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I am currently in the process of going through Ticciati's Quantum Field Theory for Mathematicians, which states the following (Theorem 13.7.11):

"Let $g$ be a differentiable function from $S^3$ to a [connected] simple group $G$. Then the winding number of $g$ is given by $$ \frac{1}{24\pi ^2}\int \epsilon ^{rst}\mathrm{tr}\, (A_rA_sA_t), $$ where $A_r=-(\partial _rg)g^{-1}$."

What is he referring to here by "winding number"?

Presumably he is referring to the degree of the map, but to the best of my knowledge, there is no notion of degree of a map unless the dimensions of the domain and co-domain manifolds are the same. He mentions previously that every such map is homotopic to a map from $S^3$ to either an $SU(2)$ or $SO(3)$ subgroup of $G$, in which case the notion of degree would make sense. What result is he referring to here and how does one make precise sense of the notion of a "winding number" in this case (he provides no precise definition)?

As a side (and perhaps irrelevant) comment, I don't really think this text deserves the "for Mathematicians" qualifier. I've found it to be better than most other texts, but its level of rigor is nowhere near that found in typical mathematics texts, and as such, I suspect that this 'Theorem' might be a bit tainted by the usual 'physicist slop' that comes with the territory (i.e. quantum field theory).

Andrews
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Jonathan Gleason
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    This reference says this comes from a theorem of Bott http://books.google.com/books?id=BCE_ey8wTOkC&lpg=PA408&ots=eXI3IKUzXe&dq=%22winding%20number%22%20%22lie%20group%22&pg=PA408#v=onepage&q&f=false – Eric O. Korman Apr 27 '13 at 18:09
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    Every nontrivial simple Lie group has $\pi_3=\mathbb{Z}$, so he could be referring to the induced map on $\pi_3$... – Jason DeVito - on hiatus Apr 27 '13 at 18:15
  • Dear Jonathan, It probably wouldn't hurt to present the actual formula. Then someone might be able to confirm that it is the map on $\pi_3$ which the author is referring to. Regards, – Matt E Apr 27 '13 at 21:28
  • @MattE Updated. – Jonathan Gleason Apr 27 '13 at 21:36

1 Answers1

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Amplifying Jason DeVito's comment above, you may want to look at this MO page (which includes a nice answer by Jason), and also this one, where the term index is used, rather than degree.

Community
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Matt E
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  • Thanks for the compliment! I didn't post my comment as an answer because I have almost no familiarity with QFT, so I was hesitant to assume that I had the right idea. – Jason DeVito - on hiatus Apr 27 '13 at 20:18
  • Is $\pi_3 G = \mathbb Z$ enough to imply the statement that any map (up to homotopy) $S^3 \to G$ factors through an $SU(2)$ or $SO(3)$ subgroup? I can see how to get this for something like $SU(n)$ (since $\pi_k SU(n) = \pi_k SU(n-1)$) but what about in general? Does the inclusion of an $SU(2)$ or $SO(3)$ subgroup always induce an isomorphism on $\pi_3$? – Eric O. Korman Apr 27 '13 at 20:28