In the previous post, we defined local zeta functions and meromorphically extended them to the complex plane. We now work towards defining the global zeta function. Although local zeta functions are supposed to be local factors of the zeta function, we are going to allow multiplying local zeta functions \( \zeta _ v(f _ v, c _ v) \) under some restrictive conditions. But let us first develop some general theory.
4. Fourier analysis on the adeles#
We now start investigating the global analogue of the local theory we have developed so far. For \( K \) a number field, we have different places \( v \) and the corresponding completions \( K _ v \), which are local fields.
Definition 1. We define the adeles \( \mathbb{A} _ K \) and the ideles \( \mathbb{A} _ K^\times \) as \[ \displaystyle \begin{aligned} \mathbb{A} _ K & \displaystyle= \{ (x _ v) _ v : x _ v \in K _ v, x _ v \in \mathcal{O} _ v \text{ for almost all } v \}, \\ \mathbb{A} _ K^\times & \displaystyle= \{ (x _ v) _ v : x _ v \in K _ v^\times, x _ v \in \mathcal{O} _ v^\times \text{ for almost all } v \}, \end{aligned} \] with the topologies generated by, for \( S \) finite, \( \prod _ {v \in S}^{} U _ v \times \prod _ {v \notin S}^{} \mathcal{O} _ v \) in \( \mathbb{A} _ K \) and \( \prod _ {v \in S}^{} U _ v \times \prod _ {v \notin S}^{} \mathcal{O} _ v^\times \) in \( \mathbb{A} _ K^\times \). (Note that the topology of \( \mathbb{A} _ K^\times \) is not the subspace topology inherited from \( \mathbb{A} _ K \).)
For all but finitely many places \( v \), the local field \( K _ v \) is going to be unramified over \( \mathbb{Q} _ p \) or \( \mathbb{F} _ p \). This implies that \( \mathcal{D} _ v = \mathcal{O} _ v \) for almost all \( v \). Then on \( \mathbb{A} _ K \) we may define the measure \[ \displaystyle \mu\biggl( \prod _ {v \in S}^{} U _ v \times \prod _ {v \notin S}^{} \mathcal{O} _ v \biggr) = \prod _ {v \in S}^{} \mu _ v(U _ v) \times \prod _ {v \notin S}^{} \mu _ v(\mathcal{O} _ v) \] because for all but finitely \( v \), we have \( \mu _ v(\mathcal{O} _ v) = (N\mathcal{D} _ v)^{-1/2} = 1 \). We may further multiply all the local characters to get a character \[ \displaystyle \chi : \mathbb{A} _ K \rightarrow S^1; \quad \chi(x) = \prod _ {v}^{} \chi _ v(x). \] This indeed is continuous because on the finite \( v \), the kernel contains \( \prod _ {v \text{ fin}}^{} \mathcal{D} _ v \) and it is open.
Theorem 2. If \( K \) is a number field, then for \( G = \mathbb{A} _ K \), the map \[ \displaystyle \Phi : \mathbb{A} _ K \rightarrow \widehat{\mathbb{A}} _ K; \quad x \mapsto (y \mapsto \chi(xy)) \] is an isomorphism of topological groups.
Proof. First we show that it is an isomorphism of groups. For \( c \) a character, the inverse image of \( V = \{ e^{i \theta} : \lvert \theta \rvert < 0.1 \} \) should contain some open subgroup \( U = \prod _ {v \in S}^{} U _ v \times \prod _ {v \notin S}^{} \mathcal{O} _ v^\times \), but because there is no subgroup of \( S^1 \) contained in \( V \), the image of \( U \) should be just \( \{1\} \). Then we see that \( c \) should just come from multiplying characters \( c _ v \in \widehat{K} _ v \cong K _ v \), with \( c _ v \) vanishing on \( \mathcal{O} _ v \) for almost all \( v \).
Now we show that it is a homeomorphism. Continuity is immediate from continuity of the multiplication map. In the other direction, the image of \( U = \prod _ {v \in S}^{} B _ {r _ v}(0) \times \prod _ {v \notin S}^{} \mathcal{O} _ v \) contains the open set of characters consisting of those sending the compact set \( \prod _ {v \in S}^{} \mathcal{D} _ v \overline{B} _ {0.01 / r _ v}(0) \times \prod _ {v \notin S}^{} \mathcal{D} _ v) \) into \( V = \{ e^{i \theta} : \lvert \theta \rvert < 0.1 \} \). ▨
Note that the same argument shows that, if \( G _ v \) are locally compact abelian groups with \( H _ v \subseteq G _ v \) open and compact in \( G _ v \), then the dual of the restricted product \( \prod _ {}^{\prime} G _ v \) is topologically isomorphic to the restricted product \( \prod _ {}^{\prime} \widehat{G} _ v \) along \( H _ v^\ast \), where \( H _ v^\ast = \ker(\widehat{G} _ v \rightarrow \widehat{H} _ v) \).
The next step is to check if our measure is correctly normalized with respect to the identification \( \mathbb{A} _ K \cong \widehat{\mathbb{A}} _ K \). Consider \( f = \prod _ {v}^{} f _ v \) for \( f _ v \) Schwartz–Bruhat functions \( K _ v \rightarrow \mathbb{C} \), with \( f _ v = \hat{f} _ v = \mathbf{1} _ {\mathcal{O} _ v} \) for \( v \notin S \) (so that \( f \) itself is a Schwartz–Bruhat function on \( \mathbb{A} _ K \)). Its Fourier transform is \[ \displaystyle \begin{aligned} \hat{f}(\xi) & \displaystyle= \int _ {\prod _ {v \in S}^{} K _ v}^{} \int _ {\prod _ {v \notin S}^{} \mathcal{O} _ v}^{} f(x) \overline{\chi(x\xi)} d\mu(x) \\ & \displaystyle= \prod _ {v \in S}^{} \int _ {K _ v}^{} f(x _ v) \overline{\chi(x _ v \xi _ v)} d\mu(x _ v) \prod _ {v \notin S}^{} \int _ {\mathcal{O} _ v}^{} f(x _ v) \overline{\chi(x _ v \xi _ v)} d\mu(x _ v) = \prod _ {v}^{} \hat{f} _ v(\xi _ v). \end{aligned} \] It follows that \( \lVert f \rVert _ {L^2} = \prod _ {v}^{} \lVert f _ v \rVert _ {L^2} = \prod _ {v}^{} \lVert \hat{f} _ v \rVert _ {L^2} = \lVert \hat{f} \rVert _ {L^2} \) in this case, so we see that the normalization is correct.
Corollary 3. For \( f \in \mathcal{S}(\mathbb{A} _ K) \), we have the Fourier inversion formula \[ \displaystyle \hat{f}(\xi) = \int _ {\mathbb{A} _ K}^{} f(x) \overline{\chi(x \xi)} d\mu(x), \quad f(x) = \int _ {\mathbb{A} _ K}^{} \hat{f}(\xi) \chi(x \xi) d\mu(\xi). \] Moreover, \( \lVert \hat{f} \rVert _ {L^2} = \lVert f \rVert _ {L^2} \).
For an \( x \in \mathbb{A} _ K^\times \) we have a multiplication map \( \times x^{-1} : \mathbb{A} _ K \rightarrow \mathbb{A} _ K \) and the push-forward measure \( S \mapsto \mu(x S) \) is a Haar measure. This means that it is a constant times \( \mu(S) \), and comparing at \( \prod _ {v} \overline{B} _ 1(0) \), we easily see that this constant should be \( \lvert x \rvert = \prod _ {v}^{} \lvert x _ v \rvert _ v \), which is sometimes called the content of \( x \). That is, \[ \displaystyle \mu(x S) = \lvert x \rvert \mu(S) \] for any measurable set \( S \).
The field \( K \) sits inside \( \mathbb{A} _ K \) as a "full rank lattice", with the diagonal embedding \[ \displaystyle K \hookrightarrow \mathbb{A} _ K; \quad x \mapsto (x _ v = x). \] That is, \( K \) is a discrete subgroup of \( \mathbb{A} _ K \), and moreover the quotient \( \mathbb{A} _ K / K \) is compact and thus has finite volume. This suggest that there should be some analogue of the Poisson summation formula. (If you have seen the derivation of the functional equation for the Riemann zeta function, you might recall that this was used at some point. So the Poisson summation formula is really the main analytic input we will need.)
Lemma 4. \( \mu(\mathbb{A} _ K / K) = 1 \), where this means the measure of a fundamental domain.
Proof. Consider \( \mathcal{O} _ K \hookrightarrow \prod _ {v \text{ inf}}^{} K _ v \), and take \( D \) a fundamental domain of this lattice. Then \[ \displaystyle \prod _ {v \text{ fin}}^{} \mathcal{O} _ v \times D \subseteq \mathbb{A} _ K \] is a fundamental domain of \( K \hookrightarrow \mathbb{A} _ K \). Moreover, its measure on the finite component is \( \prod _ {v}^{} (N \mathcal{D} _ v)^{-1/2} \) and the measure on the infinite component is \( \sqrt{\lvert D \rvert} = \prod _ {v}^{} (N\mathcal{D} _ v)^{1/2} \). So the total measure is \( 1 \). ▨
To prove Poisson summation, we should look at Pontryagin duality on \( \mathbb{A} _ K / K \). We can put a measure on \( \mathbb{A} _ K / K \) using the measure on the fundamental domain \( D \). Also, we can put a discrete measure on \( K \) as \( \mu(S) = \lvert S \rvert \).
Lemma 5. For \( x \in \mathbb{A} _ K \), the character \( y \mapsto \chi(xy) \) on \( K \) is trivial if and only if \( x \in K \).
This is saying that the dual lattice of \( K \) is itself, if we identify \( \mathbb{A} _ K \cong \widehat{\mathbb{A}} _ K \).
Proof. If \( x \in K \), then \( xy \in K \) for all \( y \in K \). So we check that \( \chi(x) = 0 \) for all \( x \in K \). To see this, we note that the contributions from infinite places are \( -\mathrm{Tr} _ {K/\mathbb{Q}} (x) \in \mathbb{Q}/\mathbb{Z} \). The contributions from finite places are also rational, so \( \chi(x) \in \mathbb{Q}/\mathbb{Z} \) for all \( x \in K \). Now we check that \( \chi(x) \) is integral at all primes \( p \). To do this, we note that \( \chi _ v(x) \) is defined as the rational number such that \( \mathrm{Tr}(x) - \chi _ v(x) \in \mathbb{Z} _ p \). Then because \( K \otimes _ \mathbb{Q} \mathbb{Q} _ p \cong \prod _ {v \mid p}^{} K _ v \), the trace map can be described in two equivalently ways: \[ \displaystyle \mathrm{Tr} _ {K/\mathbb{Q}} \otimes \mathrm{id} _ {\mathbb{Q} _ p} : K \otimes _ \mathbb{Q} \mathbb{Q} _ p \rightarrow \mathbb{Q} _ p, \quad \prod _ {v \mid p}^{} K _ v \xrightarrow{\mathrm{Tr}} \prod _ {v \mid p}^{} \mathbb{Q} _ p \xrightarrow{\mathrm{add}} \mathbb{Q} _ p. \] It is then clear that \( \mathrm{Tr} _ {K/\mathbb{Q}}(x) - \sum _ {v \mid p}^{} \chi _ v(x) \in \mathbb{Z} _ p \), and hence \( \chi(x) = 0 \) for \( x \in K \).
Now consider \( K^\ast \subseteq \mathbb{A} _ K \cong \widehat{\mathbb{A}} _ K \) the set of characters on \( \mathbb{A} _ K \) that vanish on \( K \). We have just proved that \( K^\ast \supseteq K \), but \( K^\ast \) is discrete because \( \mathbb{A} _ K / K \) is compact. Thus \( [K^\ast : K] \) should have finite index but \( K^\ast \) is a vector space over \( K \). This shows that \( K^\ast = K \). ▨
This implies that the Pontryagin dual of \( \mathbb{A} _ K / K \) is \( K \) with the identification given by \[ \displaystyle \mathbb{A} _ K / K \rightarrow \widehat{K}; \quad x \mapsto (y \mapsto \chi(xy)). \] Here, \( K \) is given a discrete topology. Moreover, if we give \( K \) the discrete measure \( \mu(S) = \lvert S \rvert \), then \[ \displaystyle \int _ {K}^{} \delta _ 0(x) \overline{\chi(x \xi)} d\mu(x) = \mathbf{1} _ {\mathbb{A} _ K/K} (\xi) \] and \( \lVert \delta _ 0 \rVert _ {L^2(K)} = \lVert \mathbf{1} \rVert _ {L^2(\mathbb{A} _ K/K)} = 1 \) implies that we are using the correct normalization for the measures.
Theorem 6 (Poisson summation). For any \( f \in \mathcal{S}(\mathbb{A} _ K) \), we have \[ \displaystyle \sum _ {x \in K}^{} f(x) = \sum _ {\xi \in K}^{} \hat{f}(\xi). \]
Proof. For \( f \in \mathcal{S}(\mathbb{A} _ K) \), we can define a function \( g \) on \( \mathbb{A} _ K / K \) by \[ \displaystyle g(x) = \sum _ {y \in K}^{} f(x + y), \] and this becomes a Schwartz–Bruhat function on \( \mathbb{A} _ K / K \). Now its Fourier transform is \[ \displaystyle \hat{g}(\xi) = \int _ {\mathbb{A} _ K/K}^{} g(x) \overline{\chi(x \xi)} dx = \int _ {\mathbb{A} _ K/K}^{} \sum _ {y \in K}^{} f(x+y) \overline{\chi(x \xi)} dx = \int _ {\mathbb{A} _ K}^{} f(x) \overline{\chi(x\xi)} dx \] for \( \xi \in K \). Note that this is just the restriction of \( \hat{f} \) to \( K \). Then the inverse Fourier transform of \( \hat{g} \) evaluated at \( 0 \) is \[ \displaystyle \sum _ {y \in K}^{} f(y) = g(0) = \int _ {K}^{} \hat{g}(\xi) d\mu(\xi) = \sum _ {\xi \in K}^{} \hat{f}(\xi). \] This proves the theorem. ▨
If we use a different lattice, we get a different dual lattice and a different normalizing constant.
Theorem 7. For \( a \in \mathbb{A} _ K^\times \) and \( f \in \mathcal{S}(\mathbb{A} _ K) \), we have \[ \displaystyle \sum _ {x \in K}^{} f(a x) = \frac{1}{\lvert a \rvert} \sum _ {\xi \in K}^{} \hat{f}(a^{-1} \xi). \]
Proof. Similar. ▨
5. Multiplicative quasi-characters#
Before defining the zeta function, let us develop some structure theory for multiplicative characters and quasi-characters on \( \mathbb{A} _ K^\times \). By the exactly same argument as for additive characters on \( \mathbb{A} _ K \), we see that all quasi-characters look like \[ \displaystyle x = (x _ v) _ v \mapsto \prod _ {v}^{} c _ v(x _ v) \] for quasi-characters \( c _ v : K _ v^\times \rightarrow \mathbb{C} \) such that \( c _ v \vert _ {\mathcal{O} _ v^\times} = 1 \) for almost all \( v \). Also, we can define measures on \( \mathbb{A} _ K^\times \) as \[ \displaystyle d\mu^\times(x) = \prod _ {v \in S}^{} d\mu _ v^\times(x _ v) \] because \( \mu _ v^\times(\mathcal{O} _ v^\times) = (N\mathcal{D} _ v)^{-1/2} = 1 \) for almost all \( v \).
Definition 8. We define \( \mathbb{A} _ K^1 = \{ x \in \mathbb{A} _ K^\times : \lvert x \rvert = 1 \} \).
The product formula states that \( K^\times \subseteq \mathbb{A} _ K^1 \). We shouldn’t expect \( \mathbb{A} _ K^\times / K^\times \) to be compact because \( \lvert - \rvert \) gives a surjective map to \( \mathbb{R} _ {>0} \), but when we restrict to content \( 1 \) elements \( \mathbb{A} _ K^1 / K^\times \) is compact.
This motivates us to define a measure on \( \mathbb{A} _ K^1 \). This should be given so that \[ \displaystyle 1 \rightarrow \mathbb{A} _ K^1 \rightarrow \mathbb{A} _ K^\times \rightarrow \mathbb{R} _ {>0} \rightarrow 1 \] is a short exact sequence of abelian groups with measure (so that a splitting gives an measure-preserving isomorphism), where \( \mathbb{R} _ {>0} \) is given the measure \( dt / t \).
We can explicitly find a fundamental domain for \( \mathbb{A} _ K^1 / K^\times \). On the infinite part, there is a log embedding \( \prod _ {v \mid \infty}^{} K _ v^\times \rightarrow \mathbb{R}^{r _ 1+r _ 2-1} \) whose image is a lattice, so we can find a fundamental domain \( E _ \infty \) for this. Then we may take \( E _ 0 = \prod _ {v \nmid \infty}^{} \mathcal{O} _ v^\times \times E _ \infty \). If \( b _ 1, \ldots, b _ h \in \mathbb{A} _ K^1 \) are are chosen so that \( \prod _ {\mathfrak{p}}^{} \mathfrak{p}^{v _ \mathfrak{p}(b _ j)} \) are representatives of the class group \( \mathrm{Cl}(K) \), then it is not hard to see that \[ \displaystyle E = b _ 1 E _ 0 \cup \cdots \cup b _ h E _ 0 \] is a fundamental domain for \( K^\times \) inside \( \mathbb{A} _ K^1 \). Using this, we can explicitly compute the measure of the fundamental domain.
Proposition 9. The measure of \( \mathbb{A} _ K^1 / K^\times \) is \[ \displaystyle \mu^\times(\mathbb{A} _ K^1 / K^\times) = \frac{2^{r _ 1} (2\pi)^{r _ 2} h _ K \mathrm{reg} _ K}{w _ K \sqrt{\lvert D \rvert}}, \] where \( r _ 1, r _ 2 \) are the number of real and complex embeddings, \( h _ K \) is the class number, \( w _ K \) is the number of roots of unity in \( K \), \( \mathrm{reg} _ K \) is the regulator.
This will become the residue of the zeta function at \( s = 1 \).
Proof. If you compute the volume of \( E \), by computing the volume of \( E _ 0 \), you get the right hand side. ▨
From now on, we are going to only consider quasi-characters that are trivial on \( K^\times \). Equivalently, these are quasi-characters on \( \mathbb{A} _ K^\times / K^\times \). But because \( \mathbb{A} _ K^1 / K^\times \) is compact, any quasi-character \( c \) will send \( \mathbb{A} _ K^1 / K^\times \) to \( S^1 \). Therefore any quasi-character looks like \[ \displaystyle c(x) = \tilde{c}(x) \lvert x \rvert^\sigma \] with \( \tilde{c} : \mathbb{A} _ K^\times / K^\times \rightarrow S^1 \) and some real number \( \sigma \). Again, such \( \sigma \) is called the exponent of the quasi-character \( c \).
If we use the previous fact that every quasi-character on \( \mathbb{A} _ K^\times \) is a product of local quasi-characters on \( K _ v^\times \), we can further write the arbitrary quasi-character \( c \) as \[ \displaystyle c(x) = \lvert x \rvert^\sigma \prod _ {v}^{} \tilde{c} _ v(x _ v) \] for characters \( \tilde{c} _ v : K _ v^\times \rightarrow S^1 \) with \( \tilde{c} _ v \vert _ {\mathcal{O} _ v} = 1 \) for almost all \( v \). However, the condition \( c(K^\times) = 1 \) gives an additional condition that is tricky to interpret. It turns out that the space of quasi-characters \( \mathbb{A} _ K^\times / K^\times \rightarrow \mathbb{C}^\times \) is a disjoint union of \( \mathbb{R}^{r _ 1 + r _ 2 + 1} \) topologically.
6. Global zeta function#
Let \( c \) be a quasi-character on \( \mathbb{A} _ K^\times / K^\times \). For \( f \in \mathcal{S}(\mathbb{A} _ K) \), we define \[ \displaystyle \zeta(f, c) = \int _ {\mathbb{A} _ K^\times}^{} f(x) c(x) d\mu^\times(x). \] Note that although \( f \) is defined on \( \mathbb{A} _ K \), we are only integrating on \( \mathbb{A} _ K^\times \), which is a very small subset of \( \mathbb{A} _ K \) because most component of \( (x _ v) \in \mathbb{A} _ K^\times \) satisfies \( x _ v \in \mathcal{O} _ v^\times \).
To see why we are calling this the zeta function, we look at the case when \( f \) looks like \[ \displaystyle f(x) = \prod _ {v}^{} f _ v(x _ v) \] for \( f _ v \in \mathcal{S}(K _ v) \) and \( f _ v = \mathbf{1} _ {\mathcal{O} _ v} \) for almost all \( v \). To compute the zeta function, we factor out all \( K _ v^\times \) with \( \tilde{c} _ v \vert _ {\mathcal{O} _ v} \neq 1 \) and \( f _ v \neq \mathbf{1} _ {\mathcal{O} _ v} \) and \( N \mathcal{D} _ v \neq 1 \). Then it is not hard to see that \[ \displaystyle \zeta(f, c) = \prod _ {v}^{} \zeta _ v(f _ v, c _ v). \] Set \( f _ \mathfrak{p} = \mathbf{1} _ {\mathcal{O} _ \mathfrak{p}} \) for all finite \( \mathfrak{p} \), and also set \( c = \lvert - \rvert^s \). Then we recover \[ \displaystyle \zeta(f, \lvert - \rvert^s) = \prod _ {v \mid \infty}^{} \zeta _ v(f _ v, \lvert - \rvert^s) \prod _ {\mathfrak{p}}^{} \frac{(N\mathcal{D} _ \mathfrak{p})^{-1/2}}{1 - (N\mathfrak{p})^{-s}} = \zeta _ K(s) \cdot \frac{1}{\sqrt{\lvert D \rvert}} \prod _ {v \mid \infty}^{} \zeta _ v(f _ v, \lvert - \rvert^s), \] which is the Dedekind zeta function \[ \displaystyle \zeta _ K(s) = \sum _ {\mathfrak{a} \subseteq \mathcal{O} _ K}^{} \frac{1}{(N\mathfrak{a})^s} = \prod _ {\mathfrak{p} \subseteq \mathcal{O} _ K}^{} \frac{1}{1 - (N\mathfrak{p})^{-s}} \] only up to a non-interesting meromorphic function.
Theorem 10. For \( c \) a quasi-character and \( f \in \mathcal{S}(\mathbb{A} _ K) \) a Schwartz–Bruhat function, \( \zeta(f, c \lvert - \rvert^s) \) (which only converges on \( \Re(s) > 1 - \sigma \)) can be extended to a meromorphic function on the entire complex plane. Moreover, it satisfies the functional equation \[ \displaystyle \zeta(f, c) = \zeta(\hat{f}, \hat{c}) \] where \( \hat{c} = \lvert - \rvert / c \) as before. This function \( \zeta(f, c \lvert - \rvert^s) \) is moreover a holomorphic function on the entire plane if \( c \) is not of the form \( \lvert - \rvert^s \). For \( \zeta(f, \lvert - \rvert^s) \), there are simple poles at \( s = 0 \) and \( s = 1 \), each with residue \( -\mu^\times(\mathbb{A} _ K^1 / K^\times) \hat{f}(0) \) and \( +\mu^\times(\mathbb{A} _ K^1 / K^\times) f(0) \).
Let us prove this. We first write \[ \displaystyle \zeta _ t(f, c) = \int _ {\mathbb{A} _ K^t}^{} f(x) c(x) d\mu^\times(x) \] where \( \mathbb{A} _ K^t = \{ x \in \mathbb{A} _ K^\times : \lvert x \rvert = t \} \), considered as a coset for \( \mathbb{A} _ K^1 \) with the same measure. Then we can write \[ \displaystyle \zeta(f, c) = \int _ {\mathbb{R} _ {>0}}^{} \zeta _ t(f, c) \frac{dt}{t}, \] because we defined the measure on \( \mathbb{A} _ K^1 \) so that the measure on \( \mathbb{A} _ K^\times \) decomposes as the measure on \( \mathbb{A} _ K^1 \) times \( \frac{dt}{t} \). Here, \( \zeta _ t(f, c) \) is always finite for each fixed \( t \) because \( f \) is Schwartz–Bruhat. If \( f \) looks like \( \prod _ {v}^{} f _ v \), then \( f _ v \) is compactly supported and bounded for finite \( v \). Moreover, if the exponent of \( c \) is \( \sigma \), then you can see that \( \zeta _ t(f, c) = O(t^{\sigma-1}) \) as \( t \rightarrow 0 \) and \( \zeta _ t(f, c) \) decays faster than any polynomial as \( t \rightarrow \infty \). This shows that \( \zeta(f, c) \) absolutely converge for \( \sigma > 1 \). Then \[ \displaystyle s \mapsto \zeta(f, c \lvert - \rvert^s) \] is a holomorphic function on \( \Re(s) > 1 - \sigma \).
To get a functional equation, consider a fundamental domain \( E^t \) for \( \mathbb{A} _ K^t / K^\times \). The idea is that \( c \) takes \( K^\times \) to \( 1 \), and so \( c(xy) = c(x) \) for \( y \in K^\times \). Then we can write \[ \displaystyle \zeta _ t(f, c) = \int _ {E^t}^{} \sum _ {y \in K^\times}^{} f(xy) c(xy) d\mu^\times(x) = \int _ {E^t}^{} c(x) \biggl( \sum _ {y \in K}^{} f(xy) - f(0) \biggr) d\mu^\times(x) \] and apply Poisson summation. Then we immediately get \[ \displaystyle \begin{aligned} \zeta _ t(f, c) + f(0) \int _ {E^t}^{} c(x) d\mu^\times(x) & \displaystyle= \int _ {E^t}^{} \frac{c(x)}{\lvert x \rvert} \sum _ {y \in K}^{} \hat{f}\Bigl( \frac{y}{x} \Bigr) d\mu^\times(x) \\ & \displaystyle= \int _ {E^{1/t}}^{} \frac{\lvert x \rvert}{c(x)} \sum _ {y \in K}^{} \hat{f}(xy) d\mu^\times(x) \\ & \displaystyle= \zeta _ {1/t}(\hat{f}, \hat{c}) + \hat{f}(0) \int _ {E^{1/t}}^{} \hat{c}(x) d\mu^\times(x). \end{aligned} \]
This additional term \( \int _ {E^t}^{} c(x) d\mu^\times(x) \) is annoying, but if you think about it, it vanishes for almost all \( c \). Because \( E^t \) is essentially like \( \mathbb{A} _ K^1 / K^\times \), this integral is essentially integrating a character on \( \mathbb{A} _ K^1 / K^\times \), which is a compact group. This can be thought of as evaluating the Fourier transform of \( \mathbf{1} _ {\mathbb{A} _ K^1 / K^\times} \), which is a delta function at the origin. Note that compactness of \( \mathbb{A} _ K^1 / K^\times \) implies discreteness of its dual group, and so the delta function makes sense as a Schwartz–Bruhat function.
Proposition 11. If \( c \) is a quasi-character on \( \mathbb{A} _ K^\times \), then \[ \displaystyle \int _ {E^t}^{} c(x) d\mu^\times(x) = \begin{cases} \mu^\times(\mathbb{A} _ K^1 / K^\times) t^s & \displaystyle c = \lvert - \rvert^s, \\ 0 & \displaystyle \text{otherwise}. \end{cases} \]
Proof. This is essentially what we said above. At \( c = \lvert - \rvert^s \), we can explicitly evaluate the integral on the left hand side, and in other cases it is zero. ▨
If \( c \) is not of the form \( \lvert - \rvert^s \), then we immediately get \[ \displaystyle \zeta _ t(f, c) = \zeta _ {1/t}(\hat{f}, \hat{c}) \] and so \[ \displaystyle \zeta(f, c) = \int _ {1}^{\infty} \zeta _ t(f, c) \frac{dt}{t} + \int _ {0}^{1} \zeta _ {1/t}(\hat{f},\hat{c}) \frac{dt}{t} = \int _ {1}^{\infty} (\zeta _ t(f, c) + \zeta _ t(\hat{f}, \hat{c})) \frac{dt}{t}. \] Because \( \zeta _ t(f, c) \) and \( \zeta _ t(\hat{f}, \hat{c}) \) both decays rapidly as \( t \rightarrow \infty \), this converges nicely and we get that \[ \displaystyle s \mapsto \zeta(f, c \lvert - \rvert^s) \] extends holomorphically to the entire complex plane. Moreover, we clearly have \( \zeta(f, c) = \zeta(\hat{f}, \hat{c}) \) and so \( \zeta(f, c \lvert - \rvert^s) = \zeta(\hat{f}, \hat{c} \lvert - \rvert^{-s}) \).
If \( c = \lvert - \rvert^s \), \[ \displaystyle \zeta _ t(f, \lvert - \rvert^s) + \mu^\times(\mathbb{A} _ K^1 / K^\times) f(0) t^s = \zeta _ {1/t}(\hat{f}, \lvert - \rvert^{1-s}) + \mu^\times(\mathbb{A} _ K^1 / K^\times) \hat{f}(0) t^{1-s}. \] Then \[ \displaystyle \begin{aligned} \zeta(f, \lvert - \rvert^s) & \displaystyle= \int _ {1}^{\infty} \zeta _ t(f, \lvert - \rvert^s) dt + \int _ {0}^{1} (\zeta _ {1/t}(\hat{f}, \lvert - \rvert^s) + \mu^\times(\mathbb{A} _ K^1 / K^\times) \hat{f}(0) t^{1-s}) \frac{dt}{t} \\ & \displaystyle\qquad - \frac{\mu^\times(\mathbb{A} _ K^\times / K^\times) f(0)}{s} \\ & \displaystyle= \int _ {1}^{\infty} (\zeta _ t(f, \lvert - \rvert^s) + \zeta _ t(\hat{f}, \lvert - \rvert^{1-s})) \frac{dt}{t} + \mu^\times(\mathbb{A} _ K^1 / K^\times) \Bigl( \frac{\hat{f}(0)}{s-1} - \frac{f(0)}{s} \Bigr). \end{aligned} \] The integral converges to a holomorphic function in \( s \) because \( \zeta _ t(f, c) \) decays rapidly as \( t \rightarrow \infty \). Therefore \( \zeta _ t(f, \lvert - \rvert^s) \) has a meromorphic extension to the entire complex plane, and it has two poles at \( s = 0, 1 \). The residues at \( s = 1 \) and \( s = 0 \) are \( \pm \mu^\times(\mathbb{A} _ K^1 / K^\times) \), and satisfies the functional equation \( \zeta _ t(f, \lvert - \rvert^s) = \zeta _ t(\hat{f}, \lvert - \rvert^{1-s}) \). This finishes the proof of the theorem.
Corollary 12. The Dedekind zeta function \( \zeta _ K(s) \) satisfies the functional equation \[ \displaystyle \zeta _ K(1-s) = \frac{1}{\lvert D \rvert} \cdot \zeta _ K(s) \rho _ \mathbb{R}(s)^{r _ 1} \rho _ \mathbb{R}(s)^{r _ 2} \] where \( r _ 1, r _ 2 \) are the number of real and complex embeddings, and the functions \( \rho _ \mathbb{R} \) and \( \rho _ \mathbb{C} \) are defined as \[ \displaystyle \rho _ \mathbb{R}(s) = 2^{1-s} \pi^{-s} \cos\Bigl( \frac{\pi s}{2} \Bigr) \Gamma(s), \quad \rho _ \mathbb{C}(s) = (2\pi)^{1-2s} \frac{\Gamma(s)}{\Gamma(1 - s)}. \] Moreover, the residue of \( \zeta _ K(s) \) at \( s = 1 \) is \[ \displaystyle \mathrm{Res} _ {s=1} \zeta _ K(s) = \mu^\times(\mathbb{A} _ K^1 / K^\times) = \frac{2^{r _ 1} (2\pi)^{r _ 2} h _ K \mathrm{reg} _ K}{w _ K \sqrt{\lvert D \rvert}}. \]
Proof. We have seen that for \( f = \prod _ {v}^{} f _ v \) with \( f _ \mathfrak{p} = \mathbf{1} _ {\mathcal{O} _ \mathfrak{p}} \), \[ \displaystyle \zeta(f, \lvert - \rvert^s) = \frac{1}{\sqrt{\lvert D \rvert}} \zeta _ K(s) \prod _ {v \mid \infty}^{} \zeta _ v(f _ v, \lvert - \rvert^s). \] Then \( \hat{f} = \prod _ {v}^{} \hat{f} _ v \) and then functional equation gives us that \[ \displaystyle \begin{aligned} \zeta(f, \lvert - \rvert^s) & \displaystyle= \zeta(\hat{f}, \lvert - \rvert^{1-s}) = \prod _ {\mathfrak{p}}^{} \zeta _ \mathfrak{p}(\mu(\mathcal{O} _ \mathfrak{p}) \mathbf{1} _ {\mathcal{D} _ \mathfrak{p}^{-1}}, \lvert - \rvert^{1-s}) \prod _ {v \mid \infty}^{} \zeta _ p(\hat{f} _ v, \lvert - \rvert^{1-s}) \\ & \displaystyle= \sqrt{\lvert D \rvert} \zeta _ K(1-s) \prod _ {v \mid \infty}^{} \frac{\zeta _ v(f _ v, \lvert - \rvert^{s})}{\rho _ v(\lvert - \rvert^s)}. \end{aligned} \] Comparing the two gives the functional equation. For the residue, we note \( \zeta _ v(f _ v, \lvert - \rvert) = \hat{f} _ v(0) \) and hence \( \prod _ {v \mid \infty}^{} \zeta _ v(f _ v, \lvert - \rvert) \) is \( \prod _ {v \mid \infty}^{} \hat{f} _ v(0) = \sqrt{\lvert D \rvert} \hat{f}(0) \). Therefore the residue of \( \zeta _ K(s) \) at \( s = 1 \) is \( \mu^\times(\mathbb{A} _ K^1 / K^\times) \). ▨
To get a symmetric form of the functional equation, what you can do is to find explicitly a function \( f \) such that \( \hat{f} \). Then you get \( \zeta(f, \lvert - \rvert^s) = \zeta(f, \lvert - \rvert^{1-s}) \). This will immediately give, for instance, the symmetric functional equation of the Riemann zeta function.
References#
[Tat67] J. T. Tate, Fourier analysis in number field, and Hecke’s zeta-functions, Algebraic Number Theory (Proc. Instructional Conf., Brighton, 1965), Thompson, Washington, D.C., 1967, pp. 305–347.