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Conformally Invariant Variational Problems. - SAM

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Observe that ⎧⎪ ⎨<br />

⎪ ⎩<br />

〈⃗e z ,⃗e z 〉 = 0<br />

〈⃗e z ,⃗e z 〉 = 1 2<br />

(X.99)<br />

⃗e z ∧⃗e z = i 2 ⃗e 1 ∧⃗e 2<br />

Introduce moreover the Weingarten Operator expressed in our<br />

conformal coordinates (x 1 ,x 2 ) :<br />

⃗H 0 := 1 [<br />

⃗I(e1 ,e 1 )−<br />

2<br />

⃗ I(e 2 ,e 2 )−2i ⃗ I(e 1 ,e 2 )]<br />

.<br />

With these notations the following lemma holds<br />

Lemma X.3. Let Φ ⃗ be a conformal immersion of D 2 into R m<br />

π T (∂ zH)−i ⃗ ⋆(∂z ⃗n∧ H) ⃗ 〈 〉<br />

= −2 ⃗H, H0 ⃗ ∂ zΦ<br />

⃗ (X.100)<br />

and hence<br />

∂ zH ⃗ −3π⃗n (∂ zH)−i ⃗ ⋆(∂z ⃗n∧ H) ⃗<br />

〈 〉<br />

= −2 ⃗H, H0 ⃗ ∂ zΦ−2π⃗n ⃗ (∂ zH)<br />

⃗<br />

(X.101)<br />

Remark X.2. Observe that with these complex notations the<br />

identity (X.85), which reads in conformal coordinates<br />

[<br />

− e−2λ<br />

2 div ∇H ⃗ −3π ⃗n (∇H)+⋆(∇ ⃗ ⊥ ⃗n∧ H) ⃗ ]<br />

(X.102)<br />

= ∆ ⊥H ⃗ + Ã( H)−2| ⃗ H| ⃗ 2 H ⃗ ,<br />

becomes<br />

〈 〉<br />

4e −2λ R<br />

(∂ z<br />

[π ⃗n (∂ zH)+ ⃗ ⃗H, H0 ⃗<br />

∂ z<br />

⃗ Φ<br />

])<br />

= ∆ ⊥<br />

⃗ H + Ã( ⃗ H)−2 | ⃗ H| 2 ⃗ H ,<br />

(X.103)<br />

149<br />

✷<br />

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