A tensor whose components in an
orthonormal basis are given by the Levi-Civita symbol (a tensor of
covariant rank ) is sometimes called a
permutation tensor. Under the ordinary transformation rules for tensors the Levi-Civita symbol is unchanged under pure rotations, consistent with that it is (by definition) the same in all coordinate systems related by orthogonal transformations. However, the Levi-Civita symbol is a
pseudotensor because under an
orthogonal transformation of
Jacobian determinant −1, for example, a
reflection in an odd number of dimensions, it
should acquire a minus sign if it were a tensor. As it does not change at all, the Levi-Civita symbol is, by definition, a pseudotensor. As the Levi-Civita symbol is a pseudotensor, the result of taking a cross product is a
pseudovector, not a vector. Under a general
coordinate change, the components of the permutation tensor are 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. In index-free tensor notation, the Levi-Civita symbol is replaced by the concept of the
Hodge dual. Summation symbols can be eliminated by using
Einstein notation, where an index repeated between two or more terms indicates summation over that index. For example, :\varepsilon_{ijk} \varepsilon^{imn} \equiv \sum_{i=1,2,3} \varepsilon_{ijk} \varepsilon^{imn}. In the following examples, Einstein notation is used.
Two dimensions In two dimensions, when all each take the values 1 and 2: {{NumBlk|: | \varepsilon_{ij} \varepsilon^{mn} = {\delta_i}^m {\delta_j}^n - {\delta_i}^n {\delta_j}^m | }} {{NumBlk|: | \varepsilon_{ij} \varepsilon^{in} = {\delta_j}^n | }} {{NumBlk|: | \varepsilon_{ij} \varepsilon^{ij} = 2. | }}
Three dimensions Index and symbol values In three dimensions, when all each take values 1, 2, and 3: {{NumBlk|: | \varepsilon_{ijk} \varepsilon^{pqk}=\delta_i{}^{p}\delta_j{}^q - \delta_i{}^q\delta_j{}^p | }} {{NumBlk|: | \varepsilon_{jmn} \varepsilon^{imn}=2{\delta_j}^i | }} {{NumBlk|: | \varepsilon_{ijk} \varepsilon^{ijk}=6. | }}
Product The Levi-Civita symbol is related to the
Kronecker delta. In three dimensions, the relationship is given by the following equations (vertical lines denote the determinant): :\begin{align} \varepsilon_{ijk}\varepsilon_{lmn} &= \begin{vmatrix} \delta_{il} & \delta_{im} & \delta_{in} \\ \delta_{jl} & \delta_{jm} & \delta_{jn} \\ \delta_{kl} & \delta_{km} & \delta_{kn} \\ \end{vmatrix} \\[6pt] &= \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). \end{align} A special case of this result occurs when one of the indices is repeated and summed over: :\sum_{i=1}^3 \varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} In Einstein notation, the duplication of the index implies the sum on . The previous is then denoted . If two indices are repeated (and summed over), this further reduces to: :\sum_{i=1}^3 \sum_{j=1}^3 \varepsilon_{ijk}\varepsilon_{ijn} = 2\delta_{kn}
n dimensions Index and symbol values In dimensions, when all take values : {{NumBlk|: | \varepsilon_{i_1 \dots i_n} \varepsilon^{j_1 \dots j_n} = \delta^{j_1 \dots j_n}_{i_1 \dots i_n} | }} {{NumBlk|: | \varepsilon_{i_1 \dots i_k~i_{k+1} \dots i_n} \varepsilon^{i_1 \dots i_k~j_{k+1} \dots j_n} = \delta_{ i_1 \ldots i_k~i_{k+1} \ldots i_n}^{i_1 \dots i_k~j_{k+1}\ldots j_n} = k!~\delta^{j_{k+1} \dots j_n}_{i_{k+1} \dots i_n} | }} {{NumBlk|: | \varepsilon_{i_1 \dots i_n}\varepsilon^{i_1 \dots i_n} = n! | }} where the exclamation mark () denotes the
factorial, and is the
generalized Kronecker delta. For any , the property :\sum_{i, j, k, \dots = 1}^n \varepsilon_{ijk\dots}\varepsilon_{ijk\dots} = n! follows from the facts that • every permutation is either even or odd, • , and • the number of permutations of any -element set number is exactly . The particular case of () with k=n-2 is \varepsilon_{i_1\dots i_{n-2}jk}\varepsilon^{i_1\dots i_{n-2}lm} = (n-2)!(\delta_j^l\delta_k^m - \delta_j^m\delta^l_k)\,.
Product In general, for dimensions, one can write the product of two Levi-Civita symbols as: \varepsilon_{i_1 i_2 \dots i_n} \varepsilon_{j_1 j_2 \dots j_n} = \begin{vmatrix} \delta_{i_1 j_1} & \delta_{i_1 j_2} & \dots & \delta_{i_1 j_n} \\ \delta_{i_2 j_1} & \delta_{i_2 j_2} & \dots & \delta_{i_2 j_n} \\ \vdots & \vdots & \ddots & \vdots \\ \delta_{i_n j_1} & \delta_{i_n j_2} & \dots & \delta_{i_n j_n} \\ \end{vmatrix}.
Proof: Both sides change signs upon switching two indices, so without loss of generality assume i_1 \leq \cdots\leq i_n, j_1 \leq \cdots \leq j_n. If some i_c = i_{c+1} then left side is zero, and right side is also zero since two of its rows are equal. Similarly for j_c = j_{c+1}. Finally, if i_1 , then both sides are 1.
Proofs For (), both sides are antisymmetric with respect of and . We therefore only need to consider the case and . By substitution, we see that the equation holds for , that is, for and . (Both sides are then one). Since the equation is antisymmetric in and , any set of values for these can be reduced to the above case (which holds). The equation thus holds for all values of and . Using (), we have for () : \varepsilon_{ij} \varepsilon^{in} = \delta_i{}^i \delta_j{}^n - \delta_i{}^n \delta_j{}^i = 2 \delta_j{}^n - \delta_j{}^n = \delta_j{}^n \,. Here we used the
Einstein summation convention with going from 1 to 2. Next, () follows similarly from (). To establish (), notice that both sides vanish when . Indeed, if , then one can not choose and such that both permutation symbols on the left are nonzero. Then, with fixed, there are only two ways to choose and from the remaining two indices. For any such indices, we have :\varepsilon_{jmn} \varepsilon^{imn} = \left(\varepsilon^{imn}\right)^2 = 1 (no summation), and the result follows. Then () follows since and for any distinct indices taking values , we have :\varepsilon_{ijk} \varepsilon^{ijk} = 1(no summation, distinct ) ==Applications and examples==