194 M. Yoshino
where |λ
1
| 1 |λ
2
|. Hence we have e
iθ
− λ
2
ue
iθ
≡ 1
− λ
1
e
− iθ
− 1
f = f
+
+ f
−
≡ f
+
modulo negative powers. Here f
+
resp. f
−
consists of Fourier coefficients of non- negative resp. negative part. Hence, we have
e
iθ
− λ
2
ue
iθ
= f
+
. The solution is given by ue
iθ
= e
iθ
− λ
2 −
1
f
+
. Hence the cokernel vanishes. This ends the proof.
6. Index formula of an ordinary differential operator
We will give an elementary proof of an index formula. Cf. Malgrange, Komatsu, Ramis. Let
⊂ C be a bounded domain satisfying the following condition. A.1 There exists a conformal map ψ : D
w
= {|z| w} 7→ such that ψ can be extented in some neighborhood of D
w
= {|z| ≤ w} holomorphically. Let w 0, µ
≥ 0, and define G
w
µ = {u =
X
n
u
n
x
n
; kuk
2
: =
X
n
|u
n
| w
n
n n
− µ
2
∞}, where n
− µ = 1 if n − µ ≤ 0. Clearly, G
w
µ is a Hilbert space. Define A
w
µ as
the totality of holomorphic functions ux on such that uψz ∈ G
w
µ .
Consider an N × N N ≥ 1 matrix-valued differential operator
Px , ∂
x
= p
i j
x , ∂
x
, where p
i j
is holomorphic ordinary differential operator on . For simplicity, we as- sume that there exist real numbers ν
i
, µ
j
i, j = 1, . . . , N such that
or d p
i j
≤ µ
j
− ν
i
, or d p
ii
= µ
i
− ν
i
. Hence
1 Px , ∂
x
:
N
Y
j =1
A
w
−µ
j
−→
N
Y
j =1
A
w
−ν
j
. If we write
p
i j
x , ∂
x
=
µ
j
− ν
i
X
k=0
a
k
x ∂
k x
, a
k
x ∈ O
Riemann–Hilbert problem 195
we obtain, by the substitution x = ψz
˜p
i j
z, ∂
z
= X
k=µ
j
− ν
i
a
k
ψ zψ
′
z
− k
∂
k z
+ · · · . Here the dots denotes terms of order µ
j
− ν
i
, which are compact operators. Define a Toeplitz symbol Q
z by Q
z : = q
i j
z. Here 2
q
i j
z = a
µ
j
− ν
i
ψ zzψ
′
z
ν
i
− µ
j
. Then we have
T
HEOREM
3. Suppose A.1. Then the map 1 is a Fredholm operator if and only if
3 det Q
z 6= 0
for ∀z ∈ C, |z| = w.
If 3 holds the Fredholm index of 1, χ : = dim
C
Ker P − codim
C
Im P is given by the following formula
4 −χ =
1 2π
I
| z|=w
dlog det Q
z, where the integral is taken in counterclockwise direction.
Proof. Suppose 3. We want to show the Fredholmness of 1. For the sake of simplicity, we suppose that µ
j
− ν
i
= m, i.e., ord p
i j
= m. If we lift P onto the torus and we multiply the lifted operator on torus with
hD
θ
i
− m
we obtain an operator π Q
on H
2
modulo compact operators. It is easy to show that π Q
on H
2
is a Fredholm operator. cf. [3]. Because the difference of these operators are compact operators the
lifted operator is a Fredholm operator. In order to see the Fredholmness of 1 we note that the kernel of the operator on
the boundary coincides with that of the operator inside under trivial analytic extension because of a maximal principle. The same property holds for a cokernel. Therefore the
Fredholmness of the lifted operator implies the Fredholmness of 1. Conversely, assume that 1 is a Fredholm operator. We want to show 3. By the
argument in the above we may assume that the operator π Q
on H
2
is a Fredholm operator. For the sake of simplicity, we prove in the case N
= 1, a single case. We denote π Q
by T . Let K be a finite dimensional projection onto K er T . Then there exists a constant c 0 such that
kT f k + kK f k ≥ ck f k, ∀ f ∈ H
2
. It follows that
kπ Q
π g
k + kπ K πgk + ck1 − πgk ≥ ckgk, ∀g ∈ L
2
.
196 M. Yoshino
Let U be a multiplication operator by e
iθ
. Then we have kπ Q
π U
n
g k + kπ K πU
n
g k + ck1 − πU
n
g k ≥ ckU
n
g k,
∀g ∈ L
2
. Because U preserves the distance we have
kU
− n
π Q
π U
n
g k + kπ K πU
n
g k + ckU
− n
1 − πU
n
g k ≥ ckgk,
∀g ∈ L
2
. The operator U
− n
π U
n
is strongly bounded in L
2
uniformly in n. We have U
− n
π U
n
g → g
strongly in L
2
for every trigonometric polynomial g. Therefore it follows that
U
− n
π U
n
g → g strongly in L
2
. Thus U
− n
1 − πU
n
g converges to 0 strongly, and U
− n
π Q
π U
n
g = U
− n
π U
n
Q
U
− n
π U
n
g → Q
in the strong sense. On the other hand, because U
n
converges to 0 weakly π K πU
n
g tends to 0 strongly by the compactness of K . It follows that
kQ
g k ≥ ckgk
for every g ∈ L
2
. If Q
vanishes at some point t , there exists g with support in some
neighborhood of t with norm equal to 1. This contradicts the above inequality. Hence
we have proved the assertion. Next we will show the index formula 4. For the sake of simplicity, we assume that
w = 1 and Q
z is a rational polynomial of z, namely Q
z = cz − λ
1
· · · z − λ
m
z − λ
m+1
· · · z − λ
m+n
z
− k
. Here
|λ
1
| ≤ · · · ≤ |λ
m
| 1 |λ
m+1
| ≤ · · · ≤ |λ
m+n
|. We can easily see that the right-hand side of 4 is equal to m
− k. We will show that the Fredholm index of the operator
π Q
: H
2
→ H
2
is equal to k − m. Because z − λ
m+1
· · · z − λ
m+n
does not vanish on the unit disk the multiplication operator with this function is one-to-one on H
2
. We may assume that Q
z = z − λ
1
· · · z − λ
m
z
− k
. We can calculate the kernel and the cokernel of this operator by constructing a
recurrence relation. Let us first consider the case Q
z = z − λz
− k
|λ| 1. By substituting u
= P
∞ n=0
u
n
z
n
into π
z − λz
− k
u = 0
Riemann–Hilbert problem 197
we obtain z
− λz
− k
∞
X
n=0
u
n
z
n
=
∞
X
n=0
u
n−1
− λu
n
z
n−k
≡ 0, modulo negative powers of z. By comparing the coefficients we obtain the following
recurrence relation u
k−1
− u
k
λ = 0,
u
k
− λu
k+1
= 0, . . . Here u
, u
1
, . . . u
k−2
are arbitrary. Suppose that u
k−1
= c 6= 0. Then we have u
k
= cλ, u
k+1
= cλ
2
, . . . Because the radius of convergence of the function u constructed from this series is 1,
u is not in the kernel. Therefore, the kernel is k − 1 dimensional.
Next we want to show that the cokernel is trivial, namely the map is surjective. Consider the following equation
π z
− λz
− k
u = f =
∞
X
n=0
f
n
z
n
. By the same arguement as in the above we obtain
u
k−1
− u
k
λ = f
, u
k
− λu
k+1
= f
1
, u
k+1
− λu
k+2
= f
2
, . . . By setting
u = u
1
= · · · = u
k−2
= 0, we obtain, from the above recurrence relations
u
k−1
= λu
k
+ f = f
+ λ f
1
+ λ
2
u
k+1
= f + λ f
1
+ λ
2
f
2
+ λ
3
u
k+2
+ · · · = f
+ λ f
1
+ λ
2
f
2
+ λ
3
f
3
+ · · · The series in the right-hand side converges because
|λ| 1. Similarly we have u
k
= λu
k+1
+ f
1
= f
1
+ λ f
2
+ λ
2
u
k+2
= f
1
+ λ f
2
+ λ
2
f
3
+ λ
3
u
k+3
+ · · · = f
1
+ λ f
2
+ λ
2
f
3
+ λ
3
f
4
+ · · · . The series also converges. In the same way we can show that u
j
j = k − 1, k, k +
1, . . . can be determined uniquely. Hence the map is surjective. It follows that Ind =
k − 1. This proves the index formula. The general case can be treated in the same way
by solving a recurrence relation. We give an alternative proof of this fact. We recall the following facts.
The operator π z
− k
has exactly k dimensional kernel given by the basis 1, z, . . . , z
k−1
. The map πz − λ |λ| 1 has one dimensional cokernel. Indeed,
the equation z − λ
P u
n
z
n
= 1 does not have a solution in H
2
because we have
198 M. Yoshino
u = −1λ, u
1
= −1λ
2
, u
2
= −1λ
3
, ..., which does not converge on the torus.
These facts show the index formula for particular symbols. In order to show the index formula for general symbols we recall the following
theorems. T
HEOREM
4 A
TKINSON
. If A : H
2
→ H
2
and B : H
2
→ H
2
are Fredholm operators B A is a Fredholm operator with the index
Ind B A = Ind B + Ind A.
T
HEOREM
5. For the Toeplitz operators π q : H
2
→ H
2
and π p : H
2
→ H
2
the operator π pq
− π pπq is a compact operator. These theorems show that the index formula for Q
is reduced to the one with symbols given by every factor of the factorization of Q
.
7. Riemann-Hilbert problem - Case of 2 variables
