# Tensor identity in L&L book 2

How does the identity $$\epsilon^{prst}A_{ip}A_{kr}A_{ls}A_{mt}=-A\epsilon_{iklm}$$ with the Levi Civita symbol $$\epsilon$$ and the determinant A of the matrix $$A_{ik}$$ follow from the equation $$\epsilon^{iklm}\epsilon_{prst}=-\begin{vmatrix} \delta^i_p & \delta^i_r & \delta^i_s & \delta^i_t \\ \delta^k_p & \delta^k_r & \delta^k_s & \delta^k_t \\ \delta^l_p & \delta^l_r & \delta^l_s & \delta^l_t \\ \delta^m_p & \delta^m_r & \delta^m_s & \delta^m_t \\ \end{vmatrix}$$ where the $$\delta$$ are Kronecker deltas and the vertical lines represent the determinant?
(i found it in §6 of L&L book 2 where it says that the first equation follows directly from the second and cannot really see how to draw the connection)

• Perhaps this can be of help? Commented May 4 at 14:55
• what happened to $j$? Also, who know NIST was keeping standards on isotropic tensors up to rank-8? The section on Capelli's Identity may address your question: nvlpubs.nist.gov/nistpubs/jres/79b/jresv79bn1-2p49_a1b.pdf ...wait, never mind, you're in 4-D. NIST is 3...are we still Euclidean, since you're using latin super/super scripts? Note also: NIST skips "l" because it looks like "1", but it keeps "j" around.
– JEB
Commented May 4 at 16:34
• It's literally just the Leibniz formula en.wikipedia.org/wiki/Leibniz_formula_for_determinants Commented May 8 at 12:30
• Hmm, I wouldn't think too hard about the second identity. The first identity is almost self evident from the fact that $\det A = \epsilon^{ijkl} A_{0i}A_{1j}A_{2k}A_{3l}$ and the antisymmetry of the determinant when changing rows (you just have to take care of the sign-convention). Commented May 8 at 15:01

These trivialities were developed simply by checking both sides for simple entries from $$0,1$$. The full n dimensional LeviCivita symbol is the coefficient tensor of the n-dimensional signed volume element, normal notation today $$d^4x = dx_1\wedge \dots dx_4$$ as a multilinear antisymmetric function of 4 variables.

Since the four indices $$i,k,l,m$$ in the result are fixed, one can simply contract the LeviCivita tensor with four vectors differently named vectors

$$\text{tc}=\sum_{(i_1,i_5),(i_2,i_6),(i_3,i_7),(i_4,i_8)}\left(\epsilon ^{(4)} \otimes \vec a \otimes \vec b \otimes \vec c \otimes \vec d \right)$$

e.g. explicitely in Mathematica

   tc = TensorContract[
TensorProduct[LeviCivitaTensor[4],
(Subscript[a, #] &) /@ Range[4],
Subscript[b, #] & /@ Range[4],
Subscript[c, #] & /@ Range[4],
Subscript[d, #] & /@ Range[4]] ,
{{1, 5}, {2, 6}, {3, 7}, {4, 8}}]


$$a_4 b_3 c_2 d_1-a_3 b_4 c_2 d_1-a_4 b_2 c_3 d_1+a_2 b_4 c_3 d_1+a_3 b_2 c_4 d_1-a_2 b_3 c_4 d_1$$ $$-a_4 b_3 c_1 d_2+a_3 b_4 c_1 d_2+a_4 b_1 c_3 d_2-a_1 b_4 c_3 d_2-a_3 b_1 c_4 d_2+a_1 b_3 c_4 d_2$$ $$+a_4 b_2 c_1 d_3-a_2 b_4 c_1 d_3-a_4 b_1 c_2 d_3+a_1 b_4 c_2 d_3+a_2 b_1 c_4 d_3-a_1 b_2 c_4 d_3$$ $$-a_3 b_2 c_1 d_4+a_2 b_3 c_1 d_4+a_3 b_1 c_2 d_4-a_1 b_3 c_2 d_4-a_2 b_1 c_3 d_4+a_1 b_2 c_3 d_4$$

This is apparently equal to the determinant of the four vectors, that is the alternating sum of the diagonal product $$a_1 b_2 c_3 d_4$$ over all $$4!$$ permutations.

The negative sign is the effect of lowering or lifting all four indices with negative signature of the Lorentz metric