User srilakshmi - MathOverflow

Image of a Galois representation

2012-01-20T07:35:45Z

Notation:

$E$ is a non-CM Elliptic curve over $\mathbb{Q}$.
$p$ is an ordinary prime.
$f$ - cuspidal eigenform of weight $k$ = 2 attached with $E$.
$\rho_f$ - the global 2-dimensional $p$-adic Galois representation attached with $f$. $\rho_f$ : $G_S$ $\rightarrow$ $\mathrm{GL}_2({\mathbb{Z}}_p)$.
$G_S:= \mathrm{Gal}(\mathbb{Q}_S/{\mathbb{Q}})$, where $\mathbb{Q}_S$ - maximal unramified extension outside the set $S=\{\text{ bad primes of } E \} \cup\{ p, \infty \}$ .

Assume that the residual representation $\overline{\rho}_f$ is $p$-split.

The prime $p$ is an ordinary prime. So the the image of $\rho_f$ restricted to the decomposition group $G_p:=\mathrm{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$ will be of the form $\rho_f$ $\mid$ $G_p$ $\sim$ $\begin{pmatrix} a & * \\ 0 & d \end{pmatrix}$ . The residual representation $\overline{\rho}_f$ is $p$-split. So, $\overline{\rho}_f$ $\sim$ $\begin{pmatrix} \omega \lambda_p^{-1}(\overline{a}_p) & 0 \\ 0 & \lambda_p(\overline{a}_p) \end{pmatrix}$ , where $\lambda_p$ is an unramified character which sends $\mathrm{Frob}_p$ to $\overline{a}_p$, $\overline{a}_p \in \mathbb{F}_p$ is the mod $p$ reduction of the $p$-th coefficent $a_p$ of $f$, and $\omega$ is the $p$-adic cyclotomic character.

Question: What is the image of the representation $\rho_f: G_p \rightarrow \mathrm{GL}_2(\mathbb{Z}/p^n \mathbb{Z})$, where $(\mathbb{Z}/p^n\mathbb{Z})^2 \simeq (E[p^n])$, the $p^n$-torsion points of $E$, for some fixed $n \geq 2$? Is it possible to compute it (or atleast it's order) by using MAGMA/SAGE/PARI?

Answer by Srilakshmi for Pairing on elliptic curve

2013-03-23T10:51:29Z

The groups $G_1$ and $G_2$ are generated by $P$ and $Q$ respecively. For every $s$, $f_{s,P}$ is the Miller function associated with $s$. In Optimized (lower degree Miller functions) pairing, one imposed condition on $s$ is : $s \equiv q \pmod r$. The integer $k$ is minimal such that $r$ $\mid$ ${q^k-1}$, which implies that $r$ $\mid$ $(s^k-1)$.

Answer by Srilakshmi for About equivalent statements of the Birch and Swinnerton-Dyer Conjecture

2013-03-20T13:41:38Z

Weaker version of BSD (Parity Conjecture):

$$(-1)^{\mathrm{rank}(E/K)} = w(E/K),$$ where $w(E/K)$ ( +1 or -1) is the global root number of $E/K$.

Standard statement:

BSD I:

If $K$ is a number field and $E$ is an elliptic curve over $K$, then $$\mathrm{ord}_{s=1} L(E/K,s) = \mathrm{rank} (E/K), $$ where $\mathrm{rank}(E/K)$ := Analytic rank of $E$ over $K$ := Mordell-Weil rank of $E$ over $K$.

BSD II: The order of $Ш$ is finite and the leading coeffcient of $L(E/K,s)$ at $s=1$ is given by $$\lim_{s \to 1} \frac{L(E/K,s)}{ (s-1)^r} = \frac{R.|Ш|.C}{\sqrt{{\triangle}_K} {|T|}^2 },$$ where $r$ is the Mordell-Weil rank of $E/K$, $R$ is the regulator of $E/K$ (with respect to the Neron-Tate height pairing), $|Ш|$ is the order of the Tate-Shafarevich group, $|T|$ is the order of the torsion group, $\triangle_K$ is the discriminant of $K$ and $C$ = $\prod_{v} c_{v} $ is the product of the local tamagawa numbers ($v$ varies over places of $K$).

When $K$ is a function field over a finite field of +ve characteristic,

$\mathrm{ord}_{s=1} L(E/K,s) = \mathrm{rank} (E/K) \iff |Ш| < \infty \iff |Ш_l^{\infty}| < \infty$ for some $l \iff \mathrm{ord}_{s=1} L(E/K,s) \leq \mathrm{rank} (E/K)$

Answer by Srilakshmi for Deducing BSD from Gross-Zagier and Kolyvagin

2013-03-12T08:26:49Z

Some results can be found in this survey: A survey on development of the Gross- Zagier formulas and their applications: Elliptic curves, L-functions, and CM-points by Shou-Wu Zhang.

The Doi-Naganuma Lift

2013-02-12T12:07:34Z

Let $F$=$\mathbb{Q}(\sqrt{D})$ be a real quadratic field and $\mathcal{O}_F$ be the ring of integers of $F$. The generating series $\Omega^{(k)}(z_1, z_2, \tau )$ = $\sum^{\infty}_{m=1} m^{k-1} \omega^{(k)}_m(z_1, z_2) e^{2\pi i m \tau}$ ($z_1$, $z_2$, $\tau$ $\in$ $\mathbb{H}$, the upper half plane) is both a Hilbert modular form (with respect to $z_1$ and $z_2$) and a classical modular form (with respect to $\tau$), where $\omega^{(k)}_m(z_1, z_2) = \sum_{(a,b,\lambda)}^{\prime} \frac{1}{(a z_1 z_2 + \lambda z_1 + \lambda^{\prime} z_2 + b)^k}$, the summation is over all triples $(a, b, \lambda)$ satisfying the conditions $a$, $b$ $\in$ $\mathbb{Z}$ and Norm of $\lambda$ - $ab$ = $\frac{m}{D}$, $\lambda$ $\in$ $\delta^{-1}$, $\delta$ is the principal ideal $(\sqrt{D})$. The notation $\lambda^{\prime}$ denote the conjugate of $\lambda$.

Zagier constructed $\Omega^{(k)}(z_1, z_2, \tau )$ (Reference: ``Modular forms associated to real quadratic fields"). The petersson inner product $\langle . , \Omega^{(k)} \rangle$ defines a linear map from classical cusp forms to Hilbert cusp forms, $S_k(\Gamma_0(D), \chi_D)$ to $S_{(k,k)}(SL_2(\mathcal{O}_F))$.

I'd like to know the history behind. My question is : What was the motivation for the construction of $\Omega^{(k)}(z_1, z_2, \tau )$? Thanks!

Answer by Srilakshmi for Who first noticed that the Hilbert symbol is a Steinberg symbol ?

2012-08-29T07:52:31Z

Matsumoto's computation of $K_2$ of a field in "Introduction to Algebraic K-Theory" by John Milnor.

Special value of $L$-function

2012-08-27T05:20:23Z

Let $p$ be a prime number. Let $f$ be a newform of weight 2 on $Γ_0(p)$, and $E_f$ denote the associated newform quotient of $J_0 (N)$ over $\mathbb{Q}$. Is there a way to express the algebraic part of the special $L$-value of $E_f$ over $K$, a totally real quadratic field, in terms of the conjugacy classes of maximal orders in the definite quaternion algebra over $\mathbb{Q}$ ramified at $p$ and $\infty$?. p.s: when $K$ is a quadratic imaginary field, there is such an expression.

Answer by Srilakshmi for Interesting mathematical documentaries

2012-06-20T11:03:22Z

PBS documentary A Brilliant Madness that looks at the life of Nobel-prize winning mathematician, John Nash

http://www.youtube.com/watch?v=chXIfhJ36Iw

Answer by Srilakshmi for Tamagawa Number of Elliptic Curves over $\mathbb{Q}$

2012-05-10T06:35:48Z

Look at http://mathoverflow.net/questions/71044/intuition-behind-the-tamagawa-number : Tate's article in Antwerp IV (Springer Lecture Notes in Mathematics 476), Modular Functions of One Variable IV.

Elliptic subfields of a function field

2012-05-09T10:37:59Z

Let $C$ be a curve and $K(C)$ be its function field of genus 2, where $K$ = $\mathbb{C}$.

The number of essential elliptic subfields of $K(C)$ is 0 or 2 or $\infty$.

Edit: I am looking for a proof. Thanks!

Conductor of an elliptic curve

2012-05-09T09:44:23Z

Given any elliptic curve over $\mathbb{Q}$ of conductor $N$, by modularity of elliptic curves, there exists a surjective morphism from $X_0(N)$ $\rightarrow$ $E$.There may be several such 'N' and morphisms from $X_0(N)$ $\rightarrow$ $E$. One such $N$ is the conductor of $E$.

If there exists an elliptic curve $E$ and a morphism $f$: $X_0(p)$ $\rightarrow$ $E$ for a fixed prime $p$, then "will it imply that the conductor of $E$ is $p$".

Order of Ш (Sha)

2012-04-30T06:08:26Z

To prove the BSD conjecture, one has to know about 'the finiteness of the Shafarevich Tate group'. But, an example of an elliptic curve of rank 2 (whose Sha group $Ш(E/\mathbb{Q})$ is finite) is not yet known.

Is there any example of an elliptic curve of rank 2 such that $p$-primary components of Ш are trivial for $p$ outside a finite set of primes?. In particular, $Ш(E/\mathbb{Q})[p]$ is trivial for $p$ $\neq$ 2, 3, 5, 7.

Bound for the number of rational points on the modular curve

2012-02-17T06:24:58Z

By Mazur's theorems (Reference:- Modular curves and the Eisenstein ideal, 1977), we know that the only rational points of X_0(N) for N any prime > 163 are the two cusps (o) and (oo) (|X_0(N)(Q)| = 2 ).

Is there any bound for |X_0(N)(Q)|, where N is an arbitrary +ve integer?.

More Generally,

Is there a bound on the number of rational points on the modular curve i.e. for |X_0(N)(K)|, where K is some number field. We know that |X_0(N)(K)| is finite.

Trace of Frobenius over $F_q$

2012-03-18T14:43:29Z

Let $q_0$ be a prime and $q$ = $q_0^n$. Let $a(F_q/F_{q_0})$ denote any integer which is trace of Frobenius over the field $F_q$ for some elliptic curve which can be defined over $F_{q_0}$.

It is stated in Mazur's paper (Rational isogenies of Primes degree, Inventiones mathematicae, 1978) that $a(F_3^{12}/F_3)$ = 658, -1358, +1458.

I can get only $\pm$ 1458 or $\pm$ 729 or 0 if i use the following thm.

(It might be a simple answer. But, I don't see that how 658, - 1358 occurs).

Theorem: Let $q_0$ be a prime and $q$ = $q_0^n$. Then there exists an elliptic curve $E$ dened over $F_q$ such that the trace of the Frobenius equals to $\beta$ ($\beta$ $\leq$ $\lfloor 2\sqrt{q} \rfloor$) if and only if one of the following cases occur:

(i) $q_0$ does not divide $\beta$,

(ii). $q$ is a square (i.e. $n$ is even) and

$\beta$ = $\pm 2 \sqrt{q}$ or

$\beta$ = $\pm \sqrt{q}$ ($q_0$ $\not\equiv$ 1 (mod 3)) or

$\beta$= 0 and ($q_0$ $\not\equiv$ 1 (mod 4)),

(iii) $q$ is not a square (i.e. $n$ is odd) and

$\beta$ = 0 or $\beta$ = $\pm$ $q_0^{n+1/2}$ and $q_0$ = 2,3.

Answer by Srilakshmi for Galois Cohomology maps

2012-01-20T13:57:23Z

Note that $\mathrm{dim}(H^{1}(G_S,\mathrm{Ad}^{0}))$ = 2. So, if your map is injective, then the image will be of dimension 2. Otherwise, dimension 1 or 0. Finding the injectiveness of the map is a seperate problem (not known yet). See for more details: Recent paper by E.Ghate and V.Vatsal (Locally indecomposable Galois representations).

Comment by Srilakshmi

2013-04-03T15:41:02Z

I meant the induced map is injective.

Comment by Srilakshmi

2013-03-20T20:02:14Z

Is there any relation between $s$ and $q$? What are $P$ and $Q$?

Comment by Srilakshmi

2013-03-16T06:23:40Z

mathoverflow.net/questions/123813/…

Comment by Srilakshmi

2013-02-12T17:03:21Z

Thanks for removing the tag. There is a geometric interpretation of the Doi-Naganuma lifting and it's adjoint. I amn't sure arithmetic-geometry tag would be suitable (Probably you can retag).

Comment by Srilakshmi

2012-08-29T09:35:48Z

Dear Sir, I thought Matsumoto was the first person.

Comment by Srilakshmi

2012-08-19T06:15:42Z

When i came to know abt. this conjecture, i thought first row p1, p2, p3, p4,....and second row p2-p1, p3-p2, p4-p3,... and third row p3-2p2+p1...and i ended up with pascal's triangle. i.e. To Prove (n-1)C0 p_n - (n-1)C1 p_(n-1) +...+(-1)^(n-1) (n-1)C1 p1 = 1. This might be done by applying a formula for p_n (for eg. paper by willans ). Then i realised that i forgot about the absolute values of the differences.

Comment by Srilakshmi

2012-06-20T07:16:37Z

@Marvis: I gave the link for the film "A beautiful mind" ( biographical drama film based on the life of John Nash). It is not a documentary. i.e. why down vote.

Comment by Srilakshmi

2012-06-18T05:28:35Z

@Joseph: Which software did you use to draw these curves?

Comment by Srilakshmi

2012-06-08T08:43:09Z

Dear Francois, Thanks for your comment. I also thought that the statement can be generalized.

Comment by Srilakshmi

2012-05-23T06:21:28Z

What is special about genus 2 here? Can we generalize for higher genus too (looking at the jacobian decomposition of Jac(C)).

Comment by Srilakshmi

2012-05-11T07:53:43Z

It is a standard proof for your question (using unique-prime-factorization theorem).

Comment by Srilakshmi

2012-05-10T05:58:29Z

Dear Francois, Thanks for your comments.

Comment by Srilakshmi

2012-05-09T12:36:12Z

Essential subfield: A subfield of +ve genus and also "maximal" in the sense that it is not contained in any other subfield of same genus. Elliptic subfield: Genus 1 subfield of K(C).

Comment by Srilakshmi

2012-05-09T11:29:52Z

Dear David, Thanks for your comment.

Comment by Srilakshmi

2012-05-01T13:59:57Z

Dear Alex, Thanks for your comment.