TITLE: Triviality of Associated Bundles
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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.