Levi-Civita symbol: Definition from Answers.com

The Levi-Civita symbol, also called the permutation symbol, antisymmetric symbol, or alternating 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 Tensor density
6 See also
7 References

Definition

Visualization of the Levi-Civita symbol.

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

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

i.e. $\varepsilon_{ijk}$ 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.

The formula for the three dimensional Levi-Civita symbol is:

$\varepsilon_{ijk} = \frac{\left( j-i \right)\left( k-i \right)\left( k-j \right)}{2}$

The formula in four dimensions is:

$\varepsilon_{ijkl} = \frac{\left( j-i \right)\left( k-i \right)\left( l-i \right)\left( k-j \right)\left( l-j \right)\left( l-k \right)}{12}$

Visualization of the Levi-Civita-Symbol

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

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

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

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

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

or more simply:

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

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

The tensor whose components in an orthonormal basis 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 acquires a minus sign. Because the Levi-Civita symbol is a pseudotensor, the result of taking a cross product is a pseudovector, not a vector.

Note that under a general coordinate change, the components of the permutation tensor get multiplied by the jacobian of the transformation matrix. This implies that in coordinate frames different from the one in which the tensor was defined, its components can differ from those of the Levi-Civita symbol by an overall factor. If the frame is orthonormal, the factor will be ±1 depending on whether the orientation of the frame is the same or not.

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: $\varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}$ )

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

Generalization to n dimensions

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

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

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

The generalized formula is:

$\varepsilon_{a_1 a_2 a_3 \ldots a_n} = \prod_{i=1}^{n-1} \left( \frac{1}{i!} \ \prod_{j=i+1}^n ( a_j-a_i ) \right)$

where n is the dimension (rank).

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.

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)

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).

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

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

Notation

Main article: Antisymmetric tensor

A shorthand notation for anti-symmetrization is denoted by a pair of square brackets. For example, in arbitrary dimensions, for a rank 2 covariant tensor M,

$M_{[ab]} = \frac{1}{2!}(M_{ab} - M_{ba}) \,,$

and for a rank 3 covariant tensor T,

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

In three dimensions, these are equivalent to

$M_{[ab]} = \varepsilon_{abc} \, \frac{1}{2!} \, \varepsilon^{dec} M_{de} \,,$

$T_{[abc]} = \varepsilon_{abc} \, \frac{1}{3!} \, \varepsilon^{def} T_{def} \,.$

While in four dimensions, these are equivalent to

$M_{[ab]} = \frac{1}{2!} \, \varepsilon_{abcd} \, \frac{1}{2!} \, \varepsilon^{efcd} M_{ef} \,,$

$T_{[abc]} = \varepsilon_{abcd} \, \frac{1}{3!} \, \varepsilon^{efgd} T_{efg} \,.$

Tensor density

The Levi-Civita symbol may be considered to be a tensor density in two different ways. It may be regarded as a contravariant tensor density of weight +1 or as a covariant tensor density of weight -1. In four dimensions,

$\epsilon^{\alpha \beta \gamma \delta} = \epsilon_{\alpha \beta \gamma \delta} \,.$

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 Creative Commons Attribution/Share-Alike License.

This entry is from Wikipedia, the leading user-contributed encyclopedia. It may not have been reviewed by professional editors (see full disclaimer)

Donate to Wikimedia

	What does Civitas Dei mean? Read answer...
	What does Princeps Civitas mean? Read answer...
	What does the Latin word 'civitas' mean? Read answer...

	What is Tripolis Syric Civitas mean?
	What is levies?
	What are Levi's?

		Sci-Tech Dictionary. McGraw-Hill Dictionary of Scientific and Technical Terms. Copyright © 2003, 1994, 1989, 1984, 1978, 1976, 1974 by McGraw-Hill Companies, Inc. All rights reserved. Read more
		Wikipedia. This article is licensed under the Creative Commons Attribution/Share-Alike License. It uses material from the Wikipedia article "Levi-Civita symbol". Read more

Levi-Civita symbol

Contents