To find some fixes to the initial core growth problem we presented last time, let’s refine the split lemma as our first step. We show:

Lemma 1 (\(r\)-Split Lemma): let \( r \in [m] \) divide both \( m \in [n] \) and \( n=|X| \). For any \( {\mathcal F} \subset {X \choose m} \), there exists an \( r \)-split \( {\boldsymbol X}=(X_1, X_2, \ldots, X_r) \) and subfamily \( {\mathcal F}’ \) of cardinality at least \( {n/r \choose m/r}^r |{\mathcal F}| \Big/ {n \choose m} \) such that \[  |U \cap X_j| = \frac{m}{r}, \quad \textrm{for all}~ U \in {\mathcal F}’ ~\textrm{and}~ X_j.  \]

Here an \( r \)-split is a natural generalization of an \( m \)-split, so \( X_j \) are mutually disjoint \( n/r \)-sets as defined last time. The claim generalizes the original split lemma that finds \( |{\mathcal F}’|  \ge  \left( \frac{n}{m} \right)^m  |{\mathcal F}| \Big/ {n \choose m} > |{\mathcal F}| e^{-m}, \) which is given on p.2 of  Paper 1 ver. 1 on the left. Let’s say such an \( {\mathcal F}’ \) is on the \(r\)-split \({\boldsymbol X} \) as well.

Corollary 3.2 on p. 11 of Paper 3 further generalizes this to a primitively weighted \( X \). There, the primitive norm \( \| \cdot \| \) is defined just as the 1/15/24 post by a one-parameter weight function \( w_* \). The whole Section 3 of the paper is about splitting the \(X \) according to a given \( {\mathcal F} \). This could be a tool set for finding some topological properties of \(2^X \).

Our proof of the lemma uses the same technique as the original split lemma, described on the extra page of the 3/3/23 post. The claim given there finds the next \( X_j \) such that the incident subfamily \( {\mathcal F}_{\boldsymbol X} \subset {\mathcal F} \) is large enough meeting \( |U \cap X_i| = 1 \) for all \( i \le j \). For Lemma 1, we change its RHS into \( m/r \).

Let’s see how Paper 3 finds the first \( n/r \)-set \( X_1 \) in the last paragraph of p. 10, where \[ q= \frac{m}{r}, \quad d= \frac{n}{r}, \quad j=0, \quad {\boldsymbol X} = ( \emptyset ), \quad {\mathcal F}_{\boldsymbol X} = {\mathcal F},  \quad \textrm{and} \quad \gamma_{\boldsymbol X} = e^{-\kappa({\mathcal F})}. \]

It is rephrased as follows:

The factor \( \prod_{i=0}^{r-1} {n-di \choose d } \) is the number of \(r \)-splits \({\boldsymbol X} \), so there must be a subfamily \( {\mathcal F}’ \subset {\mathcal F} \) on some \({\boldsymbol X} \) such that \( |{\mathcal F}’ | \ge {d \choose q}^r e^{-\kappa({\mathcal F})} = {d \choose q}^r |{\mathcal F}| \Big/ {n \choose m}. \)

We’ll detail the induction step for any \( j \) to finish proving Lemma 1 in the next post.

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Having finished our proof of Proposition 2.1 of Paper S4 last time, we’ve now seen the details of the rank-\( g\) construction. If we convince ourselves with the other two topics given in the 11/13/23 post, we’ll be able to verify our current target property \( \Gamma(m^{\frac{1}{2}+\epsilon}) \)-condition \( \Rightarrow \) 3 mutually disjoint set in \( {\mathcal F} \), through Proposition 3 in the same post. The claim is explicitly stated as Theorem 1.1 in Paper 3 on the left as well.

Let’s think about how we can prove it. If the given \( {\mathcal F} \) is on an \(m\)-split as in the 3/3/23 post, we can already initiate the rank-\(g\) construction on the family: let \( {\boldsymbol X} = (X_1, X_2, \ldots, X_m) \) be the \(m\)-split, so \( X_i \) are \( m \) mutually disjoint \( n/m \)-sets whose union is the universal set \( X \). Considering the \( \Gamma \)-condition of \( {\mathcal F} \), put \[ m_* = m^{\frac{1}{2}+\epsilon^2}, \quad \textrm{and} \quad X_* = \bigcup_{i=1}^{m_*} X_i.\] This is a situation just the same as the 11/27/23 post. By the following arguments we’ve seen, there are \( (1-\epsilon) {n_* \choose m_*} \) sets \( Y \in {X_* \choose n_*} \), each \( {\mathcal F}_Y \) of which meets the density requirement and exported collective \( \Gamma \)-condition.

Then we select \( k \) mutually disjoint sets \( Y_1, Y_2, \ldots, Y_k \) among those \( Y \) to finish the first step like the 11/6/23 post. Because of it, we achieve the desired disjointness within \( X_* \) for Prop. 3 on 11/13/23. So we can choose the next \( X_* \) to continue the construction.

Those are described in the 7 steps on p. 15-22 of Paper 3, with which we are mostly familiar through our recent discussions. We need to see, however, how the push-up lemma works in Step 6. It is the second topic on 11/13/23. 

When starting proof of Prop 3. on 11/13/23, we are not given such an \( {\boldsymbol X} \), but just an \( {\mathcal F} \) satisfying the \( \Gamma (c m^{\frac{1}{2} +\epsilon} k \ln^2 k) \)-condition. If we apply the split lemma, we can get a subfamily \( {\mathcal F}’ \subset {\mathcal F} \) on an \( {\boldsymbol X} \) such that \( |{\mathcal F}’| > e^{-m} |{\mathcal F}| \). But it doesn’t help here because it may not satisfy a necessary \( \Gamma \)-condition.

We will have to go inside the split lemma to construct a working \(r \)-split \( (X_1, X_2, \ldots, X_r) \) from scratch. Here each \( |X_i| \) should be around \( n m_* / m \) so \( r \simeq \frac{m}{m_*}= m^{\frac{1}{2}- \epsilon^2} \) if \(k=3\). We’ll need Lemma A.1 on p. 10 of Paper S4 to discuss like the last three paragraphs on p. 12. This will give us such an \( r \)-split and \( k \)-mutually disjoint sets \( C_1, C_2, \ldots, C_k \) such that \( {\mathcal F}[C_i] \) includes enough \(m\)-sets \( U \) with \( |U \cap X_i| \simeq m/r  \) for all strips \( X_i \) of \({\boldsymbol X} \), still satisfying a good \( \Gamma \)-condition. Selecting those \(U \) into \( {\mathcal F}_i[C_i] \), we’ll be able to start the rank-\( g\) construction on \( {\mathcal F}_i[C_i] \).

This is about partitioning families, universal sets, and/or included m-sets where we can present some interesting facts. To discuss them, I’ll soon restart publishing weekly.

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We are in the last step to prove Proposition 2.1 of Paper S4. Step 3 on p. 9-10 derives the desired exported collective \( \Gamma \)-condition of most \( {\mathcal F}_Y \), which we’ll confirm in this post as well. Please recall that in the 11/23/24 post and also in Step 1, we checked the truth of the density requirement \( |{\mathcal F}_Y | > \frac{1}{2} {l \choose m_*} {n_* \choose m_*}^{-1} |{\mathcal F}| \) for more than \( (1-\epsilon^2) { n_* \choose \epsilon n_*} \) sets \( Y \in {X_* \choose l} \).

So we can apply the \( g^{th}\) mark lemma to \( {\mathcal H} \) with the \( \Gamma_g \)-condition. Since \( \sqrt c \) is much larger than \( \epsilon^{-2} \):

With both i) and ii) on the bottom of p. 7 confirmed, we’ve now finished our proof of Proposition 2.1. By this, we demonstrated that the collective \( \Gamma \)-condition can be exported to most \( {\mathcal F}_Y \). The method based on the rank-g construction can be effective to prove Proposition 3 given in the 11/13 post. Its confirmation is our current goal that implies \( \Gamma (m^{\frac{1}{2} + \epsilon} ) \) of \( {\mathcal F} \) \( \Rightarrow \) three disjoint sets in \( {\mathcal F} \).

We’ll start our next topic next time.

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I’ve changed my topic plan slightly. Please open the index page. We’ll finish the review of Paper S4 in the next post. To fully convince ourselves of the main claim of Paper 3, we’ll discuss the two topics other than the rank-g construction as mentioned in the 11/13/23 post: 2. the push-up lemma, and 3. initial core growth. The third topic is about splitting sets and families involving Lemma A.1 of Paper S4 and the split lemma given in the 3/3/23 post, appropriate for the last topic of the current chapter.

Then I’ll explain topic 2 and main recursive steps in the next chapter to finish presenting all the materials in Paper 3. I’ll publish weekly again during summer.

Back to Step 2 of the main construction for Proposition 2.1, please recall the terminology defined last time. We confirm the \( \Gamma_g(b_*, h) \)-condition in the rest of this post.

Note about C) that \[ m_*= \frac{m n_*}{n}, \quad b>m^{1+\epsilon} \frac{n_*}{n}=m_* m^{\epsilon}, \quad \textrm{and} \quad \frac{1}{g} < \epsilon^2, \] by definition. So \( b_\dagger= \frac{b^{1-1/g}}{m^{1/g}} \) is greater than \( m_* m^{\epsilon/2} \), which makes sense for us to consider the \( \Gamma( b_\dagger, h_T ) \)-condition there. Also \( b_* = \frac{b_\dagger}{2^g (g-1) m_*} \) in D) is greater than any constant because so is m.

Next time, we’ll apply the obtained \( \Gamma_g\)-condition to the \( g^{th} \)-mark lemma to confirm the desired exported collective \( \Gamma \)-condition.

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Having finished our proof of the conversion lemma last time, we are done with the first three tasks of the four listed in the 12/11/23 post. Also recalling our proof scenario of Proposition 2.1 discussed in the 11/27/23 post, it remains to carry out the last task 4 correctly before completing all of the proof: given \( \Gamma(2b) \) of \({\mathcal F} \), we are trying to show that most \( {\mathcal F}_Y \) meet the density requirement, and the exported collective \( \Gamma \)-condition. By this driving force, we can show Proposition 3 given in the 11/13/23 post, which implies \( \Gamma( m^{\frac{1}{2} +\epsilon} ) \) of \( {\mathcal F} \) \( \Rightarrow \) three mutually disjoint sets in it.

The two key properties are shown by the three steps described on p. 8-10 of Paper S4. Step 1 checks the density requirement, and as the 11/27/23 post figured out, it’s our familiar application of the weighted EGT. We just recall here that the EGT is applied to \( {\mathcal H}={X_* \choose m_*} \) for the given \( n_* \)-set \( X_* \), then most \( {\mathcal F}_Y = {\mathcal F} \cap {X – X_* \cup Y \choose m} \) meet the density requirement.

For the second property, we define the following weight with the given sufficiently large constant g:

This is our outcome for the fourth task. In the next two posts, we’ll check the truth of the \( \Gamma_g(b_*, h) \)-condition, and that the \(g^{th} \) mark lemma with it implies the desired collective \( \Gamma\)-condition as in Step 3.

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Let’s finish proving the conversion lemma in this post. We’re confirming (22) given last time. Assume true for g-1 and prove for g. By induction hypothesis:

The second line is justified by (1′) and the figure: the induction hypothesis considers \( {\boldsymbol U}=(U_1, U_2, \ldots, U_{g-1} ) \in {\mathcal P}_{k, g-1} \), so \( Union ({\boldsymbol U} ) = (g-1)m-k \) by definition. Add the last element \( U_g \in {\mathcal F} \) to \( {\boldsymbol U}\) such that \( |Union({\boldsymbol U}) \cap U_g| = j-k \). There are \( {(g-1)m – k \choose j-k}\) or less choices of such (j-k)-intersection, each producing the weight at most \( h b^{j-k} \| {\mathcal F} \|_* \) by the primitive \( \Gamma(b, h) \)-condition of \({\mathcal F} \). So we get the second line.

The third line is due to \( b_g=2^{g-2} b \) and the note below it.

We have proven the conversion lemma.

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We clarified the role of the conversion lemma in the rank-g construction last time. It’ll be used to covert a \( \Gamma(b, h) \)-condition given on one-parameter weight \( w_* \) to a \( \Gamma_g \)-condition on the g-parameter weight w primitive with \( w_* \), i.e., w: \( (U_1, U_2, \ldots, U_g ) \mapsto \prod_{i=1}^g w_*(U_i) \). Then we can apply it to \( g^{th} \) mark lemma in order to export the collective \( \Gamma \)-condition.

The lemma on p. 8 is rephrased into:

The Conversion Lemma: if \( \| \cdot \| \) is primitive with \( \| \cdot \|_* \), \[ \Gamma(b, h)~\textrm{of}~ {\mathcal F} ~\textrm{on}~ \| \cdot \|_* ~\textrm{with}~h \ge 1, \qquad \Rightarrow \qquad \Gamma_g \left[ \frac{b}{2^{g-2} (g-1) m} ,~h^{g-1} \right]~\textrm{of}~ {\mathcal F} ~\textrm{on}~ \| \cdot \|. \]

Let’s start its proof. We verify \[ (22) \qquad \| {\mathcal P}_{j, g} \| < h^{g-1} b_g^{-j} {(g-1)m \choose j} \| {\mathcal F} \|^g, \quad \forall j \in [(g-1)m], \quad \textrm{where}~b_g=2^{-g+2} b, \] by induction on \( g \ge 2\) noting that:

• (22) means the target \( \Gamma_g \)-condition because \( b_g^{-j} {(g-1)m \choose j} \le \left[ \frac{b_g}{(g-1)m} \right]^{-j} \).
• Paper S4 shows the same for \( g’=[2, g] \). We just confirm it by induction on g here without g’. This simplifies the notation slightly.
• For a given fixed \( w_* \) inducing \( \| \cdot \|_* \), we show (22) for the \( w \) primitive with \( w_* \), for which we may add the subscript g to write \( w_g \) and \( \| \cdot \|_g \). It’s necessary to distinguish it from the norms for g-1 in the induction step.
• To distinguish between the original \( \Gamma \) and the \( \Gamma_g \)-condition, we call the former primitive \(\Gamma \)-condition.
• As said before, the \( \Gamma_g \)-condition must exclude any \( U \not \in {\mathcal F} \) from the norm calculation. It’s achieved by \( U \not \in {\mathcal F} \Rightarrow w_*(U)=0 \) implied by the primitive \( \Gamma \)-condition.

Then prove (22) for the basis \( g=2 \) as follows.

We’ll prove the induction step next time.

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Last time, we proved the \( g^{th} \) mark lemma on p. 2 of Paper S4. It is the generalization of the double mark lemma for the rank-g construction. By this we completed the second task of the four listed in the 12/11/23 post.

The \( g^{th} \) mark lemma is stronger implying the whole statement of the double mark lemma for g=2. It is shown by the inequality \( {n \choose m}^g {n-gm \choose l-gm} \le {n \choose l}{l \choose m}^g \) tautologically true for every possible l, m, n and g. The proof is simple not counting \( g^{th} \)-marks along a combinatorial identity like we did for the double mark lemma. The equation given in Remark C) on p.3 might lead to something else that is interesting.

Despite the simplicity of the new proof, it is interesting to find the identity (6) in the 8/25/23 post to count double marks along its algebraic proof, with the accurate approximation of \( {x-y \choose y-j} \) shown in the 9/9/23 post. I think both ways are equally good to know.

Before moving to the third task to prove the conversion lemma, let’s look at Section 1.3. It’s to connect the rank-g weight \( w: (2^X)^g \rightarrow {\mathbb R}_{\ge 0} \) to the g=1 weight \( 2^X \rightarrow {\mathbb R}_{\ge 0} \) we dealt with in Paper S3. We say that \( w \) is primitive if it is defined by \[ (U_1, U_2, \ldots, U_g ) \mapsto \prod_{i=1}^g w_*(U_i), \qquad \textrm{for some}~~ w_* :~2^X \rightarrow {\mathbb R}_{\ge 0}. \]

If so, \( w \) is essentially the same as \( w_* \) because for any family \( {\mathcal G} \subset 2^X \), \[ \| {\mathcal G} \|^g = \sum_{\boldsymbol{U} \in {\mathcal G}^g } w( \boldsymbol{U} ) = \sum_{(U_1, U_2, \ldots, U_g) \in {\mathcal G}^g } ~~\prod_{i=1}^q w_*( U_i) = \left[ \sum_{U \in {\mathcal G}} w_*(U) \right]^g = \| {\mathcal G} \|_*^g, \] meaning \( \| {\mathcal G} \| = \| {\mathcal G} \|_* \) where \( \| \cdot \|_* \) is the norm induced by \( w_* \).

Then we can define the weight of a mark \( (U, Y) \) to be \( w_*(U) \), so the family \( {\mathcal M} \) of marks has the norm \[ \| {\mathcal M} \| = \sum_{{Y \choose l}} \left\| {Y \choose m} \right \| = \| {\mathcal F} \| {n-m \choose l-m} = {n \choose m}{n-m \choose l-m} e^{-\kappa({\mathcal F})} = {n \choose m}{l \choose m} e^{-\kappa({\mathcal F})} \] as before and G) on p. 5 due to \( \| \cdot \| = \| \cdot \|_* \).

By this we can make the same quadratic optimization arguments for g=2 started at the 9/17/23 post. So we have Theorem 1.3 on p. 4 that is almost the same as Theorem 2.1 of Paper S3 we proved before. It says that if \( {\mathcal F} \) satisfies the \( \Gamma_2( \frac{4 \gamma n}{l}, 1 ) \)-condition, almost all \( l \)-sets Y have approximate norm \( {l \choose m} \| {\mathcal F} \| \Big/ {n \choose m} \).

We’ll do the next task to prove Lemma 1.4 on primitive weight. Although such weight is similar to the \( w_* \), we can accumulate them to make a non-primitive weight. That’s what we did for g=2: in the the 11/20/23 post , we designed the 2-parameter weight \(w \) such that \( w ( {\mathcal F}_Y ) = \sum_{T \subset X – X_,~T \ne \emptyset} ~~ \frac{|{\mathcal F}_Y[T]|^2}{b^{-|T|} {m \choose |T|}} \). You can see that the summand for each T is primitive, but the total weight may not be. We’ll covert the given plain \( \Gamma \)-condition to a \( \Gamma_g \)-condition for each primitive part of the total weight by the lemma. Step 2 on p. 8 accumulates them to show the \( \Gamma_g \)-condition as a whole. Then apply it to the \( g^{th} \) mark lemma to export the collective \( \Gamma \)-condition.

the 11/20/23 post .

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Recently, my teaching schedule changed and I’m preparing materials for a new course. So, I probably don’t publish new posts during the coming spring semester as often as before, but at least once a couple of weeks. I’ll publish weekly again after the semester is over.

Also, I’m working on fixing Paper 1 having found a new way to understand the \( \hat \Gamma \)-family. I’m planning to change Paper 2 into an overview of Paper 3 with the same organization as Paper S4 adding slightly stronger results. I hope I can upload the second versions of Papers 1 and 2 on arXiv soon.

Back to our review of Paper S4, let’s prove the \( g^{th} \) mark lemma on p. 2 in this post. Given the \( \Gamma_g(b, h) \) condition \( \| {\mathcal P}_{j, g} \| < h b^{-j} \| {\mathcal F} \|^g \), we want to tightly bound the number of \( g^{th} \) marks \( ( {\boldsymbol U}, Y ) \) close to the maximum \( {n \choose l} {l \choose m}^g e^{-g \kappa({\mathcal F} )} \), or their weight sum.

As in Remark A) we saw last time, each \( {\boldsymbol U} \in {\mathcal P}_j \) independently produces the weight \( w( {\boldsymbol U} ) {n – gm+ j \choose l-gm+j} \) in the domain of all \( g^{th} \) marks: by the definition of \( {\mathcal P}_j \), the union of \( U_i \), i.e., \( Union( {\boldsymbol U}) = U_1 \cup U_2 \cup \cdots \cup U_g \), has the cardinality \( gm-j \), so it’s included in \( {n – gm+ j \choose l-gm+j} \) sets \( Y \in {X \choose l} \).

By the \(\Gamma_g \)-condition, the total weight of the \( g^{th} \) marks \( ( {\boldsymbol U}, Y ) \) with j> 0 is \[ \sum_{j=1}^{(g-1)m} \| {\mathcal P}_j \| {n – gm+j \choose l – gm+j} \le \sum_{j=1}^{(g-1)m} h b^{-j} \| {\mathcal F} \|^g {n – gm+j \choose l – gm+j} \] \[ = h \sum_{j=1}^{(g-1)m} b^{-j} {n \choose m}^g e^{- g\cdot \kappa({\mathcal F})} {n – gm+j \choose l – gm+j}, \] since \( \| {\mathcal F} \| = {n \choose m} e^{- \kappa({\mathcal F})} . \)

In there:

• \( {n – gm+ j \choose l – gm+j} < \left( \frac{2n}{l} \right)^j { n – gm \choose l – gm} \) as in Remark E) on p. 3.
  • Because the LHS equals \( {n -gm \choose l-gm} \prod_{j’=1}^j \frac{n-gm+j’}{l-gm+j’} \) by the identity \( {x \choose y} = \frac{x}{y} {x-1 \choose y-1} \).
  • \( \frac{n-gm+j’}{l-gm+j’} < \frac{2n}{l} \) since \( n \) and \( l \) are both sufficiently larger than \( gm \) as assumed by the lemma.
• \( {n \choose m}^g {n-gm \choose l-gm} \le {n \choose l}{l \choose m}^g \) as in C).
  • C) shows a stronger result that might be useful for another claim. (We’ll check this much later.)
  • However, the inequality itself is proven by the footnote of p. 3 that counts the total number of \( g^{th} \) marks when \( {\mathcal F} = {X \choose m} \).
  • We did this in the 9/2/23 post for g=2. The argument is essentially the same here.
  • This is a tautological inequality holding whenever \( 0 \le m \le l \le n \) and \( g \ge 1 \).

Therefore with \( b= 4 \gamma n / l \):

as on p.4 proving (1.1) above it, and the lemma.

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After proving Proposition 2.1 of Paper S4 for g=2, we found a way to generalize the method to a larger rank g last time. On a g-parameter weight \( w \) with the associated norm \( \| \cdot \| \) and sparsity \( \kappa (\cdot ) \), we want to consider \( g^{th} \) marks \( ( {\boldsymbol U}, Y) \), the neighbor tuple families \( {\mathcal P}_{j, g} \), and the new \( \Gamma_g(b, h) \)-condition defined on p. 1-2. These will support our method of the rank-g construction for proving Prop. 2.1 completely.

Among the four tasks presented, we’ve finished the first one with the defined \( \Gamma_g(b, h) \)-condition that mainly cares about the union of the \( U_i \) in \( {\boldsymbol U} \) for manageable proof complexity. The notation \( Union ({\boldsymbol U} ) = \bigcup_{i=1}^g U_i \) used there is something we should keep in mind afterward. The family \( {\mathcal P}_{j, g} \) collects \( {\boldsymbol U} = (U_1, U_2, \ldots, U_g) \in {\mathcal F}^g \) such that \( Union({\boldsymbol U}) = gm-j \), consistently with the case g=2. The \( \Gamma_2 \)-condition requires \( \| {\mathcal P}_{j, g} \| < h b^{-j} \| {\mathcal F} \|^g \) for \( j \in [m] \). Its RHS has no binomial coefficient such as \( {(g-1)m \choose j} \), instead of which the considered b value is \( 4 \gamma n / l \) unlike \( b= 5 \gamma n m / l \) so far.

Let’s move to the second task to generalize the double mark lemma. We want to show that the \( \Gamma_g \)-condition means $$ \| {\mathcal D}_g \| < \frac{\left( 1+ \frac{h}{\gamma} \right){n \choose l}{l \choose m}^g}{{n \choose m}^g} \| {\mathcal F} \|^g = \left( 1+ \frac{h}{\gamma} \right){n \choose l}{l \choose m}^g e^{-g \cdot \kappa ({\mathcal F})}, $$ where \( {\mathcal D}_g \) is the family of \( g^{th} \) marks \( ( {\boldsymbol U}, Y) \), and its norm is the sum of \( w( {\boldsymbol U} ) \) for all \( ( {\boldsymbol U}, Y) \). Lemma 1.1 on p. 2 states this.

We may drop the subscript g when its value is clear, so we can write \( {\mathcal D} \), \( {\mathcal P}_j \), etc. just as before. Below the description of the \( g^{th} \) mark lemma, the paper says:

Those are common with g=2 being clear except for \( j \le (g-1)m \): the subscript j of \( {\mathcal P}_{j, g} \) comes from \( Union({\boldsymbol U}) = gm-j \), so j must not exceed \( (g-1) m\).

Since \( \| {\mathcal D} \| \) is the sum of \( \left\| {Y \choose l} \right\|^g =w \left( {\mathcal F}_Y \right) \) for all l-sets Y, the \( g^{th} \) mark lemma implies that most Y meet \( w \left( {\mathcal F}_Y \right) < \epsilon^{-2} \left( 1+ h \gamma^{-1} \right) {l \choose m}^g {n \choose m}^{-g} \| {\mathcal F} \|^g \).

By the weighted EGT, most Y satisfy the density requirement \( w( |{\mathcal F}_Y ) > \frac{1}{2} {l \choose m_*} {n_* \choose m_*}^{-1} |{\mathcal F}| \) with \( n \rightarrow n_* \) and \( m \rightarrow m_* \) as the 11/27 post. So if the g-parameter weight \( w \) is set as Step 2 on p. 8, we will have most Y to meet both the density requirement and exported \(g^{th} \) collective \( \Gamma \)-condition. That’s what Step 3 concludes on p. 10 finishing the proof of Prop. 2.1.

We’ll prove the \( g^{th} \) mark lemma next time.

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