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I am reading Theorem 9.20 (Flowout Thoerem) from Lee's Introduction to Smooth Manifolds, Second edition.

A part of the theorem states the following:

Let $M$ be a smooth manifold and $S$ be a $k$-dimensional embedded submanifold in $M$. Let $V$ be a smooth vector field on $M$ which is nowhere tangent to $S$. Let $\theta:\mathcal D\to M$ be the maximal flow associated with $V$, and let $\mathcal O=(\mathbf R\times S)\cap D$. Let $\Phi=\theta|_{\mathcal D}$.

Then $\partial/\partial t\in \mathfrak X(\mathcal O)$ is $\Phi$-related to $V$.

First Question: What is $\partial/\partial t$? Is it defined as $$ d(id\times \varphi)_p\left(\frac{\partial}{\partial t}\Big|_p\right)=(1; 0,\ldots, 0)\in \mathbf R\times \mathbf R^k $$ where $id\times \varphi$ is a smooth chart on $\mathcal O$? If yes then we should also check that this is well-defined.

Second Question: What is the theorem trying to say geometrically? I can verify the details given in the proof but am having no insight as to what is really being said.

Thanks.

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The vector field $\partial/\partial t$ is just the standard coordinate vector field on $\mathbb R$. By means of the canonical identification of each tangent space to a product manifold with the direct sum of tangent spaces of the factors, we can view it as a vector field on $\mathbb R\times S$. It's canonically defined, so there's no need to check that it's well defined.

As for what the theorem means geometrically, I tried to suggest the geometric meaning in the illustration that comes with it (see below). The idea is that you start with a $k$-dimensional submanifold $S$, and then flowing out along integral curves of $V$ gives you a certain $(k+1)$-dimensional submanifold containing $S$, to which $V$ is tangent.

(BTW, there's a typo in your transcription of the theorem statement -- the definition of $\mathcal O$ should be $(\mathbb R\times S)\cap \mathcal D$, not $(\mathbb R\times S)\times \mathcal D$.)

enter image description here

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  • $\begingroup$ You made a typo in the same too! It is $(\mathbf R\times S)\cap \mathcal D$, not $(\mathbf R\times S)\cup \mathcal D$. $\endgroup$ – caffeinemachine Jul 12 '15 at 22:14
  • $\begingroup$ I see. This has a nice geometric meaning. My main hindrance was a lack of understanding of what $\partial/\partial t$ means. This clarifies it. $\endgroup$ – caffeinemachine Jul 12 '15 at 22:18
  • $\begingroup$ Good grief. Fixed now. Thanks. $\endgroup$ – Jack Lee Jul 12 '15 at 22:37

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