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The standard torus $\mathbb{S}^1 \times \mathbb{S}^1$ can be given a flat metric if we construct the torus as a quotient space of a rectangle with opposite sides identified and declare the Riemannian metric on the torus as the pullback of the metric on the flat rectangle. By construction this metric is flat.

I want to generalize this construction to construct metrics of constant (sectional) curvature on the $n$-holed torus. So in particular let's start with the $2$-holed torus: as seen here, we can construct it as an octagon with opposite sides identified. So why does the same construction I mentioned for the standard torus not generalize to give a flat metric on the $2$-holed torus? I have heard that if we do such a construction starting out with a hyperbolic octagon instead, we can get a metric of constant negative curvature on the $2$-holed torus. But I fail to see where my construction of the flat metric fails. It has to fail because of Gauss-Bonnet, but I don't see how (I know the angles don't add up in the flat construction - whereas they do add up for a hyperbolic octagon with sides of angle $\frac{\pi}{4}$ - but I'm not seeing why this is crucial).

Furthermore, how can we construct an $n$-holed torus as a quotient space of a polygon? Do we take a $4n$-polygon and identify its sides in some manner (how?)? Do they admit constant negative curvature metrics as well?

Matheus Andrade
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    A Riemannian metric allows you to calculate angles between tangent vectors, and if you have a bunch of tangent vectors on a surface and calculate the consecutive angles between them they have to add up to $2\pi$. The proposed flat metric for the octagon will fail to have this property. And yes, for the surface of genus $g$ you identify sides of a $4g$-gon, and hyperbolic tilings give them hyperbolic metrics. – Qiaochu Yuan Jul 27 '22 at 00:27
  • @QiaochuYuan but why do consecutive angles have to add up to $2 \pi$? Obviously this is true in the Euclidian case but I don't see why it has to be true for an arbitrary Riemannian manifold. – Matheus Andrade Jul 27 '22 at 00:29
  • @QiaochuYuan Ah, I think this follows from Hopf's "turning tangents" theorem. Am I correct? – Matheus Andrade Jul 27 '22 at 00:39
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    No theorem needed. In the end there are coordinates in a neighborhood of a vertex. The edges have to be smooth curves ending at the vertex. There has to be a well defined metric tensor at the vertex. This all implies the angles have to add up to $2\pi$. – Deane Jul 27 '22 at 01:34
  • @Deane I think I see what you mean. Since in the quotient space all the original vertices collapse to a single point, the sum of the angles has to be the angle around that collapsed point, and the angle around a point is $2 \pi$. This still feels a bit too informal to me. Is this what you meant or did you think of another approach? Thanks for replying. – Matheus Andrade Jul 27 '22 at 02:03
  • That's more or less it. My explanation can be made rigorous. In the tangent space at the vertex, you have the tangent vector to each edge. The angles between consecutive tangent vectors add up to $2\pi$ simply because if you go around the origin once, you go through an angle of $2\pi$. – Deane Jul 27 '22 at 02:25
  • @Deane Ah, I think I got it! Rigorously, using coordinates at the collapsed point and using the fact that the eight lines corresponding to the eight edges in the original octagon must intersect at exactly the collapsed point, we see that the angles have to add up to $2 \pi$ simply because of high school geometry. Do you agree? – Matheus Andrade Jul 27 '22 at 02:28
  • An angle is the distance between two points on the unit circle. If you have a finite number of points on the unit circle, then the complement consists of a finite number of disjoint open arcs. It follows that the sum of the lengths of the arcs equals the length of the circle. – Deane Jul 27 '22 at 02:35
  • @Deane that's another way of seeing it I hadn't thought about. Thanks! If you want to make an answer out of these comments I'll accept it. – Matheus Andrade Jul 27 '22 at 02:39
  • Sorry but I'm too lazy. Please feel free to write your own detailed answer. By the way, the fact that all the angles add up to $2\pi$ at a vertex is a crucial fact in the proof of Gauss-Bonnet. – Deane Jul 27 '22 at 02:41
  • @Deane No problem, that's okay! You've helped a lot already anyway. It's been a long time since I studied Gauss-Bonnet so I didn't remember this fact. I don't have the time to write the details right now so I'll wait and see if someone else wants to write a detailed answer. If that doesn't happen I'll do as you said. Thanks! – Matheus Andrade Jul 27 '22 at 02:43
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    Not that you asked, but edge-gluing a Euclidean octagon results in a flat metric away from the vertex, with incident angle $8 \times (3\pi/4) = 6\pi$ at the vertex. The "total curvature" of the vertex is $2\pi - 6\pi = -4\pi$, consistent with Gauss-Bonnet. – Andrew D. Hwang Jul 27 '22 at 02:47
  • @AndrewD.Hwang thanks for the observation! That's very interesting. But what would happen to the metric at the vertex? I'm not sure I understand the construction you mentioned. – Matheus Andrade Jul 27 '22 at 02:50
  • There's a "saddle cone singularity" with incident angle $6\pi$. One way to make a model is to take three disks of paper, slit each radially from the center, and join the cut edges cyclically. – Andrew D. Hwang Jul 27 '22 at 22:41

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