# Nilpotence of the stable Hopf map via framed cobordism

The Pontryagin-Thom construction shows that the stable homotopy groups of spheres are the same as the groups of stably framed manifolds up to cobordism. Specifically the Hopf map corresponds to the circle with its Lie group framing. It is well known that this element $\eta$ in stable homotopy (like all elements of positive degree) is nilpotent in the stable homotopy ring, more specifically $\eta^4$ is trivial. On the cobordism side, multiplication in the ring is just the product of manifolds. So the corresponding statement about manifolds is that the 4-dimensional torus with its Lie group framing is nullcobordant. Is there a geometric way to understand this? What makes the 4d torus so different from the 2d and 3d ones?

• A closely related fact, which quickly implies this one, is that the 3-torus is cobordant to a disjoint union of four copies of the 3-sphere (with its unit quaternion framing). Since that's just 3-dimensional it might be easier to see. – Noah Snyder Mar 31 '15 at 21:19
• Really? I thought it was $12$ copies. $\eta^3$ is $2$-torsion, so isn't it $12$ times the generator of stable $\pi_3$? – Qiaochu Yuan Mar 31 '15 at 21:26
• @QiaochuYuan: You're right, I forgot that there were primes other than 2. (Or rather, I was looking at the picture for just the 2-part.) – Noah Snyder Mar 31 '15 at 21:41
• @Chris: I don't follow. The signature isn't sensitive to a choice of framing, and there are other framings on tori in all dimensions with respect to which they're framed nullcobordant. Also, the signature of the $4$-torus is zero. – Qiaochu Yuan Mar 31 '15 at 21:59
• The 12 seems to big to be easily geometrically explained, but it's enough to just show that the 3-torus is cobordant to $M \coprod M$ for some 3-manifold M. – Noah Snyder Mar 31 '15 at 22:08

Let $\eta$ be the framed 1-manifold which is the Lie group framing on the circle and let $\nu$ be the Lie group framing on $S^3 = Spin(3)$. I am probably going to conflate these framed manifolds with their classes in frame cobordism. I hope you forgive me.

There are lots of geometric ways to see that $2 \eta = 0$ in stably framed bordism so I will take that as given.

One thing which makes dimension four different from dimensions two and three is the existence of the K3 surface.

What I will show below is that there is a way to cut the K3 surface in half to get a 4-manifold, which I will call $X$, with boundary the 3-torus $T^3$. We will see that if we remove 12 points from $X$ then it admits an (unstable) framing, which moreover restricts to the Lie group framing $\nu$ at each of the 12 points and restricts to the Lie group framing $\eta^3$ on the 3-torus.

This is a geometric incarnation of the relation $\eta^3 = 12 \nu$, which, as you observed, immediately shows that $\eta^4 = 0$ as well since $2 \eta = 0$.

This is closely related to the fact that $24 \nu = 0$ and I will start there.

## 24 Torsion in the 3rd stable stem, via geometry

The 3rd stable stem ( = the 3rd stable homotopy group of spheres = group of stably framed 3-manifolds up to cobordism) is $\mathbb{Z}/24$ and it is a very natural question to ask for a geometric description for the source of this 24. In fact this has been asked and answered before on MO. What I am about to describe is really just a synthesis of the answers given by Tilman and Mrowka for that previous question, which gives a geometric reason why $24 \nu = 0$.

The key geometric ingredient is the existence of the K3 surface. This complex surface is famous for its many interesting and useful properties. The two salient facts about this 4-manifold are the following:

1. The cohomology groups of the K3 surface are: $\mathbb{Z}, 0, \mathbb{Z}^{22}, 0, \mathbb{Z}$. In particular it is simply connected and the Euler characteristic is 24.
2. The K3 surface is hyperkahler. This means that there is not just one complex structure on the tangent bundle of the K3 but that there are three different ones $I, J, K$ and they generate a quaternion algebra bundle.

This second property is critical because it means that if we are given a non-zero vector $v$ in the tangent space of the K3, then we actually get a frame $(v, Iv, Jv, Kv)$. On a K3 surface (and submanifolds thereof) we can turn non-vanishing vector fields into 3-framings.

The Euler class obstructs the existence of a non-vanishing section of the tangent bundle, however since the Euler characteristic is 24, if we puncture the K3 surface 24 times (introducing 24 3-sphere boundary components) then we can find a non-vanishing vector field. Applying the hyperkahler structure to this gives us a 3-framing of the punctured K3 surface. This is now a framed bordism which shows that the sum of those 24 framed 3-spheres is zero in the framed bordism group.

In fact I think we could find a non-vanishing vector field on the K3 after puncturing just one time. However by puncturing 24 times we can arrange for each of the punctures to be a simple source for the vector field. This is important because then when we apply the hyperkahler structure near the source we see that it does indeed give the Lie group framing on each of those 24 3-spheres.

We conclude that $24 \nu = 0$.

## Half the K3 surface and $\eta^3 = 12 \nu$, via geometry

There is an important famous way to construct the K3 surface called the Kummer construction. I will just discuss the topological aspects first. In this construction you first consider the 4-torus $T^4$, which we will think of as a Lie group.

Next we take the quotient by the involution $\sigma: a \mapsto -a$. This results in a quotient space/orbifold $T^4/ \sigma$ which has 16 singularities. To get the k3 surface we "blow-up" those 16 points. The result is the K3 surface.

To get half the K3 surface we will modify this construction. Consider one of the circle factors of the 4-torus. The $Z/2$ action comes from a product of an action on each of these factors. It is simply reflecting the circle through the x-axis. It has two fixed points in the circle at (1,0) and (-1,0). The product of these fixed points in each circle give us the 16 fixed points in $T^4$.

We will take one circle factor and consider cutting it in half into the semicircles with either positive or negative x-coordinate. The action is compatible with this splitting (it is an equivariant decomposition) and each semicircle now has only one critical point.

This decomposition gives us a decomposition of of the 4-torus $$T^4 = (I \times T^3) \cup_{\partial I \times T^3} (I \times T^3)$$

The boundary of each half consists of two disjoint $T^3$s, and the $\mathbb{Z}/2$-action exchanges them. The quotient $$(I \times T^3)/\sigma$$ is an orbifold with boundary $T^3$ and with eight singular points on the interior. If we blow-up those eight points, then we get a manifold $X$ which is half the K3 surface in the sense that $$X \cup_{T^3} X = K3.$$

Since the Euler characteristic of $T^3$ is zero, this means that $\chi(X) = \frac{1}{2} \chi(K3) = 12$.

Now just as the Euler characteristic obstructs the existence of a non-vanishing vector field on a closed manifold, on a manifold with boundary it obstructs the existence of a non-vanishing vector field which further restricts to the inward pointing normal vector field near the boundary.

Thus if we take the 4-manifold $X$ and puncture it 12 times, then we may choose a non-vanishing vector field which is pointing inward at each of the punctures (i.e. they are sources) and which is inward pointing along the boundary 3-torus. Then we apply the hyperkahler structure to extend this to a 3-framing.

I already mentioned that near each of the 3-spheres we get the Lie group framing. This is believable because we just need to understand what is going on near an isolated point and so some standard flat model of $R^4$ is a good enough approximation, at least to identify the homotopy class of the induced framing.

However I claim that the induced framing on the 3-torus agrees with $\eta^3$. If so then this bordism witnesses the relation $\eta^3 = 12 \nu$, and we are done.

## Why is this framed 3-torus $\eta^3$?

There might be an easy way to see this, but I at least found one way to do this, by looking more closely a the geometric structure on the K3 surface. Much of this comes from D. Joyce's book Compact Manifolds with Special Holonomy.

One of the claimed properties of the K3 surface is that it has a hyperkahler structure, but how can we construct such a structure explicitly? Here specifically I mean a Kahler metric which has special holonomy given by $SU(2)$. We know that the moduli space of such metrics is 20 dimensional, how to we construct points in there?

One method is to go back to the Kummer construction. On the 4-torus we have a flat Kahler metric, viewing it as the quotient of $\mathbb{C}^2$ by a lattice. This is invariant under $\sigma$ and so descends to give a flat metric on $T^4/\sigma$, which we can think about as being a singular degenerate metric on the K3 surface. We can think of this as being a point on the boundary of some compactification of the space of hyperkahler metrics and the idea is to deform this metric to get the desired one. Away from the singular points the flat metric is a good approximation, but near the singular points it is bad, so you replace it with the "Eguchi-Hanson" metric. Then you have to interpolate. The details of this seem difficult and Joyce attributes it to Page, LeBrun-Singer, and Topiwala.

So in the end we see that the metric (and hence the hyperkahler structure) looks like the flat metric when you are far away from the singular points. The 3-torus is, by construction, just about as far away from these singular points as possible. We can also choose the punctures to be in the "flat region" as well.

Once the torus and the punctures have a nearby hyperkahler structure which is approximately flat, it is easy to see that our construction of a framing produces $\nu$ near each of the punctures and produces $\eta^3$ on the 3-torus.