4. Proof of Theorem 1.8, the Slicing Theorem
It is the goal of this Section to prove the Slicing Theorem (Theorem 1.8). Recall the statement:
For each there exists such that if
satisfies and if is a harmonic
-splitting map, then there exists a subset
which satisfies the following:
- (1)
.
- (2)
If then is nonempty.
- (3)
For each and there exists a lower
triangular matrix
such that is an -splitting map.
The main tool will be the Transformation theorem (Theorem 1.11) proved in Section 3 and its applications. Recall first from Definition 1.10 the singular radius , where such that for the conclusions of Theorem 1.11 hold for . It is clear from the definition that if , then there exists such that
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(4.1) |
where in the second inequality, we have used that is an -splitting to conclude that , for sufficiently small.
Put
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Let denote the -dimensional measure of the image .
In view of the Transformation theorem,
to conclude the proof of
the Slicing theorem, it suffices to show
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(4.2) |
where as .
Let us denote by , the measure such that
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(4.3) |
Note that is a probability measure on .
In the arguments to come we will need have
a certain doubling property. While in principle, it is too much to ask that is actually a doubling measure,
next we observe that has a partial doubling property which will suffice for our purposes.
Lemma 4.1.
For each and
we have the doubling condition
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(4.4) |
Proof.
By Theorem 1.11, there exists a
lower triangular
matrix such that
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(4.5) |
is an -splitting. Let denote the Riemannian measure
and set . Define
the measure by . Then
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(4.6) |
In particular this gives us
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(4.7) |
and it is equivalent to show the ratio bound for . Now since is an -splitting
we have the estimate
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(4.8) |
Hence, we also have the estimate
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(4.9) |
which of course uses the doubling property for the Riemannian measure. By combining the previous
two estimates we get
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(4.10) |
Finally, by using the definition of we arrive at:
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(4.11) |
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(4.12) |
which by (4.7) completes the proof.
∎
By a standard covering lemma, let us choose a collection of disjoint balls, such that
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(4.13) |
For each such ball let be such that
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(4.14) |
Now since the balls are mutually disjoint,
Theorem 1.9, together with (4.14) and Lemma 4.1 (the doubling property of ) gives
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(4.15) |
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(4.16) |
The proof of the Slicing theorem (Theorem 1.8) requires that the image of
under have small measure. If in (4.15) the measure were instead the
usual riemannian measure,
then
since is Lipschitz, standard estimates could be used to show just that.
On the face of it, however, the -content estimate is much weaker, since for balls where the determinant
of is small we have .
On the other hand, in the spirit of Sard’s theorem, we will see in the next lemma
that at least for balls with , we recover
this loss because the volume of the image is correspondingly small.
Lemma 4.2.
If , then
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(4.17) |
Proof.
As in Lemma 4.1, choose a lower triangular matrix such that
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(4.18) |
is an -splitting and define the measure as in Lemma 4.1.
Then as in (4.5), .
Since is an -splitting, we have the estimates
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(4.19) |
By the first estimate above,
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(4.20) |
where in the last step we have used volume monotonicity for the Riemannian measure. On the other hand, by the second
estimate of (4.19),
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(4.21) |
Combining these gives the estimate
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(4.22) |
To relate these back to the original function , we observe that
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(4.23) |
which immediately gives
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(4.24) |
This completes the proof.
∎
We can now finish the proof of Theorem 1.8. Indeed, we have by (4.13), (4.15), (4.17), that
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(4.25) |
By taking sufficiently small, this suffices to complete the proof.