The proof of this tensor identity

Question

I found that $\epsilon^{ijk}g_{jm}g_{kn}=\det(g)g^{il}\epsilon_{lmn}$, is it correct? What's proof of it? This formula was found when I was studying tensor cross product, I tried to prove that if $A\times B=C$ holds for $A^i,B^i,C^i$, it will also holds for $A_i,B_i,C_i$.
For example:

Answer 1

Presumably, you are dealing with a three-dimensional space. If you are dealing with a $d$-dimensional space, then I think this is true:
$$\epsilon^{i_1i_2\ldots i_d}g_{i_2j_2}g_{i_3j_3}\cdots g_{i_dj_d}=\det(g)\,g^{i_1j_1}\,\epsilon_{j_1j_2\ldots j_d}\,.$$ The proof for the general case is essentially the same.

Back to the original problem, let $(x^l)$, $(y^m)$, and $(z^n)$ be arbitrary ($3$-dimensional) vectors. We want to show that $$\epsilon^{ijk}g_{jm}g_{kn}x_iy^mz^n=\det(g)\,g^{il}\epsilon_{lmn}x_iy^mz^n\,.$$ However, $x_i=g_{il}x^l$, so $$\epsilon^{ijk}g_{jm}g_{kn}x_iy^mz^n=\epsilon^{ijk}g_{il}g_{jm}g_{kn}x^ly^mz^n\,,$$ whereas $$\det(g)\,g^{il}\epsilon_{lmn}x_iy^mz^n=\det(g)\,\epsilon_{lmn}x^ly^mz^n\,.$$ Thus, it suffices to prove that $$\epsilon^{ijk}g_{il}g_{jm}g_{kn}=\det(g)\,\epsilon_{lmn}\,.\tag{*}$$

In the right-hand side of (*), if $l$, $m$, and $n$ are not pairwise distinct, then it equals $0$. If $l=m$, for instance, then the left-hand side is $$\epsilon^{ijk}g_{il}g_{jm}g_{kn}=\epsilon^{ijk}g_{im}g_{jl}g_{kn}=\epsilon^{jik}g_{jm}g_{il}g_{kn}=-\epsilon^{ijk}g_{il}g_{jm}g_{kn}\,.$$ Therefore, the left-hand side is $0$ as well.

Now, suppose that $(l,m,n)$ is a permutation of $(1,2,3)$ with sign $s\in\{-1,+1\}$. Thus, the right-hand side of (*) is numerically $s\,\det(g)$. On the other hand, by a direct expansion of determinant, we get $$\epsilon^{ijk}g_{il}g_{jm}g_{kn}=\det\left(\begin{bmatrix}g_{1l}&g_{1m}&g_{1n}\\g_{2l}&g_{2m}&g_{2n}\\g_{3l}&g_{3m}&g_{3n} \end{bmatrix}\right)\,.$$ Because $$\begin{bmatrix}g_{1l}&g_{1m}&g_{1n}\\g_{2l}&g_{2m}&g_{2n}\\g_{3l}&g_{3m}&g_{3n} \end{bmatrix}=\begin{bmatrix}g_{11}&g_{12}&g_{13}\\g_{21}&g_{22}&g_{23}\\g_{31}&g_{32}&g_{33} \end{bmatrix}P=gP\,,$$ where $P$ is the permutation matrix corresponding to the permutation $(1,2,3)\mapsto(l,m,n)$, we get that the numerical value of the left-hand side of (*) is $$\epsilon^{ijk}g_{il}g_{jm}g_{kn}=\det(gP)=\det(g)\,\det(P)=s\,\det(g)\,,$$ as $\det(P)=s$.