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:
$\epsilon^{ijk}\epsilon_{pqr}=\delta_p^i(3\delta_q^j-\delta_q^j)-\delta_q^i(3\delta_p^j-\delta_p^j)+(\delta_p^j\delta_q^i-\delta_p^i\delta_q^j)=\delta_p^i\delta_q^j-\delta_p^j\delta_q^i$
Best Answer
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 $$\delta^k{}_k = \delta^1{}_1 + \delta^2{}_2 + \delta^3{}_3 = 1 + 1 + 1 = 3 .$$