November 20, 2017

Warlpiri Groups

Warlpiri Kinship Relations and the Dihedral Group
Robert F Szalapski, PhD
www.CallMeDrRob.com

§ 1. Introduction

While pretty much every person that might see this article has been exposed to Algebra, not so many have been exposed to Group Theory. When I was first exposed to Group Theory my impression was that it is less complicated than Linear Algebra and vector spaces, and yet it sometimes seems a bit more abstract and challenging to fully grasp. Yet, it is utterly pervasive in Geometry, Physics and Abstract Algebra among other places. As you shall see, it arises in some unexpected places!

I took a graduate course in Ethnomathematics at Nazareth College working towards my New York State certifications in Education. Professor Heather Lewis chose the book by Marcia Ascher, Ethnomathematics, A Multicultural View of Mathematical Ideas, CRC Press (2010), Chapman & Hall/CRC (1991). In Chapter 3, “The Logic of Kin Relations”, we are introduced to the kinship relations of the Warlpiri of Australia. Every person is a member of one of eight different groups based upon maternal lineage. Suitable marriages are determined by the group to which each man and woman belongs. These relations are described in terms of an abstract representation of the dihedral group. For some reason I felt inclined to work out a concrete representation of this abstract group, and my notes are presented here.

In particular I will use a representation in terms of 8\times 8 real matrices and column vectors of length 8. The language is a bit strange since group refers to the dihedral group, but it also refers to groups of people. I will refer to these latter “groups” as “states”, but the word set might have been just as good. This is more consistent with other fields, but more importantly we avoid the confusion between the abstract mathematical group and a population of people. The type of notation I use looks something like the Dirac bra-ket notation used in Quantum Mechanics. The figures were created on my SmartBoard.

If a reader is looking for a basic introduction to Group Theory, this wouldn’t be the best article. The basic notion of a group and also group representation as a set of matrices closed under the group operation is essential. However, this is a nice example of working out the relationships and representations from a concrete example.

§ 2. The Abstract Representation

The population is split into eight “states”. An individual is placed in a state based upon the state to which his or her mother belonged. To introduce some notation for the abstract view, let’s represent the states this way.

\displaystyle|1>,\;|2>,\;|3>,\;|4>,\;|5>,\;|6>,\;|7>,\;|8> (1)

Now consider the mother operator, m. If m operates on some state |i>, it returns the state |j> with the understanding that all of the people in state |i> were born to mothers in state |j>. The cyclical digram

\displaystyle 1\rightarrow 3\rightarrow 2\rightarrow 4\rightarrow 1 (2)

is the same as writing

\displaystyle m\,|1> \displaystyle= \displaystyle|3> (3)
\displaystyle m\,|3> \displaystyle= \displaystyle|2> (4)
\displaystyle m\,|2> \displaystyle= \displaystyle|4> (5)
\displaystyle m\,|4> \displaystyle= \displaystyle|1> (6)

Similarly the cyclical digram

\displaystyle 5\rightarrow 8\rightarrow 6\rightarrow 7\rightarrow 5 (7)

is the same as writing

\displaystyle m\,|5> \displaystyle= \displaystyle|8> (8)
\displaystyle m\,|8> \displaystyle= \displaystyle|6> (9)
\displaystyle m\,|6> \displaystyle= \displaystyle|7> (10)
\displaystyle m\,|7> \displaystyle= \displaystyle|5> (11)

This is summarized by Figure 1.

Figure 1.

The mother relationships are summarized. The state |j> of the mother of an individual in state |i> is found by following the arrow labeled with the mother operator m. The eight states are divided into two cycles of four states each.

We must also consider the father operator, f, which acts on the state of an individual and returns the state of the father. To understand this we must consider marriages and understand that marriages take place only as follows:

  • A |1> marries a |5> (symmetric).

  • A |2> marries a |6> (symmetric).

  • A |3> marries a |7> (symmetric).

  • A |4> marries an |8> (symmetric).

See Figure 2 where marriage is indicated by the dashed lines connecting the inner and outer contours.

Figure 2.

The two distinct four-cycles of mother relationships are repeated, but marriage is indicated with dashed lines connecting the inner and outer contours. Father relationships have been added as well using arrows labeled with the father operator f. Notice that these arrows are bidirectional indicating four distinct two-cycles.

Suppose that an individual is in state |1>. We can see that the mother of that individual is in state |3>, and following the dotted line to the other contour we find the father in state |7>. In turn we can find the mother of an individual in state |7> residing in state |5> who is connected to the father in state |1>. The father of a |1> is a |7>, and the father of a |7> is a |1>. This two-cycle is indicated in the figure with a bidirectional arrow. We discover four additional two cycles which are also indicated with bidirectional arrows.

The two-cycle

\displaystyle 1\rightarrow 7\rightarrow 1 (12)

states that

\displaystyle f\,|1> \displaystyle= \displaystyle|7> (13)
\displaystyle f\,|7> \displaystyle= \displaystyle|1> (14)

while

\displaystyle 2\rightarrow 8\rightarrow 2 (15)

states that

\displaystyle f\,|2> \displaystyle= \displaystyle|8> (16)
\displaystyle f\,|8> \displaystyle= \displaystyle|2> (17)

Similarly

\displaystyle 3\rightarrow 6\rightarrow 3 (18)

states that

\displaystyle f\,|3> \displaystyle= \displaystyle|6> (19)
\displaystyle f\,|6> \displaystyle= \displaystyle|3> (20)

and

\displaystyle 4\rightarrow 5\rightarrow 4 (21)

states that

\displaystyle f\,|4> \displaystyle= \displaystyle|5> (22)
\displaystyle f\,|5> \displaystyle= \displaystyle|4> (23)

Up to this point I have only introduced some notation. While m and f are two members of the dihedral group, they operate on these collections of people, the states, which we represent by |1>,\cdots,|8>.

§ 3. The Matrix Representation

What I am going to do now is to replace the collections of people, the states, enumerated in Eqn. (1) with column vectors of length eight. The column vectors are

\displaystyle|1>\Rightarrow\left(\begin{array}[]{c}1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right),\;\;|2>\Rightarrow\left(\begin{array}[]{c}0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right),\;\;|3>\Rightarrow\left(\begin{array}[]{c}0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right),\;\;|4>\Rightarrow\left(\begin{array}[]{c}0\\ 0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 0\end{array}\right), (24)
\displaystyle|5>\Rightarrow\left(\begin{array}[]{c}0\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0\\ 0\end{array}\right),\;\;|6>\Rightarrow\left(\begin{array}[]{c}0\\ 0\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0\end{array}\right),\;\;|7>\Rightarrow\left(\begin{array}[]{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ 0\end{array}\right),\;\;|8>\Rightarrow\left(\begin{array}[]{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 1\end{array}\right) (25)

We will represent the abstract operation m with an 8\times 8 matrix M that will multiply these column vectors to realize the operations of Eqn. (2) through Eqn. (11). Most matrix elements of M will be zeros, but there will a 1 in every row and column for a total of 8 non-zero elements. I can tell which elements will have a value of 1 by looking at those equations. If this is not obvious to you, don’t be concerned. You may simply verify that the results are correct by explicit matrix-vector multiplication. I can see that

\displaystyle M_{{3,1}} \displaystyle= \displaystyle 1 (26)
\displaystyle M_{{2,3}} \displaystyle= \displaystyle 1 (27)
\displaystyle M_{{4,2}} \displaystyle= \displaystyle 1 (28)
\displaystyle M_{{1,4}} \displaystyle= \displaystyle 1 (29)

and

\displaystyle M_{{8,5}} \displaystyle= \displaystyle 1 (30)
\displaystyle M_{{6,8}} \displaystyle= \displaystyle 1 (31)
\displaystyle M_{{7,6}} \displaystyle= \displaystyle 1 (32)
\displaystyle M_{{5,7}} \displaystyle= \displaystyle 1 (33)

Now we can write down the matrix M as

\displaystyle M \displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\end{array}\right) (34)

By explicit multiplication

\displaystyle\left(\begin{array}[]{cccccccc}0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\end{array}\right)\left(\begin{array}[]{c}1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right) \displaystyle= \displaystyle\left(\begin{array}[]{c}0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right) (35)

which is the same statement as Eqn. (3); the mother of a |1> is a |3>. In a similar fashion we can verify the behavior of mother operator M on all of the other column vectors.

Notice that the matrix M has a block diagonal form of four 4\times 4 matrices. The upper-left sub-matrix contains non-zero elements because it connects column vectors of Eqn. (24) to other column vectors from the same set; this corresponds to the cycle on the left-hand side of Fig. 1 and in Eqn. (2). Similarly the lower-right sub-matrix contains non-zero elements because it connects column vectors of Eqn. (25) to other column vectors from that same set; this corresponds to the cycle on the right-hand side of Fig. 1 and in Eqn. (7). However, the upper-right and lower-left 4\times 4 sub-matrices contain only zeros because the mother operator never relates column vectors of Eqn. (24) to column vectors of Eqn. (25), it never mixes the cycles on the left-hand side of Fig. 1 with those on the right-hand side.

We can obtain m^{2} and m^{3} by multiplying M with itself. We should verify that m^{2} and m^{3} have the same block-diagonal structure as M, and we should also verify that M^{4} returns the identity matrix.
By explicit calculation,

\displaystyle M^{2} \displaystyle= \displaystyle MM (36)
\displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&1&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&1&0\end{array}\right) (37)
\displaystyle M^{3} \displaystyle= \displaystyle MM^{2} (38)
\displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\end{array}\right) (39)
\displaystyle M^{4} \displaystyle= \displaystyle M^{2}M^{2} (40)
\displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\end{array}\right) (41)

All have the predicted block structure, and M^{4} produces the identity as expected. For further confirmation we can compute, as an example,

\displaystyle\left(\begin{array}[]{c}1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right) \displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\end{array}\right)\left(\begin{array}[]{c}0\\ 0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right) (42)

which states that the great-grandmother of a |3> is a |1>, which may be cross-checked using either if the diagrams..

We will find an 8\times 8 matrix F to represent the father operator, f, using Eqn. (12) through Eqn. (23). Most matrix elements of F will be zeros, but there will a 1 in every row and column for a total of 8 non-zero elements. I can tell which elements will have a value of 1 by looking at those equations. If this is not obvious to you, don’t be concerned. You may simply verify that the results are correct by explicit matrix-vector multiplication. I can see that

\displaystyle F_{{7,1}} \displaystyle= \displaystyle 1 (43)
\displaystyle F_{{1,7}} \displaystyle= \displaystyle 1 (44)
\displaystyle F_{{8,2}} \displaystyle= \displaystyle 1 (45)
\displaystyle F_{{2,8}} \displaystyle= \displaystyle 1 (46)
\displaystyle F_{{6,3}} \displaystyle= \displaystyle 1 (47)
\displaystyle F_{{3,6}} \displaystyle= \displaystyle 1 (48)
\displaystyle F_{{5,4}} \displaystyle= \displaystyle 1 (49)
\displaystyle F_{{4,5}} \displaystyle= \displaystyle 1 (50)

Now we can write down the matrix F as

\displaystyle F \displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\end{array}\right) (51)

Again notice the block diagonal structure. In this case the upper-left and lower-right 4\times 4 sub-matrices are all zeros while the upper-right and lower-left sub-matrices contain the non-zero entries. This is because the father operation takes us back and forth between the states of Eqn. (24) and those of Eqn. (25). Also refer to Fig. 2. By explicit computation

\displaystyle F^{2} \displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\end{array}\right) (52)

as expected.

Our complete set of operators is

\displaystyle\left\{ 1,M,M^{2},M^{3},F,MF,M^{2}F,M^{3}F\right\} (53)

It is completely straightforward to find the remaining operators by matrix multiplication. We can then use matrix multiplication to verify the 8\times 8 multiplication table corresponding to these operators. It is by constructing the complete table that we verify closure of the set in Eqn. (53). By explicit computation,

\displaystyle MF \displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\end{array}\right)\left(\begin{array}[]{cccccccc}0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\end{array}\right)
\displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\end{array}\right)

and

\displaystyle M^{2}F \displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&1&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&1&0\end{array}\right)\left(\begin{array}[]{cccccccc}0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\end{array}\right)
\displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\end{array}\right)

and

\displaystyle M^{3}F \displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\end{array}\right)\left(\begin{array}[]{cccccccc}0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\end{array}\right)
\displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&1&0\\ 0&1&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\end{array}\right)

We can multiply any state by one of these matrix operators to see what state is returned. The results may easily be checked against Fig. 2.

§ 4. Illustration that fmf=m^{3}.

Eight copies of Figure. 2 are presented in Figure 3, and the individual instances are labeled with the numerals 1 through 8, that label designating the starting state within the diagram. The relations fmf and m^{3} are then traced by hand, and in every instance the two paths arrive at the same final state. This demonstrates clearly that the two operations are the same, i.e. fmf=m^{3}.

Figure 3.

Starting from each of the states |i> as labeled above the individual diagram, the operations fmf and m^{3} are traced with the hand-drawn contours demonstrating that fmf=m^{3}.

Alternatively we can verify the relation in a single step using our concrete representation as 8\times 8 matrices.

\displaystyle FMF \displaystyle= \displaystyle F(MF) (54)
\displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\end{array}\right)\left(\begin{array}[]{cccccccc}0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\end{array}\right) (55)
\displaystyle= \displaystyle\left(\begin{array}[]{cccccccc}0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\end{array}\right) (56)
\displaystyle= \displaystyle M^{3} (57)

Comparison with Eqn. (39) verifies that indeed FMF=M^{3}, as expected. For the abstract group fmf=m^{3}, but we can demonstrate this with a single matrix multiplication as opposed to the process of exhaustion used in Figure 3.

 

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