Let $g_{ij}$ be a Riemannian metric tensor on an open subset $U\subseteq \mathbb{R}^n$, and let $p\in U$.

I would guess the following is true:

for $\epsilon$ sufficiently small, the $g$-geodesic ball $B^g(p,\epsilon)$ is a Euclidean-convex subset of $U$.

Is this true? Especially, does anyone know a reference for this?

Even better, suppose that $g_{ij}$ is scale-$C^{k}$-controlled in the usual way: $U=B(p,r)$ (the Euclidean ball), and for some $Q>1$,

  • $Q^{-1}\delta_{ij}\leq g_{ij}\leq Q\delta_{ij}$
  • for multi-indices $1\leq |\beta|\leq k$, we have $r^{|\beta|}|D^{\beta}g_{ij}|\leq Q-1$

Can one estimate explicitly an $\epsilon=\epsilon(r,Q)$ such that the above is true?

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    $\begingroup$ I am not sure if the geodesic balls have to be convex, but you can always take the normal coordinates centered at any point of $p\in U$ and in that system of coordinates small geodesic balls centered at $p$ are just the Euclidean balls. Perhaps it will help. $\endgroup$ – Piotr Hajlasz Feb 6 at 0:39
  • $\begingroup$ That's true, but I'm interested in general co-ordinates! $\endgroup$ – macbeth Feb 6 at 8:28
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    $\begingroup$ I don't know the details, but is it maybe possible to prove this by a blowup argument? If there exists a sequence of non-convex geodesic balls with their radii converging to zero, then if they're rescaled to radius $1$, then you get a sequence of balls converging to the flat ball in standard coordinates but the limiting ball is not strictly convex. Something like that? $\endgroup$ – Deane Yang Feb 6 at 9:33
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    $\begingroup$ Just as a follow up to Deane Yang's suggestion, if you perform the blow-up argument in your coordinates, the geodesic balls should converge to an ellipsoid in Euclidean space. To make this argument work, it seems necessary to assume the metric is scale-$C^k$-controlled. I'm not sure how to prove an explicit estimate for epsilon; the issue seems to be that the extrinsic curvature of the metric sphere in the coordinates needs to be positive. As such, I'm guessing that for a uniform estimate, you need scale-$C^2$-control, as that's the regularity at which curvature estimates are possible. $\endgroup$ – Gabe K Feb 6 at 16:28

This is certainly true. If you choose $\epsilon>0$ smaller than half the injectivity radius of $g$ at $p$ (in particular, sufficiently small that the exponential map $\exp_p:T_pU\to U$ is well-defined for all $v\in T_pM$ of $g$-length less than $\epsilon$), then, on $B^g(p,\epsilon)$ the squared $g$-distance from $p$ is a smooth function $\rho:B^g(p,\epsilon)\to\mathbb{R}$ that can be written in the form $$ \rho = (y^1)^2 + \cdots + (y^n)^2 $$ where $y = (y^i)$ is a smooth coordinate system on $B^g(p,\epsilon)$ that is centered on $p$.

Hence $\rho$ has an isolated proper local minimum at $p$, with a positive definite Hessian at $p$. This implies, by an elementary argument, that the sublevel sets $\rho \le \delta$, for all $\delta\in(0,\epsilon)$ sufficiently close to zero, are convex in $\mathbb{R}^n$.

Added remark about estimates: Actually, if one assumes, for simplicity, that $p=0\in\mathbb{R}^n$, then, for $g = g_{ij}(x)\mathrm{d}x^i\mathrm{d}x^j$, one has a Taylor expansion with remainder of the form $$ \rho = g_{ij}(0)x^ix^j + g_{ijk}(x)x^ix^jx^k $$ where the functions $g_{ijk}$ and their first and second derivatives can be bounded in $U$ in terms of $g_{ij}(0)$ and uniform bounds on the first, second, and third derivatives (with respect to the coordinates $x^i$) of the $g_{ij}(x)$ in $U$. Once one has this, the local convexity of the function $\rho$ (which will imply that its sublevel sets near $p=0$ are convex in the usual sense) is a matter of checking whether $$ H = (H_{ij}) = \left(\frac{\partial^2\rho}{\partial x^i\partial x^j}\right) $$ is positive semidefinite in the region where $\rho(x)\le\epsilon^2$. Again, this can be checked in terms of the explicit bounds that can be worked out from these formulae, so there certainly will be an estimate of how large one can take $\epsilon$ based on the data that the OP wants to use. Probably, one can do better than this by examining the $g_{ijk}(x)$ more carefully. I would expect that one only needs uniform bounds on $g_{ij}$ and its first (and maybe second) derivatives to get such a bound.

  • $\begingroup$ Can you say how close is sufficient? $\endgroup$ – Matt F. Feb 6 at 18:50
  • $\begingroup$ @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 $\mathbb{R}^n$. $\endgroup$ – Robert Bryant Feb 6 at 19:45
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    $\begingroup$ 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? $\endgroup$ – Gabe K Feb 6 at 20:21
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    $\begingroup$ @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. $\endgroup$ – Robert Bryant Feb 6 at 21:30
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    $\begingroup$ 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? $\endgroup$ – macbeth Feb 6 at 21:50

Unfortunately I can't get explicit estimates on $\epsilon(r,Q)$, but here's an argument to show such an $\epsilon$ exists. As a notational remark, I'm going to denote the Riemannian distance between $p$ and $q$ in the metric $g$ by $d(p,q)$ and the distance from $p$ to $q$ with respect to the inner product $g(0)$ (extended in the natural way to all of $U$) by $|p-q|_g$.

Suppose that $g$ is $C^2$-scaled controlled with precision $Q$. Then, whenever $|p-q|_g$ is small, $d(p,q)$ and $|p-q|_g$ are close in $C^2$ norm. If I'm remembering the implicit function theorem correctly, it can be used to show that if two functions are close in $C^2$ norm, their level sets are close in $C^2$ norm as well. The level sets of $|p-q|_g$ are ellipsoids that are uniformly convex (by the $Q$-precision) and since any surface that is $C^2$-close to a strictly-convex surface is itself convex, for small $r$ the level set $d(p,q) = r$ is also convex.

In fact, we could have done even better. Since any surface that is $C^1$ close to a strictly-convex surface is also convex, if $g$ is merely $C^1$-scale controlled, we get the same result. It's a good question whether this can be made explicit and I'll add details if I think of anything towards that end.


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