# There is it an easier proof that if $g(X,X)$ is a constant $\nabla_X X=0$?

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.

• Exactly the same as this? – user10354138 Jun 9 at 14:40
• Oh yes, it's that. Would you say that I should erase the question for not having it duplicated? – elescararriba Jun 9 at 14:46
• 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. – user10354138 Jun 9 at 14:49
• ok, thanks for the reference! – elescararriba Jun 9 at 14:50