7.3 Basis representation of tensors

中文

We now construct a basis of $V^{(r,s)}$ from any given basis $\left\{{\mathbf{e}}_i\right\}$ of $V$ and its dual basis $\{{\epsilon }^j\}$ and display tensors of type $(r,s)$ with respect to this basis. While the expressions that arise often turn out to have a rather complicated appearance as multicomponented objects, this is simply a matter of becoming accustomed to the notation. It is still the representation of tensors most frequently used by physicists.

Tensor product

If $T$ is a tensor of type $(r,s)$ and $S$ is a tensor of type $(p,q)$ then define $T\otimes S$ , called their tensor product, to be the tensor of type $(r+p,s+q)$ defined by

$$\begin{array}{c}(T\otimes S)({\omega }^1,\dots ,{\omega }^r,{\rho }^1,\dots ,{\rho }^p,u_1,\dots ,u_s,v_1,\dots ,v_q)\\ =T({\omega }^1,\dots ,{\omega }^r,u_1,\dots ,u_s)S({\rho }^1,\dots ,{\rho }^p,v_1,\dots ,v_q).\end{array}\text{\hspace{1em}(7.15)}$$

This product generalizes the definition of tensor products of vectors and covectors in the previous section. It is readily shown to be associative

$$T\otimes (S\otimes R)=(T\otimes S)\otimes R,$$

so there is no ambiguity in writing expressions such as $T\otimes S\otimes R$

If $e_1,\dots ,e_n$ is a basis for $V$ and ${\epsilon }^1,\dots ,{\epsilon }^n$ the dual basis of $V^{\ast }$ then the tensors

$$e_{i_1}\otimes \cdots \otimes e_{i_r}\otimes {\epsilon }^{j_1}\otimes \cdots \otimes {\epsilon }^{j_s}(i_1,i_2,\dots ,j_1,\dots ,j_s=1,2,\dots ,n)$$

form a basis of $V^{(r,s)}$ , since every tensor $T$ of type $(r,s)$ has a unique expansion

$$T=T_{j_1\dots j_s}^{i_1\dots i_r}{e}_{i_1}\otimes \cdots \otimes e_{i_r}\otimes {\epsilon }^{j_1}\otimes \cdots \otimes {\epsilon }^{j_s}\text{\hspace{1em}(7.16)}$$

where

$$T_{j_1\dots j_s}^{i_1\dots i_r}=T({\epsilon }^{i_1},\dots ,{\epsilon }^{i_r},e_{j_1},\dots ,e_{j_s})\text{\hspace{1em}(7.17)}$$

are called the components of the tensor $T$ with respect to the basis $\{e_i\}$ of $V$.

Exercise

Prove these statements in full detail. Despite the apparent complexity of indices the proof is essentially identical to that given for the case of $V^{(0,2)}$ in Theorem 7.1.

The components of a linear combination of two tensors of the same type are given by

$$\begin{array}{rl}(T+aS{)}_{kl\dots }^{ij\dots }& =(T+aS)({\epsilon }^i,{\epsilon }^j,\dots ,e_k,e_l,\dots )\\ & =T({\epsilon }^i,{\epsilon }^j,\dots ,e_k,e_l,\dots )+aS({\epsilon }^i,{\epsilon }^j,\dots ,e_k,e_l,\dots )\\ & =T_{kl\dots }^{ij\dots }+a{S}_{kl\dots }^{ij\dots }.\end{array}$$

The components of the tensor product of two tensors $T$ and $S$ are given by

$$(T\otimes S{)}_{kl\dots mn\dots }^{ij\dots pq\dots }=T_{kl\dots }^{ij\dots }{S}_{mn\dots }^{pq\dots }$$

The proof follows from Eq. (7.15) on setting ${\omega }^1={\epsilon }^i,{\omega }^2={\epsilon }^j,\dots ,{\rho }^1={\epsilon }^k,\dots ,u_1=e_p,u_2=e_q,\mathsf{etc}$.

Exercise

Show that in components, a multilinear map $T$ has the expression

$$T(\omega ,\rho ,\dots ,u,v,\dots )=T_{kl\dots }^{ij\dots }{w}_i{r}_j\dots u^k{v}^l\dots \text{\hspace{1em}(7.18)}$$

where $\omega =w_i{\epsilon }^i,\rho =r_j{\epsilon }^j,u=u^k{e}_k$, etc.

Change of basis

Let $\{e_i\}$ and $\{e_j^{\prime }\}$ be two bases of $V$ related by

$$e_i=A_i^j{e}_j^{\prime },e_j^{\prime }=A_j^{\prime k}{e}_k\text{\hspace{1em}(7.19)}$$

where the matrices $[A^i{}_j]$ and $[{{\mathbf{A}}^{\prime }}^i{}_j]$ are inverse to each other,

$$A_j^{\prime k}{A}_i^j=A_j^k{A}_i^{\prime j}={\delta }_i^k.\text{\hspace{1em}(7.20)}$$

As shown in Chapter 3 the dual basis transforms by Eq. (3.32),

$${\epsilon }^{\prime j}=A_k^j{\epsilon }^k\text{\hspace{1em}(7.21)}$$

and under the transformation laws components of vectors $v=v^i{e}_i$ and covectors $\omega =w_j{\epsilon }^j$ are

$$v^{\prime j}=A_i^j{v}^i,w_i^{\prime }=A_i^{\prime j}{w}_j.\text{\hspace{1em}(7.22)}$$

The terminology ‘contravariant’ and ‘covariant’ transformation laws used in Chapter 3 is motivated by the fact that vectors and covectors are tensors of contravariant degree 1 and covariant degree 1 respectively.

If $\mathcal{T}={\mathcal{T}}^{ij}{e}_i\otimes e_j$ is a tensor of type (2, 0) then

$$\begin{array}{rl}& T=T^{ij}{e}_i\otimes e_j\\ & =T^{ij}{A}_i^k{e}_k^{\prime }\otimes A_j^l{e}_l^{\prime }\\ & =T^{\prime kl}{e}_k^{\prime }\otimes e_l^{\prime }\end{array}$$

where

$$T^{\prime kl}=T^{ij}{A}_i^k{A}_j^l.\text{\hspace{1em}(7.23)}$$

Exercise

Alternatively, show this result from Eq. (7.21) and

$$T^{\prime kl}=T({\epsilon }^{\prime k},{\epsilon }^{\prime l}).$$

Similarly the components of a covariant tensor $T=T_{ij}{\epsilon }^i\otimes {\epsilon }^j$ of degree 2 transform as

$$T_{kl}^{\prime }=T_{ij}{A}_k^{\prime i}{A}_l^{\prime j}.\text{\hspace{1em}(7.24)}$$

Exercise

Show Eq. (7.24) (i) by transformation of ${\epsilon }^i$ using Eq. (7.21), and (ii) from $T_{ij}^{\prime }=T(e_i,e_j)$ using Eq. (7.19).

In the same way, the components of a mixed tensor $T=T_j^i{e}_i\otimes {\epsilon }^j$ can be shown to have the transformation law

$$T_l^{\prime k}=T_j^i{A}_i^k{A}_l^{\prime j}.\text{\hspace{1em}(7.25)}$$

Exercise

Show Eq. (7.25).

Before giving the transformation law of components of a general tensor it is useful to establish a convention known as the kernel index notation. In this notation we denote the indices on the transformed bases and dual bases by a primed index, $\{e_{i^{\prime }}^{\prime }|i^{\prime }=1,\dots ,n\}$ and $\{{\epsilon }^{\prime j^{\prime }}|j^{\prime }=1,\dots ,n\}$ . The primes on the ‘kernel’ letters $e$ and $\epsilon$ are essentially superfluous and little meaning is lost in dropping them, simply writing $e_{i^{\prime }}$ and ${\epsilon }^{j^{\prime }}$ for the transformed bases. The convention may go even further and require that the primed indices range over an indexed set of natural numbers $i^{\prime }=1^{\prime },\dots ,n^{\prime }$ . These practices may seem a little bizarre and possibly confusing. Accordingly, we will only follow a ‘half-blown’ kernel index notation, with the key requirement that primed indices be used on transformed quantities. The main advantage of the kernel index notation is that it makes the transformation laws of tensors easier to commit to memory.

Instead of Eq. (7.19) we now write the basis transformations as

$$e_i=A_i^{i^{\prime }}{e}_{i^{\prime }}^{\prime },e_{j^{\prime }}^{\prime }=A_{j^{\prime }}^j{e}_j\text{\hspace{1em}(7.26)}$$

where the matrix array $\mathsf{A}=[A_i^{j^{\prime }}]$ is always written with the primed index in the superscript position, while its inverse ${\mathsf{A}}^{-1}=[A^{\prime k}{}_{j^{\prime }}]$ has the primed index as a subscript. The relations Eq. (7.20) between these are now written

$$A_{j^{\prime }}^{\prime k}{A}_i^{j^{\prime }}={\delta }_i^k,A_k^{j^{\prime }}{A}_{i^{\prime }}^{\prime k}={\delta }_{i^{\prime }}^{j^{\prime }},\text{\hspace{1em}(7.27)}$$

which take the place of Eq. (7.20).

The dual basis satisfies

$${\epsilon }^{\prime i^{\prime }}(e_{j^{\prime }}^{\prime })={\delta }_{j^{\prime }}^{i^{\prime }}$$

and is related to the original basis by Eq. (7.21), which read

$${\epsilon }^i=A_{j^{\prime }}^{\prime i}{\epsilon }^{\prime j^{\prime }},{\epsilon }^{\prime i^{\prime }}=A_j^{i^{\prime }}{\epsilon }^j.\text{\hspace{1em}(7.28)}$$

The transformation laws of vectors and covectors Eq. (7.22) are replaced by

$$v^{\prime i^{\prime }}=A_j^{i^{\prime }}{v}^j,w_{j^{\prime }}^{\prime }=A_{j^{\prime }}^{\prime i}{w}_i.\text{\hspace{1em}(7.29)}$$

Exercise

If $e_1=e_1^{\prime }+e_2^{\prime },e_2=e_2^{\prime }$ are a basis transformation on a two-dimensional vector space $V$, write out the matrices $[A_j^{i^{\prime }}]$ and $[A^{\prime i}]$ and the transformation equation for the components of a contravariant vector $v^i$ and a covariant vector $w_j$

The tensor transformation laws Eq. (7.23), Eq. (7.24) and (7.25) can be replaced by

$$\begin{array}{r}{T}^{\prime i^{\prime }{j}^{\prime }}=A_i^{i^{\prime }}{A}_j^{j^{\prime }}{T}^{ij},\\ T_{i^{\prime }{j}^{\prime }}^{\prime }=A_{i^{\prime }}^{\prime i}{A}_{j^{\prime }}^{\prime j}{T}_{ij},\\ T_{k^{\prime }}^{\prime i^{\prime }}=A_i^{i^{\prime }}{A}_{k^{\prime }}^{\prime k}{T}_k^i.\end{array}$$

When transformation laws are displayed in this notation the placement of the indices immediately determines whether $A_i^{i^{\prime }}$ or ${{\mathbf{A}}^{\prime }}_{j^{\prime }}^j$ is to be used, as only one of them will give rise to a formula obeying the conventions of summation convention and kernel index notation.

Exercise

Show that the components of the tensor $\delta$ are the same in all bases by

(a) showing $e_i\otimes {\epsilon }^i=e_{i^{\prime }}^{\prime }\otimes {\epsilon }^{\prime i^{\prime }}$ , and

(b) using the transformation law Eq. (7.25).

Now let $T$ be a general tensor of type $(r,s)$

$$T=T_{j_1\dots j_s}^{i_1\dots i_r}{e}_{i_1}\otimes \cdots \otimes e_{i_r}\otimes {\epsilon }^{j_1}\otimes \cdots \otimes {\epsilon }^{j_s}$$

where

$$T_{j_1\dots j_s}^{i_1\dots i_r}=T({\epsilon }^{i_1},\dots ,{\epsilon }^{i_r},e_{j_1},\dots e_{j_s}).$$

The separation in spacing between contravariant indices and covariant indices is not strictly necessary but has been done partly for visual display and also to anticipate a further operation called ‘raising and lowering indices’, which is available in inner product spaces. The transformation of the components of $T$ is given by

$$\begin{array}{rl}& T_{j_1^{\prime },\dots ,j_s^{\prime }}^{\prime i_1^{\prime },\dots ,i_r^{\prime }}=T({\epsilon }^{\prime i_1^{\prime }},\dots ,{\epsilon }^{\prime i_r^{\prime }},e_{j_1^{\prime }}^{\prime },\dots ,e_{j_s^{\prime }}^{\prime })\\ & =T(A_{i_1}^{i_i^{\prime }}{\epsilon }^{i_1},\dots ,A_{i_r}^{i_r^{\prime }}{\epsilon }^{i_r},A_{j_1^{\prime }}^{\prime j_1},\dots ,A_{j_s^{\prime }}^{\prime j_s}{e}_{j_s})\\ & =A_{i_1}^{i_1^{\prime }}\dots A_{i_r}^{i_r^{\prime }}{A}_{j_1^{\prime }}^{\prime j_1}\dots A_{j_s^{\prime }}^{\prime j_s}{T}_{j_1\dots j_s}^{i_1\dots i_r}.\end{array}\text{\hspace{1em}(7.30)}$$

The general tensor transformation law of components merely replicates the contravariant and covariant transformation law given in Eq. (7.29) for each contravariant and covariant index separately. The final formula Eq. (7.30) compactly expresses a multiple summation that can represent an enormous number of terms, even in quite simple cases. For example in four dimensions a tensor of type (3, 2) has $4^{3+2}=1024$ components. Its transformation law therefore consists of 1024 separate formulae, each of which has in it a sum of 1024 terms that themselves are products of six indexed entities. Including all indices and primes on indices, the total number of symbols used would be that occurring in about 20 typical books.

Problems

Problem 7.10

Let $e_1,e_2$ and $e_3$ be a basis of a vector space $V$ and $e_{i^{\prime }}^{\prime }$ a second basis given by

$$\begin{array}{l}{e}_1^{\prime }=e_1-e_2,\\ e_2^{\prime }=e_3,\\ e_3^{\prime }=e_1+e_2.\end{array}$$

(a) Display the transformation matrix ${\mathsf{A}}^{\prime }=[A_{i^{\prime }}^{\prime i}]$

(b) Express the original basis $e_i$ in terms of the $e_{i^{\prime }}^{\prime }$ and write out the transformation matrix $\mathsf{A}=[A_j^{j^{\prime }}]$

(c) Write the old dual basis ${\epsilon }^i$ in terms of the new dual basis ${\epsilon }^{\prime }{i}^{\prime }$ and conversely.

(d) What are the components of the tensors $T=e_1\otimes e_2+e_2\otimes e_1+e_3\otimes e_3$ and $S=e_1\otimes {\epsilon }^1+3e_1\otimes {\epsilon }^3-2e_2\otimes {\epsilon }^3-e_3\otimes {\epsilon }^1+4e_3\otimes {\epsilon }^2$ in terms of the basis $e_i$ and its dual basis?

(e) What are the components of these tensors in terms of the basis $e_{i^{\prime }}^{\prime }$ and its dual basis?

Problem 7.11

Let $V$ be a vector space of dimension 3, with basis $e_1,e_2,e_3$ . Let $T$ be the contravariant tensor of rank 2 whose components in this basis are $T^{ij}={\delta }^{ij}$ , and let $S$ be the covariant tensor of rank 2 whose components are given by $S_{ij}={\delta }_{ij}$ in this basis. In a new basis $e_1^{\prime },e_2^{\prime },e_3^{\prime }$ defined by

$$\begin{array}{l}{e}_1^{\prime }=e_1+e_3\\ e_2^{\prime }=2e_1+e_2\\ e_3^{\prime }=3e_2+e_3\end{array}$$

calculate the components $T^{\prime i^{\prime }{j}^{\prime }}$ and $S_{i^{\prime }{j}^{\prime }}^{\prime }$

Problem 7.12

Let $T:V\to V$ be a linear operator on a vector space $V$. Show that its components $T_j^i$ given by Eq. (3.6) are those of the tensor $\hat{T}$ defined by

$$\hat{T}(\omega ,v)=⟨\omega ,Tv⟩.$$

Prove that they are also the components with respect to the dual basis of a linear operator $T^{\ast }:V^{\ast }\to V^{\ast }$ defined by

$$⟨T^{\ast }\omega ,v⟩=⟨\omega ,Tv⟩.$$

Show that tensors of type $(r,s)$ are in one-to-one correspondence with linear maps from $V^{(s,0)}$ to $V^{(r,0)}$ , or equivalently from $V^{(0,r)}$ to $V^{(0,s)}$

Problem 7.13

Let $T:V\to V$ be a linear operator on a vector space $V$. Show that its components $T_j^i$ defined through Eq. (3.6) transform as those of a tensor of rank (1,1) under an arbitrary basis transformation.

Problem 7.14

Show directly from Eq. (7.14) and the transformation law of components $g_{ij}$

$$g_{j^{\prime }{k}^{\prime }}^{\prime }=g_{jk}{A}_{j^{\prime }}^{\prime j}{A}_{k^{\prime }}^{\prime k},$$

that the components of an inverse metric tensor $g^{ij}$ transform as a contravariant tensor of degree 2,

$$g^{\prime i^{\prime }{k}^{\prime }}=A_l^{i^{\prime }}{g}^{lk}{A}_k^{k^{\prime }}.$$