Degenerate Clifford Algebras and the Radical

Introduction

A quadratic form can fail to be non-degenerate, and when it does the Clifford algebra acquires a nilpotent factor that is invisible in the non-degenerate theory. The failure is concentrated in the radical of the form, the submodule of vectors orthogonal to everything; the radical is totally isotropic, its Clifford generators square to zero and anticommute, and the algebra they generate is the exterior algebra on the radical. The Clifford algebra of a degenerate form is therefore the graded tensor product of the Clifford algebra of the reduced non-degenerate form and an exterior algebra, and the radical generates a nilpotent ideal that is the ring-theoretic radical of the algebra.

This article proves the reduction, identifies the nilpotent factor, computes the rank, and gives the criterion for semisimplicity. The base is a commutative ring $R$ in which $2$ is invertible, the module is free of finite rank, and the form may be degenerate. The construction of the Clifford algebra, the fundamental relation and the graded tensor product are from Clifford Algebras; the radical and non-degeneracy of a form are from Bilinear Forms; the exterior algebra is from The Exterior Algebra. The non-degenerate case is developed, whose basis, filtration and centre are cited rather than repeated.

The Radical of a Quadratic Form

Definition and First Properties

Definition. The radical of a quadratic form $q$ on $V$ with polar form $B$ is the submodule

$$ \operatorname{rad}(q) = \{v \in V : B(v, w) = 0 \text{ for all } w \in V\}. $$

Equivalently, $\operatorname{rad}(q)$ is the kernel of the linear map $V \to V^*$, $v \mapsto B(v, \cdot)$. The form is non-degenerate exactly when this map is an isomorphism; over a field that is the same as $\operatorname{rad}(q) = 0$, while over a general commutative ring the map must also be surjective.

Proposition. The radical is a totally isotropic subspace, and $q$ vanishes on it:

$$ v \in \operatorname{rad}(q) \implies q(v) = B(v, v) = 0. $$

Proof. Take $w = v$ in the defining condition.

Proposition. The radical is orthogonal to all of $V$ and contains every vector orthogonal to all of $V$; hence it is the largest totally isotropic subspace orthogonal to the whole space. For a subspace $W$ complementary to the radical, the restriction $q|_W$ is non-degenerate.

Proof. Orthogonality to all of $V$ is the definition. If $W$ is a complement, $V = W \oplus \operatorname{rad}(q)$, then a vector $w \in W$ orthogonal to all of $W$ is orthogonal to all of $V$, hence lies in $W \cap \operatorname{rad}(q) = 0$.

The Reduced Form

Proposition. The form $q$ induces a quadratic form $\bar{q}$ on the quotient $V/\operatorname{rad}(q)$ by $\bar{q}(v + \operatorname{rad}(q)) = q(v)$, and the polar form of $\bar{q}$ is the induced bilinear form on the quotient; when $\operatorname{rad}(q)$ is a direct summand with complement $W$, the quotient is identified with $W$ and $\bar{q}$ with $q|_W$.

Proof. The form $q$ is constant on the cosets of the radical because for $r \in \operatorname{rad}(q)$ one has $q(v + r) = q(v) + 2B(v, r) + q(r) = q(v)$, using $B(v, r) = 0$ and $q(r) = 0$. The polar form descends by the same computation applied to $B$.

Remark. Over a field every subspace is a direct summand, so a complement $W$ always exists and $\bar{q}$ is a form on $W$. Over a general commutative ring the existence of a complement is an additional hypothesis, and it is assumed throughout the decomposition below.

The Reduction to the Non-Degenerate Case

The Splitting

Let $W$ be a complement of the radical, so that

$$ V = W \oplus \operatorname{rad}(q), \qquad q|_W = \bar{q} \text{ non-degenerate.} $$

Because every element of $\operatorname{rad}(q)$ is orthogonal to every element of $V$, and in particular to every element of $W$, the decomposition is an orthogonal decomposition of quadratic spaces:

$$ (V, q) = (W, \bar{q}) \perp (\operatorname{rad}(q), 0), $$

where $0$ denotes the zero form on the radical. The second summand is the zero form on a space that is its own radical.

The Tensor Product Decomposition

Theorem. With $W$ a complement of $\operatorname{rad}(q)$, there is an isomorphism of $\mathbb{Z}/2$-graded algebras

$$ \mathrm{Cl}(V, q) \cong \mathrm{Cl}(W, \bar{q}) \,\hat{\otimes}\, \Lambda(\operatorname{rad}(q)), $$

where $\Lambda(\operatorname{rad}(q))$ is the exterior algebra on the radical and $\hat\otimes$ is the graded tensor product of Clifford Algebras.

Proof. The orthogonal decomposition above and the splitting theorem of Clifford Algebras give

$$ \mathrm{Cl}(V, q) \cong \mathrm{Cl}(W, \bar{q}) \,\hat{\otimes}\, \mathrm{Cl}(\operatorname{rad}(q), 0), $$

and $\mathrm{Cl}(\operatorname{rad}(q), 0) \cong \Lambda(\operatorname{rad}(q))$ because the defining relation $v^2 = 0$ on the second factor is exactly the defining relation of the exterior algebra.

Proof by relations. The same statement can be read off from the fundamental relation. Write $V = W \oplus \operatorname{rad}(q)$ and let $v, w \in W$ and $r, s \in \operatorname{rad}(q)$. The fundamental relation gives three families:

• $vw + wv = 2B(v, w)\cdot 1$ for $v, w \in W$, the relations of $\mathrm{Cl}(W, \bar{q})$;
• $vr + rv = 2B(v, r)\cdot 1 = 0$ for $v \in W$, $r \in \operatorname{rad}(q)$, so the generators of $W$ and of the radical anticommute;
• $rs + sr = 2B(r, s)\cdot 1 = 0$ for $r, s \in \operatorname{rad}(q)$, so the radical generators anticommute and each squares to zero, the relations of $\Lambda(\operatorname{rad}(q))$.

The first family generates the first factor, the third generates the second, and the second family is precisely the sign rule of the graded tensor product.

Dependence on the Complement

Proposition. The isomorphism depends on the choice of complement $W$. Two complements $W, W'$ of the radical give algebras $\mathrm{Cl}(W, \bar{q})$ and $\mathrm{Cl}(W', \bar{q})$ that are isomorphic — the two projections of $W'$ onto $W$ along the radical define an isometry and hence an isomorphism of Clifford algebras — but the isomorphism between the two tensor product decompositions of $\mathrm{Cl}(V, q)$ depends on that choice.

Proof. The projection $W' \to W$ along $\operatorname{rad}(q)$ is an isometry for $\bar{q}$ because $\bar{q}$ is constant on cosets of the radical; it is bijective by symmetry of the decomposition. The induced isomorphism of Clifford algebras is the one from functoriality.

The Nilpotent Factor and the Rank

The Exterior Algebra on the Radical

Let $r_0 = \operatorname{rank}\operatorname{rad}(q)$ and let $r_1, \ldots, r_{r_0}$ be a basis of the radical. Then

$$ \Lambda(\operatorname{rad}(q)) = \bigoplus_{k \geq 0} \Lambda^k(\operatorname{rad}(q)), $$

with basis the products $r_{i_1}\cdots r_{i_k}$ for $i_1 < \cdots < i_k$, and the generators satisfy

$$ r_i^2 = 0, \qquad r_ir_j = -r_jr_i \quad (i \neq j). $$

Every element of $\bigoplus_{k \geq 1}\Lambda^k(\operatorname{rad}(q))$ is nilpotent, with nilpotency index at most $r_0 + 1$: a product of more than $r_0$ of the generators vanishes because some index repeats.

The Rank

Theorem. Let $V$ be free of rank $n$ and $r_0 = \operatorname{rank}\operatorname{rad}(q)$. Then $\mathrm{Cl}(V, q)$ is free of rank

$$ \operatorname{rank}\mathrm{Cl}(V, q) = 2^{n - r_0} \cdot 2^{r_0} = 2^n. $$

Pro. The graded tensor product of two free modules of ranks $2^{n - r_0}$ and $2^{r_0}$ is free of rank $2^{n - r_0} \cdot 2^{r_0} = 2^n$. The rank of the non-degenerate factor is the dimension count.

So the presence of a radical changes the structure but not the rank: it replaces part of the algebra by an exterior algebra on the same number of generators.

The Nilpotent Ideal

Theorem. The image of $\operatorname{rad}(q)$ generates a two-sided ideal

$$ \mathrm{N} = \operatorname{rad}(q)\cdot \mathrm{Cl}(V, q) = \mathrm{Cl}(V, q)\cdot \operatorname{rad}(q), $$

which under the decomposition is $\mathrm{Cl}(W, \bar{q}) \otimes \bigoplus_{k\geq 1}\Lambda^k(\operatorname{rad}(q))$. The ideal $\mathrm{N}$ is nilpotent, and the quotient $\mathrm{Cl}(V, q)/\mathrm{N}$ is $\mathrm{Cl}(W, \bar{q})$. For a field $F$ of characteristic not $2$, $\mathrm{N}$ is the Jacobson radical of $\mathrm{Cl}(V, q)$.

Proof. For $r \in \operatorname{rad}(q)$ and $x \in \mathrm{Cl}(V, q)$ the products $rx$ and $xr$ lie in $\mathrm{N}$, so $\mathrm{N}$ is a two-sided ideal; the decomposition identifies it with the positive part of the exterior factor tensored with the non-degenerate factor. Its $(r_0+1)$-st power vanishes because a product of $r_0 + 1$ radical generators vanishes, so $\mathrm{N}^{r_0+1} = 0$. The quotient is obtained by setting the radical generators to zero, which is the defining presentation of $\mathrm{Cl}(W, \bar{q})$. A nilpotent ideal is contained in the Jacobson radical, and when $\mathrm{Cl}(W, \bar{q})$ is semisimple the quotient has zero Jacobson radical, so the two coincide.

Corollary. For a field $F$ of characteristic not $2$, $\mathrm{Cl}(V, q)$ is semisimple if and only if $q$ is non-degenerate.

Proof. If $q$ is non-degenerate then $\mathrm{N} = 0$ and the algebra is $\mathrm{Cl}(W, \bar{q})$, which is semisimple as a finite-dimensional algebra over a field that is either simple or a product of two simple algebras. If $q$ is degenerate then $\mathrm{N} \neq 0$ is nilpotent, so the algebra is not semisimple.

The Nilpotency Index

Theorem. With the notation of the decomposition,

$$ \mathrm{N}^k = \mathrm{Cl}(W, \bar{q}) \otimes \bigoplus_{j \geq k} \Lambda^j(\operatorname{rad}(q)), $$

and if the radical is free of rank $r_0$ the nilpotency index of $\mathrm{N}$ is exactly $r_0 + 1$.

Proof. A product of $k$ elements of $\mathrm{N}$ involves at least $k$ radical generators in each of its terms after expansion in a basis of $\operatorname{rad}(q)$, so $\mathrm{N}^k$ is contained in the displayed submodule. Conversely that submodule is spanned by the products $x\,r_{i_1}\cdots r_{i_j}$ with $x \in \mathrm{Cl}(W, \bar q)$ and $j \geq k$, each of which is a product of $k$ elements of $\mathrm{N}$ when $j \geq k$: write it as $\bigl(x r_{i_1}\cdots r_{i_{j-k}}\bigr)\bigl(r_{i_{j-k+1}}\bigr)\cdots\bigl(r_{i_j}\bigr)$. Hence the two submodules coincide. For the index, $\Lambda^{r_0}(\operatorname{rad}(q)) \cong R$ with generator the product of a basis, which is nonzero in the free case; therefore $\mathrm{N}^{r_0} = \mathrm{Cl}(W,\bar q)\otimes\Lambda^{r_0}(\operatorname{rad}(q))$ is nonzero as a free $R$-module, while $\mathrm{N}^{r_0+1} = 0$ since there is no exterior degree above $r_0$.

Thus the nilpotency index grows linearly with the rank of the radical; in the non-degenerate case $\mathrm{N} = 0$ and the algebra has no nilpotent factor at all.

The Filtration by the Radical Ideal

The ideal $\mathrm{N}$ is generated by the image of $\operatorname{rad}(q)$ in $\mathrm{Cl}(V,q)$, so it is independent of any choice of complement, and so is the decreasing filtration

$$ \mathrm{Cl}(V, q) = \mathrm{N}^0 \supseteq \mathrm{N}^1 \supseteq \mathrm{N}^2 \supseteq \cdots \supseteq \mathrm{N}^{r_0+1} = 0. $$

Proposition. The successive quotients have rank

$$ \operatorname{rank}\bigl(\mathrm{N}^k/\mathrm{N}^{k+1}\bigr) = 2^{\,n - r_0}\binom{r_0}{k}, $$

so the associated graded has total rank $\sum_k 2^{n-r_0}\binom{r_0}{k} = 2^{n-r_0}\cdot 2^{r_0} = 2^n$, which is the rank of the algebra.

Proof. Choosing a complement identifies $\mathrm{N}^k$ with $\mathrm{Cl}(W,\bar q)\otimes\bigoplus_{j\geq k}\Lambda^j(\operatorname{rad}(q))$ by the theorem above, and $\Lambda^k(\operatorname{rad}(q))$ is free of rank $\binom{r_0}{k}$; hence the quotient $\mathrm{N}^k/\mathrm{N}^{k+1}$ is free of the stated rank, and the identification of the associated graded is independent of the complement because the filtration is.

So the filtration detects the rank of the radical through the length of the filtration and the ranks of its quotients: the rank of $\mathrm{N}/\mathrm{N}^2$ is $2^{n-r_0}r_0$, which together with the length $r_0 + 1$ of the filtration determines $r_0$, and then $2^n$ determines $n - r_0$. This is the precise sense in which the Clifford algebra of a degenerate form retains the rank of the radical, even though it does not retain the form on it.

The Split Extension

The decomposition also describes the algebra as an extension of a semisimple algebra by a nilpotent ideal.

Theorem. Let $\mathrm{G} = \mathrm{Cl}(W, \bar q) \otimes 1$ be the image of the reduced Clifford algebra in $\mathrm{Cl}(V, q)$. Then $\mathrm{G}$ is a subalgebra, $\mathrm{N}$ is a two-sided ideal, $\mathrm{G} \cap \mathrm{N} = 0$ and $\mathrm{G} + \mathrm{N} = \mathrm{Cl}(V,q)$; hence

$$ \mathrm{Cl}(V, q) = \mathrm{G} \oplus \mathrm{N} $$

as a direct sum of $R$-modules, the quotient map restricts to an isomorphism $\mathrm{G} \to \mathrm{Cl}(V, q)/\mathrm{N}$, and $\mathrm{N}$ is the largest nilpotent two-sided ideal of the algebra.

Proof. The image of $\mathrm{Cl}(W,\bar q)$ under $x \mapsto x \otimes 1$ is a subalgebra because the map is an algebra homomorphism; the image of the positive part of the exterior algebra is $\mathrm{N}$ by the theorem above, and the intersection is zero because the two factors meet only in degree zero on the second side: an element $x \otimes b$ with $b$ of positive degree is not of the form $y \otimes 1$. The ranks add, $2^{n - r_0} + 2^{n-r_0}(2^{r_0} - 1) = 2^n$, so the sum is everything, and the restriction of the quotient map is injective with image of full rank, hence an isomorphism. Finally $\mathrm{N}$ is nilpotent and the Jacobson radical of the algebra, so it contains every nilpotent ideal.

Remark. This is the split extension of the semisimple algebra $\mathrm{G} \cong \mathrm{Cl}(V,q)/\mathrm{N}$ by the nilpotent bimodule $\mathrm{N}$: the products of an element of $\mathrm{G}$ with an element of $\mathrm{N}$ lie in $\mathrm{N}$, and the multiplication of the whole algebra is determined by the multiplication in $\mathrm{G}$, the bimodule structure of $\mathrm{N}$ and the multiplication inside $\mathrm{N}$. When $r_0 = 1$ the interior product vanishes, so $\mathrm{N}^2 = 0$ and the extension is a split null extension. The complement $\mathrm{G}$ is not canonical, since it comes from the choice of $W$, while $\mathrm{N}$ and the isomorphism class of the quotient are.

The Degenerate and Non-Degenerate Cases Compared

The table records how each feature of the algebra behaves as the radical grows.

The last line is the content of the reduction: the Clifford algebra of a degenerate form sees the form only through its non-degenerate part, and loses all information about the radical except its rank.

In the centre row $\omega = e_1 \cdots e_n$ denotes the volume element of an orthogonal basis of $V$, and $\Lambda^{\mathrm{ev}}(\operatorname{rad}(q))$ is the even part of the exterior algebra on the radical, spanned by the products of an even number of radical generators. Moving $e_i$ past the $n - i$ generators to its right in $\omega e_i$ and past the $i - 1$ generators to its left in $e_i\omega$ gives $\omega e_i = (-1)^{n-i}a_i\,e_1\cdots\hat e_i\cdots e_n$ and $e_i\omega = (-1)^{i-1}a_i\,e_1\cdots\hat e_i\cdots e_n$, so $\omega e_i = (-1)^{n-2i+1}e_i\omega = (-1)^{n-1}e_i\omega$: for odd $n$ the volume element commutes with every generator and is therefore central. For the non-degenerate column this is the classical statement: the centre is $R$ for even $n$ and $R \oplus R\omega$ for odd $n$, which is the case $\operatorname{rad}(q) = 0$ of the degenerate entry, where $\Lambda^{\mathrm{ev}}(0) = R$. In the degenerate column the graded tensor product with $\Lambda(\operatorname{rad}(q))$ enlarges the even part of the centre, and two verifications give the count. First, an element $1 \otimes b$ with $b$ even commutes with every element of $\mathrm{Cl}(V, q)$: even elements of a supercommutative algebra commute with all elements of it, so $b$ passes the generators of $\mathrm{Cl}(W, \bar q)$ without a sign; this produces the subspace $\Lambda^{\mathrm{ev}}(\operatorname{rad}(q))$, of rank $2^{r_0 - 1}$ when $r_0 \geq 1$ and of rank $1$ when $r_0 = 0$. Second, no central elements other than these and the multiples of $\omega$ occur. A homogeneous central term $a \otimes b$ must be graded-central in each of the two factors with parity twists dual to one another; the elements of the reduced non-degenerate factor that can occur are the scalars and the multiples of its volume element $\omega_W$, and the only even elements of $\Lambda(\operatorname{rad}(q))$ that can occur as the second factor are the even elements themselves, while an odd second factor must be a multiple of the radical volume element; so a term not of the form $1 \otimes b$ with $b$ even has $a$ a multiple of $\omega_W$ and $b$ a multiple of $\omega_{\operatorname{rad}}$, that is, it is a multiple of $\omega$, and such a term occurs exactly when $\dim W$ and $r_0$ have opposite parity, that is, when $n$ is odd. Hence the centre is $\Lambda^{\mathrm{ev}}(\operatorname{rad}(q))$ for even $n$ and $\Lambda^{\mathrm{ev}}(\operatorname{rad}(q)) \oplus R\omega$ for odd $n$. When $q = 0$ the reduced factor is $R$, and the formula reads $\Lambda^{\mathrm{ev}}(V)$ for even $n$ and $\Lambda^{\mathrm{ev}}(V) \oplus R\omega$ for odd $n$, which is the centre of the exterior algebra.

Examples

The Zero Form

If $q = 0$ then every vector is in the radical and $\operatorname{rad}(q) = V$; the complement is $0$ and $\bar{q}$ is the empty form. The theorem gives

$$ \mathrm{Cl}(V, 0) \cong \Lambda(V), $$

so the Clifford algebra of the zero form is the exterior algebra, as it must be, since the defining relation $v^2 = 0$ is the defining relation of $\Lambda(V)$. This is the extreme degenerate case, and it is the one in which the nilpotent ideal is the whole positive part.

The Dual Numbers

Let $R = \mathbb{R}$ and $V = \mathbb{R}$ with the zero form, $q(x) = 0$. Then $\operatorname{rad}(q) = V$ and

$$ \mathrm{Cl}(\mathbb{R}, 0) \cong \mathbb{R}[\varepsilon]/(\varepsilon^2), \qquad \varepsilon^2 = 0, $$

which is the algebra of dual numbers $\mathbb{D}'$. This is the one-dimensional degenerate Clifford algebra, and its unit $\varepsilon$ is the radical generator; the algebra is local with maximal ideal $(\varepsilon)$, and it is the smallest example of a Clifford algebra with a nonzero radical. It is recorded in the number-system dictionary of the companion entry with this article.

A Degenerate Form in Two Variables

Let $V = \mathbb{R}^2$ with $q(x, y) = x^2$. The polar form is $B((x, y), (x', y')) = xx'$, whose kernel is the line $\{(0, y)\} = \operatorname{rad}(q)$; a complement is the line spanned by $e_1$ with $\bar{q}(e_1) = 1$, so $\mathrm{Cl}(W, \bar{q}) \cong \mathbb{R}[t]/(t^2 - 1) \cong \mathbb{R}\times\mathbb{R}$, the split complex algebra $\mathbb{D}$. Write $e$ for the generator of the complement, $e^2 = 1$, and $\varepsilon$ for the generator of the radical, $\varepsilon^2 = 0$. The decomposition is

$$ \mathrm{Cl}(\mathbb{R}^2, x^2) \cong \mathbb{D}\,\hat{\otimes}\,\Lambda(\mathbb{R}), \qquad e^2 = 1, \quad \varepsilon^2 = 0, \quad e\varepsilon = -\varepsilon e, $$

a four-dimensional algebra. The grading of $\mathbb{D}$ is nontrivial — its generator $e$ is odd — so the graded tensor product is not the ordinary tensor product, and the algebra is not the product of two copies of $\mathbb{R}[\varepsilon]/(\varepsilon^2)$: the idempotents $(1 \pm e)/2$ are not central, since $e\varepsilon = -\varepsilon e$, and the centre is only $\mathbb{R}$. The radical generator spans the nonzero nilpotent ideal $\mathrm{N} = \operatorname{span}\{\varepsilon, e\varepsilon\}$ with $\mathrm{N}^2 = 0$, and the reduced form is the one-dimensional form of square $+1$, in agreement with the rank count $2^2 = 4$.

A Radical of Rank Two

Let $V = \mathbb{R}^3$ with $q(x, y, z) = x^2$, so that $\operatorname{rad}(q)$ is spanned by $e_2, e_3$ and the complement $W = \langle e_1\rangle$ carries the form of square $1$, whose Clifford algebra is the split complex algebra $\mathbb{D}$. The decomposition reads

$$ \mathrm{Cl}(\mathbb{R}^3, x^2) \cong \mathbb{D} \,\hat{\otimes}\, \Lambda(\mathbb{R}^2), \qquad \text{rank } 2 \cdot 4 = 8 = 2^3, $$

with generators $e$ of square $1$ and $\varepsilon, \eta$ of square $0$, all three odd and mutually anticommuting. The nilpotent ideal has rank $6$, spanned by the elements containing a radical generator; its square is spanned by the two elements $\varepsilon\eta$ and $e\varepsilon\eta$ and is nonzero, while its cube vanishes. So the nilpotency index is $3 = r_0 + 1$, the value predicted by the theorem, and the nilpotent factor has index greater than $2$. The centre is $\Lambda^{\mathrm{ev}}(\operatorname{rad}(q)) \oplus \mathbb{R}\omega = \operatorname{span}\{1, \varepsilon\eta, e\varepsilon\eta\}$, of rank $3$, in agreement with the formula for odd $n$.

A Non-Degenerate Reduced Factor

Let $V = \mathbb{R}^3$ with $q(x, y, z) = x^2 - y^2$, the hyperbolic plane on the first two coordinates and the zero form on the last. The radical is spanned by $e_3$, the reduced form is the hyperbolic plane, and $\mathrm{Cl}(W, \bar q) \cong M_2(\mathbb{R})$, the Clifford algebra of a hyperbolic plane being a matrix algebra. Hence

$$ \mathrm{Cl}(\mathbb{R}^3, x^2 - y^2) \cong M_2(\mathbb{R}) \,\hat{\otimes}\, \Lambda(\mathbb{R}), \qquad \text{rank } 4 \cdot 2 = 8, $$

and the ideal $\mathrm{N} = M_2(\mathbb{R}) \otimes \mathbb{R}\varepsilon$ has rank $4$ with $\mathrm{N}^2 = 0$, so the nilpotency index is $2 = r_0 + 1$ with $r_0 = 1$. The example shows that the nilpotency index is governed by the rank of the radical alone and is independent of the size of the reduced algebra.

Summary

The radical $\operatorname{rad}(q) = \{v : B(v, w) = 0 \ \forall w\}$ of a quadratic form is a totally isotropic subspace on which $q$ vanishes, and it is the kernel of the map $V \to V^*$ induced by the polar form; the form is non-degenerate exactly when the radical is zero. The form descends to the quotient by the radical, and when the radical is a direct summand with complement $W$, the reduced form $q|_W$ is non-degenerate and $(V, q) = (W, q|_W) \perp (\operatorname{rad}(q), 0)$.

The decomposition theorem states that for such a complement

$$ \mathrm{Cl}(V, q) \cong \mathrm{Cl}(W, q|_W) \,\hat{\otimes}\, \Lambda(\operatorname{rad}(q)), $$

the graded tensor product of the Clifford algebra of the reduced non-degenerate form and the exterior algebra on the radical. It follows both from the splitting theorem for orthogonal sums and from the three families of fundamental relations among vectors of $W$ and of the radical. The isomorphism depends on the choice of complement, although different complements give isomorphic factors.

The generator $r$ of the radical satisfies $r^2 = 0$, so the exterior algebra is nilpotent in positive degree. In fact $\mathrm{N}^k = \mathrm{Cl}(W, \bar{q}) \otimes \bigoplus_{j\geq k}\Lambda^j(\operatorname{rad}(q))$ and the nilpotency index is exactly $\operatorname{rank}\operatorname{rad}(q) + 1$ when the radical is free. The radical generates a nilpotent two-sided ideal $\mathrm{N}$, the quotient by which is the non-degenerate factor; over a field of characteristic not $2$, $\mathrm{N}$ is the Jacobson radical and $\mathrm{Cl}(V, q)$ is semisimple if and only if $q$ is non-degenerate. The centre is the even part of the exterior algebra on the radical, of rank $2^{r_0 - 1}$ for $r_0 \geq 1$, together with the span of the volume element when $n$ is odd, which reduces to the classical $R$ or $R \oplus R\omega$ when the radical vanishes. The rank of the algebra is $2^n$ whether or not the form is degenerate, since the dimension lost by the reduced form is recovered by the exterior factor. The extreme case is the zero form, for which $\mathrm{Cl}(V, 0) = \Lambda(V)$; the one-dimensional case is the algebra of dual numbers $\mathbb{D}'$.

Summary of Notation

Further Reading

• Max-Albert Knus, Quadratic and Hermitian Forms over Rings, Grundlehren der mathematischen Wissenschaften 294 (Springer, 1991), for the radical of a form and the Clifford algebra over a ring.
• H. Blaine Lawson and Marie-Louise Michelsohn, Spin Geometry (Princeton University Press, 1989), for the algebraic preliminaries on Clifford algebras of possibly degenerate forms.
• Pertti Lounesto, Clifford Algebras and Spinors (Cambridge University Press, 2nd ed. 2001), for the exterior-algebra factor and worked degenerate examples.
• T. Y. Lam, Introduction to Quadratic Forms over Fields, Graduate Studies in Mathematics 67 (American Mathematical Society, 2005), for the radical, non-degeneracy and the reduction of a form.
• Nicolas Bourbaki, Algebra I (Springer, 1998), for the filtration and the exterior-algebra comparison without a non-degeneracy hypothesis.