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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.

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    $\begingroup$ Not a complete answer, but such Killing vectors are twist-free; although this condition seems stronger. Recall that a Killing vector $X$ is twist-free if $X^\flat \wedge dX^\flat = 0$, where $X^\flat = g(X,\cdot)$ in your notation. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Mar 2, 2011 at 13:00

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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.

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  • $\begingroup$ Great! I figured there was a more convenient way to think about things. $\endgroup$
    – Rbega
    Commented Mar 2, 2011 at 20:45
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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 $p$.

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  • $\begingroup$ As it ended up answering the question your approach makes the most sense. The way I thought about things was: Fix a point $p$ so that $X(p)\neq 0$ Let $U$ be a small neighborhood of $p$ so that in $p$ there is a smooth surface $\Sigma$ through $p$ and normal to $X$. Let $\phi_s$ be the family of isometries generated by $X$ clearly for $U$ small enough $\phi_s(\Sigma)$ foliate $U$ and all the leaves are orthogonal to $X$. This allows one to think of $s$ as a function on $U$. One has $ds=\frac{X}{|X|^2}$. $\endgroup$
    – Rbega
    Commented Mar 2, 2011 at 20:50

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