Conformally Invariant Variational Problems. - SAM

Assuming now that (X.240) holds we can then go backwards
in the equivalences in order to obtain
]]
I
[∂ z
[B⃗e z −2iπ ⃗n (∂ zH) ⃗ = 0 (X.260)
where B is given by (X.253). Observe now that
I[∂ z α] = 0 ⇐⇒ ∂ x2 α 1 +∂ x1 α 2 = 0
where α = α 1 +iα 2 and α i ∈ R. Hence
I[∂ z α] = 0 ⇐⇒ ∃a ∈ R s.t. α = ∂ z a .
Thus (X.260) is equivalent to the existence of a map ⃗ L from D 2
into R m such that
∂ z
⃗ L = B⃗ez −2iπ ⃗n (∂ z
⃗ H) .
This is exactly (X.252) for which we have proved that this is
equivalent to (X.239). This finishes the proof of theorem X.18.
✷
Observe that we have just established the following lemma
which gives some new useful formula.
Lemma X.7. Let ⃗ Φ be a conformal immersion of the disc D 2 .
⃗Φ is conformal Willmore on D 2 if and only there exists a
smooth map ⃗ L from D 2 into R m and an holomorphic function
f(z) such that
∂ z ( L−2i ⃗ H) ⃗ = 2i | H| ⃗ 2 ∂ zΦ+[e ⃗ −2λ f(z)−4i < H, ⃗ H ⃗ 0 >] ∂ zΦ ⃗ .
(X.261)
In particular the following system holds
⎧
⎨ < ∂ zΦ,∂z ⃗ ( L−2i ⃗ H) ⃗ >= i | H| ⃗ 2 e 2λ
⎩
< ∂ z
⃗ Φ,∂z ( ⃗ L−2i ⃗ H) >= 2 −1 f(z)−2i e 2λ < ⃗ H, ⃗ H 0 >
(X.262)
✷
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