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I am trying to find an explicit way to view global holomorphic sections of $\Omega^{1} \otimes \mathcal{O} (2)$ over $\mathbb{CP}^{2}$. I guess what I mean by "explicit" would be a formulation over an affine open $U_i \subset \mathbb{CP}^{2}$. According to what I found in Okoneck, Schneider and Spindler, there is a 3-dimensional space of such sections, but I want this for a computation in differential geometry.

flag	asked Nov 29 2010 at 3:57 David Johnson 13●2

Sasha · Accepted Answer · 2010-11-29 05:39:17Z UTC

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If $x,y,z$ are coordinates on $P^2$ then the 3 sections of $\Omega(2)$ are given by $xdy - ydx$, $ydz-zdy$, and $zdx-xdz$.

answered Nov 29 2010 at 5:39

Sasha
10.1k●1●7●17

	That is elegant, and the sort of answer I wanted. Can you give me a reference as to why this is true? – David Johnson Nov 29 2010 at 5:43
	It follows from Yuhao's answer plus the Koszul complex exactness. – Torsten Ekedahl Nov 29 2010 at 6:08
	I think it is necessary to clarify the notations here: as $x,y,z$ are only sections of $\mathcal{O}(1)$, "d" of them doesn't make sense in the usual way, in fact, as sections of $\Omega(2)$ the above sections are $x^2d(\dfrac{y}{x})$, etc. which can be written formally as $xdy-ydx$. – Yuhao Huang Nov 29 2010 at 6:22
	If $P^2 = (V - \{0\})/C^*$ then you can think about $x$, $y$, and $z$ as about the coordinates on $V$. Then $dx$, $dy$ and $dz$ are forms on $V$. – Sasha Nov 29 2010 at 6:27
	Between Sasha's answers and Torsten and Yuhao's explanations, I believe this makes sense to me now. To Yuhao, as a section of $\mathcal{O}(1)$ is a linear polynomial in the homogeneous coordinates, I think that the formulation of Sasha is correct, and dx has meaning in terms of the pullback via a local section of $\mathbb{C}^3\\{(0,0,0)}\to \mathbb{P}^2$ over a coordinate neighborhood. Thanks for the answers. – David Johnson Nov 30 2010 at 4:36

Yuhao Huang · Answer 2 · 2010-11-29 05:10:21Z UTC

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Use the Euler sequence:

$0 \to \Omega^1_{\mathbb{P}^n_A/A} \to \mathcal{O}_{\mathbb{P}^n_A}(-1)^{\oplus n+1} \to \mathcal{O}_{\mathbb{P}^n_A} \to 0.$

Everything can be seen explicitly from here. Tensoring the sequence with $\mathcal{O}(2)$ gives an exact sequence, which is still exact if you take global sections because every monomial of degree 2 is a multiple of a monomial of degree 1.:) So the dim you want = 9-6 =3.

answered Nov 29 2010 at 5:10

Yuhao Huang
3,067●5●29

The dimension I can see, but what I am looking for is explicit formulas, like Sasha's --- which is probably exactly what I need. – David Johnson Nov 29 2010 at 5:45

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Well, the 3 sections Sasha write down is precisely a basis for the kernel of the map obtained by taking global sections. The wikipedia page of Euler sequence contains a description of the maps of the Euler sequence, which is precisely what you missed, I guess. – Yuhao Huang Nov 29 2010 at 6:01

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You get the kernel of the map of global sections $\mathcal O_{\mathbb P^n_A}(1)^{\oplus n+1}\to \mathcal O_{\mathbb P^n_A}(2)$ by using the exactness of the Koszul complex of the $x_0,\dots,x_n$. – Torsten Ekedahl Nov 29 2010 at 6:07

Global sections of $\Omega^{1} \otimes \mathcal{O} (2)$ over $\mathbb{CP}^{2}$

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