I Expressing Vectors of Dual Basis w/Metric Tensor

[AndersF]

TL;DR

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

 

[vanhees71]

Have a look at this posting. I hope that helps!

https://www.physicsforums.com/threads/dual-vector.994347/post-6399833

 

[AndersF]

none, because they're not right

Oh, why not? They are written like this in my textbook

 

[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 $\langle \hat{\boldsymbol{e}}_i, \boldsymbol{e}_j \rangle = \boldsymbol{g}(\boldsymbol{e}_i, \boldsymbol{e}_j) = g_{ij}$

 

[vanhees71]

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
$$L_{\vec{v}}(\vec{w})=g(\vec{v},\vec{w}).$$
In your notation $L_{\vec{v}}=\hat{\vec{v}}$.

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.

 

[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

 

[vanhees71]

Well, in principle yes. But if you identify the vectors with their duals via the fundamental form in a pseudo-Euclidean (or Euclidean) vector space as explained by the above introduced corresponding isomorphism it makes sense. E.g., then you get from the above formula
$$\boldsymbol{e}^i(\boldsymbol{e}_k)=g^{ij} \boldsymbol{e}_j(\boldsymbol{e}_k)=g^{ij} g_{jk}=\delta_k^i,$$
as it must be.

BTW the $g^{ij}$ are the contravariant components of the pseudo-metric not the pseudo-metric itself. Of course physicists use simply "metric" for both the pseudo-metric and its (various) components all the time. This confused me a lot when I started to learn the subject, but one gets used to it.