Proof of Theorem 5.1.
We will prove the result by contradiction. So let us assume it is false. Then there exists a sequence of
Riemannian manifolds satisfying , and such that
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(5.3) |
with and a vertex.
Note first that by the noncollapsing assumption we have .
Now by Lemma 1.7, there exists -splitting maps
with . Fix some sequence which is
tending to zero so slowly compared to , that Theorem 1.8 holds
for with . Let be
the corresponding good values of , and let
be fixed regular values.
Note that is smooth outside of the singular set . In particular on
we have , where is the harmonic
radius as in Section 1 and denotes distance. By the standard -regularity theorem, it follows
that the convergence of is in away from , for every
and . Let be the
-Gromov Hausdorff maps, and let us denote .
Then by the previous statements, for every , all sufficiently large, and
, we have .
Consider again the submanifold . Define the scale
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(5.4) |
By the considerations of the previous paragraph, this minimum is actually obtained at some
, with . Moreover, since , the cross-section of
the cone factor, satisfies , it follows that . According to Theorem 1.8,
there exists a lower triangular matrix such that
is an -splitting map.
Note that we have renormalized so that each of our regular values is the zero level set.
Now let us consider the sequence . After passing to a subsequence if necessary,
which we will continue to denote by , have
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(5.5) |
in the pointed Gromov-Hausdorff sense, where splits off isometrically.
We begin by observing that by our noncollapsing
assumption we have , and hence, in the rescaled spaces, we have
for all . In particular, has Euclidean volume
growth at i.e.
for all .
After possibly passing to another subsequence, we can limit the functions to a function .
Note that by our normalization, we have
are -splittings, and that by
Theorem 1.11, we have for each that
are -splittings. In particular, we can conclude that
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(5.6) |
where is the projection map and .
Now by construction, in the rescaled spaces we have for any that .
Therefore, the limit is in a neighborhood of , and hence
is a nonsingular surface. Thus, since it follows that is
at least a manifold with . Since the Ricci curvature is uniformly
bounded, in fact tending to zero, we have by the standard -regularity theorem that the
convergence is in
. Because the convergence is in we have that
converges continuously; [A90]. In particular, we have that and so .
On the other hand, since and is it follows that is a
smooth Ricci flat manifold.
This is easiest to see by writing directly in harmonic coordinates on , see [A90]
for the argument. Now since , we can conclude that is smooth and Ricci flat, hence flat.
In particular, we have that is flat. Since we have already shown that
has Euclidean volume growth, this implies that .
However, we have also already concluded that , which gives us our desired contradiction.
โ