Conformally Invariant Variational Problems. - SAM

Since [ ⃗ V,⃗e s ] is tangent to ⃗ Φ t (Σ 2 ), by composing the previous
identity with the orthogonal projection π T on the tangent space
to ⃗ Φ t (Σ 2 ) we obtain
[ ⃗ V,⃗e s ] = π T (d⃗e s ·∂ t )−π T (d ⃗ V ·e s ) = ∇ ∂t ⃗e s −∇ es
⃗ V . (X.76)
Using (X.73) we have
D es (∇ ∂t ⃗e s ) = D ∇∂t ⃗e s
⃗e s
Since ∇ ∂t ⃗e s = 0 along the curve [0,1]×{p} one deduces from
the previous identity that
D es (∇ ∂t ⃗e s )(t,p) ≡ 0 .
(X.77)
Combining (X.75) with (X.76) and (X.77) we obtain
D ∂
⃗H = 1
∂t 2
2∑ [D es (dV ⃗ ]
·e s )−2D es (∇ esV) ⃗
s=1
. (X.78)
We decompose ⃗ V = ⃗ V N + ⃗ V T where ⃗ V N = π ⃗n ( ⃗ V) and ⃗ V T =
π T ( ⃗ V) and afterwrittingd ⃗ V ·e s = ∇ es
⃗ V +Des ⃗ V, (X.78)becomes
D ∂
⃗H = 1
∂t 2
Observe that
and
+ 1 2
2∑
s=1
2∑
s=1
[
]
D es (D esV ⃗ N )−D es (∇ esV ⃗ N )
[
]
D es (D esV ⃗ T )−D es (∇ esV ⃗ T )
(X.79)
2∑
D es (D esV ⃗ N ) = ∆ ⊥V ⃗ N , (X.80)
s=1
D es (∇ es
⃗ V N ) = ⃗ I(e s ,∇ es
⃗ V N ) .
(X.81)
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