$$ \newcommand{\wh}{\widehat} \newcommand{\R}{\mathbf R} \newcommand{\mr}{\mathscr} \newcommand{\set}[1]{\{#1\}} \newcommand{\inclusion}{\hookrightarrow} \newcommand{\vp}{\varphi} $$

I am trying to understand the following theorem:

Theorem. Let $M$ be a smooth manifold and $S$ be a subset of $M$. There is a unique smooth structure on $S$, if one exists, which makes it into a smooth manifold such that $i:S\inclusion M$ is a smooth embedding.

The only way I could prove this is via the lemma proved below whose proof is very long. I think the proof of the above theorem should be fairly straightforward and clear and should not require to do what I have done. Does somebody have a short proof?

Lemma. Let $S$ be a $k$-dimensional embedded submanifold of a smooth manifold $M$. Then for each $p\in S$, there exist a smooth charts $(U_p, \vp_p)$ and $(V_p, \psi_p)$ about $p$ on $S$ and $M$ respectively such that $U_p=V_p\cap S$, and $\psi_p\circ\vp_p^{-1}:\wh U_p\to \wh V_p$ (the `hat' denotes the image of the open set under the corresponding chart) \begin{equation*} \psi_p\circ\vp_p^{-1}(x_1 , \ldots, x_k) = (x_1 , \ldots, x_k, 0 , \ldots, 0) \end{equation*} for all $(x_1 , \ldots, x_k)\in \wh U_p$. Therefore $\vp_p=(\pi\circ \psi_p\circ i)|_{V_p\cap S}$, where $\pi:\R^n\to \R^k$ is the projection on the first $k$ coordinates, and the collection of smooth charts $\mr U=\set{V_p\cap S,\ (\pi\circ \psi_p\circ i)|_{V_p\cap S}}_{p\in S}$ is a smooth atlas on $S$.

Proof: Since $i:S\inclusion M$ is an immersion (it's more than that), by the Constant Rank Theorem, we know that there exists a smooth chart $(U,\vp)$ on $S$ containing the point $p$, and a smooth chart $(V,\psi)$ on $M$, again containing $p$, such that $U\subseteq V$ and $\psi\circ\vp^{-1}(x_1,\ldots,x_k)=(x_1,\ldots,x_k,0,\ldots,0)$ for all $(x_1,\ldots,x_k)\in \wh U$. Since $i:S\inclusion M$ is in particular a topological embedding, we know that $U$ is open in $M$ and thus we may WLOG assume that $U=V\cap S$.
We now show that $\vp=(\pi\circ \psi\circ i)|_{V\cap S}$. For take $q\in V\cap S$, and say $\vp(q)=(x_1 , \ldots, x_k)$. Now we have $\psi\circ \vp^{-1}(x_1 , \ldots, x_k)=(x_1, , \ldots, x_k, 0 , \ldots, 0)$, giving $(\pi\circ \psi\circ i)(q)=(x_1 , \ldots, x_k)$.

So we have shown that for each $p\in S$, there exist a smooth charts $(U_p, \vp_p)$ and $(V_p, \psi_p)$ about $p$ on $S$ and $M$ respectively such that $U_p=V_p\cap S$, and $\psi_p\circ\vp_p^{-1}:\wh U_p\to \wh V_p$ \begin{equation*} \psi_p\circ\vp_p^{-1}(x_1 , \ldots, x_k) = (x_1 , \ldots, x_k, 0 , \ldots, 0) \end{equation*} for all $(x_1 , \ldots, x_k)\in \wh U_p$. It remains to show that $\mr U=\set{U_p, \vp_p}_{p\in S}$ is a smooth atlas on $S$. To see this, consider note that \begin{equation*} \vp_q\circ \vp_p^{-1} = \vp_q\circ \psi_p^{-1} \circ \psi_p\circ \vp_p^{-1} \end{equation*} and this map sends $(x_1 , \ldots, x_k)$ to $\pi\circ\psi_q^{-1}\circ\psi_p(x_1 , \ldots, x_k, 0 , \ldots, 0)$ for all $(x_1 , \ldots, x_k)\in \vp_p(U_p\cap U_q)$, and hence is smooth.


1 Answer 1


I would be very surprised if you could avoid using some form of the lemma or the Constant Rank Theorem for maps $\mathbb{R}^k \to \mathbb{R}^n$. Just as we define smooth $n$-manifolds by relating them to $\mathbb{R}^n$, the property of being a smooth $k$-submanifold is inherently connected to the way $\mathbb{R}^k$ sits in $\mathbb{R}^n$. But we can replace the "lemma" with a special case of the Constant Rank Theorem: If $f: \mathbb{R}^k \to \mathbb{R}^n$ is smooth and $df_p$ is injective for some $p$, then there is a diffeomorphism $g: \mathbb{R}^n \to \mathbb{R}^n$ such that, near $p$, the composition $g \circ f$ is the canonical immersion $(x_1,\ldots,x_k) \mapsto (x_1,\ldots,x_k,0,\ldots,0)$.

Now for the uniqueness theorem:

Proof. We want to show that there is a unique maximal smooth atlas for $S$ such that the inclusion $i: S \hookrightarrow M$ is a smooth embedding. Therefore it suffices to show the following: If $\, \phi_1,\phi_2 : U \to \mathbb{R}^k$ are charts on $U \subset S$ such that $$i \circ \phi_1^{-1}, i \circ \phi_2^{-1}: \mathbb{R}^k \to M$$ are smooth, then $\phi_1$ and $\phi_2$ are compatible, i.e. $\phi_2 \circ \phi_1^{-1}$ is smooth. To show this, fix a chart $\psi: V\subset M \to \mathbb{R}^n$ where $V \cap S=U$. We know that $\psi \circ i \circ \phi_1^{-1}: \mathbb{R}^k \to \mathbb{R}^n$ is an immersion, so for any point $x \in\mathbb{R}^k$ there is a diffeomorphism $g: \mathbb{R}^n \to \mathbb{R}^n$ such that $g \circ( \psi \circ i \circ \phi_1^{-1})$ is the canonical immersion near $x$. Letting $\pi: \mathbb{R}^n \to \mathbb{R}^k$ be the canonical submersion, it follows that $$ \pi \circ g \circ \psi \circ i \circ \phi_1^{-1}=\operatorname{id}. \tag{1}$$ Finally, we have $$\phi_1 \circ \phi_2^{-1}= \operatorname{id} \circ (\phi_1 \circ \phi_2^{-1})=\pi \circ g \circ \psi \circ i \circ \phi_1^{-1} \circ (\phi_1 \circ \phi_2^{-1}) = \pi \circ g \circ \psi \circ i \circ \phi_1^{-1}. \tag{2}$$ The map on the right is smooth because $\pi$, $g$, $\psi$, and $i \circ \phi_1^{-1}$ are smooth. $\square$

Here's another perspective that uses the same ideas:

Lemma. Consider maps $X \overset{f}{\to} Y^k \overset{g}{\to} Z^n$ such that $f$ is continuous, $g$ is a smooth immersion, and $g\circ f$ is smooth. Then $f$ is also smooth.

Proof. Fix a point $x \in X$ and charts $(V,\phi)$ around $y=f(x) \in Y$ and $(W,\psi)$ around $z=g(y) \in Z$ in which $g$ is locally represented by the canonical immersion. Letting $\pi: \mathbb{R}^{n} \to \mathbb{R}^{k}$ be the canonical submersion, it follows that $\pi \circ \psi \circ g|_V \circ \phi^{-1}: \mathbb{R}^{k} \to \mathbb{R}^{k}$ is the identity. Since $f$ is continuous, we can find a neighborhood $U$ of $x$ such that $f(U)\subset V$. Then the composition $$ \phi \circ f|_U=\pi \circ \psi \circ g|_V \circ \phi^{-1} \circ \phi \circ f|_U =\pi \circ \psi \circ g|_V \circ f|_U$$ is smooth since $\pi$, $\psi$, and $g|_V \circ f|_U = (g \circ f)|_U$ are smooth. Since $\phi$ is a diffeomorphism, $f$ is smooth. $\square$

This also proves the uniqueness result you're after: If $i : S \hookrightarrow M$ is a smooth embedding and you have another smooth embedding $i': S' \to M$ with $i'(S')=i(S)$, then there is an obvious continuous bijection $f:S' \to S$ such that $i\circ f=i'$. The preceding lemma implies that $f$ and $f^{-1}$ are smooth, hence $S'$ and $S$ are diffeomorphic.

  • $\begingroup$ Could you please give some more details. Where exactly did we use the inverse function theorem? And how do we get the right inverses $f_i$? $\endgroup$ May 14, 2015 at 13:20
  • 1
    $\begingroup$ The trouble with this argument is that you have to decide which smooth structure you're using on $S$ when you compute the composition. If it's $S_1$, how do you know $f_2$ is smooth? If it's $S_2$, how do you know $\iota_1$ is smooth into $S$? $\endgroup$
    – Jack Lee
    May 14, 2015 at 14:39
  • $\begingroup$ @JackLee: I completely agree. Circularly, the argument I had in mind uses results which are established by the slice chart "lemma" and its friends. Seems to me like you need some version of that result, at least the Inverse Function Theorem. $\endgroup$
    – Kyle
    May 14, 2015 at 14:54
  • 1
    $\begingroup$ Unfortunately, your lemma is not true as stated. Here's a counterexample: Let $X=\mathbb R$, $Y=(-\pi,\pi)$, and $Z= \mathbb R^2$, and define $g: Y\to Z$ by $g(t)=(\sin 2t,\sin t)$. Then $g$ is an injective immersion but not an embedding, and its image is a figure-$8$ curve called a lemniscate. If we define $G: X\to Z$ by $G(t)=(\sin 2t,\sin t)$, then the image of $G$ is the same as that of $g$, and we can let $f=g^{-1}\circ G$ because $g$ is bijective onto its image. Now $g$ and $g\circ f = G$ are both smooth, but $f$ is not even continuous. See my Intro to Smooth Manifolds for more. $\endgroup$
    – Jack Lee
    May 15, 2015 at 17:14
  • $\begingroup$ @JackLee: Thanks for the observation and counterexample! I have (hopefully) fixed the lemma by requiring $f$ to be continuous. I guess the lemniscate foils my former argument because, for example, there is no way to simultaneously (1) find neighborhoods $V$ of $0 \in (-\pi,\pi)$ and $W$ of $(0,0)$ in which $g$ is equivalent to the canonical immersion and (2) find a neighborhood $U$ of $\pi \in \mathbb{R}$ such that $f(U)$ lies inside $V$. Then we're out of luck because we can't study the composition $\pi \circ \psi \circ g|_V \circ f$ near $x$. $\endgroup$
    – Kyle
    May 19, 2015 at 17:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .