About the Yang-Mills equation Perhaps this is a sick question. How do we know that Yang-Mills equation does not have a non-trivial classical solution in Minkowski spacetime in vacuum? By non-trivial solution, I mean one that connects a pure gauge to another with a different Chern-Simons charge. We know that there are solutions in Euclidean spacetime, the BPST instantons, that can do this. And I am aware of the existence of barriers between two pre-vacua with different Chern-Simons charges. But I do not know how to prove (or show) it explicitly.
 A: Answered shortly. It has to do with compactness. Boundary conditions are very relevant for instanton-like solutions since by definition they must have a finite action. This immediately implies that such solutions must have a particular decaying behavior. In this process what is done mathematically is, compactify space-time. However this procedure makes sense only if your norm is euclidean because the quadratic form
$$x_\mu x^\mu = x_1^2 + x_2^2 + x_3^2 + x_4^2$$ 
is positive definite (technically $SO(4)$ symmetric which is a compact group), so a single point labeled as infinity can be defined, it representing all coordinates increasing and getting farther and farther from the origin. And then a solution can be found since expressions such as $|x|^{-n}$ make sense at infinity. This is not the case in the Minkowski case since its symmetry group is $SO(3,1)$ is not compact one cannot write solutions that fall-off consistently and give out finite actions.
Note: The above statements are not sufficient (just necessary) to have instanton solutions. Even in Euclidean signature in $\mathbb{R}^4$ with $U(1)$ symmetry, one does not have not trivial solutions because of gauge-equivalence.
