I am reading through a section on vector calculus in an electromagnetism book and it has started to use suffix notation and the Levi-Civita alternating tensor in order to prove some identities. Some of the identities I am familiar with and others I am not. The notation is new to me as are the concept of tensors and I am struggling to apply both to do things which I can already do.

As an example it is stated in the book that it is much simpler and more concise to write the product rule as


and I know to work out the product rule I can use the determinant formula. What I am having trouble formulating in my head is how to actually read the above definition and get the correct expansion for the vector-product. So to make sure I understood the notation I expanded it out on paper but ended up getting the wrong answer. My interpretations and assumptions are as below:

$$\left(\mathbf{A}\times\mathbf{B}\right)_i = \epsilon_{ijk}A_jB_k$$ $$ = \epsilon_{ijk}A_yB_z\mathbf{i} + \epsilon_{jki}A_zB_x\mathbf{j} + \epsilon_{kij}A_xB_y\mathbf{k} + \epsilon_{jik}A_yB_z\mathbf{j} + \epsilon_{kji}A_zB_x\mathbf{k} + \epsilon_{ikj}A_xB_y\mathbf{i} $$

From here it can be seen that the first three terms are correct, but the last three are not. I don't understand how the notation links back to the correct unit vector. Here I have just take the first subscript letter in $\epsilon_{ijk}$ to also represent the relevant unit vector. So I could really from getting the first three terms correct using the notation correctly write out the last three terms but that is just because I know what it should be. What I do not understand is how the subscripts of the tensor link up to the subscripts of the two vectors. I hope that is clear, if it isn't please leave a comment and I will try to remove anything confusing.

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    $\begingroup$ What you written down is wrong. Note in suffix notation, you sum things which appear twice. So you sum the right hand for j = 1 to 3 and k = 1 to 3. There is no y or z... $\endgroup$ – Lost1 Jan 4 '14 at 21:56
  • $\begingroup$ The book I am using introduced it as $A_i$ represents the summation of $i$ from 1 to 3 and also representing the vector $A_i=A_x+A_y+A_z$. I had took it to mean that when $i=1$ it referred to the $x$ component of $\mathbf{A}$. So for example the dot product could be written as $\mathbf{A}.\mathbf{B}=A_iB_i$. $\endgroup$ – Aesir Jan 5 '14 at 11:56
  • $\begingroup$ So xyz are basically 123. Okay but what you wrote down is still wrong... A_i is not the sum of those things... You only sum things when there are two of the same index.... $\endgroup$ – Lost1 Jan 5 '14 at 11:58
  • $\begingroup$ Here you need to replace the levi civita symbol with the appropriate 1 or -1... From its definitions. $\endgroup$ – Lost1 Jan 5 '14 at 12:01
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    $\begingroup$ Also, to add to Ivo Terek's answer, you can write the vector resulting from the cross product itself as $\epsilon_{ijk} A_j B_k \mathbf{\hat{e}_i}$. That form is useful if you want to perform further vector operations, eg, $A \times B \cdot C$. $\endgroup$ – user_of_math Jul 11 '14 at 4:17

Actually, when you write $(A \times B)_i$, this is no longer a vector, but just one of its components. The vector itself is $$(A \times B) = ((A \times B)_1, (A \times B)_2, (A \times B)_3)$$ Having this in mind, the expression $$(A \times B)_i = \epsilon_{ijk}A_j B_k$$ means three equations, for $i$ ranges from $1$ to three. The equations are: $$(A \times B)_1 = \epsilon_{1jk}A_j B_k \\ (A \times B)_2 = \epsilon_{2jk}A_j B_k \\ (A \times B)_3 = \epsilon_{3jk}A_j B_k $$ For a better understanding of it, let's write everything for $(A \times B)_1$, say. I'll begin with the sum in $j$. So: $$\begin{align} (A \times B)_1 &= \epsilon_{\color {red}{11}k}A_1B_k + \epsilon_{12k}A_2B_k + \epsilon_{13k}A_3B_k \\ &= \epsilon_{12k} A_2B_k + \epsilon_{13k} A_3 B_k \end{align}$$ Notice that in the indices, repeats are zeros. Now, let's do the sum on $k$: $$\begin{align} (A \times B)_1 &= \epsilon_{\color{red}{1}2\color{red}{1}}A_2 B_1 + \epsilon_{1\color {red}{22}}A_2B_2 + \epsilon_{123}A_2B_3 + \epsilon_{\color{red}{1}3\color{red}{1}}A_3B_1 + \epsilon_{132}A_3B_2 + \epsilon_{1\color{red}{33}}A_3B_3 \\ &= \epsilon_{123}A_2B_3 + \epsilon_{132}A_3B_2 \\ &= A_2B_3 - A_3B_2 \end{align}$$

You can do the same for the other ones, to get used to it. Also, I reccomend getting a few basic properties of the cross product, and prove them using only this notation. I once asked a question related to it, you might find it helpful.


Lost 1 is correct what you wrote is totally absurd....it is utter nonsense. Recall :

$$\epsilon_{ijk} = -\epsilon_{ikj} = -\epsilon_{jki} = -\epsilon_{kji}.$$

We realize that product vanishes, i.e. it becomes zero....which I must say it's wrong for we don't have enough info to justify that.

So we rewrite the product as follows.

$$C_x = (\vec{A} \times \vec{B}) = \epsilon_{ijk} A_j B_k = \epsilon_{xyz} A_y B_z + \epsilon_{xzy} A_z B_y$$ I have ignored the other expressions since they lead to zero, i.e.

$$\epsilon_{yzx} = \epsilon_{yxz} = \epsilon_{zyx} =\epsilon_{zxy} = 0,$$

since in our expression we have $C_x$ and for the other components:

$$C_y = \epsilon_{yzx} - \epsilon_{yxz} \text{ and } C_z = \epsilon_{zxy} - \epsilon_{zyx}.$$


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