Take the 2-minute tour ×

I was reading the lemma 4.1 in "J.M.Lee - Introduction to smooth manifolds" which says that given a smooth $n$-manifold $M$, then the tangent bundle $TM$ is a smooth $2n$-manifold.

If $\pi: TM\rightarrow M$ is the natural projection, given an atlas $\mathcal A=\{(U_i,\phi_i)\}$ on $M$, we can define a collection of sets $\{\pi^{-1}(U_i)\}$ and a collection of functions $\widetilde\phi_i:\pi^{-1}(U_i)\rightarrow \mathbb R^{2n}$ which both satisfy the lemma 1.23 namely the construcion lemma for smooth manifolds.

More precisely, on $TM$ is defined the topology having as base the set
$$B=\{\widetilde{\phi_i}^{-1}(V)\; \textrm{for all $i$}:\,\textrm{$V$ is open in $\mathbb R^{2n}$}\}$$

I have some problems to prove that $TM$ is Hausdorff: clearly it is enough to show that given two distinct point $P=(p,X),Q=(q,Y)\in TM$, either then exist some $\pi^{-1}(U_i)$ containing both $P$ and $Q$ or there exist disjoint sets $\pi^{-1}(U_i)$ and $\pi^{-1}(U_j)$ with $P\in \pi^{-1}(U_i)$ and $Q\in \pi^{-1}(U_j)$. If $P$ and $Q$ lie in the same fiber of $\pi$ it is all clear, but if they lie in different fibres than $\pi (P)=p\neq q=\pi(Q)$ and J.M. Lee says

there exist disjoint smooth coordinate domains $U,V$ for $M$ such that $p\in U$ and $q\in V$

Why is this true? $M$ is Hausdorff, but open sets of $M$ are not the smooth coordinate domains (the latter form a subset of the former)!

share|improve this question
Given disjoint open sets $U, V \subset M$ with $p \in U$ and $q \in V$, you can find subsets $U' \subseteq U$ and $V' \subseteq V$ which are smooth coordinate domains for $M$ and $p \in U'$, $q \in V'$. Let $U$ and $V$ be such open sets and let $X$ be a coordinate neighbourhood for $p$ and $Y$ a coordinate neighbourhood for $q$. Then set $U' = U \cap X$ and $V' = V \cap Y$. –  Michael Albanese Oct 13 '12 at 15:41

1 Answer 1

up vote 1 down vote accepted

First, let me fix some terminology (taken from Lee's book).

Let $M$ be a smooth $n$-dimensional manifold. A coordinate chart on $M$ is a pair $(U, \varphi)$ where $U$ is an open subset of $M$ and $\varphi : U \to \tilde{U}$ is a homeomorphism from $U$ to an open subset $\tilde{U} = \varphi(U) \subset \mathbb{R}^n$. The set $U$ is called a coordinate domain.

Now to the situation at hand. As $M$ is Hausdorff, and $p \neq q$, there are disjoint open sets $U, V \subset M$ with $p \in U$ and $q \in V$. As you note, there is no guarantee that $U$ and $V$ will be coordinate domains. While that is true, we can take them to be coordinate domains without loss of generality. Let $(X, \varphi)$, $(Y, \psi)$ be charts (i.e. $X, Y$ are coordinate domains) with $p \in X$ and $q \in Y$ (note, we do not require $X$ and $Y$ to be disjoint).

Claim: The sets $U' = U\cap X$, $V' = V\cap Y$ are coordinate domains with $p \in U'$ and $q \in V'$.

As $U'$ and $V'$ are open, and the restriction of a homeomorphism to an open subset is a homeomorphism, $(U', \varphi|_{U'})$ and $(V', \psi|_{V'})$ are charts; that is, $U'$ and $V'$ are coordinate domains. As $p \in U$ and $p\in X$, $p \in U'$; likewise, as $q \in V$ and $q \in Y$, $q \in V'$.

So, without loss of generality, we can take $U$ and $V$ to be coordinate domains (if they aren't, pass to subsets $U' \subseteq U$, $V' \subseteq V$ which are, and call these sets $U$ and $V$ respectively).

share|improve this answer

Your Answer


