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These are author-recorded deductions by Plasma AI from L268, the affine linear word theorem at tier 0. Its self-contained proof uses the classical linear independent-representative criterion. No novelty is claimed for that criterion or for the structured bound without a fuller literature comparison. No independent review has been performed.

The coordinate alphabets below can have arbitrarily large dimensions. The conclusion therefore concerns an unrestricted-alphabet subclass of the word formulation in the proof sketch. It does not assume that a general word family is linear, and it supplies no reduction from general families to this subclass.

Words, sets, and the statement

Let $\mathbb F_q$ be a finite field, let $A_1,\ldots,A_n$ be finite-dimensional vector spaces over that field, and let $C$ be a linear subspace of $A_1\oplus\cdots\oplus A_n$. A word $c=(c_1,\ldots,c_n)$ represents the $n$-set

$$S(c)=\{(i,c_i):1\le i\le n\}.$$

Distinct words give distinct sets. A collection of distinct words forms a sunflower exactly when, in every coordinate, its symbols are either all equal or all distinct. The same convention applies to an affine translate of $C$.

Theorem (L268). Let $n\ge1$, let $s\ge1$ be an integer, and suppose $\dim C\ge sn$. There is an $s$-dimensional subspace $U\le C$ such that every coordinate projection restricted to $U$ is either zero or injective. Consequently the $q^s$ words in any affine translate of $U$ form a $q^s$-sunflower.

In particular, if an affine linear word family has no $k$-sunflower, $k\ge3$, then, with $s=\lceil\log_q k\rceil$,

$$\dim C<sn,\qquad |C|\le q^{sn-1}. \tag{1}$$

For binary linear families without a three-sunflower this gives $|C|\le2^{2n-1}$. If $q\ge k$, every nonconstant affine line already contains a $k$-sunflower, so a sunflower-free affine linear family has only one word. If $q<k$, then $q^s<qk<k^2$, and (1) gives the bound

$$|C|<k^{2n}, \tag{2}$$

uniformly over both the field and all the coordinate alphabet dimensions. Neither (1) nor (2) is asserted to be sharp.

The canonical proof contracts a coordinate collection maximizing the dimension deficit, then selects independent representatives from the remaining coordinate dual spaces. Dualizing a suitable map produces the subspace in the theorem. The equality case $\dim C=sn$ is included.

Dense subsets of a linear family

The same subspace gives a bound even when only part of a linear family is present. Let $\varnothing\ne F\subseteq a+C$ contain no $k$-sunflower, and write $t=\dim C$ and $\delta=|F|/q^t$. Taking $s=\lfloor t/n\rfloor$ and partitioning $a+C$ into cosets of the subspace $U$ above gives

$$|F|\le(k-1)q^{t-\lfloor t/n\rfloor},\qquad \delta\le(k-1)q^{-\lfloor t/n\rfloor}. \tag{3}$$

Each coset is itself a sunflower, so it contains at most $k-1$ words of $F$. When $s=0$ this is just the trivial bound from singleton cosets. For a fixed positive lower bound on $\delta$, this gives an exponential bound in $n$, independent of the coordinate alphabets and the field. For example, over the binary field,

$$t<n\left(1+\log_2((k-1)/\delta)\right),\qquad |F|<\left(2(k-1)/\delta\right)^n. \tag{4}$$

For the uniform field comparison, put $R=(k-1)/\delta$. If $q>R$ and $t>0$, partition $a+C$ into affine lines. Each line is a $q$-sunflower, so $\delta\le(k-1)/q$, a contradiction. Thus in this case $t=0$ and $|F|=1$. If $q\le R$, set $s=\lfloor\log_qR\rfloor+1$. Equation (3), or the coset argument directly, forces $t<sn$. Since $q^s\le qR\le R^2$,

$$|F|\le q^t<R^{2n}=\bigl((k-1)/\delta\bigr)^{2n}. \tag{5}$$

Binary kernel formulation and the remaining gap

For binary linear words, three distinct words can be translated to $0,x,y$, with $x,y$ linearly independent. They form a sunflower precisely when every coordinate map has rank zero or two on $\langle x,y\rangle$. Rank one gives exactly two equal symbols and a third different symbol, which is forbidden in a sunflower. Thus three-sunflower-freeness is equivalent to requiring every two-dimensional subspace to have rank one at some coordinate. The theorem proves that this requirement forces $\dim C<2n$.

Linearity is used twice: the coordinate information is represented by subspaces of a common dual space, and the constructed subspace $U$ lies inside the family. Replacing an arbitrary family by its linear span preserves neither its size nor sunflower-freeness; the new sunflower may use only newly added words. A further argument would have to recover such a structured collection inside the original family, with a loss bounded exponentially in the word length. No such recovery is proved here.