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I was pondering on the symmetries that \begin{equation} \epsilon_{ijk}\delta_{\ell m} \end{equation} might have upon interchanging indices of the Kronecher delta and Levi-Civita symbol, e.g. the interchange of $i$ and $m$ or $\ell$.

Following this answer I get (see the proof below) \begin{equation} \epsilon_{ijk}\delta_{\ell m} = \epsilon_{\boldsymbol{m}jk}\delta_{\ell \boldsymbol{i}} + \epsilon_{\boldsymbol{\ell} jk}\delta_{\boldsymbol{i} m} \end{equation} This seems nice, and it's something I am looking for. But if I contract both sides with $\epsilon_{ijk}$ to make sure it holds well (sanity check!), I get the obvious contradiction \begin{equation} 6\delta_{\ell m} = 2 \delta_{im}\delta_{i\ell} + 2 \delta_{i\ell} \delta_{im} = 4 \delta_{\ell m} \end{equation} What could possibly have gone wrong? Also, and more importantly, what would be the symmetries anyway?

Edit: Proof:

Set $\hat x_i$ as the sum (see this answer)

$$\hat x_i=\delta_{i\ell}\hat x_\ell+\delta_{im}\hat x_m$$

(no summation is implied over repeated indices) so $$\begin{align} \epsilon_{ijk}\delta_{\ell m}&=\left(\hat x_i\cdot(\hat x_j \times \hat x_k)\right)\left(\hat x_{\ell} \cdot\hat x_m\right)\\\\ &=\left([\delta_{i\ell}\hat x_\ell+\delta_{im}\hat x_m]\cdot(\hat x_j \times \hat x_k)\right)\left(\hat x_{\ell} \cdot\hat x_m\right)\\\\ &=\delta_{i\ell}(\hat x_j \times \hat x_k)\cdot \hat x_{\ell}\hat x_{\ell}\cdot(\hat x_m)\\\\ &+\delta_{im}(\hat x_j \times \hat x_k)\cdot \hat x_{m}\hat x_{m}\cdot(\hat x_{\ell})\\\\ &=\delta_{i\ell}(\hat x_j \times \hat x_k)\cdot \hat x_m +\delta_{im}(\hat x_j \times \hat x_k)\cdot \hat x_{\ell}\\\\ &=\delta_{i\ell} \epsilon_{mjk} + \delta_{im}\epsilon_{\ell jk} \end{align}$$

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  • $\begingroup$ Nowhere in the linked answer is there a delta times an epsilon. So how did you get your equation? $\endgroup$
    – coiso
    Commented Jun 5, 2023 at 17:43
  • $\begingroup$ @elemelons I added the proof. Yet as I type, I reckon I violated the completeness of $\hat{x}_i$'s as a basis. This must be the problem. $\endgroup$
    – Bjaam
    Commented Jun 5, 2023 at 18:00
  • $\begingroup$ In Einstein notation, we have $\hat{x}_i=\delta_i^\ell\hat{x}_\ell$. Note that's being summed over $\ell$. I don't know why you wrote the RHS out twice with two different indices, that's wrong. But then in $(\hat{x}_i\cdot(\hat{x}_j\times\hat{x}_k))(\hat{x}_\ell\cdot\hat{x}_m)$, you replaced $\hat{x}_i$ with something that used $\ell$ as an index, even though $\ell$ was already being used as an index outside of that summation. You've confused yourself by overusing indices and not keepin track of when they are being summed over. $\endgroup$
    – coiso
    Commented Jun 5, 2023 at 18:15
  • $\begingroup$ Compare to e.g. $3x=(\sum_{x=1}^2x)x=\sum_{x=1}^2x^2=5$, which proves three times anything is $5$. Not to mention, your current final line includes repeated indices, but the equation I asked how you got does not. $\endgroup$
    – coiso
    Commented Jun 5, 2023 at 18:16
  • $\begingroup$ Sorry, the last line had a typo. Nothing is summed over. In the link the answerer to that post provides, he specifically mentions that in $\hat x_i=\delta_{i\ell}\hat x_\ell+\delta_{im}\hat x_m$ the summation convention is NOT intended. So, it's not an expansion in the basis. Apart from that heinous typo in the last line of my proof, all indices are free. $\endgroup$
    – Bjaam
    Commented Jun 5, 2023 at 18:41

2 Answers 2

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Let's first relabel the indices for better tracking: \begin{equation} \epsilon_{abc}\delta_{ij} \end{equation} Following this answer, let's assume none of the indices $(a,b,c)$ is equal to either of the other two, and hence the number $i$ is equal to one of $a$, $b$, or $c$. So, using the property of the Kronecker delta we can write

$$\hat x_i=\delta_{ia}\hat x_a+\delta_{ib}\hat x_b + \delta_{ic}\hat x_c$$

Hence, $$\begin{align} \epsilon_{abc}\delta_{ij}&=\left(\hat x_a\cdot(\hat x_b \times \hat x_c)\right)\left(\hat x_{i} \cdot\hat x_j\right)\\\\ &=[\hat x_a\cdot(\hat x_b \times \hat x_c)][(\delta_{ia}\hat x_a+\delta_{ib}\hat x_b + \delta_{ic}\hat x_c) \cdot\hat x_j)]\\\\ &=\delta_{ia}(\hat x_b \times \hat x_c)\cdot \hat x_{a}\hat x_{a}\cdot(\hat x_j)\\\\ &+\delta_{ib}(\hat x_c \times \hat x_a)\cdot \hat x_{b}\hat x_{b}\cdot(\hat x_{j})\\\\ &+\delta_{ic}(\hat x_a \times \hat x_b)\cdot \hat x_{c}\hat x_{c}\cdot(\hat x_{j})\\\\ &=\delta_{ia}(\hat x_b \times \hat x_c)\cdot \hat x_j + \delta_{ib}(\hat x_c \times \hat x_a)\cdot \hat x_j + \delta_{ic}(\hat x_a \times \hat x_b)\cdot \hat x_j\\\\ &= \epsilon_{jbc} \delta_{ia} + \epsilon_{ajc} \delta_{ib}+ \epsilon_{abj}\delta_{ic} \end{align}$$

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Using the properties of the generalized Kronecker delta \begin{align} 3!\epsilon_{i[jk}\delta_{l]m} &= \delta^{pqr}_{jkl}\epsilon_{ipq}\delta_{rm} \\ &= \epsilon_{jkl}\epsilon^{pqr}\epsilon_{ipq}\delta_{rm} \\ &= \epsilon_{jkl}\delta^{pqr}_{ipq}\delta_{rm} \\ &= \epsilon_{jkl}\delta^{rpq}_{ipq}\delta_{rm} \\ &= 2\epsilon_{jkl}\delta^{r}_{i}\delta_{rm} \\ &= 2\epsilon_{jkl}\delta_{im} \\ \end{align} where the $[\dots]$ denote antisymmetrization. Now the left hand side is $$3!\epsilon_{i[jk}\delta_{l]m} = 2(\epsilon_{ijk}\delta_{lm} +\epsilon_{ikl}\delta_{jm} +\epsilon_{ilj}\delta_{km}) $$ so dividing by $2$ you get $$\epsilon_{ijk}\delta_{lm} +\epsilon_{ikl}\delta_{jm} +\epsilon_{ilj}\delta_{km} =\epsilon_{jkl}\delta_{im}. $$

I want to remark that I translated the first calculation in my post from Penrose graphical notation. Here is my original calculation. I hope it convinces you to at least take a look at the wiki article :D it's very nice for tensor calculations.

penrose graphical notation

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  • $\begingroup$ It's brilliant! I'm going to learn the crap out of it! I looked, and I'm going to start with these suggestions. What else would you recommend? $\endgroup$
    – Bjaam
    Commented Jun 6, 2023 at 20:08
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    $\begingroup$ @Bjaam yup those resources look allright. Also take a look at the apendices in the first part of Penrose's book together with Rindler: Spinors and Space-time. $\endgroup$ Commented Jun 7, 2023 at 5:58

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