What is the spin connection on the spin bundle $S$ over $R^3$? Let metric be $dx_1^2+dx_2^2+dx_3^2$ with orientation $dx_1 \wedge dx_2 \wedge dx_3$. From my understanding, the spin bundle over $R^3$ is a trivial rank 4 bundle equipped with anticommuting involutions I, J e.g. $I=\begin{pmatrix} 0 & 1 &0 &0 \\ -1 &0 &0 &0 \\ 0 & 0 &0 &1 \\0 &0 &-1&0\end{pmatrix}, J=\begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}$.
We need a representation $\Gamma$ of the Clifford algebra of $TR^3$ onto the space $R^4$ that commutes with I,J. Let $\frac{\partial}{\partial x_1}, \frac{\partial}{\partial x_2}, \frac{\partial}{\partial x_3}$ be frame of $TR^3$. I suppose $\Gamma$ could be given by $\Gamma(\frac{\partial}{\partial x_1})=\begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}$, $\Gamma(\frac{\partial}{\partial x_2})=\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\end{pmatrix}$, $\Gamma(\frac{\partial}{\partial x_3})=\begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}$.
Then what remains is to define a connection on $S$ that is both compatiable with the Levi-Civita connection on $R^3$ and with the inner product on $R^4$, i.e. if $s,s'$ are sections of $S$ then $\nabla_X(\Gamma(Y)s)= \Gamma(Y) \nabla_X s + \Gamma(\nabla_X Y) s$ for all vector fields $X,Y$ on $R^3$ and $X<s,s'>=<\nabla_X s,s'>+<s,\nabla_X s'>$
For our spin connection, I believe there are 4x4x3=48 Christoffel symbols to calculate. Are we to use compatability conditions to generate 48 equations and solve for the Christoffel symbols? How does one get the spin connection? I am looking for explicit form of this connection and calculation, not abstract definition of this connection like "unique lift of Levi-Civita connection" etc.
This is an old question but an entertaining one, so I will offer the following algorithm for finding the spin connection. First, note that since $\operatorname{Spin}^+(s,t)$ is the double cover of the orientation and time orientation preserving special orthogonal group, their Lie algebras are isomorphic. Any oriented and time oriented orthonormal basis induces a basis for the Clifford algebra associated to $\mathbb {R}^{t+s}$ equipped with the inner product of signature $(s,t)$ by mapping $e_i\mapsto \gamma_i$, where $\gamma_i$ are mathematical gamma matrices.
Now, as you should know if we have a Levi-Civita connection on the tangent bundle of a pseudo Riemannian manifold $M$, we can write $\nabla$ in any local frame orthonormal frame as: \begin{align} \nabla e_a=\xi_{ab}\eta^{bc}\otimes e_c \end{align} for some one forms $\xi_{ab}$, and the matrix $\eta^{bc}$ which is the inner product of split signature. This induces a unique connection one form $A_{SO^+}$ in the principal bundle of oriented, and time oriented, orthonormal frames, such that for any local frame (i.e. section of this principal bundle) we have that: \begin{align} (e^*(A_{SO}))^c_a=\xi_{ab}\eta^{bc} \end{align} If we $M$ comes equipped with a spin structure, there is a $2$ to $1$ $\lambda:\operatorname{Spin}^+(t,s)\rightarrow SO^+(t,s)$ equivariant homomorphism: \begin{align} \Lambda:\operatorname{Spin}^+(M)\longrightarrow SO^+(M) \end{align} where $\lambda$ is the covering map. The spin connection $A$ is defined by \begin{align} A=(\lambda_{*))^{-1}\circ \Lambda^*(A_{SO}) \end{align} So locally, the spin covariant derivative $\nabla^s$ is given by: \begin{align} \nabla^s\psi=d\psi+\kappa_*(\epsilon^*A^s)\psi \end{align} where $\epsilon^*$ is a section of the $\operatorname{Spin}^+(M)$, $\kappa_*$ is the induced representation of the Lie algebra on the vector space of Dirac spinors $\Delta_{s+t=n}$, and $\psi$ is a smooth map $U\rightarrow \Delta_n$. It follows that: \begin{align} \kappa_*(\epsilon^*A^s)= \kappa_*\circ (\lambda_*)^{-1}\circ (\Lambda\circ \epsilon)^*A_{SO} \end{align} Now note that $\Lambda\circ \epsilon$ is a section of $SO^+(M)$, i.e. an oriented and time oriented orthonormal local frame, so we can write this as: \begin{align} \kappa_*(\epsilon^*A^s)= \kappa_*\circ (\lambda_*)^{-1}\circ (e^*A_{SO}) \end{align} It remains to determine an explicit formula for the map $\kappa_*\circ (\lambda_*)^{-1}:\mathfrak{so}^+(t,s)\rightarrow \operatorname{End}(\Delta_n)$. Let $A\in \mathfrak{so}^+(t,s)$, and write the components as $A_a^c=w_{ab}\eta^{bc}$ for a fixed orthonormal basis of $e_a$. This orthonormal basis induces gamma matrices $\gamma_a$, then one can show that: $$\kappa_*\circ (\lambda_*)^{-1}(A)=\frac{1}{4}w_{ab}\gamma^{ab}$$, where: $$\gamma^{ab}=\frac{1}{2}[\gamma_c\eta^{ca}, \gamma_d\eta^{db}]$$ It follows that an explicit formula for the Spin covariant derivative in any local frame is given by: \begin{align} \nabla^s\psi=d\psi+\frac{1}{4}\xi_{ab}\gamma^{ab}\cdot \psi \end{align} So, you just need to fix gamma matrices on, and find the one forms $\xi_{ab}$ to explicitly calculate the spin covariant derivative. In particular, since the Levi-Civita connection on $\mathbb R^3$ is trivial, we have that $\xi_{ab}=0$, so the spin covariant derivative is just $d\psi$.