# Covariant Exterior Derivative action on the $\mathrm{End}(E)$-valued p-forms

Suppose I define an operator $$d_A$$ by its action on sections $$s\in \Gamma(E)=\Omega_M^0(E)$$ of some vector bundle $$\Pi:E\rightarrow M$$ in a trivializing neighbourhood $$U\subset M$$ as $$d_As|_U=(ds+A\wedge s)|_U$$ for some connection $$A=(A^i_j)$$ with $$A^i_j\in \Omega^1(U)$$. By writing $$\omega\in\Omega^r_M(E)$$ as $$\omega=e_i\wedge \omega^i$$ where $$\omega^i\in \Omega^r(M)$$ and $$e_i\in \Gamma(E)$$ are local basis sections, and imposing, $$d_A(\eta\wedge \omega^i)=d_A\eta\wedge \omega^i+(-1)^{\mathrm{deg}(\eta)}\eta\wedge d\omega^i\qquad \forall \eta\in \Omega^r_M(E)\quad (1)$$ I've shown, $$d_A\omega=d_A(e_i\wedge \omega^i)=d_Ae_i\wedge \omega^i+(-1)^0e_i\wedge d\omega^i\\ =de_i\wedge\omega^i+A\wedge e_i\wedge \omega^i+e_i\wedge d\omega^i\\ =d(\omega)-e_i\wedge d\omega^i+A\wedge\omega+e_i\wedge d\omega^i\\ =d\omega+A\wedge \omega$$ Now suppose I have $$\sigma\in \Gamma(\mathrm{End}(E))=\Omega^0_M(\mathrm{End}(E))$$, I have been told that one can extend $$d_A$$ as a map, $$d_A:\Omega^r_M(\mathrm{End}(E))\rightarrow \Omega_M^{r+1}(\mathrm{End}(E))$$ by first imposing, $$d_A(\sigma)s=d_A(\sigma s)-\sigma d_As\quad (2)$$ to show, $$d_A(\mu\wedge \alpha)=d_A\mu\wedge \alpha+(-1)^{\mathrm{deg}(\mu)}(\mu\wedge d_A\alpha)\qquad\forall \mu\in \Omega_M^p(\mathrm{End}(E)),\alpha\in\Omega_M^q(E)$$ then it follows that, $$d_A(\mu\wedge \beta)=d_A\mu\wedge \beta+(-1)^{\mathrm{deg}(\mu)}(\mu\wedge d_A\beta)\qquad\forall \mu\in \Omega_M^p(\mathrm{End}(E)),\beta\in\Omega_M^q(\mathrm{End}(E))$$ It also apparently follows that if $$\mu\in \Omega_M^2(\mathrm{End}(E))$$ then, $$d_A\mu=d\mu+A\wedge \mu-\mu\wedge A$$ Here is my confused attempt, maybe consider them more as ideas:

From (2), if I ASSUME that I can do the following,

$$d_A(\sigma\wedge s)=d(\sigma\wedge s)+A\wedge\sigma\wedge s =(d\sigma)\wedge s+\sigma\wedge ds+A\wedge\sigma\wedge s$$

Combining with (2) we have,

$$(d_A\sigma)\wedge s=(d\sigma)\wedge s+\sigma\wedge ds+A\wedge\sigma\wedge s-\sigma\wedge ds-\sigma\wedge A\wedge s\\ =d\sigma\wedge s+A\wedge\sigma\wedge s-\sigma\wedge A\wedge s\\ =(d\sigma+A\wedge\sigma-\sigma\wedge A)\wedge s\\ \implies d_A\sigma=d\sigma+A\wedge\sigma-\sigma\wedge A$$

Then, if we write $$\omega\in \Omega_M^p(\mathrm{End}(E))$$ as $$\omega=f_i\wedge \omega^i$$ where $$f^i\in \Gamma(\mathrm{End}(E))=\Omega_M^0(\mathrm{End}(E))$$, and if I also ASSUME that (1) holds with for $$\eta\in \Omega_M^r(\mathrm{End}(E))$$ then,

$$d_A\omega=d_A(f_i\wedge \omega^i)=d_Af_i\wedge \omega^i+(-1)^0f_i\wedge d\omega^i\\ =df_i\wedge \omega^i+A\wedge f^i\wedge\omega^i-f^i\wedge A\wedge \omega^i+f_i\wedge d\omega^i \\ =d(\omega)-f^i\wedge d\omega^i+A\wedge\omega-(-1)^{(r)}f^i\wedge\omega^i\wedge A+f_i\wedge d\omega^i\\ =d\omega+A\wedge\omega-(-1)^r\omega\wedge A$$

which gives the $$\Omega^2_M(\mathrm{End}(E))$$ as a corollary.

I've had some other thoughts, none of them well enough formulated to write down, any help with the above would be very much appreciated. There is a reference: https://www.dpmms.cam.ac.uk/~agk22/vb.pdf page 35.