Low dimensional structures in a high dimensional space
$begingroup$
Let $n ge 3$ and $x(t) in mathbb{R}^n$ be a vector-valued function of $t in I = [0,M]$.
Suppose $x_1(t) = [t,cdots,t]^T = textbf{1}t$ which is a line, a 1-dimensional manifold.
Then $$x_1(I) = {x(t)|t in I}$$ can be viewed as a subset in 1-dimension $mathbb{R}$.
Similarly, if
$$x_2(t) = [1,1,1,0,cdots,0]^Tsqrt{1-(t/M)^2} + [0,0,0,1,cdots,1]^T(t/M),$$ $x_2(I)$ can be viewed as a half circle, as a subset in 2-dimenision $mathbb{R}^2$.
Now let consider
$$
x_3(t) = [cos(t-theta_1), cos(t-theta_2), cdots, cos(t-theta_{n})]^T.
$$
In this case, it is unclear that what dimensional object $x_3(I)$ is.
I am wondering if there is a mathematical notion or terminology which describing the dimensionality of $x(I)$ in general (e.g. $x_3(I)$).
But I don't want to (even continuously) deform $x(I)$ to identify the dimensionality. If I allow any deformation on $x(I)$, I believe $x_2(I)$ and $x_3(I)$ become a 1-dimensional "manifold" as they only depend on a single variable $t$.
I can see that $x_1(I)$ is isomorphic to $I$ (as a set) and
$x_2(I)$ is isomorphic to the half circle ${(x,y) | x^2 + y^2 = 1, 0 le x, y}$ (as a set). It seems that these might be related to some isomorphic properties, however, I am not sure.
Any comments/suggestions/answers will be very appreciated.
reference-request manifolds low-dimensional-topology
asked Dec 19 '18 at 23:03
induction601induction601
1,307314
$endgroup$
add a comment |
$begingroup$
Let $n ge 3$ and $x(t) in mathbb{R}^n$ be a vector-valued function of $t in I = [0,M]$.
Suppose $x_1(t) = [t,cdots,t]^T = textbf{1}t$ which is a line, a 1-dimensional manifold.
Then $$x_1(I) = {x(t)|t in I}$$ can be viewed as a subset in 1-dimension $mathbb{R}$.
Similarly, if
$$x_2(t) = [1,1,1,0,cdots,0]^Tsqrt{1-(t/M)^2} + [0,0,0,1,cdots,1]^T(t/M),$$ $x_2(I)$ can be viewed as a half circle, as a subset in 2-dimenision $mathbb{R}^2$.
Now let consider
$$
x_3(t) = [cos(t-theta_1), cos(t-theta_2), cdots, cos(t-theta_{n})]^T.
$$
In this case, it is unclear that what dimensional object $x_3(I)$ is.
I am wondering if there is a mathematical notion or terminology which describing the dimensionality of $x(I)$ in general (e.g. $x_3(I)$).
But I don't want to (even continuously) deform $x(I)$ to identify the dimensionality. If I allow any deformation on $x(I)$, I believe $x_2(I)$ and $x_3(I)$ become a 1-dimensional "manifold" as they only depend on a single variable $t$.
I can see that $x_1(I)$ is isomorphic to $I$ (as a set) and
$x_2(I)$ is isomorphic to the half circle ${(x,y) | x^2 + y^2 = 1, 0 le x, y}$ (as a set). It seems that these might be related to some isomorphic properties, however, I am not sure.
Any comments/suggestions/answers will be very appreciated.
reference-request manifolds low-dimensional-topology
asked Dec 19 '18 at 23:03
induction601induction601
1,307314
$endgroup$
$begingroup$
$x_2$ is still a 1-dimensional object.
$endgroup$
– Dog_69
Dec 19 '18 at 23:09
$begingroup$
Sorry I didn't explain this clearly. What I am trying to say is that even the half circle is a line with the curvature, the only way we can realize whether this line is a half circle or not is to put this object into the 2-dimensional space. In this sense, I would like to see this as a 2-dimensional object.
$endgroup$
– induction601
Dec 19 '18 at 23:11
$begingroup$
@Dog_69 For example, $x_2(I)$ is isomorphic to HC (HC = Half Circle) where HC is a subset in "2D", not in 1D.
$endgroup$
– induction601
Dec 19 '18 at 23:13
add a comment |
$begingroup$
Let $n ge 3$ and $x(t) in mathbb{R}^n$ be a vector-valued function of $t in I = [0,M]$.
Suppose $x_1(t) = [t,cdots,t]^T = textbf{1}t$ which is a line, a 1-dimensional manifold.
Then $$x_1(I) = {x(t)|t in I}$$ can be viewed as a subset in 1-dimension $mathbb{R}$.
Similarly, if
$$x_2(t) = [1,1,1,0,cdots,0]^Tsqrt{1-(t/M)^2} + [0,0,0,1,cdots,1]^T(t/M),$$ $x_2(I)$ can be viewed as a half circle, as a subset in 2-dimenision $mathbb{R}^2$.
Now let consider
$$
x_3(t) = [cos(t-theta_1), cos(t-theta_2), cdots, cos(t-theta_{n})]^T.
$$
In this case, it is unclear that what dimensional object $x_3(I)$ is.
I am wondering if there is a mathematical notion or terminology which describing the dimensionality of $x(I)$ in general (e.g. $x_3(I)$).
But I don't want to (even continuously) deform $x(I)$ to identify the dimensionality. If I allow any deformation on $x(I)$, I believe $x_2(I)$ and $x_3(I)$ become a 1-dimensional "manifold" as they only depend on a single variable $t$.
I can see that $x_1(I)$ is isomorphic to $I$ (as a set) and
$x_2(I)$ is isomorphic to the half circle ${(x,y) | x^2 + y^2 = 1, 0 le x, y}$ (as a set). It seems that these might be related to some isomorphic properties, however, I am not sure.
Any comments/suggestions/answers will be very appreciated.
reference-request manifolds low-dimensional-topology
asked Dec 19 '18 at 23:03
induction601induction601
1,307314
$endgroup$
Let $n ge 3$ and $x(t) in mathbb{R}^n$ be a vector-valued function of $t in I = [0,M]$.
Suppose $x_1(t) = [t,cdots,t]^T = textbf{1}t$ which is a line, a 1-dimensional manifold.
Then $$x_1(I) = {x(t)|t in I}$$ can be viewed as a subset in 1-dimension $mathbb{R}$.
Similarly, if
$$x_2(t) = [1,1,1,0,cdots,0]^Tsqrt{1-(t/M)^2} + [0,0,0,1,cdots,1]^T(t/M),$$ $x_2(I)$ can be viewed as a half circle, as a subset in 2-dimenision $mathbb{R}^2$.
Now let consider
$$
x_3(t) = [cos(t-theta_1), cos(t-theta_2), cdots, cos(t-theta_{n})]^T.
$$
In this case, it is unclear that what dimensional object $x_3(I)$ is.
I am wondering if there is a mathematical notion or terminology which describing the dimensionality of $x(I)$ in general (e.g. $x_3(I)$).
But I don't want to (even continuously) deform $x(I)$ to identify the dimensionality. If I allow any deformation on $x(I)$, I believe $x_2(I)$ and $x_3(I)$ become a 1-dimensional "manifold" as they only depend on a single variable $t$.
I can see that $x_1(I)$ is isomorphic to $I$ (as a set) and
$x_2(I)$ is isomorphic to the half circle ${(x,y) | x^2 + y^2 = 1, 0 le x, y}$ (as a set). It seems that these might be related to some isomorphic properties, however, I am not sure.
Any comments/suggestions/answers will be very appreciated.
reference-request manifolds low-dimensional-topology
reference-request manifolds low-dimensional-topology
asked Dec 19 '18 at 23:03
induction601induction601
1,307314
asked Dec 19 '18 at 23:03
induction601induction601
1,307314
edited Dec 20 '18 at 5:51
induction601
asked Dec 19 '18 at 23:03
induction601induction601
1,307314
asked Dec 19 '18 at 23:03
induction601induction601
1,307314
asked Dec 19 '18 at 23:03
induction601induction601
1,307314
1,307314
$begingroup$
$x_2$ is still a 1-dimensional object.
$endgroup$
– Dog_69
Dec 19 '18 at 23:09
$begingroup$
Sorry I didn't explain this clearly. What I am trying to say is that even the half circle is a line with the curvature, the only way we can realize whether this line is a half circle or not is to put this object into the 2-dimensional space. In this sense, I would like to see this as a 2-dimensional object.
$endgroup$
– induction601
Dec 19 '18 at 23:11
$begingroup$
@Dog_69 For example, $x_2(I)$ is isomorphic to HC (HC = Half Circle) where HC is a subset in "2D", not in 1D.
$endgroup$
– induction601
Dec 19 '18 at 23:13
add a comment |
$begingroup$
$x_2$ is still a 1-dimensional object.
$endgroup$
– Dog_69
Dec 19 '18 at 23:09
$begingroup$
Sorry I didn't explain this clearly. What I am trying to say is that even the half circle is a line with the curvature, the only way we can realize whether this line is a half circle or not is to put this object into the 2-dimensional space. In this sense, I would like to see this as a 2-dimensional object.
$endgroup$
– induction601
Dec 19 '18 at 23:11
$begingroup$
@Dog_69 For example, $x_2(I)$ is isomorphic to HC (HC = Half Circle) where HC is a subset in "2D", not in 1D.
$endgroup$
– induction601
Dec 19 '18 at 23:13
$begingroup$
$x_2$ is still a 1-dimensional object.
$endgroup$
– Dog_69
Dec 19 '18 at 23:09
$begingroup$
$x_2$ is still a 1-dimensional object.
$endgroup$
– Dog_69
Dec 19 '18 at 23:09
$begingroup$
Sorry I didn't explain this clearly. What I am trying to say is that even the half circle is a line with the curvature, the only way we can realize whether this line is a half circle or not is to put this object into the 2-dimensional space. In this sense, I would like to see this as a 2-dimensional object.
$endgroup$
– induction601
Dec 19 '18 at 23:11
$begingroup$
Sorry I didn't explain this clearly. What I am trying to say is that even the half circle is a line with the curvature, the only way we can realize whether this line is a half circle or not is to put this object into the 2-dimensional space. In this sense, I would like to see this as a 2-dimensional object.
$endgroup$
– induction601
Dec 19 '18 at 23:11
$begingroup$
@Dog_69 For example, $x_2(I)$ is isomorphic to HC (HC = Half Circle) where HC is a subset in "2D", not in 1D.
$endgroup$
– induction601
Dec 19 '18 at 23:13
$begingroup$
@Dog_69 For example, $x_2(I)$ is isomorphic to HC (HC = Half Circle) where HC is a subset in "2D", not in 1D.
$endgroup$
– induction601
Dec 19 '18 at 23:13
add a comment |
0
active
oldest
votes
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%2f3046965%2flow-dimensional-structures-in-a-high-dimensional-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3046965%2flow-dimensional-structures-in-a-high-dimensional-space%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
$begingroup$
$x_2$ is still a 1-dimensional object.
$endgroup$
– Dog_69
Dec 19 '18 at 23:09
$begingroup$
Sorry I didn't explain this clearly. What I am trying to say is that even the half circle is a line with the curvature, the only way we can realize whether this line is a half circle or not is to put this object into the 2-dimensional space. In this sense, I would like to see this as a 2-dimensional object.
$endgroup$
– induction601
Dec 19 '18 at 23:11
$begingroup$
@Dog_69 For example, $x_2(I)$ is isomorphic to HC (HC = Half Circle) where HC is a subset in "2D", not in 1D.
$endgroup$
– induction601
Dec 19 '18 at 23:13