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I'm trying to follow the answer to this post, that cites the identity $$ \epsilon_{i_{1} \ldots i_{n}} A_{j_{1}}^{i_{1}} \cdots A_{j_{n}}^{i_{n}}=\operatorname{det} A \epsilon_{j_{1} \ldots j_{n}} $$ as a proof of the invariance of the Levi-Civita tensor. How would I prove this identity? It seems like just the definition of the determinant, but I'm not sure about the second Levi-Civita symbol on the right hand side. I don't know if the definition of determinant is trivial, or if I'm missing a critical step?

Once I have this identity, then the invariance follows obviously because rotations are in $SU(2)$, but I'm stuck on this identity, any help would be appreciated.

Qmechanic
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Andrew Hardy
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1 Answers1

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Consider the following expression : $$\epsilon_{i_1i_2...i_n}A^{i_1}_{1}A^{i_2}_{2}...A^{i_n}_{n}$$

Now THIS is exactly the definition of the determinant of $A$. Why ? Well the determinant is defined as $$detA=\sum_{\sigma\in S_n} \epsilon(\sigma)A^{\sigma(1)}_{1}A^{\sigma(2)}_{2}...A^{\sigma(n)}_{n}$$

Which, if you look hard enough, is exactly the same as the above expression, you can see the sum over all permutations is handled by the contraction with the Levi-Civita tensor, and the sign $\epsilon(\sigma)$ comes from the sign of the Levi-Civita.

But now we have all we need. Indeed, consider the tensor defined by $T_{j_1j_2...j_n}=\epsilon_{i_1i_2...i_n}A^{i_1}_{j_1}A^{i_2}_{j_2}...A^{i_n}_{j_n}$. You can easily show that it is completely anti-symmetric in its indices, try it !

But totally anti-symmetric tensors of rank n are necessarily proportional to the Levi-Civita tensor !(So here, $T_{j_1j_2...j_n}=C\epsilon_{j_1j_2...j_n}$) Indeed, once you know one of the elements, all the other elements of the tensor are determined using the antisymmetry. That's great, because we know : $T_{12..n}=detA=C\epsilon_{12...n}=C$.

Plugging back in yields $$T_{j_1j_2...j_n}=detA\epsilon_{j_1j_2...j_n}$$ And this is exactly as we wished to show.

Frotaur
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