Let $M$ be a Riemannian manifold with the Levi-Civita connection. For $x \in M$ and $X \in T_x(M)$ we can consider geodesic $\gamma$ with the properties that $\gamma(0)=x$ and $\gamma'(0)=X$. Such geodesic exists and is unique: existence is guaranteed provided we take vectors $X$ in some suitable neighborhood of $0$ in $T_x(M)$. If we define $y:=\gamma(1)$ we arrive at the definition of exponential map $\exp_x:T_xM \to M$ (defined only on some small neighborhood of $0$ in $T_xM$).
From the other hand, we can consider $GL(n,\mathbb{R})$ as an open set in $\mathbb{R}^{n^2}$: therefore it becomes manifold on its own right. I would like to understand why in this situation $\exp_{I}(A)=e^{A}$ where on the left hand side is the exponential map from riemannian manifold at identity matrix and the right hand side is just the exponential of the matrix. Here we identify $T_I(Gl(n))$ with $\mathbb{R}^{n^2}$.

  • $\begingroup$ A curve $\gamma(t) = \exp{(At)}$ has properties: 1) $\gamma(0) = I$, 2) $\gamma'(t) = A \exp{(At)}$ and hence $\gamma'(0) = A$. But I struggle to show for which Levi-Civita connection $\exp{(At)}$ is also a geodesic (because I never really had a differential geometry as a subject). $\endgroup$
    – Evgeny
    Mar 20, 2017 at 20:31

1 Answer 1


You haven't specified which Riemannian metric you want to consider on $\operatorname{GL}_n(\mathbb{R})$. I don't know if you can put a Riemannian metric on $\operatorname{GL}_n(\mathbb{R})$ such that the corresponding geodesics are given by $\exp(tA)$ but you can show the following:

  1. There exists natural connection $\nabla$ on $\operatorname{GL}_n(\mathbb{R})$ whose associated parallel transport is given by left translations. This connection has torsion so it is not the Levi-Civita connection of a Riemannian metric on $\operatorname{GL}_n(\mathbb{R})$. Still, one can consider the geodesics of $\nabla$ and the geodesics starting at the identity are given by $t \mapsto \exp(tA)$. The same holds for the natural connection whose associated parallel transport is given by right translations.
  2. There exists a natural torsion-free connection $\nabla$ on $\operatorname{GL}_n(\mathbb{R})$ called sometimes the Cartan connection whose geodesics starting at the identity are given by $t \mapsto \exp(tA)$. It is not metric (for any metric) so it is not a Levi-Civita connection. For details, see here and here.
  3. If $G \subseteq \operatorname{GL}_n(\mathbb{R})$ is a compact Lie subgroup and you endow it with any bi-invariant Riemannian metric $g$ then $\exp^g_I(tA) = \exp(tA)$ for all $A \in T_p(G) \subseteq M_n(\mathbb{R})$. Here, the left hand side is the exponential map of Riemannian geometry and the right hand side is the exponential of matrices.

The problem with generalizing $(3)$ from $G$ to the whole of $\operatorname{GL}_n(\mathbb{R})$ is that $\operatorname{GL}_n(\mathbb{R})$ has no bi-invariant metric.

  • $\begingroup$ So, geodesic exponential is just inspired by Lie group exponential (or vice versa?), not a particular example of it? $\endgroup$
    – Evgeny
    Mar 21, 2017 at 10:48
  • $\begingroup$ @Evgeny: Well, you don't really need a metric to define geodesics and an exponential map, only a connection on the tangent bundle. In this sense, the exponential map of the Levi-Civita connection yields the Riemannian geodesic exponential map and the exponential map of a different connection yields the exponential map of Lie groups. $\endgroup$
    – levap
    Mar 21, 2017 at 11:31

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