# Equivariant Venice Lemma

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?

• Btw, why is it called "Venice lemma"? – Qfwfq Oct 16 '19 at 18:20