# Recognising Chern-Weil forms

Given a smooth vectorbundle $$E\to B$$ with connection $$\nabla$$, the (real or complex) characteristic classes of $$E$$ are the cohomology classes of the Chern-Weil forms associated to $$\nabla$$.

Suppose $$E$$ is complex, and that we have a form $$\omega\in \bigoplus_i \Omega^{2i}(B;\mathbb R)$$ which represent $$ch(E)$$. Is there a connection $$\nabla$$ on $$E$$ such that $$\omega=ch(\nabla)$$? I'm pretty sure the answer is no; if I take the Chern-character forms of one connection in some degrees, and of another connection in other degrees, it seems unreasonable to expect there to be a third connection with the resulting combination of Chern-character forms. I would love to see a concrete example though. Also, it clearly sometimes happens that a form is the Chern-Weil form of a connection, and I wonder when:

Are there any known conditions we can impose on $$\omega$$ to ensure the existence of a connection $$\nabla$$ with $$ch(\nabla)=\omega$$?

I was aware of differential K-theory from before. The novelty of Simons-Sullivan, which I had not fully appreciated, is that they represent classes as $$(E,\{\nabla\})$$, where $$\{\nabla\}$$ is an equivalence class of connections under the equivalence relation $$\nabla\sim \nabla'$$ whenever the Chern-Simons form $$\widetilde{ch}(\nabla,\nabla')$$ is exact. (This is stronger than $$ch(\nabla)=ch(\nabla')$$.) Other authors (such as Freed-Lott, Klonoff, Bunke-Schick, Karoubi) use triples $$(E,\nabla,\phi)$$, where $$\phi$$ is a form. Then $$(E,\nabla,\phi)\sim (E',\nabla',\phi')$$ if $$E=E'$$ and $$\phi=\widetilde ch(\nabla,\nabla')+\phi'$$. This way it seems we are getting more forms as $$ch(E,\nabla,\phi):=ch(\nabla)+d\phi$$ than just the Chern-Weil forms. The Simons-Sullivan paper implies that this is not the case.