recovering the information of the vertices from a simplex Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The simplest $Delta$-complex structure on $S^2$Uniformly distributed points over the surface of the standard simplexEdge identification implies vertex identification (but not vice versa)?How to allow simplicial maps with “degeneracies” in the case of simplicial complexUpper Bound for Vertices of Intersection of Subspace and SimplexWhy does the definition of 2-cycle as $kerpartial_2$ work so magically?Question about theorem in Hatcher's Algebraic TopologyUnderstanding the n-dimensional Simplex in TopologyExplanation needed for the representation of simplex as a polyhedronBarycentric subdivision preserves geometric realization
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recovering the information of the vertices from a simplex
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The simplest $Delta$-complex structure on $S^2$Uniformly distributed points over the surface of the standard simplexEdge identification implies vertex identification (but not vice versa)?How to allow simplicial maps with “degeneracies” in the case of simplicial complexUpper Bound for Vertices of Intersection of Subspace and SimplexWhy does the definition of 2-cycle as $kerpartial_2$ work so magically?Question about theorem in Hatcher's Algebraic TopologyUnderstanding the n-dimensional Simplex in TopologyExplanation needed for the representation of simplex as a polyhedronBarycentric subdivision preserves geometric realization
$begingroup$
It seems quite obvious that, when given a simplex, its set of vertices is uniquely determined by the simplex. The formal formulation of this intuition is as follows:
Suppose that the points $v_0,v_1,dots,v_k$ and
$w_0,w_1,dots,w_l$ are affinely independent sets of points of
$mathbbR^n$. If the $k$-simplex $sigma=[v_0,dots,v_k]$ and the $l$-simplex $tau=[w_0,dots, w_l]$
are equal, then $k=l$ and
$v_0,v_1,dots,v_k=w_0,w_1,dots,w_l$.
We clearly have $k=l$ because an $m$-simplex is homeomorphic to the closed ball of dimension $m$ and no two closed ball of different dimensions are homeomorphic to each other. But I cannot prove that $v_0,v_1,dots,v_k=w_0,w_1,dots,w_l$.
I think I must be overlooking something very obvious... Can anyone help me? Thanks in advance.
general-topology algebraic-topology simplex simplicial-complex
asked Mar 27 at 3:32
KenKen
191110
$endgroup$
add a comment |
$begingroup$
It seems quite obvious that, when given a simplex, its set of vertices is uniquely determined by the simplex. The formal formulation of this intuition is as follows:
Suppose that the points $v_0,v_1,dots,v_k$ and
$w_0,w_1,dots,w_l$ are affinely independent sets of points of
$mathbbR^n$. If the $k$-simplex $sigma=[v_0,dots,v_k]$ and the $l$-simplex $tau=[w_0,dots, w_l]$
are equal, then $k=l$ and
$v_0,v_1,dots,v_k=w_0,w_1,dots,w_l$.
We clearly have $k=l$ because an $m$-simplex is homeomorphic to the closed ball of dimension $m$ and no two closed ball of different dimensions are homeomorphic to each other. But I cannot prove that $v_0,v_1,dots,v_k=w_0,w_1,dots,w_l$.
I think I must be overlooking something very obvious... Can anyone help me? Thanks in advance.
general-topology algebraic-topology simplex simplicial-complex
asked Mar 27 at 3:32
KenKen
191110
$endgroup$
$begingroup$
What does it mean for two simplices to be equal?
$endgroup$
– Connor Malin
Mar 27 at 3:46
$begingroup$
@ConnorMalin Two simplices are equal when they are equal as sets.
$endgroup$
– Ken
Mar 27 at 4:04
1
$begingroup$
Given a simplex, you can recover its vertices by its extreme points. Maybe this helps.
$endgroup$
– JHF
Mar 27 at 16:33
add a comment |
$begingroup$
It seems quite obvious that, when given a simplex, its set of vertices is uniquely determined by the simplex. The formal formulation of this intuition is as follows:
Suppose that the points $v_0,v_1,dots,v_k$ and
$w_0,w_1,dots,w_l$ are affinely independent sets of points of
$mathbbR^n$. If the $k$-simplex $sigma=[v_0,dots,v_k]$ and the $l$-simplex $tau=[w_0,dots, w_l]$
are equal, then $k=l$ and
$v_0,v_1,dots,v_k=w_0,w_1,dots,w_l$.
We clearly have $k=l$ because an $m$-simplex is homeomorphic to the closed ball of dimension $m$ and no two closed ball of different dimensions are homeomorphic to each other. But I cannot prove that $v_0,v_1,dots,v_k=w_0,w_1,dots,w_l$.
I think I must be overlooking something very obvious... Can anyone help me? Thanks in advance.
general-topology algebraic-topology simplex simplicial-complex
asked Mar 27 at 3:32
KenKen
191110
$endgroup$
It seems quite obvious that, when given a simplex, its set of vertices is uniquely determined by the simplex. The formal formulation of this intuition is as follows:
Suppose that the points $v_0,v_1,dots,v_k$ and
$w_0,w_1,dots,w_l$ are affinely independent sets of points of
$mathbbR^n$. If the $k$-simplex $sigma=[v_0,dots,v_k]$ and the $l$-simplex $tau=[w_0,dots, w_l]$
are equal, then $k=l$ and
$v_0,v_1,dots,v_k=w_0,w_1,dots,w_l$.
We clearly have $k=l$ because an $m$-simplex is homeomorphic to the closed ball of dimension $m$ and no two closed ball of different dimensions are homeomorphic to each other. But I cannot prove that $v_0,v_1,dots,v_k=w_0,w_1,dots,w_l$.
I think I must be overlooking something very obvious... Can anyone help me? Thanks in advance.
general-topology algebraic-topology simplex simplicial-complex
general-topology algebraic-topology simplex simplicial-complex
asked Mar 27 at 3:32
KenKen
191110
asked Mar 27 at 3:32
KenKen
191110
asked Mar 27 at 3:32
KenKen
191110
asked Mar 27 at 3:32
KenKen
191110
asked Mar 27 at 3:32
KenKen
191110
191110
$begingroup$
What does it mean for two simplices to be equal?
$endgroup$
– Connor Malin
Mar 27 at 3:46
$begingroup$
@ConnorMalin Two simplices are equal when they are equal as sets.
$endgroup$
– Ken
Mar 27 at 4:04
1
$begingroup$
Given a simplex, you can recover its vertices by its extreme points. Maybe this helps.
$endgroup$
– JHF
Mar 27 at 16:33
add a comment |
$begingroup$
What does it mean for two simplices to be equal?
$endgroup$
– Connor Malin
Mar 27 at 3:46
$begingroup$
@ConnorMalin Two simplices are equal when they are equal as sets.
$endgroup$
– Ken
Mar 27 at 4:04
1
$begingroup$
Given a simplex, you can recover its vertices by its extreme points. Maybe this helps.
$endgroup$
– JHF
Mar 27 at 16:33
$begingroup$
What does it mean for two simplices to be equal?
$endgroup$
– Connor Malin
Mar 27 at 3:46
$begingroup$
What does it mean for two simplices to be equal?
$endgroup$
– Connor Malin
Mar 27 at 3:46
$begingroup$
@ConnorMalin Two simplices are equal when they are equal as sets.
$endgroup$
– Ken
Mar 27 at 4:04
$begingroup$
@ConnorMalin Two simplices are equal when they are equal as sets.
$endgroup$
– Ken
Mar 27 at 4:04
1
1
$begingroup$
Given a simplex, you can recover its vertices by its extreme points. Maybe this helps.
$endgroup$
– JHF
Mar 27 at 16:33
$begingroup$
Given a simplex, you can recover its vertices by its extreme points. Maybe this helps.
$endgroup$
– JHF
Mar 27 at 16:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Trickier than I expected. Suppose that $v_0,dots,v_k$ does not equal $w_0,dots,w_k$. Then without loss of generality we can assume $v_0 = Sigma r_i w_i$ where multiple $r_i$ are nonzero. Again, without loss of generality assume that these are $r_0,r_1$. Then there is a function $(-epsilon, epsilon) rightarrow sigma$ defined by $t rightarrow (r_0 +t)w_0 + (r_1-t)w_1 + r_2 w_2 +dots + r_k w_k$. Such a function cannot exist because at $t=0$ this passes through $v_0$, and no such line segment passes through $v_0$ that is also contained in $sigma$ because such a thing would necessarily contain points that when written as a sum of the $v_i$ would have negative coefficients.
answered Mar 27 at 4:26
Connor MalinConnor Malin
649111
$endgroup$
add a comment |
$begingroup$
As JHF had pointed out in the comment, the vertices of a given simplex can be characterized as the extreme points of the simplex. (If $S$ is a convex set, a point $x$ in $S$ is called its extreme point if for each $y$ in $Ssetminus x$, any open line segment containing $x$ and $y$ always contains a point not in $S$.) This can be checked directly by using the barycentric coordinates.
Although the underlying idea is the same as that of Connor Malin's answer, and despite this answer being not my idea, I will post this because I feel like this is the most "natural" way to approach the problem. Big thanks to Connor Malin and JHF!
answered Mar 28 at 2:14
KenKen
191110
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Trickier than I expected. Suppose that $v_0,dots,v_k$ does not equal $w_0,dots,w_k$. Then without loss of generality we can assume $v_0 = Sigma r_i w_i$ where multiple $r_i$ are nonzero. Again, without loss of generality assume that these are $r_0,r_1$. Then there is a function $(-epsilon, epsilon) rightarrow sigma$ defined by $t rightarrow (r_0 +t)w_0 + (r_1-t)w_1 + r_2 w_2 +dots + r_k w_k$. Such a function cannot exist because at $t=0$ this passes through $v_0$, and no such line segment passes through $v_0$ that is also contained in $sigma$ because such a thing would necessarily contain points that when written as a sum of the $v_i$ would have negative coefficients.
answered Mar 27 at 4:26
Connor MalinConnor Malin
649111
$endgroup$
add a comment |
$begingroup$
Trickier than I expected. Suppose that $v_0,dots,v_k$ does not equal $w_0,dots,w_k$. Then without loss of generality we can assume $v_0 = Sigma r_i w_i$ where multiple $r_i$ are nonzero. Again, without loss of generality assume that these are $r_0,r_1$. Then there is a function $(-epsilon, epsilon) rightarrow sigma$ defined by $t rightarrow (r_0 +t)w_0 + (r_1-t)w_1 + r_2 w_2 +dots + r_k w_k$. Such a function cannot exist because at $t=0$ this passes through $v_0$, and no such line segment passes through $v_0$ that is also contained in $sigma$ because such a thing would necessarily contain points that when written as a sum of the $v_i$ would have negative coefficients.
answered Mar 27 at 4:26
Connor MalinConnor Malin
649111
$endgroup$
add a comment |
$begingroup$
Trickier than I expected. Suppose that $v_0,dots,v_k$ does not equal $w_0,dots,w_k$. Then without loss of generality we can assume $v_0 = Sigma r_i w_i$ where multiple $r_i$ are nonzero. Again, without loss of generality assume that these are $r_0,r_1$. Then there is a function $(-epsilon, epsilon) rightarrow sigma$ defined by $t rightarrow (r_0 +t)w_0 + (r_1-t)w_1 + r_2 w_2 +dots + r_k w_k$. Such a function cannot exist because at $t=0$ this passes through $v_0$, and no such line segment passes through $v_0$ that is also contained in $sigma$ because such a thing would necessarily contain points that when written as a sum of the $v_i$ would have negative coefficients.
answered Mar 27 at 4:26
Connor MalinConnor Malin
649111
$endgroup$
Trickier than I expected. Suppose that $v_0,dots,v_k$ does not equal $w_0,dots,w_k$. Then without loss of generality we can assume $v_0 = Sigma r_i w_i$ where multiple $r_i$ are nonzero. Again, without loss of generality assume that these are $r_0,r_1$. Then there is a function $(-epsilon, epsilon) rightarrow sigma$ defined by $t rightarrow (r_0 +t)w_0 + (r_1-t)w_1 + r_2 w_2 +dots + r_k w_k$. Such a function cannot exist because at $t=0$ this passes through $v_0$, and no such line segment passes through $v_0$ that is also contained in $sigma$ because such a thing would necessarily contain points that when written as a sum of the $v_i$ would have negative coefficients.
answered Mar 27 at 4:26
Connor MalinConnor Malin
649111
edited Mar 27 at 4:51
answered Mar 27 at 4:26
Connor MalinConnor Malin
649111
answered Mar 27 at 4:26
Connor MalinConnor Malin
649111
answered Mar 27 at 4:26
Connor MalinConnor Malin
649111
649111
add a comment |
add a comment |
$begingroup$
As JHF had pointed out in the comment, the vertices of a given simplex can be characterized as the extreme points of the simplex. (If $S$ is a convex set, a point $x$ in $S$ is called its extreme point if for each $y$ in $Ssetminus x$, any open line segment containing $x$ and $y$ always contains a point not in $S$.) This can be checked directly by using the barycentric coordinates.
Although the underlying idea is the same as that of Connor Malin's answer, and despite this answer being not my idea, I will post this because I feel like this is the most "natural" way to approach the problem. Big thanks to Connor Malin and JHF!
answered Mar 28 at 2:14
KenKen
191110
$endgroup$
add a comment |
$begingroup$
As JHF had pointed out in the comment, the vertices of a given simplex can be characterized as the extreme points of the simplex. (If $S$ is a convex set, a point $x$ in $S$ is called its extreme point if for each $y$ in $Ssetminus x$, any open line segment containing $x$ and $y$ always contains a point not in $S$.) This can be checked directly by using the barycentric coordinates.
Although the underlying idea is the same as that of Connor Malin's answer, and despite this answer being not my idea, I will post this because I feel like this is the most "natural" way to approach the problem. Big thanks to Connor Malin and JHF!
answered Mar 28 at 2:14
KenKen
191110
$endgroup$
add a comment |
$begingroup$
As JHF had pointed out in the comment, the vertices of a given simplex can be characterized as the extreme points of the simplex. (If $S$ is a convex set, a point $x$ in $S$ is called its extreme point if for each $y$ in $Ssetminus x$, any open line segment containing $x$ and $y$ always contains a point not in $S$.) This can be checked directly by using the barycentric coordinates.
Although the underlying idea is the same as that of Connor Malin's answer, and despite this answer being not my idea, I will post this because I feel like this is the most "natural" way to approach the problem. Big thanks to Connor Malin and JHF!
answered Mar 28 at 2:14
KenKen
191110
$endgroup$
As JHF had pointed out in the comment, the vertices of a given simplex can be characterized as the extreme points of the simplex. (If $S$ is a convex set, a point $x$ in $S$ is called its extreme point if for each $y$ in $Ssetminus x$, any open line segment containing $x$ and $y$ always contains a point not in $S$.) This can be checked directly by using the barycentric coordinates.
Although the underlying idea is the same as that of Connor Malin's answer, and despite this answer being not my idea, I will post this because I feel like this is the most "natural" way to approach the problem. Big thanks to Connor Malin and JHF!
answered Mar 28 at 2:14
KenKen
191110
answered Mar 28 at 2:14
KenKen
191110
answered Mar 28 at 2:14
KenKen
191110
answered Mar 28 at 2:14
KenKen
191110
191110
add a comment |
add a comment |
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$begingroup$
What does it mean for two simplices to be equal?
$endgroup$
– Connor Malin
Mar 27 at 3:46
$begingroup$
@ConnorMalin Two simplices are equal when they are equal as sets.
$endgroup$
– Ken
Mar 27 at 4:04
1
$begingroup$
Given a simplex, you can recover its vertices by its extreme points. Maybe this helps.
$endgroup$
– JHF
Mar 27 at 16:33