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This question relates to this continuum mechanics online course. So far it has excellent lectures and problem sets but no solutions.

Problem 3.

Show that the permutation symbol, $\epsilon_{ijk}$ can be expressed as: $$ \epsilon_{ijk} = \begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}$$

I am also posting my answer. I am working on my rigorous mathematical writing as well as the continuum mechanics, so I would appreciate feedback of my proof-writing style as well.

Oğuzhan Kılıç
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Eric
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  • Your expression is correct and this answer might also be of interest: https://math.stackexchange.com/questions/1874812/levi-civita-and-kronecker-delta-identity/1882044#1882044 – Léreau Sep 21 '21 at 21:54
  • Closely related: https://math.stackexchange.com/q/919094 – Jean Marie Sep 22 '21 at 08:45

1 Answers1

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Starting with the conjecture:

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

It should be true that:

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

We can expand the left hand side of the statement to be the following, keeping in mind that even though there are 27 permutations of $ijk$, only the following six of them are non-zero.

$$\epsilon_{ijk}A_iB_jC_k = A_1B_2C_3 + A_2B_3C_1 + A_3B_1C_2 - A_3B_2C_1 - A_2B_1C_3 - A_1B_3C_2$$

Likewise expanding the right hand side of statement:

$$\begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}A_iB_jC_k = (\delta_{i1}\delta_{j2}\delta_{k3} + \delta_{i2}\delta_{j3}\delta_{k1} + \delta_{i3}\delta_{j1}\delta_{k2} - \delta_{i3}\delta_{j2}\delta_{k1} - \delta_{i2}\delta_{j1}\delta_{k3} - \delta_{i1}\delta_{j3}\delta_{k2})A_iB_jC_k$$

$$\begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}A_iB_jC_k = A_1B_2C_3 + A_2B_3C_1 + A_3B_1C_2 - A_3B_2C_1 - A_2B_1C_3 - A_1B_3C_2$$

$$\therefore$$

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

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

QED

Eric
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