Do we have to use the metric constructed from the tangent basis to form a line element?

Question

I'm working through Guidry's Modern General Relativity, and there is a problem where it asks to construct the metric using the tangent and dual basis vectors provided. I have done this, and my metrics match the answer key. However, when I went to construct the line element, I used the metric derived from the dual basis vectors. The solution only has the line element using the tangent basis vectors. Since I expected the line element to be invariant, I was wondering why they aren't the same and if the tangent basis one is the correct one?

Answer 1

If I understand the question properly, the idea is the following. We first consider Cartesian coordinates $(x,y,z)$ which induce the orthonormal coordinate basis vectors $\hat e_x,\hat e_y,\hat e_z$. We then choose new coordinates $(u,v,w)$ which are defined via $$\matrix{x=u+v\\y=u-v\\z=2uv+w}$$ The vector transformation law yields $$\hat e_u = \frac{\partial x}{\partial u}\hat e_x + \frac{\partial y}{\partial u}\hat e_y + \frac{\partial z}{\partial u}\hat e_z = \hat e_x + \hat e_y+2v\hat e_z$$ and similarly for the others. As a result, the components of the metric become e.g. $$\matrix{g_{uu} \equiv \hat e_u\cdot \hat e_u = 2+4v^2\\g_{uv}\equiv \hat e_u\cdot \hat e_v=4uv}$$ and so on. In this framework, this is the definition of the components of the metric.

However, when I went to construct the line element, I used the metric derived from the dual basis vectors.

I'm not entirely sure what you mean by this, but I am guessing it's something like the following. One could also define dual basis vectors by inverting the coordinate relations to obtain $$\matrix{u = \frac{x+y}{2}\\ v=\frac{x-y}{2}\\ w=z-\frac{x^2-y^2}{2}}$$ From there, one could define a dual basis $$\hat\epsilon^u = \frac{\partial u}{\partial x} \hat e_x + \frac{\partial u}{\partial y} \hat e_y + \frac{\partial u}{\partial z} \hat e_z = \frac{1}{2}\hat e_x +\frac{1}{2}\hat e_y$$ and similarly for the rest. This basis has the special property that $\hat \epsilon^i \cdot \hat e_j = \delta^i_j$, where $i,j\in\{u,v,w\}$. My guess is that you were under the impression that $$\hat e_i \cdot \hat e_j \equiv g_{ij} \overset{?}{=} \hat \epsilon^i\cdot\hat\epsilon^j$$ but in fact this is not true (as you have seen). Instead, $\hat \epsilon^i\cdot \hat \epsilon^j\equiv \gamma^{ij}$, where $\boldsymbol\gamma$ is called the dual metric. It has the property that its components are the matrix inverse of the components of the metric $g_{ij}$, i.e.

$$\mathbf g \boldsymbol \gamma = \boldsymbol \gamma \mathbf g = \mathbb I \iff \sum_j \gamma^{ij}g_{jk} = \delta^i_j \iff \boldsymbol \gamma = \mathbf g^{-1}$$

In the Cartesian basis, the components of $\mathbf g$ are simply the identity matrix (which is its own inverse), so $g_{ij}=\gamma^{ij}$; however, this is not a general feature because $\mathbf g$ and $\boldsymbol \gamma$ are not the same in a non-orthonormal basis.

The solution only has the line element using the tangent basis vectors.

The line element is, by definition, $$\mathrm ds^2 = \sum_{i,j} g_{ij}\mathrm dx^i \mathrm dx^j$$ It's not hard to see that we can't simply substitute $g_{ij}\leftrightarrow \gamma^{ij}$ and get the same result.

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why doesn't moving from tangent to dual basis qualify as change of basis?

It does - but in that case the components of the vector must change as well as the basis you choose. Explicitly, we have that $$d\mathbf r = \mathrm du \hat e_u + \mathrm dv \hat e_v + \mathrm dw \hat e_w$$ $$ds^2 = d\mathbf r \cdot d\mathbf r = g_{uu} du^2 + g_{vv} dv^2 + g_{ww} dw^2 + 2g_{uv} dudv + 2g_{vw} dvdw + 2g_{uw}dudw$$

If you want to change to the basis $\{\hat \epsilon^\mu,\hat \epsilon^v,\hat \epsilon^w\}$, then the components of $d\mathbf r$ are no longer $du,dv,$ and $dw$. Explicitly, we may write $$d\mathbf r = du \hat e_u + dv \hat e_v + dw\hat e_w = A\hat \epsilon^u + B \hat \epsilon^v + C\hat \epsilon^w$$ $$\implies \hat e_u \cdot d\mathbf r = du g_{uu} + dv g_{uv} + dw g_{uw} = A$$ and similarly for $B$ and $C$. Computing the line element in this basis would yield $$ds^2 = A^2 \gamma^{uu} + B^2\gamma^{vv} + C^2 \gamma^{ww} + 2AB \gamma^{uv} + 2BC \gamma^{vw} + 2AC \gamma^{uw}$$

which is perfectly valid.

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All that being said, I should make a point to note that this formulation of differential geometry - in which "dual vectors" and "tangent vectors" live in the same vector space - is antiquated and conceptually muddy. I far prefer the more modern treatment I describe e.g. here, as well as in several related answers elsewhere. Since the latter is both conceptually cleaner and in quite widespread use, I would personally suggest you consider learning it (though of course, I'm just a person on the internet, so you may do whatever feels natural and comfortable for you).