Parallel propagator along a single coordinate:
${P_{\mu }(x_L^{\mu },x_R^{\mu }{)}^{\alpha }}_{\nu }=exp(-{\int }_{x^{\prime \mu }=x_L^{\mu }}^{x^{\prime \mu }=x_R^{\mu }}{\mathrm{\Gamma }}_{\mu }(x^{\prime \mu })d{x}^{\prime \mu })$
...where ${\mathrm{\Gamma }}_{\mu }$ is a matrix with components ${{\mathrm{\Gamma }}^{\alpha }}_{\mu \nu }$

Parallel propagator linearly interpolated between coordinates:
$P_{\mu }(x,y)=exp(-{\int }_{\lambda =0}^{\lambda =1}{\mathrm{\Gamma }}_{\mu }(x+\lambda v)v^{\mu }d\lambda )$

Matrix exponential:
$exp(A)=exp(R\cdot \mathrm{\Lambda }\cdot L)=R\cdot exp(\mathrm{\Lambda })\cdot L$
$R^{-1}\cdot exp(A)\cdot L^{-1}=exp(\mathrm{\Lambda })$
$\mathrm{\Lambda }=log(R^{-1}\cdot exp(A)\cdot L^{-1})$
$A=R\cdot \mathrm{\Lambda }\cdot L=R\cdot log(R^{-1}\cdot exp(A)\cdot L^{-1})\cdot L$
$A=log(exp(A))=log(exp(R\cdot \mathrm{\Lambda }\cdot L))=log(R\cdot exp(\mathrm{\Lambda })\cdot R)$

TODO for $A=R_A\cdot {\mathrm{\Lambda }}_A\cdot L_A$, prove $log(A)=R_A\cdot log({\mathrm{\Lambda }}_A)\cdot L_A$
Let $B=log(A)$, so $exp(B)=A$
and $B=R_B\cdot {\mathrm{\Lambda }}_B\cdot L_B$
and (by argument of diagonalization of $exp(A_{ij})={\delta }_{ij}exp(A_{ij})$) we know that $R_B=R_A,L_B=L_A$
so ${\mathrm{\Lambda }}_B=log({\mathrm{\Lambda }}_A)$
so $log(A)=R_A\cdot log({\mathrm{\Lambda }}_A)\cdot L_A$
where, for diagonal matrix, $(log({\mathrm{\Lambda }}_A){)}_{ij}=0$ for $i\ne j$, or $log(A_{ij})$ for $i=j$.

Flux
Jacobian Eigendecomposition:
$F^{\alpha }=$ flux.
${A^{\alpha }}_{\mu }=\frac{\partial F^{\alpha }}{\partial U^{\mu }}=$ flux Jacobian.
$A=R\cdot \mathrm{\Lambda }\cdot L$

Let $-\int {\mathrm{\Gamma }}_{\mu }{v}^{\mu }d\lambda =-\int \mathrm{\Gamma }(\lambda ,v)d\lambda =A$ ... notice the first $\mathrm{\Gamma }$ parameter is its dependent variable $\lambda$, while the second is the vector $v^{\mu }$ that the one-form $\mathrm{\Gamma }$ is actiong upon. Maybe it would be more proper math to write $\mathrm{\Gamma }(\lambda )(v)$?
So $A=R\cdot \mathrm{\Lambda }\cdot L=-\int {\mathrm{\Gamma }}_{\mu }{v}^{\mu }d\lambda$
And $exp(-\int {\mathrm{\Gamma }}_{\mu }{v}^{\mu }d\lambda )=R\cdot \mathrm{\Lambda }\cdot L$
And $-\int {\mathrm{\Gamma }}_{\mu }{v}^{\mu }d\lambda =R\cdot log(\mathrm{\Lambda })\cdot L$

How about if we say ${\mathrm{\nabla }}_{e_{\mu }}{e}_{\nu }={{\mathrm{\Gamma }}^{\alpha }}_{\mu \nu }{e}_{\alpha }$ is equivalent to $\frac{\partial F^{\alpha }}{\partial U^{\mu }}$?

How about if $-{{\mathrm{\Gamma }}^{\alpha }}_{\mu \nu }{\hat{x}}^{\mu }={A^{\alpha }}_{\nu }$?
$\int -{{\mathrm{\Gamma }}^{\alpha }}_{\mu \nu }{v}^{\mu }d\lambda =\int {A^{\alpha }}_{\nu }d\lambda$

Then there's Dullemond, Wang's Hydrodynamics notes:
Start with ${\partial }_{t}U+{\partial }_{x^i}{F}^i=0$
(In arbitrary curvilinear coordiantes ${\partial }_{t}U+{\mathrm{\nabla }}_i{F}^i=0$)
eqn. 7.36:
$F(U(x_R))-F(U(x_L))={\int }_0^1\frac{\partial }{\partial \lambda }F(U(\lambda ))d\lambda$
$=({\int }_0^1\frac{\partial F}{\partial U}d\lambda )\cdot (U(x_R)-U(x_L))$
$={\int }_0^1(R\cdot \mathrm{\Lambda }\cdot L)d\lambda \cdot (U(x_R)-U(x_L))$
...
$={\int }_0^1(R\cdot exp(log(\mathrm{\Lambda }))\cdot L)d\lambda \cdot (U(x_R)-U(x_L))$
...
$exp(F(U(x_R))-F(U(x_L)))=exp(\left({\int }_0^1\frac{\partial F}{\partial U}d\lambda \right)\cdot (U(x_R)-U(x_L)))$ ... or does that go outside the exp?
$=exp\left({\int }_0^1\frac{\partial F}{\partial U}d\lambda \right)(U(x_R)-U(x_L))$
$=exp({\int }_0^{1}R\cdot \mathrm{\Lambda }\cdot Ld\lambda )(U(x_R)-U(x_L))$
$exp(F(U(x_R)))\cdot exp(-F(U(x_L)))=exp({\int }_0^{1}R\cdot \mathrm{\Lambda }\cdot Ld\lambda )(U(x_R)-U(x_L))$
so the flux $F$ is the log of the connection ${\mathrm{\Gamma }}_x$ that is parallel-propagated along the x-coordiante.

Parallel propagator:
$e(x_R)=e(x_L)\cdot P^{-1}(x_L,x_R)$
$=e(x_L)\cdot exp(-{\int }_0^1{\mathrm{\Gamma }}_{v}d\lambda )$
$e(x_L{)}^{-1}e(x_R)=exp(-{\int }_0^1{\mathrm{\Gamma }}_{v}d\lambda )$
$exp(log(e(x_L{)}^{-1}e(x_R)))=exp(-{\int }_0^1{\mathrm{\Gamma }}_{v}d\lambda )$
$exp(log(e(x_R))-log(e(x_L)))=exp(-{\int }_0^1{\mathrm{\Gamma }}_{v}d\lambda )$
$log(e(x_R))-log(e(x_L))={\int }_0^1-{\mathrm{\Gamma }}_{v}d\lambda$
...substitute...
$F(U(x_R))\leftrightarrow log(e(x_R))$
$F(U(x_L))\leftrightarrow log(e(x_L))$
$\frac{\partial F}{\partial U}\cdot (U(x_R)-U(x_L))\leftrightarrow -{\mathrm{\Gamma }}_v$
...to get...
$F(U(x_R))-F(U(x_L))={\int }_0^1\frac{\partial F}{\partial U}(U(x_R)-U(x_L))d\lambda$
$F(U(x_R))-F(U(x_L))={\int }_0^1\frac{\partial F}{\partial U}\frac{\partial U}{\partial \lambda }d\lambda$
$F(U(x_R))-F(U(x_L))={\int }_0^1\frac{\partial F}{\partial \lambda }d\lambda$

Then there is Misner, Thorn, Wheeler "Gravitation" exercise 16.6 showing how ${\partial }_{t}U+{\mathrm{\nabla }}_{x^i}{F}^i=0$ is the same as ${\mathrm{\nabla }}_{\nu }{T}^{\mu \nu }=0$
where ${\partial }_t\rho +{\mathrm{\nabla }}_{x^i}(\rho v^i)=0$ becomes ${\mathrm{\nabla }}_{\mu }{T}^{t\mu }=0$
and ${\partial }_t(\rho v^j)+{\mathrm{\nabla }}_{x^i}(\rho v^i{v}^j+g^{ij}P)=0$ becomes ${\mathrm{\nabla }}_{\mu }{T}^{j\mu }=0$