For a given topological group $G$ there are natural transformations $$K^\* \leftarrow K^\*\_G \overset a\to H^{\*\*}(EG \times\_G -;\mathbb Q)$$ from equivariant K-theory, the first forgetting the $G$-structure of a bundle and $a$ inducing from an equivariant bundle over $X$ a nonequivariant one over $EG \times\_G X$ and taking the Chern character. These maps do respectably at detecting equivariant K-theory classes in nice cases, so I'm interested in how they can fail. 


Let us simplify matters by rationalizing and taking $X$ and $G$ compact. Then the forgetful map factors as $$K^\*\_G(X;\mathbb Q) \overset a\to H^{\*\*}(EG \times\_G X;\mathbb Q) 
\to H^{\*\*}(X;\mathbb Q) \overset\cong\to K^\*(X;\mathbb Q),$$ so we really only want the kernel of $a$. As $a$ can be viewed as completion of the $R(G) \otimes \mathbb Q$-module $K\_G^\*(X;\mathbb{Q})$ at the augmentation ideal $I(G;\mathbb Q)$, its kernel consists of classes annihilated by some element of $1 + I(G;\mathbb Q)$. 
Such classes exist when $G$ is discrete (already in $R(G) \otimes \mathbb Q$), but I don't know any where $G$ is connected.


**Does anyone have an explicit example of this happening when $G$ is connected**?



Equivariant bundles invisible in K-theory and Borel cohomology
The simplest example of what you are looking for occurs when $G = S^1$ and $X=S^1/C\_6$, where $C\_6$ is the group of 6th roots of unity. Then the map
$$ K\_G(X) \rightarrow K(EG\times\_G X)$$
identifies with 
$$ R(C\_6) \rightarrow K(BC\_6),$$
which has free abelian kernel of rank 2 (corresponding to the two conjugacy classes in $C\_6$ of order not a power of a prime).


