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Apr 18, 2019 at 15:23 vote accept tparker
Apr 16, 2019 at 17:29 answer added Alex Gavrilov timeline score: 17
Sep 15, 2018 at 15:38 comment added Mizar The same argument shows that Gauss-Bonnet holds for compact surfaces with boundary: observe that at each boundary component you have a well-defined inner normal vector field, hence the geodesic curvature $k_g$ is well defined.
Sep 15, 2018 at 15:24 comment added Mizar This should hold also for closed non-orientable surfaces $S$, viewing $\iint K\,dA$ as the integral of the Gaussian curvature $K$ with respect to the area measure (speaking about measures completely avoids orientability). Proof: if you pass to the oriented double cover $\widehat S$, both sides $\iint K\,dA$ and $2\pi\chi(S)$ double their value; since they are equal for $\widehat S$, they coincide also for $S$.
Sep 15, 2018 at 15:11 history rollback tparker
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Sep 14, 2018 at 21:11 history edited tparker CC BY-SA 4.0
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Sep 13, 2018 at 7:27 comment added Sofie Verbeek @tparker Both spellings are usually fine! See grammar-monster.com/lessons/hyphens_in_prefixes.htm
Sep 13, 2018 at 6:51 comment added DLIN GB-formula holds holds for closed non-orientable surface, here $dA$ means the volum-measure. For one example, $\mathbb RP^2$.
Sep 13, 2018 at 6:05 comment added Ryan Budney The theorem holds perfectly well for non-orientable surfaces. You just have to be a little careful when computing the integral. Depending on what formula you use for $K dA$, you might be using a formalism that assumes orientability -- something like a differential form formalism. But you can make perfect sense of both the left and right hand side of the Gauss-Bonnet formula for non-orientable surfaces, and you can prove it in a variety of ways. The double cover is one.
Sep 13, 2018 at 4:41 comment added tparker In researching this, I found another question about orientable surfaces on which the experts seem split exactly 50-50: whether or not the word "non-orientable" is hyphenated.
Sep 13, 2018 at 4:40 history asked tparker CC BY-SA 4.0