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Consider a compact surface $\mathcal{S} \subset \mathbb{R}^3$ without boundary and assume we are given the Gaussian curvature $K(x), x\in \mathcal{S}$. It is know that Gaussian curvature does not fully determine the Riemannian metric, as for example discussed in

Does the curvature determine the metric?

But for a given Gaussian curvature, does one have the freedom to choose a Riemannian metric such that the Gauss-Bonnet theorem holds?:

$$\int_{\mathcal{S}} K(x) d\mu_g(x) = 2\pi\; \chi(\mathcal{S}),$$

where $\mu_g$ is the Riemmanian volume measure induced by the metric $g$ and $\chi(\mathcal{S})$ is the Euler characteristic.

Is there even a constructive method for a Riemmanian metric $g$ under the conditions that the resulting Gaussian curvature is $K(x),x\in \mathcal{S}$ and the Gauss-Bonnet theorem is fulfilled?

(maybe this question is too general as stated above, but then might become more meaningful with assumptions on $K$.)

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madison54
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    If the surface $S$ is diffeomorphic to a sphere or torus and the function $K$ is everywhere negative, clearly this is not possible. Similarly, if $S$ is diffeomorphic to a torus or higher genus surface and $K>0$ everywhere, this is not possible. – Ben McKay Apr 04 '14 at 13:11
  • a related MO question could be http://mathoverflow.net/questions/157420/a-converse-to-the-gauss-bonnet-theorem – Ali Taghavi Apr 06 '14 at 08:18

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If $K$ is to be the Gauss curvature of some metric on a compact two-dimensional manifold $M$ with Euler characteristic $χ(M)$, then from the Gauss-Bonnet theorem one has the following conditions: (a) $K$ is positive somewhere if $χ(M)>0$, (b) $K$ changes sign or $K≡0$ if $χ(M)=0$, (c) $K$ is negative somewhere if $χ(M)<0$. Theorem: $K∈C^∞(M)$ is the Gauss curvature of some Riemannian metric on the two-dimensional, compact, connected manifold $M$ if and only if $K$ satisfies the above sign condition. MR0375153 (51 #11349) Kazdan, Jerry L.; Warner, F. W. Existence and conformal deformation of metrics with prescribed Gaussian and scalar curvatures. Ann. of Math. (2) 101 (1975), 317–331. 53C20 (58G99)

Ben McKay
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    Interesting result! I guess the minor question that remains then, is which Riemannian surfaces are embeddable in $\mathbb{R}^3$, since for example the flat torus is not. – Jaap Eldering Apr 04 '14 at 16:50
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    Not a minor question at all. Embeddability in $\mathbb{R}^3$ is a very complicated problem, even locally. There are abstract Riemannian metrics on surfaces with marked point, so that no neighborhood of the marked point admits any isometric embedding into $\mathbb{R}^3$: MR2393070 (2009k:53177) Reviewed Nadirashvili, Nikolai(F-PROV-APT); Yuan, Yu(1-WA) Improving Pogorelov's isometric embedding counterexample. (English summary) Calc. Var. Partial Differential Equations 32 (2008), no. 3, 319–323. 53C45 – Ben McKay Apr 04 '14 at 17:11