A question on Lee's Proof of the Global Frobenius Theorem (Lemma 19.22)

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I'm afraid this is a stupid question — I'm not a mathematician, so please correct me when I'll be saying something wrong — but I've been stuck at this point for so long that I thought it would be wise to ask for help.

Hereafter is an excerpt from Introduction to Smooth Manifold by John Lee. The last line is where I'm lost.

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Essentially, $D$ is an involutive (smooth) distribution of rank $k$ on a (smooth) manifold $\mathcal M$ of dimension $m$, $\mathcal N_1$ and $\mathcal N_2$ are integral manifolds for $D$, with $\emptyset\ne\mathcal N_1 \cap \mathcal N_2$, and we want to show that $\mathcal N_1 \cap \mathcal N_2$ is open with respect to, say, $\mathcal N_1$ (whose topology may be finer than the subspace topology induced by $\mathcal M$).

To this end, let $q\in \mathcal N_1 \cap \mathcal N_2$, and let us consider a chart $(W,\varphi)$ of $\mathcal M$ in $q$ that is flat for $D$ – this means just that $\varphi(W)$ is a cube in $\mathbb R^m$, and that $\partial/\partial x^1,\dots,\partial/\partial x^k$ is a (local) frame for $D$ over $W$. We call $V_i$ ($i=1,2$) the connected component (I guess with respect to the topology of $\mathcal N_i$) of $\mathcal N_i \cap W$ containing $q$. From previous results, we know that our $V_i$'s are open sets of the same slice $S=\{p\in W \,\vert\, \varphi^{k+1}(p)=c^{k+1},\dots,\varphi^{m}(p)=c^{m}\}$ of $W$, therefore $V_1\cap V_2$ is open in $S$ with respect to the subspace topology.

Here, as far as I understand, Lee is saying: since $V_1\cap V_2$ is open in $S$, then it is open in $\mathcal N_1$. What is the line of reasoning one should be following?

PS: my sketch argument would be along the following lines. Since $V_1$ is a connected component of $\mathcal N_1\cap W$, it should be open in $\mathcal N_1$. Also, for a subset of $V_1$, being open in $V_1$, or in $\mathcal N_1$ should be the same. From the fact that $V_1\cap V_2$ is open in $S$, we have $V_1\cap V_2= G\cap S$, for some open set $G$ of $\mathcal M$. Then $V_1\cap V_2= G\cap V_1$, hence is open in the subspace topology induced by $\mathcal M$, hence in $V_1$, hence in $\mathcal N_1$. But this seems to be too convoluted, I feel like I'm missing some very basic stuff.