$\newcommand{\ep}{\epsilon}$
Let $GL_n^+$ be the Lie group of invertible $n \times n$ matrices with positive determinant. In particular it's a connected open submanifold of the Euclidean space $\mathbb{R}^{n^2}$.
Now consider it with the induced metric from $\mathbb{R}^{n^2}$. (So it's a Riemannian submanifold of $\mathbb{R}^{n^2}$).
For clarity, this means endowing $GL_n^+ $ with the pullback metric of the Euclidean metric $e$ along the inclusion $ i:GL_n^+ \to (\mathbb{R}^{n^2},e) $. Explicitly: $g_z(X,Y) =tr(X^TY)$ , where $X,Y \in T_z(GL_n^+)$.
Questions:
(1) Is there an explicit formula for the Riemannian distance between two matrices $A,B \in GL_n^+$?
Conjecture: The Riemannian distance equals the Euclidean one.
Attempted proof:
Since $GL_n^+$ is open in $\mathbb{R}^{n^2}$, it follows that it's a totally geodesic submanifold. (That is, all it's geodesics are geodesics in $(\mathbb{R}^{n^2},e)$, i.e they are the usual straight lines in Euclidean space).
$GL^+(\mathbb{R}^n)$ is open $\Rightarrow$ for any $A \in GL^+(\mathbb{R}^n)$, there is an Euclidean ball centered around it which is contained in $GL_n^+$. Hence, for all matrices close enough to $A$ their distance from $A$ is just the Euclidean one. (since the straight line beteen them is in our submanifold).
Now consider $A,B \in GL_n^+$. Let $\alpha:[0,1] \to GL_n^+$ be the straight line path between them. Then:
$$ \det(\alpha(t))=\det(A+t(B-A)) $$ is a polynomial in $t$ of degree $\le n$ . Hence, it has only finitely many zeroes. This implies there are no more than $n$ points $t_i$ where $\alpha(t_i)$ is not invertible.
Hence, we only need to show we can make arbitrary small perturbations around each such 'bad' non-invertible matrix. This would imply the Riemannian distance equals the Euclidean one.
It would be nice if someone could find a neat argument to show this maneuver is indeed possible.
Update: This conjecture is false. The key point (as noted by Jason DeVito and loup blanc) is that the determinant is negative for non-negligible parts of the straight-line path $\alpha$. Now, by continuity argument, any path which approximates too closely the straight path must enter a region of negative determinant.
It turns out that the behaviour depends on the number of sign changes of the determinant.
Example for a case where the distance is Euclidean (a "jump" is possible): take $n=2$,$A,B=\pm Id$. Start with a path from $Id$ to $\begin{bmatrix}\ep & 0 \\ 0 & \ep\end{bmatrix}$. Then go via
(1) $t \to \begin{bmatrix}\ep & -t \\ t & \ep\end{bmatrix}$ to $\begin{bmatrix}\ep & -\ep \\ \ep & \ep\end{bmatrix}$ ($t$ goes $0 \to \ep)$ .
(2) $t \to \begin{bmatrix}t & -\ep \\ \ep & t\end{bmatrix}$ to $\begin{bmatrix}-\ep & -\ep \\ \ep & -\ep\end{bmatrix}$ ($t$ goes $\ep \to -\ep)$.
(3) $t \to \begin{bmatrix}-\ep & -t \\ t & -\ep\end{bmatrix}$ to $\begin{bmatrix}-\ep & 0 \\ 0 & -\ep\end{bmatrix}$ ($t$ goes $\ep \to 0)$ .
Now continue with straight line until reaching $-Id$.
How much this maneuver cost us?
The derivatives of the 3 broken straight paths we took were: $\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}, Id , \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$. their norms are $\sqrt 2$. Hence, the total lenght is $\sqrt2 \cdot 4\ep$ which is arbitrarily small, as required.
(Also, note that the determinant was always $t^2 + \ep^2 > 0$ so we stayed in $GL_n^+$).
(2) Can we compute explicitly for a given $A \in GL_n^+$, it's distance from $SO(n)$?
$dist(A,SO(n)) =\underset{X \in SO(n)}{\text{min}} d(A,X)$
($SO(n)$ is the special orthogonal group and the minimum exists since $SO(n)$ is compact and $d$ is continuous)
And who is the minimizer (the closest matrix to $A$ in $SO(n)$)? Is it unique?
Note: Right or Left multiplication by elements of $SO(n)$ are isometries of $GL_n^+$ with the induced metric. Thus, $d$ is left (right)-$SO(n)$ invariant.
In particular, if $ A = U\Sigma V^T $ is the SVD-dscomposition of $A$, then: $dist(A,SO(n)) = dist(\Sigma,SO(n)) $ , where $\Sigma$ is a square, diagonal matrix whose diagonal elements are the (strictly positive) singular values of $A$.
So for the question of computing the distance from $SO(n)$ (and the minimizer) is reduced to matrices of this type. (i.e diagonal + positive entries).