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3. In ''n'' dimensions, when all <math>i_1,...,i_n,j_1,...,j_n</math> are in <math>\{1,...,n\},</math>:
::<math>\begin{align}& \varepsilon_{i_1 \dots i_n} \varepsilon^{j_1 \dots j_n} = n! \delta^{j_1}_{[ i_1} \dots \delta^{j_n}_{i_n ]} &&(7)\\& \varepsilon_{i_1 \dots i_k~i_{k+1}\dots i_n} \varepsilon^{i_1 \dots i_k~j_{k+1}\dots j_n}= k!(n-k)!~\delta^{j_{k+1}}_{[ i_{k+1}} \dots \delta^{j_n}_{i_n ]} &&(8)\\& \varepsilon_{i_1 \dots i_n}\varepsilon^{i_1 \dots i_n} = n! &&(9)\end{align}</math>
===Proofs===
For equation 1, both sides are [[antisymmetric]] with respect of <math>ij</math> and <math>mn</math>. We therefore only need to consider the case <math>i\neq j</math> and <math>m\neq n</math>. By substitution, we see that the equation holds for <math>\varepsilon_{12} \varepsilon^{12}</math>, i.e., for <math>i=m=1</math> and <math>j=n=2</math>. (Both sides are then one). Since the equation is antisymmetric in <math>ij</math> and <math>mn</math>, any set of values for these can be reduced to the above case (which holds). The equation thus holds for all values of <math>ij</math> and <math>mn</math>. Using equation 1, we have for equation 2
<math> \varepsilon_{ij}\varepsilon^{in}</math> <math> =</math> <math> \delta_i^i \delta_j^n - \delta^n_i \delta^i_j</math>
::<math> =</math> <math> 2 \delta_j^n - \delta^n_j</math>
::<math> =</math> <math> \delta_j^n</math>.
Here we used the [[Einstein summation convention]] with <math>i</math> going from <math>1</math> to <math>2</math>. Equation 3 follows similarly from equation 2. To establish equation 4, let us first observe that both sides vanish when <math>i\neq j</math>. Indeed, if <math>i\neq j</math>, then one can not choose <math>m</math> and <math>n</math> such that both permutation symbols on the left are nonzero. Then, with <math>i=j</math> fixed, there are only two ways to choose <math>m</math> and <math>n</math> from the remaining two indices. For any such indices, we have <math>\varepsilon_{jmn} \varepsilon^{imn} = (\varepsilon^{imn})^2 = 1</math> (no summation), and the result follows. Property (5) follows since <math>3!=6</math> and for any distinct indices <math>i,j,k</math> in <math>\{1,2,3\}</math>, we have <math>\varepsilon_{ijk} \varepsilon^{ijk}=1</math> (no summation).
==Examples==
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