I need to evaluate this multiplication of kronecker deltas
$$\delta_{ij}\delta_{km}\delta_{jk}\delta_{im}$$
seems to be a very simple exercise, but my question is: Can I change the order of the deltas to reduce the terms? I mean to use this property: $$\sum_{k} \delta_{ik}\delta_{kj} = \delta_{ij}.$$
thanks
You can reorder the deltas. This follows from basic properties of arithmetic. Changing the order of factors in a product is allowed because of commutativity of multiplication. With the summation convention in place, however, you have to think a bit harder about why certain rearrangements are allowed, since what looks like a product is actually a sum of products. Consider an arbitrary function $f$ that takes two elements of $\{1,2\}$ as arguments, and consider the expression $$ \sum_{i=1}^2\sum_{j=1}^2f_{ij}. $$ If you adopt the sensible convention that the inner sum is to be performed first, this represents $f_{11}+f_{12}+f_{21}+f_{22}$, which can be thought of as the row-first summation of the array elements in $$ \begin{array}{cc} f_{11} & f_{12}\\ f_{21} & f_{22}. \end{array} $$ If you change the order of the summations, $$ \sum_{j=1}^2\sum_{i=1}^2f_{ij}. $$ you get $f_{11}+f_{21}+f_{12}+f_{22}$, which is equal to the previous expression because of commutativity of addition. This is like doing a column-first summation of the array above.
Now consider the double summation $$ \sum_{j=1}^2\sum_{k=1}^2f_{ij}f_{jk}f_{k\ell}. $$ The distributive law allows use to write this as $$ \sum_{j=1}^2f_{ij}\sum_{k=1}^2f_{jk}f_{k\ell}. $$ This is because, for fixed $i$, $j$, and $\ell$, every term in the inner sum over $k$ has the factor $f_{ij}$, which may therefore be pulled out, e.g. (with $i=j=1$, $\ell=2$) we have $$ f_{11}f_{11}f_{12}+f_{11}f_{12}f_{22}=f_{11}(f_{11}f_{12}+f_{12}f_{22}). $$ Now if $f$ were the delta function, we could then simplify the inner sum to $\delta_{12}$.
Thinking in these terms should allow you to decide for yourself when an rearrangement is allowed.