The Spinor Norm and the Structure of the Orthogonal Group with Inner Conjugation

Introduction

A rotation and its negative look alike as linear maps but not as versors, and the orthogonal group carries an invariant that records the difference. It is the spinor norm, a homomorphism from the orthogonal group to the square classes of the field, defined on a reflection in a vector by the value of the form on that vector. Over the real numbers with a positive definite form every value of the form is a positive number, hence a square, so the invariant is invisible; over the rational numbers, and over any field in which a positive quantity need not be a square, it cuts the orthogonal group into pieces and is the invariant that governs which rotations are products of reflections of a prescribed length.

The article defines the spinor norm, proves that it is a well-defined homomorphism, and describes the resulting chain of subgroups of the orthogonal group: the spinorial kernel, the products of two reflections with square product, and the commutator subgroup. The identification of the last two is the structure theorem of Dieudonné, quoted here; the proof of the part that follows from the Clifford algebra is given. The group so described is exactly the image of the spin group, and that is the sense in which the spinor norm measures the failure of the double cover to be a cover of the whole orthogonal group.

The Clifford algebra, the generating relation $v^2=q(v)$ and the universal property are from Clifford Algebras; the reflection $\rho_u(v)=-uvu^{-1}$, the Clifford group, the signed inner conjugation action, the Clifford norm, the generation of the Clifford group by non-isotropic vectors, Cartan–Dieudonné and the kernel of the signed inner conjugation are from The Clifford, Pin and Spin Groups with Signed Inner Conjugation; the versors, their lengths and their parity are from Versors, Rotors and the Sandwich Action with Signed Inner Conjugation; the bivector description of the rotations and the quaternion case are from The Clifford Algebra as a Lie Algebra and Quaternion Rotations and Reflections; the arithmetic for which the spinor norm is used is from The Clifford Invariant and the Spinor Genus with Inner Conjugation. Nothing owned by those entries is re-derived. The base is a field $F$ of characteristic not $2$, with $q$ non-degenerate on a finite-dimensional space $V$ of dimension $n$.

Reflections and Versors

Reflections

Theorem. For a non-isotropic vector $u\in V$ the map

$$ \rho_u:V\longrightarrow V, \qquad \rho_u(v)=v-2B(v,u)q(u)^{-1}u=-uvu^{-1}, $$

is an isometry of determinant $-1$ fixing $u$ and reversing the orthogonal complement of $u$.

Proof. This is the reflection of The Clifford, Pin and Spin Groups with Signed Inner Conjugation, where its properties are established.

Versors of a Given Length

Theorem. Every element $\sigma\in O(V,q)$ is of the form $\sigma=\mathrm{Ad}^{\alpha}_x$ for some versor $x$, and if $x'$ is another versor with $\mathrm{Ad}^{\alpha}_{x'}=\sigma$ then $x'=\lambda x$ for some $\lambda\in F^\times$.

Proof. By Cartan–Dieudonné every isometry is a product of reflections, $\sigma=\rho_{v_1}\cdots\rho_{v_k}=\mathrm{Ad}^{\alpha}_{v_1\cdots v_k}$, and $x=v_1\cdots v_k$ is a versor; each $v_i$ is non-isotropic because a reflection is defined only in such a vector. If $\mathrm{Ad}^{\alpha}_{x'}=\mathrm{Ad}^{\alpha}_x$ then $\mathrm{Ad}^{\alpha}_{x^{-1}x'}$ is the identity on $V$, so $x^{-1}x'$ acts as a scalar on $V$ and hence, being central on the algebra generated by $V$, is a scalar $\lambda$; the kernel of the signed inner conjugation on the Clifford group is the group of nonzero scalars.

Definition. The length of a versor $x=v_1\cdots v_k$ is the number $k$, well defined modulo two, and its parity is the parity of $k$.

Proposition. For a versor $x$ of length $k$ one has $\det\mathrm{Ad}^{\alpha}_x=(-1)^k$ and $N(x)=(-1)^kq(v_1)\cdots q(v_k)$.

Proof. Each reflection has determinant $-1$, and $\mathrm{Ad}^{\alpha}_x$ is the product of the $k$ reflections. The norm is multiplicative on the Clifford group with $N(u)=-q(u)$ for a vector, by the conventions of The Clifford, Pin and Spin Groups with Signed Inner Conjugation, and it is a scalar on the versor.

So the parity of the length is determined by the determinant of the action, and the product of the values of the form on the factors is determined by the norm; the two differ by the sign $(-1)^k$, which is why the definition below uses the norm and the parity together.

The Spinor Norm

Definition

Definition. Let $\sigma\in O(V,q)$ and let $x$ be a versor with $\sigma=\mathrm{Ad}^{\alpha}_x$ and length $k$. The spinor norm of $\sigma$ is

$$ \theta(\sigma)=(-1)^kN(x)\,(F^\times)^2\in F^\times/F^{\times2}. $$

Theorem. The spinor norm is well defined: the class $(-1)^kN(x)$ depends only on $\sigma$, and not on the versor or on the length by which it is expressed.

Proof. Let $x$ and $x'$ be two versors representing $\sigma$. By the theorem above $x'=\lambda x$ for some $\lambda\in F^\times$; the norm is multiplicative and a scalar commutes with everything, so $N(x')=N(\lambda x)=\lambda^2N(x)$, and the two values differ by the square $\lambda^2$. The length of an expression is defined only modulo two, and the parity of the length is determined by the determinant of the isometry it defines: $\det\sigma=(-1)^k=(-1)^{k'}$ by the proposition above, so $(-1)^k=(-1)^{k'}$. Hence $(-1)^kN(x)$ and $(-1)^{k'}N(x')$ differ by the square $\lambda^2$, and their classes in $F^\times/F^{\times2}$ agree.

Theorem. $\theta$ is a homomorphism: for $\sigma,\tau\in O(V,q)$,

$$ \theta(\sigma\tau)=\theta(\sigma)\,\theta(\tau). $$

Proof. Let $\sigma=\mathrm{Ad}^{\alpha}_x$ and $\tau=\mathrm{Ad}^{\alpha}_y$ with lengths $j$ and $l$. Then $\sigma\tau=\mathrm{Ad}^{\alpha}_{xy}$ and $xy$ is a versor of length $j+l$, so

$$ \theta(\sigma\tau)=(-1)^{j+l}N(xy)=(-1)^j(-1)^lN(x)N(y)=\theta(\sigma)\theta(\tau), $$

using the multiplicativity of the norm and the fact that the product of the classes is the class of the product in $F^\times/F^{\times2}$.

Theorem. On a reflection the spinor norm is the value of the form:

$$ \theta(\rho_u)=q(u)\,(F^\times)^2 . $$

Proof. Take $x=u$, of length $1$ and norm $N(u)=-q(u)$; then $\theta(\rho_u)=(-1)^1(-q(u))(F^\times)^2=q(u)(F^\times)^2$.

The Spinorial Kernel and the Reduced Group

The Kernel

Definition. The spinorial kernel is $O'(V,q)=\ker\theta$; the spinorial kernel of the rotations is $SO'(V,q)=SO(V,q)\cap\ker\theta$.

Theorem. The image of the spin group under the signed inner conjugation action lies in $O'(V,q)$; that is, $\theta(\mathrm{Ad}^{\alpha}_R)=1$ for every rotor $R$.

Proof. A rotor is an even versor of norm $N(R)=1$, so its length is even and $\theta(\mathrm{Ad}^{\alpha}_R)=(-1)^{2m}N(R)=1$.

Corollary. $\mathrm{Ad}^{\alpha}(\mathrm{Spin}(V,q))\subseteq SO'(V,q)$, and the double cover of the special orthogonal group of Versors, Rotors and the Sandwich Action with Signed Inner Conjugation is a cover of $SO'(V,q)$ by the rotors.

Proof. The image lies in the kernel by the theorem and in the special orthogonal group because a rotor has even length. The kernel of $\mathrm{Ad}^{\alpha}$ on the rotors is $\{\pm1\}$, by the cited entry.

Products of Two Reflections

Theorem. Let $u,v\in V$ be non-isotropic and suppose $q(u)q(v)$ is a square, $q(u)q(v)=c^2$ with $c\in F^\times$. Then $\rho_u\rho_v$ lies in the image of the spin group; explicitly

$$ R=\frac{uv}{c}\ \text{ is a rotor},\qquad \mathrm{Ad}^{\alpha}_R=\rho_u\rho_v . $$

Proof. The element $R$ is even and a versor, and $N(R)=N(uv)c^{-2}=N(u)N(v)c^{-2}$; now $N(u)N(v)=q(u)q(v)=c^2$, using $N(u)=-q(u)$ and the two signs, so $N(R)=1$ and $R$ is a rotor. The action is $\mathrm{Ad}^{\alpha}_{uv}=\mathrm{Ad}^{\alpha}_u\mathrm{Ad}^{\alpha}_v=\rho_u\rho_v$, and multiplying by the central scalar $c^{-1}$ does not change the signed inner conjugation.

The Chain of Subgroups

Theorem. The following inclusions hold:

$$ [O(V,q),O(V,q)]\ \subseteq\ SO'(V,q)\ \subseteq\ SO(V,q)\ \subseteq\ O(V,q), $$

and the quotient $SO(V,q)/SO'(V,q)$ embeds in $F^\times/F^{\times2}$, so that it is abelian of exponent two.

Proof. The spinor norm takes values in an abelian group, so its kernel contains every commutator; a commutator of isometries has determinant $1$, so the commutator subgroup lies in $SO$. The second inclusion is the definition. The spinor norm restricted to $SO$ has image $SO/SO'$ in $F^\times/F^{\times2}$ by the first isomorphism theorem, and every square class has square $1$, so the quotient has exponent two.

Theorem (Dieudonné). Let $q$ be non-degenerate and $n\ge3$. Then the inclusions above are equalities at the first place,

$$ [O(V,q),O(V,q)]=SO'(V,q)=SO(V,q)\cap\ker\theta, $$

and the group $SO'(V,q)$ is generated by the products $\rho_u\rho_v$ of two reflections with $q(u)q(v)$ a square. Equivalently, the image of the spin group in $SO(V,q)$ is the commutator subgroup of $O(V,q)$.

Proof. The proof of the reverse inclusions is the structure theorem of Dieudonné for the orthogonal group, quoted here from the literature cited below; the direction proved above is the inclusion $[O,O]\subseteq SO'$, and the generation by products of two reflections with square product is the statement that the subgroup they generate, which lies in the image of the spin group by the theorem of the previous subsection, is all of $SO'$.

Remark (small dimensions). For $n\le2$ the theorem fails as stated: in dimension two the orthogonal group is abelian, its commutator subgroup is trivial, and $SO'(V,q)$ need not be trivial, as the example of the hyperbolic plane over $\mathbb{Q}$ shows, where $\rho_u\rho_v$ has spinor norm $q(u)q(v)$ which can be a nonsquare. The statement for $n\ge3$ is the one used in the arithmetic of The Clifford Invariant and the Spinor Genus with Inner Conjugation.

Worked Cases

Definite Real Forms

Let $F=\mathbb{R}$ and let $q$ be positive definite. For every nonzero $u$ the value $q(u)$ is positive, hence a square in $\mathbb{R}$, so $\theta$ is the trivial homomorphism and $$ O'(V,q)=O(V,q),\qquad SO'(V,q)=SO(V,q),\qquad \Omega(V,q)=SO(V,q). $$

The spinor norm carries no information over the reals with a positive definite form, and the spin group covers the whole of $SO(V,q)$: the double cover $\mathrm{Spin}(n)\to SO(n)$ is onto. This is the reason the spinor norm is met in the arithmetic literature and not in the theory of the classical compact groups.

Remark (negative definite forms). If instead $-q$ is positive definite then $q(u)<0$ for every nonzero $u$ and the class of $q(u)$ is the class of $-1$, which is not a square in $\mathbb{R}$; a reflection then has spinor norm $-1$ and $O'(V,q)$ has index two in $O(V,q)$. On the rotations nothing changes, because a rotation is a product of an even number of reflections and the product of an even number of negative values is positive: $\theta$ is still trivial on $SO(V,q)$, and the double cover still reaches all of it. It is only the reflections whose spinor norm depends on the sign convention of the form, and the definite case of the table above is the positive definite one.

The Sum of Four Squares over $\mathbb{Q}$

Let $F=\mathbb{Q}$ and $q=\langle1,1,1,1\rangle$, so that $q(y)=y_1^2+y_2^2+y_3^2+y_4^2$.

Theorem. The image of the spinor norm $\theta:O(V,q)\to\mathbb{Q}^\times/\mathbb{Q}^{\times2}$ is the set of positive square classes, which has index two in $\mathbb{Q}^\times/\mathbb{Q}^{\times2}$ and is infinite; hence $O'(V,q)$ is a proper subgroup of $O(V,q)$, of infinite index, and $SO'(V,q)$ is of infinite index in $SO(V,q)$.

Proof. The form is positive definite and its coefficients are all $1$, so $q(y)>0$ for every nonzero $y$, and $\theta(\rho_y)=q(y)$ modulo squares is therefore a positive square class; the sign class is not attained, so the image lies in the subgroup of positive classes, of index two. Conversely, by the theorem of Lagrange every positive rational is a sum of four rational squares, so every positive square class is a value of the form and the image is the whole of that subgroup. The positive square classes are an infinite elementary abelian $2$-group, generated by the classes of the primes, so both the image and the quotient $O/O'$ it is isomorphic to are infinite. A reflection in a vector with $q(y)=2$ has spinor norm the class of $2$, which is not a square, so $O'$ is proper.

Corollary. For this form the quotient $SO(V,q)/SO'(V,q)$ is infinite, and so is the quotient by the commutator subgroup, by Dieudonné's theorem.

Proof. The quotient embeds in the image of the spinor norm on the rotations, which is a quotient of the infinite group of positive square classes.

Indefinite Forms

Let $F=\mathbb{R}$ and $q=\langle1,-1,1\rangle$, of signature $(2,1)$, so that $\mathbb{R}^\times/\mathbb{R}^{\times2}=\{\pm1\}$ with the sign as the only square class.

Theorem. $\theta(\rho_u)$ is the sign of $q(u)$; hence $\theta$ is surjective, $O'(V,q)$ has index two in $O(V,q)$, and the reflections in spacelike and in timelike vectors fall on different sides.

Proof. The square classes of $\mathbb{R}^\times$ are represented by $\pm1$, and $q(u)$ is a nonzero real, so its class is its sign.

Remark. The Lorentzian case is the one in which the two-sidedness of the spinor norm is physically visible, since a reflection in a spacelike vector and a reflection in a timelike vector have different spinor norms; the reader is referred to Biquaternion Rotations and Lorentz Transformations for the realisation of these groups inside the biquaternion algebra, where the image of the spin group is the identity component of the Lorentz group.

Summary

The spinor norm is the homomorphism

$$ \theta:O(V,q)\longrightarrow F^\times/F^{\times2},\qquad \theta(\rho_u)=q(u)\,(F^\times)^2, $$

defined on an isometry $\sigma=\mathrm{Ad}^{\alpha}_x$ by $\theta(\sigma)=(-1)^kN(x)(F^\times)^2$ for a versor $x$ of length $k$. It is well defined because a second versor representing $\sigma$ differs from the first by a scalar, whose norm is a square, and because the parity of the length is the determinant of $\sigma$; it is a homomorphism because the norm is multiplicative and the lengths add. On a reflection it returns the value of the form on the vector, so it records the length of the reflection in the square-class sense.

Its kernel $O'(V,q)$ contains the image of the spin group, since a rotor has even length and norm one, and the resulting chain is

$$ [O,O]\ \subseteq\ SO'(V,q)=SO\cap\ker\theta\ \subseteq\ SO\ \subseteq\ O, $$

the first inclusion holding because the values lie in an abelian group, and the identification of the commutator subgroup with the spinorial kernel of the rotations being the structure theorem of Dieudonné for $n\ge3$, quoted here. The subgroup $SO'(V,q)$ is generated by the products $\rho_u\rho_v$ of two reflections with $q(u)q(v)$ a square, and the rotor $R=uv/c$ realises such a product when $q(u)q(v)=c^2$; the quotient $SO/SO'$ embeds in $F^\times/F^{\times2}$ and so is abelian of exponent two. The image of the spin group is therefore the reduced orthogonal group, and the double cover fails to reach the whole of $SO$ exactly by the spinor norm.

Over $\mathbb{R}$ with a positive definite form every value of the form is a square and the invariant is trivial, so the spin group covers all of $SO(n)$; with a negative definite form the reflections take the value $-1$ while the rotations still take the value $1$, so the rotations are unaffected and only the reflections carry the sign. Over $\mathbb{Q}$ with the sum of four squares the values of the form are positive and every positive rational is a sum of four rational squares, so the image of the spinor norm is the group of positive square classes, of index two in the square-class group and itself infinite; the spinorial kernel and the reduced group are therefore proper, and the reduced group is of infinite index in $SO$. Over $\mathbb{R}$ with the form of signature $(2,1)$ the only square classes are the two signs, the invariant is the sign of the form on the vector, and its kernel has index two. The arithmetic for which the invariant is used, the classification of lattices by their genera and their spinor genera, is the subject of The Clifford Invariant and the Spinor Genus with Inner Conjugation.

Summary of Notation

Symbol	Meaning
$\rho_u$	Reflection in the non-isotropic vector $u$
$x=v_1\cdots v_k$	Versor, $k$ its length
$N(x)$	Clifford norm, $N(u)=-q(u)$ on a vector
$\mathrm{Ad}^{\alpha}_x=\alpha(x)(\cdot)x^{-1}$	Signed inner conjugation action
$\theta(\sigma)=(-1)^kN(x)$	Spinor norm, in $F^\times/F^{\times2}$
$\theta(\rho_u)=q(u)$	Value on a reflection, modulo squares
$O'(V,q)=\ker\theta$	Spinorial kernel
$SO'(V,q)=SO\cap\ker\theta$	Spinorial kernel of the rotations
$[O,O]$	Commutator subgroup, $=SO'$ for $n\ge3$
$R=uv/c$, $q(u)q(v)=c^2$	Rotor realising a product of two reflections
$SO/SO'\hookrightarrow F^\times/F^{\times2}$	The quotient, abelian of exponent two