Precise connection between Poincare Duality and Serre Duality

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).

I'd like to point out that there is a generalization of Poincare duality which lives purely in the land of smooth manifolds and looks like Serre duality. Let $M$ be a compact connected smooth $n$-manifold. Let $E$ be a vector bundle on $M$ equipped with a flat (some people say integrable) connection $\nabla$. Let $T^{\ast}$ be the cotangent bundle to $M$. For a vector bundle $V$ on $M$, let $C^{\infty}(V)$ be the sheaf of smooth sections of $V$. So $C^{\infty}(V)(U)$ is smooth sections of $V$ over $U$.

The connection $\nabla$ induces maps $C^{\infty}(E \otimes \bigwedge^k T^{\ast}) \to C^{\infty}(E \otimes \bigwedge^{k+1} T^{\ast})$. These maps form a complex $$0 \to C^{\infty}(E)(M) \to C^{\infty}(E \otimes T^{\ast})(M) \to \cdots \to C^{\infty}(E \otimes \bigwedge\nolimits^n T^{\ast})(M) \to 0.$$

Define $H_{DR}^i(M, E, \nabla)$ (not standard notation), to be the cohomology groups of this complex. Then we have:

Relation to sheaf cohomology

Let $E_0$ be the subsheaf of $C^{\infty}(E)$ given by the kernel of $\nabla$. (The so-called flat sections of $E$.) Then $$H^i_{sheaf}(M, E_0) \cong H^i_{DR}(M, E, \nabla).$$ Proof sketch: $$E_0 \to C^{\infty}(E) \to C^{\infty}(E \otimes T^{\ast}) \to \cdots \to C^{\infty}(E \otimes \bigwedge\nolimits^n T^{\ast})\to 0$$ is a resolution of $E_0$ by acyclic sheaves.

Duality We have $H^n(M, \bigwedge^n T^{\ast}) \cong \mathbb{R}$ and the cup product pairing $$H_{DR}^q(M, E, \nabla) \otimes H_{DR}^{n-q}(M, E^{\vee} \otimes \bigwedge\nolimits^n T^{\ast}, \nabla') \longrightarrow H_{DR}^n(M, \bigwedge\nolimits^n T^{\ast}) \cong \mathbb{R}$$ is perfect. Here $\nabla'$ is the connection on $E^{\vee} \otimes \bigwedge^n T^{\ast}$ which is adjoint to $\nabla$, in a sense I don't want to define.

Note that, if $M$ is orientable, then $\bigwedge^n T^{\ast}$ is trivial, which makes the statement simpler but look a bit less like Serre duality.

If $E$ is the trivial one dimensional bundle, and $\nabla$ is the standard connection $f \mapsto df$, then $H^i_{DR}(M, E, \nabla)$ is the standard DeRham cohomology $H^i_{DR}(M)$. So, if $E$ and $\nabla$ are as above, and $M$ is orientable, we recover Poincare duality.

I recall a good discussion of this in Voisin's book Hodge Theory and Complex Algebraic Geometry, volume I, chapter 5.3.2. I talked about this in my Hodge Theory course.