# 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?

• Just to highlight the issue, Stokes' theorem doesn't give that this is 0 if $\partial M$ is nonzero. The answer below nicely expands on this. Jul 7, 2020 at 16:02
• Other users think that you are leaving out a factor of $1/24\pi^2$, is this the case? Jul 7, 2020 at 16:36
• Jul 7, 2020 at 16:47

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 $$\pi_3(G)$$.