Timeline for Is there a free cohomology ring space functor?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 30, 2012 at 10:41 | comment | added | Jesper Grodal | The problem of exactly which polynomial rings can be realized was known as the Steenrod problem, and has a complete solution. It turns out that e.g., for algebras over Z/2 the only ones possible are those arising form classifying spaces of compact Lie groups, and one extra one with generators in degree 15, 14, 12, and 8 related to the exotic 2-compact group DI(4). See "The Steenrod problem of realizing polynomial cohomology rings. J. Topology 1 (2008), 747-760" and "The classification of 2-compact groups. J. Amer. Math. Soc. 22 (2009), 387-436" for much more info. | |
| Jan 24, 2012 at 4:04 | vote | accept | Will Sawin | ||
| Jan 24, 2012 at 2:23 | comment | added | Tom Goodwillie | A polynomial ring on one generator in dimension $3$ is impossible. In fact, the squaring map $H^3\to H^6$ in mod $2$ cohomology must be zero if $H^5=0$ because (1) for $x\in H^n$ with mod $2$ coefficients the Steenrod square $Sq^nx$ is the cup product $x\cup x$ and (2) $Sq^3=Sq^1\circ Sq^2$. | |
| Jan 24, 2012 at 2:01 | comment | added | Will Sawin | Which ones aren't, and why? A reference would of course work as well. | |
| Jan 24, 2012 at 0:27 | history | answered | Peter May | CC BY-SA 3.0 |