Let $c_v, c_w$ be two geodesics starting at a point $p\in M$, where M is a nonpositively curved, complete, smooth Riemannian manifold. Say $c_v(\varepsilon) = \exp_p(\varepsilon v)$ and $c_w(\varepsilon) = \exp_p(\varepsilon w)$ for some tangent vectors $v,w\in T_p(M)$. Edit 3: With the counterexamples in the answers in mind, we should assume the geodesics will need to form a triangle $\triangle(p, c_v(1),c_w(1))$ whose side lengths are within some (nonzero) lower and upper bounds.

Let $L(\varepsilon)=\mathrm{dist}(c_v(\varepsilon),c_w(\varepsilon)$) be the distance between corresponding points $\varepsilon$ out on the geodesics.

One can use the nonpositive curvature to get convexity of $L$ and prove that if $L(\varepsilon)$ looks linear at any one point, i.e. $L(\varepsilon_0) = \varepsilon_0 L(1)$, then this in fact is true for all $\varepsilon$, not just $\varepsilon_0$. Consequently, the curvature of the $v,w$ plane at $p$ is $0$.

Does an approximate version of this hold? That is, if we know some $\varepsilon_0$ has $L(\varepsilon_0)$ "close to" $\varepsilon_0 L(1)$, does this mean the sectional curvature of the $u,v$ plane at $p$ is "close to" $0$?

This seems intuitive, at least if there's some kind of homogeneity or lots of symmetry, but I'm not very familiar with curvature, and can't seem to piece together a formal justification of this.

If anyone knows and can explain this, or has a good reference to look at that discusses this function $L$ in detail, I'd appreciate it. This seems like something that should be known or easy to someone in the area. I've been trying to learn this material, and have seen a few references talking about geodesic variations and $L$ near $0$, but I'm not sure if I'm getting anywhere. Of course, if it isn't true, I'd like to know why, or if it is true in slightly nicer spaces.


Edit To try to comment more on how I think an argument might go:

My guess is that the curvature as you move along the geodesics is nonincreasing, so the amount of convexity (i.e. 2nd derivative) of $L(\varepsilon)$ is nondecreasing. This, along with the convexity of $L$ provides some bounds on $L$ on $(0,\varepsilon_0)$. Specifically, it must be under the line through $(0,0)$ and $(\varepsilon_0, L(\varepsilon_0))$ in order to be convex at all, and above the quadratic through $(0,0)$, $(\varepsilon_0, L(\varepsilon_0))$, and $(1,L(1))$ in order to satisfy the 2nd derivative requirement. This seems to control the Riemannian angle and curvature at $p$.

In what spaces would we know the curvature is nonincreasing and that argument makes some sense?

Edit 2 The only answer below shows that this won't work as generally as I originally discussed this problem. So, I'm interested in when it does hold. I'm particularly interested in symmetric spaces, and my intuition is still firmly that this should be true in that setting.


No, you should not expect such an estimate. The assumption that $L(\epsilon_0)\approx \epsilon_0 L(1)$ can be interpreted in the sense that on scale $\epsilon_0$ and larger, the space looks flat on average. It does not control curvature pointwise; in particular, you can have very negative curvature on a scale $\delta\ll \epsilon_0$ around $p$.

Example. Consider metrics of the form $ds^2= e^{2u} (dx^2+dy^2)$. The curvature is nonpositive if and only if $\Delta u\ge 0$. Choose $u$ radially symmetric, i.e., $u =u(r)$ where $r=\sqrt{x^2+y^2}$. Specifically, $$ u(r)=\begin{cases} \ln \delta+\dfrac{r^2-\delta^2}{2\delta^2} ,\quad & 0<r\le \delta \\ \ln r ,\quad & r\ge \delta \end{cases} $$ It is easy to see that $ru'$ is increasing, hence $u$ is subharmonic and the surface is nonpositively curved. (Technically, the curvature is discontinuous at $r=\delta$, which can be fixed by cooking up a more complicated subharmonic function that matches the logarithm at $u=\delta$ with its higher derivatives at $\delta$.) Since $u$ is radially symmetric, geodesics starting at the origin are lines.

The curvature is zero for $r>\delta$; indeed, this part of the surface can be mapped by $z\mapsto z^2/2$ (in complex notation) isometrically onto the annulus $\{\delta^2/2< r< \infty\}$ covered twice. In particular, $L(\epsilon)$ is an affine function of $\epsilon$ there, thus asymptotic to linear on large scale.

But near $0$, the curvature is massively negative, due to $\Delta u = 2r/\delta^2$. Another way to see this is to observe that the perimeter of geodesic disk of radius $R$ is $\approx 4\pi R$ for large $R$ (due to the annulus being doubly-covered). By the Gauss-Bonnet theorem, the integral curvature of this surface is equal to $-2\pi$. Which is also consistent with the fact that the surface is made from a cone with the angle excess of $2\pi$ by smoothing out the vertex.

  • $\begingroup$ Ah, nice counterexample. Do you think it helps if we bound the lengths to avoid arbitrarily large or small triangles? For example, only look at side length 1 and compare at halfway along them? This is more what I had in mind. Or, do you have any insight into why it might be true for particularly nice spaces? Symmetric spaces in particular. $\endgroup$ – matt Dec 28 '13 at 13:01

The point of this answer is to show that even in the nicest possible Riemannian manifolds, approximate linearity of $L(\epsilon)$ does not imply that the sectional curvature $K(p, Span(u,v))$ is close to zero. Namely, start with your favorite nonpositively curved Riemannian manifold $M$ (say, hyperbolic plane, if you wish) and a fixed point $p\in M$ such that the sectional curvature $K(p,\sigma)=k$ is nonzero for some tangent plane $\sigma\subset T_pM$. Suppose for a moment that the (unit) vectors $u, v\in \sigma$ are equal. Then the function $L(\epsilon)$ is identically zero. Now, hold the vector $u$ and consider a sequence of unit vectors $v_i\in \sigma$ converging to $v$. Let $L_i(\epsilon)$ denote the corresponding distance functions between the geodesics $\exp_p({\mathbb R}_+ u), \exp_p({\mathbb R}_+ v_i)$. Then for each fixed $\epsilon_0>0$, for all sufficiently large $i$, the restriction of the function $L_i$ to the interval $[0,\epsilon_0]$ is close to zero and, hence, is approximately linear. On the other hand, the sectional curvature of the tangent plane through $u, v_i$ is of course, still equal to $k\ne 0$, independently of $i$.

  • $\begingroup$ Please see my comment on @PostNoBulls answer. I interpret your answer similarly as indicating that bounds on the side lengths are needed. This avoids sequences of triangles becoming degenerate as in your answer, or "too global" as in his answer. My intuition is that avoiding those and/or having enough homogeneity/symmetry will make it work. Constant curvature spaces are a bit silly of me to ask this question in. Maybe something like $\mathbb{H}^2 \times \mathbb{R}^2$ makes more sense as a potential example. $\endgroup$ – matt Jan 6 '14 at 15:35

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