Mixing
On embedding of a triangulation
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Let $X$ be a triangulable topological space, i.e, a topological space which is homeomorphic with a finite simplicial complex. Suppose $X$ can be embedded in $mathbbR^d$ for some $d$. Is it possible to find a simplicial complex inside $mathbbR^d$ that triangulates $X$?
As a motivation for this question: Fáry's theorem states that any simple planar graph can be drawn without crossings so that its edges are straight line segments. That is, the ability to draw graph edges as curves instead of as straight line segments does not allow a larger class of graphs to be drawn
algebraic-topology
asked Jul 25 at 8:10
123...
379113
add a comment |Â
up vote
3
down vote
favorite
Let $X$ be a triangulable topological space, i.e, a topological space which is homeomorphic with a finite simplicial complex. Suppose $X$ can be embedded in $mathbbR^d$ for some $d$. Is it possible to find a simplicial complex inside $mathbbR^d$ that triangulates $X$?
As a motivation for this question: Fáry's theorem states that any simple planar graph can be drawn without crossings so that its edges are straight line segments. That is, the ability to draw graph edges as curves instead of as straight line segments does not allow a larger class of graphs to be drawn
algebraic-topology
asked Jul 25 at 8:10
123...
379113
2
What do you mean by "simplicial complex inside $mathbbR^d$"? If you know the simplicial structure of $X$ then this is the simplicial complex? Or are you saying "how to construct the triangulation assuming it exists"? I'm quite lost here.
– freakish
Jul 25 at 9:21
2
@freakish OP has a topological embedding and wants a piecewise linear embedding.
– Mike Miller
Jul 25 at 10:35
@MikeMiller just take a geometric realization of $X$'s abstract simplicial complex. Or am I missing something?
– freakish
Jul 25 at 10:48
2
@freakish This is a space that does not come equipped with a piecewise linear embedding into $Bbb R^d$. You may embed it into Euclidean space by ensuring the vertices are affinely independent, but usually the dimension of the Euclidean space will be $Bbb R^2n+1$, where $n$ is the dimension of your complex. OP wants to keep the dimension fixed. (As an example, you can of course embed every graph in $Bbb R^3$ piecewise linearly, but it is not obvious why every planar graph has a piecewise linear planar embedding; Fary's theorem is not trivial.)
– Mike Miller
Jul 25 at 11:00
Oops, I now see that you always come equipped with an embedding of a simplicial complex with $k$ vertices into $Bbb R^k$. But other than this correction I think everything else in my comment is correct.
– Mike Miller
Jul 25 at 11:02
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $X$ be a triangulable topological space, i.e, a topological space which is homeomorphic with a finite simplicial complex. Suppose $X$ can be embedded in $mathbbR^d$ for some $d$. Is it possible to find a simplicial complex inside $mathbbR^d$ that triangulates $X$?
As a motivation for this question: Fáry's theorem states that any simple planar graph can be drawn without crossings so that its edges are straight line segments. That is, the ability to draw graph edges as curves instead of as straight line segments does not allow a larger class of graphs to be drawn
algebraic-topology
asked Jul 25 at 8:10
123...
379113
Let $X$ be a triangulable topological space, i.e, a topological space which is homeomorphic with a finite simplicial complex. Suppose $X$ can be embedded in $mathbbR^d$ for some $d$. Is it possible to find a simplicial complex inside $mathbbR^d$ that triangulates $X$?
As a motivation for this question: Fáry's theorem states that any simple planar graph can be drawn without crossings so that its edges are straight line segments. That is, the ability to draw graph edges as curves instead of as straight line segments does not allow a larger class of graphs to be drawn
algebraic-topology
asked Jul 25 at 8:10
123...
379113
edited Jul 25 at 8:19
asked Jul 25 at 8:10
123...
379113
asked Jul 25 at 8:10
123...
379113
asked Jul 25 at 8:10
123...
379113
379113
2
What do you mean by "simplicial complex inside $mathbbR^d$"? If you know the simplicial structure of $X$ then this is the simplicial complex? Or are you saying "how to construct the triangulation assuming it exists"? I'm quite lost here.
– freakish
Jul 25 at 9:21
2
@freakish OP has a topological embedding and wants a piecewise linear embedding.
– Mike Miller
Jul 25 at 10:35
@MikeMiller just take a geometric realization of $X$'s abstract simplicial complex. Or am I missing something?
– freakish
Jul 25 at 10:48
2
@freakish This is a space that does not come equipped with a piecewise linear embedding into $Bbb R^d$. You may embed it into Euclidean space by ensuring the vertices are affinely independent, but usually the dimension of the Euclidean space will be $Bbb R^2n+1$, where $n$ is the dimension of your complex. OP wants to keep the dimension fixed. (As an example, you can of course embed every graph in $Bbb R^3$ piecewise linearly, but it is not obvious why every planar graph has a piecewise linear planar embedding; Fary's theorem is not trivial.)
– Mike Miller
Jul 25 at 11:00
Oops, I now see that you always come equipped with an embedding of a simplicial complex with $k$ vertices into $Bbb R^k$. But other than this correction I think everything else in my comment is correct.
– Mike Miller
Jul 25 at 11:02
add a comment |Â
2
What do you mean by "simplicial complex inside $mathbbR^d$"? If you know the simplicial structure of $X$ then this is the simplicial complex? Or are you saying "how to construct the triangulation assuming it exists"? I'm quite lost here.
– freakish
Jul 25 at 9:21
2
@freakish OP has a topological embedding and wants a piecewise linear embedding.
– Mike Miller
Jul 25 at 10:35
@MikeMiller just take a geometric realization of $X$'s abstract simplicial complex. Or am I missing something?
– freakish
Jul 25 at 10:48
2
@freakish This is a space that does not come equipped with a piecewise linear embedding into $Bbb R^d$. You may embed it into Euclidean space by ensuring the vertices are affinely independent, but usually the dimension of the Euclidean space will be $Bbb R^2n+1$, where $n$ is the dimension of your complex. OP wants to keep the dimension fixed. (As an example, you can of course embed every graph in $Bbb R^3$ piecewise linearly, but it is not obvious why every planar graph has a piecewise linear planar embedding; Fary's theorem is not trivial.)
– Mike Miller
Jul 25 at 11:00
Oops, I now see that you always come equipped with an embedding of a simplicial complex with $k$ vertices into $Bbb R^k$. But other than this correction I think everything else in my comment is correct.
– Mike Miller
Jul 25 at 11:02
2
2
What do you mean by "simplicial complex inside $mathbbR^d$"? If you know the simplicial structure of $X$ then this is the simplicial complex? Or are you saying "how to construct the triangulation assuming it exists"? I'm quite lost here.
– freakish
Jul 25 at 9:21
What do you mean by "simplicial complex inside $mathbbR^d$"? If you know the simplicial structure of $X$ then this is the simplicial complex? Or are you saying "how to construct the triangulation assuming it exists"? I'm quite lost here.
– freakish
Jul 25 at 9:21
2
2
@freakish OP has a topological embedding and wants a piecewise linear embedding.
– Mike Miller
Jul 25 at 10:35
@freakish OP has a topological embedding and wants a piecewise linear embedding.
– Mike Miller
Jul 25 at 10:35
@MikeMiller just take a geometric realization of $X$'s abstract simplicial complex. Or am I missing something?
– freakish
Jul 25 at 10:48
@MikeMiller just take a geometric realization of $X$'s abstract simplicial complex. Or am I missing something?
– freakish
Jul 25 at 10:48
2
2
@freakish This is a space that does not come equipped with a piecewise linear embedding into $Bbb R^d$. You may embed it into Euclidean space by ensuring the vertices are affinely independent, but usually the dimension of the Euclidean space will be $Bbb R^2n+1$, where $n$ is the dimension of your complex. OP wants to keep the dimension fixed. (As an example, you can of course embed every graph in $Bbb R^3$ piecewise linearly, but it is not obvious why every planar graph has a piecewise linear planar embedding; Fary's theorem is not trivial.)
– Mike Miller
Jul 25 at 11:00
@freakish This is a space that does not come equipped with a piecewise linear embedding into $Bbb R^d$. You may embed it into Euclidean space by ensuring the vertices are affinely independent, but usually the dimension of the Euclidean space will be $Bbb R^2n+1$, where $n$ is the dimension of your complex. OP wants to keep the dimension fixed. (As an example, you can of course embed every graph in $Bbb R^3$ piecewise linearly, but it is not obvious why every planar graph has a piecewise linear planar embedding; Fary's theorem is not trivial.)
– Mike Miller
Jul 25 at 11:00
Oops, I now see that you always come equipped with an embedding of a simplicial complex with $k$ vertices into $Bbb R^k$. But other than this correction I think everything else in my comment is correct.
– Mike Miller
Jul 25 at 11:02
Oops, I now see that you always come equipped with an embedding of a simplicial complex with $k$ vertices into $Bbb R^k$. But other than this correction I think everything else in my comment is correct.
– Mike Miller
Jul 25 at 11:02
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
As it was mentioned in the question any planar graph
has a planar drawing where all edges are straight segments. But, a higher-dimensional analog of this assertion fails! Brehm and Sarkaria showed that for every $dgeq 2$ and every $k$, $d+1leq kleq 2d$, there exist a
finite $d$-dimensional simplicial complexes $K$ that can be embedded in
$mathbbR^k$ but not linearly. For more information, please see
"Brehm, Ulrich, and Karanbir S. Sarkaria. "Linear vs. piecewise-linear embeddability of simplicial complexes." MPI-[Ber.]/Max-Planck-Inst. f@: ur Mathematik; 92-52 (1992)."
answered Aug 4 at 22:06
123...
379113
add a comment |Â
up vote
0
down vote
I think what you are looking for is the Simplicial Approximation Theorem:
If $K$ is a finite simplicial complex and $L$ is an arbitrary simplicial
complex, then any map $f :Kto L$ is homotopic to a map that is simplicial with respect to some iterated barycentric subdivision of $K$.
You can find the proof in Hatcher Algebraic Topology book.
The proof makes it clear that the simplicial approximation $g$ can be chosen not
just homotopic to $f$ but also close to $f$ (If you Choose a metric on $K$ that restricts to the standard Euclidean metric
on each simplex of $K$) if we allow subdivisions of $L$ as well as $K$.
answered Jul 29 at 8:32
Elad
52429
Thank you very much for your time and comment. I know this theorem. Actually, you can interpret my question as follows. Suppose we have a triangluabale subset of of some R^d, say for example d=3, we know taht there is an abdtract simplicial complex that can be realized in the dimension at most 7 and its polyhedron is homeomorphic with the given subset. My qurstion now is that : Is there a simplicial complex that its polyhedron is homeomorphic with the given subset and can be realized in dimension 4.
– 123...
Jul 29 at 12:10
Sadly I don’t think I can do any better then you. If it’s a manifold you can use Whitney embedding theorems and get sum results
– Elad
Jul 29 at 19:28
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As it was mentioned in the question any planar graph
has a planar drawing where all edges are straight segments. But, a higher-dimensional analog of this assertion fails! Brehm and Sarkaria showed that for every $dgeq 2$ and every $k$, $d+1leq kleq 2d$, there exist a
finite $d$-dimensional simplicial complexes $K$ that can be embedded in
$mathbbR^k$ but not linearly. For more information, please see
"Brehm, Ulrich, and Karanbir S. Sarkaria. "Linear vs. piecewise-linear embeddability of simplicial complexes." MPI-[Ber.]/Max-Planck-Inst. f@: ur Mathematik; 92-52 (1992)."
answered Aug 4 at 22:06
123...
379113
add a comment |Â
up vote
2
down vote
accepted
As it was mentioned in the question any planar graph
has a planar drawing where all edges are straight segments. But, a higher-dimensional analog of this assertion fails! Brehm and Sarkaria showed that for every $dgeq 2$ and every $k$, $d+1leq kleq 2d$, there exist a
finite $d$-dimensional simplicial complexes $K$ that can be embedded in
$mathbbR^k$ but not linearly. For more information, please see
"Brehm, Ulrich, and Karanbir S. Sarkaria. "Linear vs. piecewise-linear embeddability of simplicial complexes." MPI-[Ber.]/Max-Planck-Inst. f@: ur Mathematik; 92-52 (1992)."
answered Aug 4 at 22:06
123...
379113
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As it was mentioned in the question any planar graph
has a planar drawing where all edges are straight segments. But, a higher-dimensional analog of this assertion fails! Brehm and Sarkaria showed that for every $dgeq 2$ and every $k$, $d+1leq kleq 2d$, there exist a
finite $d$-dimensional simplicial complexes $K$ that can be embedded in
$mathbbR^k$ but not linearly. For more information, please see
"Brehm, Ulrich, and Karanbir S. Sarkaria. "Linear vs. piecewise-linear embeddability of simplicial complexes." MPI-[Ber.]/Max-Planck-Inst. f@: ur Mathematik; 92-52 (1992)."
answered Aug 4 at 22:06
123...
379113
As it was mentioned in the question any planar graph
has a planar drawing where all edges are straight segments. But, a higher-dimensional analog of this assertion fails! Brehm and Sarkaria showed that for every $dgeq 2$ and every $k$, $d+1leq kleq 2d$, there exist a
finite $d$-dimensional simplicial complexes $K$ that can be embedded in
$mathbbR^k$ but not linearly. For more information, please see
"Brehm, Ulrich, and Karanbir S. Sarkaria. "Linear vs. piecewise-linear embeddability of simplicial complexes." MPI-[Ber.]/Max-Planck-Inst. f@: ur Mathematik; 92-52 (1992)."
answered Aug 4 at 22:06
123...
379113
answered Aug 4 at 22:06
123...
379113
answered Aug 4 at 22:06
123...
379113
answered Aug 4 at 22:06
123...
379113
379113
add a comment |Â
add a comment |Â
up vote
0
down vote
I think what you are looking for is the Simplicial Approximation Theorem:
If $K$ is a finite simplicial complex and $L$ is an arbitrary simplicial
complex, then any map $f :Kto L$ is homotopic to a map that is simplicial with respect to some iterated barycentric subdivision of $K$.
You can find the proof in Hatcher Algebraic Topology book.
The proof makes it clear that the simplicial approximation $g$ can be chosen not
just homotopic to $f$ but also close to $f$ (If you Choose a metric on $K$ that restricts to the standard Euclidean metric
on each simplex of $K$) if we allow subdivisions of $L$ as well as $K$.
answered Jul 29 at 8:32
Elad
52429
Thank you very much for your time and comment. I know this theorem. Actually, you can interpret my question as follows. Suppose we have a triangluabale subset of of some R^d, say for example d=3, we know taht there is an abdtract simplicial complex that can be realized in the dimension at most 7 and its polyhedron is homeomorphic with the given subset. My qurstion now is that : Is there a simplicial complex that its polyhedron is homeomorphic with the given subset and can be realized in dimension 4.
– 123...
Jul 29 at 12:10
Sadly I don’t think I can do any better then you. If it’s a manifold you can use Whitney embedding theorems and get sum results
– Elad
Jul 29 at 19:28
add a comment |Â
up vote
0
down vote
I think what you are looking for is the Simplicial Approximation Theorem:
If $K$ is a finite simplicial complex and $L$ is an arbitrary simplicial
complex, then any map $f :Kto L$ is homotopic to a map that is simplicial with respect to some iterated barycentric subdivision of $K$.
You can find the proof in Hatcher Algebraic Topology book.
The proof makes it clear that the simplicial approximation $g$ can be chosen not
just homotopic to $f$ but also close to $f$ (If you Choose a metric on $K$ that restricts to the standard Euclidean metric
on each simplex of $K$) if we allow subdivisions of $L$ as well as $K$.
answered Jul 29 at 8:32
Elad
52429
Thank you very much for your time and comment. I know this theorem. Actually, you can interpret my question as follows. Suppose we have a triangluabale subset of of some R^d, say for example d=3, we know taht there is an abdtract simplicial complex that can be realized in the dimension at most 7 and its polyhedron is homeomorphic with the given subset. My qurstion now is that : Is there a simplicial complex that its polyhedron is homeomorphic with the given subset and can be realized in dimension 4.
– 123...
Jul 29 at 12:10
Sadly I don’t think I can do any better then you. If it’s a manifold you can use Whitney embedding theorems and get sum results
– Elad
Jul 29 at 19:28
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think what you are looking for is the Simplicial Approximation Theorem:
If $K$ is a finite simplicial complex and $L$ is an arbitrary simplicial
complex, then any map $f :Kto L$ is homotopic to a map that is simplicial with respect to some iterated barycentric subdivision of $K$.
You can find the proof in Hatcher Algebraic Topology book.
The proof makes it clear that the simplicial approximation $g$ can be chosen not
just homotopic to $f$ but also close to $f$ (If you Choose a metric on $K$ that restricts to the standard Euclidean metric
on each simplex of $K$) if we allow subdivisions of $L$ as well as $K$.
answered Jul 29 at 8:32
Elad
52429
I think what you are looking for is the Simplicial Approximation Theorem:
If $K$ is a finite simplicial complex and $L$ is an arbitrary simplicial
complex, then any map $f :Kto L$ is homotopic to a map that is simplicial with respect to some iterated barycentric subdivision of $K$.
You can find the proof in Hatcher Algebraic Topology book.
The proof makes it clear that the simplicial approximation $g$ can be chosen not
just homotopic to $f$ but also close to $f$ (If you Choose a metric on $K$ that restricts to the standard Euclidean metric
on each simplex of $K$) if we allow subdivisions of $L$ as well as $K$.
answered Jul 29 at 8:32
Elad
52429
answered Jul 29 at 8:32
Elad
52429
answered Jul 29 at 8:32
Elad
52429
answered Jul 29 at 8:32
Elad
52429
52429
Thank you very much for your time and comment. I know this theorem. Actually, you can interpret my question as follows. Suppose we have a triangluabale subset of of some R^d, say for example d=3, we know taht there is an abdtract simplicial complex that can be realized in the dimension at most 7 and its polyhedron is homeomorphic with the given subset. My qurstion now is that : Is there a simplicial complex that its polyhedron is homeomorphic with the given subset and can be realized in dimension 4.
– 123...
Jul 29 at 12:10
Sadly I don’t think I can do any better then you. If it’s a manifold you can use Whitney embedding theorems and get sum results
– Elad
Jul 29 at 19:28
add a comment |Â
Thank you very much for your time and comment. I know this theorem. Actually, you can interpret my question as follows. Suppose we have a triangluabale subset of of some R^d, say for example d=3, we know taht there is an abdtract simplicial complex that can be realized in the dimension at most 7 and its polyhedron is homeomorphic with the given subset. My qurstion now is that : Is there a simplicial complex that its polyhedron is homeomorphic with the given subset and can be realized in dimension 4.
– 123...
Jul 29 at 12:10
Sadly I don’t think I can do any better then you. If it’s a manifold you can use Whitney embedding theorems and get sum results
– Elad
Jul 29 at 19:28
Thank you very much for your time and comment. I know this theorem. Actually, you can interpret my question as follows. Suppose we have a triangluabale subset of of some R^d, say for example d=3, we know taht there is an abdtract simplicial complex that can be realized in the dimension at most 7 and its polyhedron is homeomorphic with the given subset. My qurstion now is that : Is there a simplicial complex that its polyhedron is homeomorphic with the given subset and can be realized in dimension 4.
– 123...
Jul 29 at 12:10
Thank you very much for your time and comment. I know this theorem. Actually, you can interpret my question as follows. Suppose we have a triangluabale subset of of some R^d, say for example d=3, we know taht there is an abdtract simplicial complex that can be realized in the dimension at most 7 and its polyhedron is homeomorphic with the given subset. My qurstion now is that : Is there a simplicial complex that its polyhedron is homeomorphic with the given subset and can be realized in dimension 4.
– 123...
Jul 29 at 12:10
Sadly I don’t think I can do any better then you. If it’s a manifold you can use Whitney embedding theorems and get sum results
– Elad
Jul 29 at 19:28
Sadly I don’t think I can do any better then you. If it’s a manifold you can use Whitney embedding theorems and get sum results
– Elad
Jul 29 at 19:28
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862163%2fon-embedding-of-a-triangulation%23new-answer', 'question_page');
);
Post as a guest
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
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
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
2
What do you mean by "simplicial complex inside $mathbbR^d$"? If you know the simplicial structure of $X$ then this is the simplicial complex? Or are you saying "how to construct the triangulation assuming it exists"? I'm quite lost here.
– freakish
Jul 25 at 9:21
2
@freakish OP has a topological embedding and wants a piecewise linear embedding.
– Mike Miller
Jul 25 at 10:35
@MikeMiller just take a geometric realization of $X$'s abstract simplicial complex. Or am I missing something?
– freakish
Jul 25 at 10:48
2
@freakish This is a space that does not come equipped with a piecewise linear embedding into $Bbb R^d$. You may embed it into Euclidean space by ensuring the vertices are affinely independent, but usually the dimension of the Euclidean space will be $Bbb R^2n+1$, where $n$ is the dimension of your complex. OP wants to keep the dimension fixed. (As an example, you can of course embed every graph in $Bbb R^3$ piecewise linearly, but it is not obvious why every planar graph has a piecewise linear planar embedding; Fary's theorem is not trivial.)
– Mike Miller
Jul 25 at 11:00
Oops, I now see that you always come equipped with an embedding of a simplicial complex with $k$ vertices into $Bbb R^k$. But other than this correction I think everything else in my comment is correct.
– Mike Miller
Jul 25 at 11:02