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.


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