2

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.

Arctic Char
  • 16,007

1 Answers1

1

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 $$[e_r,e_{\theta}] = \nabla_{e_r}e_{\theta}-\nabla_{e_{\theta}}e_e.$$

Gibbs
  • 8,230
  • Thanks for your help. I will try to follow your hints. However, how should I compute the norm $\lVert e_r \rVert_g$ ? Is it $\lVert e_r \rVert_g = \sqrt{g(e_r, e_r)}$ ? I am not sure. –  Jun 07 '18 at 13:10
  • Yes exactly what you say. – Gibbs Jun 07 '18 at 13:43
  • Your hints are fine, but until now I do not really see what to do. I understand how you determined $g(e_r, e_{\theta})$. What is $g(e_r, e_r)$ then ? If I plug in $e_r$ and $e_{\theta}$, I would get $0 = e_r(g(e_{\theta}, Z)) - g(\triangledown_{e_r} e_{\theta}, Z) - g(e_{\theta}, \triangledown_{e_r} Z)$ and how to continue ? Sorry, I really feel ashamed that I am not able to do the computations, but I never did something like this before and think it is better to ask someone than giving up. If you would like to help me and find the time, it is better if you could give me more details, please. –  Jun 07 '18 at 14:04
  • If you want to compute $g(e_r, e_r)$ you just need to use the new expression of $g$: $g(e_r, e_r) =\frac{4} {1+r^2} dr^2(e_r,e_r)=\frac{4} {1+r^2} \left(\frac{1+r^2}{2} \right)^2$. – Gibbs Jun 07 '18 at 15:00
  • Ok ! Thanks ! I see it now. But I still don´t see how to compute $\nabla_{e_r} e_r, \nabla_{e_r}e_{\theta},...$ using the relation $\nabla g = 0$. Sorry that I ask so much. In fact, this no homework question. It is for exam preparation. Of course, I thank you that I can learn from you. –  Jun 07 '18 at 15:37
  • I edit my answer with some more information. – Gibbs Jun 08 '18 at 09:45
  • Thanks a lot ! It is now clear to me ;). –  Jun 08 '18 at 15:54