6.4 Grassmann algebras

Multivectors

Hermann Grassmann took a completely different direction to generalize Hamilton’s quaternion algebra (1844), one in which there is no need for an inner product. Grassmann’s idea was to regard entire subspaces of a vector space as single algebraic objects and to define a method of multiplying them together. The resulting algebra has far-reaching applications, particularly in differential geometry and the theory of integration on manifolds (see Chapters 16 and 17). In this chapter we present Grassmann algebras in a rather intuitive way, leaving a more formal presentation to Chapter 8.

Let $V$ be any real vector space. For any pair of vectors $u$ , $v \in V$ define an abstract quantity $u \land v$ . , subject to the following identifications for all vectors $u$ , $v$ , $w \in V$ and scalars $a$ , $b \in R$

(a u + b v) \land w = a u \land w + b v \land w, (6.8)

u \land v = - v \land u . (6.9)

Any quantity $u \land v$ will be known as a simple 2-vector or bivector. Taking into account the identities Eq. (6.8) and Eq. (6.9), we denote by $Λ^{2} (V)$ the vector space generated by the set of all simple 2-vectors. Without fear of confusion, we denote vector addition in $Λ^{2} (V)$ by the same symbol as for the vector space $V$ . Every element $A$ of $Λ^{2} (V)$ , generically known as a 2-vector, can be written as a sum of bivectors,

A = i = 1 \sum r a_{i} u_{i} \land v_{i} = a_{1} u_{1} \land v_{1} + a_{2} u_{2} \land v_{2} + \dots + a_{r} u_{r} \land v_{r} . (6.10)

From Eq. (6.8) and Eq. (6.9) the wedge operation is obviously linear in the second argument,

u \land (a v + b w) = a u \land v + b u \land w . (6.11)

Also, for any vector $u$ ,

u \land u = - u \land u ⟹ u \land u = 0.

As an intuitive aid it is useful to think of a simple 2-vector $u \land v$ as representing the subspace or ‘area element’ spanned by $u$ and $v$ . If $u$ and $v$ are proportional to each other, $v = a u$ , the area element collapses to a line and vanishes, in agreement with $u \land v = a u \land u = 0$ . No obvious geometrical picture presents itself for non-simple 2-vectors, and a sum of simple bivectors such as that in Eq. (6.10) must be thought of in a purely formal way.

If $V$ has dimension $n$ and ${e_{1}, \dots, e_{n}}$ is a basis of $V$ then for any pair of vectors $u = u^{i} e_{i}, v = v^{j} e_{j}$

u \land v = u^{i} v^{j} e_{i} \land e_{j} = - v^{j} u^{i} e_{j} \land e_{i} .

Setting

e_{ij} = e_{i} \land e_{j} = - e_{j i} (i, j = 1, 2, \dots, n)

it follows from Eqs. Eq. (6.10) and Eq. (6.8) that all elements of $Λ^{2} (V)$ can be written as a linear combination of the bivectors $e_{ij}$

A = A^{ij} e_{ij} .

Furthermore, since $e_{ij} = - e_{j i}$ , each term in this sum can be converted to a sum of terms

A = i < j \sum (A^{ij} - A^{j i}) e_{ij},

and the space of 2-vectors, $Λ^{2} (V)$ , is spanned by the set

E_{2} = {e_{ij} ∣ 1 \leq i < j \leq n} .

As $E_{2}$ consists of $(2 n)$ elements it follows that

dim (Λ^{2} (V)) \leq (2 n) = \frac{n ( n - 1 )}{2} .

In the tensorial approach of Chapter 8 it will emerge that the set $E_{2}$ is linearly independent, whence

dim (Λ^{2} (V)) = (2 n) .

The space of $r$ -vectors, $Λ^{r} (V)$ , is defined in an analogous way. For any set of $r$ vectors $u_{1}, u_{2}, \dots, u_{r}$ , let the simple $r$ -vector spanned by these vectors be defined as the abstract object $u_{1} \land u_{2} \land \dots \land u_{r}$ , and define a general $r$ -vector to be a formal linear sum of simple $r$ -vectors,

A = J = 1 \sum N a_{J} u_{J 1} \land u_{J 2} \land \dots \land u_{J r} where a_{J} \in R, u_{J i} \in V .

In forming such sums we impose linearity in the first argument,

(a u_{1} + b u_{1}^{'}) \land u_{2} \land \dots \land u_{r} = a u_{1} \land u_{2} \land \dots \land u_{r} + b u_{1}^{'} \land u_{2} \land \dots \land u_{r}, (6.12)

and skew symmetry in any pair of vectors,

u_{1} \land \dots \land u_{i} \land \dots \land u_{j} \land \dots \land u_{r} = - u_{1} \land \dots \land u_{j} \land \dots \land u_{i} \land \dots \land u_{r} . (6.13)

As for 2-vectors, linearity holds on each argument separately

u_{1} \land \dots \land (a u_{i} + b u_{i}^{'}) \land \dots \land u_{r} = a u_{1} \land \dots \land u_{i} \land \dots \land u_{r} + b u_{1} \land \dots \land u_{i}^{'} \land \dots \land u_{r},

and for a general permutation $π$ of $1, 2, \dots, r$

u_{1} \land u_{2} \land \dots \land u_{r} = (- 1)^{π} u_{π (1)} \land u_{π (2)} \land \dots \land u_{π (r)} . (6.14)

If any two vectors among $u_{1}, \dots, u_{r}$ are equal then $u_{1} \land u_{2} \land \dots \land u_{r}$ vanishes,

u_{1} \land \dots \land u_{i} \land \dots \land u_{i} \land \dots \land u_{r} = - u_{1} \land \dots \land u_{i} \land \dots \land u_{i} \land \dots \land u_{r} = 0. (6.15)

Again, it is possible to think of a simple $r$ -vector as having the geometrical interpretation of an $r$ -dimensional subspace or volume element spanned by the vectors $u_{1}, \dots, u_{r}$ . The general $r$ -vector is a formal sum of such volume elements.

If $V$ has dimension $n$ and ${e_{1}, \dots, e_{n}}$ is a basis of $V$ , it is convenient to define the $r$ -vectors

e_{i_{1} i_{2} \dots i_{r}} = e_{i_{1}} \land e_{i_{2}} \dots \land e_{i_{r}} .

For any permutation $π$ of $1, 2, \dots, r$ Eq. (6.14) implies that

e_{i_{1} i_{2} \dots i_{r}} = (- 1)^{π} e_{i_{π (1)} i_{π (2)} \dots i_{π (r)}},

while if any pair of indices are equal, say $i_{a} = i_{b}$ for some $1 \leq a < b \leq r$ , then $e_{i_{1} i_{2} \dots i_{r}} = 0$ For example

e_{123} = - e_{321} = e_{231}, etc. e_{112} = e_{233} = 0, etc.

By a permutation of vectors $e_{j}$ the $r$ -vector $e_{i_{1} i_{2} \dots i_{r}}$ may be brought to a form in which $i_{1} < i_{2} < \dots < i_{r}$ , to within a possible change of sign. Since the simple $r$ -vector spanned by vectors $u_{1} = u_{1}^{i} e_{i}, \dots, u_{r} = u_{r}^{i} e_{i}$ is given by

u_{1} \land u_{2} \land \dots \land u_{r} = u_{1}^{i_{1}} u_{2}^{i_{2}} \dots u_{r}^{i_{r}} e_{i_{1} i_{2} \dots i_{r}},

the vector space $Λ^{r} (V)$ is spanned by the set

E_{r} = {e_{i_{1} i_{2} \dots i_{r}} ∣ 1 \leq i_{1} < i_{2} < \dots < i_{r} \leq n} .

As for the case $r = 2$ , every $r$ -vector $A$ can be written, using the summation convention,

A = A^{i_{1} i_{2} \dots i_{r}} e_{i_{1} i_{2} \dots i_{r}}, (6.16)

which can be recast in the form

A = i_{1} < i_{2} < \dots < i_{r} \sum \tilde{A}^{i_{1} i_{2} \dots i_{r}} e_{i_{1} i_{2} \dots i_{r}} (6.17)

where

\tilde{A}^{i_{1} i_{2} \dots i_{r}} = σ \sum (- 1)^{σ} A^{i_{σ (1)} i_{σ (2)} \dots i_{σ (r)}} .

When written in the second form, the components $\tilde{A}^{i_{1} i_{2} \dots i_{r}}$ are totally skew symmetric,

\tilde{A}^{i_{1} i_{2} \dots i_{r}} = (- 1)^{π} \tilde{A}^{i_{π (1)} i_{π (2)} \dots i_{π (r)}}

for any permutation $π$ of $(1, 2, \dots, r)$ . As there are no further algebraic relationships present with which to simplify the $r$ -vectors in $E_{r}$ , we may again assume that the $e_{i_{1} i_{2} \dots i_{r}}$ are linearly independent. The dimension of $Λ^{r} (V)$ ) is then the number of ways in which $r$ values can be selected from the $n$ index values ${1, 2, \dots, n}$ , i.e.

dim Λ^{r} (V) = (r n) = \frac{n !}{r ! ( n - r )!} .

For $r > n$ the dimension is zero, $dim Λ^{r} (V) = 0$ , since each basis $r$ -vector $e_{i_{1} i_{2} \dots i_{r}} = e_{i_{1}} \land e_{i_{2}} \land \dots \land e_{i_{r}}$ must vanish by Eq. (6.15) since some pair of indices must be equal.

Exterior product

Setting the original vector space $V$ to be $Λ^{1} (V)$ and denoting the field of scalars by $Λ^{0} (V) = R$ , we define the vector space

Λ (V) = Λ^{0} (V) \oplus Λ^{1} (V) \oplus Λ^{2} (V) \oplus \dots \oplus Λ^{n} (V) .

The elements of $Λ (V)$ , called multivectors, can be uniquely written in the form

A = A_{0} + A_{1} + A_{2} + \dots + A_{n} where A_{r} \in Λ^{r} (V) .

The dimension of $Λ (V)$ is found by the binomial theorem,

dim (Λ (V)) = r = 0 \sum n (r n) = (1 + 1)^{n} = 2^{n} . (6.18)

Define a law of composition $A \land B$ for any pair of multivectors $A$ and $B$ , called exterior product, satisfying the following rules:

(EP1) If $A = a \in R = Λ^{0} (V)$ and $B \in Λ^{r} (V)$ the exterior product is defined as scalar multiplication, $a \land B = B \land a = a B$

(EP2) If $A = u_{1} \land u_{2} \land \cdot \cdot \cdot \land u_{r}$ is a simple $r$ -vector and $B = v_{1} \land v_{2} \land \cdot \cdot \cdot \land v_{s}$ a simple $s$ -vector then their exterior product is defined as the simpl $(r + s)$ -vector

A \land B = u_{1} \land \dots \land u_{r} \land v_{1} \land \dots \land v_{s} .

(EP3) The exterior product $A \land B$ is linear in both arguments,

(a A + b B) \land C = a A \land C + b B \land C, A \land (a B + b C) = a A \land B + b A \land C .

Property (EP3) makes $Λ (V)$ into an algebra with respect to exterior product, called the Grassmann algebra or exterior algebra over $V$ .

By (EP2), the product of a basis $r$ -vector and basis $s$ -vector is

e_{i_{1} \dots i_{r}} \land e_{j_{1} \dots j_{s}} = e_{i_{1} \dots i_{r} j_{1} \dots j_{s}}, (6.19)

and since

e_{i_{1} i_{2} \dots i_{r}} \land (e_{j_{1} j_{2} \dots j_{s}} \land e_{k_{1} k_{2} \dots k_{t}}) = (e_{i_{1} i_{2} \dots i_{r}} \land e_{j_{1} j_{2} \dots j_{s}}) \land e_{k_{1} k_{2} \dots k_{t}} = e_{i_{1} i_{2} \dots i_{r} j_{1} j_{2} \dots j_{s} k_{1} k_{2} \dots k_{t}}

the associative law follows for all multivectors by the linearity condition (EP3),

A \land (B \land C) = (A \land B) \land C for all A, B, C \in Λ (V) .

Thus $Λ (V)$ is an associative algebra. The property that the product of an $r$ -vector and an $s$ -vector always results in an $(r + s)$ -vector is characteristic of what is commonly called a graded algebra.

Example 6.5

General products of multivectors are straightforward to calculate from the exterior products of basis elements. Some simple examples are
$e_{i} \land e_{j} = e_{ij} = - e_{j i} = - e_{j} \land e_{i},$ $e_{i} \land e_{i} = 0 for all i = 1, 2, \dots, n,$ $e_{1} \land e_{23} = e_{123} = - e_{132} = - e_{213} = - e_{2} \land e_{13},$ $e_{14} \land e_{23} = e_{1423} = - e_{1324} = - e_{13} \land e_{24} = e_{1234} = e_{12} \land e_{34},$ $e_{24} \land e_{14} = e_{2414} = 0,$ $(a e_{1} + b e_{23}) \land (a^{'} + b^{'} e_{34}) = a a^{'} e_{1} + a^{'} b e_{23} + a b^{'} e_{134} .$

Properties of exterior product

If $A$ is an $r$ -vector and $B$ an $s$ -vector, they satisfy the ‘anticommutation rule

A \land B = (- 1)^{r s} B \land A . (6.20)

Since every $r$ -vector is by definition a linear combination of simple $r$ -vectors it is only necessary to prove Eq. (6.20) for simple $r$ -vectors and $s$ -vectors

A = x_{1} \land x_{2} \land \dots \land x_{r}, B = y_{1} \land y_{2} \land \dots \land y_{s} .

Successively perform $r$ interchanges of positions of each vector $y_{i}$ to bring it in front of $x_{1}$ , and we have

A \land B = x_{1} \land x_{2} \land \dots \land x_{r} \land y_{1} \land y_{2} \land \dots \land y_{s} = (- 1)^{r} y_{1} \land x_{1} \land x_{2} \land \dots \land x_{r} \land y_{2} \land \dots \land y_{s} = (- 1)^{2 r} y_{1} \land y_{2} \land x_{1} \land x_{2} \land \dots \land x_{r} \land y_{3} \land \dots \land y_{s} = \dots = (- 1)^{sr} y_{1} \land y_{2} \land \dots \land y_{s} \land x_{1} \land x_{2} \land \dots \land x_{r} = (- 1)^{r s} B \land A,

as required. Hence an $r$ -vector and an $s$ -vector anticommute, $A \land B = - B \land A$ , if both $r$ and $s$ are odd. They commute if either one of them has even degree.

The following theorem gives a particularly quick and useful method for deciding whether or not a given set of vectors is linearly independent.

Theorem 6.2 · Linear dependence and vanishing exterior products

Vectors $u_{1}, u_{2}, \dots, u_{r}$ are linearly dependent if and only if their wedge product vanishes,
$u_{1} \land u_{2} \land \dots \land u_{r} = 0.$

Proof

If the vectors are linearly dependent then without loss of generality we may assume that $u_{1}$ is a linear combination of the others,

u_{1} = a^{2} u_{2} + a^{3} u_{3} + \dots + a^{r} u_{r} .

Hence

u_{1} \land u_{2} \land \dots \land u_{r} = \sum_{i = 2}^{r} a^{i} u_{i} \land u_{2} \land \dots \land u_{r} = \sum_{i = 2}^{r} \pm a^{i} u_{2} \land \dots \land u_{i} \land u_{i} \land \dots \land u_{r} = 0.

This proves the only if part of the theorem.

Conversely, suppose $u_{1}, \dots, u_{r} (r \leq n)$ are linearly independent. By Theorem 3.7 there exists a basis ${e_{j}}$ of $V$ such that

e_{1} = u_{1}, e_{2} = u_{2}, \dots, e_{r} = u_{r} .

Since $e_{1} \land e_{2} \land \dots \land e_{r}$ is a basis vector of $Λ^{r} (V)$ it cannot vanish.

Example 6.6

If $e_{1}, e_{2}$ and $e_{3}$ are three basis vectors of a vector space $V$ then the vectors $e_{1} + e_{3}, e_{2} + e_{3}, e_{1} + e_{2}$ are linearly independent, for
$(e_{1} + e_{3}) \land (e_{2} + e_{3}) \land (e_{1} + e_{2}) = (e_{12} + e_{13} + e_{32}) \land (e_{1} + e_{2}) = e_{132} + e_{321} = 2 e_{132} \neq = 0.$
On the other hand, the vectors $e_{1} - e_{3}, e_{2} - e_{3}, e_{1} - e_{2}$ are linearly dependent since
$(e_{1} - e_{3}) \land (e_{2} - e_{3}) \land (e_{1} - e_{2}) = (e_{12} - e_{13} - e_{32}) \land (e_{1} - e_{2}) = e_{132} - e_{321} = 0.$
We return to the subject of exterior algebra in Chapter 8.

Problems

Problem 6.9

Let ${e_{1}, e_{2}, \dots, e_{n}}$ be a basis of a vector space $V$ of dimension $n \geq 5$ . By calculating their wedge product, decide whether the following vectors are linearly dependent or independent
$e_{1} + e_{2} + e_{3}, e_{2} + e_{3} + e_{4}, e_{3} + e_{4} + e_{5}, e_{1} + e_{3} + e_{5} .$
Can you find a linear relation among them?

Problem 6.10

Let $W$ be a vector space of dimension 4 and ${e_{1}, e_{2}, e_{3}, e_{4}}$ a basis. Let $A$ be the 2-vector on $W$
$A = e_{2} \land e_{1} + a e_{1} \land e_{3} + e_{2} \land e_{3} + c e_{1} \land e_{4} + b e_{2} \land e_{4} .$
Write out explicitly the equations $A \land u = 0$ where $u = u^{1} e_{1} + u^{2} e_{2} + u^{3} e_{3} + u^{4} e_{4}$ and show that they have a non-trivial solution if and only if $c = ab$ . In this case find two vectors $u$ and $v$ such tha $A = u \land v$ .

Problem 6.11

Let $U$ be a subspace of $V$ spanned by linearly independent vectors ${u_{1}, u_{2}, \dots, u_{p}}$

(a) Show that the $p$ -vector $E_{U} = u_{1} \land u_{2} \land \cdot \cdot \cdot \land u_{p}$ is defined uniquely up to a factor by the subspace $U$ in the sense that if ${u_{1}^{'}, u_{2}^{'}, \dots, u_{p}^{'}}$ is any other linearly independent set spanning $U$ then the $p$ -vector $E_{U}^{'} \equiv u_{1}^{'} \land \cdot \cdot \cdot \land u_{p}^{'}$ is proportional to $E_{U}; i . e ., E_{U}^{'} = c E_{U}$ for some scalar $c$ .

(b) Let $W$ be a $q$ -dimensional subspace of $V$ , with corresponding $q$ -vector $E_{W}$ . Show that $U \subseteq W$ if and only if there exists $a (q - p)$ -vector $F$ such that $E_{W} = E_{U} \land F$

(c) Show that if $p > 0$ and $q > 0$ then $U \cap W = {0}$ if and only if $E_{U} \land E_{W} \neq = 0$

Szekeres · 数学物理

章节

6.4 Grassmann algebras

Multivectors

Exterior product

Properties of exterior product

Problems

关系图谱

目录

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