Mixing
area formula for integration on manifolds
$begingroup$
Let $g:mathbb{R}^nmapstomathbb{R}^n$ be a smooth function, $M$ be an $m$-dimensional manifold embedded in $mathbb{R}^n$ ($m<n$) with a parameterization $f:Umapsto M$, where $Usubsetmathbb{R}^m$. Let $h:mathbb{R}^nmapsto mathbb{R}$ be a smooth function. Let $Df$ and $Dg$ be the Jacobian matrices and assume them to be nonsingular. I would like to understand how to correctly apply the area formula for $int_{g(M)} h(x)dx$.
(a) Does the form of integration on $M$ exist for $int_{g(M)} h(x)dx$ in general? I do not quite believe $int_{g(M)} h(x)dx = int_{M} h(g(x))|text{det}[Dg(x)]|dx$ is correct.
(b) What about the integration on $U$? It seems $Dg(f(x))Df(x)$ is the Jacobian for $gcirc f$, so do we have $int_{g(M)} h(x)dx = int_{U} h(g(f(x)))sqrt{|text{det}[Df(x)^TDg(f(x))^TDg(f(x))Df(x)}]|dx$?
integration differential-geometry smooth-manifolds
asked Feb 2 at 20:14
Matt SMatt S
103
$endgroup$
add a comment |
$begingroup$
Let $g:mathbb{R}^nmapstomathbb{R}^n$ be a smooth function, $M$ be an $m$-dimensional manifold embedded in $mathbb{R}^n$ ($m<n$) with a parameterization $f:Umapsto M$, where $Usubsetmathbb{R}^m$. Let $h:mathbb{R}^nmapsto mathbb{R}$ be a smooth function. Let $Df$ and $Dg$ be the Jacobian matrices and assume them to be nonsingular. I would like to understand how to correctly apply the area formula for $int_{g(M)} h(x)dx$.
(a) Does the form of integration on $M$ exist for $int_{g(M)} h(x)dx$ in general? I do not quite believe $int_{g(M)} h(x)dx = int_{M} h(g(x))|text{det}[Dg(x)]|dx$ is correct.
(b) What about the integration on $U$? It seems $Dg(f(x))Df(x)$ is the Jacobian for $gcirc f$, so do we have $int_{g(M)} h(x)dx = int_{U} h(g(f(x)))sqrt{|text{det}[Df(x)^TDg(f(x))^TDg(f(x))Df(x)}]|dx$?
integration differential-geometry smooth-manifolds
asked Feb 2 at 20:14
Matt SMatt S
103
$endgroup$
add a comment |
$begingroup$
Let $g:mathbb{R}^nmapstomathbb{R}^n$ be a smooth function, $M$ be an $m$-dimensional manifold embedded in $mathbb{R}^n$ ($m<n$) with a parameterization $f:Umapsto M$, where $Usubsetmathbb{R}^m$. Let $h:mathbb{R}^nmapsto mathbb{R}$ be a smooth function. Let $Df$ and $Dg$ be the Jacobian matrices and assume them to be nonsingular. I would like to understand how to correctly apply the area formula for $int_{g(M)} h(x)dx$.
(a) Does the form of integration on $M$ exist for $int_{g(M)} h(x)dx$ in general? I do not quite believe $int_{g(M)} h(x)dx = int_{M} h(g(x))|text{det}[Dg(x)]|dx$ is correct.
(b) What about the integration on $U$? It seems $Dg(f(x))Df(x)$ is the Jacobian for $gcirc f$, so do we have $int_{g(M)} h(x)dx = int_{U} h(g(f(x)))sqrt{|text{det}[Df(x)^TDg(f(x))^TDg(f(x))Df(x)}]|dx$?
integration differential-geometry smooth-manifolds
asked Feb 2 at 20:14
Matt SMatt S
103
$endgroup$
Let $g:mathbb{R}^nmapstomathbb{R}^n$ be a smooth function, $M$ be an $m$-dimensional manifold embedded in $mathbb{R}^n$ ($m<n$) with a parameterization $f:Umapsto M$, where $Usubsetmathbb{R}^m$. Let $h:mathbb{R}^nmapsto mathbb{R}$ be a smooth function. Let $Df$ and $Dg$ be the Jacobian matrices and assume them to be nonsingular. I would like to understand how to correctly apply the area formula for $int_{g(M)} h(x)dx$.
(a) Does the form of integration on $M$ exist for $int_{g(M)} h(x)dx$ in general? I do not quite believe $int_{g(M)} h(x)dx = int_{M} h(g(x))|text{det}[Dg(x)]|dx$ is correct.
(b) What about the integration on $U$? It seems $Dg(f(x))Df(x)$ is the Jacobian for $gcirc f$, so do we have $int_{g(M)} h(x)dx = int_{U} h(g(f(x)))sqrt{|text{det}[Df(x)^TDg(f(x))^TDg(f(x))Df(x)}]|dx$?
integration differential-geometry smooth-manifolds
integration differential-geometry smooth-manifolds
asked Feb 2 at 20:14
Matt SMatt S
103
asked Feb 2 at 20:14
Matt SMatt S
103
edited Feb 3 at 15:27
Matt S
asked Feb 2 at 20:14
Matt SMatt S
103
asked Feb 2 at 20:14
Matt SMatt S
103
asked Feb 2 at 20:14
Matt SMatt S
103
103
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Welcome to MSE. Let me see if I can help get towards your answer.
First, integration is defined (for the kinds of things I'm guessing you're studying -- roughly "manifolds" at the level of Spivak's "Calculus on Manifolds," or maybe Guillemin and Pollack's book) only over manifolds. That is to say
$$
int_B w
$$
is defined only if $B$ is a manifold. And if $B$ is an $m$-dimensional manifold, then $w$ should be an $m$-form. Assuming $B$ is an orientable manifold with a volume-form $omega$, I think that what you're trying to do is integrate $homega$. It's just that you've called $omega$ by the name $dx$.
But the main point is that this all makes sense only if $B$ is a manifold. When you write
$$
int_{g(M)} h(x) ~dx,
$$
you need to know that $g(M)$ is indeed a manifold, and has a volume form. For that, you probably need a restriction on $g$. You might, for instance, say that $g$ is a diffeomorphism, in which case $g(M)$ is sure to be a manifold.
Once you do that, things are relatively easy: if you have a parameterization $f: U to M$ (one that resembles, say, latitude-longitude coordinates on a sphere, and hence is a diffeomorphism almost everywhere, and is surjective), then $g circ f$ will be a parameterization of $g(M)$. And now your approach, in part "b", says how to compute the integral over the manifold by instead integrating over a nice open subset of $Bbb R^n$.
I can never remember the exact formula for the part b thing, but it sure looks as if you have it right.
Just to go back to the earlier "is g(M) a manifold" issue, think of something like a nice mobius-band shaped thing, sitting on the $xy$ plane in 3-space. But instead of a half-twist, put in a full twist, so that this is really just an embedding of a cylinder in 3-space. Call that $M$.
Now let's let $g: Bbb R^3 to Bbb R^3: (x, y, z) to (x, y, 0)$. That's a smooth function, but $g(M)$ is not a smooth manifold: depending on how you oriented your Mobius band, it looks line an annulus pinched at a couple of points, or like a circle that fattens out at a couple of points. In either case, it's not a manifold, so integrating over it doesn't really make sense.
answered Feb 3 at 15:48
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
Thank you very much for your help. Indeed I forgot to say $g$ is a diffeomorphism and $g(M)$ is a manifold. I was trying to verify the formula in a very nice setting. Now I also believe part (b) is correct.
$endgroup$
– Matt S
Feb 3 at 16:10
$begingroup$
One of the tough jobs when using MSE is formulating a question clearly and completely, so that you don't waste the time of the people answering. I've found that when I've managed to do so, sometimes I already know the answer. :) Anyhow, glad to have helped out.
$endgroup$
– John Hughes
Feb 3 at 16:14
$begingroup$
Any thoughts on part (a) will also be appreciated.
$endgroup$
– Matt S
Feb 3 at 16:21
$begingroup$
I was mostly addressing part "a", saying it made no sense unless $g$ was nice, and if $g$ WAS nice, then the integral is given by the usual rule for integrating when there's a a parameterization (which I think you got right in part b). Part b gives a formula for the LHS. The RHS is just the integration of a function (albeit a complicated one -- namely $h(g(x))|text{det}[Dg(x)]|$ -- over $M$, and you've got a formula for computing such things. Plug in and compare the two!
$endgroup$
– John Hughes
Feb 3 at 16:33
add a comment |
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$begingroup$
Welcome to MSE. Let me see if I can help get towards your answer.
First, integration is defined (for the kinds of things I'm guessing you're studying -- roughly "manifolds" at the level of Spivak's "Calculus on Manifolds," or maybe Guillemin and Pollack's book) only over manifolds. That is to say
$$
int_B w
$$
is defined only if $B$ is a manifold. And if $B$ is an $m$-dimensional manifold, then $w$ should be an $m$-form. Assuming $B$ is an orientable manifold with a volume-form $omega$, I think that what you're trying to do is integrate $homega$. It's just that you've called $omega$ by the name $dx$.
But the main point is that this all makes sense only if $B$ is a manifold. When you write
$$
int_{g(M)} h(x) ~dx,
$$
you need to know that $g(M)$ is indeed a manifold, and has a volume form. For that, you probably need a restriction on $g$. You might, for instance, say that $g$ is a diffeomorphism, in which case $g(M)$ is sure to be a manifold.
Once you do that, things are relatively easy: if you have a parameterization $f: U to M$ (one that resembles, say, latitude-longitude coordinates on a sphere, and hence is a diffeomorphism almost everywhere, and is surjective), then $g circ f$ will be a parameterization of $g(M)$. And now your approach, in part "b", says how to compute the integral over the manifold by instead integrating over a nice open subset of $Bbb R^n$.
I can never remember the exact formula for the part b thing, but it sure looks as if you have it right.
Just to go back to the earlier "is g(M) a manifold" issue, think of something like a nice mobius-band shaped thing, sitting on the $xy$ plane in 3-space. But instead of a half-twist, put in a full twist, so that this is really just an embedding of a cylinder in 3-space. Call that $M$.
Now let's let $g: Bbb R^3 to Bbb R^3: (x, y, z) to (x, y, 0)$. That's a smooth function, but $g(M)$ is not a smooth manifold: depending on how you oriented your Mobius band, it looks line an annulus pinched at a couple of points, or like a circle that fattens out at a couple of points. In either case, it's not a manifold, so integrating over it doesn't really make sense.
answered Feb 3 at 15:48
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
Thank you very much for your help. Indeed I forgot to say $g$ is a diffeomorphism and $g(M)$ is a manifold. I was trying to verify the formula in a very nice setting. Now I also believe part (b) is correct.
$endgroup$
– Matt S
Feb 3 at 16:10
$begingroup$
One of the tough jobs when using MSE is formulating a question clearly and completely, so that you don't waste the time of the people answering. I've found that when I've managed to do so, sometimes I already know the answer. :) Anyhow, glad to have helped out.
$endgroup$
– John Hughes
Feb 3 at 16:14
$begingroup$
Any thoughts on part (a) will also be appreciated.
$endgroup$
– Matt S
Feb 3 at 16:21
$begingroup$
I was mostly addressing part "a", saying it made no sense unless $g$ was nice, and if $g$ WAS nice, then the integral is given by the usual rule for integrating when there's a a parameterization (which I think you got right in part b). Part b gives a formula for the LHS. The RHS is just the integration of a function (albeit a complicated one -- namely $h(g(x))|text{det}[Dg(x)]|$ -- over $M$, and you've got a formula for computing such things. Plug in and compare the two!
$endgroup$
– John Hughes
Feb 3 at 16:33
add a comment |
$begingroup$
Welcome to MSE. Let me see if I can help get towards your answer.
First, integration is defined (for the kinds of things I'm guessing you're studying -- roughly "manifolds" at the level of Spivak's "Calculus on Manifolds," or maybe Guillemin and Pollack's book) only over manifolds. That is to say
$$
int_B w
$$
is defined only if $B$ is a manifold. And if $B$ is an $m$-dimensional manifold, then $w$ should be an $m$-form. Assuming $B$ is an orientable manifold with a volume-form $omega$, I think that what you're trying to do is integrate $homega$. It's just that you've called $omega$ by the name $dx$.
But the main point is that this all makes sense only if $B$ is a manifold. When you write
$$
int_{g(M)} h(x) ~dx,
$$
you need to know that $g(M)$ is indeed a manifold, and has a volume form. For that, you probably need a restriction on $g$. You might, for instance, say that $g$ is a diffeomorphism, in which case $g(M)$ is sure to be a manifold.
Once you do that, things are relatively easy: if you have a parameterization $f: U to M$ (one that resembles, say, latitude-longitude coordinates on a sphere, and hence is a diffeomorphism almost everywhere, and is surjective), then $g circ f$ will be a parameterization of $g(M)$. And now your approach, in part "b", says how to compute the integral over the manifold by instead integrating over a nice open subset of $Bbb R^n$.
I can never remember the exact formula for the part b thing, but it sure looks as if you have it right.
Just to go back to the earlier "is g(M) a manifold" issue, think of something like a nice mobius-band shaped thing, sitting on the $xy$ plane in 3-space. But instead of a half-twist, put in a full twist, so that this is really just an embedding of a cylinder in 3-space. Call that $M$.
Now let's let $g: Bbb R^3 to Bbb R^3: (x, y, z) to (x, y, 0)$. That's a smooth function, but $g(M)$ is not a smooth manifold: depending on how you oriented your Mobius band, it looks line an annulus pinched at a couple of points, or like a circle that fattens out at a couple of points. In either case, it's not a manifold, so integrating over it doesn't really make sense.
answered Feb 3 at 15:48
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
Thank you very much for your help. Indeed I forgot to say $g$ is a diffeomorphism and $g(M)$ is a manifold. I was trying to verify the formula in a very nice setting. Now I also believe part (b) is correct.
$endgroup$
– Matt S
Feb 3 at 16:10
$begingroup$
One of the tough jobs when using MSE is formulating a question clearly and completely, so that you don't waste the time of the people answering. I've found that when I've managed to do so, sometimes I already know the answer. :) Anyhow, glad to have helped out.
$endgroup$
– John Hughes
Feb 3 at 16:14
$begingroup$
Any thoughts on part (a) will also be appreciated.
$endgroup$
– Matt S
Feb 3 at 16:21
$begingroup$
I was mostly addressing part "a", saying it made no sense unless $g$ was nice, and if $g$ WAS nice, then the integral is given by the usual rule for integrating when there's a a parameterization (which I think you got right in part b). Part b gives a formula for the LHS. The RHS is just the integration of a function (albeit a complicated one -- namely $h(g(x))|text{det}[Dg(x)]|$ -- over $M$, and you've got a formula for computing such things. Plug in and compare the two!
$endgroup$
– John Hughes
Feb 3 at 16:33
add a comment |
$begingroup$
Welcome to MSE. Let me see if I can help get towards your answer.
First, integration is defined (for the kinds of things I'm guessing you're studying -- roughly "manifolds" at the level of Spivak's "Calculus on Manifolds," or maybe Guillemin and Pollack's book) only over manifolds. That is to say
$$
int_B w
$$
is defined only if $B$ is a manifold. And if $B$ is an $m$-dimensional manifold, then $w$ should be an $m$-form. Assuming $B$ is an orientable manifold with a volume-form $omega$, I think that what you're trying to do is integrate $homega$. It's just that you've called $omega$ by the name $dx$.
But the main point is that this all makes sense only if $B$ is a manifold. When you write
$$
int_{g(M)} h(x) ~dx,
$$
you need to know that $g(M)$ is indeed a manifold, and has a volume form. For that, you probably need a restriction on $g$. You might, for instance, say that $g$ is a diffeomorphism, in which case $g(M)$ is sure to be a manifold.
Once you do that, things are relatively easy: if you have a parameterization $f: U to M$ (one that resembles, say, latitude-longitude coordinates on a sphere, and hence is a diffeomorphism almost everywhere, and is surjective), then $g circ f$ will be a parameterization of $g(M)$. And now your approach, in part "b", says how to compute the integral over the manifold by instead integrating over a nice open subset of $Bbb R^n$.
I can never remember the exact formula for the part b thing, but it sure looks as if you have it right.
Just to go back to the earlier "is g(M) a manifold" issue, think of something like a nice mobius-band shaped thing, sitting on the $xy$ plane in 3-space. But instead of a half-twist, put in a full twist, so that this is really just an embedding of a cylinder in 3-space. Call that $M$.
Now let's let $g: Bbb R^3 to Bbb R^3: (x, y, z) to (x, y, 0)$. That's a smooth function, but $g(M)$ is not a smooth manifold: depending on how you oriented your Mobius band, it looks line an annulus pinched at a couple of points, or like a circle that fattens out at a couple of points. In either case, it's not a manifold, so integrating over it doesn't really make sense.
answered Feb 3 at 15:48
John HughesJohn Hughes
65.5k24293
$endgroup$
Welcome to MSE. Let me see if I can help get towards your answer.
First, integration is defined (for the kinds of things I'm guessing you're studying -- roughly "manifolds" at the level of Spivak's "Calculus on Manifolds," or maybe Guillemin and Pollack's book) only over manifolds. That is to say
$$
int_B w
$$
is defined only if $B$ is a manifold. And if $B$ is an $m$-dimensional manifold, then $w$ should be an $m$-form. Assuming $B$ is an orientable manifold with a volume-form $omega$, I think that what you're trying to do is integrate $homega$. It's just that you've called $omega$ by the name $dx$.
But the main point is that this all makes sense only if $B$ is a manifold. When you write
$$
int_{g(M)} h(x) ~dx,
$$
you need to know that $g(M)$ is indeed a manifold, and has a volume form. For that, you probably need a restriction on $g$. You might, for instance, say that $g$ is a diffeomorphism, in which case $g(M)$ is sure to be a manifold.
Once you do that, things are relatively easy: if you have a parameterization $f: U to M$ (one that resembles, say, latitude-longitude coordinates on a sphere, and hence is a diffeomorphism almost everywhere, and is surjective), then $g circ f$ will be a parameterization of $g(M)$. And now your approach, in part "b", says how to compute the integral over the manifold by instead integrating over a nice open subset of $Bbb R^n$.
I can never remember the exact formula for the part b thing, but it sure looks as if you have it right.
Just to go back to the earlier "is g(M) a manifold" issue, think of something like a nice mobius-band shaped thing, sitting on the $xy$ plane in 3-space. But instead of a half-twist, put in a full twist, so that this is really just an embedding of a cylinder in 3-space. Call that $M$.
Now let's let $g: Bbb R^3 to Bbb R^3: (x, y, z) to (x, y, 0)$. That's a smooth function, but $g(M)$ is not a smooth manifold: depending on how you oriented your Mobius band, it looks line an annulus pinched at a couple of points, or like a circle that fattens out at a couple of points. In either case, it's not a manifold, so integrating over it doesn't really make sense.
answered Feb 3 at 15:48
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 15:48
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 15:48
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 15:48
John HughesJohn Hughes
65.5k24293
65.5k24293
$begingroup$
Thank you very much for your help. Indeed I forgot to say $g$ is a diffeomorphism and $g(M)$ is a manifold. I was trying to verify the formula in a very nice setting. Now I also believe part (b) is correct.
$endgroup$
– Matt S
Feb 3 at 16:10
$begingroup$
One of the tough jobs when using MSE is formulating a question clearly and completely, so that you don't waste the time of the people answering. I've found that when I've managed to do so, sometimes I already know the answer. :) Anyhow, glad to have helped out.
$endgroup$
– John Hughes
Feb 3 at 16:14
$begingroup$
Any thoughts on part (a) will also be appreciated.
$endgroup$
– Matt S
Feb 3 at 16:21
$begingroup$
I was mostly addressing part "a", saying it made no sense unless $g$ was nice, and if $g$ WAS nice, then the integral is given by the usual rule for integrating when there's a a parameterization (which I think you got right in part b). Part b gives a formula for the LHS. The RHS is just the integration of a function (albeit a complicated one -- namely $h(g(x))|text{det}[Dg(x)]|$ -- over $M$, and you've got a formula for computing such things. Plug in and compare the two!
$endgroup$
– John Hughes
Feb 3 at 16:33
add a comment |
$begingroup$
Thank you very much for your help. Indeed I forgot to say $g$ is a diffeomorphism and $g(M)$ is a manifold. I was trying to verify the formula in a very nice setting. Now I also believe part (b) is correct.
$endgroup$
– Matt S
Feb 3 at 16:10
$begingroup$
One of the tough jobs when using MSE is formulating a question clearly and completely, so that you don't waste the time of the people answering. I've found that when I've managed to do so, sometimes I already know the answer. :) Anyhow, glad to have helped out.
$endgroup$
– John Hughes
Feb 3 at 16:14
$begingroup$
Any thoughts on part (a) will also be appreciated.
$endgroup$
– Matt S
Feb 3 at 16:21
$begingroup$
I was mostly addressing part "a", saying it made no sense unless $g$ was nice, and if $g$ WAS nice, then the integral is given by the usual rule for integrating when there's a a parameterization (which I think you got right in part b). Part b gives a formula for the LHS. The RHS is just the integration of a function (albeit a complicated one -- namely $h(g(x))|text{det}[Dg(x)]|$ -- over $M$, and you've got a formula for computing such things. Plug in and compare the two!
$endgroup$
– John Hughes
Feb 3 at 16:33
$begingroup$
Thank you very much for your help. Indeed I forgot to say $g$ is a diffeomorphism and $g(M)$ is a manifold. I was trying to verify the formula in a very nice setting. Now I also believe part (b) is correct.
$endgroup$
– Matt S
Feb 3 at 16:10
$begingroup$
Thank you very much for your help. Indeed I forgot to say $g$ is a diffeomorphism and $g(M)$ is a manifold. I was trying to verify the formula in a very nice setting. Now I also believe part (b) is correct.
$endgroup$
– Matt S
Feb 3 at 16:10
$begingroup$
One of the tough jobs when using MSE is formulating a question clearly and completely, so that you don't waste the time of the people answering. I've found that when I've managed to do so, sometimes I already know the answer. :) Anyhow, glad to have helped out.
$endgroup$
– John Hughes
Feb 3 at 16:14
$begingroup$
One of the tough jobs when using MSE is formulating a question clearly and completely, so that you don't waste the time of the people answering. I've found that when I've managed to do so, sometimes I already know the answer. :) Anyhow, glad to have helped out.
$endgroup$
– John Hughes
Feb 3 at 16:14
$begingroup$
Any thoughts on part (a) will also be appreciated.
$endgroup$
– Matt S
Feb 3 at 16:21
$begingroup$
Any thoughts on part (a) will also be appreciated.
$endgroup$
– Matt S
Feb 3 at 16:21
$begingroup$
I was mostly addressing part "a", saying it made no sense unless $g$ was nice, and if $g$ WAS nice, then the integral is given by the usual rule for integrating when there's a a parameterization (which I think you got right in part b). Part b gives a formula for the LHS. The RHS is just the integration of a function (albeit a complicated one -- namely $h(g(x))|text{det}[Dg(x)]|$ -- over $M$, and you've got a formula for computing such things. Plug in and compare the two!
$endgroup$
– John Hughes
Feb 3 at 16:33
$begingroup$
I was mostly addressing part "a", saying it made no sense unless $g$ was nice, and if $g$ WAS nice, then the integral is given by the usual rule for integrating when there's a a parameterization (which I think you got right in part b). Part b gives a formula for the LHS. The RHS is just the integration of a function (albeit a complicated one -- namely $h(g(x))|text{det}[Dg(x)]|$ -- over $M$, and you've got a formula for computing such things. Plug in and compare the two!
$endgroup$
– John Hughes
Feb 3 at 16:33
add a comment |
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