3
$\begingroup$

In my reading of the (excellent!) paper of Grabowski and Rotkiewicz on higher vector bundles (https://arxiv.org/abs/math/0702772), I have encountered the following argument which I do not understand. See proof of Theorem 2.1 in the above reference, paragraph starting with "Hence, due to the Implicit function theorem....", they use the following:

Let $N \subseteq F$ be a connected closed embedded submanifold and $\mathcal{V}: F \rightarrow E$ a smooth map (of connected smooth manifolds without boundary), such that:

1) $\mathcal{V}$ restricted to $N$ defines an embedding $\mathcal{V}|_{N} \rightarrow E$ whose image is a closed embedded submanifold of $E$;

2) $\mathcal{V}$ is has a one-to-one tangent map $D_{n}\mathcal{V}$ at all points $n \in N$, that is it is a local diffeomorphism at all points $n \in N$.

3) Maybe it is not very important, maybe it is: The manifold $E$ is in fact a vector bundle over $N$ and the image $\mathcal{V}(N)$ is preciesely the (image of) the zero section $0_{N} \subseteq E$.

The claim: There is an open neighborhood $U_{N} \subseteq F$ of $N$ and an open neighborhood $W_{0} \subseteq E$ of $0_{N} \equiv \mathcal{V}(N)$, such that $\mathcal{V}: U_{N} \rightarrow W_{0}$ is a (global) diffeomorphism.

Remark: They specifically say "Since $\mathcal{V}|_{N}$ is an embedding, we can say...", so the point 1) is supposedly crucial in this statement.

My ideas: Due to 2), for every $n \in N$, there is an open neighborhood $U_{n}$ of $n$ and $W_{n}$ of $\mathcal{V}(n)$, such that $\mathcal{V}: U_{n} \rightarrow W_{n}$ is a diffeomorphism. Then take $U_{N} = \cup_{n \in N} U_{n}$ and $W_{0} = \cup_{n \in N} W_{n}$. Then $\mathcal{V}: U_{N} \rightarrow W_{0}$ is clearly surjective. However, I struggle to prove that it is injective. Clearly one has to use 1) somehow, or maybe even 3). The only thing I was able to show was that one can choose $U$ and $V$ to be connected.

Any help, someone? I know that in general, surjective local diffeomorphisms are not diffeomorphisms (not even covering maps).

Thanks, Jan Vysoký

$\endgroup$

1 Answer 1

3
$\begingroup$

Lemma 1: Let $N\subset F$ be a compact embedded submanifold of a smooth manifold $F$, and let $\nu: F \rightarrow E$ be a smooth map which is injective on $N$ and a local diffeomorphism at every $n\in N$. Then there is a neighborhood $U\subset F$ of $N$ such that $\nu: U \rightarrow E$ is injective.

Proof: Let $T(N)\subset F$ be a tubular neighborhood of $N$. It is diffeomorphic to the normal bundle of $N$ in $F$, and hence we can pick a bundle metric on it. Also, we denote the smooth base point projection by $\pi: T(N)\rightarrow N$. For every $n\in\mathbb{N}$, set $T_n:=\{v\in T(N) \mid |v|<\frac{1}{n}\}$; this is an open neighborhood of $N$. For a contradiction, suppose that for every $n\in\mathbb{N}$ there are $u_n$, $v_n\in T_n$ such that $u_n\neq v_n$ and $\nu(u_n)=\nu(v_n)$ (i.e., $\nu$ is not injective on $T_n$). Since $N$ is compact, we can assume that $\pi(u_n)\to u \in N$ and $\pi(v_n)\to v\in N$ in $N$ as $n\to \infty$ (possibly after picking a subsequence). By construction, we have $|u_n|$, $|v_n| \to 0$, and hence $u_n\to u$ and $v_n\to v$ in $F$ (using the product structure of $T(N)$ in a neighbothood of $u$ and $v$, respectively). Because $\nu$ is continuous and $\nu(u_n)=\nu(v_n)$ for all $n$, we get $\nu(u) = \nu(v)$. Because $\nu$ is injective on $N$, it follows that $u=v$. We denote $w:= u = v\in N$. Let $W\subset F$ be an open neighborhood of $w$ such that $\nu(W)\subset E$ is open and $\nu: W \rightarrow \nu(W)$ is a diffeomorphism. There is an $n_0\in \mathbb{N}$ such that $u_{n_0}$, $v_{n_0}\in W$. By construction, we have $\nu(u_{n_0})\neq \nu(v_{n_0})$, which is a contradiction. QED.

Intersecting your $U_N$ and my $U$, we get a neighborhood of $N$ such that $\nu: U_N\cap U \rightarrow \nu(U_N\cap U)$ is a diffeomorphism (it is a bijective local diffeomorphism).

EDIT: It holds for non-compact manifolds as well using some topological tricks with compact exhaustions. I wonder if there is a better geometrical construction...

Lemma 2: Let $\nu: F \rightarrow E$ be a smooth map which is injective on a compact subset $K\subset N$ and a local diffeomorphism at every $k\in K$. Then there is a neighborhood $U\subset F$ of $K$ such that $\nu: U \rightarrow E$ is injective.

Proof: This is a variation of Lemma 1. In fact, one does not need the embedded submanifold $N$. One picks a system of neighborhoods $U_n$ of $K$ in $F$ such that $\bar{U}_n$ is compact, $\bar{U}_{n+1} \subset U_n$ and $\bigcap_n U_n = K$. This is possible since $F$ is a metric space. Suppose, for the contradiction, that there are $u_n\neq v_n$ in $U_n$ such that $\nu(u_n)= \nu(v_n)$. Because $\bar{U}_1$ is compact, we get $\pi(u_n)\to u$ and $\pi(v_n)\to v$ for some subsequence and some $u$, $v\in \bar{U}_1$. Because of $(\bar{U}_{n+1})$ being nested in $(U_{n})$, we have $u$, $v\in \bigcap \bar{U}_n = \overline{\bigcap U_n} = K$. One then proceeds as in the proof of Lemma 1. QED.

Lemma 3: Let $\nu: F \rightarrow E$ be a continuous map which restricts to a homeomorphism of embedded submanifolds $N\subset F$ and $\nu(N)\subset E$. Let $(N_n)$ be a compact exhaustion of $N$, i.e., $N_n$ for $n\in \mathbb{N}$ are compact sets such that $N_n \subset \mathrm{int}(N_{n+1})$ and $\bigcup_n N_n = N$. We set $S_n:= N_n\backslash \mathrm{int}(N_{n-1})$, where $N_0 := \emptyset$. Then there are neighborhoods $U_n$ of $S_n$ in $F$ such that $$ \nu(U_n) \cap \nu(\bigcup_{|m-n|>1} U_m) = \emptyset. $$

Proof: Firstly, because $\nu$ is a homeomorphism, it maps the compact exhaustion of $N$ to a compact exhaustion of $\nu(N)$, and it intertwines the construction of $S_n$. Now, let $T(\nu(N))$ be a tubular neighborhood of $\nu(N)$ in $E$ isomorphic to the normal bundle. Because the manifold $N$, resp. $\nu(N)$ is a normal topological space, we can inductively construct neighborhoods $V_n$ of $\nu(S_n)$ such that $V_n \cap \bigcup_{|m-n|>1} V_m = \emptyset$ (we construct them such that $\bar{V}_n\subset \mathrm{int}(\nu(N_{n+1}))$ in every step). Now, let $W_n:= T(V_n)$, where $T(V_n)$ is the restriction of the normal bundle, resp. tubular neighborhood to $V_n$. Clearly, the family $(W_n)$ also satisfy the intersection property. It is easy to see that $U_n:= \nu^{-1}(W_n)$ have the desired properties. QED.

Lemma 4 (non-compact version of Lemma 1): Let $N\subset F$ be an embedded submanifold of a smooth manifold $F$, and let $\nu: F \rightarrow E$ be a smooth map which restricts to an embedding of $N$ and which is a local diffeomorphism at every $n\in N$. Then there is a neighborhood $U\subset F$ of $N$ such that $\nu: U \rightarrow E$ is injective.

Proof: Pick an exhaustion of $N$ by compact sets as in Lemma 3. Let $U_n$ be the neighborhoods of $S_n$ such that $\nu(U_n)\cap \nu(\bigcup_{|n-m|>1}U_m) = \emptyset$ for every $n$. By Lemma 2, we can find beighborhoods $U_n'$ of $S_{n-1}\cup S_n \cup S_{n+1}$ such that the restriction of $\nu$ to $U_n'$ is injective. We set $W_n := U_n \cap U_n'$. It is easy to check that $\nu$ is injective on $U:= \bigcup W_n$. QED

Now, as in the compact case, one intersects my $U$ with your $U_N$ and obtains the following:

PROPOSITION: Let $\nu: F\rightarrow E$ be a smooth map, and let $N$ be an embedded submanifold of $F$ such that $\nu$ restricts to an embedding of $N$ and such that $\nu$ is a local diffeomorphism at every $n\in N$. Then $\nu$ extends to a diffeomorphism of neighborhoods of $N$ and $\nu(N)$. (All manifolds are assumed to be Hausdorff and paracompact.)

$\endgroup$
9
  • $\begingroup$ That is very nice, thank you! I understand why you need compactness in your proof, and I see not way around this at first glance. Unfortunately, I need it without this restriction - but I will definitely take note of your approach, thanks. $\endgroup$
    – Jan Vysoky
    May 10, 2019 at 20:57
  • $\begingroup$ Ah, I see, I misunderstood "closed". $\endgroup$
    – Pavel
    May 10, 2019 at 21:00
  • $\begingroup$ Btw., I think that 3) is irrelevant because you can always shrink E to a tubular neighborhood $T$of $\nu(N)$ and set $F=\nu^{-1}(T)$. $\endgroup$
    – Pavel
    May 10, 2019 at 21:20
  • $\begingroup$ Yeah, I also think it is not important, that is why added this separately. However, it is entirely possible that the statement holds only for the particular $\mathcal{V}$ they are considering (although I don't think so), so it does not hold in full generality... $\endgroup$
    – Jan Vysoky
    May 10, 2019 at 21:36
  • 1
    $\begingroup$ I have found a reference with a very similar ideas here: staff.ustc.edu.cn/~wangzuoq/Courses/18F-Manifolds/Notes/… $\endgroup$
    – Jan Vysoky
    May 13, 2019 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.