Remember the Christoffel symbols This might be a little bit different from what is asked normally on this page, but does anybody here know a good way to remember the definition of the Christoffel symbol?
\[ \Gamma^k_{ij} = \frac 12 \sum_{l=1}^2 g^{kl} \left(\frac{\partial g_{il}}{\partial u^j} + \frac{\partial g_{jl}}{\partial u^i} - \frac{\partial g_{ij}}{\partial u^l}\right)
\]
I have to use this quite often recently, but I always have to look all the indices up, because I have no intuition for this beast.
 A: Here are what helped me to remember these formulas:
(1) using Einstein summation notation $A_i B^i:=\sum_{i=1}^2 A_i B^i$, $A^i B_i:=\sum_{i=1}^2 A^i B_i$.
(2) define $f_{,i}:=\frac{\partial f}{\partial u^i}$.
(3) $i,j$ are symmetric in $\Gamma^k_{ij}$.  $i,j$ are symmetric in $g^{ij}$ and $g_{ij}$.
Now the Christoffel symbols becomes:
$$\Gamma^k_{ij} = \frac 12 g^{kl} \left(g_{li,j} + g_{lj,i} - g_{ij,l}\right)\tag{1}$$
If you define $\Gamma_{lij} = g_{lk}\Gamma^k_{ij}$, then you have:
$$\Gamma_{lij} = \frac 12 \left(g_{li,j} + g_{lj,i} - g_{ij,l}\right)\tag{2}$$
And now look at the positions of these indices $k,i,j$ or $l,i,j$ in (1) and (2) and try to remember them.
Hope it helps!
A: It is better not to regard the formula for the Christoffel symbols as "the definition", but rather as a consequence of a theorem. The theorem is that given a metric $g$ there is a unique torsion-free connection $\nabla$ (called the Levi-Civita connection of $g$) such that $\nabla g = 0$. The proof of this theorem gives a derivation of the formula, and remembering the argument is the best way to remember the formula.
Fix any torsion-free affine connection $D$. We look for a torsion-free connection $\nabla = \nabla + \Pi$ such that $\nabla_{i}g_{jk} = 0$. This gives the equation
$$ 0 = \nabla_{i}g_{jk} = D_{i}g_{jk} - \Pi_{ijk} - \Pi_{ikj},$$
where $\Pi_{ijk} = \Pi_{ij}\,^{p}g_{pk}$ and the Einstein summation and abstract index conventions are in use (so indices are formal labels indicating tensor symmetries and not depending on any choice of frame, although they can be interpreted as the components with respect to a frame if a frame is chosen). We can rewrite the equation as
\begin{equation}
 \Pi_{ijk} + \Pi_{ikj} = D_{i}g_{jk} .
\end{equation}
Now we try to solve this equation for $\Pi_{ijk}$. We note that $\Pi_{ijk} = \Pi_{jik}$ because $D$ is torsion-free. Consider the cyclic permutations:
$$D_{i}g_{jk}  = \Pi_{ijk} + \Pi_{ikj} = \Pi_{ijk} + \Pi_{kij},$$
$$D_{j}g_{ki}  = \Pi_{jki} + \Pi_{jik} = \Pi_{jki} + \Pi_{ijk},$$ 
$$D_{k}g_{ij}  = \Pi_{kij} + \Pi_{kji} = \Pi_{kij} + \Pi_{jki}.$$
We want an expression for $\Pi_{ijk}$. Adding the first two equations and subtracting the last gives
$$ D_{i}g_{jk} + D_{j}g_{ki} - D_{k}g_{ij} = 2\Pi_{ijk}.$$
Dividing by $2$ and raising the index $k$ with the inverse tensor $g^{kp}$ gives
$$ \Pi_{ij}\,^{k} = \tfrac{1}{2}g^{kp}(D_{i}g_{jp} + D_{j}g_{ip} - D_{p}g_{ij}).$$
The formula in terms of the partial derivatives of the components of the metric with respect to a coordinate frame is a special case of the preceding. Choose coordinates $x^{1}, \dots, x^{n}$ and define a connection $\partial$ by the requirement that the coordinate coframe $dx^{1}, \dots, dx^{n}$ be parallel. Then $\partial_{i}g_{jk}$ can be interpreted as the partial derivative $\tfrac{\partial g_{jk}}{\partial x^{k}}$ and the formulas above yield the formula in coordinates for the Christoffel symbols.
