Computations with respect to a Riemannian metric I have the following Riemannian metric on $\mathbb{R}^2$ : $$g = 4\frac{1}{(1+r^2)^2}(dx^2+dy^2),\text{ with }r^2 = x^2 + y^2.$$ At every point of $\mathbb{R}^2\backslash \{(0,0)\}$, let $$e_r = \frac{1+r^2}{2} \partial_r, \quad e_{\theta} = \frac{1+r^2}{2r}\partial_{\theta},$$
where $(r, \theta)$ are the polar coordinates on $\mathbb{R}^2$. I didn´t know how to determine $g(e_r, e_{\theta})$ and to compute $\lVert e_r \rVert_g$. I thank Gibbs for the explanations. I am still unsure how to calculate $\nabla_{e_r} e_r, \nabla_{e_r}e_{\theta}$,etc, where $\nabla$ denotes the Levi-Civita connection. I was recommended to use that the Levi-Civita connection satisfies $$0 = X(g(Y,Z))-g(\nabla_XY,Z)-g(Y,\nabla_XZ).$$
However, can I also use that $\nabla$ is torsion-free, i.e. $\nabla_X Y - \nabla_Y X - [X, Y] = 0$ ?
Thanks in advance.
 A: First of all, to compute $g(e_r,e_{\theta})$ you should write $g$ in terms of $dr^2$ and $d\theta^2$. If you try to differentiate the two equations defining polar coordinates
$$
\begin{cases}
x = r\cos \theta \\
y = r\sin \theta
\end{cases}
$$
you find $dx = \cos \theta dr - r\sin \theta d\theta$ and $ dy = \sin \theta dr + r\cos \theta d\theta$. Then $dx^2+dy^2 = dr^2+r^2d\theta^2$. So 
$$g = \frac{4}{1+r^2}(dr^2+r^2d\theta^2).$$
Remember that here $dr^2$ is a shorthand for $dr^{\otimes 2}$. 
Now, for $X,Y,Z$ vector fields on $\mathbb{R}^2$, the Levi-Civita connection satisfies
$$0 = X(g(Y,Z))-g(\nabla_XY,Z)-g(Y,\nabla_XZ) $$
because $\nabla g = 0$. So, for example, if $X=Y=Z=e_{\theta}$ you get
$$0 = e_{\theta}(g(e_{\theta},e_{\theta})) = 2g(\nabla_{e_{\theta}}e_{\theta},e_{\theta})$$
which means that $\nabla_{e_{\theta}} e_{\theta}$ is proportional to $e_r$. You can try all the eight possibilities (i.e. $X=Y=Z=e_r, X=Y=e_r$ and $Z=e_{\theta}$, etc.) and find relations between the fields you have. Once you have the equations defining $\nabla_{e_r}e_r, \nabla_{e_r}e_{\theta}$, etc., use
$$[e_r,e_{\theta}] = \nabla_{e_r}e_{\theta}-\nabla_{e_{\theta}}e_e.$$
