First of all, I don't know much about Riemannian geometry. The following statements are in a paper about $G$-spaces that I'd like to understand, or at least I'd like to find some references.

Let $(Y,d_1)$ be a metric space. We endow $Y\times\mathbb{R}$ with the metric $$d((y_1,t_1),(y_2,t_2))=d_1(y_1,y_2)+\min\{|t_1-t_2|,1\}.$$

This is ok.

Now, let $Y$ be a $2$-dimensional torus with a flat Riemannian metric.

I had understood that a Riemannian metric isn't really a metric, for a Riemannian metric is an inner product on the tangent space, so I don't get how is that we have $Y$ as a metric space. What is next is more confusing to me:

Let $g(t),t\in\mathbb{R}$ a dense $1$-parameter subgroup of $Y$ and let $H\subseteq Y\times\mathbb{R}$, $H=\{(g(t),t):t\in\mathbb{R}\}$.

Then we have that in $H$, with the metric of subspace of $Y\times\mathbb{R}$: $$d((g(s),s),(g(t),t))=(1+\| \dot{g}(0)\|)|t-s|$$for small $|t-s|$, where $\dot{g}(0)$ is the tangent of the $1$-parameter group $g(t),t\in\mathbb{R}$ and $\|\cdot \|$ is the norm of the tangent space of $Y$ at the identity element derived from the Riemannian tensor.

Two questions about it:

1) What does it mean small $|t-s|$? I mean, how small it needs to be?

2) Is there any book or reference where I can get to know what is really $\dot{g}(0)$? And what is really the metric on the torus $Y$? It would be nice if we just can give them explicitly.

Thank you.


First of all, the metric you put on a complete Riemannian manifold is usually the one given by $d(p,q) = $ the minimal length of a geodesic joining $p$ and $q$.

I don't know what the good notion of "small" is in this context.

Finally, $g$ is a map $\mathbb{R}\to M$, so $\dot g$ denotes the velocity vector field (of which you then take the norm using the Riemannian metric.

  • $\begingroup$ Thank you very much, Daniel. Do you have any clue how to get the equality $d((g(s),s),(g(t),t))=(1+\| \dot{g}(0)\|)|t-s|$? $\endgroup$ – Talexius Aug 3 '16 at 19:33
  • $\begingroup$ It should be an equation holding "infinitesimally" (in the sense that in reality you should have a $+O(|t-s|^2)$ in the equation), and it is basically Pythagora's theorem. Try drawing $g(t)$ as a straight vertical line and the time parameter as a horizontal line to get an idea of what is happening. $\endgroup$ – Daniel Robert-Nicoud Aug 4 '16 at 1:18

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.