How are Christoffel symbols explicitly computed?

Question

In Tu's book on differential geometry he says:

Let $\nabla$ be an affine connection on a manifold $M$ and let $(U, x^1, \ldots, x^n)$ be a coordinate open set in $M$. Denote by $\partial_i$ the coordinate vector field $\partial/\partial x^i$. Then $\nabla_{\partial_i} \partial_j$ is a linear combination of $\partial_1, \ldots, \partial_n$, so there exist numbers $\Gamma_{ij}^k$ at each point such that $$\nabla_{\partial_i}\partial_j = \sum_{k=1}^n \Gamma_{ij}^k \partial_k. \tag{1}$$ These $n^3$ functions $\Gamma_{ij}^k$ are called the Christoffel symbols of the connection $\nabla$ on the coordinate open set $(U, x^1, \ldots, x^n)$. By the $\mathcal{F}$-linearity in the first argument of a connection, the Christoffel symbols completely describe a connection on $U$.

I have two questions about this passage,

1. Wouldn't we also need the linearity of the second argument of a connection to completely determine a connection on $U$? This allows us to write the connection of a vector field $Y$ along a vector field $X$ as $$\nabla_X Y = \sum_{i,j} \nabla_{x^i\partial_i}{y^j \partial_j}$$ so that each term in this sum has its own expansion (1).

2. How are the Christoffel symbols computed in practice? I originally thought that the Christoffel symbol $\Gamma$ is a tensor with the notation $$\Gamma_{ij} \equiv \Gamma(\partial_i, \partial_j).$$ However based on other questions I have read on here, this seems to be the incorrect formula on how $\Gamma$ is computed. How does $\Gamma$ explicitly depend on $i$ and $j$?

Answer 1

This is Levi-Civita connection.

Given a coordinate system, $\{x^\alpha\}$ you can represent the vector space having basis vectors $\partial _\alpha$. This is short hand for $\frac{\partial \vec{r}}{\partial x^a}$ where $\vec{r}$ is the position vector.

Another common representation is $\vec{e_\alpha}=\frac{\partial \vec{r}}{\partial x^\alpha}$

Vectors at some position can be presented as the sum of products of the components of the vector with the basis vectors.

So $\vec{V}=\sum_{\alpha =0}^n V^\alpha\vec{e_\alpha}$

The basis vectors change with position, so do vector components, so it isn't sufficient just to take the derivative of the components.

$\frac{\partial \vec{V}}{\partial x^ \beta}=\sum_{\alpha =0}^n \frac{\partial V^\alpha}{\partial x^\beta}\vec{e_\alpha}+\sum_{\alpha =0}^nV^\alpha\frac{\partial \vec{e_\alpha}}{\partial x^\beta}$

You need the derivative of the components and the derivative of the basis vectors.

Derivatives of the basis vectors are themselves vectors that can be expressed as components times the same basis vetors.

$\frac{\partial \vec{e_\alpha}}{\partial x^\beta}=\Gamma^\gamma_{\alpha \beta}\vec{e_\gamma}$

By this definition, the Christtofel Symbol is the $\gamma$ component of the $\beta$ derivative of the $\alpha$ basis vector .

But keep in mend $\vec{e_\alpha}=\frac{\partial \vec{r}}{\partial x^\alpha}$ so $\frac{\partial \vec{e_\alpha}}{\partial x^\beta}=\frac{\partial^2 \vec{r}}{\partial x^\alpha \partial x^\beta}=\frac{\partial ^2 \vec{r}}{\partial x^\beta \partial x^\alpha}$

So $\Gamma^\gamma_{\alpha \beta}=\Gamma^\gamma_{\beta \alpha}$

So the full vector derivative is:

$\frac{\partial \vec{V}}{\partial x^\beta} = \sum_{\alpha =0}^n \frac{\partial V^\alpha}{x^\beta}\vec{e_\alpha}+\sum_{\alpha=0}^n V^\alpha \frac{\vec{e_\alpha}}{x^\beta}=\sum_{\alpha =0}^n \frac{\partial V^\alpha}{\partial x^\beta}\vec{e_\alpha}+\sum_{\alpha=0}^n\sum_{\gamma=0}^nV^\alpha \Gamma^\gamma_{\alpha \beta}\vec{e_\gamma}$

Regarding calculating the Christoffel Symbols, which it should be noted are not tensor coordinates, you can use the definition.

$\frac{\partial \vec{e_\alpha}}{\partial \beta}=\sum _{\gamma=0}^n \Gamma^\gamma_{\alpha\beta}\vec{e_\gamma}= \frac{\partial^2 \vec{r}}{\partial \alpha \partial \beta}$

$\frac{\partial \vec{e_\alpha}}{\partial \beta}\cdot \vec{e_\mu}=\sum _{\gamma=0}^n \Gamma^\gamma_{\alpha\beta}\vec{e_\gamma}\cdot \vec{e_\mu}= \frac{\partial^2 \vec{r}}{\partial \alpha \partial \beta} \cdot \vec{e_\mu}$

$g_{\alpha \beta}=\vec{e_\alpha}\cdot \vec{e_\beta}$ is the metric.

$\sum_{\gamma=0}^n \Gamma^\gamma_{\alpha \beta}g_{\mu \gamma}=\frac{\partial ^2 \vec{r}}{\partial \alpha \partial \beta}\cdot \vec{e_\mu}$

And this implies a series of linear equatiosn from which you can figure out the Christoffel Symbols.

So you can use the definition, or that linear relation.

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Here's some more intro details: Christoffel Symbols

Answer 2

That, while $$\nabla_XY=\nabla_{\sum_iX^i\partial_i}\left(\sum_jY^j\partial_j\right),$$ $$=\sum_iX^i\left(\nabla_{\partial_i}\left(\sum_jX^j\partial_j\right)\right),$$ in the other hand, being a Leibnizian's operation, it complies
$$\nabla_XY=\nabla_{X}\left(\sum_jY^j\partial_j\right),$$ $$=\sum_j\nabla_{X}\left(Y^j\partial_j\right),$$ $$=\sum_j(\nabla_{X}Y^j)\partial_j+\sum_jY^j(\nabla_{X}\partial_j).$$

Now, if you can asure that the tangent space can be imbued with a differentiable metric (bilinear, symmetric, positive defined and non-degenerated) map $g:T_pM\times T_pM\to\mathbb R$, then for the coordinated basis $\partial_i$ one constructs the metric tensor $g_{ij}=g(\partial_i,\partial_j)$. This matrix, at $p$ covered by $U$, would be invertible, having inverse matrix $[g^{ij}]$. Then, at least in the coordinated patch $$\Gamma^k{}_{ij}=\frac{1}{2}\sum_sg^{ks} (\partial_j g_{is}+\partial_ig_{sj}-\partial_sg_{ij}),$$ accurately.