0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ Apr 2 at 23:37

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Apr 3 at 15:28
  • $\begingroup$ ...there should be equivalence between the last question and the fact that... $\endgroup$ Apr 3 at 15:31
  • $\begingroup$ @Willie: Thank you very much for such a thorough answer! I will try to understand this. $\endgroup$
    – ZZZ
    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. $\endgroup$ Apr 9 at 3:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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