On Kronecker terms over global function fields

Abstract

We establish a general Kronecker limit formula of arbitrary rank over global function fields with Drinfeld period domains playing the role of upper-half plane. The Drinfeld–Siegel units come up as equal characteristic modular forms replacing the classical $\Delta$. This leads to analytic means of deriving a Colmez-type formula for “stable Taguchi height” of CM Drinfeld modules having arbitrary rank. A Lerch-Type formula for “totally real” function fields is also obtained, with the Heegner cycle on the Bruhat–Tits buildings intervene. Also our limit formula is naturally applied to the special values of both the Rankin–Selberg L-functions and the Godement–Jacquet L-functions associated to automorphic cuspidal representations over global function fields.

CM theory of Drinfeld modules

Let $\rho$ be a Drinfeld A-module of rank r over ${\mathbb C}_\infty$. For $f \in {\text {End}}_A(\rho /{\mathbb C}_\infty ) \subset {\mathbb C}_\infty \{\tau \}$, let $d_f$ be the constant term of f. Put

$$\begin{aligned} \mathcal {O}:= \{d_f : f \in {\text {End}}_A(\rho /{\mathbb C}_\infty ) \} \subset {\mathbb C}_\infty . \end{aligned}$$

Then $(f \mapsto d_f)$ gives a ring isomorphism from ${\text {End}}_A(\rho /{\mathbb C}_\infty )$ to $\mathcal {O}$ (cf. [26, Theorem 13.25]). In particular, it is known that:

• Let $\Lambda _\rho \subset {\mathbb C}_\infty$ be the A-lattice associated to $\rho$. Then $\mathcal {O}= \{ c \in {\mathbb C}_\infty : c \Lambda _\rho \subset \Lambda _\rho \}$;

• The field K of fractions of $\mathcal {O}$ is imaginary (i.e. $\infty$ is not split in K) with $[K:k] \mid r$.

For $c \in \mathcal {O}$, we let $\rho _c$ be the endomorphism of $\rho$ with $d_{\rho _c} = c$.

The A-lattice $\Lambda _\rho$ can be viewed as an $\mathcal {O}$-module. We may say that two Drinfeld A-modules $\rho _1$ and $\rho _2$ of rank r over ${\mathbb C}_\infty$ with ${\text {End}}_A(\rho _1/{\mathbb C}_\infty ) = {\text {End}}_A(\rho _2/{\mathbb C}_\infty ) \cong \mathcal {O}$have the same genus if $\Lambda _{\rho _1}\otimes _A O_v \cong \Lambda _{\rho _2}\otimes _A O_v$ as $\mathcal {O}\otimes _A O_v$-modules for each finite place v of k.

Suppose $\rho$ is CM (i.e. $[K:k] = r$). Then as an $\mathcal {O}$-module, the lattice $\Lambda _{\rho }$ is isomorphic to an ideal $\mathfrak {I}$ of $\mathcal {O}$ with

$$\begin{aligned} {\text {End}}_\mathcal {O}(\mathfrak {I}):= \{ \alpha \in K : \alpha \mathfrak {I}\subset \mathfrak {I}\} = \mathcal {O}. \end{aligned}$$

We say that $\rho$ has principal genus if $\Lambda _\rho \otimes _A O_v \cong \mathcal {O}\otimes _A O_v$ for each finite place v of k, or equivalently, $\mathfrak {I}$ is an invertible ideal of $\mathcal {O}$.

In this section, we extend the work of Hayes in [15, Theorem 8.5] on the CM theory of Drinfeld modules having principal genus to the case of arbitrary genus. More precisely, we shall prove the following theorem:

Theorem A.1

Let $\rho$ be a CM Drinfeld A-module of rank r over ${\mathbb C}_\infty$. Identify the endomorphism ring ${\text {End}}_A(\rho /{\mathbb C}_\infty )$ with an A-order $\mathcal {O}$ of an imaginary field K with $[K:k] = r$.

• (1) The Drinfeld A-module $\rho$ is isomorphic (over ${\mathbb C}_\infty$) to a CM Drinfeld A-module of rank r defined over $H_\mathcal {O}$, where $H_\mathcal {O}$ is the ring class field of $\mathcal {O}$.

• (2) Suppose the Drinfeld module $\rho$ is defined over $H_\mathcal {O}$. Then ${\text {End}}_A(\rho /{\mathbb C}_\infty ) = {\text {End}}_A(\rho /H_\mathcal {O})$. Moreover, for an integral ideal $\mathcal {A}$ of $\mathcal {O}$ which is invertible, let ${\text {Frob}}_\mathcal {A}\in {\text {Gal}}(H_\mathcal {O}/K)$ be the Frobenius automorphism associated to $\mathcal {A}$ via the Artin map. Then

  $$\begin{aligned} {\text {Frob}}_\mathcal {A}(\rho ) \cong \mathcal {A}* \rho . \end{aligned}$$

Here $\mathcal {A}* \rho$ is the unique Drinfeld A-module satisfying

$$\begin{aligned} \rho _\mathcal {A}\cdot \rho _a = (\mathcal {A}* \rho )_a \cdot \rho _\mathcal {A}, \quad \forall a \in A, \end{aligned}$$

where $\rho _\mathcal {A}\in H_\mathcal {O}\{ \tau \}$ is the monic generator of the left ideal of $H_\mathcal {O}\{ \tau \}$ generated by endomorphisms $\rho _c$ for all $c \in \mathcal {A}\subset \mathcal {O}$.

Remark A.2

Let $\rho$ be a Drinfeld module of rank r over ${\mathbb C}_\infty$ with CM by $\mathcal {O}$. Let $\Lambda _\rho \subset {\mathbb C}_\infty$ be the A-lattice associated to $\rho$, which is equipped with an $\mathcal {O}$-module structure. Then for an invertible ideal $\mathcal {A}$ of $\mathcal {O}$, the A-lattice associated to $\mathcal {A}* \rho$ is homothetic to $\mathcal {A}^{-1} \cdot \Lambda _\rho$ (cf. [15, Proposition 5.10 and the equation (5.18)]). Suppose $\rho$ is defined over $H_\mathcal {O}$ (via a fixed embedding $H_\mathcal {O}\hookrightarrow {\mathbb C}_\infty$). Then Theorem A.1 (2) tells us that the A-lattice associated to $\text {Frob}_\mathcal {A}(\rho )$ also lies in the homothety class of $\mathcal {A}^{-1} \cdot \Lambda _\rho$.

We first recall the needed properties in the explicit class field theory over global function fields. Further details are referred to [13, Chapter 7] and [15].

1.1 Explicit class field theory

Let $\rho ^o$ be a CM Drinfeld A-module of rank r over ${\mathbb C}_\infty$ so that ${\text {End}}(\rho ^o)$ is identified with $O_K$, the integral closure of A in the imaginary field K. Viewing $\rho$ as a Drinfeld $O_K$-module of rank 1, suppose $\rho ^o$ is sign-normalized (cf. [13, Theorem 7.2.15]). Then $\rho ^o$ is actually defined over $O_{H^+}$, the integral closure of A in $H^+$, where $H^+$ is the “narrow” Hilbert class field of $O_K$ (cf. [13, Section 7.4]). Given an ideal $\mathfrak {A}$ of $O_K$, let ${\text {Frob}}_\mathfrak {A}\in {\text {Gal}}(H^+/K)$ be the Frobenius automorphism associated to $\mathfrak {A}$. Then (cf. [13, Theorem 7.4.8]):

$$\begin{aligned} {\text {Frob}}_\mathfrak {A}(\rho ^o) = \mathfrak {A}* \rho ^o. \end{aligned}$$

For an integral ideal $\mathfrak {C}$ of $O_K$, let

$$\begin{aligned} \rho ^o[\mathfrak {C}] := \{ \lambda \in \overline{K} : \rho ^o_a (\lambda ) = 0,\ \forall a \in \mathfrak {C}\}, \end{aligned}$$

and put $H^+_{\mathfrak {C}} := H^+(\rho ^o[\mathfrak {C}])$. Then (cf. [13, Proposition 7.5.4 and Corollary 7.5.5]):

Theorem A.3

The field extension $H^+_\mathfrak {C}/ K$ is abelian with

$$\begin{aligned} {\text {Gal}}(H^+_\mathfrak {C}/K ) \cong \mathcal {I}_{O_K}(\mathfrak {C})/\mathcal {P}^+_\mathfrak {C}. \end{aligned}$$

Here $\mathcal {I}_{O_K}(\mathfrak {C})$ is the group generated by ideals of $O_K$ coprime to $\mathfrak {C}$, and $\mathcal {P}^+_\mathfrak {C}$ is the subgroup generated by principal ideals $\alpha O_K$, where $\alpha \in O_K$ is positive and $\alpha \equiv 1 \bmod \mathfrak {C}$. Moreover, for an integral ideal $\mathfrak {A}$ of $O_K$ coprime to $\mathfrak {C}$, one has

$$\begin{aligned} {\text {Frob}}_\mathfrak {A}(\lambda ) = \rho ^o_\mathfrak {A}(\lambda ), \quad \forall \lambda \in \rho ^o[\mathfrak {C}]. \end{aligned}$$

Remark A.4

• (1) Let $\infty _K$ be the unique place of K lying above $\infty$ and ${\mathbb F}_{\infty _K}$ the residue field at $\infty _K$. Then $H^+_\mathfrak {C}/K$ is tamely ramified at $\infty _K$ with ramification index $\#({\mathbb F}_{\infty _K})-1$ (cf. [13, Proposition 7.5.8 and Corollary 7.5.9]).

• (2) Let $\Lambda ^o = \Lambda _{\rho ^o}\subset {\mathbb C}_\infty$ be the A-lattice corresponding to $\rho ^o$. Then we may write $\rho ^o[\mathfrak {C}]$ as

  $$\begin{aligned} \rho ^o[\mathfrak {C}] = \left\{ \exp _{\Lambda ^o}(\alpha ) : \alpha \in \frac{\mathfrak {C}^{-1} \Lambda ^o}{\Lambda ^o} \right\} . \end{aligned}$$

Given an integral ideal $\mathfrak {A}$ of $O_K$ coprime to $\mathfrak {C}$, recall that $\rho ^o_\mathfrak {A}\in H^+\{\tau \}$ is the monic generator of the left ideal generated by $\rho ^o_c$ for all $c \in \mathfrak {A}$. let $d_\mathfrak {A}$ be the constant term of $\rho ^o_\mathfrak {A}$. Then we have

$$\begin{aligned} d_\mathfrak {A}= \prod _{0 \ne \lambda \in \rho ^o[\mathfrak {A}]} \lambda \quad \quad \text { and } \quad \quad \Lambda _{{\text {Frob}}_\mathfrak {A}(\rho ^o)} = \Lambda _{\mathfrak {A}* \rho ^o} = d_\mathfrak {A}\cdot \mathfrak {A}^{-1} \Lambda ^o. \end{aligned}$$

Theorem A.3 implies that

$$\begin{aligned} {\text {Frob}}_\mathfrak {A}\big (\exp _{\Lambda ^o}(\alpha )\big )= & {} \rho _\mathfrak {A}^o\big (\exp _{\Lambda ^o}(\alpha )\big ) = \exp _{d_\mathfrak {A}\mathfrak {A}^{-1} \Lambda ^o} ( d_\mathfrak {A}\alpha ), \quad \forall \alpha \in \frac{\mathfrak {C}^{-1} \Lambda ^o}{\Lambda ^o}. \nonumber \\ \end{aligned}$$

(A.1)

1.1.1 Frobenius action on Drinfeld $\mathcal {O}$-modules

Let $\Lambda \subset {\mathbb C}_\infty$ be an A-lattice of rank r so that $\mathcal {O}:= \{ c \in {\mathbb C}_\infty : c\Lambda \subset \Lambda \}$ is an A-order of an imaginary field K with $[K:k] = r$. Let $\mathfrak {C}\lhd O_K$ be the conductor of $\mathcal {O}$. Take $\Lambda ^o := \mathfrak {C}\cdot \Lambda$. Then

$$\begin{aligned} O_K \cdot \Lambda = \mathfrak {C}^{-1} \cdot \Lambda ^o \supset \Lambda \supset \Lambda ^o, \end{aligned}$$

and the Drinfeld A-module $\rho ^o$ corresponding to $\Lambda ^o$ is CM by $O_K$.

Assume that $\rho ^o$ is sign-normalized (thus defined over $H^+$). Take

$$\begin{aligned} u(x) = u_{\Lambda /\Lambda ^o}(x) := \prod _{\alpha \in \frac{\Lambda }{\Lambda ^o}} \big (x - \exp _{\Lambda ^o}(\alpha )\big ) \quad \text {and} \quad d_u:= \prod _{0 \ne \alpha \in \frac{\Lambda }{\Lambda ^o}}\exp _{\Lambda ^o}(\alpha ). \end{aligned}$$

Then u corresponds to a twisted polynomial $u(\tau ) \in H^+_\mathfrak {C}\{ \tau \}$ with constant term $d_u$. Moreover, let $\rho ^{d_u \Lambda }$ be the Drinfeld A-module corresponding to the A-lattice $d_u \Lambda$. Then

$$\begin{aligned} u \cdot \rho ^o = \rho ^{d_u\Lambda } \cdot u, \end{aligned}$$

which tells us that $\rho ^{d_u\Lambda }$ is defined over $H^+_\mathfrak {C}$. In fact, we have:

Proposition A.5

The Drinfeld A-module $\rho = \rho ^{d_u \Lambda }$ is defined over $H^+_{\mathcal {O}}$, the “narrow” ring class field of $\mathcal {O}$. Moreover, given an invertible ideal $\mathcal {A}$ of $\mathcal {O}$, let ${\text {Frob}}_\mathcal {A}\in {\text {Gal}}(H^+_{\mathcal {O}}/K)$ be the Frobenius automorphism associated to $\mathcal {A}$. Then

$$\begin{aligned} {\text {Frob}}_\mathcal {A}(\rho ) = \mathcal {A}* \rho . \end{aligned}$$

Proof

Let $\mathcal {I}_\mathcal {O}$ (resp. $\mathcal {I}_\mathcal {O}(\mathfrak {C})$) be the group generated by invertible ideals of $\mathcal {O}$ (resp. coprime to $\mathfrak {C}$), and $\mathcal {P}_\mathcal {O}^+$ (resp. $\mathcal {P}_\mathcal {O}^+(\mathfrak {C})$) is the subgroup generated by principal ideals $\alpha \mathcal {O}$, where $\alpha$ is positive (resp. and coprime to $\mathfrak {C}$). Then

$$\begin{aligned} {\text {Gal}}(H^+_\mathcal {O}/K) \cong {\text {Pic}}^+(\mathcal {O}) := \mathcal {I}_\mathcal {O}/ \mathcal {P}_\mathcal {O}^+\cong \mathcal {I}_\mathcal {O}(\mathfrak {C}) / \mathcal {P}_\mathcal {O}^+(\mathfrak {C}), \end{aligned}$$

and we have the following commutative diagram:

Here the vertical map on the right hand side is induced from $\mathfrak {A}\mapsto \mathfrak {A}\cap \mathcal {O}$ for every integral ideal $\mathfrak {A}$ of $O_K$ coprime to $\mathfrak {C}$. Thus to prove that $\rho$ is defined in $H_\mathcal {O}^+$, it suffices to show that ${\text {Frob}}_\mathfrak {A}(\rho ) = \rho$ for every integral ideal $\mathfrak {A}$ of $O_K$ coprime to $\mathfrak {C}$ and $\mathfrak {A}\cap \mathcal {O}\in \mathcal {P}_\mathcal {O}^+(\mathfrak {C})$.

Let $\mathfrak {A}$ be an integral ideal of $O_K$ coprime to $\mathfrak {C}$ and ${\text {Frob}}_\mathfrak {A}\in {\text {Gal}}(H_{\mathfrak {C}}^+/K)$ be the Frobenius element corresponding to $\mathfrak {A}$. We have

$$\begin{aligned} {\text {Frob}}_\mathfrak {A}(u) \cdot {\text {Frob}}_\mathfrak {A}(\rho ^o) = {\text {Frob}}_\mathfrak {A}(\rho ) \cdot {\text {Frob}}_\mathfrak {A}(u). \end{aligned}$$

Note that $O_K\cdot \Lambda$ is a projective $O_K$-module of rank 1. Since $\mathfrak {A}$ and $\mathfrak {C}$ are relatively prime, one gets $O_K \cap \mathfrak {A}^{-1} \mathfrak {C}= \mathfrak {C}$ and

$$\begin{aligned}&\Lambda ^o \subseteq \Lambda \cap \mathfrak {A}^{-1}\Lambda ^o \subseteq (O_K\Lambda ) \cap \mathfrak {A}^{-1}\mathfrak {C}\cdot (O_K\Lambda ) \\&\quad = (O_K \cap \mathfrak {A}^{-1}\mathfrak {C}) \cdot (O_K\Lambda ) = \mathfrak {C}\cdot (O_K\Lambda ) = \Lambda ^o. \end{aligned}$$

Thus $\Lambda \cap \mathfrak {A}^{-1}\Lambda ^o = \Lambda ^o$ and we have the following isomorphism

$$\begin{aligned} \frac{\Lambda }{\Lambda ^o} \cong \frac{\Lambda + \mathfrak {A}^{-1}\Lambda ^o}{\mathfrak {A}^{-1}\Lambda ^o} \quad \text { by sending } \alpha + \Lambda ^o\hbox { to }\alpha + \mathfrak {A}^{-1}\Lambda ^o\hbox { for all } \alpha \in \Lambda . \end{aligned}$$

The equality (A.1) then implies

$$\begin{aligned} {\text {Frob}}_\mathfrak {A}(u) (x) = \prod _{\alpha ' \in \frac{d_\mathfrak {A}(\Lambda + \mathfrak {A}^{-1} \Lambda ^o)}{d_\mathfrak {A}\mathfrak {A}^{-1} \Lambda ^o}} \Big (x-\exp _{d_\mathfrak {A}\mathfrak {A}^{-1} \Lambda ^o}(\alpha ')\Big ). \end{aligned}$$

Since ${\text {Frob}}_\mathfrak {A}(\rho ^o) = \mathfrak {A}* \rho ^o$, which corresponds to the lattice $d_\mathfrak {A}\mathfrak {A}^{-1} \Lambda ^o$, we obtain that ${\text {Frob}}_\mathfrak {A}(\rho )$ actually corresponds to the lattice

$$\begin{aligned} (d_{{\text {Frob}}_\mathfrak {A}(u)} d_\mathfrak {A})\cdot (\Lambda + \mathfrak {A}^{-1} \Lambda ^o) = (d_{{\text {Frob}}_\mathfrak {A}(u)} d_\mathfrak {A}) \cdot \mathcal {A}^{-1} \Lambda , \end{aligned}$$

where $d_{{\text {Frob}}_\mathfrak {A}(u)}$ is the constant term of ${{\text {Frob}}_\mathfrak {A}(u)}$ and $\mathcal {A}:= \mathfrak {A}\cap \mathcal {O}$ is an invertible ideal of $\mathcal {O}$ coprime to $\mathfrak {C}$. On the other hand, the lattice corresponding to $\mathcal {A}* \rho$ is $(d_\mathcal {A}\cdot d_u) \cdot \mathcal {A}^{-1} \Lambda$, where $d_\mathcal {A}$ is the constant term of $\rho _\mathcal {A}$. It is straightforward that

$$\begin{aligned} d_\mathcal {A}\cdot d_u = d_{{\text {Frob}}_\mathfrak {A}(u)} d_\mathfrak {A}. \end{aligned}$$

Therefore

$$\begin{aligned} {\text {Frob}}_\mathfrak {A}(\rho )= & {} \mathcal {A}* \rho . \end{aligned}$$

(A.2)

Suppose $\mathfrak {A}\cap \mathcal {O}= \mathcal {A}= \alpha \cdot \mathcal {O}$, where $\alpha$ is positive. Then $\rho _{\alpha \mathcal {O}} = \rho _\alpha$ and

$$\begin{aligned} {\text {Frob}}_\mathfrak {A}(\rho ) = (\alpha \mathcal {O}) * \rho = \rho . \end{aligned}$$

Therefore $\rho$ is defined over $H_\mathcal {O}^+$.

Moreover, for an invertible ideal $\mathcal {A}$ of $\mathcal {O}$ coprime to $\mathfrak {C}$, we put $\mathfrak {A}:= \mathcal {A}\cdot O_K$, and let ${\text {Frob}}_\mathcal {A}\in {\text {Gal}}(H_\mathcal {O}^+/K)$ and ${\text {Frob}}_\mathfrak {A}\in {\text {Gal}}(H_\mathfrak {C}^+/K)$ be the Frobenius elements corresponding to $\mathcal {A}$ and $\mathfrak {A}$, respectively. Then ${\text {Frob}}_\mathcal {A}= {\text {Frob}}_\mathfrak {A}\big |_{H_\mathcal {O}^+}$, and the Eq. (A.2) says

$$\begin{aligned} {\text {Frob}}_\mathcal {A}(\rho ) = {\text {Frob}}_\mathfrak {A}(\rho ) = \mathcal {A}* \rho , \end{aligned}$$

which completes the proof. $\square$

1.2 Proof of Theorem A.1

Let $\rho$ be a CM Drinfeld A-module of rank r. Identifying ${\text {End}}(\rho )$ with an A-order $\mathcal {O}$ of an imaginary field K, suppose $\mathcal {O}$ has conductor $\mathfrak {C}$. Let $H_\rho$ be the field of invariants of $\rho$ (cf. [15, Theorem 6.6]). Then it is known that $H_\rho$ is contained in $K_\infty$ (cf. [15, Proposition 6.2 and 6.4]). Therefore Proposition A.5 implies that

$$\begin{aligned} H_\rho \subset H^+_\mathcal {O}\cap K_\infty = H_\mathcal {O}. \end{aligned}$$

Let $\Lambda _\rho \subset {\mathbb C}_\infty$ be the A-lattice corresponding to $\rho$. Viewing $\Lambda _\rho$ as an $\mathcal {O}$-module, it is the fact that for two invertible ideals $\mathcal {A}, \mathcal {B}$ of $\mathcal {O}$, we have

$$\begin{aligned} \mathcal {A}\cdot \Lambda _\rho \cong \mathcal {B}\cdot \Lambda _\rho \quad \text { if and only if } \quad \mathcal {A}^{-1} \mathcal {B}\text { is a principal ideal of }\mathcal {O}. \end{aligned}$$

The explicit description of Frobenius actions in Proposition A.5 then assures that

$$\begin{aligned}{}[H_\rho : K] = \#{\text {Pic}}(\mathcal {O}) = [H_\mathcal {O}:K]. \end{aligned}$$

Therefore $H_\rho = H_\mathcal {O}$ and the Theorem holds. $\square$