#13253: galois_action on cusps has a bug and incorrect documentation
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Reporter: mderickx | Owner: craigcitro
Type: defect | Status: closed
Priority: major | Milestone: sage-5.3
Component: modular forms | Resolution: fixed
Keywords: | Work issues:
Report Upstream: N/A | Reviewers: Marco Streng
Authors: Maarten Derickx | Merged in: sage-5.4.beta0
Dependencies: | Stopgaps:
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Comment (by cremona):
I was using this function recently and unfortunately the documentation
makes incorrect claims for its applicability! Being based on Steven's
book, it works for *some* congruence subgroups of level N, but *not all*
of them.
The example I have is this. I have a subgroup of level 13, index 91,
consisting of matrices in PSL(2,Z) whose mod-13 reduction lie in a
subgroup of PSL(2,13) isomorphic to A4. Under the action of A4 the 84
cusps of Gamma(13) form 7 orbits of size 12 each. But the action of (t
mod 13) is only well-defined for t=1,5,8,12 (i.e. the cubes mod 13). I
could give examples of 2 cusps c1,c2 which are A4-equivalent but the
results of c.galois_action(2,13) for c=c1,c2 are not A4-equivalent. This
can be explained by looking carefully at Stevens' proof of his
proposition, which relies on the field of modular functions for the group
in question being generated by by functions whose Q-expansions have
rational coefficients. This is true for Gamma(N), Gamma0(N), Gamma1(N),
but not in general. In my case the field of coefficients required is the
cubic subfield of Q(zeta13), which explains why Stevens's formula is only
valid when t is a cube.
I think that the way to fix this is to change the documentation so that
the function does not claim to do more than it does. A complete fix would
require something really new, with input more than just the level N and an
invertible residue class t mod N. I was able to work out my specific
example, but a general implementation would be quite hard. [For the
record, the 7 cusps consist of 2 Galois orbits, of size 3 and 4. Using
Sage's function restricted to t=5 (which generates the cubes mod 13) I
found a 4-cycle and 3 fixed points, including infinity, and as I already
knew that infinity was in an orbit of size 3 that was sufficient!]
--
Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/13253#comment:11>
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