# Proof the Levi-Civita symbol is a tensor

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?

• What is $R$, what is $\epsilon_{ijk}'$? And what is Levi-Cevita symbol?? – user99914 Apr 11 '17 at 10:30
• Sorry, edited, is it more clear now? – Christopher Turnbull Apr 11 '17 at 10:38
• It is actually a pseudotensor; the Levi-Civita symbol only behaves as a honest affine tensor under proper orthogonal transformations (those with determinant +1). – Alex Provost Apr 11 '17 at 10:42
• FYR – user99914 Apr 11 '17 at 10:45
• 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 – 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.