# Two notes on line bundles

Two fun diversions about line bundles while I procrastinate finishing my post on adeles:

First, a classic example which demonstrates how geometric vector bundles (equivalently locally free sheaves) on a scheme do not quite correspond to topological vector bundles. It is well-known that the only two topological real line bundles over ${S^1}$ are the trivial bundle and the Möbius bundle with one twist. Any other bundle, with a number of other twists, is homeomorphic to one of these two, since we can cancel out pairs of twists – even isotopically, if we embed in five(?) dimensions.

But if we take the scheme ${\mathbb{RP}^1}$ (as e.g. ${\text{Proj }\mathbb{R}[x,y]}$, which has underlying space homeomorphic to ${S^1}$, it is clear we get many line bundles: specifically, the twisting sheaves ${\mathcal{O}(n)}$ for each ${n\in \mathbb{Z}}$. (Here we implicitly use the equivalence of categories between locally free sheaves and schemes over ${\mathbb{RP}^1}$ which are geometric vector bundles.) What gives?

It’s not that there’s something fundamentally geometrically different about the scheme structure: a geometric vector bundle over ${\mathbb{RP}^1}$ really does have underlying topological space a line bundle over the same space; that’s clear from typical definitions as given in, e.g. Hartshorne chapter 2. What’s different is the morphisms in the category of topological line bundles versus the category of geometric line bundles over a scheme. The latter morphisms are, of course, simply morphisms of schemes over ${\mathbb{RP}^1}$, and so have to be given locally by regular functions. But the twist-canceling homeomorphisms can’t be given this way (intuitively, think about having to twist the lines out to infinity all the way around; these involve poles, algebraically), so they are not isomorphisms in the category of geometric vector bundles.

Intuitively, if we “allowed topological morphisms,” then the even ${\mathcal{O}(n)}$ would all be trivial whereas the odd ${\mathcal{O}(n)}$ would all be the Möbius bundle. This also explains the odd behavior of global sections (i.e. that all the nonnegative ${n}$ have global sections while all the negative ones don’t): the negative evens do have nonvanishing globals; it’s just that, again, they can’t be given by functions which formally satisfy the local regularity conditions of the twisting sheaves. On the other hand, the odd positive ones don’t really have nonvanishing global sections, since there are points at which the odd-degree polynomials corresponding to global sections of that twisting sheaf vanish.

Exercise. Someone should rigorize this line of thinking to create a gorgeous proof that every odd-degree polynomial has a real zero.

Second, a while back, I was trying to gain intuition for the Picard group by reading through different perspectives on the invertibility of line bundles. There’s of course the inverse transition maps, tensor-hom adjunction, and all that jazz, but I also stumbled on an amusing one: consider the classifying space for line bundles, the infinite Grassmannian ${\mathbb{CP}^\infty}$. An isomorphism class of line bundles on a space ${X}$ is equivalently a homotopy class of maps into the classifying space, so we have reduced our problem to the much simpler and more intuitive one of putting a group structure on ${\text{Hom}(X,\mathbb{CP}^\infty)}$ in the homotopy category!

A little Schubert calculus gives us the following: first, the cohomology ring ${H^*(\mathbb{CP}^\infty)}$ is concentrated in even dimension, so it is a genuine commutative ring. Second, since the product is given by the cup product, which corresponds to intersection product, the closed points of ${\text{Spec } H^*(\mathbb{CP}^\infty)}$ correspond to maximal ideals generated by Schubert cycles associated to 0-dimensional subspaces, points. In fact, it turns out that ${\text{Spec } H^*(\mathbb{CP}^\infty)}$ is the formal affine line, ${\widehat{\mathbb{A}}^1}$, whose canonical additive structure on its closed points is induced by tensor product of the line bundles associated to the points of the Grassmannian (as a moduli space) associated to the closed points.

This structure can then be pulled back to ${\text{Hom}(X,\mathbb{CP}^\infty)}$, using the techniques in this paper. Theorem 1.1 implies that the well-known induced map in cohomology ${\text{Hom}(X,\mathbb{CP}^\infty)\rightarrow \text{Hom}(H^*(\mathbb{CP}^\infty),H^*(X))}$ is injective. We can then compose it with the isomorphism $\text{Hom}(H^*(\mathbb{CP}^\infty),H^*(X))=\text{Hom}(\text{Spec }H^*(X),\text{Spec }H^*(\mathbb{CP}^\infty))$. As a hom into an abelian scheme, this last object naturally inherits a group structure. This can then be pulled back to ${\text{Hom}(X,\mathbb{CP}^\infty)}$ by naturality, giving us our group structure on isomorphism classes of line bundles.

The best part is that this doesn’t even work, since ${H^*(Y)}$ isn’t actually a strictly commutative ring if it doesn’t have vanishing odd cohomology, so it doesn’t have a spectrum. To really complete this elegant line of reasoning, I think it is necessary to develop the methods of noncommutative geometry. Alain Connes still dreams at night.