Mixing
Multivariable Substitution Rule VS Pullback of Volume Forms
Clash Royale CLAN TAG#URR8PPP
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Let $dmathbfx = dx^1wedge cdots wedge dx^n$ be the volume form on $UsubsetmathbbR^n$ and smooth $c : [a_1,b_1]timescdotstimes [a_n,b_n] to U$, $mathbft = (t_1,dots,t_n) mapsto mathbfx = mathbfc(mathbft) = (c^1(t_1,dots,t_n),dots,c^n(t_1,dots,t_n))$.
I’ve encountered the following two formulae:
Multivariable Substitution Rule:
$$dmathbfx = dmathbfc = vert detleft((Dmathbfc)_mathbftright) vert dmathbft$$
Pullback of Volume Forms:
$$c^*(dmathbfx) = dmathbfc = detleft((Dmathbfc)_mathbftright) dmathbft $$
$(Dmathbfc)_mathbft$ is the total derivative of $mathbfc$ at $mathbft$.
My question is, are these two formulae referring to different things? It feels like these two equations are talking about the same thing, but then why does one have an absolute value on the Jacobian determinant?
I’m trying to understand this so I can prove that integrals of differential forms only flip signs under a re-parameterisation that reverses orientation.
differential-geometry differential-forms
asked Jul 18 at 5:56
user577413
838
add a comment |Â
up vote
0
down vote
favorite
Let $dmathbfx = dx^1wedge cdots wedge dx^n$ be the volume form on $UsubsetmathbbR^n$ and smooth $c : [a_1,b_1]timescdotstimes [a_n,b_n] to U$, $mathbft = (t_1,dots,t_n) mapsto mathbfx = mathbfc(mathbft) = (c^1(t_1,dots,t_n),dots,c^n(t_1,dots,t_n))$.
I’ve encountered the following two formulae:
Multivariable Substitution Rule:
$$dmathbfx = dmathbfc = vert detleft((Dmathbfc)_mathbftright) vert dmathbft$$
Pullback of Volume Forms:
$$c^*(dmathbfx) = dmathbfc = detleft((Dmathbfc)_mathbftright) dmathbft $$
$(Dmathbfc)_mathbft$ is the total derivative of $mathbfc$ at $mathbft$.
My question is, are these two formulae referring to different things? It feels like these two equations are talking about the same thing, but then why does one have an absolute value on the Jacobian determinant?
I’m trying to understand this so I can prove that integrals of differential forms only flip signs under a re-parameterisation that reverses orientation.
differential-geometry differential-forms
asked Jul 18 at 5:56
user577413
838
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $dmathbfx = dx^1wedge cdots wedge dx^n$ be the volume form on $UsubsetmathbbR^n$ and smooth $c : [a_1,b_1]timescdotstimes [a_n,b_n] to U$, $mathbft = (t_1,dots,t_n) mapsto mathbfx = mathbfc(mathbft) = (c^1(t_1,dots,t_n),dots,c^n(t_1,dots,t_n))$.
I’ve encountered the following two formulae:
Multivariable Substitution Rule:
$$dmathbfx = dmathbfc = vert detleft((Dmathbfc)_mathbftright) vert dmathbft$$
Pullback of Volume Forms:
$$c^*(dmathbfx) = dmathbfc = detleft((Dmathbfc)_mathbftright) dmathbft $$
$(Dmathbfc)_mathbft$ is the total derivative of $mathbfc$ at $mathbft$.
My question is, are these two formulae referring to different things? It feels like these two equations are talking about the same thing, but then why does one have an absolute value on the Jacobian determinant?
I’m trying to understand this so I can prove that integrals of differential forms only flip signs under a re-parameterisation that reverses orientation.
differential-geometry differential-forms
asked Jul 18 at 5:56
user577413
838
Let $dmathbfx = dx^1wedge cdots wedge dx^n$ be the volume form on $UsubsetmathbbR^n$ and smooth $c : [a_1,b_1]timescdotstimes [a_n,b_n] to U$, $mathbft = (t_1,dots,t_n) mapsto mathbfx = mathbfc(mathbft) = (c^1(t_1,dots,t_n),dots,c^n(t_1,dots,t_n))$.
I’ve encountered the following two formulae:
Multivariable Substitution Rule:
$$dmathbfx = dmathbfc = vert detleft((Dmathbfc)_mathbftright) vert dmathbft$$
Pullback of Volume Forms:
$$c^*(dmathbfx) = dmathbfc = detleft((Dmathbfc)_mathbftright) dmathbft $$
$(Dmathbfc)_mathbft$ is the total derivative of $mathbfc$ at $mathbft$.
My question is, are these two formulae referring to different things? It feels like these two equations are talking about the same thing, but then why does one have an absolute value on the Jacobian determinant?
I’m trying to understand this so I can prove that integrals of differential forms only flip signs under a re-parameterisation that reverses orientation.
differential-geometry differential-forms
asked Jul 18 at 5:56
user577413
838
asked Jul 18 at 5:56
user577413
838
asked Jul 18 at 5:56
user577413
838
asked Jul 18 at 5:56
user577413
838
838
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
Yes, they essentially refer to the same thing. As you can guess, this has something to do with keeping signs consistent. Consider the effect of switching the order of variables in integration. Fubini says this is okay, right?
Let's just simply integrate with the variables being $x,y in mathbbR^2$. In the case where I do regular Riemann integration with substitution, $dxdy = dydx$ so there's no sign change from the "differential". But if we took only the determinant, then notice that changing the order of $x$ and $y$ multiplies the determinant by $-1$ (because we switch two columns which changes the sign of the determinant). So to keep the value the same, we take the absolute value of the determinant.
How does this work in the case of differential forms? We know that $dx wedge dy = - dy wedge dx$ so we have one sign change from here. To "counteract" it, we don't take the absolute value of the determinant. So we have two sign switches per variable switch, so the overall integral remains the same.
answered Jul 18 at 6:31
Osama Ghani
1,027312
Then am I correct to say that the two refer to the same thing, except the first is in Riemann integration and the latter in integration of differential forms?
– user577413
Jul 18 at 8:31
Also, your claim that integrals of differential forms remain the same under change of variables, correct me if I’m wrong but don’t integrals flip signs under a re-parameterisation of the domain that reverses orientation?
– user577413
Jul 21 at 16:40
I didn’t say the integral stays the same under an arbitrary change of variables, just a switch in the order of integration. A change of variables changes the integral for both Riemann and differential forms (but by different means)
– Osama Ghani
Jul 22 at 4:17
I’m not understanding what you mean by “changes the integralâ€� Are you referring to the expression of the integral or its value?
– user577413
Jul 22 at 5:39
My bad, I'm referring to the value of the integral. To summarize what I mean, a switch in the order of integration doesn't affect the value of either the Riemann integral or the integral of differential forms. An orientation-reversing parametrization however will change the sign of both integrals, by affecting different things. In the "regular" Riemann integral, the bounds reverse to give a negative number. In the integration of differential forms, it drops out from the negative sign in the Jacobian determinant.
– Osama Ghani
Jul 22 at 9:03
add a comment |Â
up vote
0
down vote
If both formulas are correct, but lead to different results, they have to be about different things. The crux is in the notation: In the first formula $dbf x$ and $dbf t$ denote the unsigned "volume element" associated to Lebesgue measure in $mathbb R^n$, whereas in the second formula your $dbf x$ as well as $dbf t=dt_1wedgeldotswedge dt_n$ denote the volume form defined in terms of exterior algebra. The latter comes with a sign, and your formula takes care of that.
For myself I use the letter $d$ for signed things and $rm d$ for unsigned ones. E.g., I write
$$rm vol(B)=int_B 1>rm d(bf x) .$$
answered Jul 18 at 8:48
Christian Blatter
164k7107306
(Sigh) I thought differential forms are meant to be an alternative approach to multivariable calculus. So why do all the texts I read keep using results from other approaches (Ordinary Multivariable, Lebesgue Integrals)? Can an exposition not be done independently?
– user577413
Jul 21 at 15:18
Differential forms are not an "alternative" approach to multivariable calculus, but a geometrically uptuned and more sophisticated one. Students have to go through different stages of sophistication in all sorts of topics.
– Christian Blatter
Jul 21 at 15:41
I see that ‘alternative’ is not the right word. What I meant was: I thought differential forms are meant to be a reformulation of multivariable calculus that is more general, from which theorems of multivariable calculus are derived as special cases rather than using these special cases to prove more general results?
– user577413
Jul 21 at 16:36
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes, they essentially refer to the same thing. As you can guess, this has something to do with keeping signs consistent. Consider the effect of switching the order of variables in integration. Fubini says this is okay, right?
Let's just simply integrate with the variables being $x,y in mathbbR^2$. In the case where I do regular Riemann integration with substitution, $dxdy = dydx$ so there's no sign change from the "differential". But if we took only the determinant, then notice that changing the order of $x$ and $y$ multiplies the determinant by $-1$ (because we switch two columns which changes the sign of the determinant). So to keep the value the same, we take the absolute value of the determinant.
How does this work in the case of differential forms? We know that $dx wedge dy = - dy wedge dx$ so we have one sign change from here. To "counteract" it, we don't take the absolute value of the determinant. So we have two sign switches per variable switch, so the overall integral remains the same.
answered Jul 18 at 6:31
Osama Ghani
1,027312
Then am I correct to say that the two refer to the same thing, except the first is in Riemann integration and the latter in integration of differential forms?
– user577413
Jul 18 at 8:31
Also, your claim that integrals of differential forms remain the same under change of variables, correct me if I’m wrong but don’t integrals flip signs under a re-parameterisation of the domain that reverses orientation?
– user577413
Jul 21 at 16:40
I didn’t say the integral stays the same under an arbitrary change of variables, just a switch in the order of integration. A change of variables changes the integral for both Riemann and differential forms (but by different means)
– Osama Ghani
Jul 22 at 4:17
I’m not understanding what you mean by “changes the integralâ€� Are you referring to the expression of the integral or its value?
– user577413
Jul 22 at 5:39
My bad, I'm referring to the value of the integral. To summarize what I mean, a switch in the order of integration doesn't affect the value of either the Riemann integral or the integral of differential forms. An orientation-reversing parametrization however will change the sign of both integrals, by affecting different things. In the "regular" Riemann integral, the bounds reverse to give a negative number. In the integration of differential forms, it drops out from the negative sign in the Jacobian determinant.
– Osama Ghani
Jul 22 at 9:03
add a comment |Â
up vote
0
down vote
Yes, they essentially refer to the same thing. As you can guess, this has something to do with keeping signs consistent. Consider the effect of switching the order of variables in integration. Fubini says this is okay, right?
Let's just simply integrate with the variables being $x,y in mathbbR^2$. In the case where I do regular Riemann integration with substitution, $dxdy = dydx$ so there's no sign change from the "differential". But if we took only the determinant, then notice that changing the order of $x$ and $y$ multiplies the determinant by $-1$ (because we switch two columns which changes the sign of the determinant). So to keep the value the same, we take the absolute value of the determinant.
How does this work in the case of differential forms? We know that $dx wedge dy = - dy wedge dx$ so we have one sign change from here. To "counteract" it, we don't take the absolute value of the determinant. So we have two sign switches per variable switch, so the overall integral remains the same.
answered Jul 18 at 6:31
Osama Ghani
1,027312
Then am I correct to say that the two refer to the same thing, except the first is in Riemann integration and the latter in integration of differential forms?
– user577413
Jul 18 at 8:31
Also, your claim that integrals of differential forms remain the same under change of variables, correct me if I’m wrong but don’t integrals flip signs under a re-parameterisation of the domain that reverses orientation?
– user577413
Jul 21 at 16:40
I didn’t say the integral stays the same under an arbitrary change of variables, just a switch in the order of integration. A change of variables changes the integral for both Riemann and differential forms (but by different means)
– Osama Ghani
Jul 22 at 4:17
I’m not understanding what you mean by “changes the integralâ€� Are you referring to the expression of the integral or its value?
– user577413
Jul 22 at 5:39
My bad, I'm referring to the value of the integral. To summarize what I mean, a switch in the order of integration doesn't affect the value of either the Riemann integral or the integral of differential forms. An orientation-reversing parametrization however will change the sign of both integrals, by affecting different things. In the "regular" Riemann integral, the bounds reverse to give a negative number. In the integration of differential forms, it drops out from the negative sign in the Jacobian determinant.
– Osama Ghani
Jul 22 at 9:03
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes, they essentially refer to the same thing. As you can guess, this has something to do with keeping signs consistent. Consider the effect of switching the order of variables in integration. Fubini says this is okay, right?
Let's just simply integrate with the variables being $x,y in mathbbR^2$. In the case where I do regular Riemann integration with substitution, $dxdy = dydx$ so there's no sign change from the "differential". But if we took only the determinant, then notice that changing the order of $x$ and $y$ multiplies the determinant by $-1$ (because we switch two columns which changes the sign of the determinant). So to keep the value the same, we take the absolute value of the determinant.
How does this work in the case of differential forms? We know that $dx wedge dy = - dy wedge dx$ so we have one sign change from here. To "counteract" it, we don't take the absolute value of the determinant. So we have two sign switches per variable switch, so the overall integral remains the same.
answered Jul 18 at 6:31
Osama Ghani
1,027312
Yes, they essentially refer to the same thing. As you can guess, this has something to do with keeping signs consistent. Consider the effect of switching the order of variables in integration. Fubini says this is okay, right?
Let's just simply integrate with the variables being $x,y in mathbbR^2$. In the case where I do regular Riemann integration with substitution, $dxdy = dydx$ so there's no sign change from the "differential". But if we took only the determinant, then notice that changing the order of $x$ and $y$ multiplies the determinant by $-1$ (because we switch two columns which changes the sign of the determinant). So to keep the value the same, we take the absolute value of the determinant.
How does this work in the case of differential forms? We know that $dx wedge dy = - dy wedge dx$ so we have one sign change from here. To "counteract" it, we don't take the absolute value of the determinant. So we have two sign switches per variable switch, so the overall integral remains the same.
answered Jul 18 at 6:31
Osama Ghani
1,027312
answered Jul 18 at 6:31
Osama Ghani
1,027312
answered Jul 18 at 6:31
Osama Ghani
1,027312
answered Jul 18 at 6:31
Osama Ghani
1,027312
1,027312
Then am I correct to say that the two refer to the same thing, except the first is in Riemann integration and the latter in integration of differential forms?
– user577413
Jul 18 at 8:31
Also, your claim that integrals of differential forms remain the same under change of variables, correct me if I’m wrong but don’t integrals flip signs under a re-parameterisation of the domain that reverses orientation?
– user577413
Jul 21 at 16:40
I didn’t say the integral stays the same under an arbitrary change of variables, just a switch in the order of integration. A change of variables changes the integral for both Riemann and differential forms (but by different means)
– Osama Ghani
Jul 22 at 4:17
I’m not understanding what you mean by “changes the integralâ€� Are you referring to the expression of the integral or its value?
– user577413
Jul 22 at 5:39
My bad, I'm referring to the value of the integral. To summarize what I mean, a switch in the order of integration doesn't affect the value of either the Riemann integral or the integral of differential forms. An orientation-reversing parametrization however will change the sign of both integrals, by affecting different things. In the "regular" Riemann integral, the bounds reverse to give a negative number. In the integration of differential forms, it drops out from the negative sign in the Jacobian determinant.
– Osama Ghani
Jul 22 at 9:03
add a comment |Â
Then am I correct to say that the two refer to the same thing, except the first is in Riemann integration and the latter in integration of differential forms?
– user577413
Jul 18 at 8:31
Also, your claim that integrals of differential forms remain the same under change of variables, correct me if I’m wrong but don’t integrals flip signs under a re-parameterisation of the domain that reverses orientation?
– user577413
Jul 21 at 16:40
I didn’t say the integral stays the same under an arbitrary change of variables, just a switch in the order of integration. A change of variables changes the integral for both Riemann and differential forms (but by different means)
– Osama Ghani
Jul 22 at 4:17
I’m not understanding what you mean by “changes the integralâ€� Are you referring to the expression of the integral or its value?
– user577413
Jul 22 at 5:39
My bad, I'm referring to the value of the integral. To summarize what I mean, a switch in the order of integration doesn't affect the value of either the Riemann integral or the integral of differential forms. An orientation-reversing parametrization however will change the sign of both integrals, by affecting different things. In the "regular" Riemann integral, the bounds reverse to give a negative number. In the integration of differential forms, it drops out from the negative sign in the Jacobian determinant.
– Osama Ghani
Jul 22 at 9:03
Then am I correct to say that the two refer to the same thing, except the first is in Riemann integration and the latter in integration of differential forms?
– user577413
Jul 18 at 8:31
Then am I correct to say that the two refer to the same thing, except the first is in Riemann integration and the latter in integration of differential forms?
– user577413
Jul 18 at 8:31
Also, your claim that integrals of differential forms remain the same under change of variables, correct me if I’m wrong but don’t integrals flip signs under a re-parameterisation of the domain that reverses orientation?
– user577413
Jul 21 at 16:40
Also, your claim that integrals of differential forms remain the same under change of variables, correct me if I’m wrong but don’t integrals flip signs under a re-parameterisation of the domain that reverses orientation?
– user577413
Jul 21 at 16:40
I didn’t say the integral stays the same under an arbitrary change of variables, just a switch in the order of integration. A change of variables changes the integral for both Riemann and differential forms (but by different means)
– Osama Ghani
Jul 22 at 4:17
I didn’t say the integral stays the same under an arbitrary change of variables, just a switch in the order of integration. A change of variables changes the integral for both Riemann and differential forms (but by different means)
– Osama Ghani
Jul 22 at 4:17
I’m not understanding what you mean by “changes the integralâ€� Are you referring to the expression of the integral or its value?
– user577413
Jul 22 at 5:39
I’m not understanding what you mean by “changes the integralâ€� Are you referring to the expression of the integral or its value?
– user577413
Jul 22 at 5:39
My bad, I'm referring to the value of the integral. To summarize what I mean, a switch in the order of integration doesn't affect the value of either the Riemann integral or the integral of differential forms. An orientation-reversing parametrization however will change the sign of both integrals, by affecting different things. In the "regular" Riemann integral, the bounds reverse to give a negative number. In the integration of differential forms, it drops out from the negative sign in the Jacobian determinant.
– Osama Ghani
Jul 22 at 9:03
My bad, I'm referring to the value of the integral. To summarize what I mean, a switch in the order of integration doesn't affect the value of either the Riemann integral or the integral of differential forms. An orientation-reversing parametrization however will change the sign of both integrals, by affecting different things. In the "regular" Riemann integral, the bounds reverse to give a negative number. In the integration of differential forms, it drops out from the negative sign in the Jacobian determinant.
– Osama Ghani
Jul 22 at 9:03
add a comment |Â
up vote
0
down vote
If both formulas are correct, but lead to different results, they have to be about different things. The crux is in the notation: In the first formula $dbf x$ and $dbf t$ denote the unsigned "volume element" associated to Lebesgue measure in $mathbb R^n$, whereas in the second formula your $dbf x$ as well as $dbf t=dt_1wedgeldotswedge dt_n$ denote the volume form defined in terms of exterior algebra. The latter comes with a sign, and your formula takes care of that.
For myself I use the letter $d$ for signed things and $rm d$ for unsigned ones. E.g., I write
$$rm vol(B)=int_B 1>rm d(bf x) .$$
answered Jul 18 at 8:48
Christian Blatter
164k7107306
(Sigh) I thought differential forms are meant to be an alternative approach to multivariable calculus. So why do all the texts I read keep using results from other approaches (Ordinary Multivariable, Lebesgue Integrals)? Can an exposition not be done independently?
– user577413
Jul 21 at 15:18
Differential forms are not an "alternative" approach to multivariable calculus, but a geometrically uptuned and more sophisticated one. Students have to go through different stages of sophistication in all sorts of topics.
– Christian Blatter
Jul 21 at 15:41
I see that ‘alternative’ is not the right word. What I meant was: I thought differential forms are meant to be a reformulation of multivariable calculus that is more general, from which theorems of multivariable calculus are derived as special cases rather than using these special cases to prove more general results?
– user577413
Jul 21 at 16:36
add a comment |Â
up vote
0
down vote
If both formulas are correct, but lead to different results, they have to be about different things. The crux is in the notation: In the first formula $dbf x$ and $dbf t$ denote the unsigned "volume element" associated to Lebesgue measure in $mathbb R^n$, whereas in the second formula your $dbf x$ as well as $dbf t=dt_1wedgeldotswedge dt_n$ denote the volume form defined in terms of exterior algebra. The latter comes with a sign, and your formula takes care of that.
For myself I use the letter $d$ for signed things and $rm d$ for unsigned ones. E.g., I write
$$rm vol(B)=int_B 1>rm d(bf x) .$$
answered Jul 18 at 8:48
Christian Blatter
164k7107306
(Sigh) I thought differential forms are meant to be an alternative approach to multivariable calculus. So why do all the texts I read keep using results from other approaches (Ordinary Multivariable, Lebesgue Integrals)? Can an exposition not be done independently?
– user577413
Jul 21 at 15:18
Differential forms are not an "alternative" approach to multivariable calculus, but a geometrically uptuned and more sophisticated one. Students have to go through different stages of sophistication in all sorts of topics.
– Christian Blatter
Jul 21 at 15:41
I see that ‘alternative’ is not the right word. What I meant was: I thought differential forms are meant to be a reformulation of multivariable calculus that is more general, from which theorems of multivariable calculus are derived as special cases rather than using these special cases to prove more general results?
– user577413
Jul 21 at 16:36
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If both formulas are correct, but lead to different results, they have to be about different things. The crux is in the notation: In the first formula $dbf x$ and $dbf t$ denote the unsigned "volume element" associated to Lebesgue measure in $mathbb R^n$, whereas in the second formula your $dbf x$ as well as $dbf t=dt_1wedgeldotswedge dt_n$ denote the volume form defined in terms of exterior algebra. The latter comes with a sign, and your formula takes care of that.
For myself I use the letter $d$ for signed things and $rm d$ for unsigned ones. E.g., I write
$$rm vol(B)=int_B 1>rm d(bf x) .$$
answered Jul 18 at 8:48
Christian Blatter
164k7107306
If both formulas are correct, but lead to different results, they have to be about different things. The crux is in the notation: In the first formula $dbf x$ and $dbf t$ denote the unsigned "volume element" associated to Lebesgue measure in $mathbb R^n$, whereas in the second formula your $dbf x$ as well as $dbf t=dt_1wedgeldotswedge dt_n$ denote the volume form defined in terms of exterior algebra. The latter comes with a sign, and your formula takes care of that.
For myself I use the letter $d$ for signed things and $rm d$ for unsigned ones. E.g., I write
$$rm vol(B)=int_B 1>rm d(bf x) .$$
answered Jul 18 at 8:48
Christian Blatter
164k7107306
answered Jul 18 at 8:48
Christian Blatter
164k7107306
answered Jul 18 at 8:48
Christian Blatter
164k7107306
answered Jul 18 at 8:48
Christian Blatter
164k7107306
164k7107306
(Sigh) I thought differential forms are meant to be an alternative approach to multivariable calculus. So why do all the texts I read keep using results from other approaches (Ordinary Multivariable, Lebesgue Integrals)? Can an exposition not be done independently?
– user577413
Jul 21 at 15:18
Differential forms are not an "alternative" approach to multivariable calculus, but a geometrically uptuned and more sophisticated one. Students have to go through different stages of sophistication in all sorts of topics.
– Christian Blatter
Jul 21 at 15:41
I see that ‘alternative’ is not the right word. What I meant was: I thought differential forms are meant to be a reformulation of multivariable calculus that is more general, from which theorems of multivariable calculus are derived as special cases rather than using these special cases to prove more general results?
– user577413
Jul 21 at 16:36
add a comment |Â
(Sigh) I thought differential forms are meant to be an alternative approach to multivariable calculus. So why do all the texts I read keep using results from other approaches (Ordinary Multivariable, Lebesgue Integrals)? Can an exposition not be done independently?
– user577413
Jul 21 at 15:18
Differential forms are not an "alternative" approach to multivariable calculus, but a geometrically uptuned and more sophisticated one. Students have to go through different stages of sophistication in all sorts of topics.
– Christian Blatter
Jul 21 at 15:41
I see that ‘alternative’ is not the right word. What I meant was: I thought differential forms are meant to be a reformulation of multivariable calculus that is more general, from which theorems of multivariable calculus are derived as special cases rather than using these special cases to prove more general results?
– user577413
Jul 21 at 16:36
(Sigh) I thought differential forms are meant to be an alternative approach to multivariable calculus. So why do all the texts I read keep using results from other approaches (Ordinary Multivariable, Lebesgue Integrals)? Can an exposition not be done independently?
– user577413
Jul 21 at 15:18
(Sigh) I thought differential forms are meant to be an alternative approach to multivariable calculus. So why do all the texts I read keep using results from other approaches (Ordinary Multivariable, Lebesgue Integrals)? Can an exposition not be done independently?
– user577413
Jul 21 at 15:18
Differential forms are not an "alternative" approach to multivariable calculus, but a geometrically uptuned and more sophisticated one. Students have to go through different stages of sophistication in all sorts of topics.
– Christian Blatter
Jul 21 at 15:41
Differential forms are not an "alternative" approach to multivariable calculus, but a geometrically uptuned and more sophisticated one. Students have to go through different stages of sophistication in all sorts of topics.
– Christian Blatter
Jul 21 at 15:41
I see that ‘alternative’ is not the right word. What I meant was: I thought differential forms are meant to be a reformulation of multivariable calculus that is more general, from which theorems of multivariable calculus are derived as special cases rather than using these special cases to prove more general results?
– user577413
Jul 21 at 16:36
I see that ‘alternative’ is not the right word. What I meant was: I thought differential forms are meant to be a reformulation of multivariable calculus that is more general, from which theorems of multivariable calculus are derived as special cases rather than using these special cases to prove more general results?
– user577413
Jul 21 at 16:36
add a comment |Â
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