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3 Fixed matrix

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 $Gl_2(k)$?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 $(1 \\begin{pmatrix} 1 & 1$1 \\$ 0 \& 1)$1 \end{pmatrix}$. 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.

2 added 27 characters in body; added 1 characters in body

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 $(1 \& 1$ \ \ $0 \& 1)$. 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 with 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.

1

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 $(1 \& 1$ \ \ $0 \& 1)$. 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 $(1,-1)$ 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 with 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.