# Explicit computation of spinor norm

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.

• 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. (I hope self reference is OK if not, don't read next comment.) They're … – LSpice Aug 3 at 20:11
• … described in §5.1 of Fintzen, Kaletha, and Spice - On certain sign characters … in the form that's of interest to us, but most of them come from Scharlau's book Quadratic and Hermitian forms. – LSpice Aug 3 at 20:12
• @LSpice Thank you a lot for this decent set of references. I've checked Scharlau's book and it already contains quite a lot, even if not exactly what I've hoped for. I'll accept if you post it as an answer. – lisyarus Aug 4 at 7:25
• @LSpice Yes, was thinking of exactly the same. Thank you! – lisyarus Aug 4 at 10:33
• In terms of an algorithm (not a polynomial formula) to compute the spinor norm, you can decompose your orthogonal transformation as a product of reflections. Then each reflections has a simple lift to the Clifford group and you can take the corresponding product. – Aurel Aug 4 at 14:11