ScalingStacks

Proof of Estimate ( 1.9 ) of Theorem 1.5 . [01ZH]

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Proof of Estimate (1.9) of Theorem 1.5.

Let MnM^{n} satisfy |RicM4|≤3|{\rm Ric}_{M^{4}}|\leq 3 and Vol⁡(B1​(p))>v>0{\rm Vol}(B_{1}(p))>{\rm v}>0. We will prove the estimate for the harmonic radius rhr_{h}. The same argument works in the Einstein case to control the regularity scale.

So let 0<ϵ<<10<\epsilon<<1 be fixed with δ⁡(n,ϵ)\delta(n,\epsilon) chosen to satisfy Theorem 8.3. Consider the set

{x∈B1​(p):rh​(x)<r}.\displaystyle\{x\in B_{1}(p):r_{h}(x)<r\}\,. (8.19)

In view of the doubling condition implied by the Bishop-Gromov inequality, we have by a standard construction that there exists a covering {Br​(xj)}1N\{B_{r}(x_{j})\}_{1}^{N} with

{x∈B1​(p):rh​(x)<r}⊆⋃jBr​(xj),\displaystyle\{x\in B_{1}(p):r_{h}(x)<r\}\subseteq\bigcup_{j}B_{r}(x_{j})\,, (8.20)

but such that {Br/4​(xj)}\{B_{r/4}(x_{j})\} are disjoint. Such coverings, which we will term “efficient”, will be constructed several times below. Note that

Tr​({x∈B1​(p):rh​(x)<r})⊆⋃jB2​r​(xj),\displaystyle T_{r}\big(\{x\in B_{1}(p):r_{h}(x)<r\}\big)\subseteq\bigcup_{j}B_{2r}(x_{j})\,, (8.21)

and thus

Vol⁡(Tr​({x∈B1​(p):rh​(x)<r}))≤∑1NVol⁡(B2​r​(xj))≤C⁡(n)​N⋅r4.\displaystyle{\rm Vol}\Big(T_{r}\big(\{x\in B_{1}(p):r_{h}(x)<r\}\big)\Big)\leq\sum_{1}^{N}{\rm Vol}(B_{2r}(x_{j}))\leq C(n)N\cdot r^{4}\,. (8.22)

Hence, our goal is to control the number of balls NN in the covering. Denote by 𝒞≡{xj}1N\mathcal{C}\equiv\{x_{j}\}_{1}^{N} this collection of points.

Now note the following: if xjx_{j} is one of our ball centers and Tα​(xj)=0T_{\alpha}(x_{j})=0, then by Theorem 8.3 we have for every x∈Arα/2,2​rα​(xj)x\in A_{r_{\alpha}/2,2r_{\alpha}}(x_{j}) that rh​(x)>r¯​(v)⋅rαr_{h}(x)>\bar{r}({\rm v})\cdot r_{\alpha}. In particular, if rα>r¯−1​rr_{\alpha}>\bar{r}^{-1}r, this implies that

xk∉Arα/2,2​rα​(x).\displaystyle x_{k}\not\in A_{r_{\alpha}/2,2r_{\alpha}}(x)\,. (8.23)

Now let us inductively build a sequence of decreasing subsets 𝒞k+1⊆𝒞k⊆⋯⊆𝒞\mathcal{C}^{k+1}\subseteq\mathcal{C}^{k}\subseteq\cdots\subseteq\mathcal{C} and associated radii sk=rαk>0s_{k}=r_{\alpha_{k}}>0 with diam⁡(𝒞k)<4​sk{\rm diam}(\mathcal{C}^{k})<4s_{k}. There are three key inductive properties that will be proved about these sets:

  1. (1)

    There exists C⁡(n)>0C(n)>0 such that the cardinality of 𝒞k\mathcal{C}^{k} satisfies

    |#​𝒞k|≥C−k​|#​𝒞|=C−k​N.\displaystyle\big|\#\mathcal{C}^{k}\big|\geq C^{-k}\big|\#\mathcal{C}\big|=C^{-k}N\,. (8.24)
  2. (2)

    For every xjk∈𝒞kx^{k}_{j}\in\mathcal{C}^{k} we have

    ∑0≤α≤αkTαδ​(xjk)≥k.\displaystyle\sum_{0\leq\alpha\leq\alpha_{k}}T^{\delta}_{\alpha}(x^{k}_{j})\geq k\,. (8.25)
  3. (3)

    If |#​𝒞k|>1\big|\#\mathcal{C}^{k}\big|>1 and sk>r¯−1​rs_{k}>\bar{r}^{-1}r then 𝒞k+1≠∅\mathcal{C}^{k+1}\neq\emptyset.

Before constructing the sequence of sets, let us see that once the construction is complete, we will have proved our desired estimate on NN. Indeed, let kk be the largest index such that 𝒞k≠∅\mathcal{C}^{k}\neq\emptyset. By the third property we must have either |#​𝒞k|=1|\#\mathcal{C}^{k}|=1 or sk≤r¯−1​rs_{k}\leq\bar{r}^{-1}r, at which point we get by a covering argument that |#​𝒞k|<C⁡(n)|\#\mathcal{C}^{k}|<C(n). By Lemma 8.5 and the second property we have that k≤k⁡(n,v,δ)=k⁡(n,v)k\leq k(n,{\rm v},\delta)=k(n,{\rm v}), and thus by the first property we have

N≤C​(n)k⁡(n,v)⋅|#​𝒞k|≤C⁡(n,v),\displaystyle N\leq C(n)^{k(n,{\rm v})}\cdot|\#\mathcal{C}^{k}|\leq C(n,{\rm v})\,, (8.26)

which proves the result.

Now let 𝒞0≡𝒞\mathcal{C}^{0}\equiv\mathcal{C} with s0=1s_{0}=1. Clearly, the inductive properties hold for 𝒞0\mathcal{C}^{0}. Assume we have built 𝒞k⊆𝒞\mathcal{C}^{k}\subseteq\mathcal{C} with sk>0s_{k}>0 satisfying the inductive properties, and let us build 𝒞k+1\mathcal{C}^{k+1}. First note that if |#​𝒞k|=1|\#\mathcal{C}^{k}|=1 or sk≤r¯−1​rs_{k}\leq\bar{r}^{-1}r, then we let 𝒞k+1=∅\mathcal{C}^{k+1}=\emptyset. Our construction will otherwise give us a nonempty 𝒞k+1\mathcal{C}^{k+1}, so that the third inductive property will automatically be satisfied. So let us denote sk′=diam⁡(𝒞k)⋅2−10s^{\prime}_{k}={\rm diam}(\mathcal{C}^{k})\cdot 2^{-10}. Choose an efficient covering {Bsk′​(xjk)}\{B_{s^{\prime}_{k}}(x^{k}_{j})\}, where xjk∈Skx^{k}_{j}\in S^{k}, so that the balls in {Bsk′/4​(xjk)}\{B_{s^{\prime}_{k}/4}(x^{k}_{j})\} are disjoint. Note that because diam⁡(𝒞k)<4​sk{\rm diam}(\mathcal{C}^{k})<4s_{k}, the usual doubling estimates imply that there are at most C⁡(n)C(n) balls in this covering. We choose the ball Bsk′​(y)B_{s^{\prime}_{k}}(y) such that 𝒞k∩Bsk′​(y)\mathcal{C}^{k}\cap B_{s^{\prime}_{k}}(y) has the largest cardinality of any ball from the covering. Then we define 𝒞k+1=𝒞k∩Bsk′​(y)\mathcal{C}^{k+1}=\mathcal{C}^{k}\cap B_{s^{\prime}_{k}}(y).

By that by our choice of ball, Bsk′​(y)B_{s^{\prime}_{k}}(y), we have

|#​𝒞k+1|\displaystyle\big|\#\mathcal{C}^{k+1}\big| =|#​𝒞k∩Bsk′​(y)|≥C​(n)−1​|#​𝒞k|≥C−(k+1)​N,\displaystyle=\big|\#\mathcal{C}^{k}\cap B_{s^{\prime}_{k}}(y)\big|\geq C(n)^{-1}\big|\#\mathcal{C}^{k}\big|\geq C^{-(k+1)}N\,, (8.27)

so that 𝒞k+1\mathcal{C}^{k+1} satisfies the first inductive property. To find sk+1s_{k+1} and prove the second inductive property, let us define the following. For each xjk+1∈𝒞k+1x^{k+1}_{j}\in\mathcal{C}^{k+1} if

Tαk+7δ​(xjk+1)=1,\displaystyle T^{\delta}_{\alpha_{k}+7}(x^{k+1}_{j})=1\,, (8.28)

then let us set βj=αk+7\beta_{j}=\alpha_{k}+7, and otherwise let βj≥αk+8\beta_{j}\geq\alpha_{k}+8 be the largest integer such that Tβj−1δ​(xjk+1)=0T^{\delta}_{\beta_{j}-1}(x^{k+1}_{j})=0 but Tβjδ​(xjk+1)=1T^{\delta}_{\beta_{j}}(x^{k+1}_{j})=1. Note that Bsk′​(y)⊆Brαk+7​(xjk)B_{s^{\prime}_{k}}(y)\subseteq B_{r_{\alpha_{k}+7}}(x^{k}_{j}). Let αk+1≡max⁡{βj,⌈−log2⁡(r¯​r−1)⌉}\alpha_{k+1}\equiv\max\{\beta_{j},\lceil-\log_{2}\big(\bar{r}r^{-1}\big)\rceil\} with xk+1∈𝒞k+1x^{k+1}\in\mathcal{C}^{k+1} the associated element which attains the maximum, and note by (8.23) that

𝒞k+1=𝒞k∩Bsk′​(y)=𝒞k∩B2−αk+1+1​(xk+1).\displaystyle\mathcal{C}^{k+1}=\mathcal{C}^{k}\cap B_{s^{\prime}_{k}}(y)=\mathcal{C}^{k}\cap B_{2^{-\alpha_{k+1}+1}}(x^{k+1})\,. (8.29)

In particular, with sk+1=rαk+1s_{k+1}=r_{\alpha_{k+1}} then diam⁡(𝒞k+1)<4​sk+1{\rm diam}(\mathcal{C}^{k+1})<4s_{k+1}, and the second inductive property holds, which completes the induction step of the construction, and hence, the proof.

∎

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