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.

Answer 1

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.

Answer 2

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".

Answer 3

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 $A\circ B$.

Answer 4

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 $$\begin{aligned} &(2 \odot 0) \odot \infty = 0 \odot \infty = 1 \\ \neq& 2 \odot (0 \odot \infty) = 2 \odot 1 = 2 \text{.} \end{aligned}$$

Answer 5

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 $a\ne c$.