A subset $S$ of a manifold $N$ of dimension $n$ is a regular submanifold of dimension $k$ if for every $p \in S$ there is a coordinate neighbourhood $(U,\phi)$ of $p$ in the maximal atlas of $N$ such that $U \cap S$ is defined by the vanishing of $n-k$ of the coordinate functions.

This is Tu's definition, An introduction to Manifolds, 2nd ed. Why do we demand equality between the set $U \cap S$ and the zero set of the vanishing coordinates? Tu stresses this fact in the example on the following page, where he excludes the chart corresponding to $V=(-2,0) \times (-1,1)$ in relation to the regular submanifold $S=(-1,1) \subset \mathbb{R}^2$. The inverse image of such a chart would include some points in the zero set outside of $U \cap S$, but is this a problem? We seem to restrict to $U \cap S$ anyhow. Were some similar questions in the past.

Added comment: So the author continues by introducing the restriction of the first $k$ components of a chart $(U,\phi)=(U,x^1,...,x^n)$ to $U \cap S$, $\phi_S: U \cap S \to \mathbb{R}^k$, $\phi_S=(x^1,...,x^k$), and shows that if $\{(U,\phi)\}$ is a collection of charts with the desired properties then $\{(U \cap S, \phi_S)\}$ is an atlas for $S$. To do so, he shows that the corresponding transition functions are smooth, which seems to hold in either case. In a subsequent section he also shows that the map $g:N \to S$ induced by a smooth map $f: N \to M$, whose image $f(N)$ lies in a regular submanifold $S$ of dimension $s$, is again smooth. In the proof, he notes that for a suitable chart $(V,\phi)$, relative to $S$, about a point $f(p) \in M$, he can pick an open set about $p \in U$ with $f(U) \subset V \cap S$, such that $\phi(f(q))=(y^1(f(q)),...,y^s(f(q)),0,...,0)$. From this he concludes, what seems to be nothing but trivial and from here on a matter of definition, with the words "It follows that on $U$, $\phi_S \circ g=(y^1 \circ f,...,y^s \circ f)"$, sounding as if there were anything more to it. The latter also seems to be a recurring theme for a build up that initially seems to be a matter of finding suitable definitions (excepting that such abstractions may be far from obvious) to go in accord with the comparatively deep results from Calculus and the Implicit function theorem.


1 Answer 1


Tu requires that for every $p \in S$ there exists a chart $(U,\phi)$ on $N$ with $p \in U$ such that $\phi^{-1}(\tilde{\mathbb R}^k) = U \cap S$. Here $\tilde{\mathbb R}^k = \{(x_1,\ldots,x_k,0,\ldots, 0 ) \mid x_1,\ldots,x_k \in \mathbb R \}$.

It seems that you ask why he doesn't relax this to the requirement that $U \cap S \subset \phi^{-1}(\tilde{\mathbb R}^k)$.

Such a chart would include some points in the zero set outside of $U∩S$, but is this a problem?

If we agree that $S$ should inherit from $N$ a natural structure of a smooth $k$-manifold, then this relaxed requirement causes indeed a problem.

As an example take $N = \mathbb R^2$. In the relaxed sense any subset $S$ of the $x$-axis would be a regular submanifold of dimension $1$. I think this is not a reasonable approach.

What we could do is to require that $U \cap S$ is an open subset of $\phi^{-1}(\tilde{\mathbb R}^k)$. This is the case in Tu's example in Fig. 9.1.

In fact, under this assumption there is an open subset $U^* \subset U$ such that $(U^*, \phi^* = \phi \mid_{U^*})$ is adapted chart relative to $S$, and this means that both definitions are equivalent.

How to find $U^*$? We know that $\tilde{\mathbb R}^k$ is closed in $\mathbb R^n$, thus $C = \phi^{-1}(\tilde{\mathbb R}^k)$ is a closed subset of $U$. Since $U \cap S$ is open in $C$, we see that $C^* = C \setminus (U \cap S)$ is closed in $C$, thus closed in $U$. Now take $U^* = U \setminus C^*$.

  • $\begingroup$ That sounds most reasonable, absolutely. Great answer, thanks. $\endgroup$
    – undefined
    Commented Aug 1, 2023 at 13:57

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