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




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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](https://encyclopediaofmath.org/wiki/Orthogonal_group) 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.



Explicit computation of spinor norm
Posted from the comments ([1](https://mathoverflow.net/questions/368195/explicit-computation-of-spinor-norm#comment929028_368195) [2](https://mathoverflow.net/questions/368195/explicit-computation-of-spinor-norm#comment929029_368195) [3](https://mathoverflow.net/questions/368195/explicit-computation-of-spinor-norm#comment929030_368195)), by [request](https://mathoverflow.net/questions/368195/explicit-computation-of-spinor-norm#comment929133_368195).


Of course, the spinor formula itself arguably is an 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. It's assembled from some facts described in §5.1 of [Fintzen, Kaletha, and Spice - On certain sign characters …](https://arxiv.org/abs/1912.03286v1) in the form that's of interest to us, but most of them come from Scharlau's book [Quadratic and Hermitian forms](https://link.springer.com/book/10.1007/978-3-642-69971-9).


Although it turned out not to be most useful for us, §2 of [Zassenhaus - On the spinor norm](https://doi.org/10.1007/BF01650092), and specifically (2.1), might be closer to what you want. It's a nice paper, but not the last word on the subject; you may like to look at [who cites it](https://mathscinet.ams.org/mathscinet/search/publications.html?refcit=148760).


