User srilakshmi - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:24:10Z http://mathoverflow.net/feeds/user/20754 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86179/image-of-a-galois-representation Image of a Galois representation Srilakshmi 2012-01-20T07:35:45Z 2013-04-03T11:27:51Z <p>Notation:</p> <ul> <li>$E$ is a non-CM Elliptic curve over $\mathbb{Q}$.</li> <li>$p$ is an ordinary prime.</li> <li>$f$ - cuspidal eigenform of weight $k$ = 2 attached with $E$.</li> <li>$\rho_f$ - the global 2-dimensional $p$-adic Galois representation attached with $f$. $\rho_f$ : $G_S$ $\rightarrow$ $\mathrm{GL}_2({\mathbb{Z}}_p)$.</li> <li>$G_S:= \mathrm{Gal}(\mathbb{Q}_S/{\mathbb{Q}})$, where $\mathbb{Q}_S$ - maximal unramified extension outside the set <code>$S=\{\text{ bad primes of } E \} \cup\{ p, \infty \}$</code>.</li> </ul> <p>Assume that the residual representation $\overline{\rho}_f$ is $p$-split. </p> <p>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 $\rho_f$ $\mid$ $G_p$ $\sim$ <code>$\begin{pmatrix} a &amp; * \\ 0 &amp; d \end{pmatrix}$</code>. The residual representation $\overline{\rho}_f$ is $p$-split. So, $\overline{\rho}_f$ $\sim$ <code>$\begin{pmatrix} \omega \lambda_p^{-1}(\overline{a}_p) &amp; 0 \\ 0 &amp; \lambda_p(\overline{a}_p) \end{pmatrix}$</code>, 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.</p> <blockquote> <p><strong>Question</strong>: 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?</p> </blockquote> http://mathoverflow.net/questions/124976/pairing-on-elliptic-curve/125364#125364 Answer by Srilakshmi for Pairing on elliptic curve Srilakshmi 2013-03-23T10:51:29Z 2013-03-23T10:51:29Z <p>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 $r$ $\mid$ $(s^k-1)$. </p> http://mathoverflow.net/questions/125050/about-equivalent-statements-of-the-birch-and-swinnerton-dyer-conjecture/125067#125067 Answer by Srilakshmi for About equivalent statements of the Birch and Swinnerton-Dyer Conjecture Srilakshmi 2013-03-20T13:41:38Z 2013-03-20T15:35:17Z <p>Weaker version of BSD (Parity Conjecture): </p> <p>$$(-1)^{\mathrm{rank}(E/K)} = w(E/K),$$ where $w(E/K)$ ( +1 or -1) is the global root number of $E/K$.</p> <p>Standard statement: </p> <p>BSD I: </p> <p>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 $K$.</p> <p>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 $K$).</p> <p>When $K$ is a function field over a finite field of +ve characteristic,</p> <p>$\mathrm{ord}_{s=1} L(E/K,s) = \mathrm{rank} (E/K) \iff |Ш| &lt; \infty \iff |Ш_l^{\infty}| &lt; \infty$ for some $l \iff \mathrm{ord}_{s=1} L(E/K,s) \leq \mathrm{rank} (E/K)$</p> http://mathoverflow.net/questions/124202/deducing-bsd-from-gross-zagier-and-kolyvagin/124298#124298 Answer by Srilakshmi for Deducing BSD from Gross-Zagier and Kolyvagin Srilakshmi 2013-03-12T08:26:49Z 2013-03-12T08:33:45Z <p>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.</p> http://mathoverflow.net/questions/121589/the-doi-naganuma-lift The Doi-Naganuma Lift Srilakshmi 2013-02-12T12:07:34Z 2013-02-12T17:02:06Z <p>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 $\lambda$.</p> <p>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 $S_{(k,k)}(SL_2(\mathcal{O}_F))$. </p> <p>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! </p> http://mathoverflow.net/questions/105791/who-first-noticed-that-the-hilbert-symbol-is-a-steinberg-symbol/105802#105802 Answer by Srilakshmi for Who first noticed that the Hilbert symbol is a Steinberg symbol ? Srilakshmi 2012-08-29T07:52:31Z 2012-08-29T07:52:31Z <p>Matsumoto's computation of $K_2$ of a field in "Introduction to Algebraic K-Theory" by John Milnor. </p> http://mathoverflow.net/questions/105600/special-value-of-l-function Special value of $L$-function Srilakshmi 2012-08-27T05:20:23Z 2012-08-28T00:14:24Z <p>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. </p> http://mathoverflow.net/questions/100033/interesting-mathematical-documentaries/100114#100114 Answer by Srilakshmi for Interesting mathematical documentaries Srilakshmi 2012-06-20T11:03:22Z 2012-06-20T11:03:22Z <p>PBS documentary A Brilliant Madness that looks at the life of Nobel-prize winning mathematician, John Nash</p> <p><a href="http://www.youtube.com/watch?v=chXIfhJ36Iw" rel="nofollow">http://www.youtube.com/watch?v=chXIfhJ36Iw</a></p> http://mathoverflow.net/questions/96542/tamagawa-number-of-elliptic-curves-over-mathbbq/96543#96543 Answer by Srilakshmi for Tamagawa Number of Elliptic Curves over $\mathbb{Q}$ Srilakshmi 2012-05-10T06:35:48Z 2012-05-10T06:45:27Z <p>Look at <a href="http://mathoverflow.net/questions/71044/intuition-behind-the-tamagawa-number" rel="nofollow">http://mathoverflow.net/questions/71044/intuition-behind-the-tamagawa-number</a> : Tate's article in Antwerp IV (Springer Lecture Notes in Mathematics 476), Modular Functions of One Variable IV.</p> http://mathoverflow.net/questions/96426/elliptic-subfields-of-a-function-field Elliptic subfields of a function field Srilakshmi 2012-05-09T10:37:59Z 2012-05-10T04:57:48Z <p>Let $C$ be a curve and $K(C)$ be its function field of genus 2, where $K$ = $\mathbb{C}$. </p> <p>The number of essential elliptic subfields of $K(C)$ is 0 or 2 or $\infty$.</p> <p>Edit: I am looking for a proof. Thanks!</p> http://mathoverflow.net/questions/96421/conductor-of-an-elliptic-curve Conductor of an elliptic curve Srilakshmi 2012-05-09T09:44:23Z 2012-05-09T11:33:09Z <p>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 $E$.</p> <p>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 $p$".</p> http://mathoverflow.net/questions/95555/order-of-sha Order of Ш (Sha) Srilakshmi 2012-04-30T06:08:26Z 2012-04-30T12:50:24Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/88694/bound-for-the-number-of-rational-points-on-the-modular-curve Bound for the number of rational points on the modular curve Srilakshmi 2012-02-17T06:24:58Z 2012-03-19T14:36:17Z <p>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 ).</p> <p>Is there any bound for |X_0(N)(Q)|, where N is an arbitrary +ve integer?.</p> <p>More Generally,</p> <p>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.</p> http://mathoverflow.net/questions/91537/trace-of-frobenius-over-f-q Trace of Frobenius over $F_q$ Srilakshmi 2012-03-18T14:43:29Z 2012-03-18T15:16:50Z <p>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 $F_{q_0}$. </p> <p>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. </p> <p>I can get only $\pm$ 1458 or $\pm$ 729 or 0 if i use the following thm.</p> <p>(It might be a simple answer. But, I don't see that how 658, - 1358 occurs).</p> <p>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:</p> <p>(i) $q_0$ does not divide $\beta$,</p> <p>(ii). $q$ is a square (i.e. $n$ is even) and</p> <p>$\beta$ = $\pm 2 \sqrt{q}$ or </p> <p>$\beta$ = $\pm \sqrt{q}$ ($q_0$ $\not\equiv$ 1 (mod 3)) or</p> <p>$\beta$= 0 and ($q_0$ $\not\equiv$ 1 (mod 4)),</p> <p>(iii) $q$ is not a square (i.e. $n$ is odd) and</p> <p>$\beta$ = 0 or $\beta$ = $\pm$ $q_0^{n+1/2}$ and $q_0$ = 2,3. </p> http://mathoverflow.net/questions/68390/galois-cohomology-maps/86207#86207 Answer by Srilakshmi for Galois Cohomology maps Srilakshmi 2012-01-20T13:57:23Z 2012-01-20T13:57:23Z <p>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).</p> http://mathoverflow.net/questions/126339/generators-for-a-certain-congruence-subgroup-of-sln-z/126370#126370 Comment by Srilakshmi Srilakshmi 2013-04-03T15:41:02Z 2013-04-03T15:41:02Z I meant the induced map is injective. http://mathoverflow.net/questions/124976/pairing-on-elliptic-curve Comment by Srilakshmi Srilakshmi 2013-03-20T20:02:14Z 2013-03-20T20:02:14Z Is there any relation between $s$ and $q$? What are $P$ and $Q$? http://mathoverflow.net/questions/124659/analytic-rank-of-an-elliptic-curve-with-algebraic-rank-0 Comment by Srilakshmi Srilakshmi 2013-03-16T06:23:40Z 2013-03-16T06:23:40Z <a href="http://mathoverflow.net/questions/123813/inequality-relating-rank-and-analytic-rank" rel="nofollow" title="inequality relating rank and analytic rank">mathoverflow.net/questions/123813/&hellip;</a> http://mathoverflow.net/questions/121589/the-doi-naganuma-lift Comment by Srilakshmi Srilakshmi 2013-02-12T17:03:21Z 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). http://mathoverflow.net/questions/105791/who-first-noticed-that-the-hilbert-symbol-is-a-steinberg-symbol/105802#105802 Comment by Srilakshmi Srilakshmi 2012-08-29T09:35:48Z 2012-08-29T09:35:48Z Dear Sir, I thought Matsumoto was the first person. http://mathoverflow.net/questions/34669/is-there-any-progress-toward-solving-gilbreaths-conjecture Comment by Srilakshmi Srilakshmi 2012-08-19T06:15:42Z 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. http://mathoverflow.net/questions/100033/interesting-mathematical-documentaries/100081#100081 Comment by Srilakshmi Srilakshmi 2012-06-20T07:16:37Z 2012-06-20T07:16:37Z @Marvis: I gave the link for the film &quot;A beautiful mind&quot; ( biographical drama film based on the life of John Nash). It is not a documentary. i.e. why down vote. http://mathoverflow.net/questions/99850/rank-of-x-x2-1-c-c2-1-y2-over-mathbbq-for-given-rational-va/99866#99866 Comment by Srilakshmi Srilakshmi 2012-06-18T05:28:35Z 2012-06-18T05:28:35Z @Joseph: Which software did you use to draw these curves? http://mathoverflow.net/questions/96426/elliptic-subfields-of-a-function-field/96444#96444 Comment by Srilakshmi Srilakshmi 2012-06-08T08:43:09Z 2012-06-08T08:43:09Z Dear Francois, Thanks for your comment. I also thought that the statement can be generalized. http://mathoverflow.net/questions/96426/elliptic-subfields-of-a-function-field/96444#96444 Comment by Srilakshmi Srilakshmi 2012-05-23T06:21:28Z 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)). http://mathoverflow.net/questions/96647/a-proof-for-irrationality-of-sqrt2 Comment by Srilakshmi Srilakshmi 2012-05-11T07:53:43Z 2012-05-11T07:53:43Z It is a standard proof for your question (using unique-prime-factorization theorem). http://mathoverflow.net/questions/96426/elliptic-subfields-of-a-function-field/96444#96444 Comment by Srilakshmi Srilakshmi 2012-05-10T05:58:29Z 2012-05-10T05:58:29Z Dear Francois, Thanks for your comments. http://mathoverflow.net/questions/96426/elliptic-subfields-of-a-function-field Comment by Srilakshmi Srilakshmi 2012-05-09T12:36:12Z 2012-05-09T12:36:12Z Essential subfield: A subfield of +ve genus and also &quot;maximal&quot; in the sense that it is not contained in any other subfield of same genus. Elliptic subfield: Genus 1 subfield of K(C). http://mathoverflow.net/questions/96421/conductor-of-an-elliptic-curve/96433#96433 Comment by Srilakshmi Srilakshmi 2012-05-09T11:29:52Z 2012-05-09T11:29:52Z Dear David, Thanks for your comment. http://mathoverflow.net/questions/95555/order-of-sha/95562#95562 Comment by Srilakshmi Srilakshmi 2012-05-01T13:59:57Z 2012-05-01T13:59:57Z Dear Alex, Thanks for your comment.