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In Witten's paper Three Dimensional Gravity Revisited and Quantization of Chern-Simons Theory with Complex Gauge Group, he used a fact that for a principal $G$-bundle, the quantization of the Chern number is reduced to the case of its maximal compact subgroup.

For example, the reason why the Chern-level $k$ of the $\mathrm{SL}(2,\mathbb{C})$-Chern-Simons theory

$$S[A]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right)$$

takes integer values ($k\in\mathbb{Z}$) is reduced to the case of the $\mathrm{SU}(2)$-Chern-Simons theory, because the group manifold $\mathrm{SL}(2,\mathbb{C})$ can be continuously shunk to $\mathrm{SU}(2)$.

Can anybody tell me how to prove this? Are there any references showing the proof that are easy to understand for physics students?

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    $\begingroup$ What is the mathematical translation or definition of "the quantization of (...) is reduced to the case of..." into maths? $\endgroup$
    – Qfwfq
    Commented Aug 3, 2018 at 17:37
  • $\begingroup$ @Qfwfq it is saying that the reason why $k\in\mathbb{Z}$ is reduced to... $\endgroup$
    – Xenomorph
    Commented Aug 3, 2018 at 17:38

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I don't quite understand the quantization part of the question.

However, the characteristic numbers are always given by pairing products of the (characteristic) cohomology classes with the fundamental class of the manifold $M$, say. Such classes are pullbacks of classes in the cohomology of $BG$, where $G$ is the structure group (where $M\to BG$ classifies the bundle).

Now, The cohomology of $BG$ is homotopy invariant in the sense that if $G\to H$ is a homomorphism which is also a homotopy equivalence of underlying spaces, then the map $H^\ast(BH) \to H^\ast(BG)$ is an isomorphism.

If we take $H := SL_n(\Bbb C)$ and $G = SU_n(\Bbb C)$, we get the statement you are looking for.

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  • $\begingroup$ Thank you John. Do you think it is true for $H=SL(2,\mathbb{R})$ and $G=U(1)$? I asked this question in another post mathoverflow.net/q/307352/120604 $\endgroup$
    – Xenomorph
    Commented Aug 3, 2018 at 17:54
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    $\begingroup$ yes, it is. The map $SO(2) \to SL_2(\Bbb R)$ is a homotopy equivalence. They are both homotopy equivalent to a circle, as is $U(1)$. $\endgroup$
    – John Klein
    Commented Aug 3, 2018 at 17:55
  • $\begingroup$ Would you please give me a reference where I can study the details? $\endgroup$
    – Xenomorph
    Commented Aug 3, 2018 at 17:57
  • $\begingroup$ So, in the end, what was the statement the OP was looking for? $\endgroup$
    – Qfwfq
    Commented Aug 3, 2018 at 18:00
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    $\begingroup$ More generally, every Lie group $G$ with finitely many components has a maximal compact subgroup $K$ and the inclusion $K \to G$ is a deformation retraction and therefore a homotopy equivalence. $\endgroup$
    – Ben McKay
    Commented Aug 3, 2018 at 23:55

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