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Conformally Invariant Variational Problems. - SAM

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order to hope to construct energy controlled liftings. However,<br />

if one removes the requirement to control the energy, every map<br />

⃗n ∈ W 1,2 (D 2 ,Gr m−2 (R m )) admits a W 1,2 lifting. The following<br />

theorem is proved in [He] chapter 5.2.<br />

Theorem X.9. Let⃗n in W 1,2 (D 2 ,Gr m−2 (R m )), then thereexists<br />

⃗e 1 ,⃗e 2 ∈ W 1,2 (D 2 ,S m−1 ) such that<br />

⃗e 1 ∧⃗e 2 = ⋆⃗n .<br />

Wecanthenmakethefollowingobservation. Let⃗n ∈ W 1,2 (D 2 ,S m−1 )<br />

and ⃗e 1 ,⃗e 2 ∈ W 1,2 (D 2 ,S m−1 ) given by the previous theorem.<br />

Consider θ ∈ W 1,2 to be the unique solution of<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

∆θ = −div in D 2<br />

〈<br />

∂θ<br />

∂ν = − ⃗e 1 , ∂⃗e 〉<br />

2<br />

∂ν<br />

then the lifting ⃗ f = ⃗ f 1 +i ⃗ f 2 given by<br />

⃗f 1 +i ⃗ f 2 = e iθ (⃗e 1 +i⃗e 2 ) ,<br />

∫<br />

is Coulomb. Let λ such that −∇ ⊥ λ =< f ⃗ 1 ,∇f ⃗ 2 > and such that<br />

∂D<br />

λ = 0, one easily sees that λ satisfies<br />

2<br />

⎧<br />

⎨ −∆λ =< ∇ ⊥ f1 ⃗,∇f ⃗ 2 > in D 2<br />

(X.153)<br />

⎩<br />

λ = 0 on ∂D 2<br />

Observe that since ∇⃗e 1 is perpendicular to ⃗e 1 and since ∇⃗e 2 is<br />

perpendicular to ⃗e 2 one has<br />

< ∇ ⊥ ⃗ f1 ,∇ ⃗ f 2 >=< π ⃗n (∇ ⊥ ⃗ f1 ),π ⃗n (∇ ⃗ f 2 ) > (X.154)<br />

✷<br />

168

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