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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":

enter image description here

user267839
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1 Answers1

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Let $U\subset\mathbb{CP}^1$ be an open subset.

Sections $U\to p^{-1}(U)$ of $p\colon X\to\mathbb{CP}^1$ look like $[x:y]\mapsto(f_1/g_1,f_2/g_2)$ where $f_1$ and $g_1$ are homogeneous polynomials of the same degree, and the same holds for $f_2$ and $g_2$. We moreover require $f_1/g_1\cdot x+f_2/g_2\cdot y=0$, and such that $g_1$ and $g_2$ have no zeroes on $U$. This corresponds to the rational function $f_1/g_1\cdot y^{-1}=-f_2/g_2\cdot x^{-1}$ of homogeneous degree $-1$, so gives a well-defined section of $\mathcal O_{\mathbb{CP}^1}(-1)$. Indeed, at $[x:y]\in U$ with $y\ne0$ the expression $f_1/g_1\cdot y^{-1}$ is well-defined and when $x\ne0$ the expression $f_2/g_2\cdot x^{-1}$ is well-defined.

Conversely, suppose $F/G$ is a section of $\mathcal O_{\mathbb{CP}^1}(-1)$, i.e., where $F$ and $G$ are homogeneous and $\deg(G)=\deg(F)+1$ and $G$ has no zeroes on $U$. Then $[x:y]\mapsto (Fy/G,-Fx/G)$ is a section $U\to p^{-1}(U)$.

Kenta S
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  • probably you want the condition $f_1/g_1\cdot y-f_2/g_2\cdot x=0$? Let me try to understand your argument: the local section $U\to p^{-1}(U)= X \vert _U={(z,l) \in \Bbb C^2 \times U: z \in l }$ is given in full terms as $[x:y]\mapsto (((f_1/g_1)(x,y),(f_2/g_2)(x,y)), [x:y])$. we abbreviate $ (f_i/g_i)(x,y) $ to simply $f_i/g_i$ keeping in mind that we consider it as evaluated in $[x:y]$. In order to land in $ X$ the ratios of $(f_1/g_1,f_2/g_2)$ and $c \cdot (x,y) $ should coinside for every non zero constant $ c$. So we conclude $f_1/g_1\cdot (f_2/g_2)^{-1}= x \cdot y^{-1}$ and this is – user267839 Jan 05 '23 at 19:07
  • equivalent to the condition $f_1/g_1\cdot y-f_2/g_2\cdot x=0$, isn't it? – user267839 Jan 05 '23 at 19:09
  • Moreover it is known that in contrast to $\mathcal O_{\mathbb{CP}^1}(1)$ the tautological bundle $X =\mathcal O_{\mathbb{CP}^1}(-1)$ as it was defined above has no global sections. – user267839 Jan 05 '23 at 19:12
  • Right, requiring the condition $f_1/g_1\cdot y-f_2/g_2\cdot x=0$ is also ok; it is a matter of convention. I chose the (possibly non-standard) convention that $[a:b]\in\mathbb P^1$ corresponds to the line $ax+by=0$. – Kenta S Jan 05 '23 at 19:20
  • And yes, I agree that $\mathcal O_{\mathbb{CP}^1}(-1)$ has no global sections. – Kenta S Jan 05 '23 at 19:22
  • But the way you approach it is exactly that one I was interested in as I posed this question providing exactly as you did a bridge between topological and abgebraic descriptions of the tautological bundle. I think here the problm with $\mathcal {O}(-1)$ and wrong conclusion is that it has non trivial global sections came from some swapped denominator, since the right conclusion should be that no local sections of $\mathcal {O}(-1)$ should be able to glue to global ones. In contrast I think that if we apply exactly the same methods of your analysis to $\mathcal {O}(1)$ this should give correct – user267839 Jan 05 '23 at 19:32
  • result. I try to borrow some time next days to think carefully about it – user267839 Jan 05 '23 at 19:33
  • I think that topologically we can describe $\mathcal {O}(-1)$ in analogy to $X$ as $X^{\vee}:={(z,[v]) \ \vert \ : [v] \in \Bbb CP^1, z \in ( \mathbb{C} \cdot v)^{\vee} }$ and do similar analysis as you did above. If we fix an open set $U\subset\mathbb{CP}^1$, do you see how in this case the local sections over $ U $ should look like? Say for $ U := \mathbb{CP}^1 - V(L) $ for certain linear polynomial $ L$. Do they allow similar description as you did for $ X $ in terms of fractions of polynomials? – user267839 Jan 05 '23 at 20:01