In almost all places I look, the generalized Kronecker-delta symbol is defined as:
$$\varepsilon_{i_1...i_p}\varepsilon^{j_1...j_p}=\delta _{j_1...j_p}^{i_1...i_p}$$
Which is the way I know it to be defined. Howver in the answers here we have the person answering writing that:
$$\varepsilon_{b_1...b_n}\varepsilon^{a_1...a_n}=\text{sgn}(g)\delta _{b_1...b_n}^{a_1...a_n}$$
Where $g$ is the metric. I don't even know how to formulate my question here. Is this sign just a part of a different definition for the generalized Kronecker delta? Does it arrise naturally somehow? Or even maybe those two equalities are the same (I don't see how that would be the case). The metric is a structure which is defined separately. So if we don't have a defined metric it would follow that we couldnt define the generalized Kronecker-delta. And the thing is that there has to be this sign in order for the answer to give the correct proof to the linked question. Can you help me understand?
