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The infamous K3 surface has many constructions in many fields ranging from algebraic geometry to algebraic topology. Its many properties are well known. For this question I am really interested in the K3 surface from a algebraic topology perspective (hence I view it as a particular smooth 4-manifold).

As we have seen in this MO question, and also this one, the K3 surface plays an important role in the third stable homotopy group of spheres. It can be viewed as the source of the 24 in this group $\pi_3^{st} = \mathbb{Z}/24$. Here is a brief review of how that goes: the stable homotopy groups (in degree n) of spheres are the same as cobordism classes of stably framed manifolds (of dimension n). In dimension three the generator is given by $\nu = (S^3, \text{Lie})$, the 3-sphere with its Lie group framing (where we think of $S^3 \subseteq \mathbb{H}$ as the group of unit quaterions).

The K3 surface is responsible for the relation $24 \nu = 0$ which comes from the existence of a certain framed 4-manifold whose boundary is a disjoint union of 24 copies of the framed 3-sphere $\nu$. The two important properties of the K3 surface which give rise to this are:

  1. The K3 surface is an example of an almost hyperkahler manifold, which means that its tangent bundle admits a reduction to an $S^3 = SU(2)$ vector bundle. In fact it is integrable (so a hyperkahler manifold), but that is not important for this. What is important is that it admits three different complex structures $I,J$ and $K$ which generate a quaterionic algebra. This means that given a vector field $v$ on the K3 surface we get a quadruple of vector fields $$(v, Iv, Jv, Kv)$$ which form a frame whereever $v$ is non-zero.
  2. The Euler characteristic of the K3 surface, the obstruction to the existence of a non-vanishing vector field, is 24.

This means that if we cut out 24 small balls from the K3 surface, we get a manifold $X$ with boundary $\partial X = \sqcup_{24} S^3$, 24 copies of the 3-sphere. $\chi(X)= 0$, and so $X$ admits a non-vanishing vector field, which restricts to the inward pointing normal vector field at each boundary component. Using the hyperkahler structure this gives us an (unstable) framing of $X$ which restricts to the Lie group framing at each of the boundary 3-spheres. Hence $24 \nu = 0$.

What I would like to understand is if there is an analog of this story which replaces the quaternions with the octonions?

The $\pi_7^{st} = \mathbb{Z}/240$ and is generated by the framed 7-sphere $\sigma$ which is $S^7$ with its "Lie group framing" coming from its embedding as the unit octonions $S^7 \subseteq \mathbb{O}$. It is not actually a Lie group, because the multiplication is non-associative, but there is still enough structure to obtain a framing (by, say, left-invariant vectors), and it is the generator.

I am wondering if there is an octonionic analog of the K3 surface which, in a similar way, leads us to the conclusion $240 \sigma = 0$? In other words I am looking for a smooth 8-manifold such that:

  1. Its tangent bundle admits a multiplication by the octonions, in the same way K3's tangent bundle admits a multiplication by the quaternions. In particular given a vector field $v$ we should get an octuple: $$(v, Iv, Jv, Kv, Ev, IEv, JEv, KEv)$$ which is a frame wherever $v$ is non-zero; and
  2. The Euler characteristic should be 240.

Is there such a manifold? and how can we construct it? Does this sort of octonionic multiplication on the tangent bundle have a name? If there isn't such a manifold is there perhaps some replacement which still allows us to see, geometrically, that $240 \sigma = 0$?

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    $\begingroup$ If $I,J,K,E,IE,JE,KE$ are maps from the tangent bundle to itself than their multiplication is necessarily associative. $\endgroup$ – Will Sawin Sep 14 '15 at 15:52
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    $\begingroup$ @WillSawin Maybe one could utilize triality setup - there are three 8-dimensional spaces (presumably the tangent, "left" half-spinors and "right" half-spinors) with actions of these $I$, $J$, $K$, etc. jumping between them in such a way that it is impossible to come up with a compatible choice of a single system of isomorphisms between these three spaces which would produce associative actions. I realize this is more than vague, I just want to throw into the discussion the triality thing somehow :D $\endgroup$ – მამუკა ჯიბლაძე Sep 14 '15 at 16:24
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    $\begingroup$ Can you do a kummer-esque construction to the 8-torus and get anything nice? $\endgroup$ – Dylan Wilson Sep 14 '15 at 17:57
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    $\begingroup$ Probably you are already aware of all this, but this book review written by John Baez contains a lot of interesting numerology that suggests an interesting answer to your question. math.ucr.edu/home/baez/octonions/conway_smith $\endgroup$ – Sam Gunningham Sep 15 '15 at 3:07
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    $\begingroup$ Someone should mention G_2 manifolds at some point, surely... $\endgroup$ – David Roberts Sep 15 '15 at 12:40
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Yes, such M exists. The boundary connected sum of 28 copies of the Milnor plumbing has boundary diffeomorphic to $S^7$ so it can be closed off with $D^8$ and you can let $M$ be the connected sum of the resulting closed manifold and 8 copies 7 copies of $S^4 \times S^4$.

Why does that work? Let $f: M \to \mathbb{O}P^1 = S^8$ have degree 240 and let $L$ denote the canonical bundle over $\mathbb{O}P^1$ whose euler class generates $H^8(S^8;\mathbb{Z})$. Since $M$ has signature $28 \cdot 8 = 224$ and since $$\mathcal{L}_2 = \frac{7}{45}p_2 = \frac{42}{45}e \in H^8(\mathbb{O}P^1;\mathbb{Q})$$ we see from the signature formula that $p_2(f^*(L)) = p_2(TM)$ from which we deduce that $TM$ and $f^*(L)$ are stably isomorphic (they are both stably trivial away from the top cell of $M$ so $p_2$ is the only obstruction). Since the two bundle have the same euler class we deduce that they are also unstably isomorphic, i.e. $$f^*(L) \cong TM,$$ which should give the bundle condition you're looking for.

Remark: Maybe it follows from the results in Kreck's paper "Surgery and duality" that up to diffeomorphism this is the only possible 3-connected $M$ which is parallelizable away from a point. (EDIT: No, see this comment.)

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  • $\begingroup$ This is really really great, thank you! I have lots of questions. When you say the Milnor pluming, I take it you mean the pluming on the $E_8$-graph, yes? I am confused about two points. The first is: did you really mean 8 copies of $S^4 \times S^4$? or did you mean 7 copies? When I tried to compute the Euler characteristic of M I am off by 2, which could mean I made an arithmetic mistake, or that we really want 7 copies. More importantly, I see how the argument works if we know that $p_1(TM) = 0$. Can you remind me how we can see that? $\endgroup$ – Chris Schommer-Pries Sep 15 '15 at 12:08
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    $\begingroup$ Also for others who might read this and be confused like I was, one key point is that for this bundle L on $\mathbb{OP}^2$, we have $e(L)$ is the generator but $p_2(L) = 6 e(L)$. This follows from the calculation that the map $S^8 \to BSO$ which is the generator in $\pi_8$ induces multiplication by 6 in cohomology (generated by $p_2$ on the RHS). This is also why on String manifolds there is a $p_2/6$ class. $\endgroup$ – Chris Schommer-Pries Sep 15 '15 at 12:11
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    $\begingroup$ To see that $p_1(TM)$ is zero note that it is enough to check this on the obvious generators for the 4-dimensional homology (the $S^4$'s). The tangent bundle of $M$ on these spheres is the sum of two copies of $TS^4$ and hence stably trivial so $p_1$ is zero. Nice question by the way. It would be nice if there was a nicer answer. Along this line there is Hirzebruch's question of 24-d manifold with an action of the Monster Group. $\endgroup$ – Tom Mrowka Sep 15 '15 at 14:42
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    $\begingroup$ Yes, 7 sounds right now. $\endgroup$ – user80296 Sep 15 '15 at 17:33
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    $\begingroup$ It would be nice if there was a construction of this manifold that didn't require knowing a priori that $bP_8 = \mathbb{Z}/28$. I was hoping to use this manifold to prove that $240 \sigma = 0$, but since the computation that $bP_8 = \mathbb{Z}/28$ presupposes $\pi_7^{st} = \mathbb{Z}/240$, this is circular reasoning. $\endgroup$ – Chris Schommer-Pries Sep 16 '15 at 8:35

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