On Ternary Clifford Algebras on Two Generators Defined by Extra-Special 3-Groups of Order 27

Abstract

The main objective of this work is to show how to construct a ternary ${\mathbb {Z}}_3$-graded Clifford algebra on two generators by using a group algebra of an extra-special 3-group G of order 27. The approach used is an extension of the method implemented to classify ${\mathbb {Z}}_2$-graded Clifford algebras as images of group algebras of Salingaros 2-groups [2]. We will show how non-equivalent irreducible representations of the ${\mathbb {Z}}_3$-graded Clifford algebra are determined by two distinct irreducible characters of G of degree 3. We comment on applying this approach to defining p-ary Clifford-like algebras on two generators and finding their irreducible representations on the basis of extra-special p-groups of order $p^3$ for $p > 3.$ Finally, we will comment on possibly using this approach to define p-ary Clifford-like algebras on three and more generators by using group central products and their group algebras.

Appendices

Appendix A: Multiplication Table in $C \ell _{2}^{\frac{1}{3}}$

A multiplication table for $C \ell _{2}^{\frac{1}{3}}= \langle \mathbf {e}_1, \mathbf {e}_2 \mid \mathbf {e}_1^3=\mathbf {e}_2^3=1, \mathbf {e}_1\mathbf {e}_2 =\omega \mathbf {e}_2\mathbf {e}_1\rangle$ with elements shown in (4) can be obtained from the multiplication table of the matrix ternary algebra P shown in Table 3 in Appendix B by making these replacements:

$$\begin{aligned} P_1\mapsto & {} 1,\quad P_2 \mapsto \mathbf {e}_1\mathbf {e}_2^2,\quad P_3 \mapsto \mathbf {e}_2\mathbf {e}_1^2,\quad P_4 \mapsto \mathbf {e}_1,\quad P_5 \mapsto \mathbf {e}_2,\nonumber \\ P_6\mapsto & {} \mathbf {e}_1^2\mathbf {e}_2^2,\quad P_7 \mapsto \mathbf {e}_1^2,\quad P_8 \mapsto \mathbf {e}_2^2,\quad P_9 \mapsto \mathbf {e}_1\mathbf {e}_2. \end{aligned}$$

(56)

Appendix B: Multiplication Table in $P \cong C \ell _{2}^{\frac{1}{3}}$

Table 3 A multiplication table for $P \cong C \ell _{2}^{\frac{1}{3}}$

where, in the notation from [4], we have

$$\begin{aligned} P_1 = I_3&= \left[ \begin{array}{rrr} 1 &{} 0 &{} 0\\ 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 1\end{array}\right] ,&P_2 = E_1E_2^2 =\left[ \begin{array}{rrr} \omega ^2 &{} 0 &{} 0\\ 0 &{} \omega &{} 0\\ 0 &{} 0 &{} 1\end{array}\right] , \\ P_3 = E_2E_1^2&= \left[ \begin{array}{rrr} \omega &{} 0 &{} 0\\ 0 &{} \omega ^2 &{} 0\\ 0 &{} 0 &{} 1\end{array}\right] ,&\quad \, P_4 = E_1 = \left[ \begin{array}{rrr} 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 1\\ 1 &{} 0 &{} 0\end{array}\right] ,\\ P_5 = E_2&= \left[ \begin{array}{rrr} 0 &{} \omega &{} 0\\ 0 &{} 0 &{} \omega ^2\\ 1 &{} 0 &{} 0\end{array}\right] ,&P_6 = E_1^2E_2^2 = \left[ \begin{array}{rrr} 0 &{} \omega &{} 0\\ 0 &{} 0 &{} 1\\ \omega ^2 &{} 0 &{} 0\end{array}\right] ,\\ P_7 = E_1^2&= \left[ \begin{array}{rrr} 0 &{} 0 &{} 1\\ 1 &{} 0 &{} 0\\ 0 &{} 1 &{} 0\end{array}\right] ,&\quad \, P_8 = E_2^2 = \left[ \begin{array}{rrr} 0 &{} 0 &{} 1\\ \omega ^2 &{} 0 &{} 0\\ 0 &{} \omega &{} 0\end{array}\right] \\ P_9 = E_1E_2&= \left[ \begin{array}{rrr} 0 &{} 0 &{} \omega ^2\\ 1 &{} 0 &{} 0\\ 0 &{} \omega &{} 0\end{array}\right] . \end{aligned}$$

with $\omega = e^{2 \pi i/3},$ $\omega ^2=-1-\omega .$

Appendix C: A Complete Character Table of $G_3$ and $H_3$

Like $G_2=D_8$ and $H_2=Q_8$, for each prime $p>2$, the groups $G_p$ and $H_p$ have the same character table but different presentations.

1.1 C.1. Group $G_3$

We use the following presentation for $G=G_3$:

$$\begin{aligned} G = \langle a,b,z : a^3=b^3=z^3=1,\, ab=baz,\, az=za, \, bz=zb \rangle . \end{aligned}$$

(57)

Hence, G is of exponent 3 (cf. Theorem 3). Generators for G can be represented by permutations in disjoint cycle notation used in SymGroupAlgebra as follows²²

$$\begin{aligned} a= & {} p([[1,2,3], [4, 16, 17], [5, 20, 21],[6, 10, 14], [7, 11, 15], [8, 24, 19],\nonumber \\&[9, 26, 22], [12, 18, 25], [13, 23, 27]]), \end{aligned}$$

(58a)

$$\begin{aligned} b= & {} p([[1, 4, 5], [2, 8, 9], [3, 12, 13],[6, 18, 22], [7, 19, 23], [10, 16, 27],\nonumber \\&[11, 25, 20], [14, 24, 21], [15, 17, 26]]), \end{aligned}$$

(58b)

$$\begin{aligned} z= & {} p\left( \left[ [1, 6, 7], [2, 10, 11], [3, 14, 15], [4, 18, 19], [5, 22, 23], [8, 16, 25],\right. \right. \nonumber \\&\left. \left. [9, 27, 20], [12, 24, 17], [13, 21, 26]\right] \right) \end{aligned}$$

(58c)

From Sect. 2 we know that each element in G can be written in the form $a^rb^sz^t$ with $r,s,t=0,1,2.$ Then, to simplify our notation and to perform computations in SymGroupAlgebra, we use the following names for the elements of G while the identity element 1 of G we denote as $\epsilon$:

$$\begin{aligned} \begin{aligned} g_1&=\epsilon , \;\;g_2 =a, \;\; g_3 =a^2, \;\; g_4 =b, \;\; g_5 =b^2,\\ g_6&=z, \;\;g_7 =z^2, \;\; g_8 =abz^2, \;\; g_9 =ab^2z, \;\; g_{10} = az,\\ g_{11}&=az^2, \;\;g_{12} =a^2bz, \;\; g_{13} =a^2b^2z^2, \;\; g_{14} =a^2z, \;\; g_{15} =a^2z^2,\\ g_{16}&=ab, \;\;g_{17} =a^2b, \;\; g_{18} =bz, \;\; g_{19} =bz^2, \;\; g_{20} =ab^2,\\ g_{21}&=a^2b^2, \;\; g_{22} =b^2z, \;\; g_{23} =b^2z^2,\;\; g_{24} =a^2bz^2,\;\; g_{25} =abz,\\ g_{26}&=a^2b^2z, \;\; g_{27} =ab^2z^2. \end{aligned} \end{aligned}$$

(59)

Since G is extra-special, the center $Z(G) = \{\epsilon ,z,z^2\}$. G has eleven conjugacy classes which we denote by $K_i,$ $i=1,\ldots ,11.$ The conjugacy classes have the following elements and class representatives:

$$\begin{aligned} \begin{aligned} K_1&= (\epsilon )^{G_3} =\{\epsilon \},\quad K_2 = (z^2)^G=\{g_{7}\},\quad K_3 = (z)^G=\{g_{6}\},\\ K_4&= (a)^G=\{g_{2}, g_{10}, g_{11}\},\quad K_5 = (a^2)^G = \{g_{3}, g_{14}, g_{15}\},\\ K_6&= (b)^G = \{g_{4}, g_{18}, g_{19}\},\quad K_7 = (b^2)^G =\{g_{5}, g_{22}, g_{23}\},\\ K_8&= (abz^2)^G = \{g_{8}, g_{16}, g_{25}\},\quad K_9 = (ab^2z)^G = \{g_{9}, g_{20}, g_{27}\},\\ K_{10}&= (a^2bz)^G = \{g_{12}, g_{17}, g_{24}\},\quad K_{11} = (a^2b^2z^2)^G = \{g_{13}, g_{21}, g_{26}\}. \end{aligned} \end{aligned}$$

(60)

The class sums $\{\epsilon , g_7,g_6,g_{2}+g_{10}+g_{11},g_{3}+g_{14}+g_{15}, \ldots , g_{13}+g_{21}+g_{26}\}$ provide a basis for the center $Z(\mathbb {C}[G])$ of the group algebra $\mathbb {C}[G].$ We are now ready to present a complete character table of $G_3$ and $H_3$ (see Table 4). The table has been computed by finding all irreducible (inequivalent) representations of $G_3$. To find the nine linear characters $\chi _1,\ldots ,\chi _9,$ it is sufficient to completely decompose the reducible $\mathbb {C}G$-module $\mathbb {C}Ge_1$ given in (18) into a direct sum of nine irreducible submodules, each with the appropriate linear character. The nonlinear characters $\rho _{10}$ and $\rho _{11}$ were found by the method discussed in Sect. 3.1. Of course, to find the character table, it is much more convenient to use Theorem 2 and formulas (13), especially for $p>3$. However, in this paper we are also interested in the non-linear representations $\rho _{10}$ and $\rho _{11}$. In the first column of Table 4, we also show each character in the notation of Theorem 2.

Table 4 Character table for $G_3$ and $H_3,$ $\omega =e^{2 \pi i/3}$

1.2 C.2. Group $H_3$

We use the following presentation for $H=H_3$:

$$\begin{aligned} H = \langle a,b,z : a^{9}=b^3=1, \, z=a^3,\, ab = baz,\, az=za,\, bz=zb \rangle \end{aligned}$$

(61)

and H is of exponent 9 (cf. Theorem 3). Generators for H can be represented by permutations in SymGroupAlgebra as follows²³:

$$\begin{aligned} a= & {} p([[1, 2, 8, 6, 11, 15, 7, 12, 3], [4, 16, 13, 18, 25, 24, 19, 9, 17],\nonumber \\&[5, 20, 27, 22, 10, 14, 23, 26, 21]]), \end{aligned}$$

(62a)

$$\begin{aligned} b= & {} p([[1, 4, 5], [2, 9, 10], [3, 13, 14], [6, 18, 22], [7, 19, 23], [8, 24, 21],\nonumber \\&[11, 16, 26], [12, 25, 20], [15, 17, 27]]), \end{aligned}$$

(62b)

$$\begin{aligned} z= & {} p([[1, 6, 7], [2, 11, 12], [3, 8, 15], [4, 18, 19], [5, 22, 23], [9, 16, 25],\nonumber \\&[10, 26, 20], [13, 24, 17], [14, 21, 27]]). \end{aligned}$$

(62c)

We use similar names for the elements of $H_3$ as those for the elements of $G_3$ shown in (59) except we call them $h_1=\epsilon , h_2=a, \ldots ,h_{27}=ab^2z^2$. Then, the elements $h_4=b, h_5=b^2, h_6=z, h_7=z^2, h_{18}=bz, h_{19}=bz^2, h_{22}=b^2z, h_{23}=b^2z^2$ are all of order 3 whereas all the remaining non-trivial elements in $H_3$ are of order 9. Of course, $Z(H)=\{\epsilon ,z,z^2\}$ and H also has 11 conjugacy classes that formally look the same as the conjugacy classes of G shown in (60) (except that the orders of the elements may be different).

Appendix D: Multiplication Table in $\mathbb {C}[G]/\mathcal {J}$

We have split the multiplication table of $\mathbb {C}[G]/\mathcal {J}$ into three parts:

• (1) Part 1 shows columns 1–6, that is, products in $\mathbb {C}[G]^{(i)}\mathbb {C}[G]^{(0)}$ for $i=0,1,2;$

• (2) Part 2 shows columns 7–12, that is, products in $\mathbb {C}[G]^{(i)}\mathbb {C}[G]^{(1)}$ for $i=0,1,2;$

• (3) Part 3 shows columns 13–18, that is, products in $\mathbb {C}[G]^{(i)}\mathbb {C}[G]^{(2)}$ for $i=0,1,2$.

In each table, we show the graded components of $\mathbb {C}[G]/\mathcal {J}$ (Tables 5, 6, 7).

Table 5 Part 1: A multiplication table in the ${\mathbb {Z}}_3$-graded algebra $\mathbb {C}[G]/\mathcal {J}$ (columns 1–6)

Table 6 Part 2: A multiplication table in the ${\mathbb {Z}}_3$-graded algebra $\mathbb {C}[G]/\mathcal {J}$ (columns 7–12)

Table 7 Part 3: A multiplication table in the ${\mathbb {Z}}_3$-graded algebra $\mathbb {C}[G]/\mathcal {J}$ (columns 13–18)

Appendix E: Matrix Representation of $\mathbb {C}[G]/\mathcal {J}$ Based on $\rho _{10}$

In the representation the basis elements of

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are represented by the following matrices:

$$\begin{aligned}&K_1=\eta _{\rho _{10}}(\epsilon ) =\left[ \begin{array}{ccc} 1 &{} 0 &{} 0\\ 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 1 \end{array}\right] , \;&~ ~~~\quad K_2=\eta _{\rho _{10}}(z) =\left[ \begin{array}{ccc} \omega ^2 &{} 0 &{} 0\\ 0 &{} \omega ^2 &{} 0\\ 0 &{} 0 &{} \omega ^2 \end{array}\right] ,\\&K_3=\eta _{\rho _{10}}(b) =\left[ \begin{array}{ccc} 1 &{} 0 &{} 0\\ 0 &{} \omega &{} 0\\ 0 &{} 0 &{} \omega ^2 \end{array}\right] ,\;&~~~\quad K_4=\eta _{\rho _{10}}(bz) =\left[ \begin{array}{ccc} \omega ^2 &{} 0 &{} 0\\ 0 &{} 1 &{} 0\\ 0 &{} 0 &{} \omega \end{array}\right] ,\\&K_5=\eta _{\rho _{10}}(b^2) =\left[ \begin{array}{ccc} 1 &{} 0 &{} 0\\ 0 &{} \omega ^2 &{} 0\\ 0 &{} 0 &{} \omega \end{array}\right] , \,&\quad K_6=\eta _{\rho _{10}}(b^2z) =\left[ \begin{array}{ccc} \omega ^2 &{} 0 &{} 0\\ 0 &{} \omega &{} 0\\ 0 &{} 0 &{} 1 \end{array}\right] ,\\&K_7=\eta _{\rho _{10}}(a) =\left[ \begin{array}{ccc} 0 &{} 0 &{} 1\\ 1 &{} 0 &{} 0\\ 0 &{} 1 &{} 0 \end{array}\right] ,\,&~~~\quad K_8=\eta _{\rho _{10}}(az) =\left[ \begin{array}{ccc} 0 &{} 0 &{} \omega ^2\\ \omega ^2 &{} 0 &{} 0\\ 0 &{} \omega ^2 &{} 0 \end{array}\right] ,\\&K_9=\eta _{\rho _{10}}(ab) =\left[ \begin{array}{ccc} 0 &{} 0 &{} \omega ^2\\ 1 &{} 0 &{} 0\\ 0 &{} \omega &{} 0 \end{array}\right] ,\,&~~~\, K_{10}=\eta _{\rho _{10}}(abz) =\left[ \begin{array}{ccc} 0 &{} 0 &{} \omega \\ \omega ^2 &{} 0 &{} 0\\ 0 &{} 1 &{} 0 \end{array}\right] ,\\&K_{11}=\eta _{\rho _{10}}(ab^2) =\left[ \begin{array}{ccc} 0 &{} 0 &{} \omega \\ 1 &{} 0 &{} 0\\ 0 &{} \omega ^2 &{} 0 \end{array}\right] ,\,&~ \, K_{12}=\eta _{\rho _{10}}(ab^2z) =\left[ \begin{array}{ccc} 0 &{} 0 &{} 1\\ \omega ^2 &{} 0 &{} 0\\ 0 &{} \omega &{} 0 \end{array}\right] ,\\&K_{13}=\eta _{\rho _{10}}(a^2) =\left[ \begin{array}{ccc} 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 1\\ 1 &{} 0 &{} 0 \end{array}\right] ,\,&\quad K_{14}=\eta _{\rho _{10}}(a^2z) =\left[ \begin{array}{ccc} 0 &{} \omega ^2 &{} 0\\ 0 &{} 0 &{} \omega ^2\\ \omega ^2 &{} 0 &{} 0 \end{array}\right] ,\\&K_{15}=\eta _{\rho _{10}}(a^2b) =\left[ \begin{array}{ccc} 0 &{} \omega &{} 0\\ 0 &{} 0 &{} \omega ^2\\ 1 &{} 0 &{} 0 \end{array}\right] ,\,&~~\, K_{16}=\eta _{\rho _{10}}(a^2bz) =\left[ \begin{array}{ccc} 0 &{} 1 &{} 0\\ 0 &{} 0 &{} \omega \\ \omega ^2 &{} 0 &{} 0 \end{array}\right] ,\\&K_{17}=\eta _{\rho _{10}}(a^2b^2) =\left[ \begin{array}{ccc} 0 &{} \omega ^2 &{} 0\\ 0 &{} 0 &{} \omega \\ 1 &{} 0 &{} 0 \end{array}\right] ,\,&~K_{18}=\eta _{\rho _{10}}(a^2b^2z) =\left[ \begin{array}{ccc} 0 &{} \omega &{} 0\\ 0 &{} 0 &{} 1\\ \omega ^2 &{} 0 &{} 0 \end{array}\right] . \end{aligned}$$

Notice that these 18 matrices are, as expected, linearly dependent. In particular,

$$\begin{aligned} K_{2i-1}=\omega K_{2i}\quad (1 \le i \le 9). \end{aligned}$$

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