# Christoffel symbols for Poincare metric in terms of non-coordinate frame

I just want to verify that I am calculating the Christoffel symbols for the Levi-Civita connection correctly on the upper half plane $H = \{(u,v) \in \mathbb{R}^2 | v > 0\}$ with the Poincare metric $h = v^{-2}(du^2 + dv^2)$, using the $h$-orthonormal vector fields $E_1 = v\partial_u$ and $E_2 = v\partial_v$. I have the following expression $$\langle \nabla_{E_i}E_j, E_l \rangle = \frac{1}{2}(E_i\langle E_j, E_l \rangle + E_j \langle E_l, E_i \rangle - E_l \langle E_i, E_j \rangle$$ $$- \langle E_j, [E_i, E_l] \rangle - \langle E_l, [E_j, E_i] \rangle + \langle E_i, [E_l, E_j] \rangle)$$ which simplifies to $\Gamma^k_{ij}\langle E_k, E_l \rangle$ where I think that $\langle E_k, E_l \rangle$ is equal to the identity matrix $I_2$, since $E_1$ and $E_2$ are $h$-orthonormal. This yields $$\Gamma^k_{ij} = \frac{1}{2}I_2^{kl}(E_i\langle E_j, E_l \rangle + E_j \langle E_l, E_i \rangle - E_l \langle E_i, E_j \rangle$$ $$- \langle E_j, [E_i, E_l] \rangle - \langle E_l, [E_j, E_i] \rangle + \langle E_i, [E_l, E_j] \rangle)$$ Given that $[E_1, E_2] = -v\partial_u = -[E_2, E_1] = v\partial_u$, we can calculate the Christoffel symbols. Here's one to verify that I am calculating correctly: $$\Gamma^2_{11} = \frac{1}{2}I^{2l}(E_1\langle E_1, E_l \rangle + E_1 \langle E_l, E_1 \rangle - E_l\langle E_1, E_1 \rangle - \langle E_1, [E_1, E_l] \rangle - \langle E_l, [E_1, E_1]\rangle + \langle E_1, [E_l, E_1]\rangle)$$ for $l \neq 2$, all terms vanish, and for $l = 2$ the first three terms vanish, leaving $$\Gamma^2_{11} = \frac{1}{2}(-\langle E_1, [E_1, E_2]\rangle - \langle E_2, 0 \rangle + \langle E_1, [E_2, E_1] \rangle) = \frac{1}{2}(\langle E_1, -E_1 \rangle + \langle E_1, E_1 \rangle = 1$$

If someone could help me verify whether this is correct, that would be appreciated.

Instead of saying that $\left<E_k, E_l \right>$ is equal to the identity matrix, I would write $\left<E_k, E_l \right> = \delta_{kl}$. Since this is a constant function (independent of $(u,v)$), the first three terms in your formula always vanish. Thus, we are left with

$$\Gamma_{ij}^k = \frac{1}{2} \left( -\left< E_j, [E_i, E_k] \right> - \left< E_k, [E_j, E_i] \right> + \left< E_i, [E_k, E_j] \right>\right).$$

As you calculated, we have

$$[E_1, E_2] = -E_1, [E_2, E_1] = E_1, [E_1, E_1] = [E_2, E_2] = 0.$$

Thus, we get for example

$$\Gamma_{11}^2 = \frac{1}{2} \left( -\left< E_1, [E_1, E_2] \right> - \left< E_2, [E_1, E_1] \right> + \left< E_1, [E_2, E_1] \right> \right) = \frac{1}{2} \left( - \left< E_1, -E_1 \right> + \left<E_1, E_1 \right> \right) = \left< E_1, E_1 \right> = 1.$$

Thus, your calculation is correct although you missed a sign in the last formula.