Timeline for Representablity of Cohomology Ring
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 20, 2009 at 18:22 | comment | added | Eric Wofsey | Dinakar: The cup product is represented by the tensor product of the two universal classes in H^*(S^1 \times S^1)=H^*(S^1) \otimes H^*(S^1). This is just a generator of H^2. Now the 2-skeleton of CP^\infty is just CP^1=S^2, and the inclusion of the 2-skeleton gives an isomorphism on H^2. Thus the universal class on CP^\infty pulls back to the fundamental class on S^2, which will then pull back to the desired class on the torus via any degree 1 map. So you can get your map by taking S^1 \times S^1 \to S^2=CP^1 \to CP^\infty, where the first map has degree 1. | |
| Oct 20, 2009 at 15:21 | comment | added | Dinakar Muthiah | So the cup product H^1 x H^1 -> H^2, should come from a homotopy class of maps S^1 x S^1 -> CP^inf ? Is it clear what this map should be? | |
| Oct 20, 2009 at 15:19 | vote | accept | Dinakar Muthiah | ||
| Oct 20, 2009 at 14:05 | comment | added | Andrew Stacey | You lose the action of those operations that are unstable but not stable. For example, in K-theory you lose all but two of the Adams operations. The problem is that when you put your representing objects into a whole then your morphisms have to be globally defined (otherwise you aren't really putting your objects into one object). There are lots of interesting operations (such as Adams operations) which cannot be globally defined. | |
| Oct 20, 2009 at 10:42 | comment | added | Sam Derbyshire | Could you explain what you have in mind when saying that putting a cohomology theory into a single object loses unstable operations? Which information precisely is lost, and how come? | |
| Oct 20, 2009 at 7:40 | history | answered | Andrew Stacey | CC BY-SA 2.5 |