Operator which is symmetric but not associative?

Question

Addition and multiplication are symmetric and associative. But I have no idea about operators which are symmetric but not associative. Please help me listing any such 2-arity operators.

score 5 · Answer 1 · answered Jun 11 '13 at 12:33

One standard example is the set of all $n\times n$ real matrices, with a product $\circ$ defined by $$ A\circ B = \frac12(AB+BA). $$ This is commutative but not associative. This is related to Jordan algebras which arise in a number of places. Note that if $A$ and $B$ are symmetric, then so is $A\circ B$.

score 5 · Answer 2 · answered Jun 22 '13 at 17:58

Consider the average of two real numbers, i.e., $a*b=\frac{a+b}2$. (Or you can take midpoint of two points in a plane, it works all the same.) Clearly, $a*b=b*a$ but
$$a*(b*c)=\frac{a+\frac{b+c}2}2=\frac{2a+b+c}4$$ and
$$(a*b)*c=\frac{\frac{a+b}2+c}2=\frac{a+b+2c}4$$
are different whenever $a\ne c$.

score 3 · Accepted Answer · answered Jun 11 '13 at 12:46

Symmetric operations are easy to come by, aren't they? Just write down a symmetric polynomial at random, like $x^2+y^2$, or make up a symmetric operation table haphazardly. It might be associative, but the odds are it's not.

score 2 · Answer 4 · answered Jun 11 '13 at 12:58

Here's another example. Define the following multiplication operator on $\overline{\mathbb{R}} = \mathbb{R} \cup \{\infty\}$ $$ a \odot b = \begin{cases} a\cdot b &\text{if $a,b \in \mathbb{R}$} \\ 1 &\text{if $a=0$, $b=\infty$ or $a=\infty$, $b=0$} \\ \infty &\text{if $a\neq 0$, $b=\infty$ or $a=\infty$, $b\neq 0$} \end{cases} $$ Then $$\begin{aligned} &(2 \odot 0) \odot \infty = 0 \odot \infty = 1 \\ \neq& 2 \odot (0 \odot \infty) = 2 \odot 1 = 2 \text{.} \end{aligned}$$

score 1 · Answer 5 · answered Aug 09 '15 at 15:27

Every commutative (symmetric) quasigroup operation. In particular, for example, the one associated to a Steiner Triple System $S_{\nu}$:

$$ a\circ b:=\left\{ \begin{array}{ccc} a & , & b=a\\ c\colon\left\{a,b,c\right\}\in S_{\nu} & , & b\ne a \end{array} \right. $$

over a set $X$ with $\nu > 3$ elements.

Avitus · Answer 6 · 2013-06-11T11:54:03.870

1

I do not know if it is exactly what you are searching for, but $A_\infty$-algebras are deformation of associative algebras; their composition is associative up to "higher homotopies". In other words, they are not associative, in general.

They have been introduced by Stasheff in algebraic topology and nowadays they are widely used in quantum algebra. A great survey is Keller's: http://arxiv.org/pdf/math/0510508v3.pdf

Symmetry in this context is a relaxed concept, leading to "cyclic $A_\infty$-structures".

edited Jun 11 '13 at 11:54

answered Jun 11 '13 at 11:48

Avitus

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Thank much for avitus, your material is very sophisticated, I'd rather want any operators in elementary level. – logician Jun 11 '13 at 11:58
Then take a set $A={a,b}$ with $a$ and $b$ distinct. Define the product $$ as follows: $aa=b$, $bb=a$ and $ab=ba=a$. It is symmetric but not associative, as $a(ab)=aa=b$, while $(aa)b=b*b=a$. – Avitus Jun 11 '13 at 12:12

Operator which is symmetric but not associative?

6 Answers6