All I know is that a Kleinian group is defined as a discrete subgroup of the group PSL(2,C). Here C = the field of a complex numbers. PSL(2,C) is defined as the group of linear fractional transformations PSL(2,C) = {z —> (az+b)/(cz+d) | a,b,c,z in C with ad-bc ≠ 0} PSL(2,C) can be identified with the group of 2x2 complex matrices a b { } c d such that ad-bc = 1 under matrix multiplication, IF matrices M and -M are identified with each other. I.e., PSL(2,C) = SL(2,C) / {±I}. plane PSL(2,C) is of particular interest because it is exactly the group of conformal maps of the sphere S^2 to itself. Conformal means angle- and sense-preserving. The rotations of the sphere, which form the group SO(3), are certainly all conformal. So the rotation groups of the regular polyhedra are discrete subgroups of PSL(2,C), i.e., Kleinian groups. Note_1: The group of conformal transformations of the open disk, which is also the group of isometries of the hyperbolic plane, can be identified with the subgroup PSL(2,R) of PSL(2,C), defined as PSL(2,R) = {z —> (az+b)/(cz+d) | a,b,c,d in R with ad-bc ≠ 0}. Here the hyperbolic plane H^2 is identified with the upper half plane, i.e., the set H+ = {z in C | Im(z) > 0}. (S^2 is the unique simply connected surface with constant curvature = 1, and H^2 is the unique simply connected surface with constant curvature = -1.) (When a discrete subgroup of PSL(2,C) lies entirely within PSL(2,R), it is usually called a "Fuchsian" group, and the term Kleinian group is usually reserved for discrete subgroups of PSL(2,C) that are not Fuchsian.) Note_2: PSL(2,C) can also be identified with the isometry group of 3-dimensional hyperbolic space H^3, which makes it important in the study of 3-manifolds. This can also be identified with the conformal transformations of the open unit ball in R^3. Finite groups of rotations of S^2 can be the three isometry groups of regular solids, or a finite cyclic group, or else a dihedral group (the rotation group in 3D of a regular polygon). But a finite Kleinian group need not consist of only rotations of S^2. For instance, the element z —> 1 - 1/z is of order 3 in PSL(2,C) but it's not a rotation. But: Any finite subgroup of PSL(2,C) is known to be *conjugate* to a cyclic or dihedral group or the tetrahedral or octahedral or icosahedral group. (Theorem 6.3.1 in http://www.matcuer.unam.mx/~jseade/Libro-Klein.pdf .) —Dan Allan Wechsler wrote: ----- Because I'm just an amateur, I've never really engaged with the concept of Kleinian groups before, so I'm only just starting with elementary concepts. So I have a couple of really basic questions. I think I read that all the finite spherical groups (including the symmetry groups of all polyhedra) occur as Kleinian groups. Is that correct? If so, is it because PSL(2,C) has a subgroup that is equivalent to the rotation group of a sphere? Are any _other_ finite groups Kleinian? That is, is there a finite Kleinian group that is _not_ the symmetry group of a polyhedron? (When I say "is", I hope it's clear that I mean "is isomorphic to". When I say "finite", I mean a finite number of elements, not the concept of "geometrically finite" that I haven't yet wrapped my head around in the context of Kleinian groups.) -----