A tensor of rank $n$ has components $T_{ij\cdots k}$ (with $n$ indices) with respect to each basis $\{\mathbf{e}_i\}$ or coordinate system $\{x_i\}$, and satisfies the following rule of change of basis: $$ T_{ij\cdots k}' = R_{ip}R_{jq}\cdots R_{kr}T_{pq\cdots r}. $$

Define the Levi-Civita symbol as: $$ \varepsilon_{ijk} = \begin{cases} +1 & ijk \text{ is even permutation}\\ -1 & ijk\text{ is odd permutation}\\ 0 & \text{otherwise (ie. repeated suffices)} \end{cases} $$

Show that $ \varepsilon_{ijk} $ is a rank 3 tensor.

I actually have a proof but I can't understand it! Can someone help me out?

$$ \varepsilon_{ijk}' = R_{ip}R_{jq}R_{kr}\varepsilon_{pqr} = (\det R)\varepsilon_{ijk} = \varepsilon_{ijk}, $$

This shows that $\varepsilon_{ijk}$ obeys the transformation law, so sure... but I don't follow what happened after the second equals sign

EDIT: Does this only hold for Cartesian coordinate systems, because then $R$ would be an orthogonal matrix with det 1 or -1?

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    $\begingroup$ What is $R$, what is $\epsilon_{ijk}'$? And what is Levi-Cevita symbol?? $\endgroup$ – user99914 Apr 11 '17 at 10:30
  • $\begingroup$ Sorry, edited, is it more clear now? $\endgroup$ – Christopher Turnbull Apr 11 '17 at 10:38
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    $\begingroup$ It is actually a pseudotensor; the Levi-Civita symbol only behaves as a honest affine tensor under proper orthogonal transformations (those with determinant +1). $\endgroup$ – Alex Provost Apr 11 '17 at 10:42
  • $\begingroup$ FYR $\endgroup$ – user99914 Apr 11 '17 at 10:45
  • $\begingroup$ Thanks Alex , could you show me where $\det R$ came from, I thought $\det(A) = \varepsilon_{j_1j_2\cdots j_n}A_{j_11}A_{j_22}\cdots A_{j_nn}$... I don't quite see the jump here $\endgroup$ – Christopher Turnbull Apr 11 '17 at 10:50

If $R^i_j$ are the components of the matrix of an orthogonal linear transformation in Euclidean $3$-space, then the general transformation rule for an affine tensor $\epsilon_{ijk}$ should read $$\epsilon'_{ijk} = R^p_iR^q_j R^r_k \epsilon_{pqr}.$$

Since here $\epsilon$ is the Levi-Civita symbol, whose values depend not on the coordinate system but only on the numerical indices $i,j,k$ (i.e., $\epsilon' = \epsilon$), this is the same thing as $$\epsilon_{ijk} = R^p_iR^q_j R^r_k \epsilon_{pqr}.$$

Note that since $\epsilon_{pqr}$ vanishes on degenerate multi-indices, the right-hand side only consists of six terms (one for each proper multi-index) and looks like $R^1_iR^2_jR^3_k - R^2_iR^1_jR^3_k + \cdots$. To evaluate this, there are three cases to consider:

  • If any two of $i,j,k$ share the same value, then the terms cancel each other pairwise and we get $0$.
  • If $ijk$ is an even permutation, then this is just the Leibniz formula for the determinant, and the right-hand side is $\det(R)$.
  • If $ijk$ is an odd permutation, then we may swap components pairwise in every term to recover $-\det(R)$ from the previous computation.

Therefore: if $R$ is a proper ($\det = 1$) orthogonal transformation, we find that the right-hand side coincides with the left-hand side, and $\epsilon$ behaves like a honest affine tensor of rank $3$. On the other hand, if $R$ is improper ($\det = -1$), we find that $$R^p_iR^q_j R^r_k \epsilon_{pqr} = -\epsilon_{ijk}.$$ This says that the Levi-Civita symbol is not a proper affine tensor but rather a pseudotensor.

  • $\begingroup$ My understanding is, that the Levi-Civita-symbol is only a pseudotensor with respect to orthogonal transformations. As soon as we admit arbitrary transformations with det(R) not necessarily being equal to +-1, we get this determinant to be the factor in your last equation. But this would violate the definition of a pseudotensor. I think in this general case the Levi-Civita symbol is only a tensor density, that is, a relative tensor of weight 1. See e.g. Anadijiban Das, "Tensors", p.39. Is my understanding correct? $\endgroup$ – Roland Salz Aug 2 '20 at 20:01
  • $\begingroup$ @RolandSalz Sounds correct to me! $\endgroup$ – Alex Provost Aug 2 '20 at 23:57
  • $\begingroup$ Thanks for looking into that after all these years. In fact, I tried to prove the general case $ \epsilon'_{a_1 \dots a_n} = (\det R)^{-1} R^{b_1}_{a_1} \dots R^{b_n}_{a_n} \epsilon_{b_1 \dots b_n} $, but I havn't been able to. That's why I put up another question: Proof that the Levi-Civita symbol is a tensor density. $\endgroup$ – Roland Salz Aug 4 '20 at 16:00

I met this problem today and this is what I am trying: $$ \epsilon_{ijk}=det(e_i\ e_j \ e_k) $$ Let A be an orthogonal transformation, then: $$ \begin{aligned} \epsilon'_{ijk}&\equiv\epsilon_{lmn}=det( e_l \ e_m \ e_n)\\ &=det(Ae_i\ Ae_j \ Ae_k)\\ &=(det(A))^3 det(e_i e_j e_k)\\ &=det(A)\cdot \epsilon_{ijk}\\ &=\pm \epsilon_{ijk} \end{aligned} $$ I think this can show that the Levi-Civita is a tensor (or pseudotensor).

  • $\begingroup$ Sorry, I suddenly found that I am not sure whether I can let $\epsilon'_{ijk}\equiv\epsilon_{lmn}$ straightforwardly :) $\endgroup$ – Peter_H Apr 1 '20 at 14:26

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