1
$\begingroup$

Let $u$ be a radially decreasing function defined on $\mathbb{R}^n$. We consider the metric $g=e^{2u}\delta$ where $\delta$ is the standard Euclidean metric on $\mathbb{R}^n$. Let $B_r$ be the ball centered at the origin with Euclidean radius $r$. Then for any $x \in \partial B_r$, by direct computation we know that the line segment connecting the origin and $x$ must be a geodesic. My question is that, is it true that the line segment joining the origin and $x$ is the unique geodesic connecting these two points?

Intuitively this is correct, because such line segment seems to be length minimizing. However, unfortunately I'm not be able to prove this....

I'm actually not very familiar with Riemannian geometry. Any ideas, comments or references will be really appreciated!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
+50
$\begingroup$

It must be length minimizing, because it is the unique geodesic from the origin to $x$. In particular, any ray from the origin is geodesic so we know all the geodesics from the origin. Since the shortest path from the origin to $x$ is a geodesic and there is only one such geodesic, it must be length minimizing.

Note that this does not mean that the shortest path from $x$ to $-x$ is necessarily a line segment through the origin. One could imagine the surface being shaped like a light bulb with the origin being the top of the bulb. If $x$ and $-x$ are on the threading, it's much shorter to go around the threading rather than go to the top and back down.

Improve this answer
$\endgroup$
9
  • $\begingroup$ In particular, this is a property of radially symmetric metrics. It doesn't depend on them being conformally flat or the conformal factor decreasing. $\endgroup$
    – Gabe K
    Commented Apr 24, 2020 at 14:30
  • 1
    $\begingroup$ Could you show why it is the unique geodesic from the origin to $x$? I know it is a geodesic, but I don't know how to show uniqueness. $\endgroup$
    – student
    Commented Apr 24, 2020 at 17:27
  • $\begingroup$ Via the exponential map, there is a correspondence between geodesics emanating from the origin and unit tangent vectors. For each unit tangent vector, there is a geodesic given by the ray based at the origin in that direction. As such, those are all of the geodesics and there is only one that reaches $x$. $\endgroup$
    – Gabe K
    Commented Apr 24, 2020 at 18:19
  • $\begingroup$ Sorry I still could not get it. Why there cannot exist other unit tangent vectors along the direction of which the corresponding geodesic reaches $x$? Also how would you exclude cut locus? $\endgroup$
    – student
    Commented Apr 24, 2020 at 18:24
  • $\begingroup$ I suppose I should say why the shortest path from the origin to $x$ is given by a geodesic at all, since we don't assume that the metric is complete. It's an ugly argument, but a proof of this can be found here. mathoverflow.net/questions/317523/… $\endgroup$
    – Gabe K
    Commented Apr 24, 2020 at 18:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .