Vanishing of quartic and sextic twists of L-functions

Abstract

Let E be an elliptic curve over ${{\mathbb {Q}}}$. We conjecture asymptotic estimates for the number of vanishings of $L(E,1,\chi )$ as $\chi$ varies over all primitive Dirichlet characters of orders 4 and 6, subject to a mild hypothesis on E. Our conjectures about these families come from conjectures about random unitary matrices as predicted by the philosophy of Katz-Sarnak. We support our conjectures with numerical evidence. Compared to earlier work by David, Fearnley and Kisilevsky that formulated analogous conjectures for characters of any odd prime order, in the composite order case, we need to justify our use of random matrix theory heuristics by analyzing the equidistribution of the squares of normalized Gauss sums. To do this, we introduce the notion of totally order $\ell$ characters to quantify how quickly the quartic and sextic Gauss sums become equidistributed. Surprisingly, the rate of equidistribution in the full family of quartic (resp., sextic) characters is much slower than in the sub-family of totally quartic (resp., sextic) characters. We provide a conceptual explanation for this phenomenon by observing that the full family of order $\ell$ twisted elliptic curve L-functions, with $\ell$ even and composite, is a mixed family with both unitary and orthogonal aspects.

1 Introduction

Vanishings of elliptic curve L-functions at the value $s=1$ (normalized so that the functional equation relates s and $2-s$) is central to a great deal of modern number theory. For instance, if an L-function associated to an elliptic curve vanishes at $s=1$, then the Birch–Swinnerton-Dyer Conjecture predicts that the curve will have infinitely many rational points.

Additionally, statistical questions about how often L-functions within a family vanish at the central value have also been of broad interest. For example, it is expected (as first conjectured by Chowla [7]) that, for all primitive Dirichlet characters $\chi$, we have that the central value of the Dirichlet L-function of $\chi$ never vanishes.

A fruitful way of studying such questions has been to model L-functions using random matrices. For example, Conrey, Keating, Rubinstein and Snaith [8] consider the family of twisted L-functions $L(f,s, \chi _d)$ associated to a fixed modular form f of weight k and varying quadratic characters $\chi _d$. They show that the random matrix theory model predicts that infinitely many values $L(f,s, \chi _d)$ are zero when the weight of f is 2 or 4, but that only finitely many of the values are zero when the weight is at least 6.

Another example, due to David, Fearnley and Kisilevsky [10, 11], instead uses the random matrix model to give conjectural asymptotics for the number of vanishings of elliptic curve L-functions twisted by families of Dirichlet characters of a fixed order. In particular, they predict that for an elliptic curve E over ${{\mathbb {Q}}}$, the values $L(E,1,\chi )$ are zero infinitely often if $\chi$ has order 3 or 5, but for characters $\chi$ with a fixed prime order $\ell \ge 7$, only finitely many values $L(E,1,\chi )$ are zero. The elliptic curves we consider in this paper are all defined over ${{\mathbb {Q}}}$.

Other authors have also considered the vanishings of central values of higher order twists of elliptic curve L-functions. For example, Mazur and Rubin [23] use statistical properties of modular symbols to heuristically estimate the probability that $L(E,1,\chi )$ vanishes. Their Conjecture 11.1 implies that for an elliptic curve E over ${{\mathbb {Q}}}$, there should be only finitely many characters $\chi$ of a fixed order $\ell$ such that $L(E,1,\chi )=0$ and $\varphi (\ell ) > 4$. This further implies the following: Let E be an elliptic curve and let $F/{{\mathbb {Q}}}$ be an infinite abelian extension such that $\text {Gal}(F/{{\mathbb {Q}}})$ has only finitely many characters of orders 2, 3 and 5. Then E(F) is finitely generated. Finally, their Proposition 3.2 relates the (order of) vanishing of $L(E,1,\chi )$ to the growth in rank of E over a finite abelian extension $F/{{\mathbb {Q}}}$. In particular, if BSD holds for E over both ${{\mathbb {Q}}}$ and F, then

$$\begin{aligned} \text {rank}(E(F)) = \text {rank}(E({{\mathbb {Q}}})) + \sum _{\chi :\text {Gal}(F/{{\mathbb {Q}}})\rightarrow \mathbb {C}^\times }\text {ord}_{s=1}L(E,s,\chi ). \end{aligned}$$

Even more recently, Kisilevsky and Kuwata [20] show that for any elliptic curve E over ${{\mathbb {Q}}}$ there are infinitely many primitive sextic characters for which the central value of the twisted L-function vanishes.

1.1 Notation and statement of the main conjecture

We fix the following notation. See Definition 1 for the definition of totally order $\ell$ characters but, roughly speaking, these are order $\ell$ characters that, when factored into characters of prime power conductors, have all their factors also of order $\ell$. We make the definitions for any $\ell \in {{\mathbb {Z}}}_{>0}$ but in what follows we focus our attention on the quartic and sextic cases; i.e., $\ell =4$ or $\ell =6$. Set

$$\begin{aligned} \Psi _\ell&= \{ \text {primitive Dirichlet characters}\, \chi \, \text {of order}\, \ell \}\\ \Psi ^\textrm{tot}_\ell&= \{\chi \in \Psi _\ell \, \text {that are totally order}\, \ell \}\\ \Psi ^\prime _\ell&= \{ \chi \in \Psi _\ell \, \text {with}\, \textrm{cond}(\chi )\, \text {prime}\}. \end{aligned}$$

Note that $\Psi _{\ell }' \subseteq \Psi _{\ell }^{\textrm{tot}} \subseteq \Psi _{\ell }$.

Along the way we will need to estimate the number of characters in each family and so we define:

$$\begin{aligned} \Psi _\ell (X)&= \{\chi \in \Psi _\ell : \textrm{cond}(\chi )\le X\}\\ \Psi ^\textrm{tot}_\ell (X)&= \{\chi \in \Psi ^\textrm{tot}_\ell : \textrm{cond}(\chi )\le X\}\\ \Psi ^\prime _\ell (X)&= \{\chi \in \Psi ^\prime _\ell : \textrm{cond}(\chi )\le X\}. \end{aligned}$$

For an elliptic curve E over ${{\mathbb {Q}}}$ we also define:

$$\begin{aligned} {{\mathcal {F}}}_{\Psi _\ell ,E}&= \{L(E,s,\chi ):\, \chi \in \Psi _\ell \}\\ {{\mathcal {F}}}_{\Psi _\ell ,E}(X)&= \{L(E,s,\chi )\in {{\mathcal {F}}}_{\Psi _{\ell }, E}:\, \chi \in \Psi _\ell (X)\}. \end{aligned}$$

We also define ${{\mathcal {F}}}_{\Psi ^\textrm{tot}_\ell ,E}$ and ${{\mathcal {F}}}_{\Psi ^\textrm{tot}_\ell ,E}(X)$ analogously for $\Psi ^\textrm{tot}_\ell$ in place of $\Psi _\ell$; we do the same with $\Psi ^\prime _\ell$, as well. Finally, let

$$\begin{aligned} V_{\Psi _\ell ,E}(X)&= \{L(E,s,\chi ) \in {{\mathcal {F}}}_{\Psi _\ell ,E}(X):\, L(E,1,\chi ) = 0\}\\ V_{\Psi ^\textrm{tot}_\ell ,E}(X)&= \{L(E,s,\chi ) \in {{\mathcal {F}}}_{\Psi ^\textrm{tot}_\ell ,E}(X):\, L(E,1,\chi ) = 0\}\\ V_{\Psi ^\prime _\ell ,E}(X)&= \{L(E,s,\chi ) \in {{\mathcal {F}}}_{\Psi ^\prime _\ell ,E}(X):\, L(E,1,\chi ) = 0\}. \end{aligned}$$

With this notation, we make the following conjecture.

Conjecture 1

Let E be an elliptic curve over ${{\mathbb {Q}}}$.

• 1. There exist positive constants $b^\prime _{E,4}$ and $b^\prime _{E,6}$ such that

  $$\begin{aligned} |V_{\Psi _4^\prime ,E}(X)| \sim b^\prime _{E,4} X^{1/2} \log ^{-3/4} X \end{aligned}$$

  and

  $$\begin{aligned} |V_{\Psi _6^\prime ,E}(X)| \sim b^\prime _{E,6} X^{1/2} \log ^{-3/4} X \end{aligned}$$

  as $X \rightarrow \infty$.

• 2. Suppose that E is not isogenous to a curve with a rational point of order d with

  • $d=2$ in the quartic case

  • $d=2$ or $d=3$ in the sextic case.

  Then, there exist positive constants $b_{E,4}$ and $b_{E,6}$ so that

  $$\begin{aligned} |V_{\Psi _4,E}(X)| \sim b_{E,4} X^{1/2} \log ^{5/4} X \end{aligned}$$

  and

  $$\begin{aligned} |V_{\Psi _6,E}(X)| \sim b_{E,6} X^{1/2} \log ^{9/4} X \end{aligned}$$

  as $X \rightarrow \infty$.

• 3. Moreover, if we restrict only to those twists by totally quartic or totally sextic characters, then there exist positive constants $b^\textrm{tot}_{E,4}$ and $b^\textrm{tot}_{E,6}$ such that

  $$\begin{aligned} |V_{\Psi _4^\textrm{tot},E}(X)| \sim b^\textrm{tot}_{E,4} X^{1/2} \log ^{1/4} X \end{aligned}$$

  and

  $$\begin{aligned} |V_{\Psi _6^\textrm{tot},E}(X)| \sim b^\textrm{tot}_{E,6} X^{1/2} \log ^{1/4} X \end{aligned}$$

  as $X \rightarrow \infty$.

Remark 1

To assist the reader in comparing the powers of $\log {X}$ in the above asymptotics, we point out here that for $\ell = 4$, $|\Psi _{4}(X)|$ is roughly $\log X$ times as large as $|\Psi _{4}^{\textrm{tot}}(X)|$, which in turn is roughly $\log X$ times as large as $|\Psi _{4}'(X)|$. For $\ell = 6$, then $|\Psi _{6}(X)|/|\Psi _{6}^{\textrm{tot}}(X)| \asymp (\log X)^2$, and $|\Psi _{6}^{\textrm{tot}}(X)|/|\Psi _{6}'(X)| \asymp \log X$. Hence, in each of the three families with a given value of $\ell$, the proportion of vanishing twists has the same order of magnitude. See Proposition 7, Lemma 8, Proposition 9, and Lemma 10 below for asymptotics of the underlying families of characters.

In particular, we conjecture that families of elliptic curve L-functions twisted by quartic and sextic characters vanish infinitely often at the central value. The mild conditions placed on E for twists by characters of composite conductor are similar to those found in [10]. Roughly speaking, with each prime factor of the conductor of the twisting character, some extra divisibility in the discretization parameter might arise (see Sect. 2.1 for more information about the discretization). The conditions are not necessary for twists by characters of prime conductor because we can only gain at most an extra factor of some fixed integer, which should affect the constant term $b_{E,\ell }'$ but not the power of $\log X$.

1.2 Outline of the paper

There are two main ingredients needed to be able to apply random matrix theory predictions to our families of twists. The first is a discretization for the central values. As described in Sect. 2.1 this can be done for curves E satisfying certain technical conditions as described in [36]. We need this discretization in order to approximate the probability that $L(E,1,\chi )$ vanishes.

The second ingredient is a proper identification of the symmetry type of the family, which is largely governed by the distribution of the sign of the functional equation within the family (see Sect. 4 of [9]). This directly leads to an investigation around the equidistribution of squares of Gauss sums of quartic and sextic characters, which has connections to the theory of metaplectic automorphic forms [30]. See Sect. 3.1 for a thorough discussion.

It is a subtle feature that the families of twists of elliptic curve L-functions by the characters in $\Psi _{\ell }^{\textrm{tot}}$ and $\Psi _{\ell }'$ have unitary symmetry type, but for composite even values of $\ell$, the twists by $\Psi _{\ell }$ should be viewed as a mixed family. To elaborate on this point, consider the case that $\ell =4$, and first note that a character $\chi \in \Psi _{4}$ factors uniquely as a totally quartic character times a quadratic character of relatively prime conductors. The totally quartic family has a unitary symmetry, but the family of twists of an elliptic curve by quadratic characters has orthogonal symmetry. This tension between the totally quartic aspect and the quadratic aspect is what leads to the mixed symmetry type. The situation is analogous to the family $L(E, 1+it, \chi _d)$; if $t=0$ and d varies then one has an orthogonal family, while if d is fixed and t varies, then one has a unitary family. See [34] for more discussion on this family.

Another interesting feature of these families is that $\Psi _{\ell }(X)$ is larger than $\Psi _{\ell }^{\textrm{tot}}(X)$ by a logarithmic factor. For instance, when $\ell =4$, then $\Psi _{4}^{\textrm{tot}}(X)$ grows linearly in X (see Proposition 7 below), and of course $\Psi _{2}(X)$ also grows linearly in X. Similarly to how the average size of the divisor function is $\log {X}$, this indicates that $|\Psi _{4}(X)|$ grows like $X \log {X}$ (see Lemma 8 below).

The rest of the paper is organized as follows. In the next section we give the necessary background and notation for L-functions and their central values and discuss the discretization we use in the paper. In the subsequent section we estimate some sums involving quartic and sextic characters and discuss totally quartic and sextic characters in more detail. In the final section, we motivate the asymptotics in Conjecture 1 and provide numerical evidence that supports them.

2 L-functions and central values

Let E be an elliptic curve defined over ${{\mathbb {Q}}}$ of conductor $N_E$. The L-function of E is given by the Euler product

$$\begin{aligned} L(E,s) = \prod _{p\not \mid N_E} \left( 1 - \tfrac{a_p}{p^s} + \tfrac{1}{p^{2s-1}}\right) ^{-1} \prod _{p\mid N_E} \left( 1-\tfrac{a_p}{p^s} \right) ^{-1} = \sum _{n\ge 1} \frac{a_n}{n^s}. \end{aligned}$$

The modularity theorem [6, 35, 37] implies that L(E, s) has an analytic continuation to all of $\mathbb {C}$ and satisfies the functional equation

$$\begin{aligned} \Lambda (E,s) = \left( \tfrac{\sqrt{N_E}}{2\pi }\right) ^s \Gamma (s) L(E,s) = w_E\Lambda (E,2-s) \end{aligned}$$

where the sign of the functional equation is $w_E = \pm 1$ and is the eigenvalue of the Fricke involution. Let $\chi$ be a primitive character and let $\textrm{cond}(\chi )$ be its conductor and suppose that $\textrm{cond}(\chi )$ is coprime to the conductor $N_E$ of the curve. The twisted L-function has Dirichlet series

$$\begin{aligned} L(E,s,\chi ) = \sum _{n\ge 1} \frac{a_n\chi (n)}{n^s} \end{aligned}$$

and the functional equation (cf. [17, Prop. 14.20])

$$\begin{aligned} \Lambda (E,s,\chi )&= \left( \tfrac{\textrm{cond}(\chi )\sqrt{N_E}}{2\pi }\right) ^s \Gamma (s) L(E,s,\chi )\nonumber \\&= \tfrac{w_E \chi (N_E)\tau (\chi )^2}{\textrm{cond}(\chi )} \Lambda (E,2-s,\overline{\chi }), \end{aligned}$$

(1)

where $\tau (\chi )=\sum _{r\in {{\mathbb {Z}}}/m{{\mathbb {Z}}}}\chi (r) e^{2\pi i r/m}$ is the Gauss sum and $m = \textrm{cond}(\chi )$.

2.1 Discretization

To justify our Conjecture 1, we need a condition that allows us to deduce that $L(E,1,\chi )=0$, for a given E and $\chi$ of order $\ell$. In particular, we show that $L(E,1,\chi )$ is discretized (see Lemma 25) and so there exists a constant $c_{E,\ell }$ such that $|L(E,1,\chi )| < c_{E,\ell }/\sqrt{\textrm{cond}(\chi )}$ implies $L(E,1,\chi )=0$. In this section we prove the results necessary for the discretization.

Let E be an elliptic curve over ${{\mathbb {Q}}}$ with conductor $N_E$. Let $\chi$ be a nontrivial primitive Dirichlet character of conductor m and order $\ell$. Let $\Omega _{+}(E)$ and $\Omega _{-}(E)$ denote the real and imaginary periods of E, respectively, with $\Omega _{+}(E) > 0$ and $\Omega _{-}(E) \in i {{\mathbb {R}}}_{>0}$.

The algebraic L-value is defined by

$$\begin{aligned} L^{{{\,\textrm{alg}\,}}}(E,1, \chi ) := \frac{L(E,1,\chi ) \cdot m}{\tau (\chi ) \Omega _\pm {}(E)} = \chi (-1) \frac{L(E,1,\chi ) \tau (\overline{\chi })}{\Omega _\pm {}(E)}, \end{aligned}$$

(2)

where $\Omega _\pm {}(E)$ is determined by whether $\chi$ is an even or odd character, that is, by $\chi (-1)$.

While it has been known for some time that algebraic L-values are algebraic numbers, recent work of Weirsema and Wuthrich [36] characterizes conditions on E and $\chi$ which guarantee integrality. In particular, under the assumption that the Manin constant $c_0(E) =1$, if the conductor m is not divisible by any prime of additive reduction for E, then $L^{{{\,\textrm{alg}\,}}}(E,1,\chi ) \in {{\mathbb {Z}}}[\zeta _\ell ]$ is an algebraic integer [36, Theorem 2]. For a given curve E, we will avoid the finitely many characters $\chi$ for which $L^{{{\,\textrm{alg}\,}}}(E,1, \chi )$ fails to be integral.

The following propositions provide a discretization of the special values $L(E,1,\chi )$. These results are similar to the case of cubic characters considered in [10] since ${{\mathbb {Q}}}(\zeta _3)^+ = {{\mathbb {Q}}}$, as opposed to characters of prime order $\ell \ge 5$ where further steps were needed to find an appropriate discretization [11].

Proposition 2

Let $\chi$ be a primitive Dirichlet character of odd order $\ell$ and conductor m. Then

$$\begin{aligned} L^{{{\,\textrm{alg}\,}}}(E,1, \chi ) = {\left\{ \begin{array}{ll} \chi (N_E)^{(\ell +1)/2} \, n_E(\chi ), &{} \text {if}\, w_E = 1, \\ (\zeta _\ell - \zeta _\ell ^{-1})^{-1} \, \chi (N_E)^{(\ell +1)/2} \, n_E(\chi ) &{} \text {if}\, w_E = -1, \end{array}\right. } \end{aligned}$$

for some algebraic integer $n_E(\chi ) \in {{\mathbb {Z}}}[\zeta _\ell + \zeta _\ell ^{-1}] = {{\mathbb {Z}}}[\zeta _\ell ] \cap {{\mathbb {R}}}$.

Proposition 3

Let $\chi$ be a primitive Dirichlet character of even order $\ell$ and conductor m. Then $L^{{{\,\textrm{alg}\,}}}(E,1,\chi ) = k_E \, n_E(\chi )$ where $n_E(\chi )$ is some algebraic integer in ${{\mathbb {Z}}}[\zeta _\ell + \zeta _\ell ^{-1}] = {{\mathbb {Z}}}[\zeta _\ell ] \cap {{\mathbb {R}}}$ and $k_E$ is a constant depending only on the curve E. In particular, when $w_E = 1$ we have

$$\begin{aligned} k_E = {\left\{ \begin{array}{ll} (1 + \chi (N_E)) &{} \text {if}\, \chi (N_E) \ne -1 \\ \zeta _\ell ^{\ell /4}, &{} \text {if}\, 4 \mid \ell \, \text {and}\, \chi (N_E) = -1 \\ (\zeta _\ell - \zeta _\ell ^{-1}) &{} \text {if}\, 4 \not \mid \ell \, \text {and}\, \chi (N_E) = -1. \end{array}\right. } \end{aligned}$$

Proof of Prop 2 and Prop 3

Since E is defined over ${{\mathbb {Q}}}$, we have $\overline{L(E,1,\chi )} = L(E,1, \overline{\chi })$. Using the functional equation and the fact that $\overline{\tau (\chi )} = \chi (-1) \tau (\overline{\chi })$, we obtain

$$\begin{aligned} L^{{{\,\textrm{alg}\,}}}(E,1,\chi )&= \chi (-1) \cdot \frac{L(E,1,\chi ) \tau (\overline{\chi })}{ \Omega _\pm {}(E)} \\&= \chi (-1) \cdot \frac{w_E \, \chi (N_E) \, \tau (\overline{\chi }) \tau (\chi )^2}{m \cdot \Omega _\pm {}(E)} L(E,1,\overline{\chi }) \\&= \frac{w_E \, \chi (N_E) \, \tau (\chi )}{\Omega _\pm {}(E)}L(E,1,\overline{\chi }) \\&= w_E \chi (N_E) \, \overline{ \frac{\chi (-1) \, \tau (\overline{\chi }) L(E,1, \chi )}{\Omega _\pm {}(E)}} \\&= w_E \chi (N_E) \, \overline{L^{{{\,\textrm{alg}\,}}}(E,1,\chi )} \end{aligned}$$

Thus $L^{{{\,\textrm{alg}\,}}}(E,1,\chi )$ is a solution to the equation $z = w_E \chi (N_E) \overline{z}$. Note that if $z_1, z_2 \in {{\mathbb {Z}}}[\zeta _\ell ]$ are two distinct solutions to this equation, then $z_1/\overline{z}_1 = z_2/\overline{z}_2$ so that $z_1/z_2 = \overline{z}_1/\overline{z}_2 = \overline{(z_1/z_2)}$, hence $z_1/z_2 \in {{\mathbb {R}}}$. Thus $L^{{{\,\textrm{alg}\,}}}(E,1,\chi ) = \alpha z$ with $\alpha \in {{\mathbb {Z}}}[\zeta _\ell ] \cap {{\mathbb {R}}}= {{\mathbb {Z}}}[\zeta _\ell + \zeta _\ell ^{-1}]$ and $z \in {{\mathbb {Z}}}[\zeta _\ell ]$.

Suppose that $w_E =1$. When $\ell$ is odd, we can take $z = \chi (N_E)^\frac{\ell +1}{2}$. Now suppose that $\ell$ is even. If $\chi (N_E) \ne -1$, since $\chi (N_E) = \zeta _\ell ^r$ for some $1 \le r \le \ell$, we may take $z = (1 + \chi (N_E))$. Indeed, we have $w_E \chi (N_E) \overline{z}=\zeta _\ell ^r(1 + \zeta _\ell ^{\ell -r}) = \zeta _\ell ^r + 1 = z$. If $4 \mid \ell$ and $\chi (N_E) = -1 = \zeta _\ell ^{\ell /2}$, we take $z = \zeta _\ell ^{\ell /4}$. Finally, if $4 \not \mid \ell$ and $\chi (N_E) = -1$ take $z = \zeta _\ell - \zeta _\ell ^{-1} = 2i \, \text {Im}(\zeta _\ell )$.

When $w_E = -1$ and $\ell$ is odd, we may take $z = (\zeta _\ell - \zeta _\ell ^{-1})^{-1} \chi (N_E)^\frac{\ell +1}{2}$. When $\ell$ is even, if $\chi (N_E) = -1$ then we may take $z = \zeta _\ell + \zeta _\ell ^{-1} = 2 {\text {Re}}(\zeta _\ell )$, and if $\chi (N_E) \ne -1$ then we make take $z = 1 - \chi (N_E)$. $\square$

Remark 2

We note that for $\ell$ even, $|k_E| \le 2$. It is clear that $|\zeta _\ell ^{\ell /4}| = 1$ and $|2i \, \text {Im}(\zeta _\ell )| \le 2$. Observe $|(1 + \chi (N_E)| \le 2$, by the triangle inequality.

Since $L(E,1,\chi )$ vanishes if and only if $n_E(\chi )$ does, we may interpret the integers $n_E(\chi )$ as a discretization of $L(E,1,\chi )$. Note that in [10] it was conjectured that if E is isogenous to a curve with a 3-torsion point and $\chi$ is a cubic character, the integer $n_E(\chi )$ is likely to be divisible by a power of 3. Since these powers of 3 could change the power of the log in Conjecture 12, we added the hypothesis restricting the isogeny class of E.

3 Estimates for Dirichlet characters

In this section we discuss various aspects of Dirichlet characters of order 4 and 6. A necessary condition for a family of L-functions to be modeled by the family of unitary matrices is that the signs must be uniformly distributed on the unit circle (see, e.g., [12]). From (1), $L(E, s, \chi )$ has sign $w_E \chi (N_{E}) \frac{\tau (\chi )^2}{\textrm{cond}(\chi )}$; we will largely focus on the distribution of the square of the Gauss sums, since the techniques may handle the extra factor $\chi (N_E)$ without difficulty (see Sect. 3.1 and the comments after Corollary 20 for more details). To obtain our estimates for the number of vanishings $|V_{\Psi _\ell ,E}(X)|$ (respectively, $|V_{\Psi _\ell ^\prime ,E}(X)|$ and $|V_{\Psi _\ell ^\textrm{tot},E}(X)|$) we must estimate the size of $\Psi _\ell (X)$ (respectively, $\Psi _\ell ^\prime (X)$ and $\Psi _\ell ^\textrm{tot}(X)$) as well as the size of an associated sum. We also discuss the family of totally quartic and sextic characters to explain some phenomena we observed in our computations.

3.1 Distributions of Gauss sums

In view of (1), the distribution of the root number of $L(E, s, \chi )$ is largely controlled by the distribution of $\tau (\chi )^2$. Indeed, the techniques for understanding the latter problem naturally generalize to handle the former, so we will focus on the distribution of squares of Gauss sums.

Patterson [30], building on work of Heath-Brown and Patterson [15] on the cubic case, showed that the normalized Gauss sum $\tau (\chi )/\sqrt{\textrm{cond}(\chi )}$ is uniformly distributed on the circle for $\chi$ varying in each of $\Psi _{\ell }^{\textrm{tot}}$ and $\Psi _{\ell }'$. This result was first announced in [28]; see [4] for an excellent summary of this and other work related to the distributions of Gauss sums. Patterson’s method moreover shows that the argument of $\tau (\chi ) \chi (k)$ is equidistributed for any fixed nonzero integer k, and hence so is the argument of $\tau (\chi )^2 \chi (k)$.

For the case of quartic and sextic characters with arbitrary conductors, there do not appear to be any results in the literature that imply their Gauss sums are uniformly distributed. In Fig. 1 we see the distributions of Gauss sums of characters of orders 3 through 9 of arbitrary conductor up to 200000. We included characters of order 4 and 6 since those examples are the focus of the paper; we included characters of orders 3, 5, and 7 as consistency checks (in [10, 11] the authors rely on them being uniformly distributed); and we included composite orders 8 and 9 to see if something similar happens in those cases as happens in the quartic case. In all cases but the quartic case, we see that the distributions of the angles of the signs appear to be uniformly distributed. The quartic distribution has two obvious peaks that we discuss below, in Remark 4.

Fig. 1

Each histogram represents the distribution of the argument of the $\tau (\chi )^2/\textrm{cond}(\chi )$ for characters of order 3 through 9, from top left to bottom right. Each histogram is made by calculating the Gauss sums of characters in $\Psi _{\ell }$ of each conductor up to 200000. The argument of $\tau (\chi )^2/\textrm{cond}(\chi )$ ranges from $-\pi$ to $\pi$ and this range has been divided into 10 bins

The images in Fig. 1 suggest that the family of matrices that best models the vanishing of $L(E,1,\chi )$ is unitary in every case except possibly the case of quartic characters. Nevertheless, in Sect. 3.4 we show that the squares of the quartic Gauss sums are indeed equidistributed, despite what the data suggest. Indeed, we prove that the squares of the sextic and quartic Gauss sums are equidistributed, allowing us to apply the heuristics from random matrix theory as in Sect. 4.

3.2 Totally quartic and sextic characters

Much of the background material in this section can be found with proofs in [16, Ch. 9].

Definition 1

Let $\chi$ be a primitive Dirichlet character of conductor q and order $\ell$. For prime p, let $v_p$ be the p-adic valuation, so that $q = \prod _{p} p^{v_p(q)}$. We correspondingly factor $\chi = \prod _{p} \chi ^{(p)}$, where $\chi ^{(p)}$ has conductor $p^{v_p(q)}$. We say that $\chi$ is totally order $\ell$ if each $\chi ^{(p)}$ is exact order $\ell$. By convention we also consider the trivial character to be totally order $\ell$ for every $\ell$.

Remark 3

In the proofs and theorems below, we will often be considering factorizations of characters into products of characters. The notational conventions we have adopted from here on are:

• $\chi _2,\chi _3,\chi _4,\chi _6$ indicate that the character is totally order 2, 3, 4, 6, respectively.

• otherwise a subscript on a character will either be a primary prime (see Sects. 3.2.1 and 3.2.2) or a product of primary primes.

• if we need to identify both the order and the conductor we write the conductor as a superscript: e.g., $\chi _2^{(p)}$ would be a quadratic character of conductor p.

• we use $\psi$ for a character whose order or conductor is not essential to the context.

3.2.1 Quartic characters

The construction of quartic Dirichlet characters uses the arithmetic in ${\mathbb {Z}}[i]$. The ring ${\mathbb {Z}}[i]$ has class number 1, unit group $\{ \pm 1, \pm i \}$, and discriminant $-4$. We say $\alpha \in {\mathbb {Z}}[i]$ with $(\alpha , 2) = 1$ is primary if $\alpha \equiv 1 \pmod {(1+i)^3}$. Any odd element in ${\mathbb {Z}}[i]$ has a unique primary associate, which comes from the fact that the unit group in the ring ${\mathbb {Z}}[i]/(1+i)^3$ may be identified with $\{ \pm 1, \pm i\}$. An odd prime p splits as $p=\pi \overline{\pi }$ if and only if $p \equiv 1 \pmod {4}$. Given $\pi$ with $N(\pi ) = p$, define the quartic residue symbol $[\frac{\alpha }{\pi }]$ for $\alpha \in {\mathbb {Z}}[i]$ with $(\alpha , \pi ) = 1$, by $[\frac{\alpha }{\pi }] \in \{ \pm 1, \pm i\}$ and $[\frac{\alpha }{\pi }] \equiv \alpha ^{\frac{p-1}{4}} \pmod {\pi }$. The map $\chi _{\pi }(\alpha ) = [\frac{\alpha }{\pi }]$ from $({\mathbb {Z}}[i]/(\pi ))^{\times }$ to $\{ \pm 1, \pm i\}$ is a character of order 4. If $\alpha \in {\mathbb {Z}}$, then $[\frac{\alpha }{\pi }]^2 \equiv \alpha ^{\frac{p-1}{2}} \equiv (\frac{\alpha }{p}) \pmod {\pi }$. Therefore, $\chi _{\pi }^2(\alpha ) = (\frac{\alpha }{p})$, showing in particular that the restriction of the quartic residue symbol to ${\mathbb {Z}}$ defines a primitive quartic Dirichlet character of conductor p.

Lemma 4

Every primitive totally quartic Dirichlet character of odd conductor is of the form $\chi _{\beta }$, where $\beta = \pi _1 \dots \pi _k$ is a product of distinct primary primes, $(\beta , 2 \overline{\beta }) = 1$, and where

$$\begin{aligned} \chi _{\beta }(\alpha ) = \Big [\frac{\alpha }{\beta }\Big ] = \prod _{i=1}^{k} \Big [\frac{\alpha }{\pi _i}\Big ]. \end{aligned}$$

(3)

Each totally quartic primitive character of even conductor is of the form $\psi \chi _{\beta }$ where $\psi$ is one of four quartic characters of conductor $2^4$, and $\chi _{\beta }$ is totally quartic of odd conductor as in (3).

Proof

We begin by classifying the quartic Dirichlet characters of odd prime-power conductor. If $p \equiv 3 \pmod {4}$, there is no quartic character of conductor $p^a$, since $\phi (p^a) = p^{a-1}(p-1) \not \equiv 0 \pmod {4}$. Since $\phi (p) = p-1$, if $p \equiv 1 \pmod {4}$, there are two distinct quartic characters of conductor p, namely, $\chi _{\pi }$ and $\chi _{\overline{\pi }}$, where $p = \pi \overline{\pi }$. There are no primitive quartic characters modulo $p^j$ for $j \ge 2$. To see this, suppose $\chi$ is a character of conductor $p^j$, and note that $\chi (1+p^{j-1}) \ne 1$, while $\chi (1+p^{j-1})^p = \chi (1+p^j) = 1$, so $\chi (1+p^{j-1})$ is a nontrivial pth root of unity. Since p is odd, $\chi (1+p^{j-1})$ is not a 4th root of unity, so $\chi$ cannot be quartic and primitive.

By the above classification, a primitive totally quartic Dirichlet character $\chi$ of odd conductor must factor over distinct primes $p_i \equiv 1 \pmod {4}$, and the p-part of $\chi$ must be $\chi _{\pi }$ or $\chi _{\overline{\pi }}$, where $\pi \overline{\pi } = p$. We may assume that $\pi$ and $\overline{\pi }$ are primary primes. Hence $\chi$ factors as $\prod _i \chi _{\pi _i}$. The property that $\beta := \pi _1 \dots \pi _k$ is squarefree (meaning, squarefree as an element in ${\mathbb {Z}}[i]$) is equivalent to the condition that the $\pi _i$ are distinct. Moreover, the property $(\beta , \overline{\beta }) = 1$ means that for each $p_i$, precisely one of $\pi _i$ and $\overline{\pi _i}$ occurs in $\beta$. Hence, every quartic character of odd conductor arises uniquely in the form (3).

Next we treat $p=2$. There are four primitive quartic characters of conductor $2^4$, since $({\mathbb {Z}}/(2^4))^{\times } \simeq {\mathbb {Z}}/(2) \times {\mathbb {Z}}/(4)$. We claim there are no primitive quartic characters of conductor $2^j$, with $j \ne 4$. For $j \le 3$ or $j=5$ this is a simple finite computation. For $j \ge 6$, one can show this as follows. First, $\chi (1+2^{j-1}) = -1$, since $\chi ^2(1+2^{j-1}) = \chi (1+2^j) = 1$, and primitivity shows $\chi (1+2^{j-1}) \ne 1$. By a similar idea, $\chi (1+2^{j-2})^2 = \chi (1+2^{j-1}) = -1$, so $\chi (1+2^{j-2}) = \pm i$. We finish the claim by noting $\chi ^2(1+2^{j-3}) = \chi (1+2^{j-2}) = \pm i$, so $\chi (1+2^{j-3})$ is a square-root of $\pm i$, and hence $\chi$ is not quartic. With the claim established, we easily obtain the final sentence of the lemma. $\square$

Example 1

The first totally quartic primitive character of composite conductor has conductor 65. While there are 8 quartic primitive characters of conductor 65, the LMFDB labels of the totally quartic ones are 65.18, 65.47, 65.8, and 65.57.

3.2.2 Sextic characters

The construction of sextic Dirichlet characters uses the arithmetic in the Eisenstein integers ${\mathbb {Z}}[\omega ]$, where $\omega = e^{2 \pi i/3}$. The ring ${\mathbb {Z}}[\omega ]$ has class number 1, unit group $\{ \pm 1, \pm \omega , \pm \omega ^2 \}$, and discriminant $-3$. We say $\alpha \in {\mathbb {Z}}[\omega ]$ with $(\alpha , 3) = 1$ is primary¹ if $\alpha \equiv 1 \pmod { 3}$. Warning: our usage of primary is consistent with [15], but conflicts with the definition of [16]. However, it is easy to translate since $\alpha$ is primary in our sense if and only if $-\alpha$ is primary in the sense of [16]. Any element in ${\mathbb {Z}}[\omega ]$ coprime to 3 has a unique primary associate, which comes from the fact that the unit group in the ring ${\mathbb {Z}}[\omega ]/(3)$ may be identified with $\{ \pm 1, \pm \omega , \pm \omega ^2 \}$. An unramified prime $p \in {\mathbb {Z}}$ splits as $p=\pi \overline{\pi }$ if and only if $p \equiv 1 \pmod {3}$. Given $\pi$ with $N(\pi ) = p$, define the cubic residue symbol $(\frac{\alpha }{\pi })_3$ for $\alpha \in {\mathbb {Z}}[\omega ]$ by $(\frac{\alpha }{\pi })_3 \in \{1, \omega , \omega ^2 \}$ and $(\frac{\alpha }{\pi })_3 \equiv \alpha ^{\frac{p-1}{3}} \pmod {\pi }$. The map $\chi _{\pi }(\alpha ) = (\frac{\alpha }{\pi })_3$ from $({\mathbb {Z}}[\omega ]/(\pi ))^{\times }$ to $\{ 1, \omega , \omega ^2 \}$ is a character of order 3. The restriction of $\chi _{\pi }$ to ${\mathbb {Z}}$ induces a primitive cubic Dirichlet character of conductor p. Note that $\chi _{\pi } = \chi _{-\pi }$.

Motivated by the fact that a sextic character factors as a cubic times a quadratic, we next discuss the classification of cubic characters.

Lemma 5

Every primitive cubic Dirichlet character of conductor coprime to 3 is of the form $\chi _{\beta }$, where $\beta = \pi _1 \dots \pi _k$ is a product of distinct primary primes, $(\beta , 3 \overline{\beta }) = 1$, and where

$$\begin{aligned} \chi _{\beta }(\alpha ) = \Big (\frac{\alpha }{\beta }\Big )_3 = \prod _{i=1}^{k} \Big (\frac{\alpha }{\pi _i}\Big )_3. \end{aligned}$$

(4)

The cubic primitive characters of conductor divisible by 3 are of the form $\psi \chi _{\beta }$ where $\psi$ is one of two cubic characters of conductor $3^{2}$, and $\chi _{\beta }$ is cubic of conductor coprime to 3.

Proof

The classification of such characters with conductor coprime to 3 is given by [1, Lemma 2.1], so it only remains to treat cubic characters of conductor $3^j$. The primitive character of conductor 3 is not cubic. Next, the group $({\mathbb {Z}}/(9))^{\times }$ is cyclic of order 6, generated by 2. There are two cubic characters, determined by $\chi (2) = \omega ^{\pm 1}$. Next we argue that there is no primitive cubic character of conductor $3^j$ with $j \ge 3$. For this, we first observe that $\chi (1+3^{j-1}) = \omega ^{\pm 1}$, since primitivity implies $\chi (1+3^{j-1}) \ne 1$, and $\chi (1+3^{j-1})^3 = \chi (1+3^j) = 1$. Next we have $\chi (1+3^{j-2})^3 = \chi (1+3^{j-1}) = \omega ^{\pm 1}$, so $\chi (1+3^{j-2})$ is a cube-root of $\omega ^{\pm 1}$. Therefore, $\chi$ cannot be cubic. $\square$

3.3 Counting characters

To start, we count all the quartic and sextic Dirichlet characters of conductor up to some bound and in each family. Such counts were found for arbitrary order in [13] by Finch, Martin and Sebah, but since we are interested in only quartic and sextic characters, in which case the proofs simplify, we prove the results we need. Moreover, we need other variants for which we cannot simply quote [13], so we will develop a bit of machinery that will be helpful for these other questions as well. Similar counts have appeared throughout the literature: see, for example [21] and [26].

We begin with a lemma based on the Perron formula.

Lemma 6

Suppose that a(n) is a multiplicative function such that $|a(n)| \le d_k(n)$, the k-fold divisor function, for some $k \ge 0$. Let $Z(s) = \sum _{n \ge 1} a(n) n^{-s}$, for $\textrm{Re}(s) > 1$. Suppose that for some integer $j \ge 0$, $(s-1)^j Z(s)$ has a analytic continuation to a region of the form $\{\sigma + it: \sigma > 1-\frac{c}{\log (2+|t|)} \}$, for some $c>0$. In addition, suppose that Z(s) is bounded polynomially in $\log {(2+|t|)}$ in this region. Then

$$\begin{aligned} \sum _{n \le X} a(n) = X P_{j-1}(\log {X}) + O(X (\log X)^{-100}), \end{aligned}$$

(5)

for $P_{j-1}$ some polynomial of degree $\le {j-1}$ (interpreted as 0, if $j=0$).

The basic idea is standard, yet we were unable to find a suitable reference.

Proof sketch

One begins by a use of the quantitative Perron formula, for which a convenient reference is [25, Thm. 5.2]. This implies

$$\begin{aligned} \sum _{n \le X} a(n) = \frac{1}{2 \pi i} \int _{\sigma _0 - iT}^{\sigma _0 + iT} Z(s) X^s \frac{ds}{s} + R, \end{aligned}$$

(6)

where R is a remainder term, and we take $\sigma _0 = 1 + \frac{c}{\log {X}}$. Using [25, Cor. 5.3] and standard bounds on mean values of $d_k(n)$, one can show $R \ll \frac{X}{T} \textrm{Poly}(\log {X})$. Next one shifts the contour of integration to the line $1- \frac{c/2}{\log {T}}$. The pole (if it exists) of Z(s) leads to a main term of the form $X P_{j-1}(\log {X})$, as desired. The new line of integration is bounded by

$$\begin{aligned} \textrm{Poly}(\log {T}) X^{1-\frac{c/2}{\log {T}}}. \end{aligned}$$

(7)

Choosing $\log {T} = (\log X)^{1/2}$ gives an acceptable error term. $\square$

3.3.1 Quartic characters

Let $\Psi _{4}^{\textrm{tot}, \text {odd}}(X) \subseteq \Psi _4^{\textrm{tot}}(X)$ denote the subset of characters with odd conductor.

Proposition 7

For some constants $K^\textrm{tot}_4, K_4^{\textrm{tot},\textrm{odd}} > 0$, we have

$$\begin{aligned} | \Psi _4^{\textrm{tot}}(X)| \sim K_4^\textrm{tot}X, \qquad \text {and} \qquad | \Psi _4^{\textrm{tot}, \textrm{odd}}(X)| \sim K_4^{\textrm{tot},\textrm{odd}} X. \end{aligned}$$

(8)

Moreover,

$$\begin{aligned} | \Psi _4'(X)| \sim \frac{X}{\log {X}}. \end{aligned}$$

(9)

Proof

By Lemma 4,

$$\begin{aligned} | \Psi _4^{\textrm{tot}, \text {odd}}(X)| = \sum _{\begin{array}{c} 0 \ne (\beta ) \subseteq {\mathbb {Z}}[i] \\ (\beta , 2\overline{\beta }) = 1\\ \beta \text { squarefree} \\ N(\beta ) \le X \end{array}} 1, \end{aligned}$$

(10)

and

$$\begin{aligned} | \Psi _4^{\textrm{tot}}(X)| =| \Psi _4^{\textrm{tot}, \text {odd}}(X)| + 4 | \Psi _4^{\textrm{tot}, \text {odd}}(2^{-4} X)|. \end{aligned}$$

(11)

To show (8), it suffices to prove the asymptotic formula for $| \Psi _4^{\textrm{tot}, \text {odd}}(X)|$. In view of Lemma 6, it will suffice to understand the Dirichlet series

$$\begin{aligned} Z_4(s) = \sum _{\begin{array}{c} 0 \ne (\beta ) \subseteq {\mathbb {Z}}[i] \\ (\beta , 2\overline{\beta }) = 1\\ \beta \text { squarefree} \end{array}} \frac{1}{N(\beta )^s} = \prod _{\begin{array}{c} \pi \ne \overline{\pi } \\ (\pi , 2)=1 \end{array}} ( 1+ N(\pi )^{-s}) = \prod _{p \equiv 1 \pmod {4}} (1+p^{-s})^2. \end{aligned}$$

(12)

Let $\psi$ be the primitive character modulo 4, so that $\zeta (s) L(s, \psi ) = \zeta _{{\mathbb {Q}}[i]}(s)$. Then

$$\begin{aligned} Z_4(s) = \zeta _{{\mathbb {Q}}[i]}(s) \prod _p (1-p^{-s})(1-\psi (p)p^{-s}) \prod _{p \equiv 1 \pmod {4}} (1+p^{-s})^2, \end{aligned}$$

(13)

which can be simplified as

$$\begin{aligned} Z_4(s) = \zeta _{{\mathbb {Q}}[i]}(s) \zeta ^{-1}(2s) (1+2^{-s})^{-1} \prod _{p \equiv 1 \pmod {4}}(1-p^{-2s}). \end{aligned}$$

(14)

Therefore, $Z_4(s)$ has a simple pole at $s=1$, and its residue is a positive constant. Moreover, the standard analytic properties of $\zeta _{{\mathbb {Q}}[i]}(s)$ let us apply Lemma 6, giving the result.

The asymptotic on $\Psi _4'(X)$ follows from the prime number theorem in arithmetic progressions, since there are two quartic characters of prime conductor $p \equiv 1 \pmod {4}$, and none with $p \equiv 3 \pmod {4}$. $\square$

Lemma 8

We have

$$\begin{aligned} |\Psi _4(X)| = K_4 X \log {X} + O(X), \end{aligned}$$

(15)

for some $K_4 > 0$

Proof

Every primitive quartic Dirichlet character factors uniquely as $\chi _4 \chi _2$ with $\chi _4$ totally quartic of conductor $q_4 > 1$ and $\chi _2$ quadratic of conductor $q_2$, with $(q_4, q_2) = 1$. It is convenient to drop the condition $q_4 > 1$, thereby including the quadratic characters; this is allowable since the number of quadratic characters is O(X), which is acceptable for the claimed error term.

The Dirichlet series for $|\Psi _4(X)|$, modified to include the quadratic characters, is

$$\begin{aligned} Z_{4}^{\text {all}}(s) = \sum _{\begin{array}{c} 0 \ne (\beta ) \subseteq {\mathbb {Z}}[i] \\ (\beta , 2\overline{\beta }) = 1\\ \beta \text { squarefree} \end{array}} \frac{1}{N(\beta )^s} \sum _{\begin{array}{c} q_2 \in {\mathbb {Z}}_{\ge 1} \\ (q_2, 2N(\beta )) = 1 \end{array}} \frac{1}{q_2^s}. \end{aligned}$$

(16)

A calculation with Euler products shows $Z_{4}^{\text {all}}(s) = \zeta _{{\mathbb {Q}}[i]}(s) \zeta (s) A(s)$, where A(s) is given by an absolutely convergent Euler product for $\textrm{Re}(s) > 1/2$. Since $Z_{4}^{\text {all}}(s)$ has a double pole at $s=1$, this shows the claim, using Lemma 6. $\square$

3.3.2 Sextic characters

Next we turn to the sextic case. The proof of the following proposition is similar to the proof of Proposition 7 and so we omit it here.

Proposition 9

For some $K^\textrm{tot}_6 > 0$, we have

$$\begin{aligned} |\Psi _{6}^{\text {tot}}(X)| \sim K_6^\textrm{tot}X, \qquad \text {and} \qquad |\Psi _{6}'(X)| \sim \frac{X}{\log {X}}. \end{aligned}$$

(17)

A primitive totally sextic Dirichlet character factors uniquely as a primitive cubic character (with odd conductor, since $2 \not \equiv 1 \pmod {3}$), times the Jacobi symbol of the same modulus as the cubic character. In general, a primitive sextic character factors uniquely as $\chi _6 \chi _3 \chi _2$ of modulus $q_6 q_3 q_2$, pairwise coprime, with $\chi _6$ totally sextic of conductor $q_6$, $\chi _3$ cubic of conductor $q_3$, and $\chi _2$ quadratic of conductor $q_2$.

Lemma 10

We have $|\Psi _{6}(X)| = K_6 X (\log {X})^{2} + O(X \log {X})$, for some $K_6 > 0$.

Proof

Write $\chi = \chi _6 \chi _3 \chi _2$ as above. Note that membership in $\Psi _6(X)$ requires $q_6 > 1$, which is an unpleasant condition when working with Euler products. However, the number of $\chi = \chi _3 \chi _2$, i.e., with $\chi _6 = 1$, is $O(X \log {X})$, so we may drop the condition $q_6 > 1$ when estimating $|\Psi _6(X)|$.

For simplicity, we count the characters with $q_2$ odd and $(q_6 q_3, 3) = 1$; the general case follows similar lines. The Dirichlet series for this counting function is

$$\begin{aligned} Z_{6}^{\text {all}}(s) = \sum _{\begin{array}{c} 0 \ne (\beta _6) \subseteq {\mathbb {Z}}[\omega ] \\ (\beta _6, 3\overline{\beta _6}) = 1\\ \beta _6 \text { squarefree} \end{array}} \frac{1}{N(\beta _6)^s} \sum _{\begin{array}{c} 0 \ne (\beta _3) \subseteq {\mathbb {Z}}[\omega ] \\ (\beta _3, 3\overline{\beta _3}) = 1\\ \beta _3 \text { squarefree} \\ (N(\beta _3), N(\beta _6) = 1 \end{array}} \frac{1}{N(\beta _3)^s} \sum _{\begin{array}{c} q_2 \in {\mathbb {Z}}_{\ge 1} \\ (q_2, 2N(\beta _3 \beta _6)) = 1 \end{array}} \frac{1}{q_2^s}. \end{aligned}$$

A calculation with Euler products shows $Z_{6}^{\text {all}}(s) = \zeta _{{\mathbb {Q}}[\omega ]}(s)^2 \zeta (s) A(s)$, where A(s) is given by an absolutely convergent Euler product for $\textrm{Re}(s) > 1/2$. Since $Z_{6}^{\text {all}}(s)$ has a triple pole at $s=1$, this shows the claim, using Lemma 6. $\square$

3.4 Equidistribution of Gauss sums

We first focus on the quartic case, and then turn to the sextic case.

3.4.1 Quartic characters

The following standard formula can be found as [17, (3.16)].

Lemma 11

Suppose that $\psi = \psi _1 \psi _2$ has conductor $q = q_1 q_2$, with $(q_1, q_2) = 1$, and $\psi _i$ of conductor $q_i$. Then

$$\begin{aligned} \tau (\psi _1 \psi _2) = \psi _2(q_1) \psi _1(q_2) \tau (\psi _1) \tau (\psi _2). \end{aligned}$$

(18)

Corollary 12

Suppose that $\psi = \psi _1 \psi _2$ has odd conductor $q = q_1 q_2$, with $(q_1, q_2) = 1$, and $\psi _i$ of conductor $q_i$. Then

$$\begin{aligned} \tau (\psi _1 \psi _2)^2 = \tau (\psi _1)^2 \tau (\psi _2)^2. \end{aligned}$$

(19)

Proof

By Lemma 11, we will obtain the formula provided $\psi _2^2(q_1) \psi _1^2(q_2) = 1$. Note that $\psi _i^2$ is the Jacobi symbol, so $\psi _2^2(q_1) \psi _1^2(q_2) = (\frac{q_1}{q_2})(\frac{q_2}{q_1}) =1$, by quadratic reciprocity, using that $q_1 \equiv q_2 \equiv 1 \pmod {4}$. $\square$

Lemma 13

Suppose $\pi \in {\mathbb {Z}}[i]$ is a primary prime, with $N(\pi ) = p \equiv 1 \pmod {4}$. Let $\chi _{\pi }(x) = [\frac{x}{\pi }]$ be the quartic residue symbol. Then

$$\begin{aligned} \tau (\chi _{\pi })^2 = - \chi _{\pi }(-1) \sqrt{p} \pi . \end{aligned}$$

(20)

More generally, if $\beta$ is primary, squarefree, with $(\beta , 2 \overline{\beta }) = 1$, then

$$\begin{aligned} \tau (\chi _{\beta })^2 = \mu (\beta ) \chi _{\beta }(-1) \sqrt{N(\beta )} \beta , \end{aligned}$$

(21)

where $\mu (\pi _1 \dots \pi _k) = (-1)^k$ is the Möbius function defined on primary squarefree elements of ${\mathbb {Z}}[i]$.

Proof

The formula for $\chi _{\pi }$ follows from [16, Thm.1 (Chapter 8), Prop. 9.9.4]. The formula for general $\beta$ follows from Corollary 12 and Lemma 4. $\square$

Lemma 14

Suppose that $\chi = \chi _2 \chi _4$ is a primitive quartic character with odd conductor q, with $\chi _2$ quadratic of conductor $q_2$, $\chi _4$ totally quartic of conductor $q_4$, and with $q_2 q_4 = q$.

Then

$$\begin{aligned} \tau (\chi )^2 = \Big (\frac{-q_4}{q_2}\Big ) q_2 \tau (\chi _4)^2. \end{aligned}$$

(22)

Proof

By Lemma 11, we have $\tau (\chi )^2 = \chi _2(q_4)^2 \chi _4(q_2)^2 \tau (\chi _2)^2 \tau (\chi _4)^2$. To simplify this, note $\chi _2(q_4)^2 = 1$, $\chi _4^2(q_2) = (\frac{q_2}{q_4}) = (\frac{q_4}{q_2})$, and $\tau (\chi _2)^2 = \epsilon _{q_2}^2 q_2 = (\frac{-1}{q_2}) q_2$. $\square$

Our next goal is to express $\tau (\chi _{\beta })^2$ in terms of a Hecke Grossencharacter. Define

$$\begin{aligned} \lambda _{\infty }(\alpha ) = \frac{\alpha }{|\alpha |}, \qquad \alpha \in {\mathbb {Z}}[i], \alpha \ne 0. \end{aligned}$$

(23)

Next define a particular character $\lambda _{1+i} : R^{\times } \rightarrow S^1$, where $R = {\mathbb {Z}}[i]/(1+i)^3$, by

$$\begin{aligned} \lambda _{1+i}(i^k) = i^{-k}, \qquad k \in \{0,1,2,3 \}. \end{aligned}$$

(24)

This indeed defines a character since $R^{\times } \simeq {\mathbb {Z}}/4{\mathbb {Z}}$, generated by i. For $\alpha \in {\mathbb {Z}}[i]$, $(\alpha , 1+i) = 1$, define

$$\begin{aligned} \lambda ((\alpha )) = \lambda _{1+i}(\alpha ) \lambda _{\infty }(\alpha ). \end{aligned}$$

(25)

For this to be well-defined, we need that the right hand side of (25) is constant on units in ${\mathbb {Z}}[i]$. This is easily seen, since $\lambda _{\infty }(i^k) = i^k = \lambda _{1+i}(i^k)^{-1}$. Therefore, $\lambda$ defines a Hecke Grossencharacter, as in [17, Sect. 3.8]. Moreover, we note that

$$\begin{aligned} \frac{\tau (\chi _{\beta })^2}{N(\beta )} = \mu (\beta ) \Big (\frac{2}{N(\beta )} \Big ) \lambda ((\beta )) \end{aligned}$$

(26)

since this agrees with (21) for $\beta$ primary, and is constant on units.

According to [17, Theorem 3.8], the Dirichlet series

$$\begin{aligned} L(s, \lambda ^{k}) = \sum _{0 \ne (\beta ) \subseteq {\mathbb {Z}}[i]} \frac{\lambda ((\beta ))^{k}}{N(\beta )^s}, \qquad (k \in {\mathbb {Z}}), \end{aligned}$$

(27)

defines an L-function having analytic continuation to $s \in {\mathbb {C}}$ with no poles except for $k = 0$. The same statement holds when twisting $\lambda ^{k}$ by a finite-order character.

For $k \in {\mathbb {Z}}$, define the Dirichlet series

$$\begin{aligned} Z(k,s) = \sum _{\begin{array}{c} 0 \ne (\beta ) \subseteq {\mathbb {Z}}[i] \\ (\beta , 2\overline{\beta }) = 1\\ \beta \text { squarefree} \end{array}} \frac{(\tau (\chi _{\beta })^2/N(\beta ))^{k}}{N(\beta )^s}, \qquad \textrm{Re}(s) > 1. \end{aligned}$$

(28)

Proposition 15

Let $\delta _{k} = -1$ for k odd, and $\delta _{k} = +1$ for k even. We have

$$\begin{aligned} Z(k, s) = A(k,s) L(s, (\lambda \cdot \psi )^{k} )^{\delta _{k}}, \quad \text {where} \quad \psi (\beta ) = \Big (\frac{2}{N(\beta )}\Big ), \end{aligned}$$

(29)

and where A(k, s) is given by an Euler product absolutely convergent for $\textrm{Re}(s) > 1/2$.

In particular, the zero free region (as in [17, Theorem 5.35]) implies that Z(k, s) is analytic in a region of the type postulated in Lemma 6. Moreover, the proof of [25, Theorem 11.4] shows that Z(k, s) is bounded polynomially in $\log (2+|t|)$ in this region.

Proof

The formula (26) shows that Z(k, s) has an Euler product of the form

$$\begin{aligned} Z(k,s) = \prod _{(\pi ) \ne (\overline{\pi })} (1 + (-1)^{k} \frac{\psi ^{k}(\pi ) \lambda ^{k}((\pi )) }{N(\pi )^s}). \end{aligned}$$

(30)

This is an Euler product over the split primes in ${\mathbb {Z}}[i]$. We extend this to include the primes $p \equiv 3 \pmod {4}$ as well, with $N(\pi ) = p^2$. It is convenient to define $\psi (1+i) = 0$, so we can freely extend the product to include the ramified prime $1+i$. In all, we get

$$\begin{aligned} Z(k, s) = \Big [\prod _{\mathfrak {p}} (1 - \frac{\psi ^{k}(\mathfrak {p}) \lambda ^{k}(\mathfrak {p}) }{N(\mathfrak {p})^s}) \Big ]^{-\delta _{k}} \prod _{p} (1 + O(p^{-2s})). \end{aligned}$$

(31)

Note the product over $\mathfrak {p}$ is $L(s, (\lambda \cdot \psi )^{k})^{\delta _{k}}$, as claimed. $\square$

According to Weyl’s equidistribution criterion [17, Ch. 21.1], a sequence of real numbers $\theta _n$, $1 \le n \le N$ is equidistributed modulo 1 if and only if $\sum _{n \le N} e(k \theta _n) = o(N)$ for each integer $k \ne 0$. We apply this to $e(\theta _n) = (\tau (\chi )^2/q)$, whence $e(k \theta _n) = (\tau (\chi )^2/q)^k$. Due to the twisted multiplicativity formula (18), the congruence class in which 2k lies modulo $\ell$ may have a simplifying effect on $\tau (\chi )^{2k}$. For instance, when $\ell =4$, then k even leads to a simpler formula than k odd. This motivates treating these cases separately. As a minor simplification, below we focus on the sub-family of characters of odd conductor. The even conductor case is only a bit different.

Corollary 16

The Gauss sums $\tau (\chi )^2/q$ for $\chi$ totally quartic of odd conductor q, equidistribute on the unit circle.

Proof

The complex numbers $\tau (\chi )^2/q$ lie on the unit circle. Weyl’s equidistribution criterion says that these normalized squared Gauss sums equidistribute on the unit circle provided

$$\begin{aligned} \sum _{\begin{array}{c} 0 \ne (\beta ) \subseteq {\mathbb {Z}}[i] \\ (\beta , 2\overline{\beta }) = 1\\ \beta \text { squarefree} \\ N(\beta ) \le X \end{array}} (\tau (\chi _{\beta })^2/N(\beta ))^{k} = o(X), \end{aligned}$$

(32)

for each nonzero integer k. In turn, this bound is implied by Proposition 15, using the zero-free region for the Hecke Grossencharacter L-functions in [17, Theorem 5.35]. $\square$

To contrast this, we will show that the normalized Gauss sums $\tau (\chi )^2/q$, with $\chi$ ranging over all quartic characters, equidistribute slowly. More precisely, we have the following result.

Proposition 17

Let $k \in 2 {\mathbb {Z}}$, $k \ne 0$. There exists $c_{k} \in {\mathbb {C}}$ such that

$$\begin{aligned} \sum _{\begin{array}{c} q \le X \\ (q,2) = 1 \end{array}} \sum _{\begin{array}{c} \chi : \chi ^4 = 1 \\ \textrm{cond}(\chi ) = q \end{array}} (\tau (\chi )^2/q)^{k} = c_{k} X + o(X). \end{aligned}$$

(33)

Fig. 2

This histogram represents the distribution of the argument of the $\tau (\chi )^2/\textrm{cond}(\chi )$ for totally quartic characters. The histogram is made by calculating the Gauss sums of totally quartic characters of conductor up to 300000. The argument of $\tau (\chi )^2/\textrm{cond}(\chi )$ ranges from $-\pi$ to $\pi$ and this range has been divided into 20 bins

Remark 4

Recall from Lemma 8 that the total number of such characters grows like $X \log {X}$, so Proposition 17 shows that the rate of equidistribution is only $O((\log {X})^{-1})$ here. In contrast, in the family of totally quartic characters, the GRH would imply a rate of equidistribution of the form $O(X^{-1/2+\varepsilon })$. This difference in rates of equidistribution is supported by Fig. 2 in which we see that the arguments of squares of the Gauss sums of totally quartic characters quickly converge to being uniformly distributed, as compared to the Gauss sums of all quartic characters.

In addition, one can derive a similar result when restricting to $\chi \in \Psi _4(X)$, simply by subtracting off the contribution from the quadratic characters alone.

Proof

As in Lemma 14, write $\chi = \chi _2 \chi _4$, with $\chi _2$ quadratic and $\chi _4$ totally quartic. Then $\tau (\chi )^4/(q_1 q_2)^2 = \tau (\chi _4)^4/q_2^2$. The analog of Z(k, s), using k even to simplify, is

$$\begin{aligned} Z^{\text {all}}(k,s) = \sum _{\begin{array}{c} 0 \ne (\beta ) \subseteq {\mathbb {Z}}[i] \\ (\beta , 2\overline{\beta }) = 1\\ \beta \text { squarefree} \end{array}} \frac{\tau (\chi _{\beta })^{2 k} /N(\beta )^{k}}{N(\beta )^s} \sum _{\begin{array}{c} q_2 \in {\mathbb {Z}}_{\ge 1} \\ (q_2, 2N(\beta )) = 1 \end{array}} \frac{1}{q_2^s}. \end{aligned}$$

(34)

Referring to the calculation in Proposition 15, we obtain

$$\begin{aligned} Z^{\text {all}}(k,s) = \zeta (s) L(s,\lambda ^{k}) A(s), \end{aligned}$$

(35)

where $A_{}(s)$ is an Euler product absolutely convergent for $\textrm{Re}(s) > 1/2$. Since this generating function has a simple pole at $s=1$, we deduce Proposition 17. $\square$

As mentioned above, in order to deduce equidistribution, by Weyl’s equidistribution criterion, we also need to consider odd values of k in (33). This is more technical than the case for even k, so we content ourselves with a conjecture.

Conjecture 18

For each odd k, there exists $\delta > 0$ such that

$$\begin{aligned} \sum _{\begin{array}{c} q \le X \\ (q,2) = 1 \end{array}} \sum _{\begin{array}{c} \chi : \chi ^4 = 1 \\ \textrm{cond}(\chi ) = q \end{array}} (\tau (\chi )^2/q)^{k} \ll _{k, \delta } X^{1-\delta }. \end{aligned}$$

(36)

Remark 5

By Lemma 14 and (26), this problem reduces to understanding sums of the rough shape

$$\begin{aligned} \sum _{\begin{array}{c} \beta , q_2 \\ q_2 N(\beta ) \le X \end{array}} \Big (\frac{-N(\beta )}{q_2}\Big ) \mu (\beta ) \Big (\frac{2}{N(\beta )}\Big ) \lambda ((\beta ))^k, \end{aligned}$$

where we have omitted many of the conditions on $\beta$ and $q_2$. In the range where $q_2$ is very small, the GRH gives cancellation in the sum over $\beta$. Conversely, in the range where $N(\beta )$ is very small, the GRH gives cancellation in the sum over $q_2$. This discussion indicates that Conjecture 18 follows from GRH, with any $\delta <1/4$.

Unconditionally, one can deduce some cancellation using the zero-free region for the $\beta$-sum (with $q_2$ very small), and a subconvexity bound for the $q_2$-sum (with $N(\beta )$ very small). In the range where both $q_2$ and $N(\beta )$ have some size, then Heath-Brown’s quadratic large sieve [14] gives some cancellation. Since we logically do not need an unconditional proof of equidistribution, we omit the details for brevity.

Remark 6

Conjecture 18 and Proposition 17 together imply that the squares of the quartic Gauss sums do equidistribute in the full family $\Psi _4(X)$.

3.4.2 Sextic characters

Now we turn to the sextic Gauss sums.

Lemma 19

Suppose that $\chi$ is totally sextic of conductor q, and write $\chi = \chi _2 \chi _3$ with $\chi _2$ quadratic and $\chi _3$ cubic, each of conductor q. Suppose that $\chi _3$ is given as $\chi _{\beta }$ with $\beta$ a product of primary primes, as in Lemma 5. Then

$$\begin{aligned} \tau (\chi ) = \mu (q) \chi _3(2) \tau (\chi _2) \tau (\chi _3) \overline{\beta } q^{-1}. \end{aligned}$$

(37)

Proof

By [17, (3.18)], $\tau (\chi _2) \tau (\chi _3) = J(\chi _2, \chi _3) \tau (\chi )$, where $J(\chi _2, \chi _3)$ is the Jacobi sum. It is easy to show using the Chinese remainder theorem that if $\chi _2 = \prod _p \chi _2^{(p)}$ and $\chi _3 = \prod _p \chi _3^{(p)}$, then

$$\begin{aligned} J(\chi _2, \chi _3) = \prod _p J(\chi _2^{(p)}, \chi _3^{(p)}). \end{aligned}$$

(38)

The Jacobi sum for characters of prime conductor can be evaluated explicitly using the following facts. By [22, Prop. 4.30],

$$\begin{aligned} J(\chi _2^{(p)}, \chi _3^{(p)}) = \chi _3^{(p)}(2^2) J(\chi _3^{(p)}, \chi _3^{(p)}). \end{aligned}$$

(39)

Suppose that $\chi _3^{(p)} = \chi _{\pi }$, where $\pi \overline{\pi } = p$, and $\pi$ is primary. Then [16, Ch. 9, Lem. 1] implies $J(\chi _{\pi }, \chi _{\pi }) = - \pi$. (Warning: they state the value $\pi$ instead of $-\pi$, but recall their definition of primary is opposite our convention. Also recall that $\chi _{\pi } = \chi _{-\pi }$.) Gathering the formulas, we obtain

$$\begin{aligned} \tau (\chi _2) \tau (\chi _3) = \tau (\chi ) \chi _3(2)^2 \prod _{\pi _i | \beta } (- \pi _i) = \tau (\chi ) \chi _3(2)^2 \mu (q) \beta . \end{aligned}$$

(40)

Rearranging this and using $\beta \overline{\beta } = q$ completes the proof. $\square$

Corollary 20

Suppose that $\chi$ is totally sextic of conductor q, and write $\chi = \chi _2 \chi _3$ with $\chi _2$ quadratic and $\chi _3$ cubic, each of conductor q. Suppose that $\chi _3$ is given as $\chi _{\beta }$ with $\beta$ a product of primary primes, as in Lemma 5.

$$\begin{aligned} \tau (\chi )^2/q = \chi _3(4) \Big (\frac{-1}{q}\Big ) \tau (\chi _{\beta })^2 \overline{\beta }^2/q^2. \end{aligned}$$

(41)

Patterson [29] showed that $\tau (\chi _{\beta })/\sqrt{q}$ is uniformly distributed on the unit circle, as $\chi _{\beta }$ ranges over primitive cubic characters. The same method gives equidistribution after multiplication by a Hecke Grossencharacter, and so similarly to the quartic case above, we deduce:

Corollary 21

(Patterson) The Gauss sums $\tau (\chi )^2/q$, for $\chi$ totally sextic of conductor q, equidistribute on the unit circle.

In light of Corollary 20, Proposition 17, and Conjecture 18, it seems reasonable to conjecture that the points $\tau (\chi )^2/q$ are equidistributed on the unit circle, as $\chi$ varies over all sextic characters. To see a limitation in the rate of equidistribution, it is convenient to consider $\tau (\chi )^6/q^3$, which is multiplicative for $\chi$ sextic. For $q \equiv 1 \pmod {4}$, and $\chi = \chi _2$ quadratic, we have $\tau (\chi _2)^2/q = 1$, so the quadratic part is constant. For $\chi$ cubic and $q \equiv 1 \pmod {4}$,

$$\begin{aligned} \tau (\chi _{\beta })^6/q^3 = \mu (\beta ) \tau (\overline{\chi _{\beta }})^3 \overline{\beta }^3 = q^{-1} \overline{\beta }^2, \end{aligned}$$

(42)

which is nearly a Hecke Grossencharacter. A similar formula holds for $\chi$ totally sextic, namely

$$\begin{aligned} \tau (\chi )^6/q^3= q^{-4} \overline{\beta }^8. \end{aligned}$$

(43)

Therefore, carrying out the same steps as in Proposition 17 shows that

$$\begin{aligned} \sum _{\begin{array}{c} q \le X \\ q \equiv 1 \pmod {4} \end{array}} \sum _{\begin{array}{c} \chi \in \Psi _6 \\ \textrm{cond}(\chi ) = q \end{array}} \Big (\tau (\chi )^6/q^3\Big )^k = C_k X + o(X). \end{aligned}$$

(44)

This is less of an obstruction than in the quartic case, since here the rate of equidistribution is $O((\log X)^{-2})$ instead of $O((\log X)^{-1})$, due to the fact that $|\Psi _{6}^{}(X)|$ is approximately $\log X$ times as large as $|\Psi _4^{}(X)|$.

Similarly to the discussion of the quartic case in Remarks 5 and 6, we make the following conjecture without further explanation.

Conjecture 22

The Gauss sums $\tau (\chi )^2/q$, for $\chi$ ranging in $\Psi _6(X)$, equidistribute on the unit circle.

3.5 Estimates for quartic and sextic characters

In order to apply the random matrix theory conjectures, we need variants on Proposition 7, Lemma 8, Proposition 9, and Lemma 10, as follows.

Lemma 23

For primitive Dirichlet characters $\chi$ of order $\ell$ we have for $\ell = 4$ and $\ell =6$ that

$$\begin{aligned} \sum _{\chi \in \Psi _{\ell }(X)}\frac{1}{\sqrt{\textrm{cond}{(\chi })}} \sim 2 K_{\ell } \sqrt{X} (\log X)^{d(\ell ) -2}, \end{aligned}$$

(45)

and

$$\begin{aligned}{} & {} \sum _{\chi \in \Psi ^\textrm{tot}_{\ell }(X)}\frac{1}{\sqrt{\textrm{cond}{(\chi })}} \sim 2 K_{\ell }^{\textrm{tot}} \sqrt{X}, \nonumber \\{} & {} \sum _{\chi \in \Psi ^\prime _{\ell }(X)}\frac{1}{\sqrt{\textrm{cond}{(\chi })}} \sim 2\frac{\sqrt{X}}{\log X}. \end{aligned}$$

(46)

Proof

These estimates follow from a straightforward application of partial summation or from a minor modification of Lemma 6 since the generating Dirichlet series for one of these sums has its pole at $s=1/2$ instead of at $s=1$. $\square$

4 Random matrix theory: conjectural asymptotic behavior

This section closely follows the exposition of §3 of [10] and §4 of [11].

Let U(N) be the set of unitary $N \times N$ matrices with complex coefficients which forms a probability space with respect to the Haar measure.

For a family of L-functions with symmetry type U(N), Katz and Sarnak conjectured that the statistics of the low-lying zeros should agree with those of the eigenangles of random matrices in U(N) [18]. Let $P_A(\lambda ) = \det (A - \lambda I)$ be the characteristic polynomial of A. Keating and Snaith [19] suggest that the distribution of the values of the L-functions at the critical point is related to the value distribution of the characteristic polynomials $|P_A(1)|$ with respect to the Haar measure on U(N).

For any $s \in \mathbb {C}$ we consider the moments

$$\begin{aligned} M_U(s,N) := \int _{U(N)} |P_A(1)|^s \, d\text {Haar} \end{aligned}$$

for the distribution of $|P_A(1)|$ in U(N) with respect to the Haar measure. In [19], Keating and Snaith proved that

$$\begin{aligned} M_U(s,N) = \prod _{j=1}^N \frac{\Gamma (j) \Gamma (j+s)}{\Gamma ^2(j+s/2)}, \end{aligned}$$

(47)

so that $M_U(s,N)$ is analytic for $\text {Re}(s) > -1$ and has meromorphic continuation to the whole complex plane. The probability density of $|P_A(1)|$ is given by the Mellin transform

$$\begin{aligned} p_U(x,N) = \frac{1}{2 \pi i } \int _{\text {Re}(s) = c} M_U(s,N) x^{-s-1} \, ds, \end{aligned}$$

for some $c > -1$.

In the applications to the vanishing of twisted L-functions we consider in this paper, we are only interested in small values of x where the value of $p_U(x,N)$ is determined by the first pole of $M_U(s,N)$ at $s = -1$. More precisely, for $x \le N^{-1/2}$, one can show that

$$\begin{aligned} p_U(x,N) \sim G^2(1/2) N^{1/4} \qquad \text {as } N \rightarrow \infty , \end{aligned}$$

where G(z) is the Barnes G-function with special value [2]

$$\begin{aligned} G(1/2) = \text {exp}\left( \frac{3}{2} \zeta ^\prime (-1) - \frac{1}{4} \log \pi + \frac{1}{24}\log 2\right) . \end{aligned}$$

We will now consider the moments for the special values of twists of L-functions. We then define, for any $s \in \mathbb {C}$, the following sum of evaluations at $s=1$ of L-functions primitive order $\ell$ characters of conductor less than X:

$$\begin{aligned} M_E(s,X) = \frac{1}{\#{\mathcal {F}}_{\Psi _\ell ,E}(X)} \sum _{L(E,s,\chi ) \in {\mathcal {F}}_{\Psi _\ell ,E}(X)} |L(E,1,\chi )|^s. \end{aligned}$$

(48)

Then, since the families of twists of order $\ell$ are expected to have unitary symmetry, we have

Conjecture 24

(Keating and Snaith Conjecture for twists of order $\ell$) With the notation as above,

$$\begin{aligned} M_E(s,X) \sim a_E(s/2) M_U(s,N) \qquad \text {as } N = 2 \log X \rightarrow \infty , \end{aligned}$$

where $a_E(s/2)$ is an arithmetic factor depending only on the curve E.

From Conjecture 24, the probability density for the distribution of the special values $|L(E,1,\chi )|$ for characters of order $\ell$ is

$$\begin{aligned} p_E(x,X)= & {} \frac{1}{2 \pi i} \int _{\text {Re}(s) = c} M_E(s,X) x^{-s-1} \, ds \end{aligned}$$

(49)

$$\begin{aligned}\sim & {} \frac{1}{2 \pi i} \int _{\text {Re}(s) = c} a_E(s/2) M_U(s,N) x^{-s-1} \, ds \end{aligned}$$

(50)

as $N = 2 \log X \rightarrow \infty$. As above, when $x \le N^{-1/2}$, the value of $p_E(x,X)$ is determined by the residue of $M_U(s,N)$ at $s = -1$, thus (50) suggests that for $x \le (2 \log X)^{-1/2}$,

$$\begin{aligned} p_E(x,X) \sim 2^{1/4} a_E(-1/2) G^2(1/2) \log ^{1/4}(X) \end{aligned}$$

(51)

as $X \rightarrow \infty$.

We now use the probability density of the random matrix model with the properties of the integers $n_E(\chi )$ to obtain conjectures for the vanishing of the L-values $|L(E,1,\chi )|$. When $\chi$ is either quartic or sextic, the discretization $n_E(\chi )$ is a rational integer since ${{\mathbb {Z}}}[\zeta _\ell ] \cap {{\mathbb {R}}}= {{\mathbb {Z}}}$ when $\ell =4$ or 6.

Lemma 25

Let $\chi$ be a primitive Dirichlet character of order $\ell = 4$ or 6. Then

$$\begin{aligned} |L(E,1, \chi )| = \frac{c_{E,\ell }}{\sqrt{\textrm{cond}(\chi )}} |n_E(\chi )|, \end{aligned}$$

where $c_{E,\ell }$ is a nonzero constant which depends only on the curve E and $\ell$.

Proof

By rearranging equation (2) we obtain

$$\begin{aligned} |L(E,1,\chi )| = \left| \frac{\Omega _\epsilon (E) \,\tau (\chi ) \, k_E \, n_E(\chi )}{\textrm{cond}(\chi )}\right| = \frac{|\Omega _\epsilon (E) \, k_E \, n_E(\chi )|}{\sqrt{\textrm{cond}(\chi )}} = \frac{ c_{E,\ell }|n_E(\chi )|}{\sqrt{\textrm{cond}(\chi )}}, \end{aligned}$$

where the nonzero constant $k_E$ is that of Proposition 3. $\square$

Fig. 3

Verification of Conjecture 1 for quartic twists of 11.a.1

Fig. 4

Verification of Conjecture 1 for sextic twists of 11.a.1

Fig. 5

Verification of parts of Conjecture 1 for twists of 37.a.1

We write

$$\begin{aligned} \text {Prob} \{|L(E,1,\chi )| = 0 \} = \text {Prob} \{|L(E,1,\chi )| < B(\textrm{cond}(\chi )) \}, \end{aligned}$$

(52)

for some function $B(\textrm{cond}(\chi ))$ of the character. By Lemma 25 we may take $B(\textrm{cond}(\chi )) = \dfrac{c_{E,\ell }}{\sqrt{\textrm{cond}(\chi )}}$. Note that since $c_{E,\ell } \ne 0$, if

$$\begin{aligned}\frac{|n_E(\chi )| c_{E,\ell }}{\sqrt{\textrm{cond}(\chi )}} < \frac{c_{E,\ell }}{\sqrt{\textrm{cond}(\chi )}},\end{aligned}$$

then $|n_E(\chi )| <1$ and hence must vanish since $|n_E(\chi )| \in {{\mathbb {Z}}}_{\ge 0}$.

Using (51), we have

$$\begin{aligned} \text {Prob}\{|L(E,1,\chi )| = 0\}= & {} \int _0^{B(\textrm{cond}(\chi ))} 2^{1/4} a_E(-1/2) G^2(1/2) \log ^{1/4}(X) \, dx \\= & {} 2^{1/4} a_E(-1/2) G^2(1/2) \log ^{1/4}(X) B(\textrm{cond}(\chi )) \end{aligned}$$

Summing the probabilities gives

$$\begin{aligned} |V_{\Psi _\ell ,E}(X)| = 2^{1/4} c_{E,k} a_E(-1/2) G^2(1/2) \log ^{1/4}(X) \sum _{\textrm{cond}(\chi ) \le X} \frac{1}{\sqrt{\textrm{cond}(\chi )}}. \end{aligned}$$

Thus, by the analysis in §3.3, we have

$$\begin{aligned} |V_{\Psi _4,E}(X)|&\sim 2^{5/4} c_{E,4} K_4 a_E(-1/2) G^2(1/2) \log ^{1/4}(X) \sqrt{X} \log X \\&\sim b_{E,4} X^{1/2} \log ^{5/4} X \end{aligned}$$

and

$$\begin{aligned} |V_{\Psi _6,E}(X)|&\sim 2^{5/4} c_{E,6} K_6 a_E(-1/2) G^2(1/2) \log ^{1/4}(X) \sqrt{X} (\log X)^2 \\&\sim b_{E,6} X^{1/2} \log ^{9/4} X \end{aligned}$$

as $X \rightarrow \infty$.

Moreover, if we restrict to those characters that are totally quartic or sextic, we get the following estimates

$$\begin{aligned} |V_{\Psi _4^\textrm{tot},E}(X)|&\sim 2^{5/4} c_{E,4} K_4^\textrm{tot}a_E(-1/2) G^2(1/2) \log ^{1/4}(X) \sqrt{X} \\&\sim b^\textrm{tot}_{E,4} X^{1/2} \log ^{1/4} X \end{aligned}$$

and

$$\begin{aligned} |V_{\Psi _6^\textrm{tot},E}(X)|&\sim 2^{5/4} c_{E,6} K_6^\textrm{tot}a_E(-1/2) G^2(1/2)\\&\sim b^\textrm{tot}_{E,6} X^{1/2} \log ^{1/4} X \end{aligned}$$

as $X \rightarrow \infty$.

Finally, if we restrict only to those twists by characters of prime conductor, we conclude

$$\begin{aligned} |V_{\Psi _4^\prime ,E}(X)|&\sim 2^{5/4} c_{E,4} a_E(-1/2) G^2(1/2) \log ^{1/4}(X) \frac{\sqrt{X}}{\log X} \\&\sim b^\prime _{E,4} X^{1/2} \log ^{-3/4} X \end{aligned}$$

and

$$\begin{aligned} |V_{\Psi _6^\prime ,E}(X)|&\sim 2^{5/4} c_{E,6} a_E(-1/2) G^2(1/2) \log ^{1/4}(X) \frac{\sqrt{X}}{\log X} \\&\sim b^\prime _{E,6} X^{1/2} \log ^{-3/4} X \end{aligned}$$

as $X \rightarrow \infty$.

4.1 Computations

Here we provide numerical evidence for Conjecture 1. The computations of the Conrey labels for the characters were done in SageMath [32] and the computations of the L-functions were done in PARI/GP [27]. We used the built-in way PARI/GP computes L-functions; that is, roughly speaking, the computations were done via a slight modification, due to Booker [5] and Molin [24], of Poisson summation. The L-function computations were done in a distributed way on the Open Science Grid. For each curve, we generated a PARI/GP script to calculate a twisted L-function for each primitive character of order 4 and 6, and then combined the results into one file at the end. The combined wall time of all the computations was more than 50 years. The code and data are available at [3].

In Fig. 3 we plot the points

$$\begin{aligned} (X,\tfrac{ X^{1/2} \log ^{5/4} X}{|V_{\Psi _4,\texttt {11.a.1}}(X)|}),(X,\tfrac{X^{1/2}\log ^{-3/4} X}{|V_{\Psi _4^\prime ,\texttt {11.a.1}}(X)|}), (X,\tfrac{X^{1/2}\log ^{1/4} X}{|V_{\Psi _4^\textrm{tot},\texttt {11.a.1}}(X)|}) \end{aligned}$$

that provides a comparison between the predicted vanishings of $L(E,1,\chi )$ for quartic characters and for the curve 11.a.1. In Fig. 4 we plot the analogous points for the same curve but for sextic twists. In Fig. 5 we plot the points

$$\begin{aligned} (X,\tfrac{X^{1/2}\log ^{-3/4} X}{|V_{\Psi _4^\prime ,\texttt {37.a.1}}(X)|}), (X,\tfrac{X^{1/2}\log ^{-3/4} X}{|V_{\Psi _6^\prime ,\texttt {37.a.1}}(X)|}) \end{aligned}$$

Even though we are most interested in the families of all quartic and sextic twists, we include the families of twists of prime conductor because there are far fewer such characters and so we can calculate the number of vanishings up to a much larger X. We include the families of twists by totally quartic and sextic characters to highlight the transition between the family of prime conductors and the family of all conductors.