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A submanifold (of $\mathbb{R}^n$), it appears, can be defined in several equivalent ways. One definition, paraphrased from Amann and Escher's Analysis II, is as follows:

A subset $M$ of $\mathbb{R}^n$ is said to be a smooth $m$-dimensional submanifold of $\mathbb{R}^n$ if for every $x \in M$ there exists an open neighborhood $U$ of $x$, an open set $V$ in $\mathbb{R}^n$ and a smooth diffeomorphism $\phi:U \rightarrow V$ such that $\phi(U \cap M) = V \cap (\mathbb{R}^m \times \{0\})$

Everything about this definition makes sense to me except the intersection of $V$ with the Cartesian product of $\mathbb{R}^m$ and $\{0\}$ instead of just $\mathbb{R}^m$. Why not just require $\phi(U \cap M) = V \cap (\mathbb{R}^m)$?

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This is just to indicate how exactly $\mathbb{R}^m$ sits inside $\mathbb{R}^n$, namely as the first $m$ coordinates, as opposed to e.g. the last $m$ coordinates which would be written as $\{0\} \times \mathbb{R}^{m}$ by Amann-Escher. –  t.b. May 18 '11 at 19:56
    
@Theo: so that {0} actually means {0,...,0} (n-m zeroes) ? –  Weltschmerz May 18 '11 at 20:56
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@Theo: That seems reasonable; I was not interpreting the $0$ as the author intended. I thought he was crossing with the set containing the number $0$ which seemed very mysterious @Weltschmerz: Yes, that's my interpretation –  ItsNotObvious May 18 '11 at 21:03
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@Weltschmerz: Yes. More precisely, write $\mathbb{R}^{n} = \mathbb{R}^{m} \times \mathbb{R}^{m-n}$ (by definition $\mathbb{R}^{n}$ is the product of $n$ copies of $\mathbb{R}$. Since $V \subset \mathbb{R}^{n}$ above, the set $\mathbb{R}^{m} \times \{0\}$ is viewed as a subset of $\mathbb{R}^{n}$ and the $0$ stands for the vector $(0,\ldots,0)$ with $n-m$ entries. In other words a vector $(x_1,\ldots,x_m) \in \mathbb{R}^{m}$ is viewed as $(x_1,\ldots,x_m,0,\ldots,0) \in \mathbb{R}^{n}$. –  t.b. May 18 '11 at 21:07

1 Answer 1

up vote 4 down vote accepted

On Willie's order, I'm copy-pasting the comments into an answer:

This is just to indicate how exactly $\mathbb{R}^m$ sits inside $\mathbb{R}^n$, namely as the first $m$ coordinates, as opposed to e.g. the last $m$ coordinates which would be written as $\{0\} \times \mathbb{R}^m$ by Amann-Escher. – Theo Buehler May 18 at 19:56

@Theo: so that $\{0\}$ actually means $(0,...,0)$ ($n-m$ zeroes) ? – Weltschmerz May 18 at 20:56

@Theo: That seems reasonable; I was not interpreting the $0$ as the author intended. I thought he was crossing with the set containing the number $0$ which seemed very mysterious @Weltschmerz: Yes, that's my interpretation – 3Sphere May 18 at 21:03

@Weltschmerz: Yes. More precisely, write $\mathbb{R}^n = \mathbb{R}^m \times \mathbb{R}^{m−n}$ (by definition $\mathbb{R}^n$ is the product of $n$ copies of $\mathbb{R}$. Since $V\subset \mathbb{R}^n$ above, the set $\mathbb{R}^m \times \{0\}$ is viewed as a subset of $\mathbb{R}^n$ and the $0$ stands for the vector $(0,\ldots,0)$ with $n−m$ entries. In other words: a vector $(x_1,\ldots,x_m) \in \mathbb{R}^m$ is viewed as $(x_1,\ldots,x_m,0,\ldots,0) \in \mathbb{R}^n$. – Theo Buehler May 18 at 21:07


/me prods @Theo with a cow-prod. "Come on, copy what you wrote and paste it into an answer." – Willie Wong 19 mins ago

@Willie: :))) aye, aye, Sir! – Theo Buehler 11 mins ago

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