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I am trying to prove the following identity:

$\epsilon^{ijk}\epsilon_{pqk}=\delta_p^i\delta_q^j-\delta_p^j\delta_q^i$

Starting from the following identity:

$\epsilon^{ijk}\epsilon_{pqr}=\begin{vmatrix} \delta^i_p & \delta^i_q & \delta^i_r \\ \delta^j_p & \delta^j_q & \delta^j_r \\ \delta^k_p & \delta^k_q & \delta^k_r \\ \end{vmatrix}$

But, when I expand out the matrix, I end up with:

$\epsilon^{ijk}\epsilon_{pqr}=\delta_p^i(\delta_q^j\delta_r^k-\delta_q^k\delta_r^j)-\delta_q^i(\delta_p^j\delta_r^k-\delta_p^k\delta_r^j)+\delta_r^i(\delta_p^j\delta_q^k-\delta_p^k\delta_q^j)$

Contracting by setting r=k, I obtained:

$\epsilon^{ijk}\epsilon_{pqr}=\delta_p^i(\delta_q^j\delta_k^k-\delta_q^k\delta_k^j)-\delta_q^i(\delta_p^j\delta_k^k-\delta_p^k\delta_k^j)+\delta_k^i(\delta_p^j\delta_q^k-\delta_p^k\delta_q^j)$

Then:

$\epsilon^{ijk}\epsilon_{pqr}=\delta_p^i(\delta_q^j-\delta_q^j)-\delta_q^i(\delta_p^j-\delta_p^j)+(\delta_p^j\delta_q^i-\delta_p^i\delta_q^j)$

So finally:

$\epsilon^{ijk}\epsilon_{pqr}=\delta_p^j\delta_q^i-\delta_p^i\delta_q^j$

Which is the identity I'm trying to prove, except that the right side is multiplied by -1 for some reason. I must be doing something wrong along the way, but I haven't been able to find it. Any ideas?

EDIT:

A similar question has been answered elsewhere, but it is only when expanding the determinant that I arrive at this problem. For that reason, I decided to make a separate post about this.

EDIT 2:

Added intermediate steps.

FINAL EDIT:

As Travis pointed out, I incorrectly replaced $\delta_k^k$ by $1$ instead of $3$. After correcting this, I arrived at the correct solution:

$\epsilon^{ijk}\epsilon_{pqr}=\delta_p^i(3\delta_q^j-\delta_q^j)-\delta_q^i(3\delta_p^j-\delta_p^j)+(\delta_p^j\delta_q^i-\delta_p^i\delta_q^j)=\delta_p^i\delta_q^j-\delta_p^j\delta_q^i$

Rafael Hurtado
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  • I arrived at the same problem when I worked through the solution given there. Since I couldn't find anything referring specifically to my problem, I decided to repost here. – Rafael Hurtado Mar 05 '19 at 19:06
  • Have a look at this https://math.stackexchange.com/questions/1874812/levi-cevita-and-kronecker-delta-identity – mfl Mar 05 '19 at 19:15
  • Thank you for the link, but what I am interested in is the expansion of the determinant into the answer. I understand why the identity makes sense, it's just the proof of it that is not working out for me. – Rafael Hurtado Mar 05 '19 at 19:17

1 Answers1

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It's hard to say for sure without seeing the intermediate steps of the contraction, but when carrying it out did you perhaps replace $\delta^k{}_k$ with $1$ instead of $3$? The latter is correct, because $$\delta^k{}_k = \delta^1{}_1 + \delta^2{}_2 + \delta^3{}_3 = 1 + 1 + 1 = 3 .$$

Travis Willse
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