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(V0, . . . , Vk), by V ′ the (k ′ +1)-uple (V ′
0
′ , . . .,V k ′), and by S the set {s1, . . . , sk, s ′ 1 +i, . . .,s′ k ′ +i}.
The X p and X 1,p norms on Σ + are defined as in section 5.3, and so are the spaces X p
S (Σ+ ,�n ) and
X 1,p
S (Σ+ 1,p
,�n ). Let X
satisfying the boundary conditions
S ,W,V ,V ′(Σ + ,�n ) be the completion of the space of maps u ∈ C ∞ S ,c (Σ + ,�n )
u(it) ∈ N ∗ W ∀t ∈ [0, 1], u(s) ∈ N ∗ Vj ∀s ∈ [sj, sj+1], u(s + i) ∈ N ∗ V ′
j ∀s ∈ [s ′ j, s ′ j+1],
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.
5.21. Theorem. Assume that Φ + (1)N ∗ Vk ∩ N ∗ V ′
k ′ = (0). Then theÊ-linear bounded operator
is Fredholm of index
∂A : X 1,p
S ,W,V ,V ′(Σ + ,�n ) → X p
S (Σ+ ,�n ), ∂Au = ∂u + Au,
ind∂A = n
2 − µ(Φ+ N ∗ Vk, N ∗ V ′ 1
k ′) −
2 (dimV0 + dimW − 2 dimV0 ∩ W)
− 1
k�
′
′ 1
(dimV 0 + dimW − 2 dimV 0 ∩ W) − (dimVj−1 + dimVj − 2 dimVj−1 ∩ Vj)
2 2
− 1
2
j=1
�
(dim V ′
k ′
j=1
j−1
′
′ ′
+ dim V j − 2 dimV j−1 ∩ V j ).
(118)
Proof. The proof of the fact that ∂A is semi-Fredholm is analogous to the case of the full strip,
treated in section 5.4. There remains compute the index. By an additivity formula analogous
to (113), it is enough to prove (118) in the case with no jumps, that is k = k ′ = 0, V = (V0),
V ′ = (V ′
0 ). In this case, we have a formula of the type
ind ∂A = −µ(Φ + N ∗ V0, N ∗ V ′
0 ) + c(W; V0; V ′
0 ),
and we have to determine the correction term c.
Assume W = (0), so that N ∗W = iÊn . Let us compute the correction term c when V0
and V ′
0 are either (0) orÊn . We can choose the map A to be the constant map A(s, t) = αI, for
α ∈]0, π/2[, so that Φ + (t) = eiαt . The Kernel and co-kernel of ∂αI are easy to determine explicitly,
by separating the variables in the corresponding boundary value PDE’s:
(i) If V0 = V ′
0 =Ên , the kernel and co-kernel of ∂αI are both (0). Since µ(e iαtÊn ,Ên ) = −n/2,
we have c((0);Ên ;Ên ) = −n/2.
(ii) If V0 = V ′
0 = (0), the kernel of ∂αI is iÊn e −αs , while its co-kernel is (0). Since µ(e iαt iÊn , iÊn ) =
−n/2, we have c((0); (0); (0)) = n/2.
(iii) If either V0 =Ên and V ′
0 = (0), or V0 = (0) and V ′
0 =Ên , the kernel and co-kernel of ∂αI are
both (0). Since µ(e iαtÊn , (0)) = µ(e iαt (0),Ên ) = 0, we have c((0);Ên ; (0)) = c((0); (0);Ên ) =
0.
Now let W, V0, and V ′
0 be arbitrary (with W partially orthogonal to both V0 and V ′
0 ). Let
U ∈ U(n) be such that UN ∗ ∗ W = iÊn . Then UN V0 = N ∗W0 and UN ∗V ′
0 = N ∗W ′ 0 , where
W0 = (V0 ∩ W) ⊥ ∩ (V0 + W), W ′ 0
70
′
= (V 0 ∩ W)⊥ ∩ (V ′
0 + W).