# Expressing the vectors of the dual basis with the metric tensor

• I
AndersF
TL;DR Summary
Why can we express the dual basis vectors in terms of the original basis vectors through the dual metric tensor in this way? ##\mathbf{e}^i=g^{ij}\mathbf{e}_j##
I'm trying to understand why it is possible to express vectors ##\mathbf{e}^i## of the dual basis in terms of the vectors ##\mathbf{e}_j## of the original basis through the dual metric tensor ##g^{ij}##, and vice versa, in these ways:

##\mathbf{e}^i=g^{ij}\mathbf{e}_j##

##\mathbf{e}_i=g_{ij}\mathbf{e}^j##

What would be the mathematical justification for these expressions?

ergospherical
none, because they're not right

AndersF
none, because they're not right
Oh, why not? They are written like this in my textbook

vanhees71
ergospherical
well you can tell immediately
##\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}##

Last edited:
AndersF
Gold Member
2022 Award
Well, yes, but usually in spaces with a fundamental form you identify the vectors and the dual vectors through the canonical mapping ##V \rightarrow V^*## via ##\vec{v} \mapsto L_{\vec{v}}##, where
$$L_{\vec{v}}(\vec{w})=g(\vec{v},\vec{w}).$$

Then of course ##\hat{e}^i=g^{ij} \hat{e}_j=e^i##. Usually you don't distinguish between ##\hat{e}_i## and ##e_i## anymore. It's somewhat sloppy notation though.

cianfa72, Orodruin and AndersF
AndersF
Okay, thanks, I see that it was the notation that confused me.

ergospherical
yea but what I mean is that you can't raise and lower the indices attached to the basis vectors with the metric like this "##\boldsymbol{e}^i = g^{ij} \boldsymbol{e}_j##", because then you are equating objects which live in different spaces 😥

AndersF
$$\boldsymbol{e}^i(\boldsymbol{e}_k)=g^{ij} \boldsymbol{e}_j(\boldsymbol{e}_k)=g^{ij} g_{jk}=\delta_k^i,$$