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Gluing $2n$-gons

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1

$begingroup$

I am confused about this how I would visualize/prove this.

Given that we have a regular $2n$-gon that preserves orientation (the surface is orientable because the $2n$-gon's opposite sides are glued in pairings), what would the resulting surface be, up to homeomorphism?

general-topology geometry algebraic-topology topological-groups geometric-topology

share|cite|improve this question

edited Mar 12 at 15:47

David G. Stork

11.1k41432

asked Mar 12 at 15:38

DatSci13DatSci13

358

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $n=2$: torus.
  $endgroup$
  – David G. Stork
  Mar 12 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know this. Sorry, I reworded my question to make it more clear what I am thinking about/looking for.
  $endgroup$
  – DatSci13
  Mar 12 at 15:47
  

  
  
  
  

add a comment | 

1

$begingroup$

I am confused about this how I would visualize/prove this.

Given that we have a regular $2n$-gon that preserves orientation (the surface is orientable because the $2n$-gon's opposite sides are glued in pairings), what would the resulting surface be, up to homeomorphism?

general-topology geometry algebraic-topology topological-groups geometric-topology

share|cite|improve this question

edited Mar 12 at 15:47

David G. Stork

11.1k41432

asked Mar 12 at 15:38

DatSci13DatSci13

358

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $n=2$: torus.
  $endgroup$
  – David G. Stork
  Mar 12 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know this. Sorry, I reworded my question to make it more clear what I am thinking about/looking for.
  $endgroup$
  – DatSci13
  Mar 12 at 15:47
  

  
  
  
  

add a comment | 

$begingroup$

I am confused about this how I would visualize/prove this.

Given that we have a regular $2n$-gon that preserves orientation (the surface is orientable because the $2n$-gon's opposite sides are glued in pairings), what would the resulting surface be, up to homeomorphism?

general-topology geometry algebraic-topology topological-groups geometric-topology

share|cite|improve this question

edited Mar 12 at 15:47

David G. Stork

11.1k41432

asked Mar 12 at 15:38

DatSci13DatSci13

358

$endgroup$

share|cite|improve this question

edited Mar 12 at 15:47

David G. Stork

11.1k41432

asked Mar 12 at 15:38

DatSci13DatSci13

358

share|cite|improve this question

edited Mar 12 at 15:47

David G. Stork

11.1k41432

asked Mar 12 at 15:38

DatSci13DatSci13

358

edited Mar 12 at 15:47

David G. Stork

11.1k41432

edited Mar 12 at 15:47

David G. Stork

11.1k41432

asked Mar 12 at 15:38

DatSci13DatSci13

358

asked Mar 12 at 15:38

DatSci13DatSci13

358

1 Answer
1

active

oldest

votes

3

$begingroup$

Suppose that one glues the opposite sides of a $2n$-gon in way that preserves orientation, to produce a closed oriented surface $S$.

If $n=2k$ is even then $S$ is a closed oriented surface of genus $k$.

If $n=2k+1$ is odd then $S$ is also obtains a closed oriented surface of genus $k$.

One proof is simply to apply the classification of surfaces: by using the gluing diagram, you count the numbers

• 
  $V = #text{vertices of $S$}$


• $E = #text{edges of $S$}$


• $F = #text{faces of $S$}$

You then compute
$$chi(S)=V-E+F
$$
and you use the theorem that a closed, oriented surface $S$ has genus $g$ if and only if $chi(S)=2-2g$.

So, let's start counting. Two of the terms are easy. First, $E = n$ because $2n$ edges of the polygon are glued in pairs to give $n$ edges of $S$. Also, $F=1$ because the one polygon itself gives $1$ face of $S$.

To count $V$, you must count the "vertex cycles" of the gluing polygon (which I'm sure you learned to do, but if not then I can add an explanation). If you do this you'll discover that there are two outcomes:

• If $n=2k$ is even then all of the $2n=4k$ vertices of the polygon are in a single vertex cycle, hence $V=1$. We get
  $$chi(S)=1-n+1=2-n=2-2k
  $$
  and so $S$ has genus $k$.


• If $n=2k+1$ is even then the $2n=4k+2$ vertices of the polygon fall into exactly $2$ vertex cycles, which alternate around the boundary of the polygon. Thus $V=2$. We get
  $$chi(S) = 2 - n + 1 = 3 - n = 3 - (2k+1) = 2-2k
  $$
  and so $S$ has genus $k$.

Here's a picture for gluing the hexagon ($n=3$) to produce a torus, which I found by googling "hexagon gluing to give a torus".

https://www.researchgate.net/figure/Gluing-the-edges-of-a-hexagon-into-a-torus_fig5_324889547

share|cite|improve this answer

edited Mar 12 at 16:16

answered Mar 12 at 16:09

Lee MosherLee Mosher

50.9k33888

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP asks for visualization. Please draw such a surface or point to one online.
  $endgroup$
  – David G. Stork
  Mar 12 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lee Mosher, Thanks again! :)
  $endgroup$
  – DatSci13
  Mar 12 at 16:14
  
  
  

  
  
  
  

add a comment | 

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3

$begingroup$

Suppose that one glues the opposite sides of a $2n$-gon in way that preserves orientation, to produce a closed oriented surface $S$.

If $n=2k$ is even then $S$ is a closed oriented surface of genus $k$.

If $n=2k+1$ is odd then $S$ is also obtains a closed oriented surface of genus $k$.

One proof is simply to apply the classification of surfaces: by using the gluing diagram, you count the numbers

• 
  $V = #text{vertices of $S$}$


• $E = #text{edges of $S$}$


• $F = #text{faces of $S$}$

You then compute
$$chi(S)=V-E+F
$$
and you use the theorem that a closed, oriented surface $S$ has genus $g$ if and only if $chi(S)=2-2g$.

So, let's start counting. Two of the terms are easy. First, $E = n$ because $2n$ edges of the polygon are glued in pairs to give $n$ edges of $S$. Also, $F=1$ because the one polygon itself gives $1$ face of $S$.

To count $V$, you must count the "vertex cycles" of the gluing polygon (which I'm sure you learned to do, but if not then I can add an explanation). If you do this you'll discover that there are two outcomes:

• If $n=2k$ is even then all of the $2n=4k$ vertices of the polygon are in a single vertex cycle, hence $V=1$. We get
  $$chi(S)=1-n+1=2-n=2-2k
  $$
  and so $S$ has genus $k$.


• If $n=2k+1$ is even then the $2n=4k+2$ vertices of the polygon fall into exactly $2$ vertex cycles, which alternate around the boundary of the polygon. Thus $V=2$. We get
  $$chi(S) = 2 - n + 1 = 3 - n = 3 - (2k+1) = 2-2k
  $$
  and so $S$ has genus $k$.

Here's a picture for gluing the hexagon ($n=3$) to produce a torus, which I found by googling "hexagon gluing to give a torus".

https://www.researchgate.net/figure/Gluing-the-edges-of-a-hexagon-into-a-torus_fig5_324889547

share|cite|improve this answer

edited Mar 12 at 16:16

answered Mar 12 at 16:09

Lee MosherLee Mosher

50.9k33888

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP asks for visualization. Please draw such a surface or point to one online.
  $endgroup$
  – David G. Stork
  Mar 12 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lee Mosher, Thanks again! :)
  $endgroup$
  – DatSci13
  Mar 12 at 16:14
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Suppose that one glues the opposite sides of a $2n$-gon in way that preserves orientation, to produce a closed oriented surface $S$.

If $n=2k$ is even then $S$ is a closed oriented surface of genus $k$.

If $n=2k+1$ is odd then $S$ is also obtains a closed oriented surface of genus $k$.

One proof is simply to apply the classification of surfaces: by using the gluing diagram, you count the numbers

• 
  $V = #text{vertices of $S$}$


• $E = #text{edges of $S$}$


• $F = #text{faces of $S$}$

You then compute
$$chi(S)=V-E+F
$$
and you use the theorem that a closed, oriented surface $S$ has genus $g$ if and only if $chi(S)=2-2g$.

So, let's start counting. Two of the terms are easy. First, $E = n$ because $2n$ edges of the polygon are glued in pairs to give $n$ edges of $S$. Also, $F=1$ because the one polygon itself gives $1$ face of $S$.

To count $V$, you must count the "vertex cycles" of the gluing polygon (which I'm sure you learned to do, but if not then I can add an explanation). If you do this you'll discover that there are two outcomes:

• If $n=2k$ is even then all of the $2n=4k$ vertices of the polygon are in a single vertex cycle, hence $V=1$. We get
  $$chi(S)=1-n+1=2-n=2-2k
  $$
  and so $S$ has genus $k$.


• If $n=2k+1$ is even then the $2n=4k+2$ vertices of the polygon fall into exactly $2$ vertex cycles, which alternate around the boundary of the polygon. Thus $V=2$. We get
  $$chi(S) = 2 - n + 1 = 3 - n = 3 - (2k+1) = 2-2k
  $$
  and so $S$ has genus $k$.

Here's a picture for gluing the hexagon ($n=3$) to produce a torus, which I found by googling "hexagon gluing to give a torus".

https://www.researchgate.net/figure/Gluing-the-edges-of-a-hexagon-into-a-torus_fig5_324889547

share|cite|improve this answer

edited Mar 12 at 16:16

answered Mar 12 at 16:09

Lee MosherLee Mosher

50.9k33888

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP asks for visualization. Please draw such a surface or point to one online.
  $endgroup$
  – David G. Stork
  Mar 12 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lee Mosher, Thanks again! :)
  $endgroup$
  – DatSci13
  Mar 12 at 16:14
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Suppose that one glues the opposite sides of a $2n$-gon in way that preserves orientation, to produce a closed oriented surface $S$.

If $n=2k$ is even then $S$ is a closed oriented surface of genus $k$.

If $n=2k+1$ is odd then $S$ is also obtains a closed oriented surface of genus $k$.

One proof is simply to apply the classification of surfaces: by using the gluing diagram, you count the numbers

• 
  $V = #text{vertices of $S$}$


• $E = #text{edges of $S$}$


• $F = #text{faces of $S$}$

You then compute
$$chi(S)=V-E+F
$$
and you use the theorem that a closed, oriented surface $S$ has genus $g$ if and only if $chi(S)=2-2g$.

So, let's start counting. Two of the terms are easy. First, $E = n$ because $2n$ edges of the polygon are glued in pairs to give $n$ edges of $S$. Also, $F=1$ because the one polygon itself gives $1$ face of $S$.

To count $V$, you must count the "vertex cycles" of the gluing polygon (which I'm sure you learned to do, but if not then I can add an explanation). If you do this you'll discover that there are two outcomes:

• If $n=2k$ is even then all of the $2n=4k$ vertices of the polygon are in a single vertex cycle, hence $V=1$. We get
  $$chi(S)=1-n+1=2-n=2-2k
  $$
  and so $S$ has genus $k$.


• If $n=2k+1$ is even then the $2n=4k+2$ vertices of the polygon fall into exactly $2$ vertex cycles, which alternate around the boundary of the polygon. Thus $V=2$. We get
  $$chi(S) = 2 - n + 1 = 3 - n = 3 - (2k+1) = 2-2k
  $$
  and so $S$ has genus $k$.

Here's a picture for gluing the hexagon ($n=3$) to produce a torus, which I found by googling "hexagon gluing to give a torus".

https://www.researchgate.net/figure/Gluing-the-edges-of-a-hexagon-into-a-torus_fig5_324889547

share|cite|improve this answer

edited Mar 12 at 16:16

answered Mar 12 at 16:09

Lee MosherLee Mosher

50.9k33888

$endgroup$

share|cite|improve this answer

edited Mar 12 at 16:16

answered Mar 12 at 16:09

Lee MosherLee Mosher

50.9k33888

answered Mar 12 at 16:09

Lee MosherLee Mosher

50.9k33888

answered Mar 12 at 16:09

Lee MosherLee Mosher

50.9k33888