Difference of the curvature forms of two Chern connections is exact

Question

Let $h$, $h'$ denote Hermitian metrics on a line bundle $\pi: L \to M$.

Then the curvature from associated to each of these metrics (via the Chern connection) is $$\omega = \bar{\partial} \partial ( \text{log}(h)) $$ $$\omega' = \bar{\partial} \partial ( \text{log}(h')) $$

Now we want to show that the difference of these two curvature forms is an exact form. The proof of this I am trying to understand (produced in a lecture) starts by letting $h_t = (1-t)h + th'$ which is itself a Hermitian metric.

Then given $\omega_t = \bar{\partial} \partial ( \text{log}(h_t))$, we get \begin{equation} \frac{\partial \omega_t}{\partial t} = \bar{\partial} \partial ( \text{log}(h') - \text{log}(h))\\ \end{equation}

I don't understand how they got this though. I would have thought \begin{equation} \begin{split} \frac{\partial \omega_t}{\partial t} &= \bar{\partial} \partial \frac{\partial}{\partial t} \Big( \text{log}((1-t)h + th')\Big)\\ &= \bar{\partial} \partial \Big(\frac{-h + h'}{h_t} \Big) \end{split} \end{equation}

What am I doing wrong?

Answer 1

For this you don't need the homotopy. Just note $\bar\partial\partial \log h' - \bar\partial\partial\log h = d(\partial \log(h'/h))$.