I made recent discovery that civita symbols product can be proved in easy way. I still struggle to finish off the proof.
Here:
[https://math.stackexchange.com/questions/3827709/product-of-two-levi-civita-permutation-symbols][1] and here:[https://math.stackexchange.com/questions/1442326/kronecker-delta-and-levi-civita-symbol/1442347?newreg=fe0f7435bd3b401db705589e94db2eda][2]
are proofs. Have to prove
$$\begin{align}
\varepsilon_{abc}\varepsilon_{k\ell m}&=\left(\left(\delta_{ak}\hat x_k+\delta_{a\ell}\hat x_\ell+\delta_{am}\hat x_m\right)\cdot(\hat x_b\times \hat x_c)\right)(\hat x_k\cdot(\hat x_\ell\times \hat x_m))\\\\
\tag3
\end{align}$$
starting from
$$\begin{align}
\varepsilon_{abc}\varepsilon_{k\ell m}&=(\hat x_a\cdot(\hat x_b\times \hat x_c))(\hat x_k\cdot(\hat x_\ell\times \hat x_m))\\\\
\tag1
\end{align}$$
I make
$$\varepsilon_{abc}\varepsilon_{k\ell m}=(\hat{x_1}+\hat{x_2}+\hat{x_3})(\hat{x_1}+\hat{x_2}+\hat{x_3})$$ because civita $\epsilon$ gather terms in each direction.
So I get $$(\hat{x_1}+\hat{x_2}+\hat{x_3})(\hat{x_1}+\hat{x_2}+\hat{x_3})=\hat{x_1}\cdot\hat{x_1}+\hat{x_2}\cdot\hat{x_2}+\hat{x_3}\cdot\hat{x_3}\\+
\hat{x_1}\cdot\hat{x_2}+\hat{x_2}\cdot\hat{x_3}+\hat{x_3}\cdot\hat{x_1}\\+
\hat{x_1}\cdot\hat{x_3}+\hat{x_2}\cdot\hat{x_1}+\hat{x_3}\cdot\hat{x_2}$$
because $(1+2+3)(1+2+3)=\color{red}{1*1}+\color{green}{1*2}+1*3+2*1+\color{red}{2*2}*2+\color{green}{2*3}+\color{green}{3*1}+3*2+\color{red}{3*3}=\\
\color{red}{1*1}+\color{red}{2*2}+\color{red}{3*3}+\color{green}{1*2}+\color{green}{2*3}+\color{green}{3*1}+1*3+2*1+3*1$
so $$\hat{x_1}\cdot\hat{x_1}+\hat{x_2}\cdot\hat{x_2}+\hat{x_3}\cdot\hat{x_3}\\+
\hat{x_1}\cdot\hat{x_2}+\hat{x_2}\cdot\hat{x_3}+\hat{x_3}\cdot\hat{x_1}\\+
\hat{x_1}\cdot\hat{x_3}+\hat{x_2}\cdot\hat{x_1}+\hat{x_3}\cdot\hat{x_2}=1+1+1\\
+0+0+0+0+0+0$$
I cannot understand how rewriting civita in following way would make zero terms became other number than zero?
$$\hat{x_1}\cdot\hat{x_1}+\hat{x_2}\cdot\hat{x_2}+\hat{x_3}\cdot\hat{x_3}\ makes\ \delta_{ak}\hat x_k \\$$
$$\hat{x_1}\cdot\hat{x_2}+\hat{x_2}\cdot\hat{x_3}+\hat{x_3}\cdot\hat{x_1}\ makes\ \delta_{a\ell}\hat x_\ell\\$$
$$\hat{x_1}\cdot\hat{x_3}+\hat{x_2}\cdot\hat{x_1}+\hat{x_3}\cdot\hat{x_2}=1+1+1\ makes\ \delta_{am}\hat x_m\\$$
So how from zero we get $\delta_{a\ell}\hat x_\ell $ making equation?
We would normally get
$\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot \hat x_{\ell}\hat x_{k}\cdot(\hat x_l\times \hat x_m)\right)$
but civita product
$$\hat x_k\cdot (\hat x_\ell\times \hat x_m)=\hat x_\ell\cdot (\hat x_m\times \hat x_k)=\hat x_m\cdot (\hat x_k\times \hat x_\ell)$$
makes the trick to return
$\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot \hat x_{\ell}\hat x_{\ell}\cdot(\hat x_m\times \hat x_k)\right)$
How normally terms would be zero but once civita is applied it gives different results from zero?
How $\hat{x_l}\hat{x_k}$ would make normally zero but after appling civita it makes $\hat{x_l}\hat{x_l}$ so it gives non zero result?
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
IN SHORT
$$\begin{align}
\varepsilon_{abc}\varepsilon_{k\ell m}&=\left(\left(\delta_{ak}\hat x_k+\delta_{a\ell}\hat x_\ell+\delta_{am}\hat x_m\right)\cdot(\hat x_b\times \hat x_c)\right)(\hat x_k\cdot(\hat x_\ell\times \hat x_m))\\\\
\tag3
\end{align}$$
MAKE
$$\delta_{a\ell}\hat x_\ell(\hat x_k\cdot(\hat x_\ell\times \hat x_m)) $$EQUAL TO ZERO BECAUSE OF DOT PRODUCT OF $\hat{x_l}\hat{x_k}=0$
but after appling civita we get $\hat x_{\ell}\hat x_{\ell}\cdot(\hat x_m\times \hat x_k)$ which is non zero.
How appling civita make normally zero terms become nonzero?
[1]: Product of two Levi-Civita permutation symbols
[2]: Kronecker delta and Levi-Civita symbol
##############################
If I am right in understanding that triple cross product does not
return unit vector direction so two civita symbols are just multplication but not dot product so that there are no zeros.
Am I right in understanding?
