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I'm completely new with the sum notation and Levi-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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    $\begingroup$ 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. $\endgroup$
    – yoyo
    Feb 22, 2011 at 21:18
  • $\begingroup$ 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. $\endgroup$
    – a1337q
    Feb 22, 2011 at 21:19
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    $\begingroup$ This proof looks fine. $\endgroup$
    – mjqxxxx
    Feb 22, 2011 at 22:11
  • $\begingroup$ Assuming the first equality is your definition of wedge product, or is a property you have proved already, this is fine. $\endgroup$
    – Ross Millikan
    Feb 23, 2011 at 4:39
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    $\begingroup$ 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 $\epsilon_{ijk} = -\epsilon_{kji}$ $\endgroup$
    – a1337q
    Feb 24, 2011 at 13:50

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