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Can there be a manifold with uncountable genus? Can they be compact?
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Just some random thoughts I had after coming across this paper on Riemann Surfaces of Infinite Genus: [http://www.math.ubc.ca/~feldman/papers/allriem.pdf]
I'm reasonable convinced that $mathbbR^n$ cannot have uncountably many regions (of non-zero measure) removed, but I don't quite see that this ruins the potential for manifolds which can be sometimes a lot more whacky than $mathbbR^n$.
Do the Homotopy Groups or Holonomy Groups or the Cohomology Groups still make any sense in this context? I think they'd have infinitely many generators, which would get strange.
differential-geometry algebraic-topology manifolds infinity
asked 2 days ago
DPatt
112
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up vote
1
down vote
favorite
Just some random thoughts I had after coming across this paper on Riemann Surfaces of Infinite Genus: [http://www.math.ubc.ca/~feldman/papers/allriem.pdf]
I'm reasonable convinced that $mathbbR^n$ cannot have uncountably many regions (of non-zero measure) removed, but I don't quite see that this ruins the potential for manifolds which can be sometimes a lot more whacky than $mathbbR^n$.
Do the Homotopy Groups or Holonomy Groups or the Cohomology Groups still make any sense in this context? I think they'd have infinitely many generators, which would get strange.
differential-geometry algebraic-topology manifolds infinity
asked 2 days ago
DPatt
112
1
A compact manifold has finitely generated (co)homology groups, so a well-defined, finite Euler characteristic.
– Lord Shark the Unknown
2 days ago
The pdf link isn't working for some reason.
– CaptainAmerica16
2 days ago
@CaptainAmerica16 Try going here: math.ubc.ca/~feldman/papers and looking for the link: allriem.pdf
– DPatt
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Just some random thoughts I had after coming across this paper on Riemann Surfaces of Infinite Genus: [http://www.math.ubc.ca/~feldman/papers/allriem.pdf]
I'm reasonable convinced that $mathbbR^n$ cannot have uncountably many regions (of non-zero measure) removed, but I don't quite see that this ruins the potential for manifolds which can be sometimes a lot more whacky than $mathbbR^n$.
Do the Homotopy Groups or Holonomy Groups or the Cohomology Groups still make any sense in this context? I think they'd have infinitely many generators, which would get strange.
differential-geometry algebraic-topology manifolds infinity
asked 2 days ago
DPatt
112
Just some random thoughts I had after coming across this paper on Riemann Surfaces of Infinite Genus: [http://www.math.ubc.ca/~feldman/papers/allriem.pdf]
I'm reasonable convinced that $mathbbR^n$ cannot have uncountably many regions (of non-zero measure) removed, but I don't quite see that this ruins the potential for manifolds which can be sometimes a lot more whacky than $mathbbR^n$.
Do the Homotopy Groups or Holonomy Groups or the Cohomology Groups still make any sense in this context? I think they'd have infinitely many generators, which would get strange.
differential-geometry algebraic-topology manifolds infinity
asked 2 days ago
DPatt
112
asked 2 days ago
DPatt
112
asked 2 days ago
DPatt
112
asked 2 days ago
DPatt
112
112
1
A compact manifold has finitely generated (co)homology groups, so a well-defined, finite Euler characteristic.
– Lord Shark the Unknown
2 days ago
The pdf link isn't working for some reason.
– CaptainAmerica16
2 days ago
@CaptainAmerica16 Try going here: math.ubc.ca/~feldman/papers and looking for the link: allriem.pdf
– DPatt
2 days ago
add a comment |Â
1
A compact manifold has finitely generated (co)homology groups, so a well-defined, finite Euler characteristic.
– Lord Shark the Unknown
2 days ago
The pdf link isn't working for some reason.
– CaptainAmerica16
2 days ago
@CaptainAmerica16 Try going here: math.ubc.ca/~feldman/papers and looking for the link: allriem.pdf
– DPatt
2 days ago
1
1
A compact manifold has finitely generated (co)homology groups, so a well-defined, finite Euler characteristic.
– Lord Shark the Unknown
2 days ago
A compact manifold has finitely generated (co)homology groups, so a well-defined, finite Euler characteristic.
– Lord Shark the Unknown
2 days ago
The pdf link isn't working for some reason.
– CaptainAmerica16
2 days ago
The pdf link isn't working for some reason.
– CaptainAmerica16
2 days ago
@CaptainAmerica16 Try going here: math.ubc.ca/~feldman/papers and looking for the link: allriem.pdf
– DPatt
2 days ago
@CaptainAmerica16 Try going here: math.ubc.ca/~feldman/papers and looking for the link: allriem.pdf
– DPatt
2 days ago
add a comment |Â
1 Answer
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There's nothing particularly strange about homotopy groups or homology groups having a countable infinity of generators (unless the manifold is compact, as said in the comment of @LordSharktheUnknown).
For example the ladder surface is the boundary of a regular neighborhood in $mathbb R^3 = mathbb R^2 times mathbb R$ of
$$bigl((mathbb R times 0,1) cup (mathbb Z times [0,1])bigr) times 0
$$
The ladder surface has countably generated $pi_1$ and $H_1$, but neither is finitely generated. Perhaps an even simpler example is $mathbb R^2$ minus the radius $1/3$ balls centered at the points of $mathbb Z times 0$.
Here's a general fact along these lines. Connected smooth manifolds are usually required by definition to be paracompact (this is unnecessary for Riemann surfaces, which are proved to be paracompact by Rado's Theorem). One can then use a partition of unity argument to prove that there is a "good open covering" by open sets homeomorphic to balls which is locally finite and such that any finite subset of the covering intersects in either the empty set or a subset homeomorphic to a ball. By applying combination theorems (e.g. Van Kampen's theorem for $pi_1$ and the Mayer Vietoris theorem for $H_n$, perhaps coupled with direct limit arguments) one can then conclude that $pi_1$ and each $H_n$ have countable generating sets (any of them could, nonetheless, still be finitely generated). By applying somewhat deeper combination theorems one can also prove that the higher homotopy groups $pi_n$ have countable generating sets.
answered 10 hours ago
Lee Mosher
45.2k33478
It's nice to know that for a surface, being 'tame' (homeomorphic to the interior of a compact manifold with boundary) is equivalent to $pi_1$ being finitely generated is equivalent to $H_1$ being finitely generated. But it is a nice exercise to try to think about $pi_1$ of $S^2$ minus a Cantor set.
– Mike Miller
9 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
There's nothing particularly strange about homotopy groups or homology groups having a countable infinity of generators (unless the manifold is compact, as said in the comment of @LordSharktheUnknown).
For example the ladder surface is the boundary of a regular neighborhood in $mathbb R^3 = mathbb R^2 times mathbb R$ of
$$bigl((mathbb R times 0,1) cup (mathbb Z times [0,1])bigr) times 0
$$
The ladder surface has countably generated $pi_1$ and $H_1$, but neither is finitely generated. Perhaps an even simpler example is $mathbb R^2$ minus the radius $1/3$ balls centered at the points of $mathbb Z times 0$.
Here's a general fact along these lines. Connected smooth manifolds are usually required by definition to be paracompact (this is unnecessary for Riemann surfaces, which are proved to be paracompact by Rado's Theorem). One can then use a partition of unity argument to prove that there is a "good open covering" by open sets homeomorphic to balls which is locally finite and such that any finite subset of the covering intersects in either the empty set or a subset homeomorphic to a ball. By applying combination theorems (e.g. Van Kampen's theorem for $pi_1$ and the Mayer Vietoris theorem for $H_n$, perhaps coupled with direct limit arguments) one can then conclude that $pi_1$ and each $H_n$ have countable generating sets (any of them could, nonetheless, still be finitely generated). By applying somewhat deeper combination theorems one can also prove that the higher homotopy groups $pi_n$ have countable generating sets.
answered 10 hours ago
Lee Mosher
45.2k33478
It's nice to know that for a surface, being 'tame' (homeomorphic to the interior of a compact manifold with boundary) is equivalent to $pi_1$ being finitely generated is equivalent to $H_1$ being finitely generated. But it is a nice exercise to try to think about $pi_1$ of $S^2$ minus a Cantor set.
– Mike Miller
9 hours ago
add a comment |Â
up vote
0
down vote
There's nothing particularly strange about homotopy groups or homology groups having a countable infinity of generators (unless the manifold is compact, as said in the comment of @LordSharktheUnknown).
For example the ladder surface is the boundary of a regular neighborhood in $mathbb R^3 = mathbb R^2 times mathbb R$ of
$$bigl((mathbb R times 0,1) cup (mathbb Z times [0,1])bigr) times 0
$$
The ladder surface has countably generated $pi_1$ and $H_1$, but neither is finitely generated. Perhaps an even simpler example is $mathbb R^2$ minus the radius $1/3$ balls centered at the points of $mathbb Z times 0$.
Here's a general fact along these lines. Connected smooth manifolds are usually required by definition to be paracompact (this is unnecessary for Riemann surfaces, which are proved to be paracompact by Rado's Theorem). One can then use a partition of unity argument to prove that there is a "good open covering" by open sets homeomorphic to balls which is locally finite and such that any finite subset of the covering intersects in either the empty set or a subset homeomorphic to a ball. By applying combination theorems (e.g. Van Kampen's theorem for $pi_1$ and the Mayer Vietoris theorem for $H_n$, perhaps coupled with direct limit arguments) one can then conclude that $pi_1$ and each $H_n$ have countable generating sets (any of them could, nonetheless, still be finitely generated). By applying somewhat deeper combination theorems one can also prove that the higher homotopy groups $pi_n$ have countable generating sets.
answered 10 hours ago
Lee Mosher
45.2k33478
It's nice to know that for a surface, being 'tame' (homeomorphic to the interior of a compact manifold with boundary) is equivalent to $pi_1$ being finitely generated is equivalent to $H_1$ being finitely generated. But it is a nice exercise to try to think about $pi_1$ of $S^2$ minus a Cantor set.
– Mike Miller
9 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There's nothing particularly strange about homotopy groups or homology groups having a countable infinity of generators (unless the manifold is compact, as said in the comment of @LordSharktheUnknown).
For example the ladder surface is the boundary of a regular neighborhood in $mathbb R^3 = mathbb R^2 times mathbb R$ of
$$bigl((mathbb R times 0,1) cup (mathbb Z times [0,1])bigr) times 0
$$
The ladder surface has countably generated $pi_1$ and $H_1$, but neither is finitely generated. Perhaps an even simpler example is $mathbb R^2$ minus the radius $1/3$ balls centered at the points of $mathbb Z times 0$.
Here's a general fact along these lines. Connected smooth manifolds are usually required by definition to be paracompact (this is unnecessary for Riemann surfaces, which are proved to be paracompact by Rado's Theorem). One can then use a partition of unity argument to prove that there is a "good open covering" by open sets homeomorphic to balls which is locally finite and such that any finite subset of the covering intersects in either the empty set or a subset homeomorphic to a ball. By applying combination theorems (e.g. Van Kampen's theorem for $pi_1$ and the Mayer Vietoris theorem for $H_n$, perhaps coupled with direct limit arguments) one can then conclude that $pi_1$ and each $H_n$ have countable generating sets (any of them could, nonetheless, still be finitely generated). By applying somewhat deeper combination theorems one can also prove that the higher homotopy groups $pi_n$ have countable generating sets.
answered 10 hours ago
Lee Mosher
45.2k33478
There's nothing particularly strange about homotopy groups or homology groups having a countable infinity of generators (unless the manifold is compact, as said in the comment of @LordSharktheUnknown).
For example the ladder surface is the boundary of a regular neighborhood in $mathbb R^3 = mathbb R^2 times mathbb R$ of
$$bigl((mathbb R times 0,1) cup (mathbb Z times [0,1])bigr) times 0
$$
The ladder surface has countably generated $pi_1$ and $H_1$, but neither is finitely generated. Perhaps an even simpler example is $mathbb R^2$ minus the radius $1/3$ balls centered at the points of $mathbb Z times 0$.
Here's a general fact along these lines. Connected smooth manifolds are usually required by definition to be paracompact (this is unnecessary for Riemann surfaces, which are proved to be paracompact by Rado's Theorem). One can then use a partition of unity argument to prove that there is a "good open covering" by open sets homeomorphic to balls which is locally finite and such that any finite subset of the covering intersects in either the empty set or a subset homeomorphic to a ball. By applying combination theorems (e.g. Van Kampen's theorem for $pi_1$ and the Mayer Vietoris theorem for $H_n$, perhaps coupled with direct limit arguments) one can then conclude that $pi_1$ and each $H_n$ have countable generating sets (any of them could, nonetheless, still be finitely generated). By applying somewhat deeper combination theorems one can also prove that the higher homotopy groups $pi_n$ have countable generating sets.
answered 10 hours ago
Lee Mosher
45.2k33478
answered 10 hours ago
Lee Mosher
45.2k33478
answered 10 hours ago
Lee Mosher
45.2k33478
answered 10 hours ago
Lee Mosher
45.2k33478
45.2k33478
It's nice to know that for a surface, being 'tame' (homeomorphic to the interior of a compact manifold with boundary) is equivalent to $pi_1$ being finitely generated is equivalent to $H_1$ being finitely generated. But it is a nice exercise to try to think about $pi_1$ of $S^2$ minus a Cantor set.
– Mike Miller
9 hours ago
add a comment |Â
It's nice to know that for a surface, being 'tame' (homeomorphic to the interior of a compact manifold with boundary) is equivalent to $pi_1$ being finitely generated is equivalent to $H_1$ being finitely generated. But it is a nice exercise to try to think about $pi_1$ of $S^2$ minus a Cantor set.
– Mike Miller
9 hours ago
It's nice to know that for a surface, being 'tame' (homeomorphic to the interior of a compact manifold with boundary) is equivalent to $pi_1$ being finitely generated is equivalent to $H_1$ being finitely generated. But it is a nice exercise to try to think about $pi_1$ of $S^2$ minus a Cantor set.
– Mike Miller
9 hours ago
It's nice to know that for a surface, being 'tame' (homeomorphic to the interior of a compact manifold with boundary) is equivalent to $pi_1$ being finitely generated is equivalent to $H_1$ being finitely generated. But it is a nice exercise to try to think about $pi_1$ of $S^2$ minus a Cantor set.
– Mike Miller
9 hours ago
add a comment |Â
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1
A compact manifold has finitely generated (co)homology groups, so a well-defined, finite Euler characteristic.
– Lord Shark the Unknown
2 days ago
The pdf link isn't working for some reason.
– CaptainAmerica16
2 days ago
@CaptainAmerica16 Try going here: math.ubc.ca/~feldman/papers and looking for the link: allriem.pdf
– DPatt
2 days ago