12.07.2015 Views

Ivancevic_Applied-Diff-Geom

Ivancevic_Applied-Diff-Geom

Ivancevic_Applied-Diff-Geom

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.

248 <strong>Applied</strong> <strong>Diff</strong>erential <strong>Geom</strong>etry: A Modern Introductioni, j, k = 1, ..., r, such that for each i, j,[v i , v j ] = h k ij · v k .Let v ≠ 0 be a right–invariant vector–field on a Lie group G. Then theflow generated by v through the identity e, namelyg ε = exp(εv) e ≡ exp(εv),is defined for all ε ∈ R and forms a one–parameter subgroup of G, withg ε+δ = g ε · g δ , g 0 = e, g −1ε = g −ε ,isomorphic to either R itself or the circle group SO(2). Conversely, anyconnected 1D subgroup of G is generated by such a right–invariant vector–field in the above manner [Olver (1986)].For example, let G = GL(n) with Lie algebra gl(n), the space of alln × n matrices with commutator as the Lie bracket. If A ∈ gl(n), then thecorresponding right–invariant vector–field v A on GL(n) has the expression[Olver (1986)]v A = a i kx k j ∂ x ij.The one–parameter subgroup exp(εv A ) e is found by integrating the systemof n 2 ordinary differential equationsdx i jdε = ai kx k j , x i j(0) = δ i j, (i, j = 1, ..., n),involving matrix entries of A. The solution is just the matrix exponentialX(ε) = e εA , which is the one–parameter subgroup of GL(n) generated bya matrix A in gl(n).Recall that the exponential map exp : g → G is get by setting ε = 1 inthe one–parameter subgroup generated by vector–field v :Its differential at 0,is the identity map.exp(v) ≡ exp(v) e.d exp : T g| 0 ≃ g → T G| e ≃ g

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

Saved successfully!

Ooh no, something went wrong!