Is there any relationship between gauge field and spin connection? For a spinor on curved spacetime, $D_\mu$ is the covariant derivative for fermionic fields is
$$D_\mu = \partial_\mu - \frac{i}{4} \omega_{\mu}^{ab} \sigma_{ab}$$
where $\omega_\mu^{ab}$ are the spin connection.
And the transformation of spin connection is very similar to gauge field.
So is there any relationship between them. If there is any good textbook or reference containing this area, please cite it. Thanks!
 A: A gauge field for a particular group $G$ can be thought of as a connection, or a $G$ Lie algebra valued differential form. If we recall the Riemann curvature,
$$R(u,v)w = \left( \nabla_u \nabla_v - \nabla_v \nabla_u -\nabla_{[u,v]}\right)w$$
If $[u,v]=0$ the expression simplifies to the usual tensor in general relativity. Similarly, we may think of the field-strength of a gauge field as a curvature - it's essentially a commutator of covariant derivatives and attempts to quantify the affect of parallel transportation on tensorial objects. For a $U(1)$ field,
$$F=\mathrm{d}A $$
with no additional terms, because the analogue of the $\nabla_{[u,v]}$ term vanishes as $U(1)$ is abelian and all structure constants of the group vanish. The relation to the curvature tensor becomes even clearer as we express the field-strength in explicit index notation,
$$F=\partial_\mu A_\nu - \partial_\nu A_\mu$$
In gravitation, the gauge group is the group of diffeomorphisms $\mathrm{Diff}(M)$, infinitesimally these are vector fields which shift the coordinates; the binary operation of the group is the Lie bracket, and the metric changes by a Lie bracket, namely,
$$g_{ab}\to g_{ab}+\mathcal{L}_\xi g_{ab}$$
where $\xi$ is our vector field. The Lorentz group $SO(1,3)$ is a subgroup of the diffeomorphism group. In addition, the Killing vectors are those which produce no gauge perturbation of the metric, i.e.
$$\nabla_\mu X_\nu -\nabla_\nu X_\mu=0$$
These Killing vector commutators may form a Lie algebra of a Lie group $G$; the generators $T_a$ of a Lie group $G$ allow us to define the structure constants,
$$[T_a,T_b]=f^{c}_{ab}T_c$$
where $f$ are the structure constants, modulo some constants according to convention.
A: As Slereah mentioned in the comments, gravity in the Palatini formulation (i.e. the one with vielbeins and the spin connection) can easily be understood as the gauge theory of the Lorentz group. 
Not only do you have a correspondance between $\omega_\mu = \omega_\mu^{ab}$ as a Lie-Algebra (i.e. Lorentz-algebra) valued vector field (which becomes more obvious when you supress the gauge group indices, as one often does in "regular" gauge theories), you also have the Riemannian $R_{\mu \nu}^{ab} = R_{\mu \nu}$ as a gauge curvature, i.e. the "$F_{\mu \nu}$" of gravity.
A: Sure enough, any gauge field has a similar connection as well as a curvature tensor. This also leaves room for a geometric interpretation, similar to the conventional approach to general relativity. A well-known example is the electromagnetic $\left(\ U(1)\ \right)$ gauge field $A_\mu$, which can be introduced by requiring local $U(1)$ invariance of the complex scalar field Lagrangian (see, for instance, Ryder's book on QFT). The curvature tensor is 
$$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$
and the covariant derivative acts as 
$$D^\mu\psi=\left(\partial^\mu + ieA^\mu\right)\psi\hspace{2cm} D^\mu\psi^*=\left(\partial^\mu-ie A^\mu\right)\psi^*$$
As mentioned by others, we can understand understand gravity in the same way, with the Christoffel connection fulfilling the same role as the spin connection you mentioned or the $ieA^\mu$ term for the electromagnetic field, while the Riemann tensor $R^\rho_{\mu\sigma\nu}$ is the curvature tensor. The curvature tensor is actually intimately related to the covariant derivative; it acts as the commutator. In general relativity, for example, we have
$$[\nabla_\mu,\nabla_\nu]V^\rho=R^\rho{}_{\sigma\mu\nu}V^\sigma$$
