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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?

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  • $\begingroup$ 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. $\endgroup$ Commented May 5, 2020 at 4:57
  • $\begingroup$ I found it here matrixcalculus.org/matrixcalculus.pdf $\endgroup$
    – JDoe2
    Commented May 5, 2020 at 14:35
  • $\begingroup$ Sorry what is meant by repeated indices in pairs? $\endgroup$
    – JDoe2
    Commented May 5, 2020 at 14:36
  • $\begingroup$ 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. $\endgroup$ Commented May 6, 2020 at 12:32
  • $\begingroup$ You mean that $\delta^{ii}$ is just a sum of ones? $\endgroup$
    – JDoe2
    Commented May 6, 2020 at 12:34

1 Answer 1

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The notation is not standard. In the usual notation, one would write

$$(B')^i{}_j = B^k{}_l \delta^{il}\delta_{kj}$$

Which by Einstein summation convention is just

$$(B')^i{}_j = \sum_k\sum_l B^k{}_l \delta^{il}\delta_{kj}$$.

The symbols $\delta_{ij}$ and $\delta^{ij}$ are defined by $\delta_{ij}=\delta^{ij}=\begin{cases}1&i=j\\0&i\neq j\end{cases}$

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