Mixing
Product of two polytopes is a polytope
$begingroup$
Please have a look at my attempt for this problem.
Let $x = begin{pmatrix} x_1\ x_2 \ end{pmatrix}, x_1 in P_1, x_2 in P_2$.
I want to show that $x in conv{P_1 times P_2}$, i.e. $x$ can be represented as the convex combination of some points of $P_1 times P_2$.
Without loss of generality, suppose that $d_1 geq d_2$. By Caratheodory's theorem, $x_1$ can be represented by at most $d_1 + 1$ points of $P_1$, i.e. there exists ${v_1,...,v_{d_1+1}} subset P_1$, and $alpha_1,...,alpha_{d_1+1}$; $ alpha_i geq 0$, $ sum alpha_i=1$, such that:
$$x_1 = alpha_1v_1+...+alpha_{d_1+1}v_{d_1+1}$$
$x_2$ can be represented similarly, with points ${w_1,...,w_{d_2+1},...,w_{d_1+1}} subset P_2$, and coefficients $beta_i$'s, such that $beta_{d_2+2},...,beta_{d_1+1}$(if exist) are all $0$'s:
$$x_2 = beta_1w_1+...+beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1}$$
So now we have $x$ as:
$$x = begin{pmatrix} x_1\ x_2 \ end{pmatrix} = begin{pmatrix} alpha_1v_1+...+alpha_{d_2+1}v_{d_2+1}+alpha_{d_2+2}v_{d_2+2}+...+alpha_{d_1+1}v_{d_1+1}\ beta_1w_1+...+beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1} \ end{pmatrix}$$
It is here that I got stuck. The points $begin{pmatrix} v_i\ w_i \ end{pmatrix}$ above are certainly in $P_1 times P_2$, but are the coefficients right? Shouldn't the coefficients be in $mathbb{R}$, instead of $mathbb{R}^2$?
===================================
Edit: the polytopes here are all convex polytopes.
Edit 2: Caratheodory's theorem: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)
convex-geometry polytopes discrete-geometry convex-hulls
asked Jan 1 at 15:22
ensbanaensbana
237113
$endgroup$
This question has an open bounty worth +50
reputation from ensbana ending tomorrow.
The current answers do not contain enough detail.
|
show 2 more comments
$begingroup$
Please have a look at my attempt for this problem.
Let $x = begin{pmatrix} x_1\ x_2 \ end{pmatrix}, x_1 in P_1, x_2 in P_2$.
I want to show that $x in conv{P_1 times P_2}$, i.e. $x$ can be represented as the convex combination of some points of $P_1 times P_2$.
Without loss of generality, suppose that $d_1 geq d_2$. By Caratheodory's theorem, $x_1$ can be represented by at most $d_1 + 1$ points of $P_1$, i.e. there exists ${v_1,...,v_{d_1+1}} subset P_1$, and $alpha_1,...,alpha_{d_1+1}$; $ alpha_i geq 0$, $ sum alpha_i=1$, such that:
$$x_1 = alpha_1v_1+...+alpha_{d_1+1}v_{d_1+1}$$
$x_2$ can be represented similarly, with points ${w_1,...,w_{d_2+1},...,w_{d_1+1}} subset P_2$, and coefficients $beta_i$'s, such that $beta_{d_2+2},...,beta_{d_1+1}$(if exist) are all $0$'s:
$$x_2 = beta_1w_1+...+beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1}$$
So now we have $x$ as:
$$x = begin{pmatrix} x_1\ x_2 \ end{pmatrix} = begin{pmatrix} alpha_1v_1+...+alpha_{d_2+1}v_{d_2+1}+alpha_{d_2+2}v_{d_2+2}+...+alpha_{d_1+1}v_{d_1+1}\ beta_1w_1+...+beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1} \ end{pmatrix}$$
It is here that I got stuck. The points $begin{pmatrix} v_i\ w_i \ end{pmatrix}$ above are certainly in $P_1 times P_2$, but are the coefficients right? Shouldn't the coefficients be in $mathbb{R}$, instead of $mathbb{R}^2$?
===================================
Edit: the polytopes here are all convex polytopes.
Edit 2: Caratheodory's theorem: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)
convex-geometry polytopes discrete-geometry convex-hulls
asked Jan 1 at 15:22
ensbanaensbana
237113
$endgroup$
This question has an open bounty worth +50
reputation from ensbana ending tomorrow.
The current answers do not contain enough detail.
$begingroup$
Why can't $d_1=d_2$?
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:31
$begingroup$
Could you please define polytope? Because from what I know, polytopes need not necessarily be convex.
$endgroup$
– toric_actions
Jan 1 at 16:14
$begingroup$
@LordSharktheUnknown $d_1$ might or might not be equal to $d_2$. But I addressed this by representing $x_2$ with some extra terms of coefficients $0$, so that the representation of $x_1$ and $x_2$ both consist of $d_1+1$ terms.
$endgroup$
– ensbana
Jan 1 at 20:23
$begingroup$
@toric_actions this is from a class about convex polytopes, so the convexity of the polytopes is implied. Sorry for the confusion! Also, in our class we have two definitions for convex polytopes: (1) as the convex hull of finitely many points, and (2) as finite intersection of closed half-spaces.
$endgroup$
– ensbana
Jan 1 at 20:26
$begingroup$
Could you provide a definition of polytope you use? If you define it as a bounded polyhedron, the statement is trivial.
$endgroup$
– LinAlg
Jan 2 at 15:21
|
show 2 more comments
$begingroup$
Please have a look at my attempt for this problem.
Let $x = begin{pmatrix} x_1\ x_2 \ end{pmatrix}, x_1 in P_1, x_2 in P_2$.
I want to show that $x in conv{P_1 times P_2}$, i.e. $x$ can be represented as the convex combination of some points of $P_1 times P_2$.
Without loss of generality, suppose that $d_1 geq d_2$. By Caratheodory's theorem, $x_1$ can be represented by at most $d_1 + 1$ points of $P_1$, i.e. there exists ${v_1,...,v_{d_1+1}} subset P_1$, and $alpha_1,...,alpha_{d_1+1}$; $ alpha_i geq 0$, $ sum alpha_i=1$, such that:
$$x_1 = alpha_1v_1+...+alpha_{d_1+1}v_{d_1+1}$$
$x_2$ can be represented similarly, with points ${w_1,...,w_{d_2+1},...,w_{d_1+1}} subset P_2$, and coefficients $beta_i$'s, such that $beta_{d_2+2},...,beta_{d_1+1}$(if exist) are all $0$'s:
$$x_2 = beta_1w_1+...+beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1}$$
So now we have $x$ as:
$$x = begin{pmatrix} x_1\ x_2 \ end{pmatrix} = begin{pmatrix} alpha_1v_1+...+alpha_{d_2+1}v_{d_2+1}+alpha_{d_2+2}v_{d_2+2}+...+alpha_{d_1+1}v_{d_1+1}\ beta_1w_1+...+beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1} \ end{pmatrix}$$
It is here that I got stuck. The points $begin{pmatrix} v_i\ w_i \ end{pmatrix}$ above are certainly in $P_1 times P_2$, but are the coefficients right? Shouldn't the coefficients be in $mathbb{R}$, instead of $mathbb{R}^2$?
===================================
Edit: the polytopes here are all convex polytopes.
Edit 2: Caratheodory's theorem: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)
convex-geometry polytopes discrete-geometry convex-hulls
asked Jan 1 at 15:22
ensbanaensbana
237113
$endgroup$
Please have a look at my attempt for this problem.
Let $x = begin{pmatrix} x_1\ x_2 \ end{pmatrix}, x_1 in P_1, x_2 in P_2$.
I want to show that $x in conv{P_1 times P_2}$, i.e. $x$ can be represented as the convex combination of some points of $P_1 times P_2$.
Without loss of generality, suppose that $d_1 geq d_2$. By Caratheodory's theorem, $x_1$ can be represented by at most $d_1 + 1$ points of $P_1$, i.e. there exists ${v_1,...,v_{d_1+1}} subset P_1$, and $alpha_1,...,alpha_{d_1+1}$; $ alpha_i geq 0$, $ sum alpha_i=1$, such that:
$$x_1 = alpha_1v_1+...+alpha_{d_1+1}v_{d_1+1}$$
$x_2$ can be represented similarly, with points ${w_1,...,w_{d_2+1},...,w_{d_1+1}} subset P_2$, and coefficients $beta_i$'s, such that $beta_{d_2+2},...,beta_{d_1+1}$(if exist) are all $0$'s:
$$x_2 = beta_1w_1+...+beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1}$$
So now we have $x$ as:
$$x = begin{pmatrix} x_1\ x_2 \ end{pmatrix} = begin{pmatrix} alpha_1v_1+...+alpha_{d_2+1}v_{d_2+1}+alpha_{d_2+2}v_{d_2+2}+...+alpha_{d_1+1}v_{d_1+1}\ beta_1w_1+...+beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1} \ end{pmatrix}$$
It is here that I got stuck. The points $begin{pmatrix} v_i\ w_i \ end{pmatrix}$ above are certainly in $P_1 times P_2$, but are the coefficients right? Shouldn't the coefficients be in $mathbb{R}$, instead of $mathbb{R}^2$?
===================================
Edit: the polytopes here are all convex polytopes.
Edit 2: Caratheodory's theorem: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)
convex-geometry polytopes discrete-geometry convex-hulls
convex-geometry polytopes discrete-geometry convex-hulls
asked Jan 1 at 15:22
ensbanaensbana
237113
asked Jan 1 at 15:22
ensbanaensbana
237113
edited Jan 1 at 20:33
ensbana
asked Jan 1 at 15:22
ensbanaensbana
237113
asked Jan 1 at 15:22
ensbanaensbana
237113
asked Jan 1 at 15:22
ensbanaensbana
237113
237113
This question has an open bounty worth +50
reputation from ensbana ending tomorrow.
The current answers do not contain enough detail.
This question has an open bounty worth +50
reputation from ensbana ending tomorrow.
The current answers do not contain enough detail.
$begingroup$
Why can't $d_1=d_2$?
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:31
$begingroup$
Could you please define polytope? Because from what I know, polytopes need not necessarily be convex.
$endgroup$
– toric_actions
Jan 1 at 16:14
$begingroup$
@LordSharktheUnknown $d_1$ might or might not be equal to $d_2$. But I addressed this by representing $x_2$ with some extra terms of coefficients $0$, so that the representation of $x_1$ and $x_2$ both consist of $d_1+1$ terms.
$endgroup$
– ensbana
Jan 1 at 20:23
$begingroup$
@toric_actions this is from a class about convex polytopes, so the convexity of the polytopes is implied. Sorry for the confusion! Also, in our class we have two definitions for convex polytopes: (1) as the convex hull of finitely many points, and (2) as finite intersection of closed half-spaces.
$endgroup$
– ensbana
Jan 1 at 20:26
$begingroup$
Could you provide a definition of polytope you use? If you define it as a bounded polyhedron, the statement is trivial.
$endgroup$
– LinAlg
Jan 2 at 15:21
|
show 2 more comments
$begingroup$
Why can't $d_1=d_2$?
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:31
$begingroup$
Could you please define polytope? Because from what I know, polytopes need not necessarily be convex.
$endgroup$
– toric_actions
Jan 1 at 16:14
$begingroup$
@LordSharktheUnknown $d_1$ might or might not be equal to $d_2$. But I addressed this by representing $x_2$ with some extra terms of coefficients $0$, so that the representation of $x_1$ and $x_2$ both consist of $d_1+1$ terms.
$endgroup$
– ensbana
Jan 1 at 20:23
$begingroup$
@toric_actions this is from a class about convex polytopes, so the convexity of the polytopes is implied. Sorry for the confusion! Also, in our class we have two definitions for convex polytopes: (1) as the convex hull of finitely many points, and (2) as finite intersection of closed half-spaces.
$endgroup$
– ensbana
Jan 1 at 20:26
$begingroup$
Could you provide a definition of polytope you use? If you define it as a bounded polyhedron, the statement is trivial.
$endgroup$
– LinAlg
Jan 2 at 15:21
$begingroup$
Why can't $d_1=d_2$?
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:31
$begingroup$
Why can't $d_1=d_2$?
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:31
$begingroup$
Could you please define polytope? Because from what I know, polytopes need not necessarily be convex.
$endgroup$
– toric_actions
Jan 1 at 16:14
$begingroup$
Could you please define polytope? Because from what I know, polytopes need not necessarily be convex.
$endgroup$
– toric_actions
Jan 1 at 16:14
$begingroup$
@LordSharktheUnknown $d_1$ might or might not be equal to $d_2$. But I addressed this by representing $x_2$ with some extra terms of coefficients $0$, so that the representation of $x_1$ and $x_2$ both consist of $d_1+1$ terms.
$endgroup$
– ensbana
Jan 1 at 20:23
$begingroup$
@LordSharktheUnknown $d_1$ might or might not be equal to $d_2$. But I addressed this by representing $x_2$ with some extra terms of coefficients $0$, so that the representation of $x_1$ and $x_2$ both consist of $d_1+1$ terms.
$endgroup$
– ensbana
Jan 1 at 20:23
$begingroup$
@toric_actions this is from a class about convex polytopes, so the convexity of the polytopes is implied. Sorry for the confusion! Also, in our class we have two definitions for convex polytopes: (1) as the convex hull of finitely many points, and (2) as finite intersection of closed half-spaces.
$endgroup$
– ensbana
Jan 1 at 20:26
$begingroup$
@toric_actions this is from a class about convex polytopes, so the convexity of the polytopes is implied. Sorry for the confusion! Also, in our class we have two definitions for convex polytopes: (1) as the convex hull of finitely many points, and (2) as finite intersection of closed half-spaces.
$endgroup$
– ensbana
Jan 1 at 20:26
$begingroup$
Could you provide a definition of polytope you use? If you define it as a bounded polyhedron, the statement is trivial.
$endgroup$
– LinAlg
Jan 2 at 15:21
$begingroup$
Could you provide a definition of polytope you use? If you define it as a bounded polyhedron, the statement is trivial.
$endgroup$
– LinAlg
Jan 2 at 15:21
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Try first to visualize the effect for low dimensions.
Let be $P_1$ and $P_2$ both be points along the x- resp. y-axis. Then $P_1times P_2$ obviously is nothing but the point $(P_1, P_2)$.
Let $P_1$ still be some point on the x-axis, while $P_2$ be the line segment from $a$ to $b$ along the y-axis. Then $P_1times P_2$ happens to be the line segment between the points $(P_1, a)$ and $(P_1, b)$.
Then let $P_1$ be such a line segment from $a$ to $b$ along the x-axis, and $P_2$ be the line segment from $c$ to $d$ along the y-axis. Then $P_1times P_2$ happens to be the rectangle with vertices $(a, c)$, $(a, d)$, $(b, c)$, and $(b, d)$.
In fact, the Cartesian polytopal product is nothing but a brique product. The outcome $P_1times P_2$ is just the $(P_1, P_2)$-duoprism. As seen from the above examples, a single point is the neutral element of that product. And for the dimensions you'd get the sum formula: $dim(P_1times P_2)=dim(P_1)+dim(P_2)$.
Eg. take $P_1$ to be some regular $n$-gonal polygon of side length being unity and $P_2$ to be a line segment, also of unit size. Then $P_1times P_2$ happens to be nothing but the Archimedean (uniform) $n$-prism. But you well can consider eg. the duoprism from a stellated dodecahedron and a great icosahedron, if you'd like. That one then happens to be $3+3=6$ dimensional!
--- rk
answered Jan 6 at 16:53
Dr. Richard KlitzingDr. Richard Klitzing
1,47616
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058574%2fproduct-of-two-polytopes-is-a-polytope%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try first to visualize the effect for low dimensions.
Let be $P_1$ and $P_2$ both be points along the x- resp. y-axis. Then $P_1times P_2$ obviously is nothing but the point $(P_1, P_2)$.
Let $P_1$ still be some point on the x-axis, while $P_2$ be the line segment from $a$ to $b$ along the y-axis. Then $P_1times P_2$ happens to be the line segment between the points $(P_1, a)$ and $(P_1, b)$.
Then let $P_1$ be such a line segment from $a$ to $b$ along the x-axis, and $P_2$ be the line segment from $c$ to $d$ along the y-axis. Then $P_1times P_2$ happens to be the rectangle with vertices $(a, c)$, $(a, d)$, $(b, c)$, and $(b, d)$.
In fact, the Cartesian polytopal product is nothing but a brique product. The outcome $P_1times P_2$ is just the $(P_1, P_2)$-duoprism. As seen from the above examples, a single point is the neutral element of that product. And for the dimensions you'd get the sum formula: $dim(P_1times P_2)=dim(P_1)+dim(P_2)$.
Eg. take $P_1$ to be some regular $n$-gonal polygon of side length being unity and $P_2$ to be a line segment, also of unit size. Then $P_1times P_2$ happens to be nothing but the Archimedean (uniform) $n$-prism. But you well can consider eg. the duoprism from a stellated dodecahedron and a great icosahedron, if you'd like. That one then happens to be $3+3=6$ dimensional!
--- rk
answered Jan 6 at 16:53
Dr. Richard KlitzingDr. Richard Klitzing
1,47616
$endgroup$
add a comment |
$begingroup$
Try first to visualize the effect for low dimensions.
Let be $P_1$ and $P_2$ both be points along the x- resp. y-axis. Then $P_1times P_2$ obviously is nothing but the point $(P_1, P_2)$.
Let $P_1$ still be some point on the x-axis, while $P_2$ be the line segment from $a$ to $b$ along the y-axis. Then $P_1times P_2$ happens to be the line segment between the points $(P_1, a)$ and $(P_1, b)$.
Then let $P_1$ be such a line segment from $a$ to $b$ along the x-axis, and $P_2$ be the line segment from $c$ to $d$ along the y-axis. Then $P_1times P_2$ happens to be the rectangle with vertices $(a, c)$, $(a, d)$, $(b, c)$, and $(b, d)$.
In fact, the Cartesian polytopal product is nothing but a brique product. The outcome $P_1times P_2$ is just the $(P_1, P_2)$-duoprism. As seen from the above examples, a single point is the neutral element of that product. And for the dimensions you'd get the sum formula: $dim(P_1times P_2)=dim(P_1)+dim(P_2)$.
Eg. take $P_1$ to be some regular $n$-gonal polygon of side length being unity and $P_2$ to be a line segment, also of unit size. Then $P_1times P_2$ happens to be nothing but the Archimedean (uniform) $n$-prism. But you well can consider eg. the duoprism from a stellated dodecahedron and a great icosahedron, if you'd like. That one then happens to be $3+3=6$ dimensional!
--- rk
answered Jan 6 at 16:53
Dr. Richard KlitzingDr. Richard Klitzing
1,47616
$endgroup$
add a comment |
$begingroup$
Try first to visualize the effect for low dimensions.
Let be $P_1$ and $P_2$ both be points along the x- resp. y-axis. Then $P_1times P_2$ obviously is nothing but the point $(P_1, P_2)$.
Let $P_1$ still be some point on the x-axis, while $P_2$ be the line segment from $a$ to $b$ along the y-axis. Then $P_1times P_2$ happens to be the line segment between the points $(P_1, a)$ and $(P_1, b)$.
Then let $P_1$ be such a line segment from $a$ to $b$ along the x-axis, and $P_2$ be the line segment from $c$ to $d$ along the y-axis. Then $P_1times P_2$ happens to be the rectangle with vertices $(a, c)$, $(a, d)$, $(b, c)$, and $(b, d)$.
In fact, the Cartesian polytopal product is nothing but a brique product. The outcome $P_1times P_2$ is just the $(P_1, P_2)$-duoprism. As seen from the above examples, a single point is the neutral element of that product. And for the dimensions you'd get the sum formula: $dim(P_1times P_2)=dim(P_1)+dim(P_2)$.
Eg. take $P_1$ to be some regular $n$-gonal polygon of side length being unity and $P_2$ to be a line segment, also of unit size. Then $P_1times P_2$ happens to be nothing but the Archimedean (uniform) $n$-prism. But you well can consider eg. the duoprism from a stellated dodecahedron and a great icosahedron, if you'd like. That one then happens to be $3+3=6$ dimensional!
--- rk
answered Jan 6 at 16:53
Dr. Richard KlitzingDr. Richard Klitzing
1,47616
$endgroup$
Try first to visualize the effect for low dimensions.
Let be $P_1$ and $P_2$ both be points along the x- resp. y-axis. Then $P_1times P_2$ obviously is nothing but the point $(P_1, P_2)$.
Let $P_1$ still be some point on the x-axis, while $P_2$ be the line segment from $a$ to $b$ along the y-axis. Then $P_1times P_2$ happens to be the line segment between the points $(P_1, a)$ and $(P_1, b)$.
Then let $P_1$ be such a line segment from $a$ to $b$ along the x-axis, and $P_2$ be the line segment from $c$ to $d$ along the y-axis. Then $P_1times P_2$ happens to be the rectangle with vertices $(a, c)$, $(a, d)$, $(b, c)$, and $(b, d)$.
In fact, the Cartesian polytopal product is nothing but a brique product. The outcome $P_1times P_2$ is just the $(P_1, P_2)$-duoprism. As seen from the above examples, a single point is the neutral element of that product. And for the dimensions you'd get the sum formula: $dim(P_1times P_2)=dim(P_1)+dim(P_2)$.
Eg. take $P_1$ to be some regular $n$-gonal polygon of side length being unity and $P_2$ to be a line segment, also of unit size. Then $P_1times P_2$ happens to be nothing but the Archimedean (uniform) $n$-prism. But you well can consider eg. the duoprism from a stellated dodecahedron and a great icosahedron, if you'd like. That one then happens to be $3+3=6$ dimensional!
--- rk
answered Jan 6 at 16:53
Dr. Richard KlitzingDr. Richard Klitzing
1,47616
answered Jan 6 at 16:53
Dr. Richard KlitzingDr. Richard Klitzing
1,47616
answered Jan 6 at 16:53
Dr. Richard KlitzingDr. Richard Klitzing
1,47616
answered Jan 6 at 16:53
Dr. Richard KlitzingDr. Richard Klitzing
1,47616
1,47616
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058574%2fproduct-of-two-polytopes-is-a-polytope%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Why can't $d_1=d_2$?
$endgroup$
– Lord Shark the Unknown
Jan 1 at 15:31
$begingroup$
Could you please define polytope? Because from what I know, polytopes need not necessarily be convex.
$endgroup$
– toric_actions
Jan 1 at 16:14
$begingroup$
@LordSharktheUnknown $d_1$ might or might not be equal to $d_2$. But I addressed this by representing $x_2$ with some extra terms of coefficients $0$, so that the representation of $x_1$ and $x_2$ both consist of $d_1+1$ terms.
$endgroup$
– ensbana
Jan 1 at 20:23
$begingroup$
@toric_actions this is from a class about convex polytopes, so the convexity of the polytopes is implied. Sorry for the confusion! Also, in our class we have two definitions for convex polytopes: (1) as the convex hull of finitely many points, and (2) as finite intersection of closed half-spaces.
$endgroup$
– ensbana
Jan 1 at 20:26
$begingroup$
Could you provide a definition of polytope you use? If you define it as a bounded polyhedron, the statement is trivial.
$endgroup$
– LinAlg
Jan 2 at 15:21