Spin Factors and the Clifford Envelope with Inner Conjugation

Introduction

This article treats the spin factor, the Jordan algebra of degree two built from a module with a quadratic form, and the Clifford envelope that makes it special. The base structure is a commutative ring $R$ with identity $1 \neq 0$, and the Jordan conventions are those of Jordan Algebras: a commutative product $\bullet$, the square $x^2 = x\bullet x$, and the identity $[L_x, L_{x^2}] = 0$.

The spin factor is the simplest Jordan algebra after the base ring itself. It carries the same data as a quadratic form, and its linear algebra is the linear algebra of that form. What makes it interesting is the way it is forced to be special: there is a canonical embedding of the spin factor into the symmetrisation of the Clifford algebra of the form, and the Clifford relations are exactly the equations that make the embedding a Jordan homomorphism. The Clifford algebra is constructed here explicitly as a quotient of the tensor algebra, and only the properties needed for the envelope are used.

The article defines spin factors and derives their degree-two structure, constructs the Clifford algebra from the tensor algebra, proves the Clifford-envelope embedding, and identifies the degree-two algebras of the classification with the spin factors, including the Hermitian $2 \times 2$ matrix algebras over $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$.

Spin Factors

Definition

Let $V$ be an $R$-module and let $q : V \to R$ be a quadratic form, that is, a map with $q(\lambda v) = \lambda^2 q(v)$ for $\lambda \in R$ and with the associated map

$$ B : V \times V \to R, \qquad B(v, w) = q(v + w) - q(v) - q(w), $$

bilinear; when $2$ is invertible one divides by $2$ and writes $B(v,w) = \tfrac12\bigl(q(v+w) - q(v) - q(w)\bigr)$, so that $B(v,v) = q(v)$. We use the normalised form and assume $2$ invertible from here on; $B$ is the polar form of $q$. The spin factor of $(V,q)$ is the $R$-module

$$ JSpin(V) = R \cdot 1 \oplus V $$

with the product

$$ (\alpha, v) \bullet (\beta, w) = \bigl(\alpha\beta + B(v, w),\ \alpha w + \beta v\bigr). $$

The element $1 = (1,0)$ is the unit, and $V = (0, V)$ is the vector part. When $V = R^n$ with the standard form write $JSpin_n$; then $\dim_R JSpin_n = n + 1$.

Remark. The product is the symmetrisation of the product in the Clifford algebra, and it is the unique commutative product on $R\oplus V$ with unit $1$, restricting to the given quadratic form in the sense that the square of $(0,v)$ is $(q(v), 0)$ and the products of the unit with $V$ are the scalar action. This is shown in the envelope theorem below.

The Degree-Two Property

Proposition. Every element $x = (\alpha, v)$ of $JSpin(V)$ satisfies the quadratic relation

$$ x^2 - 2\alpha\, x + \bigl(\alpha^2 - q(v)\bigr) 1 = 0 , $$

and consequently $JSpin(V)$ has degree two: the subalgebra generated by $x$ is at most two-dimensional, spanned by $1$ and $x$.

Proof. Compute $x^2 = (\alpha, v)\bullet(\alpha, v) = (\alpha^2 + B(v,v), 2\alpha v) = (\alpha^2 + q(v), 2\alpha v)$. Then

$$ x^2 - 2\alpha x + (\alpha^2 - q(v))1 = (\alpha^2 + q(v) - 2\alpha^2 + \alpha^2 - q(v),\ 2\alpha v - 2\alpha v) = (0, 0). $$

The quadratic form appearing in the relation is the generic norm $N(x) = \alpha^2 - q(v)$ of $JSpin(V)$, and the linear form $T(x) = 2\alpha$ is the trace; this is the degree-two case of the trace and norm of the previous article, where the characteristic polynomial is $x^2 - T(x)x + N(x)1 = 0$ and no separate quadratic coefficient $S$ occurs. The linear form $T$ is not the bilinear trace form $T(x,y) = \operatorname{tr}(L_{x\bullet y})$ of Jordan Algebras: writing $L_z$ in the basis $1, e_1, \ldots, e_n$ of $JSpin_n$ gives $\operatorname{tr}(L_z) = (n+1)\gamma$ for $z = (\gamma, w)$, so that

$$ T(x, y) = (n+1)\bigl(\alpha\beta + B(v, w)\bigr), \qquad x = (\alpha, v),\ y = (\beta, w), $$

which is $\frac{n+1}{2}$ times the generic trace of the product, $T(x\bullet y) = 2(\alpha\beta + B(v,w))$. In these terms $x$ is invertible in $JSpin(V)$ exactly when $N(x)$ is a unit of $R$, with $x^{-1} = N(x)^{-1}(T(x)1 - x) = N(x)^{-1}(\alpha, -v)$; over a field this is the condition $N(x) \neq 0$.

Idempotents and the Peirce Decomposition

Assume that $V$ contains a vector $e$ with $q(e) = 1$; over $\mathbb{R}$ with $q(e) > 0$ this is achieved by rescaling $e$ by $q(e)^{-1/2}$, and the normalisation is kept from here on. Then

$$ u_+ = \tfrac12(1 + e), \qquad u_- = \tfrac12(1 - e) $$

are idempotents with $u_+ + u_- = 1$ and $u_+ \bullet u_- = 0$. They give the Peirce decomposition of Jordan Algebras with respect to $u_+$:

$$ JSpin(V) = J_1(u_+) \oplus J_{1/2}(u_+) \oplus J_0(u_+), $$

$$ J_1(u_+) = R\,u_+, \qquad J_0(u_+) = R\,u_-, \qquad J_{1/2}(u_+) = e^{\perp} = \{v \in V : B(e, v) = 0\}. $$

The middle space has dimension $\dim V - 1$, which is the degree-two shape: two lines plus the orthogonal complement of a non-isotropic vector. For $V = R^2$ the Peirce decomposition has $\dim J_{1/2} = 1$, and the spin factor is three-dimensional.

Example. The smallest spin factor is $JSpin_2 = R \oplus R^2$, of dimension $3$, with the quadratic form $q(v) = v_1^2 + v_2^2$. The idempotents $u_\pm = \tfrac12(1 \pm e_1)$ and the element $e_2$ satisfy $e_2 \bullet e_2 = 1$, $u_\pm \bullet e_2 = \tfrac12 e_2$, and $u_+ \bullet u_- = 0$; this is the three-dimensional algebra identified in Jordan Algebras as $H_2(\mathbb{R})$.

The Clifford Algebra

Construction from the Tensor Algebra

Let $T(V) = \bigoplus_{n\geq0} V^{\otimes n}$ be the tensor algebra of $V$, as in Tensor Powers and the Free Algebra and Modules, §13. The Clifford algebra of the quadratic form $q$ is the quotient

$$ \mathrm{Cl}(V, q) = T(V) \big/ I_q , $$

where $I_q$ is the two-sided ideal generated by the elements

$$ v \otimes v - q(v)\, 1, \qquad v \in V . $$

We write $vw$ for the product in $\mathrm{Cl}(V,q)$, so that the defining relations read

$$ v^2 = q(v), \qquad vw + wv = 2B(v, w), $$

the second obtained by polarising the first. The algebra inherits from the tensor algebra a filtration , from the parity of the number of tensor factors, a $\mathbb{Z}/2\mathbb{Z}$-grading; it is associative and unital. The construction and its universal property are treated in Clifford Algebras of this part; what is used here is only that $\mathrm{Cl}(V,q)$ is an associative algebra with these relations.

The Universal Property

Theorem. Let $A$ be an associative unital $R$-algebra and let $f : V \to A$ be $R$-linear with $f(v)^2 = q(v)1$ for all $v \in V$. Then $f$ extends uniquely to an algebra homomorphism $F : \mathrm{Cl}(V,q) \to A$.

Proof. The linear map $f$ extends uniquely to an algebra homomorphism $T(V) \to A$ by the universal property of the tensor algebra (Tensor Powers and the Free Algebra). The element $v \otimes v - q(v)1$ maps to $f(v)^2 - q(v)1 = 0$, so the homomorphism kills the generators of $I_q$ and descends to $\mathrm{Cl}(V,q)$. Uniqueness holds because $V$ generates $\mathrm{Cl}(V,q)$.

Corollary. If $V$ is free with basis $e_1, \ldots, e_n$, then $\mathrm{Cl}(V,q)$ is free of rank $2^n$ with basis the products $e_{i_1}e_{i_2}\cdots e_{i_k}$ for $1 \leq i_1 < i_2 < \cdots < i_k \leq n$, together with $1$ for $k = 0$. In particular

$$ \dim_R \mathrm{Cl}(V,q) = 2^{\dim_R V}. $$

Pro. The relations rewrite any word in the $e_i$ as an alternating word, so the displayed products span; independence follows from a normal-form argument, or from the standard structure theorem.

Remark. The sign convention is the one fixed by $v^2 = q(v)$: the Clifford relation is $vw + wv = 2B(v,w)$. With the other common convention $vw + wv = -2B(v,w)$ one replaces $q$ by $-q$. The shared notation of the corpus reserves $q$ for the quadratic form and $g$ for a bilinear form, and $\mathrm{Cl}(V,q)$ is written accordingly.

The Even Part

The grading of $T(V)$ by the number of tensor factors descends to a $\mathbb{Z}/2\mathbb{Z}$-grading

$$ \mathrm{Cl}(V,q) = \mathrm{Cl}^0(V,q) \oplus \mathrm{Cl}^1(V,q), $$

with $\mathrm{Cl}^0$ the even part, spanned by products of an even number of vectors, and $\mathrm{Cl}^1$ the odd part, containing $V$. The even part is a subalgebra; the odd part is a bimodule. The main involution is the algebra automorphism $\alpha$ acting as $+1$ on $\mathrm{Cl}^0$ and $-1$ on $\mathrm{Cl}^1$. The reversal (or transpose) is the anti-automorphism reversing the order of a product, the unique anti-automorphism fixing $V$ pointwise.

The Clifford Envelope

The Embedding

The inclusion $V \hookrightarrow \mathrm{Cl}(V,q)$ is $R$-linear and satisfies $v^2 = q(v)1$, so $v \mapsto v$ is exactly a map of the type considered by the universal property. The Clifford envelope of the spin factor is the Clifford algebra $\mathrm{Cl}(V,q)$ itself, the associative algebra generated by $1$ and $V$; inside it, the linear span of $1$ and $V$ is a copy of the spin factor.

Theorem. The map

$$ \phi : JSpin(V) \longrightarrow \mathrm{Cl}(V,q)^+, \qquad \phi(\alpha, v) = \alpha\, 1 + v , $$

is an injective Jordan homomorphism. Consequently $JSpin(V)$ is a Jordan algebra, and it is special.

Proof. The map is $R$-linear, and injective because $1$ and the elements of $V$ are independent in $\mathrm{Cl}(V,q)$: the corollary above exhibits a basis of $\mathrm{Cl}(V,q)$ containing $1$ and the basis vectors of $V$, and in general the filtration of $\mathrm{Cl}(V,q)$ has associated graded the exterior algebra, so $R \oplus V$ embeds. For the Jordan property compute, for $x = (\alpha, v)$ and $y = (\beta, w)$,

$$ \tfrac12\bigl(\phi(x)\phi(y) + \phi(y)\phi(x)\bigr) = \tfrac12\bigl((\alpha + v)(\beta + w) + (\beta + w)(\alpha + v)\bigr) $$

$$ = \alpha\beta + \tfrac12(\alpha w + \beta v + v w + \beta v + \alpha w + w v) = \alpha\beta + \tfrac12(vw + wv) + \alpha w + \beta v $$

$$ = \bigl(\alpha\beta + B(v,w)\bigr) 1 + \alpha w + \beta v = \phi\bigl((\alpha, v) \bullet (\beta, w)\bigr), $$

where the Clifford relation $vw + wv = 2B(v,w)$ was used in the third line. Hence $\phi$ is a Jordan homomorphism. The ambient algebra $\mathrm{Cl}(V,q)^+$ is a Jordan algebra by the theorem of Jordan Algebras on the symmetrisation of an associative algebra, and the image of an injective Jordan homomorphism is a Jordan subalgebra; therefore $JSpin(V)$ is a Jordan algebra, isomorphic to that image. It is special by construction.

Remark. The theorem reduces the Jordan identity of $JSpin(V)$ to associativity in $\mathrm{Cl}(V,q)$: once the product of two vectors is symmetrised by $vw + wv = 2B(v,w)$, the identity holds automatically in any associative algebra. The compensating identity of the Jordan algebra is bought at the price of the anticommutation relation in the envelope.

The Envelope is Not the Spin Factor

The Clifford algebra is much larger than the spin factor: $\dim \mathrm{Cl}(V,q) = 2^n$ against $\dim JSpin(V) = n + 1$ for $V$ free of rank $n$. The even part $\mathrm{Cl}^0(V,q)$ contains $1$ and the products $vw$ of two vectors; the span of $1$ and $V$ generates the full Clifford algebra by multiplying, since products of odd numbers of vectors sweep out $\mathrm{Cl}^1$, but the linear span of $1$ together with $V$ is only the spin factor. The envelope is therefore a genuine enlargement, and the spin factor is its degree-two part.

Example. For $V = R^2$ with $q(v) = v_1^2 + v_2^2$ and orthogonal basis $e_1, e_2$, the Clifford algebra has basis $1, e_1, e_2, e_1e_2$ with $e_1^2 = e_2^2 = 1$, $e_1e_2 = -e_2e_1$, and $(e_1e_2)^2 = -1$; so $\mathrm{Cl}(V,q) \cong M_2(R)$, the algebra of $2 \times 2$ matrices, and $JSpin_2$ is the span of $1, e_1, e_2$ inside it, which is the space of symmetric $2\times2$ matrices under the identification $e_1 \mapsto \begin{pmatrix}1&0\\0&-1\end{pmatrix}$, $e_2 \mapsto \begin{pmatrix}0&1\\1&0\end{pmatrix}$. The general identification of Clifford algebras with matrix algebras over $\mathbb{R}$ and $\mathbb{C}$ is the content.

Degree-Two Algebras and Hermitian Matrices

The Classification of Degree-Two Algebras

Over a field of characteristic not $2$, a simple Jordan algebra of degree two with a nondegenerate trace form is exactly a spin factor $JSpin(V,q)$ with $q$ nondegenerate; the quadratic form is determined by the algebra up to isometry, as the next proposition shows. This is the degree-two part of the classification quoted in Jordan Algebras.

The Hermitian $2 \times 2$ Algebras

The Hermitian matrix algebras of Special and Exceptional Jordan Algebras of size two are spin factors.

Theorem. For $D \in \{\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}\}$ there is an isomorphism of real Jordan algebras

$$ H_2(D) \cong JSpin_{1 + \dim_{\mathbb{R}} D} . $$

Proof. In $H_2(D)$ the diagonal idempotents $e_{11}, e_{22}$ satisfy $e_{11} + e_{22} = 1$ and $e_{11}\bullet e_{22} = 0$, and the Peirce space $J_{1/2}(e_{11})$ is the space of off-diagonal entries, isomorphic to $D$ as a real module, of dimension $\dim_{\mathbb{R}}D$. A spin factor $JSpin_n$ has, by the computation above, middle Peirce space of dimension $n - 1$. Matching dimensions gives $n = 1 + \dim_{\mathbb{R}}D$; the product on the middle space is the symmetrised product of off-diagonal entries, which is the scalar product of $D$ up to the normalisation of the form, so the two algebras have the same product and are isomorphic.

Corollary. $H_2(\mathbb{R}) \cong JSpin_2$, $H_2(\mathbb{C}) \cong JSpin_3$, $H_2(\mathbb{H}) \cong JSpin_5$, $H_2(\mathbb{O}) \cong JSpin_9$. In particular all of these algebras are special and of degree two, and the exceptional phenomenon of Special and Exceptional Jordan Algebras is confined to size at least three.

Remark. The case $D=\mathbb{C}$ is the biquaternion case of the corpus. $H_2(\mathbb{C})\cong JSpin_3$ is the spin factor of a three-dimensional definite form, so its Clifford envelope is $\mathrm{Cl}_{3,0}\cong M_2(\mathbb{C})$, the biquaternion algebra $\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$, equivalently $\mathbb{B}\cong\mathrm{Cl}^{+}_{3,1}$. The identification $\mathbb{B}\cong\mathrm{Cl}_{3,0}$, with the competing labelling and its sign consequence, is stated, and the algebra itself is not covered here; the biquaternion vocabulary of this corpus is used here unchanged, and none of it is re-derived.

The Isometry Class of the Form

The spin factor determines its form exactly, up to isometry.

Proposition. Let $q, q'$ be two quadratic forms on the same free module $V$ of finite rank over a base ring $R$ that is a field of characteristic not $2$. Then $JSpin(V,q) \cong JSpin(V,q')$ as Jordan algebras if and only if $q$ and $q'$ are isometric, that is, there is $L \in GL_R(V)$ with $q'(Lv) = q(v)$ for all $v \in V$. The forward implication uses only that $2$ is invertible in $R$; the converse uses that $R$ is a field.

Proof. If $L$ is such an isometry then $B'(Lv, Lw) = B(v, w)$ for all $v, w$, and the map

$$ \Phi : JSpin(V,q) \longrightarrow JSpin(V,q'), \qquad \Phi(\alpha, v) = (\alpha, Lv) $$

is a Jordan isomorphism, since its scalar part is $\alpha\beta + B'(Lv,Lw) = \alpha\beta + B(v,w)$ and its vector part is $\alpha Lw + \beta Lv$. Conversely, let $\Phi$ be an isomorphism. The unit $1$ is the unique element $u$ with $u \bullet z = z$ for all $z$, so $\Phi(1) = 1$. The vector part is intrinsically characterised as the set of elements $z$ with $z^2 \in R\cdot 1$ that do not themselves lie in $R\cdot 1$: for $z = (\alpha, v)$ the square is $(\alpha^2 + q(v),\, 2\alpha v)$, which lies in $R\cdot1$ precisely when $\alpha v = 0$, and over a field this means $\alpha = 0$ or $v = 0$. Hence $\Phi$ maps $V$ to $V$, say $v \mapsto Lv$ with $L$ linear. Comparing the vector parts of $\Phi(x)\bullet'\Phi(y)$ and $\Phi(x\bullet y)$ gives $\alpha Lw + \beta Lv$ for both, and comparing the scalar parts gives $B'(Lv, Lw) = B(v,w)$; polarising, $L$ is an isometry of $q$ onto $q'$.

It follows that the spin factor determines the quadratic form up to isometry and nothing finer, and that $JSpin(V,q)$ remembers exactly the invariants of the form — its rank, its discriminant and its signature over $\mathbb{R}$. In particular the spin factor of a definite form is never isomorphic to the spin factor of an indefinite form of the same rank.

The Structure Group and Conformal Transformations

The structure group of a Jordan algebra $J$ with generic norm $N$ is

$$ \operatorname{Str}(J) = \{s \in GL_R(J) : N(sx) = \nu(s) N(x) \text{ for some } \nu(s) \in R^{\times}\}, $$

the linear maps preserving the norm up to a scalar; it contains the automorphism group and acts on $J$ preserving the Jordan structure up to a factor. For a spin factor it is the conformal orthogonal group of the norm.

Theorem. Let $q$ be a nondegenerate quadratic form on a free module $V$ of finite rank $n$ over a field $R$ of characteristic not $2$, and let $N(\alpha, v) = \alpha^2 - q(v)$ be the generic norm on $JSpin(V,q)$. Then the structure group of $JSpin(V,q)$ is the group of similitudes of $N$ on $R \oplus V$,

$$ \operatorname{Str}(JSpin(V)) \cong CO(R \oplus V, N) = \{s \in GL_R(R \oplus V) : N(sx) = \nu(s) N(x) \text{ for some } \nu(s) \in R^{\times}\}, $$

whose Lie algebra is the structure algebra, of dimension $\binom{n}{2} + n + 1 = 1 + \binom{n+1}{2}$. The subgroup of similitudes with multiplier $1$ is the orthogonal group $O(R \oplus V, N)$ of the norm, which contains $O(V,q)$ acting on the vector part and fixing the unit; the map $(\alpha, v) \mapsto (\lambda\alpha, \lambda v)$ has multiplier $\lambda^2$, so when every multiplier is a square in $R$ — for instance when $R$ is quadratically closed, as $\mathbb{C}$ is — every similitude is a scalar multiple of an isometry of $N$ and the group is generated by $O(R\oplus V, N)$ together with the dilations. Over $\mathbb{R}$ the multiplier need not be a square: for $JSpin_1$ the swap $(\alpha, v) \mapsto (v, \alpha)$ multiplies the norm $\alpha^2 - v^2$ by $-1$, so it is a similitude that is not a scalar multiple of an isometry.

Proof. An invertible $s$ satisfies $N(sx) = \nu(s)N(x)$ for all $x$ if and only if it satisfies $B_N(sx, sy) = \nu(s)B_N(x,y)$ for all $x, y$, because a quadratic form is recovered from its polar form when $2$ is invertible; this is the defining property of the conformal orthogonal group of $N$ on $R \oplus V$. Infinitesimally, $s$ lies in the structure algebra if and only if $B_N(sx, y) + B_N(x, sy) = \lambda(s)B_N(x,y)$ for some linear functional $\lambda$. Writing $s$ in blocks $(a, \varphi, b, c)$ with respect to the decomposition $R \oplus V$, this condition forces $a$ to be a scalar, $b$ to be the adjoint of $\varphi$ for $B$, and $c$ to be $a$ times the identity plus a $B$-skew endomorphism of $V$; the solutions are therefore parametrised by a scalar, a vector and a skew endomorphism, of dimension $1 + n + \binom{n}{2}$, which is exactly the dimension of $\operatorname{Der}(JSpin(V)) \oplus L(JSpin(V)) = \mathrm{SO}(V,q) \oplus (R \oplus V)$. The maps fixing the unit and preserving $N$ are the isometries of $q$ on $V$, and a similitude is a scalar multiple of an isometry of $N$ as soon as its multiplier is a square in $R$.

Remark. The norm is $\langle 1\rangle \perp (-q)$ on the $(n+1)$-dimensional module $R \oplus V$, so the structure group acts on a space one dimension larger than $V$; its Lie algebra has dimension $\binom{n+1}{2} + 1$, larger by $n$ than the dimension $\binom{n}{2} + 1$ of the conformal group $CO(V,q)$ of $q$ itself, whose action on $V$ does not extend to the algebra.

The same description applies to the split forms, with the orthogonal group replaced by the indefinite orthogonal group of the same signature; the structure group is then the conformal orthogonal group of the split norm, and the spin factor is the coordinate algebra of the corresponding quadric. The structure group acts transitively on the interior of the cone of squares, and this is the link between the spin factors of this article and the symmetric cones.

The Spin Group and the Vector Representation

The Spin Group

Let $q$ be nondegenerate on a free module $V$ of finite rank over a field of characteristic not $2$. The Clifford group is

$$ \Gamma(V,q) = \{s \in \mathrm{Cl}(V,q)^{\times} : s V s^{-1} \subseteq V\}, $$

and the spin group is the subgroup of elements of the even part of spinor norm one,

$$ \operatorname{Spin}(V,q) = \{s \in \Gamma(V,q)^0 : s \bar s = 1\}, $$

where $\bar s$ is the reversal and $\Gamma(V,q)^0 = \Gamma(V,q) \cap \mathrm{Cl}^0(V,q)$ the even part. The vector representation is

$$ \rho : \operatorname{Spin}(V,q) \longrightarrow SO(V,q), \qquad \rho(s)(v) = s v s^{-1}. $$

Theorem. Let $q$ be definite and of rank at least $2$. The vector representation is a surjective group homomorphism onto $SO(V,q)$ with kernel $\{\pm 1\}$, so $\operatorname{Spin}(V,q)$ is the double cover of $SO(V,q)$; it preserves the quadratic form, and its image is the identity component of $O(V,q)$. For an indefinite form the same statement holds with $SO(V,q)$ replaced by its identity component.

The construction and the classification of the groups $\operatorname{Spin}_{p,q}$ are.

Action on the Spin Factor

Every element $s \in \operatorname{Spin}(V,q)$ acts on $\mathrm{Cl}(V,q)$ by conjugation $x \mapsto s x s^{-1}$; since $s$ is even, this preserves the subspace $R \cdot 1 \oplus V$ and acts trivially on the unit. The resulting map

$$ \operatorname{Spin}(V,q) \longrightarrow \operatorname{Aut}(JSpin(V)), \qquad s \longmapsto \mathrm{id}_R \oplus \rho(s), $$

is a homomorphism of groups, and for $q$ definite of rank at least $2$ it maps $\operatorname{Spin}(V,q)$ onto the identity component of the automorphism group of the spin factor with kernel $\{\pm 1\}$, so that $\operatorname{Spin}(V,q)$ is a double cover of $SO(V,q) \cong \operatorname{Aut}(JSpin(V))^0$; the full automorphism group of $JSpin(V)$ is $O(V,q)$ acting on the vector part and fixing the unit, in accordance with the row $\operatorname{Der}(JSpin_n) = \mathrm{SO}(n)$ of the table in Special and Exceptional Jordan Algebras.

Summary

A spin factor is the Jordan algebra $JSpin(V) = R\cdot 1 \oplus V$ with product

$$ (\alpha, v)\bullet(\beta, w) = (\alpha\beta + B(v,w),\ \alpha w + \beta v), $$

where $q$ is a quadratic form on $V$ and $B$ its polar form, normalised by $B(v,v) = q(v)$. Every element satisfies $x^2 - 2\alpha x + (\alpha^2 - q(v))1 = 0$, so the degree is two, and the Peirce decomposition relative to a rank-one idempotent is two lines together with the orthogonal complement of the idempotent vector. The Clifford algebra $\mathrm{Cl}(V,q) = T(V)/\langle v\otimes v - q(v)1\rangle$ is associative, satisfies $v^2 = q(v)$ and $vw + wv = 2B(v,w)$, has universal property and rank $2^{\dim V}$, and splits into even and odd parts. The Clifford envelope is the embedding

$$ \phi : JSpin(V) \hookrightarrow \mathrm{Cl}(V,q)^+, \qquad \phi(\alpha, v) = \alpha 1 + v , $$

which is an injective Jordan homomorphism precisely because of the Clifford relation $vw + wv = 2B(v,w)$; it exhibits the spin factor as special. The Hermitian $2\times2$ matrix algebras are spin factors, $H_2(D) \cong JSpin_{1+\dim_{\mathbb{R}}D}$, and the spin group acts on the spin factor through the vector representation, a double cover of $SO(V,q)$.

Summary of Notation

Symbol	Meaning
$R$	Commutative ring with identity $1 \neq 0$
$q$	Quadratic form on $V$
$B(v,w)$	Polar form of $q$, normalised so that $B(v,v) = q(v)$
$JSpin(V) = R\cdot1 \oplus V$	Spin factor
$JSpin_n$	Spin factor of $V = R^n$, of dimension $n+1$
$u_\pm = \tfrac12(1\pm e)$	Idempotents of a spin factor, $q(e) = 1$
$T(V) = \bigoplus_n V^{\otimes n}$	Tensor algebra of $V$
$\mathrm{Cl}(V,q) = T(V)/I_q$	Clifford algebra, $v^2 = q(v)$, $vw + wv = 2B(v,w)$
$\mathrm{Cl}^0, \mathrm{Cl}^1$	Even and odd parts
$\phi(\alpha,v) = \alpha 1 + v$	Clifford envelope of the spin factor
$T(x) = 2\alpha$, $N(x) = \alpha^2 - q(v)$	Trace and generic norm of a spin factor
$\Gamma(V,q)$	Clifford group
$\operatorname{Str}(J)$	Structure group of $J$, norm-preserving up to scalar
$\mathrm{STR}(J)$	Structure algebra $\operatorname{Der}(J) \oplus L(J)$, its Lie algebra
$\operatorname{Spin}(V,q)$	Spin group, double cover of $SO(V,q)$
$\bar s$	Reversal (transpose) anti-automorphism of $\mathrm{Cl}(V,q)$, fixing $V$ pointwise
$\rho(s)(v) = s v s^{-1}$	Vector representation
$H_2(D)$	Hermitian $2\times2$ matrices, $\cong JSpin_{1+\dim_{\mathbb{R}}D}$