Expansions and Summations in Clifford Algebra · Expansions and Summations in Clifford Algebra Hongbo Li1) Abstract. Clifford algebra is an important tool in theoretical physics,

MM Research Preprints, 112–154MMRC, AMSS, Academia, Sinica, BeijingNo. 21, December 2002

Expansions and Summations in Clifford Algebra

Hongbo Li1)

Abstract. Clifford algebra is an important tool in theoretical physics, computationalgeometry and engineering applications, and coordinate-free computation is a salient fea-ture of the geometric algebraic version of Clifford algebra. In this paper we establishsome fundamental formulas on coordinate-free expansions of the geometric product ofvectors and blades into the inner products and outer products of vectors, and on thereverse procedure of computing the combinatorial sum of vector and blade expressions.These formulas should serve as the foundations of coordinate-free symbolic computationswith Clifford algebra.

1. Introduction

Clifford algebra is an important tool in both mathematics and physics. The version ofClifford algebra in Hestenes and Sobczyk (1984) is also called geometric algebra, which hasimportant applications in theoretical physics, computational geometry and engineering, seeHestenes (1966, 1987), Crippen and Havel (1988), Doran et al. (1993), Li (1997, 2001),Ashdown et al. (1998), Sommer (2000), etc.

A fundamental task in geometric algebra is to compute the geometric product of vectorsand blades, the latter being the outer products of vectors. The computation generally in-volves representing the geometric product with simpler products, typically the inner productand the outer product of vectors. This procedure is called the expansion of the geometricproduct. The reverse procedure is summation.

In the literature, expansions are often carried out with respect to various coordinatesystems. On the other hand, in many applications it is crucial to do expansions and sum-mations in a coordinate-free way, at least on the symbolic level. Geometric algebra providesa convenient computational environment for this task.

In this paper, we establish some fundamental formulas on expanding the geometric prod-uct of vectors and blades, and study the properties of these expansions. We then establisha series of formulas on combinatorial summations of some tensor product and geometricproduct expressions. In details, section 3 is on dimension-free expansions and summationsof the geometric product of vectors, section 4 is on expansions and summations of vectorswith given dimension, section 5 is on null vectors, section 6 is on expansions of the geometricproduct of blades, and section 7 is on summations of some expressions of blades and vectors.

We believe that the formulas developed here should prove to be fundamental to variouscoordinate-free computing tasks involving Clifford algebra.

1)Academy of Mathematics and System Sciences, Chinese Academy of Sciences, Beijing 100080, China

Expansions and Summations in Clifford Algebra 113

2. Clifford algebra: definition and associated products

We start with a brief introduction of the definition and some associated products in Clif-ford algebra in a coordinate-free manner. For a thorough investigation we refer to Hestenesand Sobczyk (1984), Hestenes (1987). The following is a definition of Clifford algebra com-mon in mathematics literature:

Let K be a field whose characteristic is not 2. Let V be a finite dimensional inner productspace over K. The Clifford algebra G(V) generated by V is the quotient of the tensor algebra⊗(V) generated by V modulo the two-sided ideal generated by elements of the form a⊗a−a·a,where a ∈ V.

The product in G(V) is called the geometric product and is denoted by juxtaposition.While this product is associative, it is neither commutative, nor anticommutative, nor divis-ible. For example, for any two vectors a,b in V,

ab = −ba + 2a · b. (2.1)

For any vector a, if a · a = a2 6= 0, then a is invertible in that vector b = a/a2 satisfiesab = ba = 1. A vector is called a null vector if its square is zero.

An element of G(V) is called a multivector. When the dimension of V is n, the grades ofa general multivector A range from 0 up to n:

A =n∑

i=0

〈A〉i. (2.2)

Here 〈A〉i is the i-vector part of A. If A = 〈A〉i for some i, it is called an i-vector. The gradeof an i-vector is i. The i-vectors form a subspace Gi(V) of G(V), and the vector space G(V),called the Grassmann space generated by V, is the direct sum of the i-vector subspaces fori from 0 to n.

When i = 0, the following notation is often used: for any multivector A, 〈A〉 = 〈A〉0. Abasic property is that for any multivectors A,B,

〈AB〉 = 〈BA〉. (2.3)

The grade of a scalar is set to be 0, and the grade of a vector in V is set to be 1. Theouter product of a sequence of r vectors a1, . . . ,ar is defined by

a1 ∧ · · · ∧ ar =1r!

∑σr

sign(σr)aσ1r· · ·aσr

r. (2.4)

Here the summation is over all permutations σr = σ1r , . . . , σ

rr of 1, . . . , r. In particular, for

any two vectors a,b in V,ab = ba + 2a ∧ b. (2.5)

A multivector is called an r-blade if it is the outer product of r vectors. An r-vector is alinear combination of r-blades. The following are basic properties of the geometric product:

114 H. LI

let a1, . . . ,ak be vectors, let Ar be an r-vector and Bs be an s-vector, then

a1 · · ·ak =[ k2]∑

i=0

〈a1 · · ·ak〉k−2i,

ArBs =min(r,s)∑

i=0

〈ArBs〉r+s−2i.

(2.6)

The outer product of vectors can be extended in a multilinear and associative manner tomultivectors, when defining the outer product of a scalar with any multivector to be the scalarmultiplication. The outer product in Clifford algebra is identical with that in Grassmannalgebra. By definition, the Grassmann algebra (also called exterior algebra) generated by Vis the quotient of the tensor algebra ⊗(V) modulo the two-sided ideal generated by elementsof the form a ⊗ a, where a ∈ V. The product in Grassmann algebra is called the outerproduct, and satisfies

a1 ∧ · · · ∧ ar =1r!

∑σr

sign(σr)aσ1r⊗ · · · ⊗ aσr

r(2.7)

for vectors a1, . . . ,ar. The outer product of an r-vector Ar and an s-vector Bs satisfies

Ar ∧Bs = (−1)rsBs ∧Ar. (2.8)

The outer product of blades Ar1 , . . . , Arkis either zero or of grade r1 + · · ·+ rk.

The inner product of two vectors can be extended to two multivectors in a bilinear manneras follows: (1) the inner product of a scalar with any multivector is set to be zero; (2) forany r-vector Ar and s-vector Bs,

Ar ·Bs = (−1)rs−min(r,s)Bs ·Ar; (2.9)

(3) for vectors a,a1, . . . ,ar and s-vector Bs, where r ≤ s,

a · (a1 ∧ · · · ∧ ar) =r∑

i=1

(−1)i+1a · ai (a1 ∧ · · · ∧ ai ∧ · · · ∧ ar),

(a1 ∧ · · · ∧ ar) ·Bs = (a1 ∧ · · · ∧ ar−1) · (ar ·Bs).(2.10)

Here ai denotes that ai does not occur in the product. The extended inner product has theproperty that for vectors a1, . . . ,ar,b1, . . . ,br,

(a1 ∧ · · · ∧ ar) · (b1 ∧ · · · ∧ br) = (−1)r(r−1)

2 det(ai · bj)i,j=1..r. (2.11)

The inner product of an r-vector Ar and an s-vector Bs is either zero or of grade |r− s|.In particular, for any vector a and multivector A,

aA = a ·A + a ∧A. (2.12)


3. Vectors

In this section, we study the expansions of the geometric product of a sequence of vec-tors in an inner product space with unknown dimension. The following notations are usedthroughout this paper.

Let α be a sequence of indices. A set of subsequences σi1 , . . . , σir , where ij is the size ofthe subsequence σij = σ1

ij, . . . , σ

ijij

, is called a partition of α, if the σ’s are pairwise disjointand the union of them is α. In particular, if r = 2, then σi1 is called the remainder of σi2 inα, and is often denoted by σi2 .

Let a1, . . . ,ak be a sequence of vectors. Let σi1 , . . . , σir be a partition of 1, . . . , k. Thenthe subsequence of vectors aσ1

i1, . . . ,a

σi1i1

is denoted by σi1(a). When r = 2, for a function f ,

the summation ∑

1≤σi1≤k

sign(σi1 , σi1)f(σi1(a), σi1(a))

is over all partitions σi1 , σi1 of 1, . . . , k, and sign(σi1 , σi1) is the sign of permutation. Likewise,the summation ∑

{σi1}⊆{α}

sign(σi1 , σi1)f(σi1(a), σi1(a))

is over all partitions σi1 , σi1 of α. The number i1 is often omitted when it is too long. Wheni1 = 0 or #(α), the size of α, set

∑

{σ0}⊆{α}sign(σ0, σ0)f(σ0(a), σ0(a)) = f(1, α(a)),

∑

{σ#(α)}⊆{α}sign(σ#(α), σ#(α))f(σ#(α)(a), σ#(α)(a)) = f(α(a), 1).

(3.1)

The geometric product of the sequence of vectors σi1(a) is denoted by the same symbol.The tensor product and the wedge product of the sequence are denoted by

⊗σi1(a) and

Λσi1(a) respectively.

Theorem 3.1. [Fundamental theorem on the geometric product of vectors] For any 1 ≤ l ≤k/2,

〈a1 · · ·ak〉k−2l =∑

1≤σ2l≤k

sign(σ2l, σ2l)〈σ2l(a)〉〈σ2l(a)〉k−2l. (3.2)

Theorem 3.2. For any l > 0,

〈a1 · · ·a2l〉 =2l∑

i=2

(−1)ia1 · ai〈a1ai〉. (3.3)

Here 〈a1ai〉 is a shorthand notation of 〈a2 · · ·ai−1ai+1 · · ·a2l〉.Proof. We prove (3.2) and (3.3) at the same time.

First, assume that l = 1. Then (3.3) is obvious. When k = 2, (3.2) is obvious. Assumethat for k = m− 1, (3.2) is true. When k = m,

116 H. LI

〈a1 · · ·am〉m−2 = (a1 ∧ · · · ∧ am−1) · am + 〈a1 · · ·am−1〉m−3 ∧ am

=m−1∑

i=1

(−1)i+m+1(ai · am)a1 ∧ · · · ∧ ai ∧ · · · ∧ am−1

+∑

1≤i<j≤m−1

(−1)i+j+1(ai · aj)a1 ∧ · · · ∧ ai ∧ · · · ∧ aj ∧ · · · ∧ am

=∑

1≤i<j≤m

(−1)i+j+1ai · aj〈aiaj〉m−2.

So (3.2) is true for l = 1 and any k ≥ 2.Now assume that for l = n− 1 and any k ≥ 2n− 2, (3.2), (3.3) are true. For l = n,

〈a1 · · ·a2n〉 = a1 · 〈a2 · · ·a2n〉1 = a1 ·(

2n∑

i=2

(−1)i〈a1ai〉ai

).

So (3.3) is true for l = n.For l = n, (3.2) is true for k = 2n. Assume that (3.2) is true for l = n and k = m−1 ≥ 2n.

When k = m, let φ2n−2, τ = τm+1−2n be a partition of 1 . . . ,m−1, and let ψ2n, ω = ωm−1−2n

be another partition. Let i be an element of τ , and let i, τ i be a partition of τ . Denote theascending sequence formed by elements in {i,m, φ2n−2} by π2n. Then

〈a1 · · ·am〉m−2n

= 〈a1 · · ·am−1〉m−2n+1 · am + 〈a1 · · ·am−1〉m−2n−1 ∧ am

=∑

1≤φ2n−2≤m−1

sign(φ2n−2, τ)〈φ2n−2(a)〉〈τ(a)〉m+1−2n · am

+∑

1≤ψ2n≤m−1

sign(ψ2n, ω)〈ψ2n(a)〉〈ω(a)〉m−1−2n ∧ am

=∑

1≤φ2n−2≤m−1

∑

i∈{τ}sign(φ2n−2, τ) sign(τ i, i)ai · am〈φ2n−2(a)〉〈τ i(a)〉m−2n

+∑

1≤ψ2n≤m−1

sign(ψ2n, ψ2n)〈ψ2n(a)〉〈ψ2n(a)am〉m−2n

=∑

1≤π2n≤m, m∈{π2n}

∑

i∈{π2n}−{m}sign(τ i, π2n)sign(φ2n−2, i)ai · am〈τ i(a)〉m−2n 〈φ2n−2(a)〉

+∑

1≤ψ2n≤m, m/∈{ψ2n}sign(ψ2n, ψ2n)〈ψ2n(a)〉〈ψ2n(a)〉m−2n

=∑

1≤π2n≤m

sign(π2n, π2n)〈π2n(a)〉〈π2n(a)〉m−2n.

So (3.2) is true for l = n and any k ≥ 2n. Here we have used the following identity

sign(φ2n−2, τ) sign(τ i, i) = sign(φ2n−2, τi, i, m)

= sign(τ i, φ2n−2, i, m)= sign(τ i, π2n) sign(φ2n−2, i).


Remark. If σr = i1, . . . , ir is a subsequence of 1, . . . , k, then

sign(σr, σr) = (−1)r(r+1)

2+i1+···+ir . (3.4)

Lemma 3.3. For any 0 ≤ m ≤ l,

〈a1 · · ·a2l〉 =1

Cml

∑

1≤σ2m≤2l

sign(σ2m, σ2m) 〈σ2m(a)〉〈σ2m(a)〉. (3.5)

Proof. The cases m = 0 and m = l are obvious. When m = 1, by (2.3) and (3.3),

〈a1 · · ·a2l〉 =12l

2l∑

i=1

〈aiai+1 · · ·a2la1a2 · · ·ai−1〉

=12l

∑

1≤i6=j≤2l

(−1)i+j+1ai · aj 〈aiaj〉

=1l

∑

1≤i<j≤2l

(−1)i+j+1ai · aj 〈aiaj〉.

So (3.5) is true for m = 1.Assume that (3.5) is true for m−1 where m > 1. Let σ2m−2, π = π2l−2m+2 be a partition

of 1, . . . , 2l, and let ω2, ψ2l−2m be a partition of π. Denote the ascending sequence formedby elements in {ω2, σ2m−2} by τ2m. Then

〈a1 · · ·a2l〉 =1

Cm−1l

∑

1≤σ2m−2≤2l

sign(σ2m−2, π) 〈σ2m−2(a)〉〈π(a)〉

=1

Cm−1l (l −m + 1)

∑

1≤σ2m−2≤2l

∑

{ω2}⊆{π}sign(σ2m−2, π)sign(ω2, ψ2l−2m)

〈σ2m−2(a)〉〈ω2(a)〉〈ψ2l−2m(a)〉

=(m− 1)!(l −m)!

l!

∑

1≤τ2m≤2l

sign(τ2m, ψ2l−2m) 〈ψ2l−2m(a)〉 ∑

{ω2}⊆{τ2m}sign(σ2m−2, ω2)〈ω2(a)〉〈σ2m−2(a)〉

=m!(l −m)!

l!

∑

1≤τ2m≤2l

sign(τ2m, τ2m) 〈τ2m(a)〉〈τ2m(a)〉.

Corollary 3.4. For any 1 ≤ l ≤ k/2, any 0 ≤ m ≤ l,

〈a1 · · ·ak〉k−2l =1

Cml

∑

1≤σ2m≤k

sign(σ2m, σ2m) 〈σ2m(a)〉〈σ2m(a)〉k−2l. (3.6)

Lemma 3.5. For any 1 ≤ l ≤ k/2, any 0 ≤ m ≤ k − 2l,

〈a1 · · ·ak〉k−2l =1

Cmk−2l

∑

1≤σm≤k

sign(σm, σm) 〈σm(a)〉m ∧ 〈σm(a)〉k−2l−m. (3.7)

118 H. LI

Proof. The cases m = 0 and m = k − 2l are obvious. For other cases, let σ2l, πk−2l be apartition of 1, . . . , k, and let τm, ω = ωk−2l−m be a partition of πk−2l. Let φk−m be theascending sequence formed by elements in {σ2l, ω}. By (3.2) and the following identity

a1 ∧ · · · ∧ ak =1

Cmk

∑

1≤σm≤k

sign(σm, σm) 〈σm(a)〉m ∧ 〈σm(a)〉k−m (3.8)

for any 0 < m < k, we get

〈a1 · · ·ak〉k−2l

=∑

1≤σ2l≤k

sign(σ2l, πk−2l)〈σ2l(a)〉〈πk−2l(a)〉k−2l

=1

Cmk−2l

∑

1≤σ2l≤k

∑

{τm}⊆{πk−2l}sign(σ2l, πk−2l)sign(τm, ω)〈σ2l(a)〉〈τm(a)〉m ∧ 〈ω(a)〉k−2l−m

=1

Cmk−2l

∑

1≤τm≤k

sign(τm, φk−m)〈τm(a)〉m ∧∑

{σ2l}⊆{φk−m}sign(σ2l, ω)〈σ2l(a)〉〈ω(a)〉k−2l−m

=1

Cmk−2l

∑

1≤τm≤k

sign(τm, τm)〈τm(a)〉m ∧ 〈τm(a)〉k−2l−m.

The following theorem is a direct consequence of (3.6) and (3.7).

Theorem 3.6. [General theorem on the geometric product of vectors] For any 1 ≤ l ≤ k/2,any 0 ≤ m ≤ k − 2l, and any 0 ≤ n ≤ l,

〈a1 · · ·ak〉k−2l =1

Cnl Cm

k−2l

∑

1≤σm+2n≤k

sign(σm+2n, σm+2n)〈σm+2n(a)〉m ∧ 〈σm+2n(a)〉k−2l−m.

(3.9)

Corollary 3.7. Let Ai be an i-vector. Then for any 0 ≤ m ≤ l,

Ai · 〈a1 · · ·an+2l+i〉n+i

=1

Cml

∑

1≤σ2m+i≤n+2l+i

sign(σ2m+i, σ2m+i) Ai · 〈σ2m+i(a)〉i 〈σ2m+i(a)〉n.(3.10)

Proof. Let σ2l, πn+i be a partition of 1, . . . , n + 2l + i, let φ2m, ψ2l−2m be a partition of σ2l,and let τi, ωn be a partition of πn+i. Let χ2m+i be the ascending sequence formed by elementsin {φ2m, τi}. By (3.2) and (3.5),

Ai · 〈a1 · · ·an+2l+i〉n+i

=∑

1≤σ2l≤n+2l+i

sign(σ2l, πn+i) 〈σ2l(a)〉Ai · 〈πn+i(a)〉n+i

=1

Cml

∑

1≤σ2l≤n+2l+i

∑

{φ2m}⊆{σ2l}

∑

{τi}⊆{πn+i}sign(σ2l, πn+i) sign(φ2m, ψ2l−2m)

sign(τi, ωn) 〈φ2m(a)〉〈ψ2l−2m(a)〉Ai · 〈τi(a)〉i 〈ωn(a)〉n


=1

Cml

∑

1≤χ2m+i≤n+2l+i

sign(χ2m+i, χ2m+i) Ai · ∑

{φ2m}⊆{χ2m+i}sign(φ2m, τi)

〈φ2m(a)〉〈τi(a)〉i

∑

{ωn}⊆{χ2m+i}sign(ψ2l−2m, ωn) 〈ψ2l−2m(a)〉〈ωn(a)〉n

=1

Cml

∑

1≤χ2m+i≤n+2l+i

sign(χ2m+i, χ2m+i) Ai · 〈χ2m+i(a)〉i 〈χ2m+i(a)〉n.

Lemma 3.8.

a1 ∧ · · · ∧ ak =[ k2]∑

i=0

∑

1≤σ2i≤k

(−1)isign(σ2i, σ2i)〈σ2i(a)〉 σ2i(a). (3.11)

Proof. The conclusion is obvious for k = 1, 2. Assume that it is true for k < m where m > 2.Let 1 ≤ j ≤ m − 1, and let j, πm−2 be a partition of 1, . . . , m − 1. Let ω2l, χ = χm−2l−2 bea partition of πm−2. Let φ2l+2 be the ascending sequence formed by elements in {ω2l, j, m}.By the induction hypothesis and (3.3),

a1 ∧ · · · ∧ am = (a1 ∧ · · · ∧ am−1)am − (a1 ∧ · · · ∧ am−1) · am

=[m−1

2]∑

i=0

∑

1≤σ2i≤m−1

(−1)isign(σ2i, σ2i)〈σ2i(a)〉 σ2i(a)am −m−1∑

j=1

[m2

]−1∑

l=0∑

{ω2l}⊆{πm−2}(−1)lsign(πm−2, j)sign(ω2l, χ)aj · am 〈ω2l(a)〉χ(a)

=[m−1

2]∑

i=0

∑

1≤σ2i≤m, m/∈{σ2i}(−1)isign(σ2i, σ2i)〈σ2i(a)〉 σ2i(a)

−[m

2]−1∑

l=0

∑

1≤φ2l+2≤m, m∈{φ2l+2}(−1)lsign(φ2l+2, χ)〈φ2l+2(a)〉χ(a)

=[m−1

2]∑

i=0

∑

1≤σ2i≤m, m/∈{σ2i}(−1)isign(σ2i, σ2i)〈σ2i(a)〉 σ2i(a)

+[m

2]∑

i=1

∑

1≤φ2i≤m, m∈{φ2i}(−1)isign(φ2i, χ)〈φ2i(a)〉χ(a)

=[m

2]∑

i=0

∑

1≤σ2i≤m

(−1)isign(σ2i, σ2i)〈σ2i(a)〉 σ2i(a).

120 H. LI

Corollary 3.9.

〈a1 · · ·ak〉k−2l =[ k2]−l∑

i=0

∑

1≤σ2l+2i≤k

(−1)iC ll+i sign(σ2l+2i, σ2l+2i)〈σ2l+2i(a)〉 σ2l+2i(a). (3.12)

The following theorem is a direct consequence of Lemma 3.8. It establishes a connectionbetween the inner product of two blades of the same grade and the scalar part of the geometricproduct of the sequence of vectors representing the two blades.

Theorem 3.10. For any 0 ≤ m ≤ 2l,

〈a1 · · ·a2l〉 = 〈(a1 ∧ · · · ∧ am)〈am+1 · · ·a2l〉m〉

+[m

2]∑

i=1

∑

1≤σ2i≤m

(−1)i+1sign(σ2i, σ2i) 〈σ2i(a)〉〈σ2i(a)am+1 · · ·a2l〉.(3.13)

It can also be written in the following form: for any 0 < m < 2l,

(a1∧· · ·∧am) · 〈am+1 · · ·a2l〉m =[m

2]∑

i=0

∑

1≤σ2i≤m

(−1)isign(σ2i, σ2i) 〈σ2i(a)〉〈σ2i(a)am+1 · · ·a2l〉.

(3.14)

Lemma 3.11. The classical combinatorial symbol C lk is defined as follows: for any 0 ≤ l ≤ k,

C lk =

k!l!(k − l)!

. (3.15)

Now we extend it to any integers k, l so that the following rules are always satisfied:

• For any l > k, C lk = 0.

• For any integers k, l,C l

k = C lk−1 + C l−1

k−1. (3.16)

Then

1. C lk = 0, for any l < 0 ≤ k.

2. For any l ≥ k > 0,C−l−k = (−1)k+lCk−1

l−1 . (3.17)

Proof. For any k ≥ 0, from C−1k + C0

k = C0k+1, we get C−1

k = 0. Assume that for some l > 1,

C−(l−1)k = 0 for any k ≥ 0. From C−l

k + C−(l−1)k = C

−(l−1)k+1 , we get C−l

k = 0. This proves thefirst property.

For the second property, first we prove C−l−1 = (−1)l+1 for any l > 0. From C−1

−1 + C0−1 =C0

0 , we get C−1−1 = 1. Assume that C

−(l−1)−1 = (−1)l for some l > 1. From C−l

−1 + C−(l−1)−1 =

C−(l−1)0 = 0, we get C−l

−1 = (−1)l+1.


Now assume that for some k > 1, C−l−(k−1) = (−1)k+l−1Ck−2

l−1 for any l ≥ k.

From C−k−1−k + C−k

−k = C−k−(k−1), we get C

−(k+1)−k = −Ck−2

k−1 − 1 = −k. So (3.17) is true

for k and l = k + 1. Assume that for some l > k, C−(l−1)−k = (−1)k+l−1Ck−1

l−2 . From

C−l−k + C

−(l−1)−k = C

−(l−1)−(k−1), we get C−l

−k = (−1)k+lCk−2l−2 + (−1)k+lCk−1

l−2 = (−1)k+lCk−1l−1 . So

(3.17) is true for k and any l ≥ k + 1.

Theorem 3.12. Let k, l ≥ 0, and let a1, . . . ,a2k+2l be vectors. Then for any 0 ≤ m ≤ 2l,∑

m+1≤σ2l−m≤2k+2l

sign(σ2l−m, σ2l−m)〈a1 · · ·amσ2l−m(a)〉〈σ2l−m(a)〉

=[m

2]∑

i=0

∑

1≤π2i≤m

Ck−ik+l−msign(π2i, π2i)〈π2i(a)〉〈π2i(a)am+1 · · ·a2k+2l〉.

(3.18)

where the combinatorial symbol is extended in Lemma 3.11.

Proof. The case l = 0 is obvious. When m = 0, (3.18) is just (3.5). When m = 2l, the rightside of (3.18) has only one nonzero term, which corresponds to i = l and which equals theleft side.

Assume that (3.18) is true for any 0 ≤ m ≤ r − 1 where 0 < r < 2l. Let π2i, ωr−2i

be a partition of 1, . . . , r, and let φ2j , ψ = ψr−2i−2j be a partition of ωr−2i. Let χ2i+2j bethe ascending sequence formed by elements in {π2i, φ2j}. Let σ2l−r, τ2k be a partition ofr + 1, . . . , 2k + 2l. By (3.10) (3.13), (3.14) and the induction hypothesis,

122 H. LI

∑

r+1≤σ2l−r≤2k+2l

sign(σ2l−r, τ2k)〈a1 · · ·arσ2l−r(a)〉〈τ2k(a)〉

=∑


sign(σ2l−r, τ2k)(a1 ∧ · · · ∧ ar) · 〈σ2l−r(a)〉r 〈τ2k(a)〉

+[ r2]∑

i=1

∑

1≤π2i≤r

∑


(−1)i+1sign(π2i, ωr−2i)sign(σ2l−r, τ2k)

〈π2i(a)〉〈ωr−2i(a)σ2l−r(a)〉〈τ2k(a)〉

= Ckk+l−r(a1 ∧ · · · ∧ ar) · 〈ar+1 · · ·a2k+2l〉r +

[ r2]∑

i=1

∑

1≤π2i≤r

(−1)i+1sign(π2i, ωr−2i)〈π2i(a)〉

[ r2]−i∑

j=0

Ck−jk+l−r+i

∑

{φ2j}⊆{ωr−2i}sign(φ2j , ψ)〈φ2j(a)〉〈ψ(a)ar+1 · · ·a2k+2l〉

= Ckk+l−r〈a1 · · ·a2k+2l〉+ Ck

k+l−r

[ r2]∑

i=1

∑

1≤π2i≤r

(−1)isign(π2i, ωr−2i)〈π2i(a)〉

〈ωr−2i(a)ar+1 · · ·a2k+2l〉+

[ r2]∑

i=1

[ r2]−i∑

j=0

∑

1≤χ2i+2j≤r

∑

{π2i}⊆{χ2i+2j}{(−1)i+1Ck−j

k+l−r+i

sign(χ2i+2j , ψ)sign(π2i, φ2j)〈π2i(a)〉〈φ2j(a)〉〈ψ(a)ar+1 · · ·a2k+2l〉}

= Ckk+l−r〈a1 · · ·a2k+2l〉+

[ r2]∑

h=1

∑

1≤χ2h≤r

sign(χ2h, χ2h)〈χ2h(a)〉

〈χ2h(a)ar+1 · · ·a2k+2l〉(

(−1)hCkk+l−r +

h∑

i=1

(−1)i+1CihCk−h+i

k+l−r+i

).

Now we only need to prove that for any h > 0,

(−1)hCkk+l−r +

h∑

i=1


k+l−r+i = Ck−hk+l−r. (3.19)

When h = 1, (3.19) is just −Ckk+l−r +Ck

k+l−r+1 = Ck−1k+l−r. Assume that (3.19) is true for

some h− 1, where h > 1. By (3.16) and the induction hypothesis,


(−1)hCkk+l−r +

h∑

i=1


k+l−r+i

= (−1)hCkk+l−r +

h∑

i=1

(−1)i+1(Ci

h−1Ck−h+i+1k+l−r+i+1 − Ci

h−1Ck−h+i+1k+l−r+i

+Ci−1h−1C

k−h+i+1k+l−r+i − Ci−1

h−1Ck−h+i+1k+l−r+i+1

)

= (−1)hCkk+l−r +

h−1∑

i=1

(−1)i+1Cih−1C

k−(h−1)+ik+(l−r+1)+i −

h−1∑

i=1

(−1)i+1Cih−1C

k−(h−1)+ik+l−r+i

+h−1∑

i=0

(−1)i+1Cih−1C

(k+1)−(h−1)+i(k+1)+(l−r)+i −

h−1∑

i=0

(−1)i+1Cih−1C

(k+1)−(h−1)+i(k+1)+(l−r+1)+i

= (−1)hCkk+l−r + C

k−(h−1)k+l−r+1 + (−1)hCk

k+l−r+1 − (Ck−(h−1)k+l−r + (−1)hCk

k+l−r)

+ (−1)hCk+1k+l−r+1 − (−1)hCk+1

k+l−r+2

= Ck−hk+l−r.

4. n-dimensional vectors

In this section we study the expansions of the geometric product of n-dimensional vectors.

Lemma 4.1. [Cramer’s rule] Let dim(V) = n. Then

n+1∑

i=1

(−1)i+1〈a1 · · · ai · · ·an+1〉nai = 0. (4.1)

Corollary 4.2. [Generalized Cramer’s rule] Let dim(V) = n. Then

n+2l+1∑

i=1

(−1)i+1〈a1 · · · ai · · ·an+2l+1〉nai = 0. (4.2)

Proof. Let σi be the remainder of element i in 1, . . . , n + 2l + 1. Let τ2l, πn be a partition ofσi. Let ωn+1 be the ascending sequence formed by elements in {i, πn}. By (3.2) and (4.1),

n+2l+1∑

i=1

(−1)i+1〈a1 · · · ai · · ·an+2l+1〉n ai

=n+2l+1∑

i=1

∑

{τ2l}⊆{σi}sign(i, σi)sign(τ2l, πn)〈τ2l(a)〉〈πn(a)〉n ai

=∑

1≤τ2l≤n+2l+1

sign(τ2l, ωn+1)〈τ2l(a)〉 ∑

i∈{ωn+1}sign(i, πn)〈πn(a)〉n ai

= 0.

124 H. LI

Proposition 4.3. Let dim(V) = n. Let a1, . . . ,an+2l be vectors. Then for any 1 ≤ i ≤ l,∑

2≤σ2i−1≤n+2l

sign(σ2i−1, σ2i−1)〈a1σ2i−1(a)〉〈σ2i−1(a)〉n = 0. (4.3)

Proof. By (3.10), the left side of (4.3) equals Ci−1l−1a1 · 〈a2 · · ·an+2l〉n+1 = 0.

Corollary 4.4. Let dim(V) = n. Then for any 0 ≤ k ≤ l,

〈a1 · · ·an+2l〉n =1

Ckl

∑

1≤σ2k≤n+2l−1

sign(σ2k, σ2k)〈σ2k(a)〉〈σ2k(a)an+2l〉n. (4.4)

Proof. The case k = 0 is obvious. By (3.6),

〈a1 · · ·an+2l〉n=

1Ck

l

∑

1≤σ2k≤n+2l−1

sign(σ2k, σ2k)〈σ2k(a)〉〈σ2k(a)an+2l〉n

+∑

1≤σ2k−1≤n+2l−1

(−1)nsign(σ2k−1, σ2k−1)〈σ2k−1(a)an+2l〉〈σ2k−1(a)〉n .

(4.5)

By (4.3), the last term on the right side of (4.5) equals zero.

Lemma 4.5. Let dim(V) = n. Then for any 0 ≤ m ≤ n + 2l,

〈a1 · · ·an+2l〉n =1l

∑

1≤σ2≤m

sign(σ2, σ2)〈σ2(a)〉〈σ2(a)am+1 · · ·an+2l〉n

+∑

m+1≤τ2≤n+2l

sign(τ2, τ2)〈τ2(a)〉〈a1 · · ·amτ2(a)〉n .

(4.6)

Proof. By (3.6), the equality holds for m = 0. Assume that (4.6) is true for m = k − 1. By(4.3), ∑

k≤τ2≤n+2l

sign(τ2, τ2)〈τ2(a)〉〈a1 · · ·ak−1τ2(a)〉n

=∑

k+1≤τ2≤n+2l

sign(τ2, τ2)〈τ2(a)〉〈a1 · · ·akτ2(a)〉n

+n+2l∑

i=k+1

(−1)i+k+1ak · ai 〈a1 · · · ak · · · ai · · ·an+2l〉n

=∑

k+1≤τ2≤n+2l

sign(τ2, τ2)〈τ2(a)〉〈a1 · · ·akτ2(a)〉n

+k−1∑

i=1

(−1)i+k+1ak · ai 〈a1 · · · ai · · · ak · · ·an+2l〉n.


By this and the induction hypothesis, we have

〈a1 · · ·an+2l〉n=

1l

∑

1≤σ2≤k−1

sign(σ2, σ2)〈σ2(a)〉〈σ2(a)ak · · ·an+2l〉n

+∑

k≤τ2≤n+2l

sign(τ2, τ2)〈τ2(a)〉〈a1 · · ·ak−1τ2(a)〉n

=1l

∑

1≤σ2≤k, k/∈{σ2}sign(σ2, σ2)〈σ2(a)〉〈σ2(a)ak+1 · · ·an+2l〉n

+∑

k+1≤τ2≤n+2l

sign(τ2, τ2)〈τ2(a)〉〈a1 · · ·akτ2(a)〉n

+∑

1≤σ2≤k, k∈{σ2}sign(σ2, σ2)〈σ2(a)〉〈σ2(a)ak+1 · · ·an+2l〉n

.

Thus (4.6) is true for m = k.

Theorem 4.6. Let dim(V) = n. Then for any 0 ≤ m ≤ n + 2l, any 0 ≤ k ≤ l,

〈a1 · · ·an+2l〉n

=1

Ckl

∑

1≤σ2k≤m

sign(σ2k, σ2k)〈σ2k(a)〉〈σ2k(a)am+1 · · ·an+2l〉n

+k∑

i=1

(−1)i+1 Cik

Cil

∑

m+1≤τ2i≤n+2l

sign(τ2i, τ2i)〈τ2i(a)〉〈a1 · · ·amτ2i(a)〉n

=1

Ckl

∑

m+1≤τ2k≤n+2l

sign(τ2k, τ2k)〈τ2k(a)〉〈a1 · · ·amτ2k(a)〉n

+k∑

i=1

(−1)i+1 Cik

Cil

∑

1≤σ2i≤m

sign(σ2i, σ2i)〈σ2i(a)〉〈σ2i(a)am+1 · · ·an+2l〉n .

(4.7)

It can also be written as∑

1≤σ2k≤m

sign(σ2k, σ2k)〈σ2k(a)〉〈σ2k(a)am+1 · · ·an+2l〉n

=k∑

i=0

∑

m+1≤τ2i≤n+2l

(−1)iCk−il−i sign(τ2i, τ2i)〈τ2i(a)〉〈a1 · · ·amτ2i(a)〉n.

(4.8)

Proof. By Ckl Ci

k = Cil C

k−il−i , (4.8) is identical to the first equality in (4.7). For the two

equalities in (4.7), by symmetry we only need to prove the first one.The case k = 0 is obvious. By (4.6), the equality holds for k = 1 and any 0 ≤ m ≤ n+2l.

When m = n+2l, according to (3.6), the equality holds for any 1 ≤ k ≤ l; when m = n+2l−1,according to (4.4), the equality holds for any 1 ≤ k ≤ l.

126 H. LI

Assume that the equality holds for k = r − 1 and any 0 ≤ m ≤ n + 2l, where 2 ≤ r ≤ l.Below we prove it for k = r and any 0 ≤ m ≤ n + 2l − 2. There are three cases to consider:

Case 1. 2r ≤ m.Let σ2r−2, χ = χm−2r+2 be a partition of 1, . . . , m, and let τ2, π = πn+2l−m−2 be a

partition of m + 1, . . . , n + 2l. Let φ2i, ψ = ψn+2l−m−2−2i be a partition of π. By theinduction hypothesis,

〈a1 · · ·an+2l〉n

=1

Cr−1l

∑

1≤σ2r−2≤m

sign(σ2r−2, χ)〈σ2r−2(a)〉〈χ(a)am+1 · · ·an+2l〉n

+r−1∑

i=1

(−1)i+1 Cir−1

Cil

∑

m+1≤τ2i≤n+2l

sign(τ2i, π)〈τ2i(a)〉〈a1 · · ·amπ(a)〉n

=1

Cr−1l

∑

1≤σ2r−2≤m


+r−1∑

i=2

(−1)i+1 Cir−1

Cil

∑

m+1≤τ2i≤n+2l

sign(τ2i, π)〈τ2i(a)〉〈a1 · · ·amπ(a)〉n

+r

l

∑

m+1≤τ2≤n+2l

sign(τ2, π)〈τ2(a)〉〈a1 · · ·amπ(a)〉n

−1l

∑

m+1≤τ2≤n+2l

sign(τ2, π)〈τ2(a)〉〈a1 · · ·amπ(a)〉n,

(4.9)

and

〈a1 · · ·amπ(a)〉n

=1

Cr−1l−1

∑

1≤σ2r−2≤m

sign(σ2r−2, χ)〈σ2r−2(a)〉〈χ(a)π(a)〉n

+r−1∑

i=1

(−1)i+1 Cir−1

Cil−1

∑

{φ2i}⊆{π}sign(φ2i, ψ)〈φ2i(a)〉〈a1 · · ·amψ(a)〉n

.

(4.10)


First, let ω2i+2 be the ascending sequence formed by elements in {τ2, φ2i}. Then

−1l

r−1∑

i=1

(−1)i+1 Cir−1

Cil−1

∑

m+1≤τ2≤n+2l

∑

{φ2i}⊆{π}sign(τ2, π)sign(φ2i, ψ)〈τ2(a)〉

〈φ2i(a)〉〈a1 · · ·amψ(a)〉n

= −1l

r−1∑

i=1

(−1)i+1 Cir−1

Cil−1

∑

m+1≤ω2i+2≤n+2l

∑

{τ2}⊆{ω2i+2}sign(τ2, φ2i)sign(ω2i+2, ψ)

〈τ2(a)〉〈φ2i(a)〉〈a1 · · ·amψ(a)〉n

=r−1∑

i=1

(−1)i(i + 1)Ci

r−1

lCil−1

∑

m+1≤ω2i+2≤n+2l

sign(ω2i+2, ω2i+2)〈ω2i+2(a)〉

〈a1 · · ·amω2i+2(a)〉n

=r∑

i=2

(−1)i+1Ci−1

r−1

Cil

∑

m+1≤τ2i≤n+2l

sign(τ2i, τ2i)〈τ2i(a)〉〈a1 · · ·amτ2i(a)〉n.

(4.11)

Second, let ξ2, ηm−2r be a partition of χ, and let ζ2r be the ascending sequence formedby elements in {σ2r−2, ξ2}. By (4.6),

∑

m+1≤τ2≤n+2l

sign(τ2, π)〈τ2(a)〉〈χ(a)π(a)〉n = (l − r + 1)〈χ(a)am+1 · · ·an+2l〉n

−∑

{ξ2}⊆{χ}sign(ξ2, ηm−2r)〈ξ2(a)〉〈ηm−2r(a)am+1 · · ·an+2l〉n.

(4.12)

By (3.6),

1lCr−1

l−1

∑

1≤σ2r−2≤m

∑

{ξ2}⊆{χ}sign(σ2r−2, χ)sign(ξ2, ηm−2r)〈σ2r−2(a)〉〈ξ2(a)〉

〈ηm−2r(a)am+1 · · ·an+2l〉n

=1

lCr−1l−1

∑

1≤ζ2r≤m

∑

{ξ2}⊆{ζ2r}sign(ζ2r, ηm−2r)sign(σ2r−2, ξ2)〈σ2r−2(a)〉〈ξ2(a)〉

〈ηm−2r(a)am+1 · · ·an+2l〉n

=1

Crl

∑

1≤ζ2r≤m

sign(ζ2r, ζ2r)〈ζ2r(a)〉〈ζ2r(a)am+1 · · ·an+2l〉n.

(4.13)

128 H. LI

Using (4.10), (4.11), (4.12), (4.13), we get

−1l

∑

m+1≤τ2≤n+2l

sign(τ2, π)〈τ2(a)〉〈a1 · · ·amπ(a)〉n

= − 1

lCr−1l−1

∑

1≤σ2r−2≤m

∑

m+1≤τ2≤n+2l

sign(σ2r−2, χ)sign(τ2, π)〈σ2r−2(a)〉〈τ2(a)〉

〈χ(a)π(a)〉n −1

l

r−1∑

i=1

(−1)i+1 Cir−1

Cil−1

∑

m+1≤τ2≤n+2l

∑

{φ2i}⊆{π}sign(τ2, π)

sign(φ2i, ψ)〈τ2(a)〉〈φ2i(a)〉〈a1 · · ·amψ(a)〉n

= − l − r + 1

lCr−1l−1

∑

1≤σ2r−2≤m


+1

lCr−1l−1

∑

1≤σ2r−2≤m

∑

{ξ2}⊆{χ}sign(σ2r−2, χ)sign(ξ2, ηm−2r)〈σ2r−2(a)〉〈ξ2(a)〉

〈ηm−2r(a)am+1 · · ·an+2l〉n

+r∑

i=2

(−1)i+1Ci−1

r−1

Cil

∑

m+1≤τ2i≤n+2l

sign(τ2i, τ2i)〈τ2i(a)〉〈a1 · · ·amτ2i(a)〉n

= − 1

Cr−1l

∑

1≤σ2r−2≤m


+1

Crl

∑

1≤ζ2r≤m

sign(ζ2r, ζ2r)〈ζ2r(a)〉〈ζ2r(a)am+1 · · ·an+2l〉n

+r∑

i=2

(−1)i+1Ci−1

r−1

Cil

∑

m+1≤τ2i≤n+2l

sign(τ2i, τ2i)〈τ2i(a)〉〈a1 · · ·amτ2i(a)〉n.

(4.14)

Substituting (4.14) into the last term of (4.9), we get the first equality of (4.7) for k = r.Case 2. 2r − 2 ≤ m < 2r.In this case, the right side of (4.12) contains only the first term, and (4.13) does not occur.

The middle term on the right side of the last equality in (4.14) vanishes. The previous proofis still valid.

Case 3. 2r − 2 > m.In this case, the first term on the right side of the last equality in (4.9) vanishes, so does

the first term on the right side of (4.10). The proof can be finished by substituting (4.10),(4.11) into the last term of (4.9).


Corollary 4.7. [The dual of (3.13)] When dim(V) = n, for any 0 ≤ m ≤ n + 2l,

〈a1 · · ·an+2l〉n =∑

m+1≤τ2l≤n+2l

sign(τ2l, τ2l)〈τ2l(a)〉〈a1 · · ·amτ2l(a)〉n

+l∑

i=1

∑

1≤σ2i≤m

(−1)i+1sign(σ2i, σ2i)〈σ2i(a)〉〈σ2i(a)am+1 · · ·an+2l〉n.

(4.15)

Theorem 4.8. When dim(V) = n, let k, l ≥ 0, and let u = n + 2k + 2l. Define

Tk,l(a1 . . .am) =∑

m+1≤σ2k−m≤u

sign(τ2k−m, τ2k−m)〈a1 · · ·amσ2k−m(a)〉〈σ2k−m(a)〉n (4.16)

for any 0 ≤ m ≤ 2k. Then

Tk,l(a1 . . .am) =

{0, if m = 2i + 1, 0 ≤ i ≤ k − 1;C l

k+l−i〈a1 · · ·a2i〉〈a2i+1 · · ·au〉n, if m = 2i, 0 ≤ i ≤ k.

(4.17)

Proof. By (3.10),

Tk,l(a1 . . .am)

=∑

m+1≤σ2k−m≤u

[m2

]∑

i=0

sign(σ2k−m, σ2k−m)〈〈a1 · · ·am〉m−2i〈σ2k−m(a)〉m−2i〉〈σ2k−m(a)〉n

=[m

2]∑

i=0

Ck−m+ik+l−m+i 〈〈a1 · · ·am〉m−2i〈am+1 · · ·au〉n+m−2i〉n

=

{0, if m is odd;C l

k+l−h 〈a1 · · ·a2h〉〈a2h+1 · · ·au〉n, if m = 2h.

Lemma 4.9. For any 0 ≤ t ≤ r,∑

1≤σt≤r

sign(σt, σt)〈σt(a)〉t · 〈σt(a)〉r−t = 0. (4.18)

Proof. The conclusion is obvious for t = 0 and t = r. When 0 < t < r, let σt, τr−t be apartition of 1, . . . , r. Let i be an element in σt, and let its remainder in σt be σi. Let j be anelement in τr−t, and let its remainder in τr−t be τ j . Let π2 be the ascending sequence formedby {i, j}, and let ωr−2 be the ascending sequence formed by elements in {σi, τ j}. Then

130 H. LI

∑

1≤σt≤r

sign(σt, σt)〈σt(a)〉t · 〈σt(a)〉r−t

=1t

∑

1≤σt≤r

∑

i∈{σt}

∑

j∈{τr−t}sign(σt, τr−t)sign(σi, i)sign(j, τ j)ai · aj

〈〈σi(a)〉t−1〈τ j(a)〉r−t−1〉|r−2t|

=1t

∑

1≤π2≤r

∑

{σi}⊆{ωr−2}sign(π2, ωr−2)sign(σi, τ j)〈〈σi(a)〉t−1〈τ j(a)〉r−t−1〉|r−2t|

∑

j∈{π2}sign(i, j)ai · aj

= 0.

Corollary 4.10. For any s, t, k, l ≥ 0,∑

1≤σs+2k≤s+t+2k+2l

sign(σs+2k, σs+2k)〈σs+2k(a)〉s · 〈σs+2k(a)〉t = 0. (4.19)

Theorem 4.11. When dim(V) = n, let k, l ≥ 0, and let u = 2n + 2k + 2l. Define

Uk,l(a1 . . .am) =∑

m+1≤σ=σn+2k−m≤u

sign(σ, σ)〈a1 · · ·amσ(a)〉n 〈σ(a)〉n (4.20)

for any 0 ≤ m ≤ n + 2k. Then

Uk,l(a1 . . .am) =

0, if m < n;0, if m = n + 2i− 1, 1 ≤ i ≤ k;C l

k+l−i〈a1 · · ·an+2i〉n 〈an+2i+1 · · ·au〉n, if m = n + 2i, 0 ≤ i ≤ k.

(4.21)

Proof. By (4.19) and (3.6),

Uk,l(a1 . . .am)

=∑


[m2

]∑

i=0

sign(σ, σ)(〈a1 · · ·am〉m−2i ∧ 〈σ(a)〉n−m+2i)〈σ(a)〉n

=[m

2]∑

i=max(0,[m−n+12

])

∑


sign(σ, σ)〈〈a1 · · ·am〉m−2i 〈〈σ(a)〉n−m+2i〈σ(a)〉n〉m−2i〉

=

0, if m− n is odd or negative;C l

k+l−h 〈a1 · · ·an+2h〉n 〈an+2h+1 · · ·au〉n, if m− n = 2h.


Lemma 4.12. Let i, j, k, l, s, t be nonnegative integers in which i ≤ s + 2k and j ≤ t + 2l.Let u = s + t + 2k + 2l and v = s + 2k − i. Let {σi}, {τj} be non intersecting subsets of1, . . . , u of size i, j respectively. Denote by σiτj the ascending sequence formed by elementsin {1, . . . , u} − {σi, τj}. Define

Ss,tk,l (σi(a), τj(a)) =

∑

{π=πs+2k−i}⊆{σiτj}sign(π, π)〈σi(a)π(a)〉s〈τj(a)π(a)〉t. (4.22)

Then for any permutation σi of j + 1, . . . , j + i, if i + j ≤ s + 2k, then

Ss,tk,l (σi(a),a1 . . .aj) =

j∑

r=0

∑

1≤φr≤j

(−1)j(s+i)+rsign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a), ). (4.23)

Here φ†r is the reverse of the sequence φr. If i + j > s + 2k, then

Ss,tk,l (σi(a),a1 . . .aj) =

v−1∑

r=0

∑

1≤φr≤j

(−1)j(s+i)+rsign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a), )

+∑

1≤φv≤j

(−1)(j+1)(s+i)sign(φ†v, φv)〈σi(a)φ†v(a)〉s 〈φv(a)ai+j+1 · · ·au〉t.(4.24)

Proof. First, we prove that for any 0 ≤ h ≤ min(j, v),

Ss,tk,l (σi(a),a1 . . .aj) =

h∑

r=0

∑

j−h+1≤φr≤j

(−1)h(s+i)+rsign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a),a1 . . .aj−h).

(4.25)

Ss,tk,l (σi(a),a1 . . .aj−1)

=∑

{πv}⊆{j,i+j+1,i+j+2,...,u}sign(πv, πv)〈σi(a)πv(a)〉s〈a1 · · ·aj−1πv(a)〉t

= (−1)s+i∑

i+j+1≤πv≤u

sign(πv, πv)〈σi(a)πv(a)〉s〈a1 · · ·aj πv(a)〉t

+∑

i+j+1≤ωv−1≤u

sign(ωv−1, ωv−1)〈σi(a)ajωv−1(a)〉s〈a1 · · ·aj−1ωv−1(a)〉t

= (−1)s+iSs,tk,l(σi(a),a1 . . .aj) + Ss,t

k,l(σi(a)aj ,a1 . . .aj−1),

i.e.,

Ss,tk,l(σi(a),a1 . . .aj) = (−1)s+i

{Ss,t

k,l(σi(a),a1 . . .aj−1)− Ss,tk,l(σi(a)aj ,a1 . . .aj−1)

}. (4.26)

This proves (4.25) for h = 1. When h = 0, (4.25) is obvious.Assume that (4.25) is true for h = m− 1, i.e.,

Ss,tk,l (σi(a),a1 . . .aj) =

m−1∑

r=0

∑

j−m+2≤φr≤j

(−1)(m−1)(s+i)+rsign(φ†r, φr)

Ss,tk,l(σi(a)φ†r(a),a1 . . .aj−m+1).

(4.27)

132 H. LI

Substituting

Ss,tk,l(σi(a)φ†r(a),a1 . . .aj−m+1) = (−1)s+i+r{Ss,t

k,l(σi(a)φ†r(a),a1 . . .aj−m)

− Ss,tk,l(σi(a)φ†r(a)aj−m+1,a1 . . .aj−m)} (4.28)

into (4.27), we get

Ss,tk,l(σi(a),a1 . . .aj)

=m−1∑

r=0

∑

j−m+1≤φr≤j, j−m+1/∈{φr}(−1)m(s+i)+rsign(φ†r, φr)S

s,tk,l(σi(a)φ†r(a),a1 . . .aj−m)

+m∑

r=1

∑

j−m+1≤φr≤j, j−m+1∈{φr}(−1)m(s+i)+rsign(φ†r, φr)S

s,tk,l(σi(a)φ†r(a),a1 . . .aj−m)

=m∑

r=0

∑

j−m+1≤φr≤j

(−1)m(s+i)+rsign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a),a1 . . .aj−m).

This proves (4.25).When j ≤ v, setting h = j in (4.25), we get (4.23). When j > v, setting h = v in (4.25),

we get

Ss,tk,l (σi(a),a1 . . .aj) =

v∑

r=0

∑

j−v+1≤φr≤j

(−1)s+i+rsign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a),a1 . . .aj−v).

(4.29)Now we prove that for any 0 ≤ h ≤ j − v,

Ss,tk,l (σi(a),a1 . . .aj)

=v−1∑

r=0

∑

j−v−h+1≤φr≤j

(−1)(h−1)(s+i)+rsign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a),a1 · · ·aj−v−h)

+∑

j−v−h+1≤φv≤j

(−1)h(s+i)sign(φ†v, φv)〈σi(a)φ†v(a)〉s 〈a1 · · ·aj−v−hφv(a)ai+j+1 · · ·au〉t.

(4.30)When h = 0, (4.30) is just (4.29). Assume that (4.30) is true for some h ≥ 0. Substituting

(4.28) into (4.30) for m = h + v + 1, and using the following arguments,

(1)∑


(−1)h(s+i)sign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a),a1 · · ·aj−v−h−1)

=∑

j − v − h ≤ φr ≤ j,j − v − h /∈ φr

(−1)h(s+i)+rsign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a),a1 · · ·aj−v−h−1),

(2)v−1∑

r=0

∑


(−1)h(s+i)+1sign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a)aj−v−h,a1 · · ·aj−v−h−1)

=v−1∑

r=0

∑

j − v − h ≤ φr+1 ≤ j,j − v − h ∈ φr+1

(−1)h(s+i)+r+1sign(φ†r+1, φr+1)

Ss,tk,l(σi(a)φ†r+1(a),a1 · · ·aj−v−h−1)


=v−1∑

r=1

∑

j − v − h ≤ φr ≤ j,j − v − h ∈ φr

(−1)h(s+i)+rsign(φ†r, φr)Ss,tk,l(σi(a)φ†r(a),a1 · · ·aj−v−h−1)

+∑

j − v − h ≤ φv ≤ j,j − v − h ∈ φv

(−1)(h+1)(s+i)sign(φ†v, φv)Ss,tk,l(σi(a)φ†v(a),a1 · · ·aj−v−h−1),

(3)∑

j−v−h+1≤φv≤j

(−1)h(s+i)sign(φ†v, φv)〈σi(a)φ†v(a)〉s 〈a1 · · ·aj−v−hφv(a)ai+j+1 · · ·au〉t

=∑

j − v − h ≤ φv ≤ j,j − v − h /∈ φv

(−1)(h+1)(s+i)sign(φ†v, φv)Ss,tk,l(σi(a)φ†v(a),a1 · · ·aj−v−h−1),

we get (4.30) for h + 1. This proves (4.30). Setting h = j − v in (4.30), we get (4.24).

Theorem 4.13. Let dim(V) = n. Let i, j, k, l ≥ 0.

1. Let i ≤ 2k and j ≤ n + 2l. Let u = n + 2k + 2l − i − j. Then for any vectorsb1, . . . ,bj ,a1, . . . ,ai, c1, . . . , cu,

∑

1≤σ2k−i≤u

sign(σ2k−i, σ2k−i)〈a1 · · ·aiσ2k−i(c)〉〈b1 · · ·bj σ2k−i(c)〉n

= (−1)i(j+1)

min(k, [ i+j2

])∑

m=[ i+12

]

∑

1≤τ2m−i≤j

C lk+l−m sign(τ †2m−i, τ2m−i)〈a1 · · ·aiτ

†2m−i(b)〉

〈τ2m−i(b)c1 · · · cu〉n.

(4.31)

2. Let i ≤ n + 2k and j ≤ n + 2l. Let u = 2n + 2k + 2l − i − j. Then for any vectorsb1, . . . ,bj ,a1, . . . ,ai, c1, . . . , cu,

∑

1≤σ=σn+2k−i≤u

sign(σ, σ)〈a1 · · ·aiσ(c)〉n 〈b1 · · ·bj σ(c)〉n

= (−1)(n+i)(j+1)

min(k, [ i+j−n2

])∑

m=max(0, [ i+1−n2

])

∑

1≤τ=τn+2m−i≤j

C lk+l−m sign(τ †, τ)

〈a1 · · ·aiτ†(b)〉n 〈τ(b)c1 · · · cu〉n.

(4.32)

In particular, (4.32) equals zero when i + j < n.

Proof. 1. The left side of (4.31) equals S0,nk,l (a1 · · ·ai,b1 · · ·bj). By Lemma 4.12, if i+j ≤ 2k,

then

S0,nk,l (a1 · · ·ai,b1 · · ·bj) =

j∑

r=0

∑

1≤φr≤j

(−1)ij+rsign(φ†r, φr)S0,nk,l (a1 · · ·aiφ

†r(b), )

=j∑

r=0

∑

1≤φr≤j

(−1)ij+rsign(φ†r, φr)Tk,l(a1 · · ·aiφ†r(b)).

134 H. LI

Let i + r = 2m. Theni

2≤ m ≤ i + j

2. By (4.17),

S0,nk,l (a1 · · ·ai,b1 · · ·bj) =

[ i+j2

]∑

m=[ i+12

]

∑

1≤τ2m−i≤j

(−1)i(j+1)C lk+l−m sign(τ †2m−i, τ2m−i)

〈a1 · · ·aiτ†2m−i(b)〉〈τ2m−i(b)c1 · · · cu〉n.

If i + j > 2k, let i + r = 2m. Then

S0,nk,l (a1 · · ·ai,b1 · · ·bj) =

k−1∑

m=[ i+12

]

∑

1≤τ2m−i≤j

(−1)i(j+1)C lk+l−m sign(τ †2m−i, τ2m−i)

〈a1 · · ·aiτ†2m−i(b)〉〈τ2m−i(b)c1 · · · cu〉n

+∑

1≤φ2k−i≤j

(−1)i(j+1)sign(φ†2k−i, φ2k−i)

〈a1 · · ·aiφ†2k−i(b)〉〈φ2k−i(b)c1 · · · cu〉n

= (−1)i(j+1)k∑

m=[ i+12

]

∑

1≤τ2m−i≤j

C lk+l−m sign(τ †2m−i, τ2m−i)

〈a1 · · ·aiτ†2m−i(b)〉〈τ2m−i(b)c1 · · · cu〉n.

2. The proof is similar.

5. Null vectors

Null vectors play an important role in the space-time algebra (Hestenes 1966) and theconformal algebra (Sommer 2000), both of which are geometric algebras. In this section weprovide some special expansion and summation formulas on null vectors.

Lemma 5.1. (Hestenes and Sobczyk, 1984) For any vectors a1, . . . ,ak,

a1 · · ·ak + (−1)ka2 · · ·aka1 = 2a1 · (a2 · · ·ak) = 2k∑

i=2

(−1)ia1 · ai(a2 · · · ai · · ·ak). (5.1)

Proposition 5.2. [Fundamental formula on the geometric product of null vectors] If a21 = 0,

then

a1a2 · · ·aka1 = 2k∑

i=2

(−1)ia1 · ai (a2 · · · ai · · ·aka1). (5.2)

Corollary 5.3. If a21 = 0, then

a1a2a3a1 = −a1a3a2a1 = 2〈a1a2a3〉1a1. (5.3)

Corollary 5.4. If a21 = 0, then

〈a1 · · ·aka1〉k−2l−1 = 2(a1 · 〈a2 · · ·ak〉k−2l−1) ∧ a1 = 2a1 · (〈a2 · · ·ak〉k−2l−1 ∧ a1). (5.4)


In particular,〈a1a2 · · ·a2la1〉1 = 2〈a1 · · ·a2l〉a1, (5.5)

and if dim(V) = n, then

〈a1a2 · · ·an+2la1〉n−1 = 2〈a1 · · ·an+2l〉n a1. (5.6)

Proposition 5.5. If dim(V) = n and a21 = a2

n+2l−1 = 0, then

a1 · an+2l−1〈a1 · · ·an+2l−1〉n−1 = a1 ∧ an+2l−1 ∧ 〈a1 · · ·an+2l−1〉n−3. (5.7)

Proof. We have

0 = 〈a1a1a2 · · ·an+2l−1an+2l−1〉n−1

= a1 ∧ 〈a1 · · ·an+2l−1〉n−3 ∧ an+2l−1 + 〈a1〈a1 · · ·an+2l−1〉n−1an+2l−1〉n−1.

On the other hand, by

a1 ∧ 〈a1 · · ·an+2l−1〉n−1 = an+2l−1 ∧ 〈a1 · · ·an+2l−1〉n−1 = 0, (5.8)

we have

〈a1〈a1 · · ·an+2l−1〉n−1an+2l−1〉n−1 = a1 ∧ (〈a1 · · ·an+2l−1〉n−1 · an+2l−1)= (−1)na1 · an+2l−1〈a1 · · ·an+2l−1〉n−1.

Theorem 5.6. Let a21 = · · · = a2

2l+1 = 0. Denote 〈ai〉 = 〈a1 · · · ai · · · a2l+1〉. Then for any0 ≤ k ≤ 2l + 1,

∑

1≤i<j≤k

+∑

k<i<j≤2l+1

(−1)i+j+1ai · aj 〈ai〉〈aj〉 = −1

2〈a1 · · ·a2l+1〉21. (5.9)

Proof. Fix an element j of 1, . . . , 2l + 1, by (3.3) and (5.2), we get

〈aja1 · · ·a2l+1〉 =

j−1∑

i=1

+2l+1∑

i=j+1

(−1)i+1ai · aj〈ai〉 = 2

j−1∑

i=1

(−1)i+1ai · aj〈ai〉.

Soj−1∑

i=1

(−1)i+1ai · aj〈ai〉 =2l+1∑

i=j+1

(−1)i+1ai · aj〈ai〉. (5.10)

Now let j vary from k+1 to 2l+1, then sum up the corresponding equality (5.10) multipliedby (−1)j+1〈aj〉, we get

∑

i<j, j>k

(−1)i+jai · aj〈ai〉〈aj〉 =∑

i>j>k

(−1)i+jai · aj〈ai〉〈aj〉.

136 H. LI

By (3.2),

〈a1 · · ·a2l+1〉21 =∑

1≤i, j≤2l+1

(−1)i+jai · aj〈ai〉〈aj〉 = 2∑

1≤i<j≤2l+1

(−1)i+jai · aj〈ai〉〈aj〉

= 2

∑

i<j≤k

+∑

i<j, j>k

(−1)i+jai · aj〈ai〉〈aj〉.

Theorem 5.7. Let dim(V) = n, and let a21 = · · · = a2

n+2l+1 = 0. Denote 〈ai〉n =〈a1 · · · ai · · ·an+2l+1〉n. Then for any 0 ≤ k ≤ n + 2l + 1,

∑

1≤i<j≤k

−∑

k<i<j≤n+2l+1

(−1)i+j+1ai · aj 〈ai〉n 〈aj〉n = 0. (5.11)

Proof. Fix an element j of 1, . . . , n + 2l + 1, by (4.2) we get

0 = aj ∧(

n+2l+1∑

i=1

(−1)i+1ai〈ai〉n)

=

j−1∑

i=1

+n+2l+1∑

i=j+1

(−1)i+1ai · aj〈ai〉n. (5.12)

Let j vary from k+1 to n+2l+1, then sum up the geometric product of the correspondingequality (5.12) with (−1)j+1〈aj〉n, we get

∑

i<j, j>k

(−1)i+jai · aj〈ai〉n 〈aj〉n = −∑

i>j>k

(−1)i+jai · aj〈ai〉n 〈aj〉n.

On the other hand, by (4.2),

0 = (−1)n−1

(n+2l+1∑

i=1

(−1)i+1ai〈ai〉n)2

= 2∑

i<j

(−1)i+jai · aj〈ai〉n 〈aj〉n

= 2

∑

i<j≤k

+∑

i<j, j>k

(−1)i+jai · aj〈ai〉n 〈aj〉n

= 2

∑

i<j≤k

−∑

i>j>k

(−1)i+jai · aj〈ai〉n 〈aj〉n.

Proposition 5.8. Let dim(V) = n be even, and let a21 = · · · = a2

n+2l+1 = 0. Denote〈ai〉g = 〈a1 · · · ai · · ·an+2l+1〉g. Then for any 0 ≤ k ≤ n + 2l + 1,

∑

1≤i<j≤k

(−1)i+j+1ai · aj (〈ai〉n〈aj〉+ 〈ai〉〈aj〉n)

=∑

k<i<j≤n+2l+1

(−1)i+j+1ai · aj (〈ai〉n〈aj〉 − 〈ai〉〈aj〉n).(5.13)


Proposition 5.9. If a21 = a2

m+1 = 0 for some 1 ≤ m ≤ 2l − 1, then

a1 · am+1 〈a1 · · ·a2l〉

=[m−1

2]∑

i=1

∑

2≤σ2i≤m

(−1)i+1sign(σ2i, σ2i) 〈a1σ2i(a)am+1〉〈a1σ2i(a)am+1 · · ·a2l〉

+ (−1)m(a1 ∧ · · · ∧ am+1) · 〈am+1 · · ·a2la1〉m+1.

(5.14)

Proof. Apply (3.13) to 〈a1 · · ·am+1am+1am+2 · · ·a2la1〉 = 0, we get

〈a1 · · ·am+1am+1am+2 · · ·a2la1〉= (a1 ∧ · · · ∧ am+1) · 〈am+1 · · ·a2la1〉m+1

+[m+1

2]∑

i=1

∑

1≤σ2i≤m+1

(−1)i+1sign(σ2i, σ2i) 〈σ2i(a)〉〈σ2i(a)am+1 · · ·a2la1〉.

On the other hand,∑

1≤σ2i≤m+1

(−1)i+1sign(σ2i, σ2i) 〈σ2i(a)〉〈σ2i(a)am+1 · · ·a2la1〉

=∑

1≤σ2i≤m+1, {1,m+1}⊆{σ2i}(−1)i+1sign(σ2i, σ2i)〈σ2i(a)〉〈a1σ2i(a)am+1 · · ·a2l〉

=∑

2≤σ2i−2≤m

(−1)i+msign(σ2i−2, σ2i−2) 〈a1σ2i−2(a)am+1〉〈a1σ2i−2(a)am+1 · · ·a2l〉,

and in particular,∑

1≤σ2≤m+1

sign(σ2, σ2)〈σ2(a)〉〈σ2(a)am+1 · · ·a2la1〉 = (−1)m+1a1 · am+1 〈a1 · · ·a2l〉.

Lemma 5.10. When dim(V) = n,

〈a1 · · ·an+2l〉n = (−1)n−1〈a2 · · ·an+2la1〉n. (5.15)

Proof. The left side equals

a1 ∧ 〈a2 · · ·an+2l〉n−1 = (−1)n−1〈a2 · · ·an+2l〉n−1 ∧ a1 = (−1)n−1〈a2 · · ·an+2la1〉n.

Proposition 5.11. [The dual of (5.14)] If dim(V) = n and a21 = a2

n = 0, then

a1 · an 〈a1 · · ·an+2l〉n=

l∑

i=1

∑

2≤σ2i≤n−1

(−1)i+1sign(σ2i, σ2i) 〈a1σ2i(a)an〉〈a1σ2i(a)an · · ·an+2l〉n

+ 〈a1an · · ·an+2l〉〈a1 · · ·an〉n.

(5.16)

138 H. LI

Proof. Apply (4.15) to 〈a1 · · ·ananan+1 · · ·an+2la1〉n = 0 for m = n, we get

〈a1 · · ·ananan+1 · · ·an+2la1〉n= 〈an · · ·an+2la1〉〈a1 · · ·an〉n

+l+1∑

i=1

∑

1≤σ2i≤n

(−1)i+1sign(σ2i, σ2i)〈σ2i(a)〉〈σ2i(a)an · · ·an+2la1〉n.

On the other hand,∑

1≤σ2i≤n

(−1)i+1sign(σ2i, σ2i)〈σ2i(a)〉〈σ2i(a)an · · ·an+2la1〉n

=∑

1≤σ2i≤n, {1,n}⊆{σ2i}(−1)i+nsign(σ2i, σ2i)〈σ2i(a)〉〈a1σ2i(a)an · · ·an+2l〉n

=∑

2≤σ2i−2≤n−1

(−1)isign(σ2i−2, σ2i−2)〈a1σ2i−2(a)an〉〈a1σ2i−2(a)an · · ·an+2l〉n,

and in particular,∑

1≤σ2≤n

sign(σ2, σ2)〈σ2(a)〉〈σ2(a)an · · ·an+2la1〉n = −a1 · an 〈a1 · · ·an+2l〉n.

6. Blades

In this section, we study the dimension-free expansions of the geometric product of asequence of blades.

Lemma 6.1. [Laplace expansion of determinant] For any 1 ≤ l ≤ r − 1,

(a1 ∧ · · · ∧ ar) · (b1 ∧ · · · ∧ br)

=∑

1≤σl≤r

sign(σl, σl) Λσl(a) · (b1 ∧ · · · ∧ bl) Λσl(a) · (bl+1 ∧ · · · ∧ br)

=∑

1≤σl≤r

sign(σl, σl) (ar−l+1 ∧ · · · ∧ ar) · Λσl(b) (a1 ∧ · · · ∧ ar−l) · Λσl(b).

(6.1)

Theorem 6.2. For any r, s, l ≥ 0,

〈(a1 ∧ · · · ∧ ar)(b1 ∧ · · · ∧ bs)〉r+s−2l

=∑

1≤σl≤r

∑

1≤τl≤s

sign(σl, σl)sign(τl, τl) 〈Λσl(a)Λτl(b)〉Λσl(a) ∧ Λτ(b). (6.2)

Remark. For any 0 ≤ l ≤ min(r, s),

〈ArBs〉r+s−2l = (−1)rs−l〈BsAr〉r+s−2l. (6.3)


Proof. When r = 0 or s = 0 or l = 0 or l = min(r, s), (6.2) is obvious. When l > min(r, s),both sides of (6.2) are zero.

When r = 1, (6.2) is obvious. Assume that (6.2) is true for r = m− 1 and any s, l ≥ 0,where m > 1. When r = m, for any s > 1 and 0 < l < min(r, s),

〈(a1 ∧ · · · ∧ am)(b1 ∧ · · · ∧ bs)〉m+s−2l

= 〈a1(a2 ∧ · · · ∧ am)(b1 ∧ · · · ∧ bs)〉m+s−2l

−m∑

i=2

(−1)ia1 · ai 〈(a2 ∧ · · · ∧ ai ∧ · · · ∧ am)(b1 ∧ · · · ∧ bs)〉m+s−2l

= a1 · 〈(a2 ∧ · · · ∧ am)(b1 ∧ · · · ∧ bs)〉(m−1)+s−2(l−1)

+a1 ∧ 〈(a2 ∧ · · · ∧ am)(b1 ∧ · · · ∧ bs)〉(m−1)+s−2l

−m∑

i=2

(−1)ia1 · ai 〈(a2 ∧ · · · ∧ ai ∧ · · · ∧ am)(b1 ∧ · · · ∧ bs)〉(m−2)+s−2(l−1)

=∑

2≤σl−1≤m

∑

1≤τl−1≤s

sign(σl−1, σl−1)sign(τl−1, τl−1)〈Λσl−1(a)Λτl−1(b)〉

a1 · (Λσl−1(a) ∧ Λτl−1(b))+

∑

2≤ξl≤m

∑

1≤ηl≤s

sign(ξl, ξl)sign(ηl, ηl)Λξl(a) · Ληl(b)a1 ∧ Λξl(a) ∧ Ληl(b)

−m∑

i=2

∑

{πl−1}⊆{2,...,m}−{i}

∑

1≤ωl−1≤s

(−1)isign(πl−1, πl−1)sign(ωl−1, ωl−1)

a1 · ai 〈Λπl−1(a)Λωl−1(b)〉Λπl−1(a) ∧ Λωl−1(b).

(6.4)

Let σl−1, αm−l be a partition of 2, . . . , m. Let j be an element in αm−l, whose remainder inαm−l is σj . Let ψm−2 be the ascending sequence formed by elements in {σj , σl−1}. Similarly,let τl−1, β = βs−l+1 be a partition of 1, . . . , s. Let k be an element in β, whose remainder inβ is τk. Let χl be the ascending sequence formed by elements in {τl−1, k}. Then

∑

2≤σl−1≤m

∑

1≤τl−1≤s

sign(σl−1, σl−1)sign(τl−1, τl−1)〈Λσl−1(a)Λτl−1(b)〉

a1 · (Λσl−1(a) ∧ Λτl−1(b))

=∑

2≤σl−1≤m

∑

1≤τl−1≤s

∑

j∈{αm−l}sign(αm−l, σl−1)sign(τl−1, β)sign(j, σj)

a1 · aj 〈Λσl−1(a)Λτl−1(b)〉Λσj(a) ∧ Λβ(b)

+∑

2≤σl−1≤m

∑

1≤τl−1≤s

∑

k∈{β}(−1)m−lsign(αm−l, σl−1)sign(τl−1, β)sign(k, τk)

a1 · bk 〈Λσl−1(a)Λτl−1(b)〉Λαm−l(a) ∧ Λτk(b)

=m∑

j=2

∑

{σl−1}⊆{2,...,m}−{j}

∑

1≤τl−1≤s

sign(j, ψm−2)sign(σj , σl−1)sign(τl−1, β)

a1 · aj 〈Λσl−1(a)Λτl−1(b)〉Λσj(a) ∧ Λβ(b)

+∑

1≤φl≤m, 1∈{φl}

∑

1≤χl≤s

∑

k∈{χl}sign(φl, φl)sign(χl, χl)sign(τl−1, k)

a1 · bk 〈Λσl−1(a)Λτl−1(b)〉Λφl(a) ∧ Λχl(b).

(6.5)

140 H. LI

Substituting (6.5) into (6.4), using sign(j, ψm−2) = (−1)j and (6.1), we get

〈(a1 ∧ · · · ∧ am)(b1 ∧ · · · ∧ bs)〉m+s−2l

=∑

1≤φl≤m, 1∈{φl}

∑

1≤χl≤s

sign(φl, φl)sign(χl, χl)Λφl(a) · Λχl(b) Λφl(a) ∧ Λχl(b)

+∑

1≤ξl≤m, 1/∈{ξl}

∑

1≤ηl≤s

sign(ξl, ξl)sign(ηl, ηl)Λξl(a) · Ληl(b) Λξl(a) ∧ Ληl(b).

Corollary 6.3. Let Ar = a1 ∧ · · · ∧ ar, Bs = b1 ∧ · · · ∧ bs. Then for any 1 ≤ l ≤ min(r, s),

〈ArBs〉r+s−2l =∑

1≤σl≤r

sign(σl, σl)Λσl(a) ∧ (Λσl(a) ·Bs)

=∑

1≤σl≤r

sign(σl, σl)(Ar · Λσl(b)) ∧ Λσl(b).(6.6)

Lemma 6.4. For any 0 ≤ l ≤ min(r, s), any 0 ≤ m ≤ l,

〈(a1 ∧ · · · ∧ ar)(b1 ∧ · · · ∧ bs)〉r+s−2l

=1

Cml

∑

1≤σm≤r

∑

1≤τm≤s

sign(σm, σm)sign(τm, τm)〈Λσm(a)Λτm(b)〉〈Λσm(a)Λτm(b)〉r+s−2l.

(6.7)

Proof. The cases m = 0 and m = l are obvious. Below we assume that 0 < m < l.Let σl, φr−l be a partition of 1, . . . , r, and let ζm, πl−m be a partition of σl. Let ξr−m

be the ascending sequence formed by elements in {φr−l, πl−m}. Similarly, let σbl , φ

bs−l be

a partition of 1, . . . , s, and let ζbm, πb

l−m be a partition of σbl . Let ξb

s−m be the ascendingsequence formed by elements in {φb

r−l, πbl−m}. By (6.2) and (6.1),

〈(a1 ∧ · · · ∧ ar)(b1 ∧ · · · ∧ bs)〉r+s−2l

=1

Cml

∑

1≤σl≤r

∑

1≤σbl≤s

∑

{ζm}⊆{σl}

∑

{ζbm}⊆{σb

l}sign(φr−l, σl)sign(σb

l , φbs−l)sign(πl−m, ζm)

sign(πbl−m, ζb

m)〈Λζm(a)Λζbm(b)〉〈Λπl−m(a)Λπb

l−m(b)〉Λφr−l(a) ∧ Λφbs−l(b)

=1

Cml

∑

1≤ζm≤r

∑

1≤ζbm≤s

∑

{πl−m}⊆{ξr−m}

∑

{πbl−m

}⊆{ξbr−m}

sign(φr−l, πl−m)sign(ξr−m, ζm)sign(ζbm,

ξbs−m)sign(πb

l−m, φbs−l)〈Λζm(a)Λζb

m(b)〉〈Λπl−m(a)Λπbl−m(b)〉Λφr−l(a) ∧ Λφb

s−l(b)

=1

Cml

∑

1≤ζm≤r

∑

1≤ζbm≤s

sign(ζm, ζm)sign(ζbm, ζb

m)〈Λζm(a)Λζbm(b)〉〈Λζm(a)Λζb

m(b)〉r+s−2l.


Theorem 6.5. For any 0 ≤ l ≤ min(r, s), any 0 ≤ m ≤ l, any 0 ≤ i ≤ r − l, and any0 ≤ j ≤ s− l,

〈(a1 ∧ · · · ∧ ar)(ar+1 ∧ · · · ∧ ar+s)〉r+s−2l

=1

Cml Ci

r−lCjs−l

∑

1≤σm+i≤r

∑

r+1≤τm+j≤r+s

sign(σm+i, τm+j , σm+i, τm+j)

〈Λσm+i(a)Λτm+j(a)〉i+j ∧ 〈Λσm+i(a)Λτm+j(a)〉r−i+s−j−2l.

(6.8)

Proof. Let σm, φr−m be a partition of 1, . . . , r, and let ωl−m, τr−l be a partition of φr−m. Letζi, π = πr−l−i be a partition of τr−l. Let ξm+i be the ascending sequence formed by elementsin {σm, ζi}, and let η = ηr−m−i be the ascending sequence formed by elements in {ωl−m, π}.

Similarly, let σbm, φb

s−m be a partition of r + 1, . . . , r + s, let ωbl−m, τ b

s−l be a partition ofφb

s−m, and let ζbj , π

b = πbs−l−j be a partition of τ b

s−l. Let ξbm+j be the ascending sequence

formed by elements in {σbm, ζb

j}, and let ηb = ηbs−m−j be the ascending sequence formed by

elements in {ωbl−m, πb}. By (6.7) and (6.2),

〈(a1 ∧ · · · ∧ ar)(ar+1 ∧ · · · ∧ ar+s)〉r+s−2l

=1

Cml

∑

1≤σm≤r

∑

r+1≤σbm≤r+s

sign(φr−m, σm)sign(σbm, φb

s−m)〈Λσm(a)Λσbm(a)〉

〈Λφr−m(a)Λφbs−m(a)〉(r−m)+(s−m)−2(l−m)

=1

Cml

∑

1≤σm≤r

∑

r+1≤σbm≤r+s

∑

{ωl−m}⊆{φr−m}

∑

{ωbl−m

}⊆{φbs−m}

sign(φr−m, σm)sign(σbm, φb

s−m)

sign(τr−l, ωl−m)sign(ωbl−m, τ b

s−l)〈Λσm(a)Λσbm(a)〉〈Λωl−m(a)Λωb

l−m(a)〉Λτr−l(a) ∧ Λτ b

s−l(a)

=1

Cml Ci

r−lCjs−l

∑

1≤σm≤r

∑

r+1≤σbm≤r+s

∑

{ωl−m}⊆{φr−m}

∑

{ωbl−m

}⊆{φbs−m}

∑

{ζi}⊆{τr−l}

∑

{ζbj}⊆{τb

s−l}

sign(ζi, π, ωl−m, σm, σbm, ωb

l−m, ζbj , π

b)〈Λσm(a)Λσbm(a)〉〈Λωl−m(a)Λωb

l−m(a)〉Λζi(a) ∧ Λπ(a) ∧ Λζb

j (a) ∧ Λπb(a)

=1

Cml Ci

r−lCjs−l

∑

1≤ξm+i≤r

∑

r+1≤ξbm+j≤r+s

∑

{ζi}⊆{ξm+i}

∑

{ζbj}⊆{ξb

m+j}

∑

{ωl−m}⊆{η}

∑

{ωbl−m

}⊆{ηb}

sign(ζi, σm, σbm, ζb

j , π, ωl−m, ωbl−m, πb)〈Λσm(a)Λσb

m(a)〉〈Λωl−m(a)Λωbl−m(a)〉

Λζi(a) ∧ Λζbj (a) ∧ Λπ(a) ∧ Λπb(a)

=1

Cml Ci

r−lCjs−l

∑

1≤ξm+i≤r

∑

r+1≤ξbm+j≤r+s

sign(ξm+i, ξbm+j , η, ηb)

〈Λξm+i(a)Λξbm+j(a)〉(m+i)+(m+j)−2m ∧ 〈Λη(a)Ληb(a)〉(r−m−i)+(s−m−j)−2(l−m).

142 H. LI

Here we have used

sign(ζi, π, ωl−m, σm, σbm, ωb

l−m, ζbj , π

b)Λζi(a) ∧ Λπ(a) ∧ Λζbj (a) ∧ Λπb(a)

= sign(ζi, σm, σbm, ζb

j , π, ωl−m, ωbl−m, πb)Λζi(a) ∧ Λζb

j (a) ∧ Λπ(a) ∧ Λπb(a).

Proposition 6.6. Let 1 ≤ r, s, t ≤ n be integers such that r + s + t = 2n. Then

〈(a1 ∧ · · · ∧ ar)(b1 ∧ · · · ∧ bs)(c1 ∧ · · · ∧ ct)〉=

∑

1≤σn−t≤r

∑

1≤τn−r≤s

∑

1≤πn−s≤t

sign(σn−t, σn−t)sign(τn−r, τn−r)sign(πn−s, πn−s)

〈Λσn−t(a)Λτn−r(b)〉〈Λτn−r(b)Λπn−s(c)〉〈Λπn−s(c)Λσn−t(a)〉.

(6.9)

Proof. By (6.2) and (6.1),

〈(a1 ∧ · · · ∧ ar)(b1 ∧ · · · ∧ bs)(c1 ∧ · · · ∧ ct)〉= 〈(a1 ∧ · · · ∧ ar)(b1 ∧ · · · ∧ bs)〉r+s−2(n−t) · (c1 ∧ · · · ∧ ct)

=∑

1≤σn−t≤r

∑

1≤σbn−t≤s

sign(σn−t, σn−t)sign(σbn−t, σ

bn−t) 〈Λσn−t(a)Λσb

n−t(b)〉

(Λσn−t(a) ∧ Λσbn−t(b)) · (c1 ∧ · · · ∧ ct)

=∑

1≤σn−t≤r

∑

1≤σbn−t≤s

∑

1≤τn−r≤t

sign(σn−t, σn−t)sign(σbn−r, σ

bn−r)sign(τn−r, τn−r)

〈Λσn−t(a)Λσbn−t(b)〉〈Λσn−t(a)Λτn−r(c)〉〈Λσb

n−t(b)Λτn−r(c)〉.

Theorem 6.7. Let 1 ≤ r, s, t ≤ n be integers such that r + s + t = 2n. Let 0 ≤ n′ ≤ n, andlet 0 ≤ r′, s′, t′ ≤ n′ be integers such that r′+s′+ t′ = 2n′ and 0 ≤ r−r′, s−s′, t− t′ ≤ n−n′.Then

〈(a1 ∧ · · · ∧ ar)(ar+1 ∧ · · · ∧ ar+s)(ar+s+1 ∧ · · · ∧ a2n)〉

=1

Cn′−r′n−r Cn′−s′

n−s Cn′−t′n−t

∑

1≤σr′≤r

∑

r+1≤τs′≤r+s

∑

r+s+1≤πt′≤2n

sign(σr′ , τs′ , πt′ , σr′ , τs′ , πt′)

〈Λσr′(a)Λτs′(a)Λπt′(a)〉〈Λσr′(a)Λτs′(a)Λπt′(a)〉.(6.10)

Proof. Let Ct = ar+s+1 ∧ · · · ∧ a2n. By (6.8), for any 0 ≤ m ≤ n− t, any 0 ≤ i ≤ n− s, and


any 0 ≤ j ≤ n− r,

〈(a1 ∧ · · · ∧ ar)(ar+1 ∧ · · · ∧ ar+s)Ct〉= 〈(a1 ∧ · · · ∧ ar)(ar+1 ∧ · · · ∧ ar+s)〉r+s−2(n−t) · Ct

=1

Cmn−tC

in−sC

jn−r

∑

1≤σm+i≤r

∑

r+1≤τm+j≤r+s

sign(σm+i, τm+j , σm+i, τm+j)

(〈Λσm+i(a)Λτm+j(a)〉i+j ∧ 〈Λσm+i(a)Λτm+j(a)〉t−i−j) · Ct

=1

Cmn−tC

in−sC

jn−r

∑

1≤σm+i≤r

∑

r+1≤τm+j≤r+s

∑

r+s+1≤πi+j≤2n

sign(σm+i, τm+j , σm+i, τm+j ,

πi+j , πi+j)〈Λσm+i(a)Λτm+j(a)〉i+j · Λπi+j(a) 〈Λσm+i(a)Λτm+j(a)〉t−i−j · Λπi+j(a)

=1

Cmn−tC

in−sC

jn−r

∑

1≤σm+i≤r

∑

r+1≤τm+j≤r+s

∑

r+s+1≤πi+j≤2n

sign(σm+i, τm+j , πi+j , σm+i,

τm+j , πi+j)〈Λσm+i(a)Λτm+j(a)Λπi+j(a)〉〈Λσm+i(a)Λτm+j(a)Λπi+j(a)〉.Setting r′ = m + i, s′ = m + j, t′ = i + j, we get m = n′− t′, i = n′− s′, j = n′− r′, and

n′ = m + i + j, hence (6.10).

Lemma 6.8. For 1 ≤ i ≤ k, let A(i)ri be an ri-vector. Then 〈A(1)

r1 · · ·A(k)rk 〉g = 0 for any

g < 2max(r1, . . . , rk)−∑k

i=1 ri.

Proof. Let rm = max(r1, . . . , rk). For 1 ≤ i ≤ k, let si =∑i

h=1 rh. If 2rm − sk > 0, thenrm − sm−1 > sk − sm ≥ 0. Since

A(1)r1 · · ·A(k)

rk

=

sm∑

j=rm−sm−1

〈A(1)r1· · ·A(m)

rm〉j

(sk−sm∑

l=0

〈A(m+1)rm+1

· · ·A(k)rk〉l

)

=sk∑

h=rm−sm−1−sk+sm

〈

sm∑

j=rm−sm−1

〈A(1)r1· · ·A(m)

rm〉j

(sk−sm∑

l=0

〈A(m+1)rm+1

· · ·A(k)rk〉l

)〉h,

for any g < 2rm − sk = rm − sm−1 − sk + sm, we have 〈A(1)r1 · · ·A(k)

rk 〉g = 0.

Proposition 6.9. Let r1, . . . , rk be a sequence of positive integers, where ri1 ≥ ri2 ≥ ri3 ≥ rj

for any j ∈ {1, . . . , k} − {i1, i2, i3}. Let r1 + · · · + rk = sk. For 1 ≤ i ≤ k, let A(i)ri be an

ri-blade. Let the dimension of the vector space spanned by vectors in the A’s be n. ThenBk,l = 〈A(1)

r1 · · ·A(k)rk 〉sk−2l equals zero if l is not within the following range:

ri1 + ri2 − n ≤ l ≤ min(sk − ri1 , sk + n− ri1 − ri2 − ri3). (6.11)

Proof. The conclusion is obvious for k = 1. When k = 2, if r1 + r2 > n, then for any l suchthat B2,l is nonzero, r1 + r2 − 2l ≤ n− r1 + n− r2, so

max(0, r1 + r2 − n) ≤ l ≤ min(r1, r2).

144 H. LI

This proves (6.11) for k = 2.Assume that (6.11) is true for k = m− 1. Since

Bm,l =l∑

h=0

〈〈Bm−1,h〉sm−1−2hA(m)rm〉(sm−1−2h)+rm−2(l−h),

for any nonzero term in the sum, we have

max(0, sm − 2h− n) ≤ l − h ≤ min(sm−1 − 2h, rm). (6.12)

On the other hand, by the induction hypothesis, for nonzero 〈Bm−1,h〉sm−1−2h,

max(0, ri + rj − n | 1 ≤ i < j < m) ≤ h≤ min(sm−1 − ri, sm−1 + n− ri − rj − rk | 1 ≤ i < j < k < m).

(6.13)

Combining (6.12) and (6.13), we get

max(l − rm, sm − n− l, 0, ri + rj − n | 1 ≤ i < j < m) ≤ h≤ min(l, sm−1 − l, sm−1 − ri, sm−1 + n− ri − rj − rk | 1 ≤ i < j < k < m),

from which we get

max(0,sm − n

2, rm + ri − n, ri + rj − n, rm + ri + rj + rk − 2n | 1 ≤ i < j < k < m) ≤ l

≤ min(sm

2, sm−1, sm − ri, sm + n− ri − rj − rk, sm−1 + n− ri − rj | 1 ≤ i < j < k < m),

which contains (6.11) for k = m.

Corollary 6.10. Let Ar, Bs be r-blade and s-blade respectively. If Ct is a t-blade in bothAr and Bs, then for any l < t, 〈ArBs〉r+s−2l = 0.

Theorem 6.11. [Fundamental theorem on the geometric product of blades] Let r1, . . . , rk

be positive integers. For 1 ≤ i ≤ k where k > 1, let si =∑i

h=1 rh. Let a1, . . . ,askbe vectors,

and let A(i)ri = asi−1+1 ∧ · · · ∧ asi . Then for any l ≥ 0,

〈A(1)r1· · ·A(k)

rk〉sk−2l =

∑

i1 + · · ·+ ik = 2l,0≤ij≤min(l,rj), 1≤j≤k

∑

sj−1 + 1 ≤ σ(j)ij

≤ sj ,

1≤j≤k

sign(σ(1)i1

, . . . , σ(k)ik

,

σ(1)i1

, . . . , σ(k)ik

) 〈Λσ(1)i1

(a) · · ·Λσ(k)ik

(a)〉Λσ(1)i1

(a) ∧ · · · ∧ Λσ(k)ik

(a).(6.14)

Here σ(j)ij

is a subsequence of sj−1 + 1, . . . , sj with length ij .

Theorem 6.12. In the same notations as in Theorem 6.11, let sk = 2n. Then

〈A(1)r1 · · ·A(k)

rk 〉 =∑

i2 + · · ·+ ik = r1,0≤ij≤rj , 2≤j≤k

∑

sj−1 + 1 ≤ σ(j)ij

≤ sj ,

2≤j≤k

sign(σ(2)i2

, . . . , σ(k)ik

, σ(2)i2

, . . . , σ(k)ik

)

〈Λσ(2)i2

(a) · · ·Λσ(k)ik

(a)〉A(1)r1 · (Λσ

(2)i2

(a) ∧ · · · ∧ Λσ(k)ik

(a)).

(6.15)


Remark. If r1 = · · · = rk = 1, then Theorem 6.11 and Theorem 6.12 become Theorem 3.1and Theorem 3.2 respectively.

We prove Theorem 6.11 and Theorem 6.12 at the same time.

Proof. When k = 2, (6.15) is obvious, and (6.14) is just (6.2) in which i1 = i2 = l andsign(σl, σl)sign(τl, τl) = sign(σl, τl, σl, τl). Assume that both (6.14) and (6.15) hold for k =m− 1 where m > 2. When k = m,

(6.15): for sm = 2n, since r1 = r2 + · · ·+ rm − 2(n− r1),

〈A(1)r1 · · ·A(m)

rm 〉= A

(1)r1 · 〈A(2)

r2 · · ·A(m)rm 〉r1

=∑

i2 + · · ·+ im = 2n− 2r1,0≤ij≤min(n−r1,rj), 2≤j≤m

∑

sj−1 + 1 ≤ σ(j)ij

≤ sj ,

2≤j≤m

sign(σ(2)i2

, . . . , σ(m)im

, σ(2)i2

, . . . , σ(m)im

)

〈Λσ(2)i2

(a) · · ·Λσ(m)im

(a)〉A(1)r1 · (Λσ

(2)i2

(a) ∧ · · · ∧ Λσ(m)im

(a)).

Let hj = rj − ij , τ(j)hj

= σ(j)ij

for 2 ≤ j ≤ m. Then h2 + · · · + hm = r1 and max(0, rj +r1−n) ≤ hj ≤ rj for 2 ≤ j ≤ m. If rj + r1−n > 0 and hj < rj + r1−n for some 2 ≤ j ≤ m,then by Lemma 6.8, 〈Λσ

(2)r2−h2

(a) · · ·Λσ(m)rm−hm

(a)〉 = 0. Thus we get (6.15) for k = m.(6.14): by the induction hypothesis, Lemma 6.8 and (6.6),

〈A(1)r1· · ·A(m)

rm〉sm−2l

=l∑

h=0

〈〈A(1)r1· · ·A(m−1)

rm−1 〉sm−1−2hA(m)rm〉(sm−1−2h)+rm−2(l−h)

=l∑

h=0

∑

i1 + · · ·+ im−1 = 2h,0≤ij≤min(h,rj), 1≤j≤m−1

∑

sj−1 + 1 ≤ σ(j)ij

≤ sj ,

1≤j≤m−1

sign(σ(1)i1

, . . . , σ(m−1)im−1

, σ(1)i1

, . . . , σ(m−1)im−1

)

〈Λσ(1)i1

(a) · · ·Λσ(m−1)im−1

(a)〉〈(Λσ(1)i1

(a) ∧ · · · ∧ Λσ(m−1)im−1

(a))A(m)rm 〉(sm−1−2h)+rm−2(l−h)

=l∑

h=0

∑

i1 + · · ·+ im−1 = 2h,ij≥0, 1≤j≤m−1

∑

sj−1 + 1 ≤ σ(j)ij

≤ sj ,

1≤j≤m−1

∑

sm−1+1≤ω(m)l−h

≤sm

sign(σ(1)i1

, . . . , σ(m−1)im−1

, σ(1)i1

, . . . , σ(m−1)im−1

) sign(ω(m)l−h, ω

(m)l−h)〈Λσ

(1)i1

(a) · · ·Λσ(m−1)im−1

(a)〉{(Λσ

(1)i1

(a) ∧ · · · ∧ Λσ(m−1)im−1

(a)) · Λω(m)l−h(a)

}∧ Λω

(m)l−h(a).

(6.16)

146 H. LI

For 1 ≤ j ≤ m− 1, let π(j)hj

, τ (j) = τ(j)rj−ij−hj

be a partition of σ(j)ij

. Then

(Λσ(1)i1

(a) ∧ · · · ∧ Λσ(m−1)im−1

(a)) · Λω(m)l−h(a)

=∑

h1 + · · ·+ hm−1 = l − h,hj≥0, 1≤j≤m−1

sign(τ (1), . . . , τ (m−1), π(1)h1

, . . . , π(m−1)hm−1

)

{(Λπ

(1)h1

(a) ∧ · · · ∧ Λπ(m−1)hm−1

(a)) · Λω(m)l−h(a)

}Λτ (1)(a) ∧ · · · ∧ Λτ (m−1)(a).

(6.17)

For 1 ≤ j ≤ m− 1, let mj = ij + hj . Let mm = l − h. Then∑m

i=1 mi = 2l. Let ω(j)mj be

the ascending sequence formed by elements in {σ(j)ij

, π(j)hj}. By (6.15), for k = m,

〈Λω(1)m1(a) · · ·Λω

(m)mm(a)〉

=∑

h1 + · · ·+ hm−1 = mm,hj≥0, 1≤j≤m−1

∑

{σ(j)ij}⊆{ω(j)

mj}, 1≤j≤m−1

sign(σ(1)i1

, . . . , σ(m−1)im−1

, π(1)h1

, . . . , π(m−1)hm−1

)

〈Λσ(1)i1

(a) · · ·Λσ(m−1)im−1

(a)〉 (Λπ(1)h1

(a) ∧ · · · ∧ Λπ(m−1)hm−1

(a)) · Λω(m)mm(a).

(6.18)Substitute (6.17) into (6.16), use (6.18), Lemma 6.8 and the following identity for h1 +

· · ·+ hm−1 = mm,

sign(σ(1)i1

, . . . , σ(m−1)im−1

, σ(1)i1

, . . . , σ(m−1)im−1

) sign(ω(m)mm , ω

(m)mm)sign(τ (1), . . . , τ (m−1),

π(1)h1

, . . . , π(m−1)hm−1

)

= sign(σ(1)i1

, . . . , σ(m−1)im−1

, τ (1), . . . , τ (m−1), π(1)h1

, . . . , π(m−1)hm−1

, ω(m)mm , ω

(m)mm)

= sign(σ(1)i1

, . . . , σ(m−1)im−1

, π(1)h1

, . . . , π(m−1)hm−1

, ω(m)mm , τ (1), . . . , τ (m−1), ω

(m)mm)

= sign(σ(1)i1

, . . . , σ(m−1)im−1

, π(1)h1

, . . . , π(m−1)hm−1

)sign(ω(1)m1 , . . . , ω

(m)mm , ω

(1)m1 , . . . , ω

(m)mm),

we get


〈A(1)r1 · · ·A(m)

rm 〉sm−2l

=∑

m1 + · · ·+ mm = 2l,mj≥0, 1≤j≤m

∑

h1 + · · ·+ hm−1 = mm,hj≥0, 1≤j≤m−1

∑

sj−1 + 1 ≤ ω(j)mj

≤ sj ,

1≤j≤m

∑

{σ(j)ij} ⊆ {ω(j)

mj},

1≤j≤m−1

sign(σ(1)i1

, . . . , σ(m−1)im−1

, σ(1)i1

, . . . , σ(m−1)im−1

) sign(ω(m)mm , ω

(m)mm)

sign(τ (1), . . . , τ (m−1), π(1)h1

, . . . , π(m−1)hm−1

)〈Λσ(1)i1

(a) · · ·Λσ(m−1)im−1

(a)〉{(Λπ

(1)h1

(a) ∧ · · · ∧ Λπ(m−1)hm−1

(a)) · Λω(m)mm(a)

}Λτ (1)(a) ∧ · · · ∧ Λτ (m−1)(a) ∧ Λω

(m)mm(a)

=∑

m1 + · · ·+ mm = 2l,mj≥0, 1≤j≤m

∑

sj−1 + 1 ≤ ω(j)mj

≤ sj ,

1≤j≤m

sign(ω(1)m1

, . . . , ω(m)m , ω(1)

m1, . . . , ω(m)

mm)

Λω(1)m1

(a) ∧ · · · ∧ Λω(m)mm

(a){ ∑

h1 + · · ·+ hm−1 = mm,hj≥0, 1≤j≤m−1

∑

{σ(j)ij} ⊆ {ω(j)

mj},

1≤j≤m−1

sign(σ(1)i1

, . . . , σ(m−1)im−1

,

π(1)h1

, . . . , π(m−1)hm−1

)〈Λσ(1)i1

(a) · · ·Λσ(m−1)im−1

(a)〉 (Λπ(1)h1

(a) ∧ · · · ∧ Λπ(m−1)hm−1

(a)) · Λω(m)mm(a)

}

=∑

m1 + · · ·+ mm = 2l,0≤mj≤min(l,rj), 1≤j≤m

∑

sj−1 + 1 ≤ ω(j)mj

≤ sj ,

1≤j≤m

sign(ω(1)m1

, . . . , ω(m)m , ω(1)

m1, . . . , ω(m)

mm)

〈Λω(1)m1(a) · · ·Λω

(m)mm(a)〉Λω

(1)m1(a) ∧ · · · ∧ Λω

(m)mm(a).

7. Blades and vectors

In this section, we establish some beautiful formulas on combinatorial summations oftensor products and geometric products of blades and vectors.

Lemma 7.1. [Fundamental lemma] Let G be the Grassmann space generated by vectorspace V. For any nonnegative integer t, define a mapping P∧

t : G −→ G ⊗ G as follows: forany multivectors A(1), . . . , A(k), any vectors a1, . . . ,ar,

P∧t (A(1) + · · ·+ A(k)) = P∧

t (A(1)) + · · ·+ P∧t (A(k)),

P∧t (a1 ∧ · · · ∧ ar) = 0, if r < t,

P∧t (a1 ∧ · · · ∧ ar) =

∑

1≤σt≤r

sign(σt, σt)Λσt(a)⊗ Λσt(a), if r ≥ t.(7.1)

Then P∧t is a well defined linear mapping.

Proof. We only need to consider the case when r > t > 0. First, the following mappingP⊗

t : ⊗(V) −→ ⊗(V) is linear: for any elements A(1), . . . , A(k) in ⊗(V), any vectors a1, . . . ,ar

148 H. LI

in V,

P⊗t (A(1) + · · ·+ A(k)) = P⊗

t (A(1)) + · · ·+ P⊗t (A(k)),

P⊗t (a1 ⊗ · · · ⊗ ar) = 0, if r < t,

P⊗t (a1 ⊗ · · · ⊗ ar) =

∑

1≤σt≤r

sign(σt, σt)(⊗

σt(a))⊗ (⊗

σt(a)), if r ≥ t.(7.2)

Second, let τr be a permutation of 1, . . . , r. Let σt, πr−t be a partition of τr. Let σt bethe ascending sequence formed by elements in {σt}, and let πr−t be the ascending sequenceformed by elements in {πr−t}. By (2.7),

P⊗t (a1 ∧ · · · ∧ ar)

=1r!

∑

{τr}={1,...,r}sign(τr)P⊗

t (⊗

τr(a))

=1r!

∑

{τr}={1,...,r}

∑

{σt}⊆{τr}sign(σt, πr−t)sign(σt)sign(πr−t) (

⊗σt(a))⊗ (

⊗πr−t(a))

=t!(r − t)!

r!

∑

1≤σt≤r

sign(σt, πr−t)

1

t!

∑

{σt}={σt}sign(σt)

⊗σt(a)

⊗ 1

(r − t)!

∑

{πr−t}={πr−t}sign(πr−t)

⊗πr−t(a)

=1Ct

r

P∧t (a1 ∧ · · · ∧ ar).

Thus, P∧t is a well-defined linear mapping induced by P⊗

t .

Corollary 7.2. Let G be the Grassmann space generated by vector space V. Let U be alinear space, and let L : G × G −→ U be a bilinear map. Then for any 0 ≤ t ≤ r,

∑

1≤σt≤r

sign(σt, σt)L(Λσt(a),Λσt(a)) (7.3)

is a linear function of a1 ∧ · · · ∧ ar.

Theorem 7.3. For any s, t, k, l ≥ 0, let u = s + t + 2k + 2l, then∑

1≤σs+2k≤u

sign(σs+2k, σs+2k)〈σs+2k(a)〉s〈σs+2k(a)〉t = Ckk+lC

ss+t 〈a1 · · ·au〉s+t. (7.4)

Proof. Let σs+2k, τt+2l be a partition of 1, . . . , u, let π2k, ωs be a partition of σs+2k, and letξ2l, ηt be a partition of τt+2l. Let φ2k+2l be the ascending sequence formed by elements in{π2k, ξ2l}. Let ψs+t be the ascending sequence formed by elements in {ωs, ηt}. By (3.2),(3.5) and (4.18),


∑

1≤σs+2k≤u

sign(σs+2k, τt+2l)〈σs+2k(a)〉s〈τt+2l(a)〉t

=∑

1≤σs+2k≤u

∑

{π2k}⊆{σs+2k}

∑

{ξ2l}⊆{τt+2l}sign(σs+2k, τt+2l)sign(π2k, ωs)sign(ξ2l, ηt)

〈π2k(a)〉〈ξ2l(a)〉Λωs(a) Ληt(a)

=∑

1≤φ2k+2l≤u

∑

{π2k}⊆{φ2k+2l}

∑

{ωs}⊆{ψs+t}sign(φ2k+2l, ψs+t)sign(π2k, ξ2l)sign(ωs, ηt)

〈π2k(a)〉〈ξ2l(a)〉Λωs(a) ∧ Ληt(a)

= Ckk+lC

ss+t

∑

1≤φ2k+2l≤u

sign(φ2k+2l, ψs+t)〈φ2k+2l(a)〉Λψs+t(a)

= Ckk+lC

ss+t 〈a1 · · ·au〉s+t.

Theorem 7.4. Let Ar = a1 ∧ · · · ∧ ar, Bs = b1 ∧ · · · ∧ bs. Then for any 0 ≤ t ≤ r, any0 ≤ l ≤ min(t, r + s− t),

∑

1≤σt≤r

sign(σt, σt)〈Λσt(a) (Λσt(a) ∧Bs)〉r+s−2l = Ct−lr−l〈ArBs〉r+s−2l. (7.5)

For any 0 ≤ t ≤ s, any 0 ≤ l ≤ min(t, r + s− t),∑

1≤σt≤s

sign(σt, σt)〈(Ar ∧ Λσt(b)) Λσt(b)〉r+s−2l = Ct−ls−l〈ArBs〉r+s−2l. (7.6)

Proof. Let σt, τr−t be a partition of 1, . . . , r, let πl, ωt−l be a partition of σt, and let φl−i, ψ =ψr−t−l+i be a partition of τr−t. Let αi, βl−i be a partition of πl. Let χ = χr−2l+i bethe ascending sequence formed by elements in {ωt−l, ψ}, let ζ2l−2i be the ascending sequenceformed by elements in {βl−i, φl−i}, and let γr−i be the ascending sequence formed by elementsin {ζ2l−2i, χ}.

Let ξi, ηs−i be a partition of 1, . . . , s. By (6.2),∑

1≤σt≤r

sign(σt, τr−t)〈Λσt(a) (Λτr−t(a) ∧Bs)〉r+s−2l

=∑

1≤σt≤r

∑

{πl}⊆{σt}

l∑

i=0

∑

{αi}⊆{πl}

∑

{φl−i}⊆{τr−t}

∑

1≤ξi≤s

(−1)i(r−t−l+i)sign(σt, τr−t)

sign(ωt−l, πl)sign(αi, βl−i)sign(φl−i, ψ)sign(ξi, ηs−i)〈Λαi(a)Λξi(b)〉〈Λβl−i(a)Λφl−i(a)〉Λωt−l(a) ∧ Λψ(a) ∧ Ληs−i(b)

=l∑

i=0

∑

1≤ξi≤s

∑

1≤αi≤r

∑

{ζ2l−2i}⊆{γr−i}

∑

{ωt−l}⊆{χ}(−1)i(r−i)sign(ξi, ηs−i)sign(αi, γr−i)

sign(χ, ζ2l−2i)sign(ωt−l, ψ)〈Λαi(a)Λξi(b)〉Λωt−l(a) ∧ Λψ(a) ∧ Ληs−i(b) ∑

{φl−i}⊆{ξ2l−2i}sign(βl−i, φl−i)〈Λβl−i(a)Λφl−i(a)〉

.

(7.7)

150 H. LI

By (4.18), on the right side of (7.7), only the term corresponding to i = l is nonzero. So∑

1≤σt≤r

sign(σt, σt)〈Λσt(a) (Λσt(a) ∧Bs)〉r+s−2l

=∑

1≤ξl≤s

∑

1≤αl≤r

∑

{ωt−l}⊆{χr−l}(−1)l(r−l)sign(ξl, ηs−l)sign(αl, γr−l)sign(ωt−l, ψr−t)

〈Λαl(a)Λξl(b)〉Λωt−l(a) ∧ Λψr−t(a) ∧ Ληs−l(b)

= Ct−lr−l

∑

1≤ξl≤s

∑

1≤αl≤r

sign(γr−l, αl)sign(ξl, ηs−l)〈Λαl(a)Λξl(b)〉Λγr−l(a) ∧ Ληs−l(b)

= Ct−lr−l 〈ArBs〉r+s−2l.

This proves (7.5). (7.6) can be derived from (7.5) as follows:

Ct−ls−l〈ArBs〉r+s−2l = (−1)rs−lCt−l

s−l〈BsAr〉r+s−2l

= (−1)rs−l∑

1≤σt≤s

sign(σt, σt)〈Λσt(b) (Λσt(b) ∧Ar)〉t+(s−t+r)−2l

= (−1)rs−l+t(s−t+r)−l∑

1≤σt≤s

sign(σt, σt)〈(Λσt(b) ∧Ar) Λσt(b)〉r+s−2l

=∑

1≤σt≤s

sign(σt, σt)〈(Ar ∧ Λσt(b)) Λσt(b)〉r+s−2l.

Corollary 7.5. Let Ar = a1 ∧ · · · ∧ ar, and let Bs be an s-vector. Then for any 0 < t < r,

∑

1≤σt≤r

sign(σt, σt)Λσt(a) · (Λσt(a) ∧Bs) =

0, ifr + s

2< t < r;

〈ArBs〉r+s−2t, if 1 ≤ t ≤ r + s

2.

(7.8)

Corollary 7.6. Let Ar = Ct ∧ A′r−t, Bs = Ct ∧ B′s−t, where A′r−t, B′

s−t, Ct are blades ofgrade r − t, s− t, t respectively. Then

〈ArBs〉r+s−2t = Ct(A′r−t ∧Bs). (7.9)

Theorem 7.7. Let Ar = a1 ∧ · · · ∧ ar, Bs = b1 ∧ · · · ∧ bs. Let

Q =∑

1≤σt≤r

sign(σt, σt)Λσt(a) (Λσt(a) ·Bs).

If 0 ≤ t ≤ r − s, thenQ = Ct

r−sAr ·Bs. (7.10)

If max(0, r − s) ≤ t < r, then for any 0 ≤ l ≤ min(t, s− r + t),

〈Q〉s−r+2t−2l = C lr+l−t〈ArBs〉s−r+2t−2l. (7.11)


Proof. If r − t ≥ s, then Q = 〈Q〉r−s. The reason is that when expanding Q, the followingfactor occurs, in which α2h is a subsequence of 1, . . . , r:

∑

{βh}⊆{α2h}sign(βh, βh)〈Λβh(a)Λβh(a)〉. (7.12)

According to (4.18), (7.12) equals zero unless h = 0.Let σt, τr−t be a partition of 1, . . . , r, and let πs, ω = ωr−t−s be a partition of τr−t. Let

ψr−s be the ascending sequence formed by elements in {σt, ω}. The proof of (7.10) can becompleted by the following calculation:

〈Q〉r−s =∑

1≤σt≤r

sign(σt, σt)Λσt(a) ∧ (Λσt(a) ·Bs)

=∑

1≤σt≤r

∑

{πs}⊆{τr−t}sign(σt, τr−t)sign(ω, πs)(Λπs(a) ·Bs) Λσt(a) ∧ Λω(a)

=∑

1≤πs≤r

∑

{σt}⊆{ψr−s}sign(ψr−s, πs)sign(σt, ω)(Λπs(a) ·Bs) Λσt(a) ∧ Λω(a)

= Ctr−s

∑

1≤πs≤r

sign(ψr−s, πs)(Λπs(a) ·Bs) Λψr−s(a)

= Ctr−s Ar ·Bs.

When r−t ≤ s, let φl, ψt−l be a partition of σt. Let χ = χr+l−t be the ascending sequenceformed by elements in {φl, τr−t}. Similarly, let τ b

r−t, σb = σb

s−r+t be a partition of 1, . . . , s,and let φb

l , ψb = ψb

s−r+t−l be a partition of σb. Let χb = χbr+l−t be the ascending sequence

formed by elements in {τ br−t, φ

bl}. By (6.2) and (6.7),

〈Q〉s−r+2t−2l

=∑

1≤σt≤r

∑

{φl}⊆{σt}

∑

1≤τbr−t≤s

∑

{φbl}⊆{σb}

sign(σt, τr−t)sign(ψt−l, φl)sign(τ br−t, σ

b)sign(φbl , ψ

b)

〈Λφl(a) Λφbl (b)〉〈Λτr−t(a) Λτ b

r−t(b)〉Λψt−l(a) ∧ Λψb(b)

=∑

1≤ψt−l≤r

∑

{φl}⊆{χ}

∑

1≤ψb≤s

∑

{φbl}⊆{χb}

sign(ψt−l, χ)sign(φl, τr−t)sign(χb, ψb)sign(τ br−t, φ

bl )

〈Λφl(a) Λφbl (b)〉〈Λτr−t(a) Λτ b

r−t(b)〉Λψt−l(a) ∧ Λψb(b)

= C lr+l−t

∑

1≤ψt−l≤r

∑

1≤ψb≤s

sign(ψt−l, χ)sign(χb, ψb)〈Λχ(a) Λχb(b)〉Λψt−l(a) ∧ Λψb(b)

= C lr+l−t 〈ArBs〉r+s−2(r+l−t).

Theorem 7.8. Let Ar = a1 ∧ · · · ∧ ar, Bs = b1 ∧ · · · ∧ bs. Then∑

1≤σt≤r

sign(σt, σt)〈Λσt(a) Bs Λσt(a)〉r+s−2l = (−1)stb(t, r, l)〈BsAr〉r+s−2l, (7.13)

152 H. LI

where

b(t, r, l) =min(t,l)∑

i=max(0,t+l−r)

(−1)iCil C

t−ir−l (7.14)

is the coefficient of xt in the polynomial (1− x)l(1 + x)r−l.

Proof. Let σt, τr−t be a partition of 1, . . . , r, let ξi, ηt−i be a partition of σt, and let πl−i, ω =ωr−t−l+i be a partition of τr−t. Then l + t − r ≤ i ≤ min(t, l). Let φl be the ascendingsequence formed by elements in {πl−i, ξi}, and let ψr−l be the ascending sequence formed byelements in {ηt−i, ω}.

Let β1i , β2

s−l, β3l−i be a partition of 1, . . . , s. Let χl be the ascending sequence formed by

elements in {β1i , β3

l−i}. By (6.14), (4.18), (6.6) and (6.2), we get

∑

1≤σt≤r

sign(σt, σt)〈Λσt(a) Bs Λσt(a)〉r+s−2l

=∑

1≤σt≤r

min(t,l)∑

i=max(0,l+t−r)

∑

{πl−i}⊆{τr−t}

∑

{ξi}⊆{σt}

∑

{β1i ,β2

s−l,β3

l−i}={1,...,s}

sign(σt, τr−t)

sign(πl−i, ω)sign(ηt−i, ξi)sign(β1i , β2

s−l, β3l−i)〈Λξi(a)Λβ1

i (b)〉〈Λπl−i(a)Λβ3

l−i(b)〉Ληt−i(a) ∧ Λβ2s−l(b) ∧ Λω(a)

=min(t,l)∑

i=max(0,l+t−r)

∑

1≤φl≤r

∑

{ξi}⊆{φl}

∑

{ηt−i}⊆{ψr−l}

∑

1≤χl≤s

∑

{β1i }⊆{χl}

(−1)ts+isign(φl, ψr−l)

sign(πl−i, ξi)sign(ηt−i, ω)sign(β2s−l, χl)sign(β1

i , β3l−i)〈Λξi(a)Λβ1

i (b)〉〈Λπl−i(a)Λβ3

l−i(b)〉Λβ2s−l(b) ∧ Ληt−i(a) ∧ Λω(a)

=min(t,l)∑

i=max(0,l+t−r)

∑

1≤φl≤r

∑

1≤χl≤s

(−1)ts+iCil C

t−ir−l sign(β2

s−l, χl)sign(φl, ψr−l)〈Λφl(a) Λχl(b)〉

Λβ2s−l(b) ∧ Λψr−l(a)

= (−1)stb(t, r, l)〈BsAr〉r+s−2l.

Lemma 7.9. Let r > t > 0 be integers. In the Clifford algebra G(V) generated by innerproduct space V, for any vectors a1, . . . ,ar,

∑

1≤σt≤r

sign(σt, σt)σt(a)⊗ σt(a) =[ r−t

2]+[ t

2]∑

l=0

min(l,[ t2])∑

j=max(0,l−[ r−t2

])

Cjl P∧

t−2j(〈a1 · · ·ar〉r−2l). (7.15)

Proof. Let σt, τr−t be a partition of 1, . . . , r, let π2k, ωt−2k be a partition of σt, and letφ2j , ψ = ψr−t−2j be a partition of τr−t. Let l = k + j. Let χ2l be the ascending sequenceformed by elements in {π2k, φ2j}, and let ξr−2l be the ascending sequence formed by elementsin {ωt−2k, ψ}. By (3.2) and (3.5),


∑

1≤σt≤r

sign(σt, τr−t)σt(a)⊗ τr−t(a)

=∑

1≤σt≤r

[ t2]∑

k=0

[ r−t2

]∑

j=0

sign(σt, τr−t)〈σt(a)〉t−2k ⊗ 〈τr−t(a)〉r−t−2j

=[ t2]∑

k=0

[ r−t2

]∑

j=0

∑

1≤σt≤r

∑

{π2k}⊆{σt}

∑

{φ2j}⊆{τr−t}sign(π2k, ωt−2k, φ2j , ψ)〈π2k(a)〉〈φ2j(a)〉

〈ωt−2k(a)〉t−2k ⊗ 〈ψ(a)〉r−t−2j

=[ t2]∑

k=0

[ r−t2

]∑

j=0

∑

1≤χ2l≤r

∑

{π2k}⊆{χ2l}

∑

{ωt−2k}⊆{ξr−2l}sign(χ2l, ξr−2l)sign(π2k, φ2j)sign(ωt−2k, ψ)

〈π2k(a)〉〈φ2j(a)〉Λωt−2k(a)⊗ Λψ(a)

=[ r−t

2]+[ t

2]∑

l=0

min(l,[ t2])∑

k=max(0,l−[ r−t2

])

∑

1≤χ2l≤r

Ckl sign(χ2l, ξr−2l)〈χ2l(a)〉P∧

t−2k(Λξr−2l(a))

=[ r−t

2]+[ t

2]∑

l=0

min(l,[ t2])∑

k=max(0,l−[ r−t2

])

Ckl P∧

t−2k(〈a1 · · ·ar〉r−2l).

Theorem 7.10. Let a1, . . . ,ar be vectors. Then∑

1≤σt≤r

sign(σt, σt)〈σt(a)σt(a)〉r−2l = c(t, r, l)〈a1 · · ·ar〉r−2l, (7.16)

where

c(t, r, l) =min(l,[ t

2])∑

i=max(0,l+[ t−r+12

])

Cil C

t−2ir−2l (7.17)

is the coefficient of xt in the polynomial (1 + x2)l(1 + x)r−2l.

Proof. Let f : G(V) ⊗ G(V) −→ G(V) be defined by f(A ⊗ B) = AB for any multivectorsA,B. By (7.15) and (7.4),

∑

1≤σt≤r

sign(σt, σt)σt(a)σt(a) = f(∑

1≤σt≤r

sign(σt, σt)σt(a)⊗ σt(a))

=[ r−t

2]+[ t

2]∑

l=0

min(l,[ t2])∑


])

Cjl (f ◦ P∧

t−2j)(〈a1 · · ·ar〉r−2l)

=[ r−t

2]+[ t

2]∑

l=0

min(l,[ t2])∑


])

Cjl C

t−2jr−2l 〈a1 · · ·ar〉r−2l.

154 H. LI

Corollary 7.11.

∑

1≤σt≤2l

sign(σt, σt)〈σt(a)σt(a)〉 =

{0, if t is odd;Ck

l 〈a1 · · ·a2k〉, if t = 2k.(7.18)

Proof. If t = 2k−1, then c(t, 2l, l) =∑k−1

i=k Cil = 0. If t = 2k, then c(t, 2l, l) =

∑ki=k Ci

l = Ckl .

By (7.16) we get the conclusion.

8. Acknowledgments

The work is supported partially by the NKBRSF Grant G1998030600 of China, theQiushi Science and Technology Foundations of Hong Kong and the Alexander von HumboldtFoundations of Germany. Part of the work was carried out during the author’s visit at theUniversity of Kiel, Germany.

References

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[3] C. Doran, D. Hestenes, F. Sommen, N. Acker (1993): Lie Groups as Spin Groups. J. Math.Phys. 34(8): 3642–3669.

[4] P. Doubilet, G. Rota and J. Stein (1974): On the Foundations of Combinatorial Theory IX:Combinatorial Methods in Invariant Theory. Stud. Appl. Math. 57: 185–216.

[5] D. Hestenes (1966): Space-Time Algebra. Gordon and Breach, New York.[6] D. Hestenes and G. Sobczyk (1984): Clifford Algebra to Geometric Calculus. D. Reidel, Dor-

drecht, Boston.[7] D. Hestenes (1987): New Foundations for Classical Mechanics. D. Reidel, Dordrecht, Boston.[8] H. Li (1997): Hyperbolic Geometry with Clifford Algebra. Acta Appl. Math. 48(3): 317–358.[9] H. Li (2001): Hyperbolic Conformal Geometry with Clifford Algebra. International J. of Theo-

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