Since the question has been "hanging on" for a while, I think it makes sense to give an outline of Atiyah's argument in the paper. Note that the paper is short (page 1 introduction, page 5-6 reference and further work). The main arguments in the paper are in pages 2-4.
(page2)
- First he introduced the "conformal sphere", which I guess Atiyah meant $\mathbb{S}^{6}$ as a conformal manifold with a conformal metric.
- The `standard' conformal sphere $S(\infty)$ in the Minkowski 8-space $M=(x,y,t)$ can then be viewed as the limit of smaller spheres $S(c)$, where the equation $$|x|^2+|y|^2=c^2 t^2, x\in \mathbb{R}^{4}, y\in \mathbb{R}^{3}, c\in \mathbb{R}^{+}$$
- The conformal sphere does not have a Riemannian metric but it has a conformal structure induced from $S(c)$ with hyperbolic metrics (inherited from $M$?).
- Atiyah defined a notion natural, which he defined as " Natural always means
compatible with the appropriate symmetry group, which here is the Lorentz group
of automorphisms of Minkowski space". In particular here $S(\infty)$ inherit the action of $SO(1,7)$ instead of $SO(8)$. Then Atiyah suggest this is the main source of "confusion" he wanted to dispel.
(page 3)
- Atiyah suggested to use the isomorphism involving quotient of Lie groups for the conformal sphere:
$$
\mathbb{S}^{6}\cong \textrm{Spin}^{+}(7,1)/(\mathbb{Z}/2\times \textrm{Spin}(6))
$$
- The $\mathbb{Z}/2$ factor acts by interchanging $(x, t)\rightarrow (-x,-t)$. In particular, any two points on $\mathbb{S}^{6}$ can be transformed to "standard antipodal pair" $(e,0), (-e,0), e\in \mathbb{R}^{3}$. (Not sure how to prove this)
Atiyah quotes a version of index theorem which Bott proved for homogeneous vector bundles. If I recall correctly this is an independent piece work of Bott produced when he was trying to verify the index theorem. So it can be viewed as special case of index theorem. However, it is not entirely clear what he meant exactly here.
Atiyah then claim the index in the representation ring for the round $6$-sphere lands in $R(\textrm{Spin}(7))$. For the conformal sphere it lands in $R(\mathbb{Z}/2\times \textrm{Spin}(6))$. (I guess) by a deformation argument Atiyah then reduce the case to $\Gamma=\mathbb{Q}_{8}$, the quarterion group of order 8.
It is pointed out that index theorem only needs an almost complex structure (correct). Atiyah then make a bold font claim that "$\Gamma$ acts on the conformal sphere $\mathbb{S}^{6}$ without invoking any
additional symmetry". This I do not really follow. Does he mean that no other group structure in the identification $$
\mathbb{S}^{6}\cong \textrm{Spin}^{+}(7,1)/(\mathbb{Z}/2\times \textrm{Spin}(6))
$$
is actually used???
(page 4)
Atiyah suggests this "fact" is related to CPT theorem. Further the presence of $\Gamma=\mathbb{Q}_{8}$ can be understood from the point of view of complex conjugations: We have
$$
0\rightarrow \mathbb{Z}/2\rightarrow \Gamma\rightarrow \mathbb{Z}/2\times \mathbb{Z}/2\rightarrow 0
$$
And the first $\mathbb{Z}/2$ action can be interpreted as conjugation on $\mathbb{C}^{3}$ preserving conjugation.
It is pointed out that the index would be equal to $(-1)^{F}$, which is a topological invariant. For Minkowski space it would be odd. (Not sure how relevant is this...).
(The claimed proof)
- Assume there is a hypothetical complex structure on the conformal sphere $\mathbb{S}^{6}=S(\infty)$, which he viewed as limit of $S(c)$ with hyperbolic metric. Note that this sphere is homeomorphic to the standard round sphere.
- Apply the $\Gamma$-equivariant version of the theorem to $\mathbb{S}^{6}$ with the complex structure. Since the central extension $\mathbb{Z}/2$ acts by complex conjugation , it is an abelian representation (meants a representation of $V$). However, I am not sure why the fact $\mathbb{S}^{6}$ having the hypothetical complex structure means the first $\mathbb{Z}/2$ action is trivial for the representation.
- Since the theorem does not depend on the metric, with the choice of round metric on $\mathbb{S}^{6}$ the index becomes an element in the representation ring of $\Gamma$. But then $\Gamma$ acts freely on the round sphere, so its index is the regular representation of $\Gamma$, which is not abelian. In particular it cannot be identified with a representation of $V$.
The crucial part of this paper left unaddressed (to me) are:
How the existence of the complex structure on $\mathbb{S}^{6}$ "forced" it to have an abelian index when using (the equivariant version) of index theorem?
Regardless of (1), does Atiyah implicitly assume the complex structure on $\mathbb{S}^{6}$ has to be "compatible" with its structure as a homogeneous space? In order words, how is the reduction from $\mathbb{Z}/2\times Spin(6)$ to $\Gamma$ related to the complex structure given by (purportedly some wierd metric)? What would happen if it does not?
Assume (1) and (2) are addressed. Now why an almost complex structure on $\mathbb{S}^{6}$ would be incompatible with the claims in (1) and (2)? In other words, what differentiates $i$ from $J$ for the purpose of $\mathbb{Z}/2$ actions, other than the fact that the central extension inside of $\Gamma$ can be identified with the complex conjugation? It does not seem Atiyah used theorems like Nirenberg-Newlander in any significant way.
It seems to me what Atiyah is really claiming is that a "special class of metric" on the conformal sphere $\mathbb{S}^{6}$ cannot give it a nice enough complex structure with respect to the existing homogeneous space structure. It would be interesting to check if the almost complex structure on $\mathbb{S}^{6}$ is compatible with the action of $\Gamma$, for example. Otherwise I suspect there may be an arithmetic mistake somewhere in the proof.
