Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is well known that a (real) vector bundle $\pi : E\to B$ over a topological space (or manifold) $B$ is a fibre bundle whose fibres
$$F=\pi^{-1}(x), \ \ \ x\in B $$
over any $x\in B$, are diffeomorphic to a vector space $V$. On the other hand, a principal $G$-bundle is a fibre bundle $\pi : P\to B$ over $B$ with a right free action of a Lie group $G$ on $P$ such that for any open set $U\subset B$, the locally trivial fiabrations defined by: $$ \Phi_{U} : \pi^{-1}(U)\to U\times G, \ \ \Phi_{U}(p)=(\pi(p), \varphi_{U}(p)). $$ Here $$ \ \varphi_{U} :\pi^{-1}(U)\to G $$ is a $G$-equivariant map, that is $\varphi_{U}(pg)=\varphi_{U}(p)g$, for all $p\in\pi^{-1}(U)$ and $g\in G$. In the last case the fibers are submanifolds of $P$ which are always diffeomorphic with the structure group $G$.

Although for any vector bundle $\pi : E\to B$, its fibers $F\cong V$ can be considered as Lie group with operation the vector addition, in general we do not include the vector bundles as examples of principal $G$-bunldes (Although, to every vector bundle we can associate the frame bundle which is a ${\rm GL}_{n}\mathbb{R}$-principal bundle, but I dont speak here about associated bundles).

My question is about a good explanation about the fact that IN GENERAL vector bundles (themselves) do not give examples of principal $G$-bundles. For example, the tangent bundle $TM$ of a smooth manifold is a prototype example of a vector bundle, but itself it cannot be considered as a principal bundle for a Lie group $G$, is this true? Thus I aks:

Which is the basic difference between a vector bundle an a $G$-bundle, which does not allows us (almost always??), to consider the vector bundles themselves as examples of $G$-bundles?

For example, a $G$-bundle is trivial (isomorphic to the product bundle), if and only if it admits a global section, but I think that this is not true for vector bundles.

A second question is about examples of vector bundles which can be considered the same time as $G$-bundles for some Lie group (I think that such an example is a cyllinder)

Thank you very much for your attention!

share|improve this question
I don't believe your definition of a vector bundle is correct. The local trivializations need to be compatible with the vector space structures. –  Qiaochu Yuan Apr 1 '12 at 19:30
Instead of vector bundles you can consider affine bundles (see e.g. the discussion in mathoverflow.net/questions/54502/affine-bundles-over-varieties). Now, an affine bundle could be principal, with the structure group $G\cong {\mathbb R}^n$. –  Misha Apr 1 '12 at 20:56

3 Answers 3

up vote 23 down vote accepted

The difference is that, for a vector bundle, there is usually no natural Lie group action on the total space that acts transitively on the fibers. The fact that all of the fibers are, individually Lie groups, doesn't mean that there is a Lie group that acts on the whole space, restricting to each fiber to be a simply transitive action. The simplest example of this is the nontrivial line bundle over the circle. Another example is the tangent bundle of $S^2$.

share|improve this answer

As a small comment here, with the aim to avoid confusions:

The tangent bundle $TM$ of a smooth $n$-dimensional (real) manifold $M$ can be considered as the $G$-bundle associated with the ${\rm GL}_{n}\mathbb{R}$-principal bundle $L(M)$ of linear frames, that means

$$ TM=L(M) \times_{{\rm GL}(n)}\mathbb{R}^{n} = L(M) \times_{\rho}\mathbb{R}^{n} $$

where here $\rho : {\rm GL}(n)\to {\rm GL}_{n}\mathbb{R}$ is the standard matrix representation.

share|improve this answer
I think you meant " vector bundle associated with [...]", not " $G$-bundle associated with [...]". –  Qfwfq Feb 17 '13 at 0:20
thanks you are right –  314159. Apr 1 '13 at 20:31

This is an old question, but I was just thinking about this question myself and I came to this realization:

For a vector bundle over a field $K$ of rank $n$, if it were a principal bundle, it would have to be a principal bundle for the additive group of the vector space $K^n$. However, as a vector bundle, the trivializations are glued along intersections via isomorphisms of vector spaces, ie elements of $\text{GL}_n(K)$, whereas for a principal $K^n$-bundle, the glueing is done by elements of $K^n$ (ie, translations). This means that not only is a vector bundle not a principal bundle, but also that a principal $K^n$ bundle is not a vector bundle.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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