3
$\begingroup$

Apologies if this question is trivial or has been answered before.

If we consider a Yang-Mills theory (with a simple, compact Lie group $G$) on $\mathbb{R}^4$, it is well-known that all the finite-action gauge connections can be classified by the second Chern number $n$ (see this, for example) given by $$ n = \frac{1}{8\pi^2} \int_{S^4} \text{Tr} \left( F \wedge F\right).$$

The reason why we integrate over $S^4$ is that the boundary condition imposed by finite-action requirement makes it equivalent to consider principal $G$-bundles over $S^4$ instead, and it is possible to construct non-trivial principal bundles on $S^4$. However, this does not mean that we have suddenly changed our base manifold to $S^4$, just that the compactification to $S^4$ aids our classification. This brings me to my question:

Ultimately we are still living in $\mathbb{R}^4$, and any principal bundles we can construct over it is necessarily trivial and the second Chern number vanishes. Does this not mean that the so-called $\theta$ term $$ \frac{\theta}{8\pi^2} \int_{\mathbb{R}^4} \text{Tr} \left( F \wedge F\right)$$ equals 0? And if that is the case, this $\theta$ parameter would become unphysical, in contrary to basically what every paper about the $\theta$ term says.

What am I missing?

$\endgroup$
3
  • 2
    $\begingroup$ see e.g. the first section of this answer of mine $\endgroup$
    – ACuriousMind
    Commented May 11, 2021 at 8:36
  • $\begingroup$ I see! From the post you reference, it seems that the choice of using 4-sphere instead of Euclidean 4-space is one based on physics requirements (e.g. anomaly) but not mathematics alone. I'm not sure if this can be answered in a comment, but if that is the case, would it be fair to say that having the eta prime meson being so much heavier than the eta meson implies the non-trivial topology of the underlying spacetime (if we take anomaly as the explanation to the U(1)_A problem)? $\endgroup$
    – seric
    Commented May 14, 2021 at 1:47
  • $\begingroup$ I wondered about that in this question, too. As I explain in the first answer I linked, we get to $S^4$ by considering the behaviour of the field "at infinity". Whether you consider that infinity "part of spacetime" or not is your interpretation, but I think most people would not say that us considering the compactification $S^4$ because we want well-defined behaviour at infinity means "spacetime is non-trivial" - the actual spacetime is still $\mathbb{R}^4$, the infinite point is not "physically part of it". $\endgroup$
    – ACuriousMind
    Commented May 14, 2021 at 8:54

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.