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I'm completely new with the sum notation and Leva-Civita-Symbols. Can somebody tell me if what I did was correct?

Let $a,b \in \mathbb{R}^3$.

I want to prove $$ (b \wedge a) = -(a \wedge b) $$ so I did $$(b \wedge a)_i = \epsilon_{ijk}b_ja_k = - \epsilon_{ikj}b_ja_k = -\epsilon_{ikj}a_kb_j = - (a \wedge b)_i $$

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    whats your definition of wedge product (the definitions ive used have anticommutivity built in)? also, in $\mathbb{R}^3$ it's the same as the familiar cross product.2011-02-22
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    I know this is not actually a proof, since the anti-symmetry is a deep property of wedge-product itself, but I needed this index permutation in the Levi-Civita-Symbol somewhere else too. And I would like to see how these things hold together.2011-02-22
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    This proof looks fine.2011-02-22
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    Assuming the first equality is your definition of wedge product, or is a property you have proved already, this is fine.2011-02-23
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    I think it's totally wrong because rotating indices in the epsilon tensor must not change sign, because the permutation stays either even or odd as before. The sign only changes when transposing (exchanging two) indices. So again what I mean: $\epsilon_{ijk} = \epsilon_{kij}$ and $\epsilon_{ijk} = -\epsilon_{kji}$2011-02-24
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    Yes, rotating indices (in three dimensions) preserves the sign. However, in the original post there is no rotation but a simple interchange $ijk \mapsto ikj$, meaning there **is** a flip of sign and hence the proof is valid.2018-05-19

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