I'm trying to learn about Freudenthal triple systems. Here is the definition given by Helenius [1], start of Section 5:



> 
> A Freudenthal triple system is a finite-dimensional vector space $V$
>  over a field $F$ (with characteristic not 2 or 3) such that
> 
> 
> • There is a nonzero quartic form $q$ defined on $V$. A corresponding
>  4- linear form, also called $q$, is given by linearization, with $q(x, x, x, x) = q(x)$ for all $x \in V$.
> 
> 
> • There is a nondegenerate skew-symmetric bilinear form $\langle ,
> \rangle$ defined on $V$. Thus for given $x, y, z \in V$ we may define
>  the triple product $xyz$ to be the unique vector in $V$ such that
>  $q(w, x, y, z) = \langle w, xyz \rangle$ for all $w \in V$.
> 
> 
> • The triple product satisfies the following identity: $$2(xxx)xy =
> \langle y, x \rangle xxx + \langle y, xxx \rangle x.$$
> 
> 
> 


So far so good. Helenius then remarks that other sources give slightly different definitions, that are equivalent to his definition by rescaling the bilinear and/or the quartic form by a scalar. In particular, Helenius notes, the definition of Ferrar in [2] omits the 2 in the last equality.


I looked up [2] on JSTOR and indeed, the definition (beginning of section 1) is the same as in Helenius. EXCEPT that Ferrar list the additional axiom (A1) that *"xyz is symmetric in all arguments"*.


This sounds like a big deal. I find it very weird that Helenius, who explicitly states that his definition is equivalent of that of Ferrar doesn't mention this property at all. 


Moreover, I am a bit distrustful of the symmetry claim. If the definitions are indeed equivalent then the symmetry should follow from the three bullet points in the definition above but I can't see how. 


Also, there is a third article, by Faulkner [3], in which he considers a similar situation (a vector space $V$ with a triple product, a 4-linear form $q$ and a skew symmetric bilinear form $\langle, \rangle$ satisfying $q(w, x, y, z) = \langle w, xyz \rangle$) satisfying some different axioms that make it even more explicit that the triple product in general is not totally symmetric. Now Faulkner's construction need not be equivalent to the other two, but I am pretty convinced that in one special case: Faulkner's 'Example 2' starting at page 399, Ferrar's 'Prototype FTS' starting at page 314 and Helenius' '$G = E\_8$'-case discussed in detail in chapter 10, they are all really describing the same thing: the case where the vector space $V$ is 56-dimensional and the subalgebra of $\mathfrak{gl}(V)$ preserving $q$ and $\langle, \rangle$ being a Lie algebra of type $E\_7$. So at least in this case the symmetry has to come from somewhere. However, I have a hard time explicitly localizing it in [1] (Proposition 5 comes close, though.)



> 
> In summary: is Ferrar's claim that $xyz$ is totally symmetric correct and if so, how do we see it from the definitions given by Helenius and Faulkner?
> 
> 
> 


[1] Fred Helenius, 2010: Freudenthal Triple Systems from Root System Data: <https://arxiv.org/pdf/1005.1275.pdf> 


[2] J. C. Ferrar, 1972: Strictly Regular Elements in Freudenthal Triple Systems: <http://www.jstor.org/stable/1996111?seq=1#page_scan_tab_contents>


[3] John R. Faulkner, 1971: Construction of Lie algebras from a Class of Ternary Algebras: <http://www.ams.org/journals/tran/1971-155-02/S0002-9947-1971-0294424-X/S0002-9947-1971-0294424-X.pdf> 



Is the triple product in a Freudenthal triple system fully symmetric?
You say that the 4-linear form $Q$ is obtained by 'linearization' of the quartic form $q$. (I can't stand giving them both the same name.) There will be many 4-linear forms $Q$ related to $q$ by


$$ Q(x,x,x,x) = q(x) $$


but, except perhaps in characteristic 2 or 3, there is just one that is totally symmetric, meaning that


$$ Q(x\_{\sigma(1)}, \dots, x\_{\sigma(4)}) = Q(x\_1, \dots, x\_4)$$


for all $\sigma \in S\_4$. In this case there's a formula for $Q$ in terms of $q$, and people usually say it's obtained from $q$ by 'polarization'.


I mention this because you don't say in detail how $Q$ is obtained from $q$. If it's obtained by polarization, it will be totally symmetric - and then the triple product will be symmetric in all 3 arguments, solving your mystery.


