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Let $V$ be a Killing vector field and let $s \longmapsto x^i(s)$ be a curve such that

$$\dot{x} \enspace \equiv \enspace \frac{dx^i}{dx}(s) \enspace = \enspace V^i\big(x(s)\big)$$

Show that $x^i(s)$ are geodesics iff

$$\nabla_k \, g(V,V) \enspace = \enspace 0$$

along the curve $x(s)$.


I approached it as follows:

$x^i(s)$ is geodesic if it satisfies the geodesic equation, i.e.

$$\ddot{x}^i + \Gamma^i_{jk} \,\dot{x}^j \, \dot{x}^k \enspace = \enspace 0$$

I calculated

$$ \begin{align} \ddot{x}^i + \Gamma^i_{jk} \, \dot{x}^j \, \dot{x}^k \enspace &= \enspace 0 \\ {} \\ &= \enspace \tfrac{d}{ds} \dot{x}^i + \Gamma^i_{jk} \, \dot{x}^j \, \dot{x}^k \\ {} \\ &= \enspace \tfrac{d}{ds} V^i\big(x(s)\big) + \Gamma^i_{jk} \, \dot{x}^j \, \dot{x}^k \\ {} \\ &= \enspace \frac{\partial V^i}{\partial x^k} \, \frac{dx^k}{ds} + \Gamma^i_{jk} \, \dot{x}^j \, \dot{x}^k \\ {} \\ &= \enspace \partial_k V^i \, \dot{x}^k + \Gamma^i_{jk} \, \dot{x}^j \, \dot{x}^k \\ {} \\ &= \enspace \Big( \nabla_k V^i \Big) \, \dot{x}^k \\ {} \\ &= \enspace V^k \, \nabla_k V^i \tag{1}\\ {} \\ &= \enspace 0 \end{align} $$

However, the term $\nabla_k \, g(V,V)$ should yield

$$ \nabla_k \, g(V,V) \enspace = \enspace 2 \cdot g_{ij} V^j \, \nabla_k V^i \enspace = \enspace 2 \cdot V_i \, \nabla_k V^i \enspace = \enspace 0$$

$$\Longrightarrow \quad V_i \, \nabla_k V^i \enspace = \enspace 0 \tag{2}$$

What went wrong in this calculation? How does the discrepancy arise? Is the approach just flawed or is there an additional step that somehow converts equation (1) into equation (2) and vice versa?

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Since $V$ is a Killing vector field, $\nabla_iV_j=-\nabla_jV_i$. Using this in equation 2, the desired relation follows.

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    $\begingroup$ Argh surely! The one condition for $V$ that is explicitly required to hold and I did not think of it. I am an idiot. ^^ Thank you so much for your answer! $\endgroup$ – Octavius Mar 24 at 18:29

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