# Symmetries of the multiplication of Kronecker delta and Levi-Civita symbol?

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

Following this answer I get (see the proof below) $$$$\epsilon_{ijk}\delta_{\ell m} = \epsilon_{\boldsymbol{m}jk}\delta_{\ell \boldsymbol{i}} + \epsilon_{\boldsymbol{\ell} jk}\delta_{\boldsymbol{i} m}$$$$ This seems nice, and it's something I am looking for. But if I contract both sides with $$\epsilon_{ijk}$$ to make sure it holds well (sanity check!), I get the obvious contradiction $$$$6\delta_{\ell m} = 2 \delta_{im}\delta_{i\ell} + 2 \delta_{i\ell} \delta_{im} = 4 \delta_{\ell m}$$$$ 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}

• Nowhere in the linked answer is there a delta times an epsilon. So how did you get your equation? Jun 5, 2023 at 17:43
• @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. Jun 5, 2023 at 18:00
• 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. Jun 5, 2023 at 18:15
• 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. Jun 5, 2023 at 18:16
• 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. Jun 5, 2023 at 18:41

Let's first relabel the indices for better tracking: $$$$\epsilon_{abc}\delta_{ij}$$$$ Following this answer, let's assume none of the indices $$(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$$
Hence, \begin{align} \epsilon_{abc}\delta_{ij}&=\left(\hat x_a\cdot(\hat x_b \times \hat x_c)\right)\left(\hat x_{i} \cdot\hat x_j\right)\\\\ &=[\hat x_a\cdot(\hat x_b \times \hat x_c)][(\delta_{ia}\hat x_a+\delta_{ib}\hat x_b + \delta_{ic}\hat x_c) \cdot\hat x_j)]\\\\ &=\delta_{ia}(\hat x_b \times \hat x_c)\cdot \hat x_{a}\hat x_{a}\cdot(\hat x_j)\\\\ &+\delta_{ib}(\hat x_c \times \hat x_a)\cdot \hat x_{b}\hat x_{b}\cdot(\hat x_{j})\\\\ &+\delta_{ic}(\hat x_a \times \hat x_b)\cdot \hat x_{c}\hat x_{c}\cdot(\hat x_{j})\\\\ &=\delta_{ia}(\hat x_b \times \hat x_c)\cdot \hat x_j + \delta_{ib}(\hat x_c \times \hat x_a)\cdot \hat x_j + \delta_{ic}(\hat x_a \times \hat x_b)\cdot \hat x_j\\\\ &= \epsilon_{jbc} \delta_{ia} + \epsilon_{ajc} \delta_{ib}+ \epsilon_{abj}\delta_{ic} \end{align}
Using the properties of the generalized Kronecker delta \begin{align} 3!\epsilon_{i[jk}\delta_{l]m} &= \delta^{pqr}_{jkl}\epsilon_{ipq}\delta_{rm} \\ &= \epsilon_{jkl}\epsilon^{pqr}\epsilon_{ipq}\delta_{rm} \\ &= \epsilon_{jkl}\delta^{pqr}_{ipq}\delta_{rm} \\ &= \epsilon_{jkl}\delta^{rpq}_{ipq}\delta_{rm} \\ &= 2\epsilon_{jkl}\delta^{r}_{i}\delta_{rm} \\ &= 2\epsilon_{jkl}\delta_{im} \\ \end{align} where the $$[\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}.$$