1
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ What is the mathematical translation or definition of "the quantization of (...) is reduced to the case of..." into maths? $\endgroup$
    – Qfwfq
    Aug 3, 2018 at 17:37
  • $\begingroup$ @Qfwfq it is saying that the reason why $k\in\mathbb{Z}$ is reduced to... $\endgroup$
    – Valac
    Aug 3, 2018 at 17:38

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ 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 $\endgroup$
    – Valac
    Aug 3, 2018 at 17:54
  • 2
    $\begingroup$ 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
    Aug 3, 2018 at 17:55
  • $\begingroup$ Would you please give me a reference where I can study the details? $\endgroup$
    – Valac
    Aug 3, 2018 at 17:57
  • $\begingroup$ So, in the end, what was the statement the OP was looking for? $\endgroup$
    – Qfwfq
    Aug 3, 2018 at 18:00
  • 2
    $\begingroup$ 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
    Aug 3, 2018 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.