Let $E\rightarrow X$ be a vector bundle and let $\mathcal{A}$ denote the space of connections on $E$. Pulling back $E$ by the second projection we obtain a vector bundle $\mathbb{E}=p\_2^\*E\rightarrow X$ over $\mathcal{A}\times X$. 


There exists a canonical connection $\mathbb{A}$ on $\mathbb{E}$ which is flat in the $\mathcal{A}$ direction and equal to $A$ on the slice $\{A\}\times X$. We obtain the following curvature: 
$$
\begin{cases}
 \mathbb{A}^2(v,w)= R\_A(v,w) & \text{for $v,w\in T\_xX$} \\
 \mathbb{A}^2(\alpha,v)=\alpha(v) & \text{for $\alpha\in T\_A\mathcal{A}$,$v\in T\_xX$ } \\
 \mathbb{A}^2(\alpha,\beta)=0 & \text{for $\alpha,\beta \in T\_A\mathcal{A} $}.
\end{cases}
$$
These identities can be found in Donaldson's "INFINITE DETERMINANTS, STABLE BUNDLES AND CURVATURE" p236 and Itoh and Nakajima's "Yang-Mills Connections and Einstein-Hermitian Metrics" p451. 


I do not understand the pairing of the middle line. How does it follow from the definition? 



Canonical connection on $\mathcal{A}\times X$
A vector field $v$ on $X$ and a vector field $\alpha$ on $\mathcal A$ give rise to two commuting vector fields on $\mathcal A\times X$, denoted by the same letters. Then we may regard $\alpha$ as (the pullback of) an element of $\Omega^1(X;\operatorname{End}E)$ as well. Write the connection as a covariant derivative $\nabla$, choose a section $s$ of $E\to X$ and compute for its pullback (still denoted $s$) that
$$\underbrace{\nabla\_\alpha\nabla\_v}\_{=\alpha(v)\in\operatorname{End}E}s-\nabla\_v\underbrace{\nabla\_\alpha s}\_{=0} -\nabla\_{\underbrace{[\alpha,v]}\_{=0}}s=\alpha(v)s\;.$$
Here we have used the interpretation of $\alpha$ as a variation of $\nabla$, and we have used that $\nabla$ is trivial in the $\mathcal A$-direction.


