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My familiarity with concepts related to derived categories is only tangential, and little by little I intend to get more comfortable with them. I was playing around with Caldararu's introduction to the topic, and looking up various things on the web.

Here is my question (that I have every confidence is trivial for experts):

On the wiki page on Verdier duality http://en.wikipedia.org/wiki/Verdier_duality it says the following: Let $F$ be a field, and $X$ a finite dimensional (dimension is defined here cohomologically, but for our purposes a finite dimensional manifold will do) locally compact space.

In the part about Poincare duality, it says: $H^k(X,F)=[F,X[k]]$. What is the interpretation of this notation? As I see it, $[F,X[k]]$ means $Hom(F,X[k])$ in the derived category. But this means that $X$ is seen as a complex. How? And why would $Hom(F,X[k])$ equal $H^k(X,F)$?

edited Apr 19 2012 at 1:11

David Roberts
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asked Jun 23 2011 at 19:48

James D. Taylor
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I guess that there is a typo and that $[F,X[k]]$ should be understood as $[F,F[k]]$. – DamienC Jun 23 2011 at 20:27

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Why is $[F,F[k]]=H^k(X,F)$? – James D. Taylor Jun 23 2011 at 20:30

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Very shortly, for any sheaf $S$ (i.e. a complex of sheaves concentrated in degree $0$), $[F[-k],S]=H^k(X,S)$. – DamienC Jun 23 2011 at 20:37

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Thanks! You mean $[S[-k],S]$, right? Because $H^k(X,S)$ doesn't depend on $F$. – James D. Taylor Jun 23 2011 at 20:40

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I do really mean $F$, because I ma working with the derived category of $F$-modules (as in the wiki). But you can just take $F=\mathbb{Z}$ if you like. – DamienC Jun 23 2011 at 20:54

show 1 more comment

Justin Curry · Answer 1 · 2012-04-19 00:39:00Z UTC

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I just revamped what was written. Perhaps now it is more understandable:New Wiki Entry on Verdier duality.

answered Apr 19 2012 at 0:39

Justin Curry
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DamienC · Answer 2 · 2011-07-07 09:16:07Z UTC

I think here is how one should understand the last paragraph of the wiki.

Consider $f:X\to pt$. We have (all functors are derived and my $Hom$ are sheaf $Hom$) $$ Hom_{pt}(f_!F,F)=f_*Hom_X(F,f^!F) $$

$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the dual of the cohomology with compact support of $X$.

$Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$.

EDIT : to answer precisely the question I include a summary of the comments.

there is a typo in the wiki: $[F,X[k]]$ should be understood as $[F,F[k]]$.
for any sheaf of $F$-modules $S$ (concentrated in degree 0), $[F[−k],S]=H^k(X,S)$.
Contrary to what is claimed in the wiki, there is no duality between $H^k(X,F)=[F[−k],F]$ and $H_k(X,F)=[F[−k],D_X]$ (where $D_X:=f^!F$ for $f:X\to pt$). The duality is, either between $H^k_c(X,F)$ and $H_k(X,F)$, or between $H^k(X,F)$ and $H_k^{BM}(X,F)$. And as far as I understand $H^*(X,S)$ is dual to $H_c^*(X,S^\vee)$.

Understanding (the wiki page on) Verdier duality

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