Question with differential form : What mean “for $n=3$ the exterior differentiation gives again the curling operation�
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Q1) Let $n=3$ and let $$omega =F_1dx_1+F_2dx_2+f_3dx_3.$$ Then $$d omega =K_1 dx_2wedge dx_3+K_2 dx_3wedge dx_1+K_3 dx_1wedge dx_2,$$
where $(K_1,K_2,K_3)=operatornameCurl(F)$. Therefore $domega $ gives again the operation of the curling.
What does it mean ? I don't really understand. Because the curling is indeed a differential operator but is not a differential form.
Q2) Same if $omega $ is a $p-$form we have for $fin mathcal C^1$ that $$d(fomega )=dfwedge omega +f(domega ),$$
which is equivalent to $$nabla (fg)=fnabla g+gnabla f$$
if $gin mathcal C^1$. Again, le gradient is a differential operator, what is the connection with differential form ?
derivatives differential-topology differential-forms
edited Aug 26 at 12:23
Bernard
111k635102
asked Aug 26 at 12:16
MathBeginner
732312
add a comment |Â
up vote
0
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Q1) Let $n=3$ and let $$omega =F_1dx_1+F_2dx_2+f_3dx_3.$$ Then $$d omega =K_1 dx_2wedge dx_3+K_2 dx_3wedge dx_1+K_3 dx_1wedge dx_2,$$
where $(K_1,K_2,K_3)=operatornameCurl(F)$. Therefore $domega $ gives again the operation of the curling.
What does it mean ? I don't really understand. Because the curling is indeed a differential operator but is not a differential form.
Q2) Same if $omega $ is a $p-$form we have for $fin mathcal C^1$ that $$d(fomega )=dfwedge omega +f(domega ),$$
which is equivalent to $$nabla (fg)=fnabla g+gnabla f$$
if $gin mathcal C^1$. Again, le gradient is a differential operator, what is the connection with differential form ?
derivatives differential-topology differential-forms
edited Aug 26 at 12:23
Bernard
111k635102
asked Aug 26 at 12:16
MathBeginner
732312
If your recipe is to turn a vector field into a $1$-form and vice-versa, then the missing link is to take the Hodge star of $domega$, and this turns it back into a $1$-form to which the recipe applies. In particular, $$star(domega) = K_1,dx_1 + K_2,dx_2 + K_3,dx_3.$$
– Ted Shifrin
Aug 28 at 21:31
No to Q2: The gradient statement is $d(fg)$. Your statement is analogous to the curl of $fvec F$.
– Ted Shifrin
Aug 29 at 6:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Q1) Let $n=3$ and let $$omega =F_1dx_1+F_2dx_2+f_3dx_3.$$ Then $$d omega =K_1 dx_2wedge dx_3+K_2 dx_3wedge dx_1+K_3 dx_1wedge dx_2,$$
where $(K_1,K_2,K_3)=operatornameCurl(F)$. Therefore $domega $ gives again the operation of the curling.
What does it mean ? I don't really understand. Because the curling is indeed a differential operator but is not a differential form.
Q2) Same if $omega $ is a $p-$form we have for $fin mathcal C^1$ that $$d(fomega )=dfwedge omega +f(domega ),$$
which is equivalent to $$nabla (fg)=fnabla g+gnabla f$$
if $gin mathcal C^1$. Again, le gradient is a differential operator, what is the connection with differential form ?
derivatives differential-topology differential-forms
edited Aug 26 at 12:23
Bernard
111k635102
asked Aug 26 at 12:16
MathBeginner
732312
Q1) Let $n=3$ and let $$omega =F_1dx_1+F_2dx_2+f_3dx_3.$$ Then $$d omega =K_1 dx_2wedge dx_3+K_2 dx_3wedge dx_1+K_3 dx_1wedge dx_2,$$
where $(K_1,K_2,K_3)=operatornameCurl(F)$. Therefore $domega $ gives again the operation of the curling.
What does it mean ? I don't really understand. Because the curling is indeed a differential operator but is not a differential form.
Q2) Same if $omega $ is a $p-$form we have for $fin mathcal C^1$ that $$d(fomega )=dfwedge omega +f(domega ),$$
which is equivalent to $$nabla (fg)=fnabla g+gnabla f$$
if $gin mathcal C^1$. Again, le gradient is a differential operator, what is the connection with differential form ?
derivatives differential-topology differential-forms
edited Aug 26 at 12:23
Bernard
111k635102
asked Aug 26 at 12:16
MathBeginner
732312
edited Aug 26 at 12:23
Bernard
111k635102
edited Aug 26 at 12:23
Bernard
111k635102
edited Aug 26 at 12:23
Bernard
111k635102
111k635102
asked Aug 26 at 12:16
MathBeginner
732312
asked Aug 26 at 12:16
MathBeginner
732312
asked Aug 26 at 12:16
MathBeginner
732312
732312
If your recipe is to turn a vector field into a $1$-form and vice-versa, then the missing link is to take the Hodge star of $domega$, and this turns it back into a $1$-form to which the recipe applies. In particular, $$star(domega) = K_1,dx_1 + K_2,dx_2 + K_3,dx_3.$$
– Ted Shifrin
Aug 28 at 21:31
No to Q2: The gradient statement is $d(fg)$. Your statement is analogous to the curl of $fvec F$.
– Ted Shifrin
Aug 29 at 6:21
add a comment |Â
If your recipe is to turn a vector field into a $1$-form and vice-versa, then the missing link is to take the Hodge star of $domega$, and this turns it back into a $1$-form to which the recipe applies. In particular, $$star(domega) = K_1,dx_1 + K_2,dx_2 + K_3,dx_3.$$
– Ted Shifrin
Aug 28 at 21:31
No to Q2: The gradient statement is $d(fg)$. Your statement is analogous to the curl of $fvec F$.
– Ted Shifrin
Aug 29 at 6:21
If your recipe is to turn a vector field into a $1$-form and vice-versa, then the missing link is to take the Hodge star of $domega$, and this turns it back into a $1$-form to which the recipe applies. In particular, $$star(domega) = K_1,dx_1 + K_2,dx_2 + K_3,dx_3.$$
– Ted Shifrin
Aug 28 at 21:31
If your recipe is to turn a vector field into a $1$-form and vice-versa, then the missing link is to take the Hodge star of $domega$, and this turns it back into a $1$-form to which the recipe applies. In particular, $$star(domega) = K_1,dx_1 + K_2,dx_2 + K_3,dx_3.$$
– Ted Shifrin
Aug 28 at 21:31
No to Q2: The gradient statement is $d(fg)$. Your statement is analogous to the curl of $fvec F$.
– Ted Shifrin
Aug 29 at 6:21
No to Q2: The gradient statement is $d(fg)$. Your statement is analogous to the curl of $fvec F$.
– Ted Shifrin
Aug 29 at 6:21
add a comment |Â
1 Answer
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Well, you've already written the answer yourself: if you take the coefficients of the differential forms $omega$ and $domega$ and build vectors $F$ and $K$ out of them, then $K$ is the curl of $F$.
answered Aug 26 at 14:44
Hans Lundmark
33.4k564109
ok, but $domega $ will not be a curl (it won't have the same properties, does it ?) May be differential form generalize all differential operator ?
– MathBeginner
Aug 26 at 15:39
I don't think I really understand what you mean, but $domega$ is a two-form and the curl of a vector field is a (pseudo)vector field, so that's a clearly property where they differ...
– Hans Lundmark
Aug 26 at 15:53
@HansLundmark: The "pseudo" is exactly registering the fact that it's really coming from the (Hodge star) of a $2$-form. :)
– Ted Shifrin
Aug 28 at 21:30
@TedShifrin: True! And certainly they are equivalent object, even if not strictly the same. I was just trying to figure out what the comment about “not being a curl†was supposed to mean.
– Hans Lundmark
Aug 29 at 6:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Well, you've already written the answer yourself: if you take the coefficients of the differential forms $omega$ and $domega$ and build vectors $F$ and $K$ out of them, then $K$ is the curl of $F$.
answered Aug 26 at 14:44
Hans Lundmark
33.4k564109
ok, but $domega $ will not be a curl (it won't have the same properties, does it ?) May be differential form generalize all differential operator ?
– MathBeginner
Aug 26 at 15:39
I don't think I really understand what you mean, but $domega$ is a two-form and the curl of a vector field is a (pseudo)vector field, so that's a clearly property where they differ...
– Hans Lundmark
Aug 26 at 15:53
@HansLundmark: The "pseudo" is exactly registering the fact that it's really coming from the (Hodge star) of a $2$-form. :)
– Ted Shifrin
Aug 28 at 21:30
@TedShifrin: True! And certainly they are equivalent object, even if not strictly the same. I was just trying to figure out what the comment about “not being a curl†was supposed to mean.
– Hans Lundmark
Aug 29 at 6:03
add a comment |Â
up vote
1
down vote
Well, you've already written the answer yourself: if you take the coefficients of the differential forms $omega$ and $domega$ and build vectors $F$ and $K$ out of them, then $K$ is the curl of $F$.
answered Aug 26 at 14:44
Hans Lundmark
33.4k564109
ok, but $domega $ will not be a curl (it won't have the same properties, does it ?) May be differential form generalize all differential operator ?
– MathBeginner
Aug 26 at 15:39
I don't think I really understand what you mean, but $domega$ is a two-form and the curl of a vector field is a (pseudo)vector field, so that's a clearly property where they differ...
– Hans Lundmark
Aug 26 at 15:53
@HansLundmark: The "pseudo" is exactly registering the fact that it's really coming from the (Hodge star) of a $2$-form. :)
– Ted Shifrin
Aug 28 at 21:30
@TedShifrin: True! And certainly they are equivalent object, even if not strictly the same. I was just trying to figure out what the comment about “not being a curl†was supposed to mean.
– Hans Lundmark
Aug 29 at 6:03
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Well, you've already written the answer yourself: if you take the coefficients of the differential forms $omega$ and $domega$ and build vectors $F$ and $K$ out of them, then $K$ is the curl of $F$.
answered Aug 26 at 14:44
Hans Lundmark
33.4k564109
Well, you've already written the answer yourself: if you take the coefficients of the differential forms $omega$ and $domega$ and build vectors $F$ and $K$ out of them, then $K$ is the curl of $F$.
answered Aug 26 at 14:44
Hans Lundmark
33.4k564109
answered Aug 26 at 14:44
Hans Lundmark
33.4k564109
answered Aug 26 at 14:44
Hans Lundmark
33.4k564109
answered Aug 26 at 14:44
Hans Lundmark
33.4k564109
33.4k564109
ok, but $domega $ will not be a curl (it won't have the same properties, does it ?) May be differential form generalize all differential operator ?
– MathBeginner
Aug 26 at 15:39
I don't think I really understand what you mean, but $domega$ is a two-form and the curl of a vector field is a (pseudo)vector field, so that's a clearly property where they differ...
– Hans Lundmark
Aug 26 at 15:53
@HansLundmark: The "pseudo" is exactly registering the fact that it's really coming from the (Hodge star) of a $2$-form. :)
– Ted Shifrin
Aug 28 at 21:30
@TedShifrin: True! And certainly they are equivalent object, even if not strictly the same. I was just trying to figure out what the comment about “not being a curl†was supposed to mean.
– Hans Lundmark
Aug 29 at 6:03
add a comment |Â
ok, but $domega $ will not be a curl (it won't have the same properties, does it ?) May be differential form generalize all differential operator ?
– MathBeginner
Aug 26 at 15:39
I don't think I really understand what you mean, but $domega$ is a two-form and the curl of a vector field is a (pseudo)vector field, so that's a clearly property where they differ...
– Hans Lundmark
Aug 26 at 15:53
@HansLundmark: The "pseudo" is exactly registering the fact that it's really coming from the (Hodge star) of a $2$-form. :)
– Ted Shifrin
Aug 28 at 21:30
@TedShifrin: True! And certainly they are equivalent object, even if not strictly the same. I was just trying to figure out what the comment about “not being a curl†was supposed to mean.
– Hans Lundmark
Aug 29 at 6:03
ok, but $domega $ will not be a curl (it won't have the same properties, does it ?) May be differential form generalize all differential operator ?
– MathBeginner
Aug 26 at 15:39
ok, but $domega $ will not be a curl (it won't have the same properties, does it ?) May be differential form generalize all differential operator ?
– MathBeginner
Aug 26 at 15:39
I don't think I really understand what you mean, but $domega$ is a two-form and the curl of a vector field is a (pseudo)vector field, so that's a clearly property where they differ...
– Hans Lundmark
Aug 26 at 15:53
I don't think I really understand what you mean, but $domega$ is a two-form and the curl of a vector field is a (pseudo)vector field, so that's a clearly property where they differ...
– Hans Lundmark
Aug 26 at 15:53
@HansLundmark: The "pseudo" is exactly registering the fact that it's really coming from the (Hodge star) of a $2$-form. :)
– Ted Shifrin
Aug 28 at 21:30
@HansLundmark: The "pseudo" is exactly registering the fact that it's really coming from the (Hodge star) of a $2$-form. :)
– Ted Shifrin
Aug 28 at 21:30
@TedShifrin: True! And certainly they are equivalent object, even if not strictly the same. I was just trying to figure out what the comment about “not being a curl†was supposed to mean.
– Hans Lundmark
Aug 29 at 6:03
@TedShifrin: True! And certainly they are equivalent object, even if not strictly the same. I was just trying to figure out what the comment about “not being a curl†was supposed to mean.
– Hans Lundmark
Aug 29 at 6:03
add a comment |Â
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If your recipe is to turn a vector field into a $1$-form and vice-versa, then the missing link is to take the Hodge star of $domega$, and this turns it back into a $1$-form to which the recipe applies. In particular, $$star(domega) = K_1,dx_1 + K_2,dx_2 + K_3,dx_3.$$
– Ted Shifrin
Aug 28 at 21:31
No to Q2: The gradient statement is $d(fg)$. Your statement is analogous to the curl of $fvec F$.
– Ted Shifrin
Aug 29 at 6:21