MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – José Figueroa-O'Farrill Mar 2 '11 at 13:00
up vote 1 down vote accepted

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.

share|cite|improve this answer
Great! I figured there was a more convenient way to think about things. – Rbega Mar 2 '11 at 20:45

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 $p$.

share|cite|improve this answer
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 $ds=\frac{X}{|X|^2}$. – Rbega Mar 2 '11 at 20:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.