For the very first sight it may be surprise that the ordinary torus $S^1 \times S^1$ is flat: one argument to see this is the following. One can imagine a torus as a square with opposite sides identified and the square is obviously flat. However there are surfaces with higher genus (higher numbers of "holes") which also can be represented as polygons (4$g$ polygon for a surface of genus $g$) with sides indetified. However higher genus surfaces are negatively curved. So my question is

Why higher genus surfaces are negatively curved while they can be represented as flat polygons with sides identified?

up vote 10 down vote accepted

If you try to turn those polygon identifications into flat metrics you run into a problem at the vertices: the angles don't add up. This issue can only be addressed for the square.

  • I downvoted here be cause I don't understand the mathematical meaning of the sentence the "angles don't add up" – Tsemo Aristide Feb 21 '16 at 1:59
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    (+1) Particularly, for a Euclidean $4g$-gon the total incident angle at the vertex (all $4g$ polygon vertices get identified) is the total interior angle of the polygon, $\theta = (4g - 2)\pi$. Consequently, the angular defect (a.k.a. total curvature) at the vertex is $2\pi - \theta = (4 - 4g)\pi$, in agreement with Gauss-Bonnet. The resulting metric is smooth iff $\theta = 2\pi$, iff $g = 1$, iff the polygon is four-sided. – Andrew D. Hwang Feb 21 '16 at 1:59
  • @Andrew D. Hwang This is a good explanation, again for me this does not mean: "the angles don't add up" – Tsemo Aristide Feb 21 '16 at 2:31
  • @Tsemo: As I read it, "the angles add up" means each point has incident angle $2\pi$ (which is equivalent to smoothness). For a $4g$-gon with identifications, this condition is automatic except at the vertex, and holds at the vertex if and only if $g = 1$. – Andrew D. Hwang Feb 21 '16 at 3:18
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    @Tsemo: more formally, if you try to define the obvious flat metric and try to glue it together everywhere on, say, an octagon, including at the vertices, you will attempt to force there to be $8$ tangent vectors at a point (given by the edges), each of which is at an angle of $135^{\circ}$ from the next one, and of course this is impossible: these angles need to add up to $360^{\circ}$ (which is what I meant by "the angles don't add up"). – Qiaochu Yuan Feb 21 '16 at 16:31

Theorem Benzecri. A closed surface is flat if and only it is euler number is zero.

Goldman, W. Two papers which change my life: Milnor seminal work on flat manifolds and bundles.

http://arxiv.org/pdf/1108.0216.pdf

p.4 which has a good outline of the proof.

To be more precise, a surface of genus $g>1$ than cannot be endowed with a differentiable metric whose curvature vanishes identically, since it cannot be endowed with a Koszul derivative which is flat, here flat means that the curvature and the torsion form vanish identically, since we know that the torsion form of the Levi-Civita connection vanishes, a differentiable manifold endowed with a flat differentiable metric has a flat structure; i.e it is an affine manifold. It is a well-known theorem that a Riemannian closed flat manifolds are finitely covered by the torus $T^n$, their fundamental groups are crystallographic groups, so again this theorem also implies that the only oriented surface endowed with a flat metric is the torus.

Remark that in the case of dimension 2, Milnor in his paper On the existence of a connection with curvature zero, Comm. Math. Helv. 32 (1958), 215-223 Milnor has generalized the work of Benzecri, by computing an equality which represents an obstruction for a bundle defined on a surface to be flat.

  • As far as I know, the OP's question is about flatness in the Riemannian sense, which is addressed by the Gauss-Bonnet theorem. Benzecri's theorem is about a weaker notion of flatness. In any case neither theorem really answers the OP's question. – Qiaochu Yuan Feb 21 '16 at 0:17
  • May i kno?w who downvote here – Tsemo Aristide Feb 21 '16 at 0:17
  • Really, can you tell me the difference between these notions of flatness? – Tsemo Aristide Feb 21 '16 at 0:20
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    I really do not understand your pattern of behavior. On the one hand, it's clear you know an impressive amount of mathematics. On the other hand, whenever you make a mistake, you downvote and insult me when I try to correct you, claim that I'm wrong without pointing out any particular error I'm making, then delete your comments afterwards. Why do you feel the need to be so openly antagonistic? There's absolutely no reason we can't work together to answer people's questions correctly. – Qiaochu Yuan Feb 21 '16 at 0:29
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    Tsemo Aristide I gave you upvote since I didn't knew Benzecri's theorem so I found your answer useful-however your answer don't exactly answer my question-which was not to give any argument why higher genus surfaces are not flat but to explain why this argument involving identifications of sides of polygon does not work for higher genus surfaces since I once met this argument in the case of torus thats why I accepted Qiauchu Yuan's answer. – truebaran Feb 21 '16 at 15:00

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