Action of the complex conjugation on modular Galois representations

Question

Let $p$ be an odd prime, and let $T$ be the Tate module of an elliptic curve defined over $\mathbb{Q}$, or the representation attached to a modular form or to a Hida family of modular forms. Why is it true that the $\pm 1$-eigenspaces of the residual representation of $T$ under the action of the complex conjugation are both $1$-dimensional? Why couldn't they be $0$ and $2$ dimensional, or vice-versa?

More precisely:

1. Let $E$ be an elliptic curve defined over $\mathbb{Q}$ and let $E[p]$ be the subgroup of $E$ consisting of $p$-torsion points, which is isomorphic to $(\mathbb{Z}/p\mathbb{Z})^2$. Then for example in [Gross, Kolyvagin's work on modular elliptic curves, Section 3] it is written that the $\pm 1$-eigenspaces for the action of the complex conjugation on $E[p]$ are both 1-dimensional over $\mathbb{Z}/p\mathbb{Z}$.

2. Let $V_f$ be a 2-dimensional $p$-adic representation attached to a normalized cusp form. Here for example [Nekovar, Kolyvagin's method for Chow groups of Kuga-Sato varieties, Chapter 10] suggests that the $\pm 1$-eigenspaces for the action of the complex conjugation on some finite quotients of $V_f$ are both 1-dimensional.

3. If $\mathbb{T}$ is the Galois representation attached to a Hida family of modular forms, [Nekovar-Plater, On the parity of ranks of Selmer groups, (1.5.3)] tells that the $\pm 1$-eigenspaces for the action of the complex conjugation on the residual representation attached to $\mathbb{T}$ are both 1-dimensional.

Answer 1

From the complex torus side $E[p^\infty]$ is dense in $E(\Bbb{C})$ and so the complex conjugaison $\rho$ doesn't act as $+1$ or $-1$ on the whole of $E[p^\infty]$.

So for $k$ large enough $(\rho+1)E[p^k]$ and $(\rho-1)E[p^k]$ will contain some non-$O$ elements $P,Q$.

As $(\rho-1)(\rho+1)E[p^k]=O$ it gives $(\rho-1)P=O,(\rho+1)Q=O$.

Multiplying $P,Q$ by suitable powers of $p$ it will give two points of order $p$ onto which $\rho$ acts as $+1$ and $-1$ and so $\rho \in End(E[p])$ has the two 1-dimensional eigenspaces.

Answer 2

reuns’s answer is very good, but here’s another, perhaps more arithmetic argument.

First for elliptic curves:

Let $c$ denote the complex conjugation. I claim that $\det(c\mid T)=-1$: this proves the claim, since $c^2=\mathrm{id}$.

To do that, it is enough to show that for every $n \geq 1$, $\det(c \mid E[p^n])=-1$.

But it's well known that we have the Galois-equivariant bilinear perfect Weil pairing $W: E[p^n] \times E[p^n] \rightarrow \mu_{p^n}$, such that $W(x,x)=1$. If $P,Q$ are any points in $E[p^n]$, then $W(P,Q)^{-1}=c(W(P,Q))=W(cP,cQ)=W(P,Q)^{\det(c\mid E[p^n])}$, which concludes.

This idea extends (somewhat) to representations attached to modular forms (and hence to Hida families).

Let’s take a newform $f \in \mathcal{S}_k(\Gamma_1(N))$ of nebentypus $\chi$. Then our $p$-adic representation $V_f$ is such that the characteristic polynomial of the Frobenius $F_{\ell}$ at a prime $\ell \not\mid pN$ is $X^2-a_{\ell}X+\ell^{k-1}\chi(\ell)$.

In particular, let $\psi$ be the determinant of $V_f$. Then, at every $F_{\ell}$ as above, $\psi$ agrees with $\omega_p^{k-1}\chi(\alpha_N)$, where $\omega_p$ is the $p$-adic cyclotomic character and $\alpha_N$ the mod $N$-cyclotomic character.

Now, $\psi$ and $\omega_p^{k-1}\chi(\alpha_N)$ are continuous characters: therefore, by Cebotarev, they’re equal. So $\det(c\mid T)=\omega_p^{k-1}(c)\chi(\alpha_N(c))=(-1)^{k-1}\chi(-1)=-1$ and we’re done.