Non-vanishing of L(1, χ)

To prove Dirichlet’s theorem for primes in arithmetic progressions consider the weighted sums over primes in an AP and sums over primes twisted by characters.

$\displaystyle S_{l}(x)=\sum_{p \equiv l ~\bmod q} \frac{\log p}{p}$

and

$\displaystyle A=\sum_{n \leq x} \frac{\chi(n) \Lambda(n)}{n} .$

We have the following result which relates the L-values to primes.

$\displaystyle \mbox { If } L(1, \chi) \neq 0, \mbox { then } A=O(1)$

$\displaystyle \mbox { If } L(1, \chi)=0, \mbox { then } A=-\log x+O(1).$

Proof:

Use

$\displaystyle L(1, \chi) =\sum_{m \leq x} \frac{\chi(m)} {m}+ O\left(\frac{1}{x}\right), \quad \log n = \sum_{l \mid n} \Lambda(l)$

and evaluate $\sum_{n \leq x} \frac{\chi(n) \log n}{n}.$

$\displaystyle \sum_{n \leq x} \frac{\chi(n) \log n}{n}=\sum_{n \leq x}\left(\frac{\chi(n)}{ n} \sum_{l \mid n} \Lambda(l)\right)$

$\displaystyle =\sum_{l \leq x}\left(\frac{\chi(l) \Lambda(l)}{ l} \sum_{m \leq \frac{x}{l}} \frac{\chi(m)} {m}\right) = \sum_{l \leq x} \frac{\chi(l) \Lambda(l)}{ l}\left(L(1, \chi)+O\left(\frac{l}{x}\right)\right)$

$\displaystyle =L(1, \chi) \sum_{l \leq x} \frac{\chi(l) \Lambda(l)}{ l}+O(1)$

Note that $\sum_{n \leq x} \frac{\chi(n) \log n}{n},~$ ${L(1, \chi)}$ are bounded quantities, hence $A=\sum_{n \leq x} \frac{\chi(n) \Lambda(n)}{ n}$ is bounded provided ${ L(1, \chi) \neq 0.}$

In the above calculation, if we replace $\sum_{l \mid n} \Lambda(l)$ with $\sum_{l \mid n} \mu(l) \log \frac{x}{l}$ , we get

$\displaystyle \log x+\sum_{n \leq x} \frac{\chi(n) \Lambda(n)}{ n} =\sum_{n \leq x} \frac{\chi(n)}{ n} \sum_{l \mid n} \mu(l) \log \frac{x}{l}$

$\displaystyle =\sum_{l \leq x} \frac{\mu(l) \chi(l)}{d} \log \frac{x}{l} \sum_{l \leq \frac{x}{l}} \frac{\chi(l)}{l}$

$\displaystyle =L(1, \chi) \sum_{l \leq x} \frac{\mu(l) \chi(l)}{ l} \log \frac{x}{l}+O(1)$

Hence, if ${L(1, \chi)=0}$ , we have

$\displaystyle \sum_{n \leq x} \frac{\chi(n) \Lambda(n)}{ n} =-\log x +O(1)$

Summing over the characters (orthogonality), primes in an arithmetic progression can be detected using the the above sums over primes twisted with characters.

$\displaystyle \frac{1}{\varphi(q)}\sum_{\chi \bmod q} \chi(\bar a)\sum_{n \leq x} \frac{\chi(n) \Lambda(n)} {l} =\sum_{n \equiv a \bmod q } \frac{\Lambda(n)} {n}$

Thus using the previous computations, we see that problem of showing

$\displaystyle \sum_{\substack{n \equiv a \bmod q \\ n \le x}} \frac{\Lambda(n)} {n} \rightarrow \infty$

is reduced to non-vanishing of ${L(1, \chi)}$ . In fact, there can be at most one character which vanishes because of the non-negativity of the expression. The character has to be real (hence quadratic) because complex characters occur in pairs and vanishing of one forces the conjugate to also vanish.

Note that I choose to present the relation between ${L(1, \chi)}$ , its non-vanishing and primes in APs in terms of weighted(twisted) sum over primes- But you can also see all the arguments written in terms of the L-functions (it’s easier to see the formal and qualitative relationship using the L-functions more directly, you don’t have to go though these elementary manipulations)

Let’s look at some ways to prove the non-vanishing of $L(1, \chi)$ . All the known proofs use the relationship between the $L(1, \chi)$ and arithmetic of number fields.

Dirichlet’s evaluation of the L-values in 1837 paper

Bringing in additive characters , we can see that

${L(s, \chi)=\displaystyle \frac{1}{\tau(\chi)} \sum_{a=1}^{q-1} \chi(a) \sum_{n=1}^{\infty} \frac{e^{2 \pi i a n / q}}{n^{s}},}$ where $\tau(\chi) = \sum_{a=1}^{q-1} \chi(a) e^{2 \pi i a / q}$ and therefore

$\displaystyle \displaystyle L(1, \chi)=\sum_{n \geq 1} \frac{\chi(n)}{n}=-\frac{2 \tau(\chi)}{q}\left\{\begin{array}{ll} \displaystyle 2 \sum_{1 \leq m \leq q / 2} \bar{\chi}(m) \log \left|\sin \frac{\pi m}{q}\right| & \mbox { when } \chi(-1)=+1, \\ \displaystyle i \pi \sum_{1 \leq m \leq q / 2} \bar{\chi}(m)\left(1-\frac{2 m}{q}\right) & \mbox { when } \chi(-1)=-1 . \end{array}\right.$

Dirichlet’s proof using Class number Formula for Quadratic fields.

${D}$ is ${-p}$ if ${p \equiv 3 \bmod 4}$ and ${-4 p}$ if ${p \equiv 1 \bmod 4}$ .

Quadratic reciprocity allows us to ((in fact equivalent to) write ${\zeta_K(s)}$ for a quadratic field ${K=\mathbb{Q}(\sqrt{-p})}$ as ${\zeta(s)L(s, \left( \frac{D}{\cdot}\right))}$ Now computing the residue at ${s=1}$ for both the expressions we get

$\text{Res}_{s=1}\displaystyle \zeta_K(s) = \text{Res}_{s=1}{\zeta(s)L(s, \chi)}$

$\displaystyle L(1, \chi)=\frac{2 \pi h(D)}{w \sqrt{|D|}}$

The residue computation for ${\zeta_K(s)}$ is equivalent to point ideals of a bounded norms. Splitting into ideal classes, it reduces to counting lattice points- there are $\displaystyle \sim \frac{2 \pi X}{w \sqrt{|D|}}$ lattice points of norm less than $X$ in each of the $h(D)$ ideal classes. Note the discriminant is the square of the covolume, hence the factor of $\sqrt {|D|}$ .

Thus it’s easy to see the non-vanishing, in fact $h(D) \ge 1$ gives $\displaystyle L(1, \chi)\ge \frac{2 \pi }{w \sqrt{|D|}}$

Class number Formula for Cyclotomic Fields.

Similar to the quadratic cases, computing the residue at ${s=1}$ using point counting in various lattices (by embedding the field ${F}$ inside a product ${\mathbb R^{r_1} \times \mathbb C^{r_2}}$ ), we get for any arbitrary number field ${F}$ , the class number formula

$\displaystyle \lim _{s \rightarrow 1}(s-1) \zeta_{F}(s)=\frac{2^{r_{1}} \cdot(2 \pi)^{r_{2}} \cdot \text{Reg}_{K} \cdot h_{K}}{w_{K} \cdot \sqrt{\left|D_{K}\right|}}$

Specializing to

$\displaystyle \zeta_{F}(s)=\prod_{\chi} L(\chi, s)$

where ${F=\mathbb Q(\zeta_q)}$ , we get that

$\displaystyle \prod_{\chi \neq \chi_0} L(\chi, 1) =\frac{2^{r_{1}} \cdot(2 \pi)^{r_{2}} \cdot \text{Reg}_{K} \cdot h_{K}}{w_{K} \cdot \sqrt{\left|D_{K}\right|}}$

and hence ${L(1, \chi)}$ is non-zero for all the non-principal characters.

Landau type proofs: All of them use Poussin’s method to create auxillary functions with positive coefficients.

$\displaystyle F(s) =\zeta(s)^{2} L(s, \chi) L(s, \bar{\chi})$

has non-negative Dirichlet series coefficients which implies ${F(s) \xrightarrow{s\rightarrow 1} \infty}$ .
But ${L(1, \chi)=0}$ shows that ${F(1)}$ has to be bounded.

Ingham/Bateman method: They consider

$\displaystyle G(s)=\zeta(s)^{6} L(s, \chi)^{4} L(s, \bar{\chi})^{4} L\left(s, \chi^{2}\right) L\left(s, \bar{\chi}^{2}\right)$ and look at ${\log G(s)}$ which has non-negative coefficients.

More generally

$\displaystyle G_{t}(s)=\zeta(s)^{6} L(s+i t, \chi)^{4} L(s-i t, \bar{\chi})^{4} L\left(s+2 i t, \chi^{2}\right) L\left(s-2 i t, \bar{\chi}^{2}\right)$

allows to show the non-vanishing for on the 1-line. (This also helps to get quantitative estimates, and zero-free regions)

In general for any two real characters, considering

$\displaystyle \zeta(s) L(s, \chi) L\left(s, \chi^{\prime}\right) L\left(s, \chi \chi^{\prime}\right)$

allows us to establish repulsion of zeros and obtains (ineffective) bounds on Siegel zeros.

Eisenstein proof: L-functions appeared in the constant terms of Eisenstein series. Analytic properties of the Eisenstein series can be related to non-vanishing of L-functions. For instance $\zeta(s)$ appears in the denominator for $SL_2(\mathbb Z)$ .

http://web.math.princeton.edu/sarnak/ShalikaBday2002.pdf

Another proof with quadratic fields(forms) (more explicit)

Consider the multiplicative function

$\displaystyle a(n)=\sum_{d \mid n} \chi(d)$

Note that ${a(n) \ge 0}$ and ${a(n^2) \ge 1.}$ ${a(n)}$ counts the number of ideals in a quadratic field (alternatively number of representations of ${n}$ by binary quadratic forms of discriminant D)

$\displaystyle A(x)=\sum_{n \leq x} \frac{a(n)}{ \sqrt n} > \sum_{m^{2} \leq \sqrt{x}} \frac{a(m^2)}{ m} \geq \sum_{m \leq \sqrt{x}} \frac{1}{m} > \frac{\log x}{2}$

$\displaystyle =\sum_{lm \leq x}\frac{ \chi(l)}{\sqrt {lm}}$

$\displaystyle =\sum_{l \leq \sqrt{x}} \chi(l) l^{-\frac{1}{2}} \sum_{m \leq \frac{x}{l}} m^{-\frac{1}{2}}+\sum_{m \leq \sqrt{x}} m^{-\frac{1}{2}} \sum_{\sqrt{x}<l<\frac{x}{m}} \chi(l) l^{-\frac{1}{2}}$

$\displaystyle =2 L(1, \chi) x^{\frac{1}{2}}+O(1)$

Therefore ${L(1, \chi)}$ cannot be zero , in fact we get that ${L(1, \chi) \ge \frac{\log x}{\sqrt{x}}.}$

Non-vanishing of L(1, χ)

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