Is Yang-Mills Instanton number zero? I am studying Yang-Mills instanton.
Suppose we have an action in $R^4$
\begin{equation}
S=\int_{M} Tr(F\wedge *F)
\end{equation}
where $F=dA+A\wedge A$.
The instanton number $k$ is defined as
\begin{equation}
k=\int_{M} Tr(F\wedge F)
\end{equation}
Now it can be shown
\begin{equation}
Tr(F\wedge F)=d\bigg[Tr(A\wedge dA+\frac{2}{3}A\wedge A \wedge A)\bigg]
\end{equation}
Applying Stokes' theorem, we have $k=0$?
I think I made a mistake. But what's wrong with my calculation?
 A: In QFT, we are used to ignoring boundary terms because they don't affect the perturbative dynamics, but they need not be $0$.
In fact, if one performs the boundary integral on $\partial M = S^3$ (after approprietly Wick rotating to euclidean time) it can be shown that the boundary integral
\begin{equation}
    k \propto \int_{\partial M}Tr(A\wedge dA+\frac{2}{3}A\wedge A \wedge A),
\end{equation}
is independant of the radius $R$ of the 3-sphere. It's not too hard to accept that this integral will depend only on the relationship between the gauge group $G$ and the boundary. 
I won't go into details here (an excelent reference is David Tong's lecture notes on Gauge Theory), but the instanton number is characterised by $\pi_3(G)$.
From a physical point of view, the instanton number is intepreted as the class of mapping from the boundary 3-sphere into the gauge group. Field configuration belonging in different classes cannot be continuously mapped onto one another.
Hope this helps a little.
