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6.9. Lemma. There exists a positive number α(A) such that for every α ∈]0, α(A)], for every<br />

γ1, γ3 ∈ P Λ (L1), γ2, γ4 ∈ P Λ (L2), x1, x2 ∈ P Θ (H1 ⊕ H2) satisfying (216) and (217) and for<br />

every u ∈ M P α (γ1, γ2, γ3, γ4; x1, x2) there holds<br />

for suitable i, j ∈ {1, . . ., m}.<br />

π ◦ u([0, α] × {0}) = π ◦ u([0, α] × {1}) ∈ B δ/2(qi) 2 × B δ/2(qj) 2 ,<br />

Proof. By contradiction, we assume that there are an infinitesimal sequence of positive numbers<br />

(αh) and elements (λh, uh) ∈ M P αh (γ1, γ2, γ3, γ4; x1, x2) where γ1, γ2, γ3, γ4, x1, x2 satisfy (216) and<br />

(217), and<br />

π ◦ uh(shαh, 0) = π ◦ uh(shαh, 1) /∈<br />

m�<br />

i,j=1<br />

B δ/2(qi) 2 × B δ/2(qj) 2 , (223)<br />

for some sh ∈ [0, 1]. Let ch : [0, 1] → M 4 be the closed curve defined by ch(t) = π ◦ uh(0, t).<br />

Arguing as in the Convergence paragraph above, we see that up to subsequences<br />

λh → λ ∈ [0, 1],<br />

ch → c ∈ W u (γ1) × W u (γ2) × W u (γ3) × W u (γ4) in W 1,2 ([0, 1], M 4 )<br />

uh → u ∈ M K 0 (γ1, γ2; x1) × M K 0 (γ3, γ4; x2) in C ∞ loc ([0, +∞[×[0, 1] \ {(0, 0), (0, 1)}, T ∗ M 4 ).<br />

Since the space M K 0 (γ1, γ2; x1) × M K 0 (γ3, γ4; x2) is not empty, we have the index estimates<br />

m Λ (γ1) + m Λ (γ2) − µ Θ (x1) ≥ n, m Λ (γ3) + m Λ (γ4) − µ Θ (x2) ≥ n.<br />

Together with (216) this implies that<br />

m Λ (γ1) + m Λ (γ2) − µ Θ (x1) = n, m Λ (γ3) + m Λ (γ4) − µ Θ (x2) = n. (224)<br />

Moreover, (217) and the fact that the action of every orbit is non-negative implies that<br />

ËL1(γ1) +ËL2(γ2) ≤ A,�H1⊕H2(x1) ≤ A, (225)<br />

ËL1(γ3) +ËL2(γ4) ≤ A,�H1⊕H2(x2) ≤ A. (226)<br />

Furthermore, arguing as in the proof of Lemma 6.7, we find that the curve<br />

dh : [0, 1] × M 4 , dh(s) = π ◦ uh(αhs, 0) = π ◦ uh(αhs, 1),<br />

converges uniformly to the constant c(0) = c(1). By (224), (225) and (226),<br />

c(0) = c(1) = π ◦ u(0, 0) = π ◦ u(0, 1)<br />

is of the form (qi, qi, qj, qj), for some i, j ∈ {1, . . .,m}. But then the uniform convergence of (dh)<br />

to c(0) contradicts (223).<br />

We fix some α0 ∈]0, α(A)], and we choose the generic data g1, g2, H1, H2 in such a way that<br />

transversality holds for the problems M K 0 , M K α0 , and M P α0 . Then each M P α0 (γ1, γ2, γ3, γ4; x1, x2)<br />

is a smooth manifold whose boundary - if non-empty - is precisely the intersection with the regions<br />

{λ = 0} and {λ = 1}. In particular, when (216) and (217) hold, M K α0 (γ1, γ2, γ3, γ4; x1, x2) is a<br />

one-dimensional manifold with possible boundary points at λ = 0 and λ = 1, and Lemma 6.9<br />

implies that<br />

M P α0 (γ1, γ2, γ3, γ4; x1, x2) ∩ {λ = 0} = M K 0 (γ1, γ2; x1) × M K α0 (γ3, γ4; x2), (227)<br />

M P α0 (γ1, γ2, γ3, γ4; x1, x2) ∩ {λ = 1} = M K α0 (γ1, γ2; x1) × M K 0 (γ3, γ4; x2). (228)<br />

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