# Reverse engineering a metric from its properties

I have been reading an article, and I am not clear on what he does in this one part.

He starts with a Riemmannian manifold $\Sigma$ that is $\mathbb{H}^2/\Gamma$, a quotient of the hyperbolic space by some group of isometries. He then takes an isometry $\tau:\Sigma\rightarrow\Sigma$ and defines $M=\Sigma\times[0,1]/[(x,0)\sim(\tau(x),1)]$ the mapping torus.

It is claimed in the article that if you take a geodesic $\gamma$ in $M$ then the tube around it has a flat metric on it's boundary. This (unless in am missing something) is not true for the product metric. you can take $(t,\rho,z)$ to be your coordinates when $(t,\rho)$ are the Fermi coordinates in $\mathbb{H}^2$ then $ds^2=d\rho^2+cosh^2(\rho)dt^2+dz^2$ will not be flat at the boundary.

The boundary of a tube around a geodesic is flat, if you you think of a quotient of $\mathbb{H}^3$ (and that is what I originally thought he did) but $(x,r)\rightarrow(\tau(x),r+1)$ is not an isometry of $\mathbb{H}^3$, so you cant get $M=\mathbb{H}^3/G$.

Any thought on how this can be done? I thank you all in advance

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What article would that be? – Will Jagy Mar 31 '11 at 17:01
I agree with @Will: a reference would go a long way. – Igor Rivin Apr 1 '11 at 2:16
The article is "Z2-sytolic freedom and Quantum codes" by M. Friedman. – Ethan Fetaya Apr 1 '11 at 5:38
The map has a finite order, so from Thurston the structure would be H^2xR. – Ethan Fetaya Apr 1 '11 at 5:56
Are you sure Friedman doesn't just mean flat in the induced metric on the torus? That's much weaker, and is trivially true in this case. – Dylan Thurston Apr 3 '11 at 2:26