Bourbaki

Exercise 61, § 1

Tag 02M1, permanent. Cite it as [Bourbaki, Tag 02M1].

Spectral Theories, II, p. 297

Let r⩾1r\geqslant 1 be an integer. Let A=(ai,j)A = (a_{i,j}) be a symmetric positive-definite matrix of type (r,r)(r, r) with integer coefficients, and let N⩾1N\geqslant 1 be an integer such that the matrix A∗=A^*= NA−1^{-1} also has integer coefficients. We denote by Q (resp. Q~\widetilde{Q}) the integral quadratic form associated with A (resp. with A−1A^{-1}), that is to say that one has Q(x)=txAxQ(x) =^txAx for every x∈Zrx\in \mathbf{Z}^r. We denote by (a,b)↦⟨a∣b⟩A(a, b)\mapsto \langle a|b\rangle_A the bilinear form associated with A, so that ⟨a∣b⟩A=taAb\langle a|b\rangle_A=^taAb.

We denote by H\mathbf{H} the set of complex numbers zz such that I(z)>0\mathscr{I}(z)>0. It is an open subset of C\mathbf{C}. We denote by log:H→C:\mathbf{H}\rightarrow \mathbf{C} the restriction to H\mathbf{H} of the principal determination of the logarithm (FVR, III, p. 10). Let k∈Nk\in \mathbf{N} be an odd integer. For z∈Hz\in \mathbf{H}, we denote by zk=z^k= exp(kk log(zz)).

a) The group SL2(Z)_2(\mathbf{Z}) acts transitively on H\mathbf{H} by

(abcd)⋅z=az+bcz+d,(abcd)∈(a bc d)\cdot z=\frac{az + b}{cz + d},(a bc d)\in SL2(Z)_2(\mathbf{Z})

(cf. TA, I, p. 150, exercise 5).

b) For every complex number z∈Hz\in \mathbf{H} and every x∈Zrx\in \mathbf{Z}^r, the series

∑m∈Zr\sum_{m\in\mathbf{Z}^r} exp(iπQ(m)zi\pi Q(m)z)

m≡xm\equiv x mod.N.N

converges absolutely and uniformly on compact sets.

We denote by θA(z;x)\theta_A(z;x) or θQ(z;x)\theta_Q(z;x) the sum of this series ("theta function associated with Q"). It is a holomorphic function on H\mathbf{H}. c) For every y∈Zry\in \mathbf{Z}^r, one has

∑r\sum_{_r} exp(iπQ(m+y)zi\pi Q(m+y)z) =√=\surddet(A)1 (iz)∑r/2r(\frac{i}{z})\sum^{r/2_r} exp(−iπQ~(m)z+m⋅y)(-i\pi \frac{\widetilde{Q}(m)}{z}+m\cdot y),

m∈Zm∈Zm\in \mathbf{Z}m\in \mathbf{Z}

(apply the Poisson formula and exercise 1).

d) We denote by H the subgroup of (Z/NZ)r(\mathbf{Z}/N\mathbf{Z})^r formed by the elements α∈(Z/NZ)r\alpha \in (\mathbf{Z}/N\mathbf{Z})^r such that Aα=0A\alpha = 0 in Z/NZ\mathbf{Z}/N\mathbf{Z}. The group H is of cardinal det(A). Every character of H is of the form

ψβ(α)=\psi_{\beta}(\alpha ) = exp(2iπ⟨α∣β⟩AN2)(2i\pi \frac{\langle\alpha|\beta\rangle_A}{N^2}),

where β∈H\beta \in H.

e) For every α∈H\alpha \in H and z∈Hz\in \mathbf{H}, one has θQ(z+2;α)=\theta_Q(z+ 2;\alpha ) = exp(2iπQ(α)N2)θQ(z;α)(\frac{2i\pi Q(\alpha)}{N^2})\theta_Q(z;\alpha ). f) For every α∈H\alpha \in H and z∈Hz\in \mathbf{H}, one has

θQ(−1z;α)=√\theta_Q(-\frac{1}{z};\alpha )=\surddet(A)1 (−iz)r/2∑ψβ(α)θQ(z;β)(-iz)^{r/2}\sum\psi_{\beta}(\alpha )\theta_Q(z;\beta ).

β∈H\beta \in H
(ab) (a b)

g) Let g=∈g=\in SL2(Z)_2(\mathbf{Z}) be such that c≠0,d>0c\not= 0,d >0 and b≡c≡0b\equiv c\equiv 0 mod.2.2. For

cd c d

α∈H\alpha \in H and z∈Hz\in \mathbf{H}, one has

θQ(g⋅z;α)=dr/2\theta_Q(g\cdot z;\alpha ) =d^{r/2} det(A)1 (cz+d)r/2∑Φ(α,β)θQ(z;β)(cz+d)^{r/2}\sum\Phi (\alpha , \beta )\theta_Q(z;\beta ),

β∈H\beta \in H

where

Φ(α,β)=∑ψγ(α)(∑\Phi (\alpha , \beta ) =\sum\psi_{\gamma}(\alpha )(\sum exp(iπdN2(bQ(δ)+2⟨γ∣δ⟩A−cQ(γ))))(\frac{i\pi}{dN^2}(bQ(\delta ) + 2\langle \gamma |\delta \rangle_A-cQ(\gamma ))))

γ∈Hδ\gamma \in H\delta mod.dN. dN

δ≡α\delta \equiv \alpha mod.N.N

(0−1) (0-1)

(first consider θQ(g~⋅z;α)\theta_Q(\widetilde{g}\cdot z;\alpha ) where g~=g\widetilde{g}=g ; note the formula

1 0

dg~⋅z=b−(dz−c)−1d\widetilde{g}\cdot z=b-(dz-c)^{-1}, and apply the preceding question ; note that Φ(α,β)\Phi (\alpha , \beta ) depends only on β\beta modulo N).

h) Suppose moreover that c≡0c\equiv 0 mod.2N.2N. Then

θQ(g⋅z;x)=τdr/2(cz+d)kθQ(z;aα) \theta_Q(g\cdot z;x) =\frac{\tau}{d^{r/2}}(cz+d)^k\theta_Q(z;a\alpha )

where

τ=∑\tau =\sum exp(iπbQ(δ)dN2)(\frac{i\pi bQ(\delta)}{dN^2}).

δ\delta mod.dN. dN δ≡x\delta \equiv x mod.N.N

i) Suppose moreover that dd is odd. We have τ=\tau = exp(iπabQ(α)N2)τQ(c;d)(\frac{i\pi abQ(\alpha)}{N^2})\tau_Q(c;d), where

τQ(c,d)=∑e(−iπcQ(x)d) \tau_Q(c, d) =\sum e(-\frac{i\pi cQ(x)}{d})

x∈(Z/dZ)rx\in (\mathbf{Z}/d\mathbf{Z})^r ("Gauss sum"). j) Let us write d=p1⋯pkd=p_1\cdots p_k, where p1,...pkp_1, . . .p_k are prime numbers. We denote

kk (nd)=∏ℓpi(n)(\frac{n}{d})=\prod\ell_{p_i}(n) i=1i=1

where, for every odd prime number pp, ℓp\ell_p denotes the unique character of order 2 of (Z/pZ)∗(\mathbf{Z}/p\mathbf{Z})^*, extended to Z/pZ\mathbf{Z}/p\mathbf{Z} by setting ℓp(0)=0\ell_p(0) = 0 ("Jacobi symbol"). If dd and 2c2c det(A) are relatively prime, then

τQ(c,d)=(\tau_Q(c, d) =( det(A)d)(ε‾d(2cd)√d‾)rd)(\overline{\varepsilon}_d(\frac{2c}{d})\surd \overline{d})^r,

where

{1\{1 if d≡1d\equiv 1 mod.4.4

εd= \varepsilon_d=

ii if d≡−1d\equiv -1 mod.4.4.

(Diagonalize the quadratic form and use the preceding exercise.)

k) If gg is congruent to the identity modulo 4N, then

θQ(g⋅z;α)=(2cd)r(cz+d)kθQ(z;α) \theta_Q(g\cdot z;\alpha ) =(\frac{2c}{d})^r(cz+d)^k\theta_Q(z;\alpha )

This means that the mapping z↦θq(z;α)z\mapsto \theta_q(z;\alpha ) is a holomorphic modular form of weight r/2r/2 and of level 4N. (FM, in preparation.)

Solution

No solution has been written yet.

Cited by

Nothing.

Cites

Nothing.

Machine translation of the English, by gpt-5.4. Not checked by a person.