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Finding an Antiderivative for an Exact Form
Clash Royale CLAN TAG#URR8PPP
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Let $omega$ be an exact $1$-form on a path-connected manifold $M$. For a fixed $x in M$ we can define for all $y in M$.
$$
g(y) := int_gamma gamma^*(omega),
$$
where $gamma$ is a path from $x$ to $y$. Because $omega$ is exact, it is independent of $gamma$. What I want to show is that $dg = omega$. My thought is that
$$
dleft( int_gamma gamma^*(omega) right) = int d(omega(gamma)),
$$
because $d$ commutes with integration, and $gamma^*(omega)$ is defined to be $omega(gamma(t))$. Then
$$
d(omega(gamma)) = domega(gamma) cdot dgamma
$$
by chain rule. By change of variable, we get
$$
int d omega = omega.
$$
But, someone pointed out that $domega = 0$ by exactness. My thought is that this is only true when integrating over closed paths. What's going on? Thanks.
Edit: Another thought is that $d(omega)$ and $domega$ are two different things. So, that $d(omega) = 0$, but $domega$ is a element of the cotangent bundle. Thus, $domega$ is not $0$.
integration differential-forms
asked Aug 2 at 2:08
Joe Johnson 126
13.5k32466
 |Â
show 2 more comments
up vote
0
down vote
favorite
Let $omega$ be an exact $1$-form on a path-connected manifold $M$. For a fixed $x in M$ we can define for all $y in M$.
$$
g(y) := int_gamma gamma^*(omega),
$$
where $gamma$ is a path from $x$ to $y$. Because $omega$ is exact, it is independent of $gamma$. What I want to show is that $dg = omega$. My thought is that
$$
dleft( int_gamma gamma^*(omega) right) = int d(omega(gamma)),
$$
because $d$ commutes with integration, and $gamma^*(omega)$ is defined to be $omega(gamma(t))$. Then
$$
d(omega(gamma)) = domega(gamma) cdot dgamma
$$
by chain rule. By change of variable, we get
$$
int d omega = omega.
$$
But, someone pointed out that $domega = 0$ by exactness. My thought is that this is only true when integrating over closed paths. What's going on? Thanks.
Edit: Another thought is that $d(omega)$ and $domega$ are two different things. So, that $d(omega) = 0$, but $domega$ is a element of the cotangent bundle. Thus, $domega$ is not $0$.
integration differential-forms
asked Aug 2 at 2:08
Joe Johnson 126
13.5k32466
Have you tried to write it in a more classical way like $int_x^y F(z) , dz$?
– md2perpe
Aug 2 at 20:29
$int_gamma gamma^*(omega)$ is wrong anyways as I understand the notation. Presumably $int_gamma omega$ or $int_0^1 gamma^*(omega)$ was meant. (assuming paths go from 0 to 1 in the domain)
– Hurkyl
Aug 2 at 20:57
@md2perpe Yes, Fundamental Theorem of Calculus. But, I need a rigorous argument that it works when we have exact $1$-forms. This isn't homework. I'm trying to help some students with their qualifying exams.
– Joe Johnson 126
Aug 3 at 2:00
I meant that you should not just do it in a classical way, but rewrite every expression of your post in a classical way to see what is happening, where you do something incorrect.
– md2perpe
Aug 3 at 6:02
@md2perpe If $F'(z) = f(z)$, then $int_x^y f(z) , dz$ would differentiate, with respect to $y$ as $f(y)$. But, it wouldn't make sense to move the derivative inside, as $f$ is not a function of $y$. Correct? But, I still don't see how to fix my work.
– Joe Johnson 126
Aug 3 at 11:26
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $omega$ be an exact $1$-form on a path-connected manifold $M$. For a fixed $x in M$ we can define for all $y in M$.
$$
g(y) := int_gamma gamma^*(omega),
$$
where $gamma$ is a path from $x$ to $y$. Because $omega$ is exact, it is independent of $gamma$. What I want to show is that $dg = omega$. My thought is that
$$
dleft( int_gamma gamma^*(omega) right) = int d(omega(gamma)),
$$
because $d$ commutes with integration, and $gamma^*(omega)$ is defined to be $omega(gamma(t))$. Then
$$
d(omega(gamma)) = domega(gamma) cdot dgamma
$$
by chain rule. By change of variable, we get
$$
int d omega = omega.
$$
But, someone pointed out that $domega = 0$ by exactness. My thought is that this is only true when integrating over closed paths. What's going on? Thanks.
Edit: Another thought is that $d(omega)$ and $domega$ are two different things. So, that $d(omega) = 0$, but $domega$ is a element of the cotangent bundle. Thus, $domega$ is not $0$.
integration differential-forms
asked Aug 2 at 2:08
Joe Johnson 126
13.5k32466
Let $omega$ be an exact $1$-form on a path-connected manifold $M$. For a fixed $x in M$ we can define for all $y in M$.
$$
g(y) := int_gamma gamma^*(omega),
$$
where $gamma$ is a path from $x$ to $y$. Because $omega$ is exact, it is independent of $gamma$. What I want to show is that $dg = omega$. My thought is that
$$
dleft( int_gamma gamma^*(omega) right) = int d(omega(gamma)),
$$
because $d$ commutes with integration, and $gamma^*(omega)$ is defined to be $omega(gamma(t))$. Then
$$
d(omega(gamma)) = domega(gamma) cdot dgamma
$$
by chain rule. By change of variable, we get
$$
int d omega = omega.
$$
But, someone pointed out that $domega = 0$ by exactness. My thought is that this is only true when integrating over closed paths. What's going on? Thanks.
Edit: Another thought is that $d(omega)$ and $domega$ are two different things. So, that $d(omega) = 0$, but $domega$ is a element of the cotangent bundle. Thus, $domega$ is not $0$.
integration differential-forms
asked Aug 2 at 2:08
Joe Johnson 126
13.5k32466
edited Aug 2 at 19:45
asked Aug 2 at 2:08
Joe Johnson 126
13.5k32466
asked Aug 2 at 2:08
Joe Johnson 126
13.5k32466
asked Aug 2 at 2:08
Joe Johnson 126
13.5k32466
13.5k32466
Have you tried to write it in a more classical way like $int_x^y F(z) , dz$?
– md2perpe
Aug 2 at 20:29
$int_gamma gamma^*(omega)$ is wrong anyways as I understand the notation. Presumably $int_gamma omega$ or $int_0^1 gamma^*(omega)$ was meant. (assuming paths go from 0 to 1 in the domain)
– Hurkyl
Aug 2 at 20:57
@md2perpe Yes, Fundamental Theorem of Calculus. But, I need a rigorous argument that it works when we have exact $1$-forms. This isn't homework. I'm trying to help some students with their qualifying exams.
– Joe Johnson 126
Aug 3 at 2:00
I meant that you should not just do it in a classical way, but rewrite every expression of your post in a classical way to see what is happening, where you do something incorrect.
– md2perpe
Aug 3 at 6:02
@md2perpe If $F'(z) = f(z)$, then $int_x^y f(z) , dz$ would differentiate, with respect to $y$ as $f(y)$. But, it wouldn't make sense to move the derivative inside, as $f$ is not a function of $y$. Correct? But, I still don't see how to fix my work.
– Joe Johnson 126
Aug 3 at 11:26
 |Â
show 2 more comments
Have you tried to write it in a more classical way like $int_x^y F(z) , dz$?
– md2perpe
Aug 2 at 20:29
$int_gamma gamma^*(omega)$ is wrong anyways as I understand the notation. Presumably $int_gamma omega$ or $int_0^1 gamma^*(omega)$ was meant. (assuming paths go from 0 to 1 in the domain)
– Hurkyl
Aug 2 at 20:57
@md2perpe Yes, Fundamental Theorem of Calculus. But, I need a rigorous argument that it works when we have exact $1$-forms. This isn't homework. I'm trying to help some students with their qualifying exams.
– Joe Johnson 126
Aug 3 at 2:00
I meant that you should not just do it in a classical way, but rewrite every expression of your post in a classical way to see what is happening, where you do something incorrect.
– md2perpe
Aug 3 at 6:02
@md2perpe If $F'(z) = f(z)$, then $int_x^y f(z) , dz$ would differentiate, with respect to $y$ as $f(y)$. But, it wouldn't make sense to move the derivative inside, as $f$ is not a function of $y$. Correct? But, I still don't see how to fix my work.
– Joe Johnson 126
Aug 3 at 11:26
Have you tried to write it in a more classical way like $int_x^y F(z) , dz$?
– md2perpe
Aug 2 at 20:29
Have you tried to write it in a more classical way like $int_x^y F(z) , dz$?
– md2perpe
Aug 2 at 20:29
$int_gamma gamma^*(omega)$ is wrong anyways as I understand the notation. Presumably $int_gamma omega$ or $int_0^1 gamma^*(omega)$ was meant. (assuming paths go from 0 to 1 in the domain)
– Hurkyl
Aug 2 at 20:57
$int_gamma gamma^*(omega)$ is wrong anyways as I understand the notation. Presumably $int_gamma omega$ or $int_0^1 gamma^*(omega)$ was meant. (assuming paths go from 0 to 1 in the domain)
– Hurkyl
Aug 2 at 20:57
@md2perpe Yes, Fundamental Theorem of Calculus. But, I need a rigorous argument that it works when we have exact $1$-forms. This isn't homework. I'm trying to help some students with their qualifying exams.
– Joe Johnson 126
Aug 3 at 2:00
@md2perpe Yes, Fundamental Theorem of Calculus. But, I need a rigorous argument that it works when we have exact $1$-forms. This isn't homework. I'm trying to help some students with their qualifying exams.
– Joe Johnson 126
Aug 3 at 2:00
I meant that you should not just do it in a classical way, but rewrite every expression of your post in a classical way to see what is happening, where you do something incorrect.
– md2perpe
Aug 3 at 6:02
I meant that you should not just do it in a classical way, but rewrite every expression of your post in a classical way to see what is happening, where you do something incorrect.
– md2perpe
Aug 3 at 6:02
@md2perpe If $F'(z) = f(z)$, then $int_x^y f(z) , dz$ would differentiate, with respect to $y$ as $f(y)$. But, it wouldn't make sense to move the derivative inside, as $f$ is not a function of $y$. Correct? But, I still don't see how to fix my work.
– Joe Johnson 126
Aug 3 at 11:26
@md2perpe If $F'(z) = f(z)$, then $int_x^y f(z) , dz$ would differentiate, with respect to $y$ as $f(y)$. But, it wouldn't make sense to move the derivative inside, as $f$ is not a function of $y$. Correct? But, I still don't see how to fix my work.
– Joe Johnson 126
Aug 3 at 11:26
 |Â
show 2 more comments
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Have you tried to write it in a more classical way like $int_x^y F(z) , dz$?
– md2perpe
Aug 2 at 20:29
$int_gamma gamma^*(omega)$ is wrong anyways as I understand the notation. Presumably $int_gamma omega$ or $int_0^1 gamma^*(omega)$ was meant. (assuming paths go from 0 to 1 in the domain)
– Hurkyl
Aug 2 at 20:57
@md2perpe Yes, Fundamental Theorem of Calculus. But, I need a rigorous argument that it works when we have exact $1$-forms. This isn't homework. I'm trying to help some students with their qualifying exams.
– Joe Johnson 126
Aug 3 at 2:00
I meant that you should not just do it in a classical way, but rewrite every expression of your post in a classical way to see what is happening, where you do something incorrect.
– md2perpe
Aug 3 at 6:02
@md2perpe If $F'(z) = f(z)$, then $int_x^y f(z) , dz$ would differentiate, with respect to $y$ as $f(y)$. But, it wouldn't make sense to move the derivative inside, as $f$ is not a function of $y$. Correct? But, I still don't see how to fix my work.
– Joe Johnson 126
Aug 3 at 11:26