8.5 Oriented vector spaces

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$n$-Vectors and $n$-forms

Let $V$ be a vector space of dimension $n$ with basis $\{e_1,...,e_n\}$ and dual basis $\{{\epsilon }^1,\dots ,{\epsilon }^n\}$ Since the spaces ${\Lambda }^n(V)$ and ${\Lambda }^{\ast n}(V)$ are both one-dimensional, the $n$-vector

$$E=e_{12\dots n}=e_1\wedge e_2\wedge \cdots \wedge e_n$$

forms a basis of ${\Lambda }^n(V)$ , while the $n$-form

$$\Omega ={\epsilon }^{12\dots n}={\epsilon }^1\wedge \cdots \wedge {\epsilon }^n$$

is a basis of ${\Lambda }^{\ast n}(V)$ . These will sometimes be referred to as volume elements associated with this basis.

Exercise

Show that every non-zero $n$-vector is the volume element associated with some basis of $V$.

Given a basis $\{e_1,...,e_n\}$ , every $n$-vector $A$ has a unique expansion

$$\begin{array}{l}A=aE=a{e}_1\wedge e_2\wedge \cdots \wedge e_n\\ =\frac{a}{n!}{\sum }_{\sigma }(-1{)}^{\sigma }{e}_{\sigma (1)}\otimes e_{\sigma (2)}\otimes \cdots \otimes e_{\sigma (n)}\\ =\frac{a}{n!}{ϵ}^{i_1{i}_2\dots i_n}{e}_{i_1}\otimes e_{i_2}\otimes \cdots \otimes e_{i_n}\end{array}$$

where the epsilon-symbols, or Levi–Civita symbols, ${ϵ}^{i_1{i}_2\dots i_n}$ and ${ϵ}_{i_1{i}_2\dots i_n}$ are defined by

$${ϵ}^{i_1{i}_2\dots i_n}={ϵ}_{i_1{i}_2\dots i_n}=\{\begin{array}{ll}1& \text{if }{i}_1\dots i_n\text{ is an even permutation of }1,2,\dots ,n,\\ -1& \text{if }{i}_1\dots i_n\text{ is an odd permutation of }1,2,\dots ,n,\\ 0& \text{if any pair of indices are equal.}\end{array}\text{\hspace{1em}(8.21)}$$

The epsilon-symbols are clearly antisymmetric in any pair of indices.

Exercise

Show that any $n$-form $\beta$ has a unique expansion

$$\beta =b\Omega =\frac{b}{n!}{ϵ}_{i_1\dots i_n}{\epsilon }^{i_1}\otimes \cdots \otimes {\epsilon }^{i_n}.$$

Every $n$-vector and $n$-form has tensor components proportional to the epsilon-symbols,

$$\begin{array}{l}{A}^{i_1\dots i_n}=\frac{a}{n!}{ϵ}^{i_1\dots i_n},\\ {\beta }_{i_1\dots i_n}=\frac{b}{n!}{ϵ}_{i_1\dots i_n},\end{array}$$

and setting $a=b=1$ we have

$$E^{i_1\dots i_n}=\frac{1}{n!}{ϵ}^{i_1\dots i_n},{\Omega }_{i_1\dots i_n}=\frac{1}{n!}{ϵ}_{i_1\dots i_n}.\text{\hspace{1em}(8.22)}$$

Transformation laws of $n$-vectors and $n$-forms

The transformation matrix $\mathsf{A}=[A_i^{i^{\prime }}]$ appearing in Eq. (7.26) satisfie

$$A_{i_1}^{i_1^{\prime }}{A}_{i_2}^{i_2^{\prime }}\dots A_{i_n}^{i_n^{\prime }}{ϵ}^{i_1{i}_2\dots i_n}=\det [A_i^{i^{\prime }}]{ϵ}^{i_1^{\prime }{i}_2^{\prime }\dots i_n^{\prime }}.\text{\hspace{1em}(8.23)}$$

Proof

If $i_1^{\prime }=1,i_2^{\prime }=2,...,i_n^{\prime }=n$ then the left-hand side is the usual expansion of the determinant of the matrix $A$ as a sum of products of its elements taken from different rows and columns with appropriate signs, while the right-hand side is $det$ $\mathsf{A}{ϵ}^{12\dots n}=\det \mathsf{A}$ From the antisymmetry of the epsilon symbol in any pair of indices $i$ and $j$ we have

$$\dots A_i^{i^{\prime }}\dots A_j^{j^{\prime }}\dots {ϵ}^{\dots i\dots j\dots }=-\dots A_j^{i^{\prime }}\dots A_i^{j^{\prime }}\dots {ϵ}^{\dots i\dots j\dots }=-\dots A_i^{j^{\prime }}\dots A_j^{i^{\prime }}\dots {ϵ}^{\dots i\dots j\dots },$$

and the whole expression is antismmetric in any pair of indices $i^{\prime },j^{\prime }$ . In particular, it vanishes if $i^{\prime }=j^{\prime }$ . Hence $\mathrm{if}\pi$ is any permutation of $(1,2,\dots ,n)$ ) such that $i_1^{\prime }=\pi (1),i_2^{\prime }=\pi (2),\dots ,i_n^{\prime }=\pi (n)$ then both sides ofEq. (8.23) are multiplied by the sign ofthe permutation $(-1{)}^{\pi }$ , while if any pair of indices $i_1^{\prime }\dots i_n^{\prime }$ are equal, both sides of the equation vanish.

If $A=aE$ is any $n$-vector, then from the law of transformation of tensor components, $Eq.(7.30)$

$$\begin{array}{rl}& A^{i_1^{\prime }\dots i_n^{\prime }}=A^{i_1\dots i_n}{A}_{i_1}^{i_1^{\prime }}\dots A_{i_n}^{i_n^{\prime }}\\ & =\frac{a}{n!}{ϵ}^{i_1\dots i_n}{A}_{i_1}^{i_1^{\prime }}\dots A_{i_n}^{i_n^{\prime }}\\ & =\det [A_i^{j^{\prime }}]\frac{a}{n!}{ϵ}^{i_1^{\prime }\dots i_n^{\prime }}.\end{array}$$

Setting $A=a^{\prime }{E}^{\prime }$ , the factor $a$ is seen to transform as

$$a^{\prime }=a\det [A_j^{i^{\prime }}]=a\det \mathsf{A}.\text{\hspace{1em}(8.24)}$$

If $a=1$ we arrive at the transformation law of volume elements,

$$E=\det \mathsf{A}{E}^{\prime },E^{\prime }=\det {\mathsf{A}}^{-1}E.\text{\hspace{1em}(8.25)}$$

A similar formula to Eq. (8.23) holds for the inverse matrix ${\mathsf{A}}^{\prime }=[A^{\prime }{}_{j^{\prime }}^i]$ ,

$$A_{i_1^{\prime }}^{i_1}{A}_{i_2^{\prime }}^{i_2}\dots A_{i_n^{\prime }}^{i_n}{ϵ}_{i_1{i}_2\dots i_n}=\det [A_{i^{\prime }}^i]{ϵ}_{i_1^{\prime }{i}_2^{\prime }\dots i_n^{\prime }},\text{\hspace{1em}(8.26)}$$

and the transformation law for an $n$-form $\begin{array}{r}\beta =b\Omega =\frac{b}{n!}{ϵ}^{12\dots n}=b^{\prime }{\Omega }^{\prime }\end{array}$ is

$$b^{\prime }=b\det [A_{j^{\prime }}^{\prime i}]=b\det {\mathsf{A}}^{-1}\text{ that is }b=b^{\prime }\det \mathsf{A},\text{\hspace{1em}(8.27)}$$

$$\Omega =\det A^{-1}{\Omega }^{\prime },{\Omega }^{\prime }=\det A\Omega .\text{\hspace{1em}(8.28)}$$

Exercise

Prove Eqs. Eq. (8.26)–Eq. (8.28).

Note, from Eqs. Eq. (8.23) and Eq. (8.26), that the epsilon-symbols do not transform as components of tensors under general basis transformations. They do however transform as tensors with respect to the restricted set ofunimodular transformations, having $det$ $\mathsf{A}=\det [{\mathbf{A}}_i^{j^{\prime }}]=1$ . In particular, for cartesian tensors they transform as tensors provided only proper orthogonal transformations, rotations, are permitted. The term ‘tensor density’ is sometimes used to refer to entities that include determinant factors like those in Eq. (8.23) and Eq. (8.26), while scalar quantities that transform like $a$ or $b$ in Eq. (8.24) and Eq. (8.27) are referred to as ‘densities’.

Oriented vector spaces

Two bases $\{e_i\}$ and $\{e_{i^{\prime }}^{\prime }\}$ are said to have the same orientation if the transformation matrix $\mathsf{A}=[A_j^{i^{\prime }}]$ in Eq. (7.26) has positive determinant, $det$ $\mathsf{A}>0\mathsf{j}$ otherwise they are said to be oppositely oriented. Writing $\{e_i\}o\{e_{i^{\prime }}^{\prime }\}$ iff $\{e_i\}$ and $\{e_{i^{\prime }}^{\prime }\}$ have the same orientation, it is straightforward to show that $o$ is an equivalence relation and divides the set of all bases on $V$ into two equivalence classes, called orientations. A vector space $V$ together with a choice oforientation is called an oriented vector space. Any basis belonging to the selected orientation will be said to be positively oriented, while oppositely oriented bases will be called negatively oriented.

Example 8.3

Euclidean three-dimensional space ${\mathbb{E}}^3$ together with choice ofa right-handed orthonormal basis {$i$. $i$. $k$} is an oriented vector space. The orientation consists of the se of all bases related to $i$, $j$, $k$ through a positive determinant transformation. A left-handed set of axes has opposite orientation since the basis transformation will involve a reflection, having negative determinant.

Example 8.4

Let $V$ be an $n$-dimensional vector space. Denote the set of all volume ele ments on $V$ by ${\dot{\Lambda }}^n(V)$ and ${\dot{\Lambda }}^{\ast n}(V)$ . Two non-zero $n$-vectors $A$ and $B$ can be said to have the same orientation if $A=cB$ with $c>0$ . This clearly provides an equivalence relation on ${\dot{\Lambda }}^n(V)$ , dividing it into two non-intersecting equivalence classes. A selection of one of these two classes is an alternative way of specifying an orientation on a vector space $V_{;}$ for we may stipulate that a basis $\left\{{\mathbf{e}}_i\right\}$ has positive orientation if $A=aE$ with $a>0$ for all volume elements $A$ in the chosen class. By Eqs. Eq. (8.24) and Eq. (8.25), this is equivalent to dividing the set of bases on $V$ into two classes.

Now let $V$ be an oriented $n$-dimensional real inner product space having index $t=r-s$ where $n=r+s.\mathrm{If}\{e_1,e_2,\dots ,e_n\}$ is any positively oriented orthonormal frame such that

$$e_i\cdot e_j={\eta }_{ij}=\{\begin{array}{ll}{\eta }_i=\pm 1& \text{ if }i=j,\\ 0& \text{ if }i\ne j,\end{array}$$

then $r$ is the number of 1’s and $s$ the number of 1’s among the ${\eta }_i$ . As pseudo-orthogona transformations all have determinant 1, those relating positively oriented orthonorma frames must have $det$ 1. Hence, by Eq. (8.25), the volume element $E=e_1\wedge e_2\wedge \cdot \cdot \cdot \wedge e_n$ is independent of the choice of positively oriented orthonormal basis and is entirely determined by the inner product and the orientation on $V.$

By Eq. (8.22), the components of the volume element $E$ with respect to any positively oriented orthonormal basis are

$$E^{i_1{i}_2\dots i_n}=\frac{1}{n!}{ϵ}^{i_1{i}_2\dots i_n}.$$

With respect to an arbitrary positively oriented basis $e_{i:}^{\prime }$ , not necessarily orthonormal, the components of $E$ are, by Eqs. (7.30) and Eq. (8.23),

$$E^{\prime i_1^{\prime }{i}_2^{\prime }\dots i_n^{\prime }}=\det [A_i^{i^{\prime }}]\frac{1}{n!}{ϵ}^{i_1^{\prime }{i}_2^{\prime }\dots i_n^{\prime }}.\text{\hspace{1em}(8.29)}$$

Take note that these are the components ofthe volume element $E$ determined by the origina orthonormal basis expressed with respect to the new basis, not the components of the volume element $e_1^{\prime }\wedge e_2^{\prime }\wedge \cdots \wedge e_n^{\prime }$ determined by the new basis. It is possible to arrive at a formula for the components on the right-hand side of Eq. (8.29) that is independent of the transformation matrix $A$. Consider the transformation of components of the metric tensor, defined by $u\cdot v=g_{ij}{u}^i{v}^j$

$$g_{i^{\prime }{j}^{\prime }}^{\prime }=g_{ij}{A}_{i^{\prime }}^{\prime i}{A}_{j^{\prime }}^{\prime j},$$

which can be written in matrix form

$${\mathsf{G}}^{\prime }=({\mathsf{A}}^{-1}{)}^T\mathsf{G}{\mathsf{A}}^{-1}\text{ where }\mathsf{G}=[g_{ij}],{\mathsf{G}}^{\prime }=[g_{i^{\prime }{j}^{\prime }}^{\prime }].$$

On taking determinants

$$g^{\prime }=g(\det \mathsf{A}{)}^{-2}\text{ where }g=\det \mathsf{G}=\pm 1,g^{\prime }=\det {\mathsf{G}}^{\prime },$$

and subsituting in Eq. (8.29) we have

$$E^{\prime i_1^{\prime }{i}_2^{\prime }\dots i_n^{\prime }}=\frac{1}{n!\sqrt{|g^{\prime }|}}{ϵ}^{i_1^{\prime }{i}_2^{\prime }\dots i_n^{\prime }}.$$

Eliminating the primes, it follows that the components of the volume element $E$ defined by the inner product can be written in an arbitrary positively oriented basis as

$$E^{i_1{i}_2\dots i_n}=\frac{1}{n!\sqrt{|g|}}{ϵ}^{i_1{i}_2\dots i_n}.\text{\hspace{1em}(8.30)}$$

On lowering the indices of $E$ we have

$$\begin{array}{rl}& E_{i_1{i}_2\dots i_n}=g_{i_1{j}_1}{g}_{i_2{j}_2}\dots g_{i_n{j}_n}{E}^{j_1{j}_2\dots j_n}\\ & =\frac{1}{n!\sqrt{|g|}}{g}_{i_1{j}_1}{g}_{i_2{j}_2}\dots g_{i_n{j}_n}{ϵ}^{j_1{j}_2\dots j_n}\\ & =\frac{1}{n!\sqrt{|g|}}g{ϵ}_{i_1{i}_2\dots i_n}.\end{array}$$

Since the sign of the determinant $g$ is equal to ( 1)${}^s$ we have, in any positively oriented basis,

$$E_{i_1{i}_2\dots i_n}=(-1{)}^s\frac{\sqrt{|g|}}{n!}{ϵ}_{i_1{i}_2\dots i_n}.\text{\hspace{1em}(8.31)}$$

Exercise

Show that the components of the $n$-form $\Omega ={\epsilon }^{12\dots n}$ defined by a positively oriented o.n. basis are

$${\Omega }_{i_1{i}_2\dots i_n}=\frac{\sqrt{|g|}}{n!}{ϵ}_{i_1{i}_2\dots i_n}=(-1{)}^s{E}_{i_1{i}_2\dots i_n}.\text{\hspace{1em}(8.32)}$$

Epsilon-symbol identities

The epsilon-symbols satisfy a number of fundamental identities, the most general of which is

$${ϵ}_{i_1\dots i_n}{ϵ}^{j_1\dots j_n}={\delta }_{i_1\dots i_n}^{j_1\dots j_n},\text{\hspace{1em}(8.33)}$$

where the generalized $\delta$-symbol is defined by

$${\delta }_{i_1\dots i_r}^{j_1\dots j_r}=\{\begin{array}{ll}1& \text{if }{j}_1\dots j_r\text{ is an even permutation of }{i}_1\dots i_r,\\ -1& \text{if }{j}_1\dots j_r\text{ is an odd permutation of }{i}_1\dots i_r,\\ 0& \text{otherwise.}\end{array}\text{\hspace{1em}(8.34)}$$

Total contraction of Eq. (8.33) over all indices gives

$${ϵ}_{i_1\dots i_n}{ϵ}^{i_1\dots i_n}=n!.\text{\hspace{1em}(8.35)}$$

Contracting Eq. (8.33) over the first $n$ 1 indices gives

$${ϵ}_{i_1\dots i_{n-1}j}{ϵ}^{i_1\dots i_{n-1}k}=(n-1)!{\delta }_j^k,\text{\hspace{1em}(8.36)}$$

for if $k\ne j$ each term in the summation ${\delta }_{i_1\dots i_{n-1}j}^{i_1\dots i_{n-1}k}$ vanishes since in every summand either one pair ofsuperscripts or one pair ofsubscripts must be equal, while if $k=j$ the expression is a sum of $(n-1)!$ ! terms each of value +1.

The most general contraction identity arising from Eq. (8.34) is

$${ϵ}_{i_1\dots i_{n-r}{j}_1\dots j_r}{ϵ}^{i_1\dots i_{n-r}{k}_1\dots k_r}=(n-r)!{\delta }_{j_1\dots j_r}^{k_1\dots k_r},\text{\hspace{1em}(8.37)}$$

where the $\delta$-symbol on the right-hand side can be expressed in terms of Kronecke

deltas,

$$\begin{array}{c}{\delta }_{j_1\dots j_r}^{k_1\dots k_r}={\sum }_{\sigma }(-1{)}^{\sigma }{\delta }_{j_{\sigma (1)}}^{k_1}{\delta }_{j_{\sigma (2)}}^{k_2}\dots {\delta }_{j_{\sigma (r)}}^{k_r}\\ ={\delta }_{j_1}^{k_1}{\delta }_{j_2}^{k_2}\dots {\delta }_{j_r}^{k_r}-{\delta }_{j_2}^{k_1}{\delta }_{j_1}^{k_2}\dots {\delta }_{j_r}^{k_r}+\dots ,\end{array}\text{\hspace{1em}(8.38)}$$

a sum $o$ $\setminus r!$ terms in which the $j_i$ indices run over every permutation of $j_1,j_2,\dots ,j_r$

Example 8.5

In three dimensions we have

$$\begin{array}{rl}& {ϵ}_{ijk}{ϵ}^{lmn}={\delta }_i^l{\delta }_j^m{\delta }_k^n-{\delta }_i^l{\delta }_j^n{\delta }_k^m+{\delta }_i^m{\delta }_j^n{\delta }_k^l-{\delta }_i^m{\delta }_j^l{\delta }_k^n+{\delta }_i^n{\delta }_j^l{\delta }_k^m-{\delta }_i^n{\delta }_j^m{\delta }_k^l,\\ & {ϵ}_{ijk}{ϵ}^{imn}={\delta }_j^m{\delta }_k^n-{\delta }_k^m{\delta }_j^n,\\ & {ϵ}_{ijk}{ϵ}^{ijn}=2{\delta }_k^n,\\ & {\delta }_{ijk}{ϵ}^{ijk}=6.\end{array}$$

The last three identities are particularly useful in cartesian tensors where the summation convention is used on repeated subscripts, giving

$${ϵ}_{ijk}{ϵ}_{imn}={\delta }_{mj}{\delta }_{nk}-{\delta }_{mk}{\delta }_{nj},{ϵ}_{ijk}{ϵ}_{ijn}=2{\delta }_{nk},{\delta }_{ijk}{ϵ}_{ijk}=6.$$

For example, the vector product $\mathbf{u}\times \mathbf{v}$ of two vectors $\mathbf{u}=u_i{\mathbf{e}}_i$ and $\mathbf{v}=v_i{\mathbf{e}}_i$ is defined as the vector whose components are given by

$$(\mathbf{u}\times \mathbf{v}{)}_k={ϵ}_{ijk}(u\wedge v{)}^{ij}={ϵ}_{kij}{u}_i{v}_j.$$

The vector identity

$$\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}\cdot \mathbf{w})\mathbf{v}-(\mathbf{u}\cdot \mathbf{v})\mathbf{w}$$

follows from

$$\begin{array}{c}(\mathbf{u}\times (\mathbf{v}\times \mathbf{w}){)}_i={ϵ}_{ijk}{u}_j{ϵ}_{klm}{v}_l{w}_m\\ =({\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl})u_j{v}_l{w}_m=(u_j{w}_j)v_i-(u_j{v}_j)w_i=((\mathbf{u}\cdot \mathbf{w})\mathbf{v}-(\mathbf{u}\cdot \mathbf{v})\mathbf{w}{)}_i.\end{array}$$

Exercise

Show the identity $(\mathbf{u}\times \mathbf{v}{)}^2={\mathbf{u}}^2{\mathbf{v}}^2-(\mathbf{u}\cdot \mathbf{v}{)}^2$