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.

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 — 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 $\operatorname{Chains}_\bullet$ — my formulas can be read from left to right or from right to left.

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!).

So my main question is:

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?

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?)

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.

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    $\begingroup$ Isn't David Spivak's "Derived Manifolds" close to what you're looking for? math.mit.edu/~dspivak At least, for the question in your title. $\endgroup$ Aug 8, 2011 at 21:36
  • $\begingroup$ en.wikipedia.org/wiki/… Maybe that is what you are looking for. My guess is its only half the answer you want though. $\endgroup$ Aug 8, 2011 at 21:37
  • $\begingroup$ @Charlie: Right, I should have mentioned that. Although I've argued that there's obstructions to finding something as strict as I'd like, the presence of Sullivan+Quillen's work is suggestive in the other direction. But I still haven't been able to give, say, PL forms or the minimal model thereof an "intersection copairing" dual to the intersection pairing. $\endgroup$ Aug 8, 2011 at 21:56
  • $\begingroup$ @Ryan: Derived manifolds are close to what I want, although there's details to work out. They're not perfect, though. I haven't read David's approach, so my comments might not apply --- I base them on a possibly more naive approach of "dg manifolds". The problem is that when you resolve a smooth map to a submersion, you get something that is "quasiisomorphic" but not "homotopy equivalent". This is familiar from commutative algebra, and a similar thing happens in geometry. For example, there are non-empty dg manifolds with 0 cohomology. So one should expect to pick up "extra Massey products". $\endgroup$ Aug 8, 2011 at 22:02
  • $\begingroup$ This is almost certainly not what you want, but in the category of "options to think about when pondering the 'next best thing'" you might be interested in McClure's approach to intersection pairings on PL manifolds: McClure, J. E. On the chain-level intersection pairing for PL manifolds. Geom. Topol. 10 (2006), 1391–1424 $\endgroup$ Aug 8, 2011 at 22:28

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After work on this and other problems, I have come to the conclusion that the correct answer to my question is "No, there does not exist a model of chains with simultaneously strict intersection and diagonal maps". Even in characteristic 0, the problem is obstructed in dimension 1, as I prove in http://arxiv.org/abs/1308.3423. To conclude the answer to this question from what I prove in that paper, one must decide also whether the Frobenius axiom is required to hold strictly — the paper proves that even a "homotopy Frobenius" algebra structure does not exist (in the properadic sense) on Chains(R), and so certainly there is no strict Frobenius algebra structure. But the linked paper does not imply that you can't strictify both the associativity and coassociativity while keeping the Frobenius axiom weak.


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