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Let A, B be two vectors. Then A=$a^ie_i$ B=$b^ie_i$ Then why is it in order to taking the inner product of the two I must change the summation of one vector:

$a^ie_i$ $\cdot$ $b^je_j$? The summation convention is that if there is a subscript and superscript that are of the same letter then summation is implied.

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    $\begingroup$ Do you mean the change from $i$ to $j$? $i$ and $j$ are dummy indices. $b^ie_i=b_1e_1+\ldots+b_ne_n=b^je_j$ $\endgroup$ – A. Goodier Feb 4 '19 at 17:25
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The convention is: repeated indices are summed over but each index must appear at most twice, so for example the expression

$$ (a^ie_i) \cdot (b^i e_i) $$

does not follow the rule (it has four times the symbol $i$ in it), and therefore you cannot sum over the index $i$ in this convention.

So how do you go about it? The answer is in the fact that the term $b^i e_i$ does follow the rule, so

$$ b^i e_i = b^1 e_1 + b^2 e_2 + \cdots = b^j e_j = b^k e_k = b^\alpha e_\alpha = \cdots $$

you can use whatever symbol you want to label the index, it is void of meaning. With this in mind you can write the first expression as

$$ (a^ie_i) \cdot (b^j e_j) $$

which now follows the rule of the convention

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Note that

  • $a^i e_i b^i e_i$ means $\displaystyle\sum_{i = 1}^n \left( a^i e_i \cdot b^i e_i \right)$ and has $n$ terms

  • $a^i e_i b^j e_j$ means $\left(\displaystyle\sum_{i = 1}^n a^i e_i\right) \cdot \left(\displaystyle\sum_{j = 1}^nb^j e_j\right)$ and has $n^2$ terms

This are different expressions. As A. Goodier commented, one can chose the name of the (dummy) index used in the sum. However, one has to distinguish between the two expressions listed above, so one might need two different indices.

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