# Are integral curves on a connected manifold always path-connected?

I am working out some math involving integral curves of a gradient field on a smooth, connected manifold with boundary. Let's assume the function that provides the gradient field is smooth or even analytic and has no singularities or critical points. So far I assumed that such integral curves would always start and end somehow on the boundary or be circular.

A colleague pointed out that these curves are not guaranteed to exists unless the manifold is complete (which cannot be assumed as it has a boundary). The issue he points out is that integral curves are only defined locally and it might not be possible to connect the local solution domains if their radius is not bounded from below. However, he didn't know more details on this issue, so I am now left with the task to find out if there is an actual issue.

I think what he describes is the situation that the integral curves would be connected but not be path-connected. This can maybe happen if the manifold was something like the graph of sin(1/x). However, that's not a manifold: In connected manifolds are path connected I found that every connected manifold is also path connected and in https://en.wikipedia.org/wiki/Connected_space#Path_connectedness it is stated that the graph of sin(1/x) (on (0, 1]) is connected but not path-connected.

My question: Can it happen on a connected manifold that integral curves of a vector field are not path-connected? If so, what regularity requirements from the vector field and the manifold are needed to justify the assumption that integral curves are path-connected? Does it make a difference if the vector field is a gradient field of a function? In that case, what regularity requirements are needed on the function? (obviously it must be differentiable to yield a gradient field, but what else?)

I read through these lecture notes: http://math.stanford.edu/~conrad/diffgeomPage/handouts/intcurve.pdf On page 18 it is stated: “we can “glue” all such integral curves; on the union of their open interval domains we obviously get the unique maximal integral curve of the desired sort”. I am not sure if I can conclude an answer to one of my questions from that. On the other hand I think if there was a potential pitfall with path-connectedness that would have been discussed in such notes.

• The domain of a maximally defined solution $\phi$ is an open interval $(\alpha,\beta)$ (see the top of p. 6 of the Stanford lecture notes). Take two points $x_1$ and $x_2$ lying on the integral curve. We have $x_1=\phi(t_1)$ and $x_2=\phi(t_2)$, for some $t_1, t_2 \in (\alpha,\beta)$. Assume for definiteness's sake that $t_1 < t_2$. Then the restriction $\psi:=\phi|_{[t_1,t_2]}$ gives a path connecting $x_1$ and $x_2$, contained in the integral curve. Regarding regularity assumptions, it suffices to have the vector field locally Lipschitz. – user539887 May 24 at 20:26
• Alright, locally Lipschitz makes sense to avoid the curves to be torn. I remember my colleague insisted that his concern is mainly about properties of the manifold (he's a mathematician but didn't do math for years, so this might be a false concern). To resolve this concern, I wonder if also the manifold is required to be Lipschitz (en.wikipedia.org/wiki/Lipschitz_continuity#Lipschitz_manifolds). (yes I said I consider smooth or analytic manifolds so assuming Lipschitz is safe here) So, is it true that in fact it boils down to this (indirect) Lipschitz property of the manifold as well? – stewori May 25 at 15:00
• At first sight, I do not think so: how to differentiate on a Lipschitz manifold? – user539887 May 27 at 7:22
• I thought integration might still be possible. E.g. if the vector field is just given and not coming from a gradient. – stewori May 28 at 13:03
• But what is a vector field on a Lipschitz manifold? – user539887 May 29 at 7:33