7
$\begingroup$

I have been reading about Serre Duality and am wondering: What sort of computations would motivate coming up with such a duality theorem? On the Wikipedia page, it is claimed "Serre duality is the analog for coherent sheaf cohomology of Poincaré duality in topology, with the canonical line bundle replacing the orientation sheaf."

I am searching for some clarification on what coherent sheaf cohomology of Poincare duality is and how Serre came up with importing such a result.

So really, I am asking for three things here:

  1. Where can I go read about the historical development of Serre Duality and related theorems?
  2. What is coherent sheaf cohomology for Poincare duality?
  3. What sort of computations did people do to dream up of Serre Duality?

For completeness, there is also a related question here which I missed.

$\endgroup$
1
  • 1
    $\begingroup$ Small comment: there is nothing called "coherent sheaf cohomology of Poincare duality." The sentence you quoted can be rephrased as "Serre duality is the analog of Poincare duality, transported from the setting of topology to coherent sheaf cohomology." $\endgroup$ Jun 21, 2022 at 0:23

1 Answer 1

7
$\begingroup$

This got a bit longer than I expected, and a bit rambly. It isn't completely precise, but one can piece the ideas together to a complete argument. Hopefully this suggests the motivation anyway.

Here is a way to see it. A version of Poincaré duality states that on a compact oriented $n$-manifold $M$, there is a perfect pairing $H^k_{\mathrm{dR}}(M;\Bbb{R})\otimes H^{n-k}_{\mathrm{dR}}(M;\Bbb{R})\to \Bbb{R}$ given quite simply by $$ [\omega]\otimes [\eta]\mapsto \int_M \omega\wedge\eta. $$ Obviously, this needs a little unpacking. This is also not the most general version of the theorem. Remember that de Rham cohomology is the cohomology of the complex $$C^\infty(M)\xrightarrow{d} \Omega^1(M)\xrightarrow{d}\cdots \xrightarrow{d}\Omega^n(M).$$ There is a (soft) resolution of the constant sheaf $\underline{\Bbb{R}}$ on $M$ given by $$ 0\to \underline{\mathbb{R}}\to \big[\mathcal{C}^\infty(M)\to \mathcal{A}^1(M,\Bbb{R})\to \cdots \to \mathcal{A}^n(M,\mathbb{R})\to 0\big] $$ where here $\mathcal{A}^k(M,\mathbb{R})$ denotes the sheaf of real $k$-forms on $M$. To compute the sheaf cohomology, we apply the global sections functor $\Gamma(M,-)$ to the resolution of $\underline{\Bbb{R}}$. This returns the usual de Rham complex $$C^\infty(M)\xrightarrow{d} \Omega^1(M)\xrightarrow{d}\cdots \xrightarrow{d}\Omega^n(M)$$ like above. As a direct consequence, we see that the de Rham cohomology $H^k_{\mathrm{dR}}(M;\mathbb{R})$ is the $k^{th}$ sheaf cohomology of the constant sheaf $\underline{\mathbb{R}}$ on $M$. Instead of studying constant sheaf cohomology, one can study coherent sheaf cohomology - or slightly more classically, the cohomology of vector bundles.

For this, take $X$ to be a compact complex manifold. Let $\mathcal{E}$ denote a holomorphic vector bundle. Associated to this vector bundle are sheaves $\mathcal{A}^{p,q}(\mathcal{E})$ of smooth differential $(p,q)$-forms valued in the vector bundle $\mathcal{E}$. Long story short, there is something called the Dolbeault resolution in this circumstance which gives a soft resolution $$ 0\to \mathcal{E}\to \big[ \mathcal{A}^{0,0}(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})\to \cdots \to \mathcal{A}^{0,n}(\mathcal{E})\to 0\big]. $$ So, we can compute the sheaf cohomology of $\mathcal{E}$ using this resolution, and we find that the representatives of an element of $H^k(X,\mathcal{E})$ are differential $(0,k)$-forms with values in $\mathcal{E}$. There is a line bundle, $\mathcal{K}_X$, called the canonical bundle of $X$. Its sections are holomorphic $n$-forms. Sections of this bundle determine in particular smooth $(n,0)$-forms. Let's replicate Poincaré duality. Define a pairing $$ H^k(X,\mathcal{E})\otimes H^{n-k}(X,\mathcal{E}^*\otimes \mathcal{K}_X)\to \Bbb{C} $$ as follows. Given $[\alpha] \otimes [\beta]$, note that $[\alpha]$ is represented by a $(0,k)$-form valued in $\mathcal{E}$, and $[\beta]$ is represented by a $(0,n-k)$-form valued in $\mathcal{K}_X\otimes \mathcal{E}^*$. This determines an $(n,n-k)$-form valued in $\mathcal{E}^*$. The tensor product then determines an $(n,n)$-form valued in $\mathcal{E}\otimes \mathcal{E}^*$. However, there is a natural map $\mathcal{E}\otimes \mathcal{E}^*\to \mathcal{O}_X$ (the trivial line bundle); Simply evaluate a section of $\mathcal{E}^*$ on a section of $\mathcal{E}$. Using all this, we can send $[\alpha]\otimes [\beta]$ to an $(n,n)$-form valued in $\mathcal{E}\otimes \mathcal{E}^*$ and then to an $(n,n)$-form $\omega$ valued in $\mathcal{O}_X$, i.e. an ordinary $(n,n)$-form. But this is in particular a top degree form on the $2n$ dimensional manifold $X$. So, we can simply integrate: $$ [\alpha ]\otimes [\beta]\mapsto \int_X \omega. $$ This, it turns out, is the pairing that defines the isomorphism in Serre duality. This has been souped up in many different contexts and culminates in Grothendieck's theory of duality. However, the basic intuition comes from this idea of wedging some forms and integrating. In this sense, Poincaré duality is the topological version of Serre duality.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .