Dimension increase in odd reflected configurations

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These are special augmentation results, not a proof of the target. They differ from the refuted assertion in the reflection obstruction: here the center itself, not twice the center, is included, and all points are odd.

Let $c$ be a positive odd integer and let $X$ be a dissociated set of distinct positive even integers smaller than $c$. Write

$$Q=\{c\}\cup\{c-x,c+x:x\in X\}.$$

A proved orientation for at most seven differences

Theorem. If $|X|\le7$, there are signs $\sigma_x$ such that

$$\{c\}\cup\{c+\sigma_xx:x\in X\}$$

is dissociated. In particular $d(Q)\ge |X|+1$.

A balanced relation among any oriented selection gives a ternary relation on $X$, so is impossible. For the all-upper selection, every point is in $[c,2c)$. Odd parity requires even relation support. An unbalanced relation supported on at most six points has at least four terms against at most two, or at least three against one; the interval bounds rule it out. This proves the theorem for $|X|\le6$.

Let $|X|=7$. If the all-upper selection is dependent, its relation must use all eight points and have five terms against three. The pivot $c$ lies on the five-term side: if it lay on the three-term side, that side would be less than $5c$, whereas the other side would be at least $5c$. We therefore have a partition

$$X=A\sqcup B,\qquad |A|=3,\quad |B|=4,\qquad \sum A-\sum B=2c.$$

Put $S=\sum B$. Since the three members of $A$ are less than $c$, we have $S<c$, and every $a\in A$ exceeds $S$.

Flip only $b=\min B$: set $\sigma_b=-1$ and all other signs to $1$. Suppose the resulting selection has a relation. Write it as

$$sc+\sum_{x\in X}r_xx=0, \qquad s=\alpha_0+\sum_xr_x\sigma_x,$$

where $r_x,\alpha_0\in\{-1,0,1\}$. Oddness makes $s$ even; $s=0$ is excluded by dissociation of $X$. Also $\sum X=2c+2S<4c$. Reversing the relation if needed gives

$$\sum_xr_xx=2c, \qquad \sum_xr_x\sigma_x\in\{-3,-2,-1\}. \tag{1}$$

At least two members of $A$ have coefficient $+1$, since one of them together with all of $B$ has sum less than $c+S<2c$. If all three do, every $B$-coefficient must be $-1$. The signed coefficient sum in (1) is then $1$, impossible. If two do and the third $a$ has coefficient $-1$, the total is at most $2c+2S-2a<2c$. Hence precisely two have coefficient $+1$ and the third $a$ is omitted. The equation becomes

$$a=S+\sum_{u\in B}r_u u>S. \tag{2}$$

Let $u,v$ be the numbers of positive and negative coefficients on $B$. There is at least one positive coefficient. If $r_b\le0$, counting the other three coefficients gives $\sum_xr_x\sigma_x\ge1$, contrary to (1). If $r_b=1$, that sum equals $u-v$. To be negative, with $u+v\le4$, requires $(u,v)=(1,2)$ or $(1,3)$. Equation (2) then gives, respectively, $a=2b+q$ for the omitted $q\in B\setminus\{b\}$, or $a=2b$. Both are less than $S$, since $b$ is the smallest of four distinct positive numbers. This contradicts $a>S$ and proves the theorem.

Thus a centrally symmetric odd set containing its odd center has dimension at least $d(X)+1$ whenever its positive difference set has dimension at most seven. The unrestricted version is unproved.

Increasing dimension by two is false in general

Take $X=\{2,4,8,16\}$ and $c=17$. The set $X$ is dissociated, but

$$Q=\{1,9,13,15,17,19,21,25,33\},\qquad d(Q)=5.$$

Thus the tempting stronger bound $d(Q)\ge d(X)+2$ is false.

The five-set $\{1,9,13,15,17\}$ is dissociated: its ten pair sums are distinct, and its smallest triple sum exceeds $17$. Parity excludes every relation except a two-term equality, a pair-sum equality, or a triple equaling a singleton.

Every six-subset of $Q$ containing two complete reflected pairs is dependent, since both pairs sum to $34$. Otherwise it consists of $c$, one complete pair $\{c-x,c+x\}$, and one point from each other pair. If $x\in\{2,4,8\}$, then $2x\in X$, and the selected point $c\pm2x$, together with the complete pair and $c$, gives a balanced four-term relation.

It remains to consider the complete pair $\{1,33\}$. The other three choices are $a\in\{15,19\}$, $b\in\{13,21\}$ and $z\in\{9,25\}$. If $a=15$, use $1+15+17=33$; if $a=19,b=13$, use $1+13+19=33$. In the remaining cases use

$$1+17+19+21=25+33,\qquad 1+9+19+21=17+33.$$

This covers every six-subset and proves the claimed dimension.

The obstruction persists in every dimension $k\ge4$. Set

$$X_k=\{2,4,\ldots,2^k\},\quad c=2^k+1,\quad Q_k=\{c\}\cup(c+X_k)\cup(c-X_k).$$

Then $d(Q_k)=k+1$. The all-upper selection with $c$ is dissociated: in a relation its coefficient sum at $c$ is even, whereas $\sum X_k=2^{k+1}-2<2c$, so it must be balanced and then trivial. To exclude $(k+2)$-subsets, two complete pairs give an immediate relation; otherwise the set has the center, one complete pair and one choice from every other pair. Any nonlargest pair is blocked by its doubled offset. For the largest pair $\{1,2c-1\}$, choices at offsets $2,4,8$ suffice. Successively according to those choices, use

$$\begin{split} 1+(c-2)+c&=2c-1,\\ 1+(c+2)+(c-4)&=2c-1,\\ 1+(c+2)+(c+4)+(c-8)&=c+(2c-1),\\ 1+c+(c+2)+(c+4)&=(c+8)+(2c-1). \end{split}$$

Thus a general reflection theorem could guarantee at most one additional dimension.

A two-dimension increase in an affine special case

Call a set affinely dissociated if its only balanced ternary relation is trivial. If a dissociated integer set $X$ has all elements of the same $2$-adic valuation, then

$$\{0\}\cup X\cup(-X) \quad\hbox{contains an affinely dissociated }(|X|+2)\hbox{-set}. \tag{3}$$

Scale so the entries are odd, fix a pivot $x_j$, and, for $i\ne j$, choose $\sigma_i\in\{\pm1\}$ with $\sigma_i\equiv x_i\pmod4$. Use

$$Y=\{0,-x_j,x_j\}\cup\{\sigma_i x_i:i\ne j\}.$$

A balanced ternary relation not assigning opposite nonzero coefficients to $-x_j,x_j$ reduces to a ternary relation on $X$. If those two coefficients are equal, cancellation in $X$ would instead require a coefficient of absolute value two at $0$; this is also impossible. The only remaining possibility would give

$$2x_j=\sum_{i\ne j}\varepsilon_i x_i, \qquad \left|\sum_{i\ne j}\varepsilon_i\sigma_i\right|\le1.$$

But the latter sum is $2x_j\equiv2\pmod4$, a contradiction. This proves (3).

For the odd reflected configuration above, (3) gives an actual dissociated $(|X|+2)$-subset if, additionally,

$$2c>\sum X+\max X.$$

Indeed, every relation on the translated selection has an even coefficient sum at $c$. Its offset part has absolute value at most $\sum X+\max X$, so the displayed inequality forces balance.

Statement (3) for mixed valuation layers remains unproved. The counterexample above concerns actual, not affine, dissociation and does not refute it. It does refute a two-dimension increase for arbitrary $c>\max X$.

There is also a valuation-free small affine case: (3) holds for every positive dissociated $X$ with $|X|\le5$, using $Y=\{0,-M,M\}\cup(X\setminus\{M\})$ and $M=\max X$. The only possible affine relation would give $2M=\sum_{x\ne M}\varepsilon_xx$ with $|\sum\varepsilon_x|\le1$. Among at most four other coordinates this permits at most two positive coefficients, whose weighted sum is strictly less than $2M$. This is impossible.

Neither an unrestricted largest-pivot rule nor a largest-valuation rule can extend this proof. The dissociated set $\{1,2,8,15\}$ has its maximum-valuation pivot $8$ blocked by $16=1+15=-1+2+15$: avoiding a signed coefficient sum of zero in the first representation forces the signs on $1,15$ to agree, and the second then has coefficient sum $\pm1$.

The largest pivot is blocked in $X=\{8,20,36,40,42,45\}$ by

$$90=40+42+8=36+42+20-8=36+40+42-8-20.$$

The first representation forces the signs on $40,42,8$ to agree; the second then forces those on $36,20$ to agree; the last has signed coefficient sum $\pm1$. This $X$ is dissociated: successive valuation arguments eliminate $45$ and $42$, leaving four times the dissociated set $\{2,5,9,10\}$, whose subset sums are $0,2,5,7,9,10,11,12,14,15,16,17,19,21,24,26$. These examples leave free choice of the pivot open.