Determination of the Homology and the Cohomology of a Few Groups of Isometries of the Hyperbolic Plane

Abstract

In this paper we determine the homology and the cohomology groups of two properly discontinuous groups of isometries of the hyperbolic plane having non-compact orbit spaces and the fundamental group of a graph of groups with a finite vertex groups and no trivial edges by extending Lyndon's partial free resolution for finitely presented groups. For the first two groups, we obtain partial extensions and the corresponding homology. We also compute the corresponding cohomology groups for one of these groups. Finally we obtain homology and cohomology in all dimensions for the last of the above mentioned groups by constructing a full resolution for this group.

Cohomology of Fuchsian groups and non-Euclidean crystallographic groups

“…So, we find Im(d 1 2,0 ) = Z and E 2 2,0 = 0. In light of the description of the fundamental domain in Section 2, for d 1 1,0 we have the following…”

“…It then follows that E 2 1,0 = Z 2g+k+d−1 and E 2 0,0 = Z. At this point, it is easy to see that the spectral sequence will collapse trivially once we have computed the differentials d 1 1, * . We will begin with the differential d 1 1,q where q ≡ 1 (mod 4).…”

“…The case where Γ is a cocompact Fuchsian group, so = + and d = k = s = 0, was considered by Majumdar [16], however, our computation of the ring structure is new. The case = + and d = k = 0 is a corollary of a result of Huebschmann [11] and the case = − and d = k = s = 0 was considered by Akhter and Majumdar [1]. Each of these previous results used different methods to the ideas here.…”

“…At this point, it is easy to see that the spectral sequence will collapse trivially once we have computed the differentials d 1 1, * . We will begin with the differential d 1 1,q where q ≡ 1 (mod 4). Since the edges connected to the vertices corresponding to the Z m j summands have trivial stabilisers, the Z m j summands will survive to the E 2 -page.…”

“…If the boundary component i only contains odd cycles, then γ q i,s i = s i −1 l=0 γ q i,l , so we have an order 2 element in the kernel of d 1 1,q . If the boundary component has an empty period of cycles, then we have exactly one edge γ i,0 with vertex w i,0 at each end.…”