Lee Smooth Manifolds: Fundamental Theorem of Flows (9.12) Details

Question

Theorem 9.12

For the second paragraph of the proof, I do not see how it is obvious that $\theta^{(p)}$ is the unique, maximal extension of all such integral curves. Without appealing to a Zorn's Lemma argument.

Edit March 28: The Zorn's Lemma argument is unnecessary because we can view the integral curves as a subset of the product space $\mathbb{R}\times M$. Taking the union over all the graphs gives us a maximal extension.

Zorn's Lemma

Fix $p\in M$, define

$$
\mathcal{I}_p = \Biggl\{\gamma:J\to M,\quad \substack{J\text{ is an open interval containing 0, and}\\ \gamma\text{ is an integral curve of } V \text{ starting at } p}\Biggr\}
$$

If $\gamma_1$, and $\gamma_2$ are in $\mathcal{I}_p$, we define

$$
\gamma_1\lesssim\gamma_2\iff \operatorname{domain}\gamma_1\subseteq\operatorname{domain}\gamma_2
$$

Through a 'union of connected open sets with a common point is again connected and open' argument, we obtain a maximal curve denoted by $\theta^{(p)}$ with domain $\mathcal{D}^{(p)}$, similar to the proof for Tychonoff's Theorem.

Smoothness of $\theta$

Once we have these unique maximal integral curves, on the 6th paragraph of the proof (where he introduces some $t_1>t_0-\varepsilon$), I fail to see why

$$
t_1<t_0\implies (t,p_0)\in W
$$

without the added restriction $t_1>0$.

Later in the paragraph, we define $\widetilde{\theta}: [0, t_1+\varepsilon)\times U_1\to M$, with our original flow $\theta$ being smooth on $(t_1-\delta,\: t_1+\delta)\times U_1$. Shouldn't $\widetilde{\theta}$ be defined on some $(-\delta, t_1+\varepsilon)$ (by shrinking $\delta$ if necessary) instead?

By showing $\widetilde{\theta}(\cdot,p)$ is again an integral curve starting at $p$ for any $p\in U_1$.

$$\operatorname{domain}\widetilde{\theta}(\cdot,p)=(-\delta,\: t_1+\varepsilon)\implies \widetilde{\theta}(\cdot,p)\lesssim \theta^{(p)}$$

It has to be pointwise equal to $\theta$.

$$\widetilde{\theta}=\theta\vert_{(-\delta,\:t_1+\varepsilon)\times U_1}$$

Since $\widetilde{\theta}$ is smooth on $(-\delta, t_1+\varepsilon)\times U_1$, this means $\theta$ is smooth on the same neighbourhood as well, breaking the supremum.

Excerpt from book

Errata can be found here.

Answer 1

$\theta^{(p)} : \mathcal D^{(p)} \to M$ is the unique maximal integral curve starting at $p$.

We don't need Zorn's lemma here.

$\mathcal D^{(p)}$ is defined to be the union of all open intervals containing $0$ on which some integral curve starting at $p$ is defined. $\theta^{(p)}$ is defined by setting $\theta^{(p)}(t) = \gamma(t)$, where $\gamma$ is any integral curve starting at $p$ that is defined on an open interval containing $0$ and $t$; this definition is independent of the choice of $\gamma$ by the uniqueness result from the previous paragraph in the book.

It's worth noting that $\mathcal D^{(p)}$ is non-empty, because there exists at least one integral curve through $p$, by the existence theorem for ODEs.

This $\theta^{(p)}$ is an integral curve starting at $p$. It is an integral curve because for any $t \in \mathcal D^{(p)}$, we have ${\theta^{(p)}}'(t) = \gamma'(t) = V_{\gamma(t)} = V_{\theta^{(p)}(t)}$ (where $\gamma$ is any integral curve starting at $p$ defined on an open interval containing $0$ and $t$). And clearly, $\theta^{(p)}$ starts at $p$, because $\theta^{(p)}(0) = \gamma(0) = p$ (where $\gamma$ is any integral curve starting at $p$).

The statement that $\theta^{(p)}$ is a maximal integral curve starting at $p$ is synonymous with the statement that the domain of $\theta^{(p)}$ contains the domain of any other integral curve starting at $p$. This is true by virtue of how $\mathcal D^{(p)}$ was constructed in the beginning! (Because the domain of $\theta^{(p)}$ contains the domain of any other integral curve starting at $p$, $\theta^{(p)}$ agrees with any other integral curve starting at $p$ on its domain. This follows from the uniqueness result from the previous paragraph.)

The fact that the maximal integral curve starting at $p$ is unique is another consequence of the uniqueness result.

Since $t_1 < t_0$, we have $(t_1, p_0) \in W$.

You're right. The book should have said that $\varepsilon$ is chosen to be small enough such that $t_0 - \varepsilon > 0$. This would imply that $t_1 > 0$.

[By the way, there is another minor error in the book. To ensure that $t_0$ itself is positive, $t_0$ should have been defined as $\inf \{ t \in [0, \infty) : (t, p_0) \notin W \}$.]

$\widetilde \theta : [0, t_1 + \epsilon) \times U_1 \to M$ is a smooth map. Each $t \mapsto \widetilde \theta(t, p)$ is an integral curve, and $\widetilde \theta$ is a smooth extension of $\theta$ to a neighbourhood of $(t_0, p_0)$, contradicting our choice of $t_0$.

If we want to be pedantic, then yes, the domain of $\widetilde \theta$ really ought to be something like $(-\epsilon ' , t_1 + \epsilon) \times U_1$. $\widetilde \theta$ would have to be smooth, and equal to $\theta$, on the domain $(-\epsilon ' , t_1 + \epsilon) \times U_1$. I'll leave you to think about how to construct a $\widetilde \theta$ on this domain such that these properties hold.

Given a $\widetilde \theta$ defined on $(-\epsilon ' , t_1 + \epsilon) \times U_1$ satisfying these properties, we can safely say that $\theta$ is smooth on $(-\epsilon ' , t_1 + \epsilon) \times U_1$. Therefore, $(t', p_0) \in W$ for any $t' \in [0, t_1 + \epsilon)$, contradicting the fact that $t_0 = \inf \{ t \in [0, \infty) : (t, p_0) \notin W \}$ by definition.