MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the exact relationship between principal bundles, representations, and vector bundles?

share|cite|improve this question
up vote 22 down vote accepted

Let G be an algebraic group (or, since the question was tagged as differential geometry, a Lie group). Then if we're given a principal G-bundle $E_G$ and a representation V of G, we get a vector bundle out of this through the associated bundle construction: $(E_G \times V)/G$ is a vector bundle with generic fiber V. Here, G acts on $E_G \times V$ as $g(x,v) = (xg^{-1},gv).$

This shows that fixing a G-bundle determines an exact tensor functor from the category of representations of G to the category of vector bundles. There's a converse to this which says that giving an exact tensor functor from representations of G to vector bundles is equivalent to a G-bundle.

share|cite|improve this answer
Do you happen to know of a text or set of notes that discusses this point of view? I've learned about $G$-bundles from differential geometry texts, which tend to be light on the categorical language. – ಠ_ಠ Feb 1 at 6:01

Just for fun, I'm going to give a fancy reinterpretation of Mike's answer. There's nothing new here, but it's fun to say it this way:

A principal $G$-bundle $P$ on a space $X$ is the same thing as a map $f_P: X \to BG$ to the classifying stack $BG$. We can identify $BG$ with the quotient stack $[\operatorname{pt}/G]$, so a vector bundle $V$ on $BG$ is exactly the same thing as a $G$-equivariant vector bundle on the point, i.e., $V$ is a vector space with a $G$-action, i.e. $V$ is a representation of $G$. The vector bundle associated to $P$ and $V$ is precisely the pullback bundle $f_P^*V$.

share|cite|improve this answer
We can take your fun game with stacks one step further (which again adds nothing new): A vector bundle on BG is equivalent to a map of stacks $BG \to BGL_n,$ and the associated vector bundle is obtained through the composition of maps $X \to BG \to BGL_n.$ – Mike Skirvin Nov 17 '09 at 5:19
And finally, still without adding anything new, we can make this look more suggestive if we use the stack vect of vector bundles. Then a linear representation is precisely a morphism BG->Vect and the vector bundle associated to X->BG is X->BG->Vect which is an element of Vect(X). – Urs Schreiber Nov 17 '09 at 8:48

From a $G$-principal bundle $E\to B$ and a representation $V$ of $G$ you can construct a vector bundle $E\times_GV\to B$. A vector bundle $\mathcal E$ with fiber $V$, on the other hand, gives you a $GL(V)$-principal bundle by taking the bundle $F(\mathcal E)$ of frames in $\mathcal E$, and you can reconstruct $\mathcal E$ from $F(\mathcal E)$ and the tautological representation of $GL(V)$ on $V$.

share|cite|improve this answer

Just to be pedantic, these notions are for finite dimensional groups and representations. In infinite dimensions (either group or representation) one has to be a little bit more careful, see this question.

share|cite|improve this answer

To add some small point about the converse to the above answers for discrete groups: An n-dimensional vector bundle on X equipped with a flat connection is the same thing as a representation of the fundamental groupoid of X in GLn. In this case, monodromy around paths is an invariant of homotopy class of path, and so any homotopy class of path produces an isomorphism between the vector spaces over start and end points.

share|cite|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.