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What is the definition of Killing vector field?. The one my professor told me is : a smooth vector field $V$ on $M$ is called a Killing vector field for $g$ if the flow of $V$ acts by isometries of $g$.

So what does it mean by the flow of $V$ acts by isometries?

I suspect this definition is equivalent to saying the lie derivative of $g$ along a vector field is 0, then the vector field is a Killing field. But first I need to figure out the meaning of the first definition.

$g$ is just the Riemannian Metric.

janmarqz
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1 Answers1

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Imagine an ant-grid on the manifold. Each ant can determine the distance to every other ant on the manifold by means of the metric. Let each ant begin moving according to the vector field: At each moment, the vector at each ant's location determines the ant's speed and direction of motion. If the ant-to-ant distances do not change, the field is a Killing field.

Brian Tung
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  • Thanks. I guess my real question is like this. Let $\theta_t$ denote the flow. Then this $theta_t$ is a map from M to M. So we can pullback covariant tensor field by this map. Of course, here the covariant tensor field is g, the metric. Now do we require for every t, the pullback is isometry or we only require the lie derivative is 0. If we have the former case, then the lie derivative is clearly 0. –  Dec 01 '15 at 03:27
  • I think the two descriptions are equivalent. I can't think of a counterexample. – Brian Tung Dec 01 '15 at 05:22