In the paper J. Simons and D. Sullivan. Structured vector bundles define differential K-theory, one of the key ideas is the so called Venice Lemma, which essentially can be stated as
Theorem: For every exact even differential form $\omega$ on a smooth manifold $M$, there is a trivial vector bundle $V = M\times \mathbb{C}^r$ with a connection $\nabla = \mathrm{d} + A$ such that $\omega$ is the Chern character form of this connection, i.e. $\mathrm{ch}(V,\nabla) -\mathrm{ch}(V,d) = \omega$.
Here, ch means the differential form arising from Chern-Weil theory. The proof is by explicit construction and induction on the degree. My question is about an equivariant version of this statement.
The setting that I want to work in is the following. Take $G$ a finite group and let $M$ be a smooth $G$-manifold. Then, we can define the delocalized de Rham complex of even forms $\Omega^0_G(M) = \left(\displaystyle\bigoplus_{g\in G} \Omega^{even}(M^g;\mathbb{C})\right)^G$. The action here is induced by the following space level action: An element $h\in G$ sends $x\in M^g$ to $hx\in M^{hgh^{-1}}$. A delocalized equivariant differential form is therefore nothing but the choice of a centralizer $Z(g)$-invariant differential form on $M^g$ for a representative $g$ of every conjugacy class in $G$. The cohomology of this complex is known to be the correct target of the equivariant Chern character.
Now I can pick an exact form $\mathrm{d}_G \omega = \displaystyle\bigoplus_{g\in G} \mathrm{d}\omega_g$ in the above complex, and ask for an (as a vector bundle) trivial $G$-vector bundle $E\to M$ with invariant connection $\nabla$ such that its Chern character form is equal to $\omega$, i.e., as above:
$\mathrm{ch}_g(E,\nabla)-\mathrm{ch}_g(E,\mathrm{d}) = \mathrm{tr}(g \cdot \mathrm{exp}\frac{i}{2\pi} \Omega_g) - \mathrm{ch}_g(E,\mathrm{d}) = \omega_g$
for all $g\in G$, where $\Omega_g$ is the restriction of the curvature $\Omega$ to $M^g$. Is it known if such a bundle can always be constructed?
