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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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up vote 6 down vote accepted

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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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? –  Dan Petersen Sep 3 '13 at 8:16
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