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?

  • 1
    $\begingroup$ 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. $\endgroup$ Feb 6, 2019 at 0:39
  • $\begingroup$ That's true, but I'm interested in general co-ordinates! $\endgroup$
    – macbeth
    Feb 6, 2019 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, 2019 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, 2019 at 16:28

2 Answers 2


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, 2019 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 $\mathbb{R}^n$. $\endgroup$ Feb 6, 2019 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, 2019 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. $\endgroup$ Feb 6, 2019 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, 2019 at 21:50

I hope this question is still of interest. It turned out that some coauthors and I needed this fact for a recent paper. As such, I've gone through and filled in the details (and fixed some errors) from my original answer. The paper is here and the relevant section is Lemma 18 in the second appendix, which starts on page 24.

I didn't try to get explicit estimates on $\epsilon(r,Q)$, but in theory it can be done by bounding each of the constants. Sorry in advance for the length of the argument; I wasn't able to find a cleaner approach which gets around inducting on the regularity of the distance function.

To begin, we fix a point $p$, which will be the center of the ball throughout. Consider two different Riemannian metrics on $U$. The first, (denoted $g$) is the Riemannian metric of interest. The second, $g_0$, is the flat metric satisfying $(g_0){}_{ij}(x) = g_{ij}(p)$ for all $ x \in U$ (i.e. we consider the metric at $p$ and do not change the components throughout $U$). We then consider two separate distance functions. The first, which we denote $d$, is the distance from $p$ in the $g$ metric. The second, which we denote $\delta_0$, is the distance from $p$ in the $g_0$ metric.

As a broad overview, we can break up this argument into 4 steps.

  1. Using the $C^0$-scale control, we show that $d$ and $\delta_0$ are $C^0$ close.
  2. By bounding the total acceleration of $g$-geodesics in the $x$-coordinates, we show that $d$ and $\delta_0$ are $C^1$ close.
  3. By bounding the total jerk of $g$-geodesics in the $x$-coordinates, we show that $d$ and $\delta_0$ are $C^2$ close.
  4. Using the implicit function theorem, we show that the level sets of $d$ are $C^2$ close to level sets of $\delta_0$ (which are ellipsoids with bounded eccentricity). This then implies that the level sets of $d$ are convex, at least at small enough scales.

Step 1. The first step is to show that $d$ and $\delta_0$ are $C^0$ close. To do this, consider the $g$-geodesics in the $g_0$ metric (and vice versa), and use the $C^0$-scale control. This shows that the distance functions control each other, or more precisely that $$ Q^{-1} \delta_0(p,q) \leq d(p,q) \leq Q \delta_0(p,q) $$

Step 2. We now want to show that the distance functions are close in $C^1$. To do so, consider a point $q$ which is close to $p$ (either in terms of $d$ or $\delta_0$). The geodesic from $p$ to $q$ in the $g_0$ metric is a straight line whose length (in the sense of $g_0$) is $\delta_0(q)$. Furthermore, the gradient of $\delta_0$ at $q$ is of unit norm (again in the sense of $g_0$) and points in the same direction as the line segment from $p$ to $q$. On the other hand, the unit speed geodesic for the metric $g$ from $p$ to $q$ satisfy the equations \begin{equation} \label{geodesicequations} \frac{d^2 \gamma^i}{ds^2} + \Gamma^{i}_{jk}\frac{d \gamma^j}{ds}\frac{d\gamma^k}{ds} = 0 \end{equation} where \begin{equation} \label{Christoffel}\Gamma^i_{jk} = \frac{1}{2} g^{i\ell} \left( \frac{\partial g_{j \ell}}{\partial x^k} + \frac{\partial g_{k \ell}}{\partial x^j} - \frac{\partial g_{jk}}{\partial x^\ell} \right). \end{equation}

However, from the scale-$C^1$-control, we can estimate that \begin{eqnarray*} \left \vert \frac{d^2 \gamma^i}{ds^2} \right \vert &\leq & \sum_{j,k} \frac{Q}{2} \frac{3(Q-1)}{r} \left \vert \frac{d \gamma^j}{ds} \right \vert \left \vert \frac{d \gamma^k}{ds} \right \vert \end{eqnarray*} along the entire geodesic. In order to make the estimates readable, we will absorb any constants involving $n,r$ and $Q$ using the notation $ f_1 \lesssim f_2$ whenever $f_1 \leq C f_2$ for some constant $ C$ depending only on $n,Q,$ and $r$. We will also use the notation $$ \left. \left \| \frac{d^k \gamma}{ds^k} \right \|_{L^1} \right \vert_{s=\tau} = \sum_{i=1}^n \left \vert \frac{d^k \gamma^i (\tau)}{ds^k} \right \vert $$ and the corresponding notation for the $L^2$ norm as well.

In this notation, we find that $$ \left \vert \frac{d^2 \gamma^i}{ds^2} \right \vert \lesssim \left \vert \frac{d \gamma^j}{ds} \right \vert \left \vert \frac{d \gamma^k}{ds} \right \vert. $$

Summing over the $i$ index, this implies that \begin{eqnarray*} \left. \left \| \frac{d^2 \gamma}{ds^2} \right \|_{L^1} \right \vert_{s=\tau} &\lesssim & \left. \left \| \frac{d \gamma}{ds} \right \|_{L^2}^2 \right \vert_{s=\tau} \\ & \lesssim & \left \| 1 + \int_0^\tau \left. \left \| \frac{d^2 \gamma}{ds^2} \right \|_{L^1} \right \vert_{s=t} \,dt \right \|_{L^2}^2 \\ & \leq & 2 + 2\left(\int_0^\tau \left. \left \| \frac{d^2 \gamma}{ds^2} \right \|_{L^1} \right \vert_{s=t} \,dt \right )^2 \end{eqnarray*}

We define $F(\tau) = \int_0^\tau \left. \left \| \frac{d^2 \gamma}{ds^2} \right \|_{L^1} \right \vert_{s=t} \,dt$. In other words $F(\tau)$ is the total acceleration of a $g$ geodesic in coordinates. When written in terms of $F$, the above estimate shows that \begin{eqnarray*} \frac{d F}{d \tau } &\lesssim & 1 + F^2 \end{eqnarray*} Dividing both sides by $1 + F^2$ and integrating, we find that $$\arctan{F(\tau)} \lesssim \tau. $$ For $\tau$ small, this provides an upper bound for $F$. However, since $\gamma$ is a unit speed geodesic, this shows that for small $d$ (equivalently $\delta_0$), the acceleration (in coordinates) of $\gamma$ is very small. As a result, $\gamma$ is $C^1$-close to a line segment from $p$ to $q$ (in coordinates). Therefore, the gradient of $d$ at $q$ is close to the gradient of $\delta_0$ at $q$, which implies that the functions are $C^1$-close as well.

Step 3. We now want to show that the distance functions are $C^2$ close as well. To do so, we start by taking one derivative of the geodesic equations and use the $C^2$ scale control to bound the $C^1$ norm of the Christoffel symbols. When we do so, we get an estimate of the form $$\left \| \frac{ d^3 \gamma}{ d s^3 } \right \|_{L^1} \lesssim \left \| \frac{ d^2 \gamma}{ d s^2 } \right \|_{L^1} \left \| \frac{ d \gamma}{ d s } \right \|_{L^1}^2 +\left \| \frac{ d \gamma}{ d s } \right \|_{L^1}^3.$$

From here, we can basically repeat Step 2 verbatim. There are other ways of controlling this term, but it's probably simplest to do it using a method we've already done. Using Young's inequality, this shows that \begin{eqnarray*} \left \| \frac{ d^3 \gamma}{ d s^3 } \right \|_{L^1} & \lesssim & \left \| \frac{ d^2 \gamma}{ d s^2 } \right \|_{L^1}^2 + \left \| \frac{ d \gamma}{ d s } \right \|_{L^1}^4 +\left \| \frac{ d \gamma}{ d s } \right \|_{L^1}^3 \\ \end{eqnarray*} Using the bound on $F$ from the $C^1$ estimate, for small time we can control $\left \| \frac{ d \gamma}{ d s } \right \|$ by a constant, which implies \begin{eqnarray*} \left \| \dfrac{ d^3 \gamma}{ d s^3 } \right \|_{L^1} & \lesssim \left \| \dfrac{ d^2 \gamma}{ d s^2 } \right \|_{L^1}^2 + 1. \end{eqnarray*}

Using the estimate $$ \left. \left \| \frac{ d^2 \gamma}{ d s^2 } \right \| \right \vert_{s = \tau} \leq \int_0^\tau \left . \left \| \frac{ d^3 \gamma}{ d s^3 } \right \| \right \vert_{s = t} dt, $$ we can rewrite the preceding line to obtain \begin{eqnarray*} \left. \left \| \frac{ d^3 \gamma}{ d s^3 } \right \| \right \vert_{s = \tau} & \lesssim \left( \int_0^\tau \left . \left \| \dfrac{ d^3 \gamma}{ d s^3 } \right \| \right \vert_{s = t} dt \right) ^2 + 1. \end{eqnarray*}

We again integrate out the differential inequality to show that $$ \int_0^\tau \left . \left \| \dfrac{ d^3 \gamma}{ d s^3 } \right \| \right \vert_{s = t} dt \lesssim \tan(\tau). $$ For sufficiently small times $\tau$, this provides a small bound on the total jerk of $\gamma$, which also bounds the point-wise acceleration of $\gamma$. As a result, $\gamma$ is $C^2$-close to a line segment from $p$ to $q$. Therefore, the gradient of $d$ is $C^1$-close to the gradient of $\delta_0$ at q, which implies that the two distance functions are $C^2$-close.

Step 4. Finally, we show that the ball $B_{g_0}(p,\epsilon)$ is convex in the $x$-coordinates for $\epsilon$ small. Consider a point $q$ with $d(p,q)$ small and the hyperplane (in $x$-coordinates) $V$ through $q$ which is perpendicular to the line segment from $p$ to $q$. Near $q$, both the functions $d$ and $\delta_0$ have gradients which are transverse to this hyperplane. As such, by the implicit function theorem we can locally find two functions $\ell_1, \ell_2:V \to \mathbb{R}$ so that $d(v,\ell_1(v)) = d(q)$ and $d(v,\ell_2(v)) = \delta_0(q)$. In other words, we use the implicit function theorem to express the level sets of $d$ and $\delta_0$ as graphs in a small neighborhood of $q$. Furthermore, we can write the derivatives of $\ell_1$ and $\ell_2$ in terms of the derivatives of $d(q)$ and $ \delta_0(q)$, respectively.

Since $d(q)$ and $ \delta_0(q)$ are $C^2$-close, it follows that $\ell_1$ and $\ell_2$ are also $C^2$-close. However, the level sets of $\delta_0$ are ellipsoids with bounded eccentricity, so it follows that $\ell_2$ is (uniformly) strongly convex. Since $\ell_1$ is $C^2$-close to the graph of an ellipsoid, for small enough $d$ the Hessian of $\ell_1$ must be non-negative definite. This implies that $\ell_1$ is a convex function and so secants of $\ell_1$ lie entirely above the graph of $\ell_1$. In other words, if we draw a segment between two points in $\ell_1$, the intermediate points are closer to $p$ than the endpoints. Put more simply, the ball $B_g(p,\epsilon)$ is convex (at least for $\epsilon$ small enough).

Hopefully this makes sense. In my original answer, I thought it would be possible to do a similar argument with scale-$C^1$-control, but this turns out not to be the case. It might be possible to modify the scale-$C^2$-control to allow for less smooth level sets of $d$, but the assumption needs to be strong enough to ensure that the level sets are convex. At present, I'm not sure of a good way to relax scale-$C^2$-control, and this answer is more than long enough already, so I'll leave that alone.


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