Levi-Civita tensor in tetrad/vielbein basis

Question

I am reading about tetrads in a GR textbook and a question occured to me. It seems natural to assume that the Levi Civita tensor in tetrad/vielbein basis is the Levi Civita symbol in the flat indices, i.e. in 2D:

$$\sqrt{g} \ \epsilon_{\mu \nu}=\epsilon_{ab} e^{a}_\mu e^b_\nu$$ and $$\frac{1}{\sqrt{g}} \ \epsilon^{\mu \nu}=\epsilon^{ab} e_{a}^\mu e_b^\nu$$ where g is the determinant of the metric and I assumed Euclidean signature for simplicity. And similarly for any num. of dimensions: $$\sqrt{g} \ \epsilon_{\mu_1 \dots \mu_d}=\epsilon_{a_1 \dots a_d} e^{a_1}_{\mu_1} \dots e^{a_d}_{\mu_d}$$ and $$\frac{1}{\sqrt{g}} \ \epsilon^{\mu_1 \dots \mu_d}=\epsilon^{a_1 \dots a_d} e_{a_1}^{\mu_1} \dots e_{a_d}^{\mu_d}$$ Is this true? How does one prove it?

Answer 1

Hint: $$ \epsilon^{\rm symbol}_{a_1\ldots a_d} e^{a_1}_{\mu_1}\ldots e^{a_d}_{\mu_d} ~=~ \det(e^a_{\mu})\epsilon^{\rm symbol}_{\mu_1\ldots \mu_d} ~=~ \sqrt{|\det(g_{\mu\nu})|}\epsilon^{\rm symbol}_{\mu_1\ldots \mu_d} ~=~ \epsilon^{\rm tensor}_{\mu_1\ldots \mu_d}. $$

Answer 2

Observe that the definition of the determinant and its change of sign under interchanging rows or columns tells us that $$ \epsilon^{a_1\ldots a_d} e^{\mu_1}_{a_1}\ldots e^{\mu_d}_{a_d}= \epsilon^{\mu_1\ldots\mu_d} {\rm det}[e^\mu_a], $$ and remember that there is a connection between the $e^\mu_a$ and the metric.