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Let consider the complex projective space in two ways: Topologically by setting $$\mathbb{P}_n =\mathbb{P}(\mathbb{C}^{n+1}) = \mathbb{C}^{n+1}/{\sim}$$

as space of complex lines and via

$$\mathbb{P}^n = \operatorname{Proj}(\mathbb{C}[T_0, T_1, \ldots, T_n])$$

as $\mathbb{C}$-scheme.

We have the (tautological) invertible sheaf $\mathcal{O}_{\mathbb{P}_n}(1)$ defined locally by $\mathcal{O}_{\mathbb{P}_n}(1) \vert _{D_+(T_i)} = \operatorname{Spec}(\mathbb{C}[T_0, T_1, \ldots, T_n]_{(T_i),1})= \operatorname{Spec}(\mathbb{C}[T_0, T_1, \ldots, T_n])$ where $\mathbb{C}[T_0, T_1, \ldots, T_n]_{(T_i),1}$ means the $1$-graded part of canonically graded ring $\bigoplus _i \mathbb{C}[T_0, T_1, \ldots, T_n]_{(T_i),i}$. Let $\mathcal{O}_{\mathbb{P}_n}(-1)$ be the dual sheaf of $\mathcal{O}_{\mathbb{P}_n}(1)$.

It's easy to see that $\{(l,u)\in \mathbb{P}_n \times \mathbb{C}^{n+1} \vert v \in l\}$ is a line bundle over $\mathbb{P}_n$.

My question is why holds

$$\mathcal{O}_{\mathbb{P}_n}(-1) = \{(l,u)\in \mathbb{P}_n \times \mathbb{C}^{n+1} \mid v \in l\}$$

Remark: I know that we can identify locally free sheaves with vector bundles if global sections of the structure sheaf are given as regular functions $U \to \mathbb{C}$, but I don't see how to conclude the desired equation.

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  • $\begingroup$ HINT: Remember that sections $\mathscr O(k)(U)$ are homogeneous polynomials of degree $k$ on $U$. Can you see that $\mathscr O(1)$ is the dual of the tautological line bundle? $\endgroup$ – Ted Shifrin Apr 7 '18 at 3:02
  • $\begingroup$ @TedShifrin Which command do you use to produce that calligraphic $O$? Thanks in advance. $\endgroup$ – Batominovski Apr 7 '18 at 4:48
  • $\begingroup$ $$ \begin{align} & \mathbb{C}^{n+1}/\sim \\ & \mathbb{C}^{n+1}/{\sim} \end{align} $$ I changed the first line above to the second, and did some other MathJax corrections. There's a common-sense reason why you should expect to need to do it as on the second line above. $\endgroup$ – Michael Hardy Apr 7 '18 at 6:10
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Let $V$ be a vector space, and $\Bbb P(V) = X$ be the space of its lines. Write $\mathcal O_X(-1)'$ for the line bundle $\{(l,v) \in X \times V : v \in l\}$.

We will build an isomorphism $\mathcal O_X(-1)' \times \mathcal O_X(1) \to \mathcal O_X$. If $s'\in \Gamma(U,O(-1)')$ we remark that $s'$ is the same as an algebraic function $v : U \to V$ where $v(u) \in u$ for each $u \in U$. A section $s$ $\in \Gamma(U,\mathcal O(1))$ is just a regular section on $U$ and of degree $1$, say $\varphi$.

Warning : it does not define a function, e.g the section $x_0$ has no well defined value at $[1:0:0]$. There are two ways to avoid this problem : first, working with quotients of sections gives a well defined map (but this is not regular anymore), or you can also consider $\varphi$ as a regular function on $W = \{v \in V : [v] \in U\}$.

In particular we obtain an "evaluation map" $$\text{ev} : \mathcal O_X(-1) \otimes \mathcal O_X(1) \to \mathcal O_X, (v, \varphi) \mapsto \varphi(v) : (u \mapsto \varphi(v(u)))$$ which is easily checked to be an isomorphism.

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  • $\begingroup$ Hi. Thank you for the answer. One step stays unclear: Why are the elements $s' \in \Gamma(U,O(-1)')$ identical to algebraic functions $v: U \to V$ with $v(u) \in u$ property? $\endgroup$ – KarlPeter Apr 11 '18 at 21:35
  • $\begingroup$ @KarlPeter : Hi ! It's because the total space of $\mathcal O(-1)'$ is $E' =: \{(v,l) \in V \times \Bbb P(V) : v \in l \} $ and the projection $p : E' \to \Bbb P(V)$ is exactly the restriction of the projection $V \times \Bbb P(V) \to \Bbb P(V)$. In particular, $s' \in \Gamma(U, O(-1)')$ is just a map $s' : U \to E'$ with $p \circ s' = id$. This is exactly algebraic functions $v : U \to V$ with the property that $v(u) \in u$. $\endgroup$ – Nicolas Hemelsoet Apr 12 '18 at 9:43
  • $\begingroup$ I noticed that in your definition of $ev$ there occurs following problem: By your explanations above $v \in \Gamma(U, O(-1)')$ is essentially a map $v: U \to V$ and $\varphi \in \Gamma(U, O(1))$ a regular function, especially a map $U \to F$ for $F= \mathbb{R}, \mathbb{C}$. What does then the expression $\varphi(v(u))$ here mean? Since $v(u) \in V$ we get a domain problem,right? $\endgroup$ – KarlPeter Nov 15 '18 at 0:37
  • $\begingroup$ I edited my answer. $\endgroup$ – Nicolas Hemelsoet Nov 15 '18 at 5:53

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