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
We start with D
EFINITION
4. A function aθ
1
, θ
2
= P
η
a
η
e
iηθ
on T
2
: = S × S, S = {|z| = 1}
is Riemann-Hilbert factorizable with respect to T
2
if there exist nonvanishing functions a
++
, a
−+
, a
−−
, a
+−
on T
2
with Fourier supports contained repectively in I :
= {η
1
≥ 0, η
2
≥ 0}, I I :
= {η
1
≤ 0, η
2
≥ 0}, I I I :
= {η
1
≤ 0, η
2
≤ 0}, I V :
= {η
1
≥ 0, η
2
≤ 0} such that
aθ
1
, θ
2
= a
++
a
−+
a
−−
a
+−
. T
HEOREM
6. Suppose that the following conditions are verified. A.1
σ z, ξ
6= 0 ∀z ∈ T
2
, ∀ξ ∈ R
2 +
, |ξ| = 1,
A.2 i nd
1
σ = ind
2
σ = 0,
where i nd
1
σ =
1 2πi
I
| ζ |=
1
d
z
1
log σ ζ, z
2
, ξ , and i nd
2
σ is similarly defined. Then σ z, ξ is R–H factorizable.
Here the integral is an integer-valued continuous function of z
2
and ξ , which is constant on the connected set T
2
× {|ξ| = 1}. Hence it is constant.
Riemann–Hilbert problem 199
Proof. Suppose that A1 and A.2 are verified. Then the function log aθ is well defined on T
2
and smooth. By Fourier expansion we have log aθ
= b
++
+ b
−+
+ b
−−
+ b
+−
where the supports of b
++
, b
−+
, b
−−
, b
+−
are contained in I , I I , I I I , I V , respec- tively. The factorization
aθ = expb
++
expb
−+
expb
−−
expb
+−
is the desired one. This ends the proof. R
EMARK
3. The above definition can be extended to a symbol of a pseudodiffer- ential operator a
= aθ
1
, θ
2
, ξ
1
, ξ
2
. We assume that the factors a
++
, a
−+
, a
−−
, a
+−
are smooth functions of ξ , in addition.
8. Riemann-Hilbert problem and construction of a parametrix
In this section we give a rather concrete construction of a parametrix of an operator reduced on the tori under the R–H factorizability.
Let L
2
T
2
be a set of square integrable functions, and let us define subspaces H
1
, H
2
of L
2
T
2
by H
1
: =
u
∈ L
2
; u = X
ζ
1
≥
u
ζ
e
iζ θ
,
H
2
: =
u
∈ L
2
; u = X
ζ
2
≥
u
ζ
e
iζ θ
.
We note that H
2
T
2
= H
1
∩ H
2
. We define the projections π
1
and π
2
by π
1
: L
2
T
2
−→ H
1
, π
2
: L
2
T
2
−→ H
2
. Then the projection π : L
2
T
2
→ H
2
T
2
is, by definition, equal to π
1
π
2
. We define a Toeplitz operator T
+
· and T·
+
by T
+
· := π
1
aθ, D : H
1
−→ H
1
, T·
+
: = π
2
aθ, D : H
2
−→ H
2
. If the Toeplitz symbols of these operators are Riemann-Hilbert factorizable it follows
that T
+
· and T·
+
are invertible modulo compact operators, and their inverses modulo compact operators are given by
5 T
− 1
+
· = π
1
a
− 1
++
a
− 1
+−
π
1
a
− 1
−+
a
− 1
−−
π
1
, T
− 1
·
+
= π
2
a
− 1
++
a
− 1
−+
π
2
a
− 1
+−
a
− 1
−−
π
2
, where the equality means the one modulo compact operators.
T
HEOREM
7. Let aθ, D be a pseudodifferential operator on the torus. Suppose that aθ, D is R–H factorizable. Then the parametrix R of π aθ, D is given by
6 R
= πT
− 1
+
· + T
− 1
·
+
− aθ, D
− 1
, where aθ, D
− 1
is a pseudodifferential operator with symbol given by aθ, ξ
− 1
.
200 M. Yoshino
These facts are essentially proved in [9] under slightly different situation. We give the proof for the reader’s convenience. In the following A
≡ B means that A and B are equal modulo compact operators.
Proof of 5. By comparing the principal symbol of both sides we obtain aθ, D ≡
a
++
a
−+
a
−−
a
+−
. T
+
·π
1
a
− 1
++
a
− 1
+−
π
1
a
− 1
−+
a
− 1
−−
π
1
≡ π
1
a
++
a
−+
a
−−
a
+−
π
1
a
− 1
++
a
− 1
+−
π
1
a
− 1
−+
a
− 1
−−
π
1
≡ π
1
a
−+
a
−−
a
++
a
+−
a
− 1
++
a
− 1
+−
π
1
a
− 1
−+
a
− 1
−−
π
1
+π
1
a
−+
a
−−
a
++
a
+−
I − π
1
a
− 1
++
a
− 1
+−
π
1
a
− 1
−+
a
− 1
−−
π
1
≡ π
1
a
−+
a
−−
π
1
a
− 1
−+
a
− 1
−−
π
1
, where we used
I − π
1
a
− 1
++
a
− 1
+−
π
1
= 0. Therefore, the right-hand side is equal to
π
1
a
−+
a
−−
a
− 1
−+
a
− 1
−−
π
1
+ π
1
a
−+
a
−−
I − π
1
a
− 1
−+
a
− 1
−+
π
1
and hence ≡ π
1
. Here we used π
1
a
−+
a
−−
I − π
1
= 0. Similarly, we can show π
1
a
− 1
++
a
− 1
+−
π
1
a
− 1
−+
a
− 1
−−
π
1
T
+·
≡ π
1
. This ends the proof.
Proof of 6. Noting that π = π
1
π
2
we have π
T
− 1
+
· πaπ = π T
− 1
+
· π
1
π
2
aπ = π T
− 1
+
· π
1
aπ − π T
− 1
+
· π
1
I − π
2
aπ ≡ π − πa
− 1
++
a
− 1
+−
π
1
a
− 1
−+
a
− 1
−−
π
1
I − π
2
aπ = π − πa
− 1
++
a
− 1
+−
π
1
π
2
+ π
1
I − π
2
a
− 1
−+
a
− 1
−−
π
1
I − π
2
aπ. Similarly, we have
π T
− 1
·
+
π aπ
= π T
− 1
·
+
π
1
π
2
aπ = π T
− 1
·
+
π
2
aπ − π T
− 1
·
+
π
2
I − π
1
aπ ≡ π − πa
− 1
++
a
− 1
−+
π
2
a
− 1
+−
a
− 1
−−
π
2
I − π
1
aπ = π − πa
− 1
++
a
− 1
−+
π
1
π
2
+ π
2
I − π
1
a
− 1
+−
a
− 1
−−
π
2
I − π
1
aπ. On the othe hand, since a
− 1
a ≡ I we have
−πa
− 1
π aπ
= −πa
− 1
π
1
π
2
aπ ≡ −π − πa
− 1
π
1
π
2
− I aπ. By using
π
1
π
2
− I = π
1
π
2
− I + π
1
− I π
2
− π
1
− I π
2
− I
Riemann–Hilbert problem 201
we have −πa
− 1
π aπ
≡ −π
1
π
2
− π
1
π
2
a
− 1
π
1
π
2
− I aπ −πa
− 1
π
1
− I π
2
aπ + πa
− 1
π
1
− I π
2
− I aπ. Combining these relations
RT ≡ π − πa
− 1
++
a
− 1
+−
π + π
1
I − π
2
a
− 1
−+
a
− 1
−−
π
1
I − π
2
aπ −πa
− 1
++
a
− 1
−+
π + π
2
I − π
1
a
− 1
+−
a
− 1
−−
π
2
I − π
1
aπ −πa
− 1
π
1
π
2
− I aπ − πa
− 1
π
1
− I π
2
aπ + πa
− 1
π
1
− I π
2
− I aπ. We note
π + π
1
I − π
2
= I − π
1
− I π
2
− I − π
2
I − π
1
, π
+ π
2
I − π
1
= I − π
1
− I π
2
− I − π
1
I − π
2
. It follows that
RT − π ≡ πa
− 1
++
a
− 1
+−
π
1
− I π
2
− I +π
2
I − π
1
a
− 1
−+
a
− 1
−−
π
1
I − π
2
aπ + πa
− 1
π
1
− I π
2
− I aπ +πa
− 1
++
a
− 1
−+
π
1
− I π
2
− I + π
1
I − π
2
a
− 1
+−
a
− 1
−−
π
2
I − π
1
aπ. In order to show that the right-hand side operators are compact operators we will show
that the operators π ϕπ
1
− I π
2
− I , π
2
I − π
1
ϕπ
1
I − π
2
, π
1
I − π
2
ϕπ
2
I − π
1
are compact. Here ϕ is an appropriately chosen smooth function. In order to show this let
u =
X
α
u
α
e
iαθ
∈ L
2
, ϕξ =
X
β
ϕ
β
ξ e
iβθ
be the Fourier expansion of u ∈ L
2
and ϕ ∈ C
∞
, respectively. Because ϕθ, D is order zero pseudodifferential operator the Fourier coefficients of ϕ
β
ξ is rapidly
decreasing in ξ when |β| → ∞. Therefore
π ϕπ
1
− I π
2
− I u = X
µ=α+β∈ I
X
α+β=µ,α∈ I I I
ϕ
β
µ u
α
e
iµθ
. Because µ
∈ I and −α ∈ I by the definition of I and I I I , β satisfies that |β| = |µ − α| ≥ |µ|. It follows that, for all n ≥ 1 and µ
|µ|
n
X
α+β=µ,α∈ I I I
|ϕ
β
µ ||u
α
| ≤ X
|β|
n
|ϕ
β
µ ||u
α
| ∞. Indeed,
|ϕ
β
µ ||β|
n
is bounded in µ and β. It follows that the Fourier coefficients converge uniformly in u
∈ L
2
. Thus π ϕπ
1
− I π
2
− I is a compact operator. The compactness of other operators are proved similarly. Hence R is a left regularizer. We
can similarly show that R is a right regularizer. This ends the proof.
202 M. Yoshino
9. Solvability in two dimensional case
