= Killing equation
{c}
{title2=$\nabla_{(\mu}K_{\nu)}=0$}
The <Killing equation> tests whether a candidate <vector field> generates a local <isometry>, without needing to solve for its <flow of a vector field>. For the <Levi-Civita connection> of a <metric tensor> $g$, it is
$$
\nabla_{(\mu}K_{\nu)}=\tfrac12(\nabla_\mu K_\nu+\nabla_\nu K_\mu)=0,
\qquad K_\nu=g_{\nu\rho}K^\rho.
$$
Parentheses mean averaging over the two index orders. The equation says the symmetric part of the <covariant derivative> vanishes, not that the entire <covariant derivative> vanishes. The <Lie derivative> of the <metric tensor> gives the coordinate form
$$
(\mathcal L_Kg)_{\mu\nu}
=K^\rho\partial_\rho g_{\mu\nu}
+g_{\rho\nu}\partial_\mu K^\rho+g_{\mu\rho}\partial_\nu K^\rho
=\nabla_\mu K_\nu+\nabla_\nu K_\mu.
$$
The last two terms account for how <tangent vectors> move with the field; checking only $K^\rho\partial_\rho g_{\mu\nu}$ misses this. See https://arxiv.org/pdf/gr-qc/9712019#page=147[Carroll, section 5, equations (5.33) and (5.41)-(5.42)].
For the <Euclidean Killing vector field> $K=-y\partial_x+x\partial_y$, the <Euclidean metric> in <Cartesian coordinates> makes <covariant derivatives> ordinary <partial derivatives>. Its diagonal equations give $\partial_xK_x=\partial_yK_y=0$, and its mixed equation gives $\partial_xK_y+\partial_yK_x=1-1=0$. Thus a varying <vector field> can preserve a constant <metric tensor>. In contrast, $K=x\partial_x+y\partial_y$ gives $(\mathcal L_Kg)_{xx}=2$ and fails the <Killing equation>, despite the constant metric components.
If $K=\partial_z$ is a <coordinate basis> vector, its components are constant and the coordinate form reduces to $\partial_zg_{\mu\nu}=0$. Hence independence of the metric components from $z$ supplies a <Killing vector field>. This shortcut applies to that coordinate generator, not to an arbitrary varying <vector field>.
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