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I understand where the $k$ comes from in $\varepsilon_{klm}$, however why do we need to introduce $l,m$ rather than continuing with $i,j$,

i.e $\varepsilon_{kij}b_ic_j$

usainlightning
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2 Answers2

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Each time you have $2$ indices which are the same in the same term, there is an implied summation. This is known as the Einstein summation convention. So there are two reasons why we wouldn't want to use $i$ and $j$ for the indices on the second $\varepsilon$: $(1)$ we don't we to sum over each index of both $\varepsilon$'s (i.e. we don't want $\varepsilon_{ijk}\varepsilon_{kij} = \sum_i \sum_j\sum_k \varepsilon_{ijk}\varepsilon_{kij}$) and $(2)$ you are not allowed to have more than $2$ of the same index used in a single term because then the summation would become ambiguous.

For more information on the Einstein summation convention (though in a slightly different form) see the first part of my answer to this question.

$\Large {\textbf{EDIT:}}$

Definition: A scalar is an object which is invariant (doesn't change) under isometries.

Definition: An isometry is a linear transformation with the property that for any matrix $A$ which represents the isometry, $A^TA=AA^T=I$, where $I$ is the identity matrix.

Isometries are things like rotations and reflections and Lorentz boosts -- and depending on your space, any other type of distance-preserving linear transformation.

Proof that the dot product of two vectors is a scalar:
Let $R$ be a matrix representing an isometry and let $\mathbf u, \mathbf v \in \Bbb R^n$. For simplicity I'm going to using the same symbols to denote the vectors $\mathbf u, \mathbf v$ and their coordinate vectors(As well as the isometry $R$ and the matrix $R$).

So $$(R\mathbf u)\cdot (R\mathbf v) = (R\mathbf u)^T (R\mathbf v) = \mathbf u^T R^T R\mathbf v = \mathbf u^T I \mathbf v = \mathbf u^T \mathbf v = \mathbf u \cdot \mathbf v$$ Therefore the dot product is invariant under isometry.$\ \ \ \ \ \square$

In index notation, the dot product of the vectors $\mathbf u$ and $\mathbf v$ is simply written as $\sum_{i=1}^n u_i v_i = u_i v_i$ (on the LHS I have the summation written explicitly, on the RHS I have the summation implied by the Einstein summation convention). So we could rewrite the above proof as $$(R\mathbf u) \cdot (R\mathbf v) = [R\mathbf u]_j [R\mathbf v]_j = [R_{ji}u_i][R_{jk}v_k] = R_{ji}R_{ji}u_iv_k =[R^T]_{ij}R_{jk}u_iv_k = \delta_{ik}u_iv_k = u_iv_i = \mathbf u \cdot \mathbf v$$

Now let's see why the same sort of product (written in index notation) of $3$ vectors is NOT a scalar: $$\sum_j [R\mathbf u]_j [R\mathbf v]_j[R\mathbf w]_j = \sum_j ([R\mathbf u]_j [R\mathbf v]_j)[R\mathbf w]_j = \sum_{i,j,k} (u_iv_i)R_{jk}w_k \ne \sum_i u_iv_iw_i$$

So there is a good reason why the Einstein convention implies summation only over pairs of indices.

Community
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  • Do you mind expanding on what you mean by 'you are not allowed to have more than 2 of the same index used in a single term because then the summation would become ambiguous.' (I understand what ambiguous means, i'm just wondering why it's ambiguous) – usainlightning Mar 10 '15 at 10:37
  • Say you had $a_ib_ic_id_ie_j$. Does this mean $\sum a_i b_i c_i d_i e_j$ or $\sum a_i b_i \sum c_i d_i e_j$ or $\sum a_ic_i \sum b_id_ie_j$ or... ? When Einstein invented this notation, he specifically made the point of specifying that one of the "rules" is that you can only have $2$ of the same index. Any more than that and you'll need to explicitly write the summation symbol. –  Mar 10 '15 at 12:41
  • Perhaps it would be relevant to answer why a summation like $\sum_i b_i c_i d_i$ does not transform like a scalar--why summations must involve pairs of indices, in other words. – Muphrid Mar 11 '15 at 02:46
  • @Muphrid That's a good idea. Thanks. –  Mar 11 '15 at 23:29
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Yes. It is necessary. You can't use the same indices. It's a different cross product. You already used $i$ and $j$ for the first cross product.

Ben
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