Let $(M^n,g)$ be a smooth Riemannian manifold. Consider the square of the distance function $$dist^2\colon M\times M\to \mathbb{R}$$ given by $(x,y)\mapsto dist^2(x,y)$. It is easy to see that this function is infinitely smooth near the diagonal.

Now fix a point $a\in M$. Consider the Taylor series of $dist^2$ at $(a,a)$ in some coordinate system (say normal). Can one compute explicitly its coefficients up to the third order?

  • 6
    $\begingroup$ All the coefficients of degree 3 vanish. $\endgroup$ – Anton Petrunin Aug 25 '15 at 7:32

Fix a point $x_0 \in M$. Then let $x = \exp_{x_0}(t v)$ and $y=\exp_{x_0}(t w)$, with $v,w \in T_{x_0}M$.

Then we have the following formula for the distance squared between two geodesic emanating from $x_0$

$$ d^2(\exp_{x_0}(t v),\exp_{x_0}(t w)) = |v-w|^2t^2-\frac{1}{3}R(v,w,w,v) t^4 + O(t^5)$$

whre $R$ is the Riemann curvature tensor. From here you can derive an expression where you follow the two geodesics for different times (just rescale $w \to s/t w$), that is

$$ d^2(\exp_{x_0}(t v),\exp_{x_0}(s w)) = |v|^2t^2 +|w|s^2 -2 g(v,w) ts -\frac{1}{3}R(v,w,w,v) s^2 t^2 + O(|t^2+s^2|^{5/2})$$

Take now $|v| = |w|=1$, then $(t,v)$ and $(s,w)$ (both $\in [0,\epsilon) \times \mathbb{S}_{x_0}^{n-1}$) are polar coordinates of $x,y$. More explicitly, setting $t = d(x_0,x)$ and $s = d(x_0,y)$, you have

$$ d^2(x,y) = t^2 + s^2 - 2\cos\theta t s - \tfrac{1}{3} K_\sigma (1-\cos\theta^2)t^2 s^2 + O(|t^2+s^2|^{5/2})$$

where $\cos\theta = g(v,w)$ is the angle between the two vectors and $K_\sigma$ is the sectional curvature of the plane $\sigma = \mathrm{span}(v,w)$.

As you can see, this can be used effectively as a purely metric definition of sectional curvature, with no connection or covariant derivative whatsoever.

| cite | improve this answer | |
  • $\begingroup$ Great! Thanks a lot. Is there a standard reference? $\endgroup$ – MKO Aug 25 '15 at 8:48
  • $\begingroup$ In the first formula shouldn't it be $|v-w|^2$? $\endgroup$ – user35593 Aug 25 '15 at 8:49
  • 2
    $\begingroup$ I saw this formula in the book "Old and new" by Villani (Eq. 14.1 cedricvillani.org/wp-content/uploads/2012/08/preprint-1.pdf). There is no proof there and the proof I've been able to write requires a not-so-short computation in normal coordinates (more precisely, one finds explicitly the tangent vector of the geodesic passing through $x$ and $y$ in normal coordinates around $x_0$). I would be happy to know of any other reference where that formula appears (maybe with a more conceptual proof that the one I described). $\endgroup$ – Raziel Aug 25 '15 at 8:53
  • $\begingroup$ By the way, if you are interested in the proof I have it typed down already, but maybe it's a bit too long for an answer. Just tell me if you're interested. $\endgroup$ – Raziel Aug 25 '15 at 8:59
  • 1
    $\begingroup$ It's better to consider the function $\delta(x):= d(x_0,x)$. This satisfies the Eikonal equation $|\nabla\delta| = 1$ (at points where it is smooth). As a consequence its gradient flow trajectories are unit-speed geodesics emanating from $x_0$. If you consider the squared distance you still have geodesics, but not parametrized by constant speed. The picture is basically as in $\mathbb{R}^n$, what changes in general Riemannian manifolds is the occurrence of cut points $x \neq x_0$ where the distance fails to be smooth (or even differentiable). $\endgroup$ – Raziel Feb 17 '17 at 6:37

The function $dist^2$ satisfies a Hamilton-Jacobi equation. Using this you can find its Taylor expansion at $(a,a)$ up to any order. In Appendix A of this paper I use this approach to find the degree 4 Taylor expansion of this function; see Eq. (A.17) in the paper. The procedure used in this appendix can be used to find the degree $6$ expansion as well, though the formulas become increasingly more complicated. I should mention that this approach is also used in

B. DeWitt: The Global Approach to Quantum Field Theory, Oxford University Press, 2003,

pages 281-282, though the physicists' notation may be a bit hard to follow.

| cite | improve this answer | |
  • $\begingroup$ Thanks a lot for the reference. That's how one can prove that the remainder in my first formula is indeed O(t^6) and not O(t^5)! Btw, in the second formula after (A.10), $[B_\ell]_1$ should be $[A_\ell]_1$, right? $\endgroup$ – Raziel Aug 25 '15 at 16:12
  • $\begingroup$ @Raziel Yes, you're right. $\endgroup$ – Liviu Nicolaescu Aug 25 '15 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.