Let $\overline{\mathcal E}: 0\to \overline S\to \overline E\to \overline Q\to 0$ be a short exact sequence of Hermitian vector bundles over a complex manifold $X$. Here $\overline E=(E,h^E)$ is a holomorphic vector bundle $E$ together with a Hermitian metric on $E$, and similarly for $S$ and $Q$. We do not require $h^S$ and $h^Q$ to be induced from $h^E$. Then each bundle has a Chern connection, they will be denoted $\nabla^S,\ \nabla^E$ and $\nabla^Q$. We have the Bott-Chern form $\widetilde{ch}_{BC}(\overline{\mathcal E})$ which satisfies a double transgression formula
$$
d d^c\overline{ch}_{BC}(\overline{\mathcal E}) = ch(\nabla^E)-ch(\nabla^S)-ch(\nabla^Q)
$$
where $ch(\nabla)$ is the Chern-Weil form representing the Chern-character computed from the connection $\nabla$.
Smoothly $\overline{\mathcal E}$ is split in a canonical way, namely the orthogonal projection $E\to S$. Therefore $E$ carries, in addition to the Chern connection $\nabla^E$, also the connection $\nabla^S\oplus\nabla^Q$. My question is:
How does the Chern-Simons transgression form $\widetilde{ch}_{cs}(\nabla^E, \nabla^S\oplus\nabla^Q)$ relate to the Bott-Chern double transgression form $\widetilde{ch}_{BC}(\overline{\mathcal E})$?
My conjecture is that $\widetilde{ch}_{cs}(\nabla^E, \nabla^S\oplus\nabla^Q) =d^c \widetilde{ch}_{BC}(\overline{\mathcal E})$. My only evidence, however, is that both these forms are solutions to the equation $d\omega = ch(\nabla^E)-ch(\nabla^S)-ch(\nabla^Q)$. If my conjecture is a well known formula, I would greatly appreciate a reference!
Best Answer
The formula I conjectured is almost true. The correct formula is $ch_{CS}(\overline{\mathcal E})=\overline\partial ch_{BC}$. See the last line of the proof of proposition 3.15 of "Hermitian vector bundles and the equidistribution of the zeroes of holomorphic sections", in which Bott and Chern introduce the Bott-Chern form.