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I have begun to study instantons and I have the following difficulty:

$\newcommand{tr}{\operatorname{Tr}}$

I am considering theory with $SU(2)$ gauge group: $S=\frac{1}{2g^{2}}\int \tr F_{\mu\nu}^{2} $.

I've obtained the following expression for topological number $Q$ as a degree of mapping $S^{3}\longrightarrow S^{3}$ (based on differential volume form: $deg(f)=\int_{\Omega} f^{*}\omega$):

$$ Q=\frac{1}{24\pi^{2}}\int d\sigma_{\mu}\epsilon^{\mu\nu\lambda\rho}\tr\left(\omega\partial_{\nu}\omega^{-1}\cdot\omega\partial_{\lambda}\omega^{-1}\cdot\omega\partial_{\rho}\omega^{-1}\right)\tag{1} $$ Here I integrate over the sphere $S^{3}$.

Now I want to show that this expression is equal to the following: $$ Q=-\frac{1}{16\pi^{2}}\int d^{4}x \tr\left(F_{\mu\nu}{F_{\mu\nu}^{*}}\right)\tag{2} $$

The first equation I proved strictly, but for the second one I have no ideas except Stoke's theorem, but i cant do it explicitly.

Could you please explain me how to obtain second expression from the first one?

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It's Stokes's theorem. Consider a field $F = dA + A \wedge A$ such that $A$ is pure gauge at infinity, that is, $\lim_{x\to\infty} A(x) = \omega\, d \omega^{-1}$ for some $\omega : S^3 \to SU(2) \sim S^3$ where $\omega$ is a function on the 3-sphere because the limit can depend on the direction out to infinity.

In differential forms the first expression is $\newcommand{tr}{\operatorname{tr}}\tr A \wedge A \wedge A.$ Since at infinity, $F = dA + A \wedge A$ vanishes, we can add $0$ in the form of $-3\tr F \wedge A$ to this expression.1

We have $$d \tr (F\wedge A) = d \tr (dA \wedge A + A \wedge A \wedge A) = \tr dA \wedge dA + 3\tr dA \wedge A \wedge A. $$ Now apply Stokes's theorem where we regard infinity as a 3-sphere bounding spacetime. The last part follows from the cyclic property of the trace. Thus we obtain $$\int_{S^3} \tr A \wedge A \wedge A = \int_{S^3} \tr A\wedge A \wedge A - 3 F\wedge A = \int_M \tr \big[3 dA\wedge A \wedge A - 3 dA \wedge dA - 9 dA\wedge A\wedge A\big].$$ Except for the term $A^{\wedge 4} = A\wedge A \wedge A \wedge A$, we are taking the trace of $-3 F\wedge F$. But $A^{\wedge 4}$ is traceless (use the cyclic property of the trace). Hence $$\int_{S^3} A\wedge A \wedge A = -3 \int_M F\wedge F.$$

Now $F\wedge F = 2 F^{\mu\nu} F^*_{\mu\nu}$ because the $F^*_{\mu\nu}$ tensor is usually defined with a $\frac{1}{2}$ factor, so this establishes the relative $-\frac{3}{2}$ in the two expressions for $Q$.


1 This seems pulled out of thin air because it is. I add it because I know what I want to get. Usually, one goes in the opposite direction starting with $F \wedge F = d(dA \wedge A + \frac{2}{3}A\wedge A \wedge A)$. The term we added is the part of this that vanishes at infinity. Putting it back in is a little less natural.

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  • $\begingroup$ Thank you for your answer! Also I've learned how to write "trace" in the right way from it ;) $\endgroup$ – xxxxx Oct 6 '14 at 21:26
  • $\begingroup$ Btw, I think you meant $-9 dA\wedge A\wedge A$ instead of $-9 A\wedge A\wedge A$, right? (: $\endgroup$ – xxxxx Oct 8 '14 at 19:53
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Robin Ekman Oct 8 '14 at 21:12

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