A 1-form on a smooth manifold is exact if and only if it integrates to zero on every closed curve
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I am stuck on the following problem, which comes from a old qualifying exam.
Prove that a 1-form $phi$ on $M$ is exact if and only if for every closed curve
c, $int_{c} phi =0$.
One way is an application of Stokes' theorem,
if $phi = df$ then $int_{c}phi = int_{partial C}df = 0$ since $partial B=emptyset$.
I don't know how to do the other direction. I made an attempt as follows:
Choose any $x_0 in M$ define a function, $f(x)=int_{x_0}^{x}phi$. This makes sense since the integral is path independent. Now I want to prove that $f$ is smooth and $df=phi$. I can't do either.
Thanks
integration manifolds differential-forms
asked Jul 15 '14 at 20:57
Bates
493313
add a comment |
up vote
5
down vote
favorite
I am stuck on the following problem, which comes from a old qualifying exam.
Prove that a 1-form $phi$ on $M$ is exact if and only if for every closed curve
c, $int_{c} phi =0$.
One way is an application of Stokes' theorem,
if $phi = df$ then $int_{c}phi = int_{partial C}df = 0$ since $partial B=emptyset$.
I don't know how to do the other direction. I made an attempt as follows:
Choose any $x_0 in M$ define a function, $f(x)=int_{x_0}^{x}phi$. This makes sense since the integral is path independent. Now I want to prove that $f$ is smooth and $df=phi$. I can't do either.
Thanks
integration manifolds differential-forms
asked Jul 15 '14 at 20:57
Bates
493313
You may not be able to prove $f$ is smooth -- in general, it's not -- but it's once-differentiable. The proof looks a lot like the proof of the fundamental theorem of calculus (of which this is just a generalization). Same goes for $df = phi$.
– John Hughes
Jul 15 '14 at 21:18
Is the theorem true? Do I need extra requirements?
– Bates
Jul 15 '14 at 21:28
I think I can do it, if I assume that $H^1 =0$.
– Bates
Jul 15 '14 at 21:29
7
You don't need to assume $H^1=0$. But to prove $f$ is smooth, you have to assume that $phi$ is smooth. To prove that $df=phi$ at a point $p$, you can work in coordinates centered at $p$, in which case what you need to show is that $partial f/partial x^i(p) = phi_i(p)$ for each $i$, where the coordinate representation of $phi$ is $sum_i phi_i dx^i$. You can do this by carefully choosing a path whose last part travels along the $x^i$-axis in these coordinates. The proof is carried out in detail in my Introduction to Smooth Manifolds (2nd ed.), Theorem 11.42.
– Jack Lee
Jul 15 '14 at 22:19
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I am stuck on the following problem, which comes from a old qualifying exam.
Prove that a 1-form $phi$ on $M$ is exact if and only if for every closed curve
c, $int_{c} phi =0$.
One way is an application of Stokes' theorem,
if $phi = df$ then $int_{c}phi = int_{partial C}df = 0$ since $partial B=emptyset$.
I don't know how to do the other direction. I made an attempt as follows:
Choose any $x_0 in M$ define a function, $f(x)=int_{x_0}^{x}phi$. This makes sense since the integral is path independent. Now I want to prove that $f$ is smooth and $df=phi$. I can't do either.
Thanks
integration manifolds differential-forms
asked Jul 15 '14 at 20:57
Bates
493313
I am stuck on the following problem, which comes from a old qualifying exam.
Prove that a 1-form $phi$ on $M$ is exact if and only if for every closed curve
c, $int_{c} phi =0$.
One way is an application of Stokes' theorem,
if $phi = df$ then $int_{c}phi = int_{partial C}df = 0$ since $partial B=emptyset$.
I don't know how to do the other direction. I made an attempt as follows:
Choose any $x_0 in M$ define a function, $f(x)=int_{x_0}^{x}phi$. This makes sense since the integral is path independent. Now I want to prove that $f$ is smooth and $df=phi$. I can't do either.
Thanks
integration manifolds differential-forms
integration manifolds differential-forms
asked Jul 15 '14 at 20:57
Bates
493313
asked Jul 15 '14 at 20:57
Bates
493313
edited Jul 16 '14 at 2:22
user147263
asked Jul 15 '14 at 20:57
Bates
493313
asked Jul 15 '14 at 20:57
Bates
493313
asked Jul 15 '14 at 20:57
Bates
493313
493313
You may not be able to prove $f$ is smooth -- in general, it's not -- but it's once-differentiable. The proof looks a lot like the proof of the fundamental theorem of calculus (of which this is just a generalization). Same goes for $df = phi$.
– John Hughes
Jul 15 '14 at 21:18
Is the theorem true? Do I need extra requirements?
– Bates
Jul 15 '14 at 21:28
I think I can do it, if I assume that $H^1 =0$.
– Bates
Jul 15 '14 at 21:29
7
You don't need to assume $H^1=0$. But to prove $f$ is smooth, you have to assume that $phi$ is smooth. To prove that $df=phi$ at a point $p$, you can work in coordinates centered at $p$, in which case what you need to show is that $partial f/partial x^i(p) = phi_i(p)$ for each $i$, where the coordinate representation of $phi$ is $sum_i phi_i dx^i$. You can do this by carefully choosing a path whose last part travels along the $x^i$-axis in these coordinates. The proof is carried out in detail in my Introduction to Smooth Manifolds (2nd ed.), Theorem 11.42.
– Jack Lee
Jul 15 '14 at 22:19
add a comment |
You may not be able to prove $f$ is smooth -- in general, it's not -- but it's once-differentiable. The proof looks a lot like the proof of the fundamental theorem of calculus (of which this is just a generalization). Same goes for $df = phi$.
– John Hughes
Jul 15 '14 at 21:18
Is the theorem true? Do I need extra requirements?
– Bates
Jul 15 '14 at 21:28
I think I can do it, if I assume that $H^1 =0$.
– Bates
Jul 15 '14 at 21:29
7
You don't need to assume $H^1=0$. But to prove $f$ is smooth, you have to assume that $phi$ is smooth. To prove that $df=phi$ at a point $p$, you can work in coordinates centered at $p$, in which case what you need to show is that $partial f/partial x^i(p) = phi_i(p)$ for each $i$, where the coordinate representation of $phi$ is $sum_i phi_i dx^i$. You can do this by carefully choosing a path whose last part travels along the $x^i$-axis in these coordinates. The proof is carried out in detail in my Introduction to Smooth Manifolds (2nd ed.), Theorem 11.42.
– Jack Lee
Jul 15 '14 at 22:19
You may not be able to prove $f$ is smooth -- in general, it's not -- but it's once-differentiable. The proof looks a lot like the proof of the fundamental theorem of calculus (of which this is just a generalization). Same goes for $df = phi$.
– John Hughes
Jul 15 '14 at 21:18
You may not be able to prove $f$ is smooth -- in general, it's not -- but it's once-differentiable. The proof looks a lot like the proof of the fundamental theorem of calculus (of which this is just a generalization). Same goes for $df = phi$.
– John Hughes
Jul 15 '14 at 21:18
Is the theorem true? Do I need extra requirements?
– Bates
Jul 15 '14 at 21:28
Is the theorem true? Do I need extra requirements?
– Bates
Jul 15 '14 at 21:28
I think I can do it, if I assume that $H^1 =0$.
– Bates
Jul 15 '14 at 21:29
I think I can do it, if I assume that $H^1 =0$.
– Bates
Jul 15 '14 at 21:29
7
7
You don't need to assume $H^1=0$. But to prove $f$ is smooth, you have to assume that $phi$ is smooth. To prove that $df=phi$ at a point $p$, you can work in coordinates centered at $p$, in which case what you need to show is that $partial f/partial x^i(p) = phi_i(p)$ for each $i$, where the coordinate representation of $phi$ is $sum_i phi_i dx^i$. You can do this by carefully choosing a path whose last part travels along the $x^i$-axis in these coordinates. The proof is carried out in detail in my Introduction to Smooth Manifolds (2nd ed.), Theorem 11.42.
– Jack Lee
Jul 15 '14 at 22:19
You don't need to assume $H^1=0$. But to prove $f$ is smooth, you have to assume that $phi$ is smooth. To prove that $df=phi$ at a point $p$, you can work in coordinates centered at $p$, in which case what you need to show is that $partial f/partial x^i(p) = phi_i(p)$ for each $i$, where the coordinate representation of $phi$ is $sum_i phi_i dx^i$. You can do this by carefully choosing a path whose last part travels along the $x^i$-axis in these coordinates. The proof is carried out in detail in my Introduction to Smooth Manifolds (2nd ed.), Theorem 11.42.
– Jack Lee
Jul 15 '14 at 22:19
add a comment |
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I assume we are discussing the problem in the context of differential geometry, so $M$ is a smooth manifold, $phi$ is a smooth 1-form. The smoothness of $f(x)=int_{x_0}^x phi$ is a local property so we should be able to prove that by applying fundamental theorem of calculus in $mathbf R ^n$.
answered Jun 6 '17 at 22:58
wolfking01
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I assume we are discussing the problem in the context of differential geometry, so $M$ is a smooth manifold, $phi$ is a smooth 1-form. The smoothness of $f(x)=int_{x_0}^x phi$ is a local property so we should be able to prove that by applying fundamental theorem of calculus in $mathbf R ^n$.
answered Jun 6 '17 at 22:58
wolfking01
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up vote
0
down vote
I assume we are discussing the problem in the context of differential geometry, so $M$ is a smooth manifold, $phi$ is a smooth 1-form. The smoothness of $f(x)=int_{x_0}^x phi$ is a local property so we should be able to prove that by applying fundamental theorem of calculus in $mathbf R ^n$.
answered Jun 6 '17 at 22:58
wolfking01
111
add a comment |
up vote
0
down vote
up vote
0
down vote
I assume we are discussing the problem in the context of differential geometry, so $M$ is a smooth manifold, $phi$ is a smooth 1-form. The smoothness of $f(x)=int_{x_0}^x phi$ is a local property so we should be able to prove that by applying fundamental theorem of calculus in $mathbf R ^n$.
answered Jun 6 '17 at 22:58
wolfking01
111
I assume we are discussing the problem in the context of differential geometry, so $M$ is a smooth manifold, $phi$ is a smooth 1-form. The smoothness of $f(x)=int_{x_0}^x phi$ is a local property so we should be able to prove that by applying fundamental theorem of calculus in $mathbf R ^n$.
answered Jun 6 '17 at 22:58
wolfking01
111
answered Jun 6 '17 at 22:58
wolfking01
111
answered Jun 6 '17 at 22:58
wolfking01
111
answered Jun 6 '17 at 22:58
wolfking01
111
111
add a comment |
add a comment |
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You may not be able to prove $f$ is smooth -- in general, it's not -- but it's once-differentiable. The proof looks a lot like the proof of the fundamental theorem of calculus (of which this is just a generalization). Same goes for $df = phi$.
– John Hughes
Jul 15 '14 at 21:18
Is the theorem true? Do I need extra requirements?
– Bates
Jul 15 '14 at 21:28
I think I can do it, if I assume that $H^1 =0$.
– Bates
Jul 15 '14 at 21:29
7
You don't need to assume $H^1=0$. But to prove $f$ is smooth, you have to assume that $phi$ is smooth. To prove that $df=phi$ at a point $p$, you can work in coordinates centered at $p$, in which case what you need to show is that $partial f/partial x^i(p) = phi_i(p)$ for each $i$, where the coordinate representation of $phi$ is $sum_i phi_i dx^i$. You can do this by carefully choosing a path whose last part travels along the $x^i$-axis in these coordinates. The proof is carried out in detail in my Introduction to Smooth Manifolds (2nd ed.), Theorem 11.42.
– Jack Lee
Jul 15 '14 at 22:19