Normality of the Kimura 3-parameter model

TL;DR

This paper proves that all algebraic varieties related to the Kimura 3-parameter model are projectively normal, confirming a key conjecture in algebraic statistics and advancing understanding of this fundamental phylogenetic model.

Contribution

It establishes the projective normality of algebraic varieties associated with the Kimura 3-parameter model, confirming Michalek's conjecture.

Findings

All algebraic varieties of the model are projectively normal.

Confirmed a significant conjecture in algebraic statistics.

Enhances mathematical understanding of the Kimura 3-parameter model.

Abstract

The Kimura 3-parameter model is one of the most fundamental phylogenetic models in algebraic statistics. We prove that all algebraic varieties associated to this model are projectively normal, confirming a conjecture of Michalek.

Equations

$$H_n=\{(g_1,\dots ,g_n)\in G:g_1+g_2+\cdots +g_n=0\}.$$

$$\sum _{j\in A}(x_0^j+x_{\alpha }^j)+\sum _{j\notin A}(x_{\beta }^j+x_{\gamma }^j)\ge 1,$$

$$\sum _{j\in A}(x_0^j+x_{\beta }^j)+\sum _{j\notin A}(x_{\alpha }^j+x_{\gamma }^j)\ge 1,$$

$$\sum _{j\in A}(x_0^j+x_{\gamma }^j)+\sum _{j\notin A}(x_{\alpha }^j+x_{\beta }^j)\ge 1.$$

$$F_g(A)=\{x\in P_n:S_g(x,A)=1\}.$$

$$L_n=\{m\in {\mathbb{Z}}^{4n}|\forall 1\le j,j^{\prime }\le n:\sum _{g\in G}{m}_g^j=\sum _{g\in G}{m}_g^{j^{\prime }},\sum _{\begin{array}{c}g\in G\\ 1\le j\le n\end{array}}(m_g^{j}g)=0\},$$

$$V_n=\{v(0)\}\cup \{v(g{)}_{j,n}|1\le j\le n-1,g\in \{\alpha ,\beta ,\gamma \}\}.$$

$$\{a,b,c,d\}\succ \{a^{\prime },b^{\prime },c^{\prime },d^{\prime }\}\iff (a>a^{\prime })\vee (a=a^{\prime }\wedge b>b^{\prime })\vee (a=a^{\prime }\wedge b=b^{\prime }\wedge c>c^{\prime }).$$

$$\forall j\in \{1,2,\dots ,n-1\}:\{x_0^j,x_{\alpha }^j,x_{\beta }^j,x_{\gamma }^j\}⪰\{x_0^n,x_{\alpha }^n,x_{\beta }^n,x_{\gamma }^n\}$$

$$\forall j\in \{1,2,\dots ,n-1\}:x_0^j=\max \{x_0^j,x_{\alpha }^j,x_{\beta }^j,x_{\gamma }^j\}$$

$$3(x_0^j+x_g^j)\ge 3x_0^j+x_g^j\ge x_0^j+x_{\alpha }^j+x_{\beta }^j+x_{\gamma }^j=k.$$

$$S_g(x,A)\equiv k(\mathrm{mod}2).$$

$$\phi :{\mathbb{Z}}_2\times {\mathbb{Z}}_2\to {\mathbb{Z}}_2$$

$$0,\alpha \mapsto 0,\beta ,\gamma \mapsto 1.$$

$$0=\phi (0)=\phi \left(\sum _{\begin{array}{c}g\in G\\ 1\le j\le n\end{array}}(x_g^{j}g)\right)=\sum _{\begin{array}{c}g\in G\\ 1\le j\le n\end{array}}(x_g^j\phi (g))=\sum _{j=1}^n(x_{\beta }^j+x_{\gamma }^j)\cdot 1.$$

$$\sum _{j\in A}(x_0^j+x_{\alpha }^j)+\sum _{j\notin A}(x_{\beta }^j+x_{\gamma }^j)=\sum _{j=1}^n(x_{\beta }^j+x_{\gamma }^j)+\sum _{j\in A}(x_0^j+x_{\alpha }^j-x_{\beta }^j-x_{\gamma }^j)=$$

$$=\sum _{j=1}^n(x_{\beta }^j+x_{\gamma }^j)-\sum _{j\in A}(x_0^j+x_{\alpha }^j+x_{\beta }^j+x_{\gamma }^j)+2\sum _{j\in A}(x_0^j+x_{\alpha }^j)=$$

$$\sum _{j=1}^n(x_{\beta }^j+x_{\gamma }^j)-k|A|+2\sum _{j\in A}(x_0^j+x_{\alpha }^j)\equiv k|A|\equiv k(\mathrm{mod}2).$$

$$\forall j\in \{1,2,\dots ,n\},g\in G:x_g^j<k$$

$$\text{At most two multisets from }G\text{-presentation of }x\text{ are }\{k/3,k/3,k/3,0\}.$$

$$S(x-v,A)\ge \sum _{j\in A\setminus B}(x_0^j-1)\ge |A\setminus B|\ge 2.$$

$$\sum _{j\in A}(y_0^j+y_{\alpha }^j)+\sum _{j\notin A}(y_{\beta }^j+y_{\gamma }^j)$$

$$\ge \sum _{j\in A\setminus \{n\}}(x_0^j+x_{\alpha }^j-1)$$

$$\ge \sum _{j\in A\setminus \{n\}}(\lceil k/3\rceil -1)$$

$$\ge 4\lceil k/3\rceil -4\ge k-2.$$

$$\sum _{j\in A}(y_0^j+y_{\alpha }^j)+\sum _{j\notin A}(y_{\beta }^j+y_{\gamma }^j)$$

$$\ge \sum _{j\in A}(x_0^j+x_{\alpha }^j-1)$$

$$\ge 3\lceil k/3\rceil -3\ge k-3,$$

$$S_g(x-v,A)=S_g(x,A)-3\text{ or }{S}_g(x-v,A)=S_g(x,A)-1.$$

$$\text{There exists no }h\in H_n\text{ such that the following conditions holds:}$$

$$0\text{ is the most frequent element in all multisests from }G\text{-presentation of }hx.$$

$$S_{\alpha }(x,A)=\sum _{j\in A}(x_0^j+x_{\alpha }^j)+\sum _{j\notin A}(x_{\beta }^j+x_{\gamma }^j)\ge x_0^1+x_0^2+x_0^n+x_{\alpha }^n=$$

$$=x_0^1+x_0^2+k-(x_{\beta }^n+x_{\gamma }^n)\ge x_0^1+x_0^2+k-2\min \{x_0^1,x_0^2\}\ge k.$$

Full text

1. Introduction

Phylogenetics is a science that models evolution. One of the central objects in phylogenetics is the tree model. In general, a statistical model is a parametric family of probability distributions. The tree model is based on rooted tree and finite set $B$ and gives us probability distribution on $B^{l}$ where $l$ is the number of leaves of the tree. The parameters are distribution on the root and transition matrices along the edges of the tree. A group-based model is a tree model where the set $B$ is a group which acts on itself and parameters are $G$-invariant.

Since everything is finite a distribution allowed by a tree model may be represented as a vector $(p_{1},\dots,p_{n})$ where $p_{i}$’s are nonnegative and sum to one. Thus a tree model may be regarded as a map from the parameter space to the $n$-dimensional vector space.

In algebraic phylogenetics we are interested in the geometric locus of all probability distributions allowed by a given model. Precisely, the Zariski closure of this locus is an algebraic variety and one is interested in its geometric and algebraic properties [Eri+04, Sul19].

For example one asks for polynomials defining the variety—so-called phylogenetic invariants—or properties of the singular set. In this article, we investigate the latter property, namely we show that for a well-known 3-Kimura model [Kim81], the singularities are always normal. This confirms a conjecture of Michałek [Mic13, Conjecture 9.5], [Mic15, Conjecture 12.1].

The $3$-parameter Kimura model is a group-based model given by the group $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Group-based models in general and the $3$-Kimura model in particular have been recently intensively studied within algebraic statistics [SS05, BW07, MRV17, DE15, DK09, Mic11, Mic14, MV17a, CF08, CFM15, CFM17, Don16, MV17].

Apart from the fact that it was an open conjecture, there are several important reasons to study normality of the $3$-Kimura model.

• •

Group-based models allow a monomial parametrization [SS05]. Thus, one may say that they are toric varieties. However, in pure mathematics one often requires a toric variety to be normal [Ful93]. The reason is that in such a case the variety admits a nice combinatorial description in terms of a fan [CLS11]. Our result in particular implies that the normal fan of the polytope associated to the $3$-Kimura model describes the toric variety representing the model.

• •

Not all group-based models give rise to normal toric varieties: for example for the group $\mathbb{Z}_{6}$ one obtains a nonnormal variety [DM12]. The normality also fails for the $2$-Kimura model. Thus the $3$-Kimura model is distinguished with respect to that regard.

• •

Normality played an important role in the $\mathbb{Z}_{2}$ group-based model [SX10, BW07, SS05].

• •

Normality of toric varieties provides automatic bounds on degrees of phylogenetic invariants [Stu96]. In a special case of a tree with six leaves this was used in a recent proof [MV17a] of the Sturmfels-Sullivant conjecture [SS05, Conjecture 30]. On that example normality was checked by computer using software Normaliz [Bru+]. Our proof, in particular, confirms normality in this case without the necessity to rely on computer software.

It would not be possible to obtain our theorem without many great previous results. We list the most important below.

• •

Application of Discrete Fourier Transform to unravel toric structure The DFT may be considered as a clever change of coordinates, that changes the parametrization of the phylogenetic model into one given by monomials. First such applications were made by Handy and Penny [HP89]. The toric structure was studied in detail in the work of Sturmfels and Sullivant [SS05] and Michałek [Mic15].

• •

Reduction to claw trees Recall that a claw tree is a tree with just one inner vertex. It is known that one can extend many properties that hold for claw trees to arbitrary trees. This technique is well-developped to obtain phylogenetic invariants [DK09]. Further, it is known that normality in case of claw trees implies normality for arbitrary trees. For phylogenetic group-based models it was first observed in [Mic11, Lemma 5.1]. The joining of trees is a special case of a more general construction of toric fiber products [Sul07, RS16, EKS14].

• •

Facet description The vertex description of the polytopes representing group-based models are well-known [SS05, Mic11, BW07]. However, obtaining facet description from the vertex one is hard in the general case, and for phylogenetic models in particular. For the 3-Kimura model such a description was provided in [MRV17].

First two results of the above allow us to translate the question about projective normality of the variety associated to the 3-Kimura model into a purely combinatorial statement about normality of a family of polytopes. We prove the normality using only combinatorial methods. Strong tool is the facet description of the polytope because it allows us to prove that a point lies inside of the polytope by checking inequalities.

2. The polytope of the 3-Kimura model

We start by fixing notation.

Let $G$ be the group $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Let us denote its elements by $0,\alpha,\beta,\gamma$. We also denote the elements of $\mathbb{Z}_{2}$ by $\mathbf{0},\mathbf{1}$.

Let $H_{n}$ be the set of the group-based flows of length $n$ of $G$, i.e.

[TABLE]

It is easy to see that $H_{n}$ is a subgroup of $G^{n}$.

The goal of this article is to prove normality of a family of polytopes for 3-Kimura model $P_{n}\subset\mathbb{R}^{4n}$ indexed by $n\in\mathbb{N}$. Before we formally define them, we introduce further notation.

We denote the coordinates of a point $x\in\mathbb{R}^{4n}$ by $x_{g}^{j}$ where $g\in G$ and $1\leq j\leq n$. Although we are using upper indices, there will be no ambiguity since we will not use any powers in this article.

Definition 1**.**

We say that the $G$-presentation of a point $x\in\mathbb{Z}_{\geq 0}^{4n}$ is an $n$-tuple $(G_{1},\dots,G_{n})$ of multisets of elements of $G$ such that the element $g\in G$ appears exactly $x_{g}^{j}$ times in the multiset $G_{j}$. We may identify the $n$-tuple $(g_{1},\dots,g_{n})\in G^{n}$ with the $n$-tuple of multisets $(\{g_{1}\},\dots,\{g_{n}\})$.

Definition 2**.**

The vertices of $P_{n}$ are all points of $\mathbb{R}^{4n}$ which $G$-presentations are the $n$-tuples from $H_{n}$. Therefore, $P_{n}$ is a convex hull of these points.

Equivalent characterization of $P_{n}$ is given in [MRV17]. The polytope is defined by the following inequalities:

• •

$x_{g}^{j}\geq 0$ for all $g\in G,1\leq j\leq n$,

• •

$x_{0}^{j}+x_{\alpha}^{j}+x_{\beta}^{j}+x_{\gamma}^{j}=1,\text{ for all }1\leq j\leq n$,

• •

For all $A\subseteq\{1,2,\dots,n\}$ with $|A|$ being an odd number:

[TABLE]

[TABLE]

[TABLE]

We denote the left sides of the last three inequalities by $S_{\alpha}(x,A),S_{\beta}(x,A),S_{\gamma}(x,A)$ respectively. Each inequality gives us a facet of $P_{n}$. We define

[TABLE]

The lattice generated by vertices of $P_{n}$ is

[TABLE]

where the last sum is in $G$. Alternatively, we can characterize $G$-presentations of points in $L_{n}\cap\mathbb{Z}_{\geq 0}^{n}$ as follows: Every multiset has the same size and sum of all elements in multisets is [math].

Definition 3**.**

Let $v(0)$ be the vertex corresponding to the $n$-tuple $(0,\dots,0)$ and $v(g)_{j,j^{\prime}}$ be the vertex corresponding to the $n$-tuple which has on $j$-th and $j^{\prime}$-th place $g$ and all other places 0.

Let $V_{n}$ be the following set of vertices of $P_{n}$:

[TABLE]

Our goal is to prove that $P_{n}$ is normal for every positive integer $n$. Let us recall that polytope $P_{n}$ is normal if every point in $kP_{n}\cap L_{n}$ can be written as a sum of $k$ lattice points from $P_{n}$. Normality of polytope is equivalent to the fact that the associated projective toric variety is projectively normal.

It is easy to check that $P_{1}$, $P_{2}$ and $P_{3}$ are normal. Hence, in this article we consider only $n\geq 4$.

3. symmetries of $P_{n}$

Polytope $P_{n}$ has a lot of symmetries that can be described by group actions on $\mathbb{R}^{4n}$:

• •

Action of $\mathbb{S}_{n}$:

For $\sigma\in\mathbb{S}_{n}$ and $x\in\mathbb{R}^{4n}$ we define $\sigma(x)_{g}^{\sigma(j)}=x_{g}^{j}$. Intuitively, we only permute quadruples of coordinates by the upper index.

• •

Action of $H_{n}$:

For $h=(g_{1},\dots,g_{n})\in H_{n}$ and $x\in\mathbb{R}^{4n}$ and we define $(hx)_{g}^{j}=x_{(g+g_{j})}^{j}$. Intuitively, if we look at $G$-presentation of a point in $\mathbb{Z}^{4n}$ we add $g_{j}$ to elements in $G_{j}$.

• •

Action of $\text{Aut}(G)$:

For $\varphi\in\text{Aut}(G)$ and $x\in\mathbb{R}^{4n}$ we define $\varphi(x)_{\varphi(g)}^{j}=x_{g}^{j}$. Again, if we consider $G$-presentation of $x$ this is application of the automorphism $\varphi$ to elements in multisets.

All of these actions only permute coordinates in $\mathbb{R}^{4n}$ and therefore are automorphisms of $\mathbb{R}^{4n}$ as a vector space. It can be easily verified that they map vertices of $P_{n}$ to vertices of $P_{n}$ and therefore preserve $P_{n}$. It follows that these actions restricted to $L_{n}$ are automorphisms of this lattice.

We want to prove that every point $x\in kP_{n}\cap L$ decomposes to a sum of $k$ lattice points from $P_{n}$. It is enough to prove it for an image of $x$ under any of group actions described above, since $Ax=v_{1}+\dots+v_{k}$ implies $x=A^{-1}v_{1}+\dots+A^{-1}v_{k}$ for any group action $A$.

Let us define linear ordering on multisets of four real numbers with sum equal to $k$. Consider two multisets $\{a,b,c,d\}$ and $\{a^{\prime},b^{\prime},c^{\prime},d^{\prime}\}$. Without loss of generality we may assume $a\geq b\geq c\geq d$ and $a^{\prime}\geq b^{\prime}\geq c^{\prime}\geq d^{\prime}$. We say

[TABLE]

Consider $x\in kP\cap L$. If we order multisets $\{x_{0}^{j},x_{\alpha}^{j},x_{\beta}^{j},x_{\gamma}^{j}\}$ then by acting with corresponding permutation from $\mathbb{S}_{n}$ we can ensure that multiset for $j=n$ is the smallest one in this ordering.

Let us denote $g_{j}$ the most frequent element (or one of the most frequent elements) in $j$-th multiset from $G$-presentation of $x$, i.e. $x_{g_{j}}^{j}=\max\{x_{0}^{j},x_{\alpha}^{j},x_{\beta}^{j},x_{\gamma}^{j}\}$. Then by acting with $(g_{1},g_{2},\dots,g_{n-1},g_{1}+\dots+g_{n-1})\in H_{n}$ we obtain a point $x$ in which the element [math] is the most frequent in all multisets except the last one.

This means that if we need to, for a point $x\in kP_{n}\cap L$ we may without loss of generality assume the following two facts:

[TABLE]

[TABLE]

We add another definition:

Definition 4**.**

Let $x\in kP_{n}\cap L_{n}$. The vertex $v$ of $P_{n}$ is called $x$-good if all coordinates of the point $x-v$ are non-negative.

4. Preliminary results

Lemma 1**.**

Let $x\in kP_{n}\cap L_{n}$ and $j\in\{1,2,\dots,n\}$. Suppose that $x_{0}^{j}\geq x_{\alpha}^{j},x_{\beta}^{j},x_{\gamma}^{j}$ and let $g\in\{\alpha,\beta,\gamma\}$. Then $x_{0}^{j}+x_{g}^{j}\geq\lceil k/3\rceil$ and the equality holds if and only if $x_{g}^{j}=0$ and $x_{h}^{j}=k/3$ for $h\neq g$.

Proof.

[TABLE]

We divide by 3 and realise that $x_{g}^{j}$ are integers to obtain wanted inequality. The part about the equality is obvious. ∎

Lemma 2**.**

Let $x\in L_{n}$ be a point such that $x_{0}^{j}+x_{\alpha}^{j}+x_{\beta}^{j}+x_{\gamma}^{j}=k$ for all $1\leq j\leq n$. Let $g\in\{\alpha,\beta,\gamma\}$ and $A\subseteq\{1,2,\dots,n\}$ be a set of odd cardinality. Then

[TABLE]

Proof.

We consider only the case $g=\alpha$, other cases are analogous.

Consider the homomorphism

[TABLE]

[TABLE]

For $x\in L_{n}$ we get

[TABLE]

Therefore $\sum_{j=1}^{n}(x_{\beta}^{j}+x_{\gamma}^{j})$ must be even. This implies

[TABLE]

[TABLE]

[TABLE]

∎

The following lemma implies that it is sufficient to consider only such points $x$ for which the following condition holds:

[TABLE]

Lemma 3**.**

Suppose that for every positive integers $k,m$ and every $x\in kP_{m}\cap L_{m}$ such that $x_{g}^{j}<k$ for all $g,j$ we can write $x$ as a sum of $k$ vertices of $P_{m}$. Then $P_{n}$ is normal for every positive integer $n$.

Proof.

Proof by induction on $n$. $P_{1},\ P_{2}$ and $P_{3}$ are normal.

Suppose that $P_{n-1}$ is normal. We prove that also $P_{n}$ is normal. Consider a point $x\in kP_{n}\cap L_{n}$. If $x_{g}^{j}<k$ for all $g,j$ then $x$ decomposes by assumption. Therefore we may assume that $x_{g}^{j}=k$ for some $g,j$. By acting with suitable permutation we can assume that $j=n$ and then by acting with $(g,0,\dots,0,g)$ we obtain $g=0$.

Consider now the projection $\pi:\mathbb{R}^{4n}\rightarrow\mathbb{R}^{4(n-1)}$ on the first $4(n-1)$ coordinates. Since $x\in kP_{n}$ there exist positive real numbers $\lambda_{1},\dots,\lambda_{s}$ with $\lambda_{1}+\dots+\lambda_{s}=k$ such that $\lambda_{1}v_{1}+\dots+\lambda_{s}v_{s}=x$, where $v_{1},\dots,v_{s}$ are some vertices of $P_{n}$. But $(v_{i})_{0}^{n}\leq 1$ and $x_{0}^{n}=k$ implies $(v_{i})_{0}^{n}=1$ for all $i$.

Consequently, $\pi(v_{i})$ is a vertex of $P_{n-1}$ and $\pi(x)\in kP_{n-1}$. By induction hypothesis $\pi(x)$ decomposes to $\pi(x)=u_{1}+\dots+u_{k}$, where $u_{i}$ are vertices of $P_{n-1}$.

Now we simply put $u_{i}^{\prime}\in\pi^{-1}(u_{i})$ such that $(u_{i}^{\prime})_{0}^{n}=1$ and $(u_{i}^{\prime})_{g}^{n}=0$ for $g\neq 0$. Obviously all $u_{i}^{\prime}$ are vertices of $P_{n}$ and we have $x=u_{1}^{\prime}+\dots+u_{n}^{\prime}$. ∎

Lemma 4**.**

Let $x\in 2P_{n}\cap L_{n}$. Then $x$ can be written in the form $x=v+v^{\prime}$ where $v,v^{\prime}$ are vertices of $P_{n}$.

Proof.

We assume, without loss of generality, $(1)$, $(2)$ and $(3)$. Thus $x_{0}^{j}>0$ for $1\leq j\leq n-1$. If also $x_{0}^{n}>0$ then $x=v(0)+(x-v(0))$. Further, $x-v(0)$ must be a vertex of $P_{n}$ since it has non-negative coordinates and sum of elements in $G$-presentation of $x-v(0)$ is [math] since it is [math] for both $x$ and $v_{0}$. If $x_{0}^{n}=0$ then by acting with suitable $\varphi\in\text{Aut}(G)$ we have $x_{\alpha}^{n}=x_{\beta}^{n}=1$ since $x_{g}^{n}<2$ for all $g$ by condition $(3)$.

Since $S_{\gamma}(x,\{n\})\geq 2$ at least one of the numbers $x_{g}^{j}$ for $g=\alpha,\beta$; $1\leq j\leq n-1$ is greater than 0. Then $x=v(g)_{j,n}+(x-v(g)_{j,n})$ for such $g,j$. By the same arguments as above $(x-v(g)_{j,n})$ must be a vertex of $P_{n}$. ∎

Lemma 5**.**

Let $x\in kP_{n}\cap L_{n}$ be such that there are at least three multisets $\{k/3,k/3,k/3,0\}$ in $G$-presentation of $x$. Then $x=y+v$, where $v$ is a vertex of $P_{n}$ and $y\in(k-1)P_{n}$.

Proof.

By acting with suitable permutation from $\mathbb{S}_{n}$ we may assume that these three multisets are the first three. Then by acting with suitable $(g_{1},g_{2},\dots,g_{n})\in H_{n}$ we may assume $x_{0}^{j}=x_{\alpha}^{j}=x_{\beta}^{j}=k/3$ for $j=1,2,3$. We describe the $G$-presentation of $v$ (which is a $n$-tuple $(g_{1},g_{2},\dots,g_{n})$ of elements from $G$). We pick the last $n-2$ elements arbitrarily, the only condition is that $g_{j}$ belongs to the $j$-th multiset from $G$-presentation of $x$. Then we pick $g_{1}$ and $g_{2}$ such that sum of this $n$-tuple is 0 and $g_{1},g_{2},g_{3}$ are not all equal. Since $g_{1}$ and $g_{2}$ can be any from $0,\alpha,\beta$, it is possible.

Now we need to check that $x-v=y\in(k-1)P_{n}$. We only need to check the inequalities for sets $A$. However, if we try to compute $S_{g}(y,A)$ we always get at least $k-2$ already on the first three coordinates. Therefore, due to Lemma 2 the inequalities hold. ∎

From now, we may assume that $x\in kP_{n}\cap L_{n}$ satisfies the following condition since the other case is covered by the previous lemma.

[TABLE]

Lemma 6**.**

Let $x\in 3P_{n}\cap L_{n}$ satisfy $(2)$, $(4)$. Let $A\subseteq\{1,2,\dots,n\}$ be a set with $|A|\geq 5$. Then $S_{g}(x-v,A)\geq 2$ for any $g\in\{\alpha,\beta,\gamma\}$ and any $x$-good vertex $v$ of $P_{n}$.

Proof.

Let $B\subseteq\{1,2,\dots,n\}$ be the set of those indices for which multisets in $G$-presentation of $x$ are equal to $\{1,1,1,0\}$. This together with condition $(2)$ yields $x_{0}^{j}\geq 2$ for $j\not\in B$. Condition $(4)$ implies $|B|\leq 2$. It follows that

[TABLE]

∎

5. The proof

5.1. Idea of the proof

We prove for all positive integers $k,n$ that every point $x\in kP_{n}\cap L_{n}$ can be written in the form $x=y+v$ where $y\in(k-1)P_{n}$ and $v$ is a vertex of $P_{n}$. This, of course, means that also $y\in L_{n}$ since all vertices of $P_{n}$ belong to $L_{n}$ and this implies that $P_{n}$ is normal.

Consider a point $x\in kP_{n}\cap L_{n}$. It is sufficient to consider $k\geq 3$ because the case $k=2$ is solved by Lemma 4. Without loss of generality, from now we will suppose that $x$ satisfies $(1)$, $(2)$, $(3)$ and $(4)$. To conclude we need to pick an $x$-good vertex $v$ and then check that $y=x-v$ belongs to $(k-1)P_{n}$. We prove this by checking all inequalities from facet characterization of $P_{n}$ for every set $A$ with odd cardinality.

Regarding the vertex $v$, we show we can always use some vertex $v\in V_{n}$ as in Definition 3.

5.2. Big sets $A$

Proposition 7**.**

Let $x\in kP_{n}\cap L_{n},k\geq 3$ satisfy $(1)-(4)$ and let $A\subseteq\{1,2,\dots,n\}$ be a set with odd cardinality.

• a)

If $|A|\geq 5$ then $S_{g}(x-v,A)\geq k-1$ for any $g\in\{\alpha,\beta,\gamma\}$ and any $x$-good vertex $v$ of $P_{n}$.

• b)

If $|A|=3,n\not\in A$ and $x$ satisfies $(4)$ then $S_{g}(x-v,A)\geq k-1$ for any $g\in\{\alpha,\beta,\gamma\}$ and any $x$-good vertex $v$ of $P_{n}$.

Proof.

Let $y=x-v$. Clearly, it is sufficient to prove the inequality for $g=\alpha$. We begin with part a):

[TABLE]

The last inequality holds for $k\geq 4$. Case $k=3$ is covered in Lemma 6. We also used Lemma 1 and $|A\setminus\{n\}|\geq 4$. Inequality $S_{g}(y,A)\geq k-2$ together with Lemma 2 implies $S_{g}(y,A)\geq k-1$.

Proof of part b) is similar:

[TABLE]

where we again used Lemma 1. Lemma 2 implies that $S_{g}(y,A)\neq k-2$. Therefore the only bad case is when we have an equality. This is possible only if we have equality everywhere, in particular $x_{0}^{j}=x_{\beta}^{j}=x_{\gamma}^{j}=k/3$ for all $j\in A$. But this means that $x$ does not satisfy $(4)$ which is a contradiction. ∎

Therefore it is sufficient to check inequalities for $|A|=1$ and $|A|=3$ such that $n\in A$.

5.3. Small sets $A$

Since $x\in kP_{n}$ we have the inequalities $S_{g}(x,A)\geq k$ for any $g$ and any set $A$ with odd cardinality. For big sets $A$ discussed in the section 5.2 we have not used them. However, we use them for smaller sets. Our first step is to observe how does $S_{g}(x,A)$ change when we subtract some vertex $v\in V_{n}$ from $x$.

Lemma 8**.**

Let $x\in kP_{n}\cap L_{n}$, $v\in V_{n}$, $g\in\{\alpha,\beta,\gamma\}$ and $|A|=1$ or $|A|=3$ with $n\in A$. Then

[TABLE]

Moreover, for $|A|=1$ we have $S_{g}(x-v,A)=S_{g}(x,A)-1$ if and only if one of the following conditions holds:

• •

$v=v(0)$**

• •

$v=v(g)_{j,n}$* for any $1\leq j\leq n-1$*

• •

$v=v(g^{\prime})_{j,n}$* for $g^{\prime}\neq g$ and $A=\{j\}$ or $|A|=\{n\}$*

Also $S_{g}(x-v,\{n\})\geq k-1$.

Proof.

For the first part, one checks how many summands in $S_{g}(x,A)$ will decrease by 1 when we subtract $v$. The last part is clear consequence since $S_{g}(x,\{n\})\geq k$ for $x\in kP_{n}$. ∎

Now we consider the following:

Proposition 9**.**

Let $x\in kP_{n}\cap L_{n}$ satisfy conditions $(1)-(4)$. Suppose that [math] is also the most frequent element in the $n$-th multiset from $G$-presentation of $x$. Then $x-v(0)\in(k-1)P_{n}$.

Proof.

Obviously, every multiset from $G$-presentation of $x$ contains [math] so $x-v(0)$ has non-negative coordinates and therefore $v(0)$ is $x$-good. Inequalities for sets with $|A|\geq 3$ hold for $x-v$ by Proposition 7 since for sets with $|A|=3$ and $n\in A$ we can use same arguments. Inequalities for $|A|=1$ hold by Lemma 8 since we are subtracting $v(0)$. It follows that $x-v(0)\in(k-1)P_{n}$. ∎

The previous proposition implies that we can assume that for $x\in kP_{n}\cap L_{n}$ satisfying $(1),(2),(3)$ also the following condition holds:

[TABLE]

Proposition 10**.**

Let $x\in kP_{n}\cap L_{n}$ satisfy $(1)-(5)$. Then:

• a)

$x$* does not belong to any facet $kF_{g}(A)$ for $|A|=3,n\in A$, i.e. $S_{g}(x,A)>k$ for all such $A$ and $g=\alpha,\beta,\gamma$.*

• b)

$S_{g}(x-v,A)\geq k-1$* for all $v\in V_{n}$, $g\in\{\alpha,\beta,\gamma\}$ and $|A|=3,n\in A$.*

Proof.

We prove part a) by contradiction: Suppose that we have an equality for $A=\{1,2,n\}$ and $g=\alpha$. We may get to this situation by acting with suitable $\sigma\in\mathbb{S}_{n}$ and $\varphi\in\text{Aut}(G)$. We compute $S_{\alpha}(x,A)$:

[TABLE]

[TABLE]

An equality holds only if there is equality in all inequalities. In particular, it means that $x_{0}^{1}=x_{0}^{2}=x_{\beta}^{n}=x_{\gamma}^{n}$ and $x_{\alpha}^{1}=0$. But from ordering of multisets, we get that also some $x_{g_{0}}^{1}=x_{0}^{1}$ for $g_{0}=\beta$ or $g_{0}=\gamma$. By acting with $(g_{0},0,\dots,0,g_{0})\in H$ we get to the situation where [math] is the most frequent also in $n$-th multiset and still is also most frequent on the first one. This is a contradiction with condition $(5)$.

We continue with proof of part b). Part a) together with Lemma 2 implies that $S_{g}(x,A)\geq k+2$. Consequently, Lemma 8 implies $S_{g}(x-v,A)\geq k-1$ for any $v\in V_{n}$. ∎

Proposition 11**.**

Let $x\in kP_{n}\cap L_{n}$ satisfy $(1)-(5)$ and $x_{0}^{n}>0$. Then $x-v(0)\in(k-1)P_{n}$.

Proof.

Clearly, $v(0)$ is $x$-good. Inequalities for $|A|\geq 3$ hold by Propositions 7 and 10. For $|A|=1$ we have $S_{g}(x,A)\geq k$, then by Lemma 8 we get $S_{g}(x-v(0),A)\geq k-1$. Since all inequalities hold $x-v(0)\in(k-1)P_{n}$. ∎

Therefore we are left only with the case $x_{0}^{n}=0$.

5.4. Special case $x_{0}^{n}=0$

In this case we will subtract a vertex $v(g)_{j,n}$ for a special choice of $g$ and $j$. Propositions 7 and 10 and Lemma 8 imply that it is enough to check inequalities for $|A|=1,A\neq\{n\}$. We distinguish two cases depending on whether $x$ lies or does not lie on a facet $kF_{g}(A)$ for such $A$.

Proposition 12**.**

Let $x\in kP_{n}\cap L_{n}$ satisfy $(1)-(5)$, $x_{0}^{n}=0$ and $x$ does not belong to any facet $kF_{g}(A)$ for $|A|=1,A\neq\{n\}$. Then there exists a vertex $v\in V_{n}$ such that $x-v\in(k-1)P_{n}$.

Proof.

For any $v\in V_{n}$ Lemma 8 implies that for any $A$ with $|A|=1,A\neq\{n\}$ we have $S_{g}(x-v,A)\geq S_{g}(x,A)-3\geq k-1$. We used Lemma 2 to deduce inequality $S_{g}(x,A)\geq k+2$. Therefore, inequalities for every set $A$ hold for any $x$-good vertex $v\in V_{n}$, since bigger sets are taken care of by Propositions 7 and 10. Consequently, it is sufficient to pick any $x$-good vertex $v\in V_{n}$.

At least two of the numbers $x_{g}^{n}$ for $g\in\{\alpha,\beta,\gamma\}$ must be non-zero by condition $(3)$ and the fact that $x_{0}^{n}=0$. Without loss of generality, let those two coordinates be $x_{\alpha}^{n}$ and $x_{\beta}^{n}$.

Since $S_{\gamma}(x,\{n\})\geq k$ and $x_{0}^{n}+x_{\gamma}^{n}<k$, at least one of the numbers $x_{\alpha}^{j}$ and $x_{\beta}^{j}$ for $1\leq j\leq n-1$ must be non-zero. Let it be $x_{g_{0}}^{j}$. For $v=v(g_{0})_{j,n}$ all coordinates of $x-v$ are non-negative since condition $(2)$ implies $x_{0}^{j}>0$ for $1\leq j\leq n-1$. This means we have found $x$-good vertex $v\in V_{n}$ and the proposition is proved. ∎

If $x$ belongs to a facet we prove that it belongs to only one facet and that we can as well subtract a vertex $v\in V_{n}$:

Proposition 13**.**

Let $x\in kP_{n}\cap L_{n}$ satisfy $(1)-(4)$, $x_{0}^{n}=0$ and $x$ belongs to some facet $kF_{g}(A)$ for $|A|=1,A\neq\{n\}$. Then

• a)

$x$* belongs to only one such facet.*

• b)

There exists a vertex $v\in V_{n}$ such that $x-v\in(k-1)P_{n}$.

Proof.

By acting with suitable permutation from $\mathbb{S}_{n}$ and $\varphi\in\text{Aut}(G)$ we can get to situation where $x\in kF_{\alpha}(\{1\})$. We have

[TABLE]

To get an equality, there must be an equality in all inequalities, specifically $x_{\alpha}^{1}=x_{\beta}^{j}=x_{\gamma}^{j}=0$ for all $2\leq j\leq n$ and $x_{0}^{1}=x_{\alpha}^{n}=\max_{g\in G}\{x_{g}^{n}\}$.

Assumption that $x$ belongs to a facet give us strong conditions. It is easy to see that $x$ cannot belong to some other facet $kF_{\alpha}(\{j\})$ for $j<n$ because it would imply $x_{\beta}^{1}=x_{\gamma}^{1}=0$. But this is a contradiction with condition $(3)$. Also $x$ cannot belong to some $kF_{\beta}(\{j\})$ for $1\leq j<n$ because it would imply $x_{\alpha}^{i}=x_{\beta}^{i}=x_{\gamma}^{i}$ for $i\neq 1,j,n$ which is again a contradiction with $(3)$. Same arguments hold for $kF_{\gamma}(\{j\})$. This proves part a).

For part b), by the same arguments as in the proof of Proposition 12 for any $x$-good vertex $v\in V_{n},\ g\in\{\alpha,\beta,\gamma\}$ and set $A$ we have $S_{g}(x-v,A)\geq k-1$, except the case when $g=\alpha$ and $A=\{1\}$.

Since $k\leq S_{\alpha}(x,\{n\})=x_{\beta}^{1}+x_{\gamma}^{1}+x_{\alpha}^{n}$ at least one of the numbers $x_{\beta}^{1}$, $x_{\gamma}^{1}$ must be greater than 0. Also one of the numbers $x_{\beta}^{n}$ and $x_{\gamma}^{n}$ is greater than zero by condition $(3)$.

If both numbers $x_{g}^{n}$ and $x_{g}^{1}$ are greater than zero for $g=\beta$ or $g=\gamma$ then the vertex $v=v(g)_{1,n}$ is $x$-good. By Lemma 8 also $S_{\alpha}((x-v),\{1\})\geq k-1$ and therefore $x-v\in(k-1)P_{n}$.

Suppose the opposite, i.e. $x_{\beta}^{1}=0$ and $x_{\gamma}^{n}=0$ (we can get to this case by acting with $\varphi\in\text{Aut}(G)$). Then $S_{\gamma}(x,\{n\})\geq k$ implies that at least one of the numbers $x_{\alpha}^{j}$ for $2\leq j\leq n-1$ is greater than 0. Then we can subtract $v=v(\alpha)_{j,n}$ for such $j$. Again $x-v$ has non-negative coordinates and by Lemma 8 $S_{\alpha}((x-v),\{1\})\geq k-1$.

∎

Theorem 14**.**

Polytope representing 3-Kimura model $P_{n}$ is normal for every positive integer $n$.

Proof.

Consider point $x\in kP_{n}\cap L_{n}$ for some positive integer $k$. If $k=2$ then $x$ decomposes due to Lemma 4. To prove normality of $P_{n}$ it is sufficient for $k\geq 3$ to prove that there exists a vertex $v$ of $P_{n}$ such that $x-v\in(k-1)P_{n}$. Also it is sufficient to consider only points $x$ which satisfy $(1)-(3)$. The existence of such $v$ is implied by Lemma 5 and Propositions 9, 11, 12 and 13. ∎