2. Semiabelian varieties and algebraic subgroups

This section involves the structure of algebraic subgroups concerning semiabelian varieties defined over $\overline{\mathbb{Q}}$ , which plays a crucial point for the proofs in the next section. The following lemma describes the form of a connected algebraic subgroup of the direct product of the additive group $\mathbb{G}_a$ with a semiabelian variety defined over $\overline{\mathbb{Q}}$ .

Proof.  Let $\pi _a$ and $\pi$ be the projections of $\mathbb{G}_a\times S$ on $\mathbb{G}_a$ and $S$ , respectively. Put

\begin{align*}H_a=\pi _a\big (H\cap \big(\mathbb {G}_a\times \{e\} \big)\big ),\quad R=\pi \big (H\cap \big(\{0\}\times S \big)\big ),\end{align*}

where $e$ is the identity element of $S$ . Then $H_a$ and $R$ are connected algebraic subgroups of $\mathbb{G}_a$ and $S$ defined over $\overline{\mathbb{Q}}$ , respectively. Let $I$ be the image of $H$ under the projection

\begin{align*}\mathbb {G}_a\times S\rightarrow (\mathbb {G}_a\times S)/(H_a\times R)\cong (\mathbb {G}_a/ H_a)\times (S/R).\end{align*}

Denote by $p_a$ and $p$ the projections of $(\mathbb{G}_a/H_a)\times (S/R)$ onto $\mathbb{G}_a/H_a$ and onto $S/R$ , respectively. One can show that $I\cong p_a(I)$ and $I\cong p(I)$ , hence $p_a(I)\cong p(I)$ . On the other hand, $H_a$ is either trivial or $\mathbb{G}_a$ , and this leads to the algebraic subgroup $p_a(I)$ of $\mathbb{G}_a/H_a$ is either $\mathbb{G}_a$ or trivial. We claim that the first case cannot hold. In fact, if not, then this gives $p(I)\cong \mathbb{G}_a$ . Since $p(I)$ is an algebraic subgroup of $S/R$ defined over $\overline{\mathbb{Q}}$ , there exists a connected algebraic subgroup $Q$ of $S$ containing $R$ defined over $\overline{\mathbb{Q}}$ such that $p(I)\cong Q/R$ , and then $Q/R\cong \mathbb{G}_a$ . Furthermore, it follows from [Reference Perucca10, Proposition 2.1.3] that $Q$ and $R$ are semiabelian varieties. We therefore get the short exact sequences

\begin{align*}1\rightarrow \mathbb {G}_m^q\rightarrow Q\rightarrow B\rightarrow 1\end{align*}

and

\begin{align*}1\rightarrow \mathbb {G}_m^r\rightarrow R\rightarrow C\rightarrow 1\end{align*}

of $Q$ and $R$ , respectively, for non-negative integers $q,r$ and for abelian varieties $B, C$ defined over $\overline{\mathbb{Q}}$ . Since $Q/R$ is isomorphic to $\mathbb{G}_a$ , the composition of the homomorphism of algebraic groups $\mathbb{G}_m^q\rightarrow Q$ (from the first exact sequence above) with the projection $Q\rightarrow Q/R$ must be trivial. This means that $G_m^q\hookrightarrow R$ . We now consider the cases:

If $q\gt r$ , then

\begin{align*}\mathbb {G}_m^{q-r}=\mathbb {G}_m^q/\mathbb {G}_m^r\hookrightarrow R/\mathbb {G}_m^r\cong C.\end{align*}

If $q\lt r$ , then

\begin{align*}\mathbb {G}_m^{r-q}=\mathbb {G}_m^r/\mathbb {G}_m^q\hookrightarrow Q/\mathbb {G}_m^q\cong B\end{align*}

(since $\mathbb{G}_m^r\hookrightarrow R$ from the second exact sequence and $R$ is an algebraic subgroup of $Q$ ).

If $q=r$ , then

\begin{align*}\mathbb {G}_a\cong Q/R\cong (Q/\mathbb {G}_m^q)/(R/\mathbb {G}_m^r)\cong B/C.\end{align*}

But these cases cannot hold because of the fact that there is no non-trivial algebraic homomorphism from a linear algebraic group to an abelian variety defined over $\overline{\mathbb{Q}}$ (see [Reference Conrad5, Lemma 2.3]). Hence, $p_a(I)$ must be trivial, and this shows that so is $I$ . From this, we are able to conclude that $H=H_a\times R$ , which completes the proof of the lemma.

3. Proof of the theorem

(i) Let $G$ be the commutative algebraic group defined by $G=\mathbb{G}_a\times S$ . Then $G$ is defined over $\overline{\mathbb{Q}}$ and its Lie algebra $\mathrm{Lie}(G)$ is identified with $\overline{\mathbb{Q}}\times \overline{\mathbb{Q}}^{g+1}=\overline{\mathbb{Q}}^{g+2}$ . Suppose by contradiction that the numbers $1, u_0,u_1,\ldots, u_g$ are linearly dependent over $\overline{\mathbb{Q}}$ , that is there is a non-zero linear form $L$ in $g+2$ variables with coefficients in $\overline{\mathbb{Q}}$ such that $L(1, u_0, u_1,\ldots, u_g)=0$ . Let $V$ be the $\overline{\mathbb{Q}}$ -vector space defined by the zero set of $L$ in $\overline{\mathbb{Q}}^{g+2}$ . Then one has the vector $\tilde u\,:\!=\,(1,u)$ lies in the $\mathbb{C}$ -vector space $V\otimes _{\overline{\mathbb{Q}}}{\mathbb{C}}$ , and furthermore,

\begin{align*}\exp _{G({\mathbb {C}})}(\tilde u)=(1, \exp _{S({\mathbb {C}})}(u))\in G(\overline {\mathbb {Q}}).\end{align*}

Hence, thanks to the analytic subgroup theorem (see [Reference Wüstholz16] or [Reference Wüstholz17]), there exists a connected algebraic subgroup $H$ of $G$ of positive dimension defined over $\overline{\mathbb{Q}}$ for which $\mathrm{Lie}(H)$ is contained in $V$ and $\tilde u$ lies in $\mathrm{Lie}(H({\mathbb{C}}))$ . The above lemma tells us that $H$ must be of the form $H_a\times R$ with $H_a$ an algebraic subgroup of $\mathbb{G}_a$ and $R$ an algebraic subgroup of $S$ (defined over $\overline{\mathbb{Q}}$ ). Since

\begin{align*}(1,u_0, u_1, \ldots, u_g)=\tilde u\in \mathrm {Lie}(H({\mathbb {C}}))=\mathrm {Lie}((H_a\times R)({\mathbb {C}}))=\mathrm {Lie}(H_a({\mathbb {C}}))\times \mathrm {Lie}(R({\mathbb {C}})),\end{align*}

it follows that $\mathrm{Lie}(H_a({\mathbb{C}}))$ is not trivial, and this leads to $H_a$ must be $\mathbb{G}_a$ . In particular, $\mathrm{Lie}(H_a)=\overline{\mathbb{Q}}$ and since $\mathrm{Lie}(H)\subseteq V$ , the linear form $L$ is given by $L=a_0X_0+a_1X_1+\cdots +a_gX_g$ with coefficients $a_0, a_1,\ldots, a_g\in \overline{\mathbb{Q}}$ not all zero. Let $W$ denote the $\overline{\mathbb{Q}}$ -vector space defined by

\begin{align*}W=\{(w_0,w_1,\ldots, w_g)\in \overline {\mathbb {Q}}^{g+1}\,:\, a_0w_0+a_1w_1+\cdots +a_gw_g=0\}.\end{align*}

Then one has $\mathrm{Lie}(R)\subseteq W$ . Let $\pi\,:\, S\rightarrow A$ denote the projection from the extension $1\rightarrow \mathbb{G}_m\rightarrow S\rightarrow A\rightarrow 1$ . Then we get an isomorphism of algebraic groups $R/R\cap \mathbb{G}_m\cong \pi (R).$ On the other hand, since $A$ is a simple abelian variety, it follows that either $\pi (R)$ is trivial or $\pi (R)$ is $A$ . In the first case, $R=Ker\pi \cong \mathbb{G}_m$ , and therefore $\mathrm{Lie}(R)$ is identified with the $\overline{\mathbb{Q}}$ -vector subspace $\overline{\mathbb{Q}}\times \{0\}\times \cdots \times \{0\}$ of $\mathrm{Lie}(S)=\overline{\mathbb{Q}}\times \overline{\mathbb{Q}}^{g}$ . Since $\mathrm{Lie}(R)\subseteq W$ , it follows that $a_0\alpha =0$ for all $\alpha \in \overline{\mathbb{Q}}$ . This implies that $a_0=0$ . By the expression of the exponential map $\exp _{S({\mathbb{C}})}$ given in [Reference Bertolin, Philippon, Saha and Saha1, Section 2.3], we obtain $\exp _{A({\mathbb{C}})}(u_1,\ldots, u_g)\in A(\overline{\mathbb{Q}})$ , and therefore by the analytic subgroup theorem again, one can find an algebraic subgroup $B$ of $A$ defined over $\overline{\mathbb{Q}}$ of positive dimension such that $\mathrm{Lie}(B)$ is contained in the $\overline{\mathbb{Q}}$ -vector space $\{(z_1,\ldots, z_g)\in \overline{\mathbb{Q}}^g\,:\, a_1z_1+\cdots +a_gz_g=0\}$ . Note that $A$ is simple, this gives $B$ must be equal to $A$ . But this cannot happen since $\dim B=\dim _{\overline{\mathbb{Q}}}\mathrm{Lie}(B)\le g-1$ . Thus, $\pi (R)=A$ and this means that $R/R\cap \mathbb{G}_m\cong A$ . It is clear that $R\cap G_m$ is either trivial or $\mathbb{G}_m$ , one can show that $R\cong A$ (since $R$ is a proper algebraic subgroup of $S$ ). In this case, similarly, we obtain $L$ is of the form $L=a_0X_0$ . In particular, $u_0=0$ which is a contradiction, and this completes the proof for the first part of Theorem1.1.

(ii) The proof of the second part follows directly from the $p$ -adic analytic subgroup theorem (see [Reference Fuchs and Pham6] or [Reference Matev9]) and from the same argument as in the proof of the first part above.

4. Applications to complex and $\boldsymbol{p}$ -adic Weierstrass sigma functions

This last section is devoted to discuss applications concerning the complex and $p$ -adic Weierstrass sigma functions associated with elliptic curves. To begin with, let $E$ be an elliptic curve defined over $\mathbb{C}$ characterised by the Weierstrass model

\begin{align*}Y^2Z-4X^3+g_2XZ^2+g_3Z^3=0\end{align*}

with $g_2, g_3\in \mathbb{C}$ such that $g_2^3-27g_3^2\ne 0.$ Let $\Lambda$ denote the period lattice of $E$ , and $\wp$ the Weierstrass elliptic function associated with $E$ (or relative to $\Lambda$ ), which is defined as:

\begin{align*}\wp (z)=\dfrac 1{z^2}+\sum _{w\in \Lambda ^*}\Big (\dfrac 1 {(z-w)^2}-\dfrac 1 {w^2} \Big ),\end{align*}

where $\Lambda ^*=\Lambda \setminus \{0\}$ . There are two more Weierstrass functions associated with $E$ , which are auxiliary to the function $\wp$ , called Weierstrass sigma and zeta functions, respectively. They play important roles in studying elliptic curves and are considered elliptic analogues of the classical trigonometric functions. In detail, the Weierstrass sigma function $\sigma$ associated with $E$ is defined as:

\begin{align*}\sigma (z)=z\prod _{w\in \Lambda ^*}\Big (1-\dfrac z w\Big )e^{\frac z w+\frac {z^2}{2w^2}},\end{align*}

and the Weierstrass zeta function $\zeta$ associated with $E$ is defined as:

\begin{align*}\zeta (z)=\dfrac 1 z+\sum _{w\in \Lambda ^*}\Big (\dfrac 1 {z-w}+\dfrac 1 w+\dfrac z{w^2}\Big ).\end{align*}

These functions are related by:

\begin{align*}\dfrac {d}{dz}\log \sigma (z)=\zeta (z);\quad \dfrac d{dz}\zeta (z)=-\wp (z).\end{align*}

It is convenient to recall that the corresponding Laurent expansions at the origin of these Weierstrass functions are given by:

\begin{align*}\wp (z)= \dfrac 1 {z^2}+\sum _{k=1}^{\infty }(2k+1)G_{2k+2}(\Lambda )z^{2k};\end{align*}

\begin{align*}\zeta (z)=\dfrac 1 z-\sum _{k=1}^{\infty } {G}_{2k+2}(\Lambda )z^{2k+1},\end{align*}

where $G_k(\Lambda )=\displaystyle \sum _{w\in \Lambda ^*}w^{-k}$ for $k\ge 3$ are the Eisenstein series (of weight $k$ ), and

\begin{align*}\sigma (z)=\sum _{m,n=0}^{\infty }a_{m,n}\Big (\dfrac 1 2g_2\Big )^m(2g_3)^n\dfrac {z^{4m+6n+1}}{(4m+6n+1)!}\end{align*}

where the sequence $(a_{m,n})_{m,n\ge 0}$ is defined by the recurrence relation

\begin{align*}a_{m,n}=3(m+1)a_{m+1,n+1}+\dfrac {16}3(n+1)a_{m-2,n+1}-\dfrac 1 3(2m+3n-1)(4m+6n-1)a_{m-1,n}\end{align*}

with $a_{0,0}=1$ and $a_{m,n}=0$ for either $m$ or $n$ is negative. One can show that $g_2=60G_4(\Lambda )$ and $g_3=140G_6(\Lambda )$ , and by induction, ${G}_{2k+2}(\Lambda )$ can be represented as polynomials $P_{2k+2}$ in terms of $g_2, g_3$ with rational coefficients (see [Reference Chandrasekharan4, Chapter IV]).

Now, let $E$ be an elliptic curve defined over $\overline{\mathbb{Q}}$ and $G$ an extension of $E$ by $\mathbb{G}_m$ defined over $\overline{\mathbb{Q}}$ . Recall that the group Ext $^1(E,\mathbb{G}_m)$ of extension classes of $E$ by $\mathbb{G}_m$ is isomorphic to the Picard group Pic $^0(E)=E^*$ of divisors on $E$ modulo principal divisors. This means that the extension $G$ corresponds to an algebraic point $P$ in $E(\overline{\mathbb{Q}})$ (here we identify $E$ with $E^*$ ). Let $q\in{\mathbb{C}}$ be an elliptic logarithm of $P$ , that is $\exp _{E({\mathbb{C}})}(q)=P$ , here the exponential map $\exp _{E({\mathbb{C}})}$ of the elliptic curve $E$ is given by:

\begin{align*}& \exp _{E({\mathbb {C}})}\,:\, {\mathbb {C}}\rightarrow E({\mathbb {C}})\subseteq \mathbb {P}^2({\mathbb {C}})\\ & \qquad \qquad z\mapsto [\wp (z)\,:\,\wp ^{\prime}(z)\,:\,1].\end{align*}

The exponential map of $G$ is given by:

\begin{align*} & \exp _{G({\mathbb {C}})}\,:\, {\mathbb {C}}^2\rightarrow G({\mathbb {C}})\subseteq (\mathbb {P}^2\times \mathbb {P}^1)({\mathbb {C}}) \\ & \qquad\qquad (z,t)\mapsto \big ([\wp (z)\,:\,\wp ^{\prime}(z)\,:\,1], [e^tf(z)\,:\,1]\big ),\end{align*}

where $f (z)=\dfrac{\sigma (z+q)}{\sigma (z)\sigma (q)}e^{-\zeta (q)z}$ is Serre’s function (see [Reference Bertolin, Philippon, Saha and Saha1, Section 6]). We now obtain the following theorem, which is an improvement of a previous result in transcendence given by M. Waldschmidt (see [Reference Waldschmidt13, Theorem 1] or [Reference Waldschmidt14, Theorem 3.2.10]).

Theorem 4.1. Let $E$ be an elliptic curve defined over $\overline{\mathbb{Q}}$ and $\wp, \zeta, \sigma$ the Weierstrass elliptic, zeta, sigma functions associated with $E$ , respectively. Let $\alpha, \beta$ be algebraic numbers and $u, u_0$ complex numbers such that $\wp (u), \wp (u_0)$ are algebraic numbers. Suppose that $u_0$ is not a torsion point and that $u, u+u_0$ are not in the lattice of periods of $\wp$ . Then, the number

\begin{align*}\dfrac {\sigma (u+u_0)}{\sigma (u)\sigma (u_0)}e^{(\alpha -\zeta (u_0))u+\beta }\end{align*}

is transcendental.

Proof. Consider the algebraic point

\begin{align*}\exp _{E({\mathbb {C}})}(u_0)=[\wp (u_0)\,:\,\wp ^{\prime}(u_0)\,:\,1]\in E(\overline {\mathbb {Q}}).\end{align*}

This algebraic point corresponds to a semiabelian variety $G$ in Ext $^1(E,\mathbb{G}_m)$ . Then one has

\begin{align*}\exp _{G({\mathbb {C}})}(u, \alpha u+\beta )=([\wp (u)\,:\,\wp ^{\prime}(u)\,:\,1],[e^{\alpha u+\beta }f(u)\,:\,1]),\end{align*}

with $f(u)=\dfrac{\sigma (u+u_0)}{\sigma (u)\sigma (u_0)}e^{-\zeta (u_0)u}$ , and this gives

\begin{align*}e^{\alpha u+\beta }f(u)=\dfrac {\sigma (u+u_0)}{\sigma (u)\sigma (u_0)}e^{(\alpha -\zeta (u_0))u+\beta }.\end{align*}

Hence, if this number is algebraic then $\exp _{G({\mathbb{C}})}(u,\alpha u+\beta )\in G(\overline{\mathbb{Q}})$ . Applying the first part of Theorem1.1, we deduce that the elements $1, u, \alpha u+\beta$ are linearly independent over $\overline{\mathbb{Q}}$ . This contradiction proves the theorem.

It is natural to obtain a $p$ -adic analogue of the above theorem, and in order to express such a result in the $p$ -adic setting, we first recall the $p$ -adic Weierstrass elliptic, zeta and sigma functions, respectively. By definition, the $p$ -adic Weierstrass elliptic function $\wp _p$ is the (Lutz-Weil) $p$ -adic elliptic function associated with the elliptic curve $E$ (see [Reference Lutz8] and [Reference Weil15]). This function satisfies the relation (as the complex one) $\wp _p^{\prime}(z)= 4\wp _p^3(z)-g_2\wp _p-g_3,$ but only on the neighbourhood of the origin

\begin{align*}\mathcal E_p\,:\!=\,\big \{z\in \mathbb {C}_p;\, |1/4|_p\max \{|g_2|_p^{1/4}, |g_3|_p^{1/6}\}z\in B(r_p)\big \},\end{align*}

where $B(r_p)$ is the set of all $p$ -adic numbers $x$ in $\mathbb{C}_p$ with $|x|_p\lt r_p\,:\!=\,p^{-\frac{1}{p-1}}$ . Note that $\wp _p$ is analytic on $\mathcal E_p\setminus \{0\}$ and expressed by:

\begin{align*}\wp _p(z)= \dfrac 1 {z^2}+\sum _{k=1}^{\infty }(2k+1)P_{2k+2}(g_2,g_3)z^{2k}.\end{align*}

Let $\zeta _p$ and $\sigma _p$ be the $p$ -adic Weierstrass zeta and sigma functions, respectively. By definition, $\zeta _p$ and $\sigma _p$ are meromorphic functions on the domain $\mathcal E_p\setminus \{0\}$ satisfying the followings:

\begin{align*}\dfrac {d}{dz}\log \sigma _p(z)=\zeta _p(z);\quad \dfrac d{dz}\zeta _p(z)=-\wp _p(z),\end{align*}

and their corresponding $p$ -adic expansions are also given by:

\begin{align*}\zeta _p(z)=\dfrac 1 z-\sum _{k=1}^{\infty } {P}_{2k+2}(g_2,g_3)z^{2k+1}\end{align*}

and

\begin{align*}\sigma _p(z)=\sum _{m,n=0}^{\infty }a_{m,n}\Big (\dfrac 1 2g_2\Big )^m(2g_3)^n\dfrac {z^{4m+6n+1}}{(4m+6n+1)!}.\end{align*}

We obtain the following theorem.

Theorem 4.2. Let $E$ be an elliptic curve defined over $\overline{\mathbb{Q}}$ and $\wp _p, \zeta _p, \sigma _p$ the $p$ -adic Weierstrass elliptic, zeta, sigma functions associated with $E$ , respectively. Let $\alpha$ and $\beta$ be algebraic numbers and $\mathfrak{u}, \mathfrak{u}_0$ non-zero $p$ -adic numbers in $\mathcal E_p$ such that $\wp _p(\mathfrak{u}), \wp _p(\mathfrak{u}_0)$ are algebraic numbers. If $\alpha \mathfrak{u} + \beta \in B(r_p)$ , then the number

\begin{align*}\dfrac {\sigma _p(\mathfrak{u} + \mathfrak{u} _0)}{\sigma _p(\mathfrak{u})\sigma _p(\mathfrak{u}_0)}e_p^{(\alpha -\zeta _p(\mathfrak{u}_0))\mathfrak{u}+\beta }\end{align*}

is transcendental.

Proof. As in the complex case, we first consider the algebraic point

\begin{align*}\exp _{E({\mathbb {C}}_p)}(\mathfrak{u}_0)=[\wp _p(\mathfrak{u}_0)\,:\,\wp _p^{\prime}(\mathfrak{u}_0)\,:\,1]\in E(\overline {\mathbb {Q}}),\end{align*}

and this algebraic point gives the corresponding semiabelian variety $G_p$ in Ext $^1(E,\mathbb{G}_m)$ . Since we have seen from above that the $p$ -adic functions $\wp _p, \zeta _p$ and $\sigma _p$ are represented by the same power series as the complex functions $\wp, \zeta$ and $\sigma$ , respectively, one can show that the $p$ -adic exponential map of $G_p$ is also expressed as the same type as the complex exponential map. More precisely, the map $\exp _{G({\mathbb{C}}_p)}$ is given by:

\begin{align*}\exp _{G({\mathbb {C}}_p)}\,:\, \mathcal G_p\rightarrow G_p({\mathbb {C}}_p)\subseteq (\mathbb {P}^2\times \mathbb {P}^1)({\mathbb {C}}_p)\end{align*}

\begin{align*}\;\quad \quad \quad \quad \qquad \quad (z,t)\mapsto \big ([\wp _p(z)\,:\,\wp _p^{\prime}(z)\,:\,1], [e_p^tf_p(z)\,:\,1]\big ),\end{align*}

with $\mathcal G_p=\mathcal E_p\times B(r_p)$ and $f_p(z)=\dfrac{\sigma _p(z+\mathfrak{u}_0)}{\sigma _p(z)\sigma _p(\mathfrak{u}_0)}e_p^{-\zeta _p(\mathfrak{u}_0)z}$ with $e_p$ the usual $p$ -adic exponential function, and prolonged to a function on the whole ${\mathbb{C}}_p$ (thanks to [Reference Robert12, Section 5.4.4]). Assume by contradiction as the complex case that the number

\begin{align*}\dfrac {\sigma _p(\mathfrak{u}+\mathfrak{u}_0)}{\sigma _p(\mathfrak{u})\sigma _p(\mathfrak{u}_0)}e_p^{(\alpha -\zeta _p(\mathfrak{u}_0))\mathfrak{u}+\beta }\end{align*}

is algebraic, then we also get $\exp _{G_p({\mathbb{C}}_p)}(\mathfrak{u},\alpha \mathfrak{u}+\beta )\in G_p(\overline{\mathbb{Q}}),$ and therefore we are able to deduce from the second part of Theorem1.1 that $1, \mathfrak{u}, \alpha \mathfrak{u} +\beta$ are linearly independent over $\overline{\mathbb{Q}}$ (a contradiction). This completes the proof of the theorem.