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 $x=\theta^1,y=\theta^2$)

$$ \left( \begin{array}{cc} \frac{(\cos (x)+2)^2}{(\cos (x)+2)^2 (5-4 \sin (x))-4 (\cos (x)+2)^2 \cos ^2(y) (2-2 \sin (x))^2 \sin ^2(y)} & -\frac{2 (\cos (x)+2) \cos (y) (2-2 \sin (x)) \sin (y)}{(\cos (x)+2)^2 (5-4 \sin (x))-4 (\cos (x)+2)^2 \cos ^2(y) (2-2 \sin (x))^2 \sin ^2(y)} \\ -\frac{2 (\cos (x)+2) \cos (y) (2-2 \sin (x)) \sin (y)}{(\cos (x)+2)^2 (5-4 \sin (x))-4 (\cos (x)+2)^2 \cos ^2(y) (2-2 \sin (x))^2 \sin ^2(y)} & \frac{5-4 \sin (x)}{(\cos (x)+2)^2 (5-4 \sin (x))-4 (\cos (x)+2)^2 \cos ^2(y) (2-2 \sin (x))^2 \sin ^2(y)} \\ \end{array} \right) $$

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    $\begingroup$ Yeah, you have a computation error. See win.tue.nl/~rvhassel/Onderwijs/Tensor-ConTeX-Bib/…. $\endgroup$ – levap Feb 25 '19 at 13:45
  • $\begingroup$ That looks like a very useful resource. Thank you for the link. $\endgroup$ – afightingchance Feb 25 '19 at 13:59
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    $\begingroup$ It's correct. For me, it's simpler to write (the matrix representing) $h$ as $\begin{pmatrix} \left< \frac{\partial \Phi}{\partial \theta^1}, \frac{\partial \Phi}{\partial \theta^1} \right> & \left< \frac{\partial \Phi}{\partial \theta^2}, \frac{\partial \Phi}{\partial \theta^1} \right> \\ \left< \frac{\partial \Phi}{\partial \theta^2}, \frac{\partial \Phi}{\partial \theta^1} \right> & \left< \frac{\partial \Phi}{\partial \theta^2}, \frac{\partial \Phi}{\partial \theta^2} \right> \end{pmatrix}$ where $\left< \cdot, \cdot \right>$ is the standard inner product on $\mathbb{R}^3$. $\endgroup$ – levap Feb 25 '19 at 15:20
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    $\begingroup$ This is equivalent to your expression. $\endgroup$ – levap Feb 25 '19 at 15:21
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    $\begingroup$ Thanks again. Your expression indeed looks neater. I kind of derived mine without knowing about fundamental forms (I read up on this after your first link). I mustve made a mistake typing it into mathematica then... $\endgroup$ – afightingchance Feb 25 '19 at 15:34

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).

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