well you can tell immediately
AndersF said:
##\mathbf{e}^i=g^{ij}\mathbf{e}_j##
##\mathbf{e}_i=g_{ij}\mathbf{e}^j##
cannot be true, because one side is a dual vector whilst the other side is a vector
to some basis ##\{ \boldsymbol{e}_{i} \}## of ##V## is associated a dual basis ##\{ \boldsymbol{e}^i \}## of ##V^*## defined by ##\langle \boldsymbol{e}^i, \boldsymbol{e}_j \rangle = \delta^i_j##
besides there is also metric duality, which is to say that to any ##\boldsymbol{v} \in V## there is a ##f_{\boldsymbol{g}}(\boldsymbol{v}) := \boldsymbol{\hat{v}} \in V^*## such that ##\langle \hat{\boldsymbol{v}}, \boldsymbol{u} \rangle = \boldsymbol{g}(\boldsymbol{v}, \boldsymbol{u})## for any ##\boldsymbol{u} \in V##. Then $$\hat{v}_i := \langle \hat{\boldsymbol{v}}, \boldsymbol{e}_i \rangle = \boldsymbol{g}(v^j \boldsymbol{e}_j, \boldsymbol{e}_i) = g_{ij} v^j$$which referred to as lowering the index ##j##. (Because ##\boldsymbol{g}## is bilinear and non-degenerate the function ##f_{\mathbf{g}}## is injective and further because ##V## and ##V^*## are both of equal finite dimension, ##f_{\boldsymbol{g}}## is indeed a bijective function.)
n.b. also the metric duals ##\hat{\boldsymbol{e}}_i = f_{\boldsymbol{g}}(\boldsymbol{e}_i)## of the basis elements of ##V## do not coincide with the dual basis elements ##\boldsymbol{e}^i## which is clear because ##\langle \boldsymbol{e}^i, \boldsymbol{e}_j \rangle = \delta^i_j## whilst ##\langle \hat{\boldsymbol{e}}_i, \boldsymbol{e}_j \rangle = \boldsymbol{g}(\boldsymbol{e}_i, \boldsymbol{e}_j) = g_{ij}##