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such that, setting V λ
ij := ϕij({λ} ×Ê3n ), we have that each V λ
ij is relatively closed in Bδ(qi) 4 ×
Bδ(qj) 4 and
(∆ Θ M × ∆ Θ M) ∩ (Bδ(qi) × Bδ(qj)) ⊂ V λ
ij, ∀λ ∈ [0, 1], (212)
V λ
ij ⊂ (∆M×M × ∆M×M) ∩ (Bδ(qi) × Bδ(qj)), ∀λ ∈ [0, 1], (213)
V 0
ij = (∆Θ M × ∆M×M) ∩ (Bδ(qi) × Bδ(qj)), (214)
V 1
ij = (∆M×M × ∆ Θ M ) ∩ (Bδ(qi) × Bδ(qj)). (215)
Let γ1, γ3 ∈ P Λ (L1), γ2, γ4 ∈ P Λ (L2), and x1, x2 ∈ P Θ (H1 ⊕ H2) satisfy the index identity
and the action bounds
m Λ (γ1) + m Λ (γ2) + m Λ (γ3) + m Λ (γ4) − µ Θ (x1) − µ Θ (x2) = 2n, (216)
ËL1(γ1) +ËL2(γ2) +ËL1(γ3) +ËL2(γ4) ≤ A,�H1⊕H2(x1) +�H1⊕H2(x2) ≤ A. (217)
Given α > 0, we define
M P α (γ1, γ2, γ3, γ4; x1, x2)
to be the set of pairs (λ, u) where λ ∈ [0, 1] and u : [0, +∞[×[0, 1] → T ∗ M 4 is a solution of the
equation
satisfying the boundary conditions
∂H1⊕H2⊕H1⊕H2,J(u) = 0, (218)
π ◦ u(0, ·) ∈ W u (γ1) × W u (γ2) × W u (γ3) × W u (γ4), (219)
(u(s, 0), −u(s, 1)) ∈
m�
N ∗ V λ
ij if 0 ≤ s ≤ α, (220)
i,j=1
(u(s, 0), −u(s, 1)) ∈ N ∗ (∆ Θ M × ∆ Θ M) if s ≥ α, (221)
lim
s→+∞ u(s, ·) = (x1, x2). (222)
Notice that if (0, u) belongs to M P α (γ1, γ2, γ3, γ4; x1, x2), then writing u = (u1, u2) where u1 and
u2 take values into T ∗ M 2 , we have
u1 ∈ M K 0 (γ1, γ2; x1), u2 ∈ M K α (γ3, γ4; x2).
If transversality holds, we deduce the index estimates
m Λ (γ1) + m Λ (γ2) − µ Θ (x1) ≥ n, m Λ (γ3) + m Λ (γ4) − µ Θ (x2) ≥ n.
But then (216) implies that the above inequalities are indeed identities. Similarly, if (1, u) belongs
to M P α (γ1, γ2, γ3, γ4; x1, x2), we deduce that
and
u1 ∈ M K α (γ1, γ2; x1), u2 ∈ M K 0 (γ3, γ4; x2).
m Λ (γ1) + m Λ (γ2) − µ Θ (x1) = n, m Λ (γ3) + m Λ (γ4) − µ Θ (x2) = n.
Conversly, we would like to show that pairs of solutions in M K 0 ×M K α (or M K α ×M K 0 ) correspond
to elements of M P α of the form (0, u) (or (1, u)), at least if α is small. The key step is the following:
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