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I have to write down what the following means using both longhand calculations and the reduction formulae. $$\varepsilon_{ijk}\varepsilon_{ijl}$$However I get two distinct results and I can't figure out which one is wrong. Here are my approaches:

Reduction formulae

The formula says that $$\varepsilon_{ijk}\varepsilon_{ilm} = \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$$ I use it since I have the dummy index $i$, hence $$\varepsilon_{ijk}\varepsilon_{ijl} = \delta_{jj}\delta_{kl}-\delta_{jl}\delta_{kj} = \delta_{kl}-\delta_{jl}\delta_{kj}$$ where the last equality holds because $\delta_{jj}=1$ Now this is actually equal to $$\delta_{kl}-\delta_{jl}\delta_{kj} = \delta_{kl}-\delta_{1l}\delta_{k1}-\delta_{2l}\delta_{k2}-\delta_{3l}\delta_{k3}$$ Now I can fix the hanging indeces $kl$ to find certain elements. Since I have two hanging indeces, it will be a matrix, or a rank-2 tensor, Call it $(A)_{lk}$. Hence certain elements are $$a_{11} = 1-1+0+0 = 0$$ $$a_{22} = 1-0-1-0 = 0$$ $$a_{33} = 1-0-0-1 = 0$$ and clearly the other entries are all zero because the diagonal indeces are the only ones that could give non-zero kronecker deltas in that expression. Hence the result is the zero matrix $0_{3\times 3}$.

Longhand $$\varepsilon_{ijk}\varepsilon_{ijl} = \sum_{i=1}^3\sum_{j=1}^3\varepsilon_{ijk}\varepsilon_{ijl} = \varepsilon_{11k}\varepsilon_{11l}+\varepsilon_{12k}\varepsilon_{12l}+\varepsilon_{13k}\varepsilon_{13l}+\varepsilon_{21k}\varepsilon_{21l}+\varepsilon_{22k}\varepsilon_{22l}+\varepsilon_{23k}\varepsilon_{23l}+\varepsilon_{31k}\varepsilon_{31l}+\varepsilon_{32k}\varepsilon_{32l}+\varepsilon_{33k}\varepsilon_{33l} = \varepsilon_{12k}\varepsilon_{12l}+\varepsilon_{13k}\varepsilon_{13l}+\varepsilon_{21k}\varepsilon_{21l}+\varepsilon_{23k}\varepsilon_{23l}+\varepsilon_{31k}\varepsilon_{31l}+\varepsilon_{32k}\varepsilon_{32l}$$ because the ones with repeating indeces are $0$ by definition.

However now if we fix the hanging indeces to find the diagonal entries of the rank-2 tensor we see they are different. Indeed we get $$a_{11} = 0+0+0+1\times 1 + 0+(-1)\times(-1)= 1+1=2$$ $$a_{22} = 0+(-1)\times(-1) +0+0+1\times 1 +0 = 1+1 = 2$$ $$a_{33} = 1\times 1 + 0+(-1)\times(-1)+0+0+0 = 1+1=2$$

Where did I go wrong?

EDIT

After reading the comments I decided to edit the question to try to go at least a bit further. $$\varepsilon_{ijk}\varepsilon_{ijl} = \delta_{jj}\delta_{kl}-\delta_{jl}\delta_{kj} = 3\delta_{kl}-\delta_{jl}\delta_{kj}$$ which then should give $$3\delta_{kl}-\delta_{jl}\delta_{kj} = 3\delta_{kl}-\delta_{1l}\delta_{k1}-\delta_{2l}\delta_{k2}-\delta_{3l}\delta_{k3}$$ and fixing the indexes we get

$$a_{11} = 3-1+0+0 = 2$$ $$a_{22} = 3-0-1-0 = 2$$ $$a_{33} = 3-0-0-1 = 2$$.

and for any other index I would have $0$. Is this correct?

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    $\delta_{jj} \neq 1$ but $\delta_{jj} = n$ where $n$ is the dimension you are in. – 2017-02-02
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    @kmesis , with $\sigma_{jj}$ do you mean $\varepsilon_{jj}$ ? – 2017-02-02
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    Sorry for the initial typos, the original comment is fixed. – 2017-02-02
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    Another way to see that the first computation cannot be correct is to note that the trace $A_{kk}$ of $A_{lk}$ is, by definition, $\epsilon_{ijk} \epsilon_{ijk} = 3! = 6$, whereas that computation implies that $A_{kk} = 0$. – 2017-02-02
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    @kmeis I don't understand (I'm 2 lectures old on the topic). I thought that the definition of the kronecker delta was $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij}= 0$ if $i\neq j$. In the case $\delta_{jj}$ we do have the same index, so shouldn't it be $1$? Can you please expand more? – 2017-02-02
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    Yes but under Einstein notation $\delta_{ii} = \delta_{11} + \delta_{22}+\delta_{33}$. $\delta_{ii}$ is the trace of the identity matrix. – 2017-02-02
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    @Euler_Salter: Yes $\delta_{ij}=1$ iff $i=j$ -- that is, $\delta_{11}=\delta_{22}=\delta_{33}=1$ -- but writing $\delta_{jj}$ here invokes the Einstein convention, so that means $\delta_{11}+\delta_{22}+\delta_{33}=3$. – 2017-02-02
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    Oh that makes sense! I'll edit and go on and see where I get, thank you! – 2017-02-02
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    @HenningMakholm I've edited it. Is it correct? – 2017-02-02

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