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Gauge theories are considered to live on $G$-principal bundles $P$ over the spacetime $\Sigma$. For convenience, the usual text often either compactify $\Sigma$ or assume it is already compact.

An instanton is now a "false vacuum" of the theory - a local minimum of the action functional. In four dimensions, instantons are the (anti-)self-dual configurations with $\star F = \pm F$, and the Yang-Mills action then is just the integral over $\mathrm{Tr}(F \wedge F)$, which is the characteristic class of the bundle, also called its Chern class, which are connected through Chern-Weil theory. This class is a topological invariant of the principal bundle associated to a field configuration.

As this math.SE post indicates, the isomorphism classes of bundles over a manifold are in bijection with its first Cech cohomology, which, for smooth manifolds, agrees with the usual other cohomology theories if $G$ is abelian. The question is now twofold:

  1. If $G = \mathrm{U}(1)$, does the existence of instanton solutions imply non-trivial first ordinary (singular, DeRham, whatever) cohomology of spacetime? Or is it rather the case that the instanton/non-trivial bundle configuration only indicates that the gauge theory only holds on the spacetime with points (or possibly more) removed, indicating the presence of magnetic monopoles at these points rather than anything about spacetime?

  2. If $G$ is non-abelian, does the existence of instanton solutions, and hence the non-vanishing of the "non-abelian Cech cohomology" imply anything about the topological structure of spacetime? Perhaps something about the non-abelian homotopy groups rather than the abelian homology? Or, again, does this indicate a non-abelian analogon of monopoles?

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First, the reasoning in the question about isomorphism classes of bundles is wrong, because the $\check{H}^1(M,G)$ from the linked math.SE post is not the cohomology of $M$ with coefficients in $G$, but actually the Čech cohomology of $M$ for the sheaf $\mathscr{G} : U\mapsto C^\infty(U,G)$.

However, this indeed has a relation to the cohomology of $M$ itself for $G = \mathrm{U}(1)$, via $$ 0 \to \mathbb{Z} \to \mathbb{R} \to \mathrm{U}(1) \to 0$$ which turns into $$ 0 \to C^\infty(U,\mathbb{Z})\to C^\infty(U,\mathbb{R}) \to C^\infty(U,\mathrm{U}(1))\to 0$$ since $C^\infty(M,-)$ is left exact and one may convince oneself that this particular sequence is still exact since the map $C^\infty(M,\mathbb{R})\to C^\infty(M,\mathrm{U}(1))$ works by just dividing $\mathbb{Z}$ out of $\mathbb{R}$. Considering this as a sheaf sequence $0\to \mathscr{Z}\to \mathscr{R} \to \mathscr{G} \to 0$, $\mathscr{Z} = \underline{\mathbb{Z}}$ for $\underline{\mathbb{Z}}$ the locally constant sheaf since $\mathbb{Z}$ is discrete, and the sheaf of smooth real-valued functions on a manifold is acyclic due to existence of partitions of unity, so taking the sheaf cohomology one gets $$ \dots \to 0 \to H^1(M,\mathscr{G}) \to H^2(M,\underline{\mathbb{Z}})\to 0 \to \dots$$ and thus $H^1(M,\mathscr{G}) = H^2(M,\underline{\mathbb{Z}}) = H^2(M,\mathbb{Z})$ where the last object is just the usual integral cohomology of $M$. Hence, $\mathrm{U}(1)$ bundles are indeed classified fully by their first Chern class which is physically the (magnetic!) flux through closed 2-cycles, and the existence of non-trivial $\mathrm{U}(1)$-bundles would imply non-trivial second cohomology of spacetime (or rather of one-point compactified spacetime $S^4$ since one should be able to talk about the field configuration "at infinty" and the bundle being framed at infinity). Indeed, since $H^2(S^4) = 0$, the existence of $\mathrm{U}(1)$-instantons would contradict the idea that spacetime is $\mathbb{R}^4$.

For general compact, connected $G$, it turns out the possible instantons are pretty much independent of the topology of $M$ because a generic instanton is localized around a point, as the BPST instanton construction shows - the instanton has a center, and one may indeed imagine the Chern-Simons form to be a "current" that flows out of that point, giving rise to a nontrivial $\int F\wedge F$.

Topologically, one may understand this by imagining $S^4$, and giving a bundle by giving the gauge fields on the two hemisphere, gluing by specifying a gauge transformation on the overlap of the two, which can be shrunk to $S^3$, i.e. the bundle is given by a map $S^3\to G$, and the homotopy classes of such maps are the third homotopy group $\pi_3(G)$, which is $\mathbb{Z}$ for semi-simple compact $G$. Since the "equator" can be freely moved around the $S^4$, or even shrunk arbitrarily close to a point, this construction does not in fact depend of the global properties of $S^4$, it can be done "around a point".

Thus, instantons in general do not tell us anything about the topology of spacetime.


This answer has been guided by the PhysicsOverflow answer to the same question.

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I agree with a lot of what you say, but I disagree with the conclusion you draw: topologically trivial spacetimes cannot support non-trivial G-bundles, and therefore have no instantons. So existence of instantons is always telling you that spacetime is non-trivial. People who claim that instantons exist on $\mathbb{R}^4$ are ignoring the crucial point that when we demand that the field configuration vanish at "infinity," we are really saying that spacetime is a four-sphere. (More precisely, we are saying the field configuration must be one that extends to a four-sphere, which is the same for the purpose of considering G-bundles).

I agree that if spacetime is a four-sphere or any manifold with trivial second integer cohomology, then it cannot support a non-trivial circle bundle and will not admit abelian instantons. So therefore yes, the existence of an abelian instanton would imply that spacetime has non-trivial second cohomology.

For a non-Abelian gauge group, you claim that being able to shrink the equator arbitrarily close to a point implies that the construction is independent of the global topology of spacetime. Not so. You are describing the "clutching function" construction of principal bundles over spheres, (which is already referencing the very special topology of the sphere). And the whole point is that instanton charges are valued in homotopy groups of the gauge group precisely because you can have a map from $S^3$ to your gauge group with a non-trivial winding number. Yes, you may shrink the 'equator' of $S^4$ arbitrarily close to a point, but you cannot collapse it down to a point. The nontrivial winding number of the map from the equator to the gauge group can never be undone by just shrinking the copy of the sphere down. One loose way to say what the winding number (instanton charge) is would be "the obstruction to homotoping that map $S^3 \to G$ to a map $\{pt\} \to G$." (Remember, if we're talking topology then shrinking a sphere `very close to a point' doesn't mean anything--all that matters is whether the map in question can be extended to the interior of the sphere and therefore truly shrunk to a point).

If you read your answer carefully, I think you'll find it actually argues the opposite: instantons can tell us a great deal about the topology of spacetime. Please refer to the mathematical field of Donaldson theory, where the whole business is to extract invariants of four-manifolds from the instanton moduli spaces.

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  • $\begingroup$ By the way, I would very much like to give an answer to this question that you closed, but I do not have sufficient reputation to comment on this site. Would you please re-open it? It's a good question that can be given a very sensible answer. physics.stackexchange.com/q/457822 $\endgroup$ – Todd N Jun 2 at 20:35
  • $\begingroup$ Hi, again I would like to please provide an answer to the question I linked above. Please re-open the question, I have a good answer to it. It is very frustrating not to be able to reach out directly to the user who first posted it. $\endgroup$ – Todd N Jun 10 at 19:49
  • $\begingroup$ Still here and still interested in helping the user whose question you were so eager to close. Again I ask that it be reopened: it is not a bad question at all. $\endgroup$ – Todd N Jun 18 at 5:13
  • $\begingroup$ @ACuriousMind Sorry I worry you may not have seen my previous comments because I did not tag you in them. Please reopen this question so I can provide the user with an answer. Many thanks! $\endgroup$ – Todd N Jul 11 at 10:35

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