In the <http://arxiv.org/abs/math/0606464v1> I read 


"If you want to prove existence of exotic smooth structure on $\mathbb R^4$ you can do this if you are in
possession of a knot which is topologically slice but not smoothly slice (slice means zero slice genus).
Freedman has a result stating that a knot with Alexander polynomial 1 is topologically slice. We now
have an obstruction (s being non-zero) to being smoothly slice." (p.28)


What does it mean? Does anyone know the construction of exotic $\mathbb R^4$ using slice knots? Please, give me a references, if there are several constructions.


Added: <http://arxiv.org/abs/math/0408379>. Does there exist some other construction?



Slice knots and exotic $\mathbb   R^4$
From Jacob Rasmussen's paper "Knot polynomials and knot homologies", arXiv:math/0504045, p.13 of ArXiv version:



> 
> Bob Gompf kindly pointed out another such application [of Rasmussen's $s$-invariant, a concordance invariant of knots extracted from Khovanov homology]. Namely, $s$ can also be used to give a gauge-theory free proof of the existence of an exotic $\mathbb{R}^4$. Indeed, Gompf has shown that to construct such a manifold, it suffices to exhibit a knot $K$ which is topologically but not smoothly slice (see Gompf and Stipsicz, "4-manifolds and Kirby calculus", p. 522 for a proof). By a theorem of Freedman, any knot with Alexander polynomial 1 is topologically slice, so we need only find a knot $K$ with $\Delta\_K(t)=1$ and $s(K) \neq 1$. It is not difficult to provide such a knot - for example, the $(-3,5,7)$ pretzel knot will do.
> 


