Conformally Invariant Variational Problems. - SAM

The tangential projection gives
4e −2λ π T
(
∂ z
[
π ⃗n (∂ z
⃗ H)
])
= 8e −2λ 〈
∂ z (π ⃗n (∂ z
⃗ H)),⃗ez
〉
+8e −2λ 〈 ∂ z (π ⃗n (∂ z
⃗ H)),⃗ez
〉
⃗e z
⃗e z
(X.120)
Using the fact that⃗e z and⃗e z are orthogonalto the normal plane
we have in one hand using (X.112)
〈∂ z (π ⃗n (∂ z
⃗ H)),⃗ez
〉
= −e −λ 〈π ⃗n (∂ z
⃗ H),∂z
[
e λ ⃗e z
] 〉
= − eλ
2
〈∂ z
⃗ H, ⃗ H
〉 (X.121)
and in the other hand using (X.113)
〈∂ z (π ⃗n (∂ z
⃗ H)),⃗ez
〉
= −e λ 〈 π ⃗n (∂ z
⃗ H),∂z
[
e −λ ⃗e z
] 〉
= − eλ
2
〈∂ z
⃗ H, ⃗ H0
〉 (X.122)
Combining (X.120), (X.121) and (X.122) we obtain
( [ ])
4e −2λ π T ∂ z π ⃗n (∂ zH) ⃗
= −4 e −2λ [〈
∂ z
⃗ H, ⃗ H
〉
∂ z
⃗ Φ+
〈
∂ z
⃗ H, ⃗ H0
〉
∂ z
⃗ Φ
]
(X.123)
Putting (X.119) and (X.123) together we obtain
])
4e −2λ R
(∂ z
[π ⃗n (∂ zH) ⃗
〉 〉] ] (X.124)
= ∆ ⊥H ⃗ −4e −2λ R
[[〈∂ zH, ⃗ H ⃗ +
〈∂ zH, ⃗ H0 ⃗ ∂ zΦ ⃗
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