Levi-civita Symbol

DEFINITION

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

:

arepsilon_{ijk} = 
\begin{cases}
+1 & \mbox{if } (i,j,k) \mbox{ is } (1,2,3), (2,3,1) \mbox{ or } (3,1,2), \
-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) and −1 if odd.

For example, in Linear Algebra , the Determinant of a 3×3 matrix A can be written
:

\sum_{i,j,k=1}^3 arepsilon_{ijk} a_{1i} a_{2j} a_{3k}

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

and the Cross Product of two Vector s can be written as a determinant:
:

\mathbf{a 	imes 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 arepsilon_{ijk} \mathbf{e_i} a_j b_k

or more simply:
:

\mathbf{a 	imes b} = \mathbf{c},\ c_i = \sum_{j,k=1}^3 arepsilon_{ijk} a_j b_k

This can be further simplified by using Einstein Notation .

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

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

ight.

Thus, it is the Sign Of The Permutation in the case of a permutation, and zero otherwise.

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.

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

:

arepsilon_{ijk}arepsilon_{lmn} = \delta_{il}\delta_{jm}\delta_{kn} + \delta_{im}\delta_{jn}\delta_{kl} + \delta_{in}\delta_{jl}\delta_{km} - \delta_{il}\delta_{jn}\delta_{km} - \delta_{in}\delta_{jm}\delta_{kl} - \delta_{im}\delta_{jl}\delta_{kn}

\sum_{i=1}^3 arepsilon_{ijk}arepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}

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

Furthermore, it can be shown that
:

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

is always fulfilled in ''n'' dimensions. In index-free tensor notation, the Levi-Civita symbol is replaced by the concept of the Hodge Dual .

PROPERTIES

''(superscipts should be considered equivalent with subscripts)''

1. When

n=2

, we have for all

i,j,m,n

\{1,2\}

,
::

arepsilon_{ij} arepsilon^{mn}

=

\delta_i^m \delta_j^n - \delta_i^n \delta_j^m

, (1)
::

arepsilon_{ij} arepsilon^{in}

=

\delta_j^n

, (2)
::

arepsilon_{ij} arepsilon^{ij}=2

. (3)

2. When

n=3

, we have for all

i,j,k,m,n

\{1,2,3\},

arepsilon_{jmn} arepsilon^{imn}=2\delta^i_j,

(4)
::

arepsilon_{ijk} arepsilon^{ijk}=6.

(5)

Proofs

For equation 1, both sides are Antisymmetric with respect of

ij

and

mn

. We therefore only need to consider the case

i
eq j

and

m
eq n

. By substitution, we see that the equation holds for

arepsilon_{12} arepsilon^{12}

, i.e., for

i=m=1

and

j=n=2

. (Both sides are then one). Since the equation is antisymmetric in

ij

and

mn

, any set of values for these can be reduced to the above case (which holds). The equation thus holds for all values of

ij

and

mn

. Using equation 1, we have for equation 2

arepsilon_{ij}arepsilon^{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

2

. Equation 3 follows similarly from equation 2. To establish equation 4, let us first observe that both sides vanish when

i
eq j

. Indeed, if

i
eq 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

arepsilon_{jmn} arepsilon^{imn} = (arepsilon^{imn})^2 = 1

(no summation), and the result follows. The last property follows since

3!=6

and for any distinct indices

i,j,k

\{1,2,3\}

, we have

arepsilon_{ijk} arepsilon^{ijk}=1

(no summation).

\Box

EXAMPLES

1. The determinant of an

n	imes n

matrix

A=(a_{ij})

can be written as

::

\det A = arepsilon_{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 = rac{1}{n!} arepsilon_{i_1\cdots i_n} arepsilon_{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	imes B)^i = arepsilon^{ijk} A^j B^k.

For instance, the first component of

A	imes B

A^2 B^3-A^3 B^2

. From the above expression for the cross product, it is clear that

A	imes B = -B	imes 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	imes C) = arepsilon^{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	imes C)= -B\cdot(A	imes 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

::

(
abla 	imes F)^i(x) = arepsilon^{ijk}rac{\partial}{\partial x^j} F^k(x).

NOTATION

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

M_{

{Link without Title} } = rac{1}{2}arepsilon_{ab} arepsilon^{cd} M_{dc} = \, rac{1}{2}(M_{ab} - M_{ba})

and for a rank 3 tensor T,

T_{

{Link without Title} } = \, rac{1}{3!}(T_{abc}-T_{acb}+T_{bca}-T_{bac}+T_{cab}-T_{cba})

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 ).''

	CATEGORIES ABOUT LEVI-CIVITA SYMBOL

	linear algebra

	tensors

	permutations

	articles containing proofs

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Levi-civita Symbol