Let $i,j,k,l\in\{0,1,2\}$. I am looking for a simple expression $f(i,j,k,l)$ involving only (Kronecker delta) $$\delta_{ab}=\cases{1&if $a=b$\\0&else},$$ and (Levi-Civita symbol) $$\epsilon_{ab} = \cases{1&if $a<b$\\-1&if $a>b$\\0&if $a=b$},$$ and maybe constants, such that $f$ is non-zero if, and only if $\{i,j\}\neq\{k,l\}$. It would be perfect if the values taken by $f$ were only $\pm1$ or $0$, but it's not so necessary. Does anyone have one such expression under hand? Also, general techniques for constructing them would be greatly appreciated.
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Can we nest the symbols, as in $\delta_{\delta_{ab}\delta_{cd}}$? Also, note that it's always possible using only $\delta$ since you can construct the elementary checks such as $(i,j)=(2,1)$ and similar, and then multiplication is logical AND and addition is logical OR. It's just that the expression is not going to be simple. You can then construct all ${,(i,j,k,l),{:},{i,j}={k,l},}$ because this set is finite. – yo' Jun 17 '16 at 10:31
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@yo' Yes, I know, I allowed $\epsilon$ because it might give a shorter expression than just using $\delta$. I would rather avoid nesting the symbols. – Daniel Robert-Nicoud Jun 17 '16 at 10:41
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We have $$f(i,j,k,l)=\bigl(1-[ik][jl]\bigr)\bigl(1-[il][jk]\bigr),$$ where we denote $[ij]=\delta_{ij}$ for convenience.
Proof: Examine $[ik][jl]+[il][jk]$. It is one if either $(i,j)=(k,l)$ or $(i,j)=(l,k)$. It is two if both conditions are satisfied, which means all four numbers are equal. Whence $[ik][jl]+[il][jk]-[ik][jl][il][jk]$ is the negation of what you want.
yo'
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