kolutastmr

2022-07-04

I was reading a paper on hyperbolic pascal triangle and the author stated that for Schlafli symbol $\{p,q\}$ , if $(p-2)\phantom{\rule{thickmathspace}{0ex}}(q-2)=4$ , it determines the Euclidean mosaic. For $(p-2)\phantom{\rule{thickmathspace}{0ex}}(q-2)<4$ a sphere is determined and for $(p-2)\phantom{\rule{thickmathspace}{0ex}}(q-2)>4$ a hyperbolic mosaic is defined.

On the nature of mosaic specified by Schlafli symbol $\{p,q\}$ ?

On the nature of mosaic specified by Schlafli symbol $\{p,q\}$ ?

persstemc1

Beginner2022-07-05Added 18 answers

Step 1

The general idea is to compare the angles of a regular p-gon in $\{p,q\}$ to a Euclidean p-gon to see whether there is angular defect or excess. If the Euclidean p-gon has a greater angle sum, the tiling is hyperbolic; if the Euclidean p-gon has a lesser angle sum, the tiling is spherical.

So let's compare! The angle sum in a Euclidean p-gon is $(p-2)\pi /p$ . The angle sum in a p-gon in $\{p,q\}$ is $2\pi /q$ . The difference between these angle sums is

$(p-2)\pi /p-2\pi /q=(pq-2q-2p)\pi /pq=((p-2)(q-2)-4)\pi /pq$

which has the same sign as $(p-2)(q-2)-4$ , hence verifying the statement you found in this paper.

The general idea is to compare the angles of a regular p-gon in $\{p,q\}$ to a Euclidean p-gon to see whether there is angular defect or excess. If the Euclidean p-gon has a greater angle sum, the tiling is hyperbolic; if the Euclidean p-gon has a lesser angle sum, the tiling is spherical.

So let's compare! The angle sum in a Euclidean p-gon is $(p-2)\pi /p$ . The angle sum in a p-gon in $\{p,q\}$ is $2\pi /q$ . The difference between these angle sums is

$(p-2)\pi /p-2\pi /q=(pq-2q-2p)\pi /pq=((p-2)(q-2)-4)\pi /pq$

which has the same sign as $(p-2)(q-2)-4$ , hence verifying the statement you found in this paper.

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