Given that the kervaire invariant one problem has been solved in (almost) all dimensions....my question is whether there exists an exotic sphere in all sufficently lagre dimensions? Given the KervaireMilnor theory this "quasireduces" to determining whether coker J has nontrivial stable homotopy in all large dimensions. Has anyone looked at this? I've glanced at the work of Ravenel, etc. which doesn't seem to be sufficent.

2$\begingroup$ See the abstract of Behrens talk at the Milnor Abel Prize meeting. ima.umn.edu/20112012/SW1.302.1.12/abstracts.html $\endgroup$ – Ryan Budney Jan 31 '12 at 4:54
As Ryan Budney indicated, Mark Behrens gave a talk about joint work with Mike Hill, Mike Hopkins, and Mark Mahowald at the Milnor conference a few days ago. Here's a link to the video.
The "reader's digest" version of this is that, by using periodic families of elements in the stable homotopy groups of spheres, they've established a large number of congruences where exotic spheres are always forced to exist, but it's not yet exhaustive. In addition, they've done enough lowdimensional computations to know that in dimensions less than or equal to (at least) 63, there are exotic spheres in all dimensions except 1, 2, 3, 4, 5, 6, 12, and 61. The first of these were known from work of KervaireMilnor in 1963 (as part of their enumeration of the number of exotic spheres that's indicated by the Wikipedia entry), but the nonexistence of exotic spheres in dimension 61 is a new result.
Yes. For some progressions mod 4 this follows easily.
A more general question I could ask is for any fixed prime p and any large n there is a nontrivial pelement in the stable homotopy group of dimension n and/or the group of exotic spheres of dimension n?
I could say more here about Kervaire, Milnor, Bernoulli and others but I'll desist(for now:::grin:::).

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3$\begingroup$ Probably Mike Hill and Mike Hopkins  not sure who is who tho ;) $\endgroup$ – Ian Agol Jan 31 '12 at 16:13

$\begingroup$ No, the two accounts are the same person; I've merged them now. $\endgroup$ – Scott Morrison♦ Feb 1 '12 at 21:07