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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?

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    $\begingroup$ All the coefficients of degree 3 vanish. $\endgroup$
    – Anton Petrunin
    Commented Aug 25, 2015 at 7:32

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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.

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  • $\begingroup$ Great! Thanks a lot. Is there a standard reference? $\endgroup$
    – asv
    Commented Aug 25, 2015 at 8:48
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    $\begingroup$ 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
    Commented Aug 25, 2015 at 8:53
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    $\begingroup$ 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
    Commented Aug 25, 2015 at 8:59
  • $\begingroup$ Certainly I would be interested to see a proof. $\endgroup$
    – asv
    Commented Aug 25, 2015 at 9:09
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    $\begingroup$ 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
    Commented Feb 17, 2017 at 6:37
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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.

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  • $\begingroup$ 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
    Commented Aug 25, 2015 at 16:12
  • $\begingroup$ @Raziel Yes, you're right. $\endgroup$
    – Liviu Nicolaescu
    Commented Aug 25, 2015 at 20:09

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