Complement to: Willmore immersions and Loop groups
The Euler–Lagrange equation of the Willmore functional
The derivation of the Euler–Lagrange equation of the Willmore functional was done in .
However I found this proof difficult to follow and I present here an alternative redaction.
Proof — Using the notations of  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)
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
ω23 = hde˙3, e2i+hde3,e˙2i
−˙ΩX = (dλ13∧ω32+λ23ω12∧ω32+ω13∧dλ23+λ13ω13∧ω12) + λ30(ω10∧ω23+ω31∧ω02)
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
=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
+(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
= 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
dω10+ω01∧ω00+ω21∧ω20 = 0
dω20+ω02∧ω00+ω12∧ω10 = 0 (9)
ω31∧ω20+ω01∧ω32 = d(h1ω20−h2ω10) +ω00∧(h1ω02−h2ω01)
−(p11+p22)ω01∧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)∧ω03 +λ30d(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
dλ30 =λ30ω00−λ13ω10−λ23ω20. (12) We insert this relation in (11) to get
˙ΩX = λ30(p11+p22)ω01∧ω02
+d[λ13ω32−λ23ω31+λ30(h1ω20−h2ω01)]. (13) Conclusion. We obtain that
˙ΩX = Z
λ(p11+p22)ω01∧ω02, (14) and this implies that any Willmore immersion satisfies the equation
p11+p22= 0. (15)
 R.Bryant,A duality theorem for Willmore surfaces, Journal of Differential Geometry 20 (1984), 23–53.
 F. H´elein, Willmore immersions and loop groups, Journal of Differential Geometry, Vol. 50, n. 2 (1998), 331–388.