Ken, Lacan, and group theory

Chuck Grimes cgrimes at tsoft.com
Wed Nov 3 14:53:46 PST 1999


Ken,

This will be fun. You might recognize some of this from a long ago post I did to Bad on the link between structuralism and group theory--a link created by Levi-Strauss and Jean Piaget--that was originally suggested by Ernst Cassirer.

You wrote out the following truth table for a model of scientific truth:

A B A and B

---------------------------

T T T

T F F

F T F

F F F

This is then jutxaposed to Lacan's system:

Lacan adopts the term "right polariation" for clockwise directions and "left polarization" for counterclockwise directions, terms used to describe the "orientation" of knots like his Borromean knot.

Real (R)

Imaginary (I) Symbolic (S)

Lacan never provides a detailed account of all the discourses covered by this particular combinatory. He mentions only two: religious discourses: RSI and psychoanalytic discourse: IRS.

----------

Okay, both tables are isomorphic to finite abstract groups.

The first is build from the cyclic product group on two elements: 2 x 2 x 2 x 2 = 16. One representative element is (1010) and another is (0101). The total number of permutations are 16 and the elements of the group are:

0. (0000) 1. (0001) 2. (0010) 3. (0011) 4. (0100) 5. (0101) 6. (0110) 7. (0111) 8. (1000) 9. (1001) 10. (1010) 11. (1011) 12. (1100) 13. (1101) 14. (1110) 15. (1111)

These are isomorphic to the binary operators on two elements:

F0 = 0, null F1 = x . y and, x and y F2 = xy' = x/y inhibition, x but not y F3 = x the function is equal to x F4 = x'y = y/x inhibition, y but not x F5 = y the function is equal to y F6 = x'y + xy' = x o y exclusive-or, x or y but not both F7 = x + y or, x or y F8 = (x + y)' = x v y nor, not-or F9 = x'y' + xy = (x = y) equivalence, x equals y F10 = y' complement, not y F11 = x + y' = x u y implication, if y then x F12 = x' complement, not x F13 = x' + y = x n y implication, if x then y F14 = (xy)' = x ^ y nand, not-and F15 = 1 identity

This amounts to the entire logic system available to the computer. Take the elements like (0000)-(1111) and write them out as the top row and first column of a multiplication table. Use the rules F0-F15 as the rule of combination and generate sixteen tables of 256 elements each. (I know, fat chance, chuck).

What emerges from that exercise is a demostration that the only two symmetric operators are F6 and F9, and these generate the only tables with non-repeating rows and columns. This means that the only group theoretic operators are F6 and F9. The diagonal elements in a 16 x 16 array using F6 are the null element (0000), and therefore the operator is equivalent to addition and group is Abelian. Using F9, the diagonal elements are (1111) or the identity element and the operator is equivalent to multiplication. The other operators are set-theoretic and generate asymetric and/or repeating tables of elements.

Experiment with truth tables using the operators (F0-F15) so that T and F are mapped to x and y and see the results.

Now, this is the part that would really piss-off Lacan. His system of three elements (r,i,s) is isomorphic to a dihedral group of order 3, the equilateral triangle. You gave (r,s,i) and (i,r,s).

Take an equilateral triange and label the vertices anything, say (a,b,c). There will six different permutations, or 1 x 2 x 3 = 6 or 3! The permutations that you gave are two of three possible rotations that form a subgroup, or the elements cycled in order. The remaining permutations are called the reflections and these will appear as interchanges between two of the three elements. The reflections are found by flipping the triangle to exchange the position of two vertices

The collection of six permutations is called S3, the symmetric group on three elements. It is isomorphic to an automorphic subgroup of the cyclic two group laid out above.

To show the isomorphism, draw the same triangle and connect the vertices to a center point. Label the vertices (01), (11), (10) with the center as (00). Now write out the path from the center to one vertex say, (00 10). Write the same path going the other way (10 00). Each of the other possible paths going in both directions will generate all the other sixteen elements, just as they appear above. This happens because you are taking a combination of two elements, two at a time and combining them with a binary operator. That is you are doing the same thing as before.

In other words under some linear transformation, Lacan's group is nothing more than a subgroup of the above, just as the first truth table is a subgroup! If you arrange the labeling in a certain way, you will find that all three are merely different representations of the same group! That is the first truth table, an automorphism of the symmetric operators on two elements, and the last group S3.

And it only gets worse. This little three group is the logical model for an incident geometry and forms the elementary linear space group (three non-colinear points = a plane).

I don't know what matheme Lacan thought he was communicating, but this is the one he was using.

Chuck Grimes



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