3

Prove that $$ det \left[ \begin{matrix} δ_{ak} & δ_{al} & δ_{am} \\ δ_{bk} & δ_{bl} & δ_{bm} \\ δ_{ck} & δ_{cl} & δ_{cm} \\ \end{matrix} \right] =ε_{abc}ε_{klm} $$ with $a,b,c,k,l,m \in \{1,2,3\}$ where $δ_{ij}$ is Kronecker delta and $ε_{ijk}$ is Levi-Civita symbol in dimension 3

I found this problem in a very old mathematical-competition.I would apreciate if we can find a solution for this.

Paris Lamp
  • 578
  • 1
    Use the relation between Levi-Civita and Kronecker delta. – Kushal Bhuyan Sep 19 '15 at 15:11
  • I think this was asked a few hours ago, but he already has the relation and he was trying to prove on why it's correct. – mopy Sep 19 '15 at 15:35
  • Swapping any two indices on the first $\varepsilon$ symbol is equivalent to swapping rows in the det, while the indices on the second swaps columns. The effect is to change the sign of determinant. If any two indices are equal then those rows/columns are linearly dependent and det=0. Finally, when both symbols are $\varepsilon_{123}$, then $\det(I)=1$. – greg Sep 19 '20 at 03:46
  • @ParisLamp Hi! It's been a while. I hope you're staying safe and healthy during the pandemic. I've reached out to contact you a few times, but am unsure whether you've received the notes? If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) – Mark Viola Feb 28 '21 at 19:47

1 Answers1

3

Write the Kronecker Delta as

$$\delta_{ij}=\hat x_i \cdot \hat x_j \tag 1$$

in terms of then inner product of Cartesian unit vectors. Write the Levi-Civita symbol as

$$\epsilon_{ijk}=\hat x_i\cdot(\hat x_j \times \hat x_k) \tag 2$$

in terms of the scalar triple product of Cartesian unit vectors. Using $(2)$, we have

$$\begin{align} \epsilon_{abc}\epsilon_{k\ell m}&=\left(\hat x_a\cdot(\hat x_b \times \hat x_c)\right)\left(\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)\tag3\\\\ &=\delta_{ak}\left((\hat x_b \times \hat x_c)\cdot \hat x_k\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)\tag4\\\\ &+\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot \hat x_{\ell}\hat x_{\ell}\cdot(\hat x_m\times \hat x_k)\right)\\\\ &+\delta_{am}\left((\hat x_b \times \hat x_c)\cdot \hat x_{m}\hat x_{m}\cdot(\hat x_k\times \hat x_{\ell})\right)\\\\ &=\delta_{ak}\left((\hat x_b \times \hat x_c)\cdot (\hat x_{\ell} \times \hat x_m)\right)\\\\ &+\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot (\hat x_m\times \hat x_k)\right)\\\\ &+\delta_{am}\left((\hat x_b \times \hat x_c)\cdot (\hat x_k\times \hat x_{\ell})\right)\\\\ &=\delta_{ak}\left(\delta_{b\ell}\delta_{cm}-\delta_{bm}\delta_{c\ell}\right)\\\\ &+\delta_{a\ell}\left(\delta_{bm}\delta_{ck}-\delta_{bk}\delta_{cm}\right)\\\\ &+\delta_{am}\left(\delta_{bk}\delta_{c\ell}-\delta_{b\ell}\delta_{ck}\right)\\\\ &= det \left[ \begin{matrix} δ_{ak} & δ_{al} & δ_{am} \\ δ_{bk} & δ_{bl} & δ_{bm} \\ δ_{ck} & δ_{cl} & δ_{cm} \\ \end{matrix} \right] \end{align}$$

as was to be shown!

NOTE: In going from $(3)$ to $(4)$, we used the result in This Answer.

Mark Viola
  • 184,670