Curvature form computations Let:

*

*$M$ be a differential manifold,

*$\pi:E\longrightarrow M$ a vector bundle on $M$ with $E_x\cong \mathbb{R}^k$,

*$D:\Gamma(E)\longrightarrow \Gamma(T^*M\otimes E)$ a connection on $E$,

*$d$ is the exterior derivative $d:\wedge^k T^*M\longrightarrow \wedge^{k+1}T^*M$.

Now the operator $D$ can be extended to higher cotangent sections of $\wedge^k T^*M\otimes E$ by setting $D(\omega\otimes\sigma):=d(\omega)\otimes\sigma+\omega\wedge D(\sigma)$
(the $\wedge$ symbol in the second summand is an abuse but it can be defined in a reasonable way).
Now i am trying to calculate the local expression of the curvature, given by a local expression for $D(D(s))$.
One need a local expression for $E$:
Let $\{s_1,\dots,s_k\}$ be a base for $E|_{U}$ with $U$ open in $M$.
Now $s=\sigma^k s_k$ (Einstein convention adopted and $\sigma^k$ are smooth functions).
From the property of $D$ one has $D(s)=d(\sigma^k)\otimes s_k+ \sigma^k\omega^i_k\otimes s_i$ (where $D(s_k)=\omega^i_k\otimes s_i$).
For the final step a can calculate separately the two summand becouse $D$ is linear:

*

*$D(d(\sigma^k)\otimes s_k)=d^2(\sigma^k)\otimes s_k+d(\sigma^k)\wedge D(s_k)=d(\sigma^k)\wedge\omega^j_k\otimes s_j$.

*$D(\sigma^k\omega_k^i\otimes s_i)=d(\sigma^k\omega^i_k)\otimes s_i+\sigma^k\omega^i_k\wedge D(s_i)=[d(\sigma^k)\wedge\omega^i_k+\sigma^kd(\omega^i_k)]\otimes s_i+\sigma^k\omega^i_k\wedge(\omega^j_i\otimes s_j)$.

This is not what I want, I need to find $D(D(s))= \sigma^kd(\omega^i_k)\otimes s_i+\sigma^k\omega^i_k\wedge(\omega^j_i\otimes s_j)$.
Questions:

*

*Set $\theta^i_k:=\omega^i_k$ and $\omega:=\theta$.
I don't think this is a good way to do computations because I can't see what's going on.
I tried to make sense, this is not formalized so I am gonna give the idea.
Fix the base $\{s_i\}$ so that $s$ is a vector of smooth functions on $U$, $(s_1,\dots,s_k)$ and if I fix the elements $\{1\otimes s_i\}$ i can say that $D(s)=ds+\theta s=(d+\theta)s$ indeed $ds:=d(\sigma^k)*(1\otimes s_k)$ and $\theta s=s^k\theta^i_k*(1\otimes s_i)$
(star stands for an action or something like that).
Now I look at $D$ when it goes from $T^*M\otimes E$ in $T^*M\wedge T^*M\otimes E$ and if $\theta=\theta^k\otimes s_k$ i have that $D(\omega)$ is $d+\theta$ indeed $d(\theta)=d(\theta^k*(1\otimes s_k)$ and $\theta\omega=\theta^k\wedge\omega^i_k*(1\otimes s_i)$ (again the star is something not defined).
Now $D(D(s))=D(d+\theta(s))=D(ds)+D(\theta s)=(d+\theta)(ds)+(d+\theta)(\theta s)=d^2s+\theta ds+d\theta s+\theta^2s$ in the meddle I can see due to the wedge product involved that there is a cancellation and so I get the correct form I expect.
Where is the mistake in the first calculation?
Does my observation stands for a general way to proceed and how can I formalize it?

Any hints or suggest would be appreciated, thanks.
 A: First, note that you have a small mistake. The formula for the extension of $D$ should read $$D(\omega \otimes s) = {\rm d}\omega \otimes s + (-1)^{{\rm deg}(\omega)} \omega \wedge Ds$$instead.
Let $s \in \Gamma(E) = \Omega^0(M;E)$. Then $Ds \in \Omega^1(M;E)$ and $D^2s \in \Omega^2(M; E)$. Here, I use the "shorthand" $\Omega^k(M;E) = \Gamma( (T^*M)^{\wedge k}\otimes E)$ as usual.
Claim: $(D^2s)(X,Y) = R^D(X,Y)s$.
Proof: Choose a local trivialization $(e_i)$ for the bundle and write $De_j = \omega^i_{~j}\otimes e_i$, as well as $s= \sigma^ie_i$. Then $$Ds = D(\sigma^je_j) = {\rm d}\sigma^j \otimes e_j + \sigma^j \omega^i_{~j}\otimes e_i = ({\rm d}\sigma^i + \omega^i_{~j}\sigma^j)\otimes e_i.$$Then $$\begin{align} D^2s &= D(({\rm d}\sigma^i + \omega^i_{~j}\sigma^j)\otimes e_i) \\ &= {\rm d}({\rm d}\sigma^i + \omega^i_{~j}\sigma^j) \otimes e_i - ({\rm d}\sigma^i + \omega^i_{~j}\sigma^j)\wedge De_i \\ &=  (\sigma^j {\rm d}\omega^i_{~j} + {\rm d}\sigma^j \wedge \omega^i_{~j})\otimes e_i - ({\rm d}\sigma^i + \omega^i_{~j}\sigma^j)\wedge (\omega^k_{~i}\otimes e_k) \\ &\stackrel{(\ast)}{=}  (\sigma^j {\rm d}\omega^i_{~j} + {\rm d}\sigma^j \wedge \omega^i_{~j})\otimes e_i - ({\rm d}\sigma^k + \omega^k_{~j}\sigma^j)\wedge (\omega^i_{~k}\otimes e_i) \\ &= (\sigma^j {\rm d}\omega^i_{~j} + {\rm d}\sigma^j \wedge \omega^i_{~j} 
 - {\rm d}\sigma^k \wedge \omega^i_{~k} - \sigma^j\omega^k_{~j}\wedge \omega^i_{~k})\otimes e_i \\ &= (\sigma^j {\rm d}\omega^i_{~j} + \sigma^j \omega^i_{~k}\wedge \omega^k_{~j})\otimes e_i \\ &= \sigma^j({\rm d}\omega^i_{~j} + \omega^i_{~k}\wedge \omega^k_{~j})\otimes e_i \\ &= \sigma^j \Omega^i_{~j}\otimes e_i,\end{align}$$as required. In $(\ast)$ we have renamed $i\leftrightarrow k$ in the second term (to make $\_\otimes e_i$ appear on both terms, so it could be factored) and in the last step, the second structure equation $\Omega^i_{~j} = {\rm d}\omega^i_{~j} + \omega^i_{~k}\wedge \omega^k_{~j}$. Here, the curvature $2$-forms $\Omega^i_{~j}$ are defined by the relation $R^D(\_,\_)e_j = \Omega^i_{~j} \otimes e_i$. The sign convention is $R^D(X,Y) = [D_X,D_Y] - D_{[X,Y]}$. $\quad\qquad\square$
Personally, I'm not a fan of this type of computation, and I think this covariant exterior derivative is better understood if you compare it with the usual exterior derivative in a coordinate free way. For example: $$\begin{align}{\rm d}\alpha(X,Y) &= X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y]), \qquad \alpha \in \Omega^1(M) \\ {\rm d}\alpha(X,Y) &= D_X(\alpha(Y)) - D_Y(\alpha(X)) - \alpha([X,Y]), \qquad \alpha \in \Omega^1(M; E), \end{align}$$and similarly for higher degree forms. In particular, when $E = TM$, $D({\rm Id}) = \tau$ is the torsion of $D$, given by $\tau(X,Y) = D_XY-D_YX-[X,Y]$, and $$D^2\tau(X,Y,Z) = R^D(X,Y)Z + R^D(Y,Z)X + R^D(Z,Y)X$$says that the content of the first Bianchi identity for torsion-free connections is "the derivative of zero is zero". For arbitrary $E$ again, we have the second Bianchi identity $DR^D = 0$, where $R^D \in \Omega^2(M; {\rm End}\,E)$ is regarded as a ${\rm End}\,E$-valued $2$-form. See my short notes for proofs.
