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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?

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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.

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