Revisions to Largest hyperbolic disk embeddable in Euclidean 3-space?

http -> https (the question was bumped anyway)

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edited Jan 5, 2023 at 6:55

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Hilbert proved Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?

Edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region $$\mathrm{PS}=\{z \mid \mathrm{Im} z \ge 1,\; -\pi < Re z \le π\}$$ on the upper half-plane model of $H^2$. Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.

Next, Euclidean circles drawn on in the upper half-plane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$. I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $\mathrm{PS}$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

Edit 2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?

Edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region $$\mathrm{PS}=\{z \mid \mathrm{Im} z \ge 1,\; -\pi < Re z \le π\}$$ on the upper half-plane model of $H^2$. Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.

Next, Euclidean circles drawn on in the upper half-plane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$. I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $\mathrm{PS}$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

Edit 2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?

Edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region $$\mathrm{PS}=\{z \mid \mathrm{Im} z \ge 1,\; -\pi < Re z \le π\}$$ on the upper half-plane model of $H^2$. Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.

Next, Euclidean circles drawn on in the upper half-plane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$. I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $\mathrm{PS}$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

Edit 2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

formatting

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edited Sep 23, 2020 at 16:16

YCor

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Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?

editEdit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region $PS=\{z | Im z ≥ 1, -π < Re z ≤ π\}$$$\mathrm{PS}=\{z \mid \mathrm{Im} z \ge 1,\; -\pi < Re z \le π\}$$ on the upper half plane-plane model of $H^2$. Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.

Next, Euclidean circles drawn on in the upper half-half planeplane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$. I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $PS$$\mathrm{PS}$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

edit2Edit 2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?

edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region $PS=\{z | Im z ≥ 1, -π < Re z ≤ π\}$ on the upper half plane model of $H^2$. Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.

Next, Euclidean circles drawn on in the upper-half plane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$. I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $PS$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

edit2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?

Edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region $$\mathrm{PS}=\{z \mid \mathrm{Im} z \ge 1,\; -\pi < Re z \le π\}$$ on the upper half-plane model of $H^2$. Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.

Next, Euclidean circles drawn on in the upper half-plane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$. I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $\mathrm{PS}$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

Edit 2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

formatting, removed deprecated (geometry) tag - see the tag info: http://mathoverflow.net/tags/geometry/info; if there are some other geometry-related tags which are suitable, please use some of them instead

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edited Jul 25, 2017 at 7:57

Ben McKay

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Hilbert proved that there's no complete regular (C^k$C^k$ for sufficiently large k$k$) isometric embedding of the hyperbolic plane into R^3$\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or C^2$C^2$, say) isometrically embedded in R^3$\mathbb{R}^3$?

edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region PS={z | Im z ≥ 1, -π < Re z ≤ π}$PS=\{z | Im z ≥ 1, -π < Re z ≤ π\}$ on the upper half plane model of H^2$H^2$. Let z=x+iy$z=x+iy$, so that ordered pairs (x,y)∈ H^2$(x,y)∈ H^2$ when y>0$y>0$.

Next, Euclidean circles drawn on in the upper-half plane model with center (x,y\cosh r)$(x,y\cosh r)$ and radius y\sinh r$y\sinh r$ correspond to hyperbolic circles with center (x,y)$(x,y)$ and radius r$r$. I can fit a Euclidean circle of radius π$π$ centered at (0,1+π)$(0,1+π)$ into the region PS$PS$. This corresponds to a hyperbolic disk of radius arctanh(π/(1+π)) ~ 0.993$\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

edit2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

Hilbert proved that there's no complete regular (C^k for sufficiently large k) isometric embedding of the hyperbolic plane into R^3. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or C^2, say) isometrically embedded in R^3?

edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region PS={z | Im z ≥ 1, -π < Re z ≤ π} on the upper half plane model of H^2. Let z=x+iy, so that ordered pairs (x,y)∈ H^2 when y>0.

Next, Euclidean circles drawn on in the upper-half plane model with center (x,y\cosh r) and radius y\sinh r correspond to hyperbolic circles with center (x,y) and radius r. I can fit a Euclidean circle of radius π centered at (0,1+π) into the region PS. This corresponds to a hyperbolic disk of radius arctanh(π/(1+π)) ~ 0.993.

Surely one can do better?

edit2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?

edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region $PS=\{z | Im z ≥ 1, -π < Re z ≤ π\}$ on the upper half plane model of $H^2$. Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.

Next, Euclidean circles drawn on in the upper-half plane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$. I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $PS$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

edit2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

removed deprecated (geometry) tag - see the tag info: http://mathoverflow.net/tags/geometry/info; if there are some other geometry-related tags which are suitable, please use some of them instead

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