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Given a Riemannian manifold $ (M,g) $, in a neighborhood of a fixed point $ x $, the distance function $ d^2(x,\cdot) $ is smooth. In coordinates, the taylor series starts off like: $$ d^2(x,y) = g_{ij} (x-y)^i (x-y)^j + c_{ijk} (x-y)^i (x-y)^j (x-y)^k + \cdots $$ What are the $ c_{ijk} $? I am pretty sure they are related to the curvature, but I haven't been able to find a quick answer.

user87448
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  • Surely the $d^{\color{red}2}$ should mean that the odd-order terms vanish. I'd expect some curvature to show up in the fourth-order terms. – Ted Shifrin Jul 23 '13 at 19:08
  • Hi ted, that's not true because d itself is not smooth. For example $ \sqrt{x+x^2} $. – user87448 Jul 23 '13 at 20:04
  • Well, I admit I was thinking about normal coordinates. Riemann thought about this, but more in terms of the metric tensor itself. – Ted Shifrin Jul 23 '13 at 20:32
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    I think you will not get any clean formula for $c_{ijk}$ without choosing a nice coordinate system (normal). Indeed, on the interval $(-1/3,1/3)$ with metric $(1+2x)^2,dx^2$ we have $d^2(0,x)=(x+x^2)^2=x^2+2x^3+x^4$. Here $c=2$ has no relation to curvature -- after all, the manifold is isometric to a line segment. – 40 votes Jul 23 '13 at 23:03
  • Actually by symmetry you should have $c_{ijk} = 0$. The next order term is the fourth-order one. – a06e Jan 01 '22 at 13:15
  • I'd be interested in seeing the fourth-order terms ... though. – a06e Jan 01 '22 at 13:18

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