Modular symbols: sign issue with diamond operators
daveloeffler
sage: M = ModularSymbols(11, sign=0)
sage: M.diamond_bracket_operator(1)
Hecke module morphism Diamond bracket operator <1> on Modular Symbols
space of dimension 3 for Gamma_0(11) of weight 2 with sign 0 over
Rational Field defined by the matrix
[ 1 0 0]
[ 0 -1 1]
[ 0 0 1]
Domain: Modular Symbols space of dimension 3 for Gamma_0(11) of
weight ...
Codomain: Modular Symbols space of dimension 3 for Gamma_0(11) of
weight ...
The diamond operators should give an action of (ZZ / N ZZ)^* on the
space, so in particular 1 should really act as the identity! I've
traced the issue to a rogue minus sign, which had the effect of
multiplying the answer by the star involution. This doesn't have any
effect on computing diamond operators on modular forms, because there
we only use the +1 eigenspace for the star involution; but the sign 0
case is needed for some other stuff I'm working on.
Patch is up at #8620. Any chance of a quick review?
David
William Stein
Nice find! I'll review it now.
William
daveloeffler
On a related note, there's a bug in the Atkin-Lehner eigenvalue code,
which is my fault as I wrote that code. But I don't know how to fix
it. Suppose we take some space of forms of trivial or quadratic
character (so the Atkin-Lehners do preserve the space), and odd
weight. Then the Atkin-Lehner operator on the corresponding space of
modular symbols doesn't preserve the plus submodule; it swaps plus and
minus submodules, so it's not quite clear to me how to relate it to
the corresponding operator on modular forms.
To be more precise: attached to an eigenform f there is a plus
eigensymbol and a minus eigensymbol in the modular symbols space, each
only determined up to some arbitrary scaling factor. Since the Atkin-
Lehner operator swaps the plus thing for the minus thing, we have to
normalise the plus and minus symbols in a "compatible" way, and I
don't know how to do this. Any suggestions? This is now ticket #8622.
David
On Mar 28, 11:04 pm, William Stein <wst...@gmail.com> wrote:
William Stein
> Thanks, that was very quick!
>
> On a related note, there's a bug in the Atkin-Lehner eigenvalue code,
> which is my fault as I wrote that code. But I don't know how to fix
> it. Suppose we take some space of forms of trivial or quadratic
> character (so the Atkin-Lehners do preserve the space), and odd
> weight. Then the Atkin-Lehner operator on the corresponding space of
> modular symbols doesn't preserve the plus submodule; it swaps plus and
> minus submodules, so it's not quite clear to me how to relate it to
> the corresponding operator on modular forms.
>
> To be more precise: attached to an eigenform f there is a plus
> eigensymbol and a minus eigensymbol in the modular symbols space, each
> only determined up to some arbitrary scaling factor. Since the Atkin-
> Lehner operator swaps the plus thing for the minus thing, we have to
> normalise the plus and minus symbols in a "compatible" way, and I
> don't know how to do this. Any suggestions? This is now ticket #8622.
Are you sure. My initial thoughts are:
(1) If you want to work only in the +1 or -1 quotient, then there is
no way to determine the Atkin-Lehner sign, since Atkin-Lehner doesn't
act in a well defined way on the +1 or -1 modular symbol quotient
spaces.
(2) So you have to work in the sign 0 space. Then there *is* an
eigenvector and an eigenvalue. E.g., if (to simplify) x generates the
+1 space and y generates the -1 space, then v = x+y is an eigenvector,
with sign +1 or sign -1. Anyway, I don't see why it is necessary to
normalize anything -- just because Atkin-Lehner swaps +1 and -1
spaces, doesn't mean that Atkin-Lehner doesn't have an eigenvector.
Perhaps I'm just confused, since I didn't try the above out with examples.
William
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William Stein
Associate Professor of Mathematics
University of Washington
http://wstein.org
daveloeffler
On Mar 29, 6:52 pm, William Stein <wst...@gmail.com> wrote:
> On Mon, Mar 29, 2010 at 3:38 AM, daveloeffler <dave.loeff...@gmail.com> wrote:
> > To be more precise: attached to an eigenform f there is a plus
> > eigensymbol and a minus eigensymbol in the modular symbols space, each
> > only determined up to some arbitrary scaling factor. Since the Atkin-
> > Lehner operator swaps the plus thing for the minus thing, we have to
> > normalise the plus and minus symbols in a "compatible" way, and I
> > don't know how to do this. Any suggestions? This is now ticket #8622.
>
> Are you sure. My initial thoughts are:
>
> (1) If you want to work only in the +1 or -1 quotient, then there is
> no way to determine the Atkin-Lehner sign, since Atkin-Lehner doesn't
> act in a well defined way on the +1 or -1 modular symbol quotient
> spaces.
Agreed.
> (2) So you have to work in the sign 0 space. Then there *is* an
> eigenvector and an eigenvalue. E.g., if (to simplify) x generates the
> +1 space and y generates the -1 space, then v = x+y is an eigenvector,
> with sign +1 or sign -1. Anyway, I don't see why it is necessary to
> normalize anything -- just because Atkin-Lehner swaps +1 and -1
> spaces, doesn't mean that Atkin-Lehner doesn't have an eigenvector.
But if x + y is an eigenvector then x - y will be as well, with the
opposite eigenvalue. Which is the right choice, y or -y? These two
eigenvectors correspond to the cohomology classes of the two harmonic
1-forms given by f and its conjugate, which have opposite Atkin-Lehner
signs in this case. What I don't know how to do is to tell which sign
corresponds to the holomorphic 1-form and which to the anti-
holomorphic one.
David
Georg S. Weber
> But if x + y is an eigenvector then x - y will be as well, with the
> opposite eigenvalue. Which is the right choice, y or -y? These two
> eigenvectors correspond to the cohomology classes of the two harmonic
> 1-forms given by f and its conjugate, which have opposite Atkin-Lehner
> signs in this case. What I don't know how to do is to tell which sign
> corresponds to the holomorphic 1-form and which to the anti-
> holomorphic one.
>
> David
Hi David,
let me try to rephrase your question, to see whether I understand it.
Take some modular curve C of some (higher) level N, and two elliptic
curves E1, E2, (both defined over QQ) of level N, such that both E1
and E2 are "optimal quotients" of C. In the "sign 0" modular symbols
space over ZZ associated to C, there are two subspaces S1, S2 each of
(ZZ-)dimension two "belonging" to E1, E2 respectively, i.e. having the
same eigenvalues under the Hecke algebra as E1 resp. E2. Let S1 be
generated by x1 (belonging to the plus-space) and y1 (belonging to the
minus-space); and let S2 be generated by x2 (belonging to the plus-
space) and y2 (belonging to the minus-space).
Then x1, x2, y1, y2 are all well-defined only up to sign. If I
understand you correctly, then after having chosen of x1, y1, and x2,
there should by a "canonical" or "natural" choice of y2 among {y2, -
y2}, in some "compatible" way w.r.t. the pair (x1, y1)?
One could think along the following lines:
E1, as (one-dimensional) complex Riemannian manifold, implies one
choice of sign, i.e. a choice of orientation (as orientable two-
dimensional real manifold). Changing this choice means interchanging
the notions of holomorphic, and anti-holomorphic.
After pinning down x1, this is "equivalent" to a choice between y1 an -
y1.
The same for E2.
But both E1 and E2 are optimal quotients of C, C too being an
orientable manifold -- any choice of orientation on E1 "implies" a
choice of orientation on C itself, and thus also an orientation on
E2.
(The same line of arguments may be run considering the modular forms
F1, F2, "belonging" to E1, E2, which should be "compatible" when
pulled back to the upper half plane.)
If I understood your question correctly --- then I don't know the
answer to how to compute this. The result would be a "natural"
isomorphism mu between the plus-space and the minus-space, defined on
the "integral" structure, possibly canonical/unique only up to sign.
The "holomorphic" subspace of the "sign 0" modular symbols would then
be the (saturation of the) symbols of the type "a + mu(a)" for a in
the plus-subspace, in the same vein "a - mu(a)" giving the "anti-
holomorphic" subspace. Changing mu to -mu would exchange the such
defined holomorphic and anti-holomorphic subspaces. These latter
subspaces would be invariant under the Hecke algebra, and also under
the Atkin-Lehner operators ...
This topic is definitely interesting, thanks for bringing it up!
Cheers,
Georg
William Stein
> Thanks, that was very quick!
>
> On a related note, there's a bug in the Atkin-Lehner eigenvalue code,
> which is my fault as I wrote that code. But I don't know how to fix
> it. Suppose we take some space of forms of trivial or quadratic
> character (so the Atkin-Lehners do preserve the space), and odd
> weight.
The character is *not* trivial if the weight is odd, right? Since odd
weight plus trivial character = 0 space. So you're only talking
about the case of quadratic character. This comment addresses Georg's
comment that starts with elliptic curves (which isn't relevant since
the weight is odd and the character is not trivial).
Here's a little program that produces examples like David is interested in:
def f(N):
for eps in DirichletGroup(N):
if eps.order()==2 and eps(-1) == -1:
S = ModularSymbols(eps,3).cuspidal_subspace()
if S.dimension() >= 1: return S
v = [f(N) for N in [1..15] if f(N)] # silly to do "if f(N)".
Upon running it, here is the first example:
sage: S = v[0]; S
Modular Symbols subspace of dimension 2 of Modular Symbols space of
dimension 4 and level 7, weight 3, character [-1], sign 0, over
Rational Field
sage: t7 = S.hecke_operator(7).matrix(); t7
[-7 0]
[ 0 -7]
sage: w = S.atkin_lehner_operator(7).matrix(); w
[-1 4]
[-2 1]
sage: w.charpoly()
x^2 + 7
sage: w.change_ring(SR).eigenvalues()
[-I*sqrt(7), I*sqrt(7)]
sage: s = S.star_involution().matrix(); s
[ 1 -1]
[ 0 -1]
sage: s*w == w*s
False
sage: s.charpoly()
x^2 - 1
sage: S.atkin_lehner_operator(5).fcp()
(x^2 + 5)^2
So the Atkin-Lehner operator on modular symbols has two eigenvalues
sqrt(-7) and -sqrt(-7).
What does it even mean to ask which is right? They are conjugate, so
maybe indistinguishable.
Assuming you have made some choice that nails down which is which, I'm
guessing one could distinguish them by somehow using a complex period.
To compute these in weight > 2 requires using the algorithm in
[Stein-Verrill], which isn't coded in Sage yet. And I'm not even
sure this will work (I haven't thought it through).
William
-- William
Georg S. Weber
> Assuming you have made some choice that nails down which is which, I'm
> guessing one could distinguish them by somehow using a complex period.
> To compute these in weight > 2 requires using the algorithm in
> [Stein-Verrill], which isn't coded in Sage yet. And I'm not even
> sure this will work (I haven't thought it through).
>
> William
>
> -- William
Yeah,
I think this guess is good. "Just" start with the (holomorphic)
modular form F in the q-expansion space, regard q as being e^(2 pi i
z) for z in the complex upper half plane and F as a function there,
and compute it(s) complex period(s) --- by brute force/explicit
complex analytic evaluation. Upon having made your choices w.r.t. the
extra factors in the pairing, and fixing x1 (as above in the plus
space), this should give whether (the "imaginary part") +y1 or -y1
then is the "correct" choice. (This is only a rough sketch, of
course.) E1 and E2 are "comparable" in the sense that for both, one
has to stick to one and the same choice made in advance, writing
explicitly down the pairing between paths and holomorphic
differentials.
Cheers,
Georg