Conformally Invariant Variational Problems. - SAM

We have using (X.113)
∂ z ∂ z
⃗ L = ∂z f e −λ ⃗e z +2 −1 f ⃗ H 0
−2i∂ z
[〈
⃗H, ⃗ H0
〉
∂ z
⃗ Φ+π⃗n (∂ z
⃗ H)
]
.
(X.256)
The fact that L ⃗ is real valued implies that
0 = I ( )
∂ z f e −λ ⃗e z +2 −1 I
(f H ⃗ )
0
[〈 〉 ])
−2 R
(∂ z ⃗H, H0 ⃗ ∂ zΦ+π⃗n ⃗ (∂ zH) ⃗
Using identity (X.103), the previous equality is equivalent to
0 = I ( )
2∂ z f e −3λ ⃗e z +e −2λ I
(f H ⃗ )
0
(X.257)
−∆ ⊥H ⃗ − Ã( H)+2| ⃗ H| ⃗ 2 H ⃗
Decomposing this identity into the normal and tangential parts
gives that (X.257) is equivalent to
⎧
⎨ ∆ ⊥H ⃗ + Ã( H)−2| ⃗ H| ⃗ 2 H ⃗ = e −2λ I
(f ⃗ )
H 0
(X.258)
⎩
I(∂ z f ⃗e z ) = 0 .
The second line is equivalent to
∂ z f ⃗e z −∂ z f⃗e z = 0
Taking the scalar product respectively with ⃗e z and ⃗e z , using
(X.99), gives that (X.258) is equivalent to
⎧
⎨ ∆ ⊥H ⃗ + Ã( H)−2| ⃗ H| ⃗ 2 H ⃗ = e −2λ I
(f ⃗ )
H 0
(X.259)
⎩
∂ z f = 0 .
We have then proved that (X.239) implies (X.240).
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