Are covariant derivatives of Killing vector fields symmetric?

I'm reading the Lecture Notes on General Relativity by Matthias Blau, and in section 9.1 (point 1) he writes:

Let $K^\mu$ be a Killing vector field, and ${x^\mu(\tau)}$ be a geodesic. Then the quantity $$Q_K=K_\mu\dot{x}^\mu$$ is constant along the geodesic.

Now, in equation (9.2) he writes $$\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu=\frac{1}{2}(\nabla_\nu{K}_\mu+\nabla_\mu{K}_\nu)\dot{x}^\mu\dot{x}^\nu$$ which vanishes due to Killing's equation.

The thing I don't really understand is why ${\nabla_\nu{K}_\mu-\nabla_\mu{K}_\nu=0}$. I can see that because ${\mu,\nu}$ are dummy indices, then \begin{align}\nabla_\nu{K}_\mu\dot{x}^\mu\dot{x}^\nu&=\nabla_\mu{K}_\nu\dot{x}^\nu\dot{x}^\mu\\&\Longrightarrow\,\nabla_\nu{K}_\mu-\nabla_\mu{K}_\nu=0 \tag{1}\end{align} Is this argument valid? If so, does $(1)$ hold in general for Killing vectors? As I'm almost sure it doesn't, despite the title of my question (all Killing vector fields would be gradient fields), does it happens only along the geodesic? If so, what interpretation would this condition have?

• He does not say that antisymmetric part must vanish, in fact it does not in general. What he says is that only symmetric part contributes to the expression. Indeed, if you have a vector, $V^\nu$, and a tensor $T_{\mu\nu}$, then only symmetric part of $T$ contributes to $T_{\mu\nu} V^\mu V^\nu$. In his case, $T=\nabla K$ and $V^\nu=\dot{x}^\nu$
– John
Apr 26, 2014 at 8:42

A Killing vector $K^\mu$ is defined as a vector Lie derivative of metric along which vanishes. $$\mathcal{L}_K g_{\mu\nu}=0, \quad \Longrightarrow \nabla_\mu K_\nu+\nabla_\nu K_\mu=0.$$ I guess there is no need to write derivation of this equation explicitly as you can find it everywhere.
Concerning you question about the antisymmetrisation. Lets start with the expression \begin{align}\nabla_\nu{K}_\mu\dot{x}^\mu\dot{x}^\nu&=\nabla_\mu{K}_\nu\dot{x}^\nu\dot{x}^\mu\\&\Longrightarrow\,(\nabla_\nu{K}_\mu-\nabla_\mu{K}_\nu)\dot{x}^\mu\dot{x}^\nu=0 \tag{1}\end{align} In this form the last equation is trivial as we contract an antisymmetric tensor with a symmetric one $\dot{x}^\mu\dot{x}^\nu$. However, you can not derive from here the equation $$\nabla_\mu K_\nu-\nabla_\nu K_\mu=0$$ since this is true only upon contraction with a symmetric tensor.
Hence, the Killing equation $\nabla_\mu K_\nu+\nabla_\nu K_\mu=0$ that was used by Blau actually comes from the definition at the beginning of this post. The symmetrisation in the expression $\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu$ as was already mentioned by John comes from contracting with the symmetric tensor $\dot{x}^\nu\dot{x}^\mu$. In details: \begin{aligned} \nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu&=\frac12(\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu+\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu)\\ &=\frac12(\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu+\nabla_\alpha{K}_\beta\dot{x}^\alpha\dot{x}^\beta)\\ &=\frac12(\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu+\nabla_\alpha{K}_\beta\dot{x}^\beta\dot{x}^\alpha)\\ &=\frac12(\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu+\nabla_\mu{K}_\nu\dot{x}^\nu\dot{x}^\mu)\\ &=\frac12(\nabla_\nu{K}_\mu+\nabla_\mu{K}_\nu)\dot{x}^\nu\dot{x}^\mu). \end{aligned} Here in the second line I just renamed the indices, in the third the $\dot{x}^\alpha$ and $\dot{x}^\beta$ were permuted and then I renamed the indices again.
• One question: the condition of being a Killing field implies that for vector fields $X,Y$ we will have $g(\nabla_X K, Y) + g(X,\nabla_Y K)=0$. In a Riemannian manifold (i.e. signature all +), this obviously implies picking an orthonormal basis $e_{\mu}$ and taking $X = e_\mu$, $Y = e_\nu$, that $\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0$. But in a pseud-Riemannian manifold, we have $g(e_\mu,e_\nu)=\pm \delta_{\mu\nu}$. How this doesn't affect the plus sign in Killing's equation when converting to the components version?