Fix a prime $p$ and consider everything mod $p$. Steenrod operations arise somehow from the loss of information passing from the singular complex of a space to its cohomology ring. Are they exactly this gap, i.e. can I get the singular complex back from the cohomology ring of a space and its structure as a module over the Steenrod algebra?
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$\begingroup$ This can't be literally true, $X \times \mathbb{R}$ has a much bigger complex of singular simplices than $X$... or do you mean something else by "singular complex" ? $\endgroup$– Vivek ShendeCommented Feb 18, 2011 at 12:11
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1$\begingroup$ I assume he means "up to weak equivalence". $\endgroup$– arsmathCommented Feb 18, 2011 at 12:44
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$\begingroup$ I posted an answer and then noticed that it's really an answer to a different question. Then I "deleted" it, but it's still there. $\endgroup$– Tom GoodwillieCommented Feb 18, 2011 at 14:07
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$\begingroup$ @tom you can still see it but we can't. It is still there for you or a moderator to undelete. $\endgroup$– Steven GubkinCommented Feb 18, 2011 at 15:16
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$\begingroup$ I see. Well, anyway, I undeleted it and edited it a bit. $\endgroup$– Tom GoodwillieCommented Feb 18, 2011 at 19:59
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4 Answers
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No. For instance, Massey products on the cohomology are extra information that neither the ring structure nor the Steenrod operations see. The complement of the Borromean rings, for example, and the complement of three unlinked circles in $R^3$ have the same cohomology ring and Steenrod operations, but cannot be chain equivalent because of nonzero Massey products in the former.
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This is answering a slightly different question, but here goes:
If the question is "does knowing the singular (co)chains up to quasi-isomorphism determine the space up to weak equivalence?" then of course the answer is no. To some extent the answer can be turned into a yes by altering the question, considering the (co)multiplication on (co)chains as part of the structure that a quasi-isomorphism must preserve. Mike Mandell has made this idea precise (using operads) and proved powerful theorems along these lines, but I'm not an expert.
Of course, the operad structure on chains of course determines the cup product, the Steenrod operations, and more (e.g. differentials in Adams spectral sequence). EDIT: And Massey products.
answered Feb 18, 2011 at 14:05
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$\begingroup$ @Tom: you know this, but I feel compelled anyway to add the condition that spaces are nilpotent in Mandell's theorem. $\endgroup$ Commented Feb 20, 2011 at 3:17
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It depends of what structure you want to consider on the complex of singular cochains. If you want to look at it just as a complex, then the cohomology groups are enough. If you want it as a differential graded algebra, then you would need the cohomology groups with an A-infinity algebra structure, etc. Steenrod operations are something which come from the E-infinity structure on cochains, but they are weaker.
answered Feb 19, 2011 at 0:43
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I think the answer is no, due to this answer to this question. There are Moore spaces with the same cohomology rings and module structure over the mod $p$ Steenrod algebra, which nevertheless have different homology groups and so their singular complexes cannot be weakly equivalent.
answered Feb 18, 2011 at 10:11
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1$\begingroup$ I think the OP is asking about the mod $p$ singular complexes. (See the first sentence of the question.) $\endgroup$– EmertonCommented Feb 18, 2011 at 13:09
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$\begingroup$ Having said that, there are further "secondary" Bockstein operations on the mod-p singular complex that recover the information present in the p-completed singular complex. This secondary-operation structure is different for different Moore spaces. $\endgroup$ Commented Feb 18, 2011 at 13:16