PDF (1016 kB)

Proof. By the same argument used in the proof of Theorem 5.23, the operator ∂A is Fredholm
and has the same index of the operator
∂ Ã
: X1,p
S ′ ,V0×V0,W ,W ′(Σ + ,�2n ) → X p
S ′(Σ + ,�2n ), u ↦→ ∂u + Ãu,
where S ′ := {2s1, . . . , 2sk, 2s1 + i, . . .,2sk + i}, W ′ is the (k + 1)-uple (∆Ên, . . . , ∆Ên), and
Ã(z) := 1
(A(z/2) ⊕ CA(z/2 + i)C).
2
By Theorem 5.21 and by (125), the index of this operator is
ind∂ Ã = n − µ(N ∗ Wk, graphCΦ + ) − 1
2 (dim ∆Ên + dimV0 × V0 − 2 dim∆Ên ∩ (V0 × V0))
− 1
2 (dim W0 + dimV0 × V0 − 2 dimW0 ∩ (V0 × V0))
− 1
2
k�
(dim Wj−1 + dim Wj − 2 dimWj−1 ∩ Wj)
j=1
= n − µ(N ∗ Wk, graphCΦ + ) − n 1
−
2 2 (dimW0 + 2 dimV0 − 2 dimW0 ∩ (V0 × V0))
The desired formula follows.
− 1
2
k�
(dim Wj−1 + dim Wj − 2 dimWj−1 ∩ Wj).
j=1
In the case of the left half-strip Σ − , let k, V0, W be as above, and let S = {s1, . . .,sk, s1 +
i, . . .,sk + i} with
0 = s0 > s1 > · · · > sk > sk+1 = −∞.
Let X 1,p
S ,V0,W (Σ− ,�n ) the completion of the space of all maps u ∈ C ∞ S ,c (Σ − ,�n ) such that
u(it) ∈ V0 ∀t ∈ [0, 1], (u(s), u(s + i)) ∈ N ∗ Wj, ∀s ∈ [sj+1, sj], j = 0, . . . , k,
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 let Φ − : [0, 1] → Sp(2n) be the solution of the linear Hamiltonian systems
Then we have:
d
dt Φ− (t) = iA(−∞, t)Φ + (t), Φ − (0) = I.
5.25. Theorem. Assume that graphCΦ − (1) ∩ N ∗ Wk = (0). Then for every p ∈]1, +∞[ the
Ê-linear operator
is bounded and Fredholm of index
∂A : X 1,p
S ,V0,W (Σ− ,�n ) → X p
S (Σ− ,�n ), u ↦→ ∂u + Au,
ind∂A = n
2 + µ(N ∗ Wk, graphCΦ − ) − 1
2 (dimW0 + 2 dimV0 − 2 dimW0 ∩ (V0 × V0))
− 1
2
k�
(dim Wj−1 + dim Wj − 2 dimWj−1 ∩ Wj).
j=1
75