I've asked this on math.stackexchange, unsuccessfully. I hope this question is appropriate for mathoverflow.

Let $V$ be a finite-dimensional vector space over a field $K$ with $\operatorname{char}K\neq 2$, and $Q$ a non-degenerate quadratic form on $V$. The spinor norm is a homomophism

$$sn: O(V,Q) \rightarrow K^*/(K^*)^2$$

defined as $Q(v)$ for reflections by a non-isotropic vector $v$.

Alternatively, for $g \in O(V,Q)$ let $a \in \Gamma(V,Q)$ be the element of the Clifford group that realizes $g$ via an inner graded automorphism. Then, $sn(g)$ is defined as $N(a)=a^t a$, which is a scalar if $a$ comes from the Clifford group.

I am interested in explicitly computing $sn(g)$ for a given $g\in O(V,Q)$. I know a bit in some special cases:

- For the Euclidean space and the corresponding $O(n,\mathbb R)$ group the spinor norm is trivial $sn(g)=1$, since the group is generated by reflections by vectors of unit norm
- For an algebraically-closed field $K$, the spinor norm is always trivial since $K^*/(K^*)^2$ is trivial
- For a metabolic space $V = W \oplus W^*$ with the form $Q(w,f) = f(w)$, any $g \in \operatorname{GL}(W)$ gives rise to an orthogonal transformation on $V$ by the formula $g \cdot (w,f) = \left(gw, \left(g^{-1}\right)^*f\right)$. The spinor norm of this transformation is equal to $\det g$
*(this is half-anecdotal: I've heard it in a Russian video lecture on Clifford algebras, presented without complete proof).* - In particular, for any quadratic space that has a metabolic subspace as a direct (orthogonal) summand, the spinor norm is surjective.
- Clearly, the spinor norm of $\Omega(V,Q)$ (the commutator subgroup of $O(V,Q)$) is trivial, since $K^*/(K^*)^2$ is abelian. This article states that $\Omega$ is precisely the kernel of the spinor norm, providing an injective morphism $O/\Omega \rightarrow K^*/(K^*)^2$, though I don't see how it helps in actually computing the spinor norm of a given orthogonal transformation.
- I've done some calculations with the real hyperbolic plane with orthogonal basis $\{e_1, e_2\}$ such that $Q(e_1)=1$ and $Q(e_2)=-1$ by explicitly computing the elements of the Clifford group that represent certain orthogonal transformations. It seems that the spinor norm of a matrix $A$ (which is $\pm 1$ in the real case) in this basis coincides with the sign of $A_{2,2}$.
- Having in mind the connected components of an indefinite real orthogonal group $O(p,q)$ and using that the spinor norm is a continuous map to a discrete space $\{\pm 1\}$, it has to be constant on connected components, thus it is enough to compute it for a single representative from each component. This gives a generalization of the previous result, namely the spinor norm is $+1$ iff the transformation preserves orientation of the negative-definite subspace, and the spinor norm equals the determinant of the lower-right $q\times q$ submatrix (in a basis where positive-definite vectors come before negative-definite ones).
*This is basically my own findings, and I would appreciate a reference that supports/disproves this claim.*

In general, it feels that there should be some explicit (maybe polynomial?) formula $O(V,Q) \rightarrow K^*$ implementing the spinor norm, but I failed to find any references on this. In any way, I am happy with any explicit way of computing the spinor norm of an orthogonal matrix for a general quadratic form, or otherwise an explanation of why this isn't that straightforward or even possible.

isan explicit formula, for some values of explicit, so I'll take explicit to mean polynomial in the entries, as you suggest—in which case I imagine one can prove rigorously that the answer is 'no'. If some special formulæ are instead of interest, Jessica Fintzen, Tasho Kaletha, and I recently found ourselves having to do some such computations, and found that, at least for semisimple elements, there's a reasonably easy, mostly explicit (in terms of eigenvalues) answer. (I hope self reference is OK if not, don't read next comment.) They're … $\endgroup$4more comments