In Cartesian coordinates on <Euclidean space>, the <Levi-Civita connection> and the <Riemann curvature tensor> vanish. The <second covariant derivative of a Killing vector> therefore gives
$$
\partial_\mu\partial_\rho K^\nu=0.
$$
Every first partial derivative is constant on the connected space $\mathbb R^n$. Integrating once more gives $K^\nu=A^\nu+M_\mu{}^\nu x^\mu$, with constant coefficients. The <Killing equation> now becomes
$$
0=\partial_\mu K_\nu+\partial_\nu K_\mu=M_{\mu\nu}+M_{\nu\mu},\qquad
M_{\mu\nu}=M_\mu{}^\kappa\delta_{\kappa\nu}.
$$
Thus the coefficient is an <antisymmetric matrix>. Conversely, any such constant coefficients satisfy the <Killing equation>. The complete <Euclidean Killing vector field> is therefore
$$
\boxed{K=A^\nu\frac{\partial}{\partial x^\nu}
+M_\mu{}^\nu x^\mu\frac{\partial}{\partial x^\nu},\qquad M^T=-M.}
$$
The $n$ translation parameters and $n(n-1)/2$ rotation parameters give $n(n+1)/2$ independent <Killing vector fields>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 309 / 2 / iii / Solution

{c}
{title2=$K(x)=A+Bx,\quad B^T=-B$}

Every <Killing vector field> on <Euclidean space> is the sum of a constant translation and a constant infinitesimal rotation. The <second covariant derivative of a Killing vector> vanishes in flat space, so its components are affine; the <Killing equation> makes their linear coefficient an <antisymmetric matrix>.

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 Incoming links: Euclidean Killing vector field