Math-1756 · mathematics · Combinatorics

Math-1756

Periodic Laplacian certificates for robustness and explosion

Status Partially solved · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedPartially solvedOpen

Statement

Definitions and mathematical statement

On the square lattice $\mathbb Z^2$, a sandpile configuration assigns a nonnegative integer number of grains to every vertex. A vertex with at least four grains may topple legally: it loses four grains and sends one to each of its four nearest neighbors. A configuration is stable when every height belongs to $\{0,1,2,3\}$. Call a configuration stabilizable if there exists a finite or countably infinite legal toppling sequence in which every vertex topples only finitely many times and whose limiting configuration is stable. This allows activity at infinitely many different vertices; it does not require a globally finite total number of topplings.

The input is a stable periodic configuration $c$ specified by two linearly independent integer period vectors and its finitely many heights on representatives for their lattice quotient. Let $\delta_0$ be one grain at the origin and zero elsewhere. Is there an algorithm, halting on every such finite input, that decides the assertion \[ \exists n\in\mathbb Z_{\geq1}\quad c+n\delta_0\text{ is not stabilizable}? \] The universal alternative is that $c+n\delta_0$ stabilizes for every finite $n$. The period lattice and its size are unrestricted parts of the input, rather than fixed in advance. The question concerns decidability of the existence of some destabilizing addition; direct simulation of a chosen addition alone does not provide a halting decision procedure for both alternatives. No finite boundary or absorbing sink is present in this infinite-lattice model.

Short English statement

Can one decide from a periodic square-lattice background whether adding finitely many grains at the origin can ever cause nonstabilization?

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