Is there an octonionic analog of the K3 surface, with implications for stable homotopy groups of spheres? 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:


*

*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. 

*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:


*

*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

*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 $240 \sigma = 0$?
 A: 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.)
