Conformally Invariant Variational Problems. - SAM

We have moreover
∇ es
⃗ V N =
= −
2∑
< ∇ esV ⃗ N ,⃗e k > ⃗e k =
k=1
2∑
< dV ⃗ N ·e s ,⃗e k > ⃗e k
k=1
2∑
< V ⃗ N ,d⃗e k ·e s > ⃗e k = −
k=1
2∑
< V ⃗ N ,D es ⃗e k > ⃗e k
k=1
Combining this last identity with (X.81) gives
2∑
D es (∇ esV ⃗ N ) = −
s=1
2∑
s,k=1
D es ⃗e k
⃗ I(⃗es ,⃗e k ) = −Ã(⃗ V N ) . (X.82)
Now, since Φ ⃗ 0 +tV ⃗ T preserves infinitesimallythe surface Φ ⃗ 0 (Σ 2 )
an since the Willmoreenergy is independent of the parametrization,
one has
d
dt
∫
Σ 2 | ⃗ H ⃗Φ0 +t ⃗ V T | 2 dvol ( ⃗ Φ0 +t ⃗ V T ) ∗ g R m = 0 .
In other words, tangent variations do not change the Willmore
energy at the first order. Therefore we can assume V ⃗ = V ⃗ N and
combining (X.79) with (X.80) and (X.82) gives finally if V ⃗ T = 0
D ∂
⃗H = 1
∂t 2
[
∆ ⊥
⃗ V N +Ã(⃗ V N )
]
. (X.83)
Let (gij t ) ij := ( Φ+t ⃗ V) ⃗ ∗ g R m, a straightforwardcomputationgives
in local coordinates
]
gij t = g ij +t
[< ∂ xiV,∂xj ⃗ Φ ⃗ > + < ∂xjV,∂xi ⃗ Φ ⃗ > +o(t) ,
from which we deduce
det(gij t ) 2∑
det(g ij ) −1 = 2
k,j=1
g jk < ∂ xk
⃗ V,∂xj ⃗ Φ > +o(t) .
137