Conformally Invariant Variational Problems. - SAM

Another difficulty lies in the remaining degree of freedom
given by the invariance group of E on C Γ , the so called gauge
group of our problem, which is here the Möbius group M + (D 2 )
which is three dimensional as we saw in section 1. The problem
with this gauge group is non compact : by taking for instance a
sequence a n ∈ D 2 and a n → (1,0) the sequence of maps
ψ n : z −→ z −a n
1−a n z
converges weakly to a constant map which is not in M + (D 2 )
anymore. Assuming we would have a sequence of minimizer u n
converging to a solution of (V.3) and satisfying the conclusions
of theorem V.2, by composing u n with ψ n we still have a minimizing
sequence since E(u n ) = E(u n ◦ψ n ). this new minimizing
sequence of E in C(Γ), u n ◦ ψ n , converges then to a constant
which cannot be a solution to the Plateau problem.
Thus all minimizing sequences cannot lead to a solution du
in particular to the existence of a non compact gauge group
M + (D 2 ). This group however is very small (in comparison with
Diff + (D 2 ) in particular).
In order to break this gauge invariance it suffices to fix the
images of 3 distinct points on the boundary. This is the three
point normalization method. Let P 1 , P 2 and P 3 in ∂D 2 and
threepointsinΓ: Q 1 , Q 2 andQ 3 inthesameorder(withrespect
to the monotony given by definition V.2) and we introduce the
following subspace of C Γ
C ∗ (Γ) := {u ∈ C(Γ) s.t . ∀k = 1,2,3 u(P k ) = Q k } (V.25)
The following elementary lemma, whose proof is left to the
reader, garanties that
inf E(u) = inf E(u) .
u∈C(Γ) u∈C ∗ (Γ)
(V.26)
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