# Image of complex conjugation by modular representations in characteristic 2

The question I am going to ask looks well-known, and I even may have heard things about it (but since I used to be deaf to anything in characteristic 2, whatever I heard has never been recorded in my mind) so it may be seen as a request for references. I have not been able to find a reference myself.

Let $f$ be a cuspidal eigenform of weight $k \geq 2$ for some congruence subgroup of $Sl_2(\mathbb{Z})$. Then as is well-known, for every prime $p$, there exists a unique, absolutely irreducible, Galois representation $\rho : G_{\mathbb Q} \rightarrow Gl_2(K)$ where $K$ is a suitable finite extension of $\mathbb Q_p$, odd (that is such that $\rho(c)$ is conjugate to the diagonal matrix $(1,-1)$), and satisfying the Eichler-Shimura relations.

I am interested in the case $p=2$. Let $A$ be the ring of integers of $K$, $m$ is maximal ideal, and $k=A/m$ the residue field, of characteristic $2$. I want to reduce $\rho$ mod $m$. As is still well-known, there are several way to do that, one for each choice of a stable $A$-lattice $\Lambda$ in $K^2$: one defines the representation $\bar \rho_\Lambda$ over $k$ as the action of $G_{\mathbb Q}$ on $\Lambda/m \Lambda$. The various $\bar \rho_\Lambda$ have all the same semi-simplification.

Now my question: What is the conjugacy class of $\bar \rho_\Lambda(c)$ in $GL_2(k)$?

The characteristic polynomial of $\bar \rho_\Lambda(c)$ is $X^2-1 = (X-1)^2$ in $k$, so either this matrix is the identity, or it is conjugate to the unipotent matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. I'd like to know when (that is for which $f$, $\Lambda$) we are in the first case, and when we are in the second case.

Note: the fact that $\rho(c)$ is conjugate to the diagonal matrix $(1,-1)$ does not trivially implies that $\bar \rho_\Lambda(c)$ is necessarily the diagonal matrix $(1,1)$, because the two eigen-lines of $\rho(c)$ may not be in good position w.r.t the lattice $\Lambda$, that is the sum of their intersections with $\Lambda$ may be a proper sub-lattice of $\Lambda$. For example if $\rho(c)$ is the anti-diagonal matrix in the canonical basis of $K^2$, and $\Lambda = A \oplus A$ is a stable lattice, then $\bar \rho_\Lambda(c)$ is clearly not the identity.

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Joel: I fixed your matrix :-) Loic Merel gave a talk about more or less exactly this question in the Computations with Modular Forms conference in Heidelberg this summer -- I don't know if the notes are available electronically anywhere. –  David Loeffler Oct 20 '11 at 18:12
Just a comment: if the residual, mod 2 representation is irreducible (therefore semi-simple) then the semi-simplicity matter at infinity should not depend on the choice of the lattice used for reducing. –  Tommaso Centeleghe Oct 20 '11 at 22:08

Joel -- it's difficult to work out what you're asking. Of course both possibilities can occur, as Wanax said. Furthermore both possibilities can occur even for the same modular form. For example, if you consider the 2-adic representation attached to Ramanujan $\Delta$ then there is a lattice for which $\overline{\rho}(c)=1$ and another lattice for which it is not 1 (in fact for any quadratic extension of the rationals with discriminant a power of two there is some lattice which gives rise to the reducible non-semisimple representation with kernel corresponding to this field). And already in Serre's 1987 paper on his conjecture he observes that for any $S_3$ extension of $\mathbf{Q}$, totally real or not, it will be modular, and for $A_5$ extensions, totally real or not, computationally they seem to work too (thanks to Mestre).
Kevin, when you say that "both possibilities can occur even for the same modular form" then in this case the global, mod $2$ representation must be reducible, right? –  Tommaso Centeleghe Oct 20 '11 at 21:05
Tommaso: right. If $\bar \rho_\Lambda$ is irreducible, then it is independent of $\Lambda$ by Brauer-Nesbitt, and actually but is not hard to prove that there is only one stable lattice $\Lambda$ (up to a scalar). –  Joël Oct 20 '11 at 22:21
Kevin, Thanks a lot. I am trying to understand your example with $\Delta$. I see that the semi-simplification of $\bar \rho_\Lambda$ is the trivial representation (since $\Delta \equiv \sum q^{(2n+1)^2} \pmod{2}$ (Jacobi), all Hecke eigenvalues are $0=2$). By Ribet's lemma+epsilon, there is at least 2 lattices $\Lambda$ such that $\rhob_{\Lambda}$ is a non-trivial extension of the trivial character by itself.But how do you prove the existence of one such $\Lambda$ with $\bar \rho_\Lambda(c) \neq \Id$ ? –  Joël Oct 20 '11 at 22:32
Re: the $\Delta$ function mod 2: I proved this when I was a graduate student, using work of Swinnerton-Dyer which explicitly computes the image of the 2-adic representation. I don't know of a reference other than chapter 6 of my thesis: www2.imperial.ac.uk/~buzzard/maths/research/notes/phd.dvi There is a funny historical story around why I was thinking about this, but it's more than 500 characters so I'll send it to you by email. –  Kevin Buzzard Oct 20 '11 at 22:35