Let $G \subset \mathrm{GL}(n)$ be a matrix Lie group. Does there exist an affine connection under which the matrix and manifold exponential coincide?

Question

Let $G\subset \operatorname{GL}(n)$ be a matrix Lie group. I am curious about curves $\gamma(t) = g \exp(tv)$, where $g \in G$, $v \in \mathfrak{g}$, and $\exp(.)$ is the matrix exponential. If we equip $G$ with a bi-invariant metric, then geodesics take the form $\gamma(.)$ up to a constant. Of course, many matrix Lie groups cannot be equipped with a bi-invariant metric. I was wondering if there exists an affine connection under which all geodesics look like $\gamma(.)$. In other words, the manifold and matrix exponentials coincide.

Answer 1

Yes, take the the trivial connection with respect to the left trivialisation of the tangent bundle. Then, all of your curves $\gamma$ are geodesics, but there are no further geodesics.

Some more details: the tangent bundle of every (matrix) lie group is trivial: consider a basis $e_1,..,e_n$ of you Lie algebra, and the globally well-defined frame $X_1,..,X_n$ determined by $X_k(g)=g e_k$, where the right hand side is just the matrix product (if you have a matrix Lie group). Then you can define a unique connection by declaring the frame to be parallel. Equivalently, this means that all Cristoffel symbols vanish identically. This gives you a flat connection (with non-trivial torsion unless the Lie algebra is commutative, ie the lie bracket vanishes). Then, for any $v$ in the Lie group, the vector field $g\mapsto gv$ is parallel. Hence, your curves $\gamma$ are all geodesic.

If you want a torsion free connection, just add half of the commutator as a global connection 1-form. As this is skew, you get the same geodesics, but by construction this connection has vanishing torsion.