Bourbaki

Exercise 60, § 1

Tag 02M0, permanent. Cite it as [Bourbaki, Tag 02M0].

Spectral Theories, II, p. 296

a) Let H⊂C\mathbf{H}\subset \mathbf{C} be the open set of complex numbers zz such that I(z)>0\mathscr{I}(z)>0. The function

Θ(z)=∑n∈Ze2iπn2z \Theta (z) =\sum_{n\in\mathbf{Z}}e^{2i\pi n^2z}

is holomorphic on H, and it satisfies Θ(z)=(z/i)−1/2Θ(−1/z)\Theta (z) = (z/i)^{-1/2}\Theta (-1/z) for z∈Hz\in H, where z↦(z/i)1/2z\mapsto (z/i)^{1/2} is the unique holomorphic function on H which extends the function

√ \surd

iy↦y‾iy\mapsto \overline{y}. (Use Exercise 16, b).)

b) Let mm and nn be strictly positive integers. One has

√1m‾∑m−1e2iπk2n/m=e√iπ/2n∑42n−1e−2iπk2m/(2n) \surd 1\overline{m}\sum^{m-1}e^{2i\pi k^2n/m}=e\surd^{i\pi /}2n\sum^{42n-1}e^{-2i\pi k^2m/(2n)}
k=0k=0k=0k=0

(Apply the relation of the preceding question with z=−2in/m+iyz=-2in/m+iy and let y→0y\rightarrow 0.) c) For every integer m⩾1m\geqslant 1 and every n∈Zn\in \mathbf{Z}, put

τ(n,m)=∑k=0m−1e2iπk2n/m \tau (n, m) =\sum^{m-1}_{k=0}e^{2i\pi k^2n/m}

For all strictly positive integers m1m_1 and m2m_2, one has

τ(1,m1m2)=τ(m2,m1)τ(m1,m2) \tau (1, m_1m_2) =\tau (m_2, m_1)\tau (m_1, m_2)
((m−1)/2)2√((m-1)/2)^2\surd

d) If mm is odd, one has τ(1,m)=im‾\tau (1, m) =i\overline{m}. (Use a).)

e) If pp is an odd prime number, then τ(n,p)=ℓp(n)τ(1,p)\tau (n, p) =\ell_p(n)\tau (1, p), where ℓp\ell_p is the unique character of order 2 of (Z/pZ)∗(\mathbf{Z}/p\mathbf{Z})^*, extended to Z/pZ\mathbf{Z}/p\mathbf{Z} by putting ℓp(0)=0\ell_p(0) = 0.

f) Let p≠qp\not=q be two odd prime numbers. One has

ℓp(q)ℓq(p)=τ(1,pq)τ(1,p)τ(1,q) \ell_p(q)\ell_q(p) =\frac{\tau(1, pq)}{\tau(1, p)\tau(1, q)}

g) Deduce a new proof of the quadratic reciprocity law (cf. A, V, p. 156, exerc. 23): for all odd prime numbers p≠qp\not=q, one has

ℓp(q)ℓq(p)=(−1)(p−1)(q−1)/4 \ell_p(q)\ell_q(p) = (-1)^{(p-1)(q-1)/4}
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Machine translation of the English, by gpt-5-6. Not checked by a person.