Do Carmo Riemannian geometry Christoffel symbols

Question

I have been reading through Riemannian Geometry from this amazing book from Do Carmo. I am new to Riemannian Geometry and have a question about what steps Do Carmo has taken on page 55/56 to obtain the following:

We first obtain the following equality which I understand:

$$\sum_l \Gamma^l_{ij}g_{lk} = \frac{1}{2}(\frac{\partial}{\partial x_i}g_{jk}+\frac{\partial}{\partial x_j}g_{ki}-\frac{\partial}{\partial x_k}g_{ij})$$ This equality follows from an equation we obtain during the proof of the Levi Civita connection theorem. Here $g_{ij}=\langle \frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j}\rangle$. The equation is given by:

$$\langle Z,\nabla_YX \rangle = \frac{1}{2}(X\langle Y,Z \rangle + Y\langle Z,X \rangle - Z\langle X,Y \rangle - \langle [X,Z], Y\rangle - \langle [Y,Z], X \rangle - \langle [X,Y], Z\rangle)$$

However I don't really see how we obtain the following from the fact that the matrix $(g_{ij})_{ij}$ admits a inverse given by $(g^{ij})_{ij}$: $$\Gamma^m_{ij}=\frac{1}{2}\sum_k(\frac{\partial}{\partial x_i}g_{jk}+\frac{\partial}{\partial x_j}g_{ki}-\frac{\partial}{\partial x_k}g_{ij})g^{km}$$

Can someone explain how we get to this point, because I am not really seeing it.

Answer 1

In general, if $A = (a_{ij})$ is a matrix, and $A^{-1}= (a^{ij})$ denotes it's inverse, then we can compute each entry of $E_n = A\cdot A^{-1}$ as follows: $$\sum_k a_{ik} a^{kj} = \delta_{i,j},$$ where $\delta_{i,j}$ is the Kronecker-delta, i.e. $1$ if $i=j$ and $0$ otherwise. Your identity now follows from the fact that

$$ \Gamma^m_{ij} = \sum_{l} \Gamma^l_{ij}\cdot \delta_{l,m} = \sum_l \sum_k \Gamma^l_{ij}g_{lk}g^{km}=\frac{1}{2}\sum_k(\frac{\partial}{\partial x_i}g_{jk}+\frac{\partial}{\partial x_j}g_{ki}-\frac{\partial}{\partial x_k}g_{ij})g^{km}.$$

Answer 2

From a matrix point of view, your first equality is $G \Gamma_{ij} = X_{ij}$ with $G=(g_{lk})_{lk}$ a square matrix, $\Gamma_{ij}=(\Gamma_{ij}^l)_l$ a column vector, and $X_{ij} = \left(\frac{1}{2}(\partial_i g_{jk}+\partial_jg_{ik}-\partial_kg_{ij})\right)_k$ another column vector. Since $G$ has inverse $G^{-1}=(g^{ij})$, it follows that $\Gamma_{ij} = G^{-1}G\Gamma_{ij} = G^{-1}X_{ij}$. You can check that the $m$th-entry of $G^{-1}X_{ij}$ is given by the last equality that you stated.