Does there exists a finitely presented group with Dehn function $> n^3$ and all asymptotic cones simply connected

It is well known that all asymptotic cones simply connected implies polynomial Dehn function (Gromov). It is also well known that quadratic Dehn function implies all asymptotic cones simply connected (Papasoglu). Sapir and Olshanskii shows that there exists groups with cubic Dehn function, linear isodiametric function and no simply connected cones. Do we know how to construct a class of groups with arbitrary polynomial Dehn function and simply connected cones?

Added (from comment): Do we have a clue about a sufficient condition on the relations that will ensure a finitely presented group to have simply connected cones ?

• Yes: the free $k$-step nilpotent group on $d\ge 2$ generators for $k\ge 3$, which has Dehn function $\simeq n^{k+1}$ (easy and well-known) and simply connected asymptotic cones (Pansu).
– YCor
Apr 2 '14 at 21:39
• One reference for Dehn functions of free nilpotent groups of class $c$ equal to $n^c$ is: G. Baumslag, C.F. Miller III, H. Short, Isoperimetric inequalities and the homology of groups, Invent. Math. 113 (1993), 531–560. Steve Gersten had another proof (also in 1993). They proved lower bound. the upper bound was provide by Ch. Pittet, "Isoperimetric inequalities for homogeneous nilpotent groups", 1995. Apr 2 '14 at 22:45
• Thanks for both answers. Do we have a clue about a sufficient condition on the relations that will ensure a finitely presented group to have simply connected cones ? Apr 3 '14 at 7:02

A f.g. group is a finitely presented group if and only if every large enough loop in the Cayley graph can be filled by a finite number of loops of length $$\le$$ half its length.
A f.g. group has all its asymptotic cones simply connected if and only if every large enough loop in the Cayley graph can be filled by a bounded (finite) number of length $$\le$$ half its length.