MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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