TITLE: Precise connection between Poincare Duality and Serre Duality
QUESTION [41 upvotes]: 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?

REPLY [9 votes]: 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.