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4. Proof of Theorem 1.8 , the Slicing Theorem [01YB]

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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 ϵ>0\epsilon>0 there exists δ⁡(n,ϵ)>0\delta(n,\epsilon)>0 such that if MnM^{n} satisfies RicMn≥−(n−1)​δ{\rm Ric}_{M^{n}}\geq-(n-1)\delta and if u:B2​(p)→ℝn−2u:B_{2}(p)\to\mathds{R}^{n-2} is a harmonic δ\delta-splitting map, then there exists a subset Gϵ⊆B1​(0n−2)G_{\epsilon}\subseteq B_{1}(0^{n-2}) which satisfies the following:

  1. (1)

    Vol⁡(Gϵ)>Vol⁡(B1​(0n−2))−ϵ{\rm Vol}(G_{\epsilon})>{\rm Vol}(B_{1}(0^{n-2}))-\epsilon.

  2. (2)

    If s∈Gϵs\in G_{\epsilon} then u−1​(s)u^{-1}(s) is nonempty.

  3. (3)

    For each x∈u−1​(Gϵ)x\in u^{-1}(G_{\epsilon}) and r≤1r\leq 1 there exists a lower triangular matrix A∈G​L​(n−2)A\in GL(n-2) such that A∘u:Br​(x)→ℝn−2A\circ u:B_{r}(x)\to\mathds{R}^{n-2} is an ϵ\epsilon-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 sxηs_{x}^{\eta}, where η⁡(n,ϵ)\eta(n,\epsilon) such that for δ<η2\delta<\eta^{2} the conclusions of Theorem 1.11 hold for ϵ>0\epsilon>0. It is clear from the definition that if sxη>0s^{\eta}_{x}>0, then there exists ℓ=ℓx\ell=\ell_{x} such that

(sxη)2​⨏Bsxη​(x)|Δ​|ωℓ||=η​⨏Bsxη​(x)|ωℓ|≥12​η​⨏Bsxη​(x)|ω|,(s^{\eta}_{x})^{2}\fint_{B_{s^{\eta}_{x}}(x)}|\Delta|\omega^{\ell}|\,|=\eta\fint_{B_{s^{\eta}_{x}}(x)}|\omega^{\ell}|\geq\frac{1}{2}\eta\fint_{B_{s^{\eta}_{x}}(x)}|\omega|\,, (4.1)

where in the second inequality, we have used that uu is an ϵ\epsilon-splitting to conclude that |ω|≤2​|ωℓ||\omega|\leq 2|\omega^{\ell}|, for ϵ≤ϵ⁡(n)\epsilon\leq\epsilon(n) sufficiently small.

Put

ℬη=:⋃x|sxη>0Bsxη​(x).\mathcal{B}_{\eta}=:\,\bigcup_{x\,|\,s^{\eta}_{x}>0}B_{s^{\eta}_{x}}(x)\,.

Let |u​(Br​(x))||u(B_{r}(x))| denote the (n−2)(n-2)-dimensional measure of the image u​(Br​(x))u(B_{r}(x)). In view of the Transformation theorem, to conclude the proof of the Slicing theorem, it suffices to show

|u⁡(ℬη)|≤δ′​(n,δ),|u(\mathcal{B}_{\eta})|\leq\delta^{\prime}(n,\delta)\,, (4.2)

where δ′​(n,δ)→0\delta^{\prime}(n,\delta)\to 0 as δ→0\delta\to 0.

Let us denote by μ\mu, the measure such that

𝑑μ=(∫B2​(p)|ω|)−1⋅|ω|​d​vg.\displaystyle d\mu=\left(\int_{B_{2}(p)}|\omega|\right)^{-1}\cdot|\omega|dv_{g}\,. (4.3)

Note that μ\mu is a probability measure on B2​(p)B_{2}(p). In the arguments to come we will need μ\mu have a certain doubling property. While in principle, it is too much to ask that μ\mu is actually a doubling measure, next we observe that μ\mu has a partial doubling property which will suffice for our purposes.

Lemma 4.1.

For each xx and 1/2≥r≥sxη1/2\geq r\geq s^{\eta}_{x} we have the doubling condition

μ⁡(B2​r​(x))≤C⁡(n)​μ​(Br​(x)).\mu(B_{2r}(x))\leq C(n)\mu(B_{r}(x))\,. (4.4)
Proof.

By Theorem 1.11, there exists a lower triangular matrix A∈G​L​(n−2)A\in GL(n-2) such that

u′=A∘u:B2​r​(x)→ℝn−2\displaystyle u^{\prime}=A\circ u:B_{2r}(x)\to\mathds{R}^{n-2} (4.5)

is an ϵ\epsilon-splitting. Let d​vgdv_{g} denote the Riemannian measure and set ω′≡d​u′1∧⋯∧d​u′n−2\omega^{\prime}\equiv du^{\prime 1}\wedge\cdots\wedge du^{\prime n-2}. Define the measure μ′\mu^{\prime} by μ′=(∫B2​(p)|ω|)−1​|ω′|​d​vg\mu^{\prime}=\Big(\int_{B_{2}(p)}|\omega|\Big)^{-1}|\omega^{\prime}|dv_{g}. Then

μ′=det(A)​μ.\displaystyle\mu^{\prime}=\det(A)\mu\,. (4.6)

In particular this gives us

μ′​(B2​r​(x))μ′​(Br​(x))=μ​(B2​r​(x))μ​(Br​(x)),\displaystyle\frac{\mu^{\prime}(B_{2r}(x))}{\mu^{\prime}(B_{r}(x))}=\frac{\mu(B_{2r}(x))}{\mu(B_{r}(x))}\,, (4.7)

and it is equivalent to show the ratio bound for μ′\mu^{\prime}. Now since u′u^{\prime} is an ϵ\epsilon-splitting we have the estimate

⨏B2​r​(x)||ω′|−1|≤C⁡(n)​ϵ.\displaystyle\fint_{B_{2r}(x)}|\,|\omega^{\prime}|-1|\leq C(n)\epsilon\,. (4.8)

Hence, we also have the estimate

⨏Br​(x)||ω′|−1|≤Vol​(B2​r​(x))Vol​(Br​(x))​⨏B2​r​(x)||ω′|−1|≤C⁡(n)​ϵ,\displaystyle\fint_{B_{r}(x)}|\,|\omega^{\prime}|-1|\leq\frac{{\rm Vol}(B_{2r}(x))}{{\rm Vol}(B_{r}(x))}\fint_{B_{2r}(x)}|\,|\omega^{\prime}|-1|\leq C(n)\epsilon\,, (4.9)

which of course uses the doubling property for the Riemannian measure. By combining the previous two estimates we get

(1−C​ϵ)​Vol​(Br​(x))\displaystyle\big(1-C\epsilon\big){\rm Vol}(B_{r}(x)) ≤μ′​(Br​(x))≤(1+C​ϵ)​Vol​(Br​(x))\displaystyle\leq\mu^{\prime}(B_{r}(x))\leq\big(1+C\epsilon\big){\rm Vol}(B_{r}(x))\,
(1−C​ϵ)​Vol​(B2​r​(x))\displaystyle\big(1-C\epsilon\big){\rm Vol}(B_{2r}(x)) ≤μ′​(B2​r​(x))≤(1+C​ϵ)​Vol​(B2​r​(x)).\displaystyle\leq\mu^{\prime}(B_{2r}(x))\leq\big(1+C\epsilon\big){\rm Vol}(B_{2r}(x))\,. (4.10)

Finally, by using the definition of μ′\mu^{\prime} we arrive at:

μ′​(B2​r​(x))\displaystyle\mu^{\prime}(B_{2r}(x)) =(∫B2​(p)|ω|​d​vg)−1​∫B2​r​(x)|ω′|\displaystyle=\Big(\int_{B_{2}(p)}|\omega|\,dv_{g}\Big)^{-1}\int_{B_{2r}(x)}|\omega^{\prime}|
≤(1+C⁡(n)​ϵ)​(∫B2​(p)|ω|​d​vg)−1​Vol​(B2​r​(x))\displaystyle\leq(1+C(n)\epsilon)\Big(\int_{B_{2}(p)}|\omega|\,dv_{g}\Big)^{-1}{\rm Vol}(B_{2r}(x))
≤C⁡(n)​(∫B2​(p)|ω|​d​vg)−1​Vol​(Br​(x))\displaystyle\leq C(n)\Big(\int_{B_{2}(p)}|\omega|\,dv_{g}\Big)^{-1}{\rm Vol}(B_{r}(x)) (4.11)
≤C⁡(n)​(∫B2​(p)|ω|​d​vg)−1​∫Br​(x)|ω′|\displaystyle\leq C(n)\Big(\int_{B_{2}(p)}|\omega|\,dv_{g}\Big)^{-1}\int_{B_{r}(x)}|\omega^{\prime}|
=C⁡(n)​μ′​(Br​(x)),\displaystyle=C(n)\mu^{\prime}(B_{r}(x))\,, (4.12)

which by (4.7) completes the proof. ∎

By a standard covering lemma, let us choose a collection of disjoint balls, {Bsj​(xj)}≡{Bsjη​(xj)}\{B_{s_{j}}(x_{j})\}\equiv\{B_{s^{\eta}_{j}}(x_{j})\} such that

ℬη⊂⋃jB6​sj​(xj).\mathcal{B}_{\eta}\subset\bigcup_{j}B_{6s_{j}}(x_{j})\,. (4.13)

For each such ball Bsj​(xj)B_{s_{j}}(x_{j}) let 1≤ℓj≤n−21\leq\ell_{j}\leq n-2 be such that

(sj)2​⨏Bsj​(xj)|Δ​|ωℓj||\displaystyle(s_{j})^{2}\fint_{B_{s_{j}}(x_{j})}|\Delta|\omega^{\ell_{j}}|| =η​⨏Bsj​(xj)|ωℓj|≥12​η​⨏Bsj​(xj)|ω|\displaystyle=\eta\fint_{B_{s_{j}}(x_{j})}|\omega^{\ell_{j}}|\geq\frac{1}{2}\eta\fint_{B_{s_{j}}(x_{j})}|\omega|
≥12​Vol⁡(B1​(xj))Vol⁡(Bsj​(xj))​∫B2​(p)|ω|Vol⁡(B1​(xj))​η​(∫B2​(p)|ω|)−1​∫Bsj​(xj)|ω|,\displaystyle\geq\frac{1}{2}\frac{{\rm Vol}(B_{1}(x_{j}))}{{\rm Vol}(B_{s_{j}}(x_{j}))}\frac{\int_{B_{2}(p)}|\omega|}{{\rm Vol}(B_{1}(x_{j}))}\eta\Big(\int_{B_{2}(p)}|\omega|\Big)^{-1}\int_{B_{s_{j}}(x_{j})}|\omega|\,,
≥C​(n)−1​η​μ​(Bsj​(xj)).\displaystyle\geq C(n)^{-1}\eta\,\mu(B_{s_{j}}(x_{j}))\,. (4.14)

Now since the balls {Bsj​(xj)}\{B_{s_{j}}(x_{j})\} are mutually disjoint, Theorem 1.9, together with (4.14) and Lemma 4.1 (the doubling property of μ\mu) gives

∑(6​sj)−2​μ​(B6​sj​(xj))\displaystyle\sum({6s_{j}})^{-2}\mu(B_{6s_{j}}(x_{j})) ≤C⁡(n)​∑sj−2​μ​(Bsj​(xj))\displaystyle\leq C(n)\sum s_{j}^{-2}\mu(B_{s_{j}}(x_{j}))
≤C⁡(n)​∑∫Bsj​(xj)|Δ​|ωℓj||\displaystyle\leq C(n)\sum\int_{B_{s_{j}}(x_{j})}|\Delta|\omega^{\ell_{j}}|\,| (4.15)
≤C⁡(n)​η−1​∫B3/2​(p)|Δ​|ωℓj||≤δ′​(n,δ).\displaystyle\leq C(n)\eta^{-1}\int_{B_{3/2}(p)}|\Delta|\omega^{\ell_{j}}|\,|\leq\delta^{\prime}(n,\delta)\,. (4.16)

The proof of the Slicing theorem (Theorem 1.8) requires that the image of ℬδ\mathcal{B}_{\delta} under uu have small measure. If in (4.15) the measure μ\mu were instead the usual riemannian measure, then since uu is Lipschitz, standard estimates could be used to show just that. On the face of it, however, the μ\mu-content estimate is much weaker, since for balls where the determinant |ω||\omega| of uu is small we have μ⁡(Br​(x))<<Vol⁡(Br​(x))\mu(B_{r}(x))<<{\rm Vol}(B_{r}(x)).

On the other hand, in the spirit of Sard’s theorem, we will see in the next lemma that at least for balls Br​(x)B_{r}(x) with r≥sxδr\geq s^{\delta}_{x}, we recover this loss because the volume of the image u​(Br​(x))u(B_{r}(x)) is correspondingly small.

Lemma 4.2.

If 1/2≥r≥sxη1/2\geq r\geq s^{\eta}_{x}, then

|u(Br(x)|≤C(n)⋅r−2μ(Br(x)).|u(B_{r}(x)|\leq C(n)\cdot r^{-2}\mu(B_{r}(x))\,. (4.17)
Proof.

As in Lemma 4.1, choose a lower triangular matrix A∈G​L​(n−2)A\in GL(n-2) such that

u′=A∘u:B2​r​(x)→ℝn−2\displaystyle u^{\prime}=A\circ u:B_{2r}(x)\to\mathds{R}^{n-2} (4.18)

is an ϵ\epsilon-splitting and define the measure μ′\mu^{\prime} as in Lemma 4.1. Then as in (4.5), μ′=det(A)​μ\mu^{\prime}=\det(A)\mu.

Since u′u^{\prime} is an ϵ\epsilon-splitting, we have the estimates

⨏B2​r​(x)||ω′|−1|≤C⁡(n)​ϵ,\displaystyle\fint_{B_{2r}(x)}||\omega^{\prime}|-1|\leq C(n)\epsilon\,,
u′​(Br​(x))⊆B2​r​(u′​(x)).\displaystyle u^{\prime}(B_{r}(x))\subseteq B_{2r}(u^{\prime}(x))\,. (4.19)

By the first estimate above,

μ′​(Br​(x))\displaystyle\mu^{\prime}(B_{r}(x)) =(∫B2​(p)|ω|)−1​∫Br​(x)|ω′|,\displaystyle=\Big(\int_{B_{2}(p)}|\omega|\Big)^{-1}\int_{B_{r}(x)}|\omega^{\prime}|\,,
≥(1−C⁡(n)​ϵ)​Vol​(Br​(x))Vol​(B2​(p))​⨏Br​(x)|ω′|\displaystyle\geq(1-C(n)\epsilon)\frac{{\rm Vol}(B_{r}(x))}{{\rm Vol}(B_{2}(p))}\fint_{B_{r}(x)}|\omega^{\prime}|
≥(1−C​ϵ)​Vol​(Br​(x))Vol​(B3​(x))≥C⁡(n)​rn,\displaystyle\geq(1-C\epsilon)\frac{{\rm Vol}(B_{r}(x))}{{\rm Vol}(B_{3}(x))}\geq C(n)r^{n}\,, (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),

|u′​(Br​(x))|≤C⁡(n)​rn−2.\displaystyle|u^{\prime}(B_{r}(x))|\leq C(n)r^{n-2}\,. (4.21)

Combining these gives the estimate

|u′​(Br​(x))|≤C⁡(n)​r−2​μ′​(Br​(x)).\displaystyle|u^{\prime}(B_{r}(x))|\leq C(n)r^{-2}\mu^{\prime}(B_{r}(x))\,. (4.22)

To relate these back to the original function uu, we observe that

|u′​(Br​(x))|\displaystyle|u^{\prime}(B_{r}(x))| =det(A)​|u⁡(Br​(x))|,\displaystyle=\det(A)|u(B_{r}(x))|\,,
μ′​(Br​(x))\displaystyle\mu^{\prime}(B_{r}(x)) =det(A)​|μ⁡(Br​(x))|,\displaystyle=\det(A)|\mu(B_{r}(x))|\,, (4.23)

which immediately gives

|u⁡(Br​(x))|≤C⁡(n)​r−2​μ​(Br​(x)).\displaystyle|u(B_{r}(x))|\leq C(n)r^{-2}\mu(B_{r}(x))\,. (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

|u⁡(ℬη)|≤∑|u⁡(B6​sj​(xj))|≤∑sj−2​μ​(B6​sj​(xj))≤δ′​(n,δ).\displaystyle|u(\mathcal{B}_{\eta})|\leq\sum|u(B_{6s_{j}}(x_{j}))|\leq\sum s_{j}^{-2}\mu(B_{6s_{j}}(x_{j}))\leq\delta^{\prime}(n,\delta)\,. (4.25)

By taking δ\delta sufficiently small, this suffices to complete the proof.

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