Index notation

1. The vector has 2 components so that $i$ can take two values (1 or 2). the dimension of $v_i$ is thus 2
2. The object $v_i$ has one index, and thus the order is 1
3. In $m_{ij}$, both $i$ and $j$ can take two different values (1 or 2). Hence, the dimension of $m_{ij}$ is still 2. However, there are now two indices, and the order of $m_{ij}$ is 2.

1.1. $i$ is not repeated in the same term and is a free index, $j$ is repeated and is dummy
1.2. $i$ and $j$ are not repeated in the same term and are free, $k$ and $l$ are repeated and are thus dummies
1.3. $i$ is not repeated in the same term (even if it is repeated in different terms) and is therefore a free index 1.4. $i$ and $k$ exist only once in each term: free indices. $j$ and $l$ are repeated twice in each term: dummies
2. If $b_{ij}=u_i v_j$, we have equivalently $b_{kl}=u_k v_l$, hence $a_{ij}=C_{ijkl} u_k v_l$
3. The expression $c_{kl}=a_{ij}D_{ijkl}$ already includes the indices $k$ and $l$, we must therefore change those indices in the first expression before multiplying: $a_{ij}=C_{ijmn}b_{mn}$ is equivalent as in this expression $k$ and $l$ where dummy indices and we can therefore change them to any available indices, in this case $m$ and $n$. We then have $c_{kl}=C_{ijmn}b_{mn}D_{ijkl}$.

1. Here we want to sum over $i$ and $j$ from 1 to 3. Hence we have
$[\delta_{11}]^2 + [\delta_{12}]^2 + [\delta_{13}]^2 + [\delta_{21}]^2 + [\delta_{22}]^2 + [\delta_{23}]^2 + [\delta_{31}]^2 + [\delta_{32}]^2 + [\delta_{33}]^2$
where only $\delta_{11}=\delta_{22}=\delta_{33}=1$ are nonzero. Hence, $\delta_{ij}\delta_{ij}=3$.
2. Following the definition of $\delta_{ij}$ we see that it is symmetric, i.e. $\delta_{ij}=\delta_{ji}$ and consequently $\delta_{ij}-\delta_{ji}=0_{ij}$ (the indices on the zero are often skipped for brevity)
3. The result will be a 2nd order object, e.g. $b_{li}=\delta_{kl}a_{ki}$. For each entry in $b_{li}$, we must sum over the 3 values of $k$. However, $\delta_{kl}=0$ if $k\neq l$. Only when $k=l$ is $\delta_{kl}=1$. I.e. $b_{li} = \delta_{1l}a_{1i}+\delta_{2l}a_{2i}+\delta_{3l}a_{3i}$. Only the term where $\delta_{kl}=1$ (i.e. $k=l$) will remain. It has the value $a_{li}$. Consequently, $\delta_{kl}a_{ki}=a_{li}$. In general, when we contract with $\delta_{ij}$ we can replace either $i$ or $j$ by $j$ or $i$ respectively in the factor we multiply with, and remove the $\delta_{ij}$ factor.

1. As we did earlier for the Kronecker delta, we know that $\delta_{ok}=1$ only when $o=k$ (otherwise zero), and hence, $\epsilon_{mno}\delta_{ok}=\epsilon_{mnk}$
2. $\epsilon_{mkk} = \epsilon_{m11} + \epsilon_{m22} + \epsilon_{m33} = 0$
3. $\epsilon_{jki} = \epsilon_{ijk}$ following the permutation order, hence $\epsilon_{jki}+\epsilon_{ijk}=2\epsilon_{ijk}$