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In computing the variation of the action in Chern-Simons, and in other contexts, we get the following expression that is named the winding number, where $U$ comes from a gauge transformation:

$$ W[U] = \frac{1}{24 \pi^2} \int d^3x \ \epsilon^{\mu \nu \beta} \ \text{Tr}(U \partial_{\mu} U^{-1} \ U \partial_{\nu} U^{-1} \ U \partial_{\beta} U^{-1})$$

What is a complete proof about this being an integer? All the resources I have visited explicitly avoid showing this.

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The proof to this fact has to do with a mathematical concept known as Brouwer's degree of a mapping.

If one were to consider a mapping $f:\mathcal{M}\rightarrow G$ from $\mathcal{M}$ being the space-time manifold to some gauge group $G$, then the Brouwer degree of this map is

$$\deg f = \frac{\int_{\mathcal{M}} f^* \omega}{\int_G \omega}, \tag{1}$$

for any $m$-form $\omega$, where $\dim G =m$. Here $f^* \omega$ is defined as the pull-back under the mapping $f$. The quantity (1) is in general an integer.

But let's have a look at the specific case of $\mathcal{M}= S^3$ and $G= SU(2)$.

Let's consider the function $U$, parametrizing a general element of $SU(2)$,

$$ U= e^{-i \phi\sigma_3} e^{-i \theta \sigma_2} e^{-i \psi \sigma_3},$$

where $0<\phi< 2\pi$, $0<\theta< \pi$, $0<\psi< 4\pi$ are the Euler angles. This mapping covers every point of $SU(2)$ exactly once. Then we can define a $3$-form $$ \omega = U^{-1}dU \wedge U^{-1}dU \wedge U^{-1}dU.$$

Performing the integral over $SU(2)$ yields $$ \int_{SU(2)} \omega = 24\pi^2.$$

The next step is to recognize that $$f^* \omega= f^{-1}df \wedge f^{-1}df \wedge f^{-1}df.$$

This gives us the result that the Brouwer degree of a mapping from $S^3$ into $SU(2)$ is the winding number that you speak of. Now if you believe that the Brouwer degree is an integer, this completes the proof. But let's say that you don't believe that.

Now to see that $\deg f$ is an integer, we need to show that it doesn't change under small perturbations. Let's vary $f\rightarrow f+\delta f$ and showing that the integral of the first order variation of $\deg f$, which we define to be $\delta w(g)$, vanishes. Using the fact that $(f+\delta f)^{-1} = f^{-1} - f^{-1} \delta f f^{-1}$ to first order and the cyclicity of the trace, we can show that \begin{align} \int d^3x \, \delta w(g) = \frac{1}{24 \pi^2} \epsilon^{\mu \nu \rho} \int d^3 x \, 3 \partial_{\mu}\mathrm{Tr} \left[(f^{-1} \delta f) f^{-1} \partial_{\nu} f f^{-1} \partial_{\rho} f \right]. \end{align} Using Stokes' theorem and requiring that the variation $\delta f$ vanishes at the boundary (or equivalently, considering a manifold without a boundary, which is the case for $S^3$), we see that \begin{align} \int d^3 x \, \delta w(g) =0. \end{align} Thus we see that the winding number is unaffected by small variations of the mapping.

Now, in order to complete the proof, we will use the fact that maps between $S^3$ and $SU(2)=S^3$ are classified by the third homotopy group of the sphere $\pi_3(S^3)= \mathbb{Z}$. This means that the function $f$ can only wrap around the $SU(2)$ an integer number of times and so the integral $\int_{\mathcal{M}} f^* \omega$ can yield only an integer number times $\int_{SU(2)} \omega.$

The above example only shows that

$$ W[U] = \frac{1}{24 \pi^2} \int d^3x \ \epsilon^{\mu \nu \beta} \ \text{Tr}(U \partial_{\mu} U^{-1} \ U \partial_{\nu} U^{-1} \ U \partial_{\beta} U^{-1})$$

is an integer for $SU(2)$ and indeed for other gauge groups, one gets different normalizations and in this way one can get different quantization conditions on the Chern-Simons level. See this question, for example.

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    $\begingroup$ Could you fill in a little hole in the argument. So the homotopy group just implies that all continuous functions from $S^3$ to $S^3$ can be classified up to homotopy by equivalence classes labeled by integers. However why can you immediately say that $\int_{\cal M} f^* \omega$ will be then also an integer number of times $\int_{\cal M} \omega$, I feel there is a logical leap right there. How do you connect such integral with "number of wrapping around"? $\endgroup$
    – ohneVal
    Jun 17, 2020 at 14:09
  • $\begingroup$ If $f$ is a map that winds around $G=SU(2)$ $n$ times, every point in $G$ gets mapped onto $n$ times so the integral $\int_{\mathcal{M}} f* \omega$ is effectively integrating over the volume of $G$ $n$ times. If you come up with an example of an $f$ that does not map onto $G$ exactly $n$ times but is in the same homotopy class, then we can use the invariance of the winding number under small variations and deform it so that the map $f$ indeed maps onto $G$ $n$ times. This isn't rigourous, but this is physics stackexchange. $\endgroup$
    – Stratiev
    Jun 17, 2020 at 15:29
  • $\begingroup$ A topology book can probably show you why the Brouwer degree $\deg f$ is an integer, but here I've shown you the connection between the two so that can hopefully clarify things. $\endgroup$
    – Stratiev
    Jun 17, 2020 at 15:30
  • $\begingroup$ @Stratiev, do we need to assume $f$ continuous here, or can we work with $f$ continuous a.e.? $\endgroup$ Jun 17, 2020 at 15:37
  • $\begingroup$ @AlonsoPerezLona Yes, gauge transformations need to be continuous in general and the above arguments apply to continuous functions. (Pretty much every theorem in topology concerns continuous functions). $\endgroup$
    – Stratiev
    Jun 17, 2020 at 15:40

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