The Geometric Product and the Grade Decomposition

Introduction

The Clifford algebra is generated by a space of vectors subject to the single relation $v^2=q(v)$, and that relation already mixes degrees: the product of two vectors is a scalar together with a bivector. This article analyses the product of $\mathrm{Cl}(V,q)$ grade by grade. It names the product — the geometric product — and proves that the product of a $k$-vector with an $l$-vector is the sum of the grades $|k-l|$, $|k-l|+2$, $\ldots$, $k+l$, so that the whole product is assembled from a grade-lowering and a grade-raising part. The grade-raising part is the outer product, which reconstructs the exterior algebra of $V$; the grade-lowering part is the contraction. The vocabulary fixed here — the grade projection, the multivector decomposition, the inner, outer and contracted products — is the form in which the Clifford algebra of a quadratic form is read geometrically, and it is used by the companion entries The Volume Element, Duality and the Hodge Star, Versors, Rotors and the Sandwich Action with Signed Inner Conjugation and Spinors as Minimal Left Ideals with Inner Conjugation.

The Clifford algebra, its universal property, the fundamental relation $uv+vu=2B(u,v)$ and the parity grading $\mathrm{Cl}=\mathrm{Cl}^0\oplus\mathrm{Cl}^1$ are from Clifford Algebras; the basis of products of distinct generators, the $k$-vectors and multivectors, the volume element, reversion, Clifford conjugation and the grade involution are from Clifford Algebras in Finite Dimensions; the wedge product and the graded-commutativity of the exterior algebra are from The Exterior Algebra. Nothing owned by those entries is re-derived. The base is a field $F$ of characteristic not $2$, with $q$ a non-degenerate quadratic form on a finite-dimensional space $V$ and $B$ its polar form; the conventions are $v^2=q(v)\cdot1$ and $uv+vu=2B(u,v)\cdot1$.

The Grade Projection

The Grade Subspaces

Let $e_1,\ldots,e_n$ be an orthogonal basis of $V$, so that $B(e_i,e_j)=0$ for $i\neq j$ and therefore $e_ie_j=-e_je_i$; such a basis exists because the form is diagonalisable. The products $e_{i_1}\cdots e_{i_k}$ with $i_1<\cdotsClifford Algebras in Finite Dimensions.

Definition. The grade-$k$ subspace $\mathrm{Cl}_k(V,q)$ is the linear span of the basis products containing exactly $k$ factors. Its elements are the $k$-vectors, and an element of $\mathrm{Cl}(V,q)$ written as a sum of $k$-vectors of distinct grades is a multivector. The dimension of $\mathrm{Cl}_k(V,q)$ is $\binom{n}{k}$, so that

$$ \mathrm{Cl}(V,q)=\bigoplus_{k=0}^{n}\mathrm{Cl}_k(V,q), \qquad \dim\mathrm{Cl}(V,q)=\sum_{k=0}^{n}\binom{n}{k}=2^n. $$

The subscript in $\mathrm{Cl}_k(V,q)$ is a grade and is not the signature subscript of $\mathrm{Cl}_{p,q}$. The parity parts of Clifford Algebras are the sums of the even and of the odd grades,

$$ \mathrm{Cl}^0(V,q)=\bigoplus_{k\ \mathrm{even}}\mathrm{Cl}_k(V,q), \qquad \mathrm{Cl}^1(V,q)=\bigoplus_{k\ \mathrm{odd}}\mathrm{Cl}_k(V,q), $$

so that a multivector is even when its odd-grade parts vanish.

Definition. The grade projection $\langle x\rangle_k$ of $x\in\mathrm{Cl}(V,q)$ is the component of $x$ in $\mathrm{Cl}_k(V,q)$ in the decomposition above. A multivector is thus $x=\sum_k\langle x\rangle_k$, and the grade projections are the linear maps that read off the coefficient of each degree.

Independence of the Basis

The grade of an element has been defined through one orthogonal basis. The definition does not depend on that choice, and the antisymmetrisation map exhibits the same subspaces without a basis.

Proposition. For $x=v_1\wedge\cdots\wedge v_k\in\Lambda^kV$ put

$$ \mathrm{Alt}(x)=\frac{1}{k!}\sum_{\pi\in S_k}\operatorname{sgn}(\pi)\,v_{\pi(1)}\cdots v_{\pi(k)}\ \in\ \mathrm{Cl}(V,q), $$

the antisymmetrised product. Then $\mathrm{Alt}$ is linear, $\mathrm{Alt}(\Lambda^kV)\subseteq\mathrm{Cl}_k(V,q)$, and $\mathrm{Alt}:\Lambda(V)\to\mathrm{Cl}(V,q)$ is an isomorphism of vector spaces over a field of characteristic $0$ or of characteristic greater than $n$.

Proof. Linearity is clear, and the grading statement follows because every term of the sum is a product of $k$ vectors. For the basis monomials $e_{i_1}\wedge\cdots\wedge e_{i_k}$ with $i_1<\cdots

The $k$-vectors of $\mathrm{Cl}(V,q)$ are thus the image of the $k$-fold wedges, and $\langle x\rangle_k$ is the component read in the exterior algebra through $\mathrm{Alt}$. On the basis monomials $\mathrm{Alt}$ is the identity; on a non-orthogonal wedge it is the antisymmetrisation and not the naive product.

The Geometric Product and Its Grades

The Grade Theorem

Definition. The product of $\mathrm{Cl}(V,q)$ is the geometric product. The two names denote the same operation; the second records that the product is read on the grades of the multivectors rather than only on the generating vectors.

Theorem. Let $u\in\mathrm{Cl}_k(V,q)$ and $w\in\mathrm{Cl}_l(V,q)$. Then

$$ uw=\sum_{j}\langle uw\rangle_j, \qquad j=k+l,\ k+l-2,\ \ldots,\ |k-l|, $$

so that all the surviving grades have the parity of $k+l$ and lie between $|k-l|$ and $k+l$. In particular the product of two $k$-vectors is even, and the product of an even with an odd multivector is odd.

Proof. Both sides being bilinear, it suffices to take monomials $u=e_{i_1}\cdots e_{i_k}$ and $w=e_{j_1}\cdots e_{j_l}$ with increasing indices, in an orthogonal basis. Let $S$ be the set of indices common to the two monomials, of cardinality $s$. In the concatenation $uw$ each index of $S$ occurs twice and every other index once. Moving the factors so that the repeated indices become adjacent, and replacing each $e^2$ by the scalar $q(e)$, leaves a scalar multiple of the product of the $k+l-2s$ distinct indices outside $S$; the anticommutation of distinct generators contributes only signs. That product is an element of $\mathrm{Cl}_{k+l-2s}(V,q)$. As $s$ ranges from $0$ to $\min(k,l)$, the grades $k+l-2s$ are exactly $k+l,k+l-2,\ldots,|k-l|$.

The theorem is the precise sense in which the geometric product is not homogeneous but has a controlled spread: the discrepancy between the degrees is bounded by the smaller of the two, and each step of the spread is by two.

Corollary (two vectors). For $u,v\in V=\mathrm{Cl}_1(V,q)$ the theorem leaves the grades $0$ and $2$, so

$$ uv=\langle uv\rangle_0+\langle uv\rangle_2,\qquad \langle uv\rangle_0=B(u,v),\qquad \langle uv\rangle_2=\tfrac12(uv-vu). $$

The symmetric part of the product of two vectors is the scalar given by the form and the alternating part is their bivector. Summing the relation with its reverse gives the fundamental relation $uv+vu=2B(u,v)$, and subtracting gives the identification of the alternating part.

Corollary (vector and bivector). For $a\in V$ and $B\in\mathrm{Cl}_2(V,q)$ the theorem leaves the grades $1$ and $3$:

$$ aB=\langle aB\rangle_1+\langle aB\rangle_3,\qquad Ba=-\langle aB\rangle_1+\langle aB\rangle_3, $$

because the wedge of a vector with a bivector is symmetric under interchange, the Koszul sign being $(-1)^{1\cdot2}=+1$, while the grade-one parts are exchanged by the two orders. Hence

$$ \langle aB\rangle_1=\tfrac12(aB-Ba),\qquad \langle aB\rangle_3=\tfrac12(aB+Ba). $$

The two grades of $aB$ are separated by symmetry rather than by the parity of the degrees: the grade-one part is the alternating part and the grade-three part the symmetric one, the opposite assignment to the case of two vectors. For a simple bivector $B$ the grade-three part vanishes exactly when $a$ lies in the plane of $B$, and then $aB$ is a pure vector.

Proof of the second identity. In the notation of the theorem with $k=1$ and $l=2$, the monomial computation gives $aB$ in grades $1$ and $3$. For $Ba$ the same computation with the factors exchanged assigns to the grade-one part the opposite sign, because the single interchange of the odd monomial with the even monomial reverses the order of the shared factor. The grade-three part is symmetric in $a$ and $B$ by the Koszul rule applied to $\Lambda^1$ and $\Lambda^2$.

Example. In $\mathrm{Cl}_{3,0}$ take $a=e_3$ and $B=e_1e_2$. Then $aB=e_3e_1e_2=e_1e_2e_3=\omega$, of pure grade three, while $Ba=e_1e_2e_3=\omega$ as well, the grade-one parts vanishing because $e_3$ is orthogonal to the plane of $B$. Take instead $a=e_1$ and $B=e_1e_2$: then $aB=e_1e_1e_2=e_2$, of pure grade one, and $Ba=e_1e_2e_1=-e_2$, so that the grade-three parts vanish and the two orders differ by the sign of the grade-one part.

The Three Products

The grade theorem isolates two terms of the product, the lowest grade and the highest, and they are the classical operations of the algebra.

Definition. Let $u\in\mathrm{Cl}_k(V,q)$ and $w\in\mathrm{Cl}_l(V,q)$, both homogeneous of positive grade $k,l\ge1$. The inner product is

$$ u\cdot w=\langle uw\rangle_{|k-l|}, $$

the left contraction is $u\lrcorner w=\langle uw\rangle_{l-k}$ for $k\le l$ and $0$ otherwise, the right contraction is $u\lfloor w=\langle uw\rangle_{k-l}$ for $k\ge l$ and $0$ otherwise, and the outer product is

$$ u\wedge w=\langle uw\rangle_{k+l}. $$

Product	Grade selected in $uw$	Range of application
inner product $u\cdot w$	$\vert k-l\vert$	$k,l\ge1$
left contraction $u\lrcorner w$	$l-k$	$k\le l$
right contraction $u\lfloor w$	$k-l$	$k\ge l$
outer product $u\wedge w$	$k+l$	all $k,l$

Theorem. For $u\in\mathrm{Cl}_k$ and $w\in\mathrm{Cl}_l$ the geometric product is the sum of all the grades of the theorem above, of which the outer product is the highest and the contraction is the lowest:

$$ uw=\langle uw\rangle_{|k-l|}+\langle uw\rangle_{|k-l|+2}+\cdots+u\wedge w, $$

the lowest grade being the left contraction $u\lrcorner w$ when $k\le l$ and the right contraction $u\lfloor w$ when $k\ge l$, and the displayed end terms being understood with the vanishing conventions of the definitions. When $k=1$ or $l=1$ the spread consists of two grades and the identity reads

$$ uw=u\cdot w+u\wedge w, $$

and for two vectors it is the decomposition of the first corollary, $uv=B(u,v)+u\wedge v$.

Proof. The grade theorem enumerates the grades of $uw$; the highest is $k+l$ and defines the outer product, and the lowest is $|k-l|$, which is $l-k$ for $k\le l$, the left contraction, and $k-l$ for $k\ge l$, the right contraction. If $k=1$ the list $l+1,l-1,\ldots,|l-1|$ has exactly two entries, and the two are the inner product and the outer product; the case $l=1$ is the same.

Remark. The inner product of a vector with a $k$-vector agrees with the left contraction when $k\ge1$, since $|1-k|=k-1$; the two notions differ only in the conventions at $k=0$ or $l=0$, where the inner product is set to zero and the contraction of a scalar is the scalar multiple. The distinction is a matter of bookkeeping and not of substance, and the version used below is the one that makes the two-vector identity exact.

The Recovery of the Exterior Algebra

The outer product alone carries the exterior structure of the space.

Proposition. The bilinear extension of $u\wedge w=\langle uw\rangle_{k+l}$, taken on homogeneous multivectors and extended by linearity, makes $\bigoplus_k\mathrm{Cl}_k(V,q)$ a graded-commutative algebra, with

$$ u\wedge w=(-1)^{kl}\,w\wedge u \qquad (u\in\mathrm{Cl}_k,\ w\in\mathrm{Cl}_l), $$

and the map $\mathrm{Alt}$ of the first section is an isomorphism of graded algebras from $\Lambda(V)$ with its wedge product onto that algebra.

Proof. It suffices to take monomials $u=e_{i_1}\cdots e_{i_k}$ and $w=e_{j_1}\cdots e_{j_l}$ in an orthogonal basis, bilinearity doing the rest. If the two index sets are disjoint, then $uw$ and $wu$ are both of pure grade $k+l$, and reordering the concatenation moves the $l$ factors of $w$ past the $k$ factors of $u$, a factor $(-1)^{kl}$; hence $u\wedge w=(-1)^{kl}w\wedge u$. If the index sets meet, the repeated index lowers the degree of every grade of both products by at least two, so the grade $k+l$ part of $uw$ and of $wu$ vanishes and the identity reads $0=0$. The identification of the outer product with the wedge of the exterior algebra, and the graded-commutativity of the latter, are from The Exterior Algebra; that $\mathrm{Alt}$ intertwines the two products follows from the same monomial check, where both sides give the monomial of the union of the indices.

So the Clifford algebra is a deformation of the exterior algebra: it has the same underlying graded vector space, the wedge is its top-grade part, and the remaining grades are the deformation. This is the reading of the remark of Clifford Algebras that setting $q=0$ recovers $\Lambda(V)$.

The Behaviour of the Involutions on the Grades

The three standard involutions of the Clifford algebra are homogeneous of degree zero — they preserve each grade subspace — and act on it by a sign depending only on the grade. With $x\in\mathrm{Cl}_k(V,q)$,

Involution	Sign on $\mathrm{Cl}_k$	Type
grade involution $\alpha$	$(-1)^k$	automorphism
reversion $x\mapsto x^{r}$	$(-1)^{k(k-1)/2}$	anti-automorphism
Clifford conjugation $x\mapsto x^{\natural}=\alpha(x^{r})$	$(-1)^{k(k+1)/2}$	anti-automorphism

The signs are those recorded in Clifford Algebras in Finite Dimensions; the table places them on the grades of the present article. In particular the grade involution is the identity on $\mathrm{Cl}^0$ and minus the identity on $\mathrm{Cl}^1$, and reversion fixes the vectors and sends a bivector to its negative.

Corollary. Reversion fixes the scalars and the vectors, negates the bivectors and the trivectors, and fixes the four-vectors; Clifford conjugation fixes the scalars and the four-vectors and negates the vectors and the bivectors.

Proof. Substituting $k=0,1,2,3,4$ in the signs of the table: for reversion the exponents are $0,0,1,3,6$, giving $+,+,-,-,+$; for conjugation the exponents are $0,1,3,6,10$, giving $+,-,-,+,+$.

Worked Cases

The Plane

In $\mathrm{Cl}_{2,0}$ with the orthonormal basis $e_1,e_2$, the geometric product of $a=a_1e_1+a_2e_2$ and $b=b_1e_1+b_2e_2$ is

$$ ab=(a_1b_1+a_2b_2)+(a_1b_2-a_2b_1)e_1e_2, $$

the scalar part being the inner product and the bivector part the oriented area of the parallelogram that the two vectors span. The bivector $e_1e_2$ has square $e_1e_2e_1e_2=-e_1^2e_2^2=-1$, and it anticommutes with every vector of the plane.

Three Dimensions

In $\mathrm{Cl}_{3,0}$ the volume element $\omega=e_1e_2e_3$ satisfies $\omega^2=-1$ and commutes with every element, since $n$ is odd. The product of two vectors is the sum of a scalar and a bivector, and multiplying the bivector by $\omega$ returns a vector: $e_1e_2\omega=e_1e_2e_1e_2e_3=-e_3$. That identity is the duality developed in The Volume Element, Duality and the Hodge Star, where the correspondence of a bivector with a vector is made into the Hodge star.

The Product of Two Even Multivectors

The even part $\mathrm{Cl}^0$ is a subalgebra by the parity grading, and the grade theorem explains why: the grades appearing in the product of two even multivectors are sums of two even grades reduced by even steps, hence even. The same argument shows that $\mathrm{Cl}^1$ is a module over $\mathrm{Cl}^0$ and that the product of two odd multivectors is even. The bivectors of $\mathrm{Cl}_{3,0}$ span, with the scalars, the even part, and the products $e_1e_2,e_2e_3,e_3e_1$ satisfy the relations of the quaternion units:

$$ (e_1e_2)^2=(e_2e_3)^2=(e_3e_1)^2=-1,\qquad (e_1e_2)(e_2e_3)=e_1e_3=-(e_2e_3)(e_1e_2). $$

So the even part of $\mathrm{Cl}_{3,0}$ is a copy of the quaternion algebra, of dimension four, and it is the algebra on which the rotors of the next entries live.

Summary

The Clifford algebra of a non-degenerate quadratic form decomposes as a vector space into the grade subspaces $\mathrm{Cl}_k(V,q)$, of dimension $\binom{n}{k}$, spanned by the products of $k$ distinct elements of an orthogonal basis; the component of a multivector in $\mathrm{Cl}_k$ is its grade projection $\langle x\rangle_k$. The decomposition does not depend on the orthogonal basis, because the antisymmetrisation map $\Lambda(V)\to\mathrm{Cl}(V,q)$ carries each wedge monomial to the Clifford monomial of the same indices and is an isomorphism.

The product of $\mathrm{Cl}(V,q)$, the geometric product, is not homogeneous: the product of a $k$-vector with an $l$-vector is the sum of exactly the grades $|k-l|,|k-l|+2,\ldots,k+l$, all with the parity of $k+l$. The highest of these grades is the outer product $u\wedge w$, the lowest is the contraction $u\lrcorner w$, and the intermediate grades complete the product. Two special cases are the whole of the elementary calculus: for two vectors $uv=B(u,v)+u\wedge v$, and for a vector and a bivector the product has only the grades one and three, the grade-one part being the contraction and the grade-three part the wedge.

The outer product alone gives the exterior algebra, recovered here from the Clifford product as the top grade, so that the Clifford algebra is a deformation of $\Lambda(V)$ with the same underlying graded vector space. The grade involution, reversion and Clifford conjugation preserve each grade subspace and act on it by the signs $(-1)^k$, $(-1)^{k(k-1)/2}$ and $(-1)^{k(k+1)/2}$ respectively. The even part is a subalgebra; in three dimensions it is the quaternion algebra spanned by the scalars and the bivectors.

Summary of Notation

Symbol	Meaning
$\mathrm{Cl}(V,q)$	Clifford algebra, $v^2=q(v)\cdot1$, $uv+vu=2B(u,v)$
$\mathrm{Cl}_k(V,q)$	Grade-$k$ subspace, the $k$-vectors, of dimension $\binom{n}{k}$
$\langle x\rangle_k$	Grade projection, the component of $x$ in $\mathrm{Cl}_k$
$\mathrm{Cl}^0,\ \mathrm{Cl}^1$	Even and odd parts, the sums of the even and odd grades
$uv$	Geometric product
$\Lambda(V)$	Exterior algebra, recovered as the top grade of the product
$\mathrm{Alt}$	Antisymmetrisation $\Lambda(V)\to\mathrm{Cl}(V,q)$, an isomorphism
$u\cdot w=\langle uw\rangle_{\vert k-l\vert}$	Inner product of a $k$-vector with an $l$-vector
$u\lrcorner w=\langle uw\rangle_{l-k}$	Left contraction, for $k\le l$
$u\lfloor w=\langle uw\rangle_{k-l}$	Right contraction, for $k\ge l$
$u\wedge w=\langle uw\rangle_{k+l}$	Outer product
$\alpha$	Grade involution, $(-1)^k$ on $\mathrm{Cl}_k$
$x^{r}$	Reversion, $(-1)^{k(k-1)/2}$ on $\mathrm{Cl}_k$
$x^{\natural}=\alpha(x^{r})$	Clifford conjugation, $(-1)^{k(k+1)/2}$ on $\mathrm{Cl}_k$