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Chiral anomaly is computed very elegantly by Fujikawa method, which is also presented in Section 22.2 of Weinberg QFT textbook volume 2 or wikipedia.

Here, the underlying spacetime is assumed to be $\mathbb{R}^d$ (with $d=4$ in particular), which is contractible and therefore only trivial bundles are possble on it.

However, chiral anomaly leads to (a version) of the Index Theorem as in formula (22.2.49) in the aforementioned Weinberg book, implying that nontrivial topological configurations of the gauge fields are possible.

And this PE post says that nontrivial topologies of the gauge fields on $\mathbb{R}^4$ depend on asymptotic behavior at infinity, so that we can define them on $S^4$, the one-point compactification of $\mathbb{R}^4$.

So...I think there must be (very implicit) assumptions on asymptotic behavior of the gauge fields in the Fujikawa method, but I cannot figure out what those assumptions are and where they apply.

Could anyone please clarify for me?

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    $\begingroup$ Fujikawa's heat kernel method is essentially local, even though the integral over the complete spacetime gives a global invariant. $\endgroup$
    – mike stone
    Commented May 25 at 12:37
  • $\begingroup$ @mikestone Yes, and my understanding is that asymptotics of gauge configuration at infinity yields nontrivial topologies while spacetime integral does the job you said. $\endgroup$
    – Keith
    Commented May 25 at 13:27
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    $\begingroup$ Yes. I agree..... $\endgroup$
    – mike stone
    Commented May 25 at 15:44

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Indeed, from a mathematical standpoint, all the physics that has to do with the global topological terms of gauge theory "on $\mathbb{R}^4$" really happens on $S^4$. So how is it possible that the physicists never talk about $S^4$ yet obtain results related to non-trivial bundles on $S^4$ that are impossible on $\mathbb{R}^4$?

The trick is that you have to view the computations the physicist does on $\mathbb{R}^4$ as this being one of the two patches of a trivialization of the $S^4$ (i.e. one of the poles is "infinity" and this is a patch around the opposite pole), and the other patch is "at infinity" and you assume that there $A=0$ identically, since physical fields vanish at infinity. So all non-trivial phenomena happen in the other patch, and non-triviality of the bundle manifests itself in non-triviality of the transition function between the "normal" patch and the patch at infinity.

The $G$-bundles on $S^4$ can be classified by homotopy classes of maps $S^3 \to G$ (see also this answer of mine), which is essentially the homotopy class of the transition function between the two hemispherical patches, and so the "gauge at infinity" that the physicist sees in their $\mathbb{R}^4$ patch specifies the transition function of bundle, hence a non-trivial bundle on $S^4$ if as $x\to \infty$ the pure gauge $A(x)$ is not related to $A(x) = 0$ by a small gauge transformation: the "boundary at infinity" of the $\mathbb{R}^4$ is a $S^3$, and so $A \to g^{-1}\mathrm{d}g$ gives us a $\pi_3(G)$ that classifies the implicit bundle over $S^4$, and this $g$ is indeed the transition function of the bundle because we initially fixed $A=0$ in the patch at infinity, so the difference between $A=0$ and the value of $A$ in the overlap in the other patch is exactly the effect of the transition function.

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  • $\begingroup$ 1. What do you mean by "pure gauge $A(x)$"? $\endgroup$
    – Keith
    Commented May 25 at 11:25
  • $\begingroup$ 2. I will take a very close look at your link, but in this post, I do not see why $S^3$ appears even if we deal with 4D? $\endgroup$
    – Keith
    Commented May 25 at 11:26
  • $\begingroup$ Anyway, thank you for your reply! $\endgroup$
    – Keith
    Commented May 25 at 11:26
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    $\begingroup$ @Keith: 1. When $F=0$, then $A$ is pure gauge (see physics.stackexchange.com/q/98484/50583), i.e. $A = g^{-1}\mathrm{d}g = \mathrm{d}\alpha$ for a gauge transformation $g = \mathrm{e}^{\alpha}$ ($\alpha$ is the infinitesimal algebra-valued transformation). 2. Think about the two hemispheres of an ordinary sphere $S^2$ - the hemispheres are $\mathbb{R}^2$s, and their overlap/boundary is a $S^1$ - the equator. The patches in my answer are the same, just two dimensions higher and with the "equator" moved to a tiny circle around the pole at infinity. $\endgroup$
    – ACuriousMind
    Commented May 25 at 11:51
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    $\begingroup$ @Keith The connection to the homotopy group is because on the boundary we have the map $g : S^3 \to G$ and the $n$-th homotopy group of $G$ is by definition the homotopy classes of maps $S^3\to G$ $\endgroup$
    – ACuriousMind
    Commented May 26 at 11:50

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