Is the orthogonal distribution of a Killing vector field always involutive?

Question

Let $M, g$ be a semi-Riemannian manifold and $K$ a Killing vector field on $M$. Let $D = \{X \in \Gamma(TM) | g(X, K) = 0\}$ be the orthogonal distribution to $K$. Does $D$ have to be involutive, i. e. $\forall X, Y \in D: [X, Y] \in D$ (or equivalently by Frobenius' theorem $D$ is integrable).

I believe this not true in general but haven't been able to come up with a counterexample. (apparently a counterexample would need at least 4 dimensions since in dimension $\leq 3$ you can always find locally orthogonal coordinates.)

Answer 1

In $\mathbb R^3$, $K=\partial_z + x\partial_y - y\partial_x$, $D$ is the kernel of $\alpha=g(K,\cdot)=dz + x\,dy-y\,dx$, and we have $\alpha\wedge\,d\alpha=2dz\wedge dx\wedge dy\neq0$, so $D$ is not involutive.

Answer 2

On $S^3 \subset \Bbb C^2$, the vector field $X$ defined by $$X_{(z, w)} = i(z, w) = (iz, iw)$$ is a Killing field for the standard metric, but its orthogonal distribution $$\{X\}^{\perp} = \operatorname{span}\{(w, -z), (iw, -iz)\}$$ is the standard contact distribution on $S^3$ and in particular is nonintegrable.