Let $M$ be a closed manifold, $f:M\to M$ be a homeomorphism, and $\phi_t:M\to M$ be a flow. .

The map $f$ is said to be (point)-transitive if some orbit $\lbrace f^nx:n\in\mathbb{Z}\rbrace$ is dense in $M$.

The flow $\phi_t$ is said to be (flow)-transitive if some flow line $\lbrace\phi_tx:t\in\mathbb{R}\rbrace$ is dense in $M$.

My first question is:

  • is there any sufficient condition under which the time-1 map of a transitive flow is still transitive?

In the following we assume $f$ is (point)-transitive and consider a special class of flows: suspensions of $f$.

Let $r\le R$ be positive numbers and $c:M\to[r,R]$ be a continuous suspension function. Let
$M_c=\lbrace(x,r):0\le r\le c(x),x\in M\rbrace/\sim$ where $(x,c(x))\sim(fx,0)$,
and $f_t:M_c\to M_c$ represented by the time-translation $(x,r)\mapsto(x,r+t)$.

According to our definition, the suspension flow $f_t:M_c\to M_c$ is (flow)-transitive.

If $c\equiv1$, it is easy to see that the time-1 map $f_1$ is not transitive (as a homeomorphism on $M_1$). So my second question is:

  • is the time-1 map $f_1:M_c\to M_c$ (point)-transitive whenever $c$ is not constant?

Any proof or reference are good. Thank you!

As suggested by Zarathustra, it is worth to point out that the time-1 map of constant suspension flow with $c=\sqrt{2}$ is also (point)-transitive.

  • $\begingroup$ You can take $c=\sqrt 2$ and your time one map will be transitive. $\endgroup$ – Zarathustra Jan 2 '12 at 18:48
  • $\begingroup$ Oh! Thank you! I will edit the question. $\endgroup$ – Pengfei Jan 3 '12 at 4:37
  • 2
    $\begingroup$ I don't think, in general, the time-$\sqrt{2}$ map of a suspension flow of a transitive map is transitive itself. In fact, consider the following example: let $f\colon\mathbb{T}^2\to\mathbb{T}^2$ be given by $f(x,y):=(x+\sqrt{3},y+\sqrt{2}) \mathrm{mod}\mathbb{Z}^2$. Then, $f$ is a minimal diffeomorphism. The suspension flow of $f$ is $\Phi_t\colon\mathbb{T}^3\to\mathbb{T}^3$ given by $\Phi_t(x,y,z)=(x+t\sqrt{3},y+t\sqrt{2},z+t)$. In this case $\Phi_{\sqrt{2}$ is not transitive. For me it is not clear at all that any transitive suspension flow admits a time-t transitive map. $\endgroup$ – Alejandro Feb 3 '12 at 16:12
  • $\begingroup$ @Alejandro Although the time-1 map of $\Phi_t$ is isomorphic to $f$, but I don't think $\Phi_t$ is the suspension of $f$. It should be $\phi_t:(x,y,z)\to(x,y,z+t)\text{mod}\sim$ where $(x,y,c(x,y))\sim(x+\sqrt{3},y\sqrt{2},0)$. If $c\equiv\sqrt{2}$, then $\phi_1$ is transitive. $\endgroup$ – Pengfei Feb 4 '12 at 15:23
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    $\begingroup$ @Pegnfei: I'm sorry, I changed your notation and that produced some confusion. However, I think my example can be rewritten following your notation: consider the map $f : \mathbb{T}^2\ni (x,y) \mapsto (x+\sqrt{2},y+\qrt{3})\in\mathbb{T}^2$. The suspension flow of $f$ with $c\equiv \sqrt{2}$ is isomorphic to the flow $\Phi_t(x,y,z)=(x+t,y+t\sqrt{3/2},z+\frac{t}{\sqrt{2}})$ and hence, we get $\Phi_1(x,y,z)=(x+1,y+\sqrt{3/2},z+\frac{1}{\sqrt{2}})$, which is not transitive. $\endgroup$ – Alejandro Feb 5 '12 at 7:13

I think the answer for the second question is no. We can construct a counterexample as follows:

Let $f\colon M\to M$ be an arbitrary transitive homeomorphism and $u\colon M\to\mathbb (0,1/4)$ be an arbitrary non-constant continuous function. Then, let's define $$c(x):=u(x)-u(f(x))+1,\quad\forall x\in M,$$ and consider the flow $f_t\colon M_c\to M_c$ as above.

We claim $f_1$ is not transitive. In fact, given any $x\in M$ and any $t\in (0,1/4)$ it holds: $$f_1(x,u(x)+t)=(x,u(x)+t+1)=(x,c(x)+u(f(x))+t)=(f(x),u(f(x))+t).$$ That means that every compact set $U_t:=\lbrace (x,u(x)+t) : x\in M\rbrace \subset M_c$ is $f_1$-invariant and hence it cannot be transitive. Notice the function $c$ is not constant because $f$ is transitive $u$ is not constant itself.

  • $\begingroup$ Oh Alejandro! You saved my life! This example is so great! Thank you! $\endgroup$ – Pengfei Feb 18 '12 at 12:45

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