My name is Alexander FufaeV and here I write about:

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

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.
Exercises with Solutions

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 $\class{green}{k}$:

$$ \begin{align} \varepsilon_{ \class{blue}{i} \class{red}{j} \class{green}{k} } \end{align} $$

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:

$$ \begin{align} \class{blue}{i},~ \class{red}{j},~ \class{green}{k} ~\in~ \{1,2,3\} \end{align} $$

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.

What is an even permutation?

In an even (cyclic) permutation, all indices are rotated clockwise or counterclockwise. With this permutation all indices change their position.

Hover the image!

Even permutation of the indices as clockwise rotation of the indices.

Examples of even permutations

A clockwise even permutation of ($\class{blue}{i},\class{red}{j},\class{green}{k}$) yields ($\class{green}{k},\class{blue}{i},\class{red}{j}$). Do you see how the indices have been rotated here?
A clockwise even permutation of ($\class{green}{k},\class{blue}{i},\class{red}{j}$) yields ($\class{red}{j},\class{green}{k},\class{blue}{i}$).
An even permutation of ($\class{red}{j},\class{green}{k},\class{blue}{i}$) would again produce ($\class{blue}{i},\class{red}{j},\class{green}{k}$). Remember that a counterclockwise rotation of the indices is also an even permutation.

What is an odd permutation?

In the case of an odd permutation, two indices are interchanged. With this permutation only two of the three indices change their position.

Examples of odd permutations

An odd permutation of ($\class{blue}{i},\class{red}{j},\class{green}{k}$) is ($\class{red}{j},\class{blue}{i},\class{green}{k}$). Here, $\class{blue}{i}$ and $\class{red}{j}$ have been swapped, while $\class{green}{k}$ has remained in the same place.
Another odd permutation of ($\class{blue}{i},\class{red}{j},\class{green}{k}$) is ($\class{green}{k},\class{red}{j},\class{blue}{i}$). Here, $\class{blue}{i}$ and $\class{green}{k}$ have been swapped, while $\class{red}{j}$ has remained in the same place.
And the last possible odd permutation of ($\class{blue}{i},\class{red}{j},\class{green}{k}$) is ($\class{blue}{i},\class{green}{k},\class{red}{j}$). Here $\class{blue}{i}$ was left in place, while $\class{green}{k}$ and $\class{red}{j}$ were interchanged.

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.

$$ \begin{align} \varepsilon_{\class{blue}{i}\class{red}{j}\class{green}{k}} ~=~ \begin{cases} +1, & ~(\class{blue}{i}\class{red}{j}\class{green}{k}) \text{ is an even permutation of } (123)\\ -1, & ~(\class{blue}{i}\class{red}{j}\class{green}{k}) \text{ is an odd permutation of } (123) \\ ~~~0, & \text{min. 2 equal indices} \end{cases} \end{align} $$

Examples

$\varepsilon_{112} ~=~ 0$, since the first two indices are the same.
$\varepsilon_{313} ~=~ 0$, since the first and third indices are the same.
$\varepsilon_{222} ~=~ 0$, since all three indices are equal.
$\varepsilon_{123} + \varepsilon_{213} ~=~ 1 + (-1) ~=~ 0$, because for the first epsilon the indices are in the initial position and the indices of the second epsilon are an odd permutation of it.
$\varepsilon_{123} \, \varepsilon_{231} ~=~ 1 \cdot 1 ~=~ 1$, because for the first epsilon the indices are in the initial position and the indices of the second epsilon have just been rotated counterclockwise.

Cross product in index notation

Cross product of two vectors.

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) $:

$$ \begin{align} \boldsymbol{a} ~\times~ \boldsymbol{b} ~=~ \begin{bmatrix} a_2\,b_3-a_3\,b_2 \\ a_3\,b_1-a_1\,b_3 \\ a_1\,b_2-a_2\,b_1 \end{bmatrix} \end{align} $$

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:

$$ \begin{align} \boldsymbol{a} ~\times~ \boldsymbol{b} &~=~ (a_2\,b_3-a_3\,b_2) \, \boldsymbol{\hat e}_1 \\\\
&~+~ (a_3\,b_1-a_1\,b_3) \, \boldsymbol{\hat e}_2 \\\\
&~+~ (a_1\,b_2-a_2\,b_1) \, \boldsymbol{\hat e}_3 \end{align} $$

We can write the cross product 3, represented in the cartesian basis, in index notation as follows:

$$ \begin{align} \boldsymbol{a} ~\times~ \boldsymbol{b} ~=~ \varepsilon_{\class{blue}{i}\class{red}{j}\class{green}{k}} \, \boldsymbol{\hat{e}}_{\class{blue}{i}} \, a_{\class{red}{j}} \, b_{\class{green}{k}} \end{align} $$

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:

$$ \begin{align} \left( \boldsymbol{a} ~\times~ \boldsymbol{b} \right)_{\class{blue}{i}} ~=~ \varepsilon_{\class{blue}{i}\class{red}{j}\class{green}{k}} \, a_{\class{red}{j}} \, b_{\class{green}{k}} \end{align} $$

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!

Example: Completely write out the first component

Let's check if the index notation of the cross product 5 gives the correct result. For this we write instead of the general index $ \class{blue}{i} $ the number 1, which represents the 1st component of the cross product:

$$ \begin{align} \left( \boldsymbol{a} ~\times~ \boldsymbol{b} \right)_{\class{blue}{1}} ~=~ \varepsilon_{\class{blue}{1}\class{red}{j}\class{green}{k}} \, a_{\class{red}{j}} \, b_{\class{green}{k}} \end{align} $$

So now you have set the component we want to calculate. Next we need to sum over $ \class{red}{j} $ and $ \class{green}{k} $. To go through all cases, we set $ \class{red}{j} = 1 $ and go through all cases for $ \class{green}{k} =1,2,3 $. Then we set $ \class{red}{j} = 2 $ and go through all cases for $ \class{green}{k} =1,2,3 $ and analogously for $ \class{red}{j} = 3 $. Altogether we get 3 * 3 = 9 summands:

$$ \begin{align} \varepsilon_{\class{blue}{1}\class{red}{j}\class{green}{k}} \, a_{\class{red}{j}} \, b_{\class{green}{k}} &~=~ \varepsilon_{\class{blue}{1}\class{red}{1}\class{green}{1}} \, a_{\class{red}{1}} \, b_{\class{green}{1}} ~+~ \varepsilon_{\class{blue}{1}\class{red}{1}\class{green}{2}} \, a_{\class{red}{1}} \, b_{\class{green}{2}} ~+~ \varepsilon_{\class{blue}{1}\class{red}{1}\class{green}{3}} \, a_{\class{red}{1}} \, b_{\class{green}{3}} \\\\
&~+~ \varepsilon_{\class{blue}{1}\class{red}{2}\class{green}{1}} \, a_{\class{red}{2}} \, b_{\class{green}{1}} ~+~ \varepsilon_{\class{blue}{1}\class{red}{2}\class{green}{2}} \, a_{\class{red}{2}} \, b_{\class{green}{2}} ~+~ \varepsilon_{\class{blue}{1}\class{red}{2}\class{green}{3}} \, a_{\class{red}{2}} \, b_{\class{green}{3}} \\\\
&~+~ \varepsilon_{\class{blue}{1}\class{red}{3}\class{green}{1}} \, a_{\class{red}{3}} \, b_{\class{green}{1}} ~+~ \varepsilon_{\class{blue}{1}\class{red}{3}\class{green}{2}} \, a_{\class{red}{3}} \, b_{\class{green}{2}} ~+~ \varepsilon_{\class{blue}{1}\class{red}{3}\class{green}{3}} \, a_{\class{red}{3}} \, b_{\class{green}{3}} \end{align} $$

According to the definition of the Levi-Civita symbol, out of nine terms, only two are non-zero and these are the ones with different indices:

$$ \begin{align} \varepsilon_{\class{blue}{1}\class{red}{j}\class{green}{k}} \, a_{\class{red}{j}} \, b_{\class{green}{k}} ~=~ \varepsilon_{\class{blue}{1}\class{red}{2}\class{green}{3}} \, a_{\class{red}{2}} \, b_{\class{green}{3}} ~+~ \varepsilon_{\class{blue}{1}\class{red}{3}\class{green}{2}} \, a_{\class{red}{3}} \, b_{\class{green}{2}} \end{align} $$

Where $ \varepsilon_{\class{blue}{1}\class{red}{2}\class{green}{3}} ~= 1 $ because it is an even permutation. And $ \varepsilon_{\class{blue}{1}\class{red}{3}\class{green}{2}} ~= -1 $ because it is an odd permutation. Inserting results in:

$$ \begin{align} \varepsilon_{\class{blue}{1}\class{red}{j}\class{green}{k}} \, a_{\class{red}{j}} \, b_{\class{green}{k}} ~=~ a_{\class{red}{2}} \, b_{\class{green}{3}} ~-~ \, a_{\class{red}{3}} \, b_{\class{green}{2}} \end{align} $$

You proceed analogously with the 2nd and 3rd components and thus obtain all three components of the resulting vector of the cross product.

Scalar triple product in index notation with Levi-Civita symbol

Scalar triple product illustrated.

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):

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

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:

$$ \begin{align} a_{\class{blue}{i}} \, \left( \boldsymbol{b} ~\times~ \boldsymbol{c} \right)_{\class{blue}{i}} \end{align} $$

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:

$$ \begin{align} a_{\class{blue}{i}} \, \left( \boldsymbol{b} ~\times~ \boldsymbol{c} \right)_{\class{blue}{i}} ~=~ a_{\class{blue}{i}} \, \varepsilon_{\class{blue}{i}\class{red}{j}\class{green}{k}} \, b_{\class{red}{j}} \, c_{\class{green}{k}} \end{align} $$

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:

$$ \begin{align} a_{\class{blue}{i}} \, \left( \boldsymbol{b} ~\times~ \boldsymbol{c} \right)_{\class{blue}{i}} ~=~ \varepsilon_{\class{blue}{i}\class{red}{j}\class{green}{k}} \, a_{\class{blue}{i}} \, b_{\class{red}{j}} \, c_{\class{green}{k}} \end{align} $$

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:

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

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:

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

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:

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

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:

$$ \begin{align} a_{\class{blue}{i}} \, \left( \boldsymbol{b} ~\times~ \boldsymbol{c} \right)_{\class{blue}{i}} ~&=~ c_{\class{green}{k}} \, \left( \boldsymbol{a} ~\times~ \boldsymbol{b} \right)_{\class{green}{k}} \\\\
~&=~ b_{\class{red}{j}} \, \left( \boldsymbol{c} ~\times~ \boldsymbol{a} \right)_{\class{red}{j}} \end{align} $$

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

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.

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise #1: Lorentz Force Using the Levi-Civita Symbol

In the Lorentz force, you find the cross product term $ \boldsymbol{v} \times \boldsymbol{B} $. Determine the components of the cross product $ \boldsymbol{v} \times \boldsymbol{B} $ by using the Levi-Civita symbol $ \varepsilon_{ijk} $.

Here, $ \boldsymbol{v} $ and $ \boldsymbol{B} $ are three-dimensional vectors.

Solution to Problem #1

Definition of the cross product using the Levi-Civita symbol (with Einstein summation convention) is given by: 1 \[ \boldsymbol{v} \times \boldsymbol{B} = \varepsilon_{ijk} \, \boldsymbol{\hat{e}}_i \, v_j \, B_k \]

Consider individual components of the cross product. Keep it general and denote the $ i $-th component. Here, $i$ represents the first, second, or third component: 2 \[ \left( \boldsymbol{v} \times \boldsymbol{B} \right)_i = \varepsilon_{ijk} \, v_j \, B_k \]

To calculate each component of the cross product explicitly, you need to explicitly write out each of the three components in index notation. To do this, you write out the sum. The first component $ i = 1 $ is given by: 3 \[ \left( \boldsymbol{v} \times \boldsymbol{B} \right)_1 = \varepsilon_{1jk} \, v_j \, B_k \]

You don't need to write out all possible 9 terms. According to the definition of the Levi-Civita symbol, all terms containing identical indices vanish, such as the term multiplied by $ \varepsilon_{112} = 0 $. Only two terms with three distinct indices remain: 4 \[ \left( \boldsymbol{v} \times \boldsymbol{B} \right)_1 = \varepsilon_{123} \, v_2 \, B_3 + \varepsilon_{132} \, v_3 \, B_2 \]

According to the definition of the Levi-Civita symbol, $ \varepsilon_{123} = 1 $ represents an even permutation of indices, and $ \varepsilon_{132} = -1 $ represents an odd permutation of indices: 5 \[ \left( \boldsymbol{v} \times \boldsymbol{B} \right)_1 = v_2 \, B_3 - v_3 \, B_2 \]

Thus, you have determined the first component of the cross product. Analogously, you calculate the 2nd component: 6 \[ \left( \boldsymbol{v} \times \boldsymbol{B} \right)_2 = v_3 \, B_1 - v_1 \, B_3 \]

And the 3rd component: 7 \[ \left( \boldsymbol{v} \times \boldsymbol{B} \right)_3 = v_1 \, B_2 - v_2 \, B_1 \]

And now you have all three components of the cross product, which can be combined into a single vector: 8 \[ \boldsymbol{v} \times \boldsymbol{B} = \left(\begin{array}{c}v_2 \, B_3 - v_3 \, B_2 \\ v_3 \, B_1 - v_1 \, B_3 \\ v_1 \, B_2 - v_2 \, B_1 \end{array}\right) \]

Exercise #2: Product of Levi-Civita Symbols

The product of two Levi-Civita symbols can be computed as the determinant of a 3x3 matrix, where Kronecker deltas with different indices are present. In the case of identical indices, it simplifies to a 2x2 matrix: $$ \varepsilon_{kij} \, \varepsilon_{kmn} ~=~ {\begin{vmatrix}\delta_{im}&\delta_{in} \\ \delta_{jm}&\delta_{jn} \end{vmatrix}} $$

Consider the following three cases:

The Levi-Civita symbols contain both a common index $k$. Compute the determinant of the above 2x2 matrix.
Then, set the indices $i=m$ and compute the determinant again. You can use the result from the first part.
Set the indices $j=n$ equal. Now, all indices of the two Levi-Civita symbols are equal. What result do you get for the determinant now? You can use the result from the second exercise.

Solution to Problem #2.1

To compute the determinant of a 2x2 matrix, the Laplace expansion can be used. For example, expanding with respect to the 1st row results in: 1 \[ \varepsilon_{kij} \, \varepsilon_{kmn} ~=~ \delta_{im} \, \delta_{jn} ~-~ \delta_{jm} \, \delta_{in} \]

You have derived an important identity! Here, $ \varepsilon_{kij} \, \varepsilon_{kmn} $ is the product of two Levi-Civita symbols, and the above identity arises precisely when the two Levi-Civita symbols share one common index. With this knowledge, you can easily rewrite, for example, the double cross product $ \boldsymbol{a}\times \left( \boldsymbol{b}\times\boldsymbol{c} \right) $.

Solution to Problem #2.2

If the Levi-Civita Symbols have two identical indices, you can, for example, set $ i = m $ in the above identity, replacing $ i $ with $ m $: 2 \[ \varepsilon_{kmj} \, \varepsilon_{kmn} ~=~ \delta_{mm} \, \delta_{jn} ~-~ \delta_{jm} \, \delta_{mn} \]

Here, $ \delta_{mm} = 3 $ according to the Rules of Kronecker delta. And: $ \delta_{jm} \, \delta_{mn} = \delta_{jn} $. Overall, we have: 3 \[ \varepsilon_{kmj} \, \varepsilon_{kmn} ~=~ 2 \delta_{jn} \]

Solution to Problem #2.3

If all indices are equal (that is, besides $ i = m $, also $ j = n $), then you have: 4 \[ \varepsilon_{kmn} \, \varepsilon_{kmn} ~=~ 2 \delta_{nn} \]

According to the rules of Kronecker delta, we get $ \delta_{nn} = 3 $. Thus: 5 \[ \varepsilon_{kmn} \, \varepsilon_{kmn} ~=~ 6 \]