Euler characteristic of this polyhedron?
$begingroup$
I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):
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:
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
edited Feb 18 at 13:10
J. W. Tanner
2,4831117
asked Feb 18 at 10:35
LH8LH8
1368
$endgroup$
add a comment |
$begingroup$
I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):
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:
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
edited Feb 18 at 13:10
J. W. Tanner
2,4831117
asked Feb 18 at 10:35
LH8LH8
1368
$endgroup$
$begingroup$
The leftmost image in the second figure a quadrangulation, not a triangulation!
$endgroup$
– Pedro Tamaroff♦
Feb 18 at 10:39
$begingroup$
Ok, I really notice that a triangulation can be done for any convex polygon.
$endgroup$
– LH8
Feb 18 at 10:43
add a comment |
$begingroup$
I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):
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:
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
edited Feb 18 at 13:10
J. W. Tanner
2,4831117
asked Feb 18 at 10:35
LH8LH8
1368
$endgroup$
I´m trying to obtain the Euler characteristic of this polyhedron $P$, that is homeomorphic to the torus $T$ (I think):
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:
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
edited Feb 18 at 13:10
J. W. Tanner
2,4831117
asked Feb 18 at 10:35
LH8LH8
1368
edited Feb 18 at 13:10
J. W. Tanner
2,4831117
asked Feb 18 at 10:35
LH8LH8
1368
edited Feb 18 at 13:10
J. W. Tanner
2,4831117
edited Feb 18 at 13:10
J. W. Tanner
2,4831117
edited Feb 18 at 13:10
J. W. Tanner
2,4831117
2,4831117
asked Feb 18 at 10:35
LH8LH8
1368
asked Feb 18 at 10:35
LH8LH8
1368
asked Feb 18 at 10:35
LH8LH8
1368
1368
$begingroup$
The leftmost image in the second figure a quadrangulation, not a triangulation!
$endgroup$
– Pedro Tamaroff♦
Feb 18 at 10:39
$begingroup$
Ok, I really notice that a triangulation can be done for any convex polygon.
$endgroup$
– LH8
Feb 18 at 10:43
add a comment |
$begingroup$
The leftmost image in the second figure a quadrangulation, not a triangulation!
$endgroup$
– Pedro Tamaroff♦
Feb 18 at 10:39
$begingroup$
Ok, I really notice that a triangulation can be done for any convex polygon.
$endgroup$
– LH8
Feb 18 at 10:43
$begingroup$
The leftmost image in the second figure a quadrangulation, not a triangulation!
$endgroup$
– Pedro Tamaroff♦
Feb 18 at 10:39
$begingroup$
The leftmost image in the second figure a quadrangulation, not a triangulation!
$endgroup$
– Pedro Tamaroff♦
Feb 18 at 10:39
$begingroup$
Ok, I really notice that a triangulation can be done for any convex polygon.
$endgroup$
– LH8
Feb 18 at 10:43
$begingroup$
Ok, I really notice that a triangulation can be done for any convex polygon.
$endgroup$
– LH8
Feb 18 at 10:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Feb 18 at 10:38
Dan UznanskiDan Uznanski
6,88521528
$endgroup$
$begingroup$
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
Feb 18 at 10:39
$begingroup$
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
Feb 18 at 10:48
$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
Feb 18 at 11:19
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
Feb 18 at 11:24
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
Feb 18 at 12:19
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Feb 18 at 10:38
Dan UznanskiDan Uznanski
6,88521528
$endgroup$
$begingroup$
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
Feb 18 at 10:39
$begingroup$
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
Feb 18 at 10:48
$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
Feb 18 at 11:19
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
Feb 18 at 11:24
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
Feb 18 at 12:19
|
show 1 more comment
$begingroup$
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.
answered Feb 18 at 10:38
Dan UznanskiDan Uznanski
6,88521528
$endgroup$
$begingroup$
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
Feb 18 at 10:39
$begingroup$
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
Feb 18 at 10:48
$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
Feb 18 at 11:19
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
Feb 18 at 11:24
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
Feb 18 at 12:19
|
show 1 more comment
$begingroup$
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.
answered Feb 18 at 10:38
Dan UznanskiDan Uznanski
6,88521528
$endgroup$
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.
answered Feb 18 at 10:38
Dan UznanskiDan Uznanski
6,88521528
edited Feb 18 at 10:48
answered Feb 18 at 10:38
Dan UznanskiDan Uznanski
6,88521528
answered Feb 18 at 10:38
Dan UznanskiDan Uznanski
6,88521528
answered Feb 18 at 10:38
Dan UznanskiDan Uznanski
6,88521528
6,88521528
$begingroup$
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
Feb 18 at 10:39
$begingroup$
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
Feb 18 at 10:48
$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
Feb 18 at 11:19
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
Feb 18 at 11:24
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
Feb 18 at 12:19
|
show 1 more comment
$begingroup$
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
Feb 18 at 10:39
$begingroup$
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
Feb 18 at 10:48
$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
Feb 18 at 11:19
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
Feb 18 at 11:24
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
Feb 18 at 12:19
$begingroup$
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
Feb 18 at 10:39
$begingroup$
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
Feb 18 at 10:39
$begingroup$
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
Feb 18 at 10:48
$begingroup$
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
Feb 18 at 10:48
$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
Feb 18 at 11:19
$begingroup$
So, a polyedron has different Euler characteristic depending the triangulation? I´m confuse. Thanks.
$endgroup$
– LH8
Feb 18 at 11:19
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
Feb 18 at 11:24
$begingroup$
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
Feb 18 at 11:24
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
Feb 18 at 12:19
$begingroup$
"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
Feb 18 at 12:19
|
show 1 more comment
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$begingroup$
The leftmost image in the second figure a quadrangulation, not a triangulation!
$endgroup$
– Pedro Tamaroff♦
Feb 18 at 10:39
$begingroup$
Ok, I really notice that a triangulation can be done for any convex polygon.
$endgroup$
– LH8
Feb 18 at 10:43