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 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

$[E_i, E_j]=\sum c_{ij}{}^kE_k$.

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 5-11). 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 basis-invariant way of expressing this? I can't think offhand.

  • $\begingroup$ $DL_{exp tX} \circ exp(-\frac{t}{2}\text{ad}X)Y$ $\endgroup$ – Adterram Mar 29 '17 at 21:17

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