most recent 30 from http://mathoverflow.net 2013-05-21T03:22:12Z http://mathoverflow.net/feeds/question/55833 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55833/singular-complex-cohomology-ring-steenrod-operations roger123 2011-02-18T09:11:23Z 2011-02-19T00:43:15Z

<p>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?</p>

http://mathoverflow.net/questions/55833/singular-complex-cohomology-ring-steenrod-operations/55835#55835 Mark Grant 2011-02-18T10:11:20Z 2011-02-18T10:11:20Z

<p>I think the answer is no, due to <a href="http://mathoverflow.net/questions/55365/counterexamples-in-algebraic-topology/55375#55375" rel="nofollow">this answer</a> to <a href="http://mathoverflow.net/questions/55365/counterexamples-in-algebraic-topology" rel="nofollow">this question</a>. 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.</p>

http://mathoverflow.net/questions/55833/singular-complex-cohomology-ring-steenrod-operations/55844#55844 Tilman 2011-02-18T12:25:42Z 2011-02-18T12:25:42Z

<p>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.</p>

http://mathoverflow.net/questions/55833/singular-complex-cohomology-ring-steenrod-operations/55853#55853 Tom Goodwillie 2011-02-18T14:05:00Z 2011-02-18T21:07:09Z

<p>This is answering a slightly different question, but here goes:</p> <p>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.</p> <p>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.</p>

http://mathoverflow.net/questions/55833/singular-complex-cohomology-ring-steenrod-operations/55935#55935 Fernando Muro 2011-02-19T00:43:15Z 2011-02-19T00:43:15Z

<p>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.</p>