Question

Tutorial:
Sign of a permutation and the Ricci tensor

Answer 1

Let's consider the symmetric group \(S_3\), namely the group of bijective applications, also called \emph{permutations}, of the set \(\{ 1,2,3\} \). A generic element of that group, can be written as below:
\[\sigma :\{ 1,2,3\} \to \{ 1,2,3\} ,\quad \sigma = \left( {\begin{array}{*{20}{c}}
1&2&3 \\
i&j&k
\end{array}} \right)\]
where, of course, \(i,j,k\) \(\in \{ 1,2,3\} \). Each entry of the second row, is the image of the corresponding element of the first row:
\[\sigma \left( 1 \right) = i,\quad \sigma \left( 2 \right) = j,\quad \sigma \left( 3 \right) = k\]
In general, we can write a permutation, omitting the first row, so we will write the images of a permutation \(\sigma \), like below:
\[i\quad j\quad k\]
being: \(i,j,k \in \{ 1,2,3\} \).

\\
To each permutation of ${S_3}$, can be assigned a number, namely its \emph{sign}, whose definition is:
\[\operatorname{sgn} \left( \sigma \right):\quad {S_3} \to \left\{ { + 1, - 1} \right\}\]
we say that the sign of a permutation is 1, if we need of an \emph{even} number of swaps in order to go from that permutation to the identical permutation, and the sign of a permutation is ${ - 1}$, if we need of an \emph{odd} number of swaps, in order to go from that permutation to the identical permutation. A swap is the permutation which change only two elements, here is an example of swap:
\[\left( {\begin{array}{*{20}{c}}
1&2&3 \\
1&3&2
\end{array}} \right)\]
\\
whereas, the identical permutation is denoted like below:
\[Id = \left( {\begin{array}{*{20}{c}}
1&2&3 \\
1&2&3
\end{array}} \right)\]
or, simply like this:
\[1\quad 2\quad 3\]
As we well know, the elements of the set ${S_3}$ is equal to $3! = 6$, so we have 6 permutation in total, here they are:

\[\begin{array}{*{20}{c}}
1&2&3&{}&{}&2&1&3 \\
2&3&1&{}&{}&3&2&1 \\
3&1&2&{}&{}&1&3&2
\end{array}\]
At the left we have written all permutations which whose sign is \(+ 1\), whereas at the right, we have written all permutations whose sign is \(- 1\). We can go from right to left and from left to right, by simply making one swap, as we can check immediately.

In physics, and in particular in classical mechanics, the sign of a permutation, is denoted with the \(Ricci tensor\):
\[{\varepsilon _{ijk}}\]
so we can write this:
\[\begin{gathered}
{\varepsilon _{123}} = {\varepsilon _{231}} = {\varepsilon _{312}} = + 1 \hfill \\
{\varepsilon _{213}} = {\varepsilon _{321}} = {\varepsilon _{132}} = - 1 \hfill \\
\end{gathered} \]
which are the only non-zero components of such tensor, among the {27} components of \({\varepsilon _{ijk}}\).

\(Application\)
Let \(\mathbf{a},\;\mathbf{b}\) two generic vectors, of the euclidean space \(E\), then let's consider its vector product or cross product, namely the subsequent map:
\[{\mathbf{ \times }}\;:\;E \times E \to E\quad \left( {{\mathbf{a}},\;{\mathbf{b}}} \right) \to {\mathbf{a}} \times {\mathbf{b}}\]
then, using the Ricci's tensor, we can write the \(i-th\) component of the vector \(\mathbf{a} \times \mathbf{b}\), like below:
\[{\left( {{\mathbf{a}} \times {\mathbf{b}}} \right)_i} = \sum\limits_{j,k = 1}^3 {{\varepsilon _{ijk}}\;{a_j}{b_k}} \]
and since the indexes \(j,\; k\) are repeated, we can omit the symbol of summation
(\(Einstein \; convention)\):
\[{\left( {{\mathbf{a}} \times {\mathbf{b}}} \right)_i} = {\varepsilon _{ijk}}\;{a_j}{b_k}\]
The Ricci's tensor satisfies an important identity:
\[{\varepsilon _{ijk}}\;{\varepsilon _{ilm}} = {\delta _{jl}}{\delta _{km}} - {\delta _{jm}}{\delta _{kl}}\]
where \(\delta _{ij}\) is the \(Kronecker's symbol\).

Finally, from the previous discussion, we have this other property of the Ricci's tensor:
\[\begin{gathered}
{\varepsilon _{ijk}} = - {\varepsilon _{jik}} \hfill \\
{\varepsilon _{ijk}} = - {\varepsilon _{ikj}} \hfill \\
\end{gathered} \]

Answer 2

here is the corresponding PDF file:

Answer 3

Can you prove that every permutation can be written in either all even transposes or all odd ? I was curious about that.

Answer 4

whoa

Answer 5

thanks :) :) @77777jeannie77777

Answer 6

lol youre welcome i will never understand that lol im not in that level of math and i hope i wont be anytime soon hahaha :P

Answer 7

thanks again!! :)
yes! sure one day you will understand it and you will write another tutorial, better than mine
@77777jeannie77777

Answer 8

Awesome, thanks for sharing @Michele_Laino

Answer 9

thanks!! :) @Astrophysics

Answer 10

Richi

Answer 11

I thought the Ricci tensor was a contraction of the Riemann-Christoffel tensor and this was the Levi-Civitta tensor.