By Knopp M.I. (ed.)

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9, the elements w(a), w E W(r), span IRn. Therefore not all w(a) can be orthogonal to 13. After replacing a by w(a) for some wE W(r) if necessary, we may assume that a and 13 are nonorthogonal. We may also assume that a and 13 are distinct and a i=- -13, because otherwise we are done. Hence a . 13 = ± 1. Replacing 13 if necessary by -13 = s f3 (13), we may assume that a·f3 = 1. Then sasf3sa(f3) = sasf3(f3-a) = sa(f3-a-f3) = a. 10. 10 implies that in the constructions for the lattices E7 and E6 above one can take any root v and any pair of roots Vl, V2 with Vl .

4, forcing ,\ = O. Then so that b{3 = 0 for all (3, and similarly c"! = 0 for all 'Y, as required. 6 Let bl,'" ,'Yn} be a basis of ffi. n . n with t . 'Yi > 0 for all 1 s:; i s:; n. Then there exists an element Proof. Let J~ be the orthogonal projection of 'Yi on the subspace Ui spanned by all basis vectors except 'Yi, for i = 1, ... , n. Let Ji := 'Yi - J~. Then Ji . 'Yj = 0 for 1 s:; j s:; n, j f- i, and bi . 'Yi = bi . (bi + b~) = br > O. Then t = element as required provided that all ri > O.

But = 1. IR Hence c = 1, which proves the lemma. 1. 3, we first show the following lemma. 2 Let r c jRn be a lattice. Then the series converges uniformly absolutely for all T with 1m T Proof. Let r =M . zn for M E GL(n, jR), and let €:= min(Mx)2. Ixl=l :::: Vo > O. 4 Theta Functions as Modular Forms Then c L > 0 and (MX)2 2: cx 2 for all x kriTx21 xEr = L le 7fiT (MX)21 ::; xEZ n E L JRn. This yields the following estimation e-7fVQCX2 = ( xEzn f: e-7fVoc1'2) n < 00. 1'=-00 This proves the lemma.

### Analytic Number Theory: Proceedings by Knopp M.I. (ed.)

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