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meaning of $\mathcal O(-1)$ in algebraic geometry?

+ 1 like - 0 dislike
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First, i want to know what is $\mathcal O(-1)$ bundle is. (Definition or geometric interpretation, and so on)

Second I want understand how the following is constructed.

Consider a $\phi$ as a coordinates on a copy of $Z= C^N$

Then, I know \begin{align} |\phi_1|^2 + |\phi_2|^2 + \cdots |\phi_N|^2 = r \end{align} which describe $S^{2N-1}$.

Implementing $U(1)$ condition the space of solution is described by

\begin{align} CP^N = S^{2N+1}/U(1) \end{align} Now consider slightly different case \begin{align} |\phi_1|^2 + |\phi_2|^2 - |\phi_3|^2 - |\phi_4|^2 =r \end{align} this gives $\mathcal O(-1) \oplus \mathcal O(-1)$ over $CP^1$. I want to know what this means and how to obtain.

This post imported from StackExchange Mathematics at 2015-10-31 22:13 (UTC), posted by SE-user phy_math
asked Oct 28, 2015 in Mathematics by phy_math (185 points) [ no revision ]
$\mathcal O(-1)$ is the invertible sheaf on $\mathbb P^n$ corresponding to the "tautological bundle" $\{(L, x) \in \mathbb P^n \times \mathbb A^{n+1}:x\in L\}$.

This post imported from StackExchange Mathematics at 2015-10-31 22:13 (UTC), posted by SE-user Hoot
@Hoot, i understand a $\mathcal O(-1)$ as a dual of line bundle. (But still don't know the physical or geometrical interpretation...) Can you explain why $\mathcal O(-1)$ appears in the second question that i wrote?

This post imported from StackExchange Mathematics at 2015-10-31 22:13 (UTC), posted by SE-user phy_math
I think the definition is pretty geometric already: from the very start each point in $\mathbb{P}^n$ corresponds to a line, and the bundle $\mathcal{O}(-1)$ glues all of these together over $\mathbb P^n$. In the last part of your question, what's the map to $\mathbb P^1$ and what's the group action?

This post imported from StackExchange Mathematics at 2015-10-31 22:13 (UTC), posted by SE-user Hoot
@Hoot, here i guess the group action is $U(1)$.

This post imported from StackExchange Mathematics at 2015-10-31 22:13 (UTC), posted by SE-user phy_math
That's a group. I guess you multiply each coordinate by the element. How do you map to $CP^1$?

This post imported from StackExchange Mathematics at 2015-10-31 22:13 (UTC), posted by SE-user Hoot

1 Answer

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A invertible sheaf is also defined by its transition function over open subsets. In general, an invertible sheaf $O(n)$ over $\mathbb{P}^n$ is the following:

Over each $U_i=\{[*,*,..,0,*...,*]:\text{ zero in i-th position }\}$ we define $O(n)|_{U_i}=\{\text{polynomial of degree $n$ }\in O_{P^n}(U_i)\}$ and the transition function is compatible with the structure sheaf of $\mathbb{P}^n$. We can check that under this definition $O(-1)$ is isomorphic to the canonical line bundle over $\mathbb{P}^n$.( the definition mentioned by Hoot in the comment)

This post imported from StackExchange Mathematics at 2015-10-31 22:13 (UTC), posted by SE-user Ben
answered Oct 28, 2015 by Ben (0 points) [ no revision ]

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