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Product of Two Levi-Civita Symbols in N-dimensions

by akoohpaee
Tags: levicivita, ndimensions, product, symbols
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akoohpaee
#1
Jul3-11, 01:37 PM
P: 3
Dear You,

In N-dimensions Levi-Civita symbol is defined as:

\begin{align}
\varepsilon_{ijkl\dots}=
\begin{cases}
+1 & \mbox{if }(i,j,k,l,\dots) \mbox{ is an even permutation of } (1,2,3,4,\dots) \\
-1 & \mbox{if }(i,j,k,l,\dots) \mbox{ is an odd permutation of } (1,2,3,4,\dots) \\
0 & \mbox{otherwise}
\end{cases}
\end{align}



I found the following expression for the product of two Levi-Civita symbols when there are no dummy indices (i_1,...,i_n,j_1,...,j_n are in {1,...,n}):

\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 ]} &&\\& \end{align}

But I could not find its proof through literature and also I was failed to prove it!

Can you please help me? Thanks!

Best Regards,
Ali
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g_edgar
#2
Jul4-11, 04:18 PM
P: 607
When [itex]N=2[/itex] we have for example [itex]\varepsilon_{12}=+1[/itex]. Now, you didn't define [itex]\varepsilon^{12}[/itex] for us. Also, is there some Einstein summation convention in effect here?
henry_m
#3
Jul5-11, 10:18 AM
P: 160
Firstly, notice that the RHS is antisymmetric in exchange of any pair of the i's and any pair of the j's. So all you have to check is that you get the right answer for [itex]i_1=1, i_2=2, \ldots[/itex] and the same for the j's, which is straightforward.

akoohpaee
#4
Jul6-11, 01:54 PM
P: 3
Product of Two Levi-Civita Symbols in N-dimensions

### REPLY TO g_edgar ###

Hi,

Many thanks for your reply!

Assume that the array [itex]K_{ijk}[/itex] (i,j,k are in {1,2,3}) is defined in such a way that [itex]K_{ijk}=\varepsilon_{ijk}[/itex]. It can be shown that, this array behave as a tensor under covariant and contravariant coordinate transformations. In other words:

\begin{align}
\varepsilon_{ijk}=\varepsilon^{ijk}
\end{align}

Also this is the case for N-dimensional:

\begin{align}
\varepsilon_{ijklm\dots}=\varepsilon^{ijklm\dots}
\end{align}

And regarding to the second part of your question: There is no Einstein summation notation in effect here. In fact there is no summation in this statement.

Many thanks for your reply and your attention.

Ali


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