I am currently doing some exercises to prepare for an Oral next month and have found myself stuck on computing Christoffel symbols for a certain metric $h$ defined below. The setup is as follows:
We have a chart $(\theta^1,\theta^2)$ on $T^2$ and a map $\Phi : T^2 \to \mathbb{R}^3$ \begin{align*} \Phi(\theta^1,\theta^2) = \left((2+\cos\theta^1)\cos\theta^2, (2+\cos\theta^1)\sin\theta^2,\sin \theta^1\right). \end{align*} We are asked to define a metric $h$ on $T^2$ as the pullback of the Euclidean metric $$h= \Phi^*(dx^i \otimes dx^i)$$ and then compute the Christoffel symbols for $\nabla$ on $(T^2,h)$. My progress thus far has been to compute $$ h(\xi,\eta)=\left( \xi^j \frac{\partial\Phi^m}{\partial\theta^j} \right) \left( \eta^j \frac{\partial\Phi^m}{\partial\theta^j}\right), $$ with summation over $m,j$ implicit. Writing out the matrix for $h$ is tractable. I find $$ \begin{bmatrix} 5-4\sin\theta^1& 2 (2-\sin \theta^1)(2+\cos \theta^1)\sin \theta^2 \cos \theta^2\\ 2 (2-\sin \theta^1)(2+\cos \theta^1)\sin \theta^2 \cos \theta^2 & \left(2+\cos\theta^1\right)^2 \end{bmatrix} $$ However, when I start computing the Christoffel symbols I run into what feels like a computational nightmare, mainly due to the rather ugly inverse matrix for $h$. Is there any way around this or should I just suck it up (also, did I make a computation error?)?
Edit: Let me just add that Mathematica gives me the following inverse matrix (to motivate my lack of computational morals with $x=\theta^1,y=\theta^2$)
$$ \left( \begin{array}{cc} \frac{(\cos (x)+2)^2}{(\cos (x)+2)^2 (5-4 \sin (x))-4 (\cos (x)+2)^2 \cos ^2(y) (2-2 \sin (x))^2 \sin ^2(y)} & -\frac{2 (\cos (x)+2) \cos (y) (2-2 \sin (x)) \sin (y)}{(\cos (x)+2)^2 (5-4 \sin (x))-4 (\cos (x)+2)^2 \cos ^2(y) (2-2 \sin (x))^2 \sin ^2(y)} \\ -\frac{2 (\cos (x)+2) \cos (y) (2-2 \sin (x)) \sin (y)}{(\cos (x)+2)^2 (5-4 \sin (x))-4 (\cos (x)+2)^2 \cos ^2(y) (2-2 \sin (x))^2 \sin ^2(y)} & \frac{5-4 \sin (x)}{(\cos (x)+2)^2 (5-4 \sin (x))-4 (\cos (x)+2)^2 \cos ^2(y) (2-2 \sin (x))^2 \sin ^2(y)} \\ \end{array} \right) $$
Your $h_{22}$ is right, but the mixed 12 and 21 components should be 0, since the torus coordinates are orthogonal, and $h_{11}$ is 1. This is also obvious since you are changing latitude on a "tube" with unit radius.
Since you use Mathematica you should perhaps check the xAct tensor package, www.xact.es
With the xPrint GUI for xAct, which I developed, learning to use xAct is much quicker. Please see my profile for direct link (I have problems with my mobile browser right now).