Kervaire invariant: Why dimension 126 especially difficult? Is there any resource that might help non-experts gains some understanding of why
the Kervaire invariant problem remains open now only in dimension $126$? ($126 =2^7-2=2^{j+1}-2$;
whether $\theta_j=\theta_6$ exists in the "$128$-stem"), i.e., why the celebrated Hill-Hopkins-Ravenel 
proof technique fails in this last remaining case?
I read a delightful exposition by Erica Klarreich ("Mathematicians solve 45-year-old Kevaire Invariant Puzzle,"
The Best Writing on Mathematics, 2010, 373ff, Simons Foundation link), which piqued my interest.
But Michael Hopkins' online presentation slides announcing the result
in 2009 ("Applications of algebra to a problem in topology") are beyond my ken.
Perhaps this an area too abstruse for all but the experts?
I'd appreciate pointers to expositions.  Thanks!
 A: I'll give a shot at an answer.  The relevant dimensions are of the form $2^j-2$. For 
$j\leq 4$, it is easy and classical that we can construct manifolds of Kervaire invariant one. The
problem was ``reduced'' from differential topology to pure stable homotopy
theory by Browder in 1969.  Direct calculational methods in
homotopy theory were used by Barratt, Jones, and Mahowald to construct a
cell complex that can be used to solve the homotopy theory problem and prove 
that such manifolds also exist in dimensions 30 and 62.  I believe a construction 
of such a manifold has been worked out in dimension 30, but that has certainly not 
been done in dimension 62. Periodicity phenomena play a huge role in modern stable 
homotopy theory, and a crucial feature of the Hill, Hopkins, Ravenel proof is a 
periodicity of order $2^8 = 256$.  That enables them to solve the stable homotopy problem 
and prove there is no manifold of Kervaire invariant one for $j\geq 8$.  The reasons $j=7$ 
is so hard are several. Nobody has a really good reason for guessing which way the answer 
will go. There is no reason to expect a relevant periodicity of order $2^7$. Direct calculation
of the Adams spectral sequence through dimension $126$ is just plain hard: the
calculations blow up.  There is a chance that the methodology of Barratt, Jones,
and Mahowald might extend to prove existence (if that is how the answer turns out!),
but it will probably be much harder to prove nonexistence (if that is the answer).   
A: More detail on Jones's 30 manifold and analogous constructions can be found in my  Extended powers of manifolds and the Adams spectral sequence , Cont. Math. 271, 41--51.  The basic idea in it dates from 1979 or so (immediately upon seeing Jones's construction) but didn't make it into print for another 20 years.  There are also comments about the homotopy theoretic approach to the problem in   the talk I gave in Edinburgh  in 2011 at the workshop on the Kervaire invariant.
The main point is that there is a small set of homology classes in low dimension (dimension 2 for Jones' construction) which one wants to realize as the fundamental class of a manifold with tangent bundle realized by a permutation representation of $\pi_1$.  To get the 30 dimensional Kervaire manifold from $S^7$ one only needs a two manifold and an appropriate representation in $S_4$.  ($D_8$ will do:  only the Sylow 2-subgroup matters.)  To get from $S^7$ to the 62 dimensional Kervaire manifold requires a 6-manifold with a representation of $\pi_1$ in $S_8$.  Unfortunately, no such manifold exists.  In dimension 126, one would need a 14 manifold, and presumably such also fails to exist. So, one needs to find another approach that relaxes the input data needed.
