# How do we describe maps of line bundles on $\mathbb{P}^1$?

I am trying to get a very down to earth description of the sections of the holomorphic line bundles over the complex projective line. We have that for line bundles $$\mathcal{O}(d)$$ of non-negative degree $$d\geq 0$$ one has $$H^0(\mathcal{O}(d)) \simeq \mathbb{C}[z_0,z_1]^d,$$ the space of homogeneous polynomials in two variables of degree $$d$$. For $$d<0$$, the line bundle $$\mathcal{O}(d)$$ only has the zero global section. Another relevant feature is that for different degrees $$m,n$$ we have $$\mathcal{O}(m)\otimes\mathcal{O}(n)\simeq \mathcal{O}(m+n)$$, and $$\mathcal{O}(n)^*\simeq \mathcal{O}(-n)$$. This is particularly useful so as to interpret the bundle of homomorphisms $$Hom(\mathcal{O}(n),\mathcal{O}(m)) \simeq \mathcal{O}(m)\otimes \mathcal{O}(n)^* \simeq \mathcal{O}(m-n)$$

For positive $$m\geq n\geq 0$$ my intuition tells me that a global homomorphism from $$\mathcal{O}(n)$$ to $$\mathcal{O}(m)$$ sends degree-$$n$$ polynomials to degree-$$m$$ polynomials via multiplication with a degree-$$(m-n)$$ polynomial.

However, if $$m-n\geq 0$$ but $$n<0$$ I cannot describe this homomorphism in terms of global sections of $$\mathcal{O}(n)$$, as there are no nontrivial ones!

My question: how can I describe, fiberwise, this homomorphism $$\mathcal{O}(n)\rightarrow \mathcal{O}(m)$$ in terms of the degree-$$(m-n)$$ polynomial, when $$n<0$$?

Some preliminary thoughts: for negative degree bundles, I can try to describe them as the tensor power of the hyperplane bundle, with fibers given by $$\mathcal{O}(-1)|_{[z_0:z_1]} = \{([z_0:z_1],\phi):\phi\in(\mathbb{C}^2)^*,\phi(z_0,z_1)=0\}$$ My maybe-not-so-useful guess is that for $$d<0$$ I would get fibers $$\mathcal{O}(d)|_{[z_0:z_1]} = \{([z_0:z_1],\sum_i\alpha_i \phi^i_1\otimes\dots\otimes \phi^i_{-d}):\phi^i_j\in(\mathbb{C}^2)^*,\alpha_i\in\mathbb{C},\phi^i_j(z_0,z_1)=0\}$$

• Perhaps you can consider instead the induced homomorphism on the generic point? Commented Nov 25, 2021 at 12:57
• Does it help to note that even if $m \geq n > 0$, we don't get a mapping on global sections of $\mathcal{O}(n)$ to global sections of $\mathcal{O}(m)$ sending one arbitrary section $s$ to another $s'$ without multiplying by a meromorphic section $s'/s$ of $\mathcal{O}(m-n)$? Commented Nov 25, 2021 at 12:57
• It does make sense, but how can we reconcile with the interpretation of the global mapping $\mathcal{O}(n)\rightarrow\mathcal{O}(m)$ as a degree-$(m-n)$ polynomial? Commented Nov 25, 2021 at 16:13

My question: how can I describe, fiberwise, this homomorphism $$\mathcal{O}(n)\rightarrow \mathcal{O}(m)$$ in terms of the degree-$$(m-n)$$ polynomial, when $$n<0$$?

Answer: let $$C$$ be the projective line. There is by Hartshorne, Ex II.5.1 isomorphisms

$$Hom_{\mathcal{O}_C}(\mathcal{O}(m), \mathcal{O}(n)) \cong Hom_{\mathcal{O}_C}(\mathcal{O}_C, Hom(\mathcal{O}(m), \mathcal{O}(n))) \cong$$

$$Hom_{\mathcal{O}_C}(\mathcal{O}_C, \mathcal{O}(m)^*\otimes \mathcal{O}(n))\cong Hom_{\mathcal{O}_C}(\mathcal{O}_C, \mathcal{O}(n-m)) \cong H^0(C, \mathcal{O}(n-m)).$$

Here you should check that to give a map of $$\mathcal{O}_C$$-modules

$$s: \mathcal{O}_C \rightarrow \mathcal{O}(n-m)$$

is equivalent to give a global section of $$\mathcal{O}(n-m)$$ which is equivalent to giving a homogeneous polynomial of degree $$n-m$$ in $$x_0,x_1$$.

If $$n-m < 0$$ there are no such maps. If $$n-m \geq 0$$ the vector space of such maps has dimension $$n-m+1$$, and given a map $$s$$ it corresponds 1-1 to a homogeneous polynomial

$$s(x_0,x_1) \in \mathbb{C}[x_0,x_1]^{n-m}.$$

Example: A map $$s:\mathcal{O}(m) \rightarrow \mathcal{O}(n)$$

corresponds 1-1 to a homogeneous polynomial $$s(x_0,x_n)$$ of degree $$n-m$$ from the following argument:

We get a map of graded modules

$$\phi: \mathbb{C}[x_0,x_1](m)\rightarrow \mathbb{C}[x_0,x_1](n)$$

defined by

$$\phi(f(x_0,x_1)):=s(x_0,x_1)f(x_0,x_1).$$

Sheafifying this map we get an induced map

$$s:\mathcal{O}(m) \rightarrow \mathcal{O}(n)$$

of $$\mathcal{O}_C$$-modules. This construction gives all such maps by the above argument.

Example: For any $$n,m$$ with $$n-m \geq 0$$ you get a vector space of dimension $$n-m+1$$ of such maps.

The fiber: The induced map at the fiber at $$x:=(t-\alpha) \in C(\mathbb{C})$$: Let $$t:=x_1/x_0$$. We get the following map $$\phi_x$$ between fibers $$\mathcal{O}(m)(x)\cong \mathbb{C}x_0^m:=\mathbb{C}[t]/(t-\alpha)x_0^m$$ and $$\mathcal{O}(n)(x)$$:

$$\phi_x: \mathbb{C}x_0^m \rightarrow \mathbb{C}x_0^n$$

defined by

$$\phi(x)(ux_0^m):=g(\alpha)ux_0^n$$

where

$$s(x_0,x_1)=g(t)x_0^{n-m}$$.

Here the point $$x:=(a_0:a_1)$$ has $$a_0 \neq 0$$ and by definition $$x=(1: a_1/a_0)=(1:\alpha)$$. There is an isomorphism

$$\mathcal{O}(m)(D(x_0)) \cong \mathbb{C}[t]x_0^m$$

and the point $$x$$ corresponds 1-1 to the maximal ideal $$(t-\alpha) \subseteq \mathbb{C}[t]$$ with $$t:=x_1/x_0$$. The fiber at $$x$$ is by definition

$$\mathcal{O}(m)(x)\cong \mathbb{C}[t]/(t-\alpha) x_0^m \cong\mathbb{C}x_0^m.$$

A generalization: Similar results hold for any projective scheme $$X \subseteq \mathbb{P}^n_{A}$$: There are isomorphisms

$$Hom_{\mathcal{O}_X}(\mathcal{O}(m), \mathcal{O}(n)) \cong Hom_{\mathcal{O}_X}(\mathcal{O}_X, \mathcal{O}(n-m))$$

$$\cong H^0(X, \mathcal{O}(n-m)).$$

Hence for projective space $$\mathbb{P}^n_k$$ you get an equality

$$Hom_{\mathcal{O}}(\mathcal{O}(m), \mathcal{O}(n)) \cong k[x_0,..,x_n]^{n-m}.$$

Hence a map of sheaves

$$s: \mathcal{O}(m) \rightarrow \mathcal{O}(n)$$

is in 1-1 correspondence with a homogeneous polynomial

$$s(x_0,..,x_n) \in \mathbb{C}[x_0,..,x_n]^{n-m}$$

of degree $$n-m$$.

Note: Usually we write $$k[x_i]_d$$ instead of $$k[x_i]^d$$. The notation

$$k[x_i]^d :=\oplus_{i=1}^d k[x_i]e_i$$

usually means "the direct sum of $$d$$ copies of $$k[x_i]$$".

Note moreover: There is an exercise (Ex.II.5.9 in Hartshorne) classifying coherent sheaves on a projective scheme $$X:=Proj(S) \subseteq \mathbb{P}^n_A$$ with $$A$$ a finitely generated algebra over a field $$k$$. Assume $$S$$ is generated by $$S_1$$ as $$A$$-algebra and for a given graded $$S$$-module $$M$$, let

$$M_{\geq d}:=\oplus_{n \geq d} M_n.$$

Two graded $$S$$-modules $$M,N$$ are equivalent iff there is an integer $$d\geq 0$$ and an isomorphism $$M_{\geq d} \cong N_{\geq d}$$. A module $$M$$ is "quasi finitely generated" iff it is equivalent to a finitely generated module. The functor $$F(M):=\tilde{M}$$ gives an equivalence between the category of coherent sheaves on $$X$$ and the category of quasi finitely generated $$S$$-modules modulo this equivalence.

Example: In particular if $$A:=\mathbb{C}$$ is the field of complex numbers and $$E$$ is a coherent analytic sheaf on $$X$$, there is a quasi finitely generated $$S$$-module $$M$$ and an "isomorphism" $$E \cong \tilde{M}^s$$ where $$(-)^s$$ is the analytification functor described in

Smooth algebraic varieties are complex manifolds

Hence if $$L^s, E^s$$ are coherent analytic sheaves on $$X$$ with $$L^s$$ a line bundle, there is an isomorphism

$$Hom_{\mathcal{O}_X}(L,E) \cong Hom_{\mathcal{O}_X}(\mathcal{O}_X, E\otimes L^*) \cong H^0(X, E\otimes L^*) \cong$$

$$Hom_{\mathcal{O}_X^s}(\mathcal{O}_X^s, E^s \otimes (L^*)^s) \cong H^0(X^s, E^s\otimes (L^*)^s).$$

Hence you can calculate the maps from $$L^s$$ to $$E^s$$ using $$L$$ and $$E$$. Hence if $$L^s,E^s$$ are coherent analytic sheaves on $$X^s$$ and if $$L^s$$ is invertible, the set of maps $$L^s \rightarrow E^s$$ is finite dimensional: Since $$E^s\otimes (L^*)^s$$ is coherent it follows from HH.II.5.19:

The vector space

$$\Gamma(X^s, E^s\otimes (L^*)^s):=H^0(X^s, E^s\otimes (L^*)^s)\cong H^0(X, E\otimes L^*)$$

is finite dimensional. This is Theorem AppB.2.1 in Hartshorne.

• This is a nice answer that helped me figure out some things. I come from a more differential-geometric background so let me ask this: are you identifying the base point $x$ with some polynomial $t-\alpha$ in the open set $\{(z_0:z_1)\in C|z_0\neq 0\}$? One further question: in the example you justified that all morphisms at the fiber level come from sheafifying these mappings between global sections of each bundle. However, I am most interested when the first bundle has negative degree and has no global sections (with $n-m\geq 0$ still). I'd accept the answer if you clarify this! Commented Nov 29, 2021 at 10:13
• @topolosaurus - in algebraic geometry our points are maximal ideals. The point $\alpha$ corresponds to the maximal ideal $(t-\alpha)$. Commented Nov 29, 2021 at 10:26
• Note also that any complex projective manifold $X$ "is" algebraic and that any holomorphic vector bundle $E$ on $X$ is algebraic. Hence we may view the sheaves $\mathcal{O}(m)$ as (all) holomorphic line bundles on $C$. The above construction give all maps of all holomorphic line bundles on $C$. Commented Nov 29, 2021 at 10:40