Timeline for Why is the Chern Number Invariant under A Continuously Shrinking of the Structure Group?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2018 at 0:24 | comment | added | John Klein | @NewStudent There is standard place to look: (1) Milnor and Stasheff's Characteristic classes. See mathoverflow.net/questions/201368/… for information about the statement I made. Note too that the cohomology of the infinite Grassmannian (cf. Milnor-Stasheff) coincides with the cohomology of the space $BG$ for $G=\Bbb GL_n(\Bbb C)$. | |
| Aug 3, 2018 at 23:55 | comment | added | Ben McKay | 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. | |
| Aug 3, 2018 at 19:41 | comment | added | Xenomorph | The Chern-Simons form is the secondary class of the second Chern-class. The "quantization" (meaning discretization) of its level $k$ is related to the the fact that the second Chern-class belongs to an integral cohomology. So Mr John Klein showed me that bundles are classified by that $GB$ thing, which I am still struggling with. I don't think you need to worrying about what $\alpha_{H}$ looks like. I am only asking whether $k$ is quantized (i.e. $k\in\mathbb{Z}$). | |
| Aug 3, 2018 at 19:35 | comment | added | Qfwfq | @New Student: Probably. But I think I am missing some link like: is the coh class of the Chern-Simons form $[\alpha]$ (where $\alpha$ is your integrand) a char class of some bundle? Assuming "yes", then: $[\alpha]=f^*[\alpha_G]$ for $f$ a classif map and $\alpha_G$ a cohom. class on $BG$, when you express $[\alpha_G]$ as $\pi^*([\alpha_H])$ via the isom $\pi^*$ on coh given by the retraction $\pi:G\to H$, can you say anything about $\alpha_H$? And can you say something beside $\int_M \alpha = \int_M f^*\pi^*(\alpha_H)$? I just don't follow the hypothesis, nor the thesis of the statement. | |
| Aug 3, 2018 at 18:40 | comment | added | Xenomorph | @Qfwfq John Klein already answered my question. | |
| Aug 3, 2018 at 18:00 | comment | added | Qfwfq | So, in the end, what was the statement the OP was looking for? | |
| Aug 3, 2018 at 17:57 | comment | added | Xenomorph | Would you please give me a reference where I can study the details? | |
| Aug 3, 2018 at 17:55 | comment | added | John Klein | 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 $U(1)$. | |
| Aug 3, 2018 at 17:54 | comment | added | Xenomorph | 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 | |
| Aug 3, 2018 at 17:52 | vote | accept | Xenomorph | ||
| Aug 3, 2018 at 17:51 | history | answered | John Klein | CC BY-SA 4.0 |