How does spinors on manifold transformed with coordinate

Question

Everyone knows that $SU(2)$ is a double cover of $SO(3)$.
$SL(2,\mathbb{C})$ is a double cover of Proper Lorentz Group $L^\uparrow_+$.
These groups are all associated with the spinors and tensors in 3d Euclidian space and Minkowski Spacetime. When Minkowski Spacetime doing a Lorentz transformation express like the following: $$ X'=\exp(u^ib_i+v^ir_i)X $$ Which $b_i$ and $r_i$ are 4 by 4 matrices that ${(b_i)^0}_i={(b_i)^i}_0={(r_1)^3}_2=-{(r_1)^2}_3=-{(r_2)^3}_1={(r_2)^1}_3={(r_3)^2}_1=-{(r_3)^1}_2=1$ and other components are all 0, $u^i$ and $v^i$ are 3d vectors.
The spinors also transform like this:
$$ \xi^A\to{\exp(\frac{1}{2}(u^i-iv^i)\sigma_i)^A}_B\xi^B $$ (The result of Lie algebra isomorphism)
And these can be all the way extend to $ \eta_{\mu\nu}{\sigma^\mu}_{AA'}{\sigma^\nu}_{BB'}=-\epsilon_{AB}\epsilon_{A'B'} $ (the spinor metric)
But what if this is on a manifold and can not find an orthonormal coordinate system, The coordinate transformation matrices is an element of $GL(4)$, How does the spinor transform like? Are they still in $SL(2,\mathbb{C})$? And will the Pauli Matrices' components remain unchanged? I find a group $SL(3,\mathbb{C})$ have the same amount of dimension as $GL(4)$, are they also have similar relations of spinors and tensors?
It would be better if there were examples. Thanks

Answer 1

Will, I think I have found the key to understanding spinor.
Vector space and the spinor space have similar relations between tensor space and vector space, which could be considered as two "separate" spaces only connect by their definition :
A type(k,l) tensor is a multiple linear function that maps k dual vectors and l vectors to a real number $$ T:\underbrace{V^*\times\cdots\times V^*}_{k}\times\underbrace{V\times\cdots\times V}_{l}\mathop\longrightarrow\limits^{\text{linear}}\mathbb{R} $$ Spinor space(S) and vector space(V) are also "separated". The connection between them is the ${\sigma_\mu}^{AA'}$ which is also the bases $(e_\mu)^a$ of V. When choosing the bases in V that is orthonormal, it naturally gives the bases of S, which makes ${\sigma_\mu}^{AA'}$ have components from Pauli Matrices. Give any two spinors $\{o^A,l^A\}$ which can form a set of bases in S can also construct a set of bases of V through Pauli Matrices. But give any vectors forming a set of bases in V may not able to give a set of bases in S, Like the bases of V can easily construct bases in tensor space by tensor product, But a base of tensor space may not also give out the base of V.
For this reason, when doing Lorentz transformation, we can do the "same" thing to spinor space to keep ${\sigma_\mu}^{AA'}$'s components remain unchanged, but this isn't necessary. Bases for V and S can choose individually, only need to change the ${\sigma_\mu}^{AA'}$. And this might be why the spinor connection ${\gamma_{aB}}^C$ is so special.
Needless to say, using spinor tensors to represent spin-$\frac{n}{2}$ is extremely convenient when doing coordinate transformations.