Conformally Invariant Variational Problems. - SAM

where e λ = |∂ x1Φ| ⃗ = |∂x2Φ|. ⃗ The oriented Gauss map ⃗n is then
given by
⃗n = e −2λ ∂⃗ Φ
× ∂⃗ Φ
.
∂x 1 ∂x 2
We have
⎧
⎨
⎩
= − < ∇ ⊥ ⃗n,⃗e 2 >
=< ∇ ⊥ ⃗n,⃗e 1 > .
From which we deduce
⎧
⎨ −⃗n×∂ x2 ⃗n =< ∂ x2 ⃗n,⃗e 2 > ⃗e 1 − < ∂ x2 ⃗n,⃗e 1 > ⃗e 2
⎩
⃗n×∂ x1 ⃗n = − < ∂ x1 ⃗n,⃗e 2 > ⃗e 1 + < ∂ x1 ⃗n,⃗e 1 > ⃗e 2
Thus ⎧⎨
⎩
∂ x1 ⃗n−⃗n×∂ x2 ⃗n = [< ∂ x2 ⃗n,⃗e 2 > + < ∂ x1 ⃗n,⃗e 1 >] ⃗e 1
∂ x2 ⃗n+⃗n×∂ x1 ⃗n = [< ∂ x2 ⃗n,⃗e 2 > + < ∂ x1 ⃗n,⃗e 1 >] ⃗e 2
Since H = −e −λ 2 −1 [< ∂ x2 ⃗n,⃗e 2 > + < ∂ x1 ⃗n,⃗e 1 >] we deduce
(X.91) and Lemma X.2 is proved.
✷
Proof of theorem X.7 in codimension 1.
We can again assume that ⃗ Φ is conformal. First take the
divergence of (X.91) and multiply by H. This gives
−2H 2 ∆ ⃗ Φ−2H∇H ·∇ ⃗ Φ = Hdiv [ ∇⃗n+⃗n×∇ ⊥ ⃗n ] . (X.92)
Wereplace−2H∇Φin(X.92)bytheexpressiongivenby(X.91),
⃗
moreoverwealsousetheexpressionofthemeancurvaturevector
in terms of Φ ⃗ :
∆Φ ⃗ = 2e 2λ H ⃗ . (X.93)
So (X.92) becomes
−4H 2 ⃗ H e 2λ +∇H ·[∇⃗n+⃗n×∇ ⊥ ⃗n ]
= Hdiv [ ∇⃗n+⃗n×∇ ⊥ ⃗n ] .
147
(X.94)