Proof that $\epsilon_{ijk}$ is the same as the determinant of the Kronecker delta

Question

I just learned about the Kronecker Delta function and $\epsilon_{ijk}$ for the first time, and I still can't wrap my mind around how to prove that $$\epsilon_{ijk} = det\begin{pmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \end{pmatrix}$$ I understand that the RHS is $$det\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}=1$$ but how is that related to the fact that the $\epsilon$ tensor renders 1 for even, -1 for odd permutations of $i,j,k$ and $0$ for identical ones? And how can it be proven?

The only "proof" I could think of is$$ \delta_{i1}\delta_{j2}\delta_{k3} + \delta_{i2}\delta_{j3}\delta_{k1} + \delta_{i3}\delta_{j1}\delta_{k2} - \delta_{i3}\delta_{j2}\delta_{k1} - \delta_{i2}\delta_{j1}\delta_{k3} - \delta_{i1}\delta_{j3}\delta_{k2} \\ = \begin{cases} 1 \cdot \delta_{ijk} & |\{ijk\} \in \{123\},\{312\},\{231\} \\ -1 \cdot \delta_{ijk} & |\{ijk\} \in \{213\},\{321\},\{132\} \end{cases} \\ = \epsilon_{ijk} $$ but that doesn't look sound to me.

I am aware of this question, but it's kind of like two steps ahead when I'd like to be able to explain just the first.

Can somebody explain with an Einstein mindset?

Answer 1

A property of the determinant is:

• Exchanging two rows while leaving everything else unchanged changes the sign of the determinant, but not the magnitude.

This can be proven from the formula $\det(AB) = (\det A)(\det B)$. The operation of exchanging rows in a matrix $M$ is the same as taking $EM$, where $E$ is the same row exchange on the identity matrix $I$. One can easily show all such matrices $E$ have determinant $-1$.

Now note that if two rows of a matrix $M$ are identical, then exchanging them makes no change to the matrix. Therefore $\det M = -\det M$, from which it follows that $\det M = 0$:

• If a matrix has two identical rows, then its determinant is $0$.

Now define $$d_{ijk} = \det \begin{pmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \end{pmatrix}$$

The matrix for $d_{123}$ is the identity matrix, so $d_{123} = 1$. If $i = j$ or $j = k$ or $i = k$, the matrix will have two identical rows, so $d_{ijk} = 0$ in this case. And by the exchange of row rule for matrix determinants, if you exchange the values of any two of $i, j, k$, then $d_{ijk}$ changes sign.

These three facts completely determine all values of $d_{ijk}$. But $\epsilon_{ijk}$ obeys exactly the same set of rules. So the two of them must be equal.

Answer 2

As we know, if $A = (a_{ij})$ is an $n\times n$ matrix, its determinant is given by: $$\mbox{det}A = \sum_{\sigma \in S_{n}}\epsilon_{\sigma}\prod_{i=1}^{n}a_{i,\sigma(i)}$$ where $S_{n}$ is the set of all permutations $\sigma$ of the set $I_{n}:=\{1,...,n\}$ and $\epsilon_{\sigma}$ is the sign of the permutation $\sigma$. In your case, we have $n=3$.

Let us consider the matrix: $$\Delta_{ijk} := \begin{pmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \end{pmatrix} $$ For simplicity, let us consider only the case where $i,j,k$ are all different because, if not, it is easy to see that the result follows because at least two rows of $\Delta_{ijk}$ are equal, so that $\det \Delta_{ijk} = 0$ and also is $\epsilon_{ijk}$. Now, as you pointed out, we have: $$ \epsilon_{ijk} = \mbox{det} \Delta_{ijk} = \sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{i=1}^{3}\delta_{i,\sigma(i)}$$ Now, let $\eta$ be a fixed permutation of $S_{3}$. Every permutation is a bijection from $I_{3}$ to itself, so every permutation has an inverse $\eta^{-1}$. Also, it is easy to prove that $\epsilon_{\eta^{-1}} = \epsilon_{\eta}$. Now, note that: $$\epsilon_{\eta(i),\eta(j),\eta(k)} = \sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{i=1}^{3}\delta_{\eta(i),\sigma(i)} $$ If we set $\eta(i) = k$, then $i = \eta^{-1}(k)$ and: $$\sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{i=1}^{3}\delta_{\eta(i),\sigma(i)} = \sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{k=1}^{3}\delta_{k,(\sigma\circ \eta^{-1})(k)} $$ If we sum over every permutation $\sigma$ in $S_{3}$, the composite $\sigma \circ \eta^{-1}$ also covers every permutation of $S_{3}$ so we can redefine $\tilde{\sigma} = \sigma\circ\eta^{-1}$ and: $$\sum_{\sigma \in S_{3}}\epsilon_{\sigma}\prod_{k=1}^{3}\delta_{k,(\sigma\circ \eta^{-1})(k)} = \sum_{\tilde{\sigma} \in S_{3}}\epsilon_{\eta}\epsilon_{\tilde{\sigma}}\prod_{k=1}^{3}\delta_{k,\tilde{\sigma}(k)} =\epsilon_{\eta}\mbox{det}\Delta_{ijk}$$ where I've used: $$\epsilon_{\sigma} = \epsilon_{\tilde{\sigma}}\frac{1}{\epsilon_{\tilde{\sigma}}}\epsilon_{\sigma} = \epsilon_{\tilde{\sigma}}\frac{1}{\epsilon_{\eta}\epsilon_{\sigma}}\epsilon_{\sigma} = \epsilon_{\eta}\epsilon_{\tilde{\sigma}}.$$ The conclusion is that: $$\epsilon_{\eta(i),\eta(j),\eta(k)}= \epsilon_{\eta}\overbrace{\mbox{det}\Delta_{ijk}}^{=1} = \epsilon_{\eta}$$ Thus, $\epsilon_{\eta(i),\eta(j),\eta(k)}$ coincides with the sign of the permutation $\eta$, i.e. it is $1$ if $\eta$ is an even permutation and $-1$ otherwise.

Answer 3

Both sides are $1$ if $i=1,\,j=2,\,k=3$, for the reason you gave. Both sides also multiply by $-1$ if any two indices are exchanged. (The determinant, in particular, is of a matrix that swaps two of its rows in this process.) In the special case where the indices are equal, both expressions must be originally $0$; in the special case where all three indices are unequal, the sign change on each side preserves equality, but of nonzero values. Since $\epsilon_{ijk}$ is completely specified by full antisymmetry together with $\epsilon_{123}=1$, we're done.