# Tensor contraction and associativity

I'm a tensor newbie and while solving some problems in a book, I generated the following double contraction: $$\delta^i_{k} \delta^{k}_{i} = 3$$ Multiplying both sides by $\delta^{m}_{i}$ gives me $$\delta^{m}_{i} \delta^{i}_{k} \delta^{k}_{i} = 3 \delta^{m}_{i}$$ Index renaming yields $$\delta^{m}_{k} \delta^{k}_{i} = 3 \delta^{m}_{i}$$ $$\delta^{m}_{i} = 3 \delta^{m}_{i}$$ Which, kinda seems like nonsense. What rule did I break or what misrepresentation did I make?

thanks

• "Multiplying both sides by $\delta^m_i$" is illegal. In Einstein notation, if the same index appears more than twice, then the said index is not summed. – Cave Johnson Jul 21 '17 at 14:25

$$\delta_k^i\delta_i^k \equiv \sum_i\sum_k \delta_k^i\delta_i^k = \sum_i\delta_i^i = 3$$
The problem with your second equation is that you have the index $i$ repeated three times which does not make sense in this notation
• Thanks, that makes sense. Is it an odd number that fails or any number > 2? The actual problem came up in a proof that the contravariant metric is a tensor starting from the identity. $g_{im}g^{mj}=\delta^i_j$ where I know that the covariant metric is a tensor. Throw in some Jacobians and you end up with multiple instances of indices. – Marco Dalla Gasperina Jul 21 '17 at 14:29
The summation convention really insists on repeated indices occurring only twice! Your paradox illustrates why: When you multiply $\delta^i_k\delta^k_i$ by $\delta^m_i$, you are really multiplying $\sum_{i,k}\delta^i_k\delta^k_i$ by $\delta^m_i$. The answer is not $\sum_{i,k}\delta^i_k\delta^k_i\delta^m_i$! You would have to rename the summation index $i$ to $j$ first: Then you multiply $\sum_{i,k}\delta^i_k\delta^k_i=\sum_{j,k}\delta^j_k\delta^k_j$ by $\delta^m_i$, and you get the result $\sum_{j,k}\delta^j_k\delta^k_j\delta^m_i$.
The problem occurs in the second line. You're already summing over $i$ and $k$. Try replacing $\delta_i^m$ with a different index, say $\delta_j^ m$.