I would like to know how this property of the Kronecker delta works. How exactly is $\delta_{a1}\delta_{1c} + \delta_{a2}\delta_{2c} + \delta_{a3}\delta_{3c} = \delta_{ac}$?
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Is it known that $a,c$ must be one of $1,2,3$? – lulu Mar 03 '20 at 00:39
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Use the definition. What is $\delta_{a1}$ equal to? What is the value of the product $\delta_{a1}\delta_{1c}$? – pegasus Mar 03 '20 at 00:40
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Wherever I look, this property is given without any proof, like in the picture. It isn't specified if a and c are (1,2 or 3), it is just said that one sums over the same indices. In this case b (see the picture). – ali_ runnindis Mar 03 '20 at 00:44
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@pegasus how can delta(a1)delta(1c) have a value if we don't know what a and c are? – ali_ runnindis Mar 03 '20 at 00:45
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@ali_runnindis, notice that the final expression depends on $a$ and $c$. Try to calculate the expression for each of the (9) possible combinations of $a$ and $c$. – pegasus Mar 03 '20 at 00:48
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I understand. Thank you – ali_ runnindis Mar 03 '20 at 00:49