5.1. Further discussion [02D3]
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5.1. Further discussion
We can use this detailed description of the link , in the three-dimensional case to get a more precise understanding of the “topological obstruction” of Section 3.2.2. A representation defines a covering of and it is clear that the metric completion of this is again an orbifold with a metric of Ricci curvature . It is clear then that the usual proof of Myers Theorem extends to show that is compact, so the representation maps to a finite group. Thus is also finite and the torus in the discussion of 3.2.2 is in this case trivial. (Of course the set can be assumed to be homotopy equivalent to ). Moreover it is also clear that the usual proof of the Bishop Theorem extends to this case to show that the volume of cannot exceed that of . Hence the order of the cover, is bounded by where is the volume ratio, and hence by . Let be the least integer such that all integers less than or equal to divide . Then we see that the power of any such representation must be trivial. Thus if, from the beginning of the discussion in Section 3, we consider powers we never encounter the topological obstruction. The point here of course is that is determined in a simple explicit way by which in turn, in the Fano case, is known explicitly. In many practical cases of interest is not too large.
We expect that in fact the same will be true in higher dimensions (with the same ). Of course we do not expect that the singularities will always be of orbifold type, but it seems likely that the Bishop theorem can still be extended to the metric completion of a covering, as above. There is a slightly weaker statement which should be easier to prove. Let be a point in the singular set of a -dimensional link . Let be a sufficiently small ball about and the regular set. Suppose that we have found a number such that for all such points (in all tangent cones of all limits of manifolds in ) the homology group has order bounded by . Let be a representation of as above. Then in the covering defined by the pre-image of is a disjoint union of copies of . In this situation it is straightforward to apply recent results of Colding and Naber [7] to show that the regular set in the metric completion is geodesically convex, and then to extend the Bishop argument to this case. Then we see that if, from the beginning of the discussion in Section 3, we consider powers then we never encounter the topological obstruction. Arguing by induction on dimension it seems likely that in fact the number will have the property stated above so, for this weaker statement, we would consider powers . But, in fact it seems to us most likely that these higher powers of are not required.
In this direction we make the following conjecture, which (if true) would be a substantial sharpening of Theorem 1.1.
Conjecture 5.15.
For any and there is a number such that if then for any in we have
with as above.
To put this in context, recall that for a fixed the standard asymptotics is as . This essentially follows from the fact that on we have . The conjectural lower bound here is a uniform version of this over , provided we work over multiples of . On the other hand the corresponding upper bound——almost certainly fails, because at the vertex of a cone we have where is the volume ratio. This is why we believe that the plausible upper bound should include the extra factor . In a similar way, if in fact we do encounter the topological obstruction of 3.2.2 in some limit space, then it seems it would not be true that there is a lower bound on for all sufficiently large , since the twisting of the line bundle will force to be small as we approach the singularity. This phenomenon—that near to a singularity gets larger or smaller depending on divisibility—is similar to the orbifold situation considered by Ross and Thomas in [17].