4
$\begingroup$

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)?
$\endgroup$
5
  • 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 '16 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 '16 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 '16 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 '16 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 '16 at 12:09
5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This together with comments clarified to me my confusion $\endgroup$ Mar 18 '16 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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