Why can't a vertex of a $d$-dimensional polytope be in fewer than $d$ edges?
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This is motivated by the definition of simple polytopes: if all vertices of a $d$-dimensional convex polytope $P$ are in exactly $d$ edges (i.e. $1$-dimensional faces of $P$), then $P$ is simple.
I struggle to show that no vertex of $P$ can be in fewer than $d$ edges. Clearly, if this would be the case for some vertex $v$, the (translated) cone generated by $v$ and its adjacent vertices would have a lower dimension than $d$.
So if I can only show that the cone contains $P$, I would be done. That appears obvious but I've spent some time struggling with no luck. I feel I am missing something obvious.
Edited to add: I use "polytope" to mean the convex hull of finitely many points in some Euclidean space. And, for the sake of completeness, when I speak of $P$ being $d$-dimensional, I mean that its affine hull is $d$-dimensional, not the ambient space.
combinatorics geometry convex-geometry polytopes
asked Dec 3 '18 at 10:29
StefanStefan
163
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add a comment |
$begingroup$
This is motivated by the definition of simple polytopes: if all vertices of a $d$-dimensional convex polytope $P$ are in exactly $d$ edges (i.e. $1$-dimensional faces of $P$), then $P$ is simple.
I struggle to show that no vertex of $P$ can be in fewer than $d$ edges. Clearly, if this would be the case for some vertex $v$, the (translated) cone generated by $v$ and its adjacent vertices would have a lower dimension than $d$.
So if I can only show that the cone contains $P$, I would be done. That appears obvious but I've spent some time struggling with no luck. I feel I am missing something obvious.
Edited to add: I use "polytope" to mean the convex hull of finitely many points in some Euclidean space. And, for the sake of completeness, when I speak of $P$ being $d$-dimensional, I mean that its affine hull is $d$-dimensional, not the ambient space.
combinatorics geometry convex-geometry polytopes
asked Dec 3 '18 at 10:29
StefanStefan
163
$endgroup$
$begingroup$
Any answer to this is going to depend on the details of how you've defined "polytope": could you clarify precisely what definition you're using?
$endgroup$
– user3482749
Dec 3 '18 at 10:38
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I edited my post. Happy to clarify further!
$endgroup$
– Stefan
Dec 3 '18 at 10:44
$begingroup$
At least in principle, it shouldn't be necessary: we're discussing an intrinsic property of the object, which holds without considering it as being embedded in anything else.
$endgroup$
– user3482749
Dec 3 '18 at 11:01
$begingroup$
If a polytope $P$ is the convex hull of finitely many points $x_1,dots,x_n$ in some Euclidean space, then in general not all $x_i$ are vertices of $P$. You should add the requirement that no $x_i$ is contained in the convex hull of the other $x_j$.
$endgroup$
– Paul Frost
Dec 3 '18 at 14:08
$begingroup$
@PaulFrost Thank you. My question is only about the vertices (i.e. the extreme points of $P$) and not about the elements of the point set generating $P$.
$endgroup$
– Stefan
Dec 4 '18 at 2:48
add a comment |
$begingroup$
This is motivated by the definition of simple polytopes: if all vertices of a $d$-dimensional convex polytope $P$ are in exactly $d$ edges (i.e. $1$-dimensional faces of $P$), then $P$ is simple.
I struggle to show that no vertex of $P$ can be in fewer than $d$ edges. Clearly, if this would be the case for some vertex $v$, the (translated) cone generated by $v$ and its adjacent vertices would have a lower dimension than $d$.
So if I can only show that the cone contains $P$, I would be done. That appears obvious but I've spent some time struggling with no luck. I feel I am missing something obvious.
Edited to add: I use "polytope" to mean the convex hull of finitely many points in some Euclidean space. And, for the sake of completeness, when I speak of $P$ being $d$-dimensional, I mean that its affine hull is $d$-dimensional, not the ambient space.
combinatorics geometry convex-geometry polytopes
asked Dec 3 '18 at 10:29
StefanStefan
163
$endgroup$
This is motivated by the definition of simple polytopes: if all vertices of a $d$-dimensional convex polytope $P$ are in exactly $d$ edges (i.e. $1$-dimensional faces of $P$), then $P$ is simple.
I struggle to show that no vertex of $P$ can be in fewer than $d$ edges. Clearly, if this would be the case for some vertex $v$, the (translated) cone generated by $v$ and its adjacent vertices would have a lower dimension than $d$.
So if I can only show that the cone contains $P$, I would be done. That appears obvious but I've spent some time struggling with no luck. I feel I am missing something obvious.
Edited to add: I use "polytope" to mean the convex hull of finitely many points in some Euclidean space. And, for the sake of completeness, when I speak of $P$ being $d$-dimensional, I mean that its affine hull is $d$-dimensional, not the ambient space.
combinatorics geometry convex-geometry polytopes
combinatorics geometry convex-geometry polytopes
asked Dec 3 '18 at 10:29
StefanStefan
163
asked Dec 3 '18 at 10:29
StefanStefan
163
edited Dec 3 '18 at 10:43
Stefan
asked Dec 3 '18 at 10:29
StefanStefan
163
asked Dec 3 '18 at 10:29
StefanStefan
163
asked Dec 3 '18 at 10:29
StefanStefan
163
163
$begingroup$
Any answer to this is going to depend on the details of how you've defined "polytope": could you clarify precisely what definition you're using?
$endgroup$
– user3482749
Dec 3 '18 at 10:38
$begingroup$
I edited my post. Happy to clarify further!
$endgroup$
– Stefan
Dec 3 '18 at 10:44
$begingroup$
At least in principle, it shouldn't be necessary: we're discussing an intrinsic property of the object, which holds without considering it as being embedded in anything else.
$endgroup$
– user3482749
Dec 3 '18 at 11:01
$begingroup$
If a polytope $P$ is the convex hull of finitely many points $x_1,dots,x_n$ in some Euclidean space, then in general not all $x_i$ are vertices of $P$. You should add the requirement that no $x_i$ is contained in the convex hull of the other $x_j$.
$endgroup$
– Paul Frost
Dec 3 '18 at 14:08
$begingroup$
@PaulFrost Thank you. My question is only about the vertices (i.e. the extreme points of $P$) and not about the elements of the point set generating $P$.
$endgroup$
– Stefan
Dec 4 '18 at 2:48
add a comment |
$begingroup$
Any answer to this is going to depend on the details of how you've defined "polytope": could you clarify precisely what definition you're using?
$endgroup$
– user3482749
Dec 3 '18 at 10:38
$begingroup$
I edited my post. Happy to clarify further!
$endgroup$
– Stefan
Dec 3 '18 at 10:44
$begingroup$
At least in principle, it shouldn't be necessary: we're discussing an intrinsic property of the object, which holds without considering it as being embedded in anything else.
$endgroup$
– user3482749
Dec 3 '18 at 11:01
$begingroup$
If a polytope $P$ is the convex hull of finitely many points $x_1,dots,x_n$ in some Euclidean space, then in general not all $x_i$ are vertices of $P$. You should add the requirement that no $x_i$ is contained in the convex hull of the other $x_j$.
$endgroup$
– Paul Frost
Dec 3 '18 at 14:08
$begingroup$
@PaulFrost Thank you. My question is only about the vertices (i.e. the extreme points of $P$) and not about the elements of the point set generating $P$.
$endgroup$
– Stefan
Dec 4 '18 at 2:48
$begingroup$
Any answer to this is going to depend on the details of how you've defined "polytope": could you clarify precisely what definition you're using?
$endgroup$
– user3482749
Dec 3 '18 at 10:38
$begingroup$
Any answer to this is going to depend on the details of how you've defined "polytope": could you clarify precisely what definition you're using?
$endgroup$
– user3482749
Dec 3 '18 at 10:38
$begingroup$
I edited my post. Happy to clarify further!
$endgroup$
– Stefan
Dec 3 '18 at 10:44
$begingroup$
I edited my post. Happy to clarify further!
$endgroup$
– Stefan
Dec 3 '18 at 10:44
$begingroup$
At least in principle, it shouldn't be necessary: we're discussing an intrinsic property of the object, which holds without considering it as being embedded in anything else.
$endgroup$
– user3482749
Dec 3 '18 at 11:01
$begingroup$
At least in principle, it shouldn't be necessary: we're discussing an intrinsic property of the object, which holds without considering it as being embedded in anything else.
$endgroup$
– user3482749
Dec 3 '18 at 11:01
$begingroup$
If a polytope $P$ is the convex hull of finitely many points $x_1,dots,x_n$ in some Euclidean space, then in general not all $x_i$ are vertices of $P$. You should add the requirement that no $x_i$ is contained in the convex hull of the other $x_j$.
$endgroup$
– Paul Frost
Dec 3 '18 at 14:08
$begingroup$
If a polytope $P$ is the convex hull of finitely many points $x_1,dots,x_n$ in some Euclidean space, then in general not all $x_i$ are vertices of $P$. You should add the requirement that no $x_i$ is contained in the convex hull of the other $x_j$.
$endgroup$
– Paul Frost
Dec 3 '18 at 14:08
$begingroup$
@PaulFrost Thank you. My question is only about the vertices (i.e. the extreme points of $P$) and not about the elements of the point set generating $P$.
$endgroup$
– Stefan
Dec 4 '18 at 2:48
$begingroup$
@PaulFrost Thank you. My question is only about the vertices (i.e. the extreme points of $P$) and not about the elements of the point set generating $P$.
$endgroup$
– Stefan
Dec 4 '18 at 2:48
add a comment |
1 Answer
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Any vertex figure of a $d$-polytope happens to be a $(d-1)$-polytope. The smallest known $(d-1)$-polytope is a $(d-1)$-simplex, which has exactly $d$ vertices. As those vertices in turn represent the edges of that $d$-polytope, which emanate from the vertex of consideration, you are done.
--- rk
answered Dec 3 '18 at 18:09
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
$endgroup$
$begingroup$
Thank you! I struggle only with the very first part of what you wrote: why is the vertex figure of a $d$-polytope $(d-1)$-dimensional? It's intuitively obvious but I am missing an argument for that fact. (In a sense, I am restating my confusion from the OP.)
$endgroup$
– Stefan
Dec 4 '18 at 2:26
$begingroup$
Just consider chopping off a tiny vertex pyramid. Tiny wrt. the incident edge lengths for sure. Then the base of that pyramid will be a representation of the vertex figure. And for sure, it will be a $d-1$-dimensional polytope.
$endgroup$
– Dr. Richard Klitzing
Dec 12 '18 at 18:28
add a comment |
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1 Answer
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$begingroup$
Any vertex figure of a $d$-polytope happens to be a $(d-1)$-polytope. The smallest known $(d-1)$-polytope is a $(d-1)$-simplex, which has exactly $d$ vertices. As those vertices in turn represent the edges of that $d$-polytope, which emanate from the vertex of consideration, you are done.
--- rk
answered Dec 3 '18 at 18:09
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
$endgroup$
$begingroup$
Thank you! I struggle only with the very first part of what you wrote: why is the vertex figure of a $d$-polytope $(d-1)$-dimensional? It's intuitively obvious but I am missing an argument for that fact. (In a sense, I am restating my confusion from the OP.)
$endgroup$
– Stefan
Dec 4 '18 at 2:26
$begingroup$
Just consider chopping off a tiny vertex pyramid. Tiny wrt. the incident edge lengths for sure. Then the base of that pyramid will be a representation of the vertex figure. And for sure, it will be a $d-1$-dimensional polytope.
$endgroup$
– Dr. Richard Klitzing
Dec 12 '18 at 18:28
add a comment |
$begingroup$
Any vertex figure of a $d$-polytope happens to be a $(d-1)$-polytope. The smallest known $(d-1)$-polytope is a $(d-1)$-simplex, which has exactly $d$ vertices. As those vertices in turn represent the edges of that $d$-polytope, which emanate from the vertex of consideration, you are done.
--- rk
answered Dec 3 '18 at 18:09
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
$endgroup$
$begingroup$
Thank you! I struggle only with the very first part of what you wrote: why is the vertex figure of a $d$-polytope $(d-1)$-dimensional? It's intuitively obvious but I am missing an argument for that fact. (In a sense, I am restating my confusion from the OP.)
$endgroup$
– Stefan
Dec 4 '18 at 2:26
$begingroup$
Just consider chopping off a tiny vertex pyramid. Tiny wrt. the incident edge lengths for sure. Then the base of that pyramid will be a representation of the vertex figure. And for sure, it will be a $d-1$-dimensional polytope.
$endgroup$
– Dr. Richard Klitzing
Dec 12 '18 at 18:28
add a comment |
$begingroup$
Any vertex figure of a $d$-polytope happens to be a $(d-1)$-polytope. The smallest known $(d-1)$-polytope is a $(d-1)$-simplex, which has exactly $d$ vertices. As those vertices in turn represent the edges of that $d$-polytope, which emanate from the vertex of consideration, you are done.
--- rk
answered Dec 3 '18 at 18:09
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
$endgroup$
Any vertex figure of a $d$-polytope happens to be a $(d-1)$-polytope. The smallest known $(d-1)$-polytope is a $(d-1)$-simplex, which has exactly $d$ vertices. As those vertices in turn represent the edges of that $d$-polytope, which emanate from the vertex of consideration, you are done.
--- rk
answered Dec 3 '18 at 18:09
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
answered Dec 3 '18 at 18:09
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
answered Dec 3 '18 at 18:09
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
answered Dec 3 '18 at 18:09
Dr. Richard KlitzingDr. Richard Klitzing
1,50616
1,50616
$begingroup$
Thank you! I struggle only with the very first part of what you wrote: why is the vertex figure of a $d$-polytope $(d-1)$-dimensional? It's intuitively obvious but I am missing an argument for that fact. (In a sense, I am restating my confusion from the OP.)
$endgroup$
– Stefan
Dec 4 '18 at 2:26
$begingroup$
Just consider chopping off a tiny vertex pyramid. Tiny wrt. the incident edge lengths for sure. Then the base of that pyramid will be a representation of the vertex figure. And for sure, it will be a $d-1$-dimensional polytope.
$endgroup$
– Dr. Richard Klitzing
Dec 12 '18 at 18:28
add a comment |
$begingroup$
Thank you! I struggle only with the very first part of what you wrote: why is the vertex figure of a $d$-polytope $(d-1)$-dimensional? It's intuitively obvious but I am missing an argument for that fact. (In a sense, I am restating my confusion from the OP.)
$endgroup$
– Stefan
Dec 4 '18 at 2:26
$begingroup$
Just consider chopping off a tiny vertex pyramid. Tiny wrt. the incident edge lengths for sure. Then the base of that pyramid will be a representation of the vertex figure. And for sure, it will be a $d-1$-dimensional polytope.
$endgroup$
– Dr. Richard Klitzing
Dec 12 '18 at 18:28
$begingroup$
Thank you! I struggle only with the very first part of what you wrote: why is the vertex figure of a $d$-polytope $(d-1)$-dimensional? It's intuitively obvious but I am missing an argument for that fact. (In a sense, I am restating my confusion from the OP.)
$endgroup$
– Stefan
Dec 4 '18 at 2:26
$begingroup$
Thank you! I struggle only with the very first part of what you wrote: why is the vertex figure of a $d$-polytope $(d-1)$-dimensional? It's intuitively obvious but I am missing an argument for that fact. (In a sense, I am restating my confusion from the OP.)
$endgroup$
– Stefan
Dec 4 '18 at 2:26
$begingroup$
Just consider chopping off a tiny vertex pyramid. Tiny wrt. the incident edge lengths for sure. Then the base of that pyramid will be a representation of the vertex figure. And for sure, it will be a $d-1$-dimensional polytope.
$endgroup$
– Dr. Richard Klitzing
Dec 12 '18 at 18:28
$begingroup$
Just consider chopping off a tiny vertex pyramid. Tiny wrt. the incident edge lengths for sure. Then the base of that pyramid will be a representation of the vertex figure. And for sure, it will be a $d-1$-dimensional polytope.
$endgroup$
– Dr. Richard Klitzing
Dec 12 '18 at 18:28
add a comment |
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$begingroup$
Any answer to this is going to depend on the details of how you've defined "polytope": could you clarify precisely what definition you're using?
$endgroup$
– user3482749
Dec 3 '18 at 10:38
$begingroup$
I edited my post. Happy to clarify further!
$endgroup$
– Stefan
Dec 3 '18 at 10:44
$begingroup$
At least in principle, it shouldn't be necessary: we're discussing an intrinsic property of the object, which holds without considering it as being embedded in anything else.
$endgroup$
– user3482749
Dec 3 '18 at 11:01
$begingroup$
If a polytope $P$ is the convex hull of finitely many points $x_1,dots,x_n$ in some Euclidean space, then in general not all $x_i$ are vertices of $P$. You should add the requirement that no $x_i$ is contained in the convex hull of the other $x_j$.
$endgroup$
– Paul Frost
Dec 3 '18 at 14:08
$begingroup$
@PaulFrost Thank you. My question is only about the vertices (i.e. the extreme points of $P$) and not about the elements of the point set generating $P$.
$endgroup$
– Stefan
Dec 4 '18 at 2:48