A Mathematical Daydream

1. The Hardest Conjecture; 2. The Valley; 3. A Mathematical Daydream

“Those limits in certain functor categories that cannot be computed pointwise? They don’t actually exist,” one of us declared authoritatively. We, all math grad students, sat on an apartment floor, in a circle, with cards scattered over the floor.

Blank looks abounded. “Meaning, there don’t actually exist monic natural transformations some of whose components are not monomorphisms,” he clarified.

We had invented a variant of a popular card game called “spies vs. revolutionaries” – we called it “students vs. professors” – in which, in particular, each new round was heralded by its leader’s presentation of an established mathematical truth that he or she had decided we were to overturn for good. “The Banach-Tarski paradox is still true, but requires using at best six pieces, not five,” another student later suggested. “2 isn’t actually a prime,” one student blustered, citing the integer’s pathological character in many number-theoretic environments.

Soon it was my turn. “The Hodge Conjecture is false,” I fibbed, “and a counter-example is provided in my paper.”

A good laugh and a general readiness to proceed with the game cut my monologue short. Ridiculous as it was, though, I was ready to continue. The matter was one to which I had given some thought. Indeed, my work presents an interesting testing ground for a few of the ideas surrounding the Hodge Conjecture, and in particular seems to invite a heuristic argument whereby it could be used to furnish a counter-example to the conjecture. I’ll explain this mathematical daydream, and how it can be ultimately debunked.

The first unknown case

The Hodge Conjecture is an assertion about the way in which a k-dimensional smooth complex projective algebraic variety’s i-dimensional subvarieties influence its (2k – 2i)-dimensional cohomology groups. In particular, the conjecture puts forth multiple claims about any given smooth projective variety: one for each of its dimensions.

As (for example) Pierre Deligne writes in the Clay Institute’s official statement of the problem, certain special cases are already known. The conjecture’s known cases often concern certain of these dimensions.

The assertions about 0-dimensional and k-dimensional subvarieties of a given k-dimensional variety, in particular, can be shown trivially to be true. The most prominent known case of the Hodge Conjecture involves subvarieties of dimension one below the maximum. Indeed, the Lefschetz (1,1)-Theorem asserts precisely that the Hodge conjecture is true for (k – 1)-dimensional subvarieties of a k-dimensional variety. Subvarieties of this sort behave especially nicely, and, in some sense, subvarieties become progressively more inaccessible as their dimensions drop below that of the variety itself.

On the other hand, a marvelous symmetry exists among the conjecture’s respective assertions about any given variety. Indeed, the Hard Lefschetz Theorem provides an isomorphism of Hodge structures between any k-dimensional variety’s (2i)-dimensional and (2k – 2i)-dimensional cohomology groups. This means, in layman’s terms, that if for a k-dimensional variety the Hodge Conjecture is true for its (k – i)-dimensional subvarieties, then it’s also true for its i-dimensional subvarieties. This is a symmetry about the middle dimension.

Combining the above two assertions (and thanking Solomon Lefschetz, who “plant[ed] the harpoon of algebraic topology into the body of the whale of algebraic geometry” [1, p. 1]), we see that the Hodge Conjecture is known for any k-dimensional variety’s (k – 1)-dimensional and 1-dimensional subvarieties.

The first case where this does not exhaust the available dimensions, of course, is that of 2-dimensional subvarieties in a 4-dimensional variety. This is the first unknown case of the Hodge Conjecture. This is also the subject matter of my paper.

A mathematical fantasy

The Hodge Conjecture asserts, more precisely, that a k-dimensional variety’s i-dimensional subvarieties exhaust its (2k – 2i)-dimensional Hodge classes. Crucial is whether there are “enough” subvarieties in this variety to account for all of its Hodge classes.

My work, on the other hand, asserts that in certain smooth fourfolds, the surfaces of certain sorts belong to only finitely many families, where membership in a family is defined by a certain relationship involving smooth deformation. (In general, my work considers only surfaces which arise from an ambient variety, but for certain special fourfolds, the ambient variety can be scrapped. We’ll assume that we work with one such fourfold in what follows.)

This invites the following mathematical fantasy:

Consider a smooth fourfold X to which my theory applies. Thus in this fourfold, the smooth surfaces in X whose twistedness is less than or equal to r, say, belong to only finitely many families.

Fantasy. Every surface in X belongs to the same family as some smooth surface in X whose twistedness is less than or equal to r.

Meanwhile, any smooth fourfold has infinitely many 4-dimensional Hodge classes. Therefore:

Corollary. The surfaces in X belong to finitely families, and thus cannot exhaust X‘s 4-dimensional Hodge classes. The Hodge Conjecture is therefore false for X.

A schematic depiction of this fantasy is given below.

There are infinitely many Hodge classes. What if there were only finitely many algebraic cycles?

A dose of reality

2 is prime and the Banach-Tarski paradox can be done using 5 pieces, and, of course, the fantasy I’ve described above — as any seasoned mathematician would quickly see (I’m embarrassed now) — is nonsense: a variety’s subvarieties of any given dimension must belong to infinitely many families, and so this fantasy is doomed from the get-go. In this section I’ll demonstrate why.

We’ll begin by discussing some pretty geometric constructions. Recall that our k-dimensional variety lives in an n-dimensional ambient projective space. We will use the geometry of the ambient space to construct special subvarieites in our variety.

The ambient projective space contains in particular what’s called a hyperplane, or a flat subspace which has one dimension less than that of the ambient space. (The terminology reflects its generalization of the standard 2-dimensional plane in 3-dimensional space.) Intersecting the hyperplane with our k-dimensional variety, we produce a (k – 1)-dimensional subvariety. Picking another hyperplane and intersecting again, we get a (k – 2)-dimensional subvariety. Carrying on in this way, we get a distinguished family of subvarieties of diminishing dimension, which I’ll call hyperplane sections.

Each successive hyperplane section yields a subvariety of smaller dimension.

These hyperplane sections have one particularly interesting property. The intersection of any i-dimensional hyperplane section with i hyperplanes is a finite collection of points. This number, moreover, is the same no matter what i is. Indeed, any such number is equal to the number of points yielded when our original variety is intersected with k hyperplanes. (This special number is called the variety’s degree.) This number, most importantly of all, is nonzero. A variety sitting in projective space must be “caught” somehow by these hyperplanes! (This is similar to the phenomenon whereby any two lines must meet.)

We can use hyperplane sections to prove that any k-dimensional algebraic variety has infinitely many families of i-dimensional algebraic cycles, where i is any integer. The first step is to realize that the collection of families of i-dimensional subvarieties in our variety is more than just a package of disparate objects. It is, in fact, a place in which a certain sort of arithmetic can be performed — where two families can in a sense be “added” together to yield a third. It doesn’t make sense, of course, to literally add two families together the way we do integers. An analogous operation does exist here, though, and it boasts some of the same properties as addition. (This is a motif typical of the notion of group in the modern mathematical discipline of abstract algebra.)

The idea is to pick a single family of i-dimensional subvarieties and “add” it to itself infinitely many times, automatically yielding infinitely many families. There’s one catch, though. Though the operation here shares many properties with addition of integers, it doesn’t share all: a family, once added to itself enough times, could “wrap around” and become 0 again, ultimately failing to produce the infinity we desire. (This phenomenon is called torsion.) We need to pick a family of i-dimensional varieties for which we can show that this does not happen.

This is where the i-dimensional hyperplane section comes in. Any i-dimensional subvariety can be intersected with i hyperplanes, yielding a number. If we did this to the i-dimensional hyperplane section, for example, we would yield our variety’s degree. The key property is that if we did this to double the hyperplane section, we would yield double the variety’s degree, and same with triple, and quadruple, and so on. (The operation which takes a family to its degree is a group homomorphism.) If the i-dimensional hyperplane section were torsion, then some multiple of it would be the empty “0 subvariety”. But this would mean that some multiple of the variety’s degree (a nonzero number) is 0, which is ridiculous.

Therefore, the family of the i-dimensional hyperplane section is not torsion, and its infinitely many multiples yield the infinitely many families we desire. My fantasy is bankrupt!

I still can’t remember who won that game of cards.

References

I’ll translate the assertion here into more standard mathematical terms. At stake here is essentially whether, for say a smooth fourfold $X$, the Chow group $CH^2(X)$ of codimension-2 cycles in $X$ could be a finite group. (Actually, this argument goes through for any adequate equivalence relation on cycles, including homological equivalence. In fact the argument above is sufficiently vaguely stated that it’s not clear over which equivalence relation it operates, so the flexibility of the impending argument is good.)
This flimsy prospect is easily put to rest. It is enough to find an algebraic cycle $\alpha \in CH^2(X)$ and a map $CH^2(X) \rightarrow \mathbb{Z}$ for which the image of $\alpha$ in $\mathbb{Z}$ is nonzero; this would imply that $\alpha$ is not torsion. To this effect, we introduce the map $CH^2(X) \rightarrow \mathbb{Z}$ induced by pairing any class with the square $H^2$ of the hyperplane class and then taking the resulting 0-cycle’s degree. (To continue our aside: this map actually depends only on $\alpha$‘s numerical equivalence class.)
The image of $H^2 \in CH^2(X)$ under this map is none other than the degree of $X$, which is nonzero by the projectivity of $X$.