The first set of questions can be found here: Understanding (the wiki page on) Verdier duality

I'm fairly confident that I understand something wrong, so I'll write down here clearly what my set of beliefs is about what is right, and you feel free to shoot down any falsehood:

Let $X$ be our geometric object, be it a topological space, variety, scheme, or what have you. I will do two cases, one Poincare duality, and the other Serre duality.

Serre duality

I will assume $X$ is nice (a variety, projective, smooth,... of dimension $n$). Here I will look at the (abelian) category of coherent $O_X$-modules. In this case the "dualizing module" is $\omega_X[n]$. I take that to mean that $[\mathcal{E},\mathcal{F}]$ is dual to $[\mathcal{F},\mathcal{E}\otimes \omega_X[n]]$.

I also believe $H^k(X,\mathcal{F})\cong [O_X,\mathcal{F}[k]]$. Going from there, the rest is easy: $H^k(X,\mathcal{F})\cong [O_X,\mathcal{F}[k]]\cong [$dual of $\mathcal{F},O_X[k]]$, which is dual to $[O_X[k],\omega_X[n]\otimes$ dual of $\mathcal{F}]\cong [O_X, \omega_X \otimes$ dual of $\mathcal{F}[n-k]] \cong H^{n-k}(X,\omega_X \otimes $ dual of $\mathcal{F})$.

Great! However...

Poincare duality

Again assume $X$ is nice (orientable compact smooth manifold of dimension $n$). Let $K$ be field. Here I will look at the (abelian) category of $K$-vector spaces. Here the "dualizing sheaf" is $K[-n]$. I will continue somewhat similarly to the Serre duality case. I interpret the dualizing sheaf as meaning that $[\mathcal{E},\mathcal{F}]$ is dual to $[\mathcal{F},\mathcal{E}\otimes K[-n]]$ (here $\mathcal{E}$ and $\mathcal{F}$ are $K$-vector spaces).

For whatever reason, I believe that $H_k(X,\mathcal{E})\cong[\mathcal{E}[-k],D_X]$ ($D_X$ is the dualizing sheaf in general. But what does this even mean in general? For example in the Serre duality case, what would $H_k(X,\mathcal{F})$ even mean?). So:

$H_k(X,K) \cong [K[-k],K[-n]] \cong [K,K[k-n]] \cong H^{k-n}(X,K)$. Wait. What? Makes no sense!

If we use the "duality" it still makes no sense:

$[K,K[k-n]] \cong [K[k-n],K[-n]] \cong [K,K[-k]]\cong H^{-k}(X,K)$. What? Huh?

So in conclusion, I desperately want to have a handle on this, but I clearly don't. Hopefully just a nudge in the right direction would lead me to a better understanding of this yoga.

  • $\begingroup$ You should not have different signs in the shifts for Serre and Poincaré duality, in fact the shifts should be exactly the same. The conventions used for shifts (whether a positive shift is to the right or the left) is confusing (at least to me) and my guess is that you have found sources that uses different conventions (which is wrong I think there is only one generally accepted convention only I can never remember which it is). $\endgroup$ – Torsten Ekedahl Jun 24 '11 at 4:24
  • $\begingroup$ If $D_X$ were $K[n]$ in the Poincare case, it still wouldn't make sense... I guess the only thing that would would be $H^k(X,K) \cong [K[k],K[n]]$, but I don't know how to make that fit with anything... $\endgroup$ – James D. Taylor Jun 24 '11 at 4:36
  • $\begingroup$ And even if by magic that were true, my interpretation of the dualizing sheaf would be wrong... (otherwise cohomology would be periodic) $\endgroup$ – James D. Taylor Jun 24 '11 at 4:44
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    $\begingroup$ Sorry, didn't look closely at what you said. You are also wrong about the homology, your formula for $H_k(X,\mathbb E)$ couldn't be OK as homology should be covariant in $\mathbb E$. The Verdier version of Poincaré duality is usually formulated purely in terms of cohomology just as is Serre duality. To get the usual topological version one combines this with the universal coefficient formula. $\endgroup$ – Torsten Ekedahl Jun 24 '11 at 5:10
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    $\begingroup$ Yes, that's it. $\endgroup$ – Torsten Ekedahl Jun 26 '11 at 5:55

I'm not sure if this will answer your question or not, but let $\mathbb{D}$ be the Verdier dualizing sheaf on the locally compact space $X$. If $M$ is a manifold, then $\mathbb{D}[-n]$, where $n$ is the dimension of $M$, is isomorphic in the derived category to the orientation sheaf $\omega$ on $M$ (see, for example, Borel's book on interesection cohomology, I think around section V.7). The Borel-Moore homology of $X$, which is equivalent to the ordinary homology of $X$ if $X$ is compact (otherwise it's the homology theory built from locally-finite chains) is defined to be $H^{BM}_{k}(X)=H^{-k}(X; \mathbb{D})$. If you unwind all that indexing business, it says that, for a compact manifold, $H_k(M)=H^{n-k}(M; \omega)$, which is probably the duality statement you're looking for.

Of course showing that this has anything to do with classical Poincare duality (say via the cap product) is pretty far from obvious. This is part of the content of a paper I currently have in preparation with Jim McClure.

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  • $\begingroup$ What do you mean by $H^{n-k}(M,\omega)$? $\endgroup$ – James D. Taylor Jun 24 '11 at 12:47
  • $\begingroup$ I mean the n-k cohomology group of $M$ with coefficients in the orientation sheaf $\omega$. More explicitly, this would be $H^{n-k}(\Gamma(M;I^\ast))$, where $I^\ast$ is an injective resolution of $\omega$. Or it could be the Cech cohomology with coefficients in $\omega$ if you like that better. $\endgroup$ – Greg Friedman Jun 27 '11 at 6:12
  • $\begingroup$ Sorry, I don't understand why you stressed that the comparison to classical Poincaré duality is "far from obvious" (that is to say, the others are "more obvious")? It is Theorem 4.7 in Iversen's Cohomology of Sheaves. I am struggling to establish a proof on my own, but I don't see how this one should be especially more difficult than others. $\endgroup$ – Yai0Phah Dec 27 '19 at 18:23
  • $\begingroup$ Sorry, I should have said more explicitly something like "singular chain cap product." Iversen does have a Poincare duality theorem with isomorphism given by a map he calls a cap product, but that map is defined in his book purely sheaf theoretically. So it's not obvious, at least not to me, that this is the same Poincare duality isomorphism you can find proven in a more classical algebraic topology text (e.g. Hatcher, Spanier, Munkres, Dold, etc.). $\endgroup$ – Greg Friedman Dec 28 '19 at 22:55

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