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In the book General Relativity for matematicians by R.K.Sachs and H.Wu there is a problem which say:

Let $(M,g)$ be a Lorenzian manifold,$\nabla$ the Levi-Civita conexion, $f\colon M\to \mathbb{R}$ a function and $X$ a vector field physically equivalent to $df$. Prove that if $g(X,X)$ is a constant then $\nabla_X X$=0.

I got a proof using that since $\nabla$ is compatible with $g$, then: $$Xg(Y,Z)=g(\nabla_X Y,Z)+g(Y,\nabla_X Z)$$ If I write $X$ in the place of $Y$ I get:

$$Xg(X,Z)=g(\nabla_X X, Z)+g(X,\nabla_XZ)$$

And since $df$ is the 1-for physically equivalent to $X$, then $g(X,Z)=(df) Z$. Hence, $X((df)Z)=g(\nabla_X X,Z)+ (df)\nabla_X Z$. So I have left to prove is that $X((df)Z)= (df)\nabla_X Z$ because if this is true , then $g(\nabla_X X,Z)=0$ for all $Z$ and since $g$ is non degenerate, $\nabla_X X=0$.

Since $\nabla$ is de Levi-Civita conexion, then:

$$X(Zf)-Z(Xf)=(df)[X,Z]=df\nabla_X Z-df\nabla_Z X$$

And since $Xf=df X=g(X,X)=$ is a constant, $Z(Xf)=0$ and so I have to prove that $df\nabla_Z X = 0$. Which I can get because:

$$Z(g(X,X))=g(\nabla_ZX.X)+g(X,\nabla_ZX)=2 df\nabla_ZX$$.

And since $g(X,X)$ is a constant $ df\nabla_ZX=0$.

I think that there should be a more straightforward way to solve it. Does anybody knows an alternative proof? Also, I think that I haven't used that $(M,g)$ is a Lorenzian manifold so I'm not sure if I made some mistake or if it is true for semi-Riemannian manifolds in general.

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    $\begingroup$ Exactly the same as this? $\endgroup$ – user10354138 Jun 9 at 14:40
  • $\begingroup$ Oh yes, it's that. Would you say that I should erase the question for not having it duplicated? $\endgroup$ – elescararriba Jun 9 at 14:46
  • $\begingroup$ Well, that post only mentions the Riemannian case but the proof works for semi-Riemannian too. I don't think it is necessary to delete this question. $\endgroup$ – user10354138 Jun 9 at 14:49
  • $\begingroup$ ok, thanks for the reference! $\endgroup$ – elescararriba Jun 9 at 14:50

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