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is bounded and Fredholm of index
ind∂A = n
2 + µ(Φ−N ∗ Vk, N ∗ V ′ 1
k ′) −
2 (dimV0 + dimW − 2 dimV0 ∩ W)
− 1 ′
(dimV 0 + dimW − 2 dimV
2 ′
0 ∩ W) − 1
k�
(dimVj−1 + dimVj − 2 dimVj−1 ∩ Vj)
2
− 1
2
j=1
�
(dim V ′
j−1 + dim V ′
j − 2 dimV ′
k ′
j=1
Indeed, notice that if u(s, t) = v(−s, t), then
−(∂u(s, t) + A(s, t)u(s, t)) = C(∂v(−s, t) − CA(s, t)Cv(−s, t)),
j−1 ∩ V ′
where C is denotes complex conjugacy. Then the operator (120) is obtained from the operator
∂B : X 1,p
−S ,W,V ,V ′(Σ + ,�n ) → X p
−S (Σ+ ,�n ), ∂Bv = ∂v + Bv,
j ).
(121)
where B(s, t) = −CA(−s, t)C, by left and right multiplication by isomorphisms. In particular,
the indices are the same. Then Theorem 5.22 follows from Theorem 5.21, taking into account the
fact that the solution Φ + of
is Φ + (t) = CΦ − (t)C, so that
d
dt Φ+ (t) = iB(+∞, t)Φ + (t), Φ + (0) = I,
µ(Φ + N ∗ Vk, N ∗ V ′
k ′) = µ(CΦ− CN ∗ Vk, N ∗ V ′
k ′) = −µ(Φ− N ∗ Vk, N ∗ V ′
k ′).
5.8 Non-local boundary conditions
It is useful to dispose of versions of Theorems 5.9, 5.21, and 5.22, involving non-local boundary
conditions. In the case of the full strip Σ, let us fix the following data. Let k ≥ 0 be an integer,
let
−∞ = s0 < s1 < · · · < sk < sk+1 = +∞
be real numbers, and set S := {s1, . . .,sk, s1 + i, . . .,sk + i}. Let W0, W1, . . . , Wk be linear
subspaces ofÊn ×Ên such that Wj−1 and Wj are partially orthogonal, for j = 1, . . .,k, and set
W = (W0, . . . , Wk).
The space X 1,p
S ,W (Σ,�n ) is defined as the completion of the space of all u ∈ C ∞ S ,c (Σ,�n ) such
that
(u(s), u(s + i)) ∈ N ∗ Wj, ∀s ∈ [sj, sj+1], j = 0, . . .,k,
with respect to the norm �u� X 1,p (Σ).
Let A ∈ C 0 (Ê×[0, 1], L(Ê2n ,Ê2n )) be such that A(±∞, t) ∈ Sym(2n,Ê) for every t ∈ [0, 1], and
define the symplectic paths Φ + , Φ − : [0, 1] → Sp(2n) as the solutions of the linear Hamiltonian
systems
d
dt Φ± (t) = iA(±∞, t)Φ ± (t), Φ ± (0) = I.
Denote by C the complex conjugacy, and recall from section 5.1 that Φ ∈ L(Ê2n ,Ê2n ) is symplectic
if and only if graphCΦ is a Lagrangian subspace of (Ê2n ×Ê2n , ω0×ω0). Then we have the following:
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