**eV**and here I will explain the following topic:

# Levi-Civita Symbol and How to Write Cross Product with it

**Explanation**

## Table of contents

- Even and odd permutations of indices Here you will learn two possible permutations of indices that are necessary to understand the Levi-Civita symbol.
- Definition and examples Here you will learn the definition of the Levi-Civita symbol, made clear with some examples.
- Cross product in index notation Here you will learn how to write the cross product in index notation using the Levi-Civita symbol.
- Scalar triple product in index notation with Levi-Civita symbol Here you'll learn how the Levi-Civita symbol can simplify proving equations that contain a cross product.

## Video

Together with the Kronecker delta \( \delta_{\class{blue}{i}\class{red}{j}} \), the Levi-Civita symbol is a very common symbol in all subfields of physics, from classical mechanics to quantum field theory. Therefore, it is important to understand how this symbol works.

With the Levi-Civita symbol, sometimes called the *Epsilon tensor*, you can, for example...

- Easily transform and simplify complicated vector equations, such as multiple cross products
- Write equations more compactly

In principle, using the Levi-Civita symbol, we want to write vector equations in index notation to make them easier to manipulate.

Levi-Civita symbol is denoted with a small Greek epsilon \(\varepsilon\), which carries three indices \(\class{blue}{i}\), \(\class{red}{j}\) and \(\class{green}{k}\):

How you call the indices is, of course, up to you. The indices take different values depending on the dimension considered. If you work with three-dimensional vectors, then you need a Levi-Civita symbol whose indices \(\class{blue}{i}\), \(\class{red}{j}\) and \(\class{green}{k}\) take the values from 1 to 3:

The Levi-Civita symbol \(\varepsilon_{ \class{blue}{i} \class{red}{j} \class{green}{k} }\) can take three different values: +1, 0 or -1. The symbol takes only these three values - no others! When does it take which value? This depends on *how* the indices \(\class{blue}{i} \class{red}{j} \class{green}{k}\), are *ordered*, with respect to the original order. So you have the symbol \(\varepsilon_{ \class{blue}{i} \class{red}{j} \class{green}{k} }\) and you can permute the indices with each other. Let's take a closer look.

## Even and odd permutations of indices

Before you can understand the definition of the Levi-Civita symbol, you must first understand *even* and *odd* permutations of its indices.

## Definition and examples

With this knowledge you are able to understand the definition of the Levi-Civita symbol. The permutations refer to a starting position of the indices. Here we take \((\class{blue}{i}\class{red}{j}\class{green}{k}) = (123) \) as the starting position. Then the Levi-Civita symbol behaves as follows:

- If the three different indices \(\class{blue}{i}\), \(\class{red}{j}\) and \(\class{green}{k}\), are permuted
*even*with respect to the starting position \((123) \), the resulting epsilon is equal to 1. - If the three different indices \(\class{blue}{i}\), \(\class{red}{j}\) and \(\class{green}{k}\), are permuted
*odd*with respect to the starting position \((123)\), the resulting epsilon is equal to -1. - If at least two of the indices are equal, the resulting epsilon is 0.

## Cross product in index notation

One of the advantages of the definition 1

of the Levi-Civita symbol is that it allows us to write the cross product of two vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) in index notation, because the epsilon represents exactly the properties of the cross product!

Consider the cross product \( \boldsymbol{a} ~\times~ \boldsymbol{b} \) of two vectors \( \boldsymbol{a} = (a_1, a_2, a_3)\) and \( \boldsymbol{b} = (b_1, b_2, b_3) \):

The result of the cross product is a new vector perpendicular to \( \boldsymbol{a} \) and \( \boldsymbol{b} \). This property of the cross product is very useful, for example, in the description of the Lorentz force, where the cross product \(\boldsymbol{v} ~\times~ \boldsymbol{B}\) between the velocity \( \boldsymbol{v} \) of the charge and the external magnetic field \( \boldsymbol{B} \) is relevant.

We can also represent the cross product 2

in an arbitrary basis. For example, let us choose the usual cartesian basis, with the normalized and mutually perpendicular basis vectors \( \boldsymbol{\hat{e}}_1 \), \( \boldsymbol{\hat{e}}_2 \) and \( \boldsymbol{\hat{e}}_3 \) spanning the three-dimensional space:

&~+~ (a_3\,b_1-a_1\,b_3) \, \boldsymbol{\hat e}_2 \\\\

&~+~ (a_1\,b_2-a_2\,b_1) \, \boldsymbol{\hat e}_3 \end{align} $$

We can write the cross product 3

, represented in the cartesian basis, in index notation as follows:

Take a close look at the indices. All three indices \(\class{blue}{i}\), \(\class{red}{j}\) and \(\class{green}{k}\) occur twice. We use the **Einstein summation convention** in Eq. 4

, so we sum over duplicate indices.

In 4

we have written down all three components of the cross product in a compact equation. But the crucial thing here is not the compactness, but the *commutativity* of the individual factors created by the index notation. This means: You may interchange the three factors in 4

as you like. With the representation 3

you cannot do this, of course, because of all the minus and plus signs. In other words, using the Levi-Civita symbol, we have cleverly hidden all addition and subtraction of expressions in 3

, so now we have only commutative multiplication.

We can also write the components 2

of the cross product in index notation without concretely including the basis vectors. This is the so-called component notation.

In component notation, we consider *one* component of the cross product, notated in such a general way that it is representative of all three components. For this we provide the cross product with an index \( \class{blue}{i} \), like this: \( \left( \boldsymbol{a} ~\times~ \boldsymbol{b} \right)_{\class{blue}{i}} \). This means: You consider the \( \class{blue}{i} \)-th component of the cross product - the first, second or third component. The index \( \class{blue}{i} \) is representative for 1, 2 or 3.

Remember that the component \( \left( \boldsymbol{a} ~\times~ \boldsymbol{b} \right)_{\class{blue}{i}} \) of a vector, is a *pure number*, where the basis vector \( \boldsymbol{\hat{e}}_{\class{blue}{i}} \) does not appear explicitly. In component notation, the cross product is written as follows:

Whether you want to work with the vector cross product 4

or with only one component 5

of the cross product is up to you!

## Scalar triple product in index notation with Levi-Civita symbol

Using the Levi-Civita symbol, it is easy to show that the following scalar triple products are the same (try doing this without index notation):

&~=~ \boldsymbol{b} \cdot \left( \boldsymbol{c} ~\times~ \boldsymbol{a} \right) \end{align} $$

If you look closely at the equations, you will see that the vectors are cyclically interchanged to arrive at an equivalent scalar triple product. Let's start with the left side of the equation. Since this is a scalar product between \( \boldsymbol{a} \) and \( \left( \boldsymbol{b} ~\times~ \boldsymbol{c} \right) \), you give both the cross product and the vector \( \boldsymbol{a} \) the same index, for example the index \( \class{blue}{i} \). This index is then summed up according to the Einstein summation convention:

Now you have an ordinary product of two numbers: \( a_{\class{blue}{i}} \) and \( \left( \boldsymbol{b} ~\times~ \boldsymbol{c} \right)_{\class{blue}{i}} \), which is why the scalar product dot is gone. In this way we have written the scalar product in index notation.

You also learned earlier that the \( \class{blue}{i} \)th component of the cross product can be rewritten using the Levi-Civita symbol (see Eq. 5

). We exploit this to write the cross product in index notation:

Now the scalar triple product 12

contains only multiplicative factors, which is why we can swap the factors around however we want! For example, let's place the Levi-Civita symbol at the beginning:

Next, we exploit the property of the Levi Civita symbol: With an *even* permutation of the indices, we get the same Levi-Civita symbol. (With an *odd* permutation of the indices, on the other hand, we get a minus sign in front of the Levi-Civita symbol).

For the first even permutation you 'rotate' the indices once in a circle: \( \class{blue}{i}\class{red}{j}\class{green}{k} ~\rightarrow~ \class{green}{k}\class{blue}{i}\class{red}{j} \). Since nothing changes in the result, the permuted and non-permuted versions are the same:

Next even permutation (i.e., another rotation) of the indices \( \class{green}{k}\class{blue}{i}\class{red}{j} ~\rightarrow~ \class{red}{j}\class{green}{k}\class{blue}{i} \) yields another term equal to the other two in Eq. 14

:

~&=~ \varepsilon_{\class{red}{j}\class{green}{k}\class{blue}{i}} \, a_{\class{blue}{i}} \, b_{\class{red}{j}} \, c_{\class{green}{k}} \end{align} $$

If you would rotate the indices a third time \( \class{red}{j}\class{green}{k}\class{blue}{i} ~\rightarrow~ \class{blue}{i}\class{red}{j}\class{green}{k} \), then you come back to the initial state. So there is no point in doing another even permutation.

Let us sort the Eq. 15

so that the indices at the vector components have the same order as at the Levi-Civita symbols:

~&=~ \varepsilon_{\class{red}{j}\class{green}{k}\class{blue}{i}} \, b_{\class{red}{j}} \, c_{\class{green}{k}} \, a_{\class{blue}{i}} \end{align} $$

Now we just have to apply the definition 5

of the cross product backwards to the three terms in 16

to get their vector notation:

~&=~ b_{\class{red}{j}} \, \left( \boldsymbol{c} ~\times~ \boldsymbol{a} \right)_{\class{red}{j}} \end{align} $$

Each term is summed over an index. This is therefore a scalar product. With this we get the scalar triple products:

&~=~ \boldsymbol{b} \cdot \left( \boldsymbol{c} ~\times~ \boldsymbol{a} \right) \end{align} $$

As you can see, the Levi-Civita symbol is a very useful permutation symbol when dealing with equations that contain cross products. Next, you should practice a little yourself. Try proving the BAC-CAB rule using index notation and the Levi-Civita symbol.