# How the automorphism group of an elliptic curve acts at the localization of the stack $\mathcal{M}_{1, 1, k}$ at the corresponding point

I am studying the enlightening article "The Picard Group of $$\mathcal{M}_{1, 1, S}$$", written by Fulton and Olsson, but I have some problems with a proof.

Setting Let $$\mathcal{M}_{1, 1, k}$$ denote the stack of elliptic curves over an algebraically closed field $$k$$ with $$char(k)=3$$. We have $$\begin{equation*} V = Spec(k[\lambda][1/(\lambda(\lambda - 1))]) \rightarrow \mathcal{M}_{1, 1, k} \rightarrow \mathbb{A}^{1}_{k} \end{equation*}$$ where the first map (which is étale) is the Legendre family, hence induced by the ellptic curve $$E_{V} \rightarrow V$$ $$\begin{equation*} E_{V} \colon Y^{2}Z = X(X-Z)(X - \lambda Z) \end{equation*}$$ and the second map is the morphism into the coarse moduli space induced by the $$j$$-invariant. The composition is induced by the ring map sending $$j \mapsto \frac{2^{8}(\lambda^{2}-\lambda+1)^{3}}{\lambda^{2} (1 - \lambda)^{2}}$$. Recall that in $$char(k)=3$$ the elliptic curves have automorphism group $$\mathbb{Z}/2\mathbb{Z}$$ except that for the curve with $$j$$-invariant equal to $$0$$ (which in the family corresponds to $$\lambda = -1$$) in which case the automorphism group is isomorphic to $$G = \mathbb{Z}/4\mathbb{Z} \ltimes \mathbb{Z}/3\mathbb{Z}$$.

The problem I want to understand how the group $$G$$ acts on the localization in $$\mathcal{M}_{1, 1, k}$$ at the point corresponding to the curve whith automorphism group $$G$$. Since $$V \rightarrow \mathcal{M}_{1, 1, k}$$ is étale, the problem is equivalent to understanding how it acts on the localization at $$\lambda = -1$$ in V.

What it is said in the article For comfort, we can translate and consider $$\mu = \lambda + 1$$ so the localization is at $$\mu = 0$$ and we have the morphism $$Spec(k[[\mu]]) \rightarrow Spec(k[[j]])$$ induced by $$j \mapsto \frac{\mu^{6}}{(\mu^{4}-1)}$$ (since we hare in $$char(k)=3$$). Now the group $$G$$ acts on $$k[[\mu]]$$. The group sits in an exact sequence $$\begin{equation*} 1 \rightarrow \{ \pm 1 \} \rightarrow G \rightarrow S_{3} \rightarrow 1 \end{equation*}$$ and the action on $$k[[\mu]]$$ factors through the action of $$S_{3}$$ on $$k[[\mu]]$$ given by the two automorphisms $$\begin{equation*} \alpha \colon \mu \mapsto -\mu \end{equation*}$$ and $$\begin{equation*} \beta \colon \mu \mapsto \mu/(1-\mu) = \mu(1 + \mu + \mu^{2} + \ldots). \end{equation*}$$

My question is: How the authors of the article determine the actions of $$\alpha$$ and $$\beta$$?

My attempt at a solution I think to know how $$\alpha$$ acts in such a way.

First of all, if we consider a model for an elliptic curve with $$j$$ invariant equal to 0, like $$y^{2}=x^{3} + a_{4}x + a_{6}$$, with $$a_{4} \ne 0$$, the morphism of elliptic curves are of the form $$\begin{equation*} \begin{split} x &\mapsto u^{2}x' + r \\ y &\mapsto u^{3}y' \end{split} \end{equation*}$$ and imposing the equality between the curve in $$(x, y)$$ and the curve in $$(x', y')$$, i.e. imposing the morphism being an automorphism, we obtain $$u^{4}=1$$ and $$r^{3} + a_{4}r + (1-u^{2})a_{6}=0$$. The $$12$$ elements $$(u, r)$$ which satisfy the equations are the group elements of $$G$$. In particular $$\alpha = (i, 0)$$.

To see how $$\alpha$$ acts: let $$Spec(k[x, y])/(y^{2} = x^{3} + a_{2}x^{2} + a_{4}x + a_{6})) \rightarrow Spec(k)$$ be a generic elliptic curve. On the coefficients $$(a_{2}, a_{4}, a_{6})$$, the transformation $$(u, r)=(i, 0)$$ acts via $$(i, 0) \star (a_{2}, a_{4}, a_{6}) = (i^{2}a_{2}, i^{4}a_{4}, i^{6}a_{6})$$. Since we are interested at the point $$(0, 1, 1)$$ we consider the curve $$y^{2} = x^{3} + \mu x^{2} + x + 1$$ on which the action is $$(i, 0) \star (\mu) = i^{2} \mu = - \mu$$ and taking the localization at $$\mu=0$$ is exactly what we want to see the action of $$\alpha= (i, 0)$$ at the point $$\lambda = -1$$ (i.e. $$\mu=0$$).

My problem is that the previous argument cannot be applied to $$\beta$$.

Can somone help me? (maybe also with a simpler way to compute the action of $$\alpha$$)

Thanks

From Section 3 of the paper, we see that the moduli stack of elliptic curves (over a ring $$A$$) has a quotient stack presentation $$\mathcal{M}_{1,1,A} \simeq [U/G]$$ where $$U = \operatorname{Spec} A[a_{1},a_{3},a_{2},a_{4},a_{6},\Delta^{-1}]$$ and $$G = \operatorname{Spec} A[u^{\pm},r,s,t]$$. The group law on $$G$$ and the action of $$G$$ on $$U$$ come from making the change-of-variable $$$$\begin{bmatrix} X \\ Y \\ Z \end{bmatrix} \mapsto \begin{bmatrix} u^{2} & & r \\ u^{2}s & u^{3} & t \\ & & 1 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} \tag{1}\label{eqn01}$$$$ in the Weierstrass equation $$Y^{2}Z+a_{1}XYZ+a_{3}YZ^{2} = X^{3} + a_{2}X^{2}Z + a_{4}XZ^{2} + a_{6}Z^{3}$$ (see Silverman, Arithmetic of Elliptic Curves, Section III.1, Table 3.1).
Suppose $$2 \in A^{\times}$$. Given $$a_{1},a_{3},a_{2},a_{4},a_{6} \in A$$ such that $$\Delta \in A^{\times}$$, we can change variables so that $$a_{1} = a_{3} = 0$$ (which implies $$s = t = 0$$), and furthermore we can replace $$A$$ by an etale cover $$A \to A'$$ and change variables so that \begin{aligned} a_{6} &= 0 \\ a_{2}+a_{4}+1 &= 0 \end{aligned} \tag{2}\label{eqn02} (i.e. write the equation in (dehomogenized) Legendre normal form $$$$y^{2} = x(x-1)(x-\lambda) \tag{3}\label{eqn03}$$$$ for $$\lambda = a_{4}$$). The formula \eqref{eqn01} becomes $$(x,y) \mapsto (u^{2}x+r,u^{3}y)$$. The equation \eqref{eqn03} is transformed into $$$$\textstyle y^{2} = (x+\frac{r}{u^{2}})(x+\frac{r-1}{u^{2}})(x+\frac{r-\lambda}{u^{2}}) \tag{4} \label{eqn04}$$$$ and imposing the condition \eqref{eqn02} gives \begin{aligned} r(r-1)(r-\lambda) &= 0 \\ \textstyle (1+\frac{r}{u^{2}})(1+\frac{r-1}{u^{2}})(1+\frac{r-\lambda}{u^{2}}) &= 0 \end{aligned} \tag{5} \label{eqn05} respectively.
Let $$k$$ be an algebraically closed field of $$\operatorname{char} k = 3$$, and set $$A := k[[\mu]]$$ and $$\lambda = \mu-1$$. The stabilizer $$\Gamma_{\overline{x}}$$ can be identified with the set of pairs $$(u,r) \in A^{\times} \times A$$ satisfying \eqref{eqn05}, and the action $$\Gamma_{\overline{x}} \times A \to A$$ is defined by $$$$(u,r) \cdot \lambda := \textstyle \frac{1}{u^{4}}(r(r-1) + r(r-\lambda) + (r-1)(r-\lambda))$$$$ (namely the coefficient of $$x$$ in \eqref{eqn04}, the new $$a_{4}$$). You can check that the solutions are $$$$(u,r) = (\pm 1 , 0), (\pm \sqrt{\lambda},0), (\pm \sqrt{-1},1), (\pm \sqrt{\lambda-1},1), (\pm \sqrt{-\lambda},\lambda),(\pm \sqrt{1-\lambda},\lambda)$$$$ which sends $$\mu$$ to $$$$\mu,\frac{\mu}{\mu-1},-\mu,\frac{\mu}{\mu+1},\frac{-\mu}{\mu-1},\frac{-\mu}{\mu+1}$$$$ respectively. It seems the authors don't specify what $$\alpha,\beta$$ are but if they're interested in checking a certain condition for a set of generators of $$S_{3}$$, then it is enough to choose elements of order 2 and 3.