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.
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[edit] Definition
Image:Epsilontensor.svg In three dimensions, the Levi-Civita symbol is defined as follows:
- <math> \varepsilon_{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} </math>
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
- <math>
\sum_{i,j,k=1}^3 \varepsilon_{ijk} a_{1i} a_{2j} a_{3k} </math>
(and similarly for a square matrix of general size, see below)
and the cross product of two vectors can be written as a determinant:
- <math>
\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 </math> or more simply:
- <math>
\mathbf{a \times b} = \mathbf{c},\ c_i = \sum_{j,k=1}^3 \varepsilon_{ijk} a_j b_k </math>
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:
- <math>
\varepsilon_{ijk}\varepsilon_{lmn} = \det \begin{vmatrix} \delta_{il} & \delta_{im} & \delta_{in} \\ \delta_{jl} & \delta_{jm} & \delta_{jn} \\ \delta_{kl} & \delta_{km} & \delta_{kn} \\ \end{vmatrix} </math>
- <math> = \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) \,</math>
- <math>
\sum_{i=1}^3 \varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} </math>
- <math>
\sum_{i,j=1}^3 \varepsilon_{ijk}\varepsilon_{ijn} = 2\delta_{kn} </math>
[edit] Generalization to n dimensions
The Levi-Civita symbol can be generalized to higher dimensions:
- <math>\varepsilon_{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} \right. </math>
Thus, it is the sign of the permutation in the case of a permutation, and zero otherwise.
Furthermore, it can be shown that
- <math>
\sum_{i,j,k,\dots=1}^n \varepsilon_{ijk\dots}\varepsilon_{ijk\dots} = n!
</math>
is always fulfilled in n dimensions. In index-free tensor notation, the Levi-Civita symbol is replaced by the concept of the Hodge dual.
In general <math>n</math> dimensions one can write the product of two Levi-Civita symbol as:
- <math> \varepsilon_{ijk\dots}\varepsilon_{mnl\dots} = \det \begin{vmatrix}
\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{vmatrix} </math>.
Now we can contract <math>m</math> indexes, this will add a <math>m!</math> factor to the determinant and we need to omit the relevant Kronecker delta.
[edit] Properties
(superscipts should be considered equivalent with subscripts)
1. When <math>n=2</math>, we have for all <math>i,j,m,n</math> in <math>\{1,2\}</math>,
- <math> \varepsilon_{ij} \varepsilon^{mn}</math> <math> =</math> <math> \delta_i^m \delta_j^n - \delta_i^n \delta_j^m</math>, (1)
- <math> \varepsilon_{ij} \varepsilon^{in}</math> <math> =</math> <math> \delta_j^n</math>, (2)
- <math> \varepsilon_{ij} \varepsilon^{ij}=2</math>. (3)
2. When <math>n=3</math>, we have for all <math>i,j,k,m,n</math> in <math>\{1,2,3\},</math>
- <math> \varepsilon_{jmn} \varepsilon^{imn}=2\delta^i_j,</math> (4)
- <math> \varepsilon_{ijk} \varepsilon^{ijk}=6.</math> (5)
[edit] Proofs
For equation 1, both sides are antisymmetric with respect of <math>ij</math> and <math>mn</math>. We therefore only need to consider the case <math>i\neq j</math> and <math>m\neq n</math>. By substitution, we see that the equation holds for <math>\varepsilon_{12} \varepsilon^{12}</math>, i.e., for <math>i=m=1</math> and <math>j=n=2</math>. (Both sides are then one). Since the equation is antisymmetric in <math>ij</math> and <math>mn</math>, any set of values for these can be reduced to the above case (which holds). The equation thus holds for all values of <math>ij</math> and <math>mn</math>. Using equation 1, we have for equation 2 <math> \varepsilon_{ij}\varepsilon^{in}</math> <math> =</math> <math> \delta_i^i \delta_j^n - \delta^n_i \delta^i_j</math>
- <math> =</math> <math> 2 \delta_j^n - \delta^n_j</math>
- <math> =</math> <math> \delta_j^n</math>.
Here we used the Einstein summation convention with <math>i</math> going from <math>1</math> to <math>2</math>. Equation 3 follows similarly from equation 2. To establish equation 4, let us first observe that both sides vanish when <math>i\neq j</math>. Indeed, if <math>i\neq j</math>, then one can not choose <math>m</math> and <math>n</math> such that both permutation symbols on the left are nonzero. Then, with <math>i=j</math> fixed, there are only two ways to choose <math>m</math> and <math>n</math> from the remaining two indices. For any such indices, we have <math>\varepsilon_{jmn} \varepsilon^{imn} = (\varepsilon^{imn})^2 = 1</math> (no summation), and the result follows. The last property follows since <math>3!=6</math> and for any distinct indices <math>i,j,k</math> in <math>\{1,2,3\}</math>, we have <math>\varepsilon_{ijk} \varepsilon^{ijk}=1</math> (no summation). <math>\Box</math>
[edit] Examples
1. The determinant of an <math>n\times n</math> matrix <math>A=(a_{ij})</math> can be written as
- <math> \det A = \varepsilon_{i_1\cdots i_n} a_{1i_1} \cdots a_{ni_n},</math>
where each <math>i_l</math> should be summed over <math>1,\ldots, n.</math>
Equivalently, it may be written as
- <math> \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},</math>
where now each <math>i_l</math> and each <math>j_l</math> should be summed over <math>1,\ldots, n</math>.
2. If <math>A=(A^1, A^2, A^3)</math> and <math>B=(B^1, B^2, B^3)</math> are vectors in <math>R^3</math> (represented in some right hand oriented orthonormal basis), then the <math>i</math>th component of their cross product equals
- <math> (A\times B)^i = \varepsilon^{ijk} A^j B^k.</math>
For instance, the first component of <math>A\times B</math> is <math>A^2 B^3-A^3 B^2</math>. From the above expression for the cross product, it is clear that <math>A\times B = -B\times A</math>. Further, if <math>C=(C^1, C^2, C^3)</math> is a vector like <math>A</math> and <math>B</math>, then the triple scalar product equals
- <math> A\cdot(B\times C) = \varepsilon^{ijk} A^i B^j C^k.</math>
From this expression, it can be seen that the triple scalar product is antisymmetric when exchanging any adjacent arguments. For example, <math>A\cdot(B\times C)= -B\cdot(A\times C)</math>.
3. Suppose <math>F=(F^1, F^2, F^3)</math> is a vector field defined on some open set of <math>R^3</math> with Cartesian coordinates <math>x=(x^1, x^2, x^3)</math>. Then the <math>i</math>th component of the curl of <math>F</math> equals
- <math> (\nabla \times F)^i(x) = \varepsilon^{ijk}\frac{\partial}{\partial x^j} F^k(x).</math>
[edit] Notation
A shorthand notation for anti-symmetrization is denoted by a pair of square brackets. For example, for an n x n matrix, M,
<math>M_{[ab]} = \frac{1}{2}\varepsilon_{ab} \varepsilon^{cd} M_{dc} = \, \frac{1}{2}(M_{ab} - M_{ba})</math>
and for a rank 3 tensor T,
<math>T_{[abc]} = \, \frac{1}{3!}(T_{abc}-T_{acb}+T_{bca}-T_{bac}+T_{cab}-T_{cba})</math>
[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).
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