2. Notation and strategy

The hypergeometric group $\Gamma (\alpha,\beta )$ with the pair of parameters $\alpha =(0,0,0,0,0,0)$ and $\beta$ as in Table 1, with respect to the new basis (obtained by using a change of the basis matrix $K$ , for suitable $K$ , provided in Section 3), is generated by the matrices $U$ , $T$ and $R=TU$ , where $U=K^{-1}AK$ and $T=K^{-1}A^{-1}BK$ . The matrices $U, T$ , and $R$ have the following form and they preserve a symplectic form $J$ whose matrix form is as given below:

\begin{equation*} U= \left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1&1&0&0&0&0\\[5pt] 0&1&1&0 &0&0\\[5pt] 0&0&1&0&0&0\\[5pt] -a&-a&0&1&0&0 \\[5pt] 0&d&d&-1&1&0\\[5pt] 0&0&-c&1&-1&1 \end {array} \right ],\qquad T=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}@{\quad}c@{\quad}c@{\quad}c} 1&0&0&0&0&0\\[5pt] 0&1&0&0 &0&0\\[5pt] 0&0&1&0&0&1\\[5pt] 0&0&0&1&0&0 \\[5pt] 0&0&0&0&1&0\\[5pt] 0&0&0&0&0&1 \end {array} \right ], \end{equation*}

\begin{equation*} R=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1&1&0&0&0&0\\[5pt] 0&1&1&0 &0&0\\[5pt] 0&0&1-c&1&-1&1\\[5pt] -a&-a&0&1&0 &0\\[5pt] 0&d&d&-1&1&0\\[5pt] 0&0&-c&1&-1&1 \end {array} \right ],\qquad J=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&0&0&1&0&0\\[5pt] 0&0&0&0 &1&0\\[5pt] 0&0&0&0&0&1\\[5pt] -1&0&0&0&0&0 \\[5pt] 0&-1&0&0&0&0\\[5pt] 0&0&-1&0&0&0 \end {array} \right ]. \end{equation*}

Observe that in case of the hypergeometric groups appearing in Theorem 1.2, $R^m\neq I$ for any $m\in \mathbb{Z}\setminus \{0\}$ as the characteristic polynomial of $B$ has repeated roots.

Following the idea of Brav and Thomas [Reference Brav and Thomas6], we consider first the matrix

\begin{equation*} H=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1&1&0&0&0&0\\[5pt] 0&-1&0&0 &0&0\\[5pt] 0&0&1&0&0&-1\\[5pt] 0&0&0&-1&0&0 \\[5pt] 0&-d&0&-1&1&0\\[5pt] 0&0&0&0&0&-1 \end {array} \right ] \end{equation*}

where $d$ is defined as the $(5,2)$ entry of the matrix $U$ . It follows that $H$ fixes the subspace $V$ spanned by its first, third, and fifth column vectors. A computation shows that

(2.1) \begin{equation} H^2=I, \qquad HRH=R^{-1} \qquad \mbox{ and } \qquad HT^{-1}H=T \end{equation}

where $I$ is the $6\times 6$ identity matrix.

For a $6\times 6$ unipotent matrix $F$ , we define

\begin{equation*}\log\!(F)=(F-I)-\frac {1}{2}(F-I)^2+\frac {1}{3}(F-I)^3-\frac {1}{4}(F-I)^4+\frac {1}{5}(F-I)^5.\end{equation*}

Observe that $T^{-1}R$ and $TR^{-1}$ are unipotent matrices and define $P=\log\!(T^{-1}R)$ and $Q=\log\!(TR^{-1})$ .

It follows from Equation (2.1) that

\begin{equation*} H(T^{-1}R)^j=(TR^{-1})^jH\qquad \mbox { for all } j\in \mathbb {N} \end{equation*}

and from this, it follows that $HP=QH$ , and we get

(2.2) \begin{equation} HP^j=Q^jH\qquad \mbox{ for all } j\in \mathbb{N}. \end{equation}

Now, we choose a vector $v$ in $V$ and define

\begin{equation*}C^+=\left \{\sum _{j=0}^5 t_jP^j v \;:\; t_j\in {\mathbb {R}}_{\gt 0},\ \forall \ 0\le j\le 5\right \}\end{equation*}

the open cone generated by the vectors $v, Pv, P^2v, P^3v, P^4v, P^5v$ , and

\begin{equation*}C^-=\left \{\sum _{j=0}^5 t_jQ^j v \;:\; t_j\in {\mathbb {R}}_{\gt 0},\ \forall \ 0\le j\le 5\right \}\end{equation*}

the open cone generated by the vectors $v, Qv, Q^2v, Q^3v, Q^4v, Q^5v$ inside ${\mathbb{R}}^6$ . Since $HP^j=Q^jH$ (cf. Equation (2.2)) and $Hv=v$ , it follows that

(2.3) \begin{equation} C^-=HC^+\ . \end{equation}

Now, we are ready to define the subsets $X$ and $Y$ of ${\mathbb{R}}^6$ (with the standard action of the hypergeometric group $\Gamma (\alpha,\beta )$ ) that satisfy the hypotheses of the ping-pong lemma (cf. Theorem 1.3). Let

(2.4) \begin{equation} X=\pm C^+\cup \pm C^-\qquad \mbox{ and } \qquad Y=\bigcup \limits _{j\in \mathbb{Z}\setminus \{0\}} R^jX\ . \end{equation}

Let $M$ be the matrix whose column vectors are $v, Pv, P^2v, P^3v, P^4v, P^5v$ , and let $N$ be the matrix whose column vectors are $v, Qv, Q^2v, Q^3v, Q^4v, Q^5v$ . Observe that $N=HM$ .

The hypergeometric group $\Gamma (\alpha,\beta )$ is generated by its subgroups $G_1=\langle T\rangle$ and $G_2=\langle R\rangle$ . We show that $G_1\cap G_2=\{I\}$ . Observe that $R^m=T^n$ , for some $m,n\in \mathbb{Z}$ , implies that $A^{-1}B^mA=C^n$ , where $C=A^{-1}B$ fixes the standard basis vectors $e_1,e_2,\ldots,e_5$ in ${\mathbb{R}}^6$ , and we get $A^{-1}B^mAe_j=C^ne_j=e_j$ for all $1\le j\le 5$ . This implies that $B^me_j=e_j$ for all $2\le j\le 6$ . Also, $Be_1=e_2=B^me_2=B^m Be_1$ implies that $B^me_1=e_1$ . It now follows that $B^m=I$ . Since the characteristic polynomials of $B$ associated with the parameters appearing in Theorem 1.2 have repeated roots, we get that $m=0=n$ . Hence, $G_1\cap G_2=\{I\}$ .

Now, we need to verify the following conditions of the ping-pong lemma.

1. (A) $X$ and $Y$ are disjoint subsets.

2. (B) $(G_2\setminus \{I\})X\subseteq Y$ .

3. (C) $(G_1\setminus \{I\})Y \subseteq X$ .

We will verify condition $(A)$ by verifying the following statements.

1. (A1) $R^jC^+$ and $R^jC^-$ are disjoint from $\pm C^+$ for all $j \in \mathbb{Z}\setminus \{0\}$ .

2. (A2) $R^jC^+$ and $R^jC^-$ are also disjoint from $\pm C^-$ for all $j \in \mathbb{Z}\setminus \{0\}$ .

Condition (B) follows from the construction of the subset $Y$ . We will verify condition $(C)$ by verifying the following statements.

1. (C1) $T^{-1}C^+ \subseteq C^+$ .

2. (C2) $TC^- \subseteq C^-$ .

3. (C3) $T^{-1}R^j(C^+ \cup C^-) \subseteq \pm C^+$ for all $j \in \mathbb{Z}\setminus \{0\}$ .

4. (C4) $TR^j(C^+ \cup C^-) \subseteq \pm C^-$ for all $j \in \mathbb{Z}\setminus \{0\}$ .

Now, we have the following convention.

Observe that $R^jC^+\cap C^+$ is non-empty exactly if there exist column vectors $v_1,v_2$ having all of their coordinates positive such that $R^jMv_1=Mv_2$ , that is, $M^{-1}R^jMv_1=v_2$ . Similarly, $R^jC^-\cap C^+$ is non-empty exactly if there exist column vectors $v_1,v_2$ having all of their coordinates positive such that $R^jNv_1=Mv_2$ , that is, $M^{-1}R^jNv_1=v_2$ . Therefore, to verify the statement (A1), it is sufficient to show that the matrices $M^{-1}R^jM$ and $M^{-1}R^jN$ , for all $j \in \mathbb{Z}\setminus \{0\}$ , have some non-negative and some non-positive rows (as when this is the case, $M^{-1}R^jM$ and $M^{-1}R^jN$ map a column vector $v_1$ having all of its coordinates positive to a column vector $v_2$ having some positive and some negative coordinates).

Since $N=HM$ , $H^2=I$ and $HRH=R^{-1}$ (cf. Equation (2.1)), we get

(2.5) \begin{equation} N^{-1}R^jM=(HM)^{-1}R^j(HN)=M^{-1}H^{-1}R^jHN=M^{-1}HR^jHN=M^{-1}R^{-j}N \end{equation}

(2.6) \begin{equation} N^{-1}R^jN=(HM)^{-1}R^j(HM)=M^{-1}H^{-1}R^jHM=M^{-1}HR^jHM=M^{-1}R^{-j}M. \end{equation}

It follows from Equations (2.5) and (2.6) that if the matrices $M^{-1}R^jM$ and $M^{-1}R^jN$ , for all $j \in \mathbb{Z}\setminus \{0\}$ , have some non-negative and some non-positive rows, then the matrices $N^{-1}R^jN$ and $N^{-1}R^jM$ , for all $j \in \mathbb{Z}\setminus \{0\}$ , also have some non-negative and some non-positive rows. This shows that, for all $j\in \mathbb{Z}\setminus \{0\}$ , $R^jC^+$ and $R^jC^-$ both are disjoint from $\pm C^-$ . Hence, the sufficient criterion for (A1) above also proves (A2).

The condition (C) is divided into four parts from (C1) to (C4). Observe that $T^{-1}C^+\subseteq C^+$ if for any column vector $v_1$ having all of its coordinates some positive real numbers, there exists a column vector $v_2$ having all of its coordinates some positive real numbers such that $T^{-1}Mv_1=Mv_2$ . Thus, (C1) is equivalent to showing that the matrix $M^{-1}T^{-1}M$ is non-negative (as when this is the case, $M^{-1}T^{-1}M$ will map a column vector $v_1$ having all of its coordinates positive to a column vector $v_2$ having all of its coordinates positive).

Similarly, $TC^-\subseteq C^-$ if for any column vector $v_1$ having all of its coordinates some positive real numbers, there exists a column vector $v_2$ having all of its coordinates some positive real numbers such that $TNv_1=Nv_2$ . Thus, (C2) is equivalent to showing that the matrix $N^{-1}TN$ is non-negative. Again, since $N=HM$ , $H^2=I$ and $HTH=T^{-1}$ (cf. Equation (2.1)), we get

(2.7) \begin{equation} N^{-1}TN=(HM)^{-1}T(HM)=M^{-1}H^{-1}THM=M^{-1}(HTH)M=M^{-1}T^{-1}M \end{equation}

and hence the sufficient criterion for (C1) also proves (C2).

By using similar arguments as above, to verify the statement (C3) it is enough to show that both the matrices $M^{-1}T^{-1}R^jM$ and $M^{-1}T^{-1}R^jN$ , for all $j \in \mathbb{Z}\setminus \{0\}$ , are either non-negative or non-positive.

To show (C4), it is enough to show that both the matrices $N^{-1}TR^jM$ and $N^{-1}TR^jN$ , for all $j \in \mathbb{Z}\setminus \{0\}$ , are either non-negative or non-positive. Again, since $N=HM$ , $H^2=I$ , $HRH=R^{-1}$ , and $HTH=T^{-1}$ (cf. Equation (2.1)), we get

(2.8) \begin{equation} N^{-1}TR^jM=(HM)^{-1}TR^j(HN)=M^{-1}H^{-1}TR^jHN=M^{-1}HTR^jHN=M^{-1}T^{-1}R^{-j}N \end{equation}

(2.9) \begin{equation} N^{-1}TR^jN=(HM)^{-1}TR^j(HM)=M^{-1}HTR^jHM=M^{-1}T^{-1}R^{-j}M \end{equation}

and hence the sufficient criterion for (C3) also proves (C4).

In conclusion, to apply the ping-pong lemma to the groups in Table 1 with $X$ and $Y$ obtained as above from a given $v\in{\mathbb{R}}^6$ , it suffices to check that $M$ has full rank, that $M^{-1}R^jM$ and $M^{-1}R^jN$ have both non-positive and non-negative rows for all $j\neq 0$ (A1), that $M^{-1}T^{-1}M$ is non-negative (C1), and that $M^{-1}T^{-1}R^jM$ and $M^{-1}T^{-1}R^jN$ are non-negative or non-positive for all $j\neq 0$ (C3).

3. Proof of the freeness of the hypergeometric groups of Table 1

As explained above in Section 2, to show that the hypergeometric groups associated with the parameters appearing in Table 1 are free groups (using the ping-pong lemma), we need to verify only the statements (A1), (C1), and (C3) of Section 2 for these groups.

We write below the detailed verification for Example 1 and also for Example 2 as it requires some different explanations. For other examples, we provide the needed data which helps in verifying the ping-pong conditions using the methods of Examples 1 and 2. The computations are an adaptation of the methods in ref. [Reference Brav and Thomas6] to our setting. The main new ingredient is the choice of $v$ . See Section 2 for the notations.

3.1. Case 1

$\alpha =(0,0,0,0,0,0), \quad \beta =(1/2,1/2,1/2,1/2,1/2,1/2)$ .

In this case, the corresponding polynomials $f,g$ are

\begin{align*}f(x)&={x}^{6}-6\,{x}^{5}+15\,{x}^{4}-20\,{x}^{3}+15\,{x}^{2}-6\,x+1\quad \mbox { and }\\[5pt] g(x)&={x}^{6}+6\,{x}^{5}+15\,{x}^{4}+20\,{x}^{3}+15\,{x}^{2}+6\,x+1\end{align*}

and the corresponding hypergeometric group $\Gamma (\alpha,\beta )$ is generated by the companion matrices $A, B$ of the polynomials $f,g$ . We first consider the following change of the basis matrix

\begin{equation*}K=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&48&-12&0&-1&-1\\[5pt] - 64&-144&0&-1&4&5\\[5pt] 128&208&-40&3&-6&-10 \\[5pt] -64&-112&0&-3&4&10\\[5pt] 0&0&-12&1& -1&-5\\[5pt] 0&0&0&0&0&1\end {array} \right ].\end{equation*}

Then, $U=K^{-1}AK$ , $T=K^{-1}A^{-1}BK$ and $R=TU$ have the form described in Section 2 with $a=64$ , $d=48$ and $c=12$ .

Guided by computer calculations, we pick $v=\left (-\frac{1}{10}, 0,1, 0, -\frac{1162}{225}, 0\right )$ inside the fixed subspace $V$ of the matrix $H$ . Note that in this case, $-R$ is unipotent and so we can define $Z=\log\!(\!-R)$ . Then, as required, the resulting matrix $M$ has full rank, and

\begin{equation*}-R= e^Z= I+Z+\frac {Z^2}{2!}+\frac {Z^3}{3!}+\frac {Z^4}{4!}+\frac {Z^5}{5!}\end{equation*}

(3.1) \begin{equation} R^j= (\!-\!1)^je^{jZ}= (\!-\!1)^j\left (I+jZ+\frac{j^2Z^2}{2!}+\frac{j^3Z^3}{3!}+\frac{j^4Z^4}{4!}+\frac{j^5Z^5}{5!}\right ). \end{equation}

To verify (A1), we compute the matrices $M^{-1}R^jM$ and $M^{-1}R^jN$ , for all $j \in \mathbb{Z}\setminus \{0\}$ , and show that they have some non-negative and some non-positive rows.

Denote $A_i=M^{-1}\frac{Z^i}{i!}M$ , for $i=1,2,3,4,5$ . Then,

(3.2) \begin{equation} M^{-1}R^jM= (\!-\!1)^j(I+jA_1+j^2A_2+j^3A_3+j^4A_4+j^5A_5)\ . \end{equation}

By computation, we get

(3.3) \begin{equation} A_5=\left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}{\frac{98}{3375}}&{\frac{916}{3375}}&{ \frac{304}{225}}&{\frac{256}{75}}&{\frac{64}{15}}&{\frac{128}{15}} \\[5pt] -{\frac{49}{3375}}&-{\frac{458}{3375}}&-{\frac{ 152}{225}}&-{\frac{128}{75}}&-{\frac{32}{15}}&-{\frac{64}{15}} \\[5pt] {\frac{49}{16875}}&{\frac{458}{16875}}&{\frac{152 }{1125}}&{\frac{128}{375}}&{\frac{32}{75}}&{\frac{64}{75}} \\[5pt] -{\frac{49}{202500}}&-{\frac{229}{101250}}&-{ \frac{38}{3375}}&-{\frac{32}{1125}}&-{\frac{8}{225}}&-{\frac{16}{225}} \\[5pt] {\frac{8477}{12150000}}&{\frac{39617}{6075000}}&{ \frac{3287}{101250}}&{\frac{1384}{16875}}&{\frac{346}{3375}}&{\frac{ 692}{3375}}\\[5pt] -{\frac{8477}{24300000}}&-{\frac{39617 }{12150000}}&-{\frac{3287}{202500}}&-{\frac{692}{16875}}&-{\frac{173}{ 3375}}&-{\frac{346}{3375}}\end{array} \right ]. \end{equation}

Notice that $A_5$ has some non-negative and some non-positive rows. Now, we use the Archimedean property of the real numbers to show that there exists a positive integer $n\in \mathbb{Z}$ such that the entries of the first two rows of the matrix $M^{-1}R^jM$ , for all $j \in \mathbb{Z}\setminus \{0\}$ and $|j|\geq n$ , have the same sign as that of the corresponding entries of the matrix $\mathrm{sign}(j)(\!-\!1)^{|j|}A_5$ .

A sufficient $n$ can be computed after writing

(3.4) \begin{equation} (\!-\!1)^j(I+jA_1+j^2A_2+j^3A_3+j^4A_4+j^5A_5)=(\!-\!1)^j(I+j(A_1+j(A_2+j(A_3+j(A_4+jA_5))))) . \end{equation}

By using the Archimedean property of the real numbers on the entries of the matrices $A_5$ and $A_4$ , we get a positive integer $k_5$ such that the entries of the matrix $A_4+jA_5$ have the same sign as that of the corresponding entries of the matrix $A_5$ for all $j\geq k_5$ . Again using the Archimedean property of the real numbers on the entries of the matrices $A_4+k_5A_5$ and $A_3$ , we get a positive integer $k_4$ such that the entries of the matrix $A_3+j(A_4+k_5A_5)$ have the same sign as that of the corresponding entries of the matrix $A_4+k_5A_5$ for all $j\geq k_4$ . Repeating this process, we get the positive integers $k_5,k_4,k_3,k_2,k_1$ such that for all $j\ge n=\max \{k_5,k_4,k_3,k_2,k_1\}$ the entries of the matrix $(\!-\!1)^j(1+j(A_1+j(A_2+j(A_3+j(A_4+jA_5)))))$ have the same sign as that of the corresponding entries of the matrix $(\!-\!1)^jA_5$ .

In case when $j$ is a negative integer, we take $j=-l$ where $l$ is a positive integer and write the expression for the matrix $M^{-1}R^jM$ as

(3.5) \begin{equation} (\!-\!1)^j(1+jA_1+j^2A_2+j^3A_3+j^4A_4+j^5A_5)=(\!-\!1)^l(1+l(-A_1+l(A_2+l(-A_3+l(A_4-lA_5))))) \end{equation}

and use the Archimedean property of the real numbers, as it is described above, to get a positive integer $m$ so that the matrix $M^{-1}R^jM$ , for all $j\le -m \in \mathbb{Z}\setminus \{0\}$ , has some non-negative and some non-positive rows. In this case, we get that $\max \{n,m\}$ is $5$ . So, it follows that for $|j|\ge 5$ , the matrix $M^{-1}R^jM$ has some non-negative and some non-positive rows.

For the remaining values of $j \in \mathbb{Z}\setminus \{0\}$ , that is, for $j\in \mathbb{Z}$ with $1\leq |j|\leq 4$ , we compute the matrices $M^{-1}R^jM$ individually and verify that the matrix $M^{-1}R^jM$ , also for $1\leq |j|\leq 4$ , has some non-negative and some non-positive rows.

Similarly, we verify that the matrix $M^{-1}R^jN$ , for all $j \in \mathbb{Z}\setminus \{0\}$ , has some non-negative and some non-positive rows.

To verify (C1), we need to show that the matrix $M^{-1}T^{-1}M$ is non-negative. By computation we get

\begin{equation*}M^{-1}T^{-1}M=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1&{\frac {458}{225}}&0&{\frac {128}{5}}&0 &64\\[5pt] 0&1&0&0&0&0\\[5pt] 0&{\frac {229}{ 1125}}&1&{\frac {64}{25}}&0&{\frac {32}{5}}\\[5pt] 0&0&0&1&0 &0\\[5pt] 0&{\frac {39617}{810000}}&0&{\frac {692}{1125}}&1 &{\frac {346}{225}}\\[5pt] 0&0&0&0&0&1\end {array} \right ].\end{equation*}

To verify (C3), it is sufficient to show that both the matrices $M^{-1}T^{-1}R^jM$ and $M^{-1}T^{-1}R^jN$ , for all $j \in \mathbb{Z}\setminus \{0\}$ , are either non-negative or non-positive.

Using Equation (3.1) for the expression of $R^j$ , we get

(3.6) \begin{equation} M^{-1}T^{-1}R^jM= (\!-\!1)^j(M^{-1}T^{-1}M+jD_1+j^2D_2+j^3D_3+j^4D_4+j^5D_5) \end{equation}

(3.7) \begin{equation} M^{-1}T^{-1}R^jN= (\!-\!1)^j(M^{-1}T^{-1}N+jE_1+j^2E_2+j^3E_3+j^4E_4+j^5E_5) \end{equation}

where $D_i=M^{-1}\frac{T^{-1}Z^i}{i!}M$ and $E_i=M^{-1}\frac{T^{-1}Z^i}{i!}N$ , for $i=1,2,3,4,5$ .

By computation, we get

\begin{equation*}D_5=E_5=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} -{\frac {98}{3375}}&-{\frac {916}{3375}}& -{\frac {304}{225}}&-{\frac {256}{75}}&-{\frac {64}{15}}&-{\frac {128}{15} }\\[5pt] -{\frac {49}{3375}}&-{\frac {458}{3375}}&-{\frac { 152}{225}}&-{\frac {128}{75}}&-{\frac {32}{15}}&-{\frac {64}{15}} \\[5pt] -{\frac {49}{16875}}&-{\frac {458}{16875}}&-{\frac { 152}{1125}}&-{\frac {128}{375}}&-{\frac {32}{75}}&-{\frac {64}{75}} \\[5pt] -{\frac {49}{202500}}&-{\frac {229}{101250}}&-{ \frac {38}{3375}}&-{\frac {32}{1125}}&-{\frac {8}{225}}&-{\frac {16}{225}} \\[5pt] -{\frac {8477}{12150000}}&-{\frac {39617}{6075000}} &-{\frac {3287}{101250}}&-{\frac {1384}{16875}}&-{\frac {346}{3375}}&-{ \frac {692}{3375}}\\[5pt] -{\frac {8477}{24300000}}&-{\frac {39617}{12150000}}&-{\frac {3287}{202500}}&-{\frac {692}{16875}}&-{\frac {173}{3375}}&-{\frac {346}{3375}}\end {array} \right ].\end{equation*}

Since all entries of the matrices $D_5$ and $E_5$ have the same signs, by using the Archimedean property of the real numbers, as above, we find that for $|j|\geq 7$ the matrices $M^{-1}T^{-1}R^jM$ and $M^{-1}T^{-1}R^jN$ are either non-negative or non-positive. We compute that the matrices $M^{-1}T^{-1}R^jM$ and $M^{-1}T^{-1}R^jN$ , also for $1\le |j|\le 6$ , are either non-negative or non-positive.

This completes the proof of the freeness of the hypergeometric group.

3.2. Case 2

$\alpha =(0,0,0,0,0,0),\quad \beta =(1/2,1/2,1/2,1/2,1/3,2/3)$ .

We consider the change of the basis matrix

\begin{equation*}K=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&40&-11&0&-1&-1\\[5pt] - 48&-120&4&-1&4&5\\[5pt] 96&168&-34&3&-6&-10 \\[5pt] -48&-88&4&-3&4&10\\[5pt] 0&0&-11&1&- 1&-5\\[5pt] 0&0&0&0&0&1\end {array} \right ].\end{equation*}

In this case, $a=48$ , $d=40$ , $c=11$ and $v=\left (-\frac{1}{10},0,1,0,-\frac{631}{150},0\right )$ . This time neither $R$ nor $-R$ is unipotent. But $R^6$ is unipotent, so let $Z=\log\!(R^6)$ . Note that $Z^4=0$ . Then,

\begin{equation*}R^6=e^Z=I+Z+\frac {Z^2}{2!}+\frac {Z^3}{3!},\end{equation*}

(3.8) \begin{equation} R^{6n+k}=R^kR^{6n}=R^ke^{nZ}=R^k\left (I+nZ+\frac{n^2Z^2}{2!}+\frac{n^3Z^3}{3!}\right ). \end{equation}

We write $j=6n+k$ for $n \in \mathbb{N}\cup \{0\}$ and $0\leq k \leq 5$ . If we denote $A_{i,k}=M^{-1}\frac{R^kZ^i}{i!}M$ and $B_{i,k}=M^{-1}\frac{R^kZ^i}{i!}N$ for $i=1,2,3$ , then

(3.9) \begin{equation}{} M^{-1}R^{6n+k}M=M^{-1}R^kM+nA_{1,k}+n^2A_{2,k}+n^3A_{3,k} \end{equation}

(3.10) \begin{equation} M^{-1}R^{6n+k}N=M^{-1}R^kN+nB_{1,k}+n^2B_{2,k}+n^3B_{3,k}\ . \end{equation}

By denoting $D_{i,k}=M^{-1}\frac{T^{-1}R^kZ^i}{i!}M$ and $E_{i,k}=M^{-1}\frac{T^{-1}R^kZ^i}{i!}N$ for $i=1,2,3$ and using the above Equation (3.8), we get

(3.11) \begin{equation}{} M^{-1}T^{-1}R^{6n+k}M=M^{-1}T^{-1}R^kM+nD_{1,k}+n^2D_{2,k}+n^3D_{3,k} \end{equation}

(3.12) \begin{equation}{} M^{-1}T^{-1}R^{6n+k}N=M^{-1}T^{-1}R^kN+nE_{1,k}+n^2E_{2,k}+n^3E_{3,k}\ . \end{equation}

For each $k$ , we can find a $q\in \mathbb{N}$ such that the signs of the entries of $M^{-1}R^{6n+k}M$ , $M^{-1}R^{6n+k}N$ , $M^{-1}T^{-1}R^{6n+k}M$ and $M^{-1}T^{-1}R^{6n+k}N$ are the same as that of the entries of $A_{3,k}, B_{3,k}, C_{3,k}$ and $D_{3,k}$ for all $|n| \gt q$ , and we verify the ping-pong conditions as in Case 1.

3.3. Case 3

$\alpha =(0,0,0,0,0,0),\quad \beta =(1/2,1/2,1/2,1/2,1/4,3/4)$ .

We consider the change of the basis matrix

\begin{equation*}K= \left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&32&-10&0&-1&-1\\[5pt] - 32&-96&8&-1&4&5\\[5pt] 64&128&-28&3&-6&-10 \\[5pt] -32&-64&8&-3&4&10\\[5pt] 0&0&-10&1&- 1&-5\\[5pt] 0&0&0&0&0&1\end {array} \right ]. \end{equation*}

In this case, $a=32$ , $d=32$ , $c=10$ , $v=\left (-\frac{1}{11},0,1,0,-{\frac{277}{75}},0\right )$ and $Z=\log\!(R^4)$ .

3.4. Case 4

$\alpha =(0,0,0,0,0,0),\quad \beta =(1/2,1/2,1/2,1/2,1/6,5/6)$ .

We consider the change of the basis matrix

\begin{equation*}K= \left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&24&-9&0&-1&-1\\[5pt] -16 &-72&12&-1&4&5\\[5pt] 32&88&-22&3&-6&-10 \\[5pt] -16&-40&12&-3&4&10\\[5pt] 0&0&-9&1&- 1&-5\\[5pt] 0&0&0&0&0&1\end {array} \right ]. \end{equation*}

In this case $a=16$ , $d=24$ , $c=9$ , $v=\left (-\frac{1}{10},0,1,0,-{\frac{3041}{810}},0\right )$ and $Z=\log\!(\!-R^3)$ .

3.5. Case 5

$\alpha =(0,0,0,0,0,0),\quad \beta =(1/2,1/2,1/3,2/3,1/3,2/3)$ .

We consider the change of the basis matrix

\begin{equation*}K= \left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&33&-10&0&-1&-1\\[5pt] - 36&-99&7&-1&4&5\\[5pt] 72&135&-30&3&-6&-10 \\[5pt] -36&-69&7&-3&4&10\\[5pt] 0&0&-10&1&- 1&-5\\[5pt] 0&0&0&0&0&1\end {array} \right ]. \end{equation*}

In this case, $a=36$ , $d=33$ , $c=10$ , $v=\left (-\frac{1}{6},0,1,0,-{\frac{207}{40}},0\right )$ and $Z=\log\!(R^6)$ .

3.6. Case 6

$\alpha =(0,0,0,0,0,0),\quad \beta =(1/2,1/2,1/3,2/3,1/4,3/4)$ .

We consider the change of the basis matrix

\begin{equation*}K= \left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&26&-9&0&-1&-1\\[5pt] -24 &-78&10&-1&4&5\\[5pt] 48&102&-26&3&-6&-10 \\[5pt] -24&-50&10&-3&4&10\\[5pt] 0&0&-9&1&- 1&-5\\[5pt] 0&0&0&0&0&1\end {array} \right ]. \end{equation*}

In this case, $a=24$ , $d=26$ , $c=9$ , $v=\left (-\frac{1}{6},0,1,0,-{\frac{472}{105}},0\right )$ and $Z=\log\!(R^{12})$ .

3.7. Case 7

$\alpha =(0,0,0,0,0,0),\quad \beta =(1/2,1/2,1/5,2/5,3/5,4/5)$ .

We consider the change of the basis matrix

\begin{equation*}K= \left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 0&25&-9&0&-1&-1\\[5pt] -20 &-75&11&-1&4&5\\[5pt] 40&95&-24&3&-6&-10 \\[5pt] -20&-45&11&-3&4&10\\[5pt] 0&0&-9&1&- 1&-5\\[5pt] 0&0&0&0&0&1\end {array} \right ] . \end{equation*}

In this case, $a=20$ , $d=25$ , $c=9$ , $v=\left (-\frac{1}{10},0,1,0,-{\frac{221}{60}},0\right )$ and $Z=\log\!(R^{10})$ .