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Flow Box Theorem

If $M$ is a manifold of dimension $n$ and $X$ is a vector field on $M$ such that for a certain $p\in M$ $X(p)\neq0$, then there exists a chart $(U,\phi)$ on $M$ such that $p\in U$ and $\phi_\ast X$, the pushforward of $X$ by $\phi$ is the first canonical vector.

The proof I have of this result starts by completing $X(p)$ to a basis $\{X(p),v_2,\dotsc,v_n\}$ of the tangent space $T_pM$. Then it fixes a function $g:\mathbb{R}^{n-1}\to M$ such that $g(0)=p$ and $\mathrm{d}g_0(e_{i-1})=v_i$ for $i\geq2$. From that function $g$, by composing it with the flow of $X$, it constructs a chart and then checks it is the chart the theorem states to exist.

I am not asking about the details of the proof, which I have sketched above, and which I am clear about. What I would like to know is how to prove such a function exists. This point was left to us students as an exercise, but I really do not know how to start on this. I mean, I can easily imagine a smooth $g$ such that $g(0)=p$: I just need a map taking $p$ as a value, then I compose it with a translation of $\mathbb{R}$. But how do I fix the differential? I would need to make sure $\frac{\partial(g \circ h)}{\partial x^{i-1}}=v_i(h)$ for any smooth map $h:M\to\mathbb{R}$… any help?

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MickG
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Hint: Do you know about slice charts? You are essentially trying to reverse that idea. Click below for full answer.

Let $\psi: U \to \mathbb{R}^n$ be a chart in a neighborhood $U \subset M$ of $p$ such that $\psi(p)=0$. The image of $\{v_2,\ldots,v_n\}$ under $d \psi_p$ is an $(n-1)$-dimensional subspace $W$ of $T_0 \mathbb{R}^n$. Using the identification $T_0 \mathbb{R}^n \cong \mathbb{R}^n$, we can view $W$ as a subspace of $\mathbb{R}^n$ itself. After a change of basis, $W$ (and thus $T_0 W$) is spanned by $e_1,\ldots,e_{n-1} \in \mathbb{R}^{n-1} \subset\mathbb{R}^n$ as desired. You can take $g$ to be the restriction $\psi^{-1}|_{W}: W \cong \mathbb{R}^{n-1} \to M$, since $d(\psi^{-1}_0)(e_{i-1})=(d\psi_p)^{-1}(e_{i-1})=v_i.$

Kyle
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  • Now so we are taking the chart and then composing it with a change of base, which is a diffeomorphism, so we get a new chart, invert it and restrict the inverse to the span of the first $n-1$ vectors, right. Perfect. Now I know what a chart is, but what does "slice" add to "chart"? – MickG May 14 '15 at 09:43
  • @MickG: The "slice chart" reference was more for inspiration. If you have a $k$-dimensional submanifold $N\subset M$, you can find (at any point in $N\subset M$) a chart that sends $U\subset M$ to $\mathbb{R}^n$ in such a way that $U\cap N$ is sent to $\mathbb{R}^k\times 0^{n-k}$, the set of points in $\mathbb{R}^n$ whose last $n-k$ coordinates are zero. In this sense, a $k$-dimensional submanifold of $M$ always locally looks like a $k$-dimensional slice of $\mathbb{R}^n$. – Kyle May 14 '15 at 11:16
  • Ah I see. We called them "adapted charts" :). – MickG May 14 '15 at 11:21
  • And I guess you mean "regular submanifold". At least we defined r.s. as a submanifold with an adapted chart at every point. – MickG May 14 '15 at 11:22
  • @MickG: Interesting. It looks like Lee and several other authors use this adapted/slice chart definition of submanifold, too. It is equivalent to defining a submanifold to be the image of a proper embedding of one manifold into another, which is what I had in mind. – Kyle May 14 '15 at 11:40
  • We have proved that the image of an embedding (smooth map with everywhere injectve differential which is a homeomorphism onto its image with the subspace topology from the manifold containing it) is a regular submanifold as defined with slice charts. And we proved the inclusion of a submanifold in that sense is an emhedding. So indeed they are equivalent definitions. Just for the record, and to leave the definition of embedding in sight for other readers :). – MickG May 14 '15 at 12:01
  • @MickG: Just to make sure we're on the same page: does that answer your question or did you have questions about the other steps in the proof? – Kyle May 15 '15 at 13:15
  • «I am not asking about the details of the proof, which I have sketched above, and which I am clear about.», says the question. So yes, that answered my question :). And no, no questions about the rest of the proof. Not as far as I remember, at least. When I get to repeating it, if I have trouble, I will be back :). – MickG May 15 '15 at 14:25
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    With a link to a new question, I'd say. – MickG May 15 '15 at 14:25