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I was not able to do the following exercise from Manfredo's Riemannian Geometry:

Chapter 3, Exercise 5 Let $M$ be a Riemannian manifold and $X\in \mathfrak{X}(M)$. Let $p\in M$ and let $U\subset M$ be a neighborhood of $p$. Let $\varphi:(-\varepsilon,\varepsilon)\times U\to M$ be a differentiable mapping such that for any $q\in U$ the curve $t\mapsto \varphi(t,q)$ is a trajectory of $X$ passing through $q$ at $t=0$ ($U$ and $\varphi$ are given by the fundamental theorem for ordinary differential equations, Cf. Theorem 2.2). $X$ is called a Killing field (or an infinitesimal isometry) if, for each $t_0\in (-\varepsilon,\varepsilon)$, the mapping $\varphi(t_0,\,):U\to M$ is an isometry. Prove that:

(a) (...)

(d) $X$ is a Killing field $\iff \langle \nabla_YX,Z\rangle+\langle \nabla_ZX,Y\rangle=0$, for all $Y,Z\in \mathfrak{X}(M)$ (the equation above is called the Killing equation).

Then Manfredo gives a hint for proving $(\Rightarrow)$ and therefore I suppose that $(\Leftarrow)$ must be the "easy" part. However, I could not prove it. In $(\Leftarrow)$, I want to show that $$\langle d(\varphi_{t_0})_q\cdot u,d(\varphi_{t_0})_q\cdot v\rangle=\langle u,v\rangle,$$ for all $t_0\in (-\varepsilon,\varepsilon)$, $q\in U$ and $u,v\in T_qM$. But I am not being able to relate this with the Killing equation. I know that $\frac{d}{dt}\varphi(t,q)=X(\varphi(t,q))$ but $d(\varphi_t)_q$ is not exactly $\frac{d}{dt}\varphi(t,q)$...

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A Killing vector field is one satisfying $\mathcal{L}_X g=0.$ Since \begin{align} (\mathcal{L}_Xg)(Y,Z)&=Xg(Y,Z)-g([X,Y],Z)-g(Y,[X,Z]) \\ &=g(\nabla_XY,Z)+g(Y,\nabla_XZ) -g([X,Y],Z)-g(Y,[X,Z]) \\ &=g(\nabla_YX,Z)+g(Y,\nabla_ZX), \end{align} where in the second line we used the fact that the Levi-Civita connection is Riemannian and in the third that it is torsion free, we have your equivalence.

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  • $\begingroup$ Sorry @AloizioMacedo, but what does $\mathcal{L}_Xg$ mean? $\endgroup$ – Ders Nov 4 '17 at 16:56
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    $\begingroup$ @AndersonFelipeViveiros It is the Lie derivative of a tensor field. $\endgroup$ – Aloizio Macedo Nov 4 '17 at 17:00
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    $\begingroup$ @AndersonFelipeViveiros If you're not familiar with this, look at the definition. You are computing the derivative w.r.t time of the pull-back of $g$ (in this case) along the flow of $X$. This will be $0$ if and only if the pull-back of $g$ is constant in time, which will be true if and only if the flow pull-backs the metric at $\varphi(t,x)$ to $g_x$ itself (since the flow at time $0$ is the identity), i.e., if $\varphi(t, \cdot)$ is an isometry for all $t$. $\endgroup$ – Aloizio Macedo Nov 4 '17 at 17:10
  • $\begingroup$ do you recommend a more detailed text in Lie derivatives? In the link you've sent me, are the sings in Axiom 3 wrong (shouldn't they be minus instead of plus, at least it seems you have used minus above...)? $\endgroup$ – Ders Nov 4 '17 at 21:54
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    $\begingroup$ @AndersonFelipeViveiros If I recall correctly, Warner's book talks about Lie derivative. The sign is correct: $X g(Y,Z)=\mathcal{L}_X(g(Y,Z))$, so if you pass everything with minus to the left side of the equation above you will have "Axiom 3" of wikipedia. $\endgroup$ – Aloizio Macedo Nov 5 '17 at 2:38
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The metric in this case is given by

$$\left<(d\phi(t)) u, (d\phi(t)) v\right>=\left<(d\phi(t_0)) u, (d\phi(t_0)) v\right> +O(t^2)$$

and hence

$$\lim_{t\to 0}\frac{\left<(d\phi(t)) u, (d\phi(t)) v\right>-\left<(d\phi(t_0)) u, (d\phi(t_0)) v\right>}{t}=0$$

i.e.

$$\left({\frak{L}}_Xg\right)(u,v)=0$$

Then you can use the answer of Aloizio Macedo.

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