dimension of total space of a smooth vector bundle

Question

I'm reading John Lee's Introduction to Smooth Manifold Chapter 10. I'm thinking about the definition of the vector bundle arising from the use of dimension in a particular problem.

Suppose we are given a smooth vector bundle $(E,M,\pi)$ of rank $k$, where $E$ is the total space, $M$ is a smooth manifold, and $\pi$ is the "projection" from $E$ to $M$. By definition, each fiber $E_p$ is endowed with a real k-dim v.s and each point $p$ of $M$ has a neighborhood $U$ and a local trivialization $\Phi$, which in this smooth case should also be diffeomorphic, from $\pi^{-1}(U)$ to $U\times \mathbb{R}^k$.

I think that the dimension of the total space $E$ is $m+k$, where $m$ is the dimension of $M$. The intuition is like $U$ can sufficiently shrink to be a smooth chart of $M$ and is then homeomorphic to $\mathbb{R}^m$. Then $\pi^{-1}(U)$ will be a nbd homeomorphic to $\mathbb{R}^{m+k}$. Now, if one finds this not working, I wonder if it's true when $M$ is also compact, where we see the zero section is an embedding and its image is an $m$ dimensional submanifold of $E$ with each point attaching with a $k$-dimensional real vector space.

Answer 1

I'll break this down for you so that you are able to justify your claim with a rigorous argument.

Step 1: For any open subset $U$ of the base manifold $M$, $\pi^{-1}(U)$ is an open subset of the total space $E$. (It's open because $\pi$ is continuous.) Viewing $U$ as an open submanifold of $M$, $U$ and $M$ have the same dimension, since an open submanifold always has the same dimension as the manifold it's contained in. Similarly, $\pi^{-1}(U)$ can be viewed as an open submanifold of $E$, and $\pi^{-1}(U)$ and $E$ have the same dimension.

Step 2: Let $U$ be one of those open subsets of $M$ on which the bundle trivialises, i.e. there exists a diffeomorphism $\Phi : \pi^{-1}(U) \to U \times \mathbb R^k$. $\pi^{-1}(U)$ and $U \times \mathbb R^k$ have the same dimension, because two manifolds that are diffeomorphic to one another always have the same dimension.

Step 3: The dimension of $U \times \mathbb R^k$ is equal to the sums of the dimensions of $U$ and $\mathbb R^k$, because the dimension of a product of two manifolds is always equal to the sum of the dimensions of the two individual manifolds.

Putting this together, we see that $$ \dim(E) = \dim(\pi^{-1}(U))=\text{dim}(U \times \mathbb R^k)=\text{dim}(U) + \text{dim}(\mathbb R^k) = \text{dim}(M)+k. $$

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You can also follow through these steps (in reverse order) to construct a chart around a given point $e \in E$.

Let $p = \pi(e)$, and let $U$ be an open neighbourhood of $p$ on which the bundle trivialises via a diffeomorphism $\Phi : \pi^{-1}(U) \to U \times \mathbb R^k$.

Pick a chart $(V, \phi)$ for the manifold $U$ such that $p \in V$. Let $(\mathbb R^k, \text{id})$ be the canonical chart for $\mathbb R^k$. Then $(V \times \mathbb R^k, \phi \times \text{id})$ is a chart for $U \times \mathbb R^k$.

Composing with the diffeomorphism $\Phi$, we obtain the chart $(\Phi^{-1}(V \times \mathbb R^k), (\phi \times \text{id})\circ \Phi)$ for the manifold $\pi^{-1}(U)$. Note that $\Phi^{-1}(V \times \mathbb R^k)$ is in fact the same as $\pi^{-1}(V)$, so this chart can be thought of as $(\pi^{-1}(V), (\phi \times \text{id})\circ \Phi)$.

The chart $(\pi^{-1}(V), (\phi \times \text{id})\circ \Phi)$ is a chart for the manifold $\pi^{-1}(U)$, and $\pi^{-1}(U)$ is an open submanifold of $E$, so we can equally view this chart as a chart for the manifold $E$. Since $e \in \pi^{-1}(V)$ by assumption, we've succeeded in constructing a chart for $E$ around the point $e$. It's clear that this chart has the expected dimension.