How Does Levi-Civita Symmetry Relate to Tensor Permutations?

[Jyoti6297]

I know what Levi Civita is, but I don't know how to go about such questions.
Question : (1.3)

 

[anorlunda]

Welcome to PF.

I moved your post. It does not belong in New Member Introductions.

Our rules say that you must show your effort before our homework helpers are allowed to help. So please "I don't know how" is not enough. Make a try and show us your work.

 

[PeroK]

For a question like that you need the definition and/or properties of the Levi Civita symbol and the definition and/or properties of the determinant.

Without those you can make no progress.

 

[Cryo]

I agree about the need to go back to the definition. Some useful things to remember is that

1. Levi-Civita indices must be all different for it to be non-zero.
2. When it us non-zero it can only be one or minus one.
3. Every time you swap two indices around you pick up a minus

Next, you need to go to the definition of the determinant. Wikipedia has some good information on this including the connection to Levi-Civitas https://en.wikipedia.org/wiki/Determinant.

Start with a simpler task and try writing the sum out:

$\epsilon_{\alpha\beta\gamma\kappa} A^\alpha_1 A^\beta_2 A^\gamma_3 A^\kappa_4 = ?$

Take, for example the first term as $\{\alpha\beta\gamma\kappa\}=\{1234\}$ all other terms will be permutations of this sequence. You can deal with these permutations by choosing what you keep fixed:

-> Fix $\alpha=1$ permute $\beta\gamma\kappa\}=\{234\}$:
___-> Fix $\beta=2$ permute $\{\gamma\kappa\}=\{34\}$:
______-> $\{\alpha\beta\gamma\kappa\}=\{1234\}, -\{1243\}$
__________Note that the second term has a minus because I needed a single exchange of indices compared to 1234
go up few levels
___-> Fix $\beta=3$ permute $\{\gamma\kappa\}=\{24\}$:
______-> $\{\alpha\beta\gamma\kappa\}=-\{1324\}, (-)^2\{1342\}$
go up
___-> Fix $\beta=4$ permute $\{\gamma\kappa\}=\{23\}$:
______-> $\{\alpha\beta\gamma\kappa\}=\{1423\}, -\{1432\}$

Based on this you already have:

$\epsilon_{\alpha\beta\gamma\kappa} A^\alpha_1 A^\beta_2 A^\gamma_3 A^\kappa_4 = A^1_1 \left(A^2_2 \left(A^3_3 A^4_4 - A^4_3 A^3_4 \right) + A^3_2 \left(-A^2_3 A^4_4 + A^4_3 A^2_4 \right) + A^4_2 \left(A^2_3 A^3_4 - A^3_3 A^2_4 \right) \right) \dots$

You should start seeing the hierarchical structure of the determinant appearing in this permutations (the inner brackets are 2-by-2 determinants, above them is 3-by-3 etc). If you are happy with this (i.e. you can extend the arguments to proof of $\epsilon_{\alpha\beta\gamma\kappa} A^\alpha_1 A^\beta_2 A^\gamma_3 A^\kappa_4=\det\left(A\right)$), we can move to your original question

 

[Cryo]

Provided you are happy with the determinant part the rest is simple. Consider the permutation symmetry of

$\epsilon_{\alpha\beta\gamma\kappa}A^\alpha_\mu A^\beta_\nu A^\gamma_\eta A^\kappa_\sigma$

I can permute indices $\{\mu\nu\eta\sigma\}$ and offset this by permuting the ${\alpha\beta\gamma\kappa}$, e.g.

$()_{\mu\sigma\eta\nu}=\epsilon_{\alpha\beta\gamma\kappa}A^\alpha_\mu A^\beta_\sigma A^\gamma_\eta A^\kappa_\nu = \epsilon_{\alpha\beta\gamma\kappa} A^\alpha_\mu A^\kappa_\nu A^\gamma_\eta A^\beta_\sigma= - \epsilon_{\alpha\kappa\gamma\beta} A^\alpha_\mu A^\kappa_\nu A^\gamma_\eta A^\beta_\sigma$

Now rename the dummy indices $\beta\to\kappa \, and \, \kappa \to \beta$:$()_{\mu\sigma\eta\nu}= - \epsilon_{\alpha\kappa\gamma\beta} A^\alpha_\mu A^\kappa_\nu A^\gamma_\eta A^\beta_\sigma = - \epsilon_{\alpha\beta\gamma\kappa} A^\alpha_\mu A^\beta_\nu A^\gamma_\eta A^\kappa_\sigma = -()_{\mu\nu\eta\sigma}$. So the permutation symmetry of the $()$ is the same as of the Levi-Civita, so we can write:

$()_{\mu\sigma\eta\nu}=\epsilon_{\mu\nu\eta\sigma} ()_{1234} = \epsilon_{\mu\nu\eta\sigma} \cdot \epsilon_{\alpha\beta\gamma\kappa}A^\alpha_1 A^\beta_2 A^\gamma_3 A^\kappa_4$

And as you know from the pervious post, the second term is simply the determinant of $A$

 

[Jyoti6297]

I didn't get the last part.

So the permutation symmetry of the $()$ is the same as of the Levi-Civita, so we can
$()_{\mu\sigma\eta\nu}=\epsilon_{\mu\nu\eta\sigma} ()_{1234} = \epsilon_{\mu\nu\eta\sigma} \cdot \epsilon_{\alpha\beta\gamma\kappa}A^\alpha_1 A^\beta_2 A^\gamma_3 A^\kappa_4$

Do you mean
$()_{\mu\nu\eta\sigma}$ instead of $()_{\mu\sigma\eta\nu}$ here? Or am I interpreting something wrong?

 

[Cryo]

I didn't get the last part.

So the permutation symmetry of the $()$ is the same as of the Levi-Civita, so we can
$()_{\mu\sigma\eta\nu}=\epsilon_{\mu\nu\eta\sigma} ()_{1234} = \epsilon_{\mu\nu\eta\sigma} \cdot \epsilon_{\alpha\beta\gamma\kappa}A^\alpha_1 A^\beta_2 A^\gamma_3 A^\kappa_4$

Do you mean
$()_{\mu\nu\eta\sigma}$ instead of $()_{\mu\sigma\eta\nu}$ here? Or am I interpreting something wrong?

What I meant is that $()_{\mu\nu\eta\sigma}\equiv \epsilon_{\alpha\beta\gamma\kappa}A^\alpha_\mu A^\beta_\nu A^\gamma_\eta A^\kappa_\sigma$ has the same symmetry with respect to index permutation, as a co-variant Levi-Civita, so $\epsilon_{\alpha\beta\gamma\kappa}A^\alpha_\mu A^\beta_\nu A^\gamma_\eta A^\kappa_\sigma = const\cdot \epsilon_{\mu\nu\eta\sigma}$, where $const$ happens to be the determinant