global section on $\mathcal{O}(-1)$ and $\mathcal{O}(1)$

Question

I was tring to figure out the global section on $\mathcal{O}(-1)$ and $\mathcal{O}(1)$, there is a very nice explanation given here.

Which shows since the holomorphic line bundle is determined by local non vanishing holomorphic function as frame denote it $e_i $, then any global section $s\in \Gamma(\Bbb{P}^n, \mathcal{O}(k))$ is determined under this local coordinate as $s = t_ie_i$ with the coefficient $t_i \in \mathcal{O}(U_i)$ since we know the cocyle of line bundle $\mathcal{O}(1)$ and $\mathcal{O}(-1)$, we deduced that for global section on $\mathcal{O}(1)$ (which now encoded as set of $\{U_i,t_i\}$) satisfies the equation

$$t_i=\frac{z_i}{z_j}t_j \text{ on }\ U_i\cap U_j$$

However I can't figure out a detail , why this implies $t_i = 0$, and why in the same manner on $\mathcal{O}(1)$ if $$s_i=\frac{z_j}{z_i}s_j \text{ on }\ U_i\cap U_j$$

will imply $s_i=\frac{L}{z_i}$ for some $L=a_0z_0+...+a_nz_n \in (k^{n+1})^* $

It seems not very direct?

Answer 1

Let’s use the coordinates $X_0,X_1$ on $\mathbb P^1$. Then on the usual open set $U_0$ where $X_0 \neq 0$ , we have the sections on this are the elements of $k[X_1/X_0]$ and similarly we have the sections $k[X_0/X_1]$ for $U_1$.

When trying to build a global section for $\mathcal O(1)$, let’s start with a partial section on $U_0$, i.e. an element in $k[X_1/X_0]$. Now, to transition this function from $U_0$ to $U_1$ all you’re allowed to do is multiply it by $\frac{X_0}{X_1}$ and pray that it lands in $k[X_0/X_1]$. So we’re allowed to use a little bit of $X_0$ in the denominator in our starting function because the transition function will cancel it out but no more than a single power of $X_0$.

When trying to build a global section for $\mathcal O(-1)$, let’s again start with an element in $k[X_1/X_0]$. To transition this function from $U_0$ to $U_1$ all you’re allowed to do is multiply it by $\frac{X_1}{X_0}$ and pray that it lands in $k[X_0/X_1]$. Now you see the fix we’re in. Whatever polynomial you start with (even a non-zero constant), the transition function is just going to cause more headaches by putting an extra power of $X_0$ in the denominator which is not allowed to be present in the land of $k[X_0/X_1]$. Thus there are no non-zero global sections in this case.

I will leave it to you to generalize this to both $O(m)$ and to $\mathbb P^ n$. The underlying idea is essentially the same.

Answer 2

This example is very important therefore worthwhile to understand it clear.

The general statement is :

The global section of $\mathcal{O}(k)$ on $\Bbb{P}^n$ is a homogeneous polynomial of degree $k$ with $n$ variables if $k\ge 0$ , and it's zero if $k<0$.

There is a proof which has more analytic flavor(Demailly agbook COR 15.6):

Claim: there is a canonical isomorphism: $$\Phi:S^k(\Bbb{C}^{n+1})^\star\to H^0(P(\Bbb{C}^{n+1}),\mathcal{O}(k))$$

(1)Given a homogenous polynomial of degree $k$ denote it $P$, it will one to one correspond to a symmetric k-multilinear form (the symmetrization process), and each of this will uniquely determine by $\Bbb{C}$ linear map $\Bbb{C}^{n+1}\otimes ...\otimes \Bbb{C}^{n+1} \to \Bbb{C}$ (as multilinear map factor through the tensor product).

Therefore on each fiber of $\Bbb{P}^n\times (\Bbb{C}^{n+1})^{\otimes k}$ we can define a linear form $$P:\ \{l\}\times (\Bbb{C}^{n+1})^{\otimes k} \to \Bbb{C}$$ It can restrict to the fiber of $O(-k)$. Therefore we get a rough section of $O(k)$. Only needs to check it's holomorphic, however it's constant as $l\in \Bbb{P}^{n}$ varies, it's holomorphic.

(2)Given a holomorphic section $s$ on $O(k)$, at every $x\ne 0$, we have $\mu(x) = ([x],x)\in O(-1)_{[x]}$ it's holomorphic map, and tensoring it with $k$ times $\mu^k(x) \in O(-k)_{[x]}$ is still holomorphic map. Finally define

$$f(x) = s\circ(\mu^k)(x)$$ is the holomorphic map $\Bbb{C}^{n+1}\setminus\{0\}$ by the Hartogs theorem it extends to the whole space.

However we can check it's homogenous function which is also holomorphic, by Liouville's type theorem, it's the polynomial of degree $k$ (if $k\ge 0$). Therefore it's the polynomial we want.