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Systematic approach to triangulation closed combinatorial surfaces
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there are a name for a simplicial complex that is homotopic to the clique complex of its 1-skeleton?Partitioning a triangulated 2-sphere into two triangulated discstriangulation of the closed discGenerator for homology of surface of genus $g$ - Hatcher 2.2.29If you know that a shape tiles the plane, does it also tile other surfaces?Existence of subdivision of PL manifold triangulation which is combinatorial manifoldTriangulation of torus - understanding whyTriangulation to maximize the number of triangles, including those with points insideA question on triangulation of compact surfacesClosed Surfaces are combinatorial surfaces
$begingroup$
I was wondering whether there is a systematic approach to the triangulation of closed combinatorial surfaces, which we know can be shown to be homeomorphic to polygons with complete set of side identifications on the way to the classification theorem, or just more generally to the manifolds obtained from the sphere by adding $g$ handles or $h$ crosscaps.
Well, intuitively, we should be able to triangulate a handle/crosscap alone, and then we also know of a simple triangulation of the sphere (two tetrahedrons glued together on a side with the interior of that face removed), but what concerns is me whether such cut-paste of those separate triangulations would generally result in a triangulation of the desired space?
TL;DR: knowing that triangulations of closed combinatorial surfaces exist (by definition), do we know of a simple way to write those triangulations down, knowing the class of the surface given (i. e. $M_g$ or $N_h$)?
...starting from a polygon with a complete set of side identifications, say?
general-topology algebraic-topology surfaces polygons triangulation
$endgroup$
add a comment |
$begingroup$
I was wondering whether there is a systematic approach to the triangulation of closed combinatorial surfaces, which we know can be shown to be homeomorphic to polygons with complete set of side identifications on the way to the classification theorem, or just more generally to the manifolds obtained from the sphere by adding $g$ handles or $h$ crosscaps.
Well, intuitively, we should be able to triangulate a handle/crosscap alone, and then we also know of a simple triangulation of the sphere (two tetrahedrons glued together on a side with the interior of that face removed), but what concerns is me whether such cut-paste of those separate triangulations would generally result in a triangulation of the desired space?
TL;DR: knowing that triangulations of closed combinatorial surfaces exist (by definition), do we know of a simple way to write those triangulations down, knowing the class of the surface given (i. e. $M_g$ or $N_h$)?
...starting from a polygon with a complete set of side identifications, say?
general-topology algebraic-topology surfaces polygons triangulation
$endgroup$
add a comment |
$begingroup$
I was wondering whether there is a systematic approach to the triangulation of closed combinatorial surfaces, which we know can be shown to be homeomorphic to polygons with complete set of side identifications on the way to the classification theorem, or just more generally to the manifolds obtained from the sphere by adding $g$ handles or $h$ crosscaps.
Well, intuitively, we should be able to triangulate a handle/crosscap alone, and then we also know of a simple triangulation of the sphere (two tetrahedrons glued together on a side with the interior of that face removed), but what concerns is me whether such cut-paste of those separate triangulations would generally result in a triangulation of the desired space?
TL;DR: knowing that triangulations of closed combinatorial surfaces exist (by definition), do we know of a simple way to write those triangulations down, knowing the class of the surface given (i. e. $M_g$ or $N_h$)?
...starting from a polygon with a complete set of side identifications, say?
general-topology algebraic-topology surfaces polygons triangulation
$endgroup$
I was wondering whether there is a systematic approach to the triangulation of closed combinatorial surfaces, which we know can be shown to be homeomorphic to polygons with complete set of side identifications on the way to the classification theorem, or just more generally to the manifolds obtained from the sphere by adding $g$ handles or $h$ crosscaps.
Well, intuitively, we should be able to triangulate a handle/crosscap alone, and then we also know of a simple triangulation of the sphere (two tetrahedrons glued together on a side with the interior of that face removed), but what concerns is me whether such cut-paste of those separate triangulations would generally result in a triangulation of the desired space?
TL;DR: knowing that triangulations of closed combinatorial surfaces exist (by definition), do we know of a simple way to write those triangulations down, knowing the class of the surface given (i. e. $M_g$ or $N_h$)?
...starting from a polygon with a complete set of side identifications, say?
general-topology algebraic-topology surfaces polygons triangulation
general-topology algebraic-topology surfaces polygons triangulation
asked Mar 26 at 17:04
user35443user35443
203213
203213
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There is a simple answer to your question which ignores all of your remarks about handles and crosscaps. Starting from the polygon $P$, first form the 2nd barycentric subdivision of $P$, and then glue the sides of $P$. After gluing, the 2nd barycentric subdivision of $P$ becomes a triangulation of the surface.
You may wonder: one knows what the 2nd barycentric subdivision of a triangulation is, but what is the 2nd barycentric subdivision of the polygon $P$? Let me assume for simplicity that $P$ is a regular polygon in the plane. The 1st barycentric subdivision of $P$ is the following triangulation of $P$. Its $0$-skeleton contains the original vertex set of $P$, plus some additional vertices, namely the midpoint $M_E$ of each edge of $P$ and the center $C$ of $P$. Its $1$-skeleton has two edges obtained from each edge $E$ by subdividing it at the point $M_E$, plus some additional edges, namely $CV$ for each vertex $V$ of $P$, and $CM_E$ for each edge of $E$. And now, as expected, the 2nd barycentric subdivision of $P$ is the 1st barycentric subdivision of its 1st barycentric subdivision.
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$begingroup$
There is a simple answer to your question which ignores all of your remarks about handles and crosscaps. Starting from the polygon $P$, first form the 2nd barycentric subdivision of $P$, and then glue the sides of $P$. After gluing, the 2nd barycentric subdivision of $P$ becomes a triangulation of the surface.
You may wonder: one knows what the 2nd barycentric subdivision of a triangulation is, but what is the 2nd barycentric subdivision of the polygon $P$? Let me assume for simplicity that $P$ is a regular polygon in the plane. The 1st barycentric subdivision of $P$ is the following triangulation of $P$. Its $0$-skeleton contains the original vertex set of $P$, plus some additional vertices, namely the midpoint $M_E$ of each edge of $P$ and the center $C$ of $P$. Its $1$-skeleton has two edges obtained from each edge $E$ by subdividing it at the point $M_E$, plus some additional edges, namely $CV$ for each vertex $V$ of $P$, and $CM_E$ for each edge of $E$. And now, as expected, the 2nd barycentric subdivision of $P$ is the 1st barycentric subdivision of its 1st barycentric subdivision.
$endgroup$
add a comment |
$begingroup$
There is a simple answer to your question which ignores all of your remarks about handles and crosscaps. Starting from the polygon $P$, first form the 2nd barycentric subdivision of $P$, and then glue the sides of $P$. After gluing, the 2nd barycentric subdivision of $P$ becomes a triangulation of the surface.
You may wonder: one knows what the 2nd barycentric subdivision of a triangulation is, but what is the 2nd barycentric subdivision of the polygon $P$? Let me assume for simplicity that $P$ is a regular polygon in the plane. The 1st barycentric subdivision of $P$ is the following triangulation of $P$. Its $0$-skeleton contains the original vertex set of $P$, plus some additional vertices, namely the midpoint $M_E$ of each edge of $P$ and the center $C$ of $P$. Its $1$-skeleton has two edges obtained from each edge $E$ by subdividing it at the point $M_E$, plus some additional edges, namely $CV$ for each vertex $V$ of $P$, and $CM_E$ for each edge of $E$. And now, as expected, the 2nd barycentric subdivision of $P$ is the 1st barycentric subdivision of its 1st barycentric subdivision.
$endgroup$
add a comment |
$begingroup$
There is a simple answer to your question which ignores all of your remarks about handles and crosscaps. Starting from the polygon $P$, first form the 2nd barycentric subdivision of $P$, and then glue the sides of $P$. After gluing, the 2nd barycentric subdivision of $P$ becomes a triangulation of the surface.
You may wonder: one knows what the 2nd barycentric subdivision of a triangulation is, but what is the 2nd barycentric subdivision of the polygon $P$? Let me assume for simplicity that $P$ is a regular polygon in the plane. The 1st barycentric subdivision of $P$ is the following triangulation of $P$. Its $0$-skeleton contains the original vertex set of $P$, plus some additional vertices, namely the midpoint $M_E$ of each edge of $P$ and the center $C$ of $P$. Its $1$-skeleton has two edges obtained from each edge $E$ by subdividing it at the point $M_E$, plus some additional edges, namely $CV$ for each vertex $V$ of $P$, and $CM_E$ for each edge of $E$. And now, as expected, the 2nd barycentric subdivision of $P$ is the 1st barycentric subdivision of its 1st barycentric subdivision.
$endgroup$
There is a simple answer to your question which ignores all of your remarks about handles and crosscaps. Starting from the polygon $P$, first form the 2nd barycentric subdivision of $P$, and then glue the sides of $P$. After gluing, the 2nd barycentric subdivision of $P$ becomes a triangulation of the surface.
You may wonder: one knows what the 2nd barycentric subdivision of a triangulation is, but what is the 2nd barycentric subdivision of the polygon $P$? Let me assume for simplicity that $P$ is a regular polygon in the plane. The 1st barycentric subdivision of $P$ is the following triangulation of $P$. Its $0$-skeleton contains the original vertex set of $P$, plus some additional vertices, namely the midpoint $M_E$ of each edge of $P$ and the center $C$ of $P$. Its $1$-skeleton has two edges obtained from each edge $E$ by subdividing it at the point $M_E$, plus some additional edges, namely $CV$ for each vertex $V$ of $P$, and $CM_E$ for each edge of $E$. And now, as expected, the 2nd barycentric subdivision of $P$ is the 1st barycentric subdivision of its 1st barycentric subdivision.
answered Mar 26 at 18:56
Lee MosherLee Mosher
52.4k33891
52.4k33891
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