10
$\begingroup$

Sometime back i watched the documentary of Andrew Wiles proving the Fermat's Last theorem. A truly inspiring video and i still watch it whenever i am in a depressed mood. There are certain things(infact many) which i couldn't follow and i would like it to be explained here.

The first is:

The Taniyama-Shimura conjecture. In the video it's said that that an elliptic curve is a modular form in disguise. I would want someone to explain this statement. I have seen the definition of a Modular form in Wikipedia, but i can't correlate this with an elliptic curve. The definition of an elliptic curve is simply a cubic equation of the form $y^{2}=x^{3}+ax+b$. How can it be a modular form?

Next, there was a mention of this Mathematician named "Gerhard Frey" who seems to considered this question of what could happen if there was a solution to the equation, $x^{n}+y^{n}=z^{n}$, and by considering this he constructed a curve which is not modular, contradicting Taniyama-Shimura. If he had constructed such a curve then what was the need for Prof. Ribet to actually prove the Epsilon conjecture.

Lastly, here i would like to know this answer: How many of you agree with Wiles, that possibly Fermat could have fooled himself by saying that he had a proof of this result? I certainly disagree with his statement. Well, the reason, is just instinct!

$\endgroup$
  • 2
    $\begingroup$ To be frank, Chandru, this is one of your better questions. :) $\endgroup$ – J. M. is a poor mathematician Sep 18 '10 at 13:24
  • 2
    $\begingroup$ @Chandru1: Frey did not prove that the Frey curve is not modular. That is what Ribet did. $\endgroup$ – Qiaochu Yuan Sep 18 '10 at 19:31
  • 1
    $\begingroup$ I remember someone guessing that Fermat had proved his theorem for $n = 3$ and he supposed his proof would generalize, while it turns out that it doesn't. $\endgroup$ – Gunnar Þór Magnússon Nov 22 '10 at 7:43
7
$\begingroup$

The elliptic curve is not a modular form. The idea that there is a modular form $f$ associated to $E$. It satisfies $$f(z)=\sum_{n=1}^\infty c_n q^n$$ where $q=\exp(2\pi i z)$, $c_1=1$ and (with finitely many exceptions) for prime $p$, the equation $y^2=x^3+ax=b$ has $p-c_p$ solutions $(x,y)$ considered modulo $p$. It also satisfies various other conditions that I won't spell out (that it's a "newform" for a modular group $\Gamma_0(N)$ etc.)

$\endgroup$
  • $\begingroup$ Robin, please correct my impression of the whole setup if I'm wrong: what Frey did amounted to constructing a "weird" elliptic curve that, if it did not have an associated modular form, would imply the falsity of TSC and thus FLT. Meaning, what Ribet essentially did to settle the ε conjecture was to show that the Frey curve does have an associated modular form. $\endgroup$ – J. M. is a poor mathematician Sep 18 '10 at 13:22
  • 2
    $\begingroup$ @J. M.: Ribet proved (a statement that implies) that the Frey curve is not modular. That is the contradiction that allows the Taniyama-Shimura conjecture to imply FLT. $\endgroup$ – Qiaochu Yuan Sep 18 '10 at 19:30
2
$\begingroup$

The starting point of the link between elliptic curves and modular forms is the following.

From a topological point of view, elliptic curves are just (2-dimensional) tori, i.e. products $S^1\times S^1$, where $S^1$ is a circle.

A torus has always an invariant never vanishing tangent field. Dually, one can find a non-vanishing invariant differential form $\omega$ on every elliptic curve $E$. It is a useful exercise to write $\omega$ in terms of the coordinates $x$ and $y$ when $E$ is given as a Weierstrass cubic.

If you have a modular parametrization $\pi:X_0(N)\rightarrow E$ you can pull-back the form $\omega$ to $X_0(N)$ and in terms of the coordinate $z$ in the complex upper halfplane $\pi^*(\omega)=f(z)dz$ for some holomorphic function $f(z)$. It is basically immediate that $f(z)$ is a modular form of weight $2$.

$\endgroup$
  • 2
    $\begingroup$ No, that's Weierstrass parametrization. The Weierstrass $\wp$ function and its derivative are meromorphic functions on the complex plane wich are periodic with respect to a lattice $\Lambda$. Thus, they realize an embedding of ${\Bbb C}/\Lambda$ into the projective plane whose image turns out to be a cubic. $\endgroup$ – Andrea Mori Sep 18 '10 at 13:27
  • 2
    $\begingroup$ Weierstrass parametrizations exist for all elliptic curves (where "elliptic curve"="algebraic curve of genus 1") whereas modular parametrizations are more an arithmetic phenomenon (STW conjecture dealt only with elliptic curves defined over the rationals) $\endgroup$ – Andrea Mori Sep 18 '10 at 13:30
  • 1
    $\begingroup$ @J. M.: the Weierstrass function is really a function of two things: a parameter tau which lives on the upper half plane, and another parameter which lives in the elliptic curve C/(1, tau). Generally we speak of the Weierstrass function as if tau were constant (making it an elliptic function), but it is also possible to speak of the Weierstrass function as a Laurent series in its second parameter while varying tau (making its coefficients modular forms). The Weierstrass function in its totality is therefore something more complicated called a Jacobi form. $\endgroup$ – Qiaochu Yuan Sep 18 '10 at 19:47
  • 1
    $\begingroup$ @Chandru1: compared to anything else you're going to read about elliptic curves, yes. $\endgroup$ – Qiaochu Yuan Sep 21 '10 at 13:13
  • 1
    $\begingroup$ @NumThcurious, Frey picked up this curve in Yves Hellegouarch's thesis *Courbes elliptiques et équation de Fermat", Besançon, 1972. Hell. himself has written a nice introductory book on the subject, "Invitation aux math. de Fermat-Wiles", Masson ed., 1997 (I don't know if there is an English translation). At the beginning of the 70s, Hell. had not the necessary coceptual tools to conjecture that his elliptic curve could not be modular, but he was aware that it possessed so many extraordinary properties that he had nicknamed it "Roland's mare" (Roland - Orlando in Italian - was.. $\endgroup$ – nguyen quang do Nov 26 '17 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy