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In the book Quantum field theory by Mark Srednicki (chapter 93, pages 575-576) in order to compute winding number, $n$, in a 4-dimensional space with coordinates $x = (x_1, x_2, x_3, x_4)$ and such that

$$\hat{x} = (\sin\chi \sin\psi \cos\phi, \sin\chi \sin\psi \sin\phi, \sin\chi \cos\psi, \cos\chi), \quad \sum_\mu \hat{x}_\mu \hat{x}_\mu = 1$$

$n$ is given by

$$ n = -\frac{1}{24\pi^2}\int_0^\pi d\chi\int_0^\pi d\psi \int_0^{2\pi} d\phi\ \epsilon^{\alpha\beta\gamma}tr\{(U\partial_\alpha U^\dagger) (U\partial_\beta U^\dagger) (U\partial_\gamma U^\dagger)\}, \quad \epsilon^{\chi\psi\phi} = +1 $$

Where $U$ is only dependent on $\hat{x}$, belongs to $SU(2)$ and has the winding number $n$ associated. $tr$ represents the trace.

But suddenly Srednicki says that you can write $n$ as an integral over the surface of this 4-dimensional space of the form

$$ n = \frac{1}{24\pi^2}\int dS_\mu\ \epsilon^{\mu\nu\sigma\tau}tr\{(U\partial_\nu U^\dagger) (U\partial_\sigma U^\dagger) (U\partial_\tau U^\dagger)\}, \quad \partial_\nu = \partial/\partial x^\nu\ {\rm and\ so\ on} $$

I don't understand how you can go from one expression of $n$ to the other.

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  • $\begingroup$ I asked this in Math's site but I couldn't find any answer, so taking into account that there is Physics related to Winding number and my interest on it comes from theta vacua and instantons, I have cross-posted it $\endgroup$
    – Vicky
    May 30 '19 at 5:54
  • $\begingroup$ Crossposted from math.stackexchange.com/q/3179213/11127 $\endgroup$
    – Qmechanic
    May 30 '19 at 5:59
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I don't have Sredniki to hand, but it seems to be that your "four dimensional" space of the $x^\mu$ at the beginning of the question is really the three dimensional space forming the surface of a three sphere and the integral is over this three-dimensional space. Your $U\in {\rm SU}(2)$ is also parametrized by a three-sphere, so you have a winding number of 3-sphere about 3-sphere.

In the second equation, however the $x^\mu$ are clearly intended to be an actual four dimensional space and one is choosing some general (no longer necessarily a sphere) three dimensional surface embedded in that space. If you choose a surface $x^0=0$ for example then $\epsilon^{0abc}= \epsilon^{abc}$ and you are back to the original integral. Even if the new surface is not compact, you will still get an integer value for $n$ if $U\to {\rm identity}$ at infinity.

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