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Most cosmological models in general relativity have a point of singularity: There is a point, which can be reached by a geodesic of finite total length from other points of the space-time manifold, where the metric is degenerate or otherwise irregular (for example, a point where a curvature scalar is infinite). All known dust-filled models, without cosmological constant, have singular points,' and it may be conjectured that the presence of dustlike matter filling space always leads to a singularity. This paper shows that this conjecture is true for an important class of universes. Some cosmological models, not of the dust-filled type, do not have this geometric imperfection. There exists a vacuum solution, the Taub-NUT-Misner2-4 universe (here denoted T-NUT-1\J), which is a cosmological model in that it admits closed spacelike homogeneous hypersurfaces. This universe, moreover, does not have a point of singularity in the sense described above. This model, however, is unusable as a cosmological model because of incompleteness: Some timelike geodesics are of finite total length and yet do not have any limiting end point. The T-NUT-M solution is "symmetric" under the three-dimensional Lie group of Bianchi type nine [denoted SO(3,R) ]. A symmetry group is a group of transformations of the space-time manifold: Under the action of the group elements, any single given point-of the manifold traces out a subspace-in the cases under consideration, a three-dimensional hypersurface-at all of whose points the metric tensor is the same. Some of these invariant hypersurfaces are spacelike, but not all. A manifold with this type of symmetry is said to be "spatially homogeneous." In a search for information concerning the development of singularities in realistic cosmological models, one is led to impose the same symmetry conditions. Unlike the T-NUT-M vacuum solution, however, every dust-filled universe of this kind possesses a singularity; this paper will sketch the proof of this fact. Why is it desirable to look at spatially homogeneous models? Why use the group SO(3,R) instead of another? There are four reasons: First, the introduction of any symmetry group enormously simplifies the problem of solving the field equations. (The danger, of course, is that any such simplification is too much !5) Second, this particular group stands out-it allows the introduction of three directions which are equivalent as far as group properties are concerned: The structure constants of the group SO(3,R) assume a natural form when expressed in a characteristic basis of the tangent space of the group at the unit element of the group (the group algebra). The structure constants remain unchanged when this basis, which is three-dimensional, is transformed by means of a three-dimensional orthogonal transformation of determinant one (a rotation). This means that although the directions within the three-dimensional invariant hypersurfaces may not be equivalent metrically, they are similar in their group properties, which are

SINGULARITIES IN SPATIALLY HOMOGENEOUS, DUST-FILLED COSMOLOGICAL MODELS