# Coordinate-free proof of the constancy of the scalar product of a Killing vector with the geodesic tangent vector along the geodesic

I would like to know if the following proof of the constancy of the scalar product of a Killing vector with the geodesic tangent vector along the geodesic is correct. I already found a coordinate-dependent proof but my rationale seems to be rather different from the coordinate-dependent proof.

Be $$K$$ a Killing vector and $$\gamma$$ the geodesic and $$\dot{\gamma}$$ its tangent vector. $$g(\cdot, \cdot)$$ is the scalar product based on the metric $$g$$. $$\nabla_V X$$ is the covariant derivative.

$$d g(K,\dot{\gamma})(\dot{\gamma}) = \nabla_\dot{\gamma}g( K,\dot{\gamma}) = g( \nabla_\dot{\gamma}K, \dot{\gamma}) + g( K, \nabla_\dot{\gamma}\dot{\gamma}) = g( \nabla_\dot{\gamma}K, \dot{\gamma})$$

where the second term vanishes because of $$\nabla_\dot{\gamma}\dot{\gamma}=0$$ along the geodesic. The first term also vanishes since I can conclude from the coordinate-independent form of the Killing vector equation:

$$g(\nabla_V K,W) + g(V,\nabla_W K)=0$$

that

$$g(\nabla_\dot{\gamma}K,\dot{\gamma})=0$$

So $$d g(K,\dot{\gamma})(\dot{\gamma})=0$$

which shows the constancy of the scalar product of a Killing vector with the geodesic tangent vector along the geodesic. Is this derivation correct?

• it looks alright, which step are you feeling fishy? – chichi Aug 19 at 13:46
• @chichi : the sources on differential geometry I dispose of only attest : $\nabla g(X,Y) = g(\nabla X, Y) + g(X, \nabla Y)$ instead of $\nabla_U g(X,Y) = g(\nabla_U X, Y) + g(X, \nabla_U Y)$ so I don't know if the latter is correct. – Frederic Thomas Aug 19 at 13:51
• I think the latter equation you wrote is the definition of covariant derivative, no? – chichi Aug 19 at 14:03
• Well, I would say, at least in my differential geometry book the definition is $\nabla_U (X\otimes Y) = \ldots$ (plus some addtional rules), I guess since $g$ already is an additional structure. – Frederic Thomas Aug 19 at 14:14

I'll give another proof, just for fun. We know that in a pseudo-Riemannian manifold $$(M,g)$$, geodesics are critical points of the energy functional $$E[\gamma] = \frac{1}{2}\int_I g_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))\,{\rm d}t$$Since $$K$$ is Killing, the flow of $$K$$ (consisting of isometries) leaves the Lagrangian we're integrating invariant. So Noether's Theorem says that the charge $$\mathscr{J}(x,v) = \mathbb{F}L(x,v)\left(\frac{{\rm d}}{{\rm d}s}\bigg|_{s=0} \Phi_{s,K}(x)\right) = \mathbb{F}L(x,v)K_x= \frac{\rm d}{{\rm d}t}\bigg|_{t=0} \frac{1}{2}g_x(v+tK_x,v+tK_x) = g_x(v,K_x)$$is constant along geodesics. But $$\mathscr{J}(\gamma(t),\dot{\gamma}(t)) = g_{\gamma(t)}\big(\dot{\gamma}(t), K_{\gamma(t)}\big)$$, and so we're done. Here $$(x,v) \in TM$$ (which is to say that $$x \in M$$ and $$v \in T_xM$$) and $$\mathbb{F}L$$ denotes the fiber derivative of $$L$$.
• Thank you for the interest. Actually I don't know what the fiber derivative is. Furthermore, $L(x,v)$ seems to be a Lagrangian, is actually the energy functional $E[\gamma]$ meant by this? And finally, the expression $(\frac{d}{ds}|_{s=0} \Phi_{s,K}(x))$ is this an argument of $\mathbb{F}L$ or is it supposed to be multiplied to $\mathbb{F}L$ ? Is $\Phi_{s,K}$ the flow of $K$? I am not familiar with this notation. – Frederic Thomas Oct 3 at 11:05
• 1) the fiber derivative of $L$ is $\mathbb{F}L(x,v)w = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} L(x,v+tw)$; 2) yes; 3) argument, see 1); 4) it is the flow. – Ivo Terek Oct 3 at 13:22