$\newcommand{\Cof}{\operatorname{Cof}}$
Do we know how to calculate geodesics on $\text{SL}_n$ w.r.t the Euclidean metric?
More explicitly, I consider $\text{SL}_n$ with the Riemannian metric induced by the inclusion $i:\text{SL}_n \to (\mathbb{R}^{n^2},e)$. $e$ is the Euclidean metric; We take the pullback metric $i^*e$.
It seems to me straight segments between two different matrices never have constant determinant.
Edit:
This is wrong! Examples are given in Andrew's comment. An interesting question is to characterize when a straight segment between a pair of matrices is a geodesic.
Here is a partial result:
Define $\alpha(t)=A+tB$. Then $\det \alpha(t)=\text{const}$ is equivalent to $$ 0=\frac{d \det \alpha(t)}{dt}=\langle \Cof (\alpha(t)), B \rangle.\tag{1}$$
Where $\Cof A$ is the cofactor matrix of $A$. In the $2D$-case, $\Cof \begin{pmatrix} a & b \\\ c & d \end{pmatrix} =\begin{pmatrix} d & -c \\\ -b & a \end{pmatrix}$, so if we take $A= \begin{pmatrix} a & b \\\ c & d \end{pmatrix} ,B= \begin{pmatrix} \tilde a & \tilde b \\\ \tilde c & \tilde d \end{pmatrix} $ we see that $(1)$ is equivalent to
$$ 0=\langle \Cof \begin{pmatrix} a+t\tilde a & b+t\tilde b \\\ c+t\tilde c & d+t\tilde d \end{pmatrix} , \begin{pmatrix} \tilde a & \tilde b \\\ \tilde c & \tilde d \end{pmatrix} \rangle=\langle \begin{pmatrix} d+t\tilde d & -(c+t\tilde c) \\\ -(b+t\tilde b) & a+t\tilde a \end{pmatrix} , \begin{pmatrix} \tilde a & \tilde b \\\ \tilde c & \tilde d \end{pmatrix} \rangle=\tag{2}$$
$$ (\tilde ad+a\tilde d)-(c\tilde b +b\tilde c)+2t(\tilde a \tilde d -\tilde b \tilde c)=0$$
So we must have $$ \tilde a \tilde d -\tilde b \tilde c=\det B=0, \tag{3}$$
and also $$\tilde ad+a\tilde d=c\tilde b +b\tilde c. \tag{4}$$
Condition $(3)$ means that the perturbation (or direction if the segment) must be singular. I do not know how to interpret the second condition. (If someone has a nice interpretation I would be happy to see it).