MODULAR FORMS-page118

MODULAR FORMS-page118 - 114 LECTURE 11. HECKE OPERATORS...

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Unformatted text preview: 114 LECTURE 11. HECKE OPERATORS From now on we shall identify M(Γ(1))k with Fk . So we have linear operators T (n) in each space M(Γ(1))k which also leave the subspace M(Γ(1))0 invariant. k To avoid denominators in the formulas one redefines the action of operators T (n) on the vector space M(Γ(1))k by setting X T (n)f = n2k−1 T (n)∗ (f ) = n2k−1 f |k A (11.14) A∈An These operators are called the Hecke operators. Let T (n)( ∞ X cm q m ) = m=0 ∞ X bm q m . (11.15) m=0 It follows from (11.9) that for prime n = p, we have ( cpm if p |m, bm = cmp + p2k−1 cm/p if p|m. (11.16) Also, for any n, b0 = σ2k−1 (n)c0 , b 1 = cn . (11.17) 11.4 We will be interested in common eigenfunctions of operators T (n), that is, functions f ∈ Mk (Γ(1) satisfying T (n)f = λ(n)f for all n. Lemma 11.3. Suppose f is a non-zero modular form of weight 2k with respect to P Γ(1) which is a simultaneous eigenfunction for all the Hecke operators and let cn q n be its Fourier expansion. Then c1 = 0 and T (n)f = cn f. c1 Moreover, if c0 = 0 we have cn /c1 = σ2k−1 (n). Conversely, if c0 = 0 and the coefficients cn satisfy the previous equality, then f is a simultaneous eigenfunction of Hecke operators. Proof. In the notation of (11.11) we have bm = λ(n)cm , ∀m, n. If c1 = 0, then b1 = λ(n)c1 = 0. But, by (11.12) we have cn = b1 . This shows that cn = 0 for all n = 0. Thus f is constant, contradicting the assumption. So, c1 = 0, and cn = b1 = λ(n)c1 implies λ(n) = cn /c1 . If c0 = 0, we use (11.12) to get b0 = σ2k−1 (n)c0 = λ(n)c0 . This gives λ(n) = σ2k−1 (n). ...
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This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.

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