Discretization

Theorem 4.3.1 (Improvement of [4, Theorem 1.6]) Letf :R² →Rbe a centered Gaussian function with covariance functionesatisfyinge(x, y) =k(kx−yk²)fork C¹, withk⁰(0)6= 0.

Let T be a lattice in R². Then, there exists CT,f > 0 such that for any fixed η > 0, for any n > 1, with probability at least 1−^C_n^Tη^,f, the vanishing locus of f in the box [−n, n]² cuts at most one times every edge of _n2+η¹ T. In particular, with the same probability, any continuous crossing in rectangles in the box is a discrete crossing and vice versa.

Figure 4.8: Discretization through a lattice. If the lattice is too coarse (left), the discretiza-tion is not trustful enough. The soludiscretiza-tion os too refine it, but the amount of informadiscretiza-tion on the field increases and the dependency of two restrictions of the field to two disjoint rectangles increases.

In fact, the result in [4] is less precise (the power is 9 instead of 3) and its proof was pretty complicated. Here we follow the simpler proof giving better estimates of [5], see also [29].

The proof uses the following extended version of Kac-Rice formula;

Theorem 4.3.2 (Theorem 3.2 [3]) Let f be a Gaussian field on an interval I ⊂R, such that almost surely, f isC¹ and that for any x, y∈I, Cov(f(x), f(y)) is definite. Let N be the number of zeros off on an interval I. Then,

E(N(N −1)) = Z

I²

E |f⁰(x1)||f⁰(x2)|

f(x1) =f(x2) = 0

φ_{(f(x1),f(x2))}(0,0)dx1dx2. We do not write down the proof and refer to the book [3].

Corollary 4.3.3 [5, Proposition 7] Under the hypotheses of Theorem4.3.2, and ife(x, y) = k(x−y) with k C² andk⁰⁰(0)6= 0. Then, there exists a constant C depending only onk⁰⁰(0) such that

E(N(N −1))≤C|I|³. Proof. Theorem 4.3.2 gives

E(N(N −1)) = Z

(x,y)∈I²

Z

(u,v)∈R²

|u||v|e^−12hΣ−1(u,v),(u,v)i 1 2π√

det Σφ(f(x),f(y))(0,0)dxdy.

where

is the covariance of the Gaussian vector(f(x), f(y)), S1 := Var f⁰(x), f⁰(y) is the variance matrix of he Gaussian vector(f⁰(x), f⁰(y)), and

C:= Cov centered Gaussian vector with variance matrix Σ,

E(kXk^k)≤CkΣk^k/2.

4.3. PROOF OF THEOREM4.1.6 49 Proof. Since Σis a positive real symmetric matrix, there exists an orthonormal basis(ei)i

ofRⁿ of eigenvectors ofΣand associated positive eigenvalues λ1,· · · , λn, so that is a Gaussian field satisfying the hypotheses of Corollary4.3.3. By Markov’s inequality and Corollary4.3.3,

The discretization have temporally solved the analytic continuation problem. However the problem now we have to face is the following: by theorem 4.3.1, the number of points at which we look the sign off increases in a rectangleR of sizer asr⁴. The more we know on the sign R, the more it gives informations on the sign on another rectangle R⁰ and threats the independency condition of Tassion theorem.

The second application of Proposition 4.2.9 is an quantitative dependency theorem for events that depend only on the sign of the Gaussian field restricted to a finite number of points. In our case, this will be vertices of a lattice in a large box.

Theorem 4.3.5 (Piterbarg 1982 [27, Theorem 1.1, p. 37]) Let A be an event that de-pends only on the signs of the coordinates of X ∈ R^N, and S₀ = (σ_0,ij)1≤ij≤N and S₁ =

The left hand side compares the two Gaussian vectors for events depending on their signs.

The given upper bound for it is the product which increases polynomially with the dimension of the vector, for us with the number of the sites where we look at the sign of the Gaussian field, and a second term which is bounded by the difference between the covariance matrices.

Remark 4.3.6 There exists in fact an equality, see the proof below and [29] for a more profound analysis of it which gives a far better bound.

Proof of Theorem 4.3.5. We use the notation of Proposition 4.2.9, that is St :=

(1−t)S₀+tS₁ and X_tis a centered Gaussian vector associated to S_t. The proof of

Note that heref is not C², but it was used for the integration by parts in the sequence of the proof, so it is not necessary here. Now since A depends only on the signs of the x_i’s, define

∀x∈R^N, sign(x) := (sign(xi))i ∈ {−1,+1}^N. For any couplei < j and any_i, _j ∈ {−1,+1}

Z^i,j(A) :={x∈A | sign(xi) =i,sign(xj) =j}.

The integration by part ofφ_Xt(x) gives Pr(X1 ∈A)−Pr(X0 ∈A) =

Note that we bound the probability in the sum by 1, which is a very crude bound. See [30]

and [6] for better bounds.

The following corollary measures the dependency between the positive crossings of a discretized Gaussian field over two disjoint rectangles.

4.3. PROOF OF THEOREM4.1.6 51 Corollary 4.3.7 Let V be the set of vertices of a lattice in Rⁿ, f :Rⁿ →R be a Gaussian field with covariance function e, such that e(x, x) = 1, and U0 and U1 a pair of disjoint bounded open sets ofRⁿ. Then, there exists a constant C such that for any pair (A_i)_i=0,1 of events depending on the sign of f on V ∩(Ui)i=0,1,

|Pr A₀, A₁)−Pr(A₀) Pr(A₁)

≤ 1

π|(V₀|+|V₁|)² max

x∈V₀,y∈V₁|e(x, y)|(1−e(x, y)²)^−1/2. Proof. Let

X0:= (X, Y) :=

(f(x))x∈V₀,(f(y))y∈V₁

. Then, this Gaussian vector has a covariance matrixS0 of the form

S0 := Var(X0) =

Var(X) Cov(X, Y) Cov(X, Y)^t Var(Y)

. Let

S₁=

Var(X) 0

0 Var(Y)

.

IfX1 is the associated Gaussian vector, thenX⁰ := (X1,i)1≤i≤|V₀|has the same law thanX, and is independent ofX⁰⁰ := (X1,i)_{|V0|+1≤j≤|V0|+|V1|}, the latter having the same law thanY. By Theorem4.3.5,

|Pr(X₁ ∈A₀∩A₁)−Pr(X₀ ∈A₀∩A₁)| ≤ 1

π| |V₀|+|V₁|2

x∈Vmax₀,y∈V₁|e(x, y)|(1−e(x, y)²)^−1/2. Now Pr(X₀ ∈ A₀ ∩A₁) = Pr(A₀, A₁) and Pr(X₁ ∈ A₀ ∩A₁) = Pr(A₀) Pr(A₁), which concludes. 2

Proof of the main Theorem 4.1.6 . In Corollary 4.2.8 we showed that the symmetry and positivity of correlations are satisfied for our continuous model. We must now prove the last quantitative dependence condition. Let T = (V,E) be the Union Jack lattice, see Figure 4.3. Fix η > 0, such that 12 + 4η < D. This is possible since D > 16. Then, Theorem 4.3.1 shows that there exists C > 0 such that for anyn ≥2, with probability at least 1−C(nlogn)^−η, in the box [−nlogn, nlogn]², the continuous percolations happen simultaneously with the discrete site percolations on the vertices

Vn:= 1

(nlogn)^2+ηV.

Moreover by Corollary4.3.7applied toU₀:= [−nlogn, nlogn]²\[−3n,3n]²,U₁:= [−2n,2n]²\ [−n, n]² and

Xn=f_{|Vn∩[−nlogn,nlogn]}²,

there exists a constantC > 0 depending only on the kernel esuch that for sign events A₀ onV₀ (resp. A1 onV₁),

Amax0,A1

|Pr A₀, A₁)−Pr(A₀) Pr(A₁)| ≤C(nlogn)⁴(nlogn)^8+4ηn^−D.

Since this is bounded byCn^12+4η−Dlog^4n+4η, thismax tends to 0 whenngrows to infinity.

Now let >0, and N be such that

∀n≥N, C(nlogn)⁴(nlogn)^8+4ηn^−D+C(nlogn)^−η < .

Then, by4.3.1and the last computation, the max of the sign events in Tassion theorem for the continuous percolation problem is no greater than for n≥N, so that it converges to zero, which proves Theorem4.1.6. 2

4.4 Proof of Tassion’s theorem

Figure 4.9: The events A^s_α,B^s_β andX_α^s

Figure 4.10: The eventR_a^b and the event N_a^b.

For 0 < a < b, 0 < α < s/2 Consider the events R^b_a, N_a^b, A^s_α, B_β^s and X_α^s defined by figure4.10and figure4.9respectively. Note thatX_α^s can be achieved if we have 4 symmetric copies ofA_α and a vertical crossing, so that

Pr(X_α^s)≥ 1

2Pr(A^s_α)⁴. We will use a simple useful lemma:

Lemma 4.4.1 There exists C >0, such that for any a < b, k∈N^∗, Pr(R^b+k(b−a)_a )> 1

2^kPr(R^b_a)^k and if b >2a, Pr(N_2b−a^b )≥Pr(R^b_a)⁴. Proof. This is a consequence of the FKG formula and figure4.11. 2

Proof of Theorem 4.2.1. Let0< <1/2 (in the article of Tassion, = 1/3). By FKG

4.4. PROOF OF TASSION’S THEOREM 53

Figure 4.11: Pr(R^2b−a_a )≥ ¹₂(PrR^b_a)² andPr(N_2b−a^b )≥Pr(R^b_a)⁴.

Figure 4.12: This is a horizontal crossing of the rectangle (3−)s×2s produced by 4 particular events translated or symmetrized from B_β^2s (two times) and Xα^s(1+), if β ≤ 2α andα < ₂^s(1−).

inequality, we see on figure 4.12 that the probability of crossing a rectangle (3−)s×2s satisfies

Pr(R^(3−)s_2s )≥Pr(X_α^s) Pr(B^2s_β )²≥ 1

2Pr(A^s_α)⁴Pr(B_β^2s)²

if β≤2α andα+ ^s₂(1 +)< swhich means α < ^s₂(1−). We see that we must have both large probabilities forA and B. By horizontal symmetry, we have,

Pr(A^s_α) + Pr(B^s_α)≥1/4 Now define

φs: [0, s/2] → [−1,1]

φ_s(α) := Pr(B_α^s)−Pr(A^s_α).

Then φ_s is non decreasing, φ_s(0)≤ 0 since B₀^s ⊂A^s₀ and φ_s/2 ≥ 1/4 if the extremity α of s/2×]α, s/2]is not allowed in the definition ofB_α^s. Sinceφs is continuousà voir, letαs be

such thatφ(αs) = 1/8.Then,

min Pr(A^s_αs),Pr(B_α^s_s)

≥1/8.

Now, by figure4.12we have α_s< s

2(1−) andα_2s<2α_s(1+)⇒Pr(R^(3−)s_2s )≥ 1 2¹⁹. Now, fixs₀ >0,N ∈N^∗ and ∀k∈N, s_k:=s₀(₁₊² )^k.

•Assume first that

∃k∈ {0,· · ·, N}, α_sk ≥ s_k

2 (1−).

Then α_sk ≥ s_k/4 since < 1/2 and in this case we construct large crossing without by figure4.13. By this figure, we see thatF KGand the condition implies that

Pr(R

3 2sk

sk )≥Pr(X_α^sk

sk)³≥ 1 2²⁵.

• Assume now that that

Figure 4.13: Ifα_s > s/4, by FKG we can construct a crossing of the rectangle ^3s₂ ×swith three eventsX_s/4^s .

∀k∈ {0,· · ·N}, α_2sk >2α_sk(1+) and αsk < s_k

2 (1−).

Recall that the latter condition is required for the small square being in the rectangle. Then, α₍ ²

1+)^N+1 >2^Nα_s0(1+).

However, α_s ≤ s implies 2^kα₁ ≤ (₁₊² )^k+1, so that k ≤ C. That is, for s₁ large enough, there is a positive crossing ofs1R. The first part of Theorem4.2.1 is proved.

We prove now the second assertion of Theorem4.2.1. 2 We prove now the one-arm bound.

4.4. PROOF OF TASSION’S THEOREM 55 Theorem 4.4.2 For any random centered Gaussian field f : R² → R such that e(x, y) = k(kx−yk), with e(x, y)≤C/kx−yk^D with D >12, there existsγ >0 andC >0 such that for any 1≤s < t,

Pr(one arm in A(s, t)≤C(s/t)^γ.

Proof. Fix1≤s < t, and letT be a periodic symmetric triangulation. By Theorem 4.3.1, for anyη >0, there exist a constantC >0, such that the probability of the continuous one arm event inA(s, t)is bounded by the the one of the discrete one arm over the latticet^−2+η plusCt^−η. The probability pof the latter event satisfies

p ≤ Pr

h \

i∈N, s≤5ⁱ√ st≤t/2

A^c₅i√ st

i ,

whereA_s denotes the event that there is a positive circuit in the annulus A(s,2s). Then, by Corollary4.3.7,

p ≤ Pr[A^c√

st] Prh \

1≤i≤blog₅(¹₂√

t s)c

A^c

5ⁱ√ st

i

+Ct^4+4(2+η)(√ st)^−D

≤ Y

0≤i≤blog₅(¹₂√

t s)c

Pr[A^c

5ⁱ√

st] +C⁰t^4+4(2+η)(√

st)^−Dlogt.

By Theorem 4.1.6 and the FKG condition, there exists c >0, such that PrA_s ≥ c for the continuous model, so thatPrA_s≥c−t^−η for the discrete one. This implies

p ≤ C⁰⁰(1−c+t^−η)^log

√

t/s+C⁰⁰t^4+4(2+η)(√

st)^−Dlogt

≤ C⁰⁰(s/t)^{log(1−c+t−η}+C⁰⁰(s/t)^D−12−4ηlogt.

Fixη such thatD−12−4η >0andt0 such that1−c+t^−η₀ >1−c/2.We obtain the result fort≤t₀, hence for all t. 2

Chapter 5

Annex

5.1 Probability toolbox

5.1.1 Generalities.

Definition 5.1.1 1. A probability space is a triplet (Ω,F,P), where Ω is a set, F a σ-algebra, that is a non empty subset of P(Ω) which is closed under complement and coutable unions, andPa measure over(Ω,F), that is a map : P:F →[0,∞]satisfying P(∅) = 0 and P is additive for the union of disjoint sets.

2. IfΩis topological space, the Borelianσ-algebra is theσ−algebra generated by the open sets of Ω.

3. A random variable X : Ω → R is is a F-measurable map, that is for any Borelian B ⊂R, B ∈ B(R), X⁻¹(B) ∈ F. Note if X is (Ω,{Ω,∅},P)-measurable (that is we choose the trivialσ-algebra), thenX is constant.

4. We say that events (A₁,· · · , A_n)∈ Fⁿ are independent if

∀(I_k ⊂R)_k, Pr(A₁,· · · , A_N) =

N

Y

i=1

Pr(A_k).

5. (X₁,· · ·, X_N) are independent variables iff

∀(I_k ⊂R)_k, Pr(X₁ ∈I₁,· · · , X_n∈I_N) =

N

Y

i=1

Pr(X_i ∈I_k).

5.1.2 Gaussian vectors.

Form∈Rand Σ>0, a random variableX follows the normal lawN(m,Σ) iff

∀t∈R, Pr(X ≤t) = Z t

−∞

e^{−12(x−m)2Σ} dx

√ Σ√

2π. Exercise 5.1.2 Verify that Pr(X∈R) = 1. You can use the classical

Z

R

e^−12x2dx=√ 2π.

57

Definition 5.1.3 A random vector X = (Xi)i=1,···,N ∈R^N is a non-degenerate Gaussian Σis called the variance matrix of X, and denoted byVar(X).

Notation. For any Gaussian vector X ∈ R^N, and any deterministic t ∈R^N, we denote byφX(x) the density of the law ofX at x with respect to the Lebesgue measure, that is,

Pr(X∈A) = Z

x∈A

φ_X(x)dx.

Proposition 5.1.4 1. The coefficients ofΣ can be computed by the formula Σ_i,j =E (X_i−E(X_i)(X_j−E(X_j)

, or Σ =E (X−m)(X−m)^t).

2. If X1 ∼ N(m1,Σ1) ∈ R^N1 and X2 ∼ N(m2,Σ2) ∈ R^N2 are two non-degenerate Gaussian vectors, then the random vector (X₁, X₂) has variance

Σ =

Proof. To prove this (a Fourier transform), we diagonalize Σ into an orthonormal basis and use the elementary

This equality allows to define Gaussian vectors with non negative covariance matrix, that is withΣhaving a kernel. In this cases, any vector in the kernel correspond to finir.

5.1. PROBABILITY TOOLBOX 59 5.1.3 Gaussian fields

The correlation function.

Definition 5.1.6 Let(Ω,F,Pr)be a probability space. AGaussian fieldf over a topological spaceM a measurable function on f : Ω×M →R, whereM andR are equipped with their Borelian σ-algebra, such that for any finite set of points (x1,· · ·, xn) ∈ M^k, the random vector X_ω := (f(ω, x_i))ⁿ_i=1 is Gaussian.

Example 5.1.7 Let (φi)i=1,···,N be a finite set of functions φi :M → R, and (ai)i=1···n a Gaussian vector. Then, f(x) :=Pn

i=1aiφ(x) is a Gaussian field over M.

Definition 5.1.8 The two-point correlation functionassociated to a random Gaussian func-tion is the funcfunc-tion

Example 5.1.9 In example 5.1.7, the correlation function equals e(x, y) =

N

X

i=1

Cov(ai, aj)φi(x)φj(x).

Assume thata_i are independent and follow the same normal law N(0,1). Then, e(x, y) =

N

X

i=1

φ_i(x)φ_i(y).

1. For the Kac random polynomials defined by 1.0.1, e(x, y) =

2. There is another, more natural in fact, random model for polynomial, the complex Fubini-Study, or Kostlanmeasure:

where the (a_k)_k are still independent and follow the same centered normal law. We will see later why this is a good measure to choose, which is not really clear at first sight! For the moment, let us compute the 2-point correlation:

e(x, y) = Gaussian field with covariance e, such that e can be derivated in at least k times in x and k times in y, and that these derivatives are continuous. Then, almost surely f is C^k−1. Moreover, for any differential operator P,P f is a Gaussian field whose kernel equals

E(P f(x)P f(y)) =PxPye.

Theorem 5.1.11 ([1, Lemma 12.11.12]) LetM be a manifold andf :M →Ra Gaussian field, almost surelyC¹. Then, almost surely f vanishes transversally.

5.1.4 Conditional expectation.

Assume two real random variablesX, Y are defined on the same probability space have a derivable joint density, that is

F(x, y) = Pr(X ≤x, Y ≤y) = Z

s≤u,t≤v

φ_X,Y(u, v)dudv is differentiable inxand y. Note that we can recover the law ofX by

Pr(X≤x) = Pr(X≤x, Y ≤+∞) =

Define the conditional density by φX

Y =y) the associate random variable. Then,

∆→0lim Pr(X≤x|Y ∈[y±∆]) = lim

which shows that the definition ofZ_y is coherent with the intuitive definition given by de latter limit. Now

The following proposition in the Gaussian case is very useful, and is called theregression formula in books. It gives the law of Gaussian vector conditioned that another Gaussian vector has a particular value in terms of another unconditioned Gaussian vector.

Proposition 5.1.12 (see [3, Proposition 1.2]) Let X1 ∼ N(m1,Σ1) ∈ R^N1 and X2 ∼ N(m₂,Σ₂)∈R^N2 be two Gaussian vectors, such that X₂ is non-degenerate. Then, for any x₂ ∈R^N2 and any bounded functionf :R^N1→R,

E(f(X1)|X₂=x2) =E(f(X3+Cx2)),

5.2. REMINDER FOR THE PROJECTIVE SPACES 61 where

C = Cov(X1, X2)Σ⁻¹₂ ∈M_N1,N2(R) andX₃∈R^N1 is a Gaussian vector satisfying

X3 ∼N m1−Cm2,Σ1−Cov(X1, X2)Σ⁻¹₂ Cov(X1, X2)^t. .

Remark 5.1.13 1. Note that if X1 and X2 are independent, C = 0, X3 = X1 and E(f(X1)|X₂ = y) = E(f(X1)) : knowing something on X2 has no influence on the result for X₁.

2. On the converse, if X1 = X2, then (f(X1)

X1 =x2) =f(x2). Here C =IN1, and X3 = 0 !

Proof of Proposition5.1.12. The trick is to find a matrixC such thatX₃:=X₁−CX₂ is independent ofX2. SinceX3 is Gaussian, independence is equivalent toCov(X3, X2) = 0, that is

Cov(X₁, X₂) = Cov(CX₂, X₂) =E

(C(X₂−m₂))(X₂−m₂)^t)

=CS₂, so that

C = Cov(X₁, X₂)S₂⁻¹. From Proposition5.1.4,

X3∼N m1−Cm2, S1−Cov(X1, X2)S₂⁻¹Cov(X1, X2)^t. . Now

E(f(X₁)

X₂ =x₂) =E(f(X₃+CX₂)

X₂=x₂) =E(f(X₃+Cx₂)).

2

Exercise 5.1.14 For f(x) a centered Gaussian field, Pr(f(x)>0

f(y)>0) = 1 2+ 1

πarcsine(x, y).

5.2 Reminder for the projective spaces

5.2.1 Definition and basic properties

ForK=R orC, and n≥1, then-dimensional projective space KPⁿ is defined by KPⁿ=Kⁿ⁺¹/∼,

where

(X₀,· · · , X_n)∼(Y₀,· · · , Y_n)⇔ ∃t∈K^∗,(Y₀,· · · , Y_n) =t(X₀,· · ·, X_n).

In other terms,Kⁿ is the set of lines inKⁿ⁺¹. Note that we have RPⁿ=Sⁿ/∼,

whereSⁿ:={kxk= 1},and if

π :Kⁿ⁺¹ →KPⁿ,

is the projection, then for any[X]∈RPⁿ,π⁻¹([X]) =RX and π_|⁻¹

Sⁿ([X]) ={±X}.

ForK=C, we haveπ⁻¹([X]) =CX and π_|⁻¹

S²ⁿ⁺¹([X]) ={tX|t∈C,|t|= 1}.

The setKPⁿcan be equipped with the topology induced by π, that isU ⊂KPⁿ is open iff π⁻¹(U)∈Kⁿ⁺¹ is open for the standard topology in the affine space.

Theorem 5.2.1 [20, p. 60–61] The projective spaceRPⁿ (resp. CPⁿ) is a smooth compact manifold of dimensionn(resp. 2n).

Remark 5.2.2 In fact, CPⁿ is a complex manifold, that is we can choose transition func-tions as holomorphic funcfunc-tions (locally defined onCⁿ).

5.2.2 Metric on RPⁿ.

Recall that RPⁿ can be defined as the quotient of Sⁿ ⊂Rⁿ⁺¹ by the projection π :Sⁿ → Sⁿ/∼=RPⁿ. Moreover,π is a local diffeomorphism. This allows to define a natural metric onRPⁿ by

∀x∈RPⁿ,∀u, v∈T_xRPⁿ, g(u, v) :=hdπ⁻¹(u), dπ⁻¹(v)i.

Here, dπ⁻¹ is the differential of a local inverse π⁻¹. If we choose another and different inverse π⁻¹⁰, then π⁻¹⁰ = −π⁻¹, so that dπ⁻¹⁰ = −dπ⁻¹ and the metric defined by π⁻¹⁰ equals the first one.

5.3 Volumes and coarea formula

5.3.1 Riemannian volumes

We recall here some generalities for volumes in manifolds. If(Mⁿ, g)is a smooth Riemannian manifold, that is a manifold equipped with a metric g, that is a scalar product g_x on the tangent space T_xM which is smooth in x. The definition of the latter means that in local coordinates (xi)i, for any pair of tangent vectors v, w ∈ TxM, v = Pn

i=1vi∂xi and w=Pn

i=1w_i∂_xi,

g(v, w) =X

i,j

gij(x)vivj

, where g_ij(x) is a smooth function. Now, the volume form associated to g is defined in coordinates by then-form

q

(det(g_ij(x))_ij)dx₁∧ · · · ∧dx_n.

This definition does not depend on coordinates, since if(y_j) are other coordinates, then (dx_i)_i= (∂x_i

∂y_j)_ij(dy_j)_j, so that by definition of the determinant,

dx₁∧ · · · ∧dx_n= det(∂xi

∂y_j)_ijdy₁∧ · · · ∧dy_n.

5.3. VOLUMES AND COAREA FORMULA 63 Moreover,∂_xif =P

j∂_yjf^∂y_∂x^j

i so that

∂xi =X

j

∂yj

∂yj

∂x_i, so that the transformation matrix equalsP := (^∂y_∂x^j

i)_ij.This implies

G:= (gij(x))ij = (g(∂xi, ∂xj))ij (5.3.1)

= P^tGP,˜ (5.3.2)

whereG˜ is the matrix of the scalar product in the basis(∂_yi)_i, so thatdetG= det²Pdet ˜G,

and √

detGdx1∧ · · · ∧dxn= p

det ˜Gdy1∧ · · · ∧dyn.

We denote this intrinsic (that is, independent of the coordinates) form the volume form associated to the metricgand we dente it by dvol_M.

Note that if G is the matrix of a scalar product on Rⁿ, then √

detG is the volume of the volume generated by the vectors ∂_xj. In particular, if this basis is orthonormal, then

√

detG= 1. This is why in a Riemannian manifold we define

∀U ⊂M, Vol(U) :=

Z

U

dvol_M(x).

5.3.2 The coarea formula

We will need the following useful equation:

Lemma 5.3.1 For any linear map: F : (E, g)→(E⁰, g⁰) between euclidean spaces of finite dimension, such that dimE⁰≤dimE andF is onto. Then

√

detF◦F^∗=|detF_|kerF^⊥|, where in the right-hand-side,det is computed in orthormal basis.

Proof. Let(ei)i be an ONB ofkerF and (e⁰_j)j an ONG of kerF^⊥, and (f_k)_k an ONB of imF. Then,

mat(F) = 0 matF_|kerF^⊥ , so that

matF F^∗ = 0 matF_kerF^⊥ 0 matF_|^∗_kerF⊥

!

=matF_|kerF^⊥matF_|^∗_kerF⊥, which gives the result sincedetmatF^∗ = detmatF. 2

Proof of Theorem2.2.4. By Sard’s theorem, sincef isC¹, the Lebesgue measure of the set of critical points off vanishes. Since for any such critical pointxwe havedetdf◦df^∗ = 0, we can assume that anyx in both integrals are regular for f. Fixx ∈M, and choose local coordinates(x_i)_i based on an open subsetU 3x, such that

h∂_xm−n+1,· · ·, ∂xmitkerdf(x).

For this fix first coordinates. This gives a basis of TxM; find a linear change of variables to find a new basis that split in two parts that satisfies the condition, and apply this linear change of variables to the coordinates themselves. Define

∀x∈U, φ(x) := (f(x), x⁰)∈Rⁿ×R^m−n, wherex⁰ = (x₁,· · ·xm−n). Then

kerdφ(x) = kerdf(x)∩ h∂_x⁰⁰i={0},

where x⁰⁰ = (xm−n+1,· · · , x_m), so that by the local inversion theorem, φ is a local diffeo-morphism. Note that for anyx∈M,

vx :=φ⁻¹(f(x),·) :R^m−n → M (5.3.3)

x⁰ 7→ φ⁻¹(f(x), x⁰) (5.3.4) is in fact a parametrization of the fiberf⁻¹(x), so that

imdvx(x⁰) =T_vx(f(x),x⁰₎f⁻¹(x) = kerdf(v(f(x), x⁰)).

The volume form onTxf⁻¹(x) in the coordinatesx⁰ write dvol_f⁻¹_(x)=|detdvx|dx⁰ DecomposingTxM = kerdf(x)^⊥⊕kerdf(x) gives

dφ(x) =

df_|kerdf^⊥ 0

∗ dv_x⁻¹

, so that in orthonormal basis,

|detdφ(x)|=|detdf_|kerdf^⊥||detdv_x⁻¹|.

This implies Z

U

gp

detdf◦df^∗dx = Z

U

g|detdφ(x)|detdv_x(φ(x)|dx (5.3.5)

= Z

φ(U)

g◦φ⁻¹|detdvx|dx⁰dx⁰⁰ (5.3.6)

= Z

Rⁿ

Z

φ(U)∩f⁻¹(x)

g◦φ⁻¹|detdvx(x⁰)|dx⁰dx⁰⁰ (5.3.7)

= Z

x⁰⁰

Z

φ(U)∩f⁻¹(x))

gdvol_f⁻¹_(x)dx⁰dx⁰⁰ (5.3.8) Now, using a partition of unity compatible with the chart we have chosen, we obtain the result. 2

5.4 Solutions of exercices

5.4. SOLUTIONS OF EXERCICES 65

Before going on, let us check that iff(x) is independent of f(y), then the result is 1/2. In this case,e(x, y) = 0and we see that it is true. Let us go on: let v= tanθ. Then,

Other method for the beginning. Using the conditional expectation. In this case, Σ1 = Σ2 = 1 andCov(f(x), f(y)) =e(x, y). By the former proposition, for anya∈R,

(f(x)|f(y) =a)∼N(e(x, y)a,1−e(x, y)²).

We have is an ONB made of eigenfunctions, and we choose k≤L for the answer.

3. We have

eL(x, y) = 1 + 2

N

X

k=1

cos(kx) cos(ky) + sin(kx) sin(ky)

= 1 + 2

Let us use a classical trick that begins with 1 =

5.4. SOLUTIONS OF EXERCICES 67 this problem is, in a simple case, to estimate the number of zeros off in an bounded open set of Rⁿ. We assume that there exists a Gaussian centered field g : Rⁿ → R, such that for any i ∈ {1,· · · , n}, f_i is a copy of g, and all the f_i⁰s are independent. We denote by e: (Rⁿ)²→R the covariance function associated tog, and we assume thateisC¹.

1. For anyx∈Rⁿ, write the variance matrix of the vector f(x) as related toe.

We have V arf(x) = (E(fi(x)fj(x))ij) = (δije(x, x))ij) =e(x, x)1n

2. For anyx∈Rⁿ, compute the densityφ_f(x)(0)as a function of e.

We have

φ_f(x)(0) = 1

(2πe(x, x))^n/2.

3. For any x ∈ Rⁿ, denote by Df(x) ∈ Mn(R) the Jacobian matrix of f at x in the canonical basis. Write the variance matrix of the vector Df(x) as a function of e and its derivative. In this question we consider that a matrix in M_n(R) is a vector (mij)1≤i,j≤n∈Rⁿ

2, however we keep the notations1≤i, j≤ninstead of1≤k≤n². We have

V arDf(x) = V ar M at(df(x), B_can)

= V ar (∂_jf_i|x)_i,j)

= (E(∂_jf_i(x)∂<sub>`</sub>f<sub>k</sub>(x))1≤i,j,k,`≤n) = (δ_ik∂_x²_{jy`</sub>e<sub>|x=y</sub>)<sub>ij,k`}. 4. WriteCov(f(x), Df(x)).

We have

Cov(f(x), Df(x)) = (Efi(x)∂<sub>`</sub>f<sub>k</sub>(x))1≤i,k,`≤n= (δ_ik∂<sub>`</sub>e)1≤i,k,`≤n.

Til the end of the problem we assume that there exists a smooth ρ:R⁺→R such that

∀x, y∈Rⁿ, e(x, y) =ρ(−1

2kx−yk²).

5. WriteV arf(x),V arDf(x)andCov(f(x), Df(x))as a function ofρand its derivatives.

We have V arf(x) =ρ(0)1n. Besides,

∂_xje=ρ⁰(−1

2kx−yk²)(−(x_j−y_j)) and ∂_x²_iy

`e<sub>|x=y</sub> =δ<sub>i`</sub>ρ⁰(0). so that

V arDf(x) =ρ⁰(0)1_n². Lastly, Cov(f(x), Df(x)) = 0_n³.

6. LetU ⊂Rⁿ be an open subset and

N(f, U) := #{x∈U, f(x) = 0}.

Show that

EN(f, U) =cV ol(U)^αρ(0)^βρ⁰(0)^γ,

with α, β, γ and c constants which depend only on n. Give the values of α, β and γ and write cas an integral over Mn(R).

By Kac-Rice, we have

EN(f, U) = Z

U

E(|detDf(x)| |f(x) = 0)φ_f(x)(0)dx,

5.4. SOLUTIONS OF EXERCICES 69 wheredxis the Lebesgue measure. Since Cov(Df(x), f(x)) = 0, the Gaussian vectors Df(x) and f(x) are independent, so that

E(|detDf(x)| |f(x) = 0) = E(|detDf(x)|)

8. Let M ∈ M_n(R) be random, such that its coefficients are independent and follow a normal lawN(0,1).

where for any k,Xk ∈R^k is a random vector which coefficients are independent and follow a normal lawN(0,1).

5.4.8 Proof of Lemma 4.2.10 By Lemma5.1.5,

φ_X(x) = Z

λ∈R^N

e^ihx−m,λie^{−12hΣλ,λi} dλ (2π)^N, so that fori6=j, remembering that hΣλ, λi=P

i,jσijλiλj,andσji=σij,

∂ σij

φ_X(x) = Z

λ∈R^N

e^ihx−m,λi(−1

2(2λ_iλ_j))e^{−12hΣλ,λi} dλ (2π)^N, whereas for everyi, j,

∂

∂x_i∂x_jφ_X(x) = Z

λ∈R^N

(−iλ_i)(−iλ_j)e^ihx−m,λi(−1

2(2λ_iλ_j))e^{−12hΣλ,λi} dλ (2π)^N, which gives the result fori6=j.

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Dans le document Topology of random algebraic and analytic hypersurfaces (Page 47-0)