Does a dense orbit imply topological transitivity for flows on manifolds?

Question

Let $M$ be a smooth connected manifold of dimension $\geq 2$, and let $\phi: \mathbb{R} \times M \to M$ be a complete flow on $M$.

Suppose there exists a point $x_0 \in M$ whose orbit is dense in $M$, i.e.,

$\overline{\{\phi_t(x_0) : t \in \mathbb{R}\}} = M.$

Does this imply that the flow is topologically transitive? That is, for any two non-empty open sets $U, V \subset M$, does there exist a positive time $t > 0$ such that $\phi_t(U) \cap V \neq \varnothing$?

I know that in discrete-time dynamical systems (i.e., for a homeomorphism $f: M \to M$), the existence of a dense orbit is equivalent to topological transitivity. However, for flows, the situation seems more subtle.

Some related points - A dense orbit means that for any open $V$, there exists some $t \in \mathbb{R}$ with $\phi_t(x_0) \in V$, but this $t$ could be negative. -Topological transitivity requires a positive time $t > 0$ for any pair of open sets $U, V$. - On a manifold of dimension $\geq 2$, the flow might have recurrent behavior that allows "mixing" of open sets.

I suspect the answer might be yes for dimension $\geq 2$, but I'm not sure how to prove it. Could there be a counterexample?

Any references or insights would be greatly appreciated.

Answer 1

Suppose $\dim M \ge 2$, and assume $U$ and $V$ are disjoint. Assume that $x_0\in U$ (which has already or will eventually happen by density), we wish to show that $\phi_t(x_0)\in V$ for some $t>0$.

Let $F_+=\overline{\{\phi_t(x_0)|t>0\}}$, if $F_+ \cap V\neq \varnothing$ we are done, so let us assume the converse. We then denote $t_0=\sup \{t\in \Bbb R| \phi_t(x_0) \in V\} <0$. If $\phi_t(x_0) \in U$ for some $t<t_0$ then we are done again, so we assume the converse. In particular, the closed subset $F_-=\overline{\{\phi_t(x_0)|t<t_0\}}$ contains $V$ but not $x_0$.

Let $v = \phi_{-s}(x_0)\in V$ for $-s<t_0$, then $\phi_s(v)=x_0 \in U$. Then for $W$ some tiny open neighborhood of $x_0$ in $U$, $\phi_{-s}(W)$ is an open neighborhood of $v$ in $V$, which we call $W_v$. We now use the following

Lemma: for any compact interval $[a,b]$, $\phi_{[a,b]}(x_0) = \{\phi_t(x_0)|a\le t\le b\}$ is a one-dimensional compact submanifold of $M$ with boundary.

You can prove this locally first, and then use compactness to argue that the result is indeed a manifold. In particular, for any $a<b$, the intersection $\phi_{[a,b]}(x_0) \cap W_v$ has dense complement in $W_v$ simply because $\dim W_V \ge 2$. Yet the orbit of $x_0$ is dense, this means that there exists $t<<t_0$ (namely such that $t+s<t_0$) with $\phi_t(x_0) \in W_v$. But then, applying $\phi_s$, we see that $\phi_{t+s}(x_0)\in W \cap F_-$, and for $W$ small enough to be disjoint from $F_-$ (as $x_0 \not\in F_-$), this is a contradiction.