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By using the change of variable v = Uu, we find
c(W; V0; V ′
0) = c((0); W0; W ′ 0), (119)
and we are reduced to compute the latter quantity. By an easy homotopy argument, using the fact
that the Fredholm index is locally constant in the operator norm topology, we can assume that W0
and W ′ 0 are partially orthogonal. ThenÊn has an orthogonal splittingÊn = X1 ⊕ X2 ⊕ X3 ⊕ X4,
where
from which
W0 = X1 ⊕ X2, W ′ 0 = X1 ⊕ X3,
N ∗ W0 = X1 ⊕ X2 ⊕ iX3 ⊕ iX4, N ∗ W ′ 0 = X1 ⊕ iX2 ⊕ X3 ⊕ iX4.
Then the operator ∂αI decomposes as the direct sum of four operators, whose index is computed
in cases (i), (ii), and (iii) above. Indeed,
Since
we find
c((0); W0; W ′ 1
0 ) =
2 dim X4 − 1
dim X1
2
= 1
2 codim(W0 + W ′ 1
0 ) −
2 dimW0 ∩ W ′ 1
0 =
2 (n − dimW0 − dim W ′ 0 ).
dim W0 = dim(V0 + W) − dim V0 ∩ W = dimV0 + dimW − 2 dimW0 ∩ W,
dimW ′ ′
′
′
0 = dim(V 0 + W) − dimV 0 ∩ W = dimV 0 + dim W − 2 dimW ′ 0 ∩ W,
c((0); W0; W ′ n 1
0 ) = −
2
2 (dim V0 + dimW − 2 dimW0 ∩ W) − 1
2
Together with (119), this proves formula (118).
′
(dimV 0 + dimW − 2 dimW′ 0 ∩ W).
We conclude this section by considering the case of the left half-strip Σ− . Let k, k ′ ≥ 0,
V = (V0, . . . , Vk), and V ′ = (V ′ ′
0 , . . . , V ′) be as above. Let
k
−∞ = sk+1 < sk < · · · < s1 < s0 = 0, −∞ = s ′ k ′ +1 < s ′ k ′ < · · · < s′ 1 < s ′ 0 = 0,
be real numbers, and set S = {s1, . . . , sk, s ′ 1 + i, . . .,s ′ k ′ + i}.
Let X 1,p
S ,W,V ,V ′(Σ− ,�n ) be the completion of the space of maps u ∈ C ∞ S ,c (Σ − ,�n ) satisfying
the boundary conditions
u(it) ∈ N ∗ W ∀t ∈ [0, 1], u(s) ∈ N ∗ Vj ∀s ∈ [sj+1, sj], u(s + i) ∈ N ∗ V ′
j ∀s ∈ [s ′ j+1, s ′ j],
with respect to the norm �u� X 1,p (Σ − ).
Let A ∈ C 0 ([−∞, 0]×[0, 1], L(Ê2n ,Ê2n )) be such that A(−∞, t) is symmetric for every t ∈ [0, 1],
and denote by Φ − : [0, 1] → Sp(2n) the solutions of the linear Hamiltonian system
Then we have:
d
dt Φ− (t) = iA(−∞, t)Φ − (t), Φ − (0) = I.
5.22. Theorem. Assume that Φ − (1)N ∗ Vk ∩ N ∗ V ′
k ′ = (0). Then theÊ-linear operator
∂A : X 1,p
S ,W,V ,V ′(Σ − ,�n ) → X p
S (Σ− ,�n ), ∂Au = ∂u + Au, (120)
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