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We shall apply the above lemma to the complexes<br />

Ck = � M(ËΛ L1 , g1) ⊗ M(ËΛ L2 , g2) �<br />

k+n , C′ k = F Θ k (H1 ⊕ H2, J1 ⊕ J2),<br />

and to the chain maps K Λ 0 and K Λ α0 . The tensor products KΛ 0 ⊗K Λ α0 and KΛ α0 ⊗KΛ 0 are represented<br />

by the coupling - in two different orders - of the corresponding elliptic boundary value problems.<br />

4.7. Proposition. The chain maps K Λ 0 ⊗ KΛ α0 and KΛ α0 ⊗ KΛ 0<br />

are homotopic.<br />

Constructing a homotopy between the coupled problems is easier than dealing with the original<br />

ones: we can keep α0 fixed and rotate the boundary condition on the initial part of the half-strip.<br />

This argument is similar to an alternative way, due to Hofer, to prove the gluing statements in<br />

standard Floer homology. Details of the proof of Proposition 4.7 are contained in section 6.5<br />

below.<br />

Here we just construct the cycle ǫ and the chain map δ required in Lemma 4.6. Since changing<br />

the Lagrangians L1 and L2 (and the corresponding Hamiltonian) changes the chain maps appearing<br />

in diagram (53) by a chain homotopy, we are free to choose the Lagrangians so to make the<br />

construction easier.<br />

We consider a Lagrangian of the form<br />

where the potential V1 ∈ C ∞ (Ì×M) satisfies<br />

L1(t, q, v) := 1<br />

2 〈v, v〉 − V1(t, q),<br />

V1(t, q) < V1(t, q0) = 0, ∀t ∈Ì, ∀q ∈ M \ {q0}, (58)<br />

The corresponding Euler-Lagrange equation is<br />

HessV1(t, q0) < 0, ∀t ∈Ì. (59)<br />

∇tγ ′ (t) = −gradV1(t, γ(t)), (60)<br />

where ∇t denotes the covariant derivative along the curve γ. By (58) and (59), the constant curve<br />

q0 is a non-degenerate minimizer for the action functionalËΛ L1 on the free loop space (actually, it<br />

is the unique global minimizer), so<br />

m Λ (q0, L1) = 0.<br />

Notice also that the equilibrium point q0 is hyperbolic and unstable. We claim that there exists<br />

ω > 0 such that<br />

every solution γ of (60) such that γ(0) = γ(1), other than γ(t) ≡ q0, satisfiesËL1(γ) ≥ ω. (61)<br />

Assuming the contrary, there exists a sequence (γh) of solutions of (60) with γh(0) = γh(1) and<br />

0

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