5
$\begingroup$

Is the Levi-Civita symbol a tensor?

R. A. Sharipov afirm (In "Quick Introduction to Tensor Analysis", page 30) that "...the Levi-Civita symbol is NOT a tensor..."

$\epsilon_{jkq}=\epsilon^{jkq}=\left\{\begin{array}{cc}0, & \mbox{if among $j$, $k$, $q$ there are at least two equal numbers} \\ 1 & \mbox{if $(j,k,q)$ is even permutation of numbers $(1,2,3)$} \\ -1 & \mbox{if $(j,k,q)$ is odd permutation of numbers}\end{array}\right.$

What does that phrase mean?

Thanks!

$\endgroup$
2
  • $\begingroup$ What's the relevance of $\epsilon_{jkq}$? $\endgroup$
    – Deane Yang
    May 3, 2014 at 18:52
  • $\begingroup$ I'm learning about it. In particular, $e_{jkq}$ can be used to write vectorial product (cross product). $\endgroup$
    – Esteban
    May 5, 2014 at 0:59

1 Answer 1

6
$\begingroup$

The Levi-Civita symbol is a "pseudotensor", or tensor density, because it inverses sign upon inversion. (An orthogonal transformation with Jacobian $-1$ introduces a minus sign.) As a consequence, the contraction of $\varepsilon_{ijk}$ with two vectors produces a pseudovector, or axial vector -- the cross product.

For a clear discussion see page 11 of these lecture notes by Asaf Pe'er. In particular, it is explained how to convert the Levi-Civita pseudotensor into a real tensor, at least on orientable manifolds.

$\endgroup$
3
  • $\begingroup$ Thanks @Carlo. But I need to understand the words of Sharipov in the context of his article Sharipov. ¿Does he want to say: "Despite its form or notation will not be a tensor? $\endgroup$
    – Esteban
    May 5, 2014 at 1:24
  • 1
    $\begingroup$ I guess Sharipov wants to distinguish the Levi-Civita symbol from the Levi-Civita pseudotensor (or tensor density). Then he shows how to convert the pseudotensor into a real tensor. See the link to the note I added, where this is worked out in some more detail. $\endgroup$ May 5, 2014 at 6:36
  • $\begingroup$ The lecture notes are very clear! Thank you very much! $\endgroup$
    – Esteban
    May 5, 2014 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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