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Conformally Invariant Variational Problems. - SAM

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Let k ∈ {1,...,n} and choose i ≠ k, using (II.16)<br />

∂f<br />

= ∂ ( ∂ 2 e −λ )<br />

= 0<br />

∂x k ∂x i ∂x i ∂x k<br />

f is therefore constant on the domain U and we finally easily<br />

deduce the existence of two real constants A and B and a vector<br />

x 0 ∈ R n such that<br />

e −λ = A | x−x 0 | 2 +B<br />

which concludes the proof of the Lemma II.3<br />

✷<br />

Since U is assumed to be a diffeo-<br />

Proof of Theorem II.1<br />

morphism we have<br />

taking the differential gives<br />

∀y ∈ u(U) u◦u −1 (y) = y<br />

∀y ∈ u(U) du u<br />

−1<br />

(y) ◦du −1<br />

y = I n<br />

Where I n is the identity matrix on R n<br />

For any pair of vectors Y and Y ′ in R n , using Lemma II.3, we<br />

then have denoting x = u −1 (y)<br />

< Y,Y ′ > =< du x ·(du −1<br />

y ·Y),du x·(du −1<br />

y Y ′ ) ><br />

=<br />

1<br />

(A | x−x 0 | 2 +B) 2 < du−1 y ·Y,du −1<br />

y Y ′ > (II.18)<br />

Since u is conformal, u −1 is also conformal and there exists<br />

C,D ∈ R and y 0 ∈ R n s.t.<br />

∀y ∈ u(U) < du −1<br />

y Y,du−1 y ′ Y ′ >=<br />

11<br />

1<br />

(C | y −y 0 | 2 +D) 2 < Y,Y ′ ><br />

(II.19)

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