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I am reading John Lee "Introduction to Smooth Manifolds." Flowout theorem states (Theorem $9.20$, Flowout Theorem):

Suppose $M$ is a smooth manifold, $S\subset M$ is and embedded $k$-dimensional sumbanifold,and $V$ is a smooth vector field that is nowhere tangent to $S$. Let $\theta:D\rightarrow M$ be the flow of V, let $O = (\mathbb{R} \times S) \cap D$ and let $\Phi = \theta|_O$ then there exist positive function $\delta: S \rightarrow \mathbb{R}$ such that the restriction of $\Phi$ to $\{(t,p)\in O:|t|<\delta(p)\}$ is injective. In particular $\Phi(\{(t,p)\in O:|t|<\delta(p)\})$ is an immersed sumbanifold.

My question is: why does it not an embedded one? It seems to be an embedded manifold for the case when $S$ has codimension $1$. I was trying to come up with an example but failed.

Vadim
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1 Answers1

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One can use Example 4.20 to show that there exist flowouts (where more generally the domain is an open subset of $\mathbb R \times S$ containing $\{0\} \times S$), which are not embedded:

As in the book, let $$M:= \mathbb{T}^2 = \left\{ \left( z_1, z_2 \right) \in \mathbb {C}^2 \middle\vert \, \left\lvert z_1 \right\rvert = \left\lvert z_2 \right\rvert =1 \right\}$$ be the torus and $\alpha$ be an irrational number. By identifying the tangent bundle of $\mathbb{C}^2$ with $\mathbb C^4$ in the obvious manner, we may consider the vector field $$\mathbb{C}^2 \to \mathbb{C^4} \quad\colon \quad \left( z_1, z_2\right) \to \left( z_1, z_2, 2 \pi i \, z_1, 2 \pi i \alpha \, z_2 \right) \, ,$$ which restricts to a smooth vector field $V$ on $M$. The set $S:= \{ (1,1)\}$ is a $0$-dimensional embedded submanifold of $M$. Any non-zero vector at the point is nowhere tangent to $S$ and so is $V$. Now check that $$\gamma \colon \quad \mathbb R \to M \quad\colon \quad t \to \left( e^{2 \pi i \, t}, e^{2 \pi i \alpha \, t}\right)$$ is an integral curve of $V$ and defines a flowout from $S$ along $V$ in the sense that $\gamma$, considered as a map defined on $\mathbb R \times S$, is an immersed submanifold. However, $\gamma$ is not a topological embedding, as its image is dense in $M$. Yet if you restrict its domain to say $(-1,1) \times S$, then it is indeed an embedded submanifold.

My intuition tells me that, even in the general case, making the neighborhood around S 'small enough' (i.e. choosing a different function $\delta'$ on $S$ with $0< \delta' \leq \delta$) should yield embeddedness, but I have not been able to turn this into a rigorous argument yet. The idea behind this is that embeddedness tends to fail either because of similar issues as in the example above or when the manifold `touches itself in the wrong way' (see figure 4.3 in the book).

user510186
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  • Thank you for your answer, but I am a bit confused. You've said that "one can use Example 4.20 to show that there exist flowouts which are not embedded," yet in the second paragraph you claim that it is an embedded submanifold. Are you trying to say that Example 4.20 and Figure 4.3 might be good places to start? I tried those and I can't really come up with an example. – Vadim Dec 05 '17 at 20:25
  • I said that the above flowout $(\mathbb R \times S, \gamma)$, where $\gamma$ is considered as a map on $\mathbb R \times S$, is an immersed submanifold, but not an embedded one. Example 4.20 and Figure 4.3 are good illustrations of what may prevent submanifolds from being embedded in general. However, using the constant rank theorem, one can show that every (immersed) submanifold $N$ of some manifold $M$ is "locally embedded" in the sense that every point $p \in N$ admits an open neighborhood $U$ in $N$ such that $U$ (together with the inclusion mapping) is an embedded submanifold of $M$. – user510186 Dec 07 '17 at 13:24
  • While I feel honored that you marked the question as answered, I kindly urge you to reconsider. I did give you a counterexample to the claim that all flowouts are embedded. However, the conjecture that there always exist embedded flowouts is still out there - waiting for a proof or a counterexample. I tried proving it, but failed, and I would also be very much interested in a resolution of the issue. – user510186 Dec 11 '17 at 20:26