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Morphisms between affine varieties are determined simply by specifying a fixed number of polynomials. On the other hand, morphisms between projective varieties are not as easy to deal with in general and may sometimes require construction via gluing. If one tries to understand collections of morphism, it becomes tricky to keep track of all the "gluing data". This problem is solved in a brilliant way by the use of line bundles. Here we continue the discussion of the first post, where we considered the projective curve $$C=\left\{\,[X,Y,Z]\in\mathbb{P}^2\mid X^2+Y^2=Z^2\,\right\}.$$ I claimed that there is an extension $$\Xi\colon C\to\mathbb{P}^1$$ of the assignment $$\Phi\colon\left[X,Y,Z\right]\mapsto \left[\frac{Y}{X+Z},1\right].$$ Recall that this assignment assigns to a point $P$ the slope of the line through $[-1,0,1]$ and $P$. We will look at this claim more closely and see how the modern machinery works in this concrete example. But first, we look at how the extension is constructed without it.

The first thing one can notice is that the denominator vanishes at the point $[-1,0,1]$. Furthermore, this is the only point where this assignment is not defined. Thus, we only need to extend over $[-1,0,1]$. The usual trick of clearing denominators $$\left[\frac{Y}{X+Z},1\right]=[Y,X+Z]$$ fails, because $Y=X+Z=0$ at $[-1,0,1]$. The only way to define $\Xi$ is by gluing. In other words, $\Xi$ should be defined by the above assignment on the open subset $U:= C\setminus\{[-1,0,1]\}$ and we need to find another open subset $V\subset C$ and a morphism $\Phi'\colon V\to\mathbb{P}^1$ such that $U\cup V= C$ and $\Phi=\Phi'$ on the intersection $U\cap V$.

Here's what we do. Fix a point $(x,y)$ on the unit circle such that $y\neq 0$. Recall, that $\Phi(x,y)$ is the slope of the line through $(-1,0)$ and $(x,y)$. Now, construct also the line through $(1,0)$ and $(x,y)$. Since the three points lie on the circle and $(-1,0)$ and $(1,0)$ are antipodal, the two constructed lines intersect perpendicularly at $(x,y)$. Thus, the slope of the line intersecting $(1,0)$ is $-1/\lambda$ where $\lambda=\Phi(x,y)$, so we can compute the values of $\Phi$ alternatively by constructing lines through $(1,0)$. Thus, let us take $V=C\setminus\{(1,0)\}$ and define $\Phi'$ by first calculating the slope of the line through $(1,0)$ and the inverting and negating the result. In other words, let $\Phi'$ be the composition $$[X,Y,Z]\mapsto\left[\frac{-Y}{Z-X},1\right]\mapsto\left[\frac{Z-X}{Y},1\right]=[Z-X,Y].$$ Then, $U\cup V=C$ and $\Phi=\Phi'$ on the intersection $U\cap V$ by construction (one can also verify the equality by scaling the coordinates).

In conclusion, we can define the extension $\Xi$ by $$\Xi\colon[X,Y,Z]\mapsto\begin{cases} [Y,X+Z],&[X,Y,Z]\in U,\\ [Z-X,Y],&[X,Y,Z]\in V \end{cases}$$ In an ideal world, we could find two polynomials $f, g\in\mathbb{C}[X,Y,Z]$ such that $$\Xi\colon \left[X,Y,Z\right]\mapsto\left[f(X,Y,Z),g(X,Y,Z)\right].$$ But instead, we have two such representations valid only on proper subsets of $C$, and keeping track of this gluing data is what makes the use of these naive constructions unsuitable for general study of morphisms to projective space.

The problem preventing us from finding a representation of $\Xi$ by polynomials is actually not only algebraic but also topological and we can even visualise it by restricting to the real numbers. Thus, let us replace $C$ by the real unit circle $S^1$ and try to find a representation for the function $S^1\to\mathbb{RP}^1$ that maps a point of $S^1$ to the slope of the line through it. To actually view this topologically, you should imagine the functions $\frac{y}{x+1}$ and $1$ as represented by their graphs in the cylinder $S^1\times\mathbb{R}$. You should see that the function $\frac{y}{x+1}$ escapes to infinity at the point $(-1,0)$. Here is an idea one could use to fix this: since we are mapping to projective space, we are only concerned about the ratio of the functions $\frac{y}{x+1}$ and $1$. In particular, one could scale both functions by some fixed function that approaches zero at $(-1,0)$ at a suitable speed so that $\frac{y}{x+1}$ becomes bounded after scaling. However, it will always be discontinuous, because the sign of $\frac{y}{x+1}$ is determined by the function $y$, which changes sign at $(-1,0)$. Since we cannot fix the problem by only scaling the functions, we need to also change the total space where the graphs of the functions lie. Thus, imagine cutting the cylinder along the line above $(-1,0)$. Then, twist one end of the cylinder by a half-turn and glue the ends back together, obtaining a Möbius strip $M$. Finally, take a section $\varphi\colon S^1\to M$ of the Möbius strip which approaches zero at $(-1,0)$ and doesn't vanish anywhere else on $S^1$. If $\varphi$ converges to zero at a suitable speed, then the scaled section $\varphi\cdot\frac{y}{x+1}$ is bounded and can furthermore be extended to the whole of $S^1$. Thus, we now have two global sections $\varphi$ and $\varphi\cdot\frac{y}{x+1}$ of $M\to S^1$ and they define a map $$S^1\to\mathbb{RP}^1:(x,y)\mapsto\left[\varphi(x,y)\cdot\frac{y}{x+1},\varphi(x,y)\right]$$ which clearly agrees with our map $\Phi$ from before. In conclusion: while we cannot find global polynomials that define the extension, we can find global sections of a line bundle that define it.

This fixes the topological problem, but we need to be a bit more precise about how we do the constructions geometrically. One way of constructing the Möbius strip $M\to S^1$ in this setting is to realise it as a sub-bundle of the trivial vector bundle $S^1\times\mathbb{R}^2\to S^1$ as follows. Above each point $p\in S^1$, there is a copy of $\mathbb{R}^2$. We can think of $S^1$ as sitting in this copy of $\mathbb{R}^2$ and draw the line through $(-1,0)$ and $p$. As we vary $p$, these lines define a sub-line bundle of the trivial bundle. In other words, we can write $$M=\left\{\,(p,v)\in S^1\times\mathbb{R}^2\mid v\text{ is on the line through }(-1,0)\text{ and }p.\,\right\}.$$ One good mental exercise is to try to visually confirm that this indeed is the Möbius strip. It helps to think of the trivial bundle $S^1\times\mathbb{R}^2$ as a solid, open torus sitting above $S^1$. This is a fairly faithful visualisation, because $\mathbb{R}^2$ is homeomorphic to an open disk.

This construction works perfectly well over the real numbers but we need to change it a bit for it to work over other fields as well. The problem is that over the complex numbers, the curve $C$ is not contained in the affine piece as it is with the real numbers. Now, here is one way of restating the construction of $M$. We took the fibre over $p\in S^1$ to be the line obtained by taking the projective line in $\mathbb{RP}^2$ through $[-1,0,1]$ and $p$ and removing the point at infinity. What we should do is remove the point $[-1,0,1]$ and keep the rest of the line. If we do this, we get the line bundle $$\pi\colon\mathbb{P}^2\setminus\left\{[-1,0,1]\right\}\to C$$ where $\pi(x)$ is the intersection point of $C$ with the line through $x$ and $[-1,0,1]$. One can find a trivialisation such that the graph of the zero-section is the line $\left\{\,x=1\,\right\}$. Note that the section $\varphi$ with graph $\left\{\,x=0\,\right\}$ does not touch the zero-section over the open set $U$, but they intersect above the point $[-1,0,1]$. Furthermore, one can check that $\varphi$ approaches the zero-section at a suitable speed so that $$\frac{Y}{X+Z}\cdot\varphi$$ approaches 1 above $[-1,0,1]$. Therefore, we can take $$\Xi(X,Y,Z):=\left[\varphi\cdot\frac{Y}{X+Z},\varphi\right].$$

We have reached the goal of expressing the extension $\Xi$ as a pair of sections of a line bundle, but one may wonder how this is any better than simply gluing two polynomial functions. In fact, the power of line bundles comes from extracting information just by handling them abstractly without doing concrete constructions as we did here. Firstly, new line bundles can be created from old ones by taking tensor products, duals, restrictions, etc. and the properties of the new line bundles can be derived abstractly from the properties of the old ones. For example, our line bundle on $C$ with the sections defining the extension $\Xi$ can be constructed abstractly as the line bundle $\mathscr{O}_{\mathbb{P}^2}(1)\vert_C\otimes\mathscr{O}([-1,0,1])^\vee$, i.e. it is the tensor product of a line bundle obtained from another line bundle on $\mathbb{P}^2$ by restriction and the dual of a bundle "twisted" at the point $[-1,0,1]$, whatever any of this means. Also, given a line bundle $\mathscr{L}$ on a projective $k$-variety $X$ and an open set $U\subseteq X$, the sections of $\mathscr{L}$ over $U$ form a $k$-vector space, so the local study of vector bundles reduces to algebra. Furthermore, gluing of local sections can be abstracted by cohomology. In the end, the question of what are the morphisms from a curve $C$ to projective space reduces to mainly algebra.

Using the abstract machinery of cohomology, one can prove that if a projective smooth curve $C$ of genus $g$ has a line bundle $\mathscr{L}$ such that $\deg\mathscr{L}>2g$ , then there is a closed embedding $$C\hookrightarrow\mathbb{P}^{\deg\mathscr{L}-g}$$ as a degree $\deg\mathscr{L}$ curve. For example, if $C$ is a plane curve of genus $0$, then one can construct the line bundle $\mathscr{O}_{\mathbb{P}^2}(1)\vert_{C}\otimes\mathscr{O}(-p)$, where $p\in C$, as we did above. It is easy to compute that the degree of this line bundle is $1$. This shows $C$ embeds into $\mathbb{P}^1$ as a degree 1 curve. Thus, we conclude the statement made in the previous post that all smooth genus $0$ curves are isomorphic to $\mathbb{P}^1$ (at least plane curves, but it is easy to construct degree $1$ line bundles on general genus $0$ curves). We can use the general theory to study curves of higher genera as well. For example, if $E$ has genus $1$ and $\mathscr{L}$ has degree $3$, then $\deg\mathscr{L}>2g$ so there is an embedding $$E\hookrightarrow \mathbb{P}^2$$ of $E$ as a cubic curve. One can even define the group structure on $E$ from the group structure of the group of line bundles on $E$, and derive the Weierstrass form of elliptic curves by looking at the sections of $\mathscr{L}$ more closely. These examples are from Chapter 19 of Vakil's notes and one can find many more results there that can be derived from the theory of line bundles.