# Cones with deficit angle 2$\pi$ and euler characteristics

I've managed to confuse myself with cones and deficit angles. Let's consider a conical defect in 2 dimensions. So the metric is the usual one in polar coordinates, $$ds^2 = dr^2 + r^2 d\phi^2,$$ except that now $\phi \sim \phi + 2\pi(1-\alpha)$. For $\alpha = 0$, the is just flat space. When $\alpha \neq 0$, there is a singularity in the curvature, and for example the Ricci scalar acquires a delta function: $$R(x) = 4\pi \alpha \, \delta^{(2)}_{x,x'},$$ where $x'$ is the location of the conical defect. Now if we use the Gauss-Bonnet theorem (remembering that in 2 dimensions $R=K/2$, where $K$ is the Gaussian curvature), we can relate the deficit angle to the Euler characteristic (neglecting any boundary terms) $$\chi = \alpha.$$

So my confusion now is: what does it mean to have $\alpha = 1$, which is to say that the deficit angle is $2\pi$? It seems weird that I can remove the whole angle and still have a 2 dimensional space. Since I don't have much intuition for what it means to remove $2\pi$, I looked up what manifolds have $\chi = 1$, I find things like the disk (which has a boundary), and the real projective plane, which is $S^2/\mathbb{Z}_2$ and non-orientable.

So what space is a cone with deficit angle $2\pi$? (Bonus question: what space has deficit angle $4\pi$, the Euler formula would suggest a sphere?)

• when $\alpha=1$ then you only have a line segment right? so that also has $\chi=1$ – Prastt Nov 29 '15 at 10:39
• I don't know much about this kind of stuff, but I think you have to extend the definition of Euler characteristics to noncompact manifold which you obtain after introducing deficit angle $2\pi$ here. This Math.SE post may help. – Minkyu Jun 9 '16 at 12:03