50 3: Hamiltonian formalism and zero-curvature representation
There is a freedom in the matrices U(λ) and V(λ) known as gauge invariance.
Let g = g(τ, ρ) be an arbitrary invertible matrix. The transformations
˜
U = gUg
−1
+
∂g
∂ρ
g
−1
and
V = gVg
−1
+
∂g
∂τ
g
−1
(3.3.13)
map solutions to the zero-curvature equation into new solutions: if the matri-
ces (U, V) satisfy (3.3.10) then so do the matrices (
U,
V). To see this assume
that v(ρ,τ,λ) is a solution to the linear system (3.3.9), and demand that
˜v = g(ρ,τ)v be another solution for some (
U,
V). This leads to the gauge
transformation (3.3.13).
One can develop a version of the IST which recovers U(λ) and V(λ) from
a linear-scattering problem (3.3.9). The representation (3.3.10) can also be an
effective direct method of finding solutions if we know n linearly independent
solutions v
1
,...,v
n
to the linear system (3.3.9) in the first place. Let (ρ,τ,λ)
be a fundamental matrix solution to (3.3.9). The columns of are the n
linearly independent solutions v
1
,...,v
n
. Then (3.3.9) holds with v replaced
by and we can write
U(λ)=
∂
∂ρ
−1
and V(λ)=
∂
∂τ
−1
.
In practice one assumes a simple λ dependence in , characterized by a finite
number of poles with given multiplicities. One general scheme of solving
(3.3.10), known as the dressing method, is based on the Riemann–Hilbert
problem which we shall review next.
3.3.1 Riemann–Hilbert problem
Let λ ∈ C = C ∪{∞}and let be a closed contour in the extended complex
plane. In particular we can consider to be a real line −∞ <λ<∞ regarded
as a circle in
C passing through ∞. Let G = G(λ) be a matrix-valued function
on the contour . The Riemann–Hilbert problem is to construct two matrix-
valued functions G
+
(λ) and G
−
(λ) holomorphic, respectively, inside and out-
side the contour such that on
G(λ)=G
+
(λ)G
−
(λ). (3.3.14)
In the case when is the real axis G
+
is required to be holomorphic in the
upper half-plane and G
−
is required to be holomorphic in the lower half-plane.
If (G
+
, G
−
) is a solution of the Riemann–Hilbert problem, then
G
+
= G
+
g
−1
and
G
−
= gG
−
will also be a solution for any constant invertible matrix g. This ambiguity can
be avoided by fixing a values of G
+
or G
−
at some point in their domain, for