TITLE: Simplest explicit counterexample for $Vect(BG) \ne Rep(G)$ as monoids
QUESTION [9 upvotes]: Let $G$ be a topological group, $Vect(BG)$ the monoid of complex vector bundles over its classifying space (not the stack!) and $Rep(G)$ its monoid of complex representations.
Generally $Vect(BG) \ne Rep(G)$ since by Atiyah Segal completion the grothendiek ring of one is the completion of the other. Unfortunately I don't have any understanding of the Atiyah-Segal theorem so this question is not really about it.

Question: What is the simplest example of a vector bundle over a $BG$ which doesn't come from any continuous representation of $G$? Is there such an example for $\mathbb{CP}^{\infty}=BS^1$?

Edit: I realize now there are many examples for non-compact $G$. Is there a simple example for $G$ compact? (specifically $G=S^1$).

REPLY [3 votes]: Recall that $H^*(BS^3)=\mathbb{Z}[y]$ with $|y|=4$.  For any odd integer $k>0$ there is a map $\psi^k\colon BS^3\to BS^3$ with $(\psi^k)^*(y)=k^2y$.  (I am not actually sure if $k$ needs to be odd.)  There are evident inclusions $S^1\xrightarrow{i}S^3\xrightarrow{j}SU$.  The composite $\psi^k\circ Bi$ just comes from the representation 
$$z\mapsto\left[\begin{array}{cc}z^k&0 \\ 0 & z^{-k}\end{array}\right],$$
and the composite $Bj\circ \psi^k\in[BS^3,BSU]\subset K^0BS^3$ can be described in terms of Adams operations.  However, I do not know a way to construct $\psi^k$ itself without recourse to étale homotopy theory or something like that.  Section 5 of Sullivan's Geometric Topology notes is relevant here.  Anyway, if $V$ is the tautological bundle over $BS^3$ and $k>1$ then $(\psi^k)^*(V)$ is a vector bundle over $BS^3$ which comes from a virtual representation of $S^3$ but not from an honest representation.  More recent work on this sort of thing usually has the phrase "maps between classifying spaces"; there are papers by Jackowski, McClure and Oliver, and other papers by Notbohm.