I need some help trying to prove the invariance under rotations of the Levi-Civita tensor, I know that the tensor transform using $U\in SO(3)$ from unitary group with $\det(U)=1$, so with this definition of the determinant (I need to use specifically this definition), $$ \det(U)=\frac{1}{6}\epsilon^{abc}\epsilon_{ijk}U^i_aU^j_bU^k_c\qquad(\text{for U$\in$ SO(3)})=1 $$

So I need to transform the Lev-Civita tensor,

$$ \epsilon^{abc}\rightarrow U^{a'}_iU^{b'}_jU^{c'}_k\epsilon^{abc}\overset{!}{=}\epsilon^{lmn} $$

I dont know how to relate the two definitions that I have given, but it could be done, I need help for the following steps. Thanks!

  • 1
    $\begingroup$ Isn't there any summation in the quantities after the "$\rightarrow$"? $\endgroup$ – rhesu Jan 25 at 16:18

As far as I understand you have to prove that Levi-Civita tensor is constant tensor over transformation by U in $SO(3) $ ($det U=1$) and use as a tool for proving a determinant definition as $\det(U)=\frac{1}{6}\epsilon^{abc}\epsilon_{ijk}U^i_aU^j_bU^k_c$.

We identify $\epsilon^{abc}$ as an antisymmetric tensor ($\epsilon^{abc}=0$ if any two indices are the same; for example, $\epsilon^{112}=0$), $\epsilon^{123}=1$, the rest of the tensor values are equal to $+1$ or $-1$ - depending on the parity of the permutation of the indices (for example, $\epsilon^{132}=-1$). Now you want to prove that $\epsilon^{abc}\rightarrow U^{i}_aU^{j}_bU^{k}_c\epsilon^{abc}=B(U)^{ijk}$ (some tensor depending on $U$) $=\epsilon^{ijk}$ over transformation by means of any $U$ (in this formula the summation over repeated indices $a, b, c$ is implied).

Let’s check it directly. First of all, $B(U)^{ijk}=-B(U)^{jik}$ over permutation of neighboring indices. Indeed, $U^{j}_aU^{i}_bU^{k}_c\epsilon^{abc}= U^{i}_bU^{j}_aU^{k}_c\epsilon^{abc}$ $=- U^{i}_bU^{j}_aU^{k}_c\epsilon^{bac}$ But $a$ and $b$ are just the (dumb) indices of summation; we can write $a$ instead of $b$ and $b$ instead of $a$ in our expression: $U^{j}_aU^{i}_bU^{k}_c\epsilon^{abc}=- U^{i}_bU^{j}_aU^{k}_c\epsilon^{bac} =- U^{i}_aU^{j}_bU^{k}_c\epsilon^{abc}= -B(U)^{ijk}$. In the same way we can prove that $ B(U)^{jik}$ is the antisymmetric tensor for all its indices.

Now we have to prove that $ B(U)^{123}=1$. Taking into consideration that $a, b, a$ have to be different (otherwise, we get zero due to anti-symmetry of the tensor $\epsilon^{abc}$) we get: $ B(U)^{123}= U^{1}_aU^{2}_bU^{3}_c\epsilon^{abc}= U^{1}_1(U^{2}_2U^{3}_3- U^{2}_3U^{3}_2)- U^{1}_2(U^{2}_1U^{3}_3- U^{2}_3U^{3}_1)+ U^{1}_3(U^{2}_1U^{3}_2- U^{2}_2U^{3}_1)=det(U)=1$

Tensor $B(U)^{ijk}$ is antisymmetric with values $+1$ or $-1$, therefore $B(U)^{ijk}=\epsilon^{ijk}$


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