What's the measure of a point, i.e. a 0-form?
Maybe later.
What's the definition of a line i.e. a 1-form, in $\mathbb{R}$?
$\mathbf{L} = [a, b]; a,b \in \mathbb{R}$
But let's generalize this a bit, for the sake of future definitions:
$\mathbf{L} = \{ (b - a) s + a; s \in [0,1]; a,b \in \mathbb{R} \}$
$\mathbf{L} = \{ a (1 - s) + b s; s \in [0,1]; a,b \in \mathbb{R} \}$
Let $\mathbf{p}^1 = a, \mathbf{p}^2 = b, s_1 = 1 - s, s_2 = s, V = \mathbb{R}$
$\mathbf{L} = \{ \Sigma_{i=1}^2 (\mathbf{p}^i s_i); s_i \in [0,1]; \Sigma_{i=1}^2 s_i = 1; \mathbf{p}^i \in V \}$
And for any dimension, $V = \mathbb{R}^n$, the definition holds the same, all except V changes:
$\mathbf{L} = \{ \Sigma_{i=1}^2 (\mathbf{p}^i s_i); s_i \in [0,1]; \Sigma_{i=1}^2 s_i = 1; \mathbf{p}^i \in V \}$
How about the set of a triangle?
$\mathbf{T} = \{ \mathbf{p}^1 s_1 + \mathbf{p}^2 s_2 + \mathbf{p}^3 s_3; s_1, s_2, s_3 \in [0,1]; s_1 + s_2 + s_3 = 1; \mathbf{p}^1, \mathbf{p}^2, \mathbf{p}^3 \in V \}$
$\mathbf{T} = \{ \Sigma_{i=1}^3 (\mathbf{p}^i s_i); s_i \in [0,1]; \Sigma_{i=1}^3 s_i = 1; \mathbf{p}^i \in V \}$
In fact this definition can generalize to all dimensions simplexes $\mathbf{S}_n$ in dimension $V = \mathbb{R}^n$:
$\mathbf{S}_n = \{ \Sigma_{i=1}^n (\mathbf{p}^i s_i); s_i \in [0,1]; \Sigma_{i=1}^n s_i = 1; \mathbf{p}^i \in V \}$
This is called Barycentric Coordinates.
For a line, i.e. a 1-simplex, what is the measure i.e. length?
In $\mathbb{R}^1$ the measure of a line is an interval: $|\mathbf{L}^1| = |\mathbf{S}^1_1| = |\mathbf{p}^2 - \mathbf{p}^1|$
In $\mathbb{R}^n$ the measure of a line is the L-2 norm: $|\mathbf{L}^n| = |\mathbf{S}^n_1| = \left( \Sigma_{i=1}^n ({p^2}_i - {p^1}_i)^2 \right)^{\frac{1}{2}}$
Let's offset the points so that one is at the origin.
Let $q^n_i = p^n_{i+1} - p^n_1$
Then in $\mathbb{R}^n$ the measure of a line is the L-2 norm: $|\mathbf{L}^n| = |\mathbf{S}^n_1| = \left( \Sigma_{i=1}^n ({q^1}_i)^2 \right)^{\frac{1}{2}}$
Let's revisit that line definition but for 2D:
Our line in 2D can also be defined as:
$|\mathbf{S}^2_1| = \sqrt{ ({q^1}_1)^2 + ({q^1}_2)^2 }$
As a determinant, this is also equal to:
$|\mathbf{S}^2_1| = \sqrt{\left| \begin{matrix}
q^1_1 & -q^1_2 \\
q^1_2 & q^1_1
\end{matrix} \right|}$
Notice that 2nd column is just the definition of the vector perpendicular to the 1st column.
in index notation:
$|\mathbf{S}^2_1| = \sqrt{ \epsilon^{ij} q^1_i \epsilon^{jk} q^1_k }$
...aka...
$|\mathbf{S}^2_1| = \sqrt{ \delta^{ij} q^1_i q^1_j }$
Why the double permutation? More on that later.
It just so happens to generalize to any dimension courtesy of the Hodge dual definition of the interior product:
$|\mathbf{S}^n_1| = \sqrt{ \Sigma_i ({q^1}_i)^2 }$
$|\mathbf{S}^n_1| = \sqrt{ \star (\mathbf{q}^1 \wedge \star \mathbf{q}^1) }$
$|\mathbf{S}^n_1| = \sqrt{ \langle \mathbf{q}^1, \mathbf{q}^1 \rangle }$
Let $\mathbf{Q} = \mathbf{q}^1$ for the 1D case.
Then:
$|\mathbf{S}^n_1| = \sqrt{ \langle \mathbf{Q}, \mathbf{Q} \rangle }$
For a triangle i.e. 2-simplex, what's the measure i.e. area?
In $\mathbb{R}^1$, $|\mathbf{S}^1_2|$ doesn't exist.
In $\mathbb{R}^2$ it is equal to half the length of the two sides times the sine of the angle between them.
Coincidentally this is also half the determinant of the matrix with components from the line segments of the triangle sides:
$|\mathbf{S}^2_2| = \frac{1}{2} \left| \begin{matrix}
p^2_1 - p^1_1 & p^3_1 - p^1_1 \\
p^2_2 - p^1_2 & p^3_2 - p^1_2
\end{matrix} \right|$
Using our offset coordinates, our triangle volume is:
$|\mathbf{S}^2_2| = \frac{1}{2} \left| \begin{matrix}
q^1_1 & q^2_1 \\
q^1_2 & q^2_2
\end{matrix} \right|$
$|\mathbf{S}^2_2| = \frac{1}{2} \left( q^1_1 q^2_2 - q^1_2 q^2_1 \right)$
Using the Levi-Civita permutation tensor this is equal to:
$|\mathbf{S}^2_2| = \frac{1}{2} | \epsilon^{ij} q^1_i q^2_j |$
Using differential forms this is:
$|\mathbf{S}^2_2| = \frac{1}{2} \star (\mathbf{q}^1 \wedge \mathbf{q}^2)$
Let $\mathbf{Q} = \mathbf{q}^1 \wedge \mathbf{q}^2$
Then:
$|\mathbf{S}^2_2| = \frac{1}{2} \star \mathbf{Q}$
But what about $\langle \mathbf{Q}, \mathbf{Q} \rangle$? Won't that be equal to the square of the area?
Yes.
So:
$|\mathbf{S}^2_2| = \frac{1}{2} \sqrt{\langle \mathbf{Q}, \mathbf{Q} \rangle}$
So what's the definition of a 2D triangle in 3D space? I.e. what if the triangle vertexes are 3D points?
We can still exploit the cross product definition to come up with:
$|\mathbf{S}^3_2| = \frac{1}{2} |\mathbf{q}^1 \times \mathbf{q}^2|$
This just happens to have the definition:
$|\mathbf{S}^3_2| = \frac{1}{2} \sqrt{\Sigma_{k=1}^3 \left( \epsilon^{ijk} q^1_i q^2_j \right)^2 }$
aka:
$|\mathbf{S}^3_2| = \frac{1}{2} \sqrt{ (\mathbf{q}^1 \times \mathbf{q}^2) \cdot (\mathbf{q}^1 \times \mathbf{q}^2) }$
aka, via the definition that $det(|a|b|c|) = (a \times b) \cdot c$:
$|\mathbf{S}^3_2| = \frac{1}{2} \left| \begin{matrix}
q^1_1 & q^2_1 & (\mathbf{q}^1 \times \mathbf{q}^2)_1 \\
q^1_2 & q^2_2 & (\mathbf{q}^1 \times \mathbf{q}^2)_2 \\
q^1_3 & q^2_3 & (\mathbf{q}^1 \times \mathbf{q}^2)_3
\end{matrix} \right|$
This can also be thought of as: the area of a triangle in 3D is equal to the volume of that triangle extruded along its perpendicular (which is the cross of the two base vectors of the triangle).
aka:
$|\mathbf{S}^3_2| = \frac{1}{2} \sqrt{\epsilon^{ijk} q^1_i q^2_j \epsilon^{klm} q^1_l q^2_m}$
This is also equaivalent to:
$|\mathbf{S}^3_2| = \frac{1}{2} \sqrt{
\langle
\star (\mathbf{q}^1 \wedge \mathbf{q}^2),
\star (\mathbf{q}^1 \wedge \mathbf{q}^2)
\rangle
}$
Let $\mathbf{Q} = \mathbf{q}^1 \wedge \mathbf{q}^2$
$|\mathbf{S}^3_2| = \frac{1}{2} \sqrt{
\langle
\star \mathbf{Q},
\star \mathbf{Q}
\rangle
}$
Using $\langle a, b \rangle = \sigma \langle \star a, \star b \rangle$ for metric signature $\sigma=1$ in Cartesian space:
$|\mathbf{S}^3_2| = \frac{1}{2} \sqrt{
\langle \mathbf{Q}, \mathbf{Q} \rangle
}$
3D case?
$|\mathbf{S}^3_3| = \frac{1}{3!} \left| \begin{matrix}
{q^1}_1 & {q^2}_1 & {q^3}_1 \\
{q^1}_2 & {q^2}_2 & {q^3}_2 \\
{q^1}_3 & {q^2}_3 & {q^3}_3
\end{matrix} \right|$
$|\mathbf{S}^3_3| = \frac{1}{3!} ( \mathbf{q}^1 \times \mathbf{q}^2 ) \cdot \mathbf{q}^3 $
$|\mathbf{S}^3_3| = \frac{1}{3!} \epsilon^{ijk} {q^1}_i {q^2}_j {q^3}_k$
$|\mathbf{S}^3_3| = \frac{1}{3!} \star (\mathbf{q}^1 \wedge \mathbf{q}^2 \wedge \mathbf{q}^3)$
Let $\mathbf{Q} = \mathbf{q}^1 \wedge \mathbf{q}^2 \wedge \mathbf{q}^3$
$|\mathbf{S}^3_3| = \frac{1}{3!} \star \mathbf{Q}$
Once again this is equal to the square-root of the square:
$|\mathbf{S}^3_3| = \frac{1}{3!} \sqrt{ \langle \mathbf{Q}, \mathbf{Q} \rangle }$
General Case guess:
$|\mathbf{S}^n_m| = \frac{1}{m!} \sqrt{ \star (\mathbf{Q} \wedge \star \mathbf{Q}) }$, where $\mathbf{Q} = \mathbf{q}^1 \wedge ... \wedge \mathbf{q}^m$ is a m-form.
...aka...
$|\mathbf{S}^n_m| = \frac{1}{m!} \sqrt{ \langle \mathbf{Q} , \mathbf{Q} \rangle }$
... well that looks a lot obvious, since this is the definition of the length of lines in any dimension.
$|\mathbf{S}^n_m| = \frac{1}{m!} |\mathbf{Q}|$ for the Hodge inner product (TODO what's its name?)
and in component form:
$|\mathbf{S}^n_m| = \frac{1}{m!} \sqrt{\frac{1}{m!} Q_{i_1 ... i_m} Q^{i_1 ... i_m}}$
where $\mathbf{Q} = \mathbf{q^1} \wedge ... \wedge \mathbf{q^m}$
$= {q^1}_{i_1} ... {q^m}_{i_m} dx^{i_1} \wedge ... \wedge dx^{i_m}$
$= m! {q^1}_{[i_1} ... {q^m}_{i_m]} dx^{i_1} \otimes ... \otimes dx^{i_m}$
$\mathbf{Q} = Q_{[i_1 ... i_m]} dx^{i_1} \otimes ... \otimes dx^{i_m}$
$= Q_{|i_1 ... i_m|} dx^{i_1} \wedge ... \wedge dx^{i_m}$
$= \frac{1}{m!} Q_{i_1 ... i_m} dx^{i_1} \wedge ... \wedge dx^{i_m}$
$|\mathbf{S}^n_m| = \frac{1}{m!} \sqrt{
\frac{1}{m!}
{q^1}_{[i_1} {q^2}_{i_2} ... {q^m}_{i_m]}
{q^1}_{[j_1} {q^2}_{j_2} ... {q^m}_{j_m]}
g^{i_1 j_1} ... g^{i_m j_m}
}$