Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $P\rightarrow M$ be a principal (right) $G$-bundle, where $G$ is a Lie group. Given a finite-dimensional representation of $G$, $V$ say, we can define the associated bundle $P\times_{G}V\rightarrow M$. This is a vector bundle over $M$ defined as the quotient of the (free, right) action of $G$ on $P\times V$ - $(p,v)\cdot g =(p\cdot g, g^{-1}v)$.

Hence, for a given representation $V$ of $G$ we can associate to a principal $G$-bundle $P\rightarrow M$ a vector bundle $P\times_{G} V\rightarrow M$. Moreover, this assignment is functorial and so induces a map from isomorphism classes of principal $G$-bundles to $K_{0}(M)$, the Grothendieck group of vector bundles on $M$. Call this functor (and, by abuse of notation, the map it induces) $\theta_{V}$. Furthermore, it seems (there may be problems here?) that we obtain a functor

$\theta: Rep_{G}\rightarrow Fun(Prin_{G}(M),Vec(M))$

where the left hand side is the category of (finite dimensional) representations of $G$ and the right hand side is the category of functors from $Prin_{G}(M)$ to $Vec(M)$, the categories of principal $G$-bundles on $M$ and vector bundles on $M$ (respectively).

Question 1: Which representations induce the trivial map on iso-classes? For example, the trivial representation $T$ will always give

$\theta_{T}(P\rightarrow M)=M\times T$

since we can choose linearly independent generating sections of $P\times_{G} T$ using triviality of $T$. My question is, are there other representations of $G$ which afford this property?

Question 2: What am I really discussing here? Is there a name for $\theta$? Do these ideas arise in some 'deeper' (or more natural) framework?

Question 3: Is this formulation useful? Are there any interesting results related to this construction?

I have come to these conclusions as a result of thinking about associated bundles based on knowing the basic definition only and any references/comments would be appreciated. My apologies if this is standard material to topologists, or well-known to experts - I am neither.

share|improve this question
    
You may like the discussion in Husemoller's "Fiber bundles", Chapter 14, as well as "Vector bundles and homogeneous spaces" by Atiyah-Hirzebruch. –  Igor Belegradek Nov 2 '12 at 1:35

2 Answers 2

It's a quite common to think that a principal bundle is the same thing as a monoidal functor $Rep_G\to Vect(M)$ (this is part of "Tannakian philosophy" of describing objects related to G using the category of representations). I'm not finding any good references on line, but perhaps someone else can suggest one.

There's no non-trivial representation that will give a trivial functor, since the tautological bundle $EG \to BG$ gives an equivalence of tensor categories between $Rep_G$ and $Vect(BG)$.

share|improve this answer
    
This is wrong for $G=\marhbb Z$. It should be fine for Lie groups with finitely many components. –  Will Sawin Nov 2 '12 at 2:24
    
@Will: In general, noncompact Lie groups groups are not correctly detected by their finite-dimensional representation theory. The finite-dimensional representation theory of a group detects the algebraization of a said group, which is the algebraic group that best approximates your group. For example, since $\pi_1(\mathrm{SL}(2,\mathbb R)) = \mathbb Z$, there are connected real Lie groups $G$ that nontrivially cover $\mathrm{SL}(2,\mathbb R)$; but all of them have algebraization $\mathrm{SL}(2,\mathbb R)$. –  Theo Johnson-Freyd Nov 2 '12 at 5:58
    
@Ben thanks for your comment. I'll need to look in more depth at 'tannakian philosophy' etc. cheers –  George Melvin Nov 5 '12 at 0:12

Theorem: ''Let $G$ be a compact, connected Lie group and $f: G \to U(n)$ a group homomorphism such that for each principal bundle $P \to M$ on a manifold, the induced vector bundle $P \times_{G,f} \mathbb{C}^n$ is a trivial vector bundle. Then $f$ is the constant homomorphism.''

Proof: ''For a given $k$, there exists a compact manifold $M$ and a map $M \to BG$ that is $k$-connected, $dim (M) \geq 2k+1$. This is manufactured using surgery below middle dimensions. Applying this to the assumption, you get that $f$ induces the trivial map on cohomology $H^{\ast}(BU(n)) \to H^{\ast}(BG)$ of any degree.

Now assume $f$ is zero on real cohomology $H^{\ast}(BU(n)) \to H^{\ast}(BG)$. By Chern-Weil theory, $H^{\ast}(BG) \cong Sym^{\ast}(\mathfrak{g})^G$, the algebra of Ad-invariant symmetric polynomials on the Lie algebra. There is a symmetric polynomial of degree $2$ on $\mathfrak{u}(n)$ that is nowhere zero: take an invariant scalar product. Therefore, the assumption implies that $f$ has to be zero on the Lie algebra level; hence $f$ is constant on the unit component of $G$.''

I think this is true for nonconnected $G$ and believe the argument is similar to the one by Chris Gerig and myself to this question:

Non-vanishing of group cohomology in sufficiently high degree

But I do not have time to think this through right now.

share|improve this answer
    
Thanks for your answer and helpful comments. Cheers –  George Melvin Nov 5 '12 at 0:14

Your Answer

 
discard

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.