Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(This question is a spin-off from this other question, and is largely inspired by it.)

Let $f \in S_1(\Gamma_1(800),\chi)$ be the weight-one "icosahedral" eigenform constructed by Buhler in his thesis [1]:

$$f = q - iq^3 - ijq^7 - q^9 + jq^{13} + \cdots.$$

Here $i = \sqrt{-1}$ and $j = \frac{1+\sqrt{5}}{2}$; (and if you'd like to see the first 360 Fourier coefficients, then see page 69 of loc. cit). It's "icosahedral", because the corresponding representation $\rho_f : G_\mathbb{Q} \to GL_2(\mathbb{C})$ has projective image isomorphic to $A_5$. (If you're wondering what $\chi$ is, it is the product of the character of order 2 and conductor 4 with the character of order 5 and conductor 25 sending 2 to $\zeta_5$.)

Here is my basic question:

Can I use this $f$ to construct a weight 2 cuspidal eigenform $g$ whose associated mod $\lambda$ representation $\bar{\rho}_{g,\lambda}$ (for $\lambda$ some prime ideal of the coefficient field of $g$) has projective $A_5$ image?

Here are some thoughts I've had:

An idea I first saw in section 1 of Lecture 1 of Gelbart's article in [2] gives me some hope; pick a weight one Eisenstein series $E$ that is congruent to 1 mod $l$ (some rational prime) and consider the product $fE$, which will be a weight 2 cuspform (but not an eigenform). Applying a lifting lemma of Deligne and Serre might produce an eigenform with the desired property at some $\lambda$ lying above $l$.

I say "might", because the lemma I'm looking at on page 163 in [2] is working with primes above 3, and 3 may be the only prime for which this lifting works; (and since I'm chiefly interested in characteristics 7, 19 and 61, the approach may fail).

Moreover, if I want the answer as a $q$-expansion, then perhaps this lifting is not explicit enough.

(I thought about being more demanding in the question and stipulating the coefficient field of $g$ and the characteristic of $\lambda$, but decided against it...)

Finally, I had wanted to ask this question starting with the conductor 133 "tetrahedral" form found by Tate and some of his students, because that came first historically (see the previous question), and I'd then be asking about "tetrahedral" weight 2 forms; but I was unable to write down its $q$-expansion; MAGMA gives a Runtime error when you ask it to compute a basis of the weight one forms at level 133 (though it seems to be fine at smaller levels such as 23 and 47). And since I really like $q$-expansions, I used Buhler's form.

[1]: J. Buhler: Icosahedral Galois Representations. LNM 654.

[2]: G. Cornell, J. Silverman, G. Stevens (eds): Modular Forms and Fermat's Last Theorem. Springer.

share|improve this question
1  
It will be nice if some kind soul computes the $q$-expansion of the weight-$133$ tetrahedral form and posts it below Kevin's answer (mathoverflow.net/questions/97624/…). –  Chandan Singh Dalawat May 27 '12 at 3:03
1  
If Magma is really dying when you ask it to compute the level 133 form, then you should send in a bug report -- the Magma folks are very good about fixing bugs quickly. –  David Loeffler May 27 '12 at 9:39
2  
I just ran a test of computing weight 1 modular forms for levels $\le 200$ with Magma, and found two other bugs in addition to this one (it fails for $N \in \{112, 124, 133, 136, 148, 168, 171, 180, 196\}$. I've sent them a full bug report. –  David Loeffler May 27 '12 at 10:35
2  
It's funny that the magma code doesn't work because I thought they just pasted my code in, and my code works fine for these weights (as far as I can see -- it gives answers and doesn't crash at any rate!). Barinder: if you want to see the level 133 form then download www2.imperial.ac.uk/~buzzard/char0results.m . Take a look at it -- should be self-explanatory. –  Kevin Buzzard May 27 '12 at 14:36
2  
Here is a Sage code snippet that produces the q-expansion of the level 133 non-dihedral eigenform (just a linear combination of the basis vectors for the space that Kevin gives): pastebin.com/qEcuv7EF –  David Loeffler May 28 '12 at 9:31
show 2 more comments

2 Answers 2

up vote 6 down vote accepted

(This answer is partly a precis of an answer I gave to Barinder in person, which I am posting here in case anyone else is interested.)

The Deligne--Serre lifting lemma is a completely general statement about endomorphisms of free modules over discrete valuation rings; there is no need to assume that the residue field has characteristic 3 (it need not even be finite). Googling "Deligne Serre lifting" brings up multiple detailed accounts of this lemma and its proof.

So the strategy you propose will work, as long as you can find the necessary weight 1 Eisenstein series. You say you're interested in characteristic 7; try typing

EisensteinForms(Gamma1(7), 1).integral_basis()[0].qexp(50).change_ring(Zmod(7))

into Sage. So this argument will give you a weight 2 eigenform $g$ of level 931 which is congruent to $f$ mod some prime above $7$. It's still not completely clear that the mod $7$ representations of $f$ and $g$ have projective image that is tetrahedral, because the mod 7 representation attached to $f$ might have smaller image than the characteristic 0 one; but you might be able to rule this out if you think about what the smaller images might be -- the only proper quotients of $S_4$ are cyclic and dihedral groups, and these would force rather strong conditions on the reductions of $f$ and $g$ mod 7.

Sadly, getting your hands on $g$ explicitly is rather hard work because the weight 2 space it lives in is rather large: you can pin down what the character of $g$ should be, but the space of weight 2 forms of level 931 with this character is 86-dimensional. But it should be possible to get a list of eigenforms in this space if you leave Sage (or Magma) running for long enough, and to check for each of them whether there is a prime of its coefficient field above $7$ with respect to which it is congruent to $f$.

share|improve this answer
add comment

If f is an eigenform of level 800 and charachter $\chi$, then how we have $\chi(3)=0$? (since $a_{9}=-1$ and $a_{3}=-i$, then $\chi(3)=0$)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.