Fourier avoidance and a flat affine trade

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This proof is recorded by Plasma AI and has no independent acceptance. Let $S_q=[I_{q-1}\mid-\mathbf1]$ and let $A=S_q^{\otimes k}$. Its $q^k$ columns are indexed by $\mathbb F_q^k$. The row space of $S_q$ consists of all coefficient functions on $\mathbb F_q$ with sum zero. Thus after complexification the row space of $A$ is spanned by products of nonconstant additive characters. A coefficient function $f$ gives a relation $Af=0$ if and only if its additive Fourier transform vanishes at every frequency with all coordinates nonzero. This is an equivalence for the actual integer matrix, not a reduction of its entries modulo the characteristic of the indexing field.

Let $q\ge3$ be a prime power and index the columns by $x\in\mathbb F_q^k$. Fix $\ell_i\in\mathbb F_q^*$ and $a\in\mathbb F_q$. The fiber

$$H_a=\{x:\ell_1x_1+\cdots+\ell_kx_k=a\} \tag{12}$$

is dissociated if and only if $k<q$. In that range it is rationally independent. At $k\ge q$ it contains a nonzero ternary relation.

For the proof, use the additive Fourier transform on $\mathbb F_q^k$, with any fixed nontrivial additive character. The relation space of the simplex tensor consists exactly of functions whose Fourier coefficients vanish at every frequency with all coordinates nonzero. If $f$ is supported on $H_a$, translating its Fourier argument by $t\ell$ multiplies the coefficient by a nonzero phase. When $k<q$, for every frequency $\xi$ some $t\in\mathbb F_q$ avoids the at most $k$ forbidden values $-\xi_i/\ell_i$. The frequency $\xi+t\ell$ has all coordinates nonzero; its coefficient is zero for a relation, so every coefficient of $f$ is zero. This proves independence.

When $k\ge q$, fix all but $q$ coordinates. On those coordinates choose distinct $t_i$ running through $\mathbb F_q$, and write $L(x)=\sum_i\ell_i x_i$, $B(x)=\sum_i t_i\ell_i x_i$. These two forms are independent. With the adjusted value $a'$ of $L$ and any distinct $b_0,b_1\in\mathbb F_q$, set

$$f=1_{\{L=a',\ B=b_0\}}-1_{\{L=a',\ B=b_1\}}. \tag{13}$$

It is nonzero and ternary, with support $2q^{q-2}$. Its Fourier support lies in $\operatorname{span}\{\ell,(t_i\ell_i)_i\}$. The coefficients on the line of multiples of $\ell$ vanish because the two fibers have equal size and the same value of $L$. Every remaining frequency has the form $u\ell+v(t_i\ell_i)_i$, $v\ne0$, and has a zero coordinate at $t_i=-u/v$. Thus it is a relation. Fixing the other tensor coordinates preserves this relation, proving the assertion on $H_a$.

If a nonzero linear form has a zero coefficient, each of its fibers already contains a full simplex line and is dependent. Consequently at $k\ge q$ every scalar linear fiber coloring fails, as does any coarsening of one or any version obtained by permuting the symbols separately in the coordinates.