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Let $(M,g)$ be a complete Riemannian manifold, and let $C$ be a subset of $M$. We will say $C$ is convex if for any points $p,q \in C$, there exists a unique normal minimal geodesic $\gamma$ joining $p$ and $q$, such that $\gamma \subset C$.

Now, suppose we have a smooth manifold $M$, and two metrics $g_1$ and $g_2$. Briefly, I am interested to know under what conditions $(M,g_1)$ and $(M,g_2)$ have the same convex sets, or at least a class of convex sets (e.g. convex geodesic balls) in common.

That is, if $C \subset M$ is a convex set in $(M,g_1)$, is it convex in $(M,g_2)$? Certainly this isn't generally true, so what conditions on $M,g_1,g_2,C$ are necessary/sufficient?

My guess is that very restrictive conditions are needed to ensure all convex sets are shared, so let me be more precise and describe the specific case I'm interested in:

Let $M = \mathbb R^n$ and $g_1$ be the standard Euclidean metric, and let $C = B_R(x)$ be the usual ball of radius $R$ centered at $x$, and let $g_2$ be some other metric. Is $C$ convex in $(M,g_2)$? (or what conditions on $g_2,R$ are needed? In particular I imagine that $R$ cannot be too large)

In practice, I will be assuming $g_2 = c^{-2}(x)dx^2$ (from a wave equation) and that the sectional curvature is bounded above, so that there is some control over the injectivity/convexity radius.

EDIT (some more thoughts): I suspect this is tied to curvature; does the sectional curvature bound the curvature (as a curve in Euclidean space) of a geodesic of $g_2$?

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This will be a very sketchy argument, perhaps to be elaborated on later.

First thing we may note is that a short geodesic curve $C$ joining two points $p$ and $q$ is itself a convex set. So if the two metrics are to have the same convex sets, they must also have the same geodesic curves (up to reparametrisation if by geodesic we mean a path of constant velocity). (I'll use curve to mean the set $C$ and path for the parametrised map $\gamma(t)$ to help keep these two apart.)

So now we need to demonstrate that for two metrics, say $g$ and $\tilde g$, to have the same geodesic curves, the two must be related by a linear relation $g_{ij}=a^r_i a^s_j \tilde g_{rs}$ for some constant $a^r_i$.

For a given point, $O$, we can always find a coordinate system which has $O$ at the origin, and so that at $O$ we have $\tilde g_{ij}=\delta_{ij}$ and Christoffel symbols $\tilde\Gamma^k_{ij}=0$ for $\tilde g$. Thus, a geodesic path $\gamma$ (with constant speed) through $O=\gamma(0)$ should have vanishing second derivative at $O$ (in this coordinate system). If $\gamma$ is a path along the geodesic curve (not necessarily with constant speed), instead $\gamma''(0)$ need only be parallel to the direction $\gamma'(0)$.

For the metric $g$, a geodesic path through $O$ should satisfy $\gamma''_k+\gamma'^i\gamma'^j\Gamma^k_{ij}=0$ at the origin. For this to form a geodesic path under the metric $\tilde g$, we must then for all tangent vectors $u$ have $u^iu^j\Gamma^k_{ij}=\lambda u^k$ for some $\lambda$. In dimensions $2$ and higher, this forces $\Gamma^k_{ij}=0$, which in turn means the derivative $g_{ij,k}=0$ at $O$.

Unless I'm missing something, the rest should be purely technical to show that this requires that $g_{ij}$ and $\tilde g_{rs}$ are related by a constant linear transformation. For flat metrics, any linear transformation should work, while for curved metrics I suspect linear transformations other than scaling will only be possible in "flat directions".


I'll try to address the second question of when a ball of radius $R$ in one metric will remain convex in the other. I'll ignore global issues (not sure if they will matter) and only look at the question locally. This will be even more sketch then the above, and I've probably managed to sneak in a few mistakes on the fly, to read it for ideas only.

Again, to ensure this property locally, a geodesic curve in one metric should have curvature $\kappa$ at most $1/R$ in the other. To verify this, note that a short (infinitesimal) geodesic between points $p,q$ in one metric, in the other metric needs to be inside every ball of radius $R$ with $p,q$ on the surface.

Imagine we have a geodesic curve in metric $\tilde g$ through a point $O$. Also, we select a coordinate system centered around $O$ in which $\tilde g_{ij}=\delta_{ij}$ and $\tilde\Gamma^k_{ij}=0$ at the origin. We then parametrise the curve using the path $\gamma(s)$ (or $\gamma^k$ for coordinate $k$) so that $\tilde g(\gamma',\gamma')=\tilde g_{ij} {\gamma'}^i {\gamma'}^j=1$ (where $\gamma'=d\gamma/ds$); and since the curve is geodesic and $\tilde\Gamma^k_{ij}=0$, the second derivative $\gamma''=0$.

In the coordinate system $g$, this curve will have covariant derivative $D\gamma^k={\gamma'}^i{\gamma'}^j\Gamma^k_{ij}$ since we have conveniently selected a coordinate system in which $\gamma''=0$. Subtracting the acceleration part, $D\gamma-\gamma'\cdot g(\gamma',D\gamma)/g(\gamma',\gamma')$, the curve has curvature $\kappa$ where $$ \kappa^2 =\frac {|D\gamma-\gamma'\cdot g(\gamma',D\gamma)/g(\gamma',\gamma')|^2_g} {|\gamma'|^4_g} $$ which we can then write out in terms of $\gamma'$ and $\Gamma^k_{ij}$. Since we requires $\kappa\le 1/R$ for all such curves, the derived relation becomes $$ u^iu^ju^lu^m\Gamma^k_{ij}\Gamma^n_{lm}g_{km} -\frac{\left(u^iu^ju^r\Gamma^k_{ij}g_{kr}\right)^2}{u^pu^qg_{pq}} \le\frac{\left(u^pu^qg_{pq}\right)^2}{R^2} $$ for all vectors $u^i$.

So far, things were a bit technical.

Next, although the Christoffel symbols $\Gamma^k_{ij}$ depend on the choice of coordinate system, the difference $\Delta\Gamma^k_{ij}=\Gamma^k_{ij}-\tilde\Gamma^k_{ij}$ between the two metrics is independent of coordinate system changes and thus transform like tensors. The reason for this is that under a change in coordinates from $x^i$ to $\hat x^a$, they transform as $$ \hat\Gamma^c_{ab}x^k_{,c}=x^i_{,a}x^j_{,b}\Gamma^k_{ij}+x^k_{,ij} $$ where the second part of the sum is the same for both metrics.

Since the original bound was expressed in a coordinate system with $\tilde\Gamma=0$, we can just plug in $\Delta\Gamma$ instead. To also show that geodesic curves in $g$ have curvature at most $1/R$ in $\tilde g$, all we need do is to swap the two metrics: in the inequality, $\Delta\Gamma$ changes sign (no effect there since it is squared), and $g$ must be replaced by $\tilde g$.

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  • $\begingroup$ Thank you for your thoughts so far, I think this convinces me that to have all convex sets in common is very restrictive. Do you have any thoughts about the case where only a specific class of convex set is being considered? In particular, are the unit balls, in one of the metrics (in particular, in the Euclidean metric), of radius no greater than some $R_0$ also convex in the second metric? (and generally I expect not, so then what are sufficient conditions?) $\endgroup$ – BaronVT May 15 '15 at 2:19
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    $\begingroup$ @BaronVT: I suspect the weaker condition, e.g. balls of radius $R_0$ in one metric are convex in the other, will be much less restrictive, but also more complex to analyse. Locally, I think this will mean that geodesics in one metric will have curvature at most $1/R_0$ in the other. Thus, arbitrary small and smooth perturbations of the metric are permissible, and the limitation is on the size of the "non-linearity" of the change. $\endgroup$ – Einar Rødland May 15 '15 at 8:40
  • $\begingroup$ Yes, I also suspect that is the right idea: it seems a geodesic cannot exit the ball and then return without having relatively high curvature (or being very long). Do you know of any relationship between the curvature of a geodesic (as a curve in $\mathbb R^n$) and the sectional curvature of the manifold $(\mathbb R^n, g_2)$? $\endgroup$ – BaronVT May 15 '15 at 14:48
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    $\begingroup$ @BaronVT: I've added an outline for the $R$-ball version. The added derivation also leads me to think the limitation has little to do with the curvatures of the metrics (at least not locally). I haven't double checked anything, so read with caution. $\endgroup$ – Einar Rødland May 18 '15 at 17:53
  • $\begingroup$ Great, thank you for your input. I will have to check the details of what you have outlined, but I think this is certainly deserving of the bounty. $\endgroup$ – BaronVT May 18 '15 at 19:34

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