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Mar 11, 2015 at 18:00 history edited Peter Humphries
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Nov 1, 2013 at 13:28 comment added GH from MO Typo in my comment: "no longer not right-invariant" should read "no longer right-invariant". Thanks to Qiaochu for his valuable comment.
Nov 1, 2013 at 2:21 comment added Qiaochu Yuan Some details on the lift mentioned in GH's comment can be found at mathoverflow.net/questions/20281/… .
Oct 31, 2013 at 18:27 comment added GH from MO I am too sleepy for a proper answer, but basically the reason is this: a classical holomorphic modular form is a function on H=SL(2,R)/SO(2,R) that transforms in an appropriate way with respect to SL(2,Z) or a similar discrete subgroup. In particular, it is not SL(2,Z) invariant. Lifting to SL(2,R) means converting to a function that is left-invariant by SL(2,Z), the price to pay is that it is no longer not right-invariant by SO(2,R). For a half-integral weight holomorphic modular form you cannot lift to SL(2,R) this way, you need to consider a double cover of it.
Oct 31, 2013 at 18:09 history asked Fabian Werner CC BY-SA 3.0