Levi-Civita symbol

In two dimensions, the Levi-Civita symbol is defined by:

$${\displaystyle \varepsilon _{ij}={\begin{cases}+1&{\text{if }}(i,j)=(1,2)\\-1&{\text{if }}(i,j)=(2,1)\\\;\;\,0&{\text{if }}i=j\end{cases}}}$$

The values can be arranged into a 2 × 2 antisymmetric matrix:

$${\displaystyle {\begin{pmatrix}\varepsilon _{11}&\varepsilon _{12}\\\varepsilon _{21}&\varepsilon _{22}\end{pmatrix}}={\begin{pmatrix}0&1\\-1&0\end{pmatrix}}}$$

Use of the two-dimensional symbol is common in condensed matter, and in certain specialized high-energy topics like supersymmetry[1] and twistor theory,[2] where it appears in the context of 2-spinors.

For the indices (i, j, k) in ε_ijk, the values 1, 2, 3 occurring in the   cyclic order (1, 2, 3) correspond to ε = +1, while occurring in the   reverse cyclic order correspond to ε = −1, otherwise ε = 0.

In three dimensions, the Levi-Civita symbol is defined by:[3]

$${\displaystyle \varepsilon _{ijk}={\begin{cases}+1&{\text{if }}(i,j,k){\text{ is }}(1,2,3),(2,3,1),{\text{ or }}(3,1,2),\\-1&{\text{if }}(i,j,k){\text{ is }}(3,2,1),(1,3,2),{\text{ or }}(2,1,3),\\\;\;\,0&{\text{if }}i=j,{\text{ or }}j=k,{\text{ or }}k=i\end{cases}}}$$

That is, ε_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. In three dimensions only, the cyclic permutations of (1, 2, 3) are all even permutations, similarly the anticyclic permutations are all odd permutations. This means in 3d it is sufficient to take cyclic or anticyclic permutations of (1, 2, 3) and easily obtain all the even or odd permutations.

Analogous to 2-dimensional matrices, the values of the 3-dimensional Levi-Civita symbol can be arranged into a 3 × 3 × 3 array:

where i is the depth (blue: i = 1; red: i = 2; green: i = 3), j is the row and k is the column.

Some examples:

$${\displaystyle {\begin{aligned}\varepsilon _{\color {BrickRed}{1}\color {Violet}{3}\color {Orange}{2}}=-\varepsilon _{\color {BrickRed}{1}\color {Orange}{2}\color {Violet}{3}}&=-1\\\varepsilon _{\color {Violet}{3}\color {BrickRed}{1}\color {Orange}{2}}=-\varepsilon _{\color {Orange}{2}\color {BrickRed}{1}\color {Violet}{3}}&=-(-\varepsilon _{\color {BrickRed}{1}\color {Orange}{2}\color {Violet}{3}})=1\\\varepsilon _{\color {Orange}{2}\color {Violet}{3}\color {BrickRed}{1}}=-\varepsilon _{\color {BrickRed}{1}\color {Violet}{3}\color {Orange}{2}}&=-(-\varepsilon _{\color {BrickRed}{1}\color {Orange}{2}\color {Violet}{3}})=1\\\varepsilon _{\color {Orange}{2}\color {Violet}{3}\color {Orange}{2}}=-\varepsilon _{\color {Orange}{2}\color {Violet}{3}\color {Orange}{2}}&=0\end{aligned}}}$$

In four dimensions, the Levi-Civita symbol is defined by:

$${\displaystyle \varepsilon _{ijkl}={\begin{cases}+1&{\text{if }}(i,j,k,l){\text{ is an even permutation of }}(1,2,3,4)\\-1&{\text{if }}(i,j,k,l){\text{ is an odd permutation of }}(1,2,3,4)\\\;\;\,0&{\text{otherwise}}\end{cases}}}$$

These values can be arranged into a 4 × 4 × 4 × 4 array, although in 4 dimensions and higher this is difficult to draw.

Some examples:

$${\displaystyle {\begin{aligned}\varepsilon _{\color {BrickRed}{1}\color {RedViolet}{4}\color {Violet}{3}\color {Orange}{\color {Orange}{2}}}=-\varepsilon _{\color {BrickRed}{1}\color {Orange}{\color {Orange}{2}}\color {Violet}{3}\color {RedViolet}{4}}&=-1\\\varepsilon _{\color {Orange}{\color {Orange}{2}}\color {BrickRed}{1}\color {Violet}{3}\color {RedViolet}{4}}=-\varepsilon _{\color {BrickRed}{1}\color {Orange}{\color {Orange}{2}}\color {Violet}{3}\color {RedViolet}{4}}&=-1\\\varepsilon _{\color {RedViolet}{4}\color {Violet}{3}\color {Orange}{\color {Orange}{2}}\color {BrickRed}{1}}=-\varepsilon _{\color {BrickRed}{1}\color {Violet}{3}\color {Orange}{\color {Orange}{2}}\color {RedViolet}{4}}&=-(-\varepsilon _{\color {BrickRed}{1}\color {Orange}{\color {Orange}{2}}\color {Violet}{3}\color {RedViolet}{4}})=1\\\varepsilon _{\color {Violet}{3}\color {Orange}{\color {Orange}{2}}\color {RedViolet}{4}\color {Violet}{3}}=-\varepsilon _{\color {Violet}{3}\color {Orange}{\color {Orange}{2}}\color {RedViolet}{4}\color {Violet}{3}}&=0\end{aligned}}}$$

More generally, in n dimensions, the Levi-Civita symbol is defined by:[4]

$${\displaystyle \varepsilon _{a_{1}a_{2}a_{3}\ldots a_{n}}={\begin{cases}+1&{\text{if }}(a_{1},a_{2},a_{3},\ldots ,a_{n}){\text{ is an even permutation of }}(1,2,3,\dots ,n)\\-1&{\text{if }}(a_{1},a_{2},a_{3},\ldots ,a_{n}){\text{ is an odd permutation of }}(1,2,3,\dots ,n)\\\;\;\,0&{\text{otherwise}}\end{cases}}}$$

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

Using the capital pi notation Π for ordinary multiplication of numbers, an explicit expression for the symbol is:

$${\displaystyle {\begin{aligned}\varepsilon _{a_{1}a_{2}a_{3}\ldots a_{n}}&=\prod _{1\leq i<j\leq n}\operatorname {sgn}(a_{j}-a_{i})\\&=\operatorname {sgn}(a_{2}-a_{1})\operatorname {sgn}(a_{3}-a_{1})\dotsm \operatorname {sgn}(a_{n}-a_{1})\operatorname {sgn}(a_{3}-a_{2})\operatorname {sgn}(a_{4}-a_{2})\dotsm \operatorname {sgn}(a_{n}-a_{2})\dotsm \operatorname {sgn}(a_{n}-a_{n-1})\end{aligned}}}$$

where the signum function (denoted sgn) returns the sign of its argument while discarding the absolute value if nonzero. The formula is valid for all index values, and for any n (when n = 0 or n = 1, this is the empty product). However, computing the formula above naively has a time complexity of O(n²), whereas the sign can be computed from the parity of the permutation from its disjoint cycles in only O(n log(n)) cost.

In two dimensions, when all i, j, m, n each take the values 1 and 2:[3]

${\displaystyle \varepsilon _{ij}\varepsilon ^{mn}={\delta _{i}}^{m}{\delta _{j}}^{n}-{\delta _{i}}^{n}{\delta _{j}}^{m}}$

(1)

${\displaystyle \varepsilon _{ij}\varepsilon ^{in}={\delta _{j}}^{n}}$

(2)

${\displaystyle \varepsilon _{ij}\varepsilon ^{ij}=2.}$

(3)

In three dimensions, when all i, j, k, m, n each take values 1, 2, and 3:[3]

${\displaystyle \varepsilon _{ijk}\varepsilon ^{pqk}=\delta _{i}{}^{p}\delta _{j}{}^{q}-\delta _{i}{}^{q}\delta _{j}{}^{p}}$

(4)

[6]

${\displaystyle \varepsilon _{jmn}\varepsilon ^{imn}=2{\delta _{j}}^{i}}$

(5)

${\displaystyle \varepsilon _{ijk}\varepsilon ^{ijk}=6.}$

(6)

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):[4]

${\displaystyle {\begin{aligned}\varepsilon _{ijk}\varepsilon _{lmn}&={\begin{vmatrix}\delta _{il}&\delta _{im}&\delta _{in}\\\delta _{jl}&\delta _{jm}&\delta _{jn}\\\delta _{kl}&\delta _{km}&\delta _{kn}\\\end{vmatrix}}\\[6pt]&=\delta _{il}\left(\delta _{jm}\delta _{kn}-\delta _{jn}\delta _{km}\right)-\delta _{im}\left(\delta _{jl}\delta _{kn}-\delta _{jn}\delta _{kl}\right)+\delta _{in}\left(\delta _{jl}\delta _{km}-\delta _{jm}\delta _{kl}\right).\end{aligned}}}$

A special case of this result is

${\displaystyle \sum _{i=1}^{3}\varepsilon _{ijk}\varepsilon _{imn}=\delta _{jm}\delta _{kn}-\delta _{jn}\delta _{km}}$

sometimes called the "contracted epsilon identity" [7]

In Einstein notation, the duplication of the i index implies the sum on i. The previous is then denoted ε_ijkε_imn = δ_jmδ_kn − δ_jnδ_km.

${\displaystyle \sum _{i=1}^{3}\sum _{j=1}^{3}\varepsilon _{ijk}\varepsilon _{ijn}=2\delta _{kn}}$

In n dimensions, when all i₁, ...,i_n, j₁, ..., j_n take values 1, 2, ..., n:

${\displaystyle \varepsilon _{i_{1}\dots i_{n}}\varepsilon ^{j_{1}\dots j_{n}}=\delta _{i_{1}\dots i_{n}}^{j_{1}\dots j_{n}}}$

(7)

${\displaystyle \varepsilon _{i_{1}\dots i_{k}~i_{k+1}\dots i_{n}}\varepsilon ^{i_{1}\dots i_{k}~j_{k+1}\dots j_{n}}=\delta _{i_{1}\ldots i_{k}~i_{k+1}\ldots i_{n}}^{i_{1}\dots i_{k}~j_{k+1}\ldots j_{n}}=k!~\delta _{i_{k+1}\dots i_{n}}^{j_{k+1}\dots j_{n}}}$

(8)

${\displaystyle \varepsilon _{i_{1}\dots i_{n}}\varepsilon ^{i_{1}\dots i_{n}}=n!}$

(9)

where the exclamation mark (!) denotes the factorial, and δ^α...
_β... is the generalized Kronecker delta. For any n, the property

${\displaystyle \sum _{i,j,k,\dots =1}^{n}\varepsilon _{ijk\dots }\varepsilon _{ijk\dots }=n!}$

follows from the facts that

• every permutation is either even or odd,
• (+1)² = (−1)² = 1, and
• the number of permutations of any n-element set number is exactly n!.

The particular case of (8) with ${\textstyle k=n-2}$ is

$${\displaystyle \varepsilon _{i_{1}\dots i_{n-2}jk}\varepsilon ^{i_{1}\dots i_{n-2}lm}=(n-2)!(\delta _{j}^{l}\delta _{k}^{m}-\delta _{j}^{m}\delta _{l}^{k})\,.}$$

In general, for n dimensions, one can write the product of two Levi-Civita symbols as:

$${\displaystyle \varepsilon _{i_{1}i_{2}\dots i_{n}}\varepsilon _{j_{1}j_{2}\dots j_{n}}={\begin{vmatrix}\delta _{i_{1}j_{1}}&\delta _{i_{1}j_{2}}&\dots &\delta _{i_{1}j_{n}}\\\delta _{i_{2}j_{1}}&\delta _{i_{2}j_{2}}&\dots &\delta _{i_{2}j_{n}}\\\vdots &\vdots &\ddots &\vdots \\\delta _{i_{n}j_{1}}&\delta _{i_{n}j_{2}}&\dots &\delta _{i_{n}j_{n}}\\\end{vmatrix}}.}$$

Proof: Both sides change signs upon switching two indices, so without loss of generality assume ${\displaystyle i_{1}\leq \cdots \leq i_{n},j_{1}\leq \cdots \leq j_{n}}$. If some ${\displaystyle i_{c}=i_{c+1}}$ then left side is zero, and right side is also zero since two of its rows are equal. Similarly for ${\displaystyle j_{c}=j_{c+1}}$. Finally, if ${\displaystyle i_{1}<\cdots <i_{n},j_{1}<\cdots <j_{n}}$, then both sides are 1.

For (1), both sides are antisymmetric with respect of ij and mn. We therefore only need to consider the case i ≠ j and m ≠ n. By substitution, we see that the equation holds for ε₁₂ε¹², that is, 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 (1), we have for (2)

${\displaystyle \varepsilon _{ij}\varepsilon ^{in}=\delta _{i}{}^{i}\delta _{j}{}^{n}-\delta _{i}{}^{n}\delta _{j}{}^{i}=2\delta _{j}{}^{n}-\delta _{j}{}^{n}=\delta _{j}{}^{n}\,.}$

Here we used the Einstein summation convention with i going from 1 to 2. Next, (3) follows similarly from (2).

To establish (5), notice that both sides vanish when i ≠ j. Indeed, if i ≠ 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

${\displaystyle \varepsilon _{jmn}\varepsilon ^{imn}=\left(\varepsilon ^{imn}\right)^{2}=1}$

(no summation), and the result follows.

Then (6) follows since 3! = 6 and for any distinct indices i, j, k taking values 1, 2, 3, we have

${\displaystyle \varepsilon _{ijk}\varepsilon ^{ijk}=1}$ (no summation, distinct i, j, k)

In linear algebra, the determinant of a 3 × 3 square matrix A = [a_ij] can be written[8]

${\displaystyle \det(\mathbf {A} )=\sum _{i=1}^{3}\sum _{j=1}^{3}\sum _{k=1}^{3}\varepsilon _{ijk}a_{1i}a_{2j}a_{3k}}$

Similarly the determinant of an n × n matrix A = [a_ij] can be written as[5]

${\displaystyle \det(\mathbf {A} )=\varepsilon _{i_{1}\dots i_{n}}a_{1i_{1}}\dots a_{ni_{n}},}$

where each i_r should be summed over 1, ..., n, or equivalently:

${\displaystyle \det(\mathbf {A} )={\frac {1}{n!}}\varepsilon _{i_{1}\dots i_{n}}\varepsilon _{j_{1}\dots j_{n}}a_{i_{1}j_{1}}\dots a_{i_{n}j_{n}},}$

where now each i_r and each j_r should be summed over 1, ..., n. More generally, we have the identity[5]

${\displaystyle \sum _{i_{1},i_{2},\dots }\varepsilon _{i_{1}\dots i_{n}}a_{i_{1}\,j_{1}}\dots a_{i_{n}\,j_{n}}=\det(\mathbf {A} )\varepsilon _{j_{1}\dots j_{n}}}$

Let ${\displaystyle (\mathbf {e_{1}} ,\mathbf {e_{2}} ,\mathbf {e_{3}} )}$ a positively oriented orthonormal basis of a vector space. If (a¹, a², a³) and (b¹, b², b³) are the coordinates of the vectors a and b in this basis, then their cross product can be written as a determinant:[5]

${\displaystyle \mathbf {a\times b} ={\begin{vmatrix}\mathbf {e_{1}} &\mathbf {e_{2}} &\mathbf {e_{3}} \\a^{1}&a^{2}&a^{3}\\b^{1}&b^{2}&b^{3}\\\end{vmatrix}}=\sum _{i=1}^{3}\sum _{j=1}^{3}\sum _{k=1}^{3}\varepsilon _{ijk}\mathbf {e} _{i}a^{j}b^{k}}$

hence also using the Levi-Civita symbol, and more simply:

${\displaystyle (\mathbf {a\times b} )^{i}=\sum _{j=1}^{3}\sum _{k=1}^{3}\varepsilon _{ijk}a^{j}b^{k}.}$

In Einstein notation, the summation symbols may be omitted, and the ith component of their cross product equals[4]

${\displaystyle (\mathbf {a\times b} )^{i}=\varepsilon _{ijk}a^{j}b^{k}.}$

The first component is

${\displaystyle (\mathbf {a\times b} )^{1}=a^{2}b^{3}-a^{3}b^{2}\,,}$

then by cyclic permutations of 1, 2, 3 the others can be derived immediately, without explicitly calculating them from the above formulae:

${\displaystyle {\begin{aligned}(\mathbf {a\times b} )^{2}&=a^{3}b^{1}-a^{1}b^{3}\,,\\(\mathbf {a\times b} )^{3}&=a^{1}b^{2}-a^{2}b^{1}\,.\end{aligned}}}$

From the above expression for the cross product, we have:

${\displaystyle \mathbf {a\times b} =-\mathbf {b\times a} }$.

If c = (c¹, c², c³) is a third vector, then the triple scalar product equals

${\displaystyle \mathbf {a} \cdot (\mathbf {b\times c} )=\varepsilon _{ijk}a^{i}b^{j}c^{k}.}$

From this expression, it can be seen that the triple scalar product is antisymmetric when exchanging any pair of arguments. For example,

${\displaystyle \mathbf {a} \cdot (\mathbf {b\times c} )=-\mathbf {b} \cdot (\mathbf {a\times c} )}$.

If F = (F¹, F², F³) is a vector field defined on some open set of ${\displaystyle \mathbb {R} ^{3}}$ as a function of position x = (x¹, x², x³) (using Cartesian coordinates). Then the ith component of the curl of F equals[4]

${\displaystyle (\nabla \times \mathbf {F} )^{i}(\mathbf {x} )=\varepsilon _{ijk}{\frac {\partial }{\partial x^{j}}}F^{k}(\mathbf {x} ),}$

which follows from the cross product expression above, substituting components of the gradient vector operator (nabla).

In Minkowski space (the four-dimensional spacetime of special relativity), the covariant Levi-Civita tensor is

${\displaystyle E_{\alpha \beta \gamma \delta }=\pm {\sqrt {\left|\det[g_{\mu \nu }]\right|}}\,\varepsilon _{\alpha \beta \gamma \delta }\,,}$

where the sign depends on the orientation of the basis. The contravariant Levi-Civita tensor is

${\displaystyle E^{\alpha \beta \gamma \delta }=g^{\alpha \zeta }g^{\beta \eta }g^{\gamma \theta }g^{\delta \iota }E_{\zeta \eta \theta \iota }\,.}$

The following are examples of the general identity above specialized to Minkowski space (with the negative sign arising from the odd number of negatives in the signature of the metric tensor in either sign convention):

${\displaystyle {\begin{aligned}E_{\alpha \beta \gamma \delta }E_{\rho \sigma \mu \nu }&=-g_{\alpha \zeta }g_{\beta \eta }g_{\gamma \theta }g_{\delta \iota }\delta _{\rho \sigma \mu \nu }^{\zeta \eta \theta \iota }\\E^{\alpha \beta \gamma \delta }E^{\rho \sigma \mu \nu }&=-g^{\alpha \zeta }g^{\beta \eta }g^{\gamma \theta }g^{\delta \iota }\delta _{\zeta \eta \theta \iota }^{\rho \sigma \mu \nu }\\E^{\alpha \beta \gamma \delta }E_{\alpha \beta \gamma \delta }&=-24\\E^{\alpha \beta \gamma \delta }E_{\rho \beta \gamma \delta }&=-6\delta _{\rho }^{\alpha }\\E^{\alpha \beta \gamma \delta }E_{\rho \sigma \gamma \delta }&=-2\delta _{\rho \sigma }^{\alpha \beta }\\E^{\alpha \beta \gamma \delta }E_{\rho \sigma \theta \delta }&=-\delta _{\rho \sigma \theta }^{\alpha \beta \gamma }\,.\end{aligned}}}$