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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.

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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.

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  • $\begingroup$ 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

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