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Let us consider a 2-dimensional unit disk $U_0$ with a puncture at $0$ and some Riemannian metric $g$ on it, which is $\textbf{flat}$ near the puncture. ( Remark: $g$ might not be extendedable to the metric on the whole unit disk $U$). As an example think about the ice-cream cone.

We also require that $0$ is a puncture and not a removed disk or a point at infinity. Frankly speaking I am not quite sure how to express this condition correctly, but let us formally express it this way: $g$ is locally (near removed $0$) isometric to some metric $G=G_{ij}$ on the punctured $xy$-plane where $G$ is a matrix, representing the metric tensor in the punctured neighbourhood of $0$ and satisfying $|G|< C$ and $|G^{-1}|< C$ for any $(x,y)\neq (0,0)$.

We call any two such metrics $g_1$ and $g_2$ locally equivalent if there are two punctured neighbourhoods of $0$ ( small enough) $U^1$ and $U^2$ so that $(U^1,g_1)$ is isometric to $(U^2, g_2)$.

Question: what is the moduli space of such equivalence classes of flat punctured metrics?

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up vote 8 down vote accepted

The moduli space is $(0,\infty)$.

Pass to the completion, you get a point for $0$. The metric in the neighborhood of $0$ has a natural cone structure. The total angle around $0$ is the only invariant.

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Anton, thank you very much, in fact this is quite what I expected. Could you please provide some more detailed argument ( sketch of the proof) or give some references? – Axel Nov 6 '12 at 5:57
Look at the geodesics in the completion which come from $0$. If points slide along pair of such close geodesics then the $($distance$)^2$ between them is a quadratic polynomial. You may define angle between the geodesics to make this polynomial look like cosine rule. Once it is done you get a polar coordinates in your disc and can talk about total angle. – Anton Petrunin Nov 6 '12 at 17:54
I see now, thank you! – Axel Nov 7 '12 at 6:57

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