<strong>IN</strong>HOMOGENEOUS QUADRATIC FORMS 145 Proof We may and will assume that L is of the first type. We continue to use the notations as in Lemma 5.7. In particular, let M = pL, andlet{v1,v2} be the standard basis for M. Then Lemma 5.7 implies that {〈vi, pξ〉B} ≤(cδ 1−η2 )/(T 1/2−τ3 ). Recall that v1 = (m, 0,n,0) and v2 = (0,m,0,n), where gcd(m, n) = 1, and we have T 1−τ3 ≤v M =m 2 + n 2 ≤ T 1+τ3 , (60) where τ3 is a fixed multiple of the constant τ2 appearing in Theorem 5.6. We also note that 〈v1 , pξ〉B = m(pξ)4 − n(pξ)2 and that 〈v2 , pξ〉B(pξ)1 − m(pξ)3. Let τ ′ 1 ≪ 1/32κ be chosen. Then choose τ2 in Theorem 5.6 such that τ2 T τ ′ 1. Taking τ2 even smaller, we may and will assume that τ3
146 MARGULIS and MOHAMMADI Proof Let η1 be as in Proposition 5.3, andletη2 δ η2 for some kθ ∈ p1(A qL t (δ, ε)) is O(T 1−τ1 ). On the other hand, using Theorem 5.6, we see that the number of quasi-null subspaces with T/2 ≤v L ≤T and A L t (δ, ε) = ∅which are not in Qt(δ, ε, τ2) is O(T 1−τ2 ). The corollary follows with η = η1/2 and τ = min{τ1,τ2}. 6. Proof of Theorem 4.1 We complete the proof, of Theorem 4.1 in this section. Before proceeding to the proof, we need the following two statements. LEMMA 6.1 If Qξ is an irrational (2, 2) form, then the number of 2-dimensional null subspaces, say L,ofQ such that ξ ∈ v + L for some v ∈ Z 4 is at most four. Proof First, note that using [EMM2, Lemma 10.3], we may and will assume that Q is a rational form. In this case, we actually show that there are at most two such subspaces. In order to see this, suppose that there are two null rational subspaces of the same type, say Li, for i = 1, 2 such that ξ ∈ vi + Li, where vi ∈ Z 4 . Now since the Li are of the same type, they are transversal. Let {w i 1 ,wi 2 } be an integral basis for Li. Then 〈ξ,w i j 〉B ∈ Z for i, j = 1, 2. This (thanks to the transversality of the Li) implies that ξ is a rational vector, which is a contradiction. The following is essential to the proof of Theorem 4.1 and proved in Section 7. PROPOSITION 6.2 The number of quasi-null subspaces with norm between T/2 and T is O(T ). As we mentioned, this is proved in Section 7. However, for the time being, let us remark that in the case where Q is a split integral form, this is immediate. Indeed, in that case we are dealing with null subspaces, and hence we need to show this for Q = B. As we observed, however, the null subspaces of B are classified by primitive integral vectors (m, n). Now if L is a null subspace of either type which corresponds to (m, n), then v L =m 2 + n 2 . Thus the result is obvious. We now turn to the proof of Theorem 4.1.
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30 CASIM ABBAS Recall that the Reeb
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32 CASIM ABBAS that Here the energy
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36 CASIM ABBAS punctured surfaces w
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38 CASIM ABBAS that is, the central
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46 CASIM ABBAS and Jδε2 = xγ1(r)
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54 CASIM ABBAS formula (2.2) for th
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56 CASIM ABBAS Restricting any of t
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58 CASIM ABBAS where fτ (∞) = li
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64 CASIM ABBAS THEOREM 3.7 Let µ,
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66 CASIM ABBAS Since w2 − w1C 1,
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68 CASIM ABBAS be the conformal tra
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70 CASIM ABBAS so that, for z ∈ B
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72 CASIM ABBAS and, for every R>0,
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74 CASIM ABBAS W 1,p loc (C) since
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78 CASIM ABBAS is finite, the image
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80 CASIM ABBAS Converting from coor
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82 CASIM ABBAS [27] D. MCDUFF and D
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84 GUILLARMOU and TZOU where ∂ν
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88 GUILLARMOU and TZOU where F ⊂
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90 GUILLARMOU and TZOU LEMMA 2.2.4
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92 GUILLARMOU and TZOU THEOREM 2.3.
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