The following line of reasoning, apart from possible misconceptions in my part, is how instantons are usually (intuitively, at least) introduced:

(i) We look for minimum classical action solution for a pure Yang-Mills theory: these occur for pure gauge fields $A_\mu=\frac{-i}{g}U\partial_\mu U^{-1}$.

(ii) Furthermore, we work in temporal gauge and restrict ourselves to fields generated by gauge transformations which obbey $U(\mathbf{x})\to 1$ as $\mathbf{x}\to \infty$ (this appears to be a somewhat obscure and ill justified, although necessary, requirement, but is not what my question is about). This allow us to identify the infinity "(hyper)surface" as a single point and the whole patial part of spacetime $\mathbb{R}^n$ as $S^n$. Then, for each gauge configuration, we have a map from physical space to the gauge group manifold $S^n\to \mathcal{M}$, where $\mathcal{M}$ is a manifold related to some group quotient that is isomorphic to $S^{m}$. Which quotient exactly $\mathcal{M}$ represents for the general case of a $SU(N)$ theory is admitedly not clear to me and I would appreciate if something could be said about it.

(iii) If the map is surjective, then the possible minimum classical action configurations are divided into equivalence classes with different winding number. This becomes clear with the one-dimensional example of the compactified physical space being $S^1$ and the gauge groupd being $U(1)$. Maps that wind one circle over the other a different number of times cannot be continuously deformed into one another.

(iv) We then realize that the gauge field configuration related to a $m$-winding number cannot be continuously deformed into another with a $n$-winding number without ceasing to be pure gauge in the process.

I have two questions:

(1) The first one is regarding an assumption made in (iii): why does the map need to be surjective? Can't it, for instance, in the case of $U(1)$ be such that the "physical space covers only half of the group circle"? The reason for this seems to be the periodicity that results from the compactification, but I'm not sure.

(2) There are, however, another style of introduction that does not even mention the compactification of space. In this 'alternative' line of reasoning, which is less clear to me, we ask that the field be pure gauge only on the border of spacetime $\mathbb{R}^{n+1}$ (with respect to the previous space only $\mathbb{R}^n$), $x\to \infty$, which is also $S^n$, and then apply the same homotopy argument to reach the same conclusions as one would with the other procedure. My impression is that the first idea, which forces the field to be pure gauge in the whole space, actually looks for the different classical vacua and does not talk about the temporal slice because it assumes temporal gauge; while the last idea, described in this paragraph is the one which in fact looks for instanton solutions, which interpolate between vacua. Even if this is all correct, there are a lot of things that are not clear to me: Are both approaches equivalent or do they need to be taken into consideration simultaneously? How is one related to the other? Does the second approach also assumes the temporal gauge and the condition $U(\mathbf{x})\to 1$? I understand how this could be seen as "too many questions", but they are just meant to show my general confusion which could probably be settled by a few lines about the two approaches.

  • $\begingroup$ This from Tong's notes? $\endgroup$
    – JamalS
    Commented Jun 23, 2020 at 20:37
  • 1
    $\begingroup$ But gauge theory instantons are not pure gauge. If they were $\int F^2$ would be zero. They can be pure gauge at $\infty$ though. $\endgroup$
    – mike stone
    Commented Jun 23, 2020 at 20:42
  • $\begingroup$ @JamalS That is one of dozens of references I have looked at. Two other ones (each with one of the 'competing' point of views) are the books of Rajaraman and Srednicki. $\endgroup$
    – GaloisFan
    Commented Jun 23, 2020 at 20:49
  • $\begingroup$ @mikestone Yes! That is why I said in the second part of the question that the first approach, which looks for pure gauge solutions in the entire spatial slice are actually looking for classical vacua $\endgroup$
    – GaloisFan
    Commented Jun 23, 2020 at 20:50

2 Answers 2


I discuss a very similar confusion in this answer of mine.

Both approaches are trying to look for the same thing, inequivalent principal $G$-bundles on the compactified spacetime. Because it does not use the language of principal bundles, the physical literature is often very confused about what is actually going on mathematically. In the linked answer I explain why we need to look at compactified spacetime and are actually looking for inequivalent principal $G$-bundles over compactified spacetime. The first and the second approach are both attempts to physically motivate the conclusion that we really need to look at classes of maps from $S^3$ to $G$.

In the end, the point is that principal $G$-bundles, whose Chern class is the "instanton number", can be constructed on $S^4$ (compactified spacetime) by specifying the transition function on the overlap of two contractible patches covering $S^4$, and that overlap is homotopically equivalent to $S^3$, so you end up with homotopy classes $S^3 \to G$ classifying principal bundles (I explain this clutching construction a bit more in this answer of mine). This overlap is what traditional physics texts mean by the "sphere at infinity" or "border of spacetime". Whether you motivate it by looking at the "boundary" of spacetime $\mathbb{R}^4$ or by compactifying space $\mathbb{R}^3$ doesn't really matter.

I don't know what you're talking about in the part where you talk about maps $S^n\to S^m$, but it is a fact that $\pi_3(G)$ is $\mathbb{Z}$ for simple Lie groups and $\mathbb{Z}^n$ for more complicated Lie groups. $\pi_3(S^n)$, i.e. homotopy classes of maps $S^3 \mapsto S^n$, is either $0$ or $\mathbb{Z}$ depending on $n$, so this isn't a worrisome discrepancy - talking about the spheres adds nothing over looking just at $\pi_3(G)$.

  • $\begingroup$ Thank you for your answer! When I talked about "quotients of groups" and $S^m$, I was thinking of the case in which we add matter fields to the theory, when moduli spaces can appear instead of the simple $SU(N)~S^N$ (which is admittedly not directly related to the discussion). Unfortunately my differential geometry is not enough to truly appreciate your answer, but now I know where to look! $\endgroup$
    – GaloisFan
    Commented Jun 23, 2020 at 21:24

Nice question. I don't think I'd be able to answer everything but here are my thoughts. Please don't hesitate to ask if I've written things in an unclear way.

(1) Surjectivity is not necessary. The reason why it breaks down for your case is that if you consider the gauge transformation mapping onto any region of $U(1)$, which is a proper subset of $U(1)$, you'd get a map into a set that is topologically a line element and the first homotopy group of a line element is trivial $\pi_1(\mathbb{R}) \simeq 1$. If you were to choose a gauge group $\mathcal{G}$, that had a more complicated topology, such that restricting the gauge transformation to a subset of $\mathcal{G}$ still meant that the gauge transformation is classified by a non-trivial homotopy group, then you'd get an instanton (or monopole/vortex etc) with a non-trivial winding that is not surjective.

The simplest example I can come up with that would illustrate this is simply having two copies of $U(1)$, i.e. $U(1)\times U(1)$. You can define a gauge transformation that is the identity with respect to one part of the gauge group but winds around with respect to the other. I know this sounds trivial, but it captures the essence. Perhaps you can concoct a more elaborate example where the group does not factorize into the image and the "other part" so obviously, but the principles are the same - for the simple cases considered here, not covering the whole group, means you're mapping onto a different topology.

(2) As for the second part, I think that you mean that the boundary of $\mathbb{R}^n$ is $S^{n-1}$? You can see that this would mean you would have to consider a different homotopy group for the classification of the gauge transformation compared to if it was $S^{n}$. So, in general, this approach would be different. In fact, I am not familiar with any instanton solutions, which are arrived at by considering $\mathbb{R}^4$ as an $S^4$, since that would mean they are classified by $\pi_4$ (I would be happy to learn I'm wrong if you can provide me with a reference). The requirement that the gauge field is pure gauge at infinity is simply a finite energy condition, but one can choose a different gauge transformation at different direction of infinity in $\mathbb{R}^4$ and so the instantons are classified by the "boundary" of $\mathbb{R}^4$, i.e $\partial \mathbb{R}^4= S^3$.

I think the lesson to learn from this is that you don't need to compactify to get a non-trivial solution, if the boundary mapping to the gauge group is classified by a non-trivial homotopy group. However, if the homotopy group is trivial and we require finite energy configurations, then one can consider $U(x)\rightarrow \mathbb{I}$, as ${x \rightarrow \infty}$. In such a case, one is is allowed to compactify. I, too, don't know how to justify that last condition ($U(x)\rightarrow \mathbb{I}$, as ${x \rightarrow \infty}$).

  • $\begingroup$ Thank you for the thoughtful answer! I still have not had the time to read it carefuly and will just make this comment to clarify that I was, in fact, talking about $S^{n-1}$ as the boundary of $\mathbb{R}$. Thank you for pointing that out, too. I have corrected it in the question (feel free to edit the answer accordingly). $\endgroup$
    – GaloisFan
    Commented Jun 23, 2020 at 20:21

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