Parallel translation in Lie groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:41:30Z http://mathoverflow.net/feeds/question/22983 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22983/parallel-translation-in-lie-groups Parallel translation in Lie groups Alex 2010-04-29T14:04:06Z 2010-04-29T23:07:13Z <p>Let G be a Lie group with a left invariant metric. If X and Y are left invariant vector fields and [X,Y]=0, then it is easy to show that Y is parallel to exp(tX). </p> <p>But if [X,Y] is not zero, what is the parallel translate of Y along exp(tX)?</p> http://mathoverflow.net/questions/22983/parallel-translation-in-lie-groups/23053#23053 Answer by macbeth for Parallel translation in Lie groups macbeth 2010-04-29T23:07:13Z 2010-04-29T23:07:13Z <p>For a general $Y$, it will be matrix-exponential in $t$ with initial conditions determined by $Y$. Here's an explicit computation. Pick a left-invariant global frame $(E_1, \ldots E_n)$ for the group, and define structure constants</p> <p>$[E_i, E_j]=\sum c_{ij}{}^kE_k$.</p> <p>The covariant derivative of $E_i$ along the geodesic $\exp(tX)$ from 0 is the constant</p> <p>$\frac{1}{2}[X, E_i]=\frac{1}{2}\sum X^jc_{ji}{}^kE_k$</p> <p>(see eg Lee "Riemannian Manifolds" problem 5-11). Therefore a vector field</p> <p>$t\mapsto \sum f^i(t)E_i$</p> <p>along this geodesic is parallel if it is a solution to</p> <p>$0=D_t\left(\sum f^i(t)E_i\right)=\sum_k\left[(f^k)'(t)+\frac{1}{2}\sum_{i,j} f^i(t)X^jc_{ji}{}^k\right]E_k$,</p> <p>i.e., to $\forall k \ 0=(f^k)'(t)+\frac{1}{2}\sum_{i,j} f^i(t)X^jc_{ji}{}^k$.</p> <p>The parallel transport of $Y$ will be the solution to this linear system with initial value $Y$. Perhaps there's a nice basis-invariant way of expressing this? I can't think offhand.</p>