24

It seems to me that given a vector bundle, the associated principal bundle is univocally determined. In fact one has to construct a principal bundle given the base, the fibre (the group $G$ in which the transition functions of the vector bundle take values) and a local trivialization whose associated transition functions satisfy the cocycle condition.

On the other hand, it seems to me that given a principal bundle, the associated vector bundle is far from unique: first one has to specify what is the vector space $V$ constituting the typical fibre, second one has to give a representation of $G$ on $V$. Even if the principal bundle is nontrivial, by taking the trivial representation the associated vector bundle is trivial.

If what I say is correct, why is the terminology "\emph{the} associated vector bundle" so widely use when there is no such an object, even if the vector space itself is specified?

GFR
  • 5,591
  • 5
    I'm not sure where you got the idea that an associated bundle can be constructed from a principal bundle and an abstract vector space having nothing to do with $G$. An associated bundle is a vector bundle determined by a principal $G$-bundle and a representation of $G$. – Paul Siegel Jan 23 '14 at 19:57
  • 3
    Just from the fact that I have seen a few times the phrase "the associated vector bundle" being used without $V$ and/or the representation being specified. I guess that for a while I got confused into thinking of $G$ as a subgroup of $GL(n,\mathbb{K})$ and of the associated vector bundle as the vector bundle with fibre $\mathbb{K}^n$ and representation the defining representation of $G$. By the way, is it correct that a vector bundle uniquely defines the associated principal bundle like I wrote? – GFR Jan 23 '14 at 20:09
  • 2
    If $G$ happens to be a matrix group, then people do sometimes write "the associated vector bundle" to mean the vector bundle associated to the tautological representation, but otherwise the representation should be specified. – Paul Siegel Jan 23 '14 at 20:18
  • 3
    And yes, that's basically correct, though it is important to understand that the transition functions for the vector bundle really do determine transition functions for the principal bundle via the representation. In the case where $G$ is a matrix group, you might also think of the associated principal bundle as a bundle of frames. – Paul Siegel Jan 23 '14 at 20:21

1 Answers1

33

Given a topological space $X$ and a natural number $n$, to each rank $n$ real vector bundle $E\to X$ corresponds a principal $GL(n,\mathbb R)$-bundle $P(E)\to X$, called the frame bundle of $E$; the fiber $P(E)_x$ over a point $x\in X$ is the set of linear isomorphisms $p:\mathbb R^n\to E_x$. The group $GL(n,\mathbb R)$ acts on $P(E)$ on the right by $g: p\mapsto p\circ g$.

In the other direction, to a given a prinicipal $GL(n,\mathbb R)$-bundle $P\to X$ associate the vector bundle $E(P)=P\times\mathbb R^n/G$, where $G$ acts on $P\times \mathbb R^n$ diagonaly by $(p,v)\mapsto (pg,g^{-1}v)$.

You need to verify many small (and easy) details for $P(E)$ and $E(P)$ to be precisly and well defined. The upshot is that the maps $E\mapsto P(E)$ and $P\mapsto E(P)$ induce a natural bijection between the set of equivalence classes of rank $n$ real vector bundles over $X$ and the set of equivalence classes of principal $GL(n,\mathbb R)$-bundles over $X$.

You can modify the above easily to complex vector bundles and manifolds.

So we see that vector bundles (real or complex) correspond to principal $G$-bundles, where $G$ is the general linear group (real or complex). You can extend the correpondence to include other groups $G$, by adding more structure on the vector bundle (orientation, euclidean metric etc).

This may give the impression that working with vector bundles or principal bundles is "basicaly the same", perhaps a matter of taste. This is not the case, and both concepts are essential. But this will take too long to explain, perhaps the subject of a seperate question (sorry).

Gil Bor
  • 3,693
  • 3
    So, while to an abstract principal $G$-bundle there is no way to associate a vector bundle unless a representation of $G$ is given, if $G=GL(n,\mathbb{R})$, there is a canonical choice: the vector bundle with fibre $\mathbb{R}^n$ and structure group $GL(n,\mathbb{R})$, therefore we obtain a 1 to 1 correspondence between principal bundles and vector bundles,is this what you are telling me? I guess that other correspondences are: oriented vector bundle of rank $n$ <-> $GL(n,\mathbb{R})_+$ principal bundle, oriented vector bundle of rank $n$ with a metric <-> $SO(n,\mathbb{R})$ principal bundle. – GFR Jan 27 '14 at 15:10
  • 2
    By the way, I found your remark on the fact that, despite there being a bijective correspondence between the 2 objects, working with one or the other is not equivalent, so I opened a follow up question here http://math.stackexchange.com/questions/653371/when-does-the-difference-between-a-vector-bundle-and-the-associated-frame-bundle – GFR Jan 27 '14 at 15:18
  • A sort of reference is http://www-math.mit.edu/~mbehrens/18.906/prin.pdf Proposition 4.1. – Bruno Stonek Nov 30 '16 at 17:29