Simplest explicit counterexample for $Vect(BG) \ne Rep(G)$ as monoids

Question

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 $G=S^1$).

Answer 1

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.

Answer 2

Complex line bundles over $BG$ are classified by $H^1(BG, \mathbb{C}^{\times}) \cong H^2(BG, \mathbb{Z})$. On the other hand, $1$-dimensional complex representations are classified by $\text{Hom}(G, \mathbb{C}^{\times})$. There's a map from the latter to the former and it need not be surjective in general.

Explicitly, take $G = SL_2(\mathbb{R})$. This group is connected, so $BG$ is simply connected, and so the universal coefficient theorem, together with Hurewicz, gives

$$H^2(BG, \mathbb{Z}) \cong \text{Hom}(H_2(BG), \mathbb{Z}) \cong \text{Hom}(\pi_2(BG), \mathbb{Z}) \cong \text{Hom}(\pi_1(G), \mathbb{Z})$$

which is $\mathbb{Z}$ since $\pi_1(SL_2(\mathbb{R})) \cong \mathbb{Z}$. On the other hand, $G$ has no nontrivial $1$-dimensional complex representations.

I won't vouch for this being the simplest example though. I think I used to have simpler examples in mind.