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 $ch(\nabla)=\omega$?


1 Answer 1


I was reminded about the Simons-Sullivan paper "Structured vector bundles define differential K –theory". The diagram on page 2 of that article implies that in fact any form representing a Chern character class is the Chern-Weil form of a connection. (And the diagram even specifies the space of choices of such connections.)

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.


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