Timeline for Convexity in co-ordinate charts of geodesic balls
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2019 at 21:26 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Fixed some typos and careless, misleading statements
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| Feb 7, 2019 at 2:29 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added a remark about the explicit estimate for epsilon
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| Feb 6, 2019 at 22:35 | comment | added | Gabe K | In particular, any estimate on the $\epsilon$ for which metric balls are Euclidean convex needs to depend on the choice of coordinate chart, and not just on the coordinate-independent geometry of the manifold. | |
| Feb 6, 2019 at 22:24 | comment | added | Gabe K | @RobertBryant Sorry, I should have been more precise. By $C^k$-regularity, I meant a uniform scale-$C^k$-estimate. It seems that without some sort of estimate like this on the $U$-coordinates, there is no hope of having a uniform radius for which metric balls are convex. For example, I could put a coordinate chart on $R^n$ where the sphere of radius $\epsilon$, in coordinates, is a cardioid with its cusp smoothed. In order to prevent that from happening on a given length scale, it seems like there needs to be some explicit estimate on the metric tensor in those coordinates. | |
| Feb 6, 2019 at 21:54 | comment | added | macbeth | I will leave this answer un-accepted while I wait to see if someone can answer this more refined version. By your argument this can be reduced to giving an estimate $\epsilon(r,Q)$ (and probably $k=1$ suffices) such that the Riemannian distance function is convex on the Euclidean ball $B(p,\epsilon)$. This seems like it could be doable. | |
| Feb 6, 2019 at 21:50 | comment | added | macbeth | Thank you for this answer, it is a simple and satisfying argument! I am still interested in the follow-up question I mentioned (and I think this is what Gabe K was referring to): can one control the $\epsilon$ by the constants $Q$ and $r$ of a "scale-$C^{k}$-controlled" co-ordinate chart? | |
| Feb 6, 2019 at 21:36 | comment | added | Robert Bryant | @GabeK: I don't think that the OP wanted to consider low regularity metrics. The notion of "scale-$C^k$ controlled" doesn't have anything to do with regularity, it's just a way of talking about controlling the size of the first $k$ derivatives of the $g_{ij}$, not a question about whether they exist. | |
| Feb 6, 2019 at 21:30 | comment | added | Robert Bryant | @GabeK: The point is that, when the metric $g$ is smooth on $U$ (which I am assuming), then the squared $g$-distance $\rho$ from a given point $p$ is always smooth near $p$ and its Hessian at $p$ is positive definite. This is a coordinate-independent statement (I was just explaining the proof). Now you just need to know that, for any smooth function $f$ on $\mathbb{R}^n$ with a nondegenerate minimum at $p$ (i.e., the Hessian of $f$ at $p$ is positive definite), the sublevel sets $f \le f(p)+\epsilon$ near $p$ are convex for $\epsilon>0$ sufficiently small. This is a standard calculus fact. | |
| Feb 6, 2019 at 20:21 | comment | added | Gabe K | I thought this question was asking about the convexity for an arbitrary coordinate chart with $C^k$-regularity, not necessarily the coordinates in which $\rho$ has this form. Is there an easy way to translate the convexity in these nice coordinates to convexity in the arbitrary coordinates or have I misunderstood the question somehow? | |
| Feb 6, 2019 at 19:45 | comment | added | Robert Bryant | @MattF.: In terms of what? If one has appropriate bounds on the $g_{ij}$ and their first derivatives in $U$, then, in terms of these bounds, one can determine a lower estimate for how large one could take $\epsilon$ such that $B^g(p,\delta)$ will be convex in $\mathbb{R}^n$ for all $0<\delta<\epsilon$. After all, what one is comparing are the geodesics of two different affine connections, the Levi-Civita connection of $g$ and the flat affine connection of $\mathbb{R}^n$. | |
| Feb 6, 2019 at 18:50 | comment | added | user44143 | Can you say how close is sufficient? | |
| Feb 6, 2019 at 17:33 | history | answered | Robert Bryant | CC BY-SA 4.0 |