I am studying Loring W. Tu's Differential geometry, Connections, Curvature and Characteristic Classes and I am having a doubt (the same doubt I had when studying the same topic in the author's An Introduction to Manifolds).

The following definition of a vector budle is given

Definition 7.1. A $C^{\infty}$ surjection $\pi : E \to M$ is a $C^{\infty}$ vector bundle of rank $r$ if

  1. For every $p \in M$, the set $E_p:=\pi^{−1}(p)$ is a real vector space of dimension $r$;
  2. every point $p \in M$ has an open neighborhood $U$ such that there is a fiber-preserving diffeomorphism $\phi_U:\pi^{−1}(U) \to U\times \mathbb{R}^r$ that restricts to a linear isomorphism $E_p \to {p}\times \mathbb{R}^r$ on each fiber.

The following definition for bundle map is then given:

Definition 7.5. Let $\pi_{E}: E \to M$ and $π_F: F \to N$ be $C^{\infty}$ vector bundles. A $C^{\infty}$ bundle map from $E$ to $F$ is a pair of $C^{\infty}$ maps $(\phi: E \to F, \underline{\phi}: M \to N)$ such that

  1. the diagram $$ \newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lllllll} E & \ra{\phi} & F \\ \da{\pi_E} & & \da{\pi_F} \\ M & \ra{\underline{\phi}} & N \\ \end{array} $$ commutes.
  2. ${\phi}$ restricts to a linear map $\phi_p: E_{p} \to F_{\underline{\phi(p)}}$ of fibers for each $p \in M$.

A bundle map over $M$ is when $\underline{\phi}$ is the identity map. The author then says that

If there is a bundle map $\psi: F \to E$ over $M$ such that $\psi \circ \phi = \mathbb{1}_E$ and $\phi \circ \psi= \mathbb{1}_F$, then $\phi$ is called a bundle isomorphism over $M$, and the vector bundles $E$ and $F$ are said to be isomorphic over $M$.

Finally, here is the definition of trivial bundle:

Definition 7.6. A vector bundle $\phi: E \to M$ is said to be trivial if it is isomorphic to a product bundle $M \times \mathbb{R}^r \to M$ over $M$.

Here are my questions:

  1. Can we say, equivalently, that a trivial bundle is a bundle in the sense of Definition 7.1 where there exists a open $U = M$, i.e. when there is a fiber-preserving diffeomorphism with the same properties $\phi_M: \pi^{-1}(M) \to M \times \mathbb{R}^r$ ? To me $\psi \circ \phi = \mathbb{1}_E$ and $\phi \circ \psi= \mathbb{1}_F$ is equivalent to say that $\phi$ is a diffeomorphism, i.e. a bijective $C^{\infty}$ map with $C^{\infty}$ inverse.
  2. Can we say that that any manifold with a single chart has trivial tangent bundle?

Apologies in advance if my question is obvious or if, on the contrary, I am missing some macroscopic obstruction to my idea. I was wondering why the definition of trivial bundle has been given after the one of bundle isomorphism, while right after definition 7.1 the author defines trivializing open subset ($U$), trivialization ($\phi_U$ is a trivialization for $\phi^{-1}(U)$) and trivializing open cover.



1 Answer 1


Question 2 is certainly true. A single chart for $M$ is a diffeomorphism $M \xrightarrow{\psi} V \subset \mathbb R^m$, where $m = \text{dimension}(M)$ and $V$ is open. By definition of the tangent bundle of $M$, the map $D\psi : TM \to TV$ is a bundle isomorphism, and $TV$ is the restriction of $T\mathbb R^m$ which is trivial, hence $TV$ is trivial.

Regarding Question 1 and your closing paragraph, those trivialization operations that you are concerned about are only defined over individual elements of an atlas for $M$. So there is indeed a "macroscopic obstruction", but it is a simple one: Not every manifold has an atlas with a single chart, so triviality of individual charts is not yet sufficient to define what it means for $\pi_E : E \to M$ as a whole to be trivial.

You could easily proceed directly to a definition of triviality for the whole of $\pi_E : E \to M$, without first defining the concept of bundle isomorphism.

However, after later proceeding to a definition of bundle isomorphism, you would then have to prove a theorem (or lemma) saying that triviality of $\pi_E : E \to M$ is equivalent to the statement that $\pi_E : E \to M$ is isomorphic to a trivial bundle; and then your readers would object by saying "Wait, isn't that a trivial statement?" (pun intended).

Which order to do these things in becomes just an authorial decision.

  • $\begingroup$ thanks a lot. so, to sum up, I understand that there is no particular reason why triviality is defined via bundle isomorphism, just an authorial choice, and that my definition at question 1 is equivalent to the book's one, correct? $\endgroup$ Jan 29 at 21:47
  • 1
    $\begingroup$ Yes, that's basically correct. I would add that if one chooses to define vector bundle isomorphism first, in that case your definition of triviality basically translates, directly, into the statement that a vector bundle $E \mapsto M$ is trivial if and only if it is isomorphic to the product bundle $M \times R^r \mapsto M$. $\endgroup$
    – Lee Mosher
    Jan 29 at 23:58

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