By Goro Shimura
A ebook on any mathematical topic past the textbook point is of little price until it comprises new rules and new views. It is helping to incorporate new effects, only if they provide the reader new insights and are provided besides recognized outdated ends up in a transparent exposition. it's with this philosophy that the writer writes this quantity. the 2 topics, Dirichlet sequence and modular types, are conventional matters, yet the following they're taken care of in either orthodox and unorthodox methods. whatever the unorthodox therapy, the writer has made the e-book obtainable to those that will not be accustomed to such issues through together with lots of expository fabric.
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Additional info for Elementary Dirichlet Series and Modular Forms (Springer Monographs in Mathematics)
Example text
5. 19) χ∈X (2f /2 + 1)2t/f if f ∈ 2Z, (2 − 1) if f∈ / 2Z. 17). 36). 8). Special care must be given to the character of conductor 4, which is included in X but not in Y. 18) for r > 3. 18) is true even for r = 2 and 3. Remark. For odd , it often happens that 2 is a primitive root modulo m, in which case we have f = 2t, and so A = 2t + 1. 17) equals I −1 . 8. 20) B= 21−t β 2 , cχ − χ∈X α , χ(a)a with a=1 α = r − 2−1 (r β =r−1−2 r − (r + 1) r−2 r if r−1 + 1) if = 2, = 2. Here cχ is the conductor of χ.
Put ω = e(1/6), and suppose d0 − 7 ∈ 12Z. 6) can m m be written j=−m χ(j)ω −j , which equals j=1 χ(j)(ω −j − ω j ). 18) −i 3 χ0 (a) − χ0 (b) a b with a and b as in (v). 6) equals i 3χ √0 (6)hK /2. 18). Now χ0 (3) = 3 d0 , which equals −1 or 1 according as d0 ≡ 7 or 11 (mod 12). 15) in both cases as expected. 5). Put m = [d0 /6] and d0 = 2q + 1. Then the leftd0 −m−1 m (−1)a χ0 (a). 5) becomes a=−m d0 −m−1 0 as χ0 (−1) = −1, and so we only have to consider a=m+1 (−1)a χ0 (a), which equals q q a d0 −a (−1) χ0 (a) + (−1) χ0 (d0 − a) a=m+1 (−1)a χ0 (a).
2πi)−k G(χ)L(k, χ) d−1 = χ(4) χ(a)e(−da/4)Ec,k−1 (a/d). (1 + i){4k − χ(2)2k } a=1 (iii) Suppose d is prime to 6; let c = −e(1/6). (2πi)−k G(χ)L(k, χ) 6. SOME MORE FORMULAS FOR L(k, χ) 51 d−1 = χ(6) χ(a)e(−da/6)Ec,k−1 (a/d). e(1/6){2k − χ(2)}{3k − χ(3)} a=1 Proof. For simplicity, let us write F and E for Fc, k−1 and Ec, k−1 . We first prove (iii). Thus c = −e(α) with α = 1/6. 4c), we obtain m−k e(mb/d) F (6b/d) = 6k (∗) (b ∈ Z). m∈1+6Z Clearly F (6b/d) depends only on b (mod d). Also, we easily see that χ(m)m−k = 1 − χ(2)2−k (∗∗) 1 − χ(3)3−k L(k, χ).