Frobenius elements in Galois representations with SL_n image

TL;DR

This paper introduces a method for identifying Frobenius elements in Galois representations with SL_n image, leveraging the linear structure to distinguish conjugacy classes, enhancing understanding of the Galois group's structure.

Contribution

It provides a novel technique to determine Frobenius elements in Galois representations with SL_n images, distinguishing conjugacy classes more precisely than previous methods.

Findings

Method effectively distinguishes SL_n(_l)-conjugacy from GL_n(_l)-conjugacy.

Applicable to elliptic curves over number fields with specific mod l representations.

Enhances the analysis of Galois groups via linear algebraic structures.

Abstract

Suppose we have a elliptic curve over a number field whose mod $l$ representation has image isomorphic to $SL_2(\mathbb{F}_l)$. We present a method to determine Frobenius elements of the associated Galois group which incorporates the linear structure available. We are able to distinguish $SL_n(\mathbb{F}_l)$-conjugacy from $GL_n(\mathbb{F}_l)$-conjugacy; this can be thought of as being analogous to a result which distinguishes $A_n$-conjugacy from $S_n$-conjugacy when the Galois group is considered as a permutation group.

Equations

$${\rho }_{E,l}:\mathrm{Gal}(\overline{F}/F)\to {\mathrm{GL}}_2({\mathbb{F}}_l),$$

$$⟨P_1,P_2{⟩}_l\mathrm{mod}\mathfrak{p}\equiv ⟨Q_1,Q_2{⟩}_l^{k^2}\text{ for some }k\in \mathbb{Z},$$

$$m_{\mathcal{F}}(⟨Q_1,Q_2{⟩}_l^{k^2})\text{ divides }{m}_F(⟨P_1,P_2{⟩}_l)\mathrm{mod}\mathfrak{p}$$

$$T_F(B)\mathrm{mod}\mathfrak{p}\equiv T_{\mathcal{F}}(\overline{B}{)}^h\text{ for some }h\in H.$$

Full text

We now give a further criterion to distinguish between two classes in $\operatorname{SL}_{2}$ that are conjugate in $\operatorname{GL}_{2}$.

Let $\tilde{E}$ be the reduced curve at $\mathfrak{p}$ and suppose that $\{Q_{1},Q_{2}\}$ is an ordered basis of $\tilde{E}[l]$ such that the action of the Frobenius automorphism $x\mapsto x^{q}$ acts as $\sigma\in\operatorname{SL}_{2}$ on $\tilde{E}[l]$ with respect to $\{Q_{1},Q_{2}\}$.

Then $\rho_{E,l}(\operatorname{Frob}_{\mathfrak{p}})$, written with respect to a global ordered basis $\{P_{1},P_{2}\}$, is $\operatorname{SL}_{2}$-conjugate to $\sigma$ if and only if

[TABLE]

where $\langle\,,\,\rangle_{l}$ denotes the Weil pairing.

Hence there exists $B\in\operatorname{GL}_{2}$ such that $\sigma=B^{-1}\sigma_{1}B$ and by the same argument, $\det(B)$ cannot be a square. If $\det A$ is not a square, then $\det(A)/\det(B)$ is a square so $\tau$ is $\operatorname{SL}_{2}$-conjugate to $\sigma_{1}$ therefore cannot be $\operatorname{SL}_{2}$-conjugate to $\sigma$.

By the properties of the Weil pairing $\langle AP_{1},AP_{2}\rangle_{l}=\langle P_{1},P_{2}\rangle_{l}^{\det(A)}$, hence $\rho_{E,l}(\operatorname{Frob}_{\mathfrak{p}})$ is $\operatorname{SL}_{2}$-conjugate to $\sigma$ if and only if $\langle P_{1},P_{2}\rangle_{l}^{k^{2}}\mod{\mathfrak{p}}=\langle Q_{1},Q_{2}\rangle_{l}$ for some integer $k$ coprime to $l$. ∎

We don’t actually need the image to be $\operatorname{SL}_{2}$ to apply the above theorem. However, $F$ may not contain the relevant roots of unity so to combat this we should consider the minimal polynomials.

Let $m_{F}(\alpha)$ denote the minimal polynomial of $\alpha$ over $F$, for any field $F$ and algebraic number $\alpha$, and define $\operatorname{GL}_{2}^{\square}:=\{A\in\operatorname{GL}_{2}\,|\,\det A\text{ is a square }\}$.

Let $\tilde{E}$ be the reduced curve at $\mathfrak{p}$ and suppose that $\{Q_{1},Q_{2}\}$ is an ordered basis of $\tilde{E}[l]$ such that the action of the Frobenius automorphism $x\mapsto x^{q}$ acts as $\sigma$ on $\tilde{E}[l]$ with respect to $\{Q_{1},Q_{2}\}$.

Then $\rho_{E,l}(\operatorname{Frob}_{\mathfrak{p}})$, written with respect to a global ordered basis $\{P_{1},P_{2}\}$, is $G$-conjugate to $\sigma$ if and only if

[TABLE]

for some $k\in\mathbb{Z}$.

Choose a global basis $P_{1}=(\alpha_{1},\beta_{1}),P_{2}=(\overline{\alpha_{1}},\overline{\beta_{1}})\in E(\mathbb{C})$, where $\alpha_{1}\approx 0.571+1.754i,\beta_{1}\approx 0.984+2.761i$ and observe that $\langle P_{1},P_{2}\rangle_{3}=\zeta_{3}$.

Now the reduced curve $\tilde{E}$ has $18$ points so the trace of Frobenius is $2\mod{3}$, hence the image of Frobenius (with respect to $\{P_{1},P_{2}\}$) is $\operatorname{SL}_{2}$-conjugate to $\left(\begin{smallmatrix}1&n\\ 0&1\end{smallmatrix}\right)$ for some $n\in\{0,1,2\}$. These all define distinct $\operatorname{SL}_{2}$-conjugacy classes, with the non-identity elements being $\operatorname{GL}_{2}$-conjugate.

A quick check shows that $\tilde{E}(\mathbb{F}_{13})[3]\neq 9$ hence $n\neq 0$ so $\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)$ is $\operatorname{GL}_{2}$-conjugate to $\rho_{E,3}(\operatorname{Frob}_{\mathfrak{p}})$.

Now $\tilde{E}[3]$ is defined over the cubic extension $\mathbb{F}_{13}[\alpha]$, where $\alpha$ has minimal polynomial $x^{3}+2x-2$. We compute that $Q_{1}=(10,6)$, $Q_{2}=(8\alpha^{2}-\alpha+3,7\alpha^{2}+4\alpha-1)$ is a basis of $\tilde{E}[3]$ such that the Frobenius automorphism acts as $\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)$ here.

The criterion we have in this case is equivalent to checking whether $\langle P_{1},P_{2}\rangle_{3}\equiv\langle Q_{1},Q_{2}\rangle_{3}\mod{\mathfrak{p}}$. A quick calculation shows that $\langle Q_{1},Q_{2}\rangle_{3}=3$ so $\rho_{E,3}(\operatorname{Frob}_{\mathfrak{p}})$ is $\operatorname{SL}_{2}$-conjugate to $\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)$ with respect to $\{P_{1},P_{2}\}$.

Let $\mathfrak{p}=(\frac{1+5\sqrt{5}}{2})$ be a prime of $\mathbb{Q}(\sqrt{5})$ above $31$. Choose the ordered global basis $P_{1}\approx(1.69-1.54i,-1.27+2.83i),P_{2}=(1,-1)$ so $\langle P_{1},P_{2}\rangle_{5}=e^{2\pi i/5}$. This is not an element of $\mathbb{Q}(\sqrt{5})$ so we instead take its minimal polynomial $m_{\mathbb{Q}(\sqrt{5})}(e^{2\pi i/5})=x^{2}+\frac{1}{2}(1+\sqrt{5})x+1$.

One can check that $\operatorname{Frob}_{\mathfrak{p}}$ has order $5$ using the group structure of the reduced curve and so is conjugate to either $\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)$ or $\left(\begin{smallmatrix}1&2\\ 0&1\end{smallmatrix}\right)$ under the ordered basis $\{P_{1},P_{2}\}$.

Let $Q_{1}=(1,-1),Q_{2}=(26\alpha^{3}+8\alpha^{2}+23\alpha+12,16\alpha^{4}+17\alpha^{3}+29\alpha^{2}+17\alpha+2)$, where $\alpha$ has minimal polynomial $x^{5}+7x+28$. Then the Frobenius automorphism acts on $\tilde{E}[5]$ as $\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)$ with respect to the ordered basis $\{Q_{1},Q_{2}\}$.

We compute that $\langle Q_{1},Q_{2}\rangle_{5}=8$. Now $m_{\mathbb{Q}(\sqrt{5})}(e^{2\pi i/5})\equiv x^{2}+13x+1\mod{\mathfrak{p}}$ which doesn’t have $8$ as a root so $\operatorname{Frob}_{\mathfrak{p}}$ cannot be conjugate to $\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)$. Redoing the calculation with $\left(\begin{smallmatrix}1&2\\ 0&1\end{smallmatrix}\right)$, (where we take the basis $\{Q_{1},Q_{1}+2Q_{2}\}$), the Weil pairing is $2$ which is a root of $m_{\mathbb{Q}(\sqrt{5})}(e^{2\pi i/5})\mod{\mathfrak{p}}$. Hence $\operatorname{Frob}_{\mathfrak{p}}$ is $D_{10}$-conjugate to $\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)$ with respect to the basis $\{P_{1},P_{2}\}$.

[TABLE]

The final thing we wish to address is how beneficial elliptic curves were here as to the feasibility of this method for Galois representations arising from other types of objects. We can also do this for larger dimensional vector spaces, so we’ll incorporate this into our theorem.

We first recall a construction which generalises the precise properties of the Weil pairing that we want. Let $V/\mathbb{F}_{l}$ be a vector space of dimension $n$. Then the $n^{th}$ exterior power $\Lambda^{n}V^{*}$ is a one dimensional vector space of alternating multilinear forms, such that for any nonzero $T\in\Lambda^{n}V^{*}$ we have

1. (i)

$T(v_{1},\cdots,v_{n})=0$ if and only if $\{v_{1},\cdots,v_{n}\}$ are linearly dependent, 2. (ii)

$T(Av_{1},\cdots,Av_{n})=\det(A)T(v_{1},\cdots,v_{n})$ for all matrices $A\in\operatorname{GL}_{n}$.

In the case of the Weil pairing, we identified the image $\mathbb{F}_{l}$ with the $l^{th}$ roots of unity and shall do so again in our final theorem. For a field $F$, we let $\mu_{l}(F)$ denote the $l^{th}$ roots of unity in $F$.

Let $V,\overline{V}$ be two $\mathbb{F}_{l}$-vector spaces of dimension $n$. Suppose $V$ (respectively $\overline{V}$) has a faithful action of $G$ (respectively $\overline{G}$) and there exists an isomorphism $\theta:V\rightarrow\overline{V}$ such that $\theta\overline{g}=\overline{g}\theta$ for all $\overline{g}\in\overline{G}$. Furthermore, suppose that there are nonzero alternating multilinear forms $T_{F}\in\Lambda^{n}V^{*}$ and $T_{\mathcal{F}}\in\Lambda^{n}\overline{V}^{*}$ such that the diagram

{V^{n}}$${\mu_{l}(F)}$${\overline{V}^{n}}$${\mu_{l}(\mathcal{F})}$$\scriptstyle{T_{F}}$$\scriptstyle{\theta_{n}}$$\scriptstyle{\mod{\mathfrak{p}}}$$\scriptstyle{T_{\mathcal{F}}}

commutes, where $\theta_{n}(v_{1},\cdots,v_{n}):=(\theta(v_{1}),\cdots,\theta(v_{n}))$.

Suppose the $\operatorname{GL}_{n}(\mathbb{F}_{l})$-conjugacy class of $\rho(\operatorname{Frob}_{\mathfrak{p}})$ splits into $m$ classes in $\operatorname{SL}_{n}(\mathbb{F}_{l})$ and let $H\subset\mathbb{F}_{l}^{\times}$ be the unique subgroup such that $[\mathbb{F}_{l}^{\times}:H]=m$. Suppose $\sigma\in\operatorname{SL}_{n}(\mathbb{F}_{l})$ is $\operatorname{GL}_{n}(\mathbb{F}_{l})$-conjugate to $\rho(\operatorname{Frob}_{\mathfrak{p}})$ and let $\overline{B}$ be an ordered basis of $\overline{V}$ such that the Frobenius automorphism acts as $\sigma$ on $\overline{V}$ with respect to $\overline{B}$. Then $\rho(\operatorname{Frob}_{\mathfrak{p}})$, written with respect to a global ordered basis $B$, is $\operatorname{SL}_{n}(\mathbb{F}_{l})$-conjugate to $\sigma$ if and only if

[TABLE]

Note that we used that the centraliser of $\sigma$ in $\operatorname{GL}_{2}$, $C_{\operatorname{GL}_{2}}(\sigma)$, was equal to its centraliser in $\operatorname{GL}_{2}^{\square}$ when the class splits; for general $n$, the latter group should be replaced by the group $\operatorname{GL}_{n}^{(n)}(\mathbb{F}_{l})$ of matrices whose determinant is an $n^{th}$ power. Moreover, we necessarily have $(\mathbb{F}_{l}^{\times})^{n}\subset H$ and $|C_{\operatorname{GL}_{n}(\mathbb{F}_{l})}(\sigma)|=|C_{\operatorname{GL}_{n}^{(n)}(\mathbb{F}_{l})}(\sigma)|[H:(\mathbb{F}_{l}^{\times})^{n}]$, so a similar computation then proves this more general version. ∎