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).
But if [X,Y] is not zero, what is the parallel translate of Y along exp(tX)?
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). But if [X,Y] is not zero, what is the parallel translate of Y along exp(tX)? 


For a general $Y$, it will be matrixexponential in $t$ with initial conditions determined by $Y$. Here's an explicit computation. Pick a leftinvariant global frame $(E_1, \ldots E_n)$ for the group, and define structure constants $[E_i, E_j]=\sum c_{ij}{}^kE_k$. The covariant derivative of $E_i$ along the geodesic $\exp(tX)$ from 0 is the constant $\frac{1}{2}[X, E_i]=\frac{1}{2}\sum X^jc_{ji}{}^kE_k$ (see eg Lee "Riemannian Manifolds" problem 511). Therefore a vector field $t\mapsto \sum f^i(t)E_i$ along this geodesic is parallel if it is a solution to $0=D_t\left(\sum f^i(t)E_i\right)=\sum_k\left[(f^k)'(t)+\frac{1}{2}\sum_{i,j} f^i(t)X^jc_{ji}{}^k\right]E_k$, i.e., to $\forall k \ 0=(f^k)'(t)+\frac{1}{2}\sum_{i,j} f^i(t)X^jc_{ji}{}^k$. The parallel transport of $Y$ will be the solution to this linear system with initial value $Y$. Perhaps there's a nice basisinvariant way of expressing this? I can't think offhand. 

