$x^a \in \mathbb{R}^n$
$x^I \in \mathbb{R}^m$

Basis:
$e_a = {e_a}^I \frac{\partial}{\partial x^I}$

Tetrad:
${e_a}^I = e_a x^I = {e_a}^J \frac{\partial}{\partial x^J} x^I$
$= {e_a}^J \delta^J_I = {e_a}^I$

$\left[ {e_a}^I \right]_{aI} \in \mathbb{R}^{n \times m}$
$\left[ {e^a}_I \right]_{aI} \in \mathbb{R}^{n \times m}$

Co-Tetrad:
${e^a}_I = \frac{\partial x^a}{\partial x^I}$
${e_a}^I {e^b}_I = \delta^b_a$ if $dim(x^a) \le dim(x^I)$, otherwise it is equal to $ {\delta_{sub}}^b_a = diag(a \le m : 1, a \gt m : 0)$
${e_a}^I {e^a}_J = \delta^I_J$ if $dim(x^I) \le dim(x^a)$, otherwise it is equal to $ {\delta_{sub}}^I_J = diag(I \le n : 1, I \gt n : 0)$
The co-tetrad does not form an inverse with the tetrad unless the dimension of the coordinate space and the embedded space are equal.

For a holonomic basis the tetrad represents a coordinate chart:
${e_a}^I = \frac{\partial x^I}{\partial x^a}$
${e^a}_I = \frac{\partial x^a}{\partial x^I}$
${{e_a}^I}_{,b} = \frac{\partial}{\partial x^b} \frac{\partial x^I}{\partial x^a} = \frac{\partial x^I}{\partial x^b\partial x^a} = \frac{\partial x^I}{\partial x^a\partial x^b} = \frac{\partial}{\partial x^a} \frac{\partial x^I}{\partial x^b} = \frac{\partial}{\partial x^a} {e_b}^I = {{e_b}^I}_{,a}$
Therefore, for holonomic basis, ${{e_a}^I}_{,b} = {{e_{(a}}^I}_{,b)}$

Comma derivatives:
$T_{,a} = e_a (T) = {e_a}^I \frac{\partial}{\partial x^I} T$
This is not equal to $\frac{\partial}{\partial x^a}$ ... ?
or is it that it is equal, and $\frac{\partial}{\partial x^a}$ is not a partial derivative but instead a function ${e_a}$ which is a linear combination (via ${e_a}^I$) of functions of partial derivatives?
MTW uses comma derivatives and applications of $e_a$ to denote anholonomic basis operators, but I think it avoids using $\frac{\partial}{\partial x^a}$ - probably for ambiguity's sake.
Maybe I should revise/generalize my definition of $e_a$ as well?
So $T_{,a} = e_a (T)$ is not equal to $\frac{\partial}{\partial x^a} T$. In the case of an anholonomic basis should not ever be represented as this.
Though for a holonomic basis interchanging $T_{,a}$ and $\frac{\partial}{\partial x^a} T$ is fine.

Change of basis of ${e_a}^I$:
${e_a}^J {{e_b}^I}_{,J} = {e_a}^J \frac{\partial}{\partial x^J} {e_b}^I = {e_a}( {e_b}^I ) = {{e_b}^I}_{,a}$

Minkowski metric:
$[\eta_{IJ}]_{IJ} \in \mathbb{R}^{m \times m}$
$\eta_{IJ} = \eta^{IJ} = diag(-1, 1, ...)$
$\eta^{IJ}$ and $\eta_{IJ}$ raise and lower $IJK...$

Metric:
$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$
Metric inverse:
$g^{ab} = {e^a}_I {e^b}_J \eta^{IJ}$

$g^{ab}$ and $g_{ab}$ raise and lower $abc...$

The metric is, by definition, two tetrads combined.
Regardless of dimensions of coordinate or embedded space.
Therefore $\left[ g_{ab} \right]_{ab}, \left[ g^{ab} \right]_{ab} \in \mathbb{R}^{min(m,n) \times min(m,n)}$.
From this, all $a-z$ indexes in all objects that are a product of $g_{ab}$ or $g^{ab}$ are also going to be of dimension $min(m,n)$.
Nearly everything will be of dimension $min(m,n)$, with the exception of the vector $x^a$ itself, the tetrads ${e_a}^I$ and ${e^a}_I$, and any application of them via $e_a$ (i.e. "$_{,a}$").

Notice, due to the possible differences in dimensions, it is not always the case that:
${e^a}_I = g^{ab} {e_b}^J \eta_{IJ}$
${e_a}^I = g_{ab} {e^b}_J \eta^{IJ}$
This is true only when the dimension of index $a$ in $g_{ab}$, which is $min(m,n)$, is greater than or equal to the dimension of index $a$ is ${e_a}^I$, which is $n = dim(x^a)$ . So only for $m \ge n$.

Minkowski metric in relation to tetrad inverse definitions:
$(\delta_{n})^K_J = {e_b}^K {e^b}_J$
$(\delta_{n})^K_J = e^{aK} {e^b}_J g_{ab}$
$\eta_{IK} (\delta_{n})^K_J = \eta_{IK} e^{aK} {e^b}_J g_{ab}$
$\eta_{IK} (\delta_{n})^K_J = {e^a}_I {e^b}_J g_{ab}$

$(\delta_{n})^J_K = {e^a}_K {e_a}^J $
$(\delta_{n})^J_K = e_{aK} {e_a}^J g^{ab} $
$\eta^{IK} (\delta_{n})^J_K = \eta^{IK} e_{aK} {e_a}^J g^{ab} $
$\eta^{IK} (\delta_{n})^J_K = {e_a}^I {e_a}^J g^{ab} $
The Minkowski metric is only recovered from two co-tetrads if $n \ge m$, i.e. $dim(x^a) \ge dim(x^I)$

Commutation coefficients (definition from MTW):
${c_{ab}}^c e_c = [e_a, e_b]$
${c_{ab}}^c {e_c}^K \frac{\partial}{\partial x^K} = {e_a}^I \frac{\partial}{\partial x^I}( {e_b}^J \frac{\partial}{\partial x^J} ) - {e_b}^I \frac{\partial}{\partial x^I}( {e_a}^J \frac{\partial}{\partial x^J})$
${c_{ab}}^c {e_c}^K \frac{\partial}{\partial x^K} = {e_a}^I {{e_b}^J}_{,I} \frac{\partial}{\partial x^J} + {e_a}^I {e_b}^J \frac{\partial^2}{\partial x^I \partial x^J} - {e_b}^I {{e_a}^J}_{,I} \frac{\partial}{\partial x^J} - {e_b}^I {e_a}^J \frac{\partial^2}{\partial x^I \partial x^J}$
${c_{ab}}^c {e_c}^K \frac{\partial}{\partial x^K} = 2 {e_{[a}}^I {{e_{b]}}^J}_{,I} \frac{\partial}{\partial x^J}$
Apply both sides to $x^J$:
${c_{ab}}^c {e_c}^J = 2 {e_{[a}}^I {{e_{b]}}^J}_{,I}$
By substituting the definition of $e_a = _{,a}$ we find:
${c_{ab}}^c {e_c}^J = 2 {{e_{[b}}^J}_{,a]}$,
we can apply the co-tetrad to index $J$ to find:
${c_{ab}}^c {e_c}^J {e^d}_J = 2 {e^d}_J {{e_{[b}}^J}_{,a]}$,
In the case $dim(x^I) \lt dim(x^a)$ we cannot recover all of the range:
${c_{ab}}^c {\delta_{sub}}^d_c = 2 {e^d}_J {{e_{[b}}^J}_{,a]}$,
So ${c_{ab}}^c$ has $n-min(m,n)$ free parameters.
Otherwise we do:
${c_{ab}}^d = {c_{ab}}^c \delta^d_c = 2 {e^d}_J {{e_{[b}}^J}_{,a]} $
Even in the case that $dim(x^I) \lt dim(x^a)$, we can define ${c_{ab}}^c$ in the same manner and simply acknowledge that the dimension of the index $c$ is only $m = dim(x^I) \lt n = dim(x^a)$.
Therefore, for ${c_{ab}}^c$, the dimension of index $c$ is $min(m,n)$ for $m = dim(x^I)$ and $n = dim(x^a)$.

Define ${c_{ab}}^I = 2 {{e_{[b}}^I}_{,a]}$
Note that the range of index $I$ could exceed $dim(x^a)$.
We can still define ${c_{ab}}^c = {c_{ab}}^I {e^c}_I$ and, for all cases, know that the range of index $c$ is $min(m,n)$.
This is one way to define the $n-min(m,n)$ free parameters.

Christoffel symbols (definition from MTW):
$\Gamma_{abc} = \frac{1}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a} + c_{abc} + c_{acb} - c_{bca})$
Notice the difference in dimension for index $c$ between $g_{ab,c}$ and $c_{abc}$.
For $dim(x^I) \lt dim(x^a)$, the $c$ index in $g_{ab,c}$ is $dim(x^a)$, while the $c$ index in $c_{abc}$ is $dim(x^I)$.
Substitute definitions of $g_{ab}$ and $c_{abc} = {c_{ab}}^d g_{dc}$:
$\Gamma_{abc} = \frac{1}{2} \left[ ({e_a}^I {e_b}^J \eta_{IJ})_{,c} + ({e_a}^I {e_c}^J \eta_{IJ})_{,b} - ({e_b}^I {e_c}^J \eta_{IJ})_{,a} \right] + g_{cd} {e^d}_I {{e_{[b}}^I}_{,a]} + g_{bd} {e^b}_I {{e_{[c}}^I}_{,a]} - g_{ad} {e^a}_I {{e_{[c}}^I}_{,b]}$
Distribute comma derivatives:
$\Gamma_{abc} = \frac{1}{2} ( {{e_a}^I}_{,c} e_{bI} + {{e_b}^I}_{,c} e_{aI} + {{e_a}^I}_{,b} e_{cI} + {{e_c}^I}_{,b} e_{aI} - {{e_b}^I}_{,a} e_{cI} - {{e_c}^I}_{,a} e_{bI} + {{e_b}^I}_{,a} e_{cI} - {{e_a}^I}_{,b} e_{cI} + {{e_c}^I}_{,a} e_{bI} - {{e_a}^I}_{,c} e_{bI} - {{e_c}^I}_{,b} e_{aI} + {{e_b}^I}_{,c} e_{aI} )$
Cancel.
But can we cancel? Isn't there a ${\delta_{sub}}^a_b$ somewhere in there?
There is only a ${\delta_{sub}}^a_b$ within the equation defining ${c_{ab}}^c$, leaving $n-min(m,n)$ parameters free.
$\Gamma_{abc} = e_{aI} {{e_b}^I}_{,c}$

Connection coefficients (definition from MTW):
${\Gamma^a}_{bc} = g^{ad} \Gamma_{dbc}$
substitute...
${\Gamma^a}_{bc} = g^{ad} e_{dI} {{e_b}^I}_{,c}$
When $dim(x^a) \ge dim(x^I)$ we can replace the product of inverse metric and tetrad with the co-tetrad:
${\Gamma^a}_{bc} = {e^a}_I {{e_b}^I}_{,c}$
Define ${\Gamma^I}_{bc} = {{e_b}^I}_{,c}$
so ${\Gamma^a}_{bc} = g^{ad} e_{dI} {\Gamma^I}_{bc}$.
And when $m = n$:
${\Gamma^a}_{bc} = {e^a}_I {{e_b}^I}_{,c}$

Antisymmetric component of connection coefficients (MTW):
${\Gamma^a}_{[bc]} = g^{ad} e_{dI} {{e_{[b}}^I}_{,c]}$
for $dim(x^a) \ge dim(x^I)$ this equals ${e^a}_I {{e_{[b}}^I}_{,c]}$
and for $dim(x^I) \ge dim(x^a)$ as well -- i.e. only for $dim(x^a) = dim(x^I)$ -- this equals $\frac{1}{2} {c_{cb}}^a$

Spin connection definition from Wikipedia and from Yepez - "Einstein's vierbein field theory of curved space":
${\Gamma^v}_{ua} = {e^v}_I {{e_a}^I}_{,u} + {e^v}_I {{\omega_u}^I}_J {e_a}^J$
${{\omega_u}^I}_J = {e_v}^I {\Gamma^v}_{ua} {e^a}_J - {e^a}_J {{e_a}^I}_{,u}$
Writing coordinate changes into the indexes:
${\Gamma^a}_{bc} = {e^a}_I {{e_c}^I}_{,b} + {{\omega_b}^a}_c$
Rearrange:
${\Gamma^a}_{bc} - {{\omega_b}^a}_c = {e^a}_I {{e_c}^I}_{,b}$
From here I'm willing to bet that their ${\Gamma^a}_{bc}$ is symmetric on $bc$.
That would mean $\omega_{bac}$ would need to be symmetric on $bc$ as well.
However, by definition of cancelling Minkowski metric, it appears that $\omega_{bac}$ is symmetric on $ac$.
Hmm, maybe it is antisymmetric on $bc$ too?

To boot, Yepez's paper also says that torsion is the antisymmetric component of ${\Gamma^a}_{bc}$ on $bc$. So they are not purely symmetric.
In fact, in Yepez's paper, what is the spin connection with respect to the torsion?
${\Gamma^a}_{[bc]} = \frac{1}{2} {T^a}_{bc} = {e^a}_I {{e_{[c}}^I}_{,b]} + {{\omega_{[b}}^a}_{c]}$
$\frac{1}{2} {T^a}_{bc} - {{\omega_{[b}}^a}_{c]} = {e^a}_I {{e_{[c}}^I}_{,b]}$
Now the right side of this equation is equal to the above definition of ${c_{bc}}^a$.
Maybe this means that (double) the torsion and the spin connection are two separate parts of the antisymmetric component of the tetrad derivative?
Does this mean the two are arbitrarily defined?
For some arbitrary choice of spin connection, the torsion is the difference between it and the antisymmetric component of the tetrad derivative?

This isn't true, according to Wikipedia's "Torsion tensor" entry, which says (reinterpreted to match the conventions of this paper) ${T^a}_{bc} = 2 {\Gamma^a}_{[bc]} - {c_{cb}}^a$.
Notice that, above, by MTW's definition of ${\Gamma^a}_{bc}$, we find ${\Gamma^a}_{[bc]} = \frac{1}{2} {c_{cb}}^a$.
This means that, throughout MTW's book, ${T^a}_{bc}$ is always zero.
This also means that torsion is something defined distinct from ${e_a}^I$. According to Wikipedia's torsion page, that is. Yepez never mentions commutation coefficients, and therefore torsion can safely be defined in his paper as the difference between the arbitrary spin connection and the certain antisymmetric portion of the derivative of the tetrad.

And of course 'contorsion' is the value, based on the torsion, that returns us to a torsion-free connection.
Wikipedia's 'contorsion tensor' page defines this as: $K_{kij} = \frac{1}{2} ( T_{kij} + T_{jki} - T_{ijk} )$
If you add contorsion to the connection does it cancel the spin connection only, leaving you with the symmetric connection?
Or does it cancel additional terms of the symmetric connection?

Spin connection cancelling Minkowski metric:
$0 = D_a \eta_{IJ} = \eta_{IJ,a} - {{\omega_a}^K}_I \eta_{KJ} - {{\omega_a}^K}_J \eta_{KI}$
$\omega_{aIJ} = -\omega_{aJI}$
$\omega_{aIJ} = \omega_{a[IJ]}$

Defining spin connections by a covariant derivative that cancels tetrad and only applies spin connection to Minkowski indexes:
$D_b {e_a}^I = 0 = {{e_a}^I}_{,b} + {{\omega_b}^I}_J {e_a}^J$
${{\omega_b}^I}_a = -{{e_a}^I}_{,b}$
${{\omega_b}^I}_a = -{\Gamma^I}_{ab}$
$\omega_{bca} = -\Gamma_{cab}$
But if $\omega_{aIJ} = -\omega_{aJI}$
then does $\Gamma_{cab} = -\Gamma_{acb}$?
$\Gamma_{[ca]b} = {e_{[c}}^I e_{a]I,b} \ne {e_c}^I e_{aI,b}$
So this definition doesn't work with the spin connection cancelling the Minkowski metric.

Defining spin connections by a covariant derivative that cancels tetrad, applies spin connection to Minkowski indexes, and connection coefficients to parameter space indexes, considering a non-symmetric connections:
$D_b {e_a}^I = 0 = {{e_a}^I}_{,b} - {\Gamma^c}_{ab} {e_c}^I + {{\omega_b}^I}_J {e_a}^J$
${{\omega_b}^I}_a = {\Gamma^I}_{ab} - {{e_a}^I}_{,b}$
${{\omega_b}^I}_a = {{e_a}^I}_{,b} - {{e_a}^I}_{,b}$
${{\omega_b}^I}_a = 0$
This definition doesn't give us anything useful.

Spin connection, if connections were symmetric yet the tetrad was mysetriously somehow still anholonomic:
$D_b {e_a}^I = 0 = {{e_a}^I}_{,b} - {\Gamma^c}_{ab} {e_c}^I + {{\omega_b}^I}_J {e_a}^J$
${{\omega_b}^I}_a = {\Gamma^I}_{ab} - {{e_a}^I}_{,b}$
${{\omega_b}^I}_a = {{e_{(a}}^I}_{,b)} - {{e_a}^I}_{,b}$
${{\omega_b}^I}_a = {{e_{[b}}^I}_{,a]} = \frac{1}{2} {c_{ab}}^I$
This makes sense, but is redundant with the definition of the commutation coefficients.
Its origins are also a bit suspicious, with the connection being symmetric despite an anholonomic basis.
It looks like folks who use spin connection are considering a tetrad that does not communite (i.e. ${{e_a}^I}_{,b} \ne {{e_{[a}}^I}_{,b]}$), therefore does not represent a coordinate basis. This is the case for anholonomic coordinate systems, which MTW's definition of connections says anholonomic coordinate systems imply a not-necessarily-symmetric connection. However the spin-connection crowd seems to still define the connection to be symmetric (meaning the coordinate system may still yet be holonomic, but there may be a separate component to the geodesic's asymmetry -- maybe the torsion?) and puts the asymmetric portion of the geodesic into the spin connection.
In summary, where MTW would have a single no-constraints tensor ${\Gamma^a}_{bc}$, the spin-conn crowd would break that into symmetric ${\Gamma^a}_{bc}$ and antisymmetric ${{\omega_c}^a}_b$.

Torsion (Wikipedia):
${T^a}_{bc} = 2 {\Gamma^a}_{[bc]} - {c_{cb}}^a$
substituting the MTW definition of connection and commutation coefficients:
${T^a}_{bc} = {c_{cb}}^a - {c_{cb}}^a$
${T^a}_{bc} = 0$
so I'm betting torsion is missing from my MTW definition of connection coefficients.

So what is the difference between spin connections, commutation coefficients, and torsion?

${\delta^a_b}_{,c} = 0$
$({e_a}^I {e^b}_I)_{,c} = 0$
${{e_a}^I}_{,c} {e^b}_I + {e_a}^I {e^b}_{I,c} = 0$

${\delta^I_J}_{,a} = 0$
$({e_b}^I {e^b}_J)_{,a} = 0$
${{e_b}^I}_{,a} {e^b}_J + {e_b}^I {e^b}_{J,a} = 0$
${e^c}_{J,a} = -{e^c}_I {{e_b}^I}_{,a} {e^b}_J$

Riemann curvature tensor:
${R^a}_{bcd} = {\Gamma^a}_{bd,c} - {\Gamma^a}_{bc,d} + {\Gamma^a}_{ec} {\Gamma^e}_{bd} - {\Gamma^a}_{ed} {\Gamma^e}_{bc} - {\Gamma^a}_{be} {c_{cd}}^e$
Substitute definitions:
${R^a}_{bcd} = ({e^a}_I {{e_b}^I}_{,d})_{,c} - ({e^a}_I {{e_b}^I}_{,c})_{,d} + {e^a}_I {{e_e}^I}_{,c} {e^e}_J {{e_b}^J}_{,d} - {e^a}_I {{e_e}^I}_{,d} {e^e}_J {{e_b}^J}_{,c} - 2 {e^a}_I {{e_b}^I}_{,e} {e^e}_I {{e_{[d}}^I}_{,c]}$
Distribute derivatives:
${R^a}_{bcd} = {{e^a}_I}_{,c} {{e_b}^I}_{,d} + {e^a}_I {{e_b}^I}_{,dc} - {{e^a}_I}_{,d} {{e_b}^I}_{,c} - {e^a}_I {{e_b}^I}_{,cd} + {e^a}_I {{e_e}^I}_{,c} {e^e}_J {{e_b}^J}_{,d} - {e^a}_I {{e_e}^I}_{,d} {e^e}_J {{e_b}^J}_{,c} - {e^a}_I {{e_b}^I}_{,e} {e^e}_J {{e_d}^J}_{,c} + {e^a}_I {{e_b}^I}_{,e} {e^e}_J {{e_c}^J}_{,d} $
Cancel out second derivatives:
${R^a}_{bcd} = {{e^a}_I}_{,c} {{e_b}^I}_{,d} - {{e^a}_I}_{,d} {{e_b}^I}_{,c} + {e^a}_I {{e_e}^I}_{,c} {e^e}_J {{e_b}^J}_{,d} - {e^a}_I {{e_e}^I}_{,d} {e^e}_J {{e_b}^J}_{,c} - {e^a}_I {{e_b}^I}_{,e} {e^e}_J {{e_d}^J}_{,c} + {e^a}_I {{e_b}^I}_{,e} {e^e}_J {{e_c}^J}_{,d} $
Use ${{e_a}^I}_{,c} {e^b}_I = -{e_a}^I {e^b}_{I,c}$:
${R^a}_{bcd} = {e^a}_{I,c} {{e_b}^I}_{,d} - {e^a}_{I,d} {{e_b}^I}_{,c} - {e^a}_{I,c} {e_e}^I {e^e}_J {{e_b}^J}_{,d} + {e^a}_{I,d} {e_e}^I {e^e}_J {{e_b}^J}_{,c} + {e^a}_{I,e} {e_b}^I {e^e}_J {{e_d}^J}_{,c} - {e^a}_{I,e} {e_b}^I {e^e}_J {{e_c}^J}_{,d} $
Use ${e_e}^I {e^e}_J = \delta^I_J$ and cancel like terms:
${R^a}_{bcd} = {e^a}_{I,e} {e_b}^I {{e_d}^J}_{,c} {e^e}_J - {e^a}_{I,e} {e_b}^I {{e_c}^J}_{,d} {e^e}_J$
Use ${{e_a}^I}_{,c} {e^b}_I = -{e_a}^I {e^b}_{I,c}$ again:
${R^a}_{bcd} = {e^a}_{I} {{e_b}^I}_{,e} {{e_c}^J}_{,d} {e^e}_J - {e^a}_I {{e_b}^I}_{,e} {{e_d}^J}_{,c} {e^e}_J $
Factor:
${R^a}_{bcd} = {e^a}_{I} {{e_b}^I}_{,e} ( {{e_c}^J}_{,d} {e^e}_J - {{e_d}^J}_{,c} {e^e}_J)$
${R^a}_{bcd} = 2 {e^a}_{I} {{e_b}^I}_{,e} {e^e}_J {{e_{[c}}^J}_{,d]}$
...and where did all the terms go?
${R^a}_{bcd} = -{\Gamma^a}_{be} {c_{cd}}^e$ ...?
Something tells me that they wouldn't have gone away if I had defined the commutation coefficients otherwise.

$R_{ab} = {R^u}_{aub} = {{e_a}^I}_{,v} {e^u}_{I} ( {{e_u}^J}_{,b} - {{e_b}^J}_{,u} ) {e^v}_J$

$R = g^{ab} R_{ab} = {e^a}_K e^{bK} {{e_a}^I}_{,v} {e^u}_{I} ( {{e_u}^J}_{,b} - {{e_b}^J}_{,u} ) {e^v}_J$

$G_{ab} = R_{ab} - \frac{1}{2} g_{ab} R$
$