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What are some surfaces of constant integral curvature of genus $g>1?$ What are the simplest 3d forms/parameterizations?

From the Gauss-Bonnet theorem classical definition has it

$$ \Omega=\int_U K dA= 4\pi(1-g)$$

For the sphere, single and double holed tori $ (g=0,1,2)$ we have $(\Omega =4 \pi,0, -4 \pi) $ respectively.

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EDIT 1:

For example I want to see an image and its parameterization of a surface with $\Omega= -5, g\approx 1.398 ; \;$ Is that possible?

To me it makes sense to regard $\Omega $ as a geometric scalar, an isometric invariant. For example take any patch of a "circle" of constant "radial"distance from a single center C on a surface of constant arbitrary positive or negative $K$ swept by geodesic polar coordinate of arbitrarily constant radial length $u$ . This spherical/pseudospherical segmented patch has a constant $\Omega.$ This is a special, may be trivial case of constant integral curvature.

Afik KF Gauss did not define Integra Curvatura $\dfrac{\Omega}{4 \pi}$ to be valid only for integers $\mathbb Z$ in the context of Gauss-Bonnet theorem or elsewhere.

Seen that way however it also appears impossible because by definition the genus turns out fractional or non-integer. Genus is an integer, fractional holes like $ g\approx 1.398$ number of holes to me sound meaningless.

Just as fractional factorial and fractional differentiation were validated by latter day definitions, has a topological or integral calculus meaning been arrived at for a surface with e.g.,$\;g=$ three and a quarter number of holes?

Thanks!

Narasimham
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  • To clarify, the question is about explicit parametrizations of closed surfaces of genus greater than $1$? – Andrew D. Hwang Sep 16 '22 at 13:52
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    Yes... aimed to further check/verify GB thm on $g=2$ IC constant surface. – Narasimham Sep 16 '22 at 16:05
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    What does constant integral curvature mean? Be specific, – Ted Shifrin Sep 16 '22 at 19:19
  • For a sphere it is the solid angle measured in steradians... $\Omega=4 \pi$ in the specific case of a sphere.

    https://math.stackexchange.com/questions/904483/solid-angle-integration#:~:text=Element%20of%20solid%20angle%20is,full%20pseudosphere%20and%20so%20on.

     – Narasimham Sep 16 '22 at 19:44
  • @Narasimham: To clarify, you want surfaces for which $\Omega := 4\pi (1-g)$ is an integer? Well, since $g$ must be an integer, the only possibility is $g = 1$. However, it sounds like you're hoping for a definition of "genus $g$ surface" for irrational $g$. I don't think such a thing exists. – Jesse Madnick Sep 16 '22 at 23:08
  • One reason the question may be difficult to answer is, surfaces of genus $0$ and $1$ have trigonometric parametrizations, while for genus at least $2$ the natural analogs are modular functions (for the reasons in Qiaochu's answer). <> I recall seeing (but not reading) a paper in the Mathematical Intelligencer in the early-mid 1990s about an explicit parametrization of Fermat surfaces (given by $x^n+y^n+z^n=1$ in three-space), though a quick web search doesn't turn up a reference. – Andrew D. Hwang Sep 17 '22 at 12:21
  • Re: the edit, for a closed surface $g$ is an integer, while $\Omega$ is an integer multiple of $2\pi$. <> For specific qualitative drawings it's possible to see $\Omega = 2\pi(2 - 2g)$. – Andrew D. Hwang Sep 17 '22 at 12:23
  • @Andrew D. Huang: Thanks once again. So I request you (are anyone) to kindly confirm in an answer .. that non-integer $\Omega/(4 \pi)$ are inadmissible, undefined make no sense etc. in differential geometry and general topology at least in 3-space. – Narasimham Sep 17 '22 at 15:30

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There are no closed surfaces in $\mathbb{R}^3$ with with constant Gaussian curvature except for the spheres. Once $g \ge 1$ a closed surface embedded in $\mathbb{R}^3$ necessarily has positive Gaussian curvature on its "outside" (formally, at some extreme point of its convex hull) but negative Gaussian curvature on its "inside" (where the holes are).

Instead these surfaces are most easily constructed as quotients. Flat toruses can be constructed as quotients $\mathbb{R}^2/\Gamma$ of $\mathbb{R}^2$ (with the Euclidean metric) by a lattice, whereas surfaces with constant Gaussian curvature of higher genus can be constructed as quotients of the hyperbolic plane $\mathbb{H}^2$ by Fuchsian groups; this is a consequence of the uniformization theorem.

Qiaochu Yuan
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