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The classical Poincare duality is formulated in terms of cohomology groups. I am wondering if we can also formulate it in terms of complexes.

In particular, suppose $\mathcal{C}^*$ is a complex of $k$-modules for some coefficient ring $k$ such that its cohomology groups $H^*(\mathcal{C^*})$ is some Weil cohomology groups which satisfies some kind of Poincare duality. I am wondering if there is some form of Poincare duality formulated purely in terms of the complex $\mathcal{C}^*$ instead of its cohomology groups.

I understand that there is a derived version of Poincare duality, namely the Verdier duality. But Verdier duality starts with a sheaf $\mathcal{F}^*$. In my problem I already have its cohomology $R\Gamma(\mathcal{F}^*)$, so I hope to work with the cohomological complex, and not to go back to the sheaf setting.

Any thoughts or reference is highly appreciated.

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  • $\begingroup$ maybe something like this is relevant arxiv.org/pdf/math/0701309.pdf Though I do not understand whether one should expect a functorial construction at dga level or $E_{\infty}$-ring spectrum level $\endgroup$
    – user74900
    Commented Mar 12, 2019 at 4:29
  • $\begingroup$ I think the strongest form of Poincare duality possible for a complex of $k$-modules is that there is a quasi-isomorphism, canonical up to homotopy, between the complex and its dual. (Well, you could write down some explicit map, and say that this map is an isomorphism). Verdier duality guarantees this if your complex arises from $R \Gamma$ of a self-dual sheaf, say, but if you don't say what your complex arises from there's no nontrivial theorem that is useful. $\endgroup$
    – Will Sawin
    Commented Mar 12, 2019 at 5:20
  • $\begingroup$ @WillSawin What do you mean by the dual of a complex? Do you mean literally taking the dual of each $k$-modules or do you mean the dualizing complex defined by some $f^!$ functor as in the sense of Verdier duality? Thank you so much! $\endgroup$
    – yue he
    Commented Mar 13, 2019 at 0:15
  • $\begingroup$ I mean literally taking the dual of each $k$-module. Indeed, note that literally taking the dual of each $k$-module appears in the statement of Verdier duality (for the map to the point, taking the rank 1 constant sheaf on the point)! $\endgroup$
    – Will Sawin
    Commented Mar 13, 2019 at 0:22

1 Answer 1

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One way of finding a "fully derived" version of Poincaré duality is Atiyah duality. This says that for any closed manifold $M$ there is an equivalence of spectra (in the sense of algebraic topology) $$M^{-TM}\cong\mathbb{D}(\Sigma^∞_+M)$$ Here $M^{-TM}$ is the Thom spectrum for the (virtual) normal bundle, $\Sigma^\infty_+$ is the suspension spectrum and $\mathbb{D}$ is Spanier-Whitehead duality (duality in the category of spectra). Concretely this mean that for every $E_\infty$-ring spectrum $E$ such that $TM$ is $E$-orientable (i.e. such that the Thom isomorphism holds) we have an equivalence of $E$-modules $$\Sigma^{-\dim M}E\wedge \Sigma^\infty_+M\cong E\wedge M^{-TM}\cong \mathbb{D}_E(E\wedge \Sigma^∞_+M)$$

Specializing in the case of $E=HR$, the Eilenberg-MacLane spectrum associated to a ring $R$, we have that the category of $E$-modules is exactly the derived category of $R$, and $HR\wedge\Sigma^∞_+M\cong C_*(M;R)$, so the formula above gives an equivalence in $\mathscr{D}(R)$ $$C^*(M;R):=\mathbb{D}_R(C_*(M;R))\cong C_*(M;R)[-\dim M]$$

You can generalize this to your heart's content, to compact manifolds with boundaries and to Verdier duality (working with local systems of spectra, or even constructible sheaves of spectra), but I hope this is enough to show the power of working in the stable homotopy category (i.e. the category of spectra). Moreover, without the orientability assumption this gives Poincaré duality with local coefficients (since $E\wedge M^{-TM}$ can be interpreted as $E$-homology with coefficients in the local system $x\mapsto E\wedge S^{-T_xM}$).

One nice thing about this approach is that the proof of Atiyah duality is nice and geometric, as you can see from Charles Rezk's notes I linked above, and the "hard work" is outsourced to the Thom isomorphism (which is still not that hard)

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    $\begingroup$ Could you please somehow elaborate the classical analogues of maps $\eta$, $\mu$ and $\psi$ in that paper in the simplest case, and indicate how we can establish an explicit formula for a pair of inverse maps of the most classical Poincaré isomorphism? I believe that it is essentially a specialization of the proof of Proposition 2.2. It should be trivial, but seemingly I have not succeeded yet, and I have not seen the "inverse" of the Poincaré duality isomorphism being the cap product with the fundamental class written out explicitly. $\endgroup$
    – user20948
    Commented Dec 15, 2019 at 11:26
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    $\begingroup$ I mean, seemingly Rezk's notes essentially say that the duality follows algebraically from the topological input of $\eta$, $\mu$ and $\psi$ which exploits the hypothesis that $M$ is a smooth manifold and $TM$ is the tangent bundle. I guess that the structure persists after base change to $\mathbb Z$, that is, smashing with the Eilenberg-Maclane spectrum $H\mathbb Z$. The result should be a classical algebraic structure, which gives rise to explicit formulae of the most classical Poincaré isomorphism. But sorry for my ignorance, I only know one direction is a cap product, don't know the other. $\endgroup$
    – user20948
    Commented Dec 15, 2019 at 11:39
  • $\begingroup$ @Yai0Phah I'm not sure I understand your question. $\mu$ and $\eta$ corresponds to the standard algebra structure on cohomology and $\psi$ to the classical fact that homology is a module over cohomology ("the cap product"). If anything I'd have thought that the mysterious maps would be the other three (coming from the coalgebra structure on chains, which is not apparent on homology unless the coefficients are a field). $\endgroup$
    – Denis Nardin
    Commented Dec 15, 2019 at 15:40
  • $\begingroup$ Yeah, I would like to understand structures "down to earth". I chose these three because I found that in the proof of these three he took advantage of the existence of an embedding to Euclidean space, that is to say, $M$ being the a manifold and $TM$ being the associated tangent bundle are "used", while the other three seems to be formal. $\endgroup$
    – user20948
    Commented Dec 15, 2019 at 19:00
  • $\begingroup$ @Yai0Phah I don't know what I can add beyond what is in Rezk's notes. Those maps are obtained by various Pontryagin-Thom constructions, and that seems pretty geometric to me :). $\endgroup$
    – Denis Nardin
    Commented Dec 15, 2019 at 19:06

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