# Why are topological properties described by surface terms?

An example are the anomalies in abelian and non-abelian gauge quantum field theories.

For example, the abelian anomaly is $\tilde {F}_{\mu\nu}F^{\mu\nu}$ and the integral over this quantity is a topological invariant which measures a topological characteristic of the gauge field $A_\mu$.

All such quantities can be rewritten as total derivatives and then, using Gauss' law transformed into a surface integral.

What is the intuitive reason that quantities which describe topological properties can always be written as surface integrals?

• Because otherwise they would influence the equations of motion, and be non-topological by definition? – Prof. Legolasov Nov 20 '17 at 12:13
• @SolenodonParadoxus Surface terms do not influence the equations of motions and that's what makes them different. So far, so good. Now, my problem is seeing the connection to topology. Maybe a better example is the Hydrodynamical helicity en.wikipedia.org/wiki/Hydrodynamical_helicity, which is also a topological quantity and described by a surface term. It describes the "knottedness of vortex lines in the flow". Why do we describe such topological characteristic of the system with a surface integral? – jak Nov 20 '17 at 12:22
• @SolenodonParadoxus Formulated a bit differently: Why are topological properties always completely encoded in the boundary of the system? – jak Nov 20 '17 at 12:29
• -1. Not clear. Are you asking why the Divergence Theorem works? – sammy gerbil Nov 20 '17 at 12:56
• @sammygerbil no. I'm asking, why a topological quantity, such as the ""knottedness of vortex lines in the flow" or the "winding of the gauge functions" is completely determined by a surface integral, i.e. completely encoded in the boundary of the system – jak Nov 20 '17 at 12:58

I am giving you an answer from the path integral quantization point of view.

When we quantize a gauge theory, we need to sum over all the configurations of connections of a gauge group on some manifold modulo gauge transformation.

In contrast to the affine space of all connections (without dividing by gauge transformation), the latter space (of connections modulo gauge transformation) can disconnected corresponding to different sectors of bundles which cannot be deformed into each other by any combination of diffeomorphism of the base manifold or continuous deformations of the transition functions of the fiber.

In quantum theory, we absolutely need to sum over all topological sectors in the path integral. For example if we do not do that in the problem of a particle moving on a circle, we do not get the correct answer given by Schrödinger's equation.

Chern and Weil (please see the following expositionby Fecko) discovered a profound theorem that there exist topological invariants differentiating between the principal or vector bundles, with the same structure group, which can be expressed by means of certain polynomials of the curvature of the connection. These topological invariants do not depend on the connection (they are gauge invariants) nor on the field strengths but only on the bundle. Moreover, these topological invariants -named characteristic classes- can be expressed by means of cohomology classes of the base manifold (this is why they are closed).

Thus, we can use these invariants to weigh different bundles in the path integral because they do not depend on the connections nor the fields but only on the bundles.

There are other topological invariants which cannot be written in terms of differential forms, such as the Stiefel–Whitney classes, on which the existence of fermions on the manifold depends. These invariants also affect the path integral, however, more advanced techniques are needed to take them into account.

It is worthwhile to mention that not every closed form on the base manifold is an image of a characteristic class (or can be written as a combination of characteristic class). Thus, as topological terms, we can consider only special types of closed forms.

• Thanks a lot! The expostionary notes by Fecko are amazing. One question: What's the interpretation of the "different bundles" we need to sum over here? I know that we get a very different bundle e.g. when a monopole is present compared to the vacuum case. However it's not surprising that bundles are different for physically different systems and there wouldn't be a need to sum over them. Thus, I suppose you're talking about inequivalent bundles for one system, like, e.g. the vacuum. Do the inequivalent bundles you mention correspond e.g. to different instantons (winding 1, 2, etc.)? – jak Nov 20 '17 at 16:05
• Yes, The different bundles in the case of QCD are the instanton bundles. They all describe different configurations of a single system, namely QCD. – David Bar Moshe Nov 20 '17 at 16:09

Pretty much every time a physicist says "topological invariant", they mean a topological invariant of a vector bundle (the vector bundle the fields take values in, usually), the most common of which are the Chern classes.

The Chern classes can be expressed as integrals of polynomials in the curvature (it does not matter which curvature, so even if the physical theory doesn't contain a gauge field, you can simply choose/construct one ad hoc) $\int F\wedge F$, and it so happens that these polynomials in the curvature are locally total derivatives of their associated Chern-Simons forms. So, at the level of rigor of physics where one now usually disregards the "locally", the integrals of the polynomials in the curvature which yield the Chern classes are integrals of total derivatives, hence "boundary terms". The deeper mathematical story of why the curvature polynomials that represent integral cohomology classes of a bundle in DeRham cohomology should be such total derivatives is the story of secondary characteristic classes and differential cohomology.

It should be stressed, however, that once we try to be bit more rigorous than the average physicist, the Chern classes are not "surface terms". In fact, that terminology doesn't make any sense because they're invariants of vector bundles over ordinary manifolds, and ordinary manifolds do not have a boundary - any "surface terms" simply vanish on a compact manifold, and are potentially ill-defined on non-compact ones. What's really happening is that, once again, the physicist hides the global property of a bundle on something like compactified spacetime $S^4$ by just looking at it in one of the coordinate patches, pushing out all the structure of the second necessary patch "to infinity", i.e. "to the surface", precisely as in this answer of mine to your question about large gauge transformations.

Additionally, "surface term" usually carries some sort of connotation with it that the choice of function on the surface still matters - but it does not, the Chern class is independent of the choice of connection, as a true topological invariant should be, it is solely a function of the topological homeomorphism class of the bundle. Using some sort of gauge field/curvature to compute the topological invariant is merely a crutch because it's often easier than "purer" topological computations, especially for physicists who don't know such topology.

This is not a full answer but I'll give it in order to contemplate some aspects not covered in the previous (good) answers and explicitly mentioned in the OP's comment:

I'm asking, why a topological quantity, such as the ""knottedness of vortex lines in the flow" or the "winding of the gauge functions" is completely determined by a surface integral, i.e. completely encoded in the boundary of the system

In this case, the topological invariants you are referring to are related to homotopy classes. Specifically, in the case of gauge theories, we seek for finite energy (or tension in the case of a vortex line) solutions and this imposes some constraints on the asymptotic fields. Each (non negative) term in the hamiltonian density $\mathcal H$ has to vanish sufficiently fast as the fields approach the spatial infinity. In particular, the potential $V$ has to vanish asymptotically. This means that the scalar field belongs to the vacuum manifold when $r\rightarrow\infty$. This asymptotic field provides a map from the spatial infinity (which depends on the spatial dimension of the model) to the vacuum manifold and different configurations (different maps) are classified into equivalent classes according to homotopy groups. Maps belonging to different classes cannot be continuously deformed into each other and that is why non trivial maps (non trivial winding number for example) are said to be topologically stable or protected. As you can see, the topological invariants (namely the homotopy classes) in these models are encoded in the asymptotic fields or in the boundary of the system.

• thanks for your answer! However, what you point out is exactly what confuses me. The winding of the field does not happen at spatial infinity. All gauge field configurations and all gauge transformations we care about are trivial at spatial infinity. The winding happens in the bulk and not on the boundary. Suppressing the time dimension and restricting to one spatial dimension, I think this is how a gauge transformation with winding number 1 for $U(1)$ looks like: i.stack.imgur.com/mshyZ.png. A gauge trafo with winding 0 would be the one with all arrows pointing upwards etc. – jak Nov 21 '17 at 9:34
• (The picture is adapted from Frankel's Geometry of Physics page 557). When we restrict ourselves to gauge transformations that are trivial at infinity and field configurations that are trivial at infinity, spatial infinity is just a point and we can thus compactify our 3 spatial dimensions to $S^3$. Thus, I'm not sure what you mean with maps from "spatial infinity" to the vacuum manifold. – jak Nov 21 '17 at 9:36
• @JakobH Why do you say the winding of the fields happens at the bulk? Let us consider a concrete example: (3+1) Yang-Mills-Higgs. Then the winding happens at infinity, not at the bulk. Notice that what I mean by infinity is "sufficiently far from the core". The winding number of a cosmic string is counted far from its core. Same for the winding of a monopole. The point is that if two asymptotic configuration (far from the core) cannot be continuously deformed into each other, the the field for the whole space cannot as well. – Diracology Nov 21 '17 at 15:38
• @JakobH Regarding the example you mentioned, there you are mapping from $\mathrm R$ to $S^1$ and then you stereographically project $\mathrm R$ onto $S^1-\mathrm{North\, Pole}$. You end up with a map $S^1-\mathrm{North\, Pole}\rightarrow S^1$. This sounds artificial if we want to speak about winding number since this can be defined as the number of times we cover the image $S^1$ as we go around once at the domain $S^1$. In any case, the fact that this depend on the bulk is very particular to that one dimensional example. – Diracology Nov 21 '17 at 15:50
• @JakobH In the example of vortices or monopoles, the gauge transformation is not trivial at infinity at all. In fact (as discussed in the answer) the asymptotic scalar field belongs to the vacuum manifold and we obtain it by a non trivial gauge transformation (large gauge transformation) to an arbitrary point $\phi_0$. For the 't Hooft-Polyakov monopole, for example, the explicit transformation is $g(\theta,\phi)=\exp(-in\phi T_3)\exp(-i\theta T_2)\exp(in\phi T_3)$ where $T_i$ forms an $su(2)$ algebra... – Diracology Nov 21 '17 at 15:58