Non-metrizable smooth manifold? [on hold]
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Is it possible to find a smooth manifold on which it is impossible to define a metric function?
general-topology manifolds smooth-manifolds
edited Nov 7 at 13:52
amWhy
191k27223437
asked Nov 7 at 13:29
alper akyuz
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put on hold as off-topic by José Carlos Santos, Delta-u, Key Flex, TheGeekGreek, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Delta-u, Key Flex, TheGeekGreek, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
0
down vote
favorite
Is it possible to find a smooth manifold on which it is impossible to define a metric function?
general-topology manifolds smooth-manifolds
edited Nov 7 at 13:52
amWhy
191k27223437
asked Nov 7 at 13:29
alper akyuz
361112
put on hold as off-topic by José Carlos Santos, Delta-u, Key Flex, TheGeekGreek, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Delta-u, Key Flex, TheGeekGreek, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
It depends. How do you define “smooth manifold”?
– José Carlos Santos
Nov 7 at 13:31
A smooth manifold can be given a metric using a partition of unity.
– deb
Nov 7 at 13:37
2
Do you want to talk about metric tensor or distance function?
– edm
Nov 7 at 13:38
1
For example the long line?
– freakish
Nov 7 at 14:24
1
By the Whitney embedding theorem, any smooth manifold can be thought of as being inside some Euclidean space $R^N$ (you can take $N$ to be twice the dimension of the manifold). You can now define your distance function on the manifold as the restriction of the usual distance function in Euclidean space.
– Aleksandar Milivojevic
Nov 9 at 0:01
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show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is it possible to find a smooth manifold on which it is impossible to define a metric function?
general-topology manifolds smooth-manifolds
edited Nov 7 at 13:52
amWhy
191k27223437
asked Nov 7 at 13:29
alper akyuz
361112
Is it possible to find a smooth manifold on which it is impossible to define a metric function?
general-topology manifolds smooth-manifolds
general-topology manifolds smooth-manifolds
edited Nov 7 at 13:52
amWhy
191k27223437
asked Nov 7 at 13:29
alper akyuz
361112
edited Nov 7 at 13:52
amWhy
191k27223437
asked Nov 7 at 13:29
alper akyuz
361112
edited Nov 7 at 13:52
amWhy
191k27223437
edited Nov 7 at 13:52
amWhy
191k27223437
edited Nov 7 at 13:52
amWhy
191k27223437
191k27223437
asked Nov 7 at 13:29
alper akyuz
361112
asked Nov 7 at 13:29
alper akyuz
361112
asked Nov 7 at 13:29
alper akyuz
361112
361112
put on hold as off-topic by José Carlos Santos, Delta-u, Key Flex, TheGeekGreek, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Delta-u, Key Flex, TheGeekGreek, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by José Carlos Santos, Delta-u, Key Flex, TheGeekGreek, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Delta-u, Key Flex, TheGeekGreek, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
It depends. How do you define “smooth manifold”?
– José Carlos Santos
Nov 7 at 13:31
A smooth manifold can be given a metric using a partition of unity.
– deb
Nov 7 at 13:37
2
Do you want to talk about metric tensor or distance function?
– edm
Nov 7 at 13:38
1
For example the long line?
– freakish
Nov 7 at 14:24
1
By the Whitney embedding theorem, any smooth manifold can be thought of as being inside some Euclidean space $R^N$ (you can take $N$ to be twice the dimension of the manifold). You can now define your distance function on the manifold as the restriction of the usual distance function in Euclidean space.
– Aleksandar Milivojevic
Nov 9 at 0:01
|
show 2 more comments
It depends. How do you define “smooth manifold”?
– José Carlos Santos
Nov 7 at 13:31
A smooth manifold can be given a metric using a partition of unity.
– deb
Nov 7 at 13:37
2
Do you want to talk about metric tensor or distance function?
– edm
Nov 7 at 13:38
1
For example the long line?
– freakish
Nov 7 at 14:24
1
By the Whitney embedding theorem, any smooth manifold can be thought of as being inside some Euclidean space $R^N$ (you can take $N$ to be twice the dimension of the manifold). You can now define your distance function on the manifold as the restriction of the usual distance function in Euclidean space.
– Aleksandar Milivojevic
Nov 9 at 0:01
It depends. How do you define “smooth manifold”?
– José Carlos Santos
Nov 7 at 13:31
It depends. How do you define “smooth manifold”?
– José Carlos Santos
Nov 7 at 13:31
A smooth manifold can be given a metric using a partition of unity.
– deb
Nov 7 at 13:37
A smooth manifold can be given a metric using a partition of unity.
– deb
Nov 7 at 13:37
2
2
Do you want to talk about metric tensor or distance function?
– edm
Nov 7 at 13:38
Do you want to talk about metric tensor or distance function?
– edm
Nov 7 at 13:38
1
1
For example the long line?
– freakish
Nov 7 at 14:24
For example the long line?
– freakish
Nov 7 at 14:24
1
1
By the Whitney embedding theorem, any smooth manifold can be thought of as being inside some Euclidean space $R^N$ (you can take $N$ to be twice the dimension of the manifold). You can now define your distance function on the manifold as the restriction of the usual distance function in Euclidean space.
– Aleksandar Milivojevic
Nov 9 at 0:01
By the Whitney embedding theorem, any smooth manifold can be thought of as being inside some Euclidean space $R^N$ (you can take $N$ to be twice the dimension of the manifold). You can now define your distance function on the manifold as the restriction of the usual distance function in Euclidean space.
– Aleksandar Milivojevic
Nov 9 at 0:01
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
It depends on the definition of a smooth manifold $M$. Usually one requires that $M$
1) is Hausdorff,
2) is second countable,
3) has a smooth atlas.
The "minimal" requirement for a smooth manifold would be 3), but obviously 1) is a necessary condition for the existence of a metric. See https://en.wikipedia.org/wiki/Non-Hausdorff_manifold for examples of non-Hausdorff manifolds.
That 2) is necessary for the for the existence of a metric is less obvious. As a counterexample take the long line https://en.wikipedia.org/wiki/Long_line_(topology).
If 1) - 3) are satisfied, then deb's and Aleksandar Milivojevic's comments show that $M$ is metrizable.
answered Nov 12 at 17:49
Paul Frost
7,3661527
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
It depends on the definition of a smooth manifold $M$. Usually one requires that $M$
1) is Hausdorff,
2) is second countable,
3) has a smooth atlas.
The "minimal" requirement for a smooth manifold would be 3), but obviously 1) is a necessary condition for the existence of a metric. See https://en.wikipedia.org/wiki/Non-Hausdorff_manifold for examples of non-Hausdorff manifolds.
That 2) is necessary for the for the existence of a metric is less obvious. As a counterexample take the long line https://en.wikipedia.org/wiki/Long_line_(topology).
If 1) - 3) are satisfied, then deb's and Aleksandar Milivojevic's comments show that $M$ is metrizable.
answered Nov 12 at 17:49
Paul Frost
7,3661527
add a comment |
up vote
4
down vote
accepted
It depends on the definition of a smooth manifold $M$. Usually one requires that $M$
1) is Hausdorff,
2) is second countable,
3) has a smooth atlas.
The "minimal" requirement for a smooth manifold would be 3), but obviously 1) is a necessary condition for the existence of a metric. See https://en.wikipedia.org/wiki/Non-Hausdorff_manifold for examples of non-Hausdorff manifolds.
That 2) is necessary for the for the existence of a metric is less obvious. As a counterexample take the long line https://en.wikipedia.org/wiki/Long_line_(topology).
If 1) - 3) are satisfied, then deb's and Aleksandar Milivojevic's comments show that $M$ is metrizable.
answered Nov 12 at 17:49
Paul Frost
7,3661527
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
It depends on the definition of a smooth manifold $M$. Usually one requires that $M$
1) is Hausdorff,
2) is second countable,
3) has a smooth atlas.
The "minimal" requirement for a smooth manifold would be 3), but obviously 1) is a necessary condition for the existence of a metric. See https://en.wikipedia.org/wiki/Non-Hausdorff_manifold for examples of non-Hausdorff manifolds.
That 2) is necessary for the for the existence of a metric is less obvious. As a counterexample take the long line https://en.wikipedia.org/wiki/Long_line_(topology).
If 1) - 3) are satisfied, then deb's and Aleksandar Milivojevic's comments show that $M$ is metrizable.
answered Nov 12 at 17:49
Paul Frost
7,3661527
It depends on the definition of a smooth manifold $M$. Usually one requires that $M$
1) is Hausdorff,
2) is second countable,
3) has a smooth atlas.
The "minimal" requirement for a smooth manifold would be 3), but obviously 1) is a necessary condition for the existence of a metric. See https://en.wikipedia.org/wiki/Non-Hausdorff_manifold for examples of non-Hausdorff manifolds.
That 2) is necessary for the for the existence of a metric is less obvious. As a counterexample take the long line https://en.wikipedia.org/wiki/Long_line_(topology).
If 1) - 3) are satisfied, then deb's and Aleksandar Milivojevic's comments show that $M$ is metrizable.
answered Nov 12 at 17:49
Paul Frost
7,3661527
answered Nov 12 at 17:49
Paul Frost
7,3661527
answered Nov 12 at 17:49
Paul Frost
7,3661527
answered Nov 12 at 17:49
Paul Frost
7,3661527
7,3661527
add a comment |
add a comment |
It depends. How do you define “smooth manifold”?
– José Carlos Santos
Nov 7 at 13:31
A smooth manifold can be given a metric using a partition of unity.
– deb
Nov 7 at 13:37
2
Do you want to talk about metric tensor or distance function?
– edm
Nov 7 at 13:38
1
For example the long line?
– freakish
Nov 7 at 14:24
1
By the Whitney embedding theorem, any smooth manifold can be thought of as being inside some Euclidean space $R^N$ (you can take $N$ to be twice the dimension of the manifold). You can now define your distance function on the manifold as the restriction of the usual distance function in Euclidean space.
– Aleksandar Milivojevic
Nov 9 at 0:01