TITLE: Strong equivalence between intrinsic and extrinsic metrics on $GL_n^+$?
QUESTION [11 upvotes]: $\newcommand{\til}{\tilde}$
Lately, I have become interested in comparing intrinsic and extrinsic metrics on Riemannian manifolds.
Consider $GL_n^+$ (invertible matrices , $\det >0$) as an open Riemannian submanifold of $\mathbb{R}^{n^2}$.
We have two metrics on $GL_n^+$ (in the sense of metric spaces); intrinsic $d^{int}$ (when we only allow paths which stay inside $GL_n^+$) and extrinsic $d^{ext}$ (the standard Euclidean metric). 
Clarification: $d^{int}$ is the Riemann distance function on $GL_n^+$ induced by the standard Riemannian metric on $\mathbb{R}^{n^2}$ (i.e I consider $GL_n^+$ as a Riemannian submanifold of $\mathbb{R}^{n^2})$
Of course $d^{ext} \le d^{int}$. For some matrices $d^{ext}(A,B) < d^{int}(A,B)$;
If $\gamma(t)=A+t(B-A)$, then $\det(\gamma(t))$ is a polynomial which can be negative on a subinterval of $[0,1]$. (i.e the minimizing geodesic in the Euclidean space, is contained in $GL^{-}$ for some time)
(see this answer to this question for details on this specific case, and here for a cleaner proof that "nearly minimizing paths are nearly geodesic")

Note that $d^{ext},d^{int}$ both generate the same topology on $GL_n^+$.
Moreover, both are incomplete.
So, the following natural question arises:
Are $d^{ext},d^{int}$ strongly equivalent? i.e; Is it true that
$d^{int} \le C \cdot d^{ext} \text{ for some } C \in \mathbb{R}$?

REPLY [3 votes]: In two dimensions the condition $ad-bc=0$ translates into $(a+d)^2-(a-d)^2-(b+c)^2+(b-c)^2=0$ or to simplify notation $x^2+y^2=z^2+w^2$. Intersecting this with the unit sphere in $\mathbb{R}^4$ gives a flat 2-torus which decomposes the $3$-sphere into two solid tori. One solid torus consists of matrices of positive determinant and the other of matrices of negative determinant. Thus the determinant variety $ad-bc=0$ in this case is a cone on a 2-torus. 
Since the torus is smooth, the intrinsic distances in the torus are bilipschitz with ambient distance in the solid torus filling it.  In fact, one can probably get away with a Lipschitz constant not much bigger than $\frac{\pi}{2}$.  Therefore whenever one has a straight path joining two points in $GL^+(2,R)$ one can always replace its portions "dipping" into the (cone over the) "negative" solid torus by arcs following the (cone over the) flat 2-torus while retaining Lipschitz control over length of the path.  This proves that the intrinsic and the ambient distances are bilipschitz in the 2-dimensional case.
A complete answer appears at http://arxiv.org/abs/1602.01227