What is the exact relationship between principal bundles, representations, and vector bundles?
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asked Nov 17, 2009 at 3:43
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5 Answers
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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.
answered Nov 17, 2009 at 4:00
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2$\begingroup$ 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. $\endgroup$– ಠ_ಠCommented Feb 1, 2016 at 6:01
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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 $f_P^*V$.
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10$\begingroup$ 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 $X \to BG \to BGL_n.$ $\endgroup$ Commented Nov 17, 2009 at 5:19
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10$\begingroup$ 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). $\endgroup$ Commented Nov 17, 2009 at 8:48
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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 $V$.
answered Nov 17, 2009 at 3:57
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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.
answered Nov 17, 2009 at 7:42
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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.
answered Nov 17, 2009 at 6:19