# Computing the the exterior derivative of the connection matrix on a hermitian vector bundle

If $$D$$ is the Chern connection on a holomorphic hermitian vector bundle $$(E,h)$$ and $$e$$ is a holomorphic frame, then $$\vartheta=\partial h \cdot h^{-1}$$ where $$h_{ij}=h(e_i,e_j)$$. Here $$\vartheta$$ is the connection matrix and $$\partial h$$ means the del-derivative of the matrix $$h$$.

I was going through the following calculation

\begin{align*} d\vartheta &= (\partial+\bar{\partial})\vartheta =\bar{\partial}\vartheta+\partial(\partial h\cdot h^{-1})= \bar{\partial}\vartheta-\partial h \wedge \partial h^{-1}\\ &= \bar{\partial}\vartheta + \partial h\wedge h^{-1}\partial hh^{-1} = \bar{\partial}\vartheta + \partial hh^{-1}\wedge\partial hh^{-1}. \end{align*}

I don't see where the last three equalities come from, firstly why do we have

$$\bar{\partial}\vartheta+\partial(\partial h\cdot h^{-1})= \bar{\partial}\vartheta-\partial h \wedge \partial h^{-1},$$

is this an application of the Leibniz rule? Secondly why does this equal

$$\bar{\partial}\vartheta-\partial h \wedge \partial h^{-1}=\bar{\partial}\vartheta + \partial h\wedge h^{-1}\partial hh^{-1}$$

and

$$\bar{\partial}\vartheta + \partial h\wedge h^{-1}\partial hh^{-1} =\bar{\partial}\vartheta + \partial hh^{-1}\wedge\partial hh^{-1}?$$

• Perhaps it would be better if you did this calculation by writing the matrices in terms of indices and work with scalar instead of matrix-valued forms. Commented Mar 18 at 0:21

So all your questions boils down to the following sequence of identities with the reasons listed in the same line. \begin{align*} &\quad\partial(\partial h\cdot h^{-1}) \\ &= \partial(\partial h)\cdot h^{-1} - \partial h\cdot \partial h^{-1} \quad\text{graded product rule: } \partial(\alpha\wedge\omega)=\partial\alpha\wedge \omega + (-1)^{\deg \alpha}\alpha\wedge \partial\omega\\ &= 0 - \partial h\wedge \partial h^{-1}\quad\quad \partial^2=0\\ &= \partial h\wedge h^{-1}\partial hh^{-1}\quad \text{derivative of inverse matrix: } \partial h^{-1} = -h^{-1}\partial hh^{-1}\\ &=\partial hh^{-1}\wedge \partial h h^{-1}\quad \text{associativity of matrix multiplication: }A(BC) = (AB)C. \end{align*}