0
$\begingroup$

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.

$\endgroup$

2 Answers 2

4
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ That is elegant, and the sort of answer I wanted. Can you give me a reference as to why this is true? $\endgroup$ Nov 29, 2010 at 5:43
  • $\begingroup$ It follows from Yuhao's answer plus the Koszul complex exactness. $\endgroup$ Nov 29, 2010 at 6:08
  • $\begingroup$ 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 $xdy-ydx$. $\endgroup$ Nov 29, 2010 at 6:22
  • $\begingroup$ 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 $V$. $\endgroup$
    – Sasha
    Nov 29, 2010 at 6:27
  • $\begingroup$ 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. $\endgroup$ Nov 30, 2010 at 4:36
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ The dimension I can see, but what I am looking for is explicit formulas, like Sasha's --- which is probably exactly what I need. $\endgroup$ Nov 29, 2010 at 5:45
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 29, 2010 at 6:01
  • 1
    $\begingroup$ 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 $x_0,\dots,x_n$. $\endgroup$ Nov 29, 2010 at 6:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.