# Invariance of Levi-Civita tensor

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!

• Isn't there any summation in the quantities after the "$\rightarrow$"? – 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 $$B(U)^{ijk}=\epsilon^{ijk}$$