224 D. SALAMON, FLOER HOMOLOGY
Exercise 5.13. Prove that every solution u E Z(x-,x+) of fh,H(u) = 1(u) satis-
fies
E(u) = aH(x-,u-) - aH(x+,u+) + f
00
f
1
(8 8 u,1(u)) dtds,
-oo lo
where u± : B ----> M are smooth maps which satisfy u±(s^2 7rit) = x±(t) and u+ =
u-#u. Deduce that (E(u) - c)/2:::; aH(x-, u-) -aH(x+, u+):::; 2(E(u) + c) where
c is the constant of (56). Deduce further that, if ( 4) holds, then
(57) E(u):::; c + ~ (μ(u, H) + r]H(x+) - 'T]H(x-))
where 'T]H : P(H) ---->IR is defined by (11). D
With the perturbation in place let us denote the space of perturbed connecting
orbits by
M(x-, x+; H, J, r) = { u E Z(x-, x+) : Eh,H(u) E r(u)}.
For generic choices of H, J , and r , this is a finite dimensional branched manifold
of local dimension μ(u, H) near u (Exercise 2.10) and it carries a weight function
M(x-, x+; H, J, r)----> Q: u f-+ >.(u) = >.(u, 8J,H(u)).
As before we denote by M^1 (x-, x+; H , J, f) the 1-dimensional part of this moduli
space. In view of the perturbation the difficulties of Section 5.1 disappear and,
using (57), we find that the quotient
M^1 (x-, x+; H, J, r) = M^1 (x-, x+; H , J, r)/IR
is a finite set (compare with Remark 5.12). As in the case of the rational Gromov-
Witten invariants, each perturbed connecting orbit [u] E M^1 (x-, x+; H , J, r) may
belong to several branches of the zero dimensional moduli space and hence carries
a rational weight p( u) = Li EiAi, where the A i are the weights of the branches
to which u belongs, and the Ei E { ±1} are the signs determined by the coherent
orientations. The Floer chain complex is now defined as the vector space over Q,
generated by the periodic solutions of (1):
CF*(H) = EB Q(x).
xEP(H)
The boundary operator 8 = 3J,H,r is given by
3J,H,[' (y) = L p(u) (x).
xEP(H) [u)EM^1 (y,x;H, J ,f')
Theorem 5.14. Suppose that (M,w) satisfies (4) with T < 0 and let H, J, r be
chosen as above. Then 8 o 8 = 0 and the Floer homology groups
ker 3J,H,r
HF*(M,w; H , J,r) =. f)J Hr
lm ' '
are naturally isomorphic to the singular homology of M with rational coefficients.
Proof. We only sketch the proof of 8 o 8 = 0. This is equivalent to the identity
p(v)p(u) = 0
yEP(H) [v)EM'(z,y;H, J , f') [u)EM^1 (y,x;H,J , f')
