If $E$ is a complex vector bundle over a manifold $M$ then one defines the space of vector valued $p$-differential forms on them as $\Omega^p(M,E) = \Gamma ( \wedge ^p (T^\*M) \otimes E) $


The connection be defined as the map , $\nabla: \Gamma(E) \rightarrow \Omega^1(M,E)$ satisfying $\nabla(fX) = df\otimes X+f\nabla X$ where $f \in C^{\infty}(M)$ and $X \in \Gamma(E)$


* Can't the connection be thought of as an element of $\Omega^1(M,End(E))$? 
Difference of two connections though not a connection is such an element.


Curvature of the connection is defined as the map, $$R = \nabla \circ \nabla : \Gamma(E) \rightarrow \Omega^2(M,E)$$


Corresponding definition of "trace" is as a map $Tr: \Omega^p(M,End(E))\rightarrow \Omega^p(M)$ satisfying some natural conditions. 


One can show that if $A,B \in \Omega^p(M,End(E))$ then $Tr[A,B]=0$


The crucial relationship from which a version of the Chern-Weil Theorem becomes almost immediately obvious is this,


$$dTr[A] = Tr[[\nabla,A]]$$ 


The argument for this begins by choosing a different connection say $\nabla'$ and seeing that, $$Tr[[\nabla',A]] = Tr[[\nabla,A]]$$


Hence the RHS of the desired equation is independent of the connection chosen and hence it can be evaluated for any connection to get the LHS. Using the fact that the a bundle is locally trivial one can choose the "trivial" connection and this should apparently yield $dTr[A]$


* Here I can't see what is a "trivial connection". It would help if someone can write that down in local trivializing coordinates. And how for that does the evaluation of the Tr give a de-Rham derivative of the Tr. The appearance of the de-Rham derivative on the LHS looks very mysterious and thats what makes the Chern-Weil Theorem click! It would help if someone can give the intuition behind this. 


One defines the Chern form as $det(I + \frac{\sqrt{-1}}{2\pi}R)$


	+ In this definition $I$ is the identity automorphism of $E$ but $R$ takes values in $\Omega^2(M,End(E))$. Then how is the "+" defined?


What is a good reference for this approach to Chern-Weil Theory in the language of connections and curvature? 
I found the notation of Kobayashi's lectures very old to relate to and Weiping's book is too terse and Milnor-Stasheff's book does it in the language of cohomology. 



Proving the basic identity which implies the Chern-Weil theorem 
(1) No, a connection is not a section of $\Omega^1(M,\mathrm{End}(E))$: a section would act tensorially and not satisfy the Leibniz rule. The connection is $\mathbb{C}$ linear and not $C^\infty(M,\mathbb{C})$ linear. 


(2) Since you have a vector bundle over some manifold, by definition there is some complex vector space $V$ such that around a neighbourhood of some point $p$ in the base manifold $M$, $E$ splits locally as $U\times V$ (a bundle is by definition locally trivial), where $U$ is a domain in $\mathbb{R}^n$. A section of $E$ can be locally expressed as a $V$-valued function on $U$. The trivial connection is just given by component-wise partial derivation. (In coordinates, we can pick a basis of $V$ and use the standard coordinate on $\mathbb{R}^n$, then you just have a map from a domain in Euclidean space to Euclidean space, given by a collection of functions. The trivial connection acts on each of the functions like the exterior derivative.)


(3) The Chern form takes value in $H^\*(M,\mathbb{R})$... if you accept that you can add objects in $H^2$ and $H^4$, why not the expression in the determinant? If you worry about such things, perhaps you'd be happier with the definition that the k'th Chern class is given by $(i / 2\pi)^k \sigma\_k(\Omega)$, where $\sigma\_k$ is the k'th symmetric polynomial acting on $\Omega$. 


(4) Another book that may be useful is Morita's *Geometry of Differential Forms*.


