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Consider a Riemannian manifold $(M^n , g)$ and let $d_p: M^n \to [0,\infty)$ be the distance function of $p \in M^n$. Then the flow lines generated by $\nabla d_p$ are radial geodesics from $p$. Also, each geodesic sphere $S_r(p)$ is orthogonal to the flow lines.

I wonder for a general vector field $X$, can one find a "natural" section of the flow lines? For example, if we assume $X$ is a Killing vector field, can one construct $\Sigma_t$ that is orthogonal to every flow line and the flow maps each $\Sigma_t$ to another?

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  • $\begingroup$ What type of mathematical object is your 𝚺_t meant to be? (I am guessing that it is a hypersurface in M, i..e., a submanifold of dimension n-1.) Also, what are you assuming about the differentiability of the manifold M and its riemannian metric? $\endgroup$
    – Daniel Asimov
    Commented Apr 2 at 23:37

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Since you used the word "orthogonal" I assume you are working on a Riemannian manifold. So let $(M,g)$ be Riemannian, and $X$ a vector field. Your question actually has several subquestions buried in it. Let me address each one in turn.

Orthogonal hypersurface

Whether there exists (even locally) a hypersurface that is orthogonal to $X$ is a well-studied problem, and the local problem is completely solved by Frobenius's Theorem. An equivalent formulation to your question is, if we let $\omega$ be the metric dual one form to $X$ (so $\omega(Y) = g(X,Y)$), whether the kernel of $\omega$ is tangent to a hypersurface. The local obstruction is given by $\omega\wedge (d\omega)$; locally such an hypersurface exists if and only if this quantity vanishes.

Flow mapping these slices

Obviously this cannot hold in general. If $X$ is a vector field and $Y = fX$ for some scalar non-vanishing function $f$, and $X$ is hypersurface orthogonal, then obviously so is $Y$. But the flows defined by $X$ and $Y$ are in general completely different, so at most one of the two can preserve orthogonality.

If we take a step back and forget about the integrability problem, we see that you can formulate this at the infinitesimal level by asking that the flow map preserving the orthogonal complement to $X$. In terms of the dual one form $\omega$, this is $$ \mathcal{L}_X(\omega) \wedge \omega = 0 $$ Using Cartan's magic formula this expands to $$ d(\omega(X)) \wedge \omega + i_X(d\omega \wedge \omega) = 0 $$ so we see that when you already know that $\omega$ is hypersurface orthogonal, the flow maps these hypersurface to each other if and only if $$ \nabla (g(X,X)) \propto X $$

When $X$ is Killing

In this case the second result above becomes guaranteed, since $\mathcal{L}_X \omega = 0$. So the issue is only integrability. Not all Killing fields are hypersurface orthogonal. Write $\mathbb{R}^3$ in cylindrical coordinates $(r,z,\theta)$, then the vector field $\partial_z + \partial_\theta$ is Killing and nowhere vanishing. Its metric dual one form is $dz + r^2 d\theta$, whose exterior derivative is $2 r dr \wedge d\theta$, and so we see that hypersurface orthogonality fails.

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  • $\begingroup$ A very pedagogical answer. Thank a lot. But I think that for a flow mapping these slices, it suffices that the vector field be regular (nowhere vanishing) gradient field. The same conclusion should be true if there is a regular (differentiable) section (initial section) of the flow lines and the vector field is regular and orthogonal to that initial section, at least in the (one-sided?) neighborhood of that initial section. Also, there should be equivalence the last question and the fact that the vector field define riemannian flow. $\endgroup$
    – Eric Arnéo Vespira Kengne
    Commented Apr 3 at 15:28
  • $\begingroup$ ...there should be equivalence between the last question and the fact that... $\endgroup$
    – Eric Arnéo Vespira Kengne
    Commented Apr 3 at 15:31
  • $\begingroup$ @Willie: Thank you very much for such a thorough answer! I will try to understand this. $\endgroup$
    – ZZZ
    Commented Apr 4 at 14:03
  • $\begingroup$ @EricArnéoVespiraKengne: just being a gradient field is not enough. Let the manifold be $\mathbb{R}^2\setminus \{0\}$ with the Euclidean metric, and set $f(x,y) = xy$. Its gradient flow does not map level sets to level sets. // Let $f$ be a scalar function, and $X = \nabla f$ its Riemannian gradient. A necessary and sufficient conditions for the flow of $X$ to map level sets of $f$ to other level sets of $f$ is that $g(df,df)$ is constant along the level sets. This condition has other implications. $\endgroup$
    – Willie Wong
    Commented Apr 9 at 3:44

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