, We will show that it does not occur. If N ? I is simply connected, then the holonomy group must be discrete, which contradicts the preceding lemma. So we suppose, that N ? I is not simply connected. This case occurs if I is an affine subspace of real codimension 2. Let H be an affine half-space with ?H = I. By turning H around I

, The subgroup of Sim(N ) stabilizing I contains dilatations f, g with different fixed points. The lifted group Sim(I) × R containsf ,g (essentiallyf f × {0} since f = g ij does not rotate much around I) again contracting and with different fixed points. We now have a lifting of, Since I is invariant by the holonomy group ?, and since ? contains contractions (such as g ij ) it follows that I is stable by dilatations

The developing map is given by the choice of a point: take p ?M and D(p) ? N ? I, we have to choose q ? H × R such that q is send to D(p) by the covering H × R ? N ? I, denote q by D (p). The new developing map D ,

, It is sent to a path in N ? I which is covered by a pathq(t) inM since D is a covering map, and D (q(t)) = q(t) by local triviality. Since H × R is simply connected, the new holonomy group ? must be discrete

, This concludes Fried's theorem's proof since I must be constituted of a single point (which must be totally fixed by ?)

, ?H n )-manifold. If D is not surjective then it is a covering onto its image. Furthermore, D is a covering on its image if, and only if, D(M ) is equal to a connected component of ?H n ? L(?). M be a closed (PU(n, 1), ?H n )-manifold. If the developing map is not surjective

Since M is closed, it verifies the hypotheses of Fried's theorem and hence must be complete, then up to conjugation, we can suppose that ? = {?} and D(M ) ? N . Since ? is fixed by the holonomy group, we get a similarity structure (Sim(N ), N ) on M ,

, Then by minimality L(?) ? ?. If L(?) = ? then by

, Again, by Fried's theorem and since L(?) = {?}, the developing map is in fact a covering onto N , absurd. Hence, L(?) consists of at least two points and is contained in ?, Hence L(?) consists of one or at least two points. If L(?) = {a} then we can suppose it is ? and D(M ) ? N

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