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For a non-CM holomorphic modular forms of weight $k \geq 2$, the Sato–Tate conjecture is known to be true. Thanks to the work of Barnet-Lamb, Geraghty, Harris, and Taylor.

Do we have an analogous statement for CM modular forms as well? I mean, Is there a precise formulation (or a proof) of the Sato-Tate conjecture for CM modular forms of weight $k \geq 2$?

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up vote 5 down vote accepted

Since the $L$-function of a CM modular form is just that of a Hecke character, the analogue of the Sato–Tate conjecture is much simpler to prove and follows from work of Deuring, I believe. If I remember correctly the measure one uses is (proportional to) $(1-z^2)^{-1/2}$ (and, as David points out, you only consider the primes that split in the associated imaginary quadratic field since $a_p=0$ at all inert primes).

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You have to take into account the fact that $a_p = 0$ for a density 1/2 set of primes (the ones that split in the CM field). For the rest it is the measure with probability density proportional to $(1 - z^2)^{-1/2}$ (not $z^{-1/2}$ as I said in a comment earlier!) I think the nicest way of saying it is: the distribution of $a_p/2\sqrt{p}$ in the CM case is the same as the distribution of the trace of a random 2x2 real orthogonal matrix, while in the non-CM case it's the trace of a random 2x2 complex unitary matrix. – David Loeffler Nov 21 '12 at 16:04
sorry, "the ones that do not split in the CM field". – David Loeffler Nov 21 '12 at 16:04
Thanks Robert Harron and David Loeffler for your comments. – N. Kumar Nov 21 '12 at 17:11
This is of course a special case of the general Sato-Tate conjecture, where the distribution of the (weighted) conjugacy class of Frobenius is the same as the distribution of the conjugacy class of random element of the maximal compact subgroup of the Lie group whose Q_l form is the Zariski closure of the image of the Galois representation. @David: If you want that statement to work, I think you should divide by $\sqrt{p}$, not $2\sqrt{p}$. The traces of unitary matrices range from $-2$ to $2$. – Will Sawin Nov 21 '12 at 17:20
in fact, "the ones that are inert in the CM field"; $a_p$ need not be zero at a ramified prime. Crazy right? – Rob Harron Nov 21 '12 at 17:20

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