Christoffel symbols and metric

Question

For Christoffel symbol and metric, we've the following identity

$$\large \frac{1}{2}g^{\alpha\gamma}(g_{\alpha\beta,\mu}+g_{\alpha\mu,\beta}- g_{\beta\mu,\alpha})={\mathrm\Gamma^{\gamma}}_{\beta\mu}. $$

Now even though I've seen the derivation, I still can't understand what is the motivation behind the steps taken, in all the index juggling being done. Can anyone please give a motivated proof for the identity? I mean that mentioning why we are doing instead of just what we are doing. What if the identity wasn't discovered yet and you were to, what will go inside your head to come up with this useful result and why would you do the following to arrive at it, instead of just Hit and Trial.

$$\large \begin{aligned} g_{\alpha\beta,\mu}&={\mathrm\Gamma^{\nu}}_{\alpha\mu}g_{\nu\beta}+ {\mathrm\Gamma^{\nu}}_{\beta\mu}g_{\alpha\nu},\\ g_{\alpha\mu,\beta}&={\mathrm\Gamma^{\nu}}_{\alpha\beta}g_{\nu\mu}+ {\mathrm\Gamma^{\nu}}_{\mu\beta}g_{\alpha\nu},\\ -g_{\beta\mu,\alpha}&=-{\mathrm\Gamma^{\nu}}_{\beta\alpha}g_{\nu\mu}- {\mathrm\Gamma^{\nu}}_{\mu\alpha}g_{\beta\nu}. \end{aligned} $$

Answer 1

I'm not exactly sure which "derivation" you're referring to, but I've always seen Christoffel symbols introduced in the following context:

Ultimately the motivation is to define the notion of parallel transport, which means when we take a derivative of a vector field, we want our derivative to still be tangent to our manifold. So let's suppose for instance we have a manifold $M$, and we have vector fields $X, Y \in \mathcal{X}(M)$ defined locally in $M$. We want an affine connection $\nabla:\mathcal{X}(M) \times \mathcal{X}(M) \to \mathcal{X}(M)$ to act as a derivative that remains tangent to the manifold. Let $\frac{\partial}{\partial e^1}, \ldots, \frac{\partial }{\partial e^n}$ be a local basis for the tangent spaces on $M$ (defined in the same neighborhood as our vector fields), such that we can express our vector fields locally by functions $x^i, y^j: M \to \mathbb{R}$:

$$ X \;\; =\;\; x^i \frac{\partial}{\partial e^i} \;\;\; \;\;\; Y \;\; =\;\; y^j \frac{\partial}{\partial e^j}. $$

Then our covariant derivative is given locally by

$$ \nabla_XY \;\; =\;\; \nabla_{x^i\partial_i} (y^j \partial_j) \;\; =\;\; x^i \nabla_{\partial_i} (y^j\partial_j) \;\; =\;\; x^i \frac{\partial y^j}{\partial e^i} \frac{\partial }{\partial e^j} + x^iy^j \nabla_{\partial_i}\partial_j. $$

The first term is clearly in the tangent space, but we want to define $\nabla_{\partial_i}\partial_j$ to lie in the tangent space. We therefore define correction functions $\Gamma_{ij}^k: M\to \mathbb{R}$ known as the Christoffel symbols that satisfy

$$ \nabla_{\partial_i} \partial_j \;\; \equiv \;\; \Gamma_{ij}^k \frac{\partial}{\partial e^k}. $$

Now by manipulating the indices around a little bit we have that

$$ \nabla_XY \;\; =\;\; \left (x^i \frac{\partial y^k}{\partial e^i} + x^i y^j \Gamma_{ij}^k \right ) \frac{\partial}{\partial e^k}. $$

The above analysis is valid for any affine connection on a manifold, but on a Riemannian manifold we require the connection to be symmetric and compatible with the metric. The compatibility condition is that

$$ X\langle Y,Z\rangle \;\; =\;\; \langle \nabla_XY, Z\rangle + \langle Y, \nabla_XZ\rangle $$

while the symmetry condition is that

$$ \nabla_XY - \nabla_YX \;\; =\;\; [X,Y] $$

which for the basis vectors reduces to $\nabla_{\partial_i}\partial_j - \nabla_{\partial_j}\partial_i = 0$. From these equations we can in general prove that

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

If we reduce this down to the basis vectors we obtain

$$ \langle \partial_s, \nabla_{\partial_i}\partial_j\rangle \;\; =\;\; \Gamma_{ij}^k g_{sk} \;\; =\;\; \frac{1}{2} \left ( \partial_j g_{si} + \partial_i g_{sj} - \partial_s g_{ij} \right ). $$

Because we have that $g^{sm}$ is the inverse of $g_{sk}$ we have that $g_{sk} g^{sm} = \delta_k^m$ and thus we obtain

$$ \Gamma_{ij}^m \;\; =\;\; \frac{1}{2} g^{ms} \left ( \partial_j g_{si} + \partial_i g_{sj} - \partial_s g_{ij} \right ). $$

This is the intuitive explanation that I like most regarding Christoffel symbols: they are correction functions to make sure that an affine connection keeps the derivative of a vector field within the tangent space of the manifold. On a Riemannian manifold they have the given form above. To get the other equations that you have listed in your question, simply permute the indices to get different copies of the fundamental equation for $\Gamma_{ij}^s$, and then subtract/add them together to get the forms you're interested in.

Answer 2

The intuition can be explained (without any formula) in the case of a subvariety in $\bf R^n$. In the case the Levi-Cita derivative is obtained by composing the usual derivative with the projection on the tangent space. If you are in a chart $\phi (x_1,...,x_n)$, where $\phi $is a $N$vector, the tangent space is spanned by the (non orthogonal) family of vectors $X_i={\partial \phi \over \partial x_i}$. We have $g_{i,j}= <X_i,X_j>$, and writting the projection onto this space is equivalent to inverse the matrix $g_{i,j}$. In other words $\Gamma _{i,j}^k$ is the $k-$th coordinate in the base $X_i={\partial \phi \over \partial x_i}$ of the $N$ vector ${\partial ^2 \phi \over \partial x_i \partial x_j}$.The two dimensional case is enough to understand how the computation goes. For instnace if we are lucky and the base $X_i$ is othonormal, this Christoffel is just $<{\partial ^2 \phi \over \partial x_i \partial x_j},{\partial \phi \over \partial x_k} >$, and your formula is easy to check. The case of surfaces in $R^3$ already contains all difficulties (see Do Carmo)