Timeline for $E^G_\ast(E)$ tensored with the rationals
Current License: CC BY-SA 4.0
2 events
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| Oct 22, 2022 at 13:57 | comment | added | Neil Strickland | I don't really understand this comment. However, it is certainly true that $X\mathbb{Q}\wedge Y\mathbb{Q}=(X\wedge Y)\mathbb{Q}$ and that $\pi^G_*(X\mathbb{Q}\wedge Y\mathbb{Q})=\pi^G_*((X\wedge Y)\mathbb{Q})=\pi^G_*(X\wedge Y)\otimes\mathbb{Q}$, so I am not sure why you think otherwise. The value of $\pi^G_*(KU_G\wedge KU_G)\otimes\mathbb{Q}$ can easily be read off from my answer (and can be simplified using the standard isomorphism $\mathbb{Q}(\mu_n)\otimes\mathbb{Q}(\mu_n)\simeq\text{Map}((\mathbb{Z}/n)^\times,\mathbb{Q}(\mu_n))$ from Galois Theory if desired). | |
| Oct 20, 2022 at 18:23 | history | answered | Neil Strickland | CC BY-SA 4.0 |