Dedekind sum

Define the sawtooth function ${\displaystyle (\!(\,)\!):\mathbb {R} \rightarrow \mathbb {R} }$ as

${\displaystyle (\!(x)\!)={\begin{cases}x-\lfloor x\rfloor -1/2,&{\mbox{if }}x\in \mathbb {R} \setminus \mathbb {Z}$ ;\\0,&{\mbox{if }}x\in \mathbb {Z} .\end{cases}}}

We then define the Dedekind sum

${\displaystyle D:\mathbb {Z} ^{2}\times \mathbb {Z} _{>0}\to \mathbb {Q} }$

by

${\displaystyle D(a,b;c)=\sum _{n=1}^{c-1}\left(\!\!\left({\frac {an}{c}}\right)\!\!\right)\!\left(\!\!\left({\frac {bn}{c}}\right)\!\!\right).}$

For the case a = 1, one often writes

s(b, c) = D(1, b; c).

Note that D is symmetric in a and b, i.e.,

${\displaystyle D(a,b;c)=D(b,a;c),}$

and that, by the oddness of (( )),

D(−a, b; c) = −D(a, b; c).

By the periodicity of D in its first two arguments, the third argument being the length of the period for both,

D(a, b; c) = D(a+kc, b+lc; c), for all integers k,l.

If d is a positive integer, then

D(ad, bd; cd) = dD(a, b; c),

D(ad, bd; c) = D(a, b; c), if (d, c) = 1,

D(ad, b; cd) = D(a, b; c), if (d, b) = 1.

There is a proof for the last equality making use of

${\displaystyle \sum _{n=1}^{c-1}\left(\!\!\left({\frac {n+x}{c}}\right)\!\!\right)=(\!(x)\!),\qquad \forall x\in \mathbb {R} .}$

Furthermore, az = 1 (mod c) implies D(a, b; c) = D(1, bz; c).

If b and c are coprime, we may write s(b, c) as

${\displaystyle s(b,c)={\frac {-1}{c}}\sum _{\omega }{\frac {1}{(1-\omega ^{b})(1-\omega )}}+{\frac {1}{4}}-{\frac {1}{4c}},}$

where the sum extends over the c-th roots of unity other than 1, i.e., over all ${\displaystyle \omega }$ such that ${\displaystyle \omega ^{c}=1}$ and ${\displaystyle \omega \not =1}$.^[6]

Equivalently, if b and c are coprime, then

${\displaystyle s(b,c)={\frac {1}{4c}}\sum _{n=1}^{c-1}\cot \left({\frac {\pi n}{c}}\right)\cot \left({\frac {\pi nb}{c}}\right).}$

This reformulation mirrors the fact that the above cotangent function is the discrete Fourier transform of the sawtooth function.^[6]

Dedekind^[1] proved that, if b and c are coprime positive integers then

${\displaystyle s(b,c)+s(c,b)={\frac {1}{12}}\left({\frac {b}{c}}+{\frac {1}{bc}}+{\frac {c}{b}}\right)-{\frac {1}{4}}.}$

There exist several proofs from first principles, and Dedekind's reciprocity law is equivalent to quadratic reciprocity.^[7]

Rewriting the reciprocity law as

${\displaystyle 12bc\left(s(b,c)+s(c,b)\right)=b^{2}+c^{2}-3bc+1,}$

it follows that the number 6c s(b,c) is an integer.

If k = (3, c) then

${\displaystyle 12bc\,s(c,b)=0\mod kc}$

and

${\displaystyle 12bc\,s(b,c)=b^{2}+1\mod kc.}$

A relation that is prominent in the theory of the Dedekind eta function is the following. Let q = 3, 5, 7 or 13 and let n = 24/(q − 1). Then given integers a, b, c, d with ad − bc = 1 (thus belonging to the modular group), with c chosen so that c = kq for some integer k > 0, define

${\displaystyle \delta =s(a,c)-{\frac {a+d}{12c}}-s(a,k)+{\frac {a+d}{12k}}.}$

Then ${\displaystyle n\delta }$ is an even integer.

Hans Rademacher found the following generalization of the reciprocity law for Dedekind sums:^[8] If a, b, and c are pairwise coprime positive integers, then

${\displaystyle D(a,b;c)+D(b,c;a)+D(c,a;b)={\frac {1}{12}}{\frac {a^{2}+b^{2}+c^{2}}{abc}}-{\frac {1}{4}}.}$

Hence, the above triple sum vanishes if and only if (a, b, c) is a Markov triple, i.e., a solution of the Markov equation

${\displaystyle a^{2}+b^{2}+c^{2}=3abc.}$