Antiassociative algebras with R: the evitaicossa package

Antiassociative algebras

Algebras satisfying \(\mathbf{u}(\mathbf{v}\mathbf{w})=-(\mathbf{u}\mathbf{v})\mathbf{w}\) exhibit some startling behaviour. Firstly, in the vector space there are no scalars except for \(0\in\mathbb{R}\). Proof: for any \(x\in\mathbb{R}\), we have \(x^3=x(xx)=-(xx)x=-x^3\); thus \(x^3=-x^3\), so \(x=0\). Secondly, antiassociative algebras are nilpotent of order 4:

\[ (\mathbf{a}\mathbf{b})(\mathbf{c}\mathbf{d}) = -\mathbf{a}(\mathbf{b}(\mathbf{c}\mathbf{d})) = \mathbf{a}((\mathbf{b}\mathbf{c})\mathbf{d}) = -(\mathbf{a}(\mathbf{b}\mathbf{c}))\mathbf{d} = ((\mathbf{a}\mathbf{b})\mathbf{c})\mathbf{d} = -(\mathbf{a}\mathbf{b})(\mathbf{c}\mathbf{d}) \]

We see that \((\mathbf{a}\mathbf{b})(\mathbf{c}\mathbf{d})=-(\mathbf{a}\mathbf{b})(\mathbf{c}\mathbf{d})\) so \(\mathbf{a}\mathbf{b}\mathbf{c}\mathbf{d}\) (however bracketed) must be zero.

The free antiassociative algebra

We consider vector spaces generated by a finite alphabet of symbols \(\mathbf{x}_1,\ldots,\mathbf{x}_n\). These will be denoted generally by a single letter, as in \(\mathbf{a},\mathbf{b},\ldots,\mathbf{z}\). We now consider the algebra spanned by products of linear combinations of these symbols, subject only to the axioms of an algebra [and the antiassociative relation \(\mathbf{u}(\mathbf{v}\mathbf{w})=-(\mathbf{u}\mathbf{v})\mathbf{w}\)]. Given an alphabet \(\mathbf{x}_1,\ldots,\mathbf{x}_n\), the general form of an element of an antiassociative algebra will be

\[ \sum_{i}\alpha_i\mathbf{x}_i + \sum_{i,j}\alpha_{ij}\mathbf{x}_i\mathbf{x}_j+ \sum_{i,j,k}\alpha_{ijk}(\mathbf{x}_i\mathbf{x}_j)\mathbf{x}_k \]

(see Remm (2024) for a proof, but note that she uses \(\mathbf{x}_i(\mathbf{x}_j\mathbf{x}_k)\) rather than \((\mathbf{x}_i\mathbf{x}_j)\mathbf{x}_k\) for the triple products; a brief discussion is given in the appendix). In the package, the components of the first term \(\sum_{i}\alpha_i\mathbf{x}_i\) are known as “single-symbol” terms [\(\mathbf{x}_i\)] and coefficients [\(\alpha_i\)] respectively. Similarly, the components of \(\sum_{i,j}\alpha_{ij}\mathbf{x}_i\mathbf{x}_j\) are known as the “double-symbol” terms and coefficients; and the components of \(\sum_{i,j,k}\alpha_{ijk}(\mathbf{x}_i\mathbf{x}_j)\mathbf{x}_k\) are the “triple-symbol” terms and coefficients.

Addition is performed elementwise among the single-, double-, and triple- components; the result is the (formal) composition of the three results. Given

\[A= \sum_{i}\alpha_i\mathbf{x}_i + \sum_{i,j}\alpha_{ij}\mathbf{x}_i\mathbf{x}_j+ \sum_{i,j,k}\alpha_{ijk}(\mathbf{x}_i\mathbf{x}_j)\mathbf{x}_k \]

\[B= \sum_{i}\beta_i\mathbf{x}_i + \sum_{i,j}\beta_{ij}\mathbf{x}_i\mathbf{x}_j+ \sum_{i,j,k}\beta_{ijk}(\mathbf{x}_i\mathbf{x}_j)\mathbf{x}_k \]

(where the sums run from \(1\) to \(n\)), we define the sum \(A+B\) to be

\[ \sum_{i}(\alpha_i+\beta_i)\mathbf{x}_i + \sum_{i,j}(\alpha_{ij}+\beta_{ij})\mathbf{x}_i\mathbf{x}_j+ \sum_{i,j,k}(\alpha_{ijk}+\beta_{ijk})(\mathbf{x}_i\mathbf{x}_j)\mathbf{x}_k \]

Multiplication is slightly more involved. We define the product \(AB\) to be

\[ \sum_{i,j}\alpha_i\beta_{ij}\mathbf{x}_i\mathbf{x}_j +\sum_{i,j,k}\alpha_{ij}\beta_{k}(\mathbf{x}_i\mathbf{x}_j)\mathbf{x}_k -\sum_{i,j,k}\alpha_i\beta_{jk}(\mathbf{x}_i\mathbf{x}_j)\mathbf{x}_k. \]

The minus sign in front of the third term embodies antiassociativity.

The evitaicossa package

The evitaicossa package implements these relations in the context of an R-centric suite of software. I give some examples of the package in use. A good place to start is function raaa(), which returns a simple random element of the free antiassociative algebra:

raaa()

## free antiassociative algebra element:
## +1a +1c +2d +1b.c +1c.b +1c.c +2(b.a)a +1(b.b)c +1(b.c)a

(the default alphabet for this command is \(\left\lbrace\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\right\rbrace\)). We see the print method for the package which shows some of the structure of the object. This one has some single-symbol elements, some double-symbol and some triple-symbol elements.

It is possible to create elements using the aaa() or as.aaa() functions:

x  <- as.aaa(c("p","q","r"))
x1 <- aaa(s1 = c("p","r","x"),c(-1,5,6))
y <- aaa(d1 = letters[1:3],d2 = c("foo","bar","baz"),dc=1:3)
z <- aaa(
    t1 = c("bar","bar","bar"),
    t2 = c("q","r","s"),
    t3 = c("foo","foo","bar"),
    tc = 5:7)

And then apply arithmetic operations to these objects:

x

## free antiassociative algebra element:
## +1p +1q +1r

x1

## free antiassociative algebra element:
## -1p +5r +6x

x+x1

## free antiassociative algebra element:
## +1q +6r +6x

(above, note the cancellation in x+x1). Multiplication is also implemented (package idiom is to use an asterisk *):

x*(x1+y)

## free antiassociative algebra element:
## -1p.p +5p.r +6p.x -1q.p +5q.r +6q.x -1r.p +5r.r +6r.x -1(p.a)foo -2(p.b)bar
## -3(p.c)baz -1(q.a)foo -2(q.b)bar -3(q.c)baz -1(r.a)foo -2(r.b)bar -3(r.c)baz

Check:

x*(x1+y) == x*x1 + x*y

## [1] TRUE

We end with a remarkable identity:

\[ (\mathbf{a}+\mathbf{a}\mathbf{x})(\mathbf{b} + \mathbf{x}\mathbf{b})=\mathbf{a}\mathbf{b} \]

Numerically:

a <- raaa()
b <- raaa()
x <- raaa()
(a+a*x)*(b+x*b) == a*b

## [1] TRUE

Extract and replace methods

Because of the tripartite nature of antiassociative algebra, the package provides three families of extraction methods: single(), double() and triple(), which return the different components of an object:

a

## free antiassociative algebra element:
## +4a +1b +2a.b +2c.c +2d.d +1(c.d)a +4(d.b)c +4(d.d)b

single(a)

## free antiassociative algebra element:
## +4a +1b

double(a)

## free antiassociative algebra element:
## +2a.b +2c.c +2d.d

triple(a)

## free antiassociative algebra element:
## +1(c.d)a +4(d.b)c +4(d.d)b

The corresponding replacement methods are also implemented:

single(a) <- 0
a

## free antiassociative algebra element:
## +2a.b +2c.c +2d.d +1(c.d)a +4(d.b)c +4(d.d)b

double(a) <- double(b) * 1000
a

## free antiassociative algebra element:
## +2000b.b +1000b.d +2000c.d +1(c.d)a +4(d.b)c +4(d.d)b

Square bracket extraction and replacement is also implemented:

(a <- raaa(s=5))

## free antiassociative algebra element:
## +4a +9c +3b.c +1b.d +2c.b +3d.a +2d.c +2(a.d)d +3(b.a)a +3(b.c)d +1(c.d)d
## +3(d.c)b

a[s1=c("c","e"),t1="c",t2="d",t3="d"]

## free antiassociative algebra element:
## +9c +1(c.d)d

Above we pass named arguments ( et seq.) and the appropriate aaa object is returned. Zero coeffients are discarded. This mode also implements replacement methods:

(a <- raaa(s=5))

## free antiassociative algebra element:
## +8a +3b +1a.b +3b.c +4c.c +4c.d +2(a.c)c +4(b.c)a +4(b.c)d +4(c.a)c +1(c.d)d

a[s1="a",d1=c("c","w"),d2=c("d","w")] <- 888
a

## free antiassociative algebra element:
## +888a +3b +1a.b +3b.c +4c.c +888c.d +888w.w +2(a.c)c +4(b.c)a +4(b.c)d +4(c.a)c
## +1(c.d)d

The other square bracket method is to pass an (unnamed) character vector:

(a <- raaa())

## free antiassociative algebra element:
## +1a +2c +2d +3b.a +3b.b +3c.c +2(a.d)d +2(b.c)a +4(c.d)b

x

## free antiassociative algebra element:
## +5c +1d +3a.c +4d.a +1d.b +1(a.a)d +2(b.d)a +3(d.d)a

s1(x)

## A disord object with hash 059117d244e1291a3c3e4e94c5ad5bbc0ed7c254 and elements
## [1] "c" "d"
## (in some order)

sc(x)

## A disord object with hash 059117d244e1291a3c3e4e94c5ad5bbc0ed7c254 and elements
## [1] 5 1
## (in some order)

list(d1(x),d2(x),dc(x))

## [[1]]
## A disord object with hash c3d0273c48d84843b58af77622164b0c0a319848 and elements
## [1] "a" "d" "d"
## (in some order)
## 
## [[2]]
## A disord object with hash c3d0273c48d84843b58af77622164b0c0a319848 and elements
## [1] "c" "a" "b"
## (in some order)
## 
## [[3]]
## A disord object with hash c3d0273c48d84843b58af77622164b0c0a319848 and elements
## [1] 3 4 1
## (in some order)

Matrix index extraction

If square bracket extraction is given an index that is a matrix, it is interpreted rowwise:

l <- letters[1:3]
(a <- aaa(s1=l,sc=1:3, d1=l,d2=rev(l),dc=3:1,t1=l,t2=l,t3=rev(l),tc=1:3))

## free antiassociative algebra element:
## +1a +2b +3c +3a.c +2b.b +1c.a +1(a.a)c +2(b.b)b +3(c.c)a

a[cbind(l,l)]

## free antiassociative algebra element:
## +2b.b

a[cbind(rev(l),l,l)] <- 88
a

## free antiassociative algebra element:
## +1a +2b +3c +3a.c +2b.b +1c.a +1(a.a)c +88(a.c)c +88(b.b)b +88(c.a)a +3(c.c)a

Note on generalized antiassociativity

We may generalize antiassociativity to \(\mathbf{a}(\mathbf{b}\mathbf{c})=k(\mathbf{a}\mathbf{b})\mathbf{c}\). Thus associativity is recovered if \(k=1\) and antiassociativity if \(k=-1\). Then the nilpotence argument becomes:

\[ (\mathbf{a}\mathbf{b})(\mathbf{c}\mathbf{d}) = k^{-1}\mathbf{a}(\mathbf{b}(\mathbf{c}\mathbf{d})) = \mathbf{a}((\mathbf{b}\mathbf{c})\mathbf{d}) = k(\mathbf{a}(\mathbf{b}\mathbf{c}))\mathbf{d} = k^2((\mathbf{a}\mathbf{b})\mathbf{c})\mathbf{d} = k(\mathbf{a}\mathbf{b})(\mathbf{c}\mathbf{d}) \]

The value of \(k\) may be set at compile-time by editing file src/anti.h. The line in question reads:

#define K -1 // a(bc) == K(ab)c

but it is possible to change the value of K. Note that this will cause test_aac.R, one of the testthat suite, to fail R CMD check.

x <- 3
class(x) <- "foo"
`*.foo` <- function(x,y){x + y + x}
print.foo <- function(x){print(unclass(x))}
c(`(x*x)*x` = (x*x)*x,  `x*(x*x)` = x*(x*x),  `x*x*x` = x*x*x)

## (x*x)*x x*(x*x)   x*x*x 
##      21      15      21