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:
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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 $\mathbb{Z}/p\mathbb{Z}$.
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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.
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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.
Best Answer
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.