# Why is the QCD $\theta$-term aware of the topology of the space?

The QCD Lagrangian without the $$\theta$$-term $$\mathcal{L}_{QCD}=-\frac{1}{4}G_{\mu\nu}^aG^{\mu\nu a}\tag{1}$$ is not topological. However, the $$\theta$$-term $$\mathcal{L}_\theta=\frac{\theta}{32\pi^2}G^a_{\mu\nu}\tilde{G}^{\mu\nu a}\tag{2}$$ is topological!

What is so special about the $$\theta$$-term that it knows about the topology of the space?

• Assuming you've already worked through a derivation that the $\theta$-term is topological, and can see the same derivation won't apply to the Yang-Mills term, can you be more specific about what kind of answer you are looking for? As written I can't think of much to say beyond "because it has the right structure so that the argument it is topological works (total derivative, depends on holonomy, etc)." Nov 17, 2021 at 17:15
• I mean, since this term is sensitive to the topology, there must be an indication of it in the definition of the term? But all I can see is that the QCD Lagrangian and the $\theta$-term differs by a factor of $\varepsilon_{\mu\nu\sigma\rho}$. But the germ of topological information must be built in the dual field tensor. Isn't it? @Andrew
– SRS
Nov 17, 2021 at 17:35
• By derivation do you mean the relation of this term to the topological winding number? Assuming that I haven't gone through the "derivation", is there any indication that this will something to do with topology? @Andrew
– SRS
Nov 17, 2021 at 17:37
• The $\epsilon_{\mu\nu\rho\sigma}$ kind of is what makes it topological. Basically that structure enforces cancellations like $\epsilon^{\mu\nu\rho\sigma}\partial_\mu\partial_\nu \phi = 0$ for some generic field $\phi$, since partial derivatives commute and the epsilon tensor is totally antisymmetric. That guarantees the term is a total derivative, since integrating by parts you get this kind of cancellation. Nov 17, 2021 at 17:37
• The interesting thing about topological terms is not that they know about the topology, it's that they don't know about the geometry.
– d_b
Nov 17, 2021 at 20:25

1. There is nothing special about functions "knowing" about topology. In fact, things like Morse theory rely on the notion that the "typical" smooth function on a manifold always reflects some topological characteristics of the manifold in its various properties.

2. The "topological term" is a total derivative of its associated Chern-Simons current. As such, it can't have any local effect as a term in the Lagrangian, so any effect it does have would have to be called "topological" or "global" in the usual nomenclature. (Yes, this is somewhat tautological, but physicists usually do not use the mathematical notion of "topological" as precisely as mathematicians do)

3. It is crucially not the topology of spacetime that the $$\theta$$-term detects, but the topology of the principal bundle of our gauge theory. The proof that the integral over the $$\theta$$-term is a topological characteristic of this bundle is the foundation of Chern-Weil theory, and it would be hard to explain this connection any shorter than the construction of the Chern-Weil map in the Wikipedia article I linked there.

4. As AccidentalFourierTransform points out in a comment, "topological" can also mean "does not depend on the metric". The $$\theta$$-term can be computed without the use of the spacetime metric, since in the language of coordinate-free differential geometry it is just $$\mathrm{tr}(F\wedge F)$$, and the exterior product does not depend on the existence of a metric, in contrast to the standard Yang-Mills term which is $$\mathrm{tr}(F\wedge{\star}F)$$ and where the Hodge star does depend on the metric.

• Of course it is hard to say given the lack of context in the OP, but topological is also often used when the term does not depend on the spacetime metric (the standard kinetic term does, the theta term does not). In this sense, topological terms do measure the topology of the spacetime. For example, according to this definition, Chern-Simons terms are topological even though they are not total derivatives. Nov 17, 2021 at 20:59
• Can you explain how is the theta term independent of the metric but not the kinetic term? In both terms, the indices of the field strength tensor can be raised (or lowered) by bringing in two metric tensors. So implicitly both have metric tensors, right? @AccidentalFourierTransform
– SRS
Nov 19, 2021 at 13:55
• @SRS It's very simple: Starting from $F_{\mu\nu}$, you can define the $\theta$-term without ever using a metric, but you can't define the kinetic term without raising indices with a metric. Nov 19, 2021 at 13:57