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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:
as few symbols and rules as possible
brevity of the algebraic rules
phenomenological meaningfulness of symbols and rules
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.
V, the void
N, an On
[ X ], a hard boundary demarcating some entity X
< X >, a soft boundary demarcating some entity X
C = [N], a Chronon
+, a binary operation denoting coexistence
^, a binary operation denoting temporal precedence
*, a binary operation denoting interpenetration
An "entity" is a legal expression in On algebra
X, Y, etc. are variables, referring to any entities
A legal expression is
V or N
X + Y , X * Y, or X ^ Y where X and Y are legal expressions
[X] or
X(i), Y(j), etc., are also variables;
the ( and ) have
no fundamental meaning in On algebra but are
merely a notational device
S(X(i); i=1,...n = X(1) + X(2) + ... + X(n) ;
the S or "summation" operator has no fundamental
meaning in On algebra but is merely a notational
device
[ ] = < > = < V > = V =
V + X = X
N + N = V + V = V
< [X + C] + [X] + Z > = < V + Z>
[ [X] + [Y] + Z ] = [ [X+Y] + Z ]
X * X^Y = C or X^Y = V, for any arguments X and Y
S( X(i(j)); j=1,...,n) is identical for all functions
i() that permute {1,...,n}
X * Y = X + Y + X^Y
if Q = S(X(i); i=1,...,n), R= S(X(i); i=1,...,m) with Q and R as above,
then [Q]*[R] = [Q + R + S(X(i)^Y(j);i=1,...,n;j=1,...,m) ]
[Q]^[R] = [ S(X(i)^Y(j);i=1,...,n;j=1,...,m) ]
If X^Y=C then we may say equivalently that X>Y
A "temporal domain" is an pair (G,^) where G is a set of entities and ^ obeys transitivity and antisymmetry and nonidentity over G, so that
if X^Y=C, then Y^X=V
if X^Y=C and Y^Z=C then X^Z=C
X^X = V
The ^ operator may be defined in such a way as to make the set of all entities a temporal domain. Then it is called a "linear time operator."
Given an entity G = S(X(i); i=1,...,m), the "hard powerset" of G, h(G), is the collection of all entities of the form
[ S(X(i(k)); k=1,...,M) ], where each 0<=i(k)<=n.
Similarly, the "soft powerset" of G, s(G), is the collection of all entities of the form < S(X(i(k)); k=1,...,M) >, where each 0<=i(k)<=n.
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
A boundary with nothing inside it is taken to be Void, i.e.
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.
Inside a hard boundary, hard boundaries are associative, i.e.
Inside a soft boundary, associativity of hard boundaries fails,
and we instead have the rule
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
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.
For any entity X,
(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 Suppose X=[Y]. Then, [[Y]+[Y]] = [[Y+Y]] = [[V]] = [V] = V = .
This shows e.g. that [C+C] = V.
Suppose X = 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.
QED
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
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:
meets a similar element, in which case it is cancelled with the similar element using e(i) e(i) = +1, or
finds a place in between two A-elements in the proper order.
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
and we wish to compute the product
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
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:
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
To compute the sign using only bit strings, one reasons as follows. First, one lines up the two multiplicands, one under the other:
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
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.
Consider a temporal domain (G,^) where G contains n entities,
none of which is a chronon C.
: 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 a rule which holds so long as none of the X(i) in the
formula is equal to C:
where the function i() is a permutation of {1,...,n}, defined to enforce the ordering arrangement X(i(k)) < X(i(k+1)).
a rule for dealing with C, namely
What needs to be shown is that cliff is an isomorphism, i.e. that
where Q and R are elements of h(G),
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,
Observe here that, by Theorem 1, X+X and Y+Y cancel out. Given the
ordering W Now, suppose we did this computation using the DCLG rule.
In the bit string version, we would have
yielding the W and Z. Next the sign would be obtained as follows:
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
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
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.
QED
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.
and from the cancellation law
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.
QED
[] = <> = V
N + N = V
[ [X] + [Y] ] = [ [X+Y] ]
< [C + X] + X > = < V > = V =
X * Y = X + Y + X^Y
[A+B] * [C+D] = [ A + B + C + D + A^C + A^D + B^C + B^D]
etc.
4. Some Theorems
Theorem 1:
[X + X] = [V] = V =
Proof:
Corollary: For any entity G, the hard superset h(G) is finite
( 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
A = e1 e2 e3
B = e2 e3 e4
A * B = e1 e2 e3 * e2 e3 e4
eX = e1 e2 e3 e2 e3 e4 =
- e1 e2 e2 e3 e3 e4 =
- e1 e4
e1 e2 e5 * e1 e2 e4 =
- e1 e2 e1 e5 e2 e4 =
e1 e1 e2 e5 e2 e4 =
- e2 e2 e5 e4 =
e4 e5
1110 * 0111 = - 1001
1110 *
0111
11001 *
11010
Theorem 2
Then the algebra (h(G),*) is isomorphic to the Discrete Clifford
Algebra DCLG(n)
Proof
cliff( [ S(X(i);i=1,...n) ] ) =
cliff(X(i(1)) cliff(i(2)) ... cliff(i(n))
cliff( [C + X]) = -cliff(X)
cliff(Q) * cliff(R) = cliff(Q*R)
[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]
0111 *
1110
-----
1001
3 1's in 0111 coming after the 1*** in 1110
2 *1**
1 **1*
---
6
[ S(X(i);i=1,...n) ] +' [ S(Y(i);i=1,...n) ] =
[ S(X(i);i=1,...n) + S(Y(i);i=1,...n) ]
set(Q +' R) = set(Q) sym-diff set(R)
Theorem 3:
Proof:
< X+Y > * < U+V > = < < X+Y >*U + < X+Y >*V > =
X*U + X*V + Y*U + Y*V
< [X+C] + [X] > = < V >