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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

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    $\begingroup$ It seems to me that Faulkner is talking about a different algebraic structure. His four-linear form is not symmetric, while I would assume that the form associated to quartic $q$ is symmetric. That would also easily imply that the triple product is symmetric as well. $\endgroup$ May 21, 2017 at 15:29

2 Answers 2

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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.

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  • $\begingroup$ That does indeed solve the mystery! I didn't say in detail how $Q$ is obtained from $q$ because the article I was quoting doesn't, I just copied their text (as far as copying from pdf is possible at all). The only mystery that remains is how the constructions of Ferrar and Helenius relate to those of Faulkner who claims that $Q$ only satifies your identity when $\sigma$ lies in some Klein four subgroup of $S_4$. But it might well be that Faulkners $Q$ becomes totally symmetric in the special case of $E_7$ - I'll try to figure that out myself and perhaps ask a follow up question. Thank you! $\endgroup$
    – Vincent
    May 21, 2017 at 15:07
  • $\begingroup$ Also Thank You to the anonymous MSE moderator who moved my question here! $\endgroup$
    – Vincent
    May 21, 2017 at 15:20
  • $\begingroup$ For the sake of completeness: my speculation about Faulkner's $Q$ in the comment above is wrong. Faulkner's $Q$ is not totally symmetric in the special case of $E_7$, instead it is a slightly different $Q$ than the one from Ferrar/Helenius - see the details in my answer below. $\endgroup$
    – Vincent
    Jun 19, 2017 at 10:39
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I figured out the precise relationship between the 'Class of Ternary Algebras' of Faulkner and the 'Freudenthal Triple Systems' of Ferrar/Helenius and I will write it down here for the benefit of future readers (mostly my future self).

Following Helenius (as in the OP) I write $\langle . , . \rangle$ for the skew-symmetric biliear form. Following John Baez above I write $q$ for the quartic form and $Q$ for the 4-linear form. Following Faulkner I write $\langle . , . , . \rangle$ for the triple product. In all cases I use the subscript $f$ for Faulkners definitions and $h$ for Ferrar's definition (as I mostly learned his definition from Helenius). With these notational conventions we have:

$$\langle x, y \rangle_f = \langle y, x \rangle_h = - \langle x, y \rangle_h$$ $$Q_h(x_1, x_2, x_3, x_4) = \frac{1}{4!} \sum_{\pi \in S_4} Q_f(x_{\pi(1)}, x_{\pi(2)}, x_{\pi(3)}, x_{\pi(4)})$$ from which it follows that $$q_h(x) = Q_h(x, x, x, x) = Q_f(x, x, x, x) = q_f(x).$$ The relation between the triple products can be deduced from $$Q_f(x, y, z, w) = \langle \langle x, y ,z \rangle_f, w \rangle_f$$ $$Q_h(x, y, z, w) = \langle w, \langle x, y ,z \rangle_h \rangle_h = \langle \langle x, y ,z \rangle_h, w \rangle_f$$

This is not perfect. It allows us to deduce $Q_h$ from $Q_f$ but not the other way around and obtaining an explicit description of $\langle . , . , . \rangle_h$ in terms of $\langle . , . , . \rangle_f$ is messy. However, the article of Helenius provides a great way of clearing this up.

Faulkner starts with his version of the triple system and uses it to create a Lie algebra out of it; similar constructions of (the same) Lie algebra from Ferrar's version of the triple system have been given elsewhere. Helenius turns this viewpoint upside down: he starts with the Lie algebra ($\mathfrak{g}$ in his notation) identifies a special subspace $\mathfrak{g}_1$ on which he can define $\langle . , . \rangle_h, \langle . , . , . \rangle_h$ and $Q_h$ in terms of the Lie algebra structure and then shows they satisfy the Freudenthal triple system axioms. I will give here the analogous presentation of $\langle . , . \rangle_f, \langle . , . , . \rangle_f$ and $Q_f$. We may assume that the Lie-algebra $\mathfrak{g}$ is simple, apparently this is equivalent to the non-degeneracy of $\langle . , . \rangle$.

Helenius starts with picking a long root $\rho$ and defines the space $\mathfrak{g}_1$ as the sum of all the root spaces $\mathfrak{g}_\alpha$ for roots $\alpha$ satisfying $2 \frac{(\alpha, \rho)}{(\rho, \rho)} = 1$. (I try to avoid introducing another use of $\langle . , . \rangle$ here.) More generally he defines a grading of $\mathfrak{g}$ where $\mathfrak{g}_k$ is the span of the rootspaces $\mathfrak{g}_\beta$ where $2 \frac{(\beta, \rho)}{(\rho, \rho)} = k$.

We fix a generator $x_{\rho}$ of $\mathfrak{g}_\rho$ and define $x_{-\rho} \in \mathfrak{g}_{-\rho}$ in such a way that $\{x_\rho, x_{-\rho}, [x_\rho, x_{-\rho}]\}$ is a standard $\mathfrak{sl}_2$-triple.

Now we have for all $x, y, z, w \in \mathfrak{g}_1$ that

$$[x, y] = \langle y, x \rangle_f x_\rho$$ $$\langle x, y, z \rangle_f = [x, [y, [z, x_{-\rho}]]] = (ad(x)\circ ad(y) \circ ad(z)) (x_{-\rho})$$ $$Q_f(x, y, z, w) = [w, [x, [y, [z, x_{-\rho}]]]] = (ad(w) \circ ad(x)\circ ad(y) \circ ad(z)) (x_{-\rho})$$

Comparing to the equalities $$[x, y] = \langle x, y \rangle_h x_\rho$$ $$Q_h(x_1, x_2, x_3, x_4) = \frac{1}{4!} \sum_{\pi \in S_4} (ad(x_{\pi(1)}) \circ ad(x_{\pi(2)}) \circ ad(x_{\pi(3)}) \circ ad(x_{\pi(4)})) (x_{-\rho})$$ from Helenius, chapter 3, yields the above relations.

The equation $\langle x, y, z \rangle_f = [x, [y, [z, x_{-\rho}]]]$ is given explicitely in Faulkner as part of the proof of Lemma 3. Verifying that the above descriptions of $\langle . , . \rangle_f, \langle . , . , . \rangle_f$ and $Q_f$ when taken as the definitions satisfy Faulkner's axioms $T1 - T4$ is mostly a long list of application of the Jacobi-idenity together with an occasional use of the fact that the space $\mathfrak{g}_k$ is zero for $k < - 2$ and $k > 2$.

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