Levi-Civita symbol

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The Levi-Civita symbol, also called the permutation symbol or antisymmetric symbol, is a mathematical symbol used in particular in tensor calculus. It is named after the Italian mathematician and physicist Tullio Levi-Civita.

1 Definition
- 1.1 Relation to Kronecker delta
- 1.2 Generalization to n dimensions
2 Properties
- 2.1 Proofs
3 Examples
4 Notation
5 References

[edit] Definition

Visualization of a Levi-Civita symbol.

In three dimensions, the Levi-Civita symbol is defined as follows:

$\varepsilon_{ijk} = \begin{cases} +1 & \mbox{if } (i,j,k) \mbox{ is } (1,2,3), (3,1,2) \mbox{ or } (2,3,1), \\ -1 & \mbox{if } (i,j,k) \mbox{ is } (3,2,1), (1,3,2) \mbox{ or } (2,1,3), \\ 0 & \mbox{otherwise: }i=j \mbox{ or } j=k \mbox{ or } k=i, \end{cases}$

i.e. it is 1 if (i, j, k) is an even permutation of (1,2,3), −1 if it is an odd permutation, and 0 if any index is repeated.

An alternative notation for the three dimensional Levi-Civita symbol without case differentiation is

$\varepsilon_{ijk} = -[(i-j)^2%3][(i-k)^2%3][(j-k)^2%3][(j-(i%3)-\frac{1}{2})^2-\frac{5}{4}]$

with $%$ representing the modulo operator and $i,j,k \in \lbrace1,2,3\rbrace$ .

For example, in linear algebra, the determinant of a 3×3 matrix A can be written

$\det A = \sum_{i,j,k=1}^3 \varepsilon_{ijk} a_{1i} a_{2j} a_{3k}$

(and similarly for a square matrix of general size, see below)

and the cross product of two vectors can be written as a determinant:

$\mathbf{a \times b} = \begin{vmatrix} \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_3} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{vmatrix} = \sum_{i,j,k=1}^3 \varepsilon_{ijk} \mathbf{e_i} a_j b_k$

or more simply:

$\mathbf{a \times b} = \mathbf{c},\ c_i = \sum_{j,k=1}^3 \varepsilon_{ijk} a_j b_k.$

According to the Einstein notation, the summation symbol may be omitted.

The tensor whose components are given by the Levi-Civita symbol (a tensor of covariant rank n) is sometimes called the permutation tensor. It is actually a pseudotensor because under an orthogonal transformation of jacobian determinant −1 (i.e., a rotation composed with a reflection), it gets a -1. Because the Levi-Civita symbol is a pseudotensor, the result of taking a cross product is a pseudovector, not a vector.

[edit] Relation to Kronecker delta

The Levi-Civita symbol is related to the Kronecker delta. In three dimensions, the relationship is given by the following equations:

$\varepsilon_{ijk}\varepsilon_{lmn} = \det \begin{bmatrix} \delta_{il} & \delta_{im}& \delta_{in}\\ \delta_{jl} & \delta_{jm}& \delta_{jn}\\ \delta_{kl} & \delta_{km}& \delta_{kn}\\ \end{bmatrix}$

$= \delta_{il}\left( \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}\right) - \delta_{im}\left( \delta_{jl}\delta_{kn} - \delta_{jn}\delta_{kl} \right) + \delta_{in} \left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl} \right) \,$

$\sum_{i=1}^3 \varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$ ("contracted epsilon identity")

(In Einstein notation, the duplication of the i index implies the sum on i. The previous is then noted: $\varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$ )

$\sum_{i,j=1}^3 \varepsilon_{ijk}\varepsilon_{ijn} = 2\delta_{kn}$

[edit] Generalization to n dimensions

The Levi-Civita symbol can be generalized to higher dimensions:

$\varepsilon_{ijkl\dots} = \left\{ \begin{matrix} +1 & \mbox{if }(i,j,k,l,\dots) \mbox{ is an even permutation of } (1,2,3,4,\dots) \\ -1 & \mbox{if }(i,j,k,l,\dots) \mbox{ is an odd permutation of } (1,2,3,4,\dots) \\ 0 & \mbox{if any two labels are the same} \end{matrix} \right.$

Thus, it is the sign of the permutation in the case of a permutation, and zero otherwise.

Furthermore, for any n the property

$\sum_{i,j,k,\dots=1}^n \varepsilon_{ijk\dots}\varepsilon_{ijk\dots} = n!$

follows from the facts that (a) every permutation is either even or odd, (b) (+1)² = (-1)² = 1, and (c) the permutations of any n-element set number exactly n!.

In index-free tensor notation, the Levi-Civita symbol is replaced by the concept of the Hodge dual.
In general $n$ dimensions one can write the product of two Levi-Civita symbols as:

$\varepsilon_{ijk\dots}\varepsilon_{mnl\dots} = \det \begin{bmatrix} \delta_{im} & \delta_{in} & \delta_{il} & \dots \\ \delta_{jm} & \delta_{jn} & \delta_{jl} & \dots \\ \delta_{km} & \delta_{kn} & \delta_{kl} & \dots \\ \vdots & \vdots & \vdots \\ \end{bmatrix}$ .

Now we can contract $m$ indices. This will add a factor of $m!$ to the determinant and we need to omit the relevant Kronecker delta.

[edit] Properties

(in these examples, superscripts should be considered equivalent with subscripts)

1. When $n = 2$ , we have for all $i, j, m, n$ in ${1,2}$ ,

$\varepsilon_{ij} \varepsilon^{mn}$

=

$\delta_i^m \delta_j^n - \delta_i^n \delta_j^m$ , (1)

$\varepsilon_{ij} \varepsilon^{in}$

=

$\delta_j^n$ , (2)

$\varepsilon_{ij} \varepsilon^{ij}=2$ . (3)

2. When $n = 3$ , we have for all $i, j, k, m, n$ in ${1,2,3},$

$\varepsilon_{jmn} \varepsilon^{imn}=2\delta^i_j$ , (4)

$\varepsilon_{ijk} \varepsilon^{ijk}=6$ , (5)

$\varepsilon_{ijk} \varepsilon^{imn}=\delta^{m}_j\delta^{n}_k - \delta^{n}_j\delta^{m}_k$ . (6)

[edit] Proofs

For equation 1, both sides are antisymmetric with respect of $i j$ and $m n$ . We therefore only need to consider the case $i\neq j$ and $m\neq n$ . By substitution, we see that the equation holds for $\varepsilon_{12} \varepsilon^{12}$ , i.e., for $i = m = 1$ and $j = n = 2$ . (Both sides are then one). Since the equation is antisymmetric in $i j$ and $m n$ , any set of values for these can be reduced to the above case (which holds). The equation thus holds for all values of $i j$ and $m n$ . Using equation 1, we have for equation 2 $\varepsilon_{ij}\varepsilon^{in}$ $=$ $\delta_i^i \delta_j^n - \delta^n_i \delta^i_j$

=

$2 \delta_j^n - \delta^n_j$

=

$\delta_j^n$ .

Here we used the Einstein summation convention with $i$ going from $1$ to $2$ . Equation 3 follows similarly from equation 2. To establish equation 4, let us first observe that both sides vanish when $i\neq j$ . Indeed, if $i\neq j$ , then one can not choose $m$ and $n$ such that both permutation symbols on the left are nonzero. Then, with $i = j$ fixed, there are only two ways to choose $m$ and $n$ from the remaining two indices. For any such indices, we have $\varepsilon_{jmn} \varepsilon^{imn} = (\varepsilon^{imn})^2 = 1$ (no summation), and the result follows. Property (5) follows since $3! = 6$ and for any distinct indices $i, j, k$ in ${1,2,3}$ , we have $\varepsilon_{ijk} \varepsilon^{ijk}=1$ (no summation).

[edit] Examples

1. The determinant of an $n\times n$ matrix $A = (a i j)$ can be written as

$\det A = \varepsilon_{i_1\cdots i_n} a_{1i_1} \cdots a_{ni_n},$

where each $i l$ should be summed over $1,\ldots, n.$

Equivalently, it may be written as

$\det A = \frac{1}{n!} \varepsilon_{i_1\cdots i_n} \varepsilon_{j_1\cdots j_n} a_{i_1 j_1} \cdots a_{i_n j_n},$

where now each $i l$ and each $j l$ should be summed over $1,\ldots, n$ .

2. If $A = (A 1, A 2, A 3)$ and $B = (B 1, B 2, B 3)$ are vectors in $R 3$ (represented in some right hand oriented orthonormal basis), then the $i$ th component of their cross product equals

$(A\times B)^i = \varepsilon^{ijk} A^j B^k.$

For instance, the first component of $A\times B$ is $A 2 B 3 - A 3 B 2$ . From the above expression for the cross product, it is clear that $A\times B = -B\times A$ . Further, if $C = (C 1, C 2, C 3)$ is a vector like $A$ and $B$ , then the triple scalar product equals

$A\cdot(B\times C) = \varepsilon^{ijk} A^i B^j C^k.$

From this expression, it can be seen that the triple scalar product is antisymmetric when exchanging any adjacent arguments. For example, $A\cdot(B\times C)= -B\cdot(A\times C)$ .

3. Suppose $F = (F 1, F 2, F 3)$ is a vector field defined on some open set of $R 3$ with Cartesian coordinates $x = (x 1, x 2, x 3)$ . Then the $i$ th component of the curl of $F$ equals

$(\nabla \times F)^i(x) = \varepsilon^{ijk}\frac{\partial}{\partial x^j} F^k(x).$

[edit] Notation

Main article: Symmetric tensor#Properties

A shorthand notation for anti-symmetrization is denoted by a pair of square brackets. For example, for an n x n matrix, M,

$M_{[ab]} = \frac{1}{2}\varepsilon_{abc} \varepsilon^{cde} M_{de } = \, \frac{1}{2}(M_{ab} - M_{ba})$

and for a rank 3 tensor T,

$T_{[abc]} = \, \frac{1}{3!}(T_{abc}-T_{acb}+T_{bca}-T_{bac}+T_{cab}-T_{cba})$

[edit] References

Charles W. Misner, Kip S. Thorne, John Archibald Wheeler, Gravitation, (1970) W.H. Freeman, New York; ISBN 0-7167-0344-0. (See section 3.5 for a review of tensors in general relativity).

This article incorporates material from Levi-Civita permutation symbol on PlanetMath, which is licensed under the GFDL.

Levi-Civita symbol

From Wikipedia, the free encyclopedia

Contents

[edit] Definition

[edit] Relation to Kronecker delta

[edit] Generalization to n dimensions

[edit] Properties

[edit] Proofs

[edit] Examples

[edit] Notation

[edit] References

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