm {d}w, \end{aligned}$$

(A.3)

where the contours $\varGamma$ and $\gamma$ are defined as in [46, Formula (14)]. Then we have, analogous to [46, Proposition 2.1],

$$\begin{aligned} N^{(j_1 - j_2)/3} e^{\epsilon (y - x)} K'_{N, j_1, j_2}(x, y) = -\int ^{\infty }_0 H_{N, j_2}(x + t) J_{N, j_1}(y + t) \mathrm {d}t. \end{aligned}$$

(A.4)

Next, analogous to [46, Formulas (21) and (22)], define

$$\begin{aligned} H_{\infty , k_2}(x) = {}&\frac{\exp (-\epsilon x)}{2\pi } \int _{\varGamma _{\infty }} \exp \left( xz - \frac{z^3}{3} \right) \prod ^{k_2}_{i = 1} \frac{1}{z - a_i} \mathrm {d}z, \end{aligned}$$

(A.5)

$$\begin{aligned} J_{\infty , k_1}(y) = {}&\frac{\exp (\epsilon y)}{2\pi } \int _{\gamma _{\infty }} \exp \left( -yw + \frac{w^3}{3} \right) \prod ^{k_1}_{i = 1} (w - a_i) \mathrm {d}w, \end{aligned}$$

(A.6)

where the contours $\varGamma _{\infty }$ and $\gamma _{\infty }$ are defined in [46, Figure 1]. By the arguments in [46, Sections 2.1 and 2.2], we have, analogous to [46, Proposition 2.2], that for any fixed $y_0 \in {\mathbb {R}}$, there exists $C > 0$, $c > 0$, an integer $N_0 > 0$ such that

$$\begin{aligned} \left|Z_{N, j_2} H_{N, j_2}(x) - H_{\infty , k_2}(x) \right|\le {}&\frac{C \exp (-cx)}{N^{1/3}},&\text {for any } x > y_0, N \ge N_0, \end{aligned}$$

(A.7)

$$\begin{aligned} \left|Z^{-1}_{N, j_1} J_{N, j_1}(y) - J_{\infty , k_1}(y) \right|\le {}&\frac{C \exp (-cy)}{N^{1/3}},&\text {for any } y > y_0, N \ge N_0, \end{aligned}$$

(A.8)

where $Z_{N, j} = N^{(j - k)/3} \exp (NF(1))$. On the other hand, we have

$$\begin{aligned} e^{\epsilon (y - x)} {\widetilde{K}}^{k_2, k_1}_{{{\,\mathrm{Airy}\,}}, {\mathbf {a}}}(y, x) = -\int ^{\infty }_0 H_{\infty , k_2}(x + t) J_{\infty , k_1}(y + t) \mathrm {d}t. \end{aligned}$$

(A.9)

The proof is finished by using the argument in [6, Section 3.3]. $\square$

Proof of Lemma 7

By Lemma 6, for any n, the joint distribution of $\lambda ^{(N - j)}_1, \lambda ^{(N - j - 1)}_1, \lambda ^{(N - j)}_2, \lambda ^{(N - j - 1)}_2, \dotsc , \lambda ^{(N - j)}_n$ converges weakly to that of $\xi ^{(k - j)}_1, \xi ^{(k - j - 1)}_1, \xi ^{(k - j)}_2, \xi ^{(k - j - 1)}_2, \dotsc , \xi ^{(k - j)}_n$ up to a scaling transform. Hence we have that the interlacing inequality (2.6) implies the weak interlacing property

$$\begin{aligned} +\infty > \xi ^{(k - j)}_1 \ge \xi ^{(k - j - 1)}_1 \ge \xi ^{(k - j)}_2 \ge \xi ^{(k - j - 1)}_2 \ge \cdots \ge \xi ^{(k - j)}_n. \end{aligned}$$

(A.10)

On the other hand, the determinantal structure requires that the point process consisting of $\xi ^{(k - j)}_i$ and $\xi ^{(k - j - 1)}_l$ is simple, so with probability 1 the inequalities in (A.10) are all strict. So with probability 1 we have (2.7) by letting $n \rightarrow \infty$.

Proof of Lemma 8

We prove the lemma in three steps: First the right tail estimate of $\xi ^{(k)}_j$ in part 1, then the left tail estimate of $\xi ^{(k)}_j$, and at last we prove part 2 about the rigidity of $\xi ^{(k)}_n$.

Proof of the right tail estimate of $\xi ^{(k)}_j$ We note that

$$\begin{aligned} {\mathbb {P}}(\xi ^{(k)}_j> t) \le {\mathbb {P}}(\xi ^{(k)}_1 > t) \le {\mathbb {E}}( \# \text { of } \xi ^{(k)}_i \text { on } [t, +\infty )) = \int ^{+\infty }_t K^{k, k}_{{{\,\mathrm{Airy}\,}}, {\mathbf {a}}}(x, x) \mathrm {d}x.\nonumber \\ \end{aligned}$$

(A.11)

Then by (1.7),

$$\begin{aligned} \int ^{+\infty }_t K^{k, k}_{{{\,\mathrm{Airy}\,}}, {\mathbf {a}}}(x, x) \mathrm {d}x = \frac{1}{(2\pi \mathrm {i})^2} \int _{\gamma } \mathrm {d}u \int _{\sigma } \mathrm {d}v \frac{e^{\frac{u^3}{3} - tu}}{e^{\frac{v^3}{3} - tv}} \frac{\prod ^k_{j = 1} (u - a_j)}{\prod ^k_{j = 1} (v - a_j)} \frac{1}{(u - v)^2}.\nonumber \\ \end{aligned}$$

(A.12)

Let $\gamma$ and $\sigma$ be deformed into $\gamma _{{{\,\mathrm{std}\,}}}(\sqrt{t})$ and $\sigma _{{{\,\mathrm{std}\,}}}(-\sqrt{t})$ (c.f. (4.15)). By standard saddle point analysis, we find that as $t \rightarrow +\infty$, the integral (A.12) concentrates on the region $u \in \gamma _{{{\,\mathrm{std}\,}}}(\sqrt{t}) \cap \{ u - \sqrt{t} = {\mathcal {O}}(t^{-1/4}) \}$ and $v \in \sigma _{{{\,\mathrm{std}\,}}}(-\sqrt{t}) \cap \{ v + \sqrt{t} = {\mathcal {O}}(t^{-1/4}) \}$. Then we conclude that as $t \rightarrow +\infty$,

$$\begin{aligned} \int ^{+\infty }_t K^{k, k}_{{{\,\mathrm{Airy}\,}}, {\mathbf {a}}}(x, x) \mathrm {d}x = {\mathcal {O}}\left( t^{-3/2} \exp \left( -\frac{4}{3} t^{3/2} \right) \right) . \end{aligned}$$

(A.13)

Hence by choosing C properly, we have ${\mathbb {P}}(\xi ^{(k)}_j > t)< \int ^{+\infty }_t K^{k, k}_{{{\,\mathrm{Airy}\,}}, {\mathbf {a}}}(x, x) dx < C e^{-t/C}$. $\square$

Proof of the left tail estimate of $\xi ^{(k)}_j$ We note that by the interlacing property in Lemma 7, ${\mathbb {P}}(\xi ^{(k)}_j< -t)< {\mathbb {P}}(\xi ^{(0)}_j < -t)$, where $\xi ^{(0)}_j$ is the jth particle in the determinantal point process defined by the Airy kernel (1.6). Then by [53], with any $\lambda \in (0, 1)$, we have

$$\begin{aligned} {\mathbb {P}}(\xi ^{(0)}_j< -t) = \sum ^{j - 1}_{n = 0} E(n; -t) < (1 - \lambda )^{1 - j} \sum ^{\infty }_{n = 0} (1 - \lambda )^n E(n; -t) = 2^{j - 1} D(-t, \lambda ),\nonumber \\ \end{aligned}$$

(A.14)

where $E(n; -t)$ is the probability that exactly n particles are in $[-t, \infty )$ as denoted in [53, Section ID], and $D(-t, \lambda )$ is defined by [53, Formula (1.17)] as

$$\begin{aligned}&D(-t, \lambda ) = \exp \left( - \int ^{\infty }_{-t} (x + t) q(x; \lambda )^2 \mathrm {d}x \right) , \nonumber \\&\text {where} \quad \begin{aligned} \frac{\mathrm {d}q(s; \lambda )}{\mathrm {d}s^2} = {}&sq(s; \lambda ) + 2q^3(s; \lambda ), \\ q(s; \lambda ) \sim {}&\sqrt{\lambda } {{\,\mathrm{Ai}\,}}(s) \text { as } s \rightarrow \infty . \end{aligned} \end{aligned}$$

(A.15)

The function $q(s; \lambda )$ is the Ablowitz-Segur solution to the Painlevé II equation [1, 49], its asymptotics at $+\infty$ is given by the Airy function multiplied by constant $\sqrt{\lambda }$. The asymptotic behaviour of $q(s; \lambda )$ has been extensively studied, see [28] for a rigorous and systematic discussion. We then derive the upper bound of $D(-t, \lambda )$ for large t from the asymptotics of $q(s; \lambda )$, and finally justify the estimate ${\mathbb {P}}(\xi ^{(k)}_j< -t) < C e^{-t/C}$ for some properly chosen C. $\square$

Proof of the rigidity of $\xi ^{(k)}_n$ We note that by the interlacing property (2.6), for all $n > k$,

$$\begin{aligned}&{\mathbb {P}}\left( \left|\xi ^{(k)}_n + \left( \frac{3\pi n}{2} \right) ^{2/3} \right|> n^{\frac{3}{5}} \right) \nonumber \\&\quad \le {\mathbb {P}}\left( \# \text { of } \xi ^{(0)}_l \text { in } \left( -\left( \frac{3\pi n}{2} \right) ^{2/3} + n^{\frac{3}{5}}, \infty \right) \text { is } \ge n - k \right) \nonumber \\&\qquad + {\mathbb {P}}\left( \# \text { of } \xi ^{(0)}_l \text { in } \left( -\left( \frac{3\pi n}{2} \right) ^{2/3} - n^{\frac{3}{5}}, \infty \right) \text { is } < n \right) . \end{aligned}$$

(A.16)

Since $\xi ^{(0)}_n$ are the nth particle in the determinantal point process with the Airy kernel, so the problem is reduced to the rigidity of particles in this determinantal point process. The desired regidity can be deduced from the mean and variance of the number of $\xi ^{(0)}_l$ in $(-T, \infty )$ and the Markov inequality. If we denote the number of $\xi ^{(0)}_l$ in $(-T, \infty )$ as $v_1(T)$, in the notation of [51], then

$$\begin{aligned} {\mathbb {E}}(v_1(T)) = 2T^{3/2}/(3\pi ) + {\mathcal {O}}(1), \quad \text {and} \quad {{\,\mathrm{Var}\,}}(v_1(T)) = {\mathcal {O}}(\log T) \end{aligned}$$

(A.17)

as $T \rightarrow +\infty$, see [51, Theorem 1 and the paragraph above Theorem 1]⁴. That is enough to show that as $l \rightarrow \infty$,

$$\begin{aligned}&{\mathbb {P}}\left( v_1 \left( \left( \frac{3\pi n}{2} \right) ^{2/3} - n^{\frac{3}{5}} \right) \ge n - k \right) = {\mathcal {O}}\left( \frac{\log n}{n^{6/5}} \right) ,\nonumber \\&{\mathbb {P}}\left( v_1 \left( \left( \frac{3\pi n}{2} \right) ^{2/3} + n^{\frac{3}{5}} \right) < n \right) = {\mathcal {O}}\left( \frac{\log n}{n^{6/5}} \right) . \end{aligned}$$

(A.18)

By choosing the constant c properly, we obtain (2.9) for all $n \ge 2$. $\square$

Proof of Lemma 9

This lemma is analogous to Lemma 8. We prove it in four steps, with the first three steps parallel to those in the proof of Lemma 8: First, the right tail estimate of $\sigma _j$ (part 1), next the left tail estimate of $\sigma _j$ (part 1), and then the rigidity for $\sigma _n$ close to the edge (part 2), and at last the rigidity of $\sigma _n$ in the bulk (part 3). In part 1 we also need to consider $\sigma _N$, but we omit it, because the estimates for $\sigma _N$ are analogous to the estimate for $\sigma _1$.

Proof of the right tail estimate of $\sigma _j$ We use the same idea as in (A.11), and write

$$\begin{aligned}&{\mathbb {P}}(\sigma _j> 2\sqrt{N} + tN^{-\frac{1}{6}}) \le {\mathbb {P}}(\sigma _1 > 2\sqrt{N} + tN^{-\frac{1}{6}}) \nonumber \\&\quad = \int ^{\infty }_{2\sqrt{N} + t N^{-1/6}} K^{0, 0}_{{{\,\mathrm{GUE}\,}}, \varvec{\alpha }}(x, x) \mathrm {d}x = \int ^{\infty }_t K'_{N, 0, 0}(x, x) \mathrm {d}x, \end{aligned}$$

(A.19)

where $K'_{N, 0, 0}(x, x)$ is defined in (A.1). Although we can evaluate the right-hand side of (A.19) like (A.12), we prefer an indirect method that relies on result and proof of Lemma 6. We recall that as a special case of (A.4),

$$\begin{aligned} K'_{N, 0, 0}(x, x) = -\int ^{\infty }_0 H_{N, 0}(x + t) J_{N, 0}(x + t) \mathrm {d}t, \end{aligned}$$

(A.20)

and then by (A.7) and (A.8), there exists $N_0 > 0$ and $C > 0$ such that for $x > 0$, $N > N_0$,

$$\begin{aligned}&\left|Z_{N, 0} H_{N, 0}(x) - H_{\infty , k}(x) \right|\le \frac{C \exp (-cx)}{N^{1/3}}, \nonumber \\&\left|Z^{-1}_{N, j_1} J_{N, 0}(x) - J_{\infty , k}(x) \right|\le \frac{C \exp (-cy)}{N^{1/3}}, \end{aligned}$$

(A.21)

where $Z_{N, 0}= N^{-k/3} \exp (-3N/2)$ and $H_{\infty , k}, J_{\infty , k}$ are defined in (A.5) and (A.6). Hence by the very rough estimate (whose proof is omitted) that $H_{\infty , k}(x) = {\mathcal {O}}(1)$ and $J_{\infty , k}(x) = {\mathcal {O}}(1)$ for all $x > 0$, and with the help of (A.9), we have that

$$\begin{aligned} K'_{N, 0, 0}(x, x) - K^{k, k}_{{{\,\mathrm{Airy}\,}}, {\mathbf {a}}}(x, x) = {\mathcal {O}}(N^{-1/3} e^{-cx}), \quad \text {for all } x> 0 \text { and } N > N_0.\nonumber \\ \end{aligned}$$

(A.22)

Therefore, the desired right tail estimate of $\sigma _j$ is implied by the estimate (A.13) for the right tail estimate of $\xi ^{(k)}_j$. $\square$

Proof of the left tail estimate of $\sigma _j = \lambda ^{(N)}_j$ We use the same idea as in the proof of the left tail estimate of $\xi ^{(k)}_j$, that ${\mathbb {P}}(\sigma _j< 2\sqrt{N} - tN^{-1/6}) \le {\mathbb {P}}(\lambda ^{(N - k)}_j < 2\sqrt{N} - tN^{-1/6})$, so it is not hard to see that it suffices to prove that there exists $C > 0$ such that

$$\begin{aligned} {\mathbb {P}}(\lambda ^{(N - k)}_j< 2\sqrt{N - k} - t(N - k)^{-\frac{1}{6}}) < C e^{-t/C}, \quad \text {for all } 2 \le t \le 2(N - k)^{2/3}.\nonumber \\ \end{aligned}$$

(A.23)

where $\lambda ^{(N - k)}_j$ is the jth largest eigenvalue of a GUE random matrix with dimension $N - k$. The $j = 1$ case of (A.23) exists in literature, see [43, Section 5.3, especially Formula (5.16)], where a stronger version of (A.23) is derived in a very accessible way. The $j > 1$ case of (A.23) is not found in literature, to the best knowledge of the authors. However, we can extend the method in [43, Section 5.3] to solve this case. To see it, we note that like (A.14), with $\lambda \in (0, 1)$, we have

$$\begin{aligned} \begin{aligned}&{\mathbb {P}}(\lambda ^{(N - k)}_j< 2\sqrt{N - k} - t(N - k)^{-\frac{1}{6}}) \\&\quad = \sum ^{j - 1}_{n = 0} E(n; 2\sqrt{N - k} - t(N - k)^{-\frac{1}{6}}) \\&\quad < (1 - \lambda )^{1 - j} \sum ^{\infty }_{n = 0} (1 - \lambda )^n E(n; 2\sqrt{N - k} - t(N - k)^{-\frac{1}{6}}) \\&\quad = (1 - \lambda )^{1 - j} \det \left( {{\,\mathrm{Id}\,}}- \lambda K \right) , \end{aligned}\nonumber \\ \end{aligned}$$

(A.24)

where K is the $N \times N$ matrix whose (m, n) entry is

$$\begin{aligned} \langle P_{m - 1}, P_{n - 1} \rangle _{L^2((1 - \frac{t}{2}(N - k)^{-2/3}, \infty ), \mathrm {d}\mu )}, \end{aligned}$$

(A.25)

such that the meanings of $P_m$ and $\mathrm {d}\mu$ are the same as in [43, Formula (1.11)]. Then by the same arguments that leads to [43, Formula (5.14)], we have

$$\begin{aligned} \det \left( {{\,\mathrm{Id}\,}}- \lambda K \right)= & {} \prod ^N_{i = 1} (1 - \lambda \rho _i) \le e^{-\frac{1}{2} \sum ^N_{i = 1} \rho _i}\nonumber \\= & {} \exp \left( -\lambda N \mu ^N \left( (1 - \frac{t}{2}(N - k)^{-2/3}, \infty ) \right) \right) , \end{aligned}$$

(A.26)

where $\rho _i$ are the eigenvalues of K, and $\mu ^N$ is the measure defined in [43, Formula (1.4)]. We note that if we let $\lambda = 1$ in (A.26), then (A.26) is equivalent to [43, Formula (5.14)]. At last, using the estimate of $\mu ^N((1 - \frac{t}{2}(N - k)^{-2/3}, \infty ))$ given in [43, Section 5.3], we derive an estimate of $\det \left( {{\,\mathrm{Id}\,}}- \lambda K \right)$, which yields the desired estimate of ${\mathbb {P}}(\lambda ^{(N - k)}_j < 2\sqrt{N - k} - t(N - k)^{-1/6})$ and ${\mathbb {P}}(\sigma _j < 2\sqrt{N} - tN^{-1/6})$. Finally, we note that essentially the idea of the proof above is in [56]. $\square$

Proof of the rigidity of $\sigma _n$ for $n \le CN^{1/10}$ As in (2.12), we note that (2.12) is analogous to (2.9), and can be proved by an analogous argument. Instead of (A.17), we have that if $v^{(n)}_1(T)$ is the number of eigenvalues of an n-dimensional GUE random matrix in the interval $(2\sqrt{n} - n^{-1/6}T, +\infty )$, then as $n \rightarrow \infty$, $T \ge T_0$ a positive constant, and $T/n = o(1)$, by the result of [37]⁵

$$\begin{aligned} {\mathbb {E}}(v^{(n)}_1(T)) = 2T^{3/2}/(3\pi ) + {\mathcal {O}}(1), \quad \text {and} \quad {{\,\mathrm{Var}\,}}(v^{(n)}_1(T)) = {\mathcal {O}}(\log T). \end{aligned}$$

(A.27)

Then we prove (2.12) by the same argument as the proof of (2.9). $\square$

The proof of the regidity of $\sigma _n$ as in (2.13) This is a direct consequence of the interlacing property $\lambda ^{(N - k)}_n \le \sigma _n \le \lambda ^{(N - k)}_{n - k}$ and the rigidity of GUE eigenvalues in [33, Theorem 2.2], which states the rigidity of eigenvalues for Wigner matrices that of which the GUE random matrices are a special case. $\square$