The Riemannian volume form is
$$ \omega = \sqrt{|\det g|} dx^1 \wedge dx^2 \wedge \dots \wedge dx^n $$
The line element is:
$$ ds=\sqrt{g_{\mu\nu}dX^\mu dX^\nu} $$
1D:
First, the Riemannian volume form reduces to:
$$ \omega =\sqrt{|\det g|}dx $$
Second, the line element in 1D. In 1D, the quantities reduces to $g=\pmatrix{g_{xx}}$ and $\det g$= $g_{xx}$ and $\mu,\nu \in \{x\}$. Thus:
$$ \begin{align} ds&=\sqrt{g_{xx} dX^x dX^x}\\ &=\sqrt{g_{xx} } dX^x \\ &=\sqrt{|\det g| } dx \\ \end{align} $$
where $dX^x:=dx$.
Consequently, a line element in 1D is equal to the Riemannian volume form in 1D. The line element has a "richer" definition only if the line is embedded into a space of more dimensions.
2D and above:
The 2D equivalent of the line element is the parametrized surface integral (not sure of the mathematics name, but in physics, it is the same as Nambu-Goto action in general curvilinear coordinates).
First, the Riemannian volume form in 2D is
$$ \omega = \sqrt{|\det g|} dx \wedge dy $$
Now, the general notation for the p-volume element a p-surface embedded with a larger D-dimensional space is given as follows (page 19):
$$ d\mu_p = \sqrt{|\det G_{\alpha\beta}| } d^{p+1} x $$
where
$$ G_{\alpha\beta}= g_{\mu\nu} \partial_\alpha X^\mu \partial_\beta X^\nu $$
In the 2D case, this reduces to
$$ g_{\mu\nu}=\pmatrix{g_{xx} & g_{xy}\\ g_{yx} & g_{yy}} $$
and $G_{\alpha\beta} \to G_{xy}$ becomes:
$$ G_{xy}=\pmatrix { g_{xx}\partial_x X^x \partial_x X^x & g_{xy} \partial_x X^x \partial_y X^y \\ g_{yx}\partial_y X^y \partial_x X^x & g_{yy} \partial_y X^y \partial_y X^y } $$
and the determinant is:
$$ \begin{align} \det G_{xy}&=g_{xx}\partial_x X^x \partial_x X^x g_{yy} \partial_y X^y \partial_y X^y - g_{xy} \partial_x X^x \partial_y X^y g_{yx}\partial_y X^y \partial_x X^x\\ &=(g_{xx} g_{yy} - g_{xy} g_{yx}) ( \partial_x X^x \partial_y X^y)^2\\ &=\det g_{xy}( \partial_x X^x \partial_y X^y)^2 \end{align} $$
Consequently, the volume element becomes:
$$ \begin{align} d\mu_p&=\sqrt{|\det g_{xy}|}\partial_x X^x \partial_y X^y\\ &=\sqrt{|\det g_{xy}|}dxdy \end{align} $$
Based on this it certainly does seem like the idea generalizes to n-dimensions.
A few questions on the notation and exactness:
- I could use a checkup that what I have done is legit.
- For my derivation to work I need $\partial_x X^x=dx$. Can someone prove this or explain why that is? The way I see it is $\partial_x X^x=\partial X[x,y]^x / \partial x$. It is not clear how the partial derivative of a function ends up to a total derivative $dx$.
- Where does the wedge product come from? Perhaps the book I referenced abuses the notation? Should $G_{\alpha\beta}$ actually be written:
$$ G_{\alpha\beta}=g_{\mu\nu} (\partial_\alpha X^\mu \wedge \partial_\beta X^\nu) $$
edit:
For question 2, I am wondering if there is another abuse of notation in the definition of $G_{\alpha\beta}$? Should it be:
$$ G_{\alpha\beta}=g_{\mu\nu} (\partial_\alpha X^\mu d\alpha \wedge \partial_\beta X^\nu d\beta) $$
Then everything would fit neatly --- can anyone confirm?
In this case $\partial_x X^x dx = (\partial X^x /\partial x) dx= (1)dx=dx$
edit
actually I just noticed the definition by the author is
$$ d\mu_p = \sqrt{|\det G_{\alpha\beta}| } d^{p+1} x $$
thus the total differential terms are indeed included in the definition. Moving them inside the definition of $G_{\alpha\beta}$ may make the definition more intuitive because it becomes the generalization of the line element, but it is not notational abuse.
So, question 1 and 3 remains.