Section 3.3.3: Theoretical Tests

Exercise 8. ▶ [M34] (L. Carlitz.) Let

$$p(p, q, r) = 12 \sum_{0 \le j < r} \left(!\left(\frac{jp}{r}\right)!\right) \left(!\left(\frac{jq}{r}\right)!\right).$$

By generalizing the method of proof used in Lemma B, prove the following beautiful identity due to H. Rademacher: If each of $p, q, r$ is relatively prime to the other two,

$$\rho(p,q,r) + \rho(q,r,p) + \rho(r,p,q) = \frac{p}{qr} + \frac{q}{rp} + \frac{r}{pq} - 3.$$

(The reciprocity law for Dedekind sums, with $c = 0$, is the special case $r = 1$.)

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Let

$$ D(a,b;c)=\sum_{j=0}^{c-1} \left(!\left(\frac{aj}{c}\right)!\right) \left(!\left(\frac{bj}{c}\right)!\right), $$

so that

$$ \rho(a,b,c)=12D(a,b;c). $$

The statement to be proved is therefore equivalent to

$$ D(p,q;r)+D(q,r;p)+D(r,p;q)

\frac1{12} \left( \frac{p}{qr}+\frac{q}{rp}+\frac{r}{pq} \right) -\frac14 . \tag{1} $$

This is Rademacher's three-term reciprocity law.

The proof follows the same residue-calculus method used in Lemma B for the ordinary Dedekind reciprocity law.

1. Fourier expansion of the sawtooth function

For $(a,c)=1$,

$$ \left(!\left(\frac{aj}{c}\right)!\right)

-\frac1c \sum_{m=1}^{c-1} \frac{\zeta_c^{amj}}{1-\zeta_c^{m}}, \qquad \zeta_c=e^{2\pi i/c}. $$

Substituting this into the definition of $D(a,b;c)$ and using

$$ \sum_{j=0}^{c-1}\zeta_c^{(m+n)j}

\begin{cases} c,&m+n\equiv0\pmod c,\ 0,&\text{otherwise}, \end{cases} $$

gives

$$ D(a,b;c)

\frac1{4c} \sum_{m=1}^{c-1} \cot!\frac{\pi am}{c}, \cot!\frac{\pi bm}{c}. \tag{2} $$

Hence

$$ \rho(a,b,c)

\frac3c \sum_{m=1}^{c-1} \cot!\frac{\pi am}{c}, \cot!\frac{\pi bm}{c}. \tag{3} $$

Thus it suffices to prove (1).

2. The cotangent kernel

Consider

$$ f(z)=\cot(\pi pz)\cot(\pi qz)\cot(\pi rz), $$

where $p,q,r$ are pairwise relatively prime.

Because $f(z)$ is $1$-periodic, the sum of its residues in a fundamental period strip is $0$.

The poles are:

1. $z=0$;
2. $z=k/p$ $(1\le k\le p-1)$;
3. $z=k/q$ $(1\le k\le q-1)$;
4. $z=k/r$ $(1\le k\le r-1)$.

Since $p,q,r$ are pairwise coprime, these poles are all simple except for $z=0$.

3. Residues away from $0$

At $z=k/p$,

$$ \cot(\pi pz)\sim \frac1{\pi p(z-k/p)}, $$

hence

$$ \operatorname{Res}_{z=k/p}f(z)

\frac1{\pi p} \cot!\frac{\pi qk}{p} \cot!\frac{\pi rk}{p}. $$

Summing over $k$,

$$ \sum_{k=1}^{p-1} \operatorname{Res}_{z=k/p}f(z)

\frac{4}{\pi}D(q,r;p) \tag{4} $$

by (2).

Similarly,

$$ \sum_{k=1}^{q-1} \operatorname{Res}_{z=k/q}f(z)

\frac{4}{\pi}D(r,p;q), \tag{5} $$

and

$$ \sum_{k=1}^{r-1} \operatorname{Res}_{z=k/r}f(z)

\frac{4}{\pi}D(p,q;r). \tag{6} $$

4. Residue at $0$

Using

$$ \cot x

\frac1x-\frac{x}{3}+O(x^{3}), $$

we obtain

$$ \cot(\pi pz)

\frac1{\pi pz} -\frac{\pi pz}{3} +O(z^{3}), $$

and analogous expansions for $q$ and $r$.

Multiplying,

$$ f(z)

\frac{1}{\pi^{3}pqr,z^{3}} -\frac1{3\pi z} \left( \frac{p}{qr} +\frac{q}{rp} +\frac{r}{pq} \right) +O(z). $$

Therefore

$$ \operatorname{Res}_{z=0}f(z)

-\frac1{3\pi} \left( \frac{p}{qr} +\frac{q}{rp} +\frac{r}{pq} \right). \tag{7} $$

5. Sum of residues

The residue theorem for a period strip gives

$$ \operatorname{Res}{0}f +\sum{k/p}\operatorname{Res}f +\sum_{k/q}\operatorname{Res}f +\sum_{k/r}\operatorname{Res}f =0. $$

Substituting (4), (5), (6), and (7),

$$ -\frac1{3\pi} \left( \frac{p}{qr} +\frac{q}{rp} +\frac{r}{pq} \right) + \frac4\pi \Bigl( D(p,q;r)+D(q,r;p)+D(r,p;q) \Bigr) =0. $$

Hence

$$ D(p,q;r)+D(q,r;p)+D(r,p;q)

\frac1{12} \left( \frac{p}{qr} +\frac{q}{rp} +\frac{r}{pq} \right) -\frac14. $$

This is exactly Rademacher's reciprocity law. Multiplying by $12$ and recalling that $\rho=12D$, we obtain

$$ \boxed{ \rho(p,q,r)+\rho(q,r,p)+\rho(r,p,q)

\frac{p}{qr} +\frac{q}{rp} +\frac{r}{pq} -3 }. $$

For $r=1$, $D(p,q;1)=0$, and the formula reduces to the classical reciprocity law for Dedekind sums. ∎