A problem in proving a identity with tensor notation There is an exercise that has me prove the following identity $J^i_{i'j'}J^{i'}_j+J^i_{i'}J^{i'}_{jk}J^k_{j'}=0$
from the identity $J^i_{i'j'}J^{i'}_jJ^{j'}_k+J^i_{i'}J^{i'}_{jk}=0$. The problem states that I have to contract the identity with $J^k_{k'}$ and then rename $j'\rightarrow k'$. Why is it allowed to change the names of the indices? I did it anyway and got $J^i_{i'j'}J^{i'}_j\delta^{k'}_{k'}+J^i_{i'}J^{i'}_{jk}J^k_{k'}=3J^i_{i'j'}J^{i'}_j+J^i_{i'}J^{i'}_{jk}J^k_{k'}=0$ and now I am stuck here.
Edit: By these $J_{i}^{j}$s I mean $i,j$ entries of the jacobian matrix.
 A: Just as someone said in the comments, indices names are sometimes just "dummy" indices. So when indices appear repeated such as here $A_iA^i$, this means $\sum_{i=1}^n A_iA^i = \sum_{j=1}^n A_jA^j$. So the result is the same no matter what name you use for the index. This is similar to what happens with dummy variables in integration: $\int_a^bf(x)dx = \int_a^b f(u)du$. For a similar reason You can also change the name of indices which are not involved in a summation (not repeated), as long as change them in all the terms (both sides of the equation). For example, imagine I have the identity $a_i +c_i= b_i$, then it doesn't matter if I state it as $a_j +c_j= b_j$ (renaming $i$ as $j$ in all terms) but I cannot write $a_i +c_i= b_j$ or $a_i +c_j= b_j$. Using this rules you can change names while keeping consistency in the equations.
Now regarding to the identities. I don't know all the context but from what you wrote it seems that $J_{k}^{j'}J_{k'}^k = \delta_{k'}^{j'}$ right?. Then let me just remind you a classical trick with $\delta_{j}^k$. If you have $\delta_{j}^k$ multiplying with another symbol which share some index in a summation, then you can substitute indices as follows: for example $J_{i'j'}^i\delta_{k'}^{j'} = J_{i'k'}^i$, where $J_{i'j'}^i$ and $\delta_{k'}^{j'} $ share the repeated index $j'$, then $j'$ in $J_{i'j'}^i$ changes to the other index contained in $\delta_{k'}^{j'}$, resulting in $J_{i'k'}^i$ (and the Kronecker delta disappears). Why is that? Note that $\delta_{k'}^{j'}=0$ unless $k'=j'$, hence all terms in the summation $J_{i'j'}^i\delta_{k'}^{j'}$ disappear except for $J_{i'k'}^i$.
We are now ready to show the identity. First, contract with $J_{k'}^k$ just as you did:
$$
J_{i'j'}^iJ_{j}^{i'}J_k^{j'}J_{k'}^k + J_{i'}^iJ_{jk}^{i'}J_{k'}^k=0
$$
Now use the fact that $J_k^{j'}J_{k'}^k = \delta_{k'}^{j'}$:
$$
J_{i'j'}^iJ_{j}^{i'}\delta_{k'}^{j'} + J_{i'}^iJ_{jk}^{i'}J_{k'}^k=0
$$
Now, use the kronecker delta substitution trick to write $J_{i'j'}^i\delta_{k'}^{j'}  = J_{i'k'}^i$:
$$
J_{i'k'}^iJ_{j}^{i'}+ J_{i'}^iJ_{jk}^{i'}J_{k'}^k=0
$$
Note that $J_{j}^{i'}$ doesn't share any indices with $\delta_{k'}^{j'} $ so it remains the same. Now rename $k'$ as $j'$ in all the equation: change all occurrences of $k'$ to $j'$:
$$
J_{i'j'}^iJ_{j}^{i'}+ J_{i'}^iJ_{jk}^{i'}J_{j'}^k=0
$$
which finishes the proof.
