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Let $R$ be a regular submanifold of a manifold $M$ where $\dim(M)=m$ and $\dim(R)=k$. By definition, for every $r \in R$, there is a chart (in $M$) about p, $(U,\varphi) = (U,x^1,...,x^k,...,x^m)$, where $\varphi: U \to \mathbb R^m$, for which $$U \cap R = \{x^{k+1} = ... = x^{m} = 0\}$$ Define $$\varphi_R : U \cap R \to \mathbb R^k, \varphi_R = (x^1,...,x^k)$$

Why is $(U \cap R, \varphi_R)$ a chart for $R$?

We are given:

  1. $U$ is open in $M$
  2. $\varphi : U \to \varphi(U)$ is a homeomorphism.
  3. $\varphi(U)$ is open in $\mathbb R^m$

I think we must show:

  1. $U \cap R$ is open in $R$
  2. $\varphi_R : U \cap R \to \varphi(U \cap R)$ is a homeomorphism.
  3. $\varphi(U \cap R)$ is open in $\mathbb R^k$

I know how to do (1) and (2) unless (3) is assumed in proving either, but I don't know how to do (3).

If (3) is indeed what we must show, then how do we do this?

  • All I did so far is prove $\varphi(U \cap R) = \varphi_R(U \cap R) \subseteq \varphi(U) \cap$ '$\mathbb R^k$' (see 4). If the reverse inclusion $\supseteq$ is true, then we are done. Otherwise, I think this comes down to understanding $U \cap R$ (which I don't think is a variety in algebraic geometry), which I apparently do not.

If (3) is not what we must show, then what must we show instead?

$(4)$ By '$\mathbb R^k$', I mean $\mathbb R^k \times \underbrace{\{0\} \times \cdots \times \{0\}}_{\text{m-k times}}$

Thanks in advance!

user26857
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Ekhin Taylor R. Wilson
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    Here's my guess: since $\varphi$ is bijective (a homeomorphism), we know that $\varphi(U \cap R) = \varphi(U)\cap\varphi(R)$. But $\varphi(R)=\mathbb{R}^k$ by definition and $\varphi(U)$ is open in $\mathbb{R}^m$ because $\varphi$ is a coordinate chart for $M$. The rest follows from the definition of the subspace topology. Notice that $U \cap R = {x^{k+1}=\cdots=x^m=0}$ means that $\varphi(R)=\mathbb{R}^k$. – stressed out Jan 16 '19 at 05:39

1 Answers1

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I think a serious problem with the teaching methods of differential geometry, particularly manifolds, is that sometimes people use sloppy/hand-waving/difficult notations. This is currently my main issue while trying to learn about manifolds. So, even though what they intend to say is not very complicated, the sloppy notations make it really difficult to follow what they intend to say or prove.

I think what the author means to say is that

$$U \cap R = \{x^{k+1}=\cdots=x^m=0\}$$

is just a notation to say that $$\varphi(U \cap R)=\{(x_1,\cdots,x_m)\in\mathbb{R}^m:x^{k+1}=\cdots=x^m=0 \text{ and } \varphi^{-1}(x_1,\cdots,x_m) \in U\}$$

The later one is obviously isomorphic to $\mathbb{R}^k \subset \mathbb{R}^n$ intersected with $\varphi(U)$.

So, $\varphi(U \cap R)=\varphi(U)\cap\mathbb{R}^k$ where $\varphi(U)$ is open in $\mathbb{R}^m$ because $(U,\varphi)$ is a chart in $M$. By the definition of subspace topology, $\varphi(U\cap R)$ is open in $\mathbb{R}^k$.

stressed out
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  • When you say '$\varphi(R)$', are you referring to $\varphi: M \to \mathbb R^m$ instead of $\varphi: U \to \varphi(U)$? – Ekhin Taylor R. Wilson Jan 16 '19 at 06:51
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    @EkhinTaylorR.Wilson Yes. You're absolutely right. It is indeed an abuse of notation. I wanted to justify why $\varphi(U \cap R) = \varphi(U) \cap \varphi(R)$ from the set-theoretic notation and I fell into a new trap. :D – stressed out Jan 16 '19 at 06:52
  • I actually deleted the comment about abuse of notation thinking it was a different $\varphi$. So, it's $\varphi: M \to \mathbb R^m$ instead of $\varphi: U \to \varphi(U)$? Or is $\varphi(R)$ actually $\gamma(R)$ for some kind of $\gamma$ that agrees with $\varphi$ on $U$? – Ekhin Taylor R. Wilson Jan 16 '19 at 06:53
  • @EkhinTaylorR.Wilson No, your comment was indeed correct. $\varphi$ doesn't have to be defined on all of $M$, for sure. The main point, which I hope is clear, is that the set-theoretic definition of $\varphi(U \cap R)$ implies that $\varphi(U \cap R) = \varphi(U) \cap \mathbb{R}^k$. But this is of course not very set theoretic because $\mathbb{R}^k$ is not really a subset of $\mathbb{R}^n$ but only isomorphic to it. So, there's a little bit of notation issue here. – stressed out Jan 16 '19 at 06:55
  • @EkhinTaylorR.Wilson Also, for (1) and (2), note that (1) follows from the definition of subspace topology and (2) follows from the fact that the restriction of homeomorphism is itself a homeomorphism. So, I think the moral of the story is that when you have a regular submanifold $R$ in $M$, its charts are the same charts as in $M$ except that they've been restricted to $R$. So, in some sense, intuitively, they are what we expect a substructure, a submanifold in this case, to look like. Do you agree? – stressed out Jan 16 '19 at 07:22
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    So $\varphi(U \cap R) = \varphi (U) \cap Z$ where $Z \cong \mathbb R^k$? How about just $Z = \mathbb R^k \times (\underbrace{{0} \times ... \times {0})}{m-k \ \text{times}}$, so $$\varphi(U \cap R) = \varphi (U) \cap \mathbb R^k \times (\underbrace{{0} \times ... \times {0})}{m-k \ \text{times}}$$ ? – Ekhin Taylor R. Wilson Jan 16 '19 at 09:01
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    @EkhinTaylorR.Wilson Yes. You're right. $\varphi(U \cap R) = \varphi(U) \cap Z$ where $Z = \mathbb R^k \times (\underbrace{{0} \times ... \times {0})}_{m-k \ \text{times}}$. Indeed, as linear spaces they're isomorphic, i.e. $Z \cong \mathbb{R}^k$. I like the way you put it. – stressed out Jan 16 '19 at 09:05