Let’s see the main idea behind the proof of the sunflower conjecture in Paper 1. As seen last time, the given family \({\mathcal F}\) of m-sets is on an m-split \( {\boldsymbol X} =(X_1, X_2, \ldots, X_m) \), i.e., \( {\mathcal F} \subset {{\boldsymbol X} \choose m} \). We can also assume:

• \( k \ge 3 \).
• \( {\mathcal F} \) satisfies the \( \Gamma (c^c k \ln k) \)- condition meeting \( |{\mathcal F}| > (c^c k \ln k)^m \).
• \( m > c^c \ln k \).

See their reasoning in Extra Notes and Errata for Paper 1 on the left.

To explain the main intuition behind, let me present one false proof scenario of the sunflower conjecture. It uses the weighted variant of EGT that is Theorem 1.2 of Papers 2 and S1. Let’s say k=3. In the induction step of the hypothetical recursive scenario, we are given three subfamilies \( {\mathcal F}_i \subset {\mathcal F} \) for i=1, 2, 3 that are elementwise disjoint on \( {\boldsymbol X}_j = (X_1, X_2, \ldots, X_j ) \), i.e., $$ U_i \cap U_{i’} \cap Union ({\boldsymbol X}_j )= \emptyset, $$ for any \( U_i \in {\mathcal F}_i \) and \( U_{i’} \in {\mathcal F}_{i’} \) with \( i \ne i’ \). In the jth induction step, we try to show the mutual disjointness of sunflower petals on the j-subsplit \( {\boldsymbol X}_j\) of \( {\boldsymbol X} \) only. If this works, the last induction step on \( {\boldsymbol X}_m = {\boldsymbol X} \) would show a 3-sunflower included in \( {\mathcal F} \) with an empty core.

The root of the proof tries to show that any \( {\mathcal F} \) satisfying the \( \Gamma(c) \)-condition includes a 3-sunflower. The jth induction step would confirm there are three subfamilies \( {\mathcal F}_i \subset {\mathcal F} \) elementwise disjoint on \( {\boldsymbol X}_j \), by the following argument:

1. We’re given three \( {\mathcal F}_i \) elementwise disjoint on \( {\boldsymbol X}_{j-1} \) by the j-1st induction step, working to show the same on \( {\boldsymbol X}_j \). (If j=1, \( {\mathcal F}_i \) are all \( {\mathcal F} \) with an empty \( {\boldsymbol X}_0. \) )
2. The induction hypothesis also assumes each \( {\mathcal F}_i \) satisfies the \( \Gamma(c) \)-condition.
3. With the variant of EGT, the jth strip \( X_j \) of \( {\boldsymbol X} \) can be split into three mutually disjoint sets \( Y_i \) such that $$ \left| {\mathcal F}_i \cap {X – X_j \cup Y_i \choose m} \right| \simeq \frac{1}{4} |{\mathcal F}_i|. $$
   • \( X – X_j \cup Y_i \) means to exclude \( X_j – Y_i \) from consideration for \( {\mathcal F}_i \). The new \( {\mathcal F}_i \) after the jth step will be \( {\mathcal F}_i \cap {X – X_j \cup Y_i \choose m} \) that’ll be each large enough.
   • This way the three \( {\mathcal F}_i \) will be elementwise disjoint on \( {\boldsymbol X}_j\) after the jth step.
   • \( |Y_i| = |X_i|/4 = l \) and there are many such \( Y_i \in {X_j \choose l} \) in a family of l-sets found by the weighted EGT variant with the \( \Gamma(c) \)-condition.
   • Because such \( Y_i \) are a majority of \( {X_j \choose l} \), we can find three mutually disjoint l-sets \( Y_1, Y_2, Y_3 \) just as in the extra notes of this post for “finding a k-sunflower with the core S”.
4. Now each new \( {\mathcal F}_i \) is about 1/4 of the previous one. So it might not satisfy the \( \Gamma(c) \)-condition we need for the next step. Show that it does by the induction hypothesis of the root proposition.
   • Find a maximal set S such that \( |{\mathcal F}_i[S]| \ge c^{-|S|} |{\mathcal F}_i| \).
   • If S is nonempty, the family \( {\mathcal F}_i[S] \) satisfies the \( \Gamma(c) \)-condition in X-S, so it includes a 3-sunflower by the induction hypothesis of the root proposition \( \Gamma(c) \) \( \Rightarrow \) \( 3SF \subset {\mathcal F} \). (Here “3SF” means a 3-sunflower. The root is being proven by induction on m on the outer level.)
   • So there’s no nonempty S such that \( |{\mathcal F}_i[S]| \ge c^{-|S|} |{\mathcal F}_i| \), which proves the \( \Gamma(c) \)-condition.

It may sound plausible and actually has some potential, but this argument is fundamentally flawed. The error is in 4: when we look for a maximal S such that \( |{\mathcal F}_i[S]| \ge c^{-|S|} |{\mathcal F}_i| \), it’s possible that |S|=m. This means \( S \in {\mathcal F}_i \) so \( 1 = |{\mathcal F}_i[S]| \ge c^{-|S|} |{\mathcal F}_i| \) \( \Rightarrow \) \( |{\mathcal F}_i| \le c^m \), just making the \( \Gamma \)-condition false giving us no meaningful \( {\mathcal F}_i[S] \) that includes a 3-sunflower.

To prevent such a situation from happening, you must guarantee \( |{\mathcal F}_i| > c^m \) at the beginning of the jth step, but you can’t: possibly, you start the 1st step with \( |{\mathcal F}| \simeq c^m \). In the middle at the jth step, you have \( |{\mathcal F}_i| \) much smaller than \(c^m \) due to the reduction factor 1/4 for each strip.

Is this a small error we can fix with a patch? The root proposition \( \Gamma(c) \) \( \Rightarrow \) \( 3SF \subset {\mathcal F} \) is being proven by outer induction on m. You can’t help use the induction hypothesis for some \( {\mathcal F}_i[S] \), but unable to meet the requirement \( |{\mathcal F}_i[S]| > c^{m-|S|} \) when the recursion goes deeper. No, this is not a small error you can have a quick fix for. It’s a fundamental flaw.

Everything but 4 is doable. Arguments 1-3 are actually almost the same as Step 2-4 described on p.7-8 of Paper 2. This gap is a part of the elusive difficulty of proving the sunflower conjecture, and I’ve been calling it cardinality bounding problem (CBP): \( {\mathcal F}_i[S] \) for an S is too big relative to \( {\mathcal F}_i \) for an intrinsic reason so it doesn’t sustain the necessary \( \Gamma \)-condition.

To address the CBP, Paper 1 preemptively detects all maximal S such that \( {\mathcal F}[S] \) meet a \( \Gamma \)-condition in X- S, putting them into a recycle bin, say \( \hat {\mathcal F} \subset {\mathcal F} \). (Do it while \( {\mathcal F} \) is big enough to avoid CBP.) If \( \hat {\mathcal F} \) is sufficiently large, focus on the subfamily to detect a 3-sunflower in it, calling S base sets. Outside the obtained family \( {\mathcal B} \) of base sets, some \( \Gamma \)-condition of \( \hat {\mathcal F} \) is available so we can do the arguments 1-3. Inside \( {\mathcal B} \), you need to do the same operations recursively.

This is why we consider the \( \Gamma(b) \)-condition on \( {\boldsymbol X}’ \setminus S \) within \( {\mathcal B} \) as we mentioned last time. If \( \hat {\mathcal F} \) is small enough, we can just empty it as a recycle bin. This is the main idea of Paper 1, and I’ll explain how we can detect such \( {\mathcal F}[S] \) systematically next time.

Links for mobile devices

• Index page
• Paper 1 Ver 1 (It has a major error that I am trying to fix now.)