The statements of Poincare duality for manifolds and Serre Duality for coherent sheaves on algebraic varieties or analytic spaces look tantalizingly similar. I have heard tangential statements from some people that there is indeed some connection between the two. But I was never able to figure it out on myself. For instance for a naive attempt on a smooth complex manifold, the dimensions don't match. Can somebody help me out?
Algebraic Geometry – Connection Between Poincare Duality and Serre Duality
algebraic-geometrycoherent-sheavesdifferential-topologyduality-theorems
Related Solutions
Since coherence is a local property and is preserved by finite direct sums, the general question reduces to the structure sheaf. But this is coherent only in trivial cases.
If $M$ is a smooth manifold whose connected components have positive dimension, then $C^{\infty}_M$ is not coherent.
For a proof, see Prop. 7.3.8 in the course notes by Andrew Lewis on sheaf theory. By a projection argument, it suffices to deal with $M=\mathbb{R}$. If $f$ is a smooth function, which is $>0$ on $\mathbb{R}_{>0}$ and $=0$ on $\mathbb{R}_{\leq 0}$, then the kernel of the multiplication map $f : C^{\infty}_{\mathbb{R}} \to C^{\infty}_{\mathbb{R}}$ is not of finite type: Loot at the stalk at $0$. The kernel $I$ is given by those germs of smooth functions vanishing on $\mathbb{R}_{\geq 0}$. If $\mathfrak{m}$ is the maximal ideal of functions vanishing at $0$, then we have $I = \mathfrak{m} I$ by Taylor expansion. Since $I \neq 0$, Nakayama's Lemma shows that $I$ is not finitely generated.
I think you can prove this statement from (a weaker form of) the usual Serre duality statement using some compatibilities between derived functors, namely:
- For all $\mathcal{F}^{\bullet}\in D^{b}(X)$ we have a natural isomorphism $$ R\Gamma(X,-)\circ R\mathcal{Hom}(\mathcal{F}^{\bullet},-)\cong R\operatorname{Hom}(\mathcal{F}^{\bullet},-). $$
- For all $n\in \mathbb{N}$ we have a natural isomorphisms $$ \operatorname{Hom}(\mathcal{O}[-n],-)\cong H^{n}(R\Gamma(X,-)). $$
- For all $\mathcal{F}^{\bullet},\mathcal{G}^{\bullet},\mathcal{E}^{\bullet}\in D^{b}(X)$ we have an isomorphism $$ \operatorname{Hom}(\mathcal{F}^{\bullet},\mathcal{E}^{\bullet}\otimes^{L}\mathcal{G}^{\bullet})\cong \operatorname{Hom}(R\mathcal{Hom}(\mathcal{E}^{\bullet},\mathcal{F}^{\bullet}),\mathcal{G}^{\bullet}). $$ This follows from formula (3.15) in Huybrechts' book after applying derived global sections and taking cohomology on degree zero. The formula (3.15) is functorial, because it is based on a sequence of canonical isomorphisms coming from exercise II.5.1 in Hartshorne's book. Hence the previous isomorphism is functorial as well.
As a starting point consider the following statement, which is part of the usual Serre duality statement:
For all line bundles $\mathcal{L}$ on $X$ we have a natural perfect pairing $$ \operatorname{Hom}(\mathcal{L},\omega_{X})\times H^{n}(X,\mathcal{L})\to H^{n}(X,\omega_{X})\cong k $$
We generalize this statement to bounded complexes of coherent sheaves:
For all $\mathcal{F}^{\bullet}\in D^{b}(X)$ we have a natural perfect pairing $$ \operatorname{Hom}(\mathcal{F}^{\bullet},\omega_{X})\times \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{F}^{\bullet})\to {\operatorname{Hom}(\mathcal{O}_{X}[-n],\omega_{X})}\cong k $$
Proof:
We have to check that the induced morphism $$ \operatorname{Hom}(\mathcal{F}^{\bullet},\omega_{X})\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{F}^{\bullet})^{\vee} $$ is an isomorphism. We divide the proof into smaller steps:
- The result is true for finite direct sums of line bundles, because all functors are additive and we know the result for line bundles already.
- The result is true for all coherent sheaves (complexes concentrated on degree zero). To see this we use that every coherent sheaf $\mathcal{F}$ on a smooth projective variety admits a surjection from a finite direct sum of line bundles $\mathcal{E}=\oplus_{i=1}^{l}\mathcal{L}_{i}$. So we have a distinguished triangle $$ \mathcal{K} \to \mathcal{E} \to \mathcal{F} \to \mathcal{K}[1] $$ Apply the cohomological functors $\operatorname{Hom}(-,\omega_{X})$ and $\operatorname{Hom}(\mathcal{O}_{X}[-n],-)^{\vee}$ to obtain a ladder diagram with first row the exact sequence $$ 0\to \operatorname{Hom}(\mathcal{F},\omega_{X})\to \operatorname{Hom}(\mathcal{E},\omega_{X})\to \operatorname{Hom}(\mathcal{K},\omega_{X}) $$ and second row the exact sequence $$ 0\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{F})^{\vee}\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{E})^{\vee}\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{K})^{\vee}. $$ The zero on the first row corresponds to a negative ext group between coherent sheaves and the one on the second row follows from Grothendieck vanishing. By exactness of the rows and the result for finite direct sums of line bundles, the composition $\operatorname{Hom}(\mathcal{F},\omega_{X})\to \operatorname{Hom}(\mathcal{O}_{X}[-n],\mathcal{E})^{\vee}$ is injective. By commutativity of the diagram the map $\operatorname{Hom}(\mathcal{F},\omega_{X})\to\operatorname{Hom}(\mathcal{O}_{X},\mathcal{F})^{\vee}$ is injective as well. Since $\mathcal{F}$ was an arbitrary coherent sheaf on $X$, this is also true for the corresponding map for $\mathcal{K}$. Hence we can apply the 5-lemma (the version which is sometimes refered to as 4-lemma) to get the isomorphism that we wanted.
- The result is true for all bounded complexes of coherent sheaves. This can be shown by induction on the length of the complex. The result is already known for complexes of length zero, so let $\mathcal{F}^{\bullet}$ be a complex of length $n+1$ for some integer $n\geqslant 0$. For simplicity assume that this complex is concentrated between degrees $-n$ and $0$. Let $\mathcal{F}^{\bullet}_{\leqslant -1}$ be the stupid truncation (not preserving the cohomology of the complex), so that we get a distinguished triangle $$ \mathcal{F}^{\bullet}_{\leqslant -1}\to \mathcal{F}^{\bullet}\to \mathcal{F}^{0} \to \mathcal{F}^{\bullet}_{\leqslant -1}[1] $$ By induction hypothesis and arguing with the 5-lemma as before we obtain the desired isomorphism.
[End of proof]
Now we can use the compatibilities mentioned at the beginning to deduce the result in the form stated in Huybrechts' book. The conclusion is the following:
The category $D^{b}(X)$ has a Serre functor given by $$ \mathcal{F}^{\bullet} \mapsto \mathcal{F}^{\bullet}\otimes \omega_{X}[n] $$
Proof:
For all $\mathcal{F}^{\bullet}$ and $\mathcal{G}^{\bullet}$ in $D^{b}(X)$ we have the following sequence of functorial isomorphisms: $$ \operatorname{Hom}(\mathcal{F}^{\bullet},\mathcal{G}^{\bullet}) $$ $$ \cong \operatorname{Hom}(\mathcal{F}^{\bullet}\otimes \omega_{X},\mathcal{G}^{\bullet}\otimes \omega_{X}) $$ $$ \cong \operatorname{Hom}(R\mathcal{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}),\omega_{X}) $$ $$ \cong \operatorname{Hom}(\mathcal{O}_{X}[-n],R\mathcal{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}))^{\vee} $$ $$ \cong \operatorname{Hom}(\mathcal{O}_{X},R\mathcal{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}[n]))^{\vee} $$ $$ \cong H^{0}(R\Gamma(\mathcal{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}[n])))^{\vee}$$ $$ \cong H^{0}(R\operatorname{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}[n]))^{\vee}$$ $$ \cong \operatorname{Hom}(\mathcal{G}^{\bullet},\mathcal{F}^{\bullet}\otimes \omega_{X}[n])^{\vee}.$$ [End of proof]
Best Answer
As far as I know, to make a precise connection, one has to invoke Hodge theory. Suppose that $X$ is a compact smooth projective variety of dimension $d$. Then Poincare duality pairs $H^n(X,\mathbb C)$ with $H^{2d-n}(X,\mathbb C),$ for any $n$.
Now the Hodge decomposition gives $$H^n(X,\mathbb C) = \oplus_{p+q = n} H^q(X,\Omega^p)$$ and $$H^{2d-n}(X,\mathbb C) = \oplus_{p'+q' = 2 d - n} H^{q'}(X,\Omega^{p'}) = \oplus_{p + q = n}H^{d-q}(X,\Omega^{d - p}).$$
Now Serre duality gives a duality between $H^q(X,\Omega^p)$ and $H^{d - q}(X,\Omega^{d-p}),$ and the compatibility statement is that Poincare duality between $H^n$ and $H^{2 d - n}$ is induced by the direct sum of the pairings on the various summands in the Hodge decomposition given by Serre duality. (Perhaps up to signs and powers of $2 \pi i$, which I'm not brave enough to work out right now.)
Added: A good case to think about for a newcomer to Hodge theory is the case when $X$ is a compact Riemann surface (or equivalently, an algebraic curve). If the genus of $X$ is $g$, then $H^1(X,\mathbb C)$ is $2g$-dimensional, and is endowed with a symplectic pairing via Poincare duality.
Hodge theory breaks $H^1(X,\mathbb C)$ up into the sum of two $g$-dimensional subspaces, namely $H^0(X,\Omega^1)$ and $H^1(X,\mathcal O)$. These are isotropic under Poincare duality (i.e. the Poincare duality pairing vanishes when restricted to either of them), but the become dual to one another under Poincare duality, and that pairing agrees with the Serre duality pairing (up to a factor of $2\pi i$, perhaps).
The easiest part of this to understand is the inclusion $H^0(X,\Omega^1) \subset H^1(X,\mathbb C)$: a holomorphic differential gives a cohomology class just via de Rham theory (i.e. we integrate the holomorphic one form over 1-cycles); note that holomorphic 1-forms are automatically exact, because if you apply the exterior derivative, you get a holomorphic 2-form, which must vanish (because $X$ is a curve, i.e. of complex dimension one).
To see why $H^0(X,\Omega^1)$ is isotropic under Poincare duality, note that in the de Rham picture, the Poincare duality pairing corresponds to wedging forms. But wedging two holomorphic 1-forms again gives a holomorphic 2-form, which must vanish (as we already noted).