Complement to: Willmore immersions and Loop groups

(1)

Complement to: Willmore immersions and Loop groups

Frédéric Hélein

The Euler–Lagrange equation of the Willmore functional

The derivation of the Euler–Lagrange equation of the Willmore functional was done in [1].

However I found this proof difficult to follow and I present here an alternative redaction.

Proof — Using the notations of [2] we know that we may write the Willmore functional as

W(X) = Z

U

−ω₃¹∧ω₃² ≡ Z

U

ΩX,

which is precisely minus the area covered byγ. We start from an immersionX:U −→S³, not necessarily conformal, and we consider some section e of the associated bundleF_X^(γ). We perform a variation of X induced by

e₀(t) =e₀ +tλe₃, (1)

where λ : U −→ R is some smooth map with compact support in U. We can construct for each t sufficiently small the associated conformal Gauss map γ(t) = e₃(t), and the orthonormal frame (e₁(t), e₂(t)) for instance by the Gram-Schmitt orthonormalisation of (^∂e_∂x⁰(t),^∂e_∂y⁰(t)). By completing by an adequate vectore₄(t) in C₊, we construct a section e(t) ofF_t,X^(γ) fortsufficiently small. We now introduce the matrix of elements λⁱ_j depending smoothly on z such that if the dot denotes the derivatives with respect to t att = 0, we have

˙

ej =eiλⁱ_j. (2)

We have B_acλ^c_b+B_bcλ^c_a= 0 and by (1)

λ⁰₀ =λ¹₀ =λ²₀ = 0 ; λ³₀ =λ. (3) We wish to compute the first variation of R

U ΩX by this deformation.

First step. Brutal computation of ˙ΩX: Using (2) we get that

˙

ω¹₃ = hde˙₃, e₁i+hde₃,e˙₁i

= dλ¹₃+λ⁰₃ω₀¹+λ²₃ω₂¹+λ³₀ω₁⁰+λ²₁ω₃²

˙

ω²₃ = hde˙₃, e₂i+hde₃,e˙₂i

= dλ²₃+λ⁰₃ω₀²+λ¹₃ω₁²+λ³₀ω₂⁰+λ¹₂ω₃¹.

1

(2)

Thus

−˙ΩX = (dλ¹₃∧ω₃²+λ²₃ω¹₂∧ω₃²+ω¹₃∧dλ²₃+λ¹₃ω¹₃∧ω₁²) + λ³₀(ω₁⁰∧ω²₃+ω₃¹∧ω⁰₂)

+ λ⁰₃(ω₀¹∧ω²₃+ω₃¹∧ω²₀).

(4)

We then use (??) and the fact that h₁₁+h₂₂ = 0 to cancel the last term in the right hand side of (4). Moreover using the structure equations

dω₃¹+ω₀¹∧ω⁰₃+ω₂¹∧ω²₃ = 0 dω₃²+ω₀²∧ω⁰₃+ω₁²∧ω¹₃ = 0 we can transform the first term in the right hand side of (4) as

dλ¹₃∧ω²₃−λ²₃(dω¹₃+ω₀¹∧ω₃⁰) +ω¹₃∧dλ²₃+λ¹₃(dω²₃+ω₀²∧ω⁰₃)

=d(λ¹₃ω²₃−λ²₃ω¹₃) + (λ¹₃ω₀²−λ²₃ω₀¹)∧ω₃⁰. Thus we obtain that

−˙ΩX = d(λ¹₃ω₃²−λ²₃ω₃¹) + (λ¹₃ω₀²−λ²₃ω₀¹)∧ω₃⁰

+ λ³₀(ω⁰₁∧ω₃²+ω¹₃∧ω₂⁰). (5) Second step. We compute the last term in the right hand side of (5). We set ω⁰₃ = h₁ω¹₀+h₂ω₀², and we differentiate this expression. This gives

dω₃⁰ =dh₁∧ω¹₀+dh₂ ∧ω₀²+h₁dω₀¹+h₂dω²₀. We develop this relation using the structure equations and (??) as

(dh1+ 2ω₀⁰h₁−ω₁²h₂−h₁₁ω₁⁰−h₁₂ω₂⁰)∧ω₀²

+(dh₂+ 2ω₀⁰h₂−ω₂¹h₁−h₂₁ω₁⁰−h₂₂ω₂⁰)∧ω₀¹ = 0. (6) And using Cartan lemma this implies that there exist smooth functionsp₁₁, p₁₂ =p₂₁ and p₂₂ such that

dh₁+ 2ω₀⁰h₁ = ω₁²h₂+h₁₁ω₁⁰+h₁₂ω₂⁰+p₁₁ω₀¹+p₁₂ω₀² (a)

dh₂+ 2ω₀⁰h₂ = ω₂¹h₁+h₂₁ω₁⁰+h₂₂ω₂⁰+p₂₁ω₀¹+p₂₂ω₀² (b) (7) And a computation of (a)∧ω₀²−(b)∧ω¹₀ gives

dh₁∧ω₀²−dh₂∧ω₀¹+ 2ω₀⁰∧(h₁ω₀²−h₂ω¹₀)

= ω₁²∧(h₂ω²₀+h₁ω₀¹)−(ω₁³∧ω₂⁰+ω₁⁰∧ω₂³) + (p₁₁+p₂₂)ω₀¹∧ω₀². (8) Relation (8) gives us an expression for ω₁³∧ω₂⁰+ω⁰₁∧ω₂³, which we can again transform using the structure relations