The Hodge Conjecture

Get $1 million if you solve this math problem.

Matthew Ward
Feb 13 · 10 min read
Photo by jose aljovin on Unsplash

There has been a lot of talk about two of the seven Millenium Problems. These are a set of problems that have a $1 million prize attached to them. The Riemann Hypothesis is the easiest to state, and so that has a lot of popular articles on it.

The Poincare Conjecture is the only one to be solved so far, and so that has gotten some popular articles as well.

The problem I understand the best is called The Hodge Conjecture. It is a problem from algebraic geometry, which is what my degree was in (though I didn’t work anywhere near this conjecture).

This article will try to give an overview of the conjecture for a general math audience.

The Topology

Topology is basically the study of how to deform things.

Let’s think about how to measure the 1-dimensional loops on a 2-dimensional space up to deformation. We’ll use the notation X for the space we’re working on.

In our first example, we’ll let X be a sphere.

If we start with any loop (like the black one) on the surface of the sphere, we can slide it (the red ones) to a single point (the black dot). When we can do that, we’ll call the loop “equivalent to 0” because we can deform it into a trivial thing.

Since we can do that with any loop, the collection of loops up to this deforming equivalence is 0. I’ll suggestively notate this H₁(X, ℚ)=0, though, for the rigorous among you, I’ve actually described something slightly different.

One important thing to note is that the starting loop doesn’t have to be a “smooth circle.” It can be super wavy and weird or even jagged. As long as it is an honest closed loop, that is a valid starting object. The 1 that appears as a subscript is related to the fact that we started with a 1-dimensional object (the loop).

Now let’s do this with a more difficult space: a torus.

If you start with something like the red loop, it doesn’t go all the way around in either direction. So it can be deformed to a point like on the sphere example.

If you start with a wavy black loop that loops around the tube of the torus, then you can always deform it to the non-wavy black circle.

Any closed loop that goes around the central hole can be shrunk down into the middle ring.

It’s not easy to prove rigorously since I haven’t given you real definitions, but if you think about it carefully, you should be able to convince yourself that these are the only three cases.

  • You either shrink to 0.
  • You shrink to the band around the torus like you’re holding it.
  • Or you shrink to the central inner circle.

So, the only non-zero elements are generated by those two rings. In this case, we’ll suggestively write H₁(X, ℚ)=ℚ². I don’t want to get too bogged down in that. It’s called the first homology group.

The ℚ appears for technical reasons that will be mentioned in passing later. If the one loop is [A] and the second is [B], you’re allowed to make sums with these elements with rational number coefficients. So (1/2)[A] + 12[B] is such an element. This makes a lot of stuff work like making sense of intersecting.

If we call these 1-dimensional loops 1-cycles, then we’ll call the k-dimensional counterpart k-cycles. It’s a bit strange to define them precisely because we still want some notion of “being a loop” in higher dimensions.

To get a feel for this, you can think of a 3-d sphere:

If you have some 2-dimensional blob inside it, you’ll always be able to shrink it to a point. It’s kind of hard to tell from the picture, but you should note that this is “inside” the 3-dimensional sphere here. In the earlier example, we had to stay on the surface of the 2-dimensional sphere.

The k-cycles become equivalence classes in Hₖ(X, ℚ). Since all 1-cycles collapse to a point in a 3-sphere, H₁(X, ℚ)=0 and since all 2-cycles also collapse to a point, H₂(X, ℚ)=0.

In a sense, the dimension of Hₖ(X, ℚ) is going to tell you how many (n-k)-dimensional holes are in X, where n is the dimension of X. Think about the torus. H₁(X, ℚ) was 2-dimensional and that was because there were 2 “holes,” the one going around the tube and the other going around the central hole.

If you want to get a better feel for this, try doing a punctured 3-sphere on your own (where you just remove another smaller sphere from the center).

The last bit of terminology I’ll use is representative. As you see above, there are always infinitely many different k-cycles that become the same after deforming them. Each of these represents the same cycle in this fancy homology group, and so we’ll call any one of those a representative of the class.

Geometry

Geometry can mean a lot of different things, but in this article, I’ll always use it to mean “algebraic geometry.” For our purposes, this means the study of shapes given by the zero sets of polynomials.

If you’ve ever taken linear algebra, then you should be familiar with this concept. Linear algebra is the study of zero sets of linear equations. You get very simple things like planes and subspaces.

Recall that the techniques in linear algebra involved shifting back and forth between thinking about “pictures” like the null space, range space, intersections of planes and the “algebra” of actually computing things.

Linear algebra is “completely solved” in a sense, but if you let your equations have exponents so that they are polynomials, the branch of math is wide open and an active area of study.

That’s algebraic geometry: converting between the geometry of zero sets of polynomials and manipulating the algebra of these equations.

For this article, a smooth algebraic variety (in the future this will be shortened to “variety”) is a geometric space, X, given by the zero sets of polynomials (over the complex numbers ℂ) where the resulting space is “smooth” in the sense you learn about in Calculus.

There are many technical points missing here. For example, we’ll always be doing this in projective space, but that won’t matter to the broader ideas.

It turns out that we’ve already seen a few varieties in this article above.

The 2-dimensional sphere is given by a quadratic equation. The donut/torus is an elliptic curve given by a cubic equation. Hopefully, you caught the weird terminology there. An elliptic curve is a 2-dimensional torus.

Because we’re working over the complex numbers, “real” dimensions will always be even. If you think about the complex plane, it looks like ℝ² but it is only ℂ. Algebraic geometers call things by their complex dimension, so a 1-dimensional curve has 2 real dimensions and a 2-dimensional surface has 4 real dimensions.

Don’t sweat that point too much.

The last bit of terminology we need is subvariety. As you might imagine, a subvariety of X is just a subset that is also given by zero sets of polynomial equations, and hence also a variety.

Here’s the key idea that will make the Hodge Conjecture interesting: from a topologist’s viewpoint, practically nothing is a (sub-)variety.

Being the zero set of a collection of polynomials is extremely limiting. Topological things can be crazy and weird and not smooth in the slightest. If you start with a subvariety of X and you deform it as we did above in almost any way, you will end up with something that is no longer a subvariety.

Also, notice that a subvariety is formed by taking zero sets of polynomials and intersecting with X. This is inherently a “global” thing. Topologically deforming shapes is in a sense a “local” thing.

So we should be really surprised if there’s a nice connection between these two very different branches of math.

The Hodge Conjecture

The most naive version of the Hodge Conjecture we could make at this point is this: given some cycle [A] in Hₖ(X, ℚ), is there a k-dimensional subvariety Y that represents [A]? We’ll call such a subvariety an algebraic representative for [A].

Let’s unpack that for a moment. Recall that A could be super weird. It’s a fundamentally topological thing winding around holes in X. It can be jagged and complicated.

Is there a way to “deform” A into a “nice” shape defined by polynomial equations?

If you think the answer must obviously be yes, then you might not be grasping just how strong and restrictive being a subvariety is. If you think the answer must obviously be no, then you might not be grasping just how much we can deform A without changing its class.

This naive version is false by things we’ve already said. Remember that we’re over the complex numbers, so every subvariety is even-dimensional. This means that there are no algebraic representatives in H₃(X, ℚ), for example.

But even if we limit ourselves to even dimensions, there’s one more technical condition that presents a problem.

If you look up the Hodge Conjecture, the standard way to state it involves cohomology not homology, so there are superscripts, not subscripts. It turns out these are just dual concepts. If X has (real) dimension n, then we can think of [A] in Hₖ(X, ℚ) as a class in Hⁿ⁻ᵏ(X, ℚ).

One of the reasons for doing this is that cohomology has an interpretation using differential forms. This means we can use integration to make numerical computations.

There’s a beautiful decomposition of cohomology, under our assumptions on X, called the Hodge decomposition. It even uses harmonic functions. You can split the pieces up by pairs of numbers that add to the superscript. I won’t write the general pattern, but I’ll do 4 as an example:

H⁴(X, ℂ) splits up into pieces we mark with (0,4) (1,3) (2,2) (3,1) and (4,0). This doubled middle piece consists of forms called the Hodge classes. Notice that the doubling only exists if the exponent is even.

A fairly straightforward computation that uses some technical machinery shows that any subvariety [Y] must land in that middle piece. In other words, every subvariety is a Hodge class.

For example, if X is 6-dimensional (in the complex sense) and Y is a 2-dimensional subvariety, [Y] is in the (4,4) part of H⁸(X, ℚ).

If all that is a bit much, just think of this as one extra “numerical condition” that must be satisfied. Since every subvariety is a Hodge class, we know that the naive version of the Hodge conjecture isn’t true by just taking a non-zero class that isn’t a Hodge class.

The current version of the Hodge Conjecture says: every Hodge class is algebraic. In other words, we can take any Hodge class, no matter how weird, and deform it into an honest subvariety.

Some Progress

It turns out that the Hodge Conjecture is true in low dimensions due to a result of Lefschetz in 1924 from before Hodge even made the conjecture in 1950.

Lefschetz proved it for codimension 1. In other words, every Hodge class in H²(X, ℚ) is algebraic. By duality, this also shows that every Hodge class in Hⁿ⁻²(X, ℚ) is algebraic. Thus, if the dimension of X is 1, 2, or 3 the Hodge Conjecture is true.

There have been a few other cases proved over the years but only under extremely strong extra assumptions.

So, you probably know my opinion at this point considering the wording of some sections of this article. I believe it must be false just because I see no reason that we’ve put enough conditions into the conjecture yet.

The history is long and interesting, so I’m going to mischaracterize it here on purpose to make a point.

You can imagine someone making the naive conjecture and then someone comes along and points out that actually subvarieties are all Hodge classes. Then someone says, oh, well I wonder if all Hodge classes are algebraic.

Then in 1961 Atiyah and Hirzebruch showed the integral version is false. So then people were like, I wonder if we replace ℤ with ℚ if it’s true. And that’s where we’re at today.

Unlike the Riemann Hypothesis that has overwhelming evidence to support it, the Hodge Conjecture seems like a work in progress that got stuck after a couple of refinements. For goodness sake, we don’t even know if it’s true when X is 4-dimensional and given by a single polynomial equation!

It seems far more likely to me that someone will realize another necessary condition or two on algebraic classes to refine the conjecture into something that can be proved.

It should be noted that if a counterexample is found and it’s obviously pathological in some way (in the sense that a clean reformulation of the conjecture can be made that rules the example out), the million-dollar prize will not be awarded.

Cantor’s Paradise

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Matthew Ward

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Cantor’s Paradise

Medium’s #1 Math Publication!

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