# Ricci Calculus kronecker delta

I understand that in Ricci calculus $$\delta_i^j$$ represents the kronecker delta function where $$\delta^j_i=1$$ if $$i=j$$ and $$0$$ otherwise. What I am struggling with though is that I have seen it written for a matrix/(1,1)-tensor $$B$$;

$$(B')^i_j = B^j_i\delta^{ii}\delta_{jj}$$

Where $$B'$$ represents the matrix transpose of $$B$$. I am a bit confused here. What is the interpretation of $$\delta^{ii}$$? Is there an intuitive way to describe this relationship? Does anyone have any recommendations for learning more about the kronecker delta in Ricci calculus notation in situations like this?

• In the usual notation one is only allowed to have repeated indices in pairs. Where did you find this equation? If you interpret it literrally then $\delta^{ii}=\delta_{jj}=1$, so the equation says $(B')^i_j = B^j_j$, but I don't know if that is the intended meaning. Commented May 5, 2020 at 4:57
• I found it here matrixcalculus.org/matrixcalculus.pdf Commented May 5, 2020 at 14:35
• Sorry what is meant by repeated indices in pairs? Commented May 5, 2020 at 14:36
• A repeated index means it is being summed over. E.g., $a_ib^i=\sum_i a_ib^i$, $a_{ij}b^{ik}c^j_l=\sum_i\sum_ja_{ij}b^{ik}c^j_l$. The keywords are Einstein summation convention. Commented May 6, 2020 at 12:32
• You mean that $\delta^{ii}$ is just a sum of ones? Commented May 6, 2020 at 12:34

$$(B')^i{}_j = B^k{}_l \delta^{il}\delta_{kj}$$
$$(B')^i{}_j = \sum_k\sum_l B^k{}_l \delta^{il}\delta_{kj}$$.
The symbols $$\delta_{ij}$$ and $$\delta^{ij}$$ are defined by $$\delta_{ij}=\delta^{ij}=\begin{cases}1&i=j\\0&i\neq j\end{cases}$$