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:<math>f\left({az+b\over cz+d}\right) = (cz+d)^k f(z)</math>
for ''a'', ''b'', ''c'', ''d'' integers with ''ad''-''bc''=1. For example,
:<math>f(-1/z) = F(\langle 1,-1/z\rangle) = z^k F(\langle z,-1\rangle) = z^k F(\langle 1,
Functions which satisfy the modular functional equation for all matrices in a finite index subgroup of SL<sub>2</sub>('''Z''') are also counted as modular, usually with a qualifier indicating the group. Thus modular forms of ''level N'' satisfy the functional equation for matrices congruent to the identity matrix modulo ''N'' (often in fact for a larger group given by (mod ''N'') conditions on the matrix entries.)
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