Re: [sage-support] Re: modular forms of half-integer weight
William Stein
>
> Hi again,
>
> Let M_{3/2}(N) be the space of modular forms of weight 3/2, level N
> and trivial character.
>
> It seems that the Cohen-Oesterle (CO) dimensions are too small. For
> example, let
>
> f(z) = 1 + 6*q + 12*q^2 + ...
>
> be the (unique) basis element of M_{3/2}(4) and
>
> g(z) = 1 + 2*q + 4*q^2 + ...
>
> be a basis element of M_{3/2}(8). Clearly
>
> f(2*z) = 1 + 6*q^2 + 12*q^4 + ...
>
> and note that the set
>
> {f(z), f(2*z), g(z)}
>
> is linearly independent in M_{3/2}(8). Hence while dim(M_{3/2}^{CO}
> (8))=2 according to Magma/Sage, in truth we have dim(M_{3/2}(8))=3.
>
> This observation is not new; please see the undergraduate research
> paper
>
> http://www.math.clemson.edu/~kevja/REU/2004/YaraChelsea.pdf
>
> for more details. Also, it seems that, when 4|N,
>
> dim( M_{3/2}(N)) = sum_{d|N} dim( M_{3/2}^{CO}(d) )
>
> but I don't know how to prove this. In short, the half-integer
> formulas in Cohen-Oesterle need to be revisited (unless I am making a
> mistake). The implementations in both Magma & Sage would need to be
> changed or, at least, the documentation would require revision.
> Comments? Thank you,
My understanding is that there is no implementation of the
half-integer Cohen-Oesterle formulas in Sage. Isn't that right?
Magma has an implementation of some dimension formulas, evidently, but
I've never seen the code in this case.
>
> Steve
>
> P.S. Note the important word "exactly" on the third line of page 13
> in the undergraduate writeup. I'm unsure whether Cohen-Oesterle
> actually specified this and would appreciate some expert opinions!
Gonzalo Tornaria
> 2009/4/16 Steve Finch <sfi...@hotmail.com>:
>>
>> Hi again,
>>
>> Let M_{3/2}(N) be the space of modular forms of weight 3/2, level N
>> and trivial character.
>>
>> It seems that the Cohen-Oesterle (CO) dimensions are too small. For
>> example, let
>>
>> f(z) = 1 + 6*q + 12*q^2 + ...
>>
>> be the (unique) basis element of M_{3/2}(4) and
>>
>> g(z) = 1 + 2*q + 4*q^2 + ...
>>
>> be a basis element of M_{3/2}(8). Clearly
>>
>> f(2*z) = 1 + 6*q^2 + 12*q^4 + ...
Note that this form is actually in M_{3/2}(8,chi) where chi is the
(even) quadratic character of conductor 8.
In general, in half-integral weight, "lifting" a form f(z) to f(m*z)
changes the character by a quadratic character kronecker(m,*). This is
because the factor "kronecker(c,d)" which shows up in the automorphy
factor for half-integral weight.
IOW, if you use the transformation formula for f(z) to write a
transformation formula for f(m*z), you will notice that the automorphy
factor for f(m*z) under the matrix [a,b;c,d] is the same as the
automorphy factor for f(z) under the matrix [a,m*b;1/m*c,d]. For
integral weight those two are the same, but for half-integral weight
they are not unless m is a square (the details are an easy exercise).
>> This observation is not new; please see the undergraduate research
>> paper
>>
>> http://www.math.clemson.edu/~kevja/REU/2004/YaraChelsea.pdf
Thm 4.1 is good for k integral, but not for half-integral weigth.
Gonzalo
Steve Finch
Thank you for clearing up my confusion!
BTW, has a proof of the Cohen-Oesterle
dimension formula for half-integer weights
ever appeared? Much appreciation,
Steve
William Stein
For integral weight, yes, in a paper by Jordi Quer. For half
integral weight, probably not. I asked Cohen about this once, and he
said basically that anybody who could read their proof (using the
trace formula, etc.) could more easily come up with the proof
themselves. :-)
Jordi Quer's proofs in the integral weight case though are very nice
and geometric.
William