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Let $M \subset R^n$ be a $k$-dimensional smooth manifold. I want to show that if $N \subset M$ is a smooth submanifold of $M$, then $dim(N) \le k$.

I am not really sure how to start solving this question. So I'll appreciate some clues on how to start.

Gabi G
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  • What would it mean for a subset of a manifold to have dimension greater than the manifold? – user247327 Mar 23 '19 at 00:52
  • What is your definition of a submanifold? – Claire Mar 23 '19 at 13:17
  • The definition of submanifold is the same as a manifold. – Gabi G Mar 23 '19 at 14:00
  • The only thing that I have here is that for every $x \in N$ there is an open neighborhood $U \subset R^n$ such that $N \bigcap U$ is a graph of a smooth function from $R^m$. I don't really see how it helps here. – Gabi G Mar 23 '19 at 14:02
  • Maybe the second definition can help more. That for every $x \in N$ there is a open neighborhood $U \subset R^n$ and a diffeomorphism $f: U \to f(U)$ such that $f(N \bigcap U) = R^m$x${0}_{n-m}$ – Gabi G Mar 23 '19 at 14:07

1 Answers1

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A nonempty subset $M\subseteq \mathbb R^k$ is called a $m$-dimensional manifold if for every $x\in M$ there is an open neighbourhood $U\subseteq\mathbb R^k$ of $x$ and a diffeomorphims $\phi:U\to\phi(U)\subseteq\mathbb R^k$ such that $\phi(M\cap U)=\mathbb R^m\times\{0\}_{k-m}$. We will show:

If $ M\subseteq N\subseteq\mathbb R^k$ where $M$ is a $m$-dimensional and $N$ a $n$-dimensional manifold then $m\leq n$.

Note that this also implies that the rank of a manifold is well defined as you have asked here.

$\textbf{proof}:$
Let $x\in M$ and choose diffeomorphims $\phi:U\to\phi(U)$, $\psi:V\to\psi(V)$ with $x\in U\cap V$ such that $\phi(M\cap U)=\mathbb R^m\times\{0\}_{k-m}$ and $\psi(N\cap V)=\mathbb R^n\times\{0\}_{k-n}$. For $q\leq k$ let $\text{inc}_q: \mathbb R^q\to\mathbb R^k$ and $\text{pr}_q: \mathbb R^k\to\mathbb R^q$ be the canonical inclusion and projection and set $\tilde U=\text{pr}_m(\phi(U\cap V\cap M))\subseteq\mathbb R^m $.

Then we have a smooth map

$$J:\tilde U\xrightarrow{\text{inc}_m}\phi(U\cap V)\xrightarrow{\psi\circ\phi^{-1}}\psi(U\cap V)\subseteq\mathbb R^k$$

By construction $J(\tilde U)\subseteq \mathbb R^n\times\{0\}_{k-n}$ so $\text{inc}_n\circ\text{pr}_n\circ J=J$. The Jacobian $D(J)=D(\text{inc}_n)\cdot D(\text{pr}_n\circ J)$ has rank $m$ everywhere and so $D(\text{pr}_n\circ J)\in\mathbb R^{n\times m}$ has at least rank $m$ which implies $m\leq n$. $\square$

$\textbf{Edit:}$
$\phi(U\cap V\cap M)\subseteq \mathbb R^m\times\{0\}_{k-m}$, so $\text{inc}_m\,(\tilde U)=\text{inc}_m\circ\text{pr}_m\circ\phi\,(U\cap V\cap M)=\phi(U\cap V\cap M)$, $J(\tilde U) =\psi\circ\phi^{-1}\circ\text{inc}_m\,(\tilde U) =\psi\,(U\cap V\cap M) \subseteq \psi\,( V\cap N)= \mathbb R^n\times\{0\}_{k-n}$

Claire
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  • Thank you very much! Could you just explain what is "the canonical inclusion and projection"? – Gabi G Mar 23 '19 at 18:16
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    Yes maybe its not so canonical $inc_q(x_1,\dots,x_q)=(x_1,\dots,x_q,0,\dots,0)$ and $pr_q(x_1,\dots,x_k)=(x_1,\dots,x_q)$ also i have to admit that $J(\tilde U)\subseteq \mathbb R^n\times{0}_{k-n}$ is not fully obious so if you want i can add the computation. – Claire Mar 23 '19 at 18:19
  • Yes, if you can add the computation it will help – Gabi G Mar 23 '19 at 18:29
  • could you maybe clarify about the last sentence, why $D(pr_n \circ J)$ has at least rank $m$, and why does this imply $m \le n$? – Gabi G Mar 24 '19 at 13:02
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    By the chain rule $D(J)=D(\text{inc}_n)\cdot D(\text{pr}_n\circ J)$. Since $D(J)$ has rank $m$ and $rank(AB)\leq min{rank A, rank B}$ $rank(D(\text{pr}_n\circ J))\geq m$. But as $D(\text{pr}_n\circ J)$ is a $n\times m$ matrix $rank(D(\text{pr}_n\circ J))\leq n$ see for example here. – Claire Mar 24 '19 at 13:41
  • Thanks again, now I fully understand – Gabi G Mar 24 '19 at 13:52