Conformally Invariant Variational Problems. - SAM

If neither A = 0 nor B = 0
The left-hand side defines a transcendental function of τ for a
fixed s whereas the right-hand side is a rational fraction. This
is a contradiction therefore either A = 0 or B = 0
First Case A=0
In this case we have
∀i,j = 1...n < ∂u
∂x i
, ∂u
∂x j
>= 1 B δ ij
Using (II.13) we deduce that for any i,k ∈ {1...n}
∂ 2 u
∂x i ∂x k
= 0
Hence U is affine and conformal therefore there exists
S ∈ R + O(n) and y 0 ∈ u(U) s.t. u(x) = S(x−x 0 )+y 0
This proves the theorem in this case.
Second case: B=0
Introduce for y ∈ R n v(y) := y −x 0
| y −x 0 | 2 +x 0
V is conformal therefore u◦v is conformal too
∀Y,Y ′ ∈ R n ∀y ∈ v −1 (U) we have
< d(u◦v) y ·Y,d(u◦v) y Y ′ >
1
= < dv
A | v(y)−x 0 | 2 ) 2 y Y,dv y Y ′ >
1 1
=
A 2 | v(y)−x 0 | 4 | y −x 0 | = 1 4 A 2
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