Is there a good way to compute Christoffel Symbols Lets say you have a Riemannian Manifold $(M,g)$, and you have some given chart where $g = g_{ij} dx_i dx_j$ and you wish to compute the Christoffel symbols for the Riemannian connection in this chart. To do this involves:
1) Calculating the inverse matrix of $g$, which is not too bad in dimensions 2 and 3, but becomes quite painful in higher dimensions.
2) Using the formula
$\Gamma_{ij}^k = \frac{1}{2} g^{kl}(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij}) $
which if you look closely the index that is summed over on the right is $l$, so there are $n=dim M$ terms on the right that need to be evaluated, and inside each of these you must take a partial derivative of three terms in $g$, so in total on the right-side there are $3n$ terms to be computed. This is just for each value of $k$.
Luckily $\Gamma_{ij}^k$ is symmetric in $ij$, and so you only have to compute about half of the $3n^3$ quantities involved, but it is still a very involved calculation.
Does anyone know a simpler method to go about this? Or an easier way to look at the calculation of this? As $\Gamma_{ij}^k$ is an $n \times n$ symmetric matrix for each fixed $k$, maybe there is a way to write $(\Gamma_{ij}^k)$, the matrix, as a matrix product of some other matrices that would be easier to write down. That way you would not have to work with index notation for doing the calculation, which also slows things down.
 A: I'd like to expand on Hans Lundmark's answer
because the question keeps recurring.
Let $q^1,\ldots,q^n$ denote the generalized coordinates.
Introduce additional velocity variables $\dot{q}^1,\ldots,\dot{q}^n$.
If the dot accent is already reserved for other things in your theory,
use another accent to avoid confusion.
First, treat all $q^i$ and $\dot{q}^i$ as pairwise independent variables and
define, using Einstein summation convention,
$$L(q^1,\ldots,q^n;\dot{q}^1,\ldots,\dot{q}^n)
= \frac{1}{2} g_{ij}(q^1,\ldots,q^n)\,\dot{q}^i \dot{q}^j\tag{1}$$
To avoid confusion later on, I have explicitly indicated
the formal dependencies of the expressions for $g_{ij}$ and $L$.
Note that $L$ can immediately be written down when given the
first fundamental form.
Now consider a twice differentiable curve,
which makes the $q^i$ functions of some independent new parameter $\tau$,
and set $\dot{q}^i = \frac{\mathrm{d}q^i}{\mathrm{d}\tau}$.
Proposition. On the curve parameterized with $\tau$ we have
$$g^{kh}\left(\frac{\mathrm{d}}{\mathrm{d}\tau}
\left(\frac{\partial L}{\partial\dot{q}^h}\right)
- \frac{\partial L}{\partial q^h}\right)
= \ddot{q}^k + \Gamma^k_{\ ij} \dot{q}^i \dot{q}^j
\tag{2}$$
You might recognize the right-hand side as the expression
constrained by geodesics,
and you might recognize the expression wrapped around $L$ in the left-hand side
as the form of Euler-Lagrange differential equations,
which correspond to variational problems with the Lagrangian $L$.
Therefore $(2)$ hints at a variational foundation for the geodesics equation.
Such interpretations make $(2)$ easier to memorize or cross-link with
other knowledge, but all we actually need is the identity $(2)$.
The following observations should provide enough details to prove $(2)$.
We can rewrite $(2)$ to
$$\frac{\mathrm{d}}{\mathrm{d}\tau}
\left(\frac{\partial L}{\partial\dot{q}^h}\right)
- \frac{\partial L}{\partial q^h}
= g_{hk} \ddot{q}^k + \Gamma_{hij} \dot{q}^i \dot{q}^j
\tag{3}$$
The idea now is, for every $h\in\{1,\ldots,n\}$,
to take the left-hand side of $(3)$,
plug in expressions for the metric in $L$,
and rewrite the thing so that it matches the format of the right-hand side,
where dotted variables occur only in the places shown.
Then you can read off the Christoffel symbols of the first kind, $\Gamma_{h**}$,
from the coefficients of the velocity products.
To obtain $(2)$ and $\Gamma^k_{\ **}$ is then just a matter of multiplication
with the inverse metric coefficients matrix $((g^{kh}))$, or
equivalently, taking the right-hand side expressions of
$(3)$ obtained for $h\in\{1,\ldots,n\}$
and then, for each $k\in\{1,\ldots,n\}$,
finding a linear combination whose only second derivative
with respect to $\tau$ is $\ddot{q}^k$, with coefficient $1$.
This is the form given by $(2)$, so the coefficients of the velocity products
are then the Christoffel symbols of the second kind, $\Gamma^k_{\ **}$.
Remember, while doing the partial derivatives of $L$, treat the $q^i$
and the $\dot{q}^i$ as independent formal variables.
You will do that with more concrete symbol meanings and metric expressions,
but in this moderately abstract setting, you can already refine
$$\begin{align}
\frac{\partial L}{\partial\dot{q}^h}
&= g_{hj}\dot{q}^j\tag{4}
\\\frac{\partial L}{\partial q^h}
&= \frac{1}{2}\frac{\partial g_{ij}}{\partial q^h}\dot{q}^i \dot{q}^j\tag{5}
\end{align}$$
However, when doing the
$\frac{\mathrm{d}}{\mathrm{d}\tau}$ outside of $L$, stick to the curve
and apply the chain rule accordingly:
$$\frac{\mathrm{d}g_{hj}}{\mathrm{d}\tau}
= \frac{\partial g_{hj}}{\partial q^i}\,\dot{q}^i\tag{6}$$
Now $(4)$, $(5)$, $(6)$ and the Levi-Civita formula
$$\Gamma_{hij} = \frac{1}{2}\left(
\frac{\partial g_{hj}}{\partial q^i}
+ \frac{\partial g_{ih}}{\partial q^j}
- \frac{\partial g_{ij}}{\partial q^h}
\right)$$
can be used to prove $(3)$ and thereby $(2)$.
But I will focus on how to apply that proposition.
Example:
Spherical coordinates with radius $r$, longitude $\phi$, latitude $\theta$,
with $\theta=\frac{\pi}{2}$ at the equator.
At index positions, I will write coordinate names instead of digits.
The first fundamental form is
$$\mathrm{d}s^2 = \mathrm{d}r^2
+ (r^2\sin^2\theta)\,\mathrm{d}\phi^2
+ r^2\,\mathrm{d}\theta^2$$
Accordingly, the Lagrangian $L$ is
$$L = \frac{1}{2}\left(\dot{r}^2
+ (r^2\sin^2\theta)\,\dot{\phi}^2
+ r^2\,\dot{\theta}^2\right)$$
We now treat $r,\phi,\theta,\dot{r},\dot{\phi},\dot{\theta}$
as independent variables and get
$$\begin{align}
\frac{\partial L}{\partial\dot{r}} &= \dot{r}
&\frac{\partial L}{\partial r}
&= (r\sin^2\theta)\,\dot{\phi}^2 + r\,\dot{\theta}^2
\\\frac{\partial L}{\partial\dot{\phi}} &= (r^2\sin^2\theta)\,\dot{\phi}
&\frac{\partial L}{\partial\phi} &= 0
\\\frac{\partial L}{\partial\dot{\theta}} &= r^2\,\dot{\theta}
&\frac{\partial L}{\partial\theta} &= (r^2\sin\theta\cos\theta)\,\dot{\phi}^2
\end{align}$$
Now we give up the independence, consider some curve paramterized by $\tau$
and obtain
$$\begin{align}
\frac{\mathrm{d}}{\mathrm{d}\tau}
\frac{\partial L}{\partial\dot{r}} &= \ddot{r}
\\\frac{\mathrm{d}}{\mathrm{d}\tau}
\frac{\partial L}{\partial\dot{\phi}} &= (r^2\sin^2\theta)\,\ddot{\phi}
+ 2\,(r\sin^2\theta)\,\dot{r}\,\dot{\phi}
+ 2\,(r^2\sin\theta\cos\theta)\,\dot{\phi}\,\dot{\theta}
\\\frac{\mathrm{d}}{\mathrm{d}\tau}
\frac{\partial L}{\partial\dot{\theta}} &= r^2\,\ddot{\theta}
+ 2\,r\,\dot{r}\,\dot{\theta}
\end{align}$$
And so
$$\begin{align}
\frac{\mathrm{d}}{\mathrm{d}\tau}
\left(\frac{\partial L}{\partial\dot{r}}\right)
- \frac{\partial L}{\partial r}
&= \underbrace{1}_{g_{rr}}\,\ddot{r}
+ \underbrace{(-r\sin^2\theta)}_{\Gamma_{r\phi\phi}}\,\dot{\phi}^2
+ \underbrace{(-r)}_{\Gamma_{r\theta\theta}}\,\dot{\theta}^2
\\\frac{\mathrm{d}}{\mathrm{d}\tau}
\left(\frac{\partial L}{\partial\dot{\phi}}\right)
- \frac{\partial L}{\partial\phi}
&= \underbrace{(r^2\sin^2\theta)}_{g_{\phi\phi}}\,\ddot{\phi}
+ 2\,\underbrace{(r\sin^2\theta)}_{\Gamma_{\phi r\phi}
= \Gamma_{\phi\phi r}}\,\dot{r}\,\dot{\phi}
+ 2\,\underbrace{(r^2\sin\theta\cos\theta)}_{\Gamma_{\phi\phi\theta}
= \Gamma_{\phi\theta\phi}}\,\dot{\phi}\,\dot{\theta}
\\\frac{\mathrm{d}}{\mathrm{d}\tau}
\left(\frac{\partial L}{\partial\dot{\theta}}\right)
- \frac{\partial L}{\partial\theta}
&= \underbrace{r^2}_{g_{\theta\theta}}\,\ddot{\theta}
+ 2\,\underbrace{r}_{\Gamma_{\theta r\theta}
= \Gamma_{\theta\theta r}}\,\dot{r}\,\dot{\theta}
+ \underbrace{(-r^2\sin\theta\cos\theta)}_{\Gamma_{\theta\phi\phi}}
\,\dot{\phi}^2
\end{align}$$
All other Christoffel symbols of the first kind are zero.
If we had a non-diagonal metric, some right-hand side expressions would have
several second derivatives, each accompanied by a corresponding metric
coefficient.
To obtain the Christoffel symbols of the second kind, find linear combinations
of the above right-hand side expressions that leave only one second derivative,
with coefficient $1$.
Here this is easy because the metric is already in diagonal form. Therefore
$$\begin{align}
g^{rh}\left(\frac{\mathrm{d}}{\mathrm{d}\tau}
\left(\frac{\partial L}{\partial\dot{q}^h}\right)
- \frac{\partial L}{\partial q^h}\right)
&= \ddot{r}
+ \underbrace{(-r\sin^2\theta)}_{\Gamma^r_{\ \phi\phi}}\,\dot{\phi}^2
+ \underbrace{(-r)}_{\Gamma^r_{\ \theta\theta}}\,\dot{\theta}^2
\\g^{\phi h}\left(\frac{\mathrm{d}}{\mathrm{d}\tau}
\left(\frac{\partial L}{\partial\dot{q}^h}\right)
- \frac{\partial L}{\partial q^h}\right)
&= \ddot{\phi}
+ 2\,\underbrace{\left(\frac{1}{r}\right)}_{\Gamma^\phi_{\ r\phi}
= \Gamma^\phi_{\ \phi r}}\,\dot{r}\,\dot{\phi}
+ 2\,\underbrace{(\cot\theta)}_{\Gamma^\phi_{\ \phi\theta}
= \Gamma^\phi_{\ \theta\phi}}\,\dot{\phi}\,\dot{\theta}
\\g^{\theta h}\left(\frac{\mathrm{d}}{\mathrm{d}\tau}
\left(\frac{\partial L}{\partial\dot{q}^h}\right)
- \frac{\partial L}{\partial q^h}\right)
&= \ddot{\theta}
+ 2\,\underbrace{\left(\frac{1}{r}\right)}_{\Gamma^\theta_{\ r\theta}
= \Gamma^\theta_{\ \theta r}}\,\dot{r}\,\dot{\theta}
+ \underbrace{(-\sin\theta\cos\theta)}_{\Gamma^\theta_{\ \phi\phi}}
\,
\dot{\phi}^2
\end{align}$$
All other Christoffel symbols of the second kind are zero.
A: It's worth noting that if the metric is diagonal the calculations become much simpler. Explicitly,
\begin{align} &\Gamma^{\lambda}_{\;\,\mu \nu} = 0  \\
&\Gamma^{\lambda}_{\;\,\mu \mu} = -\frac{1}{2}(g_{\lambda \lambda})^{-1} \partial_{\lambda} g_{\mu \mu} \\
&\Gamma^{\lambda}_{\;\,\mu \lambda} = \partial_{\mu} (\ln \sqrt{|g_{\lambda \lambda}|}) \\
&\Gamma^{\lambda}_{\;\,\lambda \lambda} = \partial_{\lambda} (\ln \sqrt{|g_{\lambda \lambda}|}),\end{align}
where $\mu \neq \lambda \neq \nu \neq \mu$. These equations can easily be proven using the formula given in the original question.
A: You can read off the Christoffel symbols from the Euler–Lagrange equations for the Lagrangian naturally defined by the metric, $L=\frac12 g_{ij} v^i v^j$. Still messy, but at least there's not much you have to memorize in order to do the computation.
I can't recall a good reference right off the bat, but Google found this, for example:
http://count.ucsc.edu/~rmont/classes/121A/polarChristoffelE_Lag.pdf
