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Consider the basis $e_i = {e^a}_i e_a$ for some coordinate basis $e_a = \partial_a$
such that $e_i \cdot e_j = \eta_{ij}$. Therefore ${e^a}_i$ diagonalizes our metric $g_{ab} = e_a \cdot e_b$.
Mind you there are more than one diagonalization, and therefore more than one possible ${e^a}_i$ which transforms $g_{ab}$ to $\eta_{ij}$.

TODO show that $(e')_i = {\Lambda^j}_i e_j$ retains all properties for Lorentz transform ${\Lambda^j}_i$, (TODO which properties?)
so $e_i$ is gauge invariant wrt ${\Lambda^j}_i$

Let $(\hat{\nabla}_\eta)_i$ be defined as the covariant derivative associated with the metric $\eta_{ij}$ with connection possessing zero torsion, so it's the unique Levi-Civita connection of a diagonalized, (probably) non-coordinate basis.

Let ${(\hat{\Gamma}_\eta)^i}_{jk}$ be the zero-torsion Levi-Civita connection associated with the covariant derivative $(\hat{\nabla}_\eta)_i$.

${(\hat{\Gamma}_\eta)^i}_{jk}$ is defined by $(\hat\nabla_\eta)_i e_j = {(\hat{\Gamma}_\eta)^k}_{ij} e_k$.

The Levi-Civita connection with a constant metric (such as our diagonalized $e_i = {e^a}_i e_a)$:
$(\hat{\Gamma}_\eta)_{ijk} = \frac{1}{2} ( e_k (\eta_{ij}) + e_j (\eta_{ik}) - e_i (\eta_{jk}) + (c_\eta)_{ijk} + (c_\eta)_{ikj} - (c_\eta)_{kji} )$
$= \frac{1}{2} ( (c_\eta)_{ijk} + (c_\eta)_{ikj} - (c_\eta)_{kji} )$
$c_\eta$ is the commutation associated with our diagonalized $e_i$ operators.

Notice that the Levi-Civita connection of a coordinate basis is going to have structure constants $\hat{c}_{abc} = 0$
But the structure constants of the basis diagonalizing such a coordinate system are definitely not zero, as shown before:
${(c_\eta)_{ij}}^k = 2 {e^u}_{[i|} e_u ( {e^c}_{|j]} ) {e^k}_c$
${(c_\eta)_{ij}}^k = -2 {e^u}_{[i} {e^c}_{j]} e_u ( {e^k}_c )$
${(c_\eta)_{pq}}^r = -2 e_{[p} ( {e_{q]}}^k ) {e^r}_k$
${(c_\eta)_{pq}}^r = 2 e_{[p} ( e^{rk} ) e_{q]k}$
$(c_\eta)_{abc} = 2 e_{[a} ( e^{dk} ) e_{b]k} g_{dc}$

Does the Levi-Civita connection of the basis associated with the diagonalized metric equate to the Levi-Civita connection of the coordinate basis?
$(\hat{\Gamma}_\eta)_{abc} = \frac{1}{2} ( (c_\eta)_{abc} + (c_\eta)_{acb} - (c_\eta)_{cba})$
$= e_{[a} ( e^{dk} ) e_{b]k} g_{dc} + e_{[a} ( e^{dk} ) e_{c]k} g_{db} - e_{[c} ( e^{dk} ) e_{b]k} g_{da} $
$= \frac{1}{2} ( e_a ( e^{dk} ) e_{bk} g_{dc} - e_b ( e^{dk} ) e_{ak} g_{dc} + e_a ( e^{dk} ) e_{ck} g_{db} - e_c ( e^{dk} ) e_{ak} g_{db} - e_c ( e^{dk} ) e_{bk} g_{da} + e_b ( e^{dk} ) e_{ck} g_{da} )$
$= \frac{1}{2} ( - {e_c}^k e_a ( e_{bk} ) - e_{bk} e_a ( {e_c}^k ) + {e_c}^k e_b ( e_{ak} ) + e_b ( {e_a}^k ) e_{ck} + {e_b}^k e_c ( e_{ak} ) + {e_a}^k e_c ( e_{bk} ) )$
$= \frac{1}{2} ( - e_a ( {e_c}^k e_{bk} ) + e_b ( {e_a}^k e_{ck} ) + e_c ( {e_a}^k e_{bk} ) )$
$ = \frac{1}{2} (e_c (g_{ab}) + e_b(g_{ac}) - e_c(g_{ab}))$
$ = \hat{\Gamma}_{abc}$
Yup, looks like $\hat{\Gamma}_{abc} = (\hat{\Gamma}_\eta)_{abc}$
(TODO is this compatible with the change-of-basis rule of connections? Or does the change-of-basis rule even apply to anholonomic basis Levi-Civita connections?)

This can also be seen in my relation between Levi-Civita of a coordinate basis and Levi-Civita of an arbitrary non-coordinate basis:
${\hat\Gamma^a}_{bc} = {e^a}_\tilde{a} {e_b}^\tilde{b} {e_c}^\tilde{c} {\tilde{\Gamma}^\tilde{a}}_{\tilde{b}\tilde{c}} + {e^a}_\tilde{a} e_b ({e_c}^\tilde{a})$
But that would imply 'no', right? Only 'yes' when the linear transform from coordinate to non-coordinate was constant. TODO double check and fixme plz?

Levi-Civita connection in a non-coordinate basis:
If $e_a = {e_a}^{\tilde{a}} \partial_\tilde{a}$ and $e^a = {e^a}_{\tilde{a}} dx^\tilde{a}$ are linear combinations of a coordinate basis $\partial_\tilde{a}$ and dual $dx^\tilde{a}$ then we can calculate the following:
Using the identities that $\delta^a_b = {e^a}_\tilde{u} {e_b}^\tilde{u}$ and $\delta^\tilde{a}_\tilde{b} = {e_u}^\tilde{a} {e^u}_\tilde{b}$
Let $\tilde{g}_{\tilde{a}\tilde{b}} = \partial_\tilde{a} \cdot \partial_\tilde{b}$
Let $g_{ab} = e_a \cdot e_b = {e_a}^\tilde{a} \partial_\tilde{a} \cdot {e_b}^\tilde{b} \partial_\tilde{b} = {e_a}^\tilde{a} \cdot {e_b}^\tilde{b} \tilde{g}_{\tilde{a}\tilde{b}}$.
Let ${c_{ab}}^c e_c = [e_a, e_b]$, so (from the 'structure constants' worksheet on structure constants of a linear transform of a coordinate basis) ${c_{ab}}^c = 2 e_{[a} ({e_{b]}}^\tilde{c}) {e^c}_\tilde{c}$
$\hat\Gamma_{abc} = \frac{1}{2} ( e_c (g_{ab}) + e_b (g_{ac}) - e_a (g_{bc}) + c_{acb} + c_{abc} - c_{cba} )$
$= \frac{1}{2} ( e_c ({e_a}^\tilde{a} {e_b}^\tilde{b} \tilde{g}_{\tilde{a}\tilde{b}}) + e_b ({e_a}^\tilde{a} {e_c}^\tilde{c} \tilde{g}_{\tilde{a}\tilde{c}}) - e_a ({e_b}^\tilde{b} {e_c}^\tilde{c} \tilde{g}_{\tilde{b}\tilde{c}}) + 2 e_{[a} ({e_{c]}}^\tilde{d}) {e^d}_\tilde{d} g_{db} + 2 e_{[a} ({e_{b]}}^\tilde{d}) {e^d}_\tilde{d} g_{dc} - 2 e_{[c} ({e_{b]}}^\tilde{d}) {e^d}_\tilde{d} g_{da} )$
$= \frac{1}{2} ( e_c ({e_a}^\tilde{a}) {e_b}^\tilde{b} \tilde{g}_{\tilde{a}\tilde{b}} + {e_a}^\tilde{a} e_c ({e_b}^\tilde{b}) \tilde{g}_{\tilde{a}\tilde{b}} + {e_a}^\tilde{a} {e_b}^\tilde{b} e_c (\tilde{g}_{\tilde{a}\tilde{b}}) + e_b ({e_a}^\tilde{a}) {e_c}^\tilde{c} \tilde{g}_{\tilde{a}\tilde{c}} + {e_a}^\tilde{a} e_b ({e_c}^\tilde{c}) \tilde{g}_{\tilde{a}\tilde{c}} + {e_a}^\tilde{a} {e_c}^\tilde{c} e_b (\tilde{g}_{\tilde{a}\tilde{c}}) - e_a ({e_b}^\tilde{b}) {e_c}^\tilde{c} \tilde{g}_{\tilde{b}\tilde{c}} - {e_b}^\tilde{b} e_a ({e_c}^\tilde{c}) \tilde{g}_{\tilde{b}\tilde{c}} - {e_b}^\tilde{b} {e_c}^\tilde{c} e_a (\tilde{g}_{\tilde{b}\tilde{c}}) + e_a ({e_c}^\tilde{d}) e_{b\tilde{d}} - e_c ({e_a}^\tilde{d}) e_{b\tilde{d}} + e_a ({e_b}^\tilde{d}) e_{c\tilde{d}} - e_b ({e_a}^\tilde{d}) e_{c\tilde{d}} - e_c ({e_b}^\tilde{d}) e_{a\tilde{d}} + e_b ({e_c}^\tilde{d}) e_{a\tilde{d}} )$
$= \frac{1}{2} ( e_c ({e_a}^\tilde{a}) e_{b\tilde{a}} + e_c ({e_b}^\tilde{b}) e_{a\tilde{b}} + {e_a}^\tilde{a} {e_b}^\tilde{b} e_c (\tilde{g}_{\tilde{a}\tilde{b}}) + e_b ({e_a}^\tilde{a}) e_{c\tilde{a}} + e_b ({e_c}^\tilde{c}) e_{a\tilde{c}} + {e_a}^\tilde{a} {e_c}^\tilde{c} e_b (\tilde{g}_{\tilde{a}\tilde{c}}) - e_a ({e_b}^\tilde{b}) e_{c\tilde{b}} - e_a ({e_c}^\tilde{c}) e_{b\tilde{c}} - {e_b}^\tilde{b} {e_c}^\tilde{c} e_a (\tilde{g}_{\tilde{b}\tilde{c}}) + e_a ({e_c}^\tilde{d}) e_{b\tilde{d}} - e_c ({e_a}^\tilde{d}) e_{b\tilde{d}} + e_a ({e_b}^\tilde{d}) e_{c\tilde{d}} - e_b ({e_a}^\tilde{d}) e_{c\tilde{d}} - e_c ({e_b}^\tilde{d}) e_{a\tilde{d}} + e_b ({e_c}^\tilde{d}) e_{a\tilde{d}} )$
$= \frac{1}{2} ( {e_a}^\tilde{a} {e_b}^\tilde{b} {e_c}^\tilde{c} \partial_\tilde{c} (\tilde{g}_{\tilde{a}\tilde{b}}) + {e_a}^\tilde{a} {e_c}^\tilde{c} {e_b}^\tilde{b} \partial_\tilde{b} (\tilde{g}_{\tilde{a}\tilde{c}}) - {e_b}^\tilde{b} {e_c}^\tilde{c} {e_a}^\tilde{a} \partial_\tilde{a} (\tilde{g}_{\tilde{b}\tilde{c}}) ) + e_b ({e_c}^\tilde{c}) e_{a\tilde{c}} $
$\hat\Gamma_{abc} = {e_a}^\tilde{a} {e_b}^\tilde{b} {e_c}^\tilde{c} \tilde{\Gamma}_{\tilde{a}\tilde{b}\tilde{c}} + e_{a\tilde{a}} e_b ({e_c}^\tilde{a})$
${\hat\Gamma^a}_{bc} = {e^a}_\tilde{a} {e_b}^\tilde{b} {e_c}^\tilde{c} {\tilde{\Gamma}^\tilde{a}}_{\tilde{b}\tilde{c}} + {e^a}_\tilde{a} e_b ({e_c}^\tilde{a})$
...where ${\tilde{\Gamma}^\tilde{a}}_{\tilde{b}\tilde{c}}$ is the Levi-Civita connection associated with the coordinate basis $\partial_\tilde{a}$.

... now lets substitute these identities with $g_{ab} = \eta_{ab}$ and $c_{abc} = (c_\eta)_{abc}$

$\hat\Gamma_{abc} = \frac{1}{2} ( (c_\eta)_{acb} + (c_\eta)_{abc} - (c_\eta)_{cba} )$
$= \frac{1}{2} ( e_c (\eta_{ab}) + e_b (\eta_{ac}) - e_a (\eta_{bc}) + (c_\eta)_{acb} + (c_\eta)_{abc} - (c_\eta)_{cba} )$
$= \frac{1}{2} ( e_c ({e_a}^\tilde{a} {e_b}^\tilde{b} \tilde{g}_{\tilde{a}\tilde{b}}) + e_b ({e_a}^\tilde{a} {e_c}^\tilde{c} \tilde{g}_{\tilde{a}\tilde{c}}) - e_a ({e_b}^\tilde{b} {e_c}^\tilde{c} \tilde{g}_{\tilde{b}\tilde{c}}) + 2 e_{[a} ({e_{c]}}^\tilde{d}) {e^d}_\tilde{d} \eta_{db} + 2 e_{[a} ({e_{b]}}^\tilde{d}) {e^d}_\tilde{d} \eta_{dc} - 2 e_{[c} ({e_{b]}}^\tilde{d}) {e^d}_\tilde{d} \eta_{da} )$
$= \frac{1}{2} ( e_c ({e_a}^\tilde{a}) {e_b}^\tilde{b} \tilde{g}_{\tilde{a}\tilde{b}} + {e_a}^\tilde{a} e_c ({e_b}^\tilde{b}) \tilde{g}_{\tilde{a}\tilde{b}} + {e_a}^\tilde{a} {e_b}^\tilde{b} e_c (\tilde{g}_{\tilde{a}\tilde{b}}) + e_b ({e_a}^\tilde{a}) {e_c}^\tilde{c} \tilde{g}_{\tilde{a}\tilde{c}} + {e_a}^\tilde{a} e_b ({e_c}^\tilde{c}) \tilde{g}_{\tilde{a}\tilde{c}} + {e_a}^\tilde{a} {e_c}^\tilde{c} e_b (\tilde{g}_{\tilde{a}\tilde{c}}) - e_a ({e_b}^\tilde{b}) {e_c}^\tilde{c} \tilde{g}_{\tilde{b}\tilde{c}} - {e_b}^\tilde{b} e_a ({e_c}^\tilde{c}) \tilde{g}_{\tilde{b}\tilde{c}} - {e_b}^\tilde{b} {e_c}^\tilde{c} e_a (\tilde{g}_{\tilde{b}\tilde{c}}) + e_a ({e_c}^\tilde{d}) e_{b\tilde{d}} - e_c ({e_a}^\tilde{d}) e_{b\tilde{d}} + e_a ({e_b}^\tilde{d}) e_{c\tilde{d}} - e_b ({e_a}^\tilde{d}) e_{c\tilde{d}} - e_c ({e_b}^\tilde{d}) e_{a\tilde{d}} + e_b ({e_c}^\tilde{d}) e_{a\tilde{d}} )$
$= \frac{1}{2} ( e_c ({e_a}^\tilde{a}) e_{b\tilde{a}} + e_c ({e_b}^\tilde{b}) e_{a\tilde{b}} + {e_a}^\tilde{a} {e_b}^\tilde{b} e_c (\tilde{g}_{\tilde{a}\tilde{b}}) + e_b ({e_a}^\tilde{a}) e_{c\tilde{a}} + e_b ({e_c}^\tilde{c}) e_{a\tilde{c}} + {e_a}^\tilde{a} {e_c}^\tilde{c} e_b (\tilde{g}_{\tilde{a}\tilde{c}}) - e_a ({e_b}^\tilde{b}) e_{c\tilde{b}} - e_a ({e_c}^\tilde{c}) e_{b\tilde{c}} - {e_b}^\tilde{b} {e_c}^\tilde{c} e_a (\tilde{g}_{\tilde{b}\tilde{c}}) + e_a ({e_c}^\tilde{d}) e_{b\tilde{d}} - e_c ({e_a}^\tilde{d}) e_{b\tilde{d}} + e_a ({e_b}^\tilde{d}) e_{c\tilde{d}} - e_b ({e_a}^\tilde{d}) e_{c\tilde{d}} - e_c ({e_b}^\tilde{d}) e_{a\tilde{d}} + e_b ({e_c}^\tilde{d}) e_{a\tilde{d}} )$
$= \frac{1}{2} ( {e_a}^\tilde{a} {e_b}^\tilde{b} {e_c}^\tilde{c} \partial_\tilde{c} (\tilde{g}_{\tilde{a}\tilde{b}}) + {e_a}^\tilde{a} {e_c}^\tilde{c} {e_b}^\tilde{b} \partial_\tilde{b} (\tilde{g}_{\tilde{a}\tilde{c}}) - {e_b}^\tilde{b} {e_c}^\tilde{c} {e_a}^\tilde{a} \partial_\tilde{a} (\tilde{g}_{\tilde{b}\tilde{c}}) ) + e_b ({e_c}^\tilde{c}) e_{a\tilde{c}} $
$\hat\Gamma_{abc} = {e_a}^\tilde{a} {e_b}^\tilde{b} {e_c}^\tilde{c} \tilde{\Gamma}_{\tilde{a}\tilde{b}\tilde{c}} + e_{a\tilde{a}} e_b ({e_c}^\tilde{a})$
${\hat\Gamma^a}_{bc} = {e^a}_\tilde{a} {e_b}^\tilde{b} {e_c}^\tilde{c} {\tilde{\Gamma}^\tilde{a}}_{\tilde{b}\tilde{c}} + {e^a}_\tilde{a} e_b ({e_c}^\tilde{a})$


Notice that: $\partial_c (g_{ab}) + \partial_b(g_{ac}) - \partial_c(g_{ab}) = (c_\eta)_{abc} + (c_\eta)_{acb} - (c_\eta)_{cba}$

Notice that, in the diagonalized basis, since the metric is $\eta_{ij}$ is a constant, we know $(\Gamma_\eta)_{ijk}$ has the property:
$(\Gamma_\eta)_{(i|j|k)} = \frac{1}{2} e_j (\eta_{ik}) = 0$
So we know $(\Gamma_\eta)$ is antisymmetric on the 1st and 3rd indexes. This means rotation, and also means we can pick a two-form to represent this. Notice that connections are already represented as one-forms, and even on the index that isn't antisymmetric: $\omega_{ab} = \Gamma_{acb} e^c$. Now we also have the choice of representing connections as a two-form, as $(\omega_\eta)_k = \frac{1}{2} (\Gamma_\eta)_{ikj} e^i \wedge e^j$.

What happens when you calculate the coordinate-basis Levi-Civita covariant derivative of the non-coordinate basis?
$\hat\nabla_a e_j = \hat\nabla_a ({e^u}_j e_u)$
$ = \hat\nabla_a({e^u}_j) e_u + {e^u}_j \hat\nabla_a e_u$
$ = (e_a({e^c}_j) + {e^u}_j {\hat{\Gamma}^c}_{au}) e_c$

And what happens when you convert that back to the normalized coordinates?
$\hat\nabla_i e_j = \hat\nabla_{e_i} e_j = \hat\nabla_{{e^a}_i e_a} e_j = {e^a}_i \hat\nabla_{e_a} e_j$
$ = {e^a}_i \hat\nabla_a e_j$
$ = {e^a}_i (e_a({e^c}_j) + {e^u}_j {\hat{\Gamma}^c}_{au}) e_c$
$ = (e_i({e^c}_j) + {\hat{\Gamma}^c}_{ij}) e_c$
$ = (e_i({e^c}_j) {e_c}^k + {\hat{\Gamma}^k}_{ij}) e_k$

Is $\hat\nabla_i e_j = (\hat\nabla_\eta)_i e_j$?
$ = (e_i({e^c}_j) {e_c}^k + \frac{1}{2} {e_a}^k g^{au} (e_c(g_{ub}) + e_b(g_{uc}) - e_u(g_{bc})) {e^b}_i {e^c}_j) e_k$
$ = (e_i({e^c}_j) {e_c}^k + \frac{1}{2} e^{uk} {e^b}_i {e^c}_j e_c({e_u}^m e_{bm}) + \frac{1}{2} e^{uk} {e^c}_j {e^b}_i e_b({e_u}^m e_{cm}) - \frac{1}{2} {e^b}_i {e^c}_j e^{uk} e_u({e_b}^m e_{cm}) ) e_k$
$ = ( {e^a}_i e_a({e^c}_j) {e_c}^k + \frac{1}{2} e^{uk} {e^c}_j e_c(e_{ui}) + \frac{1}{2} e^{uk} {e^b}_i e_b(e_{uj}) + \frac{1}{2} {e^c}_j {e^b}_i e_b({e_c}^k) - \frac{1}{2} {e^c}_j e^{uk} e_u(e_{ci}) + \frac{1}{2} {e^b}_i {e^c}_j e_c({e_b}^k) - \frac{1}{2} {e^b}_i e^{uk} e_u(e_{bj}) ) e_k$
$ = \frac{1}{2} ( (e^{uk} {e^a}_j - {e^u}_j e^{ak}) e_a(e_{ui}) + (e^{uk} {e^a}_i - {e^u}_i e^{ak}) e_a(e_{uj}) + ({e^u}_i {e^a}_j - {e^u}_j {e^a}_i) e_a({e_u}^k) ) e_k$
$ = ( e^{uk} {e^a}_j e_{[a}(e_{u]i}) + e^{uk} {e^a}_i e_{[a}(e_{u]j}) + {e^u}_i {e^a}_j e_{[a}({e_{u]}}^k) ) e_k$
Probably yes if $\hat{\Gamma}_{abc} = (\hat{\Gamma}_\eta)_{abc}$.


What if we want to consider a connection with torsion?
Let $\nabla_\eta$ be some covariant derivative associated with operators $e_i$.
Let ${(\Gamma_\eta)^i}_{jk}$ be the connection associated with the $\nabla_\eta$ covariant derivative.
Such that $(\nabla_\eta) (T) = (e_i(T) + (\Gamma_\eta)_i \cdot (T) ) e^i$.

The connection will be defined as:
$(\Gamma_\eta)_{ijk} = \frac{1}{2} ( e_k (\eta_{ij}) + e_j (\eta_{ik}) - e_i (\eta_{jk}) + (c_\eta)_{ijk} + (c_\eta)_{ikj} - (c_\eta)_{kji} + (T_\eta)_{kij} + (T_\eta)_{jik} - (T_\eta)_{ikj} )$
$= \frac{1}{2} ( (c_\eta)_{ijk} + (c_\eta)_{ikj} - (c_\eta)_{kji} + (T_\eta)_{kij} + (T_\eta)_{jik} - (T_\eta)_{ikj} )$
$T_\eta$ is the torsion associated with the $\nabla_\eta$ operator.

TODO symmetry or antisymmetry of a connection of a diagonalized metric with torsion?


(Now I go back to the typical convention in these notes, lowercase latin for non-coordinate (in this case locally-Minkowski) basis and tilde lowercase latin for coordinate basis).
What if the metric is locally-Minkowski?
This implies more often than not that the basis will be non-coordinate.
What does the Levi-Civita connection look like?

Since the components of $g_{ab} = \eta_{ab}$ are constant, we know $e_c( g_{ab}) = 0$:
So we are left with:
$\hat\Gamma_{abc} = \frac{1}{2} ( c_{abc} + c_{acb} - c_{cba} )$

We also know now that $\hat{\Gamma}_{(a|b|c)} = \frac{1}{2} e_b (g_{ac}) = 0$

Therefore $\hat{\Gamma}_{abc}$ is antisymmetric on its 1st and 3rd indexes: $\hat{\Gamma}_{[a|b|c]} = \hat{\Gamma}_{abc}$

Verify:
$\hat{\Gamma}_{[a|b|c]} = \frac{1}{4} ( ( c_{abc} + c_{acb} - c_{cba} ) - ( c_{cba} + c_{cab} - c_{abc} ) )$
$\hat{\Gamma}_{[a|b|c]} = \frac{1}{4} ( c_{abc} + c_{acb} - c_{cba} - c_{cba} - c_{cab} + c_{abc} )$
$\hat{\Gamma}_{[a|b|c]} = \frac{1}{4} ( 2 c_{abc} + 2 c_{acb} - 2 c_{cba} )$
$\hat{\Gamma}_{[a|b|c]} = \frac{1}{2} ( c_{abc} + c_{acb} - c_{cba} )$
$ = \hat{\Gamma}_{abc}$
Correct.

Also since $g_{ab} = \eta_{ab} = \eta^{ab} = g^{ab}$, we know ${\hat{\Gamma}^a}_{bc} = \hat{\Gamma}_{abc}$ (for spatial basis at least, so long as it is a +++ signature, right?)
So for metrics with constant coefficients (TODO double check if this is the condition?), we know: ${\hat{\Gamma}^a}_{bc} = -{\hat{\Gamma}^c}_{ba}$

Thanks to the antisymmetry of the 1st and 3rd indexes of the Levi-Civita connection of a locally-Minkowski non-coordinate basis, for 3D specifically we can look at the dual of the basis:
${\Gamma^d}_b = \frac{1}{2} \epsilon^{acd} {\hat{\Gamma}}_{abc}$
Often this is depictied with the remaining non-spin index in coordinate basis, and the other two converted to the Minkowski basis:
${\Gamma^d}_{\tilde{b}} = \frac{1}{2} \epsilon^{acd} {\hat{\Gamma}}_{abc} {e^b}_{\tilde{b}}$
and because of property of duals and of our 3D antisymmetric indexes:
${\Gamma^d}_b \epsilon_{acd}$
$= \frac{1}{2} \epsilon^{efd} {\hat{\Gamma}}_{ebf} \epsilon_{acd}$
using $\epsilon^{abe} \epsilon_{cde} = \delta^{ab}_{cd}$ with + signature:
$= \frac{1}{2} {\hat{\Gamma}}_{ebf} \delta^{ef}_{ac}$
$= \frac{1}{2} {\hat{\Gamma}}_{ebf} \cdot 2 \delta^e_{[a} \delta^f_{c]}$
$= {\hat{\Gamma}}_{ebf} \cdot \delta^e_{[a} \delta^f_{c]}$
$= {\hat{\Gamma}}_{[a|b|c]}$
$= {\hat{\Gamma}}_{abc}$

So for 3D the Levi-Civita connection of a locally-Minkowski basis can be associated with an axis per-dimension.
(TODO you should be able to do this with torsion as well and relax the Levi-Civita constraint, and only constrain it to locally-Minkowski (or even only constant-component) metrics)

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