Consider a vector bundle $E$ over a manifold $M$ with flat connection, $\nabla$. From this data I can form the associative/unital differential graded algebra $\mathcal{A} = \left(\Omega^{\bullet}(M, End(E)), d\right)$ of forms with values in $End(E)$ where the differential is induced by $\nabla$ and squares to zero.
As $\mathcal{A}$ is an algebra over $\mathbb{R}$, I expect that we would say $\mathcal{A}$ is connected when $H^0(\mathcal{A})= \mathbb{R}$. However, when $M$ is simply connected, if I inspect
$$H^0(\Omega^{\bullet}(M, End(E)))= \{ s \in \Gamma(M, End(E)) \ \vert \ \nabla s = 0 \}\ne \mathbb{R}$$
I seem to be considering the $\nabla$-constant sections of $End(E)$. My thinking is that such an $s$ should be equivalent to* a choice of an endomorphism of the fiber of $E$, which is not necessarily just a choice of a real number.
What am I missing? Should I be thinking of the ring as $End(F)$? If so am I working in non-commutative algebra? Please help!
*My intuition for $\nabla s = 0$ is that $s$ does not vary from parallel transport along any choice of curve in $M$ and so given any choice in the fiber we could parallel transport that choice along $M$ (since $\nabla$ is flat this is well-defined).
Update: Note that iff $M$ is a point and the fiber is $\mathbb{R}^n$, then our complex is:
$$ \Omega^0(pt, End(E)) = End(\mathbb{R}^n) \to \Omega^1(pt, End(E))=0 \to \Omega^2(pt, End(E))=0 \to \dots $$
and so $H^0= End(\mathbb{R}^n)$ . I'm now wondering if I should be defining connectedness for some augmented dga.
