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Conformally Invariant Variational Problems. - SAM

Conformally Invariant Variational Problems. - SAM

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Example 3. We consider a map (ω ij ) i,j∈Nm in C 1 (R m ,so(m)),<br />

where so(m) is the space antisymmetricsquare m×m matrices.<br />

We impose the following uniform bound<br />

‖∇ω‖ L∞ (D 2 ) < +∞ .<br />

For maps u ∈ W 1,2 (D 2 ,R m ), we introduce the Lagrangian<br />

E ω (u) = 1 ∫ m∑<br />

|∇u| 2 + ω ij (u)∂ x u i ∂ y u j −∂ y u i ∂ x u j dxdy<br />

2 D 2 i,j=1<br />

(VI.14)<br />

The conformalinvariance of this Lagrangianarises from the fact<br />

that E ω is made of the conformally invariant Lagrangian E to<br />

whichisaddedtheintegraloverD 2 ofthe2-formω = ω i jdz i ∧dz j<br />

pulled back by u. Composing u by an arbitrary positive diffeomorphism<br />

of D 2 will not affect this integral, thereby making E ω<br />

into a conformally invariant Lagrangian.<br />

The Euler-Lagrange equation deriving from (VI.14) for variations<br />

of the form u+tξ, where ξ is an arbitrary smooth function<br />

with compact support in D 2 , is found to be<br />

∆u i −2<br />

m∑<br />

Hkl(u) i ∇ ⊥ u k·∇u l = 0<br />

k,l=1<br />

∀ i = 1,...,m. (VI.15)<br />

Here, ∇ ⊥ u l = (−∂ y u k ,∂ x u k ) 10 while Hkl i is antisymmetricin the<br />

indices k et l. It is the coefficient of the R m -valued two-form H<br />

on R m m∑<br />

H i (z) := Hkl i (z) dzk ∧dz l .<br />

k,l=1<br />

10 in our notation, ∇ ⊥ u k ·∇u l is the Jacobian<br />

∇ ⊥ u k ·∇u l = ∂ x u k ∂ y u l −∂ y u k ∂ x u l .<br />

49

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