# 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 Gold Member
2022 Award
• AndersF
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
$$\boldsymbol{e}^i(\boldsymbol{e}_k)=g^{ij} \boldsymbol{e}_j(\boldsymbol{e}_k)=g^{ij} g_{jk}=\delta_k^i,$$
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