21.06.2014 Views

Conformally Invariant Variational Problems. - SAM

Conformally Invariant Variational Problems. - SAM

Conformally Invariant Variational Problems. - SAM

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

imply<br />

D ∂<br />

⃗H = 1<br />

∂t 2<br />

2∑<br />

D es (∂ t ⃗e s )+ 1 2<br />

s=1<br />

We denote V := ∂ t<br />

⃗ Φ. Observe that<br />

2∑<br />

D [∂t ,e s ]⃗e s .<br />

s=1<br />

(X.71)<br />

∂ t ⃗e s = ∂ t (d ⃗ Φ·e s ) = d ⃗ V ·e s +d ⃗ Φ t ·∂ t e s<br />

= d ⃗ V ·e s +d ⃗ Φ·[∂ t ,e s ] = d ⃗ V ·e s +[ ⃗ V,⃗e s ] .<br />

(X.72)<br />

Observe moreover that for two tangent vector-fields X and Y<br />

on Σ 2 one has 51 D X (dΦ·Y) ⃗ = D Y (dΦ·X) ⃗ . (X.73)<br />

This last identity implies in particular that<br />

D es ([ ⃗ V,⃗e s ]) = D [∂t ,e s ]⃗e s .<br />

(X.74)<br />

Combining (X.71), (X.72) and (X.74) gives<br />

We have<br />

D ∂<br />

⃗H = 1<br />

∂t 2<br />

2∑<br />

D es (dV ⃗ ·e s )+<br />

s=1<br />

2∑<br />

D es [ V,⃗e ⃗ s ] .<br />

s=1<br />

(X.75)<br />

[ ⃗ V,⃗e s ] = d ⃗ Φ·[∂ t ,e s ] = d⃗e s ·∂ t −d ⃗ V ·e s<br />

51 Indeed in local coordinates (x 1 ,x 2 ) in Σ 2 one has<br />

D X (d ⃗ Φ·Y) =<br />

2∑<br />

k,j=1<br />

π ⃗n (∂ xk (∂ xj<br />

⃗ ΦYj )X k )<br />

= ∑ 2<br />

j,k=1 π ⃗n(∂ xj (∂ xk<br />

⃗ ΦXk )Y j )− ∑ 2<br />

j,k=1 π ⃗n(∂ xk<br />

⃗ ΦYj ∂ xj X k ) = D Y (d ⃗ Φ·X) ,<br />

where we have used the fact for all j,k = 1,2<br />

⃗n(∂ xk<br />

⃗ ΦYj ∂ xj X k ) = 0 and ⃗n(∂ xj<br />

⃗ ΦXk ∂ xk Y j ) .<br />

135

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!