Timeline for Is there a free cohomology ring space functor?
Current License: CC BY-SA 3.0
15 events
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| Jan 24, 2012 at 12:40 | comment | added | Tyler Lawson | (In both of these I'm assuming that the homology of $X$ is finitely generated in each degree. Otherwise, you have to be a lot more careful.) | |
| Jan 24, 2012 at 12:39 | comment | added | Tyler Lawson |
If, instead of either of these, you construct the free $\mathbb{Z}/2$-vector space on $X$, you get a space $Y$ where the map back on mod-2 cohomology has the following property. There exists a lift of the generators of the cohomology of $X$ to $Y$, and given any lift $\{a_i\}$ the cohomology of $Y$ is free, in the category of algebras with Steenrod operations satisfying the instability relations, on the generators $\{y_i\}$.
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| Jan 24, 2012 at 12:35 | comment | added | Tyler Lawson | For connected $X$ with a choice of basepoint, you can take $Y$ to be the infinite symmetric product $SP^\infty X$, and then the natural inclusion $X \to SP^\infty X$ gives a functorial construction with gives you the desired map on rational cohomology back. If, instead of the infinite symmetric product (the free abelian monoid on $X$) you take the free $\mathbb{Q}$-vector space (which has a canonical choice of topology), you get the same result and it provides a functorial point-set version of Vitali Kapovitch's construction. | |
| Jan 24, 2012 at 6:38 | comment | added | Will Sawin | @Vitali: Since homotopy classes are to be expected in this sort of thing, I don't think that's a problem. Cool. @Sean: A space that gives a free module for the Steenrod algebra is of separate interest. I was just clarifying a point over whether you meant, as it seemed to say, "the steenrod algebra that relates to your space". Do you know a construction for a space whose cohomology is a free module? It occurs to me that you might be able to just glue a bunch of random stuff together and then kill all the homology classes you don't want to be able to distinguish from 0. But functorial is hard. | |
| Jan 24, 2012 at 6:06 | comment | added | Sean Tilson | I guess you really wanted a ring then. What I was trying to say is that there is another object that is free with respect to the cohomology of a space and it is not a ring. And yes, of course you are right it is independent, but the map will be a map of modules like you wanted your map to induce a map of algebras. | |
| Jan 24, 2012 at 4:38 | comment | added | Vitali Kapovitch | Note that while it doesn't work over $Z_2$ by Peter May's and Tom Goodwillie's answers it does work over $\mathbb Q$ if instead of an actual map $X\to Y$ you are willing to settle for its homotopy class. just take $Y$ to be the product of $K(H^n(X,\mathbb Q),n)$ with natural map $X\to Y$ coming from $H^n(X,H^n(X,\mathbb Q))=[X, K(H^n(X,\mathbb Q),n)]$. This correspondence is clearly functorial in $X$. | |
| Jan 24, 2012 at 4:04 | comment | added | Will Sawin | Isn't the Steenrod algebra independent of the space, since it refers to cohomology operations that can be carried out on any space? | |
| Jan 24, 2012 at 4:04 | vote | accept | Will Sawin | ||
| Jan 24, 2012 at 3:26 | comment | added | Sean Tilson | What would be more likely is that you can get a space/spectrum that is free as a module over the steenrod algebra that relates to your space. | |
| Jan 24, 2012 at 0:27 | answer | added | Peter May | timeline score: 14 | |
| Jan 23, 2012 at 23:04 | comment | added | Will Sawin | I've read a bit about this spectral sequence and it does not seem obvious why it's related. Could you explain why you think it's helpful? | |
| Jan 23, 2012 at 22:50 | comment | added | Fernando Muro | You might find useful reading about the Adams spectral sequence | |
| Jan 23, 2012 at 22:50 | comment | added | Will Sawin | If it helps prove anything, yes. | |
| Jan 23, 2012 at 22:28 | comment | added | Steve D | Perhaps you'd like to restrict attention to suitably nice spaces? | |
| Jan 23, 2012 at 21:59 | history | asked | Will Sawin | CC BY-SA 3.0 |