Problem with solution: Simplify 6 Expressions with Kronecker Delta

Simplify the following expressions using the rules for calculating with Kronecker delta:

$\delta_{31}\,\delta_{33}$
$\delta_{ji}\,T_{ink}$
$\delta_{j1}\,\delta_{ji}\,\delta_{2i}$
$\delta_{ik}\,\delta_{i3}\,\delta_{3k}$
$\delta_{jj}$ with $j ~\in~ \{ 1,2,3,4 \} $
$\delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss} $ with $s ~\in~ \{ 1,2 \} $

Solution tips

Use the properties of Kronecker delta that you learned in the lesson. In all the following expressions you should keep in mind that $\delta_{ik}\,\delta_{ij}$ can be summed up to $\delta_{kj}$ and that you have to sum over equal indices: $$ \delta_{ii} ~=~ 1~+~1~+~ ... ~+~ 1 ~=~ n $$

Solution for (a)

Here we simplify the following expression: \begin{align} \delta_{31}\,\delta_{33} ~&=~ 0 \cdot 1 ~=~ 0 \end{align}

We exploited that $ \delta_{31} = 0 $ is because the indices have two different values and $ \delta_{33} = 1 $ has two equal indices.

Solution for (b)

We want to simplify the following expression: \begin{align} \delta_{ji} \, T_{ink} ~&=~ T_{jnk} \end{align}

We exploited the rule that the Kronecker delta $ \delta_{ji} $ eliminates everything except $ T_{jnk} $, where $ i = j $.

Solution for (c)

Here we simplify the following expression: \begin{align} \delta_{j1} \, \delta_{ji} \, \delta_{2i} \end{align}

For example, first combine the index $j$ in $\delta_{j1} \, \delta_{ji}$ and then combine the index $i$: \begin{align} \delta_{j1} \, \delta_{ji} \, \delta_{2i} &~=~ \delta_{i1} \, \delta_{2i} \\\\ &~=~ \delta_{12} \\\\ &~=~ 0 \end{align}

Of course, you could just as well first combine the index $i$ in $\delta_{ji} \, \delta_{2i}$: \begin{align} \delta_{j1} \, \delta_{ji} \, \delta_{2i} &~=~ \delta_{j1} \, \delta_{2j} \\\\ &~=~ \delta_{21} \\\\ &~=~ 0 \end{align}

You get the same result.

Solution for (d)

Proceed analogously to (c), except that at the end you get not 0 but 1 according to the Kronecker delta definition: \begin{align} \delta_{ik} \, \delta_{i3} \, \delta_{3k} &~=~ \delta_{k3} \, \delta_{3k} \\\\ &~=~ \delta_{33} \\\\ &~=~ 1 \end{align}

Again, the order of simplification does not matter.

Solution for (e)

Here we simplify the following expression, summing over the index $ j $ up to 4: \begin{align} \delta_{jj} \end{align} We get a sum with four summands: \begin{align} \delta_{jj} &~=~ \delta_{11} ~+~ \delta_{22} ~+~ \delta_{33} ~+~ \delta_{44} \\\\ &~=~ 1~+~1~+~1~+~1 \\\\ &~=~ 4 \end{align}

Solution for (f)

Here we simplify the following expression, summing over the index $ s $ up to 2: \begin{align} \delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss} \end{align}

Let us first combine $ \delta_{k\mu} \, \varepsilon_{kmn} $ and then write out the sum $\delta_{ss}$: \begin{align} \delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss} &~=~ \varepsilon_{\mu mn} \, \delta_{ss} \\\\ &~=~ \varepsilon_{\mu mn} \, \left( \delta_{11} ~+~ \delta_{22} \right) \\\\ &~=~ \varepsilon_{\mu mn} \, \left( 1~+~ 1 \right) \\\\ &~=~ 2\, \varepsilon_{\mu mn} \end{align}