A Quick Proof of Reciprocity for Hecke Gauss
SumsHatice Boylan∗ and Nils-Peter Skoruppa

Abstract
In this note we present a short and elementary proof of Hecke’s reciprocity law for Hecke-Gauss sums of number fields.

In Chapter VIII of his book [Hec70], Hecke introduced and studied certain Gauss sums associated to arbitrary number fields. In particular, hediscovered a reciprocity law for these sums [Hec70, Satz 163, p. 240], which
he proved by analyzing the values of suitable theta functions in the cusps.
The purpose of the present note is to give a short and elementary proof ofHecke’s reciprocity law. Our proof is based on Milgram’s formula [MH73,p. 127]X

1pe B(x, x)/2 = e(s/8),(1)
L] /L x∈L] /L
where (L, B) is an even integral lattice (i.e. L is a free Z-module of finite
rank and B a symmetric non-degenerate integer valued bilinear form on L
such that B(x, x) is even for all x in L), L] denotes the dual lattice {y ∈ L :B(y, L) ⊆ Z}, s is the signature of L, and e(x) = exp(2πix) as usual.
Hecke’s Gauss sum was defined by the formulaXC(ω) =e tr(µ2 ω)µ mod awhere K is an arbitrary number field and ω a non-zero element of K. HereN and tr denote the (absolute) norm and the trace of K and a denotes thedenominator of ωd, where d is the different of K. The sum is to be taken∗

During the preparation of the paper the first author was supported by the Scientificand Technological Research Council of Turkey TÜBİTAK
Key words and phrases. Gauss and Kloosterman sums; generalizations
AMS 2000 subject classifications. 11L05

1over a complete set representatives for the ring O of integers of K modulo a.
(Recall that the denominator of ωd is the unique integral ideal a such thatωd = b/a with an integral ideal b relatively prime to a.) It is easily checkedthat the terms of the sum C(ω) depend only on the residue class µ + a.
We state Hecke’s reciprocity law in a renormalized form that is somewhat
clearer than the original formulation. We begin with the following lemma
whose short proof will be given at the end of the paper.
Lemma. For a non-zero ω of K, the number C(ω) is non-zero if and only
if the homomorphism
ea := a/(2, a) → {±1}, µ 7→ e tr(ωµ2 )(2)is non-trivial. If this conditionis satisfied, then e (tr(ωµ2 )) depends onlypa) · [ea : a].
on µ mod ea, and |C(ω)| = N(eFor ω satisfying the condition of the lemma, we setX
1C(ω)B(ω) := pe tr(µ2 ω) =.
|C(ω)|
N(ea) µ mod ea(In fact, B(ω) is an eighth root of unity, with an explicit formula as e(s/8),
where s is the signature of a certain lattice,1 but this fact does not seem tolead to an alternative proof of the reciprocity and will not be used in thesequel.) We also setXSign(ω) =sign σ(ω),σ

where the sum runs over all real embeddings σ of K. With these notations,
Hecke’s reciprocity law can be restated as follows.
Theorem. For any non-zero ω in K such that the homomorphism (2) is
non-trivial, one hasB(ω) = e (Sign(ω)/8) B −γ 2 /4ω ,where γ denotes any number in K such that γd is integral and relatively
prime to the denominator of (4ωd)−1 .
Namely, it is easy to show that O/ea, µ + e
a 7→ tr(ωµ2 ) + Z is a non-degenerate finite
quadratic module and hence, by a theorem of Wall [Wal63, Theorem (6)], isomorphic tothe discriminant module of an even integral lattice. Then B(ω) = e(s/8) by Milgram’sformula, where s is the signature of this lattice.
12Note that under the stated hypothesis C(γ 2 /4ω) is different from 0, andhence that B(γ 2 /4ω) is defined. In fact, if tr(ωµ2 ) is integral for all µ in ea,2 2
then on setting µ = γν/2ω we see that tr(γ ν /4ω) is integral for all ν in2ωea/γ = 2b/(2, a)γd ⊆ 2b/(2, a). But the last ideal is the denominator ofγ 2 /4ω, as we will see in the course of the proof.
Proof. Using the obvious identity B(ω) = B(−ω) we can rewrite the reciprocity formula more symmetrically asB(ω)B γ 2 /4ω = e (Sign(ω)/8) .
(3)
We assume first of all that the class number of K is 1, i.e. that every idealof the ring of integers O of K is principal. Let d = δO and write ωδ = αβwith relatively prime integers α and β in K. We can then choose γ = 1/δ
and the left hand side of (3) becomesX
12 β2 αpe tr µ+ν,αδ4βδ| N(2αβ)| µ mod αν mod 2β2α
provided α is odd (so that 4β is the exact denominator of γ4ωδ = 4βandea = a = αO), which we assume for the moment. By writingµ2α(2µβ + να)21β+ ν2≡mod Oαδ
4βδ
4αβδ
δ

and on noticing that (µ, ν) 7→ 2µβ + να defines an isomorphism of O/αO ×O/2βO with O/2αβO, we see that the last double sum becomesX
τ2 1pe tr.
4αβδ| N(2αβ)| τ mod 2αβConsider the lattice L = (O, B), where B is the bilinear form on O definedby B(x, y) = tr(2αβxy/δ). It is easily checked that B is non-degenerate andtakes on even integral values. Moreover, for the dual O] of O with respect
to B we findO] = {y ∈ Q ⊗Z O : B(y, O) ⊆ Z} = (2αβ)−1 O.
Using these notations the last sum may be rewritten as
X1pe (B(x, x)/2) .
|O] /O| x∈O] /O3

But according to the formula (1) this sum equals e(s/8), where s denotes thesignature of the quadratic form B(x, x) on R ⊗Z O. It is easily checked thats = Sign(ω) which then proves (3).
To prove the general case we rewrite the left hand side of (3) as
X1γ 2ν 2 2qe tr ωµ +,(4)4ω
N(eaeb1 ) µ mod eaν mod eb1

where we write as before ωd = ba−1 with relatively prime integral ideals aand b, and where b1 denotes the denominator of γ 2 d/4ω. Recall that, for any
ideal c, we use ec = c/(2, c). Since, by definition, γd is integral and relatively
prime to the denominator of (4ωd)−1 , we find that the denominator b1 ofγ 2 d(4ω)−1 = (γd)2 (4ωd)−1 equals the denominator of (4ωd)−1 = a(4b)−1 .
From this and the fact that a and b are relatively prime, we obtain
b1 =2beb1 =.
(2, a)

4b,(4, a)b1(The second identity follows from the first one on writing e
b1 = (2,b=1)
4b/(4,a)2b2b= (4,a,2b)= (2,a).) We write(2,4b/(4,a))ωµ2 +(2ωµ + γν)2γ 2ν 2≡
mod d−1 .
4ω4ω

Now the map (µ, ν) 7→ 2ωµ + γν, O × O → 2ωO + γO induces a map
2ωO + γOφ : O/ea × O/eb1 →.
2ωea + γeb1
We claim that φ is a isomorphism. Since φ is obviously surjective it suffices
to prove thatN(2ωea + γeb1 ).
N(eaeb1 ) =N(2ωO + γO)
But this follows from:
2b2b + γad2b/(2, a) + γead1+ γO === ,eeadadadadee
2b2γbb1b1
2ωea + γeb1 =+= (O + γd) = .
(2, a)d (2, a)dd

2ωO + γO =For the last identity of the first line we use 2b/(2, a) + γdea = O since ea andeγd are relatively prime to 2b/(2, a) = b1 .
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Using this isomorphism φ we can rewrite (4) as1 Xx2 pe tr,4ω|M | x∈M]
where Mad)−1 /eb1 d−1
 . But M = L /L, where L denotes the even integral
 = (exy. Hence, we can again apply formula (1) to deduce
lattice eb1 d−1 , 2 tr 4ω
that the last sum equals e(s/8), where s is the signature of the lattice L.
Finally, to compute the signature s we note that a Gram matrix for L
is given by ∆t D∆, where D is the diagonal matrix with σi (1/2ω) on thediagonal and σi running through the embeddings of K into C, and where∆ = (σi (αj ))i,j with {αj } denoting a Z-basis of eb1 d−1 . But the signature of∆t D∆ equals Sign(1/4ω) = Sign(ω), as is obvious if K is totally real and aneasy exercise in the general case. This proves the theorem.

Proof of Lemma. Using C(ω) = C(−ω) we find that |C(ω)|2 equalsXX Xe tr(ωµ2 )e (2 tr(ωµν)) ,e tr ω(µ − ν)(µ + ν) =µ,ν mod aµ mod aν mod awhere the right hand side is obtained by substituting µ + ν 7→ µ in theleft hand side. The inner sum on the right equals N(a) if 2µωd is integral,i.e. if µ ∈ e
a = a/(2, a), and is 0 otherwise. We have thereforeX|C(ω)|2 = N(a)e tr(ωµ2 ) .
µ∈ea/aIt is easily checked that the application µ 7→ e (tr(ωµ2 )) defines a grouphomomorphism ea/a 7→ {±1}. Hence the last sum is different from 0 if and2only if tr(ωµ ) ∈ Z for all µ ∈ ea, in which case |C(ω)|2 = N(a) · [ea : a] =2eN(a) · N (2, a) . The remaining statement of the lemma is obvious.

References[Hec70] Erich Hecke. Vorlesungen über die Theorie der algebraischen Zahlen.
Chelsea Publishing Co., Bronx, N.Y., 1970. Second edition of the1923 original, with an index.
[MH73] John Milnor and Dale Husemoller. Symmetric bilinear forms.
Springer-Verlag, New York, 1973. Ergebnisse der Mathematik undihrer Grenzgebiete, Band 73.
[Wal63] C. T. C. Wall. Quadratic forms on finite groups, and related topics.
Topology, 2:281–298, 1963.
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