Complement to: Willmore immersions and Loop groups

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Complement to: Willmore immersions and Loop groups

Fr´ed´eric H´elein

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


−ω31∧ω32 ≡ Z



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 bundleFX(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) ofFt,X(γ) fortsufficiently small. We now introduce the matrix of elements λij depending smoothly on z such that if the dot denotes the derivatives with respect to t att = 0, we have


ej =eiλij. (2)

We have Bacλcb+Bbcλca= 0 and by (1)

λ001020 = 0 ; λ30 =λ. (3) We wish to compute the first variation of R

UX by this deformation.

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


ω13 = hde˙3, e1i+hde3,e˙1i

= dλ1303ω0123ω2130ω1021ω32


ω23 = hde˙3, e2i+hde3,e˙2i

= dλ2303ω0213ω1230ω2012ω31.




−˙ΩX = (dλ13∧ω3223ω12∧ω3213∧dλ2313ω13∧ω12) + λ3010∧ω2331∧ω02)

+ λ0301∧ω2331∧ω20).


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

3101∧ω0321∧ω23 = 0 dω3202∧ω0312∧ω13 = 0 we can transform the first term in the right hand side of (4) as

13∧ω23−λ23(dω1301∧ω30) +ω13∧dλ2313(dω2302∧ω03)

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

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

+ λ3001∧ω3213∧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

30 =dh1∧ω10+dh2 ∧ω02+h101+h220. 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∧ω2010∧ω23) + (p11+p2201∧ω02. (8) Relation (8) gives us an expression for ω13∧ω2001∧ω23, which we can again transform using the structure relations

1001∧ω0021∧ω20 = 0

2002∧ω0012∧ω10 = 0 (9)




ω31∧ω2001∧ω32 = d(h1ω20−h2ω10) +ω00∧(h1ω02−h2ω01)

−(p11+p2201∧ω02. (10) Third step. We inject the result of step 2 summarized in (10) in the formula for ˙ΩX given by (5). It leads to

−˙ΩX = d(λ13ω32−λ23ω31) + (λ13ω02−λ23ω01)∧ω0330d(h1ω02−h2ω10) +λ30ω00∧(h1ω20−h2ω10)


= d(λ13ω32−λ23ω31) +λ30d(h1ω20−h2ω01) +(λ30ω00−λ13ω01−λ23ω02)∧(h1ω02−h2ω10)



where we used ω30 =h1ω10+h2ω02.

Lastly we exploit a relation between the coefficients λij which we did not use before.

Since we impose the constraint ω03 = 0, for small t, we have 0 =hde˙0, e3i+hde0,e˙3i, which leads to

3030ω00−λ13ω10−λ23ω20. (12) We insert this relation in (11) to get

˙ΩX = λ30(p11+p2201∧ω02

+d[λ13ω32−λ23ω3130(h1ω20−h2ω01)]. (13) Conclusion. We obtain that



˙ΩX = Z


λ(p11+p2201∧ω02, (14) and this implies that any Willmore immersion satisfies the equation

p11+p22= 0. (15)

[1] R.Bryant,A duality theorem for Willmore surfaces, Journal of Differential Geometry 20 (1984), 23–53.

[2] F. H´elein, Willmore immersions and loop groups, Journal of Differential Geometry, Vol. 50, n. 2 (1998), 331–388.





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