Has the determinant of a involution of the first kind ever been considered as an invariant?

Question

$\DeclareMathOperator\Sym{Sym}\DeclareMathOperator\Aut{Aut}\newcommand\id{\mathrm{id}}$Let $k$ be a field of characteristic zero. Let $A, B$ be central, simple algebras over $k$ of even degrees $n,m > 1$. Let $\sigma$ be an involution on $A$, which is either symplectic or orthogonal. One way to check if an involution is of these types, is to use the fact that for the $k-$vector space $\Sym(A, \sigma) = \{ x \in A \mid \sigma(x) = x \}$ one has $\dim_k(\Sym(A,\sigma)) = n(n-1)/2$ or $n(n+1)/2$ respectively.

Since $\sigma \in \Aut(A)$, such that $\sigma \circ \sigma = \id$, we find that the determininant of $\sigma$ must be $1$ or $-1$.

It is well known that a tensorproduct of involutions induces an involution on the respective product of the algebras. If we only consider symplectic and orthogonal involutions, the resulting involution is known to be symplectic or orthogonal as well, while it is symplectic if and only if the number of symplectic factors is odd. But since $\det(\sigma \otimes \tau) = \det(\sigma)^n det(\tau)^m$ for $(A,\sigma), (B,\tau)$ and $n,m$ are even, we can conclude that if the determinant of an involution is $-1$, then it is not decomposable.

Questions: Has this property ever considered in the literature? Is an example known, where $n$ is biggen than $2$, i.e. $A$ is not a quaternion algebra?

Answer 1

There is an notion of determinant $\det(\sigma)$ of an involution $\sigma$ of the first kind on a central simple $F$-algebra $A$, which is well-known and well studied. It takes values in $F^\times/F^{\times 2}$.

It is finer than your own version, which is too coarse (the least we should expect is to recover the notion of a discriminant of a symmetric bilinear form when your c.s.a. is split, which is not the case with your version).

It is defined as the square class of the reduced norm of any skew symmetric element. If $A$ is split ans $\sigma$ is adjoint to a symmetric bilinear form $b$, then $\det(\sigma)=\det(b)$.

For symplectic involutions, it is trivial since the symplectic groups are connected groups, and thus do not admit degree one cohomological invariants. This argument shows too that your determinant will be always trivial in the symplectic case.

In degree 4, as for symmetric bilinear spaces, one may show that $(A,\sigma)$ is decomposable if and only if $\det(\sigma)$ is trivial.

You can find references and proof in the excellent Book of involutions, by Knus, Merkurjev, Rost and Tignol.

Note that, as the case of symmetric bilinear forms already shows, you cannot expect the previous result to be true for larger degrees.