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](https://mathoverflow.net/q/44866/184), and [also this one](https://mathoverflow.net/q/201587/184), 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 $240 \sigma = 0$?



Is there an octonionic analog of the K3 surface, with implications for stable homotopy groups of spheres?
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](https://mathoverflow.net/questions/218549/finding-the-octonionic-analog-of-the-k3-surface-via-almost-hyperkahler-geomet?lq=1#comment538602_218549).)


