Equivalence of Definitions of Twisted Sheaf $ \mathcal {O}(1)$

Question

Let $\mathcal {O}(-1)$ be the tautological line bundle $X$ of $ \Bbb CP^1$, where $X=\{(z,l) \in \Bbb C^2 \times \Bbb CP^1 : z \in l \}$ together with canonical projection $X \to \Bbb CP^1$ (line bundle property and co easy to prove).

Futhermore we define $ \mathcal {O}(1):= \mathcal {O}(-1)^{\vee}$, where $ \mathcal {O}(-1)^{\vee}$ can be defined in two equivalent ways :

$ \mathcal {O}(-1) \otimes \mathcal {O}(-1)^{\vee}= \mathcal {O}_{\Bbb CP^1}$ is the same as to define it as $ \mathcal {O}(-1)^V := \underline{Hom}_{\mathcal{O}_{\Bbb CP^1}}(\mathcal{O}(-1),\mathcal{O}_{\Bbb CP^1})$ (follows from evaluation map).

The other way to define $ \mathcal {O}(1)$ is the following (compare with eg Liu’s AG, page 165 or see image below):

Obviously we have $ \Bbb CP^1 = Proj (B)$ where $B = \oplus _n B_n:=\mathbb{C}[X,Y] $ is graded $\mathbb{C}$-algebra in canonical way (polynomial grade). We set $B(n)$ as a new graduated $\mathbb{C}$-algebra by defining recursively $B(n)_m := B_{n+m}$. Liu defined the $ \mathcal {O}_{\Bbb CP^1 }$ -module $ \mathcal {O}(n)$ by setting $ \mathcal {O}(n) := \widetilde{B(n)}$ .

My question is: Why this both definitions of $ \mathcal {O}(1)$ are equivalent?

Here Liu's definition of "twisting":

Answer 1

Given a geometric vector bundle $\pi \colon E \to X$, the associated sheaf $\mathscr E$ associates to an open $U \subseteq X$ the set of sections $\sigma \colon U \to E|_U$ of $\pi_U$. We want to explain why the dual of the tautological bundle $T \to \mathbb P^n$ gives $\mathcal O_{\mathbb P^n}(1)$ in the sense of algebraic geometry.

Intuitively, a local section of $T^\vee$ is a (holomorphic or algebraic) association that sends a point $p \in \mathbb P^n$ to a linear function $\ell_p \to k$, where $\ell_p \subseteq k^{n+1}$ is the line that corresponds to the element $p \in \mathbb P^n$. It should be clear that the coordinate projections $x_i \colon k^{n+1} \to k$ induce such functions globally.

What is needed to formalise and complete this argument depends on what language you want to speak (analytic or algebraic) and how much theory you are already happy with. But at least this should give you some intuition as to why the result is true.

(For example, you could proceed by trivialising $T^\vee$ on the open set $\{x_i \neq 0\}$. On this set, $x_i$ is a global generator of $T^\vee$. Compute the transition functions and compare with those of $\mathcal O_{\mathbb P^n}(1)$.)

Answer 2

Starting from the second definition, one can think of $B(1)$ as the free $B$ module on 2 generators $[x]$ and $[y]$, quotiented out by the relations:

$B(X[y]-Y[x])$.

However, the equation $X[y]-Y[x] = 0$ defines the subsheaf $\mathcal{O}(-1)$ of $\mathcal{O}\otimes\mathbb{C}^2$ via the $\tilde{}$ construction. So the first definition of the sheaf $\mathcal{O}(1)$ does indeed agree with the second definition. Thank you for this question, it made me clarify my own thoughts.

If you want, this is similar to working in $V = \mathbb{C}^2$. If you have a complex linear subspace $W \subseteq V$ given by $l = 0$, where $l$ is a linear form on $\mathbb{C}^2$, then the dual space $W^*$ is isomorphic to $V^*/(\mathbb{C}l)$. In our case, instead of working over $\mathbb{C}$, we are instead working over $B$, at least before applying the $\tilde{}$ construction. Instead of $V$, we have the free $B$-module on two generators, and instead of having $W$, we have a $B$-submodule given by one $B$-linear equation. This is to supplement my previous short description.