Mixing
Does the volume of a submanifold depend on a choice of coordinates?
$begingroup$
Let's say we have a smooth, orientable, Riemannian manifold $M$ with metric $g$, where $dim(M)=n$. Let $U$ be a submanifold contained in M such that $dim(U)=a$, where $1leq aleq n$
Let $x^1,dots,x^n$ be a set of local coordinate functions of $M$, and let $u^1,dots,u^n$ be another.
Then the $a$-dimensional volume of $U$ in the $x^n$ coordinate system is: $$int_U omega_a$$
Where, in local coordinates: $$omega_a=sqrt{|det(g)|}dx^1wedgedotswedge dx^a$$
Now, let $mu_a$ be $omega_a$ in the $u^n$ coordinate system. Then:
$$mu_a = detleft(frac{partial (x^1,dots,x^a)}{partial (u^1,dots,u^a)} right) sqrt{|det(g)|} du^1wedgedotswedge du^a$$
Thus, the $a$-dimensional volume of $U$ in the $u^n$ coordinate system is:$$int_U mu_a$$
I've heard that the volume of $U$ is dependent on our choice of coordinates, to the extent that it can be non-zero in one coordinate frame and zero in another. However, this doesn't make sense to me, since the only way this could happen is if the Jacobian of the coordinate change is equal to zero (afaik), which wouldn't make much sense.
So, I guess my question is this: is the volume of a submanifold dependent on our choice of local coordinates and, if so, is there a coordinate free formulation of submanifold volume?
differential-geometry manifolds riemannian-geometry
asked Jan 20 at 10:59
WillowWillow
85
$endgroup$
add a comment |
$begingroup$
Let's say we have a smooth, orientable, Riemannian manifold $M$ with metric $g$, where $dim(M)=n$. Let $U$ be a submanifold contained in M such that $dim(U)=a$, where $1leq aleq n$
Let $x^1,dots,x^n$ be a set of local coordinate functions of $M$, and let $u^1,dots,u^n$ be another.
Then the $a$-dimensional volume of $U$ in the $x^n$ coordinate system is: $$int_U omega_a$$
Where, in local coordinates: $$omega_a=sqrt{|det(g)|}dx^1wedgedotswedge dx^a$$
Now, let $mu_a$ be $omega_a$ in the $u^n$ coordinate system. Then:
$$mu_a = detleft(frac{partial (x^1,dots,x^a)}{partial (u^1,dots,u^a)} right) sqrt{|det(g)|} du^1wedgedotswedge du^a$$
Thus, the $a$-dimensional volume of $U$ in the $u^n$ coordinate system is:$$int_U mu_a$$
I've heard that the volume of $U$ is dependent on our choice of coordinates, to the extent that it can be non-zero in one coordinate frame and zero in another. However, this doesn't make sense to me, since the only way this could happen is if the Jacobian of the coordinate change is equal to zero (afaik), which wouldn't make much sense.
So, I guess my question is this: is the volume of a submanifold dependent on our choice of local coordinates and, if so, is there a coordinate free formulation of submanifold volume?
differential-geometry manifolds riemannian-geometry
asked Jan 20 at 10:59
WillowWillow
85
$endgroup$
add a comment |
$begingroup$
Let's say we have a smooth, orientable, Riemannian manifold $M$ with metric $g$, where $dim(M)=n$. Let $U$ be a submanifold contained in M such that $dim(U)=a$, where $1leq aleq n$
Let $x^1,dots,x^n$ be a set of local coordinate functions of $M$, and let $u^1,dots,u^n$ be another.
Then the $a$-dimensional volume of $U$ in the $x^n$ coordinate system is: $$int_U omega_a$$
Where, in local coordinates: $$omega_a=sqrt{|det(g)|}dx^1wedgedotswedge dx^a$$
Now, let $mu_a$ be $omega_a$ in the $u^n$ coordinate system. Then:
$$mu_a = detleft(frac{partial (x^1,dots,x^a)}{partial (u^1,dots,u^a)} right) sqrt{|det(g)|} du^1wedgedotswedge du^a$$
Thus, the $a$-dimensional volume of $U$ in the $u^n$ coordinate system is:$$int_U mu_a$$
I've heard that the volume of $U$ is dependent on our choice of coordinates, to the extent that it can be non-zero in one coordinate frame and zero in another. However, this doesn't make sense to me, since the only way this could happen is if the Jacobian of the coordinate change is equal to zero (afaik), which wouldn't make much sense.
So, I guess my question is this: is the volume of a submanifold dependent on our choice of local coordinates and, if so, is there a coordinate free formulation of submanifold volume?
differential-geometry manifolds riemannian-geometry
asked Jan 20 at 10:59
WillowWillow
85
$endgroup$
Let's say we have a smooth, orientable, Riemannian manifold $M$ with metric $g$, where $dim(M)=n$. Let $U$ be a submanifold contained in M such that $dim(U)=a$, where $1leq aleq n$
Let $x^1,dots,x^n$ be a set of local coordinate functions of $M$, and let $u^1,dots,u^n$ be another.
Then the $a$-dimensional volume of $U$ in the $x^n$ coordinate system is: $$int_U omega_a$$
Where, in local coordinates: $$omega_a=sqrt{|det(g)|}dx^1wedgedotswedge dx^a$$
Now, let $mu_a$ be $omega_a$ in the $u^n$ coordinate system. Then:
$$mu_a = detleft(frac{partial (x^1,dots,x^a)}{partial (u^1,dots,u^a)} right) sqrt{|det(g)|} du^1wedgedotswedge du^a$$
Thus, the $a$-dimensional volume of $U$ in the $u^n$ coordinate system is:$$int_U mu_a$$
I've heard that the volume of $U$ is dependent on our choice of coordinates, to the extent that it can be non-zero in one coordinate frame and zero in another. However, this doesn't make sense to me, since the only way this could happen is if the Jacobian of the coordinate change is equal to zero (afaik), which wouldn't make much sense.
So, I guess my question is this: is the volume of a submanifold dependent on our choice of local coordinates and, if so, is there a coordinate free formulation of submanifold volume?
differential-geometry manifolds riemannian-geometry
differential-geometry manifolds riemannian-geometry
asked Jan 20 at 10:59
WillowWillow
85
asked Jan 20 at 10:59
WillowWillow
85
edited Jan 20 at 11:26
Willow
asked Jan 20 at 10:59
WillowWillow
85
asked Jan 20 at 10:59
WillowWillow
85
asked Jan 20 at 10:59
WillowWillow
85
85
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any Riemannian manifold has a volume computed in terms of the metric, which is coordinate invariant. A submanifold acquires a Riemannian metric by restriction/pullback, so in this way it has a coordinate-invariant notion of volume.
However, this definition of the volume for $U$ is not equivalent to the one you have mentioned. This is because it defines volume as the integral of a certain multiple of $dx^1wedge cdots wedge dx^a$, but the $x^i$ are not necessarily coordinates for $U$ - they could even be orthogonal to $U$ (in which case this "volume" will be zero).
In other words, if one of the 1-forms $dx^i$ is orthogonal to the subspace $T^*U subset T^*M$ everywhere on $U$ then the differential form $dx^1wedgecdotswedge dx^a$ will vanish along $U$, and the integral $int_U omega_a$ will vanish as well. If on the other hand, all of the $dx^i$ are parallel with $T^*U$ then you will compute the correct version of volume. If the $dx^i$ are neither parallel nor orthogonal with $U$, generically you will get an incorrect but non-zero answer.
answered Jan 20 at 11:26
BenBen
4,198617
$endgroup$
$begingroup$
Ah! So the problem is my definition of the volume of U.
$endgroup$
– Willow
Jan 20 at 11:30
$begingroup$
(To be clear, by coordinate invariant maps I mean isometric change of coordinates - otherwise you can just stretch the manifold to a different size however you like.)
$endgroup$
– Ben
Jan 20 at 11:38
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%2f3080436%2fdoes-the-volume-of-a-submanifold-depend-on-a-choice-of-coordinates%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any Riemannian manifold has a volume computed in terms of the metric, which is coordinate invariant. A submanifold acquires a Riemannian metric by restriction/pullback, so in this way it has a coordinate-invariant notion of volume.
However, this definition of the volume for $U$ is not equivalent to the one you have mentioned. This is because it defines volume as the integral of a certain multiple of $dx^1wedge cdots wedge dx^a$, but the $x^i$ are not necessarily coordinates for $U$ - they could even be orthogonal to $U$ (in which case this "volume" will be zero).
In other words, if one of the 1-forms $dx^i$ is orthogonal to the subspace $T^*U subset T^*M$ everywhere on $U$ then the differential form $dx^1wedgecdotswedge dx^a$ will vanish along $U$, and the integral $int_U omega_a$ will vanish as well. If on the other hand, all of the $dx^i$ are parallel with $T^*U$ then you will compute the correct version of volume. If the $dx^i$ are neither parallel nor orthogonal with $U$, generically you will get an incorrect but non-zero answer.
answered Jan 20 at 11:26
BenBen
4,198617
$endgroup$
$begingroup$
Ah! So the problem is my definition of the volume of U.
$endgroup$
– Willow
Jan 20 at 11:30
$begingroup$
(To be clear, by coordinate invariant maps I mean isometric change of coordinates - otherwise you can just stretch the manifold to a different size however you like.)
$endgroup$
– Ben
Jan 20 at 11:38
add a comment |
$begingroup$
Any Riemannian manifold has a volume computed in terms of the metric, which is coordinate invariant. A submanifold acquires a Riemannian metric by restriction/pullback, so in this way it has a coordinate-invariant notion of volume.
However, this definition of the volume for $U$ is not equivalent to the one you have mentioned. This is because it defines volume as the integral of a certain multiple of $dx^1wedge cdots wedge dx^a$, but the $x^i$ are not necessarily coordinates for $U$ - they could even be orthogonal to $U$ (in which case this "volume" will be zero).
In other words, if one of the 1-forms $dx^i$ is orthogonal to the subspace $T^*U subset T^*M$ everywhere on $U$ then the differential form $dx^1wedgecdotswedge dx^a$ will vanish along $U$, and the integral $int_U omega_a$ will vanish as well. If on the other hand, all of the $dx^i$ are parallel with $T^*U$ then you will compute the correct version of volume. If the $dx^i$ are neither parallel nor orthogonal with $U$, generically you will get an incorrect but non-zero answer.
answered Jan 20 at 11:26
BenBen
4,198617
$endgroup$
$begingroup$
Ah! So the problem is my definition of the volume of U.
$endgroup$
– Willow
Jan 20 at 11:30
$begingroup$
(To be clear, by coordinate invariant maps I mean isometric change of coordinates - otherwise you can just stretch the manifold to a different size however you like.)
$endgroup$
– Ben
Jan 20 at 11:38
add a comment |
$begingroup$
Any Riemannian manifold has a volume computed in terms of the metric, which is coordinate invariant. A submanifold acquires a Riemannian metric by restriction/pullback, so in this way it has a coordinate-invariant notion of volume.
However, this definition of the volume for $U$ is not equivalent to the one you have mentioned. This is because it defines volume as the integral of a certain multiple of $dx^1wedge cdots wedge dx^a$, but the $x^i$ are not necessarily coordinates for $U$ - they could even be orthogonal to $U$ (in which case this "volume" will be zero).
In other words, if one of the 1-forms $dx^i$ is orthogonal to the subspace $T^*U subset T^*M$ everywhere on $U$ then the differential form $dx^1wedgecdotswedge dx^a$ will vanish along $U$, and the integral $int_U omega_a$ will vanish as well. If on the other hand, all of the $dx^i$ are parallel with $T^*U$ then you will compute the correct version of volume. If the $dx^i$ are neither parallel nor orthogonal with $U$, generically you will get an incorrect but non-zero answer.
answered Jan 20 at 11:26
BenBen
4,198617
$endgroup$
Any Riemannian manifold has a volume computed in terms of the metric, which is coordinate invariant. A submanifold acquires a Riemannian metric by restriction/pullback, so in this way it has a coordinate-invariant notion of volume.
However, this definition of the volume for $U$ is not equivalent to the one you have mentioned. This is because it defines volume as the integral of a certain multiple of $dx^1wedge cdots wedge dx^a$, but the $x^i$ are not necessarily coordinates for $U$ - they could even be orthogonal to $U$ (in which case this "volume" will be zero).
In other words, if one of the 1-forms $dx^i$ is orthogonal to the subspace $T^*U subset T^*M$ everywhere on $U$ then the differential form $dx^1wedgecdotswedge dx^a$ will vanish along $U$, and the integral $int_U omega_a$ will vanish as well. If on the other hand, all of the $dx^i$ are parallel with $T^*U$ then you will compute the correct version of volume. If the $dx^i$ are neither parallel nor orthogonal with $U$, generically you will get an incorrect but non-zero answer.
answered Jan 20 at 11:26
BenBen
4,198617
answered Jan 20 at 11:26
BenBen
4,198617
answered Jan 20 at 11:26
BenBen
4,198617
answered Jan 20 at 11:26
BenBen
4,198617
4,198617
$begingroup$
Ah! So the problem is my definition of the volume of U.
$endgroup$
– Willow
Jan 20 at 11:30
$begingroup$
(To be clear, by coordinate invariant maps I mean isometric change of coordinates - otherwise you can just stretch the manifold to a different size however you like.)
$endgroup$
– Ben
Jan 20 at 11:38
add a comment |
$begingroup$
Ah! So the problem is my definition of the volume of U.
$endgroup$
– Willow
Jan 20 at 11:30
$begingroup$
(To be clear, by coordinate invariant maps I mean isometric change of coordinates - otherwise you can just stretch the manifold to a different size however you like.)
$endgroup$
– Ben
Jan 20 at 11:38
$begingroup$
Ah! So the problem is my definition of the volume of U.
$endgroup$
– Willow
Jan 20 at 11:30
$begingroup$
Ah! So the problem is my definition of the volume of U.
$endgroup$
– Willow
Jan 20 at 11:30
$begingroup$
(To be clear, by coordinate invariant maps I mean isometric change of coordinates - otherwise you can just stretch the manifold to a different size however you like.)
$endgroup$
– Ben
Jan 20 at 11:38
$begingroup$
(To be clear, by coordinate invariant maps I mean isometric change of coordinates - otherwise you can just stretch the manifold to a different size however you like.)
$endgroup$
– Ben
Jan 20 at 11:38
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%2f3080436%2fdoes-the-volume-of-a-submanifold-depend-on-a-choice-of-coordinates%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
