7.4 Operations on tensors

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Contraction

The process of tensor product Eq. (7.15) creates tensors of higher degree from those of lower degrees,

$$\otimes :V^{(r,s)}\times V^{(p,q)}\to V^{(r+p,s+q)}.$$

We now describe an operation that lowers the degree of tensor. Firstly, consider a mixed tensor $\mathcal{T}={\mathcal{T}}_j^i{e}_i\otimes {\epsilon }^j$ of type (1, 1). Its contraction is defined to be a scalar denoted $C_1^{1}T$ given by

$$C_1^{1}T=T({\epsilon }^i,e_i)=T({\epsilon }^1,e_1)+\cdots +T({\epsilon }^n,e_n).$$

Although a basis of $V$ and its dual basis have been used in this definition, it is independent of the choice of basis, for if $e_{i^{\prime }}^{\prime }=A^{\prime }{}_{i^{\prime }}^i{e}_i$ is any other basis then

$$\begin{array}{rl}T({\epsilon }^{\prime i^{\prime }},e_{i^{\prime }}^{\prime })& =T(A_i^{i^{\prime }}{\epsilon }^i,A_{i^{\prime }}^{\prime k}{e}_k)\\ & =A_i^{i^{\prime }}{A}_{i^{\prime }}^{\prime k}T({\epsilon }^i,e_k)\\ & ={\delta }_i^{k}T({\epsilon }^i,e_k)\text{using Eq. (7.27)}\\ & =T({\epsilon }^i,e_i).\end{array}$$

In components, contraction is written

$$C_1^{1}T=T_i^i=T_1^1+T_2^2+\dots T_n^n.$$

This is a basis-independent expression since

$$T_{i^{\prime }}^{\prime i^{\prime }}=T_j^i{A}_i^{i^{\prime }}{A}_{i^{\prime }}^{\prime j}=T_j^i{\delta }_i^j=T_i^i.$$

Exercise

If $T=u\otimes \omega$, show that its contraction is $C_1^{1}T=\omega (u)$

More generally, for a tensor $T$ of type $(r,s)$ with both $r>0$ and $s>0$ one can define its $(p,q)$-contraction $(1\le p\le r,1\le q\le s)$ to be the tensor $C_q^{p}T$ of type $(r-1,s-1)$ defined by

$$\begin{array}{c}(C_q^{p}T)({\omega }^1,\dots ,{\omega }^{r-1},v_1,\dots ,v_{s-1})\\ ={\sum }_{k=1}^{n}T({\omega }^1,\dots ,{\omega }^{p-1},{\epsilon }^k,{\omega }^{p+1},\dots ,{\omega }^{r-1},\\ v_1,\dots ,v_{q-1},e_k,v_{q+1},\dots ,v_{s-1}).\end{array}\text{\hspace{1em}(7.31)}$$

Exercise

Show that the definition of $C_q^{p}T$ is independent of choice of basis. The proof is essentially identical to that for the case $r=s=1$

On substituting ${\omega }^1={\epsilon }^{i_1},\dots ,v_1=e_{j_1},\dots ,\mathrm{etc}$., we arrive at an expression for the $(p,q)$ contraction in terms of components,

$$(C_q^{p}T{)}_{j_1\dots j_{s-1}}^{i_1\dots i_{r-1}}=T_{j_1\dots j_{q-1}k{j}_{q+1}\dots j_{s-1}}^{i_1\dots i_{p-1}k{i}_{p+1}\dots i_{r-1}}.\text{\hspace{1em}(7.32)}$$

Example 7.11

Let $T$ be a tensor of type (2, 3) having components $T^{ij}{}_{klm}$ . Set $A=C_2^{1}TB=C_3^{2}T$ and $D=C_1^{1}T$ . In terms of the components of $T$ ,

$$\begin{array}{r}{A}_{km}^j=T_{kim}^{ij},\\ B_{kl}^i=T_{klj}^{ij},\\ D_{lm}^j=T_{ilm}^{ij}.\end{array}$$

Typical contraction properties of the special mixed tensor $\delta$ defined in Example 7.10 are illustrated in the following formulae:

$$\begin{array}{c}{\delta }_j^i{T}_{kl}^j=T_{kl}^i,\\ {\delta }_j^i{S}_{ik}^{lm}=S_{jk}^{lm},\\ {\delta }_i^i=\underset{n}{\underbrace{1+1+\cdots +1}}=n=\dim V.\end{array}$$

Exercise

Write these equations in $C_q^p$ form.

Raising and lowering indices

Let $V$ be a real inner product space with metric tensor $g=g_{ij}{\epsilon }^i\otimes {\epsilon }^j$ such that

$$u\cdot v=g_{ij}{u}^i{v}^j=C_1^1{C}_2^{2}g\otimes u\otimes v.$$

By Theorem 5.1 the components $g_{ij}$ are a non-singular matrix, so that $det[g_{ij}]\ne 0$ . As shown in Example 7.8 there is a tensor $g^{-1}$ whose components, written $g^{ij}=g^{ji}$ , form the inverse matrix ${\mathsf{G}}^{-1}$ . Given a vector $u=u^i{e}_i$ , the components of the covector $C_2^1(g\otimes u)$ can be written

$$u_i=g_{ij}{u}^j,$$

a process that is called lowering the index. Conversely, given a covector $\omega =w_i{\epsilon }^i$ , the vector $C_1^2{g}^{-1}\otimes \omega$ can be written in components

$$w^i=g^{ij}{w}_j,$$

and is called raising the index. Lowering and raising indices in succession, in either order, has no effect as

$$u^i={\delta }_j^i{u}^j=g^{ik}{g}_{kj}{u}^j=g^{ik}{u}_k.$$

This is important, for without this property, the convention of retaining the same kernel letter $u$ in a raising or lowering operation would be quite untenable.

Exercise

Show that lowering the index on a vector $u$ is equivalent to applying the map $\bar{g}$ in Example 7.6 to $u$, while raising the index of a covector $\omega$ is equivalent to the map ${\bar{g}}^{-1}$ of Example 7.8.

The tensors $g$ and $g^{-1}$ can be used to raise and lower indices of tensors in general, for example

$$T_i^j=g_{ik}{T}^{kj}=g^{jk}{T}_{ik}=g_{ik}{g}^{jl}{T}_l^k,etc.$$

It is strongly advised to space out the upper and lower indices of mixed tensors for this process, else it will not be clear which ‘slot’ an index should be raised or lowered into. For example

$$S_i{}^j{}_p{}^m=S_l^{kjq}{g}_{ik}{g}_{qp}{g}^{ml}.$$

If no metric tensor is specified there is no distinction in the relative ordering of covariant and contravariant indices and they can simply be placed one above the other ${}_{\mathrm{or}}$, as often done above, the contravariant indices may be placed first followed by the covariant indices. Given the capability to raise and lower indices, however, it is important to space all indices correctly. Indeed, by lowering all superscripts every tensor can be displayed in a purely covariant form. Alternatively, it can be displayed in a purely contravariant form by raising every subscript. However, unless the indices are correctly spaced we would not know where the different indices in either of these forms came from in the original ‘unlowered’ tensor.

Example 7.12

It is important to note that while ${\delta }^i{}_j$ are components of a mixed tensor the symbol ${\delta }_{ij}$ does not represent components of a tensor of covariant degree 2. We therefore try to avoid using this symbol in general tensor analysis. However, by Theorem 5.2, for a Euclidean inner product space with positive definite metric tensor $g$ it is always possible to find an orthonormal basis $\{e_1,e_2,\dots ,e_n\}$ such that

$$e_i\cdot e_j=g_{ij}={\delta }_{ij}.$$

In this case a special restricted tensor theory called cartesian tensors is frequently employed in which only orthonormal bases are permitted and basis transformations are restricted to orthogonal transformations. In this theory ${\delta }_{ij}$ can be treated as a tensor. The inverse metric tensor then also has the same components $g^{ij}={\delta }_{ij}$ and the lowered version of any component index is identical with its raised version,

$$T_{\dots i\dots }=g_{ij}{T}_{\dots }^j\cdots ={\delta }_{ij}{T}_{\dots }^j\cdots =T^{\dots i\dots }.$$

Thus every cartesian tensor may be written with all its indices in the lower position, $T_{ijk..}$ , since raising an index has no effect on the values of the components of the tensor.

In cartesian tensors it is common to adopt the summation convention for repeated indices even when they are both subscripts. For example in the standard vector theory of three dimensional Euclidean space commonly used in mechanics and electromagnetism, one adopts conventions such as

$$\mathbf{a}\cdot \mathbf{b}=a_i{b}_i\equiv a_1{b}_1+a_2{b}_2+a_3{b}_3,$$

and

$$\mathbf{a}\times \mathbf{b}={ϵ}_{ijk}{a}_j{b}_k\equiv \sum _{i=1}^3\sum _{j=1}^3{ϵ}_{ijk}{a}_j{b}_k,$$

where the alternating symbol ${ϵ}_{ijk}$ is defined by

$${ϵ}_{ijk}=\{\begin{array}{ll}1& \text{ if }ijk\text{ is an even permutation of }123,\\ -1& \text{ if it is an odd permutation of }123,\\ 0& \text{ if any pair of }ijk\text{ are equal. }\end{array}.$$

It will be shown in Chapter 8 that with respect to proper orthogonal transformations ${ϵ}_{ijk}$ is a cartesian tensor of type (0, 3)

Symmetries

A tensor $S$ of type (0, 2) is called symmetric if $S(u,v)=S(v,u)$ for all vectors $u$, $v$ in $V$ while a tensor $A$ of type (0, 2) is called antisymmetric if $A(u,v)=-A(v,u)$

Exercise

In terms of components show that $S$ is a symmetric tensor iff $S_{ij}=S_{ji}$ and $A$ is antisymmetric iff $A_{ij}=-A_{ji}$

Any tensor $T$ oftype (0, 2) can be decomposed into a symmetric and antisymmetric part, $T=\mathcal{S}(T)+\mathcal{A}(T)$ , where

$$\begin{array}{l}\mathcal{S}(T)(u,v)=\frac{1}{2}(T(u,v)+T(v,u)),\\ \mathcal{A}(T)(u,v)=\frac{1}{2}(T(u,v)-T(v,u)).\end{array}$$

It is immediate that these tensors are symmetric and antisymmetric respectively. Setting $u=e_i,v=e_j$ , this decomposition becomes

$$T_{ij}=T(e_i,e_j)=T_{(ij)}+T_{[ij]},$$

where

$$T_{(ij)}=\mathcal{S}(T{)}_{ij}=\frac{1}{2}(T_{ij}+T_{ji})$$

and

$$T_{[ij]}=\mathcal{A}(T{)}_{ij}=\frac{1}{2}(T_{ij}-T_{ji}).$$

A similar discussion applies to tensors of type (2, 0), having components $T^{ij}$ , but one cannot talk of symmetries of a mixed tensor.

Exercise

Show that $T_j^i=T_i^j$ is not a tensor equation, since it is not invariant under basis transformations.

If $A$ is an antisymmetric tensor of type (0, 2) and $S$ a symmetric tensor of type (2, 0) then their total contraction vanishes,

$$C_1^1{C}_2^{2}A\otimes S\equiv A_{ij}{S}^{ij}=0\text{\hspace{1em}(7.33)}$$

since

$$A_{ij}{S}^{ij}=-A_{ji}{S}^{ji}=-A_{ij}{S}^{ij}.$$

Problems

Problem 7.15

Let $g_{ij}$ be the components of an inner product with respect to a basis $u_1,u_2,u_3$

$$g_{ij}=u_i\cdot u_j=\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right).$$

(a) Find an orthonormal basis of the form $e_1=u_1,e_2=u_2,e_3=a{u}_1+b{u}_2+c{u}_3$ such that $a>0$ and find the index of this inner product.

(b) If $\begin{array}{r}v=u_1+\frac{1}{2}{u}_3\end{array}$ find its lowered components $v_i$

(c) Express $v$ in terms of the orthonormal basis found above, and write out its lowered components with respect to that basis.

Problem 7.16

Let $g$ be a metric tensor on a vector space $V$ and define $T$ to be the tensor

$$T=a{g}^{-1}\otimes g+\delta \otimes u\otimes \omega$$

where $u$ is a non-zero vector of $V$ and $\omega$ is a covector.

(a) Write out the components $T^{ij}{}_{kl}$ of the tensor $T$.

(b) Evaluate the components of the following four contractions:

$$A=C_1^{1}T,B=C_2^{1}T,C=C_1^{2}T,D=C_2^{2}T$$

and show that $B=C$.

(c) Show that $D=0\mathrm{iff}\omega (u)=-a$ . Hence show that if $T_{kl}^{ij}{u}^l{u}_j=0$ , then $D=0$

(d) Show that $\mathrm{if}n=\dim V>1$ then $T_{kl}^{ij}{u}^l{u}_j=0$ if and only if $\dot{a}=\omega (u)=0\mathrm{or}{u}_i{u}^i=0$.

Problem 7.17

On a vector space $V$ of dimension $n$ let $T$ be a tensor of rank (1, 1), $S$ a symmetric tensor of rank (0, 2) and $\delta$ the usual ‘invariant tensor’ of rank (1, 1). Write out the components $R^{ij}{}_{klmr}$ of the tensor

$$R=T\otimes S\otimes \delta +S\otimes \delta \otimes T+\delta \otimes T\otimes S.$$

Perform the contraction of this tensor over $i$ and $k$, using any available contraction properties of ${{\delta }^i}_j$ . Perform a further contraction over the indices $j$ and $r$.

Problem 7.18

Show that covariant symmetric tensors of rank 2, satisfying $T_{ij}=T_{ji}$ , over a vector space $V$ of dimension $n$ form a vector space of dimension $n(n+1)/2$

(a) A tensor $S$ of type $(0,r)$ is called totally symmetric if $S_{i_1{i}_2\dots i_r}$ is left unaltered by any interchange of indices. What is the dimension of the vector space spanned by the totally symmetric tensor on $V?$

(b) Find the dimension of the vector space of covariant tensors ofrank 3 having the cyclic symmetry

$$T(u,v,w)+T(v,w,u)+T(w,u,v)=0.$$