Date: Thu, 11 Mar 1999 18:19:33 +0000 (GMT)
From: Kevin Buzzard <k.buzzard@ic.ac.uk>
Subject: Re: New idea. S_2(15020)
> Totally do-able, assuming I know the character, of course.
This is 1376 we're talking about here. This might not be such a great example,
seeing as 2 is so horrible. One thing that _must_work is that once
we've found the form and proved there's a repn of A_5 type, we have to
be able to work out what the A_5 extension is. The proof should be
of the following form:
1) It's in Frey's table
2) It can't be any of them apart from this one
3) Therefore it's this one.
The step I'm worried about is (1). To do this we have to find a good
bound on the disc of a root field. All we have is a modular form.
The plan is as follows: the field will be unramified outside 5N where
N is the level of the form; for primes like 5 and odd primes it
shouldn't be too hard to explicitly work out what "type" we are,
in the sense of p46 of Buhler, using Local Langlands. [By the way,
Henniart has announced a proof of Local Langlands that is much
easier than Harris-Taylor; it doesn't need all that Shimura
variety stuff. Did RLT let on yet?!] The only thing I'm worried
about is doing this calculation at 2. Actually though... hmm... :)
Actually it seems from Buhler's tables that the max the disc can
be is 2^8 at 2, so we probably win without even thinking :) Hey
that's pretty cool. I'm even more optimistic now. The proof goes like
this: the extension is an A_5 extn unramified outside 2,5,43 and we
can also check it's unram at 5 I think; at 43 Local Langlands says
that disc=43^2; this leaves 2 but disc at 2 is <=2^8 so sqrt(disc)<=2^4*43
and 2^4*43<2083 so it's in Frey's table.
OK onto this character thing. We are "type 17" at 2, so the A_5
extension has D=A_4, I=D_2. We now have to lift these back to
GL_2(C) from PGL_2(C). Well...erm...hmm. I've forgotten how to do this :)
How does this all work? Oh yeah. I remember. Here's an example which
shows how type 2 works.
OK here's the basic principle: for any "type" there is a Dirichlet character of
(Z/p^nZ) associated to it, where n="cond" in his table. They're not too hard
to work out once you remember how to do it. Here's type 2. Let's
consider A_5 as living in PSL_2(C) (it does). We have a C_3 in this A_5
and we can think of the C_3 as being represented by the element (1 0;0 z)
where z is a primitive cube root of 1. Now we lift back to SL_2(C);
well I guess we get (z 0;0 z^2) with det=1. Now imagine a Galois
repn Gal(Q_p-bar/Q_p)-->GL_2(C) with image C_3 generated by (z 0;0 z^2)
and it's ramified and the image of inertia is this too, but p isn't
3 (in fact p is 1 mod 3) so the image of wild inertia is trivial.
What's the conductor of this repn? Well, it's abelian and conductors
add; both characters are tame so the conductor is 1 (or p^1 if you like)

so the conductor of the repn to GL_2 is 2. Well this is all well and
good but note that a twist of the repn has conductor 1: if you
twist by the inverse of one of the chars then you get a repn of
the form (1 0;0 *) where * is a tamely ramified char and this repn
has conductor 1. That's why "1" is the entry for "cond" in type 2.
But now you can do more; take the _determinant_of this repn; it will
be a character of conductor 1 i.e. a map (Z/pZ)^*-->C^*; and the
image will have order 3 (which is great because p=1 mod 3 so there
are some of these) and this is the character associated to type 2.
I guess in the input to your program this would just be denoted "3".
Now I'm brave enough to try 17. IT looks nasty though, doesn't it.
First we lift to SL_2(C). I reckon this A_4 in PSL_2(C) must become a group
of order 24 with a centre of order 2; there _is_one of these, it's
just the auts of the elliptic curve over F_2-bar with j-invariant 0.
I think it's something like SL_2(Z/3Z). The D_2 in here will probably
become a Sylow 2-subgroup of this which is probably a strange non-abelian
thing...erm..it's (1 0;0 1) (-1 0;0 -1) +/-(0 1;-1 0) +/-(1 1;1 -1)
and +/-(-1 1;1 1 ) mod 3. OK so now we have to explicitly find a 2-d
faithful complex repn of this group; ugh. Well I guess +/-(1 0;0 1) can go to
+/-(1 0;0 1). Maybe send +/-(0 1;-1 0) to themselves too. Finally
I need to find....hmmm...(algebra algebra algebra) how about
sending (1 1;1 -1) to (i 0; 0 -i), I think this works.
OK now what's the conductor of this representation? Well...hmm...this
is subtle. I don't know what's going on here. Is there now a G_2 or
something? I'm a bit confused now. I think what must be going on is that
this extra +-1 which we've introduced might make the thing more ramified,
i.e....yeah I guess it must. I think G_2 is now +-1, whereas in the A_5
extension it was trivial. *How did Buhler work out these minimal
conductors?* Is it in his book? I think that what must have happened
is something like this: maybe G_2=G_3=...=G_n=+-1 for some n. Now
the conductor of this is 2+2+(1/2)(n-1).
It's really hard to believe that twisting can make this repn any better.
But if that were the case then we'd end up showing that the lift to SL_2
was minimal, so the character would be trivial because we're going to SL_2.
And this can't be right because chi(-1) has to be not 1 for _some_prime!
I have to go out. I think something weird is happening. Maybe G_1 is
still cyclic of order 4, say generated by (1 1;1 -1). Then twisting
by a character of order 4 will do wonders, it will make the conductor
go down...maybe this is what's going on...
OK I have to go. Why don't you _try_computing at level 1376
weight 5 character of conductor 4 at 2 and of order 3 at 43?
(or weight 2 character of order 4 at 5 and the same for the other ones)
and then see if there's anything there which looks like it's giving
the repn we want. You can probably do this yourself: use the poly
given in Frey and find some p for which f factors as 2 quadratics
and a linear; then T_p will be 0 for those p. Alternatively try and
work out what's happening at 3 or something. Does this help?
Kevin

Ps I don't have time to proof-read this unfortunately. I hope it's OK.
K