7.2 Multilinear maps and tensors

中文

The dual representation of tensor product allows for an alternative definition of tensor spaces and products. Key to this approach is the observation in Section 3.7, that every finite dimensional vector space $V$ has a natural isomorphism with $V^{\ast \ast }$ whereby a vector $v$ acts as a linear functional on $V^{\ast }$ through the identification

$$v(\omega )=\omega (v)=⟨v,\omega ⟩=⟨\omega ,v⟩.$$

Multilinear maps and tensor spaces of type $(r,s)$

Let $V_1,V_2,\dots ,V_N$ be vector spaces over the field $\mathbb{R}$. A map

$$T:V_1\times V_2\times \cdots \times V_N\to \mathbb{R}$$

is said to be multilinear if it is linear in each argument separately,

$$\begin{array}{c}T(v_1,\dots ,v_{i-1},a{v}_i+b{v}_i^{\prime },\dots ,v_N)\\ =aT(v_1,\dots ,v_{i-1},v_i,\dots ,v_N)+bT(v_1,\dots ,v_{i-1},v_i^{\prime },\dots ,v_N).\end{array}$$

Multilinear maps can be added and multiplied by scalars in the usual fashion,

$$(aT+bS)(v_1,v_2,\dots ,v_s)=aT(v_1,v_2,\dots ,v_s)+bS(v_1,v_2,\dots ,v_s),$$

and form a vector space, denoted

$$V_1^{\ast }\otimes V_2^{\ast }\otimes \cdots \otimes V_N^{\ast },$$

called the tensor product of the dual spaces $V_1^{\ast },V_2^{\ast },\dots ,V_N^{\ast }$ . When $N=1$ , the word ‘multilinear’ is simply replaced with the word ‘linear’ and the notation is consistent with the concept of the dual space $V^{\ast }$ defined in Section 3.7 as the set of linear functionals on $V$. If we identify every vector space $V_i$ with its double dual $V_i^{\ast \ast }$ , the tensor product of the vector spaces $V_1,V_2,\dots ,V_N$ , denoted $V_1\otimes V_2\otimes \cdot \cdot \cdot \otimes V_N$ , is then the set of multilinear maps from $V_1^{\ast }\times V_2^{\ast }\times \cdots \times V_N^{\ast }$ to $\mathbb{R}$.

Let $V$ be a vector space of dimension $n$ over the field $\mathbb{R}$. Setting

$$V_1=V_2=\cdots =V_r=V^{\ast },V_{r+1}=V_{r+2}=\cdots =V_{r+s}=V,\text{ where }N=r+s,$$

we refer to any multilinear map

$$T:\underset{r}{\underbrace{V^{\ast }\times V^{\ast }\times \cdots \times V^{\ast }}}\times \underset{s}{\underbrace{V\times V\times \cdots \times V}}\to \mathbb{R}$$

as a tensor of type $(r,s)$ on $V$. The integer $r\ge 0$ is called the contravariant degree and $s\ge 0$ the covariant degree of $T$. The vector space of tensors of type $(r,s)$ is

denoted

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

This definition is essentially equivalent to the dual representation of the definition in Section 7.1. Both definitions are totally ‘natural’ in that they do not require a choice of basis on the vector space $V$

It is standard to set $V^{(0,0)}=\mathbb{R}$ ; that is, tensors of type (0, 0) will be identified as scalars. Tensors of type (0, 1) are linear functionals (covectors)

$$V^{(0,1)}\equiv V^{\ast },$$

while tensors of type (1, 0) can be regarded as ordinary vectors

$$V^{(1,0)}\equiv V^{\ast \ast }\equiv V.$$

Covariant tensors of degree 2

A tensor of type (0, 2) is a bilinear map $T:V\times V\to \mathbb{R}$ . In keeping with the terminology of Section 7.1, such a tensor may be referred to as a covariant tensor of degree 2 on $V$. Linearity in each argument reads

$$T(av+bw,u)=aT(v,u)+bT(w,u)\text{ and }T(v,au+bw)=aT(v,u)+bT(v,w).$$

If $\omega$, $\rho \in V^{\ast }$ are linear functionals over $V$, let their tensor product $\omega \otimes \rho$ be the covariant tensor of degree 2 defined by

$$\omega \otimes \rho (u,v)=\omega (u)\rho (v).$$

Linearity in the first argument follows from

$$\begin{array}{l}\omega \otimes \rho (au+bv,w)=\omega (au+bv)\rho (w)\\ =(a\omega (u)+b\omega (v))\rho (w)\\ =a\omega (u)\rho (w)+b\omega (v)\rho (w)\\ =a\omega \otimes \rho (u,w)+b\omega \otimes \rho (v,w).\end{array}$$

A similar argument proves linearity in the second argument $v$.

Example 7.3

Tensor product is not a commutative operation since in general $\omega \otimes \rho \ne$ $\rho \otimes \omega$ . For example, let $e_1,e_2$ be a basis of a two-dimensional vector space $V$, and let ${\epsilon }^1{\epsilon }^2$ be the dual basis of $V^{\ast }$ . If

$$\omega =3{\epsilon }^1+2{\epsilon }^2,\rho ={\epsilon }^1-{\epsilon }^2$$

then

$$\begin{array}{c}\omega \otimes \rho (u,v)=\omega (u^1{e}_1+u^2{e}_2)\rho (v^1{e}_1+v^2{e}_2)\\ =(3u^1+2u^2)(v^1-v^2)\\ =3u^1{v}^1-3u^1{v}^2+2u^2{v}^1-2u^2{v}^2\end{array}$$

and

$$\begin{array}{l}\rho \otimes \omega (u,v)=(u^1-u^2)(3v^1+3v^2)\\ =3u^1{v}^1+2u^1{v}^2-3u^2{v}^1-2u^2{v}^2\\ \ne \omega \otimes \rho (u,v).\end{array}$$

More generally, let $e_1,\dots ,e_n$ be a basis of the vector space $V$ and ${\epsilon }^1,\dots ,{\epsilon }^n$ its dual basis, defined by

$${\epsilon }^i(e_j)=⟨{\epsilon }^i,e_j⟩={\delta }_j^i(i,j=1,\dots ,n).\text{\hspace{1em}(7.6)}$$

Theorem 7.1 · A basis of the space of covariant tensors of degree two

The tensor products

$${\epsilon }^i\otimes {\epsilon }^j(i,j=1,\dots ,n)$$

form a basis of the vector space $V^{(0,2)}$ , which therefore has dimension $n^2$

Proof

The tensors ${\epsilon }^i\otimes {\epsilon }^j(i,j=1,...,n)$ are linearly independent, for if

$$a_{ij}{\epsilon }^i\otimes {\epsilon }^j\equiv \sum _{i=1}^n\sum _{j=1}^n{a}_{ij}{\epsilon }^i\otimes {\epsilon }^j=0,$$

then for each $1\le k,l\le n$ ,

$$0=a_{ij}{\epsilon }^i\otimes {\epsilon }^j(e_k,e_l)=a_{ij}{\delta }_k^i{\delta }_l^j=a_{kl}.$$

Furthermore, if $T$ is any covariant tensor of degree 2 then

$$T=T_{ij}{\epsilon }^i\otimes {\epsilon }^j\text{ where }{T}_{ij}=T(e_i,e_j)\text{\hspace{1em}(7.7)}$$

since for any pair of vector $u=u^i{e}_i,v=v^j{e}_j$ in $V$,

$$\begin{array}{rl}(T-T_{ij}{\epsilon }^i\otimes {\epsilon }^j)(u,v)& =T(u,v)-T_{ij}{\epsilon }^i(u){\epsilon }^j(v)\\ & =T(u^i{e}_i,v^j{e}_j)-T_{ij}{u}^i{v}^j\\ & =u^i{v}^j{T}_{e_i,e_j}-T_{ij}{u}^i{v}^j\\ & =u^i{v}^j{T}_{ij}-T_{ij}{u}^i{v}^j=0.\end{array}$$

Hence the $n^2$ tensors ${\epsilon }^i\otimes {\epsilon }^j$ are linearly independent and span $V^{(0,2)}$ . They therefore form a basis of $V^{(0,2)}$ -

The coefficients $T_{ij}$ in the expansion $T=T_{ij}{\epsilon }^i\otimes {\epsilon }^j$ are uniquely given by the expression on the right in Eq. (7.7), for if $T=T_{ij}^{\prime }{\epsilon }^i\otimes {\epsilon }^j$ then by linear independence of the ${\epsilon }^i\otimes {\epsilon }^j$

$$(T_{ij}^{\prime }-T_{ij}){\epsilon }^i\otimes {\epsilon }^j=0⟹T_{ij}^{\prime }-T_{ij}=0⟹T_{ij}^{\prime }=T_{ij}.$$

They are called the components of $T$ with respect to the basis $\left\{{\mathbf{e}}_i\right\}$ . For any vectors $\mathbf{u}=u^i{\mathbf{e}}_i,\mathbf{v}=v^j{\mathbf{e}}_j$

$$T(u,v)=T_{ij}{u}^i{v}^j.\text{\hspace{1em}(7.8)}$$

Example 7.4

For the linear functional $\omega$ and $\rho$ given in Example 7.3, we can write

$$\begin{array}{rl}& \omega \otimes \rho =(3{\epsilon }^1+2{\epsilon }^2)\otimes ({\epsilon }^1-{\epsilon }^2)\\ & =3{\epsilon }^1\otimes {\epsilon }^1-3{\epsilon }^1\otimes {\epsilon }^2+2{\epsilon }^2\otimes {\epsilon }^1-2{\epsilon }^2\otimes {\epsilon }^2\end{array}$$

and similarly

$$\rho \otimes \omega =3{\epsilon }^1\otimes {\epsilon }^1+2{\epsilon }^1\otimes {\epsilon }^2-3{\epsilon }^2\otimes {\epsilon }^1-2{\epsilon }^2\otimes {\epsilon }^2.$$

Hence the components of the tensor products $\omega \otimes \rho$ and $\rho \otimes \omega$ with respect to the basis tensors ${\epsilon }^i\otimes {\epsilon }^j$ may be displayed as arrays,

$$[(\omega \otimes \rho {)}_{ij}]=\left(\begin{array}{cc}3& -3\\ 2& -2\end{array}\right),[(\rho \otimes \omega {)}_{ij}]=\left(\begin{array}{cc}3& 2\\ 3& -2\end{array}\right).$$

Exercise

Using the components $(\omega \otimes \rho {)}_{12}$ , etc. in the preceding example verify the formula Eq. (7.8) for $\omega \otimes \rho \left(u,v\right)$ . Do the same for $\rho \otimes \omega \left(u,v\right)$

In general, if $\omega =w_i{\epsilon }^i$ and $\rho =r_j{\epsilon }^j$ then

$$\omega \otimes \rho =(w_i{\epsilon }^i)\otimes (r_j{\epsilon }^j)=w_i{r}_j{\epsilon }^i\otimes {\epsilon }^j,$$

and the components of $\omega \otimes \rho$ are

$$(\omega \otimes \rho {)}_{ij}=w_i{r}_j.\text{\hspace{1em}(7.9)}$$

This $n\times n$ array of components is formed by taking all possible component-by-component products of the two linear functionals.

Exercise

Prove Eq. (7.9) by evaluating $(\omega \otimes \rho {)}_{ij}=(\omega \otimes \rho )(e_i,e_j)$

Example 7.5

Let $(V,\cdot )$ be a real inner product space, as in Section 5.1. The map $g:V\times V\to \mathbb{R}$ defined by

$$g(u,v)=u\cdot v$$

is obviously bilinear, and is a covariant tensor of degree 2 called the metric tensor of the inner product. The components $g_{ij}=e_i\cdot e_j=g(e_i,e_j)$ of the inner product are the components of the metric tensor with respect to the basis $\{e_i\}$

$$g=g_{ij}{\epsilon }^i\otimes {\epsilon }^j,$$

while the inner product of two vectors is

$$u\cdot v=g(u,v)=g_{ij}{u}^i{v}^j.$$

The metric tensor is symmetric,

$$g(u,v)=g(v,u).$$

Exercise

Show that a tensor $T$ is symmetric if and only if its components form a symmetric array,

$$T_{ij}=T_{ji}\text{ for all }i,j=1,\dots ,n.$$

Example 7.6

Let $T$ be a covariant tensor of degree 2. Define the map $\bar{T}:V\to V^{\ast }$ by

$$⟨\bar{T}v,u⟩=T(u,v)\text{ for all }u,v\in V.\text{\hspace{1em}(7.10)}$$

Here $\bar{T}(v)$ has been denoted more simply by $\bar{T}\upsilon$ . The map $\bar{T}$ is clearly linear, $\bar{T}(au+bv)=a$ $\bar{T}u+b\bar{T}v$ , since

$$⟨\bar{T}(au+bv),w⟩=T(w,au+bv)=aT(w,u)+bT(w,v)=⟨a\bar{T}u+b\bar{T}v,w⟩$$

holds for all $w\in V$ . Conversely, given a linear map $\bar{T}:V\to V^{\ast }$ , Eq. (7.10) defines a tensor $T$ since $T(u,v)$ so defined is linear both in $u$ and $v$. Thus every covariant tensor of degree 2 can be identified with an element of $L(V,V^{\ast })$ .

Exercise

In components, show that $(\bar{T}v{)}_i=T_{ij}{v}^j$

Contravariant tensors of degree 2

A contravariant tensor of degree 2 on $V$, or tensor of type $(2,0)$ , is a bilinear real-valued map $S$ over $V^{\ast }$ :

$$S:V^{\ast }\times V^{\ast }\to \mathbb{R}.$$

Then, for all $a,b\in \mathbb{R}$ and all $\omega ,\rho ,\theta \in V^{\ast }$

$$S(a\omega +b\rho ,\theta )=aS(\omega ,\theta )+bS(\rho ,\theta )\text{ and }S(\omega ,a\rho +b\theta )=aS(\omega ,\rho )+bS(\omega ,\theta ).$$

If $u$ and $v$ are any two vectors in $V$, then their tensor product $u\otimes v$ is the contravariant tensor of degree 2 defined by

$$u\otimes v(\omega ,\rho )=u(\omega )v(\rho )=\omega (u)\rho (v).\text{\hspace{1em}(7.11)}$$

If $e_1,\dots ,e_n$ is a basis of the vector space $V$ with dual basis $\{{\epsilon }^j\}$ then, just as for $V^{(0,2)}$, the tensors ${\mathbf{e}}_i\otimes {\mathbf{e}}_j$ form a basis of the space $V^{(2,0)}$ , and every contravariant tensor of degree 2 has a unique expansion

$$S=S^{ij}{e}_i\otimes e_j\text{ where }{S}^{ij}=S({\epsilon }^i,{\epsilon }^j).$$

The scalars $S^{ij}$ are called the components of the tensor $S$ with respect to the basis $\{e_i\}$

Exercise

Provide detailed proofs of these statements.

Exercise

Show that the components of the tensor product of two vectors is given by

$$(u\otimes v{)}^{ij}=u^i{v}^j.$$

Example 7.7

It is possible to identify contravariant tensors of degree 2 with linear maps from $V^{\ast }$ to $V$. If $S$ is a tensor of type (0, 2) define a map $\bar{S}:V^{\ast }\to V$ by

$$(\bar{S}\rho )(\omega )\equiv \omega (\bar{S}\rho )=S(\omega ,\rho )\text{ for all }\omega ,\rho \in V^{\ast }.$$

The proof that this correspondence is one-to-one is similar to that given in Example 7.6.

Example 7.8

Let $(V,\cdot )$ be a real inner product space with metric tensor $g$ as defined in Example 7.5. Let $\bar{g}:V\to V^{\ast }$ be the map defined by $g$ using Example 7.6,

$$⟨\bar{g}v,u⟩=g(u,v)=u\cdot v\text{ for all }u,v\in V.\text{\hspace{1em}(7.12)}$$

From the non-singularity condition (SP3) of Section 5.1 the kernel of this map is $\{0\}$ , from which it follows that it is one-to-one. Furthermore, because the dimensions of $V$ and $V^{\ast }$ are identical, $\bar{g}$ is onto and therefore invertible. As shown in Example 7.7 its inverse ${\bar{g}}^{-1}:V^{\ast }\to V$ defines a tensor $g^{-1}$ of type (2, 0) by

$$g^{-1}(\omega ,\rho )=⟨\omega ,{\bar{g}}^{-1}\rho ⟩.$$

From the symmetry of the metric tensor $g$ and the identities

$$\bar{g}{\bar{g}}^{-1}={\mathrm{id}}_{V^{\ast }},{\bar{g}}^{-1}\bar{g}={\mathrm{id}}_V$$

it follows that $g^{-1}$ is also a symmetric tensor, $g^{-1}(\omega ,\rho )=g^{-1}(\rho ,\omega );$

$$\begin{array}{rl}& g^{-1}(\omega ,\rho )=⟨\omega ,{\bar{g}}^{-1}\rho ⟩\\ & =⟨\bar{g}{\bar{g}}^{-1}\omega ,{\bar{g}}^{-1}\rho ⟩\\ & =g({\bar{g}}^{-1}\rho ,{\bar{g}}^{-1}\omega )\\ & =g({\bar{g}}^{-1}\omega ,{\bar{g}}^{-1}\rho )\\ & =⟨\bar{g}{\bar{g}}^{-1}\rho ,{\bar{g}}^{-1}\omega ⟩\\ & =⟨\rho ,{\bar{g}}^{-1}\omega ⟩\\ & =g^{-1}(\rho ,\omega ).\end{array}$$

It is usual to denote the components of the inverse metric tensor with respect to any basis $\left\{{\mathbf{e}}_i\right\}$ by the symbol $g^{ij}$ , so that

$$g^{-1}=g^{ij}{e}_i\otimes e_j\text{ where }{g}^{ij}=g^{-1}({\epsilon }_i,{\epsilon }_j)=g^{ji}.\text{\hspace{1em}(7.13)}$$

From Example 7.6 we have

$$⟨\bar{g}{e}_i,e_j⟩=g(e_j,e_i)=g_{ji},$$

whence

$$\bar{g}(e_i)=g_{ji}{\epsilon }^j.$$

Similarly, from Example 7.7

$$⟨{\bar{g}}^{-1}{\epsilon }^j,{\epsilon }^k⟩=g^{-1}({\epsilon }^k,{\epsilon }^j)=g^{kj}⟹{\bar{g}}^{-1}({\epsilon }^j)=g^{kj}{e}_k.$$

Hence

$$e_i={\bar{g}}^{-1}\circ \bar{g}(e_i)={\bar{g}}^{-1}(g_{ji}{\epsilon }^j)=g_{ji}{g}^{kj}{e}_k.$$

Since $\{e_i\}$ is a basis of $V$, or from Eq. (7.6), we conclude that

$$g^{kj}{g}_{ji}={\delta }_i^k,\text{\hspace{1em}(7.14)}$$

and the matrices $[g^{ij}]$ and $[g_{ij}]$ are inverse to each other.

Mixed tensors

A tensor $R$ of covariant degree 1 and contravariant degree 1 is a bilinear map $R:V^{\ast }\times V\to \mathbb{R}$:

$$\begin{array}{l}R(a\omega +b\rho ,u)=aR(\omega ,u)+bR(\rho ,u)\\ R(\omega ,au+bv)=aR(\omega ,u)+bR(\omega ,v),\end{array}$$

sometimes referred to as a mixed tensor. Such tensors are of type (1, 1), belonging to the vector space $V^{(1,1)}$

For a vector $v\in V$ and covector $\omega \in V^{\ast }$ , define their tensor product $v\otimes \omega$ by

$$v\otimes \omega (\rho ,u)=v(\rho )\omega (u).$$

As in the preceding examples it is straightforward to show that $e_i\otimes {\epsilon }^j$ form a basis of $V^{(1,1)}$ , and every mixed tensor $R$ has a unique decomposition

$$R=R_j^i{e}_i\otimes {\epsilon }^j\text{ where }{R}_j^i=R({\epsilon }^i,e_j).$$

Example 7.9

Every tensor $R$ of type (1, 1) defines a map $\bar{R}:V\to V$ by

$$⟨\bar{R}u,\omega ⟩\equiv \omega (\bar{R}u)=R(\omega ,u).$$

The proof that there is a one-to-one correspondence between such maps and tensors of type (1, 1) is similar to that given in Examples 7.6 and 7.7. Operators on $V$ and tensors of type (1, 1) can be thought of as essentially identical, $V^{(1,1)}\cong L(V,V)$

If $\{e_i\}$ is a basis of $V$ then setting

$$\bar{R}{e}_i={\bar{R}}_i^j{e}_j$$

we have

$$⟨\bar{R}{e}_j,{\epsilon }^i⟩=R({\epsilon }^i,e_j)=R_j^i$$

and

$$⟨\bar{R}{e}_j,{\epsilon }^i⟩=⟨{\bar{R}}_j^k{e}_k,{\epsilon }^i⟩={\bar{R}}_i^k{\delta }_k^i={\bar{R}}_i^j.$$

Hence $R_i^k={\bar{R}}_i^k$ and it follows that

$$\bar{R}{e}_i=R_i^j{e}_j.$$

On comparison with Eq. (3.6) it follows that the components of a mixed tensor $R$ are the same as the matrix components of the associated operator on $V$.

Exercise

Show that a tensor $R$ of type (1, 1) defines a map $\tilde{R}:V^{\ast }\to V^{\ast }$

Example 7.10

Define the map $\delta :V^{\ast }\times V\to \mathbb{R}$ by

$$\delta (\omega ,v)=\omega (v)=v(\omega )=⟨\omega ,v⟩.$$

This map is clearly linear in both arguments and therefore constitutes a tensor of type (1, 1). $\mathrm{If}\left\{e_i\right\}$ is any basis of $V$ with dual basis $\{{\epsilon }^j\}$ , then it is possible to se

$$\delta =e_i\otimes {\epsilon }^i\equiv e_1\otimes {\epsilon }^1+e_2\otimes {\epsilon }^2+\cdots +e_n\otimes {\epsilon }^n$$

since

$$e_i\otimes {\epsilon }^i(\omega ,v)=e_i(\omega ){\epsilon }^i(v)=w_i{v}^i=\omega (v).$$

An alternative expression for $\delta$ is

$$\delta =e_i\otimes {\epsilon }^i={\delta }_j^i{e}_i\otimes {\epsilon }^j$$

from which the components of the mixed tensor $\delta$ are precisely the Kronecker delta ${\delta }^i{}_j$ . As no specific choice of basis has been made in this discussion the components of the tensor $\delta$ are ‘invariant’, in the sense that they do not change under basis transformations.

Exercise

Show that the map $\bar{\delta }:V\to V$ that corresponds to the tensor $\delta$ according to Example 7.9 is the identity map

$$\bar{\delta }={\mathrm{id}}_V.$$

Problems

Problem 7.7

Let $\bar{T}$ be the linear map defined by a covariant tensor $T$ ofdegree 2 as in Example 7.6. If $\{e_i\}$ is a basis of $V$ and $\{{\epsilon }^j\}$ the dual basis, define the matrix of components of $\bar{T}$ with respect to these bases as $[{\bar{T}}_{ji}]$ where

$$\bar{T}(e_i)={\bar{T}}_{ji}{\epsilon }^j.$$

Show that the components of the tensor $T$ in this basis are identical with the components as a map, $T_{ij}={\bar{T}}_{ij}$

Similarly if $S$ is a contravariant tensor of degree 2 and $\bar{s}$ the linear map defined in Example 7.7, show that the components ${\hat{S}}^{ij}$ are identical with the tensor components $S^{ij}$

Problem 7.8

Show that every tensor $R$ of type (1, 1) defines a map $\tilde{R}:V^{\ast }\to V^{\ast }$ by

$$⟨\tilde{R}\omega ,u⟩=R(\omega ,u)$$

and show that for a natural definition of components of this map, ${\tilde{R}}_i^k=R_i^k$

Problem 7.9

Show that the definition of tensor product of two vectors $u\times v$ given in Eq. (7.11) agrees with that given in Section 7.1 after relating the two concepts of tensor by isomorphism.