## Return to Answer

4 added 249 characters in body; added 9 characters in body

You need to write $\omega X$ as $\omega \otimes X$.

If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of ${\Omega} ^2(M,End(TM))$. 2(M,End(TM))$, because then the value of$R(X,Y)S$at a point$m\in M$depends only on$X_m,Y_m$and$S_m$and not on their derivatives (That is even more elementary result - see any introductory book on differential geometry.though we have defined$R$via differentiation). The desired identification then follows from linear algebra$Hom(\Lambda^2 V \otimes E,E) \simeq Hom(\Lambda^2 V, \otimes E^* \otimes E)$&$End(E) \simeq E^*\otimes E$. The classical formula for curvature follows directly from the definition of the action of$\nabla$on$\Omega^p(M,E)$. 3 added 3 characters in body You need to write$\omega X$as$\omega \otimes X$. If you prove that$R: \Gamma(E) \to \Omega^2(M,E)$is tensorial - i.e.$R(X,Y)(fS) = fR(X,Y)S$where$S\in\Gamma(E)$,$f\in\mathcal{C}^\infty(M)$and$X,Y\in TM$, then it follows that this mapping is indeed an element of$\Omega^2(M,End(TM))$. {\Omega} ^2(M,End(TM))$. (That is even more elementary result - see any introductory book on differential geometry.)

The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.

2 added 202 characters in body; deleted 2 characters in body; added 122 characters in body

You need to write $d\omega \omega X$ as $d\omega \omega \otimes X$, where $X$is a section of the vector bundle $E$. For the curvature - .

If you need to prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of $\Omega^2(M,End(TM))$. (That is even more elementary result - see any introductory book on differential geometry.)

The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.

1