# Triviality of Associated Bundles

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.

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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

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 $Vect(BG)$.

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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.

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