I'm trying to understand how these two perspectives on vector bundle with a $G$-action come together.

Perspective 1:

Let $P \to X$ be a principal $G$-bundle. The associated bundle construction gives an exact tensor functor from finite linear representation of $G$ to vector bundle with $G$-action: $P \times_G(-): Rep(G) \to Vect^G(X)$. How does the essential image of this functor looks like? Is it faithful? full?

Perspective 2:

Let $V \to X$ be a vector bundle with $G$-action. Under some suitable conditions on $G$ (finite will obviously do but I'm pretty sure weaker assumptions will do - perhaps semisimple is enough) we have the following characterization of $E$. Let $\{V_j\}$ be the trivial vector bundles with $G$ action over $X$ corresponding to the irreducible reprepsentations of $G$.

$$V \cong \bigoplus_j V_j \otimes Hom_{G}(V_j,V)$$

Where $G$ acts on $Hom_G(V_j,V)$ trivially. The fact that $Hom_G(V_j,V)$ is a vector bundle follows from the averaging projection operator on sections of any vector bundle with $G$-action. This is a fulll description of the objects in $Vect^G(X)$ (for when $G$ is nice enough so that it holds).

One way to spell out my confusion is this:

  • Is a vector bundle with $G$-action the same as a reduction of structure group from $GL(V)$ to $G$?

I'm pretty convinced that being a $G$-vector bundle is weaker than having structure group $G$. For example if $G$ is finite then a principal $G$ bundle will always be flat and so will any associated bundle while it looks like $G$-vector bundles mat be non-flat. I don't understand really how these POV's come together. In particular:

  • When is a $G$-vector bundle an associated bundle of some principal bundle?
  • Let $\rho : G \to GL(V)$ be a representation. How does the associated bundle $P\times_{\rho}V$ decompose via perspective 2?
  • For $G$ finite: Is every $G$-vector bundle flat (locally constant)?
  • 5
    $\begingroup$ In perspective 1, you quotient out the $G$-action to make the vector bundle, so there is no $G$-action left over. $\endgroup$
    – Ben McKay
    Mar 18, 2016 at 10:36
  • $\begingroup$ @BenMcKay So perspective 1 is actually the usual equivariant $G$-bundles on the principal bundle are the same as bundles over the base? For example if $P$ is the frame bundle then equivariant $GL_n$bundles over $P$ are the same as bundles over the base? $\endgroup$ Mar 18, 2016 at 10:40
  • 1
    $\begingroup$ Yes: if $P$ is the frame bundle of a manifold $X$ and the representation is the obvious one of $GL_n$, then the associated vector bundle is the tangent bundle of $X$, with no $GL_n$-action. $\endgroup$
    – Ben McKay
    Mar 18, 2016 at 11:15
  • 3
    $\begingroup$ Somehow your question seems to mix up the concept of a homogeneous vector bundle with the concept of the structure group of a vector bundle, so I am not sure what you mean by a G-vector bundle. One possible notion is that the base carries an action of G and you want to have a lift of this action by vector bundle homomorphisms. In the case of a homogeneous space $G/H$, this means that the bundle is associated the principal $H$-bundle $G\to G/H$. Specifying the structure group of a vector bundle is completely different issue in general, often amounting to the choice of additional structure. $\endgroup$ Mar 18, 2016 at 11:23
  • 1
    $\begingroup$ Maybe a way to avoid confusion would be to give the definition of G-vector bundle you are considering. $\endgroup$
    – Niels
    Mar 18, 2016 at 12:09

1 Answer 1


If you associate a bundle $P[V]$ to a principal $G$-bundle $P\to M$ with the help of representation $G\to GL(V)$, you also construct a mapping $\tau^V: P\times_M P[V] \to V$ which encodes the "associated bundle structure". It can be paraphrased as: it gives the coordinates of a point in $P[V]$ with respect to a frame in $P$. See 18.7 and the paragraph "Notation" after it of here. In 19.9 you find: "Recognizing induced connections". I hope that this source answers all your questions.

  • $\begingroup$ This together with comments clarified to me my confusion $\endgroup$ Mar 18, 2016 at 17:38

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