# Have you ever seen this bizarre commutative algebra?

I have encountered very strange commutative nonassociative algebras without unit, over a characteristic zero field, and I cannot figure out where do they belong. Has anybody seen these animals in any context?

For each natural $$n$$, the $$n$$-dimensional algebra $$A_n$$ with the basis $$x_1$$, ..., $$x_n$$ has the multiplication table $$x_i^2=x_i$$, $$i=1,...,n$$, and $$x_ix_j=x_jx_i=-\frac1{n-1}(x_i+x_j)$$ (for $$i\ne j$$).

The only thing I know about this algebra is its automorphism group, which is the symmetric group $$\Sigma_{n+1}$$ (for $$n>1$$). This can be seen from how I obtained the algebra in the first place.

Let $$I$$ be the linear embedding of $$A_n$$ onto the subspace of the $$(n+1)$$-dimensional space $$E_{n+1}$$ of vectors with zero coefficient sums in the standard basis, given by $$I(x_i)=\frac{n+1}{n-1}e_i-\frac1{n-1}\sum_{j=0}^ne_j,\quad i=1,...,n$$ and retract $$E_{n+1}$$ back to $$A_n$$ via the linear surjection $$P$$ given by $$P(e_0)=-\frac{n-1}{n+1}(x_1+...+x_n)$$ and $$P(e_i)=\frac{n-1}{n+1}x_i,\quad i=1,...,n.$$ Then the multiplication in $$A_n$$ is given by $$ab=P(I(a)I(b)),$$ where the multiplication in $$E_{n+1}$$ is just that of the product of $$n+1$$ copies of the base field; more precisely, it has the multiplication table $$e_ie_j=\delta_{ij}e_i,\quad i,j=0,...,n$$ (where $$\delta$$ is the Kronecker symbol).

$$A_n$$ is not Jordan either, in fact even $$(x^2)^2\ne(x^2x)x$$ in general.

Really don't know what to make of it.

I should probably explain my motivation, but unfortunately it is related to some preliminary results that have not been checked completely yet. I can only say that such structures appear on homogeneous pieces with respect to gradings of semisimple Lie algebras induced by nilpotent elements, and (somewhat enigmatically, I must admit) that all this is closely related to the answer of Noam D. Elkies to my previous question Seeking a more symmetric realization of a configuration of 10 planes, 25 lines and 15 points in projective space.

• Many thanks to @user44191 for the highly relevant tag! Don't know why it did not occur to me... – მამუკა ჯიბლაძე Dec 24 '18 at 7:17
• It vaguely reminds me of the cross product, considered as part of the quaternion product; similar to the cross product, it can be considered as the composition of "inject to larger space, take an associative product within the larger space, project back to smaller space". – user44191 Dec 24 '18 at 7:38
• @user44191 Thanks, yours is a very interesting observation since those structures I mentioned in the last (motivational) paragraph are in fact quite frequently Lie algebra structures! – მამუკა ჯიბლაძე Dec 24 '18 at 7:41

## 1 Answer

This algebra is defined on the permutation module of the symmetric group. It was studied by K. Harada and R. Griess in the 1970s and a proof that its automorphism group is the symmetric group can also be found in Haraada's paper (see also this one) and the appendix of this paper by Dong and Griess, which explains some connections with vertex operator algebras. See also remarks in Griess's paper here.

In the 1970s and 1980s there were many constructions of algebras of this spirit, in the search for model algebraic structures acted on in some canonical way by finite simple groups (I have in mind, for example, papers of A. Ryba, for example there is one constructing a commutative nonassociative algebra having the Harada-Norton group as automorphisms). My (inexpert) sense is that this literature is more or less subsumed into that about vertex operator algebras. More precisely, for vertex operator algebras whose $$0$$th graded piece is $$1$$-dimensional and $$1$$st graded piece is $$0$$-dimensional there is a commutative nonassociative algebra on the $$2$$nd graded piece that has many features in common with $$E_{n}$$ (here I have in mind papers such as the paper of Matsuo and the paper of Miyamoto on Griess algebras). My understanding of vertex operator algebras is inadequate to say more (and there are some experts who post here who can maybe say something clearer).

This algebra, call it $$E_{n}$$, has other interesting properties. Let $$\tau$$ be the Killing type form defined by $$\tau(x, y) = \operatorname{tr} L(x)L(y)$$ where $$L(x)y = xy$$ is the multiplication operator. Note that $$\operatorname{tr} L(x) =0$$ for all $$x$$ (I call such algebras exact). The symmetric bilinear form $$\tau$$ is nondegenerate and invariant in the sense that $$\tau(xy, z) = \tau(x, yz)$$ (this is somehow analogous to the properties of the Killing form of a semisimple Lie algebra). The algebra $$E_{n}$$ is determined up to isometric automorphism by the associated harmonic cubic polynomial $$\tau(xx, x)$$. This algebra is conformally associative in the sense that its associator $$[x, y, z]= (xy)z - x(yz)$$ satisfies $$\tau([x, y, z], w) =c\left(\tau(x, w)\tau(y,z) - \tau(x,z)\tau(y, w)\right)$$ where $$c$$ is the appropriate (dimension dependent) constant that I am too lazy to calculate right now. One can prove that up to isometric automorphism there is a unique $$n$$-dimensional commutative nonassociative algebra that thas the properties that a. $$\operatorname{tr} L(x) =0$$ for all x, b. its Killing type trace form $$\tau$$ is nondegenerate and invariant and c. it is conformally associative. (If you contact me directly I can provide you a preprint of mine where this is proved as part of a larger project.)

Another place these algebras $$E_{n}$$ occur is the following. Consider the Jordan algebra of real symmetric $$n\times n$$ matrices. Consider its deunitalization that comprises trace-free symmetric matrices with the product $$x \star y = \tfrac{1}{2}\left(xy + yx - \tfrac{1}{n}\operatorname{tr}(xy)I\right).$$ This is a commutative nonassociative algebra such that $$\operatorname{tr}L(x) =0$$ for all $$x$$ and the Killing type trace-form $$\tau(x, y) = \operatorname{tr}L(x)L(y)$$ is invariant (in fact it is a constant multiple of $$\operatorname{tr}(xy)$$). Although this algebra is far from being conformally nonassociative in the above sense, its subalgebra of (trace-free) diagonal matrices is isomorphic to the algebra $$E_{n-1}$$. Moreover, every every idempotent element of the deunitalized Jordan algebra is in the orbit of an idempotent of $$E_{n-1}$$. This can be used, for example, to prove that the deunitalized Jordan algebras are simple. (All this is written in my aforementioned preprint; contact me if you would like details).