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According to Deligne's "yoga of weights", the cohomology of an algebraic variety should have a weight filtration. For concreteness we can consider the rational cohomology of complex varieties, with their mixed Hodge structure.

It seems to me that in the yoga of weights there is a kind of duality between singularities and non-compactness. The simplest example should be:

  • if $X$ is smooth, then $H^n(X)$ has weights at least $n$
  • if $X$ is compact, then $H^n(X)$ has weights at most $n$

We also have the following:

  • let $X$ be a smooth variety and $X \to Y$ a smooth compactification. Then $W_n H^n(X) = \mathrm{Im}(H^n(Y) \to H^n(X))$.
  • let $X$ be a compact variety and $Y \to X$ a resolution of singularities. Then $H^n(X)/W_{n-1}H^n(X) = \mathrm{Im}(H^n(X) \to H^n(Y))$.

Are there more examples of this "duality"? Is there a unifying principle here?

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This may be already clear to you, but from my perspective, the clearest manifestation of this duality is in the setting of mixed Hodge modules (or some other version of ``mixed sheaves'').

Let $f: X \to Y$ be a morphism of complex algebraic varieties, and let $D_m(X)$ and $D_m(Y)$ refer to the derived categories of mixed Hodge modules. Then there are functors $$ f^!, f^\ast : D_m(Y) \to D_m(X) $$ $$ f_\ast, f_! : D_m(X) \to D_m(Y) $$ Let me also define the shifted functor $f^\dagger = f^! [\dim Y - \dim X]$. We also have the Verider duality functors $\mathbb D_X$ and $\mathbb D_Y$. The functors are related as follows: $ f^! \mathbb D_Y = \mathbb D_X f^\ast$, and $f_! \mathbb D_X = \mathbb D_Y f_\ast$.

We have that: $f_\ast$ and $f^!$ increase weights, whereas $f_!$ and $f^\ast$ decrease weights.

In this language:

If $f$ is smooth (i.e. submersive) then $f^\dagger$ commutes with $\mathbb D$, i.e. $\mathbb D_X f^\dagger \simeq f^\dagger \mathbb D_Y$

If f is proper then $f_\ast$ commutes with $\mathbb D$.

Thus smoothness gives a relationship between relative dualizing sheaf and constant sheaf, and properness relates relative cohomology with relative compactly supported cohomology.

For example, in the case $f: X \to pt$, we have $$ f_\ast f^\ast \mathbb Q \simeq H^\ast (X) $$ $$ f_! f^! \mathbb Q \simeq H_\ast(X)$$ $$ f_! f^\ast \mathbb Q \simeq H^\ast _c(X)$$ $$ f_\ast f^! \mathbb Q \simeq H_\ast ^{BM}(X)$$.

Smoothness of $X$ means the dualizing sheaf ($f^! \mathbb Q$) is isomorphic to the constant sheaf ($f^\ast \mathbb Q$) up to a shift. Properness means that compactly supported cohomology (with coefficients in some sheaf) is isomorphic to ordinary cohomology. This recovers your first observation.

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  • $\begingroup$ Thanks for the answer. Mostly I'm annoyed with myself - I knew perfectly well that $f_! = f_\ast$ for proper maps, that the dualizing sheaf for a smooth morphism is the constant sheaf with a degree shift, etc. but somehow I didn't put it together. Do you know if the second observation admits a proof based on general six functors principles? $\endgroup$ Sep 3, 2013 at 8:16

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