Let $A \in \mathfrak{su}(2) \otimes \mathbb R^4$, so $A$ is a collection of $4$ elements of $\mathfrak {su}(2), (A_0, \dots, A_3)$. We can consider the system of equations $$ [A_0, A_1] + [A_2, A_3] = 0 \\ [A_0, A_2] + [A_3, A_1] = 0 \\ [A_0, A_3] + [A_1, A_2] = 0, $$ which come from the anti-self-dual equations (this is saying that the element $A \in \mathfrak{su}(2) \otimes \Lambda_-$, where $\Lambda_-$ is the $-1$ eigenspace of the hodge star $* : \Lambda^2 \to \Lambda^2$).
Is there a good reason why the only solutions to these equations are $A_i = \lambda_iv$ for some choice of $v \in \mathfrak{su}(2)$? (I can prove that this is the case by identifying $(\mathbb R^3, \times)$ with $\mathfrak{su}(2)$ and considering various planes that the $A_i$ must be in, but this isn't too insightful).
Note that this is saying that if $A$ is thought of as a constant connection on $\mathbb R^4$, then $F_A = [A_i, A_j]dx^i \wedge dx^j = 0$. In other words, we have a sequence of maps $$ A \to F_A \to F_A^+ $$ and if the composition is $0$ then the first map is also $0$. I'd like to know why this is the case.
I've worked out a solution, which I'll post in case anyone might find this useful. This is similar to the argument that the ASD connections are critical points for the Yang-Mills functional
First consider $$ \text{Tr}(F_A^2) = \text{Tr}((F_A)_{ij}(F_A)_{kl}) dx^i\wedge dx^j \wedge dx^k \wedge dx^l $$ where we think of $F_A$ having matrix values in $\mathfrak{su}(2)$ and the multiplication in the above is just the multiplication of these matrices. Now we can write this as $$ \text{Tr}(F_A^2) = \sum_{\sigma \in S_4} \text{sign}(\sigma)\text{Tr}((A_{\sigma(1)}A_{\sigma(2)} - A_{\sigma(2)}A_{\sigma(1)})(A_{\sigma(3)}A_{\sigma(4)} - A_{\sigma(4)}A_{\sigma(3)})dx^1 \wedge dx^2 \wedge dx^3 \wedge dx^4 $$ this expression is equal to $0$ since $\text{Tr}(ABCD) = \text{Tr}(BCDA)$, but the cycle $(1 2 3 4)$ is odd.
If we write $F_A = F_A^+ + F_A^-$ where $F_A^{\pm}$ is the projection onto the eigenspace of the hodge star then $$ \text{Tr}(F_A^2) = \|F_A^-\|^2 - \|F_A^+\|^2. $$ This follows from the fact that for $\text{Tr}(v^2) = -\|v\|^2$ for $v \in \mathfrak {su}(2)$, where this norm is just the standard euclidean norm (perhaps multiplied by 2), and if you pick a the standard basis for the (anti-)self-dual two forms then expanding out the expression gives the result.
Hence, we have that $$ \|F_A^-\| = \|F_A^+\| $$ and so if $F_A^+ = 0$ then $F_A =0$, so $A$ is flat.