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I'm having some trouble with 2 identitys from tensor calculus. I need to proof these two guys:

  1. in euclidean 3-dimensional space, an antisymmetric matrix with entries $M_{ij}$ is equivalent to a vector $v^k=\frac{1}{2}\epsilon^{kij}\,\,M_{ij}$
  2. the inverse formula is $M_{ij}=\frac{1}{2}\epsilon_{ijk}\,\,v^k$

I know that the levi civita symbol is totally antisymmetric, and so any other totally antisymmetric object Mij will be proportional to the levi-civita symbol, but I just cant see the two informations adding up.

I appreciate any hint!

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    $\begingroup$ Have you tried writing these out explicitly? For example, what is $v^1$? What is $M_{23}$? I am a huge proponent of understanding formulas explicitly first and abstractly second. $\endgroup$
    – G. Smith
    Feb 11, 2021 at 21:49
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    $\begingroup$ For the non-explicit approach, do you know about contractions of products of Levi-Civita symbols in terms of Kronecker deltas? $\endgroup$
    – G. Smith
    Feb 11, 2021 at 21:50
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    $\begingroup$ @CaueEvangelista Think about how many independent components has an antisymmetric matrix in 3D euclidean and then how many independent components has a vector in 3D space. What those formulas are saying is that the two objects are essentially the same. $\endgroup$
    – Davide Morgante
    Feb 11, 2021 at 22:15
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    $\begingroup$ I'm having trouble in understanding the conection between this vector and the antisymmetric matrix. If you use the formulas to write out explicitly what a component like $v^1$ is in terms of components of $M$, and what $M_{23}$ is in terms of components of $v$, I think you will understand the connection. To do this you must understand the numerical values of all the components of $\epsilon$. $\endgroup$
    – G. Smith
    Feb 11, 2021 at 22:20
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    $\begingroup$ By the way, this kind of equivalence is called a duality. $\endgroup$
    – G. Smith
    Feb 11, 2021 at 22:46

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The more ways you have to convince yourself that something is correct, the better. Try G. Smith's approach, then think that $M_{ij}$ and $v^k$ have the same number of independent components, as, $$ \frac{n\left(n-1\right)}{2}=\frac{3\left(3-1\right)}{2}=3 $$ And finally, an identity to consider that will help you solve this and, e.g., prove many identities of vector calculus is, $$ \epsilon_{ijk}\epsilon^{kmn}=\left.\delta_{i}\right.^{m}\left.\delta_{j}\right.^{n}-\left.\delta_{i}\right.^{n}\left.\delta_{j}\right.^{m} $$

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