Timeline for $K$ theory and singular cohomology
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2016 at 15:35 | vote | accept | Thomas Rot | ||
| Feb 9, 2016 at 14:30 | comment | added | Sebastian Goette | @MichaelAlbanese I think this is a matter of convention. If you want to have a universal Chern character (or a universal total Chern class or whatever) on $BU$, then you have to interpret $H^\bullet$ as an infinite product. This should not come as a surprise as cohomology tends to have infinite products where homology has infinite sums. It is clear that this convention will raise problems in other places, but for characteristic classes, it seems to be the correct choice. | |
| Feb 9, 2016 at 14:19 | comment | added | Thomas Rot | @archipelago: With $K^*(X)$ I mean $K^0(X)\oplus K^1(X)$. Then $K^1(X)=\tilde K^0(SX)$ lands in $\tilde H^{2*}(SX;\mathbb{Q})$ which one can identify with $H^{2*-1}(X;\mathbb{Q})$. | |
| Feb 9, 2016 at 14:15 | history | edited | Thomas Rot | CC BY-SA 3.0 |
added 55 characters in body
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| Feb 9, 2016 at 13:48 | answer | added | Tyler Lawson | timeline score: 12 | |
| Feb 9, 2016 at 13:06 | comment | added | Michael Albanese | An example where the Chern character of a vector bundle doesn't belong to the direct sum is $X = \mathbb{CP}^{\infty}$ and $E$ the tautological line bundle. Then $\operatorname{ch}_k(E) = c_1(E)^k/k! \neq 0$, so $\operatorname{ch}(E) \not\in H^*(\mathbb{CP}^{\infty}; \mathbb{Q})$, but rather $\operatorname{ch}(E) \in \prod H^k(\mathbb{CP}^{\infty};\mathbb{Q})$. | |
| Feb 9, 2016 at 13:05 | comment | added | Gijs Heuts | It can't work for infinite complexes: consider X = BG for a finite group G. | |
| Feb 9, 2016 at 12:59 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Feb 9, 2016 at 12:58 | comment | added | Michael Albanese | If $X$ is an arbitrary cell complex, then $\operatorname{ch}(E)$ could be an infinite sum of non-zero terms, and hence not belong to $H^*(X; \mathbb{Q})$ which is the direct sum of the groups $H^k(X; \mathbb{Q})$. Instead the $\operatorname{ch}(E)$ would belong to the direct product of the groups. Note, if one assumes $X$ is a finite cell-complex, then there are only finitely many non-zero cohomology groups, there is no problem (the direct sum and direct product coincide). | |
| Feb 9, 2016 at 12:30 | comment | added | archipelago | Shouldn't it be $K^0(X)\otimes\mathbb{Q}\cong H^{2\ast}(X;\mathbb{Q})$ instead? | |
| Feb 9, 2016 at 11:18 | history | asked | Thomas Rot | CC BY-SA 3.0 |