Trivial representation of absolute Galois group, whose L-function is the Riemann zeta function.
Roots of polynomials \(f(x)\in \mathbb{Q}[x]\), we get permutation representations of \(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\), associates the Dedekind zeta function \(\zeta_K(s)\) (Number fields, Frobenius inertia), and Artin L-functions (Rep theory).
Elliptic curves, Tate module, \(L(E,s)\). (Elliptic curves, local fields)
\[\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_p \frac{1}{1-p^{-s}}\]
It encodes distribution of primes (analytic number theory), for example \(\zeta(1+)=\infty\) tells us that there are infinitely many primes.
Viewed as a function of complex variable \(s = \sigma + it\)
\[ \left|n^{-s}\right| = n^{-\sigma} \]
which converges absolutely for \(\sigma>1\).
Theorem (Riemann) \(\zeta(s)\) has meromorphic continuation to \(\mathbb{C}\) with a simple pole at \(s=1\) with residue \(1\).
The completed zeta function is defined by
\[\zeta^*(s) = \pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)\]
with the functional equation
\[\zeta^*(s) = \zeta^*(1-s).\]
The proof can be done by Poisson summation formula. For \(f:\mathbb{R}\to \mathbb{C}\), we have
\[\sum_{n\in \mathbb{Z}} f(n) = \sum_{n\in \mathbb{Z}} \hat{f}(n)\]
where
\[\hat{f}(m) = \int_{-\infty}^\infty e^{-2\pi i mx}f(x)dx\]
is the Fourier transform of \(f\).
We know that for \(f(t) = e^{-\pi t^2}\), \(\hat{f}(t) = f(t)\), so \(\widehat{f\circ c}(t) = \frac{1}{c}\hat{f}\left(\frac{t}{c}\right)\), and we get
\[\sum_n e^{-\pi n^2x} = \frac{1}{\sqrt{x}}\sum_n e^{-\frac{\pi n^2}{x}}\]
An L-function is a Dirichlet series of the form
\[L(s) = \sum_{n=1}^\infty \frac{a_n}{n^s}, \quad a_n = O(n^r)\]
which converges absolutely for \(\sigma>r+1\).
It (might) have an Euler product and degree \(d\)
\[L(s) = \prod_p \frac{1}{F_p(p^{-s})}\]
where \(F_p(T)\in \mathbb{C}[T]\) is some polynomial of degree \(d\).
For \(\zeta(s)\), \(F_p(T) = 1-T\), it is a degree \(1\) L-function.
All L-functions we are considering must have an Euler product of the above form. It is conjectured that
They have meromorphic continuation to \(\mathbb{C}\) with finitely many poles (usually none).
They satisfy certain functional equation. There is a weight \(k\), sign \(w\), conductor \(N\), \(\Gamma\)-factor \(\gamma(s)=\Gamma\left(\frac{s+\lambda_1}{2}\right)\cdots \Gamma\left(\frac{s+\lambda_d}{2}\right)\), such that
\[ L^*(s) = \left(\frac{N}{\pi^d}\right)^{\frac{s}{2}} \gamma(s) L(s)\]
satisfies \(L^*(s) = w\overline{L}^*(k-s)\), where \(\overline{L}(s) = \sum \overline{a_n}n^{-s}\).
They have a Riemann hypothesis: all non-trivial zeros lie on the line \(\Re(s)=\frac{k}{2}\).
Special value conjectures: their special values \(L(n)\) for \(n\in \mathbb{Z}\) have some special meaning.
If \(L(s)\) satisfies A+B, say with no poles.
\[L^*(s) = \int_1^\infty \Theta(\sqrt{N}x)(x^{\frac{s}{2}} + x^{\frac{k-s}{2}}) \frac{dx}{x}\]
where \(\Theta_L(x) = \sum_{n=1} a_n \phi_\gamma(n,x)\), where \(\phi_\gamma\) depends only on the gamma factors and decays exponentially with \(n\), e.g. \(e^{-\pi n^2 x}\) for \(\gamma = \Gamma(s/2)\).
\[ \Theta\left(\frac{1}{Nx}\right) = w\Theta(x)\]
This gives a way to compute \(L(s)\) numerically, need roughly \(\sqrt{N}\) terms.
There are modular forms technically new-forms of weight \(k\), level \(N\), w-eigenforms for the Artin Lehmer involution
\[f:\{\text{upper half plane}\}\to \mathbb{C}\]
such that \(\Theta(x) = f(ix)\) satisfy functional equation \(\Rightarrow\) their \(L\)-functions do satisfy A+B.
We will see two types of L-functions, there is a distinction,
sometimes we have a L-function with direct interpretation of \(a_n\), e.g. \(\zeta(s)\), \(L(\chi,s)\), \(\zeta_K(s)\) where \(a_n=\) the number of ideals of norm \(n\) in \(\mathcal{O}_K\).
for these \(L\) functions we generally know how to prove A+B.
sometimes the L-functions are given by Euler product
\(L(\rho,s)\) Artin
\(L(E,s)\) elliptic curves
For these L-functions we are currently unable to not prove A+B directly, only by reducing to C.
\(K\) a number field, \([K:\mathbb{Q}]=d\), \(\mathcal{O}=\mathcal{O}_K\) the ring of integers. We have \(K\cong \mathbb{Q}^d\) as vector space and \(\mathcal{O}\cong \mathbb{Z}^d\) as group.
For \(I\subset \mathcal{O}\), we have \(N(I) = \# \mathcal{O}/I = (\mathcal{O}:I)\) the norm of \(I\).
\(N(IJ) = NI \cdot NJ\)
\(N(n\mathcal{O}) = n^d\)
\(I\) can be written as a unique product of prime ideals
\[ I = \prod_{i=1}^r \mathfrak{P}_i^{e_i}\]
where \(\mathfrak{P}_i\) are prime ideals, \(e_i\ge 1\).
\(\mathcal{O}/\mathfrak{P}_i\) is a finite domain, so a field, and \(\mathfrak{P}_i\) is maximal.
In particular for \(I=(p)=p\mathcal{O}\) we can factor
\[p\mathcal{O}= \prod_{i=1}^g \mathfrak{P}_i^{e_i}\]
these \(\mathfrak{P}_i\) are called primes above \(p\), \(e_i(\mathfrak{P}_i,p)\) the ramification index, \(f_i(\mathfrak{P}_i,p) = [\mathcal{O}/\mathfrak{P}_i:\mathbb{F}_p]\) the inertia degree.
Taking norm we have
\[p^d = N(p\mathcal{O}) = \prod_{i=1}^g N(\mathfrak{P}_i)^{e_i} = \prod_{i=1}^r p^{e_i f_i}\]
i.e.
\[ d = \sum_{i=1}^r e_i f_i\]
\(p\nmid \Delta_K \Leftrightarrow \mathfrak{P}_i\) are distinct, i.e. \(e_i=1\).
If \(K/\mathbb{Q}\) is Galois, then \(e_i\) and \(f_i\) are all equal, denote \(e=e_i, f=f_i\) we can write \(d=efg\).
Theorem (Kummer-Dedekind) Let \(K=\mathbb{Q}[x]/(g(x))\) where \(g(x)\in \mathbb{Z}[x]\) is monic. We can compute the decomposition of \(p\mathcal{O}\) by reducing \(g(x)\) modulo \(p\).
\(\Delta_K | \Delta_g\)
For all \(p\nmid \Delta_g\), we can write \(p\mathcal{O}= \prod \mathfrak{P}_i\) as product of distinct prime ideals,
\[g(x) = g_1(x)\cdots g_r(x) \mod p\]
where \(g_i(x)\in \mathbb{F}_p[x]\) are irreducible and \(f_i = \deg g_i\).
From here we can define the Dedekind \(\zeta\)-function of \(K\),
\[a_n = \# \{\text{ideals of norm } n\}\] \[\begin{align*} \zeta_K(s) &= \sum_{n=1}^\infty \frac{a_n}{n^s} \\ &= \sum_{I\subset \mathcal{O}} \frac{1}{N(I)^s} \\ &= \sum \frac{1}{N(\mathfrak{P}_1)^{s e_1}}\cdots \frac{1}{N(\mathfrak{P}_r)^{s e_r}} \\ &= \prod_{\mathfrak{P}} \frac{1}{1-N(\mathfrak{P})^{-s}} \\ &= \prod_p \frac{1}{F_p(p^{-s})}\quad \text{exercise!} \end{align*}\]
where \(F_p(T)\in \mathbb{Z}[T]\) is degree \(d\) for \(p\nmid \Delta_K\) and degree \(<d\) for \(p|\Delta_K\).
(Solution to the exercise:
\[F_p(T) = \prod_{\mathfrak{P}|p} \left(1-T^{f_i}\right)\]
so when \(p\nmid \Delta_K\) we have \(e_i=1\), \(\deg F_p(T) = \sum f_i = d\).
When \(p|\Delta_K\), there will be some \(e_i>1\), \(\deg F_p(T) < \sum e_i f_i = d\).)
when we take \(\zeta_\mathbb{Q}(s)\), we get \(\zeta(s)\).
\(K=\mathbb{Q}(i) = \mathbb{Q}[x]/(x^2+1)\) we have \(\mathcal{O}=\mathbb{Z}[i]\) which is a PID, and \(N(a+bi) = a^2+b^2\).
\(\mathcal{O}^\times = \{\pm 1, \pm i\}\), \(\mathfrak{P}=(1+i)\) is prime, \(N(\mathfrak{P}) = 2\).
\[\zeta_K(s) = \zeta(s)\zeta(s-1)\]
where \(\zeta(s-1)\) comes from the prime \(2\).
Kummer-Dedekind for \(g(x)=x^2+1\) tells us that all \(p\neq 2\) are unramified, and
\(p\equiv 1\mod 4\), \(x^2+1\) splits, so \(p=\mathfrak{P}_1\mathfrak{P}_2\) splits into two primes
\(p\equiv 3\mod 4\), \(x^2+1\) is irreducible, so \(p=\mathfrak{P}\) is inert, \(p=N\mathfrak{P}=a^2+b^2\).
Draw a picture
Now we can compute \(\zeta_K(s)\). Since every ideal \(I=(m+ni)\) is generated unique upto units (where there are four units),
\[\begin{align*} \zeta_{\mathbb{Q}(i)}(s) &= \sum_{I\subset \mathbb{Z}[i]} \frac{1}{N(I)^s} \\ &= \frac{1}{4}\sum_{(m,n)\neq 0} \frac{1}{(m^2+n^2)^s} \\ \end{align*}\]
\(\frac{2^s}{\pi^s}\Gamma(s)\zeta_K(s)\) is the Mellin transform of \(\frac{\Theta_K(s)-1}{4}\), \[\Theta_K(x) = \sum_{(m,n)} e^{-\pi (m^2+n^2)x} = \sum_m e^{-\pi m^2 x} \sum_n e^{-\pi n^2 x} = \Theta(x)^2 = \frac{1}{x}\Theta_K\left(\frac{1}{x}\right).\]
In general, Poisson summation applies to higher dimensions, for a vector space \(V=\mathbb{R}^d\) and \(f:V\to \mathbb{C}\), define
\[\widehat{f}:V^*\to \mathbb{C}\quad \widehat{f}(\xi) = \int_V e^{-2\pi i \langle x,\xi\rangle}f(x)dx\]
where \(V^*=\mathrm{Hom}(V,\mathbb{R})\) is the dual space.
Then for any lattice \(\Lambda\subset V\) and its dual \(\Lambda^*\subset V^*\), we have
\[\sum_{\lambda\in \Lambda} f(\lambda) = \frac{1}{\mu(V/\Lambda)}\sum_{\xi\in \Lambda^*} \widehat{f}(\xi)\]
Now we can convert \(\sum_I \frac{1}{(NI)^s}\) to a sum over numbers \(\sum_{\alpha\in \mathcal{O}} \frac{1}{(N\alpha)^s}\) which is a sum on lattice, this process involves class number, units and roots of unity, then we will be ready to apply Poisson summation.
Theorem \(K\) be a number field of degree \(d=r_1+2r_2\) where \(r_1\) is the number of real embeddings and \(r_2\) is the number of complex embeddings. Then \(\zeta_K(s)\) is meromorphic with simple pole at \(s=1\) with residue \(\frac{2^{r_1}(2\pi)^{r_2} h_K R_K}{w_K \sqrt{|\Delta_K|}}\) where
\(h_K\) is the class number
\(R_K\) is the regulator
\(w_K\) is the number of roots of unity in \(K\)
this is an example of special value conjecture.
\[\zeta_K^*(s) = \left(\frac{|\Delta_K|}{\pi^d}\right)^{\frac{s}{2}} \Gamma\left(\frac{s}{2}\right)^{r_1+r_2} \Gamma\left(\frac{s+1}{2}\right)^{r_1} \zeta_K(s) \]
satisfy \(\zeta_K^*(s) = \zeta_K^*(1-s)\).
Exercise, for \(K/\mathbb{Q}\) Galois of degree \(d\), then there are infinitely many primes of \(p\in \mathbb{Z}\) that splits completely in \(K\) (those primes that \(e=f=1\)). Solution MO.218759
Exercise, for \(K=\mathbb{Q}(i)\),
\[\begin{align*} \zeta_{\mathbb{Q}(i)}(s) &= \prod_{\mathfrak{P}\subset\mathcal{O}} \frac{1}{1-N(\mathfrak{P})^{-s}} \\ &= \prod_p \frac{1} { \prod_{\mathfrak{P}|p} (1-N(\mathfrak{P})^{-s}) } \\ &= \prod_p \frac{1}{F_p(p^{-s})} \\ \end{align*}\]
where
\[F_p(T) = \begin{cases} 1-T & p=2 \\ (1-T)^2 & p\equiv 1\mod 4 \\ 1-T^2 & p\equiv 3\mod 4 \end{cases} \]
Thus
\[\begin{align*} \zeta_{\mathbb{Q}(i)}(s) &= \prod_p \frac{1}{F_p(p^{-s})} \\ &= \frac{1}{1-2^{-s}} \prod_{p\equiv 1\mod 4} \frac{1}{(1-p^{-s})^2} \prod_{p\equiv 3\mod 4} \frac{1}{(1-p^{-s})(1+p^{-s})} \\ &= \zeta(s) \prod_{p\equiv 1\mod 4} \frac{1}{1-p^{-s}} \prod_{p\equiv 3\mod 4} \frac{1}{1-(-p^{-s})} \\ &= \zeta(s)L(\chi_4,s) \end{align*}\]
This \(L(\chi_4,s)\) has no poles.
You can see that the zeros of \(\zeta_K(s)\) is the sum of the zeros of \(\zeta\) and that of \(L(\chi_4,s)\), how amazing! o.o
\(\zeta_{\mathbb{Q}(\zeta_n)}(s) = \prod \{\text{Dirichlet L-functions}\}\)
\(K\) any number field, \(\zeta_K(s) = \prod L(\rho,s)\) as Artin L-functions.