GEOMETRY, COMBINATORICS, AND ALGEBRA OF INDUCTIVELY PIERCED CODES

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Problem: For a neural code $\mathcal C\subseteq 2^{[n]}$ over a field $k$, let

and $T_{\mathcal C}=\ker(\phi_{\mathcal C})$. Reconstructed Question 5.3 asks:

If $\mathcal C$ is a nondegenerate hyperplane code, must $T_{\mathcal C}$ have a quadratic Gröbner basis? Must $T_{\mathcal C}$ be generated by quadrics?

Here “nondegenerate hyperplane code” is understood in the paper’s sense: a code realized stably by open halfspaces restricted to a convex stimulus space.

Result: The answer to both questions is no.

Use the shorthand $12=\{1,2\}$, etc. Let

This is a nondegenerate hyperplane code. Indeed take

and halfspaces

A direct inequality check gives exactly the eight codewords above. For example, if $2\notin\sigma$, then

forces $x>1/5$, hence $3\in\sigma$, $4\notin\sigma$, and $x+y<1/2$, so $1\in\sigma$; thus the only codeword without $2$ is $13$. The other cases split according to $x\le0$, $y\le x$, and $x+y<1/2$, yielding precisely

The arrangement is simple relative to $\overline X$: no triple intersection occurs in $\overline X$, and all boundary crossings are transverse and distinct. Hence the realization is stable under sufficiently small perturbations, so $\mathcal C$ is nondegenerate.

Now consider $T_{\mathcal C}$. We have

so

The $\mathbb N^4$-fiber over $(2,2,2,0)$ contains only these two monomials. Indeed, coordinate $4$ excludes all variables containing $4$, and solving

gives only $(a,b,c,d)=(1,1,1,0)$ and $(0,0,0,2)$.

Thus this fiber has exactly two monomials, with gcd $1$, and their binomial has degree $3$. No product of a quadratic binomial in $T_{\mathcal C}$ can connect them. Therefore $f$ is not in the ideal generated by quadrics, so $T_{\mathcal C}$ has no quadratic generating set. Consequently it cannot have a quadratic Gröbner basis.

Citation: The question is Lienkaemper, “Geometry, combinatorics, and algebra of inductively pierced codes,” arXiv:1811.04712, Question 5.3. The counterexample above is self-contained.