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May 27, 2021 at 6:44 comment added François Brunault @DrorSpeiser Actually with Sturm's theorem, it is sufficient to check $d$ coefficients or so, where $d$ is the dimension. So this is optimal. I have written things up here for an arbitrary congruence subgroup, since I couldn't find it in the literature.
Feb 9, 2011 at 21:02 vote accept Jill
Feb 9, 2011 at 7:56 comment added William Stein @Dror: Sturm's result in general says one has to check about $2d$ coefficients, where $d$ is the dimension. For two arbitrary cusp forms, that's pretty good (massively better than "effective Chebotarev", say), but for eigenforms it is far from optimal. Under some hypothesis, Sturm reduces the number of coefficients to roughly $2d/2^v$, for some $v$ (this $v$ is $\leq$ the number of primes dividing the level). There is a result in the Buzzard-Stein paper I mentioned that improves on Sturm's result in the case of eigenforms with nontrivial character.
Feb 8, 2011 at 20:12 comment added Dror Speiser How sharp is Sturm's result?
Feb 8, 2011 at 19:09 history answered William Stein CC BY-SA 2.5