Markov degree of the three-state toric homogeneous Markov chain model

Abstract

We consider the three-state toric homogeneous Markov chain model (THMC) without loops and initial parameters. At time $T$, the size of the design matrix is $6 \times 3\cdot 2^{T-1}$ and the convex hull of its columns is the model polytope. We study the behavior of this polytope for $T\ge 3$ and we show that it is defined by $24$ facets for all $T\ge 5$. Moreover, we give a complete description of these facets. From this, we deduce that the toric ideal associated with the design matrix is generated by binomials of degree at most $6$. Our proof is based on a result due to Sturmfels, who gave a bound on the degree of the generators of a toric ideal, provided the normality of the corresponding toric variety. In our setting, we established the normality of the toric variety associated to the THMC model by studying the geometric properties of the model polytope.

Appendix

The six cones used in defining $Q^r$ for $r=0,\ldots ,5$:

$$\begin{aligned} C^r&:= {{\mathrm{cone}}}\left( \, [1,0,1,0,0,0], [1,0,0,1,1,0], [0,1,1,0,0,1],\right. \\&\quad \left. [0,1,0,0,1,0], [0,0,0,1,0,1] \,\right) \end{aligned}$$

for $r=0,\ldots ,5$.

The six polytopes used in defining $Q^r$ for $r=0,\ldots ,5$ are given below, where the vertices are modulo the permutations of $S = \{1,2,3\}$. That is, the indexing below is $x_{12}, x_{21}, x_{13}, x_{31}, x_{23}$, and $x_{32}$. To get the full list of vertices one should use all six permutations of $\{1,2,3\}$ and permute the indices of each vertex below accordingly.

$\mathrm{{vert}}( Q^0) := [$ [0, 1, 1, 0, 0, 1], [0, 1, 2, 1/2, 1, 3/2], [0, 1, 3/2, 1, 1/2, 2], [0, 2, 2, 0, 1, 2], [0, 2, 2, 0, 2, 5], [0, 2, 2, 2/3, 0, 7/3], [0, 2, 3, 0, 2, 4], [0, 2, 4, 2, 0, 3], [0, 2, 3/2, 0, 0, 3/2], [0, 2, 7/3, 0, 2/3, 2], [0, 3, 4, 0, 0, 4], [0, 6/5, 8/5, 4/5, 2/5, 11/5], [0, 6/5, 11/5, 2/5, 4/5, 8/5], [2/3, 4/3, 4/3, 2/3, 2/3, 7/3]$]$.

${{\mathrm{vert}}}( Q^1) := [$ [0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 3, 2], [0, 0, 1, 0, 2, 3], [0, 1, 1, 0, 1, 3], [0, 1, 1, 0, 2, 2], [0, 1, 2, 0, 1, 2], [0, 1, 2, 1, 0, 2], [0, 1, 1/2, 1/2, 1/2, 1/2], [0, 1/2, 0, 1/2, 1, 1], [0, 1/2, 1, 1, 1/2, 0], [0, 1/2, 1/2, 1, 1/2, 1/2], [0, 1/2, 1/2, 1/2, 1, 1/2] $]$.

${{\mathrm{vert}}}( Q^2) := [$ [0, 1, 1, 0, 1, 2], [0, 1, 2, 1/2, 2, 5/2], [0, 1, 3, 3/2, 1, 3/2], [0, 1, 3/2, 1, 3/2, 3], [0, 1, 5/2, 2, 1/2, 2], [0, 2, 2, 0, 2, 5], [0, 2, 2, 0, 3, 4], [0, 2, 2, 1, 2, 4], [0, 2, 3, 0, 2, 4], [0, 2, 4, 2, 0, 3], [0, 2, 4, 2, 1, 2], [0, 3, 4, 0, 3, 7], [0, 3, 7, 3, 0, 4], [1/3, 2/3, 2/3, 1/3, 1/3, 2/3], [1/3, 2/3, 5/3, 5/6, 4/3, 7/6], [1/3, 2/3, 7/6, 4/3, 5/6, 5/3], [2/3, 4/3, 4/3, 2/3, 2/3, 7/3], [2/3, 4/3, 4/3, 2/3, 5/3, 4/3] $]$.

${{\mathrm{vert}}}( Q^3) := [$ [0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 1, 0], [0, 1, 1, 0, 2, 4], [0, 1, 2, 0, 2, 3], [0, 1, 3, 2, 0, 2] $]$.

${{\mathrm{vert}}}( Q^4) := [$ [0, 1, 1, 0, 0, 1], [0, 1, 2, 1/2, 1, 3/2], [0, 1, 3/2, 1, 1/2, 2], [0, 2, 2, 0, 1, 4], [0, 2, 2, 0, 2, 3], [0, 2, 2, 1, 1, 3], [0, 2, 3, 0, 1, 3], [0, 2, 3, 1, 0, 3], [0, 2, 3, 1, 1, 2], [0, 3, 4, 0, 2, 6], [0, 3, 6, 2, 0, 4], [1/3, 5/3, 5/3, 1/3, 1/3, 8/3], [1/3, 5/3, 5/3, 1/3, 4/3, 5/3] $]$.

${{\mathrm{vert}}}( Q^5) := [$ [0, 0, 0, 0, 1, 1], [0, 0, 0, 1, 4, 3], [0, 0, 1, 0, 3, 4], [0, 1, 1, 0, 2, 4], [0, 1, 1, 0, 3, 3], [0, 1, 2, 0, 2, 3], [0, 1, 3, 2, 0, 2], [0, 1, 1/2, 1/2, 3/2, 3/2], [0, 1, 3/2, 3/2, 1/2, 1/2], [0, 1/2, 0, 1/2, 2, 2], [0, 1/2, 2, 2, 1/2, 0], [0, 1/2, 1/2, 1/2, 2, 3/2], [0, 1/2, 3/2, 2, 1/2, 1/2], [1, 2, 1/2, 1/2, 1/2, 1/2] $]$.