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. I've been reading about the Cartan connection, yet I am having trouble understanding it since sources say it is not even an affine connection.

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.

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 Lie group. 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. I've been reading about the Cartan connection, yet I am having trouble understanding it since sources say it is not even an affine connection.

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. I've been reading about the Cartan connection, yet I am having trouble understanding it since sources say it is not even an affine connection.

Let $G$ 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 Lie group. 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. I've been reading about the Cartan connection, yet I am having trouble understanding it since sources say it is not even an affine connection.

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 Lie group. 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. I've been reading about the Cartan connection, yet I am having trouble understanding it since sources say it is not even an affine connection.