# Levi-Civita help

$\epsilon_{ijk}$ is the Levi-Civita tensor which is totally anti-symmetric. Let $A^{ijk}$ be a totally symmetric matrix. Is it true that $$\epsilon_{ijk}A^{ijk}=0?$$ I know this is the case for $\epsilon_{ij}A^{ij}=0$ in two dimensions. Also, do we know something about $$\epsilon_{ijk}A^{kjl}?$$

• $g_{ik}\epsilon^{kjm}x_m=h_i^j$? or $g_{ik}\epsilon^{kim}x_m=h_i^i$? – mike Nov 20 '14 at 13:07
• Sorry, I am not sure what you are asking! – Marion Nov 20 '14 at 13:11
• Can you double check the indices in $g_{ik}\epsilon^{kjm}x_m$? Should $j$ be $i$? – mike Nov 20 '14 at 13:20
• Hi! Not necessarily. In any case I found out that maybe this part (the last one) is not well asked. So you can ignore it. I will change my question anyway! Any comments on the first part? – Marion Nov 20 '14 at 13:22

$$I=\epsilon_{ijk}A^{ijk}=-\epsilon_{jik}A^{ijk}=-\epsilon_{jik}A^{jik}=-I$$.
So $I = 0$