Derivative of the Bott-Chern forms

Question

The Bott-Chern forms are constructed formally in Bismut's "Analytic Torsion and Holomorphic Determinant Bundle I" (page 74). This construction can be found as well in "Lectures on Arakelov Geometry" page 79.

These forms are used in Donaldson's "ANTI SELF-DUAL YANG-MILLS CONNECTIONS OVER COMPLEX ALGEBRAIC SURFACES AND STABLE VECTOR BUNDLES" to define a functional on the space of metrics on a holmorphic vector bundle $E\rightarrow X$. In that setup we consider the acyclic complex $$\mathcal{E}: \ 0\rightarrow (E,h_1)\xrightarrow{Id} (E,h_2) \rightarrow 0 $$ and we define $\widetilde{ch}(h_1,h_2)=\widetilde{ch}(\mathcal{E})$.

Donaldson writes the properties of the Bott-Chern forms in that constext and states that if $h_t$ is a path of metrics we have the following identity:

$$\dfrac{\partial}{\partial t}\widetilde{ch}(h_t,h_1)= -Tr(h_t^{-1}h_t'exp(i/2\pi F_{h_t})).$$

(this exact identity is taken form Itoh and Nakajima's "Yang-Mills Connections and Einstein-Hermitian Metric" p450).

I wonder how to obtain that property from the aforementioned abstract construction.

Answer 1

In order to construct the Bott-Chern class $\widetilde{ch}(h_0,h_1)$ we deform the Id complex over $\mathbb{P}^1$ \begin{equation*} 0 \rightarrow (E,\tilde{h}) \rightarrow (E,h_1) \rightarrow 0. \end{equation*} so that $i_0^*\tilde{h}=h_0$ and $i^*_{\infty}\tilde{h}\simeq h_1$. We get then \begin{equation*} \widetilde{ch}(h_0,h_1) = \int_{\mathbb{P}^1}ch(\tilde{h})\log|z|^2 \end{equation*} We now take a path of metric $h_t$ instead of $h_1$, with the corresponding $\tilde{h}_t=h_0(\tilde{g}_t\cdot,\cdot)$. Differentiating along that path we obtain \begin{eqnarray*} \dfrac{\partial}{\partial t}\widetilde{ch}(h_0,h_t) &=& \dfrac{\partial}{\partial t}\int_{\mathbb{P}^1}ch(\tilde{h}_t)\log|z|^2\\ &=& \int_{\mathbb{P}^1}\dfrac{\partial}{\partial t}ch(\tilde{h}_t)\log|z|^2\\ &=& -\int_{\mathbb{P}^1}Tr(\partial\bar{\partial}(\tilde{g}_t^{-1}\tilde{g}_t')exp(-\tilde{F}_t))\log|z|^2 \end{eqnarray*} where $\tilde{g}_t\in\Omega^0(X\times \mathbb{P}^1,End(E))$ and $\tilde{F}_t=F_{\tilde{h}_t}\in \Omega^2(X\times \mathbb{P}^1,End(E))$. Now notice that $\tilde{F}_t$ hence $exp(-\tilde{F}_t)$ are $\partial$ and $\bar{\partial}$ closed so we have \begin{equation*} Tr(\partial\bar{\partial}(\tilde{g}_t^{-1}\tilde{g}_t')exp(-\tilde{F}_t)) = \partial\bar{\partial} Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t)). \end{equation*} We can then keep going \begin{eqnarray*} \dfrac{\partial}{\partial t}\widetilde{ch}(h_0,h_t) &=& -\int_{\mathbb{P}^1}Tr(\partial\bar{\partial}(\tilde{g}_t^{-1}\tilde{g}_t')exp(-\tilde{F}_t))\log|z|^2\\ &=& -\int_{\mathbb{P}^1}\partial\bar{\partial}(Tr\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))\log|z|^2\\ &=& \int_{\mathbb{P}^1}Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))\partial\bar{\partial}\log|z|^2\\ &=& \int_{\mathbb{P}^1}Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))\partial_z\bar{\partial}_z\log|z|^2\\ &=& -2\pi i (\delta_0-\delta_{\infty})[Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))]. \end{eqnarray*} Recall that $(\tilde{h}_t)_{|0}=h_0$, so that $(\tilde{g}_t)_{|0}=Id$. We get therefore $Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))_{|0}=0$. We obtain \begin{eqnarray*} -2\pi i (\delta_0-\delta_{\infty})[Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))] &=& 2\pi i Tr(g_t^{-1}g_t'exp(-F_{h_t})). \end{eqnarray*}