Why is the differential of a differential 1-form (or the wedge product) defined as it is?
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In his Advanced Calculus of Several Variables, Edwards defines the differential $domega$ of a differential 1-form $omega=Pdx+Qdy$ to be
$$domegaequivleft(fracpartial Qpartial x-fracpartial Ppartial yright)dxdy.$$
But that definition seems cut from whole cloth. He gives some motivating examples before and after the definition, but they don't give me a good intuitive sense of what it really means.
He has mumbled a bit about the wedge product, but not by that name. He does not use it to derive the differential of a 1-form.
Why is $domega$ defined this way?
Or why is the wedge product used to multiply differential forms?
Or is there a way to picture either $domega$ or the wedge product?
I note that, if the wedge product has already been introduced as
$$dx^1wedge dx^1=dx^2wedge dx^2=0,$$
$$dx^1wedge dx^2=-dx^2wedge dx^1=1,$$
and
$$dx^idx^jequiv dx^iwedge dx^j$$
has been asserted, the differential of $omega$ can be derived as follows:
Start with the definition of differentiability
$$0=lim_Deltamathfrakxtomathfrak0fracDeltaomega_mathfrakxleft[Deltamathfrakxright]-domega_mathfrakxleft[Deltamathfrakxright]$$
$$=lim_Deltamathfrakxtomathfrak0fracDelta P_mathfrakxleft[Deltamathfrakxright]dx+Delta Q_mathfrakxleft[Deltamathfrakxright]dy-domega_mathfrakxleft[Deltamathfrakxright].$$
Take the constant objects $dx,dy$ outside the limits, but keep them in the same relative positions
$$0=lim_Deltamathfrakxtomathfrak0left[fracDelta P_mathfrakxleft[Deltamathfrakxright]right]dx+lim_Deltamathfrakxtomathfrak0left[fracDelta Q_mathfrakxleft[Deltamathfrakxright]right]dy-lim_Deltamathfrakxtomathfrak0left[fracdomega_mathfrakxleft[Deltamathfrakxright]right]$$
Apply the defintion of differentiability to P:
$$lim_Deltamathfrakxtomathfrak0left[fracDelta P_mathfrakxleft[Deltamathfrakxright]-fracdP_mathfrakxleft[Deltamathfrakxright]right]=0$$
$$nablaleft[Pleft[mathfrakxright]right]cdotmathfrakv=dP_mathfrakxleft[mathfrakvright]$$
$$=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright)left[mathfrakvright]$$
$$=fracpartial Ppartial xdxleft[mathfrakvright]+fracpartial Ppartial ydyleft[mathfrakvright]$$
$$=fracpartial Ppartial xv^x+fracpartial Ppartial yv^y.$$
So
$$dP_mathfrakx=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright).$$
The same applies to $dQ_mathfrakx$.
The differential is now:
$$domega_mathfrakx=dP_mathfrakxdx+dQ_mathfrakxdy$$
$$=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright)_mathfrakxdx+left(fracpartial Qpartial xdx+fracpartial Qpartial ydyright)_mathfrakxdy.$$
$$=left(fracpartial Qpartial x-fracpartial Ppartial yright)_mathfrakxdxdy.$$
multivariable-calculus differential-geometry definition
asked Dec 4 '17 at 23:28
Steven Hatton
662314
add a comment |Â
up vote
1
down vote
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In his Advanced Calculus of Several Variables, Edwards defines the differential $domega$ of a differential 1-form $omega=Pdx+Qdy$ to be
$$domegaequivleft(fracpartial Qpartial x-fracpartial Ppartial yright)dxdy.$$
But that definition seems cut from whole cloth. He gives some motivating examples before and after the definition, but they don't give me a good intuitive sense of what it really means.
He has mumbled a bit about the wedge product, but not by that name. He does not use it to derive the differential of a 1-form.
Why is $domega$ defined this way?
Or why is the wedge product used to multiply differential forms?
Or is there a way to picture either $domega$ or the wedge product?
I note that, if the wedge product has already been introduced as
$$dx^1wedge dx^1=dx^2wedge dx^2=0,$$
$$dx^1wedge dx^2=-dx^2wedge dx^1=1,$$
and
$$dx^idx^jequiv dx^iwedge dx^j$$
has been asserted, the differential of $omega$ can be derived as follows:
Start with the definition of differentiability
$$0=lim_Deltamathfrakxtomathfrak0fracDeltaomega_mathfrakxleft[Deltamathfrakxright]-domega_mathfrakxleft[Deltamathfrakxright]$$
$$=lim_Deltamathfrakxtomathfrak0fracDelta P_mathfrakxleft[Deltamathfrakxright]dx+Delta Q_mathfrakxleft[Deltamathfrakxright]dy-domega_mathfrakxleft[Deltamathfrakxright].$$
Take the constant objects $dx,dy$ outside the limits, but keep them in the same relative positions
$$0=lim_Deltamathfrakxtomathfrak0left[fracDelta P_mathfrakxleft[Deltamathfrakxright]right]dx+lim_Deltamathfrakxtomathfrak0left[fracDelta Q_mathfrakxleft[Deltamathfrakxright]right]dy-lim_Deltamathfrakxtomathfrak0left[fracdomega_mathfrakxleft[Deltamathfrakxright]right]$$
Apply the defintion of differentiability to P:
$$lim_Deltamathfrakxtomathfrak0left[fracDelta P_mathfrakxleft[Deltamathfrakxright]-fracdP_mathfrakxleft[Deltamathfrakxright]right]=0$$
$$nablaleft[Pleft[mathfrakxright]right]cdotmathfrakv=dP_mathfrakxleft[mathfrakvright]$$
$$=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright)left[mathfrakvright]$$
$$=fracpartial Ppartial xdxleft[mathfrakvright]+fracpartial Ppartial ydyleft[mathfrakvright]$$
$$=fracpartial Ppartial xv^x+fracpartial Ppartial yv^y.$$
So
$$dP_mathfrakx=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright).$$
The same applies to $dQ_mathfrakx$.
The differential is now:
$$domega_mathfrakx=dP_mathfrakxdx+dQ_mathfrakxdy$$
$$=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright)_mathfrakxdx+left(fracpartial Qpartial xdx+fracpartial Qpartial ydyright)_mathfrakxdy.$$
$$=left(fracpartial Qpartial x-fracpartial Ppartial yright)_mathfrakxdxdy.$$
multivariable-calculus differential-geometry definition
asked Dec 4 '17 at 23:28
Steven Hatton
662314
3
These are complicated questions and they don't have short answers. One possible answer, which is a little unsatisfying, is "to make Green's theorem true."
– Qiaochu Yuan
Dec 4 '17 at 23:30
That $dx^1wedge dx^2 = 1$ looks quite wrong. "Is there a way to picture the wedge product?" It can be thought of as a parallelogram, or a disk, or a plane weighted with a scalar. See en.wikipedia.org/wiki/Bivector
– mr_e_man
Aug 12 at 4:52
I guess that the $dx^1wedge dx^2=1$ mistake is from the fact that its integral over the unit square is $1$. But this integration does not act like a scalar; the form $dx^1wedge dx^2$ is being applied to the square's tangent bivectors. The form itself is a cotangent bivector.
– mr_e_man
Aug 12 at 6:51
add a comment |Â
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1
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up vote
1
down vote
favorite
In his Advanced Calculus of Several Variables, Edwards defines the differential $domega$ of a differential 1-form $omega=Pdx+Qdy$ to be
$$domegaequivleft(fracpartial Qpartial x-fracpartial Ppartial yright)dxdy.$$
But that definition seems cut from whole cloth. He gives some motivating examples before and after the definition, but they don't give me a good intuitive sense of what it really means.
He has mumbled a bit about the wedge product, but not by that name. He does not use it to derive the differential of a 1-form.
Why is $domega$ defined this way?
Or why is the wedge product used to multiply differential forms?
Or is there a way to picture either $domega$ or the wedge product?
I note that, if the wedge product has already been introduced as
$$dx^1wedge dx^1=dx^2wedge dx^2=0,$$
$$dx^1wedge dx^2=-dx^2wedge dx^1=1,$$
and
$$dx^idx^jequiv dx^iwedge dx^j$$
has been asserted, the differential of $omega$ can be derived as follows:
Start with the definition of differentiability
$$0=lim_Deltamathfrakxtomathfrak0fracDeltaomega_mathfrakxleft[Deltamathfrakxright]-domega_mathfrakxleft[Deltamathfrakxright]$$
$$=lim_Deltamathfrakxtomathfrak0fracDelta P_mathfrakxleft[Deltamathfrakxright]dx+Delta Q_mathfrakxleft[Deltamathfrakxright]dy-domega_mathfrakxleft[Deltamathfrakxright].$$
Take the constant objects $dx,dy$ outside the limits, but keep them in the same relative positions
$$0=lim_Deltamathfrakxtomathfrak0left[fracDelta P_mathfrakxleft[Deltamathfrakxright]right]dx+lim_Deltamathfrakxtomathfrak0left[fracDelta Q_mathfrakxleft[Deltamathfrakxright]right]dy-lim_Deltamathfrakxtomathfrak0left[fracdomega_mathfrakxleft[Deltamathfrakxright]right]$$
Apply the defintion of differentiability to P:
$$lim_Deltamathfrakxtomathfrak0left[fracDelta P_mathfrakxleft[Deltamathfrakxright]-fracdP_mathfrakxleft[Deltamathfrakxright]right]=0$$
$$nablaleft[Pleft[mathfrakxright]right]cdotmathfrakv=dP_mathfrakxleft[mathfrakvright]$$
$$=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright)left[mathfrakvright]$$
$$=fracpartial Ppartial xdxleft[mathfrakvright]+fracpartial Ppartial ydyleft[mathfrakvright]$$
$$=fracpartial Ppartial xv^x+fracpartial Ppartial yv^y.$$
So
$$dP_mathfrakx=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright).$$
The same applies to $dQ_mathfrakx$.
The differential is now:
$$domega_mathfrakx=dP_mathfrakxdx+dQ_mathfrakxdy$$
$$=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright)_mathfrakxdx+left(fracpartial Qpartial xdx+fracpartial Qpartial ydyright)_mathfrakxdy.$$
$$=left(fracpartial Qpartial x-fracpartial Ppartial yright)_mathfrakxdxdy.$$
multivariable-calculus differential-geometry definition
asked Dec 4 '17 at 23:28
Steven Hatton
662314
In his Advanced Calculus of Several Variables, Edwards defines the differential $domega$ of a differential 1-form $omega=Pdx+Qdy$ to be
$$domegaequivleft(fracpartial Qpartial x-fracpartial Ppartial yright)dxdy.$$
But that definition seems cut from whole cloth. He gives some motivating examples before and after the definition, but they don't give me a good intuitive sense of what it really means.
He has mumbled a bit about the wedge product, but not by that name. He does not use it to derive the differential of a 1-form.
Why is $domega$ defined this way?
Or why is the wedge product used to multiply differential forms?
Or is there a way to picture either $domega$ or the wedge product?
I note that, if the wedge product has already been introduced as
$$dx^1wedge dx^1=dx^2wedge dx^2=0,$$
$$dx^1wedge dx^2=-dx^2wedge dx^1=1,$$
and
$$dx^idx^jequiv dx^iwedge dx^j$$
has been asserted, the differential of $omega$ can be derived as follows:
Start with the definition of differentiability
$$0=lim_Deltamathfrakxtomathfrak0fracDeltaomega_mathfrakxleft[Deltamathfrakxright]-domega_mathfrakxleft[Deltamathfrakxright]$$
$$=lim_Deltamathfrakxtomathfrak0fracDelta P_mathfrakxleft[Deltamathfrakxright]dx+Delta Q_mathfrakxleft[Deltamathfrakxright]dy-domega_mathfrakxleft[Deltamathfrakxright].$$
Take the constant objects $dx,dy$ outside the limits, but keep them in the same relative positions
$$0=lim_Deltamathfrakxtomathfrak0left[fracDelta P_mathfrakxleft[Deltamathfrakxright]right]dx+lim_Deltamathfrakxtomathfrak0left[fracDelta Q_mathfrakxleft[Deltamathfrakxright]right]dy-lim_Deltamathfrakxtomathfrak0left[fracdomega_mathfrakxleft[Deltamathfrakxright]right]$$
Apply the defintion of differentiability to P:
$$lim_Deltamathfrakxtomathfrak0left[fracDelta P_mathfrakxleft[Deltamathfrakxright]-fracdP_mathfrakxleft[Deltamathfrakxright]right]=0$$
$$nablaleft[Pleft[mathfrakxright]right]cdotmathfrakv=dP_mathfrakxleft[mathfrakvright]$$
$$=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright)left[mathfrakvright]$$
$$=fracpartial Ppartial xdxleft[mathfrakvright]+fracpartial Ppartial ydyleft[mathfrakvright]$$
$$=fracpartial Ppartial xv^x+fracpartial Ppartial yv^y.$$
So
$$dP_mathfrakx=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright).$$
The same applies to $dQ_mathfrakx$.
The differential is now:
$$domega_mathfrakx=dP_mathfrakxdx+dQ_mathfrakxdy$$
$$=left(fracpartial Ppartial xdx+fracpartial Ppartial ydyright)_mathfrakxdx+left(fracpartial Qpartial xdx+fracpartial Qpartial ydyright)_mathfrakxdy.$$
$$=left(fracpartial Qpartial x-fracpartial Ppartial yright)_mathfrakxdxdy.$$
multivariable-calculus differential-geometry definition
asked Dec 4 '17 at 23:28
Steven Hatton
662314
edited Aug 12 at 0:50
asked Dec 4 '17 at 23:28
Steven Hatton
662314
asked Dec 4 '17 at 23:28
Steven Hatton
662314
asked Dec 4 '17 at 23:28
Steven Hatton
662314
662314
3
These are complicated questions and they don't have short answers. One possible answer, which is a little unsatisfying, is "to make Green's theorem true."
– Qiaochu Yuan
Dec 4 '17 at 23:30
That $dx^1wedge dx^2 = 1$ looks quite wrong. "Is there a way to picture the wedge product?" It can be thought of as a parallelogram, or a disk, or a plane weighted with a scalar. See en.wikipedia.org/wiki/Bivector
– mr_e_man
Aug 12 at 4:52
I guess that the $dx^1wedge dx^2=1$ mistake is from the fact that its integral over the unit square is $1$. But this integration does not act like a scalar; the form $dx^1wedge dx^2$ is being applied to the square's tangent bivectors. The form itself is a cotangent bivector.
– mr_e_man
Aug 12 at 6:51
add a comment |Â
3
These are complicated questions and they don't have short answers. One possible answer, which is a little unsatisfying, is "to make Green's theorem true."
– Qiaochu Yuan
Dec 4 '17 at 23:30
That $dx^1wedge dx^2 = 1$ looks quite wrong. "Is there a way to picture the wedge product?" It can be thought of as a parallelogram, or a disk, or a plane weighted with a scalar. See en.wikipedia.org/wiki/Bivector
– mr_e_man
Aug 12 at 4:52
I guess that the $dx^1wedge dx^2=1$ mistake is from the fact that its integral over the unit square is $1$. But this integration does not act like a scalar; the form $dx^1wedge dx^2$ is being applied to the square's tangent bivectors. The form itself is a cotangent bivector.
– mr_e_man
Aug 12 at 6:51
3
3
These are complicated questions and they don't have short answers. One possible answer, which is a little unsatisfying, is "to make Green's theorem true."
– Qiaochu Yuan
Dec 4 '17 at 23:30
These are complicated questions and they don't have short answers. One possible answer, which is a little unsatisfying, is "to make Green's theorem true."
– Qiaochu Yuan
Dec 4 '17 at 23:30
That $dx^1wedge dx^2 = 1$ looks quite wrong. "Is there a way to picture the wedge product?" It can be thought of as a parallelogram, or a disk, or a plane weighted with a scalar. See en.wikipedia.org/wiki/Bivector
– mr_e_man
Aug 12 at 4:52
That $dx^1wedge dx^2 = 1$ looks quite wrong. "Is there a way to picture the wedge product?" It can be thought of as a parallelogram, or a disk, or a plane weighted with a scalar. See en.wikipedia.org/wiki/Bivector
– mr_e_man
Aug 12 at 4:52
I guess that the $dx^1wedge dx^2=1$ mistake is from the fact that its integral over the unit square is $1$. But this integration does not act like a scalar; the form $dx^1wedge dx^2$ is being applied to the square's tangent bivectors. The form itself is a cotangent bivector.
– mr_e_man
Aug 12 at 6:51
I guess that the $dx^1wedge dx^2=1$ mistake is from the fact that its integral over the unit square is $1$. But this integration does not act like a scalar; the form $dx^1wedge dx^2$ is being applied to the square's tangent bivectors. The form itself is a cotangent bivector.
– mr_e_man
Aug 12 at 6:51
add a comment |Â
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Upon review, I discovered that Edwards gives his excuse for the definition of the product of differential forms being the implied wedge product.
Note on notation: Edwards represents the absolute value of the Jacobian by
$$left|fracdmathfrakxdmathfrakuright|.$$
After deriving the change of variables forumula for an integral; given as
$$int_Tleft[Qright]fleft[mathfrakxright]dmathfrakx=int_Qfleft[Tleft[mathfrakuright]right]left|fracdmathfrakxdmathfrakuright|dmathfrakx,$$
Edwards infroms his reader:
This observation leads to an interpretation of the change of variables formula as the result of a “mechanical†substitution procedure. Suppose $T$ is a differentiable mapping from $uv$-space to $xy$-space. In the integral $int_Tleft[Qright]fleft[x,yright]dxdy$ we want to make the substitutions
$$dx=fracpartial xpartial udu+fracpartial xpartial vdv,dx=fracpartial ypartial udu+fracpartial ypartial vdv$$
suggested by the chain rule. In formally multiplying together these two “differential forms,†we agree to the conventions
$$dudu=dvdv=0text and dudv=-dvdu,$$
for no other reason than that they are necessary if we are to get the “right†answer.
Emphasis on no is in the original. So I will emphasize the entire phrase:
for no other reason than that they are necessary if we are to get the “right†answer.
I do not like that "justification", but my reasons should be addressed in a different post.
answered Aug 11 at 23:59
Steven Hatton
662314
I don't like that reason either, nor "differential forms" in general. There is a system called "geometric calculus" which generalizes and unifies a bunch of things, and I find it easy to understand and visualize. Its fundamental integration theorem uses the geometric product instead of the wedge product (though the wedge can be defined in terms of the geometric product). One possible objection is that this product requires (or induces) a metric, unlike the wedge. See Alan Macdonald's videos youtube.com/watch?v=-JQxOYL3vhY , or Wikipedia, or geocalc.clas.asu.edu/html/UGC.html
– mr_e_man
Aug 12 at 5:41
Differential forms are a necessary "evil" for me. They are used extensively in areas of physics that interest me. I don't believe differential forms are all that bad of an idea, per se. I just think the typical exposition is misguided, and therefore misleading. Box 2.3 on Page 63 of Gravitation by Misner, Thorne and Wheeler is a good example. It sets up a straw-man about how I supposedly learned calculus, and then tells me the newfangled way is correct, whereas the way I learned it is wrong. Well, I never met their straw-man. I will follow up on your suggestion; time permitting.
– Steven Hatton
Aug 13 at 2:04
add a comment |Â
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Upon review, I discovered that Edwards gives his excuse for the definition of the product of differential forms being the implied wedge product.
Note on notation: Edwards represents the absolute value of the Jacobian by
$$left|fracdmathfrakxdmathfrakuright|.$$
After deriving the change of variables forumula for an integral; given as
$$int_Tleft[Qright]fleft[mathfrakxright]dmathfrakx=int_Qfleft[Tleft[mathfrakuright]right]left|fracdmathfrakxdmathfrakuright|dmathfrakx,$$
Edwards infroms his reader:
This observation leads to an interpretation of the change of variables formula as the result of a “mechanical†substitution procedure. Suppose $T$ is a differentiable mapping from $uv$-space to $xy$-space. In the integral $int_Tleft[Qright]fleft[x,yright]dxdy$ we want to make the substitutions
$$dx=fracpartial xpartial udu+fracpartial xpartial vdv,dx=fracpartial ypartial udu+fracpartial ypartial vdv$$
suggested by the chain rule. In formally multiplying together these two “differential forms,†we agree to the conventions
$$dudu=dvdv=0text and dudv=-dvdu,$$
for no other reason than that they are necessary if we are to get the “right†answer.
Emphasis on no is in the original. So I will emphasize the entire phrase:
for no other reason than that they are necessary if we are to get the “right†answer.
I do not like that "justification", but my reasons should be addressed in a different post.
answered Aug 11 at 23:59
Steven Hatton
662314
I don't like that reason either, nor "differential forms" in general. There is a system called "geometric calculus" which generalizes and unifies a bunch of things, and I find it easy to understand and visualize. Its fundamental integration theorem uses the geometric product instead of the wedge product (though the wedge can be defined in terms of the geometric product). One possible objection is that this product requires (or induces) a metric, unlike the wedge. See Alan Macdonald's videos youtube.com/watch?v=-JQxOYL3vhY , or Wikipedia, or geocalc.clas.asu.edu/html/UGC.html
– mr_e_man
Aug 12 at 5:41
Differential forms are a necessary "evil" for me. They are used extensively in areas of physics that interest me. I don't believe differential forms are all that bad of an idea, per se. I just think the typical exposition is misguided, and therefore misleading. Box 2.3 on Page 63 of Gravitation by Misner, Thorne and Wheeler is a good example. It sets up a straw-man about how I supposedly learned calculus, and then tells me the newfangled way is correct, whereas the way I learned it is wrong. Well, I never met their straw-man. I will follow up on your suggestion; time permitting.
– Steven Hatton
Aug 13 at 2:04
add a comment |Â
up vote
0
down vote
Upon review, I discovered that Edwards gives his excuse for the definition of the product of differential forms being the implied wedge product.
Note on notation: Edwards represents the absolute value of the Jacobian by
$$left|fracdmathfrakxdmathfrakuright|.$$
After deriving the change of variables forumula for an integral; given as
$$int_Tleft[Qright]fleft[mathfrakxright]dmathfrakx=int_Qfleft[Tleft[mathfrakuright]right]left|fracdmathfrakxdmathfrakuright|dmathfrakx,$$
Edwards infroms his reader:
This observation leads to an interpretation of the change of variables formula as the result of a “mechanical†substitution procedure. Suppose $T$ is a differentiable mapping from $uv$-space to $xy$-space. In the integral $int_Tleft[Qright]fleft[x,yright]dxdy$ we want to make the substitutions
$$dx=fracpartial xpartial udu+fracpartial xpartial vdv,dx=fracpartial ypartial udu+fracpartial ypartial vdv$$
suggested by the chain rule. In formally multiplying together these two “differential forms,†we agree to the conventions
$$dudu=dvdv=0text and dudv=-dvdu,$$
for no other reason than that they are necessary if we are to get the “right†answer.
Emphasis on no is in the original. So I will emphasize the entire phrase:
for no other reason than that they are necessary if we are to get the “right†answer.
I do not like that "justification", but my reasons should be addressed in a different post.
answered Aug 11 at 23:59
Steven Hatton
662314
I don't like that reason either, nor "differential forms" in general. There is a system called "geometric calculus" which generalizes and unifies a bunch of things, and I find it easy to understand and visualize. Its fundamental integration theorem uses the geometric product instead of the wedge product (though the wedge can be defined in terms of the geometric product). One possible objection is that this product requires (or induces) a metric, unlike the wedge. See Alan Macdonald's videos youtube.com/watch?v=-JQxOYL3vhY , or Wikipedia, or geocalc.clas.asu.edu/html/UGC.html
– mr_e_man
Aug 12 at 5:41
Differential forms are a necessary "evil" for me. They are used extensively in areas of physics that interest me. I don't believe differential forms are all that bad of an idea, per se. I just think the typical exposition is misguided, and therefore misleading. Box 2.3 on Page 63 of Gravitation by Misner, Thorne and Wheeler is a good example. It sets up a straw-man about how I supposedly learned calculus, and then tells me the newfangled way is correct, whereas the way I learned it is wrong. Well, I never met their straw-man. I will follow up on your suggestion; time permitting.
– Steven Hatton
Aug 13 at 2:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Upon review, I discovered that Edwards gives his excuse for the definition of the product of differential forms being the implied wedge product.
Note on notation: Edwards represents the absolute value of the Jacobian by
$$left|fracdmathfrakxdmathfrakuright|.$$
After deriving the change of variables forumula for an integral; given as
$$int_Tleft[Qright]fleft[mathfrakxright]dmathfrakx=int_Qfleft[Tleft[mathfrakuright]right]left|fracdmathfrakxdmathfrakuright|dmathfrakx,$$
Edwards infroms his reader:
This observation leads to an interpretation of the change of variables formula as the result of a “mechanical†substitution procedure. Suppose $T$ is a differentiable mapping from $uv$-space to $xy$-space. In the integral $int_Tleft[Qright]fleft[x,yright]dxdy$ we want to make the substitutions
$$dx=fracpartial xpartial udu+fracpartial xpartial vdv,dx=fracpartial ypartial udu+fracpartial ypartial vdv$$
suggested by the chain rule. In formally multiplying together these two “differential forms,†we agree to the conventions
$$dudu=dvdv=0text and dudv=-dvdu,$$
for no other reason than that they are necessary if we are to get the “right†answer.
Emphasis on no is in the original. So I will emphasize the entire phrase:
for no other reason than that they are necessary if we are to get the “right†answer.
I do not like that "justification", but my reasons should be addressed in a different post.
answered Aug 11 at 23:59
Steven Hatton
662314
Upon review, I discovered that Edwards gives his excuse for the definition of the product of differential forms being the implied wedge product.
Note on notation: Edwards represents the absolute value of the Jacobian by
$$left|fracdmathfrakxdmathfrakuright|.$$
After deriving the change of variables forumula for an integral; given as
$$int_Tleft[Qright]fleft[mathfrakxright]dmathfrakx=int_Qfleft[Tleft[mathfrakuright]right]left|fracdmathfrakxdmathfrakuright|dmathfrakx,$$
Edwards infroms his reader:
This observation leads to an interpretation of the change of variables formula as the result of a “mechanical†substitution procedure. Suppose $T$ is a differentiable mapping from $uv$-space to $xy$-space. In the integral $int_Tleft[Qright]fleft[x,yright]dxdy$ we want to make the substitutions
$$dx=fracpartial xpartial udu+fracpartial xpartial vdv,dx=fracpartial ypartial udu+fracpartial ypartial vdv$$
suggested by the chain rule. In formally multiplying together these two “differential forms,†we agree to the conventions
$$dudu=dvdv=0text and dudv=-dvdu,$$
for no other reason than that they are necessary if we are to get the “right†answer.
Emphasis on no is in the original. So I will emphasize the entire phrase:
for no other reason than that they are necessary if we are to get the “right†answer.
I do not like that "justification", but my reasons should be addressed in a different post.
answered Aug 11 at 23:59
Steven Hatton
662314
answered Aug 11 at 23:59
Steven Hatton
662314
answered Aug 11 at 23:59
Steven Hatton
662314
answered Aug 11 at 23:59
Steven Hatton
662314
662314
I don't like that reason either, nor "differential forms" in general. There is a system called "geometric calculus" which generalizes and unifies a bunch of things, and I find it easy to understand and visualize. Its fundamental integration theorem uses the geometric product instead of the wedge product (though the wedge can be defined in terms of the geometric product). One possible objection is that this product requires (or induces) a metric, unlike the wedge. See Alan Macdonald's videos youtube.com/watch?v=-JQxOYL3vhY , or Wikipedia, or geocalc.clas.asu.edu/html/UGC.html
– mr_e_man
Aug 12 at 5:41
Differential forms are a necessary "evil" for me. They are used extensively in areas of physics that interest me. I don't believe differential forms are all that bad of an idea, per se. I just think the typical exposition is misguided, and therefore misleading. Box 2.3 on Page 63 of Gravitation by Misner, Thorne and Wheeler is a good example. It sets up a straw-man about how I supposedly learned calculus, and then tells me the newfangled way is correct, whereas the way I learned it is wrong. Well, I never met their straw-man. I will follow up on your suggestion; time permitting.
– Steven Hatton
Aug 13 at 2:04
add a comment |Â
I don't like that reason either, nor "differential forms" in general. There is a system called "geometric calculus" which generalizes and unifies a bunch of things, and I find it easy to understand and visualize. Its fundamental integration theorem uses the geometric product instead of the wedge product (though the wedge can be defined in terms of the geometric product). One possible objection is that this product requires (or induces) a metric, unlike the wedge. See Alan Macdonald's videos youtube.com/watch?v=-JQxOYL3vhY , or Wikipedia, or geocalc.clas.asu.edu/html/UGC.html
– mr_e_man
Aug 12 at 5:41
Differential forms are a necessary "evil" for me. They are used extensively in areas of physics that interest me. I don't believe differential forms are all that bad of an idea, per se. I just think the typical exposition is misguided, and therefore misleading. Box 2.3 on Page 63 of Gravitation by Misner, Thorne and Wheeler is a good example. It sets up a straw-man about how I supposedly learned calculus, and then tells me the newfangled way is correct, whereas the way I learned it is wrong. Well, I never met their straw-man. I will follow up on your suggestion; time permitting.
– Steven Hatton
Aug 13 at 2:04
I don't like that reason either, nor "differential forms" in general. There is a system called "geometric calculus" which generalizes and unifies a bunch of things, and I find it easy to understand and visualize. Its fundamental integration theorem uses the geometric product instead of the wedge product (though the wedge can be defined in terms of the geometric product). One possible objection is that this product requires (or induces) a metric, unlike the wedge. See Alan Macdonald's videos youtube.com/watch?v=-JQxOYL3vhY , or Wikipedia, or geocalc.clas.asu.edu/html/UGC.html
– mr_e_man
Aug 12 at 5:41
I don't like that reason either, nor "differential forms" in general. There is a system called "geometric calculus" which generalizes and unifies a bunch of things, and I find it easy to understand and visualize. Its fundamental integration theorem uses the geometric product instead of the wedge product (though the wedge can be defined in terms of the geometric product). One possible objection is that this product requires (or induces) a metric, unlike the wedge. See Alan Macdonald's videos youtube.com/watch?v=-JQxOYL3vhY , or Wikipedia, or geocalc.clas.asu.edu/html/UGC.html
– mr_e_man
Aug 12 at 5:41
Differential forms are a necessary "evil" for me. They are used extensively in areas of physics that interest me. I don't believe differential forms are all that bad of an idea, per se. I just think the typical exposition is misguided, and therefore misleading. Box 2.3 on Page 63 of Gravitation by Misner, Thorne and Wheeler is a good example. It sets up a straw-man about how I supposedly learned calculus, and then tells me the newfangled way is correct, whereas the way I learned it is wrong. Well, I never met their straw-man. I will follow up on your suggestion; time permitting.
– Steven Hatton
Aug 13 at 2:04
Differential forms are a necessary "evil" for me. They are used extensively in areas of physics that interest me. I don't believe differential forms are all that bad of an idea, per se. I just think the typical exposition is misguided, and therefore misleading. Box 2.3 on Page 63 of Gravitation by Misner, Thorne and Wheeler is a good example. It sets up a straw-man about how I supposedly learned calculus, and then tells me the newfangled way is correct, whereas the way I learned it is wrong. Well, I never met their straw-man. I will follow up on your suggestion; time permitting.
– Steven Hatton
Aug 13 at 2:04
add a comment |Â
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3
These are complicated questions and they don't have short answers. One possible answer, which is a little unsatisfying, is "to make Green's theorem true."
– Qiaochu Yuan
Dec 4 '17 at 23:30
That $dx^1wedge dx^2 = 1$ looks quite wrong. "Is there a way to picture the wedge product?" It can be thought of as a parallelogram, or a disk, or a plane weighted with a scalar. See en.wikipedia.org/wiki/Bivector
– mr_e_man
Aug 12 at 4:52
I guess that the $dx^1wedge dx^2=1$ mistake is from the fact that its integral over the unit square is $1$. But this integration does not act like a scalar; the form $dx^1wedge dx^2$ is being applied to the square's tangent bivectors. The form itself is a cotangent bivector.
– mr_e_man
Aug 12 at 6:51