Antiassociative algebras with R: the `evitaicossa` package

In evitaicossa: Antiassociative Algebra

knitr::opts_chunk$set(echo = TRUE)
options(rmarkdown.html_vignette.check_title = FALSE)
library("evitaicossa")
set.seed(1)

knitr::include_graphics(system.file("help/figures/evitaicossa.png", package = "evitaicossa"))

Here I introduce the evitaicossa R package for antiassociative algebras. An algebra is a vector space in which the vectors possess a bilinear product. Formally, a vector space is a set $V$ of vectors which form an Abelian group under addition and also satisfy the following axioms:

• Compatibility, $a(b\mathbf{v})=(ab)\mathbf{v}$
• Identity, $1\mathbf{v}=\mathbf{v}$
• Distributivity WRT vector addition, $a(\mathbf{u}+\mathbf{v}) = a\mathbf{u}+a\mathbf{v}$
• Distributivity WRT field addition, $(a+b)\mathbf{u} = a\mathbf{u}+b\mathbf{u}$

Above, $\mathbf{u},\mathbf{v}\in V$ are vectors, $a,b$ are scalars [here the real numbers], and $1$ is the multiplicative identity. We also require a bilinear vector product, mapping pairs of vectors to vectors; vector multiplication is denoted using juxtaposition, as in $\mathbf{u}\mathbf{v}$, which satisfies the following axioms:

• Right distributivity, $(\mathbf{u}+\mathbf{v})\mathbf{w} = \mathbf{u}\mathbf{w}+ \mathbf{u}\mathbf{w}$
• Left distributivity, $\mathbf{w}(\mathbf{u}+\mathbf{v}) = \mathbf{w}\mathbf{u}+ \mathbf{w}\mathbf{v}$
• Compatibility, $(a\mathbf{u})(b\mathbf{v})=(ab)(\mathbf{u}\mathbf{v})$

Note the absence of commutativity and associativity. Associative algebras seem to be the most common, and examples would include multivariate polynomials [@hankin2022_mvp,@hankin2022_spray], Clifford algebras [@hankin2022_clifford], Weyl algebras [@hankin2022_weyl_arxiv], and free algebras [@hankin2022_freealg]. Non-associative algebras would include the octonions [@hankin2006_onion] and Jordan algebras [@hankin2023_jordan]. Here I consider antiassociative algebras in which the usual associativity relation $\mathbf{u}(\mathbf{v}\mathbf{w})=(\mathbf{u}\mathbf{v})\mathbf{w}$ is replaced by the relation $\mathbf{u}(\mathbf{v}\mathbf{w})=-(\mathbf{u}\mathbf{v})\mathbf{w}$.

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 @remm2024antiassociative 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()

(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
x1
x+x1

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

x*(x1+y)

Check:

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

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

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
single(a)
double(a)
triple(a)

The corresponding replacement methods are also implemented:

single(a) <- 0
a
double(a) <- double(b) * 1000
a

Square bracket extraction and replacement is also implemented:

(a <- raaa(s=5))
a[s1=c("c","e"),t1="c",t2="d",t3="d"]

Above we pass named arguments (\code{s1} et seq.) and the appropriate aaa object is returned. Zero coeffients are discarded. This mode also implements replacement methods:

(a <- raaa(s=5))
a[s1="a",d1=c("c","w"),d2=c("d","w")] <- 888
a

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

(a <- raaa())

Note on disordR discipline

If we try to access the symbols or coefficients of an aaa object [functions s1() and sc() respectively], we get a disord object [@hankin2022_disordR]. Suppose we wish to extract the single-symbol terms and the single-symbol coefficients:

x
s1(x)
sc(x)

See how the hash codes of the symbols and coeffients match. However, the double-symbol terms and coefficients, while internally matching, differ from the single-symbol stuff:

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

Above, see how the double-symbol terms and double-symbol coefficients have consistent hashes, but do not match the single-symbol objects (or indeed the triple-symbol objects).

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

Appendix

As noted above, @remm2024antiassociative 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. I chose the latter because R idiom for multiplication is left associative:

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)

Above we see that x*x*x is interpreted as (x*x)*x, which is why the sign convention in the package was adopted.

References

evitaicossa documentation built on June 28, 2024, 5:08 p.m.