Does there exist a model of chains on oriented manifolds with both a strict intersection pairing and strict functoriality for closed embeddings? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:10:56Z http://mathoverflow.net/feeds/question/72397 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72397/does-there-exist-a-model-of-chains-on-oriented-manifolds-with-both-a-strict-inter Does there exist a model of chains on oriented manifolds with both a strict intersection pairing and strict functoriality for closed embeddings? Theo Johnson-Freyd 2011-08-08T21:02:07Z 2011-08-08T21:02:07Z <p>Let $M$ be a smooth oriented $n$-dimensional manifold. My favorite model of $\operatorname{Chains}_\bullet(M) \otimes \mathbb R$ is the space of smooth compactly-supported de Rham forms on $M$, shifted in degree by $[n]$. I like it because the intersection pairing of chains is well-defined: it corresponds after the grade-shifting to the wedge product of forms. It is deeply problematic in one way, though: this model of chains is functorial only for submersions of manifolds. In particular, it is not functorial for the diagonal embedding $M \hookrightarrow M^{\times 2}$, which would generate a coproduct dual to (dg commutative) the product of forms.</p> <p>My first question is whether there exists any model of chains which has both a strict intersection pairing and a strict coproduct? I do not need a model for all manifolds &mdash; I only need something defined on, say, nice open subsets of $\mathbb R^n$, and I only need functoriality for maps as nice as embeddings. For example, does some version of "semialgebraic chains" do the trick (I'm happy working only with algebraic manifolds, and algebraic maps thereof)? For my application, I don't expect to have any problems with things like the possible difference between $\operatorname{Chains}_\bullet(M^{\times 2})$ and $\operatorname{Chains}_\bullet(M)^{\otimes 2}$. I'm also perfectly happy (possibly even more happy) to work with a model of $\operatorname{Cochains}^\bullet$ rather than $\operatorname{Chains}_\bullet$ &mdash; my formulas can be read from left to right or from right to left.</p> <p>But I do not expect such a model to exist. Naive attempts to define a push-forward along non-submersions creates chains with $\delta$-function singularities, and indeed you should expect a version of Sweedler's theorem that any cocommutative coalgebra has grouplike elements, which in my case would be "$\delta$-functions" or "points". But if I have $\delta$ functions, then in order not to create extra cohomology I would need to have step-functions among my $1$-chains, and it is hard to come up with a model that does not ultimately lead to the presence of non-(locally)-constant idempotents. But any idempotent in a dg commutative algebra is necessarily closed (Exercise!).</p> <p>So my main question is:</p> <blockquote> <p>Is the possibility of both intersection product and coproduct obstructed in some way? I.e. is there a reasonable proof that such models (where, understand, part of the proof requires making the statement of the problem precise) do not exist?</p> </blockquote> <p>Note that "take your model of $\operatorname{Chains}_\bullet$ to be $\operatorname{Homology}_\bullet$" is not an adequate answer. Here's one reason: $\operatorname{Homology}_\bullet$ has "Massey products", which is to say it is an $A_\infty$ algebra and _not_ a strictly associative algebra. (Question: is it strictly commutative?)</p> <p>My final question is for suggestions for the next best thing. For example, even with singular chains in a manifold, there is an intersection pairing that is defined up to contractible space of choices, and is homotopy-commutative: namely, you perturb your chains slightly and thereby make the intersection transverse. But this particular structure is very large, and nigh impossible for me to write explicit formulas for. So if I am right that a strict model is obstructed, then my next hope would be a model in which the homotopy-commutative structure can be given completely explicitly in a small hands-on combinatorial way.</p>