I have been reading up on the permutation tensor, and have come across the following expression (in 'Generalized Calculus with Applications to Matter and Forces' by L.M.B.C Campos page 709): $$e_{i_1,\ldots,i_n}=e^{i_1,\ldots,i_n}=\begin{cases} 0 & \text{if repeated indices} \\ 1 & \text{if ($i_1,\ldots,i_n$) is an even permutation} \\ -1 & \text{if ($i_1,\ldots,i_n$) is an odd permutation} \end{cases}$$ However when I try and prove that: $$e_{i_1,\ldots,i_n}=e^{i_1,\ldots,i_n}$$ I instead get: $$e_{i_1,\ldots,i_n}=\textrm{det}(g)e^{i_1,\ldots,i_n}$$ why is there a difference between my result and that given above? Is it to do with the way $e_{i_1,\ldots,i_n}$ has been defined?
My working $$e_{i_1,\ldots, i_n}=\sum_\sigma g_{i_{\sigma(1)}i_1}\ldots g_{i_{\sigma(n)}i_n}e^{i_{\sigma(1)}\ldots i_{\sigma(n)}}$$ where the summation is over all permutations, $\sigma$, of $(1,\ldots,n)$ $$\begin{align}e_{i_1...i_n} &=e^{i_1...i_n}\sum_\sigma \textrm{sgn}(\sigma) g_{i_{\sigma(1)}i_1}\ldots g_{i_{\sigma(n)}i_n}\\ &=\textrm{det}(g)e^{i_1,\ldots,i_n}\end{align}$$