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Why isn't $S^1$ contractible in homotopy type theory?
If the circle were contractible, then "contractible" would not be a great choice of for that property.
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Deligne finitude and finiteness of etale cohomology
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Deligne finitude and finiteness of etale cohomology
See Deligne’s theorem here which refers to $R f_*$, not $R f_!$: matematicas.unex.es/~navarro/res/sga/…
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Deligne finitude and finiteness of etale cohomology
The $X$ of that question does not refer to your $X$ in this context, but to Spec $k$, which is proper over itself.
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Deligne finitude and finiteness of etale cohomology
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Holomorphic Gauss normal map
Also, this paper (projecteuclid.org/journals/acta-mathematica/volume-223/issue-1/…) indicates that there are topological obstructions to a surface having a minimal embedding into $\mathbb{R}^3$. Every orientable surface has a complex structure and an embedding into $\mathbb{R}^3$ (mathoverflow.net/questions/112538/…), so this should give a negative answer to the noncompact case of the original question
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Holomorphic Gauss normal map
I think the book is “A Survey of Minimal Surfaces”. What Wikipedia article do you mean?
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Holomorphic Gauss normal map
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Holomorphic Gauss normal map
Right I assumed compactness. For open surfaces it sounds much more complicated. By the same argument you will need a metric of either positive or negative curvature
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Holomorphic Gauss normal map
A necessary condition is existence of a metric whose curvature has zero dimensional vanishing locus. Whether such a metric exists is an interesting question in itself
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Holomorphic Gauss normal map
Critical points of the Gauss map are where the Gaussian curvature of the surface vanishes. I think typically the vanishing locus will be a one-(real)-dimensional subset of the surface. An analytic map has isolated critical points so I think this is unlikely
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Modular forms and the cocycle condition in group cohomology
Yes— here $G$ is acting on the multiplicative group of nonvanishing analytic functions $V$ on $X$. The map $g \mapsto j_g$ defines a 1-cochain which is a cocycle. It’s a good exercise to chase this out from the formulas
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Field of fractions of etale stalk of Dedekind domain (Example from Milne's LEC)
Perhaps that should say instead: the data of the etale neighborhood determines $\sigma$ as a map $L \to K^{sep}$ (up to composition with elements of $I(\tilde{p}))$,and therefore induces a natural map $L \to (K^{sep})^{I(\tilde{p})}$. Taking the limit we get our desired isomorphism
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Field of fractions of etale stalk of Dedekind domain (Example from Milne's LEC)
so we get a natural embedding $L \to (K^{sep})^{I(\tilde{\mathfrak{p}})}$. So a choice of $\tilde{\mathfrak{p}}$ gives a canonical isomorphism to this field. which is not encoded by a geometric point over $\mathfrak{p}$ alone.
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Field of fractions of etale stalk of Dedekind domain (Example from Milne's LEC)
I think there’s a subtlety here: you are implicitly thinking of finite separable field extensions $K \to L$ as sitting inside a fixed $K^{sep}$. This is how the geometric point $\overline{x}$ over $\mathfrak{p}$ remembers the prime ideal $\tilde{\mathfrak{p}}$ which induces it: if we have a finite separable $K \to L$ and an unramified prime $\mathfrak{q}$ over $\mathfrak{p}$ in $B$, we can always conjugate by an element $\sigma$ of Gal($K^{sep}/K)$ so that $\sigma{\mathfrak{q}} = \tilde{\mathfrak{p}} \cap B$. The coset $\sigma I(\tilde{\mathfrak{p}})$ is determined by $\mathfrak{q}$,