How do you define a Riemannian metric for the unit circle. Is it $ds^2=dx^2+d\theta^2$?

I want to also measure the length of the vector from the origin. This would be a standard euclidean metric given by $ds^2=dx^2+dy^2$?

What is the difference between a euclidean and Riemannian metric?



For the first question, I think the metric for circle depends how you parametrize the circle, usually we choose $(\cos\theta,\sin\theta)$, then the metric induced from $\mathbb{R^2}$ is $g_{circle}=d\theta^2$, but if you wanna to parametrize the circle by $(x,\pm\sqrt{1-x^2})$, then the induced metric is $\frac{1}{1-x^2}dx^2$, of course, you will get different metric if you parametrize in a different way (different coordinates).

For the second question, the usual metric for $\mathbb{R}^2$ is $g=dx^2+dy^2$, write in the polar form it becomes $g=dr^2+r^2d\theta^2$, now if you restrict this metric on the unit circle, that means $r=1$, so $g_{circle}=d\theta^2$. For the difference between a Euclidean metric and Riemannian metric, I think you already know metric here means the inner product of tangent vectors of point rather than "distance metric", the Euclidean metric is just the usual inner product, when you do inner product for two vectors $<a,b>_x=a_1b_1+\cdots+a_nb_n$, this inner product is independent of starting point of your vector, but Riemannian metric is a inner product whose coefficients change pointwisely, i.e. $<a,b>_x=f(x)(a_1b_1+\cdots+a_nb_n)$, then you see the standard Euclidean metric is just a special case of Riemannian metric.

So back to the first question, for Riemannian metric on unti circle you can define
metric like $g_{cirlce}=f(\theta)d\theta^2$ or $g_{cirlce}=f(x)\frac{1}{1-x^2}dx^2$, but there are all conformal to each other if you know what conformal metric is.

  • $\begingroup$ Thanks for your answer. $\endgroup$ Sep 17 '21 at 7:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.