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Let’s consider this equation for a scalar quantity $f$ as a function of a 3D vector $a$ as:

$$ f(\vec a) = S_{ijkk} a_i a_j $$

where $S$ is a tensor of rank 4. Now, I’m not sure what to make of the index $k$ in the expression, as it doesn’t appear on the left-hand side. Is it a typo, meaning there is a $k$ missing somewhere (like $f_k$), or does it mean that it should be summed over $k$ like so:

$$f(\vec a) = \sum_i \sum_j \sum_k S_{ijkk} a_i a_j $$

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    $\begingroup$ I suppose so, but I'm no great shakes at Einstein notation. Anyway, if $S$ is of rank 4, you have to have those extra indices, so it can't be a typo. All this means is add all elements in a 'row' of the tensor, except in higher dimensions. I guess. $\endgroup$ Mar 30, 2012 at 13:21
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    $\begingroup$ Note:I am voting to close, since this seems a tad too localised. Feel free to ask this in chat if nobody has answered it in the comments by the time it gets closed :\ $\endgroup$ Mar 30, 2012 at 13:23
  • $\begingroup$ @Manishearth forget about what this equation represents, it's a general question about notation. If an index is repeated inside the same variable, does it imply summation? $\endgroup$
    – F'x
    Mar 30, 2012 at 13:24
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    $\begingroup$ I think one first has to calculate the "trace" of S indicated by the implicit Einstein sum over k, which leaves S as a second rank tensor with indices i and j. Summing over i and j as explicitely written in the second equation then gives the scalar corresponding to f on the l.h.s. $\endgroup$
    – Dilaton
    Mar 30, 2012 at 13:33
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    $\begingroup$ @Manishearth I'm pretty sure this is not the sort of thing "too localized" is for. (I think it's a fine question, actually) $\endgroup$
    – David Z
    Mar 30, 2012 at 18:50

2 Answers 2

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In the Einstein convention, pairs of equal indices to be summed over may appear at the same tensor. For example, the formula ${A_k}^k=tr~A$ is perfectly legitimate.

But your formula looks strange, as one usually sums over a lower index and an upper index, whereas you sum over lower indices only, which doesn't make sense in differential geometry unless your metric is flat and Euclidean (and then higher order tensors are very unlikely to occur).

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  • $\begingroup$ The tensor in question is the elastic stiffness tensor… $\endgroup$
    – F'x
    Mar 30, 2012 at 13:48
  • $\begingroup$ then it makes sense. $\endgroup$ Mar 30, 2012 at 13:50
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    $\begingroup$ In particle physics as well, people are frequently sloppy with the notation and use upper and lower indices interchangeably. $\endgroup$
    – David Z
    Mar 30, 2012 at 18:48
  • $\begingroup$ @DavidZaslavsky: If the two summed indices are lower, it means contract with g, so there is no ambiguity and it might not be called sloppy. $\endgroup$
    – Ron Maimon
    Mar 31, 2012 at 2:52
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You could rewrite your equation as

$$ f(\vec a) = S_{ijkl} a_i a_j \delta_{kl} $$

where $\delta_{kl}$ is the Kroneker Delta, if that helps. The last equation you've written is the right idea.

I would stress, though, that Einstein notation usually uses one upper and one lower index. This is partially so you can quickly see if your summations and indices are correct.

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