Do involutions in the symmetric group form a basis of a Jordan algebra?

Question

Let $M_d$ is a vector space spanned by the set of all the involutions in the symmetric group $S_d$ (you can treat $M_d$ as a space of function on the set of involutions endowed with standard addition and multiplication by a scalar). Define an operation $\circ\colon M_d\otimes M_d \to M_d$ as an extension by the bilinearity of the following product on the involutions.

Let $a$, $b$ are two involutions. Set $a\circ b = ab$ if $ab$ is an involution (the latter product is a usual product in $S_d$), and $0$ otherwise.

The commutativity of such an operation is straightforward. It can also be checked that such an operation is not associative for $d>2$.

I am wondering, whether $M_d$ endowed with the operation $\circ$ is a Jordan algebra?

Answer 1

Is this a counterexample?

# sage

def SGA(n):
    return SymmetricGroupAlgebra(QQ, n)

def j(a, b):
    SGA = a.parent()
    res = SGA.zero()
    for m, c in a:
        for mm, cc in b:
            lap = m.left_action_product(mm)
            if max(lap.cycle_type()) <= 2:
                res += c * cc * SGA.basis()[lap]
    return res

SGA4 = SymmetricGroupAlgebra(QQ, 4)
P = Permutations()
x = SGA4(P([2,1,3,4]))+SGA4(P([1,3,2,4]))-SGA4(P([4,3,2,1]))
y = SGA4(P([4,2,3,1]))
print j(j(x,y),j(x,x)) - j(x,j(y,j(x,x)))

> 2*[2, 1, 3, 4]

EDIT: Yes. (I have checked it by hand, as the above code is messy.)

In more standard notations: Let us write permutations in $S_4$ in cycle notation. Then, set $x = \left(1,2\right) + \left(2,3\right) - \left(1,4\right)\left(2,3\right)$ and $y = \left(1,4\right)$ in the group algebra of $S_4$. Then, using your "multiplication", $x\circ\left(y\circ\left(x\circ x\right)\right) = -5\left(2,3\right) - 2\left(1,2\right) + 5\left(1,4\right)\left(2,3\right)$, whereas $\left(x\circ y\right)\circ\left(x\circ x\right) = -5\left(2,3\right) + 5\left(1,4\right)\left(2,3\right)$.