Are framed manifolds cubulatable?
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Let's say an $n$-manifold is cubulated if it is glued out of cubes $[0,1]^n$ in a way that looks locally like the standard cubulation of $mathbb R^n$. For instance, the face $[0,1]^{k-1} times {1} times [0,1]^{n-k}$ of some cube must be glued to the face $[0,1]^{k-1} times {0} times [0,1]^{n-k}$ of some other cube, for the same value of $k$; the torus $(mathbb R/mathbb Z)^n$ is cubulated with just one cube (coming from the standard map $[0,1] to mathbb R/mathbb Z$); the Klein bottle is not cubulated in my sense.
One can imagine an equivalence relation on cubulations of $M$ in which two cubulations are equivalent if they share a common refinement. But I haven't thought through the details of exactly what "refinement" should mean.
It's clear that every cubulated manifold is framed. (A framing of an $n$-manifold $M$ is a homotopy class of trivializations of the tangent bundle, i.e. a homotopy class of vector bundle isomorphisms $TM cong mathbb R^n times M$.) Locally, a framing determines a cubulation (or rather an equivalence class of cubulations). But globally I'm not so sure. I'm worried about things like the irrational line on the torus, which could prevent a framing from arising from any finite cubulation.
Question: Does every framed manifold admit a framing-compatible cubulation?
at.algebraic-topology gt.geometric-topology smooth-manifolds condensed-matter
asked 12 hours ago
Theo Johnson-FreydTheo Johnson-Freyd
29.6k879252
$endgroup$
add a comment |
$begingroup$
Let's say an $n$-manifold is cubulated if it is glued out of cubes $[0,1]^n$ in a way that looks locally like the standard cubulation of $mathbb R^n$. For instance, the face $[0,1]^{k-1} times {1} times [0,1]^{n-k}$ of some cube must be glued to the face $[0,1]^{k-1} times {0} times [0,1]^{n-k}$ of some other cube, for the same value of $k$; the torus $(mathbb R/mathbb Z)^n$ is cubulated with just one cube (coming from the standard map $[0,1] to mathbb R/mathbb Z$); the Klein bottle is not cubulated in my sense.
One can imagine an equivalence relation on cubulations of $M$ in which two cubulations are equivalent if they share a common refinement. But I haven't thought through the details of exactly what "refinement" should mean.
It's clear that every cubulated manifold is framed. (A framing of an $n$-manifold $M$ is a homotopy class of trivializations of the tangent bundle, i.e. a homotopy class of vector bundle isomorphisms $TM cong mathbb R^n times M$.) Locally, a framing determines a cubulation (or rather an equivalence class of cubulations). But globally I'm not so sure. I'm worried about things like the irrational line on the torus, which could prevent a framing from arising from any finite cubulation.
Question: Does every framed manifold admit a framing-compatible cubulation?
at.algebraic-topology gt.geometric-topology smooth-manifolds condensed-matter
asked 12 hours ago
Theo Johnson-FreydTheo Johnson-Freyd
29.6k879252
$endgroup$
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Does the 3-sphere admit a cubulation in your sense?
$endgroup$
– Chris Schommer-Pries
10 hours ago
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@ChrisSchommer-Pries A hypercube provides one, I believe.
$endgroup$
– მამუკა ჯიბლაძე
10 hours ago
3
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Your restrictions on the cubulation seem to imply that $M$ has a flat metric and that the action of $pi_1 M$ on the universal covering space $mathbb R^n$ is an action by translation, so $pi_1 M$ is isomorphic to $mathbb Z^n$.
$endgroup$
– Lee Mosher
8 hours ago
add a comment |
$begingroup$
Let's say an $n$-manifold is cubulated if it is glued out of cubes $[0,1]^n$ in a way that looks locally like the standard cubulation of $mathbb R^n$. For instance, the face $[0,1]^{k-1} times {1} times [0,1]^{n-k}$ of some cube must be glued to the face $[0,1]^{k-1} times {0} times [0,1]^{n-k}$ of some other cube, for the same value of $k$; the torus $(mathbb R/mathbb Z)^n$ is cubulated with just one cube (coming from the standard map $[0,1] to mathbb R/mathbb Z$); the Klein bottle is not cubulated in my sense.
One can imagine an equivalence relation on cubulations of $M$ in which two cubulations are equivalent if they share a common refinement. But I haven't thought through the details of exactly what "refinement" should mean.
It's clear that every cubulated manifold is framed. (A framing of an $n$-manifold $M$ is a homotopy class of trivializations of the tangent bundle, i.e. a homotopy class of vector bundle isomorphisms $TM cong mathbb R^n times M$.) Locally, a framing determines a cubulation (or rather an equivalence class of cubulations). But globally I'm not so sure. I'm worried about things like the irrational line on the torus, which could prevent a framing from arising from any finite cubulation.
Question: Does every framed manifold admit a framing-compatible cubulation?
at.algebraic-topology gt.geometric-topology smooth-manifolds condensed-matter
asked 12 hours ago
Theo Johnson-FreydTheo Johnson-Freyd
29.6k879252
$endgroup$
Let's say an $n$-manifold is cubulated if it is glued out of cubes $[0,1]^n$ in a way that looks locally like the standard cubulation of $mathbb R^n$. For instance, the face $[0,1]^{k-1} times {1} times [0,1]^{n-k}$ of some cube must be glued to the face $[0,1]^{k-1} times {0} times [0,1]^{n-k}$ of some other cube, for the same value of $k$; the torus $(mathbb R/mathbb Z)^n$ is cubulated with just one cube (coming from the standard map $[0,1] to mathbb R/mathbb Z$); the Klein bottle is not cubulated in my sense.
One can imagine an equivalence relation on cubulations of $M$ in which two cubulations are equivalent if they share a common refinement. But I haven't thought through the details of exactly what "refinement" should mean.
It's clear that every cubulated manifold is framed. (A framing of an $n$-manifold $M$ is a homotopy class of trivializations of the tangent bundle, i.e. a homotopy class of vector bundle isomorphisms $TM cong mathbb R^n times M$.) Locally, a framing determines a cubulation (or rather an equivalence class of cubulations). But globally I'm not so sure. I'm worried about things like the irrational line on the torus, which could prevent a framing from arising from any finite cubulation.
Question: Does every framed manifold admit a framing-compatible cubulation?
at.algebraic-topology gt.geometric-topology smooth-manifolds condensed-matter
at.algebraic-topology gt.geometric-topology smooth-manifolds condensed-matter
asked 12 hours ago
Theo Johnson-FreydTheo Johnson-Freyd
29.6k879252
asked 12 hours ago
Theo Johnson-FreydTheo Johnson-Freyd
29.6k879252
asked 12 hours ago
Theo Johnson-FreydTheo Johnson-Freyd
29.6k879252
asked 12 hours ago
Theo Johnson-FreydTheo Johnson-Freyd
29.6k879252
asked 12 hours ago
Theo Johnson-FreydTheo Johnson-Freyd
29.6k879252
29.6k879252
$begingroup$
Does the 3-sphere admit a cubulation in your sense?
$endgroup$
– Chris Schommer-Pries
10 hours ago
$begingroup$
@ChrisSchommer-Pries A hypercube provides one, I believe.
$endgroup$
– მამუკა ჯიბლაძე
10 hours ago
3
$begingroup$
Your restrictions on the cubulation seem to imply that $M$ has a flat metric and that the action of $pi_1 M$ on the universal covering space $mathbb R^n$ is an action by translation, so $pi_1 M$ is isomorphic to $mathbb Z^n$.
$endgroup$
– Lee Mosher
8 hours ago
add a comment |
$begingroup$
Does the 3-sphere admit a cubulation in your sense?
$endgroup$
– Chris Schommer-Pries
10 hours ago
$begingroup$
@ChrisSchommer-Pries A hypercube provides one, I believe.
$endgroup$
– მამუკა ჯიბლაძე
10 hours ago
3
$begingroup$
Your restrictions on the cubulation seem to imply that $M$ has a flat metric and that the action of $pi_1 M$ on the universal covering space $mathbb R^n$ is an action by translation, so $pi_1 M$ is isomorphic to $mathbb Z^n$.
$endgroup$
– Lee Mosher
8 hours ago
$begingroup$
Does the 3-sphere admit a cubulation in your sense?
$endgroup$
– Chris Schommer-Pries
10 hours ago
$begingroup$
Does the 3-sphere admit a cubulation in your sense?
$endgroup$
– Chris Schommer-Pries
10 hours ago
$begingroup$
@ChrisSchommer-Pries A hypercube provides one, I believe.
$endgroup$
– მამუკა ჯიბლაძე
10 hours ago
$begingroup$
@ChrisSchommer-Pries A hypercube provides one, I believe.
$endgroup$
– მამუკა ჯიბლაძე
10 hours ago
3
3
$begingroup$
Your restrictions on the cubulation seem to imply that $M$ has a flat metric and that the action of $pi_1 M$ on the universal covering space $mathbb R^n$ is an action by translation, so $pi_1 M$ is isomorphic to $mathbb Z^n$.
$endgroup$
– Lee Mosher
8 hours ago
$begingroup$
Your restrictions on the cubulation seem to imply that $M$ has a flat metric and that the action of $pi_1 M$ on the universal covering space $mathbb R^n$ is an action by translation, so $pi_1 M$ is isomorphic to $mathbb Z^n$.
$endgroup$
– Lee Mosher
8 hours ago
add a comment |
1 Answer
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If you glue together the Riemannian metrics on the various cubes you obtain a flat metric on your cubulated manifold. So e.g. $S^3cong SU(2)$ is framed but not cubulated.
answered 10 hours ago
Dan PetersenDan Petersen
25.9k275141
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add a comment |
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$begingroup$
If you glue together the Riemannian metrics on the various cubes you obtain a flat metric on your cubulated manifold. So e.g. $S^3cong SU(2)$ is framed but not cubulated.
answered 10 hours ago
Dan PetersenDan Petersen
25.9k275141
$endgroup$
add a comment |
$begingroup$
If you glue together the Riemannian metrics on the various cubes you obtain a flat metric on your cubulated manifold. So e.g. $S^3cong SU(2)$ is framed but not cubulated.
answered 10 hours ago
Dan PetersenDan Petersen
25.9k275141
$endgroup$
add a comment |
$begingroup$
If you glue together the Riemannian metrics on the various cubes you obtain a flat metric on your cubulated manifold. So e.g. $S^3cong SU(2)$ is framed but not cubulated.
answered 10 hours ago
Dan PetersenDan Petersen
25.9k275141
$endgroup$
If you glue together the Riemannian metrics on the various cubes you obtain a flat metric on your cubulated manifold. So e.g. $S^3cong SU(2)$ is framed but not cubulated.
answered 10 hours ago
Dan PetersenDan Petersen
25.9k275141
answered 10 hours ago
Dan PetersenDan Petersen
25.9k275141
answered 10 hours ago
Dan PetersenDan Petersen
25.9k275141
answered 10 hours ago
Dan PetersenDan Petersen
25.9k275141
25.9k275141
add a comment |
add a comment |
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$begingroup$
Does the 3-sphere admit a cubulation in your sense?
$endgroup$
– Chris Schommer-Pries
10 hours ago
$begingroup$
@ChrisSchommer-Pries A hypercube provides one, I believe.
$endgroup$
– მამუკა ჯიბლაძე
10 hours ago
3
$begingroup$
Your restrictions on the cubulation seem to imply that $M$ has a flat metric and that the action of $pi_1 M$ on the universal covering space $mathbb R^n$ is an action by translation, so $pi_1 M$ is isomorphic to $mathbb Z^n$.
$endgroup$
– Lee Mosher
8 hours ago