Evaluation of \(L(1,\chi)\)

We compute \(\left|L(1,\chi)\right|\) for an odd primitive Dirichlet character \(\chi\pmod N\).

Definition 1. Fix a Dirichlet character \(\chi\pmod N\). Then we define the Dirichlet \(L\)-function \[L(s,\chi)=\sum_{n=1}^\infty\frac{\chi(n)}{n^s},\] where we recall \(\chi(n)=0\) whenever \(\gcd(n,N)>1\) by convention.

Remark 2. We see that \(L(s,\chi)\) converges absolutely and uniformly on compacts in the region \(\operatorname{Re}s>1\), so it defines a holomorphic function there. If \(\chi\) is nontrivial, then in fact the sum converges uniformly on compacts in the region \(\operatorname{Re}s>0\), but we will not need this.

Theorem 3. Fix an odd primitive Dirichlet character \(\chi\pmod N\). Then \[\left|L(1,\chi)\right|=\frac{\pi}{\left|\chi(2)-2\right|\sqrt N}\left|\sum_{1\le k\lt N/2}\chi(k)\right|.\]

Of course, there is a similar statement when \(\chi\) is even, proven by slightly altering the following argument.

The rest of the post is dedicated to the proof of Theorem 3. There are clever ways to do our evaluation, but we may as well do this directly. For now, choose \(s\) to be a real number bigger than \(1\) so that we have absolute convergence. We organize the sum \[L(s,\chi_K)=\sum_{n=1}^\infty\frac{\chi_K(n)}{n^s}\] by residue class\(\pmod N\), writing \[L(s,\chi_K)=\sum_{a=0}^{N-1}\Bigg(\chi_K(a)\sum_{n\equiv a\pmod N}\frac1{n^s}\Bigg).\] Now, we can use character sums to indicate \(n\equiv a\pmod N\), writing \[L(s,\chi_K)=\sum_{a=0}^{N-1}\Bigg(\chi_K(a)\sum_{n=1}^\infty\frac1{n^s}\sum_{k=0}^{N-1}\frac{\zeta_N^{(a-n)k}}N\Bigg).\] This rearranges into \[L(s,\chi_K)=\frac1N\sum_{k=0}^{N-1}\Bigg(\sum_{a=0}^{N-1}\chi_K(a)\zeta_N^{ak}\cdot\sum_{n=1}^\infty\frac{\zeta_N^{-nk}}{n^s}\Bigg).\] We now take \(s\to1^+\). We will handle each inner sum separately. In short, the left inner sum is a Gauss sum, and the right inner sum is \(-\log\left(1-\zeta_N^{-k}\right)\).

Let’s begin with the Gauss sum.

Definition 4. Fix a multiplicative character \(\chi\colon(\mathbb Z/N\mathbb Z)^\times\to\mathbb C^\times\) and an additive character \(\psi\colon(\mathbb Z/N\mathbb Z)\to\mathbb C^\times\). Then we define the Gauss sum \[\tau(\chi;\psi)=\sum_{a\in(\mathbb Z/N\mathbb Z)^\times}\chi(a)\psi(a).\] When \(\psi\) is the character \(\psi_1\colon a\mapsto\zeta_N^a\), we may abbreviate \(\tau(\chi;\psi_1)\) to \(\tau(\chi)\).

Remark 5. Let’s quickly classify additive characters to relate this to the usual definition. A homomorphism \(\psi\colon(\mathbb Z/N\mathbb Z)\to\mathbb C^\times\) is uniquely determined by \(\psi(1)\), which must be some \(N\)th root of unity \(\zeta_N^k\) for some \(k\). Thus, we may write each additive character \(\psi\) as \(\psi_k\colon a\mapsto\zeta_N^{ak}\) for some \(k\).

Remark 6. Many additive characters \(\psi_k\) have uninteresting Gauss sums. In particular, if \(d=\gcd(k,N)\) is bigger than \(1\), and \(\chi\) is primitive, then we claim that \(\tau(\chi;\psi_k)=0\). Well, \(\psi_k\) factors through \(\mathbb Z/(N/d)\mathbb Z\), but \(\chi\) does not, so the sum should vanish. More formally, choose some \(b\) which is \(b\equiv1\pmod{N/d}\) and \(\chi(b)\ne1\). Then \[\sum_{c\in(\mathbb Z/N\mathbb Z)^\times}\chi(c)\psi_k(c)=\sum_{c\in(\mathbb Z/N\mathbb Z)^\times}\chi(c)\psi_k(bc)=\chi(b)^{-1}\sum_{c\in(\mathbb Z/N\mathbb Z)^\times}\chi(c)\psi_k(c),\] so the sum vanishes.

Remark 7. On the other hand, even when \(\gcd(k,N)=1\), there is not too much information: one can simply re-index \[\tau(\chi;\psi_k)=\sum_{a\in(\mathbb Z/N\mathbb Z)^\times}\chi(a)\zeta_N^{ak}\] by \(a\mapsto k^{-1}a\), yielding, \(\tau(\chi;\psi_k)=\overline\chi(k)\tau(\chi)\). Note Remark 6 tells us that this equality also holds when \(\gcd(k,N)>1\), with the understanding that \(\overline\chi(k)=0\) in this case.

The previous two remarks allow us to write \[L(1,\chi)=-\frac{\tau(\chi)}N\sum_{k=0}^{N-1}\overline\chi(k)\log\left(1-\zeta_N^{-k}\right).\] We now try to evaluate the logarithm. Write \(\zeta_N^{-k}=e^{-2\pi ik/N}\), so \[1-\zeta_N^{-k}=2ie^{-\pi ik/N}\cdot\underbrace{\frac{e^{\pi ik/N}-e^{\pi ik/N}}{2i}}_{\sin\frac{\pi k}N}.\] Thus, for some choice of branch of logarithm, \[\log\left(1-\zeta_N^{-k}\right)=\log(2i)+\log\sin\frac{\pi k}N-\frac{\pi ik}N.\] When input into the sum over \(k\), we see that \(\sum_k\overline\chi(k)=0\) causes the sum with \(\log(2i)\) to vanish. (This also explains why the choice of branch is not so important.) We are left with \[L(1,\chi)=-\frac{\tau(\chi)}N\Bigg(-\frac{\pi i}N\sum_{k=0}^{N-1}\overline\chi(k)k+\sum_{k=0}^{N-1}\overline\chi(k)\log\sin\frac{\pi k}N\Bigg).\] Note that the function \(\log\sin\frac{\pi k}N\) is even (in \(k\)), so the right-hand sum is a sum of an odd function over all of \(\mathbb Z/N\mathbb Z\), so it will vanish. (We used that \(\chi\) is odd here!) It follows that \[L(1,\chi)=\frac{i\pi\tau(\chi)}{N^2}\sum_{k=0}^{N-1}\overline\chi(k)k.\] Simplifying the right-hand character sum is a little tricky, so we give it to a lemma.

Lemma 8. Let \(\chi\pmod N\) be a primitive odd Dirichlet character. Then \[\sum_{k=0}^{N-1}\chi(k)k=\frac N{\overline\chi(2)-2}\sum_{1\le k\lt N/2}\chi(k).\]

Proof. This is some technical and unenlightening manipulation. We follow Exercise 7.18 in Marcus’s Number Fields. Let \(T\) be the sum on the left-hand side, let \(T_<\) be the sum of \(\chi(k)k\) over \(k\lt N/2\), and we define \(T_>\) analogously so that \(T=T_<+T_>\). Then let \(S_<\) be the sum on the right-hand side, and define \(S_>\) to be the sum over \(k>N/2\) so that \(S_<+S_>=0\). We want to show that \((\overline\chi(2)-2)T=NS_<\).

Let’s begin with the case where \(N\) is even so that we want to show that \(-2T=NS_<\). Here, note that \(\chi(N/2+1)\) squares to \(1\), but it cannot be \(1\) by the primitivity of \(\chi\). It follows that \(\chi(N/2+1)=-1\), so \(\chi(N/2+k)=-\chi(k)\) for all \(k\). Thus, \[\begin{aligned}T_< &= \sum_{1\le k\lt N/2}\chi\left(\frac N2+k\right)(-k) \\\
&= -\sum_{1\le k\lt N/2}\chi\left(\frac N2+k\right)\left(\frac N2+k\right)+\frac N2S_> \\\
&= -T_>+NS_>.\end{aligned}\] This rearranges into the desired equality.

We now move on to the case where \(N\) is odd. For this, we combine two tricks.

• We use that \(\chi\) is odd to show that \(2T_<=T+NS_<\). Indeed, because \(\chi\) is odd, we see that \[\begin{aligned}T_< &= \sum_{1\le k\lt N/2}\chi(k)k \\\
  &=\sum_{1\le k\lt N/2}\chi(-k)(-k) \\\
  &=\sum_{1\le k\lt N/2}\chi(N-k)(N-k)+NS_<\\\
  &=T_>+NS_<.\end{aligned}\]
• We separate out the sum \(T\) into even and odd pieces to show that \(T=4\chi(2)T_<-N\chi(2)S_<\). Indeed, let \(T_0\) and \(T_1\) be the sums over the even and odd \(k\). On one hand, by dividing all even \(k\) by \(2\), we see that \(T_0=2\chi(2)T_<\). As such, it remains to show \(T_1=T_0-N\chi(2)S_<\). On the other hand, we can replace \(k\) with \(N-k\) in \(T_1\) to see \[\begin{aligned}T_1 &= \sum_{k\text{ odd}}\chi(N-k)(-k) \\\
  &= \sum_{k\text{ odd}}\chi(N-k)(N-k)-N\sum_{k\text{ odd}} \chi(N-k) \\\
  &= \sum_{k\text{ even}}\chi(k)k-N\sum_{k\text{ even}} \chi(k) \\\
  &= T_0-N\chi(2)S_<.\end{aligned}\]

To finish the proof, we solve for \(4T_<\) in both claims, which gives \[2T+2NS_<=\overline\chi(2)T+NS_<,\] which rearranges into the desired identity. \(\blacksquare\)

Thus, we see that \[L(1,\chi)=\frac{i\pi\tau(\chi)}{(\chi(2)-2)N}\sum_{1\le k\lt N/2}\overline\chi(k).\] The theorem now follows by taking absolute values with the following estimate.

Proposition 9. Fix a primitive Dirichlet character \(\chi\pmod N\). Then \(\left|\tau(\chi)\right|^2=N\).

Proof. This is a direct calculation. Observe \[\tau(\chi)\overline{\tau(\chi)}=\sum_{a,b\in(\mathbb Z/N\mathbb Z)^\times}\chi(a)\overline\chi(b)\zeta_N^{a-b}.\] Note \(\chi(a)\overline\chi(b)=\chi(a/b)\), so we may re-index the sum as \[\sum_{a,b\in(\mathbb Z/N\mathbb Z)^\times}\chi(ab)\overline\chi(b)\zeta_N^{ab-b}=\sum_{a,b\in(\mathbb Z/N\mathbb Z)^\times}\chi(a)\zeta_N^{(a-1)b}.\] We would like to isolate the sum on \(b\) so that we get a sum of roots of unity, but this does not exactly work yet because \(b\) only varies over \((\mathbb Z/N\mathbb Z)^\times\). Well, for each \(b\notin(\mathbb Z/N\mathbb Z)^\times\), we see that the term \[\sum_{a\in(\mathbb Z/N\mathbb Z)^\times}\chi(a)\zeta_N^{(a-1)b}=\zeta_N^{-b}\tau(\chi;\psi_b)\] vanishes by Remark 6. (Here is where we use that \(\chi\) is primitive!) Thus, our sum is actually \[\sum_{a\in(\mathbb Z/N\mathbb Z)^\times}\Bigg(\chi(a)\sum_{b\in\mathbb Z/N\mathbb Z}\zeta_N^{(a-1)b}\Bigg).\] Now, if \(a\ne1\), then the inner sum vanishes. Thus, we are left with \(\chi(1)\cdot N=N\). \(\blacksquare\)