Question about the Christoffel Symbol

Question

Wikipedia says that we have a Christoffel Symbol of the Second Kind, such that: $$\Gamma_{ij}^k=\frac{\partial \textbf{e}_i}{\partial x^j}\cdot\textbf{e}^k$$ Which we can convert into a Christoffel Symbol of the First Kind as so: $$\Gamma_{kij}=\frac{\partial \textbf{e}_i}{\partial x^j}\cdot\textbf{e}_k$$ I undestand the whole procedure up to that part. What I don't understand is how it say that 'rearranging' we get: $$\Gamma_{ij}^k\textbf{e}_k=\frac{\partial \textbf{e}_i}{\partial x^j}$$

Answer 1

If $\mathbf e_i$ are the given frame (basis), then $\mathbf e^i$ will be the dual basis. These covectors "eat" vectors and spit out numbers. In particular, $\langle\mathbf e^k,\mathbf e_i\rangle = \delta^k_i$. So if we apply $\mathbf e^k$ to the linear combination $\sum a^i\mathbf e_i$, we get $$\langle \mathbf e^k,\sum a^i\mathbf e_i\rangle = \sum a^i\langle\mathbf e^k,\mathbf e_i\rangle = \sum a^i\delta^k_i = a^k.$$ Thus, setting $\Gamma^k_{ij} = \langle \mathbf e^k,\dfrac{\partial\mathbf e_i}{\partial x^j}\rangle$, we do in fact get $$\frac{\partial\mathbf e_i}{\partial x^j} = \sum \Gamma^k_{ij}\mathbf e_k.$$ (Once again, I find the dot in the first equation totally confusing and misleading.)

We can identify the dual space with the original vector space using the Riemannian metric by setting $\langle\tilde{\mathbf w},\mathbf v \rangle = \mathbf w\cdot\mathbf v$. This means that $\langle\tilde{\mathbf e}_k,\mathbf e_i \rangle = \mathbf e_k\cdot \mathbf e_i = g_{ki}$. In order to make this come out equal to $\delta^k_i$, we must take the linear combination $$\mathbf e^k = \sum g^{kj}\tilde{\mathbf e}_j,$$ for then we have $$\langle \mathbf e^k,\mathbf e_i\rangle = \sum g^{kj}\mathbf e_j\cdot \mathbf e_i = \sum g^{kj}g_{ki} = \delta^k_i.$$