Spin connection

Remarks

The forms $\omega_{ab} = \eta_{ac} \omega^c{}_b$ satisfy $\omega_{ab} = -\omega_{ba}$, because $\nabla$ preserves the metric. Since $\nabla$ has no torsion, they satisfy the first structure equation of Cartan, $d\theta^a + \omega^a{}_b \wedge \theta^b = 0$. These two properties determine $\omega$ uniquely from the frame. Example: on $\R^2 \setminus \set{0}$ with polar coordinates, take $e_1 = \partial_r$ and $e_2 = r^{-1} \partial_\varphi$, so $\theta^1 = dr$ and $\theta^2 = r \, d\varphi$. Then $\omega^2{}_1 = d\varphi$ and $\omega^1{}_2 = -d\varphi$. Check: $d\theta^2 = dr \wedge d\varphi = -\omega^2{}_1 \wedge \theta^1$, and $d\theta^1 = 0 = -\omega^1{}_2 \wedge \theta^2$. Non-example: for the coordinate frame $(\partial_r, \partial_\varphi)$, which is not orthonormal, the connection matrix is not antisymmetric. Indeed $\nabla_{\partial_r} \partial_\varphi = r^{-1} \partial_\varphi$, so its diagonal entry $\omega^\varphi{}_\varphi = r^{-1} dr$ is nonzero. On a spin manifold the matrix $\omega_{ab}$ acts on spinor fields $\psi$, and in one common convention $\nabla \psi = d\psi + \tfrac14 \omega_{ab} \gamma^a \gamma^b \psi$, where $\gamma^a$ are the Clifford generators.