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Is it possible(or may be easier) to give an example of non associative algebra but commutative?

At first sight, it seems possible to prove associativity from commutativity but later realised it may no be the case.

The only example of non associative algebra which I know is Octonion but which is non-commutative. Then I started to create an artificial example. Here is my approach(same as to create an commutative): Let's $\textbf{C}$ denote the category of associative algebra. Suppose free object exists in this category say $F$ be a free associative algebra, then consider the quotient of $F$ by the ideal generated by its commutator elements. I don't know how much this work but certainly this way we will have lot of such examples.

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  • $\begingroup$ One could take the semigroup algebra (over a field, say) of the Wedderburn-Etherington monoid. See H. W. Becker, Solution to Advanced Problem 4277, Amer. Math. Monthly 56 (1949), 697-699. $\endgroup$
    – Richard Stanley
    Commented Jul 12, 2020 at 13:49
  • $\begingroup$ Isn't abelian lie algebra is associative? $\endgroup$
    – Sunny
    Commented Jul 12, 2020 at 13:57
  • $\begingroup$ In your question, it should be quotient by the ideal generated by the commutators. The result would just be an polynomial algebra, so it does not work. $\endgroup$
    – jg1896
    Commented Jul 12, 2020 at 14:03
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    $\begingroup$ But how do you know that it's polynomial algebra. $\endgroup$
    – Sunny
    Commented Jul 12, 2020 at 14:10
  • $\begingroup$ The polynomial algebra is the free object in the category of commutative associative algebras. If you begin with the free associative algebra and mod out the commutators, it is a matter of checking that the resulting algebra will also be free in the category of commutative associative algebras. Since such categories are, in fact, what is known as a variety of algebras (universal algebra parlance), the result follows. $\endgroup$
    – jg1896
    Commented Jul 12, 2020 at 14:20

3 Answers 3

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The class of Jordan algebras are the most important class of algebras in this direction.

They are defined by the two identities,

(commutativity): $xy=yx$,

(Jordan identity): $(xy)(xx)=x(y(xx))$.

They are an extremely important class of non-associative algebras (check Jacobson's Strucuture and Representation of Jordan algebras or McCrimmon's A taste of Jordan algebras)

The most easy way to produce such an algebra is by considering an associative (alternative suffices) algebra $A$ and changing the product from $.$ to $*$ as follows: $a*b = \frac{ab+ba}{2}$. Some care of course is necessary in case of characteristic two.

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    $\begingroup$ Oh nice. Thanks. $\endgroup$
    – Sunny
    Commented Jul 12, 2020 at 13:58
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The "midpoint" operation $(x,y)\mapsto\frac12(x+y)$ on the reals is commutative and not associative.

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    $\begingroup$ The same operation works in any vector space over any field of characteristic $\neq2$. $\endgroup$
    – Andreas Blass
    Commented Jul 12, 2020 at 15:25
  • $\begingroup$ simplex sigillum veri. Very nice Andreas ! $\endgroup$
    – Mirco A. Mannucci
    Commented Jul 12, 2020 at 15:29
  • $\begingroup$ How it's ring operations? What is the unit element? $\endgroup$
    – Sunny
    Commented Jul 12, 2020 at 15:45
  • $\begingroup$ Nice example. @Sunny Note this isn't bilinear so to make it into an algebra you need to linearize it, or modify it in some other way. $\endgroup$
    – Phil Tosteson
    Commented Jul 12, 2020 at 17:41
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    $\begingroup$ It's not a ring. I understood "algebra" in the sense of universal algebra. If you want it to be understood in the sense of rings and bilinearity, then you should linearize it: Take the vector space with my example as a basis, and extend the midpoint operation from basis vectors to arbitrary vectors by bilinearity. $\endgroup$
    – Andreas Blass
    Commented Jul 13, 2020 at 2:24
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To see what's going on, let's describe the free commutative, non-associative, algebra on $n$ elements $x_1, \dots, x_n$, which we call $A_n$ .

Let $X_n$ be the set of isomorphism classes of binary rooted trees $T$ with a function $f: {\rm Leaves}(T) \to \{x_1, \dots, x_n\}$, which labels the leaves. Then $A_n = k X_n$ is the free vector space with basis $X_n$. The multiplication map $A_n \otimes A_n \to A_n$ is induced by a multiplication $X_n \times X_n \to X_n$, which joins pairs of trees together at a new root.

That is, given binary labelled trees $(T_1,f_1)$ and $(T_2,f_2)$, we may form a new tree $T_1\vee T_2$, with a new root and $T_1$ above one branch of the root and $T_2$ above the other branch. (Sorry if this is unclear-- this is most easily conveyed by a picture). The new labelling is just $f_1 \sqcup f_2$.

Its not hard to check that this algebra is commutative but not associative when $n \geq 2$. E.g. $(x_1 \cdot x_2) \cdot x_2 \neq x_1 \cdot (x_2 \cdot x_2)$ since these correspond to non-isomorphic labelled trees with $3$ leaves.

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  • $\begingroup$ Alternatively, to check $(x_1x_2)x_2\neq x_1(x_2x_2)$ one can produce a small quotient where it's nonzero: the 3-dimensional (not-assumed-associative) commutative algebra with basis $(a,b,c)$ and product: $a^2=c$, $bc=cb=a$, and all other products zero. Then $(ba)a=0$ while $b(aa)=bc=d$, so $(ba)a\neq b(aa)$. $\endgroup$
    – YCor
    Commented Jul 12, 2020 at 15:21

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