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Several times, I've seen statements along the lines of "$\nabla_X Y=\nabla_Y X$ because $X$ is a Killing vector field." One example I found on Stack Exchange is here.

I have yet to understand why such arguments work, and I particularly don't understand what role the Killing structure plays in the relationship.

Could someone please explain this symmetry?

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Rawlrus Rumbleroar
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1 Answers1

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This is wrong. Taking $\mathbb{R}^2$ and the Euclidean metric, let $X=\frac{\partial}{\partial x}$ and $X$ is obvious Killing since the generated flow $\Phi(t,x,y)=(x+t,y)$ are isometric. Let $Y=x\frac{\partial}{\partial x}$, then $$D_X(Y)=\frac{\partial x}{\partial x}\frac{\partial}{\partial x}+xD_{\frac{\partial}{\partial x}}(\frac{\partial}{\partial x})=\frac{\partial}{\partial x}$$
However $$D_Y(X)=xD_\frac{\partial}{\partial x}(\frac{\partial}{\partial x})=0$$

Xipan Xiao
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  • I apologize. I guess I wasn't very clear. I'm looking for a reason why arguments such as the one linked in the question work, when the general case is not true. – Rawlrus Rumbleroar Dec 27 '14 at 22:02
  • @RawlrusRumbleroar I suspect the answer is that (at least as far as I can tell) the argument in the linked question simply doesn't work. – mollyerin Apr 17 '15 at 07:40