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I am trying to define line bundles on $\mathbb{P}^n$ as a triplet $E \xrightarrow{p} \mathbb{P}^n$. It follows the idea.

Let $m \in \mathbb{Z}$ a fixed integer. Take $E$ the quotient of $\mathbb{C}^{n+1} \times \mathbb{C}$ by the action of $\mathbb{C}^\times$ given by $$ t \cdot ((z_0,\dots,z_n),\lambda)=(t(z_0,\dots,z_n),t^m\lambda)$$

and let me denote it by $E=\mathbb{C}^{n+1} \times_{\mathbb{C}^\times} \mathbb{C}$. We have a natural $p$ which is the projection onto the first factor: $$ E=\mathbb{C}^{n+1} \times_{\mathbb{C}^\times} \mathbb{C} \rightarrow \mathbb{P}^n$$ $$ (z,l) \mapsto [z] $$

I'm wondering if it is $\mathcal{O}(m)$ or $\mathcal{O}(-m)$. Where $\mathcal{O}(m)$ is defined in terms of sheaves as the sheaf of "holomorphic maps of degree $m$".

hm2020
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wood
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    Not holomorphic maps, but homogeneous polynomials of degree $m$ (when $m>0$). For example, the only global holomorphic maps are constants. Stick with $|m|=1$. Do you get a global section with $m=1$ or with $m=-1$? – Ted Shifrin Dec 15 '22 at 18:00
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    Notice that the projection $E \to \mathbb C$ is a global section of $E^*$... – Tabes Bridges Dec 15 '22 at 18:01
  • @TedShifrin I think for m=1 I get as global sections all homogeneous polynomials of degree 1 and so, fixed m I should get O(m). However, (studying homogeneous line bundles on homogeneous spaces), I found an observation that made me doubt about this. I'm going to edit the text to provide a sketch of my argument – wood Dec 15 '22 at 19:10
  • You find a constuction of projective space using "quotients" at this link. You must do a similar construction with invertible sheaves and compare the quotient construction with the construction using a twist of the graded ring: https://math.stackexchange.com/questions/2369274/why-is-the-projection-map-mathbban1-k-setminus-0-to-mathbbpn-k/4208185#4208185 – hm2020 Dec 17 '22 at 11:55
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    @wood - In a book on algebraic geometry they define the invertible sheaf $\mathcal{O}(m)$ using the twisted coordinate ring $kx_i$. You find this in Chapter II in Hartshornes book "Algebraic geometry". – hm2020 Dec 17 '22 at 12:39
  • Thank you very much, I will look at it. However, if we look at global sections of $E$ we have $\mathbb{P}^n \rightarrow E$, $z \mapsto [z,f(z)]$ with $f$ an analytic function defined on $\mathbb{C}^{n+1}-{0}$ satisfying $f(tz)=t^mf(z)$ and so it should correspond to a homogeneous polynomials of degree $m$. Now, do you know what is wrong with this argument? – wood Dec 17 '22 at 16:33
  • There is for any finite rank locally free sheaf $E$ a canonical map from the sheaf of sections of the associated vector bundle $V(E^∗)$ to the double dual $E^{∗∗}≅E $ which is canonically isomorphic to $E$. If you do not dualize you end up making mistakes. You must define $Y:=V(E^∗):=Spec(Sym^∗(E^∗))$ as the relative spectrum of the dual $E^∗$. As you can see in the linked thread: This gives a canonical isomorphism between the sheaf of sections of $Y$ and $E$. – hm2020 Dec 18 '22 at 11:23
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    I think you meant $E=(\mathbb{C}^{n+1}\setminus 0)\times_{\mathbb{C}^\times} \mathbb{C}$ instead of just $E=\mathbb{C}^{n+1}\times_{\mathbb{C}^\times} \mathbb{C},$ right? – Grisha Taroyan Jul 12 '23 at 03:25

1 Answers1

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Question: "I'm wondering if it is O(m) or O(−m). Where O(m) is defined in terms of sheaves as the sheaf of "holomorphic maps of degree m"."

Answer: You may use schemes since the classification of holomorphic line bundles on $\mathbb{P}^n$ viewed as a complex manifold is "the same" as the classification of invertible sheaves on $\mathbb{P}^n$ viewed as a scheme. If you are a PhD student it is a good exercise to understand the relation between sheaves of sections and vector bundles mentioned in the comments.

Let $G:=Spec(k[t,1/t])$ be the multiplicative group scheme with $G(k)\cong k^*$ the units in $k$, and let $S:=Spec(k[x_0,x_1])$ and $T:=Spec(k[z])$. Define the coaction

$$G\times S \times T \rightarrow S\times T$$

by

$$\sigma: k[x_0,x_1,z] \rightarrow k[x_0,x_1,z]\otimes_k k[t,,1/t]$$

$$\sigma(x_0):=tx_0, \sigma(x_1):=tx_1$$

and

$$\sigma(z):=t^mz.$$

Let $U:=S-\{(0,0\}$ and let $V:=U \times T$ and consider the induced action $u: G\times U \rightarrow U$. You get a canonical projection map

$$\pi: E:=V/G \rightarrow U/G \cong \mathbb{P}^1$$

and it follows $\pi^{-1}(D(x_i)):=Spec(A_i)$ with

$$A_0:=k[\frac{x_1}{x_0}, (\frac{1}{x_0})^mz]$$

and

$$A_1:=k[\frac{x_0}{x_1},(\frac{1}{x_1})^mz].$$

If you consider the sheaf $A:=Sym^*_{\mathcal{O}_{\mathbb{P}^1}}(\mathcal{O}(-m))$ on $\mathbb{P}^1$ it follows there is an isomorphism of line bundles

$$ E \cong \mathbb{V}(\mathcal{O}(-m)):=Spec(A).$$

This holds in general: Your vector bundle $E$ satisfies $E \cong \mathbb{V}(\mathcal{O}(-m))\cong \mathbb{V}(\mathcal{O}(m)^*)$ where $\mathcal{O}(-m)$ is the sheaf corresponding to the graded module $k[x_0,..,x_n](-m)$ in the sense of Hartshornes book, Chapter I and II. Here $E:=V/G$ and $U/G$ are "geometric quotients" in the sense of geometric invariant theory. The global sections of the projection map

$$\pi: \mathbb{V}(\mathcal{O}(-m)) \rightarrow \mathbb{P}^n$$

are in 1-1 correspondence with $H^0(\mathbb{P}^n,\mathcal{O}(m))$, which is the vector space of homogeneous polynomials $f(x_0,..,x_n)$ of degree $m$. Hence the sheaf of sections of $\pi$ is the invertible sheaf $\mathcal{O}(m)$.

Why is the projection map $\mathbb{A}^{n+1}_k\setminus \{0\} \to \mathbb{P}^n_k$ a morphism of schemes?

In general if $L$ is an invertible sheaf on $X$, it follows $L$ is the sheaf of sections of the canonical map $\pi: \mathbb{V}(L^*) \rightarrow X$, where $\mathbb{V}(L^*):=Spec(B)$ with $B:=Sym_{\mathcal{O}_X}^*(L^*)$. Hence the map $\pi$ has $H^0(X,L)$ as global sections.

More generally: Whenever you have a linear algebraic group $G \subseteq GL_k(V)$ and a closed subgroup $H \subseteq G$ and a left $H$-module $\rho: H \rightarrow GL_k(W)$, it follows there is a quotient scheme $G/H$ that is regular of finite type and quasi projective over $k$. The $H$-module $(W,\rho)$ gives a finite rank algebraic vector bundle $\pi: E(\rho) \rightarrow G/H$. The construction is "similar" to the above example.

Example: You may construct $\mathbb{P}^n$ as follows: Let $dim_k(V)=n+1$ and let $l \subseteq V$ be a line. Let $P \subseteq SL_k(V)$ be the subgroup fixing $l$. It follows there is an isomorphism $\mathbb{P}^n \cong SL_k(V)/P$. You may choose the projection map

$$\pi: SL_k(V) \rightarrow \mathbb{P}^n$$

to map a "matrix" $g \in SL_k(V)(k)$ to its "first column".

This link answer may be of interest:

What are the algebraic features and the geometric interpretation of the symmetric algebra?

hm2020
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