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From http://www.physics.ohio-state.edu/~ntg/263/handouts/tensor_intro.pdf:

Rules of Einstein Summation Convention — If an index appears (exactly) twice, then it is summed over and appears only on one side of an equation.
A single index (called a free index) appears once on each side of the equation. So
$A_{\LARGE{i}} = B_{\LARGE{i}}C_{\LARGE{i}} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \, \, (1)$ is INvalid.
$A_{\LARGE{i}} = \epsilon_{\LARGE{ijk}}B_{\LARGE{i}}C_{\LARGE{j}} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, {(2)}$ is INvalid.

I understand (1) is invalid because there's 1 $i$ on the LHS but $2$ on the RHS. But I don't understand the rationale behind this rule? What's the problem?
$\sum_{i=1}^n A_i$ = $\sum_{i=1}^n B_iC_i $ is valid because it means $A_1 + ... + A_n = B_1C_1 + ... + B_nC_n $.

I understand (2) is invalid — On the LHS, when the summation is expanded in $i$, there's no $k$. However, on the RHS, when the 2 summations are expanded, $k$ is still there in the Levi-Civita tensor.

  • The better way to write it is as follows. For summation, write $a_ib^i$ - raising something to the $i$ here does not mean exponentiation but instead means the $i$th component. This makes it a bit clearer in my opinion because you explicitly know when something is being summed over. This notation is what is used most often in relativity for this reason. For the first one, on the left side, $i$ is a fixed number (since it only appears once) but on the right side, it is a sum - again rewrite it as $a_i = b_ic^i$ and it'll become clearer. Thus it is invalid. – Cameron Williams Jul 05 '13 at 03:17
  • $a_i = b_i c^i$ is still ambiguous, the left hand side of the equation is not being summed over while the right hand side is being summed over the index i. This leads to confusion over what is being summed and what isn't – Triatticus Jul 05 '13 at 03:19
  • @CameronWilliams I do agree that the up/down contraction is important in general, however, in the context of Euclidean space it is not needed. The covariant and contravariant components are the same since the metric and its inverse are one and the same. – James S. Cook Jul 05 '13 at 03:24
  • @JamesS.Cook you're right. I just find it to be clearer regardless of context because then there is no question. With $a_ib^i$ it is obvious that you are contracting indices. – Cameron Williams Jul 05 '13 at 03:25
  • IMHO If an index appears on both sides I would consider it free. If not, I would consider it a dummy. So I would read the first as: $A_1 = B_1 C_1$, $A_2 = B_2 C_2$, etc. No summing, because the index won't disappear. Then I would read the second as $A_i = \sum_j \sum_k \epsilon_{ijk} B_i C_j$.

    If an expression appears alone and not in an equation then the notation is indeed ambiguous.

     – Calmarius Aug 24 '13 at 15:51

1 Answers1

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Basically the point is that it is ambiguous to use an index as both a free and dummy index. $$ A_iB_{ii} \stackrel{\huge{?}}{=} A_iB_{jj} = A_i \sum_j B_{jj}$$ or $$ A_iB_{ii} \stackrel{\huge{?}}{=} A_jB_{ji} = \left(\sum_j A_jB_j\right)_i ? $$ This is a problem, we can't have both.

Added after Trapu's edit. So, to be clear, I will be non-einstein notation on the r.h.s. of the equations below: the point here is that $A_iB_{ii}$ cannot be interpreted meaningfully treating one pair of $i$ as dummies (summed over) and the other as free (not summed) $$ A_iB_{ii} = \sum_{j=1}^n A_jB_{ji} = A_1B_{1i}+A_2B_{2i}+ \cdots +A_nB_{ni} \qquad (I.) $$ verses: $$ A_iB_{ii} = \sum_{j=1}^n A_iB_{jj} = A_i\sum_{j=1}^n B_{jj} = A_i\left(B_{11}+B_{22}+ \cdots +B_{nn} \right) \qquad (II.) $$ Expressions (I.) and (II.) are two reasonable interpretations of $A_iB_{ii}$ if just one index is taken to be free. But, these are not equal.

For example, $B_{11}=1, B_{22}=-1, B_{12}=0=B_{21}$ and $A_1=1, A_2=1$, $$ (I.) \qquad A_1B_{1i}+A_2B_{2i} = B_{1i}+B_{2i} = \begin{cases} 1 & i=1 \\ -1 & i=2 \end{cases} $$ verses $$ (II.) \qquad A_i(B_{11}+B_{22}) = 0 = \begin{cases} 0 & i=1 \\ 0 & i=2 \end{cases} $$ As you can see these expressions do not agree. Therefore, we cannot use the same index for a dummy and a free index.

I would liken this problem to that I have with my calculus I students who insist they need not change the bounds in a u-substitution since they're just going to write it back in terms of $x$ at the end. However, if such a practice is made then some of the intermediate steps are wrong. We are left with the situation that what we write is insufficient to capture the precise mathemtical intent of the expression. This should be avoided since good notation ought to be unambiguous. Or, at a minimum the ambiguity should reflect a deeper mathematical structure as in the case of quotient spaces and the non-uniqueness of the representative. This is not that, this is just bad notation. It does lead to errors, trust me, I've made them.

James S. Cook
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    There is also the notion that $a_i \neq \sum \limits_i a_i$ – Triatticus Jul 05 '13 at 03:25
  • I suppose, I would write $a_i\delta_{ii}$ to sum over $i$ if I was so inclined. – James S. Cook Jul 05 '13 at 03:26
  • @Dan I'm not sure what $\delta_i$ would mean, the $\delta_{ij}$ is zero when $i \neq j$ and $1$ when $i=j$ so the symbol $\delta_{ii}$ is just $1$ and by the Einstein sum notation the expression $a_{i}\delta_{ii}$ implicits a sum over $i$. As a rule, the Kronecker delta can be used to create and destroy sums with some care. – James S. Cook Jul 05 '13 at 03:33
  • @JamesS.Cook: According to Einstein's summation convention $\delta_{ii} = n$. – user26872 Jul 05 '13 at 04:19
  • @oen but, I didn't just write $\delta_{ii}$. I wrote $a_i\delta_{ii}$ which would implicit $\sum_i a_i\delta_{ii}$ which is merely $a_1\delta_{11}+ \cdots a_n\delta_{nn} = a_1 + \cdots +a_n$. – James S. Cook Jul 05 '13 at 04:29
  • @JamesS.Cook: I believe instead of "... the symbol $\delta_{ii}$ is just $1$ ..." you mean "... the symbol $\delta_{ii}$ for fixed $i$ is $1$ ..." – user26872 Jul 05 '13 at 04:37
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    @oen Oh, so sorry, I see what you're saying now. See, this is why this notation is dangerous. It leads to these discussions. That said, it has been terribly useful to me in my life. But, yes, to be clear in the comment (I can no longer edit) I meant $\delta_{ii}=1$ for fixed $i$ as in $\delta_{11}=\delta_{22}= \cdots =\delta_{nn}=1$ – James S. Cook Jul 05 '13 at 05:13
  • @Dan: Is that because (of a rule of Einstein summation convention) — Since there's only one 1 $i$ on the LHS of eqn. (1), that $i$ must be a free index. But why's Einstein summation notation defined so that only a repeated index is summed? –  Jul 05 '13 at 08:10
  • Thanks — but I'm still lost. When I bring out the sigma notation, equation ($1$) has only the meaning that I wrote out? Where's the ambiguity? –  Jul 05 '13 at 08:23
  • @trapu your last edit is not how I would interpret $A_jB_{ji}$. It's not a component of a vector, it is the sum with a free index $i$ on each term. In particular, $A_jB_{ji}=A_1B_{1i}+A_2B_{2i}+ \cdots + A_nB_{ni}$. What you wrote does not make sense since $B_j$ is not anything. What do you mean by $B_j$? – James S. Cook Jul 05 '13 at 14:46
  • To say that $B_j$ has no meaning is not correct, in this paper it is the jth component of a vector $ \vec{B}$. For instance we know that the first component of $ (\vec{A} \times \vec{B})1 = (A_2 B_3 - A_3 B_2) = \epsilon{1jk}A_j B_k $ – Triatticus Jul 05 '13 at 20:21
  • @trapu The summation convention was just formed basically out of laziness on Einsteins part, he noted that the summations he performed so frequently had pairs of indices being summed over that he decided to not write the sigma for pairs of indices that appear on the same side of the expression, hence why it is known as Einstein summation convention. The rules were then put into place so that anyone could work backwards to the summation notation and there are no ambiguities. – Triatticus Jul 05 '13 at 20:31
  • @Dan I'm sorry, do I need to use a letter other than $B$? How would you define $B_i$ from my symbol $B_{ii}$? Notice that my comment concerned my post not the paper in question. – James S. Cook Jul 05 '13 at 21:32
  • @Dan — So if the rule had instead been — Single $ Repeated$ indices are the "new" symbol for summation — then $a_i = \sum_i a_i$ becomes true? So the equation 1 becomes $\sum_i A_i = \sum_i B_iC_i $? –  Jul 07 '13 at 07:46
  • @JamesS.Cook — Thanks a million. I'll accept and close this question after I hear back from Dan about my above comment. Abou $B_{ji}$ — I construed this as $\left(B_j\right)_i$ — the $j$th component of B in the $i$th term of AB. –  Jul 07 '13 at 07:56
  • @Dan — can you please get back to my comment 7/7? –  Oct 03 '13 at 01:09