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?

  • $\begingroup$ 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. $\endgroup$
    – 4xion
    Jul 7, 2020 at 16:02
  • $\begingroup$ Other users think that you are leaving out a factor of $1/24\pi^2$, is this the case? $\endgroup$ Jul 7, 2020 at 16:36
  • 1
    $\begingroup$ See physics.stackexchange.com/a/127880/50583 $\endgroup$
    – ACuriousMind
    Jul 7, 2020 at 16:47

1 Answer 1


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.


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