Finding the octonionic analog of the K3 surface, via (almost) hyperkahler geometry?

Question

The K3 manifold is an amazing object in mathematics which plays an important role in several fields ranging from the study of smooth 4-manifolds to algebraic geometry to differential geometry and beyond.

It also has important implications for the 3rd stable homotopy group of spheres as shown in this MO question and this one. Quickly, the fact that the K3 surface (1) is hyperkahler and (2) has Euler characteristic 24 can be viewed as the source of the 24 in $\pi_3^{st} = \Omega^{fr}_3 = \mathbb{Z}/24$. (This is the 3rd stable homotopy group of spheres, also known as stably framed 3-manifolds up to cobordism).

Motivated by this I asked in this previous MO question if there was an octonionic analog of the K3 surface which could likewise account for the 240 in $\pi_7^{st} = \Omega^{fr}_7 = \mathbb{Z}/240$. This group is generated by the octonionic hopf map after all. This octonionic version of a K3 surface would be a smooth 8-manifold with Euler characteristic 240 and possessing an octonionic analog of a hyperkahler structure (more on this later).

My question was answered in the positive! Indeed there is such a manifold. However the construction given there used the celebrated result of Kervaire and Milnor that the group of exotic 7-spheres is $\Theta_7 = bP_8 = \mathbb{Z}/28$ and is generated by the boundary of "Milnor's pluming". This is all well and good, as it answers the question I posed, however Kervaire and Milnor's computation that $bP_8 = \mathbb{Z}/28$ depends crucially on the computation (by Adams and Quillen) that the image of the J-homomorphism in this dimension is $\mathbb{Z}/240$.

One reason I was interested in this manifold is that I wanted a purely geometric reason for the 240 to appear in this dimension, just as the 24 appears for the K3 surface. However as it stands the construction of this octonionic analog of a K3 uses the 240 as input.

The K3 surface has many different constructions and I still hold out hope that this octonionic analog of the the K3 might also admit many constructions. Hopefully it admits a construction which doesn't implicitly use the fact that $\pi_7^{st} = \mathbb{Z}/240$ or that $\Theta_7 = \mathbb{Z}/28$. In fact if such a manifold can be constructed by other means, then we can probably reverse the argument and use this manifold to give a new geometric identification of these groups (or more precisely identifications of $Im\;J$ in degree 7 and the group $bP_8$).

The manifold constructed in answer to my previous question is 3-connected, and, as was suggest by the OP of that answer, I think that it is likely that a modified form of Kreck's "Surgery and duality" paper will show that it is the only 3-connected 8-manifold with the desired properties (explained below). This is in analogy with the 4-dimensional case: K3 is the only simply connected hyperkahler 4-manifold.

The K3 manifold $N$ is uniquely specified by the following properties:

1. It is simply connected (i.e. connected and $\pi_1 = 0$);
2. It is a complex manifold;
3. The first Chern class vanishes $c_1(\tau_N) = 0$.

Using the connectivity of the manifold, this last condition can be expressed in a different way. Since the manifold is complex, the classifying map of the tangent bundle factors as $N \to BU(2) \to BO(4)$. Using obstruction theory this last condition is equivalent to asking that it factors further as $$N \stackrel{f}{\to} \mathbb{HP}^1 \to BU(2)$$ where the bundle on $\mathbb{HP}^1$ is the quaternionic projective line. This means that $\tau_N \cong f^*(\gamma_\mathbb{H})$ where $\tau_N$ is the tangent bundle of $N$ and $\gamma_\mathbb{H}$ is the tautological quaternionic line bundle. In particular the tangent bundle of $M$ inherits a quternionic action.

We can try to construct a higher dimensional analog of this by replacing the complex numbers with the quaternions. In particular instead of looking for a complex surface we will be looking for a hyperkahler manifold.

Question: Is there a construction of a smooth 8-manifold $M$ (an "octonionic K3") such that

1. $M$ is 3-connected (i.e. connected and $\pi_1 = \pi_2 = \pi_3=0$);
2. $M$ is hyperkahler almost hyperkahler;
3. The first pontryagin class vanishes, $p_1(\tau_M) = 0$.

and which does not make use of the computations of exotic 7-spheres $\Theta_7 = bP_8 = \mathbb{Z}/28$ or the stable 7-stem $\pi_7^{st} = (Im \; J)_7 = \mathbb{Z}/240$?

As with the usual K3 surface, this last condition can be reformulated as saying that the classifying map of the tangent bundle factors as $$ M \stackrel{f}{\to} \mathbb{OP}^1 \to BHU(2) = BSp(2)$$ i.e. that there exists an isomorphism $\tau_M \cong f^*(\gamma_\mathbb{O})$ of the tangent bundle of $M$ with the pullback of the octonionic line bundle. In particular this gives an octonionic "action" on the tangent bundle of $M$.

Side question: Has this octonionic analog of hyperkahler, without the connectivity condition, been studied before? Does it have a name? To me it seems like a very natural generalization of the notion of (almost) hyperkahler.

If, as I suspect, there is indeed only one of these manifolds up to diffeomorphism then it will agree with the construction given here. In that case we can compute some other things about this manifold:

• The Euler characteristic is $240$;
• The integral cohomology is torsion free;
• The Betti numbers are $b^0 = b^8 = 1$, $b^4_+ = 8 * 28 + 7 = 231$, $b^4_- = 7$, and all other Betti numbers vanish;
• The signature is $\sigma = 8 * 28 = 224$;
• The second Pontryagin number is $p_2 = 6 * 240 = 1440$.

I know there has been a lot of study of hyperkahler manifolds using methods from differential geometry and algebraic geometry. Can you help me? have you seen my manifold or know how to find it?

Answer 1

The following shows that such a manifold exists, but unfortunately it uses $bP_8 \cong \mathbb Z_{28}$.

According to Corollary 5.5 of this article a highly connected $8$-manifold $M$ with $p_1(M)=0$ admits an $Sp(2)$-structure (i.e. an almost hyperkähler structure) if and only if

1. $p_2(M) -2 \chi(M)$=0
2. $\chi(M) \equiv 0 \mod 4$.

(I am identifying the Pontryagin class $p_2$ with the Pontryagin number in the obvious way)

Now let $W$ be the union of an closed $8$-disc $D^8$ with the Milnor plumbing mentioned here and consider the manifold $M= 247\#(S^4\times S^4)\#W$.

We have that $p_1(M)=0$ since the tangent bundle of $M$ is stably equal to the Whitney sum $TS^4\oplus\ldots\oplus TS^4\oplus TW$ (where we pull back the tangent bundles of $TS^4$ and $TW$ the the obvious collapsing maps $M \to S^4$ and $M \to W$) and $p_1(S^4)=p_1(W)=0$.

Furthermore we have $\chi(M) = b_4(W)+246\cdot b_4(S^4\times S^4)+2=224 +247\cdot 2+2=720(\equiv 0\mod 4)$ as well as $p_2(M) = \frac{45}{7}\sigma(M) = \frac{45}{7}\cdot 224 =1440$.

Edit: It seems to me that the example of the other post will not have an $Sp(2)$-structure if Corollary 5.5 is true. Moreover it will not be possible to construct a highly connected manifold with $Sp(2)$-structure, $p_1(M)=0$ and Euler characteristic equal to $240$ since the signature would not be an integer.