Timeline for Characteristic classes of the bundle of trace free, skew adjoint endomorphisms
Current License: CC BY-SA 4.0
6 events
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Dec 4, 2018 at 0:48 | comment | added | mme | @Warlock Yes, that is the desired property. Given any homomorphism $f: G \to H$, you have a natural way to construct a principal $H$-bundle out of any principal $G$-bundle, sending $P \mapsto P \times_G H$. In particular applying this to the universal bundle gives you a canonical map $BG \to BH$ classifying the new bundle, and it is clear that this map has the property you ask for. Actually doing the computation of what it does in cohomology usually looks like this: using known formulas for sums and tensor products and relating your construction to those. Sometimes you can run spectral seqs. | |
Dec 3, 2018 at 23:47 | vote | accept | Overflowian | ||
Dec 3, 2018 at 23:47 | comment | added | Overflowian | Thanks Qiaochu, how is the map $Bf$ defined? I think that we would like that $h_{\mathfrak{su}(2)}=Bf \circ h_{U(2)}$ ($h_{\mathfrak{su}(2)}$ and $h_{U(2)}$ are respectively the classifying maps of the frame bundles of $\mathfrak{su}(E)$ and $E$), am I right? P.S. Do you have any good book to suggest on this subject? | |
Dec 3, 2018 at 22:34 | comment | added | mme | Explicitly, if $\eta_1 \oplus \eta_2$ is the $U(2)$-bundle in question, the associated $SO(3)$-bundle is $\Bbb R \oplus (\eta_1 \otimes \eta_2^{-1})$, and your description shows how to compute the characteristic classes of this. | |
Dec 3, 2018 at 21:03 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
added 10 characters in body
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Dec 3, 2018 at 20:58 | history | answered | Qiaochu Yuan | CC BY-SA 4.0 |