One can make a model of the hyperbolic plane inside the Euclidean plane, either using the conformal model or projective model.
How does one make a model of the Euclidean plane inside the hyperbolic plane?
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3$\begingroup$ math.SE duplicate! math.stackexchange.com/questions/1347/… $\endgroup$– Qiaochu YuanCommented Aug 31, 2012 at 16:20
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$\begingroup$ There are infinitely many diffeomorphisms from the Euclidean plane to an open subset of the hyperbolic plane. You'd want to find a diffeomorphism with some nice property, to distinguish it from this enormous family. An obvious choice is to find a conformal mapping, but by Liouville's theorem there is none. Why do you want such a map? Is there any property you would like it to have? $\endgroup$– Will SawinCommented Aug 31, 2012 at 16:25
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$\begingroup$ I want to know what are the lines in the geometry, how to compute distance and angle.. $\endgroup$– i. m. soloveichikCommented Aug 31, 2012 at 17:12
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1$\begingroup$ What algebraic structure would you place on $\mathbb H^2$? $\endgroup$– Will SawinCommented Aug 31, 2012 at 20:03
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2$\begingroup$ There's the tangent space to a point in hyperbolic space. That's Euclidean. $\endgroup$– Ryan BudneyCommented Aug 31, 2012 at 20:42
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2 Answers
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I believe there is no good model of $\mathbb{E}^2$ in $\mathbb{H}^2.$ However, there is an excellent model in $\mathbb{H}^3:$ any horosphere will work.
Also This is not particularly interesting, but if you use the hyperboloid model of $\mathbb{H}^2,$ you can project it (from, e.g., the point $(2, 0, 0)$ onto the $(x, y)$ plane. This will give an algebraic model of $\mathbb{E}^2$ in $\mathbb{H}^2.$
answered Aug 31, 2012 at 16:54
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$\begingroup$ Yes I know that. That's why I want the smaller dimension. $\endgroup$ Commented Aug 31, 2012 at 17:11
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$\begingroup$ @Igor How do I see triangles have 180 degrees? $\endgroup$ Commented Aug 31, 2012 at 18:09
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$\begingroup$ The $(x, y)$ plane is a Euclidean plane? $\endgroup$ Commented Aug 31, 2012 at 19:21
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2$\begingroup$ Hyperboloid is a graph over the xy plane. $\endgroup$ Commented Aug 31, 2012 at 22:40
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This is a response to Will Sawin's comment "What algebraic structure would you place on $\mathbb{H}^2$?"
On $\mathbb{R}^n$ there is a vector space structure where
1) The metric is given by $d(\mathbf{u},\mathbf{v})=\|\mathbf{u}-\mathbf{v}\|$
and
2) given three points $U,V,W$ the angle $U\hat WV$ satisfies $\cos \theta = \frac{(-W+U)\cdot(-W+V)}{\|-W+U\|\|-W+V\|}$
Analogously, for some models of $\mathbb{H}^n$ with points identified with a subset of $\mathbb{R}^n$ there is a vector-like structure but with a noncommutative, nonassociative binary operation $\oplus$ where
1) The metric is given by $d(\mathbf{u},\mathbf{v})=\|\mathbf{u} \ominus \mathbf{v} \| $
and
2) given three points $U,V,W$ the angle $U\hat WV$ satisfies $\cos \theta = \frac{(\ominus W\oplus U)\cdot(\ominus W\oplus V)}{\|\ominus W\oplus U\|\|\ominus W\oplus V\|}$
$\| \|$ and $\cdot $ are the vector norm and dot product inherited from $\mathbb{R}^n$.
$\ominus a$ denotes the left inverse of a.
$a\ominus b$ denotes $a\oplus (\ominus b)$.
Note the use of the trig function "cos" even though this hyperbolic geometry.
For the Beltrami-Klein model the binary operation $\oplus$ is given by the formula for relativistic velocity-addition.
answered Sep 1, 2012 at 7:47