# How to show ${\epsilon^{ij}}_k {\epsilon_{ij}}^l = -2\delta^l_k$ for the Levi-Civita symbol?

I am trying to prove the following identity for the levi civita symbol $${\epsilon^{ij}}_k {\epsilon_{ij}}^l = -2\delta^l_k,$$ taken from the Ashok Das QFT book pg 153, equation 4.102.

I made use of the identity $$\epsilon_{jmn} \epsilon^{imn} = 2 \delta^i_j$$ but got a different sign from the author's result:

$${\epsilon^{ij}}_k {\epsilon_{ij}}^l = (\eta^{ia}\eta^{jb}\epsilon_{abk} )(\eta_{ic} \eta_{jd}\epsilon^{cdl}) = (\eta^{ia} \eta_{ic}) (\eta^{jb} \eta_{jd}) \epsilon_{abk} \epsilon^{cdl} = \delta^{a}_c\delta^b_d \epsilon_{abk} \epsilon^{cdl}= \epsilon_{cdk}\epsilon^{cdl} = (-\epsilon_{kdc}) (-\epsilon^{ldc}) = \epsilon_{kdc} \epsilon^{ldc} = 2\delta^l_k$$

What am I doing wrong?

• This seems to come down to conventions. – Qmechanic Jan 31 at 12:04