Conformally Invariant Variational Problems. - SAM

Denote
∂ z
⃗ L = A⃗ez +B⃗e z + ⃗ V
where A and B are complex number and ⃗ V := π ⃗n (∂ z
⃗ L) is a
complex valued normal vector to the immersed surface. The
first equation of (X.243), using (X.99), is equivalent to
IA = 0
(X.244)
Observe that if we write
∂ z
⃗ H = C ⃗ez +D ⃗e z + ⃗ W
where W ⃗ = π ⃗n (∂ zH), ⃗ one has, using (X.114) and the fact that
⃗H is orthogonal to ⃗e z
〉 〈
C = 2
〈⃗e z ,∂ zH ⃗ = −2 ∂ z (e λ ⃗e z ), H ⃗ 〉
e −λ = −e λ | H| ⃗ 2 .
Hence we deduce in particular
(X.245)
IC = 0 .
(X.246)
We have moreover using (X.113)
〉 〈
D = 2
〈⃗e z ,∂ zH ⃗ = −2 ∂ z (e −λ ⃗e z ), H ⃗ 〉
e λ = −e λ < H ⃗ 0 , H ⃗ > .
Thus combining (X.245) and (X.247) we obtain
(X.247)
∂ z
⃗ H = −| ⃗ H| 2 ∂ z
⃗ Φ− < ⃗ H0 , ⃗ H > ∂ z
⃗ Φ+π⃗n (∂ z
⃗ H)
(X.248)
The second line in the conservation law (X.243) is equivalent to
⎧
⎪⎨ I(iA−2C) = 0
( [ ]) (X.249)
⎪⎩ I ⃗e z ∧ ⃗V +2iW ⃗ = 0
[ ] [ ]
We observe that ⃗e 1 ∧ ⃗V +2iW ⃗ and ⃗e 2 ∧ ⃗V +2iW ⃗ are linearly
independent since ⃗V +2iW ⃗ is orthogonalto the
[ ]
tangent
201