Volume integral zoo: Geometric calculus, Riemannian volumes, Lagrangian densities What is the connection between these volume integrals?
The Riemannian volume-form:
$$
V=\int_M \sqrt{|g|}dx^1\wedge...\wedge dx^n \tag{1}
$$
Where $|g|$ is the determinant of the matrix representation of the metric tensor of the manifold.

The integral of geometric calculus:
$$
V=\int_M  (e_1 dx^1 \wedge ... \wedge e_n dx^n)=\int_M  (e_1 \wedge ... \wedge e_n ) d^nx  \tag{2}
$$

The integral used in Lagrangian densities found in physics, such as:
$$
S=\int \sqrt{-|g|} d^n x \tag{3}
$$

Comparing all three, one notices the following differences:


*

*For (1), the wedge product is linking the differential terms

*For (2), the wedge product is linking the basis elements of the pseudoscalar.

*For (3), the Lagrangian density has no wedge products, yet the same terms are used $\sqrt{-|g|}$ and the differential terms.


Are they all equal in some subtle way, or if they are different, how do they connect to one another?
To make sense of this zoo, it seems I am missing an equality, or at least a connection, between $dx\wedge dy$ and $dxdy$, and between $(e_1\wedge e_2)dxdy$ and $\sqrt{|g|}dxdy$?

Edit: From  Giuseppe's comment, it appears that $dxdy$ is simply a sloppy way to write $dx\wedge dy$. 
So what is left is to explain how to connect:
$$
(e_0\wedge ... e_n)dx^1 ... dx^n \to \sqrt{|g|}dx^1...dx^n
$$
I am assuming that an intermediary step involves diagonalizing the basis. For instance, it is possibly the case that :
$$
(e_0\wedge ... \wedge e_n) \stackrel{?}{=} \sqrt{|g|} (\gamma_0 \wedge ... \wedge \gamma_n) \tag{A.1}
$$
On can them eliminate the orthogonal basis $(\gamma_0 \wedge ... \wedge \gamma_n)$ by replacing it by $I$, the unit pseudoscalar:
$$
\sqrt{|g|} (\gamma_0 \wedge ... \wedge \gamma_n)=I\sqrt{|g|}
$$
Thus, the connection is $I$, as follows:
$$
\int_M (e_1 \wedge ... \wedge e_n) dx^1...dx^n = I \int_M \sqrt{|g|} dx^1...dx^n
$$
or even absorbing $I$ into the square root:
$$
\int_M (e_1 \wedge ... \wedge e_n) dx^1...dx^n =    \int_M \sqrt{-|g|} dx^1...dx^n
$$
So, the two notations are related by the unit pseudoscalar?
Can  A.1 be proven?
 A: I can partially answer this question.
In particular, it is possible to relate geometric calculus and differential forms by introducing a parameterization.
You can get the idea by considering a vector surface those span is controlled by two parameters
$$\mathbf{x} = \mathbf{x}(a , b).$$
In geometric calculus we introduce differentials that span the tangent plane at the point of evaluation
$$\begin{aligned}   dx_a &= \frac{\partial {\mathbf{x}}}{\partial {a}}\, da \\    dx_b &= \frac{\partial {\mathbf{x}}}{\partial {b}}\, db,\end{aligned}$$
so the area element for this parameterization is
$$\begin{aligned}   d^2 \mathbf{x} &= dx_a \wedge dx_b \\            &= \frac{\partial {\mathbf{x}}}{\partial {a}} \wedge \frac{\partial {\mathbf{x}}}{\partial {b}}\, da db.\end{aligned}$$
To relate this to differential forms, introduce an
orthonormal basis $ \mathbf{e}_\mu \cdot \mathbf{e}_\nu = 0, \mathbf{e}_\mu^2 = \pm 1$.  In this basis, the coordinate expansion (summation implied) of the vector $ \mathbf{x} $ is
$$   \mathbf{x} = \mathbf{e}_\mu x^\mu.$$
The coordinate expansion of the geometric area element is
$$\begin{aligned}   d^2 \mathbf{x} &=   \frac{\partial {x^\mu}}{\partial {a}} \frac{\partial {x^\nu}}{\partial {b}} \mathbf{e}_\mu \wedge \mathbf{e}_\nu\, da db \\            &=   \sum_{\mu < \nu}   \left( {      \frac{\partial {x^\mu}}{\partial {a}} \frac{\partial {x^\nu}}{\partial {b}} -      \frac{\partial {x^\nu}}{\partial {a}} \frac{\partial {x^\mu}}{\partial {b}}   } \right)      \mathbf{e}_\mu \wedge \mathbf{e}_\nu\, da db \\            &=   \sum_{\mu < \nu}      \mathbf{e}_\mu \mathbf{e}_\nu\begin{vmatrix}   \frac{\partial {x^\mu}}{\partial {a}} & \frac{\partial {x^\nu}}{\partial {a}} \\    \frac{\partial {x^\mu}}{\partial {b}} & \frac{\partial {x^\nu}}{\partial {b}}\end{vmatrix}      \, da db \\            &=   \sum_{\mu < \nu}      \mathbf{e}_\mu \mathbf{e}_\nu      \frac{\partial {(x^\mu, x^\nu)}}{\partial {(a,b)}}      \, da db.\end{aligned}$$
Each element of this sum includes a product of a pseudoscalar, a Jacobian determinant, and a scalar two parameter differential.
Now consider a two parameter differential for the same vector.  Recall that a differential (1-form) of a scalar function, again assuming two parameters, has the characteristic
$$   df  =   \frac{\partial {f}}{\partial {a}} \, da +   \frac{\partial {f}}{\partial {b}} \, db.$$
In particular, we may compute the differentials of the coordinate functions
$$\begin{aligned}   dx^\mu &= \frac{\partial {x^\mu}}{\partial {a}} \, da + \frac{\partial {x^\mu}}{\partial {b}} \, db \\    dx^\nu &= \frac{\partial {x^\nu}}{\partial {a}} \, da + \frac{\partial {x^\nu}}{\partial {b}} \, db,\end{aligned}$$
from which we can compute a 2-form
$$\begin{aligned}   dx^\mu \wedge dx^\nu   &= \left( { \frac{\partial {x^\mu}}{\partial {a}} \, da + \frac{\partial {x^\mu}}{\partial {b}} \, db } \right) \wedge \left( { \frac{\partial {x^\nu}}{\partial {a}} \, da + \frac{\partial {x^\nu}}{\partial {b}} \, db } \right) \\    &= \frac{\partial {x^\mu}}{\partial {a}} \frac{\partial {x^\nu}}{\partial {b}} \, da \wedge db + \frac{\partial {x^\mu}}{\partial {b}} \frac{\partial {x^\nu}}{\partial {a}} \, db \wedge da \\    &=\begin{vmatrix}   \frac{\partial {x^\mu}}{\partial {a}} & \frac{\partial {x^\nu}}{\partial {a}} \\    \frac{\partial {x^\mu}}{\partial {b}} & \frac{\partial {x^\nu}}{\partial {b}}\end{vmatrix}   \, da \wedge db \\    &=      \frac{\partial {(x^\mu, x^\nu)}}{\partial {(a,b)}}   \, da \wedge db.\end{aligned}$$
We have almost the same structure as with geometric algebra, however, in differential forms, the antisymmetry of the surface area element is encoded in the 2-form $ da \wedge db $ whereas in geometric calculus the required antisymmetry is encoded in a unit bivector.
Should we restrict our attention to a strictly planar subspace, the mapping between the two formalisms becomes more striking.  We now have
$$\begin{aligned}   d^2 \mathbf{x} &= \mathbf{e}_1 \mathbf{e}_2 \frac{\partial {(x^1, x^2)}}{\partial {(a,b)}} \, da db \\    dx^1 \wedge dx^2 &= \frac{\partial {(x^1, x^2)}}{\partial {(a,b)}} \, da \wedge db.\end{aligned}$$
That is, we can relate the formalisms by the mapping
$$   \mathbf{e}_1 \mathbf{e}_2 \, da db \leftrightarrow da \wedge db.$$  The 1-form has an implicit vectoral nature.
I wouldn't say that $da db$ is just a sloppy way to write $da \wedge db$, because the 2-form $da \wedge db$ from differential forms builds in an explicit antisymmetry that isn't neccessarily implied by $da db$.
It kind of looks like the $ \sqrt{\pm \begin{vmatrix} g \end{vmatrix} } $ term is probably related to the Jacobian determinant for the chosen parameterization, but I'll let somebody else elaborate on that.
