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A possibly non-associative algebra is flexible if it satisfies the identity $$(xy)x=x(yx).$$ This is clearly a very weak form of associativity —and obviously an associative algebra is flexible— but it is also a weak form of commutativity —a commutative algebra is clearly flexible— and in practice it plays both of these roles.

Flexibility is such a weak condition that it is difficult to imagine getting much out of it, but then one reads the works of Albert, Schafer, Jacobson and several others, and sees it put to use with such extraordinary artistry and skill that one ends up reflecting on the excessiveness of our standard hypotheses of commutativity and associativity :-)

But I have not found so far an explanation for the choice of the term:

Do you know why flexible algebras are called flexible?

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  • $\begingroup$ In spite of the fact that the flexibility relation involves two occurrences of $x$, is there an operad governing flexible algebras? $\endgroup$
    – Samuele Giraudo
    Commented Apr 23, 2013 at 22:21
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    $\begingroup$ The flexible identity is equivalent (over a field with more than two elements) to the multilinear identity $$(xy)z-x(yz)+(zy)x-z(yx)=0,$$ so you can express flexible algebras are algebras over an quadratic operad with a cubic relation. (Can someone see if this is Koszul?) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 23, 2013 at 23:19
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    $\begingroup$ @Samuele, if you want a definition of flexible algebras that doesn't have several occurrences of a variable in the same monomial, notice that an algebra is flexible iff $(x,y,z)+(z,y,x)=0$. The parenthesis denote associators. $\endgroup$
    – Gjergji Zaimi
    Commented Apr 23, 2013 at 23:19
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    $\begingroup$ @MarianoSuárez-Alvarez: This operad is not Koszul. The Koszul dual has the Hilbert series $f(t)=t+t^2+t^3/2+t^4/24$ (and nothing else, this operad is nilpotent), and the compositional inverse of $-f(-t)$ has negative coefficients (e.g. at $t^{12}$), contradicting the Ginzburg--Kapranov functional equation. $\endgroup$
    – Vladimir Dotsenko
    Commented Oct 16, 2014 at 9:50
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    $\begingroup$ @SamueleGiraudo: it is a symmetric operad which is not regular, so knowing the free algebra on one generator is not equivalent. In any case, to the best of my knowledge the answer to your question is no, even though it may be quite feasible. $\endgroup$
    – Vladimir Dotsenko
    Commented Oct 16, 2014 at 9:50

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I browsed through the 1948 paper by Albert where he introduced the notion of a flexible algebra. It is presented as a property that gives a certain flexibility to how algebraic operations are to be carried out. Like all good names, it is so natural it hardly needs an explanation.

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