Generalized quasispecies model on finite metric spaces: isometry groups and spectral properties of evolutionary matrices

Abstract

The quasispecies model introduced by Eigen in 1971 has close connections with the isometry group of the space of binary sequences relative to the Hamming distance metric. Generalizing this observation we introduce an abstract quasispecies model on a finite metric space X together with a group of isometries $\Gamma$ acting transitively on X. We show that if the domain of the fitness function has a natural decomposition into the union of tG-orbits, G being a subgroup of $\Gamma$, then the dominant eigenvalue of the evolutionary matrix satisfies an algebraic equation of degree at most $t\cdot \mathrm{rk}_{\mathbf {Z}} R$, where R is the orbital ring that is defined in the text. The general theory is illustrated by three detailed examples. In the first two of them the space X is taken to be the metric space of vertices of a regular polytope with the natural “edge” metric, these are the cases of a regular m-gon and of a hyperoctahedron; the final example takes as X the quotient rings $\mathbf {Z}/p^n\mathbf {Z}$ with p-adic metric.

Appendices

A Orbital rings

A.1 Orbital ring: some properties and examples

Here we provide more details on orbital rings defined in Sect. 3.2.

Lemma A.1

If the triple $(X,d,\Gamma )$ is symmetric in the sense of Definition 3.3 then matrices ${{\varvec{M}}}_A$ are symmetric and commute pairwise. Moreover, there are integer non-negative structural constants $\mu _{AB}^C=\mu _{BA}^C$, where $A, B, C \in \mathrm{Orb}$, such that

$$\begin{aligned} {{\varvec{M}}}_A{{\varvec{M}}}_B=\sum _{C\in \mathrm{Orb}} \mu _{AB}^C {{\varvec{M}}}_C. \end{aligned}$$

(A.1)

Proof

Let $(x,z)\in A$. In view of Definition 3.3 there exists an isometry $\gamma =\gamma (x,z)\in \Gamma$ such that $\gamma x=z$ and $\gamma z=x$. Thus, $(z,x)\in A$ and ${{\varvec{M}}}_A$ is symmetric.

Moreover, it follows from the definition that for any $x, z \in X$ the corresponding matrix entry

$$\begin{aligned} ({{\varvec{M}}}_A{{\varvec{M}}}_B)_{x,z}=\#\{y\in X\mid (x,y)\in A,\;(y,z)\in B\}. \end{aligned}$$

(A.2)

On the other hand, for the same transposing isometry $\gamma =\gamma (x,z)\in \Gamma$

$$\begin{aligned}&\#\{y\in X\mid (x,y)\in A,\;(y,z)\in B\}= \#\{y\in X\mid (z,y)\in B,\;(y,x)\in A\}=\\&\quad =\#\{\gamma y\in X\mid (\gamma z, \gamma y)\in B,\;(\gamma y, \gamma x)\in A\}\\&\quad =\#\{\gamma y\in X\mid (x, \gamma y)\in B,\;(\gamma y, z)\in A\}. \end{aligned}$$

It follows that $({{\varvec{M}}}_A{{\varvec{M}}}_B)_{x,z}=({{\varvec{M}}}_B{{\varvec{M}}}_A)_{x,z}$ and ${{\varvec{M}}}_A{{\varvec{M}}}_B={{\varvec{M}}}_B{{\varvec{M}}}_A$.

The pair (x, z) defines a $\Gamma$-orbit C. For $g \in \Gamma$ we have the same as in (A.2) non-negative number

$$\begin{aligned} ({{\varvec{M}}}_A{{\varvec{M}}}_B)_{g x, g z}=\#\{gy\in X\mid (g x, gy)\in A,\;(g y, g z)\in B\}. \end{aligned}$$

Thus, in the product ${{\varvec{M}}}_A{{\varvec{M}}}_B$ the matrix entries with subindices (x, z) and (gx, gz), belonging to the same $\Gamma$-orbit $C\subset X\times X$, have the same value $\mu _{AB}^C$ for any $g\in \Gamma$. It means by the definition of the orbital matrices ${{\varvec{M}}}$ and the constants $\mu$ that ${{\varvec{M}}}_A{{\varvec{M}}}_B=\cdots +\mu _{AB}^C\cdot {{\varvec{M}}}_C+\cdots$. Hence, (A.1) holds for some non-negative integer constants $\mu _{AB}^C$, depending only on the orbits A, B, C. The lemma is proved. $\square$

Moreover, we have proved

Theorem A.2

All $\mathbf {Z}$-linear combinations of ${{\varvec{M}}}_A$, $A\in \mathrm{Orb}$, compose a commutative unital ring $R=R(X,d,\Gamma )$ with unity ${{\varvec{M}}}_\Delta ={{\varvec{I}}}$ called the orbital ring associated with the symmetric triple $(X,d,\Gamma )$. As $\mathbf {Z}$-module R is free of rank $\mathrm{rk}_{\mathbf {Z}}R=|\mathrm{Orb}|=|\Gamma _0\backslash \Gamma /\Gamma _0|$.

Example A.3

Let $X=\{0,1\}^2$ be the binary square with points $x_0=[0,0]$, $x_1=[0,1]$, $x_2=[1,0]$, $x_3=[1,1]$ (binary representation of indices) with the Hamming metric d. Let $\Gamma \cong D_4$ (the dihedral group of order 8) be the group of all isometries of X. Then the triple $(X,d,\Gamma )$ is symmetric.

The set $\mathrm{Orb}$ consists of three orbits (corresponding to the three orbits, namely, d-spheres $\{x_0\}=S_0(x_0)$, $\{x_1, x_2\}=S_1(x_0)$, $\{x_3\}=S_2(x_0)$, of the stabilizer $\Gamma _0=\mathrm{St}_{\Gamma }(x_0)\cong \mathbf {Z}/2\mathbf {Z}$ acting on X) represented by matrices

$$\begin{aligned} {{\varvec{M}}_0}={{\varvec{I}}}=\left[ \begin{array}{cccc} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\\ \end{array}\right] ,\quad {{\varvec{M}}_1}=\left[ \begin{array}{cccc} 0&{}\quad 1&{}\quad 1&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 1\\ 1&{}\quad 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 1&{}\quad 1&{}\quad 0\\ \end{array}\right] ,\quad {{\varvec{M}}_2}=\left[ \begin{array}{cccc} 0&{}\quad 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0\\ \end{array}\right] . \end{aligned}$$

The multiplication table of these matrices in $R=R(X,d,\Gamma )$ is as follows:

$\times$	${{\varvec{I}}}$	${{\varvec{M}}_1}$	${{\varvec{M}}_2}$
${{\varvec{I}}}$	${{\varvec{I}}}$	${{\varvec{M}}_1}$	${{\varvec{M}}_2}$
${{\varvec{M}}_1}$	${{\varvec{M}}_1}$	$2{{\varvec{I}}}+2{{\varvec{M}}_2}$	${{\varvec{M}}_1}$
${{\varvec{M}}_2}$	${{\varvec{M}}_2}$	${{\varvec{M}}_1}$	${{\varvec{I}}}$

Remark A.4

We have not yet applied the triangle inequality $d(x,z)\le d(x,y)+d(y,z)$. It can be used for the construction of a graded ring $\mathrm{gr}\, R=\mathrm{gr}\, R(X,d,\Gamma )$. Consider the following increasing filtration on R:

$$\begin{aligned} R_{-1}=0<\mathbf {Z}=R_0<R_1<\dots <R_N=R\,\quad \text{ for } \mathbf {Z}\text{-modules }\;\;R_k=\bigoplus _{\mathrm{deg} A\le k}\mathbf {Z}{{\varvec{M}}}_A. \end{aligned}$$

It follows from the definition and the triangle inequality that $R_i\cdot R_j\subseteq R_{i+j}$. Hence, we can attach to the triple $(X,d,\Gamma )$ the graded ring

$$\begin{aligned} \mathrm{gr}\, R=\bigoplus _{ k=0}^{N} R_k/R_{k-1}. \end{aligned}$$

For instance, in the above Example A.3 (here ${{\varvec{M}}}_k\in \mathrm{gr}_k R=R_k/R_{k-1}$ is viewed as the corresponding element of $R_k$ modulo $R_{k-1}$):

$\times$	${{\varvec{I}}}$	${{\varvec{M}}_1}$	${{\varvec{M}}_2}$
${{\varvec{I}}}$	${{\varvec{I}}}$	${{\varvec{M}}_1}$	${{\varvec{M}}_2}$
${{\varvec{M}}_1}$	${{\varvec{M}}_1}$	$2{{\varvec{M}}_2}$	0
${{\varvec{M}}_2}$	${{\varvec{M}}_2}$	0	0

A.2 Orbital rings for polygonal landscapes

We use notation of Sect. 5.1. The orbital matrices ${{\varvec{M}}}_k={{\varvec{M}}}_{A_k}$ have the following entries: $({{\varvec{M}}}_k)_{ab}=1$ if $d(a,b)=k$ and $({{\varvec{M}}}_k)_{ab}=0$ otherwise (for a square $X_4$, see Example A.3).

The multiplication in the commutative ring $R_l=R(X_l,d,D_l)$ is slightly different for the cases of odd and even l. In both cases we have ${{\varvec{M}}}_0{{\varvec{M}}}_k={{\varvec{M}}}_k$ since ${{\varvec{M}}}_0={{\varvec{I}}}$ is the unity of $R_l$.

• 1. Case $l=2N+1$. It can be checked that

  $$\begin{aligned} {{\varvec{M}}}^2_k= & {} 2{{\varvec{M}}}_0+{{\varvec{M}}}_{2k}\quad \text{ if }\; 0<2k\le N,\quad {{\varvec{M}}}^2_k\\= & {} 2{{\varvec{M}}}_0+{{\varvec{M}}}_{l-2k} \quad \text{ if }\; N<2k\le 2N=l-1. \end{aligned}$$

  Also we have for $0<k<j\le N$:

  $$\begin{aligned} {{\varvec{M}}}_k{{\varvec{M}}}_j= & {} {{\varvec{M}}}_{j-k}+{{\varvec{M}}}_{j+k}\quad \text{ if }\; 0<j+k\le N, \quad {{\varvec{M}}}_k{{\varvec{M}}}_j\\= & {} {{\varvec{M}}}_{j-k}+{{\varvec{M}}}_{l-j-k} \quad \text{ if }\; N<j+k< 2N=l-1. \end{aligned}$$

• 2. Case $l=2N$. We also check that

  $$\begin{aligned} {{\varvec{M}}}^2_k=2{{\varvec{M}}}_0+{{\varvec{M}}}_{2k}\quad \text{ if }\; 0<2k< N,{{\varvec{M}}}^2_k=2{{\varvec{M}}}_0+{{\varvec{M}}}_{l-2k} \text{ if }\; N<2k< 2N=l. \end{aligned}$$

  If $2k=N$ then ${{\varvec{M}}}^2_k=2{{\varvec{M}}}_0+2{{\varvec{M}}}_{N}$. If $k=N$ then ${{\varvec{M}}}^2_N={{\varvec{M}}}_0$.

Also we have for $0<k<j< N$:

$$\begin{aligned} {{\varvec{M}}}_k{{\varvec{M}}}_j= & {} {{\varvec{M}}}_{j-k}+{{\varvec{M}}}_{j+k}\quad \text{ if }\; 0<j+k< N, \quad {{\varvec{M}}}_k{{\varvec{M}}}_j\\= & {} {{\varvec{M}}}_{j-k}+{{\varvec{M}}}_{l-j-k} \quad \text{ if }\; N<j+k< 2N=l. \end{aligned}$$

If $k+j=N$ we have ${{\varvec{M}}}_k{{\varvec{M}}}_j={{\varvec{M}}}_{j-k}+2{{\varvec{M}}}_{N}$. If $j=N$ then ${{\varvec{M}}}_k{{\varvec{M}}}_N={{\varvec{M}}}_{N-k}$.

Remark A.5

It follows that the $\mathbf {Z}$-linear mapping $\rho :R_l\longrightarrow \mathbf {Z}[2\cos (2\pi /l)]$ such that $\rho ({{\varvec{M}}}_0)=1$, $\rho ({{\varvec{M}}}_k)=2\cos (2\pi k/l)$ ($k=1,\dots , N$ for the case $l=2N+1$ and $k=1,\dots , N-1$ for the case $l=2N$) and, if $l=2N$, $\rho ({{\varvec{M}}}_N)=-\,1$, is, in fact, a ring homomorphism. Note also that $\mathbf {Z}[2\cos (2\pi /l)]=\mathbf {R}\cap \mathbf {Z}[\varepsilon ]$ is the ring of integers of the real field $\mathbf {R}\cap \mathbf {Q}[\varepsilon ]$, $Q[\varepsilon ]$ being the cyclotomic field, since $2\cos (2\pi /l)=\varepsilon +\varepsilon ^{-1}$.

A.3 Orbital rings for hyperoctahedral landscapes

Here we use notation of Sect. 5.2. The orbital matrices ${{\varvec{M}}}_k={{\varvec{M}}}_{A_k}$ have the following entries: $({{\varvec{M}}}_k)_{ab}=1$ if $d(a,b)=k$ and $({{\varvec{M}}}_k)_{ab}=0$ otherwise. For a 2-octahedron (a square) $X_2$ see Example A.3, for a 3-octahedron $X_3$ (a classical one)

$$\begin{aligned} {{\varvec{M}}_0}={{\varvec{I}}}= & {} \left[ \begin{array}{cccccc} 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 1\\ \end{array}\right] ,\, {{\varvec{M}}_1}=\left[ \begin{array}{cccccc} 0&{}\quad 1&{}\quad 1&{}\quad 1&{}\quad 1&{}\quad 0\\ 1&{}\quad 0&{}\quad 1&{}\quad 1&{}\quad 0&{}\quad 1\\ 1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 1\\ 1&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 1\\ 1&{}\quad 0&{}\quad 1&{}\quad 1&{}\quad 0&{}\quad 1\\ 0&{}\quad 1&{}\quad 1&{}\quad 1&{}\quad 1&{}\quad 0\\ \end{array}\right] ,\, {{\varvec{M}}_2}\\= & {} \left[ \begin{array}{cccccc} 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0 \end{array}\right] . \end{aligned}$$

The multiplication table of these matrices in $R_n=R(X_n,d,\Gamma _n)$ is as follows ($n\ge 2$):

$\times$	${{\varvec{I}}}$	${{\varvec{M}}_1}$	${{\varvec{M}}_2}$
${{\varvec{I}}}$	${{\varvec{I}}}$	${{\varvec{M}}_1}$	${{\varvec{M}}_2}$
${{\varvec{M}}_1}$	${{\varvec{M}}_1}$	$(2n-2)({{\varvec{I}}}+{{\varvec{M}}_2})+(2n-4){{\varvec{M}}_1}$	${{\varvec{M}}_1}$
${{\varvec{M}}_2}$	${{\varvec{M}}_2}$	${{\varvec{M}}_1}$	${{\varvec{I}}}$

B Transition matrices for the examples in Sect. 5

B.1 Transition matrix ${{\varvec{T}}}={{\varvec{T}}}_l$ and eigenpolynomials of the matrix ${{\varvec{Q}}}={{\varvec{Q}}_l}$ for the polygonal landscapes

Let the symmetric triple $(X_l,d,\Gamma )$, $\Gamma =D_l$, be as in the previous subsection and let the columns (rows) of all matrices under consideration be indexed by 0, ..., $l-1$ modulo l (since $X_l\cong \mathbf {Z}/l\mathbf {Z}$). Consider the square symmetric matrix of order l

$$\begin{aligned} {{\varvec{T}}}={{\varvec{T}}_l}=(t_{ab}):=\bigl (\cos (2\pi ab/l)-\sin (2\pi ab/l)\bigr ), \quad a,b\in X_l\cong \mathbf {Z}/l\mathbf {Z}. \end{aligned}$$

(B.1)

Theorem B.1

The matrix ${{\varvec{T}}}$ satisfies the following conditions:

1. The columns ${{\varvec{t}}}_b$ of ${{\varvec{T}}}$ compose a common eigenbasis for all orbital matrices ${{\varvec{M}}}_k$, $k=0,\dots , N$.

2. If $l=2N+1$ is odd then

$$\begin{aligned} {{\varvec{T}}}^{-1}{{\varvec{M}}}_k{{\varvec{T}}}=2{{\mathrm{diag}}}(1,\cos (2\pi k/l), \cos (4\pi k/l),\dots , \cos (2\pi (l-1)k/l)),\quad k=1,\dots , N. \end{aligned}$$

If $l=2N$ is even then

$$\begin{aligned} {{\varvec{T}}}^{-1}{{\varvec{M}}}_k{{\varvec{T}}}= & {} 2{{\mathrm{diag}}}(1,\cos (2\pi k/l), \cos (4\pi k/l),\dots , \cos (2\pi (l-1)k/l)),\quad k\\= & {} 1,\dots , N-1,\\ {{\varvec{T}}}^{-1}{{\varvec{M}}}_N{{\varvec{T}}}= & {} {{\mathrm{diag}}}(1,-1, 1,-1,\dots ,1,-1). \end{aligned}$$

3. ${{\varvec{T}}}^{-1}{{\varvec{Q}}}{{\varvec{T}}}={{\mathrm{diag}}}(P_0(q),P_1(q),\dots , P_{l-1}(q))$ where $P_0(q)=P_{X_l}(q)$ is the distance polynomial (5.1). If $l=2N+1$ is odd then

$$\begin{aligned} p_j(q)=p_{l-j}(q)=q^N+\sum _{k=1}^N 2\cos (2\pi kj/l)\,(1-q)^kq^{N-k},\quad j=1,\dots , N . \end{aligned}$$

(B.2)

If $l=2N$ is even then

$$\begin{aligned} p_j(q)= & {} p_{l-j}(q)=q^N+\sum _{k=1}^{N-1} 2\cos (2\pi kj/l)\,(1-q)^kq^{N-k}\nonumber \\&+(-\,1)^j(1-q)^N,\quad j=1,\dots , N. \end{aligned}$$

(B.3)

4. ${{\varvec{T}}}^2=l\,{{\varvec{I}}}$. In other words, ${{\varvec{T}}}^{-1}=\frac{1}{l}{{\varvec{T}}}$.

Proof

1 and 2. Consider cyclic matrices ${{\varvec{C}}}_k$ with only l non-trivial entries $({{\varvec{C}}}_k)_{a,a+k}=1$ where subindices are taken modulo l. Note that ${{\varvec{C}}}_0={{\varvec{M}}}_0={{\varvec{I}}}$ in any case and ${{\varvec{C}}}_N={{\varvec{M}}}_N$ if $l=2N$ is even. In the other cases ${{\varvec{C}}}_k+{{\varvec{C}}}_{l-k}={{\varvec{M}}}_k$.

Consider also the vectors ${{\varvec{v}}}_j=(1,\varepsilon ^j,\varepsilon ^{2j},\dots ,\varepsilon ^{(l-1)j})^T$, $j\in X_l$, $\varepsilon =e^{2\pi \mathrm {i}/l}$. Straightforward checking yields

$$\begin{aligned} {{\varvec{C}}}_k{{\varvec{v}}}_j=\varepsilon ^{kj}{{\varvec{v}}}_j. \end{aligned}$$

Note that the Vandermonde determinant $\det (\varepsilon ^{kj})\ne 0$. It follows that

$$\begin{aligned} {{\varvec{C}}}_k{{\varvec{v}}}_j=\varepsilon ^{kj}{{\varvec{v}}}_j,\quad {{\varvec{C}}}_{l-k}{{\varvec{v}}}_{j}=\varepsilon ^{-kj}{{\varvec{v}}}_{j},\quad {{\varvec{C}}}_k{{\varvec{v}}}_{l-j}=\varepsilon ^{-kj}{{\varvec{v}}}_{l-j},\quad {{\varvec{C}}}_{l-k}{{\varvec{v}}}_{l-j}=\varepsilon ^{kj}{{\varvec{v}}}_{l-j}\, \end{aligned}$$

since $\varepsilon ^l=1$. If $l=2N+1$ is odd then the real vector-columns ${{\varvec{t}}}_0={{\varvec{v}}}_0$, ${{\varvec{t}}}_j=\mathrm{Re}\,{{\varvec{v}}}_j-\mathrm{Im}\,{{\varvec{v}}}_{l-j}$, ${{\varvec{t}}}_{l-j}=\mathrm{Re}\,{{\varvec{v}}}_{l-j}+\mathrm{Im}\,{{\varvec{v}}}_{j}$ are linearly independent eigenvectors of each ${{\varvec{M}}}_k$ such that

$$\begin{aligned} {{\varvec{M}}}_0 {{\varvec{t}}}_j={{\varvec{t}}}_j,\quad {{\varvec{M}}}_k {{\varvec{t}}}_j=(\varepsilon ^{kj}+\varepsilon ^{-kj}){{\varvec{t}}}_j=2\cos (2\pi kj/l){{\varvec{t}}}_j,\quad k=1,\dots ,N. \end{aligned}$$

If $l=2N$ is even then the vector-columns ${{\varvec{t}}}_0={{\varvec{v}}}_0$ , ${{\varvec{t}}}_N=(1,-1, 1,-1,\dots ,1,-1)^\top$, ${{\varvec{t}}}_j=\mathrm{Re}\,{{\varvec{v}}}_j-\mathrm{Im}\,{{\varvec{v}}}_{l-j}$, ${{\varvec{t}}}_{l-j}=\mathrm{Re}\,{{\varvec{v}}}_{l-j}+\mathrm{Im}\,{{\varvec{v}}}_{j}$ are linearly independent eigenvectors of each ${{\varvec{M}}}_k$ such that

$$\begin{aligned} {{\varvec{M}}}_0 {{\varvec{t}}}_j= & {} {{\varvec{t}}}_j,\quad {{\varvec{M}}}_k {{\varvec{t}}}_j=\varepsilon ^{kj}+\varepsilon ^{-kj}{{\varvec{t}}}_j\\= & {} 2\cos (2\pi kj/l){{\varvec{t}}}_j,\quad k=1,\dots ,N-1,\quad {{\varvec{M}}}_N {{\varvec{t}}}_j=\cos (\pi j){{\varvec{t}}}_j. \end{aligned}$$

In any case the transition matrix ${{\varvec{T}}}$ has the form (B.1). This finishes the proof of the assertions 1 and 2.

3. Recall (3.7) that ${{\varvec{Q}}}= \sum _{k=0}^N (1-q)^k q^{N-k}{{\varvec{M}}}_k\,$. In view of the assertion 2

$$\begin{aligned} {{\varvec{T}}}^{-1}{{\varvec{Q}}}{{\varvec{T}}}=\sum _{k=0}^N (1-q)^k q^{N-k}{{\varvec{T}}}^{-1}{{\varvec{M}}}_k{{\varvec{T}}}={{\mathrm{diag}}}(p_0(q),p_1(q),\dots , p_{l-1}(q)). \end{aligned}$$

Comparing the diagonal entries we get the desired result.

4. Straightforward calculations with trigonometric sums. The theorem is proved. $\square$

Remark B.2

Note that Conjecture 3.9 is true for the triple $(X_l,d,\Gamma )$.

B.1.1 Transition matrix ${{\varvec{T}}}={{\varvec{T}}}_n$ and eigenpolynomials of the matrix ${{\varvec{Q}}}={{\varvec{Q}}_n}$ for the hyperoctahedral landscape

Let the symmetric triple $(X_n,d,\Gamma _n)$ be as in the previous subsection and let the columns (rows) of all matrices under consideration be indexed by 0, ..., $2n-1$ corresponding to $x_0,\dots , x_{2n-1}$. Consider the following four matrices of order n ($n\ge 2$):

$$\begin{aligned} {{\varvec{T}}_{00}}= & {} \left[ \begin{array}{crrcr} 1&{}\quad 1&{}\quad 1&{}\quad \dots &{}\quad 1\\ 1&{}\quad -\,1&{}\quad 0&{}\quad \dots &{}\quad 0\\ 1&{}\quad 0&{}\quad -\,1&{}\quad \dots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 1&{}\quad 0&{}\quad 0&{}\quad \cdots &{}\quad -\,1\\ \end{array}\right] ,\quad {{\varvec{T}}_{01}}=\left[ \begin{array}{rlrrr} 1&{}\quad \dots &{}\quad 1&{}\quad 1&{}\quad 1\\ 0&{}\quad \dots &{}\quad 0&{}\quad -\,1&{}\quad 1\\ 0&{}\quad \dots &{}\quad -\,1&{}\quad 0&{}\quad 1\\ \vdots &{}\quad .^{\,\displaystyle {.^{.}}}&{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ -1&{}\quad \cdots &{}\quad 0&{}\quad 0&{}\quad 1\\ \end{array}\right] ,\\ {{\varvec{T}}_{10}}= & {} \left[ \begin{array}{crrcr} 1&{}\quad 0&{}\quad 0&{}\quad \cdots &{}\quad -\,1\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad .^{\,\displaystyle {.^{.}}}&{}\quad \vdots \\ 1&{}\quad 0&{}\quad -\,1&{}\quad \dots &{}\quad 0\\ 1&{}\quad -\,1&{}\quad 0&{}\quad \dots &{}\quad 0\\ 1&{}\quad 1&{}\quad 1&{}\quad \dots &{}\quad 1\\ \end{array}\right] ,\quad {{\varvec{T}}_{11}}=\left[ \begin{array}{rrrrr} 1&{}\quad \cdots &{}\quad 0&{}\quad 0&{}\quad -\,1\\ \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 0&{}\quad \dots &{}\quad 1&{}\quad 0&{}\quad -\,1\\ 0&{}\quad \dots &{}\quad 0&{}\quad 1&{}\quad -\,1\\ -1&{}\quad \dots &{}\quad -\,1&{}\quad -\,1&{}\quad -\,1\\ \end{array}\right] . \end{aligned}$$

Here the entries of the first row and column of the matrix ${{\varvec{T}}_{00}}$ are equal to 1, the diagonal entries, except for the first one, are equal to $-1$, and the other entries are trivial. The matrices ${{\varvec{T}}_{01}}$ and ${{\varvec{T}}_{10}}$ are the horizontal and vertical mirror copies of ${{\varvec{T}}_{00}}$, the matrix ${{\varvec{T}}_{11}}$ is the horizontal mirror copy of ${{\varvec{T}}_{10}}$ multiplied by $-1$.

Consider the square symmetric matrix of order 2n

$$\begin{aligned} {{\varvec{T}}}={{\varvec{T}}_n}=\left[ \begin{array}{cc} {{\varvec{T}}_{00}}&{}\quad {{\varvec{T}}_{01}}\\ {{\varvec{T}}_{10}}&{}\quad {{\varvec{T}}_{11}}\\ \end{array}\right] . \end{aligned}$$

(B.4)

We also introduce the following four matrices of order n:

$$\begin{aligned} {{\varvec{K}}_{00}}= & {} \left[ \begin{array}{crrcr} 1&{}\quad 1&{}\quad 1&{}\quad \dots &{}\quad 1\\ 1&{}\quad 1-n&{}\quad 1&{}\quad \dots &{}\quad 1\\ 1&{}\quad 1&{}\quad 1-n&{}\quad \dots &{}\quad 1\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 1&{}\quad 1&{}\quad 1&{}\quad \cdots &{}\quad 1-n\\ \end{array}\right] ,\quad {{\varvec{K}}_{01}}=\left[ \begin{array}{rlrrr} 1&{}\quad \dots &{}\quad 1&{}\quad 1&{}\quad 1\\ 1&{}\quad \dots &{}\quad 1&{}\quad 1-n&{}\quad 1\\ 1&{}\quad \dots &{}\quad 1-n&{}\quad 1&{}\quad 1\\ \vdots &{}\quad .^{\,\displaystyle {.^{.}}}&{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 1-n&{}\quad \cdots &{}\quad 1&{}\quad 1&{}\quad 1\\ \end{array}\right] ,\\ {{\varvec{K}}_{10}}= & {} \left[ \begin{array}{crrcr} 1&{}\quad 1&{}\quad 1&{}\quad \cdots &{}\quad 1-n\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad .^{\,\displaystyle {.^{.}}}&{}\quad \vdots \\ 1&{}\quad 1&{}\quad 1-n&{}\quad \dots &{}\quad 1\\ 1&{}\quad 1-n&{}\quad 1&{}\quad \dots &{}\quad 1\\ 1&{}\quad 1&{}\quad 1&{}\quad \dots &{}\quad 1\\ \end{array}\right] ,\quad {{\varvec{K}}_{11}}=\left[ \begin{array}{rrrrr} n-1&{}\quad \cdots &{}\quad -\,1&{}\quad -\,1&{}\quad -\,1\\ \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ -1&{}\quad \dots &{}\quad n-1&{}\quad -\,1&{}\quad -\,1\\ -1&{}\quad \dots &{}\quad -\,1&{}\quad n-1&{}\quad -\,1\\ -1&{}\quad \dots &{}\quad -\,1&{}\quad -\,1&{}\quad -\,1\\ \end{array}\right] . \end{aligned}$$

Here the diagonal entries of the matrix ${{\varvec{K}}_{00}}$, except for $k_{00}=1$, are equal to $1-n$, and the other entries are equal to 1. The matrices ${{\varvec{K}}_{01}}$ and ${{\varvec{K}}_{10}}$ are the horizontal and vertical mirror copies of ${{\varvec{K}}_{00}}$, the matrix ${{\varvec{K}}_{11}}$ is the horizontal mirror copy of ${{\varvec{K}}_{10}}$ multiplied by $-1$.

Consider the square symmetric matrix of order 2n

$$\begin{aligned} {{\varvec{K}}}={{\varvec{K}}_n}=\left[ \begin{array}{cc} {{\varvec{K}}_{00}}&{}\quad {{\varvec{K}}_{01}}\\ {{\varvec{K}}_{10}}&{}\quad {{\varvec{K}}_{11}}\\ \end{array}\right] . \end{aligned}$$

(B.5)

Theorem B.3

The matrix ${{\varvec{T}}}={{\varvec{T}}}_n$, $n\ge 2$, satisfies the following conditions:

• 1. $\det ({{\varvec{T}}}_n)=(-\,1)^n\cdot 2^n\cdot n^2$, consequently, ${{\varvec{T}}}_n$ is a non-degenerate matrix.

• 2. ${{\varvec{T}}}^{-1}_n=\frac{1}{2n}{{\varvec{K}}}_n$.

• 3. The columns ${{\varvec{t}}}_b$ ($b=0,\dots ,2n-1$) of ${{\varvec{T}}}$ compose a common eigenbasis for each orbital matrix ${{\varvec{M}}}_0$, ${{\varvec{M}}}_1$, ${{\varvec{M}}}_2$.

• 4. More precisely,

  $$\begin{aligned} {{\varvec{T}}}^{-1}{{\varvec{M}}}_1{{\varvec{T}}}= & {} {{\mathrm{diag}}}(2n-2,\underbrace{-2,\dots , -2}_{n-1\;\text{ times }},\underbrace{0,\dots , 0}_{n\;\text{ times }}),\\ {{\varvec{T}}}^{-1}{{\varvec{M}}}_2{{\varvec{T}}}= & {} {{\mathrm{diag}}}(\underbrace{1,\dots , 1}_{n\;\text{ times }},\underbrace{-1,\dots , -1}_{n\;\text{ times }}). \end{aligned}$$

• 5. Let ${{\varvec{Q}}}={{\varvec{Q}}}_n=\left( (1-q)^{d(x_a,x_b)}\cdot q^{2-d(x_a,x_b)}\right)$. Then

  $$\begin{aligned} {{\varvec{T}}}^{-1}{{\varvec{Q}}}{{\varvec{T}}}= & {} {{\mathrm{diag}}}(P_0(q),P_1(q),\dots , P_{2n-1}(q))=\\= & {} {{\mathrm{diag}}}(q^2+(2n-2)(1-q)q\\&+(1-q)^2,\underbrace{(2q-1)^2,\dots , (2q-1)^2}_{n-1\;\text{ times }},\underbrace{2q-1,\dots , 2q-1}_{n\;\text{ times }}). \end{aligned}$$

Proof

1. Subtracting the row 0 from the row $2n-1$, the row 1 from the row $2n-2$, ..., the row $n-1$ from the row n (note that the subindices of matrix entries range from 0 to $2n-1$) we get

$$\begin{aligned} \det ({{\varvec{T}}})=\det \left[ \begin{array}{cc} {{\varvec{T}}_{00}}&{}\quad {{\varvec{T}}_{01}}\\ {{\varvec{T}}_{10}}&{}\quad {{\varvec{T}}_{11}}\\ \end{array}\right] =\det \left[ \begin{array}{cc} {{\varvec{T}}_{00}}&{}\quad {{\varvec{T}}_{01}}\\ {{\varvec{0}}}&{}\quad {{\varvec{2}}T_{11}}\\ \end{array}\right] =2^n\det ({{\varvec{T}}_{00}})\det ({{\varvec{T}}_{11}}). \end{aligned}$$

Adding the sum of columns $2, \ldots , n$ to the first column of the matrix ${{\varvec{T}}_{00}}$ we obtain the equality $\det ({{\varvec{T}}_{00}})=(-\,1)^{n-1}n$. In a similar way we obtain that $\det ({{\varvec{T}}_{11}})=-\,n$. Hence the desired result.

2. Straightforward checking shows that ${{\varvec{T}}}_n{{\varvec{K}}_n}=2n{{\varvec{I}}}$. Note that the matrix ${{\varvec{T}}_{00}}$ appears as a transition matrix for a simplicial landscape, see Semenov and Novozhilov (2016, Section 6) for the inverse ${{\varvec{T}}^{-1}_{00}}$ and for more details.

3 and 4. Straightforward calculations show that

$$\begin{aligned}&{{\varvec{M}}_{1}}{{\varvec{t}}}_0=(2n-2) {{\varvec{t}}}_0,\quad {{\varvec{M}}_{1}}{{\varvec{t}}}_b=-\,2 {{\varvec{t}}}_b\;\;(b=1,\dots , n-1),\\&{{\varvec{M}}_{1}}{{\varvec{t}}}_b={{\varvec{0}}}\;\;(b=n,\dots , 2n-1),\\&{{\varvec{M}}_{2}}{{\varvec{t}}}_b= {{\varvec{t}}}_b\;\;(b=0,\dots , n-1), \quad {{\varvec{M}}_{1}}{{\varvec{t}}}_b=-\,{{\varvec{t}}}_b\;\;(b=n,\dots , 2n-1). \end{aligned}$$

5. Since ${{\varvec{Q}}}=q^2{{\varvec{I}}}+q(1-q){{\varvec{M}}}_1+(1-q)^2{{\varvec{M}}}_2$ and in view of 3 we get

$$\begin{aligned} {{\varvec{T}}}^{-1}{{\varvec{Q}}}{{\varvec{T}}}= & {} q^2{{\varvec{I}}}+q(1-q){{\mathrm{diag}}}(2n-2,\underbrace{-2,\dots , -2}_{n-1\;\text{ times }},\underbrace{0,\dots , 0}_{n\;\text{ times }})\\&+\,(1-q)^2{{\mathrm{diag}}}(\underbrace{1,\dots , 1}_{n\;\text{ times }},\underbrace{-1,\dots , -1}_{n\;\text{ times }})=\\= & {} {{\mathrm{diag}}}(q^2+(2n-2)(1-q)q\\&+\,(1-q)^2,\underbrace{(2q-1)^2,\dots , (2q-1)^2}_{n-1\;\text{ times }},\underbrace{2q-1,\dots , 2q-1}_{n\;\text{ times }}). \end{aligned}$$

The theorem is proved. $\square$

Remark B.4

Note that Conjecture 3.9 is true for the triple $(X_n,d,\Gamma _n)$.

C Resulting tables

In Table 1 several known homogeneous symmetric triples $(X,d,\Gamma )$ of the landscapes are presented. Here $H_n$ is the hyperoctahedral group (the Weyl group of root system $B_n$ or $C_n$ ) of order $2^n n!$, $S_n$ is the symmetric group, $D_n$ is the dihedral group of order 2n, $A_5$ is the alternating group of order 60, $D_1\cong S_2$, $\Gamma _0=\mathrm{St}(x)$, $x\in X$. For regular polytopes P the metric space (X, d) consists of the set $X=P^{(0)}$ of vertices, the metric d is the edge metric (see Example 3.1).

Table 1 Examples of the homogeneous symmetric triples $(X,d,\Gamma )$

In Table 2 the eigenpolynomials and their multiplicities (in brackets) of the matrix ${{\varvec{Q}}}$ are given. The first one is always the (leading) distance polynomial of multiplicity 1.

Table 2 Examples of eigenpolynomials and their multiplicities for several generalized mutation matrices ${\varvec{Q}}$ corresponding to particular triples $(X,d,\Gamma )$