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These author-recorded arguments by Plasma AI develop the counting side of the proposed dichotomy in the slice-rank note. They concern words, with the conventions of the proof sketch for Problem 20. They give a sufficient condition that is uniform in the alphabets and a complete random-family calculation. They do not derive that condition from sunflower-freeness or from spread. No improvement of the unrestricted sunflower bound, novelty claim, or independent verification is asserted.

The counting extension uses the linear independent-representative lemma in the self-contained proof of L268. It counts the linear maps prescribed by that construction and retains the same author standing.

A Fourier criterion uniform in the alphabets

Fix $n\ge1$ and $k\ge2$, and let

$$\Omega=\prod_{i=1}^n\mathbb Z/B_i\mathbb Z,\qquad B_i\ge k.$$

The group structure is only an encoding of the coordinate alphabets. Injective relabeling preserves disjointness and sunflowers. Let $\mu$ be any probability measure on $\Omega$, and define its character averages and bias by

$$\phi(\xi)=\mathbb E_{x\sim\mu}\chi_\xi(x),\qquad \eta=\max_{\xi\ne0}|\phi(\xi)|.$$

Here $\chi_\xi(x)=\prod_i\exp(2\pi\mathrm i\xi_ix_i/B_i)$, and $0$ denotes the trivial character. Let $p_k(\mu)$ be the probability that $k$ independent words with law $\mu$ have pairwise distinct entries in every coordinate. These words give an empty-core $k$-sunflower. Put

$$a=\prod_{i=1}^n a_k(B_i),\qquad a_j(B)=\frac{(B)_j}{B^j},\qquad (B)_j=B(B-1)\cdots(B-j+1).$$

Counting criterion. For all such measures,

$$p_k(\mu)\ge a-\eta^2\bigl((k!)^n-a\bigr). \tag{1}$$

In particular, $\eta^2<1/(k^{kn}-1)$ guarantees a $k$-sunflower in the support of $\mu$. For three petals there is the sharper bound

$$p_3(\mu)\ge a\bigl(1-(6^n-1)\eta^2\bigr), \tag{2}$$

so $\eta<(6^n-1)^{-1/2}$ suffices, independently of the $B_i$.

Proof. Use normalized Fourier coefficients for functions on finite groups and write $\|h\|_A=\sum_\xi|\widehat h(\xi)|$. Orthogonality gives Fourier inversion. On a single alphabet let $D_B(u_1,\ldots,u_k)$ indicate that the symbols are all distinct. The matrix $([u_r=u_s])_{r,s\le k}$ is the identity when they are distinct and has repeated rows otherwise. Its determinant therefore gives

$$D_B(u_1,\ldots,u_k) =\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma) \prod_{r=1}^k[u_r=u_{\sigma(r)}]. \tag{3}$$

For each cycle of $\sigma$, the corresponding factor imposes equality of the symbols in that cycle. Equality of $s$ variables has Fourier coefficients $B^{-(s-1)}$ on the $B^{s-1}$ frequency tuples whose sum is zero, and zero elsewhere. Its Fourier norm is $1$. Distinct cycles use disjoint variables, so each summand of (3) has Fourier norm $1$, including fixed-point cycles. The triangle inequality gives $\|D_B\|_A\le k!$.

The indicator $D$ of coordinatewise disjointness on $\Omega^k$ is the product of these single-coordinate functions. Hence $\|D\|_A\le(k!)^n$, and its constant Fourier coefficient is $a$. Simultaneously translating all $k$ words leaves $D$ unchanged. Thus a nonzero Fourier coefficient at $(\xi_1,\ldots,\xi_k)$ requires $\xi_1+\cdots+\xi_k=0$. Apart from the constant term, at least two of the $\xi_j$ are nontrivial. Consequently

$$\left|p_k(\mu)-a\right| \le\eta^2\bigl(\|D\|_A-a\bigr), \tag{4}$$

which proves (1). Each $a_k(B_i)\ge k!/k^k$, since the factors $1-j/B_i$ increase with $B_i$. Therefore $(k!)^n/a\le k^{kn}$, giving the stated sufficient condition.

For $k=3$, use the exact identity

$$D_B(u,v,w)=1-[u=v]-[u=w]-[v=w]+2[u=v=w].$$

Its Fourier coefficients have the following exhaustive forms, for $B\ge3$.

Thus $\|D_B\|_A=a_3(B)+3a_3(B)+2a_3(B)=6a_3(B)$, including for composite $B$. Norms multiply over coordinates, so $\|D\|_A=6^na$. Substitution in (4) proves (2). $\square$

The proof uses only finite character orthogonality, the determinant expansion, and the triangle inequality. It is a counting argument, not a slice-rank estimate. In particular, the absence of a slice-rank saving for four or more petals does not prevent the criterion (1).

For $\mu$ uniform on a nonempty family $F$, (2) also has a precise inverse: if $F$ has no three pairwise disjoint members, then every encoding as above has a nontrivial character with $|\phi(\xi)|\ge(6^n-1)^{-1/2}$. This is only a Fourier bias conclusion. It neither supplies a dense affine subfamily nor identifies a sunflower core. The linear word theorem requires actual words inside the constructed affine structure; a bias does not provide them.

Counting in an affine space, allowing nonempty cores

Let $C\le A_1\oplus\cdots\oplus A_n$ be a linear space over $\mathbb F_2$, with arbitrary finite-dimensional coordinate spaces $A_i$. For $k\ge3$ let $H(x_1,\ldots,x_k)$ indicate that, in every coordinate, the $k$ symbols are all equal or all distinct. Write $p_C$ for its expectation on $k$ independent uniform words of any affine translate $a+C$. The value is independent of $a$.

Counting form of linear Hall. With $s=k-1$,

$$p_C\ge 2^{-s^2n}. \tag{5}$$

The assertion holds in every dimension, including $C=\{0\}$. It counts the all-equal tuples as well as the distinct sunflowers.

Proof. Put $V=C$. For each coordinate map $L_i:V\to A_i$ let $E_i=\operatorname{im}(L_i^*)\le V^*$, and write $W_I=\sum_{i\in I}E_i$. The coordinate maps are jointly injective, so $W_{[n]}=V^*$. Choose $J\subseteq[n]$ maximizing $s|I|-\dim W_I$. The empty set shows that $d=\dim W_J\le s|J|$. Unlike the subspace theorem of L268, this argument allows $J=[n]$ and imposes no lower bound on $\dim C$.

For $K\subseteq[n]\setminus J$, maximality gives

$$\dim(W_J+W_K)-\dim W_J\ge s|K|.$$

The proved linear independent-representative lemma therefore chooses $s$ vectors from each image of $E_i$ in $V^*/W_J$, for $i\notin J$, all jointly independent. Choose representatives $e_{i,j}\in E_i$. A basis of $W_J$ followed by these representatives is linearly independent in $V^*$.

Choose a uniformly random linear map $\Phi:V^*\to\mathbb F_2^s$. Require it to vanish on that basis of $W_J$ and to send $e_{i,j}$ to the $j$-th standard basis vector of $\mathbb F_2^s$. Images of the independent input vectors are independent uniform vectors, so these requirements have probability

$$2^{-sd-s^2(n-|J|)}\ge2^{-s^2n}. \tag{6}$$

A uniformly random $\Phi$ is equivalently an ordered list of $s$ independent uniform vectors $v_1,\ldots,v_s\in V$, via $\Phi(\lambda)=(\lambda(v_1),\ldots,\lambda(v_s))$. Here independent means independent samples, not necessarily linearly independent vectors. On the event (6), every coordinate in $J$ vanishes on these vectors. For $i\notin J$, the map $\mathbb F_2^s\to A_i$ sending $e_j$ to $L_i(v_j)$ is injective, because its dual has image $\Phi(E_i)=\mathbb F_2^s$. In that coordinate the $k=s+1$ vectors $0,v_1,\ldots,v_s$ have distinct symbols. Thus these $k$ vectors satisfy $H=1$.

Finally, for independent uniform $x_1,\ldots,x_k\in a+C$, the differences $x_2-x_1,\ldots,x_k-x_1$ are independent uniform vectors of $C$. Translation leaves the equality patterns unchanged, so (6) proves (5). If $J=[n]$, the prescribed map is zero and the tuples counted by this event are all equal; that case is included. $\square$

Now let $F\subseteq a+C$ be nonempty, with $M=|F|$, and let $\mu$ be the uniform measure on its translate $F-a\subseteq C$. Define its Fourier bias using only the nontrivial characters of $C$:

$$\eta_C=\max_{0\ne\lambda\in C^*} \left|\frac1M\sum_{x\in F}(-1)^{\lambda(x-a)}\right|.$$

Set $\eta_C=0$ if $C=\{0\}$. This avoids counting an ambient character that is constant on the entire affine space as a nontrivial bias. Put $K=k!+1$ and $b=2^{-(k-1)^2n}$. Then

$$\Pr_{x_1,\ldots,x_k\sim F}(H=1) \ge b-\eta_C^2(K^n-b). \tag{7}$$

Consequently the explicit conditions

$$\eta_C^2\le\frac{1}{2(K\,2^{(k-1)^2})^n},\qquad M>2^{(k-1)n+1/(k-1)} \tag{8}$$

force a $k$-sunflower in $F$, with no restriction on its core. For $k=3$ these conditions are $\eta_C^2\le1/(2\cdot112^n)$ and $M>\sqrt2\,4^n$.

Proof. On a single coordinate, the homogeneous indicator is the sum of the all-distinct indicator (3) and the all-equal indicator. Their supports are disjoint and their Fourier norms are at most $k!$ and $1$, respectively. The Fourier norm of $H$ on the full product is at most $K^n$. Restricting characters to $C^k$ can merge Fourier coefficients but cannot increase the sum of their absolute values. The constant coefficient on $C^k$ is $p_C$.

Common translation by $C$ leaves $H$ invariant. As in (4), every nonconstant Fourier term therefore uses at least two nontrivial characters of $C$, giving

$$\Pr_{F^k}(H=1)\ge p_C-\eta_C^2(K^n-p_C) \ge b-\eta_C^2(K^n-b).$$

This is (7). The same probability bound holds for any probability measure on $a+C$, with its own character averages. For such a measure the all-equal probability is $\sum_x\mu(x)^k$; for the uniform measure considered here it is $M^{1-k}$. Under (8), the probability lower bound in (7) is greater than $b/2$. If a tuple with $H=1$ repeats any whole word, no coordinate can have all distinct symbols; every coordinate is then constant, so every whole word is equal. Thus the only tuples with $H=1$ that are not distinct sunflowers are the all-equal tuples, of probability $M^{1-k}<b/2$. This proves the consequence. $\square$

Every finite word family can be encoded injectively in binary vector alphabets and placed in its affine span, so (7) has no alphabet-size restriction. It gives a necessary structural condition on a putative counterexample: if $M>2^{(k-1)n+1/(k-1)}$ and $F$ is sunflower-free, then, in every such encoding and with $C$ its affine span,

$$\eta_C^2>\frac{1}{2((k!+1)2^{(k-1)^2})^n}. \tag{9}$$

The contrapositive (9) still leaves the inverse problem open. It detects a bias of exponentially small size, not a character that is nearly constant. For a binary character of bias $\eta_C$, selecting the larger of its two affine hyperplane fibers retains only $(1+\eta_C)M/2$ words. No bound on the number of such restrictions, or on their total loss in terms of $n$ alone, follows from (9). The dimension of the ambient affine span is unrestricted. This explains why the counting extension does not itself close the general gap.

Random families: the exact exponential threshold

Here the alphabet is $[B]$ with $B\ge k$. Let $F$ be a uniformly chosen $M$-element subset of $[B]^n$, where $k\le M\le B^n$. Define

$$V_{n,k,B,M}= \sum_{\ell=2}^k \frac{\binom k\ell\binom{M-k}{k-\ell}}{\binom Mk} \left(a_\ell(B)^{-n}-1\right), \tag{10}$$

with a binomial coefficient equal to zero when its lower index exceeds its nonnegative upper index. Then

$$\Pr(F\text{ has no }k\text{ pairwise disjoint words}) \le\min(1,V_{n,k,B,M}). \tag{11}$$

For example, for $k=3$ formula (10) is

$$\frac{6}{(M)_3}\left(a_3(B)^{-n}-1\right) +\frac{18(M-3)}{(M)_3}\left(a_2(B)^{-n}-1\right). \tag{12}$$

Proof. First draw $M$ independent uniform words $W_1,\ldots,W_M$, allowing repetitions. For $S\in\binom{[M]}k$ let $I_S$ indicate that the words with indices in $S$ are pairwise disjoint, and let $X=\sum_S I_S$. Then $\mathbb E X=\binom Mk a_k(B)^n$.

If $|S\cap T|=\ell$, the shared symbols in one coordinate must be distinct. Conditional on those symbols, the two sets of unshared symbols must each be distinct and avoid the shared ones. They are independent of one another; no distinctness between the two unshared sets is required. Thus, for $0\le\ell\le k$, with $a_0(B)=a_1(B)=1$,

$$\Pr(I_S=I_T=1) =\left(\frac{(B)_\ell}{B^\ell} \left(\frac{(B-\ell)_{k-\ell}}{B^{k-\ell}}\right)^2\right)^n =a_k(B)^{2n}a_\ell(B)^{-n}. \tag{13}$$

In particular, pairs of indicators sharing exactly one index have zero covariance, just as disjoint pairs do. The number of ordered pairs $(S,T)$ with intersection size $\ell$ is $\binom Mk\binom k\ell\binom{M-k}{k-\ell}$. Summing the covariances gives the exact identity $\operatorname{Var}(X)/(\mathbb E X)^2=V_{n,k,B,M}$. Chebyshev's inequality bounds $\Pr(X=0)$ by this ratio.

To pass to distinct-word sampling, keep the set of distinct words among the $M$ draws and fill it to size $M$ by uniformly chosen unused words. The resulting set is invariant in law under every permutation of $[B]^n$, so it is a uniform $M$-subset. Every matching among the draws remains in it. This coupling proves (11). $\square$

For fixed $B\ge k\ge2$, set

$$\gamma_k(B)=a_k(B)^{-1/k}=\frac{B}{((B)_k)^{1/k}}. \tag{14}$$

If $M=\lfloor\rho^n\rfloor$ with fixed $1<\rho\le B$, then the probability of a $k$-matching tends to $1$ when $\rho>\gamma_k(B)$ and to $0$ when $\rho<\gamma_k(B)$. No assertion is made at equality.

For the upper regime, the coefficient of the $\ell$-th term of (10) is $O_k(M^{-\ell})$. The positive factors $1-j/B$ decrease with $j$, so $a_\ell(B)^{1/\ell}>a_k(B)^{1/k}$ for $1\le\ell<k$. Hence every term of (10) tends to zero when $\rho>\gamma_k(B)$. For the lower regime, the expected number of unordered $k$-matchings in the uniform $M$-subset is exactly

$$\binom Mk\frac{(B)_k^n}{(B^n)_k} =(1+o(1))\binom Mk a_k(B)^n,$$

which tends to zero; Markov's inequality finishes.

There is also a bound uniform in growing alphabets: if $k$ is fixed and $\rho>\gamma_k(k)=(k^k/k!)^{1/k}$, the upper-regime conclusion holds uniformly over $B=B(n)\ge k$ for which $\lfloor\rho^n\rfloor\le B^n$. Indeed $a_\ell(B)\ge a_\ell(k)$ makes the same finite sum decrease with $B$. In particular, rate $\rho=3$ suffices for every fixed $k$, since $k!>(k/e)^k$ gives $\gamma_k(k)<e<3$. For three petals, every fixed $\rho>(9/2)^{1/3}$ suffices uniformly over $B\ge3$.

This proves the random-family assertion left open in the single-scale discussion of the slice-rank note, and covers alphabets far beyond its counting bound. It is an average-case result. It gives no upper bound on an adversarially chosen sunflower-free family. The Fourier criterion and this second-moment argument are separate: (11) does not assert that random families satisfy the Fourier hypothesis in every group encoding.

Pair-agreement profiles do not determine sunflower behavior

For a uniform random pair $x,y\in F$, sampled independently, write

$$c_F(I)=\Pr(x_I=y_I)=\sum_u\left(\frac{|\{x\in F:x_I=u\}|}{|F|}\right)^2, \qquad I\subseteq[n].$$

Even the complete list of these numbers, together with $n$, the alphabet sizes and $|F|$, need not determine whether $F$ has a sunflower. On two ternary coordinates consider

$$F_0=\{00,01,12,22\},\qquad F_1=\{00,01,11,22\}. \tag{15}$$

In both families the six unordered pairs of distinct words have agreement sets $\varnothing,\varnothing,\varnothing,\varnothing,\{1\},\{2\}$. Thus both have

$$c_F(\varnothing)=1,\quad c_F(\{1\})=c_F(\{2\})=\frac38,\quad c_F(\{1,2\})=\frac14.$$

The three words $00,11,22$ are pairwise disjoint in $F_1$. The family $F_0$ has no three-sunflower: a triple containing both $00,01$ has exactly two equal first coordinates, and a triple containing both $12,22$ has exactly two equal second coordinates. Every three-element subset contains one of those pairs.

This distinction persists in every product power. Both $F_0^h$ and $F_1^h$ have $4^h$ words of length $2h$ and the same full pair-agreement profile, since the block profiles multiply. The family $F_1^h$ contains the three repeated diagonal words. The family $F_0^h$ is sunflower-free: three global words forming a sunflower would in each block be all equal or give three distinct sunflower words in $F_0$. Exactly two equal block words also violate the sunflower condition in a coordinate where the third differs. Every block must therefore be constant across the three words, making the global words equal.

This refutes determination of the sunflower predicate, or of its exact triple count, by the aggregate pair-agreement profile. It does not refute a sufficient inequality on that profile at a larger size rate: these examples have rate $2$. It also does not concern the fully labeled agreement relation on each pair, which does determine the sunflower triples. Joint information on three words is precisely what the aggregate profile discards.

The missing inverse step

Equations (1) and (2) give a concrete counting condition for a link or a projection whose matchings lift. Equation (7) permits nonempty cores and works relative to a binary affine span. Equation (11) verifies that typical families of exponentially many words already have matchings at constant rates. None supplies a structural alternative when the counting condition fails.

The inverse conclusions of (2) and (7) have exponentially small strength. A character correlation is not a set of points common to the members, and conditioning on it is not a link. No argument here turns that correlation into a section, a dense affine subfamily, or a restriction with only an exponential total loss. Such an argument would have to keep the original words and preserve the lifting of sunflowers. The maximal-rate obstruction also requires accounting for members outside a selected restriction.

Even favorable encoding is not a replacement for finding a core: the intersecting spread families have no matching in the empty link, so every encoding of any rainbow subfamily of them fails the matching criteria (1) and (2). The affine criterion (7) allows nonempty cores but still needs its Fourier hypothesis. A useful inverse theorem must locate another link or identify structure that supplies the missing sunflower. The equality of the pair profiles in (15) warns against replacing that inverse theorem by an exact formula in pair collision moments alone. The general core-selection and structural inverse steps remain open.