When studying Yang-Mills instantons, there are two instances where one compactifies a space.

  • When classifying vacuum states, one demands $A_\mu(\mathbf{x})$ to become a constant as $\mathbf{x} \to \infty$.
  • When finding instanton solutions, one demands $A_\mu(x)$ to become a constant as $x \to \infty$.

We may then compactify space and spacetime to $S^3$ and $S^4$ respectively. Up to small gauge transformations, we find the vacuum states are classified by $\pi_3(G)$, while the instantons are classified by topologically distinct $G$-bundles on $S^4$, which are also indexed by $\pi_3(G)$.

These assumptions are absolutely crucial for the topological arguments to work, but I haven't seen them justified. Most textbooks say these conditions are necessary for the solutions to have finite energy and finite Euclidean action, respectively, but that's simply not true. For example, I could perform a large gauge transformation in either case to give $A_\mu$ whatever dependence I want at spatial or spacetime infinity, and this does not change the energy/action, by gauge invariance.

I haven't gotten any clarification from mathematically rigorous sources either, because they tend to compactify space immediately, without any physical justification or comment. What's the real argument?

  • 1
    $\begingroup$ Large gauge transformations are a topic fraught with pitfalls, cf. physics.stackexchange.com/q/314384/50583, where my answer in passing also discusses an argument for compactified spacetime from instantons. $\endgroup$ – ACuriousMind Jul 27 '18 at 16:59
  • $\begingroup$ @ACuriousMind Actually, that answer was one of the "mathematically rigorous sources" I was talking about! As far as I can tell, you pass immediately from $\mathbb{R}^4$ to $S^4$ and argue that if we didn't do that, the mathematics would be boring, since all bundles on $\mathbb{R}^4$ would be trivial. But that doesn't tell me why, physically, one should use $S^4$. Since the choice does have physical consequences, it should have a physical justification. $\endgroup$ – knzhou Jul 27 '18 at 17:02
  • $\begingroup$ I say at the end of the first section that non-trivial bundles, i.e. instantons, are necessary for their contributions to detectable effects like the axial anomaly. On $\mathbb{R}^4$, you don't get any instanton effects. $\endgroup$ – ACuriousMind Jul 27 '18 at 17:05
  • 1
    $\begingroup$ @ACuriousMind I mean, I accept that instanton effects are real, and that you would get the wrong answer if you used $\mathbb{R}^4$, but one still has to physically justify taking $S^4$ over $\mathbb{R}^4$. For example, if you're solving for the harmonic frequencies of a string whose ends are attached to walls, you cannot just say "we postulate fixed boundary conditions, because otherwise the calculated frequencies would not match experiment". Instead, you justify those boundary conditions by saying the ends are physically held in place. $\endgroup$ – knzhou Jul 29 '18 at 11:27
  • 1
    $\begingroup$ I disagree that one has to "physically justify" anything beyond "it matches experiment". Your string analogy is misleading because there's a clear ontology of the "theory of the string" in that it corresponds clearly to the physical string, but there is no such clear uncontroversial ontology for quantum mechanics in general. $\endgroup$ – ACuriousMind Jul 29 '18 at 11:46

The justification of the compactification to $S^3$ and $S^4$ is different.

In the first case (compactification of space), the compactification can be explained as follows: (This is a plausible physical explanation, not a complete mathematical proof).

We believe that the Skyrme model explains the low energy behavior of QCD. There are plenty of experimental results supporting this assumption. In particular, this model is able to predict certain properties even for heavy Baryons within 10% of the experimental values. The Baryon number in this model is given by: $$B = \frac{1}{24\pi^2} \int_{\mathrm{space}}\mathrm{Tr} \left ( U^{-1}dU \wedge U^{-1}dU \wedge U^{-1}dU \right ) $$ If space is flat, the baryon number of any finite mass baryon vanishes. Thus flat space does not support baryons. Please, observe that this consequence is very strong physically because it tells us that space-time which is a solution of Einstein's field equations of gravitation must be compact on the spatial slices.

In the instanton case, we can still be in a physical Minkowskian space-time. The solutions in the Euclidean signature just correspond to tunneling events in the physical space time. This is the basic trick of the usage of the Euclidean signature. It happens that these solutions if restricted to finite energy must vanish at the Euclidean infinity, thus effectively describing solutions on a compactified Euclidean space-time, but the amplitudes of these solutions correspond to true tunneling amplitudes in the physical Minkowskian space time.


Mathematicians(1): Mathematicians do not have interest in a physical explanation of why space-time is compactified. They choose whatever space-time suiting the mathematical result they require. So I don't think that you can find this type of explanations in mathematically rigorous works.

Mathematicians(2): Mathematicians engaged in quantum field theory research use functorial qft machinery (especially in tqft). According to this way of thinking a quantum field theory is just a black box which accepts a manifold as an input and returns a Hilbert space in the output, i.e., the same theory is not defined on a single space-time manifold, and can be simultaneously used on compact and non-compact manifolds with Minkowskian or Euclidean signatures.

Large gauge: You cannot apply large gauge transformations on instantons, because if you do so you get another inequivalent configuration with a different instanton number. Large gauge transformations are not redundancies in the description of the field theory as are small gauge transformations. Large gauge transformations are symmetries connecting inequivalent configurations. They are singular at infinity therefor unacceptable gauge transformations. This subject was discussed here on PSE in several occasions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.