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?

  • 4
    $\begingroup$ Check out Brian Conrad's comment here: mathoverflow.net/questions/28496/… $\endgroup$ Aug 13, 2010 at 21:35
  • $\begingroup$ Ah! Thanks. But I am not able to locate the pdf he is describing. Did you get it? $\endgroup$
    – user1119
    Aug 13, 2010 at 21:51

2 Answers 2


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).

  • $\begingroup$ Thanks for explaining. This is a lot more complicated than I was suspecting. I wouldn't have known if I had not asked. $\endgroup$
    – user1119
    Aug 13, 2010 at 23:48
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    $\begingroup$ Dear George, You're welcome. To make it seem less complicated, ask yourself: what mechanism in geometry relates cohomology with complex coefficients (which is the realm of Poincare duality) to cohomology of coherent sheaves (which is the realm of Serre duality)? The answer is typically Hodge theory, and, indeed, this is one of the reasons that Hodge theory is important: it connects two worlds --- topology and algebraic geometry --- which can otherwise seem a little remote from one another. It might help to ponder the case when $X$ is a Riemann surface and $n = 1$. (See my edit for this.) $\endgroup$
    – Matt E
    Aug 14, 2010 at 0:09
  • $\begingroup$ Thanks for comments to help unravel the Hodge structure. I had earlier thought only upto the simpler connection between de Rham complex and coherent sheaves, namely that de Rham complex is actually a complex of sheaves(via the Poincare lemma). $\endgroup$
    – user1119
    Aug 14, 2010 at 0:33
  • $\begingroup$ Dear George, In fact, if you continue to think along the lines you mentioned (the de Rham complex as a complex of sheaves), you can naturally find your way to the Hodge structure on cohomology. (Perhaps comments are not the best place to elaborate on this, though.) So you certainly had some of the appropriate ingredients in mind. $\endgroup$
    – Matt E
    Aug 14, 2010 at 0:37
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    $\begingroup$ Very good answer! $\endgroup$ Feb 15, 2011 at 16:40

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.


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