7.1 Free vector spaces and tensor spaces

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Free vector spaces

If $S$ is an arbitrary set, the concept of a free vector space on $S$ over a field $\mathbb{K}$ can be thought of intuitively as the set of all ‘formal finite sums’

$$a^1{s}_1+a^2{s}_2+\cdots +a^n{s}_n\text{ where }n=0,1,2,\dots ;a^i\in \mathbb{K},s_i\in S.$$

The word ‘formal’ means that if $S$ is an algebraic structure that already has a concept of addition or scalar multiplication defined on it, then the scalar product and summation in the formal sum bears no relation to these.

More rigorously, the free vector space $F(S)$ on a set $S$ is defined as the set of all functions $f:S\to \mathbb{K}$ that vanish at all but a finite number of elements of $S$. Clearly $F(S)$ is a vector space with the usual definitions,

$$(f+g)(s)=f(s)+g(s),(af)(s)=af(s).$$

It is spanned by the characteristic functions ${\chi }_t\equiv {\chi }_{\{t\}}$ (see Example 1.7)

$${\chi }_t(s)=\{\begin{array}{ll}1& \text{ if }s=t,\\ 0& \text{ if }s\ne t,\end{array}$$

since any function having non-zero values at just a finite number of places $s_1,s_2,\dots s_n$ can be written uniquely as

$$f=f(s_1){\chi }_{s_1}+f(s_2){\chi }_{s_2}+\cdots +f(s_n){\chi }_{s_n}.$$

Evidently the elements of $F(S)$ are in one-to-one correspondence with the ‘formal finite sums’ alluded to above.

Example 7.1

The vector space ${\hat{\mathbb{R}}}^{\infty }$ defined in Example 3.10 is isomorphic with the free vector space on any countably infinite set $\mathcal{S}=\{s_1,s_2,s_3,...\}$ , since the map $\sigma :{\hat{\mathbb{R}}}^{\infty }\to F(S)$ defined by

$$\sigma \left(a^1,a^2,\dots ,a^n,0,0,\dots \right)=\sum _{i=1}^n{a}^i{\chi }_{s_i}$$

is linear, one-to-one and onto.

The tensor product $V\otimes W$

Let $V$ and $W$ be two vector spaces over a field $\mathbb{K}$. Imagine forming a product $v\otimes w$ between elements of these two vector spaces, called their ‘tensor product’, subject to the rules

$$(av+b{v}^{\prime })\otimes w=av\otimes w+b{v}^{\prime }\otimes w,v\otimes (aw+b{w}^{\prime })=av\otimes w+bv\otimes w^{\prime }.$$

The main difficulty with this simple idea is that we have no idea of what space $v\otimes w$ belongs to. The concept of a free vector space can be used to give a proper definition for this product.

Let $F(V\times W)$ be the free vector space over $V\times W$ . This vector space is, in a sense, much too ‘large’ since pairs such as $a(v,w)$ and $(av,w)$, or $(\mathbf{v}+{\mathbf{v}}^{\prime },\mathbf{w})$ and $(v,w)+(v^{\prime },w)$ are totally unrelated in $F(V\times W)$ . To reduce the vector space to sensible proportions, we define $U$ to be the vector subspace of $F(V\times W)$ generated by all elements of the form

$$(av+b{v}^{\prime },w)-a(v,w)-b\left(v^{\prime },w\right)\text{ and }(v,aw+b{w}^{\prime })-a(v,w)-b\left(v,w^{\prime }\right)$$

where, for notational simplicity, we make no distinction between a pair $(v,w)$ and its characteristic function ${\chi }_{(v,w)}\in F(V\times W)$ . The subspace $U$ contains essentially all vector combinations that are to be identified with the zero element. The tensor product of $V$ and $W$ is defined to be the factor space

$$V\otimes W=F(V\times W)/U.$$

The tensor product $v\otimes w$ of a pair of vectors $v\in V$ and $w\in W$ is defined as the equivalence class or coset in $V\otimes W$ to which $(v,w)$ belongs,

$$v\otimes w=[(v,w)]=(v,w)+U.$$

This product is bilinear,

$$(av+b{v}^{\prime })\otimes w=av\otimes w+b{v}^{\prime }\otimes w,$$ $$v\otimes (aw+b{w}^{\prime })=av\otimes w+bv\otimes w^{\prime }.$$

To show the first identity,

$$\begin{array}{l}(av+b{v}^{\prime })\otimes w=(av+b{v}^{\prime },w)+U\\ =(av+b{v}^{\prime },w)-((av+b{v}^{\prime },w)-a(v,w)-b(v^{\prime },w))+U\\ =a(v,w)+b(v^{\prime },w)+U\\ =av\otimes w+b{v}^{\prime }\otimes w,\end{array}$$

and the second identity is similar.

If $V$ and $W$ are both finite dimensional let $\{e_i|i=1,\dots ,n\}$ be a basis for $V$, and $\{f_a|a=1,\dots ,m\}$ a basis of $W$. Every tensor product $v\otimes w$ can, through bilinearity, be written

$$v\otimes w=(v^i{e}_i)\otimes (w^a{f}_a)=v^i{w}^a(e_i\otimes f_a).\text{\hspace{1em}(7.1)}$$

We will use the term tensor to describe the general element of $V\otimes W$ . Since every tensor $A$ is a finite sum of elements of the form $v\otimes w$ it can, on substituting Eq. (7.1), be expressed in the form

$$A=A^{ia}{e}_i\otimes f_a.\text{\hspace{1em}(7.2)}$$

Hence the tensor product space $V\otimes W$ is spanned by the $nm$ tensors $\{e_i\otimes f_a|i=1,\dots ,n,a=1,\dots ,m\}$

Furthermore. these tensors form a basis of $V\otimes W$ since they are linearly independent. To prove this statement, let $(\rho ,\phi )$ be any ordered pair of linear functionals $\rho \in V^{\ast },\phi \in W^{\ast }$ Such a pair defines a linear functional on $F(V\times W)$ by setting

$$(\rho ,\phi )(v,w)=\rho (v)\phi (w),$$

and extending to all of $F(V\times W)$ by linearity,

$$(\rho ,\phi )\left[\sum _r{a}^r\left(v_r,w_r\right)\right]=\sum _r{a}^r\rho \left(v_r\right)\phi \left(w_r\right).$$

This linear functional vanishes on the subspace $U$ and therefore ‘passes to’ the tensor product space $V\otimes W$ by setting

$$(\rho ,\phi )[\sum _r{a}^r{v}_r\otimes w_r]=\sum _r{a}^r(\rho ,\phi )((v_r,w_r)).$$

Let ${\epsilon }^k,{\phi }^b$ be the dual bases in $V^{\ast }$ and $W^{\ast }$ respectively; we then have

$$\begin{array}{rl}{A}^{ia}{e}_i\otimes f_a=0& ⟹({\epsilon }^j,{\phi }^b)(A^{ia}{e}_i\otimes f_a)=0\\ & ⟹A^{ia}{\delta }_i^j{\delta }_a^b=0\\ & ⟹A^{jb}=0.\end{array}$$

Hence the tensors $e_i\otimes f_a$ are l.i. and form a basis of $V\otimes W$ . The dimension of $V\otimes W$ is given by

$$\dim (V\otimes W)=\dim V\dim W.$$

Setting $V=W$ , the elements of $V\otimes V$ are called contravariant tensors of degree 2 on $V$. If $\{e_i\}$ is a basis of $V$ every contravariant tensor of degree 2 has a unique expansion

$$T=T^{ij}{e}_i\otimes e_j,$$

and the real numbers $T^{ij}$ are called the components of the tensor with respect to this basis. Similarly each element $S$ of $V^{\ast }\otimes V^{\ast }$ is called a covariant tensor of degree 2 on $V$ and has a unique expansion with respect to the dual basis ${\epsilon }^i$

$$S=S_{ij}{\epsilon }^i\otimes {\epsilon }^j.$$

Dual representation of tensor product

Given a pair of vector spaces $V_1$ and $V_2$ over a field $\mathbb{K}$, a map $T:V_1\times V_2\to \mathbb{K}$ is said to be bilinear if

$$\begin{array}{c}T(a{v}_1+b{v}_1^{\prime },v_2)=aT(v_1,v_2)+bT(v_1^{\prime },v_2),\\ T(v_1,a{v}_2+b{v}_2^{\prime },)=aT(v_1,v_2)+bT(v_1,v_2^{\prime }),\end{array}$$

for all $v_1,v_1^{\prime }\in V_1,v_2,v_2^{\prime }\in V_2$ , and $a,b\in \mathbb{K}$ . Bilinear maps can be added and multiplied by scalars in a manner similar to that for linear functionals given in Section 3.7, and form a vector space that will be denoted $(V_1,V_2{)}^{\ast }$

Every pair of vectors $(v,w)$ where $v\in V,w\in W$ defines a bilinear map $V^{\ast }\times W^{\ast }\to \mathbb{K}$. by setting

$$(v,w):(\rho ,\phi )\mapsto \rho (v)\phi (w).$$

We can extend this correspondence to all of $F(V\times W)$ in the obvious way by setting

$$\sum _r{a}^r\left(v_r,w_r\right):(\rho ,\phi )\mapsto \sum _r{a}^r\rho \left(v_r\right)\phi \left(w_r\right),$$

and since the action of any generators of the subspace $U$, such as $(av+b{v}^{\prime },w)-a(v,w)-b(v^{\prime },w)$ , clearly vanishes on $V^{\ast }\times W^{\ast }$ , the correspondence passes in a unique way to the tensor product space $V\otimes W=F(V,W)/U$ . That is, every tensor $\begin{array}{r}A={\sum }_r{a}^r{v}_r\otimes w_r\end{array}$ defines a bilinear map $V^{\ast }\times W^{\ast }\to \mathbb{K}$ , by setting

$$A(\rho ,\phi )=\sum _r{a}^r\rho (v_r)\phi (w_r).$$

This linear mapping from $V\otimes W$ into the space of bilinear maps on $V^{\ast }\times W^{\ast }$ is also one-to-one in the case of finite dimensional vector spaces. For, suppose $A(\rho ,\phi )=B(\rho ,\phi )$ for all $\rho \in V^{\ast },\phi \in W^{\ast }$ . Let $e_i,f_a$ be bases of $V$ and $W$ respectively, and ${\epsilon }^j,{\phi }^a$ be the dual bases. Writing $A=A^{ia}{e}_i\otimes f_a=0,B=B^{ia}{e}_i\otimes f_a=0$ , we have

$$A^{ia}\rho (e_i)\phi (f_a)=B^{ia}\rho (e_i)\phi (f_a)$$

for all linear functionals $\rho \in V^{\ast },\phi \in W^{\ast }$ . If we set $\rho ={\epsilon }^j,\phi ={\phi }^b$ then

$$A^{ia}{\delta }_i^j{\delta }_a^b=B^{ia}{\delta }_i^j{\delta }_a^b,$$

resulting in $A^{jb}=B^{jb}$ . Hence $A=B$ , and for finite dimensional vector spaces $V$ and $W$ we have shown that the linear correspondence between $V\otimes W$ and $(V^{\ast },W^{\ast }{)}^{\ast }$ is one-to-one,

$$V\otimes W\cong (V^{\ast },W^{\ast }{)}^{\ast }.\text{\hspace{1em}(7.3)}$$

This isomorphism does not hold for infinite dimensional spaces.

A tedious but straightforward argument results in the associative law for tensor products of three vectors

$$u\otimes (v\otimes w)=(u\otimes v)\otimes w.$$

Hence the tensor product of three or more vector spaces is defined in a unique way,

$$U\otimes V\otimes W=U\otimes (V\otimes W)=(U\otimes V)\otimes W.$$

For finite dimensional spaces it may be shown to be isomorphic with the space of maps $A:U^{\ast }\times V^{\ast }\times W^{\ast }\to \mathbb{K}$ that are linear in each argument separately.

Free associative algebras

Let $\mathcal{F}(V)$ be the infinite direct sum of vector spaces

$$\mathcal{F}(V)=V^{(0)}\oplus V^{(1)}\oplus V^{(2)}\oplus V^{(3)}\oplus \dots$$

where $\mathbb{K}=V^{(0)},V=V^{(1)}$ and

$$V^{(r)}=\underset{r}{\underbrace{V\otimes V\otimes \cdots \otimes V}}.$$

The typical member of this infinite direct sum can be written as a finite formal sum of tensors from the tensor spaces $V^{(r)}$

$$a+u+A_2+\cdots +A_r.$$

To define a product rule on $\mathcal{F}(V)$ set

$$\underset{\in V^{(r)}}{\underbrace{u_1\otimes u_2\otimes \cdots \otimes u_r}}\underset{\in V^{(s)}}{\underbrace{v_1\otimes v_2\otimes \cdots \otimes v_s}}=\underset{\in V^{(r+s)}}{\underbrace{u_1\otimes \cdots \otimes u_r\otimes v_1\otimes \cdots \otimes v_s}}$$

and extend to all of $\mathcal{F}(V)$ by linearity. The distributive law (6.1) is automatically satisfied, making $\mathcal{F}(V)$ with this product structure into an associative algebra. The algebra $\mathcal{F}(V)$ has in essence no ‘extra rules’ imposed on it other than simple juxtaposition of elements from $V$ and multiplication by scalars. It is therefore called the free associative algebra over $V$. All associative algebras can be constructed as a factor algebra of the free associative algebra over a vector space. The following example illustrates this point.

Example 7.2

If $V$ is the one-dimensional free vector space over the reals on the singleton set $S=\{x\}$ then the free associative algebra over $\mathcal{F}(V)$ is in one-to-one correspondence with the algebra of real polynomials $\mathcal{P}$ , Example 6.3, by setting

$$a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n\equiv a_0+a_{1}x+a_{2}x\otimes x+\cdots +a_n\underset{n}{\underbrace{x\otimes x\otimes \cdots \otimes x}}.$$

This correspondence is an algebra isomorphism since the product defined on $\mathcal{F}(V)$ by the above procedure will be identical with multiplication of polynomials. For example,

$$\begin{array}{c}(ax+bx\otimes x\otimes x)(c+dx\otimes x)=acx+(ad+bc)x\otimes x\otimes x+bdx\otimes x\otimes x\otimes x\otimes x\\ \equiv acx+(ad+bc)x^3+bd{x}^5\\ =(ax+b{x}^3)(c+d{x}^2),\text{etc.}\end{array}$$

Set $\mathcal{C}$ to be the ideal of $\mathcal{P}$ generated by $x^2+1$ , consisting of all polynomials of the form $f(x)(x^2+1)g(x)$ . By identifying $i$ with the polynomial class $[x]=x+\mathcal{C}$ and real numbers with the class of constant polynomials $a\to [a]$ , the algebra of complex numbers is isomorphic with the factor algebra $\mathcal{P}/\mathcal{C}$, for

$$i^2\equiv [x{]}^2=[x^2]=[x^2-(x^2+1)]=[-1]\equiv -1.$$

Grassmann algebra as a factor algebra of free algebras

The definition of Grassmann algebra given in Section 6.4 is unsatisfactory in two key aspects. Firstly, in the definition of exterior product it is by no means obvious that the rules (EP1)–(EP3) produce a well-defined and unique product on $\Lambda (V)$ . Secondly, the matter of linear independence of the basis vectors $e_{i_1{i}_2\dots i_r}(i_1<i_2<\cdot \cdot \cdot <i_r)$ had to be postulated separately in Section 6.4. The following discussion provides a more rigorous foundation for Grassmann algebras, and should clarify these issues.

Let $\mathcal{F}(V)$ be the free associative algebra over a real vector space $V$, and let $\mathcal{S}$ be the ideal generated by all elements of $\mathcal{F}(V)$ of the form $u\otimes T\otimes v+v\otimes T\otimes u$ where $u,v\in V$ and $T\in \mathcal{F}(V)$ . The general element of $\mathcal{S}$ is

$$S\otimes u\otimes T\otimes v\otimes U+S\otimes v\otimes T\otimes u\otimes U$$

where $u,v\in V$ and $S$, $T$, $U\in \mathcal{F}(V)$ . The ideal $\mathcal{S}$ essentially identifies those elements of $\mathcal{F}(V)$ that will vanish when the tensor product $\otimes$ is replaced by the wedge product $\wedge$.

Exercise

Show that the ideal $\mathcal{S}$ is generated by all elements of the form $w\otimes T\otimes w$ where $w\in V$ and $T\in \mathcal{F}(V)$ . [Hint: Set $w=u+v$.]

Define the Grassmann algebra $\Lambda (V)$ to be the factor algebra

$$\Lambda (V)=\mathcal{F}(V)/\mathcal{S},\text{\hspace{1em}(7.4)}$$

and denote the induced associative product by $\wedge$,

$$[A]\wedge [B]=[A\otimes B]\text{\hspace{1em}(7.5)}$$

where $[A]\equiv A+\mathcal{S},[B]\equiv B+\mathcal{S}$ . As in Section 6.4, the elements [$A$] of the factor algebra are called multivectors. There is no ambiguity in dropping the square brackets,

$A\equiv [A]$ and writing $A\wedge B$ for $[A]\wedge [B]$ . The algebra $\Lambda (V)$ is the direct sum of subspaces corresponding to tensors of degree $r$,

$$\Lambda (V)={\Lambda }^0(V)\oplus {\Lambda }^1(V)\oplus {\Lambda }^2(V)\oplus \dots ,$$

where

$${\Lambda }^r(V)=[V^{(r)}]=\{A+\mathcal{S}\mid A\in V^{(r)}\}$$

whose elements are called $r$-vectors. If $A$ is an $r$-vector and $B$ an $s$-vector then $A\wedge B$ is an $(r+s)$-vector.

Since, by definition, $\mathbf{u}\otimes \mathbf{v}+\mathbf{v}\otimes \mathbf{u}$ is a member of $\mathcal{S}$, we have

$$u\wedge v=[u\otimes v]=[-v\otimes u]=-v\wedge u,$$

for all $u,v\in V$ . Hence $u\wedge u=0$ for all $u\in V$

Exercise

Prove that if $A$, $B$ and $C$ are any multivectors then for all $u$, $v\in V$

$$A\wedge u\wedge B\wedge v\wedge C+A\wedge v\wedge B\wedge u\wedge C=0,$$

and $A\wedge u\wedge B\wedge u\wedge C=0$

Exercise

From the corresponding rules of tensor product show that exterior product is associative and distributive.

From the associative law

$$\left(u_1\wedge u_2\wedge \cdots \wedge u_r\right)\wedge \left(v_1\wedge v_2,\wedge \cdots \wedge v_s\right)=u_1\wedge \cdots \wedge u_r\wedge v_1\wedge \cdots \wedge v_s,$$

in agreement with (EP2) of Section 6.4. This provides a basis-independent definition for exterior product on any finite dimensional vector space $V$, having the desired properties (EP1)–(EP3). Since every $r$-vector is the sum of simple $r$-vectors, the space of $r$-vectors ${\Lambda }^r(V)$ is spanned by

$$E_r=\{e_{i_1{i}_2\dots i_r}\mid i_1<i_2<\cdots <i_r\},$$

where

$$e_{i_1{i}_2\dots i_r}=e_{i_1}\wedge e_{i_2}\cdots \wedge e_{i_r},$$

as shown in Section 6.4. It is left as an exercise to show that the set $E_r$ does indeed form a basis of the space of $r$-vectors (see Problem 7.6). Hence, as anticipated in Section 6.4, the dimension of the space of $r$-vectors is

$$\dim {\Lambda }^r(V)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!},$$

and the dimension of the Grassmann algebra $\Lambda (V)$ is $2^n$ .

Problems

Problem 7.1

Show that the direct sum $V\oplus W$ of two vector spaces can be defined from the free vector space as $F(V\times W)/U$ where $U$ is a subspace generated by all linear combinations of the form

$$(av+b{v}^{\prime },aw+b{w}^{\prime })-a(v,w)-b\left(v^{\prime },w^{\prime }\right).$$

Problem 7.2

Prove the so-called universal property of free vector spaces. Let $\phi :S\to F(S)$ be the map that assigns to any element $s\in S$ its characteristic function ${\chi }_s\in F(S)$ . If $V$ is any vector space and $\alpha :S\to V$ any map from $S$ to $V$ , then there exists a unique linear map $T:F(S)\to V$ such that $\alpha =T\circ \phi$ , as depicted by the commutative diagram

Commutative diagram for the universal property of a free vector space (Problem 7.2).

Show that this process is reversible and may be used to define the free vector space on $S$ as being the unique vector space $F(S)$ for which the above commutative diagram holds.

Problem 7.3

Let $\mathcal{F}(V)$ be the free associative algebra over a vector space $V$.

(a) Show that there exists a linear map $I:V\to \mathcal{F}(V)$ such that if $\mathcal{A}$ is any associative algebra over the same field $\mathbb{K}$ and $S:V\to \mathcal{A}$ a linear map, then there exists a unique algebra homomorphism $\alpha :\mathcal{F}(V)\to \mathcal{A}$ such that $S=\alpha \circ I$

(b) Depict this property by a commutative diagram.

(c) Show the converse: any algebra $\mathcal{F}$ for which there is a map $I:V\to \mathcal{F}$ such that the commutative diagram holds for an arbitrary linear map $S$ is isomorphic with the free associative algebra over $V$.

Problem 7.4

Give a definition of quaternions as a factor algebra of the free algebra on a three dimensional vector space.

Problem 7.5

The Clifford algebra ${\mathcal{C}}_g$ associated with an inner product space $V$ with scalar product $g(u,v)\equiv u\cdot v$ can be defined in the following way. Let $\mathcal{F}(V)$ be the free associative algebra on $V$ and the two-sided ideal generated by all elements of the form

$$A\otimes (u\otimes v+v\otimes u-2g(u,v)1)\otimes B(A,B\in \mathcal{F}(V)).$$

The Clifford algebra in question is now defined as the factor space $\mathcal{F}(V)/\mathcal{C}$ . Verify that this algebra $i$ isomorphic with the Clifford algebra as defined in Section 6.3, and could serve as a basis-independent definition for the Clifford algebra associated with a real inner product space.

Problem 7.6

Show that $E_2$ is a basis of ${\Lambda }^2(V)$ . In outline: define the maps ${\epsilon }^{kl}:V^2\to \mathbb{R}$ by

$${\epsilon }^{kl}(u,v)={\epsilon }^k(u){\epsilon }^l(v)-{\epsilon }^l(u){\epsilon }^k(v)=u^k{v}^l-u^l{v}^k.$$

Extend by linearity to the tensor space $V^{(2,0)}$ and show there is a natural passage to the factor space, ${\hat{\epsilon }}^{kl}:{\Lambda }^2(\dot{V})=V^{(2,0)}/S^2\to \mathbb{R}$. If a linear combination from $E_2$ were to vanish,

$$\sum _{i<j}{A}^{ij}{e}_i\wedge e_j=0,$$

apply the map ${\hat{\epsilon }}^{kl}$ to this equation, to show that all coefficients $A^{ij}$ must vanish separately.

Indicate how the argument may be extended to show that if $r\le n=\dim V$ then $E_r$ is a basis of ${\Lambda }^r(V)$