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Let us consider a matrix algebra $Mat_{n\times n}(K)$, where $K$ is a field, $char K \neq 2.$

It is well-known that the axiomatization of commutator operation $[A,B]=AB-BA$ on matrix algebra leads us to the theory of Lie algebras. Axiomatization of $A\circ B= \frac{1}{2}(AB+BA)$ leads us to Jordan algebras.

Let us consider an operation $A \Box B= \frac{1}{2}(AB+BA^T),$ arising, for example, in control. How we can describe a class of algebras arising from axiomatization of such an operation?

UPDATE As it was shown by Pasha Zusmanovich below considering only $\Box$ leads us to a trivial variety(of course we can try to proceed to a quasivariety..).

But, if we add a transposition to the signature situation becomes much more interesting. First of all we have $(A\Box B)^T=A\Box B^T$ and the left unit $I\Box A= A$ Really, if we consider $T$-invariant subalgebras of some matrix algebra with $\circ$, than we can note that such algebras could be decomposed (as vector spaces) to the direct sum of Jordan algebra and Lie algebra -- symmetric and antisymmetric part,respectively. Axiomatizing this decomposition we get... Commutativity for symmetric part: $$ (A+A^T)\Box (B+B^T)=(B+B^T)\Box (A+A^T), $$ Power-associativity for symmetric part: $$ (A+A^T)\Box ((A+A^T)\Box (A+A^T))= ((A+A^T)\Box (A+A^T))\Box (A+A^T) $$ Jordan identity for symmetric part: $$ ((A+A^T)\Box (B+B^T))\Box ((A+A^T)\Box (A+A^T))= (A+A^T)\Box ((B+B^T)\Box ((A+A^T)\Box (A+A^T))) $$ Anticommutativity for antisymmetric part: $$ A\Box A+A^T\Box A^T=A\Box A^T+A^T\Box A $$ Lie identity for antisymmetric part: $$ (A-A^T)\Box ((B-B^T)\Box (C-C^T))+(C-C^T)\Box ((A-A^T)\Box (B-B^T))+(B-B^T)\Box ((C-C^T)\Box (A-A^T))=0 $$

Commutativity and power-associavity for symmetric part could be seen as averaged commutativity and averaged associativity and(!) commutativity, respectively. $$ A\Box B + A\Box B^T + A^T\Box B + A^T\Box B^T= B\Box A + B^T\Box A + B\Box A^T +B^T\Box A^T $$ $$ \sum_{\sigma\in S_3}\sum_{(i,j,k)\in (\varnothing,T)^3}A_{\sigma(1)}^{i}\Box(A_{\sigma(2)}^j\Box A_{\sigma(3)}^k) =\sum_{\sigma\in S_3}\sum_{(i,j,k)\in (\varnothing,T)^3}(A_{\sigma(1)}^{i}\Box A_{\sigma(2)}^j)\Box A_{\sigma(3)}^k $$

Did anyone consider something close to varieties of algebras with identities of that type?

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Do you also want to consider the transpose in this axiomatization? (Also, it's not clear to me that Jordan algebras are the correct axiomatization of the anticommutator: it's known that exceptional Jordan algebras exist which aren't embeddable into matrix Jordan algebras.) – Qiaochu Yuan Jul 24 '11 at 19:21
No, i do not consider the transpose as a separate operation in that variety of algebras. Also i can accept an axiomatisation that is not exact, but give rise to some interesting class. Like "if you see an anticommutator then think on Jordan algebras". – probably Jul 24 '11 at 19:36
Here's a random junk idea: we work with halves throughout. Define $f(a,b) = 0.5[a,b]+ 0.5(ab+ba)$, and define $f^T$ to be the half-transpose $0.5[b,a^T]+0.5(ba^T+a^Tb)$. Then, $a\square b = 0.5(f + f^T)$. – Suvrit Jul 25 '11 at 17:59

Including the operation $A\mapsto A^T$ can be viewed, in the language of operads, in many different closely related ways: via adjoining a new unary operation $J$ that satisfies $J^2=id$ and $J(ab)=J(b)J(a)$; via considering operads over the semisimple algebra $\mathbb{C}[t]/(t^2-1)$, via splitting everything into symmetric/antisymmetric, and considering the corresponding coloured operads (this looks like what you are doing in the examples of identities you give) etc. For either approach, you will find some papers dealing with similar things, though not necessarily literally the structure you are asking about. One way to try and list the possible identities would be to use operadic Groebner bases, - this way, for example, it is possible to show that for pre-Lie algebras the symmetrised operations does not satisfy any identities (, but if there are identities, then one can detect them too, using an appropriate ordering. (It's like Groebner bases in the case of commutative algebras: if you are solving a system of polynomial equations and suspect that on all solutions one of coordinates $z_i$ has finitely many values, Groebner bases can detect that, and produce an equation in one variable that $z_i$ satisfies.)

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(This is an answer to the updated question). I have a bit trouble with your terminology (what you are talking about are definitely not direct sums of a Lie and a Jordan algebra), but I hope I understand the spirit of your question, without going to much into nitpicking. Again, being put, probably, in a bit more general framework, the question is whether the algebras with a few binary operations, one of them, for example, satisfies the Jacobi identity, another one satisfies the Jordan identity, with some compatibility conditions between them, or something like that, were considered in the literature.

The answer is "yes". In fact, there is so much papers about such sort of objects, that it is a bit difficult to point just on a few of them. Lot of such objects are studied nowadays in the framework of operadic theory, see a very interesting compendium arXiv:1101.0267 . Some authors who wrote a lot on the topic: Dzhumadil'daev (e.g., , , , ), M. Goze, Markl and Remm (look them on arXiv).

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"what you are talking about are definitely not direct sums of a Lie and a Jordan algebra" There are two subalgebras(Jordan and Lie), and my algebra is their direct sum as vector spaces(not as algebras, of course), but with non-trivial interaction of multiplications. – probably Oct 31 '11 at 7:05
Ok then. This is a commonly exploited situation in the context of (associative) rings with involutions. – Pasha Zusmanovich Oct 31 '11 at 7:32

Let me reformulate the question a bit and then provide an empirical argument in favor of the negative answer.

For an associative algebra $A$ with involution $J$, define a new binary operation on $A$ as $a$ $\square$ $b = ab + ba^J$. The question is whether this operation defines a "nice" class of algebras.

This seems to be a correct generalization of the initial question, as by analogy with Lie ($[a,b] = ab - ba$) and Jordan ($a \circ b = ab + ba$) algebras constructed from associative algebras in a similar way, "interesting" and "natural" examples of such algebras do not confined to matrix algebras $A$.

What is a "nice" class of algebras? Again, by analogy with Lie and Jordan situations we might require that it satisfies some nontrivial identity (like Jacobi or Jordan identity), i.e., form a variety of algebras. Then the answer is "no": the class of all associative algebras with involution under the opertion $\square$ do not satisfy any nontrivial identity. For, suppose such identity exist. Than each associative algebra $A$ with involution satisfies some idenity of the form $f(x_1, \dots, x_n, x_1^J, \dots, x_n^J) = 0$ with nontrivial occurences of involutions. Then by theorem of Amitsur, $A$ satisfies a power $n$ of a standard identity of degree $d$, both $n$ and $d$ are determined by $f$ (see, for example, I.N. Herstein, Rings with Involution, The Univ. of Chicago Press, 1976, $\S$ 5). As, obviously, not all algebras with involution would satisfy the latter identity, we get a contradiction.

Even if we confine ourselves with matrix algebars with transposition, as in the initial question, we can choose a matrix algebra of sufficiently large degree not satisfying a given power of a standard identity of a given degree, so even that narrower class of algebras do not form a variety.

In fact, this argument is valid for any binary operation defined in terms of addition, multiplication, and involution in an associative algebra, provided involution occurs non-trivailly, not necessary $\square$.

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What can we say if we include the transpose as a basic operation? For example, we now have the identity $(a \Box b)^T = a b^T + b^T a^T = a \Box (b^T)$. – Qiaochu Yuan Aug 8 '11 at 18:48
I don't know. Sure, this argument is valid only when we consider identitites in $\square$ only. – Pasha Zusmanovich Aug 8 '11 at 19:06
Therefore variety generated by matrix algebras from the initial question will contain an absolutely free nonassociative algebra? – probably Aug 17 '11 at 12:07
@probably: It will contain an absolutely free associative algebra (as all algebras generating the variety are associative). – Pasha Zusmanovich Aug 17 '11 at 12:15
Matrix algebras with operation $\Box$ from the initial question are nonassociative, therefore if there are no identities then this variety will contain absolutely free nonassociative algebra? – probably Aug 17 '11 at 12:41

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