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Let $\pi: E \to M$ be a fiber bundle with fiber $F$. We can define a connection to be a projection $v: TE \to VE$ satisfying $v \circ v= \text{Id}$, where $TE$ is the tangent bundle of $E$ and $VE$ is the vertical subbundle of $TE$ which consists of vectors tangent to fibers. We can think of $v$ as a $1$-form on $E$ with values on vertical vector fields, i.e. $v \in \Omega^1(E,VE)$.

The connection $v$ does not give a globally define $1$-form on $M$ but for any local trivializaton $U_{\alpha} \subseteq M$, $v$ gives a $1$-form $v_{\alpha}\in \Omega^1(U_{\alpha}, TF)$ which is obtained by pulling back $v$ by a by a trivializing section. The collection of $v_{\alpha}$ do not agree on intersections and one can write a rule showing how these transform.

I've heard that while the connection is only a locally defined $1$-form on the base with values on vertical vector fields the curvature of a connection is a globally defined $2$-form on the base with values on vertical vector fields. I would like some help in making this precise.

Lets define curvature $K \in \Omega^2(E,VE)$ by $K(X,Y)=v[X-v(X),Y-v(Y)]$, i.e. the failure of the bracket in preserving horizontal vector fields. If I do the same as above, pullback along a local section to define a collection of $2$-forms $K_{\alpha}$, will I get that all of these agree on intersections so they define a global $2$-form on $M$? How can I show this starting from the given definition of curvature? Thanks.

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I don't have time to write an answer, but I would recommend the book of Morita on Differential Forms. –  Neal Nov 20 '12 at 22:39
    
Oh I see, prove the Maurer Cartan formula locally and use this to see how the local curvature transforms. –  Manuel Nov 21 '12 at 0:24
    
See Berline, Getzler, Vergne, Heat Kernels and Dirac Operators, page 21. –  Kofi Nov 21 '12 at 21:10

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