Uniqueness of the dimension of a manifold The 2019 Stack Overflow Developer Survey Results Are...
What is this business jet?
Dropping list elements from nested list after evaluation
Are spiders unable to hurt humans, especially very small spiders?
What do these terms in Caesar's Gallic Wars mean?
How do you keep chess fun when your opponent constantly beats you?
writing variables above the numbers in tikz picture
How to charge AirPods to keep battery healthy?
Why does the nucleus not repel itself?
If climate change impact can be observed in nature, has that had any effect on rural, i.e. farming community, perception of the scientific consensus?
Did Scotland spend $250,000 for the slogan "Welcome to Scotland"?
What force causes entropy to increase?
Can withdrawing asylum be illegal?
Are there any other methods to apply to solving simultaneous equations?
Falsification in Math vs Science
Does HR tell a hiring manager about salary negotiations?
Why doesn't UInt have a toDouble()?
If I score a critical hit on an 18 or higher, what are my chances of getting a critical hit if I roll 3d20?
What is the most efficient way to store a numeric range?
How come people say “Would of”?
How to quickly solve partial fractions equation?
Mathematics of imaging the black hole
What do hard-Brexiteers want with respect to the Irish border?
Kerning for subscripts of sigma?
What do I do when my TA workload is more than expected?
Uniqueness of the dimension of a manifold
The 2019 Stack Overflow Developer Survey Results Are InDimension of submanifold is lees or equal to the dimension of the manifold.Any neighbourhood of a point $x$ in a manifold $X$ ($dim X geq 2$) has a subneighbourhood $V$ of $x$ such that $V setminus {x}$ is connectedExtending a smooth vector field on a manifoldTangent space of manifold has two unit vectors orthogonal to tangent space of its boundaryIs any orientable smooth manifold of dimension $3$ with two independent vector fields parallelizable?Proving a subset is not a submanifoldwhy there is smooth curve in any direction on manifold?On the definition of derivative between smooth manifolds in euclidean spaceWhen we construct a regular sub manifold, can we set any coordinate of the coordinate system equal to 0?Show the disjoint union of 2 n-manifolds is an n-manifoldI'm having trouble understanding manifolds.
$begingroup$
Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m ne k$. Meaning, there is no $x in M$ and $x in U_x$ a neighborhood of $x$, such that $M bigcap U_x$ has a good parametrization from $V_x subset R^m$ (where $m ne k$).
I am not really sure how to prove it. I thought that I could use the fact that for every $x in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.
However, I got stuck.
Any help would be appreciated.
calculus multivariable-calculus manifolds
$endgroup$
add a comment |
$begingroup$
Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m ne k$. Meaning, there is no $x in M$ and $x in U_x$ a neighborhood of $x$, such that $M bigcap U_x$ has a good parametrization from $V_x subset R^m$ (where $m ne k$).
I am not really sure how to prove it. I thought that I could use the fact that for every $x in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.
However, I got stuck.
Any help would be appreciated.
calculus multivariable-calculus manifolds
$endgroup$
$begingroup$
If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
$endgroup$
– reuns
Mar 22 at 1:56
1
$begingroup$
The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
$endgroup$
– o.h.
Mar 22 at 7:11
add a comment |
$begingroup$
Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m ne k$. Meaning, there is no $x in M$ and $x in U_x$ a neighborhood of $x$, such that $M bigcap U_x$ has a good parametrization from $V_x subset R^m$ (where $m ne k$).
I am not really sure how to prove it. I thought that I could use the fact that for every $x in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.
However, I got stuck.
Any help would be appreciated.
calculus multivariable-calculus manifolds
$endgroup$
Let $M$ be a $k$-dimensional manifold. I want to prove that $M$ can't be also of dimension $m$ where $m ne k$. Meaning, there is no $x in M$ and $x in U_x$ a neighborhood of $x$, such that $M bigcap U_x$ has a good parametrization from $V_x subset R^m$ (where $m ne k$).
I am not really sure how to prove it. I thought that I could use the fact that for every $x in M$ there is a neighborhood $W_x$ where $M$ is a graph of a smooth function. Then I'll get that in the same neighborhood, $M$ is graph of two functions, each of different number of variables, which will lead to a contradiction.
However, I got stuck.
Any help would be appreciated.
calculus multivariable-calculus manifolds
calculus multivariable-calculus manifolds
asked Mar 21 at 23:48
Gabi GGabi G
553210
553210
$begingroup$
If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
$endgroup$
– reuns
Mar 22 at 1:56
1
$begingroup$
The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
$endgroup$
– o.h.
Mar 22 at 7:11
add a comment |
$begingroup$
If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
$endgroup$
– reuns
Mar 22 at 1:56
1
$begingroup$
The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
$endgroup$
– o.h.
Mar 22 at 7:11
$begingroup$
If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
$endgroup$
– reuns
Mar 22 at 1:56
$begingroup$
If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
$endgroup$
– reuns
Mar 22 at 1:56
1
1
$begingroup$
The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
$endgroup$
– o.h.
Mar 22 at 7:11
$begingroup$
The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
$endgroup$
– o.h.
Mar 22 at 7:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.
$endgroup$
add a comment |
$begingroup$
For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
$$
H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
$$
where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
$$
H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
$$
This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)
Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.
$endgroup$
1
$begingroup$
Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
$endgroup$
– Gabi G
Mar 22 at 9:21
$begingroup$
@Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
$endgroup$
– o.h.
Mar 22 at 15:01
add a comment |
$begingroup$
I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?
It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.
It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.
$endgroup$
$begingroup$
Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
$endgroup$
– Gabi G
Mar 22 at 9:19
$begingroup$
@GabiG a topological manifold does not need to have a differentiable structure.
$endgroup$
– Zest
Mar 23 at 1:49
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%2f3157540%2funiqueness-of-the-dimension-of-a-manifold%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.
$endgroup$
add a comment |
$begingroup$
If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.
$endgroup$
add a comment |
$begingroup$
If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.
$endgroup$
If $M$ is a $k$-dimensional smooth manifold then each tangent space $T_pM$ is isomorphic to $mathbb R^k$. So if $M$ is also $m$-dimensional then $mathbb R^msimeqmathbb R^k$ (as vector spaces) which implies $m=k$.
edited Mar 24 at 3:16
answered Mar 22 at 3:37
triitrii
85317
85317
add a comment |
add a comment |
$begingroup$
For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
$$
H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
$$
where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
$$
H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
$$
This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)
Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.
$endgroup$
1
$begingroup$
Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
$endgroup$
– Gabi G
Mar 22 at 9:21
$begingroup$
@Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
$endgroup$
– o.h.
Mar 22 at 15:01
add a comment |
$begingroup$
For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
$$
H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
$$
where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
$$
H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
$$
This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)
Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.
$endgroup$
1
$begingroup$
Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
$endgroup$
– Gabi G
Mar 22 at 9:21
$begingroup$
@Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
$endgroup$
– o.h.
Mar 22 at 15:01
add a comment |
$begingroup$
For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
$$
H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
$$
where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
$$
H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
$$
This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)
Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.
$endgroup$
For topological manifolds, algebraic topology is your best bet (at least if you want a simple proof). For instance, if $X$ is an $n$-manifold and $xin X$ has Euclidean nbhd $Uni x$ with homeomorphism $fcolon Urightarrow mathbb R^n$, then
$$
H_i(X,Xsetminus x) xleftarrowsim H_i (U,Usetminus x) xrightarrowsim H_i(mathbb R^n,mathbb R^nsetminus f(x))
$$
where the map on the left is an excision map (i.e. inclusion) and the map on the right is the induced map $f_*$. We know that $H_i (mathbb R^n,mathbb R^nsetminus f(x))cong tilde H_{i-1}(S^{n-1})$ ($i>0$) from LES of pair $mathbb R^nsetminus f(x)subsetmathbb R^n$ and htpy equivalence $mathbb R^nsetminus f(x) simeq S^{n-1}$. In conclusion, for each $xin X$ and $i > 0$ we find that
$$
H_i (X,Xsetminus x)congtilde H_{i-1} (S^{n-1})congbegin{cases}mathbb Zquad text{if}, i=n,\ 0quadtext{else.}end{cases}
$$
This shows that the dimension of a manifold is unique when the manifold is nonempty, since then dimension is determined by local homology at $xin X$. (Uniqueness of dimension is obviously false for the manifold $X = emptyset$. Note that this is a manifold.)
Addendum. I think it may be difficult to prove uniqueness of dimension using just the tools of point-set topology. At least, I do not imagine there is an easy proof like the one given above. For instance, it is already difficult to prove that $mathbb R^mapprox mathbb R^nRightarrow m = n$ without algebraic topology.
answered Mar 22 at 7:03
o.h.o.h.
6917
6917
1
$begingroup$
Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
$endgroup$
– Gabi G
Mar 22 at 9:21
$begingroup$
@Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
$endgroup$
– o.h.
Mar 22 at 15:01
add a comment |
1
$begingroup$
Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
$endgroup$
– Gabi G
Mar 22 at 9:21
$begingroup$
@Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
$endgroup$
– o.h.
Mar 22 at 15:01
1
1
$begingroup$
Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
$endgroup$
– Gabi G
Mar 22 at 9:21
$begingroup$
Nice answer. However, I haven't studied topology yet, so I am looking for another direction. More in the sense of what I tried to do
$endgroup$
– Gabi G
Mar 22 at 9:21
$begingroup$
@Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
$endgroup$
– o.h.
Mar 22 at 15:01
$begingroup$
@Gabi G Then I recommend the answer by trii. I think jmerry was hinting at something like that answer when he mentioned "linear algebra" tools for the smooth case in his answer
$endgroup$
– o.h.
Mar 22 at 15:01
add a comment |
$begingroup$
I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?
It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.
It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.
$endgroup$
$begingroup$
Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
$endgroup$
– Gabi G
Mar 22 at 9:19
$begingroup$
@GabiG a topological manifold does not need to have a differentiable structure.
$endgroup$
– Zest
Mar 23 at 1:49
add a comment |
$begingroup$
I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?
It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.
It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.
$endgroup$
$begingroup$
Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
$endgroup$
– Gabi G
Mar 22 at 9:19
$begingroup$
@GabiG a topological manifold does not need to have a differentiable structure.
$endgroup$
– Zest
Mar 23 at 1:49
add a comment |
$begingroup$
I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?
It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.
It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.
$endgroup$
I recall several early exercises in the topological manifolds class, showing piecemeal that various small-dimensional cases weren't homeomorphic. Showing that 1-manifolds aren't higher-dimensional manifolds can be done with a connectedness argument, and showing 2-dimensional manifolds aren't higher-dimensional manifolds uses the fundamental group. Digging out my copy of Lee (Intro. to Topological Manifolds), where's the unified theorem that an $n$-manifold is a $m$-manifold only if $m=n$?
It's in Chapter 13, the last chapter of the book. We need to go all the way to homology groups to prove it.
It's much easier to prove for smooth manifolds, with the linear algebra tools that allows - but the way I learned it, manifolds have a purely topological definition. Smooth manifolds are something we define later, by taking a manifold with additional structure on it.
answered Mar 22 at 4:34
jmerryjmerry
17k11633
17k11633
$begingroup$
Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
$endgroup$
– Gabi G
Mar 22 at 9:19
$begingroup$
@GabiG a topological manifold does not need to have a differentiable structure.
$endgroup$
– Zest
Mar 23 at 1:49
add a comment |
$begingroup$
Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
$endgroup$
– Gabi G
Mar 22 at 9:19
$begingroup$
@GabiG a topological manifold does not need to have a differentiable structure.
$endgroup$
– Zest
Mar 23 at 1:49
$begingroup$
Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
$endgroup$
– Gabi G
Mar 22 at 9:19
$begingroup$
Here, manifolds and smooth manifolds are the same for me, as I haven't learned about other kinds of manifolds
$endgroup$
– Gabi G
Mar 22 at 9:19
$begingroup$
@GabiG a topological manifold does not need to have a differentiable structure.
$endgroup$
– Zest
Mar 23 at 1:49
$begingroup$
@GabiG a topological manifold does not need to have a differentiable structure.
$endgroup$
– Zest
Mar 23 at 1:49
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%2f3157540%2funiqueness-of-the-dimension-of-a-manifold%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
$begingroup$
If a manifold is a topological space where for every $x in M$ there is an open set $U_x ni x$ and a chart making it homeomorphic to an open set of $Bbb{R}^{n_x}$ then won't $n_x$ be constant on each connected component of $M$ ?
$endgroup$
– reuns
Mar 22 at 1:56
1
$begingroup$
The statement is false, since $M = emptyset$ is a manifold of dimension $n$ for all $n$. The statement is true if we require $Mneqemptyset$. This also indicates that we need to involve the points of $M$. I have given an answer below which does this, but using tools from algebraic topology.
$endgroup$
– o.h.
Mar 22 at 7:11