Euler characteristic of this polyhedron?












8












$begingroup$


I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):



enter image description here



So it should be $mathcal{X}(P)=mathcal{X}(T)=0$.



But we get $V=16, F=10, E=24$, so $mathcal{X}(P)=2$.



However, if we consider a triangulation as this two cases:



enter image description here



it is $mathcal{X}(P)=0$, because $V=C=16$ and $E=32$, and $V=16, F=32, E=48$, respectively.



So, what is it wrong?



Thanks for the support!






algebraic-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    The leftmost image in the second figure a quadrangulation, not a triangulation!
    $endgroup$
    – Pedro Tamaroff♦
    15 hours ago










  • $begingroup$
    Ok, I really notice that a triangulation can be done for any convex polygon.
    $endgroup$
    – LH8
    15 hours ago
















8












$begingroup$


I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):



enter image description here



So it should be $mathcal{X}(P)=mathcal{X}(T)=0$.



But we get $V=16, F=10, E=24$, so $mathcal{X}(P)=2$.



However, if we consider a triangulation as this two cases:



enter image description here



it is $mathcal{X}(P)=0$, because $V=C=16$ and $E=32$, and $V=16, F=32, E=48$, respectively.



So, what is it wrong?



Thanks for the support!






algebraic-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    The leftmost image in the second figure a quadrangulation, not a triangulation!
    $endgroup$
    – Pedro Tamaroff♦
    15 hours ago










  • $begingroup$
    Ok, I really notice that a triangulation can be done for any convex polygon.
    $endgroup$
    – LH8
    15 hours ago














8












8








8





$begingroup$


I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):



enter image description here



So it should be $mathcal{X}(P)=mathcal{X}(T)=0$.



But we get $V=16, F=10, E=24$, so $mathcal{X}(P)=2$.



However, if we consider a triangulation as this two cases:



enter image description here



it is $mathcal{X}(P)=0$, because $V=C=16$ and $E=32$, and $V=16, F=32, E=48$, respectively.



So, what is it wrong?



Thanks for the support!






algebraic-topology







share|cite|improve this question











$endgroup$




I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):



enter image description here



So it should be $mathcal{X}(P)=mathcal{X}(T)=0$.



But we get $V=16, F=10, E=24$, so $mathcal{X}(P)=2$.



However, if we consider a triangulation as this two cases:



enter image description here



it is $mathcal{X}(P)=0$, because $V=C=16$ and $E=32$, and $V=16, F=32, E=48$, respectively.



So, what is it wrong?



Thanks for the support!






algebraic-topology




algebraic-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 hours ago









J. W. Tanner

2,0671117




2,0671117










asked 15 hours ago









LH8LH8

1317




1317












  • $begingroup$
    The leftmost image in the second figure a quadrangulation, not a triangulation!
    $endgroup$
    – Pedro Tamaroff♦
    15 hours ago










  • $begingroup$
    Ok, I really notice that a triangulation can be done for any convex polygon.
    $endgroup$
    – LH8
    15 hours ago


















  • $begingroup$
    The leftmost image in the second figure a quadrangulation, not a triangulation!
    $endgroup$
    – Pedro Tamaroff♦
    15 hours ago










  • $begingroup$
    Ok, I really notice that a triangulation can be done for any convex polygon.
    $endgroup$
    – LH8
    15 hours ago
















$begingroup$
The leftmost image in the second figure a quadrangulation, not a triangulation!
$endgroup$
– Pedro Tamaroff♦
15 hours ago




$begingroup$
The leftmost image in the second figure a quadrangulation, not a triangulation!
$endgroup$
– Pedro Tamaroff♦
15 hours ago












$begingroup$
Ok, I really notice that a triangulation can be done for any convex polygon.
$endgroup$
– LH8
15 hours ago




$begingroup$
Ok, I really notice that a triangulation can be done for any convex polygon.
$endgroup$
– LH8
15 hours ago










1 Answer
1






active

oldest

votes


















11












$begingroup$

The problem is that your first polygon setup uses non-simple polygons. add single edge to each donut shaped face and it will work better.



Cauchy's basic proof for Euler's formula for characteristic requires that we triangulate the polygons and takes advantage of a particular property of this triangulation: adding a single edge via splitting a polygon must also add a single face. This is true of simple polygons, yes, but complex polygons like the donut shape don't have this property: adding an edge may not split the polygon into two.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, that´s what I do in the first triangulation, yeah? But what means non-simple polygons? Maybe you mean convex? In what sense it disturb a Euler characteristic? Since it´s homeomorphic to the torus, they should have same Euler characteristic, directly. Thanks!
    $endgroup$
    – LH8
    15 hours ago












  • $begingroup$
    Your polygons on the 'top' and 'bottom' goes around the hole in the torus. This is not allowed since then they are not contractable.
    $endgroup$
    – An.Ditlev
    14 hours ago










  • $begingroup$
    So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
    $endgroup$
    – LH8
    14 hours ago






  • 1




    $begingroup$
    Not at all: all triangulations of a polyhedron should have the same Euler characteristic... But in order to get an Euler characteristic that makes sense all the polygons need to be simple. A triangle within a triangle doesn't have the same layout as a hexagon, despite having technically the same number of edges and vertices.
    $endgroup$
    – Dan Uznanski
    14 hours ago






  • 3




    $begingroup$
    "Simple" here is "simply connected", meaning you can draw any loop in the polygon and shrink it down to a point without crossing any boundaries. That does not work for your top and bottom faces without triangulation, because a loop drawn around the inner boundary gets "trapped" around that boundary. Cutting the faces from inside to outside, thus excluding those offending loops, is what makes the faces simply connected again.
    $endgroup$
    – Oscar Lanzi
    13 hours ago













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

The problem is that your first polygon setup uses non-simple polygons. add single edge to each donut shaped face and it will work better.



Cauchy's basic proof for Euler's formula for characteristic requires that we triangulate the polygons and takes advantage of a particular property of this triangulation: adding a single edge via splitting a polygon must also add a single face. This is true of simple polygons, yes, but complex polygons like the donut shape don't have this property: adding an edge may not split the polygon into two.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, that´s what I do in the first triangulation, yeah? But what means non-simple polygons? Maybe you mean convex? In what sense it disturb a Euler characteristic? Since it´s homeomorphic to the torus, they should have same Euler characteristic, directly. Thanks!
    $endgroup$
    – LH8
    15 hours ago












  • $begingroup$
    Your polygons on the 'top' and 'bottom' goes around the hole in the torus. This is not allowed since then they are not contractable.
    $endgroup$
    – An.Ditlev
    14 hours ago










  • $begingroup$
    So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
    $endgroup$
    – LH8
    14 hours ago






  • 1




    $begingroup$
    Not at all: all triangulations of a polyhedron should have the same Euler characteristic... But in order to get an Euler characteristic that makes sense all the polygons need to be simple. A triangle within a triangle doesn't have the same layout as a hexagon, despite having technically the same number of edges and vertices.
    $endgroup$
    – Dan Uznanski
    14 hours ago






  • 3




    $begingroup$
    "Simple" here is "simply connected", meaning you can draw any loop in the polygon and shrink it down to a point without crossing any boundaries. That does not work for your top and bottom faces without triangulation, because a loop drawn around the inner boundary gets "trapped" around that boundary. Cutting the faces from inside to outside, thus excluding those offending loops, is what makes the faces simply connected again.
    $endgroup$
    – Oscar Lanzi
    13 hours ago


















11












$begingroup$

The problem is that your first polygon setup uses non-simple polygons. add single edge to each donut shaped face and it will work better.



Cauchy's basic proof for Euler's formula for characteristic requires that we triangulate the polygons and takes advantage of a particular property of this triangulation: adding a single edge via splitting a polygon must also add a single face. This is true of simple polygons, yes, but complex polygons like the donut shape don't have this property: adding an edge may not split the polygon into two.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, that´s what I do in the first triangulation, yeah? But what means non-simple polygons? Maybe you mean convex? In what sense it disturb a Euler characteristic? Since it´s homeomorphic to the torus, they should have same Euler characteristic, directly. Thanks!
    $endgroup$
    – LH8
    15 hours ago












  • $begingroup$
    Your polygons on the 'top' and 'bottom' goes around the hole in the torus. This is not allowed since then they are not contractable.
    $endgroup$
    – An.Ditlev
    14 hours ago










  • $begingroup$
    So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
    $endgroup$
    – LH8
    14 hours ago






  • 1




    $begingroup$
    Not at all: all triangulations of a polyhedron should have the same Euler characteristic... But in order to get an Euler characteristic that makes sense all the polygons need to be simple. A triangle within a triangle doesn't have the same layout as a hexagon, despite having technically the same number of edges and vertices.
    $endgroup$
    – Dan Uznanski
    14 hours ago






  • 3




    $begingroup$
    "Simple" here is "simply connected", meaning you can draw any loop in the polygon and shrink it down to a point without crossing any boundaries. That does not work for your top and bottom faces without triangulation, because a loop drawn around the inner boundary gets "trapped" around that boundary. Cutting the faces from inside to outside, thus excluding those offending loops, is what makes the faces simply connected again.
    $endgroup$
    – Oscar Lanzi
    13 hours ago
















11












11








11





$begingroup$

The problem is that your first polygon setup uses non-simple polygons. add single edge to each donut shaped face and it will work better.



Cauchy's basic proof for Euler's formula for characteristic requires that we triangulate the polygons and takes advantage of a particular property of this triangulation: adding a single edge via splitting a polygon must also add a single face. This is true of simple polygons, yes, but complex polygons like the donut shape don't have this property: adding an edge may not split the polygon into two.






share|cite|improve this answer











$endgroup$



The problem is that your first polygon setup uses non-simple polygons. add single edge to each donut shaped face and it will work better.



Cauchy's basic proof for Euler's formula for characteristic requires that we triangulate the polygons and takes advantage of a particular property of this triangulation: adding a single edge via splitting a polygon must also add a single face. This is true of simple polygons, yes, but complex polygons like the donut shape don't have this property: adding an edge may not split the polygon into two.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 14 hours ago

























answered 15 hours ago









Dan UznanskiDan Uznanski

6,79021528




6,79021528












  • $begingroup$
    Yes, that´s what I do in the first triangulation, yeah? But what means non-simple polygons? Maybe you mean convex? In what sense it disturb a Euler characteristic? Since it´s homeomorphic to the torus, they should have same Euler characteristic, directly. Thanks!
    $endgroup$
    – LH8
    15 hours ago












  • $begingroup$
    Your polygons on the 'top' and 'bottom' goes around the hole in the torus. This is not allowed since then they are not contractable.
    $endgroup$
    – An.Ditlev
    14 hours ago










  • $begingroup$
    So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
    $endgroup$
    – LH8
    14 hours ago






  • 1




    $begingroup$
    Not at all: all triangulations of a polyhedron should have the same Euler characteristic... But in order to get an Euler characteristic that makes sense all the polygons need to be simple. A triangle within a triangle doesn't have the same layout as a hexagon, despite having technically the same number of edges and vertices.
    $endgroup$
    – Dan Uznanski
    14 hours ago






  • 3




    $begingroup$
    "Simple" here is "simply connected", meaning you can draw any loop in the polygon and shrink it down to a point without crossing any boundaries. That does not work for your top and bottom faces without triangulation, because a loop drawn around the inner boundary gets "trapped" around that boundary. Cutting the faces from inside to outside, thus excluding those offending loops, is what makes the faces simply connected again.
    $endgroup$
    – Oscar Lanzi
    13 hours ago




















  • $begingroup$
    Yes, that´s what I do in the first triangulation, yeah? But what means non-simple polygons? Maybe you mean convex? In what sense it disturb a Euler characteristic? Since it´s homeomorphic to the torus, they should have same Euler characteristic, directly. Thanks!
    $endgroup$
    – LH8
    15 hours ago












  • $begingroup$
    Your polygons on the 'top' and 'bottom' goes around the hole in the torus. This is not allowed since then they are not contractable.
    $endgroup$
    – An.Ditlev
    14 hours ago










  • $begingroup$
    So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
    $endgroup$
    – LH8
    14 hours ago






  • 1




    $begingroup$
    Not at all: all triangulations of a polyhedron should have the same Euler characteristic... But in order to get an Euler characteristic that makes sense all the polygons need to be simple. A triangle within a triangle doesn't have the same layout as a hexagon, despite having technically the same number of edges and vertices.
    $endgroup$
    – Dan Uznanski
    14 hours ago






  • 3




    $begingroup$
    "Simple" here is "simply connected", meaning you can draw any loop in the polygon and shrink it down to a point without crossing any boundaries. That does not work for your top and bottom faces without triangulation, because a loop drawn around the inner boundary gets "trapped" around that boundary. Cutting the faces from inside to outside, thus excluding those offending loops, is what makes the faces simply connected again.
    $endgroup$
    – Oscar Lanzi
    13 hours ago


















$begingroup$
Yes, that´s what I do in the first triangulation, yeah? But what means non-simple polygons? Maybe you mean convex? In what sense it disturb a Euler characteristic? Since it´s homeomorphic to the torus, they should have same Euler characteristic, directly. Thanks!
$endgroup$
– LH8
15 hours ago






$begingroup$
Yes, that´s what I do in the first triangulation, yeah? But what means non-simple polygons? Maybe you mean convex? In what sense it disturb a Euler characteristic? Since it´s homeomorphic to the torus, they should have same Euler characteristic, directly. Thanks!
$endgroup$
– LH8
15 hours ago














$begingroup$
Your polygons on the 'top' and 'bottom' goes around the hole in the torus. This is not allowed since then they are not contractable.
$endgroup$
– An.Ditlev
14 hours ago




$begingroup$
Your polygons on the 'top' and 'bottom' goes around the hole in the torus. This is not allowed since then they are not contractable.
$endgroup$
– An.Ditlev
14 hours ago












$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
14 hours ago




$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
14 hours ago




1




1




$begingroup$
Not at all: all triangulations of a polyhedron should have the same Euler characteristic... But in order to get an Euler characteristic that makes sense all the polygons need to be simple. A triangle within a triangle doesn't have the same layout as a hexagon, despite having technically the same number of edges and vertices.
$endgroup$
– Dan Uznanski
14 hours ago




$begingroup$
Not at all: all triangulations of a polyhedron should have the same Euler characteristic... But in order to get an Euler characteristic that makes sense all the polygons need to be simple. A triangle within a triangle doesn't have the same layout as a hexagon, despite having technically the same number of edges and vertices.
$endgroup$
– Dan Uznanski
14 hours ago




3




3




$begingroup$
"Simple" here is "simply connected", meaning you can draw any loop in the polygon and shrink it down to a point without crossing any boundaries. That does not work for your top and bottom faces without triangulation, because a loop drawn around the inner boundary gets "trapped" around that boundary. Cutting the faces from inside to outside, thus excluding those offending loops, is what makes the faces simply connected again.
$endgroup$
– Oscar Lanzi
13 hours ago






$begingroup$
"Simple" here is "simply connected", meaning you can draw any loop in the polygon and shrink it down to a point without crossing any boundaries. That does not work for your top and bottom faces without triangulation, because a loop drawn around the inner boundary gets "trapped" around that boundary. Cutting the faces from inside to outside, thus excluding those offending loops, is what makes the faces simply connected again.
$endgroup$
– Oscar Lanzi
13 hours ago




















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