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Let $(M,g)$ a Riemannian manifold and $X$ a Killing field on $M$. Then, we can define the function:

$$f(p) = \|X(p)\|^2$$

The claim is:

If $p$ is a critical point of $f$ then the flow of $X$ at $p$ is a geodesic.

My attempts:

$$Xf(p) = X(\|X(p)\|^2) = 2g(\nabla_XX(p),X(p)).$$

But then, once $p$ is a critical point then $$g(\nabla_XX(p),X(p)) = 0.$$

But if I am not mistaken the condition of $X$ Killing implies the same. Right?

How to proceed?

I must conclude that $\nabla_XX(p) = 0.$

L.F. Cavenaghi
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1 Answers1

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$0=\frac{1}{2}Vf=g(\nabla_VX,X)= -g(\nabla_XX,V)$ for any $V$

That is $\nabla_XX(p)=0$

As you pointed, $X|X|^2=0$ This implies that $|X|^2$ is constant along a flow That is if $p$ is critical, flow passing through $p$ is set of critical point

HK Lee
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  • ok! The fact is: $X$ Killing implies that $g(\nabla_XX,X) = 0$ always.

    On the otherside, there is a critical point. I have to show then that the flow pass through only by critical points.

    I did not understand your hint.

     – L.F. Cavenaghi Sep 05 '16 at 21:33
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    I add some about flow passing through critical point – HK Lee Sep 06 '16 at 02:27
  • I am so sorry. Could you be more explicit? I know that $X$ Killing implies that $Xf \equiv 0.$ But, how can I conclude that $\nabla_X X = 0$ in every point? – L.F. Cavenaghi Sep 06 '16 at 17:37
  • (1) $p$ is critical Then $\nabla_XX(p)=0$ (2) $X|X|^2=0$ iff $\frac{d}{dt} |X|^2 =0$ That is $|X|^2$ is constant along a flow If $|X|^2$ has a critical pt $p$ then points on flow of $p$ are critical (3) $\nabla_XX =0$ on flow of $p$ – HK Lee Sep 06 '16 at 17:48