How to calculate cohomology $H^1(X,O_X)$,$H^2(X,O_X)$ $H^1(X,O_X^*)$ of affine plane with double origin $X=\mathbb{A}^2\cup_{\mathbb{A}^2-\{0\}}\mathbb{A}^2$? To use Cech cohomology, I cannnot find a cover whose intersection all have trivial cover. Maybe there are other ways?

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    $\begingroup$ It might help using Mayer-Vietoris sequence. I'm trying to do that $\endgroup$
    – Peter Wu
    Dec 18, 2014 at 11:40

1 Answer 1


First, calculate cohomology of the plane without the origin $\mathbb{A}^2\setminus\{0\}$. In order to do this, you may use Mayer Vietoris with respect to the covering $\mathbb{A}^2=\mathbb{A}^2_x\cup\mathbb{A}^2_y$. You get exact sequences $$0\to H^0(\mathbb{A}^2\setminus\{0\},\mathcal{O})\xrightarrow{} H^0(\mathbb{A}^2_x,\mathcal{O})\oplus H^0(\mathbb{A}^2_y,\mathcal{O})\xrightarrow{\alpha} H^0(\mathbb{A}^2_{xy},\mathcal{O})\xrightarrow{} H^1(\mathbb{A}^2\setminus\{0\},\mathcal{O})\to 0$$ $$0\to H^0(\mathbb{A}^2\setminus\{0\},\mathcal{O})\xrightarrow{} k[x,y]_x\oplus k[x,y]_y\xrightarrow{\alpha} k[x,y]_{xy}\xrightarrow{} H^1(\mathbb{A}^2\setminus\{0\},\mathcal{O})\to 0$$ $$0\to H^2(\mathbb{A}^2\setminus\{0\},\mathcal{O})\to 0$$ Since $\alpha(f,g)=g-f$, you get that $H^q(\mathbb{A}^2\setminus\{0\},\mathcal{O})$ is $k[x,y]$ for $q=0$, $\mathrm{Span}_k\{x^{-i}y^{-j}\}_{i,j>0}$ for $q=1$ and $0$ otherwise. Similarly, you have an exact sequence $$1\to H^0(\mathbb{A}^2\setminus\{0\},\mathcal{O}^*)\to H^0(\mathbb{A}^2_x,\mathcal{O}^*)\oplus H^0(\mathbb{A}^2_y,\mathcal{O}^*)\xrightarrow{\beta} H^0(\mathbb{A}^2_{xy},\mathcal{O}^*)\to H^1(\mathbb{A}^2\setminus\{0\},\mathcal{O}^*)\to 1$$ $$1\to H^0(\mathbb{A}^2\setminus\{0\},\mathcal{O}^*)\to \bigsqcup_{i\in\mathbb{Z}} k^*x^i\oplus\bigsqcup_{j\in\mathbb{Z}} k^*y^j\xrightarrow{\beta} \bigsqcup_{i,j\in\mathbb{Z}} k^*x^iy^j\to H^1(\mathbb{A}^2\setminus\{0\},\mathcal{O}^*)\to 1$$ In fact, $H^1(\mathbb{A}^2_x,\mathcal{O}^*)=H^1(\mathbb{A}^2_y,\mathcal{O}^*)=1$ since $k[x,y]_x$ is an UFD. Since $\beta(f,g)=g/f$, you have $H^0(\mathbb{A}^2\setminus\{0\},\mathcal{O}^*)=k^*$ and $H^1(\mathbb{A}^2\setminus\{0\},\mathcal{O}^*)=1$.

Now use Mayer Vietoris another time: $$0\to H^0(X,\mathcal{O})\to H^0(\mathbb{A}^2,\mathcal{O})\oplus H^0(\mathbb{A}^2,\mathcal{O})\xrightarrow{\alpha} H^0(\mathbb{A}^2\setminus\{0\},\mathcal{O})\to H^1(X,\mathcal{O})\to 0$$ $$0\to H^0(X,\mathcal{O})\to k[x,y]\oplus k[x,y]\xrightarrow{\alpha} k[x,y]\to H^1(X,\mathcal{O})\to 0$$ $$0\to H^1(\mathbb{A}^2\setminus\{0\})=\mathrm{Span}_k\{x^{-i}y^{-j}\}_{i,j>0}\to H^2(X,\mathcal{O})\to 0$$ $$1\to H^0(X,\mathcal{O}^*)\to H^0(\mathbb{A}^2,\mathcal{O}^*)\oplus H^0(\mathbb{A}^2,\mathcal{O}^*)\xrightarrow{\beta} H^0(\mathbb{A}^2\setminus\{0\},\mathcal{O}^*)\to H^1(X,\mathcal{O}^*)\to 1$$ $$1\to H^0(X,\mathcal{O}^*)\to k^*\oplus k^*\xrightarrow{\beta} k^*\to H^1(X,\mathcal{O}^*)\to 1$$ In fact, $H^1(\mathbb{A}^2,\mathcal{O}^*)=1$ since $k[x,y]$ is an UFD.

As before, $\alpha(f,g)=g-f$, hence $H^0(X,\mathcal{O})=k[x,y]$, $H^1(X,\mathcal{O})=0$ and $H^2(X,\mathcal{O})=\mathrm{Span}_k\{x^{-i}y^{-j}\}_{i,j>0}$. Similarly, $\beta(f,g)=g/f$, hence $H^0(X,\mathcal{O}^*)=k^*$ and $H^1(X,\mathcal{O}^*)=1$.

  • $\begingroup$ Caro Giulio, I took the liberty of correcting a typo in the fourth line. I also upvoted your magnificent answer and I encourage others to do so too. Could you perhaps, if you feel so inclined and have the time , explicitate some of the maps and add a reference for Mayer-Vietoris ? $\endgroup$ Dec 18, 2014 at 21:02
  • $\begingroup$ You make me blush! Anyway, in a few days I'll add some details and the reference for Mayer-Vietoris $\endgroup$ Dec 18, 2014 at 23:51
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    $\begingroup$ @Giulio Bresciani: Very nice answer. Here is a reference for Mayer-Vietoris: stacks.math.columbia.edu/download/cohomology.pdf (see Section 9) from Stack Project. $\endgroup$
    – Krish
    Dec 20, 2014 at 14:49
  • $\begingroup$ @Georges I've added details, as you asked. $\endgroup$ Dec 23, 2014 at 0:19
  • $\begingroup$ @Krish perfect, thanks. There is only one minor problem: in 9.2, they state Mayer-Vietoris only for $\mathcal{O}$-modules, and we need it also for $\mathcal{O}^*$. Anyway, it is clear that they don't really use that hypothesis, and it works perfectly for any sheaf of abelian groups. $\endgroup$ Dec 23, 2014 at 0:23

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