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Nodal Statistics of Planar Random Waves

Abstract

We consider Berry's random planar wave model (1977) for a positive Laplace eigenvalue \(E > 0\), both in the real and complex case, and prove limit theorems for the nodal statistics associated with a smooth compact domain, in the high-energy limit (\(E \rightarrow \infty\)). Our main result is that both the nodal length (real case) and the number of nodal intersections (complex case) verify a Central Limit Theorem, which is in sharp contrast with the non-Gaussian behaviour observed for real and complex arithmetic random waves on the flat 2-torus, see Marinucci et al. (2016) and Dalmao et al. (2016). Our findings can be naturally reformulated in terms of the nodal statistics of a single random wave restricted to a compact domain diverging to the whole plane. As such, they can be fruitfully combined with the recent results by Canzani and Hanin (2016), in order to show that, at any point of isotropic scaling and for energy levels diverging sufficently fast, the nodal length of any Gaussian pullback monochromatic wave verifies a central limit theorem with the same scaling as Berry's model. As a remarkable byproduct of our analysis, we rigorously confirm the asymptotic behaviour for the variances of the nodal length and of the number of nodal intersections of isotropic random waves, as derived in Berry (2002).

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Acknowledgements

The research leading to this work was supported by the Grant F1R-MTH-PUL-15CONF–CONFLUENT (Ivan Nourdin), by the Grant F1R-MTH-PUL-15STAR-STARS (Giovanni Peccati and Maurizia Rossi) and by the FNR Grant R-AGR-3376-10 (Giovanni Peccati) at the University of Luxembourg. We are grateful to Massimo Notarnicola for a number of important remarks. We also thank an anonymous referee for a careful reading of our paper, and for useful suggestions.

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Correspondence to Ivan Nourdin.

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Appendices

Appendix A

Proof of Lemma 3.1

It is a standard fact that for any \(m_1, m_2, n_1, n_2\in \mathbb {N}_{\ge 0}\)

$$\begin{aligned} {{\mathbb {E}}}\left[ \frac{\partial ^{m_1+m_2}}{\partial x_1^{m_1} \partial x_2^{m_2}} B_E(x) \frac{\partial ^{n_1+n_2}}{\partial y_1^{n_1} \partial y_2^{n_2}} B_E(y) \right]&= \frac{\partial ^{m_1 + m_2+n_1 + n_2}}{\partial x_1^{m_1}\partial y_1^{n_1} \partial x_2^{m_2}\partial y_2^{n_2}} {{\mathbb {E}}}[B_E(x) B_E(y)]\nonumber \\&= \frac{\partial ^{m_1 + m_2+n_1 + n_2}}{\partial x_1^{m_1}\partial y_1^{n_1} \partial x_2^{m_2}\partial y_2^{n_2}} r^E(x-y), \end{aligned}$$
(8.98)

where \(r^E\) is defined as in (1.9). Let us first prove that for \(x\in \mathbb {R}^2\), the covariance matrix of the centered Gaussian vector \((B_E(x), \nabla B_E(x))\) is

$$\begin{aligned} \begin{pmatrix} 1 &{}0\\ 0 &{}2\pi ^2 E\,I_2 \end{pmatrix}, \end{aligned}$$
(8.99)

where \(I_2\) denotes the \(2\times 2\)-identity matrix. Recall from (3.44) that the following integral representation holds:

$$\begin{aligned} J_0(2\pi \sqrt{E}\Vert x\Vert ) = \frac{1}{2\pi } \int _{{\mathbb {S}}^1} \mathrm{e}^{i2\pi \sqrt{E}\langle \theta , x\rangle }\,d\theta ,\qquad x\in \mathbb {R}^2, \end{aligned}$$
(8.100)

where \(d\theta \) stands for the uniform measure on the unit circle. By (8.98) and (8.100), (8.99) immediately follows. Note now that, from (8.98), in particular we have

$$\begin{aligned} {{\mathbb {E}}}[B_E(x) \partial _1 B_E(y)] = -i\sqrt{E} \int _{{\mathbb {S}}^1} \theta _1 \mathrm{e}^{2\pi \sqrt{E} i \langle \theta , x-y \rangle }\, d\theta ; \end{aligned}$$
(8.101)

in order to find an explicit expression for (8.101), let us first compute \(\int _{{\mathbb {S}}^1} \theta _1 \mathrm{e}^{i r\langle \theta , u \rangle }\, d\theta \) for \(r\in [0,+\infty )\) and any \(u\in {\mathbb {S}}^1\). Let us denote by \(r_\tau \) the rotation of angle \(\tau \) (the latter is the angle between \(\theta \) and u), then we have

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {S}}^1}\theta _1\,e^{ir\langle \theta ,u\rangle }d\theta&= \int _{-\pi }^{\pi } (r_\tau (u))_1 \,e^{ir\cos \tau }d\tau \\&= \int _{-\pi }^{\pi } (\cos \tau u_1 - \sin \tau u_2) \,e^{ir\cos \tau }d\tau \\&=- \int _{-\pi }^{\pi } (\sin \tau u_1 + \cos \tau u_2) \,e^{-ir\sin \tau }d\tau \\&=- \int _{-\pi }^{\pi } (\sin \tau u_1 + \cos \tau u_2)\big (\cos (r\sin \tau )-i\sin (r\sin \tau )\big )d\tau \\&=- u_2 \int _{-\pi }^{\pi } \cos \tau \cos (r\sin \tau )d\tau +iu_1 \int _{-\pi }^{\pi } \sin \tau \sin (r\sin \tau )d\tau \\&= -\pi u_2\big (J_1(r)+J_{-1}(r)\big ) +i \pi u_1\big (J_1(r)-J_{-1}(r)\big )\\&= 2i\pi u_1J_1(r), \end{aligned} \end{aligned}$$

where we used integral representation formulas for \(\alpha \)-order Bessel functions of the first kind \(J_\alpha \) [AS64, §9.1], so that, whenever \(x\ne y\),

$$\begin{aligned} {{\mathbb {E}}}[B_E(x){\partial }_1 B_E(y) ]= & {} 2\pi \sqrt{E} \,\frac{x_1-y_1}{\Vert x-y\Vert }\, J_1(2\pi \sqrt{E}\Vert x-y\Vert ). \end{aligned}$$
(8.102)

Analogously, we get

$$\begin{aligned} {{\mathbb {E}}}[B_E(x){\partial }_2 B_E(y) ]= & {} 2\pi \sqrt{E} \,\frac{x_2-y_2}{\Vert x-y\Vert }\, J_1(2\pi \sqrt{E}\Vert x-y\Vert ); \end{aligned}$$
(8.103)

(8.102) and (8.103) prove (3.32). For \(k,l\in \{1,2\}\) from (8.98) and (3.44) we have

$$\begin{aligned} {{\mathbb {E}}}[\partial _k B_E(x)\partial _l B_E(y) ]= & {} 2\pi E\int _{{\mathbb {S}}^1}z_kz_l\,e^{2i\pi \langle z,\sqrt{E}(x-y)\rangle }dz. \end{aligned}$$

Let us first compute \(\int _{{\mathbb {S}}^1} z_1^2\,e^{ir\langle z,u\rangle }dz \) for \((r,u)\in [0,\infty )\times {\mathbb {S}}^1\): we have, again with \(r_\tau \) denoting the rotation of angle \(\tau \),

$$\begin{aligned}&\int _{{\mathbb {S}}^1} z_1^2\,e^{ir\langle z,u\rangle }dz= \int _{-\pi }^{\pi } (r_\tau (u))_1^2 \,e^{ir\cos \tau }d\tau \\= & {} \int _{-\pi }^{\pi } (\cos \tau u_1 - \sin \tau u_2)^2 \,e^{ir\cos \tau }d\tau = \int _{-\pi }^{\pi } (\sin \tau u_1 + \cos \tau u_2)^2 \,e^{-ir\sin \tau }d\tau \\= & {} \int _{-\pi }^{\pi } \big (\sin ^2\tau \, u_1^2 + \cos ^2\tau \, u_2^2 +2\cos \tau \sin \tau u_1u_2\big ) \big (\cos (r\sin \tau )-i\sin (r\sin \tau )\big )d\tau \\= & {} \int _{-\pi }^{\pi } \big (\sin ^2\tau \, u_1^2 + \cos ^2\tau \, u_2^2\big )\cos (r\sin \tau )d\tau -i\, u_1u_2\,\int _{-\pi }^{\pi } \sin (2\tau )\,\sin (r\sin \tau )d\tau \\= & {} \frac{ 1}{2}\int _{-\pi }^{\pi } \big (1+(1-2u_1^2)\cos (2\tau )\big ) \cos (r\sin \tau )d\tau \\= & {} \pi J_0(r) + \frac{\pi }{2}(1-2u_1^2)(J_2(r)+J_{-2}(r)) = \pi J_0(r) + (1-2u_1^2)\pi J_2(r). \end{aligned}$$

Similarly

$$\begin{aligned} \int _{{\mathbb {S}}^1} z_2^2\,e^{ir\langle z,u\rangle }dz= \pi J_0(r) + (1-2u_2^2)\pi J_2(r) = \pi J_0(r) + (2u_1^2-1)\pi J_2(r), \end{aligned}$$

whereas

$$\begin{aligned}&\int _{{\mathbb {S}}^1} z_1z_2\,e^{ir\langle z,u\rangle }dz= \int _{-\pi }^{\pi } (r_\tau (u))_1(r_\tau (u))_2 \,e^{ir\cos \tau }d\tau \\= & {} \int _{-\pi }^{\pi } (\cos \tau u_1 - \sin \tau u_2)(\sin \tau u_1 + \cos \tau u_2) \,e^{ir\cos \tau }d\tau \\= & {} - \int _{-\pi }^{\pi } (\sin \tau u_1 + \cos \tau u_2)(\cos \tau u_1 - \sin \tau u_2) \,e^{-ir\sin \tau }d\tau \\= & {} \int _{-\pi }^{\pi } \big (\frac{1}{2}\sin (2\tau )\, (1-2u_1^2) -\cos (2\tau ) u_1u_2\big ) \big (\cos (r\sin \tau )-i\sin (r\sin \tau )\big )d\tau \\= & {} - u_1u_2\int _{-\pi }^{\pi } \cos (2\tau )\,\cos (r\sin \tau )d\tau \\= & {} -u_1u_2\pi (J_2(r)+J_{-2}(r)) = -2u_1u_2\pi J_2(r). \end{aligned}$$

Thus, when \(x\ne y\),

$$\begin{aligned} {{\mathbb {E}}}[{\partial }_1 B_E(x){\partial }_1 B_E(y) ]= & {} 2\pi ^2 E \left( J_0(2\pi \sqrt{E}\Vert x-y\Vert ) \right. \nonumber \\&\quad \left. + \left( 1-2\frac{(x_1-y_1)^2}{\Vert x-y\Vert ^2}\right) J_2(2\pi \sqrt{E}\Vert x-y\Vert )\right) \\ {{\mathbb {E}}}[{\partial }_2 B_E(x){\partial }_2 B_E(y) ]= & {} 2\pi ^2 E \left( J_0(2\pi \sqrt{E}\Vert x-y\Vert ) \right. \nonumber \\&\quad \left. + \left( 1-2\frac{(x_2-y_2)^2}{\Vert x-y\Vert ^2}\right) J_2(2\pi \sqrt{E}\Vert x-y\Vert ) \right) \\ {{\mathbb {E}}}[{\partial }_1 B_E(x){\partial }_2 B_E(y) ]= & {} -4\pi ^2E \frac{(x_1-y_1)(x_2-y_2)}{\Vert x-y\Vert ^2}J_2(2\pi \sqrt{E}\Vert x-y\Vert ), \end{aligned}$$

which are (3.33) and (3.34). \(\quad \square \)

The following result concerns some (known) properties of the nodal sets of \(B_E\) and its complex version.

Lemma 8.4

  1. 1.

    The value 0 is not singular for \(B_E\) a.s., i.e.

    $$\begin{aligned} {{\mathbb {P}}}(\exists x : B_E(x)=0, \nabla B_E(x)=0) = 0; \end{aligned}$$
  2. 2.

    the nodal set \(B_E^{-1}(0)\) is a smooth one dimensional submanifold of \(\mathbb {R}^2\) a.s.;

  3. 3.

    \(B_E^{-1}(0) \cap \partial {\mathcal {D}}\) consists of a finite number of points a.s.;

  4. 4.

    the nodal set \((B_E^{{\mathbb {C}}})^{-1}(0) = B_E^{-1}(0)\cap {{\widehat{B}}}_E^{-1}(0)\) consists of isolated points a.s.;

  5. 5.

    the number of nodal points \((B_E^{{\mathbb {C}}})^{-1}(0) \) in \({\mathcal {D}}\) is a.s. finite and none of them lies on \(\partial D\) a.s.

Proof

1. Proposition 6.12 in [AW] ensures that 0 is not a singular value a.s. Indeed, the hypothesis of Proposition 6.12 are satisfied, the random variables \(B_E(x)\), \(\partial _1 B_E(x)\), \(\partial _2 B_E(x)\) being independent for fixed \(x\in \mathbb {R}^2\) (Point 2. in Lemma 3.1).

2. It follows from Point 1 by Sard’s lemma.

3. Let \(\gamma \) be a unit speed parameterization of the boundary \(\partial {\mathcal {D}}\). The restriction of \(B_E\) to \(\partial {\mathcal {D}}\) is the one-dimensional Gaussian process \(t\mapsto B_E(\gamma (t))\) whose first time-derivative is

$$\begin{aligned} B_E(\gamma (t))' = \langle \nabla B_E(\gamma (t)), \dot{\gamma }(t)\rangle . \end{aligned}$$
(8.104)

From (8.104) and the arguments used in the proof of Point 1. We deduce that

$$\begin{aligned} {{\mathbb {P}}}(\exists t : B_E(\gamma (t))=B_E(\gamma (t))'=0)=0, \end{aligned}$$

i.e. the value 0 is not singular a.s. for \(B_E(\gamma )\), hence the zeros of \(B_E\) on \(\partial {\mathcal {D}}\) are isolated points a.s. (by a standard application of the inverse mapping theorem), and their number is finite (see [AT, p.269]).

4. Let us consider the two-dimensional Gaussian field on the plane \((B_E, {{\widehat{B}}}_E)\), where we recall \({{\widehat{B}}}_E\) to be an independent copy of \(B_E\). In view of Point 1., the value (0, 0) is not singular for \((B_E, {{\widehat{B}}}_E)\), hence a standard application of the inverse mapping theorem entails that the common zeros of \(B_E\) and \({{\widehat{B}}}_E\) are isolated points.

5. The value 0 being not singular for \((B_E, {{\widehat{B}}}_E)\), from [AT, p.269] the number of nodal points in \({\mathcal {D}}\) is finite a.s. We can apply Lemma 11.2.10 in [AT] to the two-dimensional random field \((B_E, {{\widehat{B}}}_E)\) restricted to the boundary \(\partial {\mathcal {D}}\) to get that \((B_E^{{\mathbb {C}}})^{-1}(0) \cap \partial {\mathcal {D}} = \emptyset \) a.s. \(\quad \square \)

Proof of Lemma 3.2

We can rewrite (3.37) by means of the co-area formula [AW, Proposition 6.13] as

$$\begin{aligned} {\mathcal {L}}_E^\varepsilon = \frac{1}{2\varepsilon } \int _{-\varepsilon }^{\varepsilon } \text {length}(B_E^{-1}(s)\cap {\mathcal {D}})\,ds, \end{aligned}$$
(8.105)

where \(B_E^{-1}(s)=\lbrace x\in \mathbb {R}^2 : B_E(x) = s\rbrace \). Theorem 3 in [APP16] ensures that the map \(s\mapsto \text {length}(B_E^{-1}(s))\) is a.s. continuous at 0, so that by the Foundamental Theorem of Calculus we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} {\mathcal {L}}_E^\varepsilon = \lim _{\varepsilon \rightarrow 0} \frac{1}{2\varepsilon } \int _{-\varepsilon }^{\varepsilon } \text {length}(B_E^{-1}(s)\cap {\mathcal {D}})\,ds = {\mathcal {L}}_E, \qquad a.s. \end{aligned}$$

In order to prove (3.40) we apply Theorem 11.2.3 in [AT], the hypothesis being satisfied.

\(\square \)

Proof of Lemma 3.3

We have \({\mathcal {L}}_E \in L^2({\mathbb {P}})\), the nodal length of \(B_E\) being a.s. bounded in \({\mathcal {D}}\) [DF88]. The collection of random variables \(\lbrace {\mathcal {L}}^\varepsilon _E \rbrace _{\varepsilon >0}\) is in \(L^2({{\mathbb {P}}})\) since

$$\begin{aligned} {\mathcal {L}}_E^\varepsilon \le \frac{1}{2\varepsilon } \int _{{\mathcal {D}}} \Vert \nabla B_E(x)\Vert \,dx, \end{aligned}$$

hence

$$\begin{aligned} \begin{aligned} {{\mathbb {E}}}[({\mathcal {L}}_E^\varepsilon )^2]&\le \frac{1}{4\varepsilon ^2} \int _{({\mathcal {D}})^2} {{\mathbb {E}}}[\Vert \nabla B_E(x)\Vert \cdot \Vert \nabla B_E(y)\Vert ]\,dx dy\\&\le {\text {area}}({\mathcal {D}}) \frac{1}{4\varepsilon ^2} \int _{{\mathcal {D}}}{{\mathbb {E}}}[\Vert \nabla B_E(x)\Vert ^2]\,dx = ({\text {area}}({\mathcal {D}}))^2 \frac{\pi ^2 E}{\varepsilon ^2} < +\infty . \end{aligned} \end{aligned}$$

In view of Lemma 3.2, in order to prove that \({\mathcal {L}}_E^\varepsilon \) converges to the nodal length in \(L^2({{\mathbb {P}}})\) it suffices to show that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} {{\mathbb {E}}}[({\mathcal {L}}_E^\varepsilon )^2] = {{\mathbb {E}}}[{\mathcal {L}}_E^2] \end{aligned}$$
(8.106)

(see also [Ros15, Lemma 7.2.1]). By Fatou’s lemma and (8.105) we get

$$\begin{aligned} {{\mathbb {E}}}[{\mathcal {L}}_E^2] \le \liminf _{\varepsilon \rightarrow 0} {{\mathbb {E}}}[({\mathcal {L}}_E^\varepsilon )^2] \le \limsup _{\varepsilon \rightarrow 0} {{\mathbb {E}}}\left[ \left( \frac{1}{2\varepsilon }\int _{-\varepsilon }^\varepsilon {\mathcal {L}}_E(s)\,ds \right) ^2\right] . \end{aligned}$$

By Jensen’s inequality

$$\begin{aligned} {{\mathbb {E}}}[{\mathcal {L}}_E^2] \le \limsup _{\varepsilon \rightarrow 0} {{\mathbb {E}}}\left[ \left( \frac{1}{2\varepsilon }\int _{-\varepsilon }^\varepsilon {\mathcal {L}}_E(s)\,ds \right) ^2\right] \le \limsup _{\varepsilon \rightarrow 0} \frac{1}{2\varepsilon }\int _{-\varepsilon }^\varepsilon {{\mathbb {E}}}\left[ {\mathcal {L}}_E(s) ^2\right] \,ds = {{\mathbb {E}}}\left[ {\mathcal {L}}_E ^2\right] , \end{aligned}$$

the last step following from the continuity of the map \(s\mapsto {{\mathbb {E}}}[{\mathcal {L}}_E(s)^2]\) at 0 which will be proven just below. Standard Kac–Rice formula [AW, Theorem 6.9] allows to write

$$\begin{aligned}&{{\mathbb {E}}}[{\mathcal {L}}_E(s)^2] \nonumber \\&= \int _{({\mathcal {D}})^2} {{\mathbb {E}}}[\Vert \nabla B_E(x)\Vert \Vert \nabla B_E(y)\Vert | B_E(x)=s, B_E(y) = s] p_{(B_E(x), B_E(y))}(s,s)\,dx dy,\nonumber \\ \end{aligned}$$
(8.107)

where \(p_{(B_E(x), B_E(y))}\) denotes the density of the random vector \((B_E(x), B_E(y))\). It suffices to show that, for \(\delta >0\), there exists a measurable function \(g=g(x,y)\) integrable on \(({\mathcal {D}})^2\) such that

$$\begin{aligned}&{{\mathbb {E}}}[\Vert \nabla B_E(x)\Vert \Vert \nabla B_E(y)\Vert | B_E(x)=s, B_E(y) = s]p_{(B_E(x), B_E(y))}(s,s) \\&\quad \le g(x,y),\qquad \forall s\in [-\delta , \delta ]. \end{aligned}$$

It is immediate that

$$\begin{aligned} p_{(B_E(x), B_E(y))}(s,s) \le p_{(B_E(x), B_E(y))}(0,0) = \frac{1}{2\pi \sqrt{1-J_0(2\pi \sqrt{E}\Vert x-y\Vert )^2}}. \end{aligned}$$

From Lemma 3.1, the vector \(\nabla B_E(x)\) conditioned to \(B_E(x)=B_E(y)=s\) is Gaussian with mean

$$\begin{aligned} s \frac{\nabla _x r^E(x-y)}{1+r^E(x-y)} \end{aligned}$$

and covariance matrix

$$\begin{aligned}&\Omega ^E(x-y) \nonumber \\&\quad := 2\pi ^2 E I_2 - \frac{1}{1-r^E(x-y)^2} \nonumber \\&\qquad \, \times \begin{pmatrix} (\partial _{x_1} r^E(x-y))^2 &{}\partial _{x_1} r^E(x-y) \partial _{x_2} r^E(x-y)\\ \partial _{x_1} r^E(x-y) \partial _{x_2} r^E(x-y) &{}(\partial _{x_2} r^E(x-y))^2 \end{pmatrix}. \end{aligned}$$
(8.108)

Jensen’s inequality yields

$$\begin{aligned}&{{\mathbb {E}}}[ \Vert \nabla B_E(x)\Vert \Vert \nabla B_E(y)\Vert | B_E(x)=s, B_E(y) = s]\nonumber \\&\quad \le {{\mathbb {E}}}[\Vert \nabla B_E(x)\Vert ^2 | B_E(x)=s, B_E(y) = s]\nonumber \\ {}&\quad = {\text{ Var }}(\partial _1 B_E(x) | B_E(x)=s, B_E(y) = s) \nonumber \\ {}&\quad + {\text{ Var }}(\partial _2 B_E(x) | B_E(x)=s, B_E(y) = s) \nonumber \\ {}&\quad + {{\mathbb {E}}}[\partial _1 B_E(x) | B_E(x)=s, B_E(y) = s]^2 + {{\mathbb {E}}}[\partial _2 B_E(x) | B_E(x)=s, B_E(y) = s]^2 \nonumber \\ {}&\quad = 4\pi ^2E -\frac{4\pi ^2E J_1(2\pi \sqrt{E} \Vert x-y\Vert )^2}{1-J_0(2\pi \sqrt{E}\Vert x-y\Vert )^2}\nonumber \\ {}&\qquad \, + s^2\frac{4\pi ^2E J_1(2\pi \sqrt{E} \Vert x-y\Vert )^2}{(1+ J_0(2\pi \sqrt{E}\Vert x-y\Vert ))^2} \nonumber \\ {}&\quad \le 2\pi ^2 E + s^2\frac{4\pi ^2E J_1(2\pi \sqrt{E} \Vert x-y\Vert )^2}{(1+ J_0(2\pi \sqrt{E}\Vert x-y\Vert ))^2} \nonumber \\ {}&\quad \le 2\pi ^2 E + \delta ^2\frac{4\pi ^2E J_1(2\pi \sqrt{E} \Vert x-y\Vert )^2}{(1+ J_0(2\pi \sqrt{E}\Vert x-y\Vert ))^2}. \end{aligned}$$
(8.109)

If we set

$$\begin{aligned} g(x,y) := \frac{1}{2\pi \sqrt{1-J_0(2\pi \sqrt{E}\Vert x-y\Vert )^2}}\left( 2\pi ^2 E + \delta ^2\frac{4\pi ^2E J_1(2\pi \sqrt{E} \Vert x-y\Vert )^2}{(1+ J_0(2\pi \sqrt{E}\Vert x-y\Vert ))^2}\right) , \end{aligned}$$

then the proof of (3.41) is concluded.

The proof of (3.42) relies on the same argument as that of (3.41). Let us first show that \({\mathcal {N}}_E\in L^2({{\mathbb {P}}})\). Theorem 6.3 in [AW] ensures that the second factorial moment of \({\mathcal {N}}_E\) has the following integral representation

(8.110)

Reasoning as in the proof of [DNPR16, Lemma 3.4], we have

$$\begin{aligned}&{{\mathbb {E}}}\left[ |\text {Jac}_{B_E, {{\widehat{B}}}_E}(x)| |\text {Jac}_{B_E, {{\widehat{B}}}_E}(y)| | B_E(x) = 0, B_E(y) = 0,\right. \nonumber \\&\quad \left. {{\widehat{B}}}_E(x) = 0, {{\widehat{B}}}_E(y) = 0 \right] \nonumber \\&\quad \ll \frac{\text {det}(\Omega ^E(x-y))}{1- J_0(2\pi \sqrt{E} \Vert x-y\Vert )^2}, \end{aligned}$$
(8.111)

where, for any \(s\in \mathbb {R}\), \(\Omega ^E(x-y)\) denotes the covariance matrix of \(\nabla B_E(x)\) conditioned to \(B_E(x) = B_E(y)= s\). Lemma 8.20 ensures that the double integral over \({\mathcal {D}}\) of the rhs of (8.111) is finite.

Let us now prove that the map \(s\mapsto {{\mathbb {E}}}[{\mathcal {N}}_E(s)^2]\) is continuous at 0. Note first that we can write

$$\begin{aligned} {{\mathbb {E}}}[{\mathcal {N}}_E(s)^2] = {{\mathbb {E}}}[{\mathcal {N}}_E(s)({\mathcal {N}}_E(s)-1)] + {{\mathbb {E}}}[{\mathcal {N}}_E(s)]. \end{aligned}$$
(8.112)

To evaluate the mean, we use Kac–Rice formula [AW, Thereom 6.2] and Lemma 3.1

$$\begin{aligned} {{\mathbb {E}}}[{\mathcal {N}}_E(s)] = \int _{{\mathcal {D}}} {{\mathbb {E}}}\left[ |\text {Jac}_{B_E, {{\widehat{B}}}_E}(x)| \right] p_{(B_E(x), {{\widehat{B}}}_E(x))}(s,s)\,dx. \end{aligned}$$
(8.113)

Since \( {{\mathbb {E}}}\left[ |\text {Jac}_{B_E, {{\widehat{B}}}_E}(x)| \right] =2\pi ^2E\) and \(p_{(B_E(x), {{\widehat{B}}}_E(x))}(s,s)\le \frac{1}{2\pi }\) for every s, then \(s\mapsto {{\mathbb {E}}}[{\mathcal {N}}_E(s)]\) is continuous.

Let us now deal with the second factorial moment, again using Kac–Rice formula [AW, Theorem 6.3].

$$\begin{aligned}&{{\mathbb {E}}}[{\mathcal {N}}_E(s) ({\mathcal {N}}_E(s)-1)] \nonumber \\&\quad = \int _{({\mathcal {D}})^2} {{\mathbb {E}}}\Big [|\text {Jac}_{B_E, {{\widehat{B}}}_E}(x)| |\text {Jac}_{B_E, {{\widehat{B}}}_E}(y)| | B_E(x)={{\widehat{B}}}_E(x)=B_E(y)= {{\widehat{B}}}_E(y)=s\Big ] \nonumber \\&\qquad \times p_{(B_E(x), {{\widehat{B}}}_E(x),B_E(y), {{\widehat{B}}}_E(y))}(s,s,s,s)\,dxdy. \end{aligned}$$
(8.114)

Jensen’s inequality yields

$$\begin{aligned} {{\mathbb {E}}}\Big [|\text {Jac}(x)|&|\text {Jac}(y)| | B_E(x)={{\widehat{B}}}_E(x)=B_E(y)= {{\widehat{B}}}_E(y)=s\Big ]\nonumber \\&\le {{\mathbb {E}}}\Big [|\text {Jac}(x)|^2 | B_E(x)={{\widehat{B}}}_E(x)=B_E(y)= {{\widehat{B}}}_E(y)=s\Big ]\nonumber \\&= 2\left( {{\mathbb {E}}}[X^2] {{\mathbb {E}}}[Y^2] - {{\mathbb {E}}}[X Y]^2 \right) , \end{aligned}$$
(8.115)

where (XY) is a random vector with the same distribution as \(\nabla B_E(x) | B_E(x) = B_E(y) =s\). Hence some straightforward computations lead to

$$\begin{aligned} {{\mathbb {E}}}[X^2] {{\mathbb {E}}}[Y^2] - {{\mathbb {E}}}[X Y]^2&= 2\pi ^2E \left( 2\pi ^2E - \frac{(\partial _{x_1} r^E(x-y))^2 + (\partial _{x_2} r^E(x-y))^2}{1 - r^E(x-y)^2} \right) \nonumber \\&+2\pi ^2E s^2 \frac{(\partial _{x_1} r^E(x-y))^2 + (\partial _{x_2} r^E(x-y))^2}{(1+r^E(x-y))^2}\nonumber \\&\le 2\pi ^2E \left( 2\pi ^2E - \frac{(\partial _{x_1} r^E(x-y))^2 + (\partial _{x_2} r^E(x-y))^2}{1 - r^E(x-y)^2} \right) \nonumber \\&+2\pi ^2E \delta ^2 \frac{(\partial _{x_1} r^E(x-y))^2 + (\partial _{x_2} r^E(x-y))^2}{(1+r^E(x-y))^2}, \end{aligned}$$
(8.116)

for \(s\in [-\delta , \delta ]\), which is integrable on \({\mathcal {D}} \times {\mathcal {D}}\).\(\square \)

Appendix B

Lemma 8.5

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} (i)\quad \mathrm{Var}(a_{1,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}} r^E(x-y)^4 dxdy \sim 9\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{1,E},a_{2,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^4\,dxdy \sim \frac{27}{2}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{1,E},a_{3,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^4\,dxdy \sim \frac{27}{2}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{1,E},a_{3,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim \frac{9}{2}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{1,E},a_{5,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{0,1}^E(x-y)^2\,dxdy \sim 3\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{1,E},a_{6,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim 3\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Proof

Let us prove (i). From Proposition 5.2,

$$\begin{aligned} {\text{ Var }}(a_{1,E})=&\, {} 24 \int _{{\mathcal {D}}} \int _{{\mathcal {D}}} r^E(x-y)^4\,dx dy \nonumber \\=&\, {} 24 \text{ area }({\mathcal {D}}) \frac{2\pi }{E} \int _1^{\sqrt{E} \cdot \text{ diam }({\mathcal {D}})} \nonumber \\ {}&\quad \times \psi \left( \frac{1}{\pi \sqrt{\psi }} \cos \left( 2\pi \psi - \frac{\pi }{4} \right) \right) ^4\, d\psi + O\left( \frac{1}{E} \right) \nonumber \\=&\, {} 24 \text{ area }({\mathcal {D}}) \frac{2}{\pi ^3 E} \int _1^{\sqrt{E} \cdot \text{ diam }({\mathcal {D}})} \frac{1}{\psi } \nonumber \\ {}&\quad \times \cos ^4 \left( 2\pi \psi - \frac{\pi }{4} \right) \, d\psi + O\left( \frac{1}{E} \right) . \end{aligned}$$
(8.117)

Thanks to (6.78) we have that, as \(E\rightarrow +\infty \),

$$\begin{aligned}&24 \text {area}({\mathcal {D}}) \frac{2}{\pi ^3 E} \int _1^{\sqrt{E} \cdot \text {diam}({\mathcal {D}})} \frac{1}{\psi } \cos ^4 \left( 2\pi \psi - \frac{\pi }{4} \right) \, d\psi \\&\quad \sim 24 \text {area}({\mathcal {D}}) \frac{2}{\pi ^3 E} \cdot \frac{3}{8} \cdot \log \sqrt{E} \\&\quad = \frac{9}{\pi ^3 E} \text {area}({\mathcal {D}}) \log E, \end{aligned}$$

that allows to conclude. The proof for the remaining terms is analogous to the proof of (i), and hence omitted. \(\quad \square \)

The proofs of the following lemmas follow from an application of Proposition 5.2, completely analogous to the one appearing in the proof of Lemma 8.5.

Lemma 8.6

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(a_{2,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^4 \,dxdy \sim \frac{315}{8}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{2,E},a_{3,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,2}^E(x-y)^4\,dxdy \sim \frac{27}{8}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{2,E},a_{4,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{45}{8}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{2,E},a_{5,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2\,dxdy \sim \frac{15}{2}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{2,E},a_{6,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{3}{2}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.7

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm {Var}(a_{3,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{2,2}^E(x-y)^4\,dxdy \sim \frac{315}{8}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(a_{3,E},a_{4,E})= & {} 24 \int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{2,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2dxdy\\&\sim \frac{45}{8}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(a_{3,E},a_{5,E})= & {} 24 \int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2 dxdy\\&\sim \frac{3}{2}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(a_{3,E},a_{6,E})= & {} 24\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2 dxdy\\&\sim \frac{15}{2}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.8

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \begin{aligned} \mathrm {Var}(a_{4,E})=&\, {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}({\widetilde{r}}_{1,1}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2+{\widetilde{r}}_{1,2}^E(x-y)^4\\ {}&+4{\widetilde{r}}_{1,1}^E(x-y) {\widetilde{r}}_{2,2}^E(x-y){\widetilde{r}}_{1,2}^E(x-y)^2)dxdy\\ {}&\quad \sim \frac{27}{8}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(a_{4,E},a_{5,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}({\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2+{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2\\ {}&+4{\widetilde{r}}_{0,1}^E(x-y){\widetilde{r}}_{0,2}^E(x-y){\widetilde{r}}_{1,1}^E(x-y){\widetilde{r}}_{1,2}^E(x-y))dxdy\\ {}&\quad \sim \frac{3}{2}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(a_{4,E},a_{6,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}({\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2+{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\\ {}&+ 4{\widetilde{r}}_{0,1}^E(x-y){\widetilde{r}}_{0,2}^E(x-y){\widetilde{r}}_{2,2}^E(x-y){\widetilde{r}}_{1,2}^E(x-y))dxdy\\ {}&\quad \sim \frac{3}{2}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned} \end{aligned}$$

Lemma 8.9

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(a_{5,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}\big (r^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2+{\widetilde{r}}^E_{0,1}(x-y)^4\\- & {} 4r^E(x-y){\widetilde{r}}^E_{1,1}(x-y){\widetilde{r}}_{0,1}^E(x-y)^2\big )\,dxdy \sim \frac{3}{2}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(a_{5,E},a_{6,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}(r^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2+{\widetilde{r}}^E_{0,2}(x-y)^4\\- & {} 4r^E(x-y){\widetilde{r}}_{0,2}^E(x-y)^2 {\widetilde{r}}^E_{2,2}(x-y))\,dxdy \sim \frac{1}{2}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.10

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(a_{6,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}\big (r^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2+{\widetilde{r}}^E_{0,2}(x-y)^4\\- & {} 4r^E(x-y) {\widetilde{r}}^E_{2,2}(x-y){\widetilde{r}}_{0,2}^E(x-y)^2\big )\,dxdy \sim \frac{3}{2}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.11

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(b_{1,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^4(u)\,dxdy \sim \frac{3}{8}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{1,E},b_{2,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{0,1}^E(x-y)^2\,dxdy \sim \frac{1}{8}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{1,E},b_{3,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim \frac{1}{8}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{1,E},b_{4,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{0,1}^E(x-y)^2\,dxdy \sim \frac{1}{8}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}\\ \mathrm{Cov}(b_{1,E},b_{5,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim \frac{1}{8}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{1,E},b_{6,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^4\,dxdy \sim \frac{9}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{1,E},b_{7,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^4\,dxdy \sim \frac{9}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{1,E},b_{8,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim \frac{3}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{1,E},b_{9,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim \frac{3}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{1,E},b_{10,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim \frac{3}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.12

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(b_{2,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2\,dxdy \sim \frac{9}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{2,E},b_{3,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{3}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{2,E},b_{4,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^4\,dxdy \sim \frac{9}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{2,E},b_{5,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim \frac{3}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{2,E},b_{6,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2\,dxdy \sim \frac{5}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{2,E},b_{7,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{2,E},b_{8,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \\ \mathrm{Cov}(b_{2,E},b_{9,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{2,E},b_{10,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}} {\widetilde{r}}^E_{0,1}(x-y) {\widetilde{r}}^E_{0,2}(x-y) {\widetilde{r}}^E_{1,1}(x-y) {\widetilde{r}}^E_{1,2}(x-y) \,dxdy\\\sim & {} \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.13

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(b_{3,E})= & {} 4 \int _{{\mathcal {D}}}\int _{{\mathcal {D}}} r^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2 \,dxdy \sim \frac{9}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{3,E},b_{4,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{0,2}^E(x-y)^2\,dxdy \sim \frac{3}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{3,E},b_{5,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^4\,dxdy \sim \frac{9}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{3,E},b_{6,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{3,E},b_{7,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2\,dxdy \sim \frac{5}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{3,E},b_{8,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{3,E},b_{9,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{3,E},b_{10,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}^E_{0,1}(x-y) {\widetilde{r}}^E_{0,2}(x-y) {\widetilde{r}}^E_{2,2}(x-y) {\widetilde{r}}^E_{1,2}(x-y) \,dxdy\\\sim & {} \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.14

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(b_{4,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}} r^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2\,dxdy \sim \frac{9}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{4,E},b_{5,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}r^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2)\,dxdy \sim \frac{3}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{4,E},b_{6,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2\,dxdy \sim \frac{5}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{4,E},b_{7,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{4,E},b_{8,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,1}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{4,E},b_{9,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{4,E},b_{10,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}^E_{0,1}(x-y) {\widetilde{r}}^E_{0,2}(x-y) {\widetilde{r}}^E_{1,1}(x-y) {\widetilde{r}}^E_{1,2}(x-y)\,dxdy\\\sim & {} \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.15

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(b_{5,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^4 \,dxdy \sim \frac{9}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{5,E},b_{6,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{5,E},b_{7,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2\,dxdy \sim \frac{5}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{5,E},b_{8,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{5,E},b_{9,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{0,1}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2\,dxdy \sim \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{5,E},b_{10,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}^E_{0,1}(x-y) {\widetilde{r}}^E_{0,2}(x-y) {\widetilde{r}}^E_{2,2}(x-y) {\widetilde{r}}^E_{1,2}(x-y)\,dxdy\\\sim & {} \frac{1}{16}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.16

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm{Var}(b_{6,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^4\,dxdy \sim \frac{105}{64}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{6,E},b_{7,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,2}^E(x-y)^4\,dxdy \sim \frac{9}{64}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{6,E},b_{8,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{15}{64}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm{Cov}(b_{6,E},b_{9,E})= & {} 4 \int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2)\,dxdy \sim \frac{15}{64}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}\\ \mathrm{Cov}(b_{6,E},b_{10,E})= & {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2\,dxdy \sim \frac{15}{64}\,\frac{\mathrm{area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Lemma 8.17

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \begin{aligned} \mathrm {Var}(b_{7,E})=&{}\, 4 \int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{2,2}^E(x-y)^4dxdy\\ {}&\quad \sim \frac{105}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(b_{7,E},b_{8,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{2,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2dxdy\\ {}&\quad \sim \frac{15}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(b_{7,E},b_{9,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{2,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2dxdy\\ {}&\quad \sim \frac{15}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(b_{7,E},b_{10,E})=&\, {} 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{2,2}^E(x-y)^2{\widetilde{r}}_{1,2}^E(x-y)^2dxdy \\ {}&\quad \sim \frac{15}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned} \end{aligned}$$

Lemma 8.18

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \begin{aligned} \mathrm {Var}(b_{8,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2dxdy \\ {}&\quad \sim \frac{9}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(b_{8,E},b_{9,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,2}^E(x-y)^4dxdy \\ {}&\quad \sim \frac{9}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(b_{8,E},b_{10,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y){\widetilde{r}}_{2,2}^E(x-y){\widetilde{r}}_{1,2}^E(x-y)^2dxdy \\ {}&\quad \sim \frac{9}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned} \end{aligned}$$

Lemma 8.19

As \(E\rightarrow +\infty \), we have

$$\begin{aligned} \mathrm {Var}(b_{9,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2dxdy\\ {}&\quad \sim \frac{9}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Cov}(b_{9,E},b_{10,E})=&{}\, 4\int _{{\mathcal {D}}}\int _{{\mathcal {D}}}{\widetilde{r}}_{1,1}^E(x-y){\widetilde{r}}_{2,2}^E(x-y){\widetilde{r}}_{1,2}^E(x-y)^2dxdy\\ {}&\quad \sim \frac{9}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E},\\ \mathrm {Var}(b_{9,E})=&{}\, \int _{{\mathcal {D}}}\int _{{\mathcal {D}}}({\widetilde{r}}_{1,1}^E(x-y)^2{\widetilde{r}}_{2,2}^E(x-y)^2\\ {}&\quad +2{\widetilde{r}}_{1,1}^E(x-y){\widetilde{r}}_{2,2}^E(x-y){\widetilde{r}}_{1,2}^E(x-y)^2\\ {}&+ {\widetilde{r}}_{1,2}^E(x-y)^4)\,dxdy \sim \frac{9}{64}\,\frac{\mathrm {area}({\mathcal {D}})}{\pi ^3}\times \frac{\log E}{E}. \end{aligned}$$

Appendix C

Proof of Lemma 7.6

Reasoning as in the proof of Proposition 5.1, we have

$$\begin{aligned} \int _{{\mathcal {D}}} \int _{{\mathcal {D}}} {\widetilde{r}}^E_{k,l}(x-y)^6\, dx dy&= \text {area}({\mathcal {D}}) \int _0^{\text {diam}({\mathcal {D}})} d\phi \, \phi \int _0^{2\pi } {\widetilde{r}}^E_{k,l}(\phi \cos \theta , \phi \sin \theta )^6\, d\theta \nonumber \\&\quad + O\left( \int _0^{\text {diam}({\mathcal {D}})} d\phi \, \phi ^2 \int _0^{2\pi } {\widetilde{r}}^E_{k,l}(\phi \cos \theta , \phi \sin \theta )^6\, d\theta \right) . \end{aligned}$$
(8.118)

Performing the change of variable \(\theta = \psi / \sqrt{E}\) in the first term on the r.h.s. of (8.118) we obtain

$$\begin{aligned} \text {area}&({\mathcal {D}}) \int _0^{\text {diam}({\mathcal {D}})} d\phi \, \phi \int _0^{2\pi } {\widetilde{r}}^E_{k,l}(\phi \cos \theta , \phi \sin \theta )^6\, d\theta \nonumber \\&= \text {area}({\mathcal {D}}) \frac{1}{E} \int _0^{\sqrt{E} \cdot \text {diam}({\mathcal {D}})} d\psi \, \psi \int _0^{2\pi } {\widetilde{r}}^1_{k,l}(\psi \cos \theta , \psi \sin \theta )^6\, d\theta . \end{aligned}$$
(8.119)

Since \(r^1(\psi \cos \theta , \psi \sin \theta )\rightarrow 1\), \({\widetilde{r}}^1_{0,i}(\psi \cos \theta , \psi \sin \theta )= O(\psi )\) and \({\widetilde{r}}^1_{i,i}(\psi \cos \theta , \psi \sin \theta )\rightarrow 1\), \({\widetilde{r}}^1_{1,2}(\psi \cos \theta , \psi \sin \theta )=O(\psi ^2)\) as \(\psi \rightarrow 0\) uniformly on \(\theta \) (\(i=1,2\)), then we can rewrite (8.119) as

$$\begin{aligned}&\text {area}({\mathcal {D}}) \frac{1}{E} \int _0^{\sqrt{E} \cdot \text {diam}({\mathcal {D}})} d\psi \, \psi \int _0^{2\pi } {\widetilde{r}}^1_{k,l}(\psi \cos \theta , \psi \sin \theta )^6\, d\theta \nonumber \\&\quad = O\left( \frac{1}{E} \right) + \text {area}({{\mathcal {D}}}) \frac{1}{E}\int _1^{\sqrt{E} \cdot \text {diam}({\mathcal {D}})} d\psi \, \psi \int _0^{2\pi } {\widetilde{r}}^1_{k,l}(\psi \cos \theta , \psi \sin \theta )^6\, d\theta . \end{aligned}$$
(8.120)

Now using (5.63) for the second term on the r.h.s. of (8.120), as \(E\rightarrow +\infty \), we have

$$\begin{aligned}&\frac{1}{E}\int _1^{\sqrt{E} \cdot \text {diam}({\mathcal {D}})} d\psi \, \psi \int _0^{2\pi } {\widetilde{r}}^1_{k,l}(\psi \cos \theta , \psi \sin \theta )^6\, d\theta \nonumber \\&\quad \ll \frac{1}{E}\int _1^{\sqrt{E} \cdot \text {diam}({\mathcal {D}})} \frac{d\psi }{\psi ^2} \sim \frac{1}{E}. \end{aligned}$$
(8.121)

For the error term on the r.h.s. of (8.118) an analogous argument yields, as \(E\rightarrow +\infty \),

$$\begin{aligned} \int _0^{\text {diam}({\mathcal {D}})} d\phi \, \phi ^2 \int _0^{2\pi } {\widetilde{r}}^E_{k,l}(\phi \cos \theta , \phi \sin \theta )^6\, d\theta \asymp \frac{\log E}{E \sqrt{E}}. \end{aligned}$$
(8.122)

Thanks to (8.121) and (8.122), (8.118) concludes the proof. \(\square \)

From (8.108), the covariance matrix of \(\nabla B_E(x)\) conditioned to \(B_E(x)=B_E(0)=0\) is

$$\begin{aligned} \Omega _E(x)= 2\pi ^2 E\, I_2 - \frac{\nabla r^E(x)^t\, \nabla r^E(x)}{1 - r^E(x)^2} , \end{aligned}$$

and its determinant is

$$\begin{aligned} \text {det}(\Omega ^E(x)) = 2\pi ^2 E \left( 2\pi ^2 E - \frac{\Vert \nabla r^E(x)\Vert ^2}{1 - r^E(x)^2} \right) . \end{aligned}$$

Lemma 8.20

As \(x\rightarrow 0\), it holds

$$\begin{aligned} \Psi _E(x) := \frac{|\mathrm{{det}}(\Omega _E(x))|}{1 - r^E(x)^2} = \frac{1}{8} (2\pi ^2 E)^2 + E^3 O(\Vert x\Vert ^2), \end{aligned}$$

where the constant involved in the “O"-notation does not depend on E.

Proof

The Taylor development of \(r^E\) centered at 0 is

$$\begin{aligned} \begin{aligned} r^E(x)&= 1 - 2\pi ^2 E\,\frac{\Vert x\Vert ^2}{2} + \frac{(2\pi ^2E)^2 \Vert x\Vert ^4}{16} + E^3 O(\Vert x\Vert ^6), \end{aligned} \end{aligned}$$
(8.123)

where, from now until the end of the proof, the constants involved in the “O"-notation do not depend on E. From (8.123) it is immediate that

$$\begin{aligned} \begin{aligned} 1-r^E(x)^2&= 2\pi ^2 E\,\Vert x\Vert ^2 - \frac{3}{8}(2\pi ^2E)^2 \Vert x\Vert ^4 + E^3 O(\Vert x\Vert ^6). \end{aligned} \end{aligned}$$
(8.124)

Analogously, we find that the Taylor development for \(\Vert \nabla r^E(x)\Vert ^2\) centered at 0 is

$$\begin{aligned} \begin{aligned} \Vert \nabla r^E(x)\Vert ^2&= 2\pi ^2 E\left( 2\pi ^2 E \Vert x\Vert ^2 + (2\pi ^2E)^2 \frac{\Vert x\Vert ^4}{2} + E^3 O(\Vert x\Vert ^6)\right) . \end{aligned} \end{aligned}$$
(8.125)

From (8.124) and (8.125) we get

$$\begin{aligned} \frac{\Vert \nabla r^E(x)\Vert ^2 }{1-r^E(x)^2}&= \frac{ 2\pi ^2 E\left( 2\pi ^2 E \Vert x\Vert ^2 + (2\pi ^2E)^2 \frac{\Vert x\Vert ^4}{2} + E^3 O(\Vert x\Vert ^6)\right) }{2\pi ^2 E\,\Vert x\Vert ^2 - \frac{3}{8}(2\pi ^2E)^2 \Vert x\Vert ^4 + E^3 O(\Vert x\Vert ^6)}\nonumber \\ {}&= \frac{ (2\pi ^2 E)^2 \Vert x\Vert ^2 \left( 1+ 2\pi ^2E\frac{\Vert x\Vert ^2}{2} + E^2 O(\Vert x\Vert ^4)\right) }{2\pi ^2 E\,\Vert x\Vert ^2\left( 1 - 2\pi ^2E \frac{3}{8}\Vert x\Vert ^2 + E^2 O(\Vert x\Vert ^4)\right) }\nonumber \\ {}&= 2\pi ^2 E \left( 1+ 2\pi ^2E\frac{\Vert x\Vert ^2}{2} + E^2 O(\Vert x\Vert ^4)\right) \nonumber \\ {}&\quad \times \left( 1 - 2\pi ^2E\frac{3}{8} \Vert x\Vert ^2 + E^2 O(\Vert x\Vert ^4)\right) \nonumber \\ {}&= 2\pi ^2 E \left( 1 - 2\pi ^2 E \frac{1}{8} \Vert x\Vert ^2 + E^2 O(\Vert x\Vert ^4) \right) . \end{aligned}$$
(8.126)

From (8.126) and using again (8.124) we can write

$$\begin{aligned} \begin{aligned} \Psi _E(x)&= \frac{\left| 2\pi ^2 E \left( 2\pi ^2 E - \frac{\Vert \nabla r^E(x)\Vert ^2}{1 - r^E(x)^2}\right) \right| }{1 - k_E(x)^2} \\&= \frac{(2\pi ^2E)^3 \frac{1}{8} \Vert x\Vert ^2 + E^4O(\Vert x\Vert ^4)}{2\pi ^2 E\,\Vert x\Vert ^2 + E^2 O( \Vert x\Vert ^4)}\\&=\frac{1}{8} (2\pi ^2 E)^2 \left( 1 + E O(\Vert x\Vert ^2)\right) \end{aligned} \end{aligned}$$

which conclude the proof. \(\quad \square \)

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Nourdin, I., Peccati, G. & Rossi, M. Nodal Statistics of Planar Random Waves. Commun. Math. Phys. 369, 99–151 (2019). https://doi.org/10.1007/s00220-019-03432-5

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