Special Killing Vector Fields - MathOverflow

Special Killing Vector Fields

2011-03-02T05:52:25Z

Consider $(M^{n},g)$ to be a Riemannian manifold and suppose that $X$ is a smooth non-trivial Killing vector field on $M$. Away from the zeros of $X$ we have a natural distribution $D$ of $(n-1)$-planes defined so that $D_p$ is orthogonal to $X_p$. If the distribution $D$ is (completely) integrable then it is straightforward to verify that the one form $\omega$ defined by $$\omega(\cdot )=\frac{1}{g(X,X)} g(X, \cdot).$$ is closed (away from $\lbrace X=0\rbrace$). Moreover, the converse also holds.

Examples in $\mathbb{R}^n$ with the euclidean metric include the the translations along the $x_i$-axis, $T_i$ and rotations around the $x_i$-axis, $R_i$. The Killing fields $T_i+R_i$ are non-examples.

My question is whether this concept already has a name and where it might appear in the literature.

Answer by Giuseppe for Special Killing Vector Fields

2011-03-02T18:03:31Z

Mine is not an answer but a question. I'll delete it if it is improper.

Why, as the questioner says, if $X$ were Killing and $D$ integrable then $\frac{1}{X^\flat(X)}.X^\flat$ should be closed on $M$? Could someone explain me the reason for this?

Here follows what I have understood:

Given a smooth non-singular vector field $X$ on a Riemannian manifold $(M,g)$, we get the smooth distribution $D$ on $M$ globally generated by the smooth non-vanishing 1-form $X^\flat=g(X,\cdot)$. By Frobenius' Theorem, $D$ is integrable iff ${X^\flat}\wedge{d{X^\flat}}=0$ on $M$. This integrability condition is at the same time necessary and sufficient for the local existence of integrating factors for $X^\flat$: i.e. for any point $p$ of $M$, there exists a function $f$ such that $f.X^\flat$ is closed in a neighborhood of $p$.

Answer by José Figueroa-O'Farrill for Special Killing Vector Fields

2011-03-02T20:38:35Z

I now think that my comment might indeed be the complete answer in the case when $X$ has no zeroes.
Guiseppe's answer has been a sort of Socratic catalyst.

Indeed, in that case the distribution $D$ defined by $\omega$ and $X^\flat$ agree. So $D$ is integrable if and only if the ideal generated by either $\omega$ or $X^\flat$ is differentiably closed, hence $dX^\flat = \alpha \wedge X^\flat$ for some one-form $\alpha$. In turn this is equivalent to $X^\flat \wedge dX^\flat = 0$, which is precisely the condition that $X$ be twist-free.