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Let $\mathbb{P}^1 $ be the complex projective line.

Using the standard affine cover, $\mathcal{U} = \lbrace U,U' \rbrace, \ \ $ we can define some quasi-coherent sheaves on $\mathbb{P}^1 $.

We can do this by defining triples $ \ (M,M',\phi)$, where $M,M'$ are $ \mathbb{C}[x] , \ \mathbb{C}[x^{-1}]$-modules respectively, and $\phi : M_{x^{-1}} \rightarrow M'_{x} $ is an isomorphism of $\mathbb{C}[x,x^{-1}] $-modules.

Now, I have defined the following sheaves (here $dx$ is just a formal generator) :

$$ \Omega^0 = (\mathbb{C}[x], \ \mathbb{C}[x^{-1}], \ x \mapsto x^{-1}), \ \ \Omega^1 = (\mathbb{C}[x] \cdot dx , \mathbb{C}[x^{-1}] \cdot dx, x \mapsto x^{-1}) $$

We then, if i have done the above correctly, have the de Rham complex on $\mathbb{P}^1 $:

$$ 0 \longrightarrow \Omega^0 \longrightarrow \Omega^1 \longrightarrow 0 $$

where the codifferentials are the derivations between the modules.

How do I go about finding a Cartan–Eilenberg resolution for this complex? I'm fairly sure I understand the definition. I've been messing about with Cech homology stuff, but I'm a bit stuck.

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    $\begingroup$ Your description of $\Omega^1$ is not correct: $\Omega^0 \ncong \Omega^1$, after all. The transition should be given by $\sum_{n \in \mathbb{Z}} a_n x^n \, \mathrm{d} x \mapsto - \sum_{n \in \mathbb{Z}} a_n x^{n+2} \, \mathrm{d} x^{-1}$. $\endgroup$
    – Zhen Lin
    Aug 3 '14 at 13:24
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Explicit injective resolutions (hence also Cartan–Eilenberg resolutions) are hard to work with concretely. (They're usually not finitely generated, for example.) For the purposes of this calculation it is easier to use Čech cohomology.

The standard affine open cover $\mathfrak{U} = \{ U, U' \}$ of $\mathbb{P}^1$ is a good open cover in the following sense:

  • The intersection of any finite set of elements of $\mathfrak{U}$ is affine, hence acyclic with respect to quasicoherent sheaf cohomology.

Let $\check{C}{}^{\bullet} (\mathfrak{U}, \mathscr{F})$ denote the Čech cochain complex of a sheaf $\mathscr{F}$ with respect to $\mathfrak{U}$. Applying $\check{C}{}^{\bullet} (\mathfrak{U}, -)$ to the de Rham complex yields a double complex: $$\begin{array}{ccccccc} \vdots & & \vdots & & \vdots & & ⋰ \\ \uparrow & & \uparrow & & \uparrow & & \\ \check{C}{}^2 (\mathfrak{U}, \Omega^0) & \to & \check{C}{}^2 (\mathfrak{U}, \Omega^1) & \to & \check{C}{}^2 (\mathfrak{U}, \Omega^2) & \to & \cdots \\ \uparrow & & \uparrow & & \uparrow & & \\ \check{C}{}^1 (\mathfrak{U}, \Omega^0) & \to & \check{C}{}^1 (\mathfrak{U}, \Omega^1) & \to & \check{C}{}^1 (\mathfrak{U}, \Omega^2) & \to & \cdots \\ \uparrow & & \uparrow & & \uparrow & & \\ \check{C}{}^0 (\mathfrak{U}, \Omega^0) & \to & \check{C}{}^0 (\mathfrak{U}, \Omega^1) & \to & \check{C}{}^0 (\mathfrak{U}, \Omega^2) & \to & \cdots \end{array}$$ I claim that since $\mathfrak{U}$ is a good open cover, the cohomology of the total complex of this double complex is the hypercohomology of the de Rham complex. (See this MO question.) In your situation, the double complex in question is not so big: $$\begin{array}{ccc} \mathbb{C}[x, x^{-1}] & \to & \mathbb{C}[x, x^{-1}] \, \mathrm{d} x \\ \uparrow & & \uparrow \\ \mathbb{C}[x] \oplus \mathbb{C}[x^{-1}] & \to & \mathbb{C}[x] \, \mathrm{d} x \oplus \mathbb{C}[x^{-1}] \, \mathrm{d} x^{-1} \end{array}$$ The total complex is then $$\mathbb{C}[x] \oplus \mathbb{C}[x^{-1}] \longrightarrow \mathbb{C}[x, x^{-1}] \oplus \mathbb{C}[x] \, \mathrm{d} x \oplus \mathbb{C}[x^{-1}] \, \mathrm{d} x^{-1} \longrightarrow \mathbb{C}[x, x^{-1}] \, \mathrm{d} x$$ where the first differential is $$f (x) + g (x^{-1}) \mapsto (f (x) - g (x^{-1})) + f' (x) \, \mathrm{d} x + g' (x^{-1}) \, \mathrm{d} x^{-1}$$ and the second differential is $$h (x) + k (x) \, \mathrm{d} x + l (x^{-1}) \, \mathrm{d} x^{-1} \mapsto (h'(x) - k (x) - x^{-2} l (x^{-1})) \, \mathrm{d} x$$ so with a little thought one sees that the cohomology groups we seek are as follows: \begin{align} H^0 & = \mathbb{C} & H^1 & = 0 & H^2 & = \mathbb{C} \end{align} Of course, this is exactly what one expects from the comparison theorem.

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  • $\begingroup$ Brilliant answer! I've been looking nearly everywhere for something like this, there's not many explicit examples like this covered in the literature I've come across. $\endgroup$
    – JC574
    Aug 3 '14 at 14:14
  • $\begingroup$ I've seen a theorem that Cech cohom. is isomorphic to hypercohom. when our space is paracompact, which doesn't seem exactly the same as what we've used here. Do you have a specific reference to the theorem you used here? $\endgroup$
    – JC574
    Aug 3 '14 at 22:23
  • $\begingroup$ See the remark about "good open covers" and the link provided above. $\endgroup$
    – Zhen Lin
    Aug 3 '14 at 22:33

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