# Flow of a vector field

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?

• What type of mathematical object is your đťšş_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? Apr 2 at 23:37

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.

• 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. Apr 3 at 15:28
• ...there should be equivalence between the last question and the fact that... Apr 3 at 15:31
• @Willie: Thank you very much for such a thorough answer! I will try to understand this.
– ZZZ
Apr 4 at 14:03
• @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. Apr 9 at 3:44