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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

$d^{int} \le C \cdot d^{ext} \text{ for some } C \in \mathbb{R}$?


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    $\begingroup$ +1 for stimulating a research paper (see below). $\endgroup$ – Sebastian Goette Feb 4 '16 at 13:15
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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

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  • $\begingroup$ Actually I think I can prove the answer is yes for $n=2$ but I don't know about $n>2$ (I cut into 2 separate comments...). $\endgroup$ – YCor Dec 29 '15 at 15:17
  • $\begingroup$ That's interesing, I would like to see the proof. @YCor $\endgroup$ – Mikhail Katz Dec 29 '15 at 15:18
  • $\begingroup$ But it's lengthy and time-consuming. The steps: 1 (for any $d$): the two distances are strongly equivalent if and only if on the 1-sphere interesected with positive determinant matrices, the usual distance and the spherical distance using paths of positive determinant matrices, are equivalent. 2) for $d=2$, on the 1-sphere the zero locus of the determinant function is not singular (for $d>2$ matrices of rank $\le d-2$ are singular points). (...) $\endgroup$ – YCor Dec 29 '15 at 15:22
  • $\begingroup$ (....) 3) (any $d$) if on the sphere the two distances (let me denote $d$ and $d^+$) are not equivalent, one can find in the 1-sphere, after extraction, converging sequences $A_n\to A$, $B_n\to B$ such that $d(A_n,B_n)/d^+(A_n,B_n)\to 0$. Note that $\det(A),\det(B)\ge 0$. If $A\neq B$ this means that $d^+(A_n,B_n)\to\infty$. Otherwise $A=B$. So in both cases, we obtain that there exists some $P$ with $\det(P)\ge 0$ at the neighborhood of which $d^+$ is not equivalent to $d$ (being unbounded in the first case, where $P=A$ or $B$, or taking $P=A=B$ in the second case). (...) $\endgroup$ – YCor Dec 29 '15 at 15:28
  • $\begingroup$ 4) If $P$ is invertible, this is obviously a contradiction. If the rank of $P$ is $d-1$, then the determinant function on the sphere has nonzero gradient at $P$ and thus locally up to a diffeomorphic change of chart, the determinant looks as a nonzero affine map vanishing at $P$, so there is no distortion at its neighborhood. 5) In particular if $d=2$, we have a contradiction, so both distances are equivalent. If $d=3$, I do not know if there is distortion at the neighbourhood of a rank 1 matrix as the determinant is singular there. But this should be doable. $\endgroup$ – YCor Dec 29 '15 at 15:32

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