Express $\delta_{ij}\delta_{jk}\delta_{ki}\delta_{mn}a_mb_n$ in terms of $\vec{a}\cdot\vec{b}$

Question

This is a question related to the Einstein summation convention. I have made an attempt at answering this question and have been incorrect. I'd really like for someone to help me understand why it is that my method fails. The following is my attempt (it should be noted here that $\delta_{ij}$ is the Kronecker-delta).

We may fix certain variables, and expand over others. For this purpose, we will start by fixing $j$, $k$, $m$ and $n$, and letting $i$ range from $1$ to $3$. We obtain \begin{equation} \delta_{ij}\delta_{jk}\delta_{ki}\delta_{mn}a_mb_n=\delta_{1j}\delta_{jk}\delta_{k1}\delta_{mn}a_mb_n+\delta_{2j}\delta_{jk}\delta_{k2}\delta_{mn}a_mb_n+\delta_{3j}\delta_{jk}\delta_{k3}\delta_{mn}a_mb_n \end{equation} Now we will let $j$ range from $1$ to $3$. We note therefore that $j$ will range from $1$ to $3$ for each term of $(4)$. Therefore, if we omit terms which equate to $0$, we obtain \begin{equation*} \begin{split} \delta_{ij}\delta_{jk}\delta_{ki}\delta_{mn}a_mb_n & =\delta_{11}\delta_{1k}\delta_{k1}\delta_{mn}a_mb_n+\delta_{22}\delta_{2k}\delta_{k2}\delta_{mn}a_mb_n+\delta_{33}\delta_{3k}\delta_{k3}\delta_{mn}a_mb_n \\ & =\delta_{1k}\delta_{k1}\delta_{mn}a_mb_n+\delta_{2k}\delta_{k2}\delta_{mn}a_mb_n+\delta_{3k}\delta_{k3}\delta_{mn}a_mb_n \end{split} \end{equation*} We now let $k$ range from $1$ to $3$, still omitting those terms which equate to $0$, and we notice that \begin{equation*} \begin{split} \delta_{ij}\delta_{jk}\delta_{ki}\delta_{mn}a_mb_n & =\delta_{11}\delta_{11}\delta_{mn}a_mb_n+\delta_{22}\delta_{22}\delta_{mn}a_mb_n+\delta_{33}\delta_{33}\delta_{mn}a_mb_n \\ & =\delta_{mn}a_mb_n+\delta_{mn}a_mb_n+\delta_{mn}a_mb_n \end{split} \end{equation*} Next, we let $m$ range from $1$ to $3$. Since all other indices have been eliminated at this point, we do not need to be concerned with them. We have that $$ \delta_{ij}\delta_{jk}\delta_{ki}\delta_{mn}a_mb_n=\delta_{1n}a_1b_n+\delta_{2n}a_2b_n+\delta_{3n}a_2b_n $$ And finally we let $n$ range from $1$ to $3$. We omit terms which equate to $0$, and hence we obtain \begin{equation*} \begin{split} \delta_{ij}\delta_{jk}\delta_{ki}\delta_{mn}a_mb_n & =\delta_{11}a_1b_1+\delta_{22}a_2b_2+\delta_{33}a_2b_3 \\ & =a_1b_1+a_2b_2+a_3b_3 \\ \end{split} \end{equation*} Hence we may introduce the dummy variable $r$, and, noting that $a_1b_1+a_2b_2+a_3b_3=a_rb_r=\vec{a}\cdot\vec{b}$, we have that $$ \delta_{ij}\delta_{jk}\delta_{ki}\delta_{mn}a_mb_n=\vec{a}\cdot\vec{b} $$

This is the answer I arrived at, although I am aware that the answer is supposed to be $3\vec{a}\cdot\vec{b}$, I am not sure why it is that my method fails. I'm fairly confident I may lack some pretty fundamental understanding when it comes to this convention, and if someone would be so kind as to aid my understanding of this convention that would be much appreciated.

Answer 1

The first three deltas simplify to $\delta_{ii}=\sum_i1$, which is the dimension, i.e. $3$.

Answer 2

Note that $\delta_{mn}\ne0$ only if $m=n$. Hence the sum is equivalent to $$\sum_{i,j,k}\delta_{ij}\delta_{jk}\delta_{ki}(a_1b_1+a_2b_2+a_3b_3)=\vec{a}\cdot\vec{b}\sum_{i,j,k}\delta_{ij}\delta_{jk}\delta_{ki}$$ where the sum is non-zero only if $i=j=k$. Hence the sum is equivalent to $$\vec{a}\cdot\vec{b}\sum_{i}\delta_{ii}\delta_{ii}\delta_{ii}=\vec{a}\cdot\vec{b}(\delta_{11}^3+\delta_{22}^3+\delta_{33}^3)=3\vec{a}\cdot\vec{b}$$ as required.

Answer 3

Your mistake came in the paragraph where you say "Next we let $m$ range from $1$ to $3$." In the equation after that you present three terms, with $m=1$, $M=2$ and $m=3$. And those three terms would all have come about from just the first of the $\delta_{mn}a_mb_n$ terms in the previous expression.

But that previous expression has $\delta_{mn}a_mb_n$ repeated three times. So you effectively left out the last two times, thus dividing the correct answer by $3$. You got $a\cdot b$, so the real answer would be $3a\cdot b$.