# Global sections of $\mathcal{O}(-1)$ and $\mathcal{O}(1)$, understanding structure sheaves and twisting.

In chapter 2 section 7 (pg 151) of Hartshorne's algebraic geometry there is an example given that talks about automorphisms of $\mathbb{P}_k^n$. In that example Hartshorne states that $\mathcal{O}(-1)$ has no global sections. However, we know that $\mathcal{O}(1)$ is generated by global sections. This is stated at the first of the section that if $\mathbb{P}_k^n = Proj k[x_0,...,x_n]$ then the $x_0,...,x_n$ give rise to global sections $x_0,...,x_n\in\Gamma(\mathbb{P}_k^n,\mathcal{O}(1))$.

I guess I don't understand this twisted structure sheaf very well, or to be honest I don't think I understand structure sheaves in general as well as I would like. The twisting part seems simple at first- you shift the grading of Proj over and then take the structure sheaf. Admittedly I don't feel comfortable using the structure sheaf other than using the basic facts about it that Hartshorne gives when it is introduced. If anyone could give some insight as to what's going on here or how I might be able to understand this better it would be much appreciated. Thanks.

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The first step, I think, is to understand how $\mathscr{O}(1)$ is the tautological line bundle over $\mathbb{P}^n$. Unfortunately that's where I fail too... –  Zhen Lin Feb 24 '12 at 19:46
@Zhen: Dear Zhen, In fact, $\mathcal O(1)$ is the dual to what is normally called the tautological line bundle. Regards, –  Matt E Feb 24 '12 at 23:39
@Matt: That explains my troubles. Thanks! –  Zhen Lin Feb 25 '12 at 0:21

There is a general technique for describing line bundles in topology, differential geometry, complex manifolds,...
We can apply it in algebraic geometry to projective space and it will give you an alternative approach which you might find appealing. Here it is.

(1)
Consider the covering $\mathcal U$ of $\mathbb P^n_k$ consisting of the open subsets $U_i=\lbrace z=[z_0:...:z_n]\in\mathbb P^n_k| z_i\neq 0\rbrace \; (i=0,...,n)$ and the functions $g_{ij}\in \Gamma(U_i\cap U_j,\mathcal O^*_{\mathbb P^n_k})$ defined by $g_{ij}(z)=\frac{z_j}{z_i}$.

These functions ("a cocycle relative to $\mathcal U$") completely characterize the sheaf $\mathcal O_{\mathbb P^n_k}(1)$ : given an arbitrary open subset $U\subset \mathbb P^n_k$, a section $s\in \Gamma(U,\mathcal O_{\mathbb P^n_k}(1))$ corresponds to a family of functions $s_i\in \Gamma(U\cap U_i,\mathcal O_{\mathbb P^n_k})$ satisfying $s_i=g_{ij}s_j$ on $U\cap U_i\cap U_j$.

Important example (1) A global section $s\in \Gamma(\mathbb P^n_k,\mathcal O_{\mathbb P^n_k}(1))$ is given by functions $s_i\in \Gamma(U_i,\mathcal O_{\mathbb P^n_k})$ satisfying $s_i=\frac{z_j}{z_i}s_j$ on $U_i\cap U_j$.
The only possible functions are of the form $s_i=\frac{L}{z_i}$ where $L=a_0z_0+...+a_nz_n \in (k^{n+1})^*$ is an arbitrary linear form on $k^{n+1}$ and we have thus $$\Gamma(\mathbb P^n_k,\mathcal O_{\mathbb P^n_k}(1))=(k^{n+1})^*$$

(-1)
Similarly, the functions $g_{ij}^{-1}(z)=\frac{z_i}{z_j}$ characterize the sheaf $\mathcal O_{\mathbb P^n_k}(-1)$
Important example (-1) A global section $t\in \Gamma(\mathbb P^n_k,\mathcal O_{\mathbb P^n_k}(-1))$ is given by functions $t_i\in \Gamma(U_i,\mathcal O_{\mathbb P^n_k})$ satisfying $t_i=\frac{z_i}{z_j}t_j$ on $U_i\cap U_j$.
Only $t_i=0$ is possible and we have thus $$\Gamma(\mathbb P^n_k,\mathcal O_{\mathbb P^n_k}(-1))=0$$

An explicit calculations for n=1 (skip if you are fed up with computing!)
In order to really understand what's going on without losing ourselves in indices , let's examine the case $n=1$ and compute $\Gamma(\mathbb P^1_k,\mathcal O(1))$.
A section $s\in \Gamma(\mathbb P^1_k,\mathcal O(1))$ is given by a function $s_0(z)=a+bz+cz^2+...\in \Gamma(U_0,\mathcal O)=k[z]\; (z=z_1/z_0)$
and a function
$s_1(w)=\alpha +\beta w+\gamma w^2+...\in \Gamma(U_1,\mathcal O) \; (w=z_0/z_1)$
satisfying (note that $w=1/z$ on $U_0\cap U_1$)
$s_0(z)=a+bz+cz^2+...=z(\alpha +\beta w+\gamma w^2+...)=\alpha z+\beta+\gamma (1/z)+...$ on $U_0\cap U_1$.
This implies that $a=\beta, b=\alpha$ and that all other coefficients are zero.
Thus, putting $L=az_0+bz_1$, we have indeed proved that $s_0=\frac{L}{z_0}, s_1=\frac{L}{z_1}$ as announced above.

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Thank you. This is very helpful. I am still digesting it, but things are clearing up slowly. –  MJoszef Feb 27 '12 at 1:02
There seems to be a typo on the second line of the important example $(1)$: Should $s_i=\frac{z_j}{z_j}s_i$ read $s_i=\frac{z_j}{z_i}s_j$ ? –  Victor Jan 29 '13 at 5:50
Dear @Victor, you are absolutely right and I have corrected my typo. A thousand thanks for your vigilance. –  Georges Elencwajg Nov 6 '13 at 20:24

$\mathcal{O}(m)$ is the bundle which is trivial on the usual open cover of $\mathbb{P}_k^n$, and with the transitional morphisms between $U_{i_1}$ and $U_{i_2}$ given by $f \to f \times (x_{i_1}/x_{i_2})^m$.

A global section of $\mathcal{O}(m)$ is the data of a family of polynomials, one polynomial $P_i \in k[x_0 / x_i \ldots x_n / x_i] = \Gamma(\mathcal{O}(m),U_i) = \Gamma(\mathcal{O}_{\mathbb{P}_k^n},U_i)$ for each of those open sets, satisfying the relations $P_{i_1}(x_0 / x_{i_1} \ldots x_n / x_{i_1}) \times (x_{i_1}/x_{i_2})^m = P_{i_2}(x_0 / x_{i_2} \ldots x_n / x_{i_2})$.

So, $P_i(x_0 / x_i \ldots x_n/ x_i) \times x_i^m$ seems to be some rational function $f \in k(x_1 \ldots x_n)$ independant of $i$, with the condition that $f(x_0 \ldots x_n)/x_i^m \in k [x_0/x_i \ldots x_n/x_i]$. This implies that the denominator of $f$ must be a power of $x_i$, for all $i$. Thus $f$ has to be a polynomial in $k[x_0 \ldots x_n]$

Now, when $m \ge 0$, the conditions $f / x_i^m \in k [x_0/x_i \ldots x_n/x_i]$ can only be realised when $f$ is a homogeneous degree $m$ polynomial in $k[x_0 \ldots x_n]$, and this gives a bijective correspondence between homogeneous degree $m$ polynomial and the global sections of $\mathcal{O}(m)$. When $m < 0$, it is not possible to find a polynomial $f$ satisfying those conditions, so there is no global section of $\mathcal{O}(m)$

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Dear mercio, your conventions seem incorrect: you don't have enough variables (you need $n+1$) in your various rings. For example you apparently didn't take into account that among your $x_1 / x_{i_1} \ldots x_n / x_{i_1}$ you will have a $x_{i_1} / x_{i_1}=1$. –  Georges Elencwajg Feb 24 '12 at 21:01
@ Georges, yes thanks I think I should have my indices go from 0 to n instead, I always get confused there. –  mercio Feb 24 '12 at 21:04