Why is the Chern Number Invariant under A Continuously Shrinking of the Structure Group?

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?

• What is the mathematical translation or definition of "the quantization of (...) is reduced to the case of..." into maths? – Qfwfq Aug 3 '18 at 17:37
• @Qfwfq it is saying that the reason why $k\in\mathbb{Z}$ is reduced to... – Libertarian Monarchist Bot Aug 3 '18 at 17:38

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.

• 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 – Libertarian Monarchist Bot Aug 3 '18 at 17:54
• 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 $U(1)$. – John Klein Aug 3 '18 at 17:55
• Would you please give me a reference where I can study the details? – Libertarian Monarchist Bot Aug 3 '18 at 17:57
• So, in the end, what was the statement the OP was looking for? – Qfwfq Aug 3 '18 at 18:00
• 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. – Ben McKay Aug 3 '18 at 23:55