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An Algebraic Foundation for Being and Time, Explaining the Emergence of Clifford Algebra Structure

Ben Goertzel
December 1997


1. Introduction

The goal of this paper is to describe what seems to be the simplest possible algebraic formalism from which the integer subset of the Clifford Algebras can be seen to emerge in a natural way. This project is inspired by the belief that the Clifford Algebras, and their quaternionic and octonionic subalgebras, are archetypal structures of fundamental importance in many areas, including physics, psychology and systems theory. It is part of an ongoing interdisciplinary investigation involving the author, Onar Aam, Kent Palmer and F. Tony Smith.

The criteria I have used in seeking a "simple" formalism are as follows:

After trying a great number of different things, the system presented here is the best I have found. Suggested simplifications are welcomed!

G. Spencer-Brown's work in Laws of Form, and Lou Kauffman's and Francisco Varela's follow-up work, have been inspirational for this work. The present formal system is not as simple and elegant as the Laws of Form system; but the difference is that, while that system is isomorphic to Boolean algebra, this one contains subsets isomorphic to Clifford Algebra, a much richer system.

The main originality of the present system in purely formal terms is the introduction of two kinds of boundaries, "hard" and "soft" boundaries. Normally in algebra we have only one kind of boundary, represented by parentheses (), and various operations that act on bounded and unbounded entities. Here I have found it most simple to restrict the number of operations to three, but to introduce two types of boundaries [] and <>, which are acted on by the operations in different ways. Basically, a soft boundary <> is a boundary that operations can distribute through, and addition can associate through; whereas a hard boundary [] prevents distribution and additive associativity, but has some special "collapse" rules of its own.

The philosophical interpretation of the formalism is only hinted at here, but has been enlarged upon in earlier writings and will be integrated with the present text in a later revision.

2. Formalism

Primitive elements of On algebra

Notation and terminology


  1. [ ] = < > = < V > = V =

  2. V + X = X

  3. N + N = V + V = V

  4. < [X + C] + [X] + Z > = < V + Z>

  5. [ [X] + [Y] + Z ] = [ [X+Y] + Z ]

  6. + =

  7. X * =

  8. X^Y = C or X^Y = V, for any arguments X and Y

  9. S( X(i(j)); j=1,...,n) is identical for all functions i() that permute {1,...,n}

  10. X * Y = X + Y + X^Y

  11. if Q = S(X(i); i=1,...,n), R= S(X(i); i=1,...,m)
    then [Q]*[R] = [Q + R + S(X(i)^Y(j);i=1,...,n;j=1,...,m) ]

  12. with Q and R as above,
    [Q]^[R] = [ S(X(i)^Y(j);i=1,...,n;j=1,...,m) ]

Further terminology

3. Interpretations and Explanations

The On, the particle of Being, is denoted N. There may be many Ons, but all are identical.

A boundary separating X from everything else is denoted [X]. The division of the boundary symbol into two parts, [ and ], is an artifact of typographical notation; in a two-dimensional symbolism, a boundary separating off X would be denoted by a box or circle drawn around X.

[N] is a Chronon, a particle of time. For convenience, a Chronon may be denoted C = [N]; this is solely for the purpose of having one symbol instead of three.

The Void, emptiness, will be denoted V. I.e.,

     V = 

A permeable or "soft" boundary separating off X is denoted . This denotes a boundary that entities can pass easily through. By contrast to , [X] will be called a "hard" boundary around X

A boundary with nothing inside it is taken to be Void, i.e.

     [] = <> = V

The "space" that an entity lives in is the hard or soft boundary surrounding it. The universe itself may be considered an implicit hard boundary, defining the space in which unbounded expressions such as X, X+Y, etc., live.

Coexistence of X and Y in a single "space" is denoted X + Y. This operation is commutative, and furthermore a sequence of entities connected by +'s in sequence is invariant with respect to permutations, e.g. X + Y + W + M = M + W + Y + X, etc.

N is an annihilator with respect to coexistence; i.e.

     N + N = V

Inside a hard boundary, hard boundaries are associative, i.e.

     [ [X] + [Y] ] = [ [X+Y] ]

Inside a soft boundary, associativity of hard boundaries fails, and we instead have the rule

    < [C + X] + X > = < V > = V = 

which is incompatible with associativity.

Temporal comparison or "temporization" of two entities X and Y is denoted by X^Y. Temporal comparison tells which of X and Y occurs first. The concept is: if X occurs before Y, X^Y=C; otherwise X^Y=V.

Interpenetration of X and Y is denoted by A*B. A*B is defined by

	X * Y = X + Y + X^Y

	[A+B] * [C+D] = [ A + B + C + D + A^C + A^D + B^C + B^D]


In general, if G and H are any two collections of coexisting entities, G*H is the collection of the following mutually coexisting entities: all entities in G, all entities in H, and all pairs X^Y where X is in G and Y is in H.

4. Some Theorems

Theorem 1:

For any entity X,

[X + X] = [V] = V = 

(i.e., X+X=V within a hard boundary)


The proof is by induction on the number n of symbols in the expression X.

We know the claim is true for n=1, as the only legal expression of length 1 is N, and [N+N] = [V] = V = .

Now, suppose the claim is true for all 0 X must be of one of the following forms: [Y], , Y+Z, Y^Z, Y*Z, where Y and Z necessarily include fewer symbols than X, and can hence be assumed by the inductive assumption to obey the claim of the theorem.

Suppose X=[Y]. Then, [[Y]+[Y]] = [[Y+Y]] = [[V]] = [V] = V = . This shows e.g. that [C+C] = V.

Suppose X =. Then, [+] = [] = [] = [V] = V =

Suppose X = Y+Z. Then, [ [Y+Z] + [Y+Z] ] = [ [ Y + Z + Y + Z ]] = [[Y + Y + Z + Z]] = [[ V + V ]] = [[V]] =

Suppose X = Y*Z. Then, [ [Y*Z] + [Y*Z] ] = [ [ Y + Z + Y^Z] + [Y + Z + Y^Z ]] = [[ Y + Y + Z + Z + Y^Z + Y^Z ]] = [[Y^Z + Y^Z]]. Now, Y^Z = either C or V, yielding [[Y^Z + Y^Z]] = [[C + C]] = [V] = V or [[Y^Z + Y^Z]] = [[V + V]] = [[V]] = V

Finally, suppose X=Y^Z. Then, by the reasoning of the previous sentence, the claim holds.


Corollary: For any entity G, the hard superset h(G) is finite

The next theorems concern two finite algebras: the Discrete Clifford Group DCLG(n) and the integer lattice of the Clifford Algebra, DCL(n).

DCLG(n) is defined in terms of a basis of elements {e(1), ..., e(n)}, and the subsets of {1,...,n}. Where A is a subset of {1,...,n} and e(i) is a basis element, the elements of DCLG(n) are of the form +e(A) or -e(A), where e(A) is the Clifford element e(j(1))e(j(2))...e(j(k)), and j(i), j(2),...,j(k) are the elements of the subset A, arranged in increasing order (the use of a standard order is important). Since we have used + and -, this yields 2n+1 elements in DCLG(n).

The multiplication rule for DCLG(n) (where multiplication is denoted by adjacency) is given by

	( x(A) e(a) ) (x(B) e(b) ) = x(A,B) e(C)

where C is the symmetric difference of the sets A and B, and x(A), x(B), and x(A,B) are +1 or -1. The sign x(A,B) of the product is determined by an algorithm using the rules.

	e(i) e(i) = +1 
      e(i) e(j) = - e(j)e(i) for i =/= j 

The algorithm works as follows. First, one creates a sequence of elements e(X) by concatenating the sequence representations e(A) and e(B) (placing e(B) at the end of e(A)). Then one performs a series of operations on eX, using the rule e(i)e(j) = - e(j)e(i) to move each of the B-elements to the left until it either:

According to this algorithm, each time a B-element is moved to the left, the sign of e(X) changes. If the total number of moves is even, then the sign of e(X) is unchanged by the algorithm, and x(A,B) = x(A) x(B). If the total number of moves is odd, then the sign of e(X) is reversed by the algorithm, and x(A,B) = - x(A)x(B). Elegantly, the algorithm also performs the symmetric difference operation: the only elements left in e(X) after the transformation process is complete, are the ones that are present in both e(A) and e(B).

Some simple examples may be helpful for understanding this algorithm. For doing example, I will adopt a simpler notation, and write e(i) = ei, e.g. e(1) = e1. Suppose we have

	A = e1 e2 e3

	B = e2 e3 e4

and we wish to compute the product

	A * B = e1 e2 e3 * e2 e3 e4

According to the DCLG algorithm, we move the elements of B over, one by one, until they meet their respective elements of A. each time we move an element of B past an element of A, we switch the sign of the product from its initial positive value. So to compute A*B, we have

	eX = e1 e2 e3 e2 e3 e4 = 

	- e1 e2 e2 e3 e3 e4 = 

	- e1 e4

The sign is negative. And what is left over after cancellation, e(1) e(4), is exactly the symmetric difference of the sets A and B.

Or, consider:

	e1 e2 e5 * e1 e2 e4 =

	- e1 e2 e1 e5 e2 e4 =

	e1 e1 e2 e5 e2 e4 =

	- e2 e2 e5 e4 =

	e4 e5

Again, each switch does a sign change, and the resultant is the signed symmetric difference of the multiplicands.

The step from DCLG(n) to the full Clifford algebra CL(n), finally, is not a large one. Clifford algebra is obtained by considering the M= 2^(N-1) elements {V(1),...,V(M)} of the DCLG to combine, not only by DCLG multiplication, but also by summation and scalar multiplication, according to standard vector space laws. Clifford algebra, as normally studied, is a continuous construction, involving either the reals or the complex numbers as a scalar field. But for many purposes, one does not require the full continuous Clifford Algebra Cl(N), only the discrete lattice subset DCL(N) obtained by considering {V1,...,VM} as a vector space with scalar coefficients drawn from the integers.

To relate DCLG to Ons algebra, it is useful to go through the intermediary representation of bit strings, strings of 0's and 1's. Suppose one writes the expansions of DCLG elements in terms of elements as bit strings, sequences of zeroes and ones, where . The length of the bit strings is n, the number of basis elements. In an algebra with four basis elements, for example, the element e(1) is represented 1000; the element e(3) is represented 0010; the element e(2) e(3) is represented 0110; the element e(1)e(2)e(3)e(4) is represented 1111.

Consider again the first eexample above, e1 e2 e3 * e2 e3 e4 = -e1 e4. In bit string notation, this result is represented

	1110 * 0111 = - 1001

To compute the sign using only bit strings, one reasons as follows. First, one lines up the two multiplicands, one under the other:

	1110 *


Then, for each 1 in the second multiplicand (B), one adds up the number of 1's in the first multiplicand (A) that come strictly after it (moving from left to right) For instance, for the 1 in the second place in B in this example, we have the one 1 in the third place in A. For the 1 in the third place in B in this example, we have no ones coming after it in A; and likewise for the 1 in the fourth place in B in this example.) Then, one takes the sum one has gotten by this procedure, and uses it to determine the sign of the product. If this sum is odd, the sign of the product is negative. If the sum is even, the sign of the product is positive.

In this example, we have moved through the 1 bits of the representation of the second multiplicand B, summing for each one the number of 1 bits of A that come after it. Equivalently, one can sum over all the 1's in the first multiplicand, A, summing for each one of these, the number of 1's in B that occur strictly before it. The two procedures always give the same result.

Similarly, the second example above, e1 e2 e5 * e1 e2 e4, translates to

	11001 *


Here, for the 1 in the first place of B, we have two 1's in A coming after it; for the 1 in the second place of B, we have one 1 in A coming after it; for the 1 in the fourth place of B, we have one 1 in A coming after it. So, the sum is four, and the sign of the product is plus.

Using this representation of the DCLG product sign rule, it is not hard to prove that the hard powerset of a temporal domain of entities is isomorphic to DCLG.

Theorem 2

Consider a temporal domain (G,^) where G contains n entities, none of which is a chronon C.
Then the algebra (h(G),*) is isomorphic to the Discrete Clifford Algebra DCLG(n)


: Let us identify the elements of G with the basis elements of a Discrete Clifford Algebra DCLG(n), supposing that the ordering of the DCLG basis elements is taken to be consistent with the linear time operator ^.

Let us call the mapping from entities into Clifford basis elements "cliff", so that e.g. w = cliff(W). For instance, if G=[W +X +Y +Z], and according to the operator ^ we have W To extend cliff into a mapping from h(G) into general DCLG elements, we use the following two rules

What needs to be shown is that cliff is an isomorphism, i.e. that where Q and R are elements of h(G),

		cliff(Q) * cliff(R) = cliff(Q*R)

where the right-hand * is the Ons interpenetration operator and the left-hand * is the DCLG operator.

Consider the example where G = [W +X +Y +Z], and consider the product,

	[X + Y + Z] * [X + Y + W] = 
		[X + Y + Z + X + Y + W + X^X + X^Y + X^W 
					     + Y^X + Y^Y + Y^W 
					     + Z^X + Z^Y + Z^W] 

Observe here that, by Theorem 1, X+X and Y+Y cancel out. Given the ordering W [X + Y + Z] * [X + Y + W] = [W + Z + V + V + C + C + V + C + C + C + C ] Since C+C=V inside a hard boundary, the result of this computation is [W].

Now, suppose we did this computation using the DCLG rule. In the bit string version, we would have

	0111 * 

yielding the W and Z. Next the sign would be obtained as follows:

		3 1's in 0111 coming after the 1*** in 1110
		2                              *1**
		1                              **1*

Since this is even, we find that there is no negative sign, consistent with the fact that our answer in Ons algebra was [W] rather than [C + W].

In general, Theorem 1 guarantees that the isomorphism will hold where sign is not considered. To see this, consider the operation

		[ S(X(i);i=1,...n) ] +' [ S(Y(i);i=1,...n) ] =

			[ S(X(i);i=1,...n) + S(Y(i);i=1,...n) ]

and the mapping set([ S(X(i);i=1,...n) ]) = {X(i);i=1,...n}. Note that the mapping set() is one-to-one because of Theorem 1: if there are multiple copies of some X(i) within ([ S(X(i);i=1,...n) ]), then they will cancel out, leaving at most a single copy of each one. It is clear that

		set(Q +' R) = set(Q) sym-diff set(R)

where "sym-diff" denotes the symmetric difference operator. This is because, if some element occurs in both Q and R, it will cancel out in the sum Q +' R by Theorem 1. The only things left in the sum will be those things that are in either Q, or R, but not both. This shows that the operation +' is isomorphic to symmetric difference; but it is already known that the DCLG multiplication operation e(A) * e(B), without signs considered, is identical to symmetric difference on the set of basis elements making up DCLG elements.


Theorem 3:

Consider a temporal domain (G,^) where G contains n entities, none of which is equal to C. Then, the algebra (s(h(G)),+,*) is isomorphic to the integer lattice DCL(n) of the real Clifford Algebra CL(n).


The elements of DCL(n) are finite sums of elements of DCLG(n).

The first claim of the proof is that the elements of DCL(n) can be expressed as finite sums of elements of G. This follows because Theorem 2 shows that G is isomorphic to DCLG(n) where the multiplication * is concerned; and the operation + in Ons algebra obeys the same rules as the operation + in DCL, i.e. it is commutative, associative and distributive with regard to the soft boundary operator <>. (The behavior of the Ons algebra + with regard to the hard boundary operator is not relevant here, because only the soft boundary operator is used in computing products in (s(h(G)),+,*). The hard boundary operator is used in computing the products X(i)*X(j) within the algebra G, but these products are a "given" in regard to DCL; they are the DCLG multiplication table.)

Next it must be shown that the elements of s(h(G)) behave under * in the same way that the elements of DCL(n) do. But this follows from the distributive law for * under the soft boundary operator, e.g.

	< X+Y > * < U+V > = < < X+Y >*U + < X+Y >*V > =
				 X*U + X*V + Y*U + Y*V

and from the cancellation law

		< [X+C] + [X] > = < V >

which shows that the h(G) correlate of a DCLG element ([X]) and its the h(G) correlate of its negative ([X+C]) cancel out.