Timeline for action of complex conjugation on Tate modules of elliptic curves
Current License: CC BY-SA 3.0
9 events
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| Oct 9, 2016 at 18:08 | comment | added | Will Chen | @user1952009 If $T_\mathbb{Z}$ is the Hecke algebra (the algebra over $\mathbb{Z}$ generated by $T_n,\langle n\rangle$ for $n\ge 1$), then for every Hecke operator $T$, let $\lambda_f(T)$ be the eigenvalue of $f$ w.r.t. $T$. Let $I_f := Ker\lambda_f$, then the sub-abelian variety mentioned above is just $J_f := J(\Gamma)/I_fJ(\Gamma)$. In the above, we usually take $\Gamma = \Gamma_1(N)$ or $\Gamma_0(N)$, such that the corresp. modular curve is defined over $\mathbb{Q}$. Finally, the Galois rep assoc. to $f$ is just the representation of $G_\mathbb{Q}$ on the Tate module of $J_f$. | |
| Oct 9, 2016 at 18:05 | comment | added | Will Chen | @user1952009 If you have a weight 2 Hecke eigenform $f$ for some congruence subgroup $\Gamma$, you can think of it as a holomorphic differential 1-form $\omega_f$ on the (compact) genus $g$ modular curve $X(\Gamma)$. Since $f$ is an eigenform, the Hecke operators stabilize the vector space generated by $f$, and since they descend to an action on the Jacobian $J(\Gamma)$ (which we think of as $\Omega^1_{hol}(X)^{\wedge}/H_1(X,\mathbb{Z})$), where they cut out a sub abelian variety of dimension equal to $[K_f,\mathbb{Q}]$, where $K_f$ is the field generated by the Fourier coefficients of $f$. | |
| Oct 9, 2016 at 9:57 | comment | added | reuns | @rtz Hi I stopped Diamond's book at the algebraic geometry chapters, but I studied the analytic chapters, and I have no problems with L-functions. Can you tell me in a few words what is a Galois representation associated to a modular form ? (possibly an example ?) | |
| Oct 9, 2016 at 8:43 | review | Close votes | |||
| Oct 9, 2016 at 11:24 | |||||
| Oct 9, 2016 at 6:50 | comment | added | stupid_question_bot | Ah fantastic! if you want to just copy/paste your comment into an answer I'd be happy to accept it. | |
| Oct 9, 2016 at 6:45 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 9, 2016 at 6:43 | history | edited | Will Chen | CC BY-SA 3.0 |
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| Oct 9, 2016 at 6:12 | comment | added | nfdc23 | The determinant is the $\ell$-adic cyclotomic character, which has value $-1$ on complex conjugation since the complex conjugate of any root of unity is its reciprocal. Since a $2 \times 2$ matrix of order 2 in characteristic 0 with determinant $-1$ must have eigenvalues $1$ and $-1$, this settles your questions for elliptic curves over any field equipped with an embedding into $\mathbf{R}$, by entirely elementary means (no need for modular forms, etc.). | |
| Oct 9, 2016 at 6:00 | history | asked | stupid_question_bot | CC BY-SA 3.0 |