Date: Fri, 12 Mar 1999 14:17:05 +0000 (GMT)
Hey! I just realised that by complete chance "type 17" is _also_the
type at 2 of Buhler's 800 repn!! So we can just look at what Buhler's
character at 2 was, and it was indeed the unique character of
conductor 4 and order 2. So this is the right char at 2. I can't
seem to do the computation at weight 5 level 1376 and this char
because my computer falls over (is the character "2 1 3"?).
Can you find the form then? It doesn't have coeffts in Z/5Z but
in F_{25}. But some twist of it will live over Z/5Z; whether
the coefft a_p lives in Z/5 or not will depend on p mod 43 :)
K
Date: Wed, 17 Feb 1999 19:02:29 +0000 (GMT)
> YES.
*sigh*.
Do you happen to know how long it took Frey et al to do the one that
we might be able to do?! Probably not as long. Aah well. I still don't
know whether we were "unlucky" or whether their method is inherently
better.
I guess it's a shame that the computation of the new one...is that 17051?...
will take a year. I've never searched beyond 17051 so I don't know what
the first one with Frob_5 having order not 3 or 5 will be.
Maybe it's time to re-trace our steps. Here's one random idea that
probably won't work :) Let's just take the second conductor in
Frey's list: it's 1376 (the first conductor is bad because I think
Frob_5 has order 5). So there should be a weight 1 form
of level 1376, with character of order 3 at 43 and fairly
nasty at 2 (can't work it out without some thought but it would
be possible, it's probably in Buhler, which I don't have to hand).
Unfortunately we note that a character of order 3 won't ever take values
in Z/5Z. So the coefficients of this form will probably take values in
something like F_{25}. But, all being well, one should be able
to twist by that character of order 3 at 43; if one does this,
one get a new weight 1 form, of character a power of 2.
Unfortunately the level has now gone up! It's now 1376*43. Presumably
this is too big to work with? I guess one would have to work at weight
5 and this level. But the thing is, this form should have coeffts
in Z/5Z (even the troublesome a_5 !!). Is there a trick by which we
can work at weight 5 and level 1376 (or even weight 2 and level 1376*5)
and find the 2-d space, and then somehow prove that the image of
Galois is not in fact all of GL_2(F_25) but that it's just a twist
of something that landed in GL_2(F_5)? This doesn't sound too
promising but it's the only idea I have at the minute.
Actually, if we were to somehow pull this off, we could maybe get

several new examples. But the problem is that how does one prove
that the image of Galois is actually in (F_25)^*.GL_2(F_5)? It
would maybe boil down to proving that a_p/chi(p) was always in F_5,
or something, where a_p is the element of F_25 that we find at
weight 5 and level 1376, and chi(p) is the appropriate cube
root of 1 depending on p mod 43. But I don't see offhand how
one could do this---even checking it for all p<=10000 doesn't
seem to be good enough, is it? Actually, maybe it is, because
a_p/chi(p) *will* be b_p for some form b_p, the problem is
that b_p will be at level 1376*43. So one would have to check
a lot more p. Maybe there is something here?
The other possibility is to find a 1-d space
at level 1376*43 and weight 5, given by the kernel of T_p-a_p.
I don't know whether or not 1376*43 is bigger than the minimal
conductor where the minimal lift actually lands in F_5...
K
Date: Wed, 17 Feb 1999 19:32:35 +0000 (GMT)
William: how about this.
There should be a weight 1 form of level 1376 and character of order
3 at 43 and something funny at 2, whose Galois repn to GL_2(C) gives
rise in the standard way to the second A_5 extension on p137 of Frey.
Say this form is sum a_nq^n. My belief is that you are now able to
"find" forms of this nature mod 5, either by looking at weight 5
and level 1376, or weight 2 and level 1376*5 and character of order 4 at 5.
However you do it, I think you can find this one. The problem is that
it won't be defined over Z/5Z, it'll probably be defined over F_25.
Having found it though, you could now look for its twist by the character
of conductor 43 and order 3 which was the local component of the form
you found. This form has got level equal to 43 * the old level, and
the same weight. But I think that one doesn't have to compute
this huge new space. You see, if one looks at the proof that
the twist by a character of a modular form is a modular form again,
you can see that in fact the twisted form is constructed using
eta and some of its generalisations. So maybe you could
1) Find the F_25 form at level 1376 and weight 5 (or wt 2 level 1376*5)
2) Compute the _twist_only, at level 1376*43, using eta etc, and
3) Check that it lives in a Hecke-invariant 1-dimensional space, and
hence the twist is defined over F_5.
and then we'd probably be done? modulo a _really_nasty calculation at 2 :)
(which is probably do-able)
This might or not might work. Raw experience has shown that usually something
unexpected happens :)
Kevin
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

Date: Mon, 22 Mar 1999 00:12:32 +0000 (GMT)
>    Aargh!!! It knows the answer but it won't tell me! It won't even tell
>    me the dimension of the space!!
>
>
> I've been frustrated by this before.  It you type "m" to turn on
> matrix display, then "c" to turn off charpoly display, before running
> the dual business it will do what you wanted.
I hadn't thought of that; thanks.
> I'll fix all of this when i get chance.  Just tell me what you want.
Rather interestingly, this comment really got me thinking: "well, what _do_
I want? Is this project going anywhere or will we get stuck at
level 2^N*43^2 which will be too big?". So I thought about what you might
have to do, on top of what you have already done, in order to get
some new examples of Artin.
And, you know: I am rather optimistic that you might not have to
do anything at all. In fact we might have a lot of new examples at our
fingertips. This is assuming that a certain calculation works: but
calculations that I've been confident will work before have failed!
Let's see what hecke can throw up against us this time!
Anyway, here's the plan. There is this A_5 extension which is unramified
outside 2 and 43; at 43 it's of type...erm...can't remember, but it's
the type which has e=3 and f=1 at 43, so the inertia group at 43 is
cyclic of order 3 and equals the decomposition group. I think this is right;
I'm at home and working from memory. Anyway, a minimal lifting of this
to GL_2(C) (minimal in the sense that the conductor is minimal) is
just 1+chi where chi is the character of order 3. This character of
order 3 doesn't reduce to Z/5Z; it forces us up to F_25 and initially
I just rejected these cases because I thought we wouldn't be able to
deal with them. But seeing as we've got stuck with the F_5 extensions
I've thought more about this F_25 one and I think we should be able to
do it.
I think I have checked that there are no problems at 5; so I think
that there should be an eigenform mod 5, call it f, of the
form sum a_n q^n, with all the a_n in F_25, giving us this minimal lift.
Let g be the conjugate
of f (i.e. g=sum b_n q^n with b_n=(a_n)^5 in F_25). Now f+g
and (f-g)/(some appropriate elt of F_25) will both have coeffts in F_5;
hence we *should* be able to find a 2-d space at level 1376 and weight 5,
mod 5, which we can prove is Hecke equivariant, and which, when you tensor
it up to F_25, will have f and g as a basis.
What will hapen in practice is that we'll find a 2-d space where we
can't diagonalise Hecke over F_5 but we can over F_25; the problem
will be to show that the two forms we have, which a priori give us
repns to GL_2(F_25), are actually twists of repns to GL_2(F_5). Here's
my plan. Firstly forget all that notation above; let's just work with
what we have. We have a 2-d space defined over F_5 and it's Hecke equivariant

and Hecke will be diagonalisable and the eigenforms f and g
will be defined over F_25. Moreover, g will have to be the conjugate
of f because that's the only way it can be; non-scalar elements of
Hecke will be forced to be diagonalisable with conjugate eigenvalues.
Now the bit of luck: I reckon that U_43 will have eigenvalues equal
to the 2 non-trivial cube roots of 1 in F_25. So by general theory
I think that one can deduce that the repn attached to f, restricted
to an inertia group at 43, will look like chi+1 where chi is
a character of order 3. Now twist f by chi^{-1} (not by chi). We get
some form, probably an oldform; but an eigenform. Take the associated newform
g'. Then my guess is that one will be able to show that the level of g' is
still 1376!! This is because the only thing that can go wrong will
be that the conductor at 43 goes up; but the conductor of the repn
at 43 will still only be 1. It's hard to believe that we'll have to
use something like Ken's level-lowering theorem but we can do this
if necessary! The point is that the conductor of 1+chi^{-1} is still 1.
Anyway, I think we'll be able to check that g' is an eigenform that
lies in our 2-d space when tensored with F_25. So g' will either
be f or g; we rule out f instantly by checking T_p for the first few
primes p and deduce that f tensor chi^{-1} is hence g=f-bar, at least
away from 43. In other words, if n is prime to 43, I think we have shown
that a_n/chi(n)=a_n-bar=a_n^5.
Now this means that a_n=a_n^5*chi(n) and let's set c_n=a_n*chi(n).
We see that (c_n)^5=a_n^5*chi(n)^2=a_n*chi(n)=c_n; so c_n is in Z/5Z.
For _all_primes n prime to 43.
So, although we haven't ever computed f*chi, and indeed it's uncomputable---
it's at level 1376*43---we have just shown that all its a_p are in Z/5Z,
for p prime to 43! So the repn goes to GL_2(Z/5Z)!!
And surely this is enough. We know exactly what's going on at 43 and
can check that the repn to PGL is only tamely ramified
at 43; the computation at 2 can't go wrong, I think I checked.
And surely there will be other forms with this property: I rejected
a _lot_of A_5 extns on the basis that the minimal lifting had character
of order a multiple of 3.
I can't imagine what can go wrong with this argument :) I think
hecke might be a bit broken; I was getting some strange "Subspace
not invariant" errors; I'll try and re-produce them. But if we get
down to a 2-d space I think we've cracked it. Basically I think
this is the computation we have to do (I'll try it Monday morning UK time);
compute the space of forms that are weight 5 mod 5 level 1376 and which
have T_5, T_19, T_31... acting as 0. The moment we get down to a 2-d
space I think we've won. We didn't win before when we were trying to
get down to a 1-d space because a_5 really didn't live in Z/5Z; but
this time it seems to me that a_2, a_5 and a_43 will all live in F_25;
in fact I think a_2=0 (not 100% sure about this), a_5=0 and a_43=omega
and as far as I can see we can't lose this time. Fingers crossed!
I might be wrong. If this works we get loads of new examples I think and you
don't even have to compute these twisting by a character maps that

I was dreading having to explain to you :) (although maybe this
can all be in hecke-v2.0, when there are routines for giving individual
forms names like Delta; and then being able to compute Delta(nz) or
Delta x chi etc etc!)
Kevin

Date: Mon, 22 Mar 1999 18:30:43 +0000 (GMT)
William: I think a lot of what I wrote yesterday was rubbish: not because
it was wrong, but because it hadn't really dawned on me that Hecke was capable
of working over finite fields other than Z/pZ; now I've finally seen
some examples of matrices representing T_p, and the first thing
I realised was they're not defined over Z/pZ! So this makes the
while thing easier: we can simply find a _1-dimensional_space
consisting of the form we want; and then look for its conjugate
and its twist, and they will both lie in another 1-d space.
It's just like that 512 problem: all you have to do is prove
that something is conjugate to its twist and you'll be done.
In our case the computations should be easy coz they're
mod 5 at level 1376 and weight 5. The only complete pain is that
either computing Hecke ops takes ages or I have to re-start with
another "f" the moment I decide I want to work out another kernel!
Is it possible to do the following: if I have an 8-d space, then
can I make Hecke compute T_p on it in a sensible way? I think this
is the best way I can currently think of for improving hecke
to do what I want it to do! The problem is that it takes a long
time to compute Hecke ops and I believe that once the space is
8-dimensional it surely shouldn't. Am I right? I want to go down
from 8 to 1; currently I'm doing this:
first use f and compute ker(T_19) and ker(T_31); then work out
fast T_p for about 5 random primes p!
Then look at the results, decide what kernel is best to compute
next (it was T_2 and then T_3-(x+2) or something); then restart
an f command with ker(T_2), ker(T_3-(x+2)), ker(T_19) and ker(T_31);
then compute some more random T_p fast!!
This is crazy I think :)
Also if one could compute fast T_p on any space, one could do the
other thing what I want to do, and that's compute fast T_p
on ker(f(T)) where f might not be monic. If for some reason the
fast T_p routine cannot be made to work on a general space,
maybe you could beef the f option up so that it includes
ker(f(T)) for T a hecke op?
By the way: does the f option compute the initial kernel like
this: it computes all of T_p and then ker(T_p-a_p) and then
all of T_q and then ker(T_q-a_q) on ker(T_p-a_p) and then
computes all of T_r and then ... or does it use faster and
faster T_p-computing the smaller the space gets?

Kevin

Date: Sat, 27 Mar 1999 21:51:09 +0000 (GMT)
> When copmputing the modular degree it is necessary to cut out the
> eigenspaces in both hom<9m,c<0 and m and then compute the discriminant of
> the natural pairing.
Interesting!
> I can work on hecke when I get to Leiden in 1.5 weeks.  The only draw back
> is that my laptop compiles code at about 1/15th the speed of my desktop!
Oh, none of this is urgent :) Why not just wait until you're back
in CA? Of course, if you have nothing at all better to do, then feel
free to continue; I'm just emphasizing that I'm not definitely putting time
pressure on you!
I have been checking more of the details of the 1376 thing and things
are still looking good. For all primes <=41, the trace of Frob_p divided
by the appropriate cube root of 1 is in Z/5Z, and I convinced myself
that Deligne-Langlands-Carayol could be used to prove that f-bar is
a twist of f, without having to do any computations at level 1376*43
(Shimura's result only says that f*chi will be modular of level 1376*43
but I think I can prove it's an oldform). Next the computations do give
us a 2-dim space where T_2,3,19 and 31 all act diagonally by the right
character, but T_5 doesn't. I think I once convinced myself that a companion
form argument showed that the repn was unramified at 5 but I can't
reproduce it and I also can't find the email I sent you about it. I'm
sure it's not too hard though.
At 43 the repn is tame by Deligne-Langlands-Carayol. At 2 we have to check
we took a minimal lift maybe but this won't be too hard, it's purely
local. If I manage to reconstruct the argument at 5 we get to deduce
that there's an extension K/Q, Galois, unramified outside 2 and 43,
discriminant of root field at 43 is just 43^2, disc at 2 will be some
reasonable power of 2, and I also had an argument that ruled out
Gal(K/Q) being A_4 or S_3 or anything, as I recall, so I think we
might well be done (assuming these arguments I never wrote down but
emailed you still exist and are right :) ). I am very optimistic
that we've done it!
K