Levi-Civita's different expansions

Question

So I've recently been introduced to the Levi-Civita and the expansion for its product. Since then we've been asked to prove some vector identities using it. What's confused me is that different websites have different expansions for the same product. I know that this is meant to be explained by the fact that even permutations don't change the value of Levi-Civita but when I use a certain expansion (e.g. the one from my notes) I'm not able to get to the right answer but if I use a different expansion (e.g. from other university notes) the answer comes out easily. I think I'm probably doing something wrong on the way that I'm not spotting.

Take this for example (I'm not sure how to make suffixes here so I've just done them as $^k$, so $^k$ is the $k$-coordinate of the vector.

$$[(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{a})]^k = ε^{kij}(\vec{a}\times\vec{b})^i (\vec{c}\times\vec{a})^j$$

$$= ε^{kij} ε^{irs} a^r b^s ε^{jmn} c^m a^n $$

$$= ε^{kij} ε^{irs} ε^{jmn} a^r b^s c^m a^n $$

At this point there's two expansions I've tried using. One gives the right answer one gives the wrong.

1st:

$$=(δ^{ir} δ^{js} - δ^{is} δ^{jr}) ε^{jmn} a^r b^s c^m a^n $$

$$=ε^{jmn} a^i b^j c^m a^n - ε^{jmn} a^j b^i c^m a^n $$

$$=a^i(\vec{a},\vec{b},\vec{c})-b^i(\vec{a},\vec{c},\vec{a})$$

$$ =a^i(\vec{a},\vec{b},\vec{c})$$

Which violates the summation convention because on one side the index is $k$ and on the other is $i$.

2nd: $$=(δ^{jr} δ^{ks} - δ^{js} δ^{kr}) ε^{jmn} a^r b^s c^m a^n$$

$$=ε^{jmn} a^j b^k c^m a^n - ε^{jmn} a^k b^j c^m a^n$$

$$=b^k(\vec{a},\vec{c},\vec{a})-a^k(\vec{b},\vec{c},\vec{a})$$

$$=-a^k(\vec{a},\vec{b},\vec{c})$$

Which is the correct identity.

What's worse is that for some other identity the 2nd might not work and the first will (e.g. when trying to prove $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{a})=(\vec{a}\cdot\vec{c})(\vec{b}\cdot\vec{a})-(\vec{b}\cdot\vec{c})(\vec{a}\cdot\vec{a}) $)

My Question is: What I'm doing wrong? Where's the problem?

Answer 1

The correct identities are:

$$\sum_{k=1}^3\epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$$

and

$$(\vec{a}\times\vec{b})_i = \epsilon_{ijk}a_jb_k$$

So let's do the $[(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{a})]$ for the $i$ component:

$$[(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{a})]_i = \epsilon_{ijk}(\vec{a}\times\vec{b})_j(\vec{c}\times\vec{a})_k = \epsilon_{ijk}\epsilon_{jlm}a_lb_m\epsilon_{krs}c_ra_s = $$

$$=\epsilon_{ijk}\epsilon_{jlm}\epsilon_{krs}a_lb_mc_ra_s$$

Now use that $\epsilon_{krs} = -\epsilon_{srk} = \epsilon_{rsk}$

$$\epsilon_{jlm}(\epsilon_{ijk}\epsilon_{rsk})a_lb_mc_ra_s = \epsilon_{jlm}a_lb_m(\delta_{ir}\delta_{js}-\delta_{is}\delta_{jr})c_ra_s = \epsilon_{jlm}a_lb_m(c_ia_j-c_ja_i)=$$ $$= (\vec{a}\cdot(\vec{a}\times\vec{b}))c_i - ((\vec{a}\times\vec{b})\cdot\vec{c})a_i = $$

Now using that $((\vec{a}\times\vec{b})\cdot\vec{c}) = ((\vec{b}\times\vec{c})\cdot\vec{a}) = ((\vec{c}\times\vec{a})\cdot\vec{b})$ then we get the answer

$$[(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{a})] = (\vec{a}\cdot(\vec{b}\times\vec{a}))\vec{c} - ((\vec{b}\times\vec{c})\cdot\vec{a})\vec{a} $$