3
$\begingroup$

$\epsilon_{ijk}$ 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 $$\epsilon_{ijk}A^{kjl}?$$

$\endgroup$
  • $\begingroup$ $g_{ik}\epsilon^{kjm}x_m=h_i^j$? or $g_{ik}\epsilon^{kim}x_m=h_i^i$? $\endgroup$ – mike Nov 20 '14 at 13:07
  • $\begingroup$ Sorry, I am not sure what you are asking! $\endgroup$ – Marion Nov 20 '14 at 13:11
  • $\begingroup$ Can you double check the indices in $g_{ik}\epsilon^{kjm}x_m$? Should $j$ be $i$? $\endgroup$ – mike Nov 20 '14 at 13:20
  • $\begingroup$ 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? $\endgroup$ – Marion Nov 20 '14 at 13:22
3
$\begingroup$

Yes. Because

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

So $I = 0$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot, this is very helpful although it confuses me since we usually define volumes in 3d like this. Would you be kind enough to help me with another question as well? In specific with this one math.stackexchange.com/questions/1029999/…. $\endgroup$ – Marion Nov 20 '14 at 13:28
  • $\begingroup$ In any case THANKS A LOT! (I cannot upvote you yet, but will return to do it when I get 15 points!) $\endgroup$ – Marion Nov 20 '14 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.