I ##\epsilon_{ijk}## in terms of ##\delta's##

[brotherbobby]

TL;DR

The Levi-Civita symbol in three dimensions $\epsilon_{ijk}$ is sometimes defined as a determinant of various kronecker symbols as under :
$$\epsilon_{ijk}\epsilon_{pqr}=\begin{vmatrix}\delta_{ip} & \delta_{iq} & \delta_{ir}\\ \delta_{jp} & \delta_{jq} & \delta_{jr}\\ \delta_{kp} & \delta_{kq} & \delta_{kr}\\ \end{vmatrix}.
$$ Can this be shown to reduce to its more usual definition where its value is $+1,\,-1,\,0$ according to whether the indices are cyclic or repeat? [equation (1) below]

The two definitions :



The Levi Civita Alternating Symbol is defined as below, taken from here. I put the relevant image above to the right to save you the trouble of having to look through the wikipage.
\begin{equation}
\epsilon_{ijk} = \begin{cases}+1, & \text{if } (i,j,k) \text{ is an even permutation of } (1,2,3) \\-1, & \text{if } (i,j,k) \text{ is an odd permutation of } (1,2,3) \\ 0, & \text{otherwise}\end{cases}
\label{usual}
\end{equation}

But here's my problem.

It is also defined via a product of itself, using a $3\times 3$ determinant of the kronecker symbols as shown to the right (same wikipage).

\begin{equation}
\epsilon_{ijk}\;\epsilon_{lmn}=\begin{vmatrix}\delta_{il} & \delta_{im} & \delta_{in}\\ \delta_{jl} & \delta_{jm} & \delta_{jn}\\ \delta_{kl} & \delta_{km} & \delta_{kn}\\ \end{vmatrix}
\label{kronecker}
\end{equation}

The issue :

I have been through the referred text on the wikipage, marked in the image as ${\color{blue}{^{[4]}}}$.

Nowhere does it say how does $\ref{kronecker}$ follow from $\ref{usual}$.

I am at a loss to prove myself, where I have tried taken values like $i=1, j=2, k=1\quad l=1, m=2, n=3$. The answers came to 1 on both sides alright, but that is not a proof. I admit I could go on and take all values from $1,2,3$ and indeed show that the second reduces to the first definition in each case. Still, won't make it a proof.

Does a proof exist whereby $\ref{kronecker}$ can be shown to reduce to $\ref{usual}$?

Request : A hint or clue would be of immense help.

 

[JimWhoKnew]

TL;DR Summary: The Levi-Civita symbol in three dimensions $\epsilon_{ijk}$ is sometimes defined as a determinant of various kronecker symbols as under :
$$\epsilon_{ijk}\epsilon_{pqr}=\begin{vmatrix}\delta_{ip} & \delta_{iq} & \delta_{ir}\\ \delta_{jp} & \delta_{jq} & \delta_{jr}\\ \delta_{kp} & \delta_{kq} & \delta_{kr}\\ \end{vmatrix}.
$$ Can this be shown to reduce to its more usual definition where its value is $+1,\,-1,\,0$ according to whether the indices are cyclic or repeat? [equation (1) below]

View attachment 362540

The two definitions :



The Levi Civita Alternating Symbol is defined as below, taken from here. I put the relevant image above to the right to save you the trouble of having to look through the wikipage.
\begin{equation}
\epsilon_{ijk} = \begin{cases}+1, & \text{if } (i,j,k) \text{ is an even permutation of } (1,2,3) \\-1, & \text{if } (i,j,k) \text{ is an odd permutation of } (1,2,3) \\ 0, & \text{otherwise}\end{cases}
\label{usual}
\end{equation}

But here's my problem.


View attachment 362541It is also defined via a product of itself, using a $3\times 3$ determinant of the kronecker symbols as shown to the right (same wikipage).

\begin{equation}
\epsilon_{ijk}\;\epsilon_{lmn}=\begin{vmatrix}\delta_{il} & \delta_{im} & \delta_{in}\\ \delta_{jl} & \delta_{jm} & \delta_{jn}\\ \delta_{kl} & \delta_{km} & \delta_{kn}\\ \end{vmatrix}
\label{kronecker}
\end{equation}

The issue :

I have been through the referred text on the wikipage, marked in the image as ${\color{blue}{^{[4]}}}$.

Nowhere does it say how does $\ref{kronecker}$ follow from $\ref{usual}$.

I am at a loss to prove myself, where I have tried taken values like $i=1, j=2, k=1\quad l=1, m=2, n=3$. The answers came to 1 on both sides alright, but that is not a proof. I admit I could go on and take all values from $1,2,3$ and indeed show that the second reduces to the first definition in each case. Still, won't make it a proof.

Does a proof exist whereby $\ref{kronecker}$ can be shown to reduce to $\ref{usual}$?

Request : A hint or clue would be of immense help.

(2) is not a definition, and is not the same as (1). It is an identity that follows from (1).

(1) is a function of 3 variables while (2) is a function of 6. How can they be identical?

A proof of (2), given (1), is outlined for n dimensions on the wikipage.


Edit:

With $i=1, j=2, k=1\quad l=1, m=2, n=3$ you should expect a zero.
Setting $l=1, m=2, n=3$ , there are 27 different inputs and by using the properties of the determinant you can show reduction to (1).

 

[brotherbobby]

Thank you @JimWhoKnew. I prefer the solution on wikipage to the one you provided in your edit above, because that was what I had in mind to do. Trying out various indices. But that is not the smarter way to it.

For reasons of completeness, I want to try out the wikipage proof for the $3\times 3$ determinant here. But let me pose the issue here again to refresh the mind.

The definition and the identity : The Levi Civita Alternating Symbol is defined as below :

\begin{equation}

\epsilon_{ijk} = \begin{cases}+1, & \text{if } (i,j,k) \text{ is an even permutation of } (1,2,3) \\-1, & \text{if } (i,j,k) \text{ is an odd permutation of } (1,2,3) \\ 0, & \text{otherwise}\end{cases}

\end{equation}

However, one can find references to an identity for the same symbol, where it appears as a product of itself and is equal to a $3\times 3$ determinant of the kronecker delta symbol as shown below :

\begin{equation}

\epsilon_{ijk}\;\epsilon_{lmn}=\begin{vmatrix}\delta_{il} & \delta_{im} & \delta_{in}\\ \delta_{jl} & \delta_{jm} & \delta_{jn}\\ \delta_{kl} & \delta_{km} & \delta_{kn}\\ \end{vmatrix}

\end{equation}

The question : How to show that equation $(4)$ above follows from $(3)$?

Attempt : Let us have $(4)$ again.

\begin{equation*}

\epsilon_{ijk}\;\epsilon_{lmn}=\begin{vmatrix}\delta_{il} & \delta_{im} & \delta_{in}\\ \delta_{jl} & \delta_{jm} & \delta_{jn}\\ \delta_{kl} & \delta_{km} & \delta_{kn}\\ \end{vmatrix}

\end{equation*}

1. Switching indices $i\leftrightarrow j$, both sides change sign.
2. If some $i=j$, both sides become 0. Same for some $p=q$.
3. Let's assume, without loss of generality, that $i<j<k\quad ,\quad p<q<r$. For all such cases, both sides are $\pm 1$.

Does this complete the proof? It looks ok to me.

 

[JimWhoKnew]

Thank you @JimWhoKnew. I prefer the solution on wikipage to the one you provided in your edit above, because that was what I had in mind to do. Trying out various indices. But that is not the smarter way to it.

For reasons of completeness, I want to try out the wikipage proof for the $3\times 3$ determinant here. But let me pose the issue here again to refresh the mind.

The definition and the identity : The Levi Civita Alternating Symbol is defined as below :

\begin{equation}

\epsilon_{ijk} = \begin{cases}+1, & \text{if } (i,j,k) \text{ is an even permutation of } (1,2,3) \\-1, & \text{if } (i,j,k) \text{ is an odd permutation of } (1,2,3) \\ 0, & \text{otherwise}\end{cases}

\end{equation}

However, one can find references to an identity for the same symbol, where it appears as a product of itself and is equal to a $3\times 3$ determinant of the kronecker delta symbol as shown below :

\begin{equation}

\epsilon_{ijk}\;\epsilon_{lmn}=\begin{vmatrix}\delta_{il} & \delta_{im} & \delta_{in}\\ \delta_{jl} & \delta_{jm} & \delta_{jn}\\ \delta_{kl} & \delta_{km} & \delta_{kn}\\ \end{vmatrix}

\end{equation}

The question : How to show that equation $(4)$ above follows from $(3)$?

Attempt : Let us have $(4)$ again.

\begin{equation*}

\epsilon_{ijk}\;\epsilon_{lmn}=\begin{vmatrix}\delta_{il} & \delta_{im} & \delta_{in}\\ \delta_{jl} & \delta_{jm} & \delta_{jn}\\ \delta_{kl} & \delta_{km} & \delta_{kn}\\ \end{vmatrix}

\end{equation*}

1. Switching indices $i\leftrightarrow j$, both sides change sign.
2. If some $i=j$, both sides become 0. Same for some $p=q$.
3. Let's assume, without loss of generality, that $i<j<k\quad ,\quad p<q<r$. For all such cases, both sides are $\pm 1$.

Does this complete the proof? It looks ok to me.

Note that
\begin{equation*}

\epsilon_{ijk}=\begin{vmatrix}\delta_{i1} & \delta_{i2} & \delta_{i3}\\ \delta_{j1} & \delta_{j2} & \delta_{j3}\\ \delta_{k1} & \delta_{k2} & \delta_{k3}\\ \end{vmatrix}

\end{equation*}
and this can be used as an alternative (equivalent) definition for $\epsilon_{ijk}$ , expressing it in terms of $\delta$'s as you wanted. The generalization to n dimensions is trivial. I would have expected the wikipage to mention it.

I wrote in #2 that the proof for the product identity is "outlined", because it claims "without loss of generality". To be rigor, you have to argue that this form can always be reached by permutations of lines and rows, and that the introduced sign changes are compatible.

In the third line of your proof, if you assume $i<j<k\quad ,\quad p<q<r$ , then (4) is +1 . If I had to submit the proof as a homework assignment, I would have tidied it up.