$i_X(dxwedge dywedge dz) $ with $X=afracpartial partial x+bfracpartial partial y+cfracpartial partial z$

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I can not understand the practical way to calculate the following:

Let $X=afracpartial partial x+bfracpartial partial y+cfracpartial partial z$
show that $dxwedge dywedge dz(X,v_1,v_2)=adywedge dz-bdxwedge dz+cdxwedge dy.$

I have this:
$i_X(dxwedge dywedge dz)(v_1,v_2)=dxwedge dywedge dz(X,v_1,v_2)=dx(X,v_1,v_2)wedge (dywedge dz)(X,v_1,v_2)\
=dx(X,v_1,v_2)wedge (dy(X)dz(v_1)-dy(v_1)dz(X)+dy(X)dz(v_2)-dy(v_2)dz(X)+dy(v_1)dz(v_2)-dy(v_2)dz(v_1))\
=[dx(v_2)dy(X)dz(v_1)-dx(v_2)dy(v_1)dz(X)]+[dx(v_1)dy(X)dz(v_2)-dx(v_1)dy(v_2)dz(X)+dx(v_2)dy(v_1)dz(v_2)-dx(X)dy(v_2)dz(v_1)]\
=[bdx(v_2)dz(v_1)-cdx(v_2)dy(v_1)]+[bdx(v_1)dz(v_2)-cdx(v_1)dy(v_2)+ady(v_1)dz(v_2)-ady(v_2)dz(v_1)]$

(The only term that gives me is $ady wedge dz$ the others do not fit me with the sign to form what I want.
How would it be? I do not handle this multiplication game well.

differential-forms

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edited Jul 30 at 6:27

asked Jul 30 at 6:14

eraldcoil

386

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up vote
1
down vote

favorite

1

I can not understand the practical way to calculate the following:

Let $X=afracpartial partial x+bfracpartial partial y+cfracpartial partial z$
show that $dxwedge dywedge dz(X,v_1,v_2)=adywedge dz-bdxwedge dz+cdxwedge dy.$

I have this:
$i_X(dxwedge dywedge dz)(v_1,v_2)=dxwedge dywedge dz(X,v_1,v_2)=dx(X,v_1,v_2)wedge (dywedge dz)(X,v_1,v_2)\
=dx(X,v_1,v_2)wedge (dy(X)dz(v_1)-dy(v_1)dz(X)+dy(X)dz(v_2)-dy(v_2)dz(X)+dy(v_1)dz(v_2)-dy(v_2)dz(v_1))\
=[dx(v_2)dy(X)dz(v_1)-dx(v_2)dy(v_1)dz(X)]+[dx(v_1)dy(X)dz(v_2)-dx(v_1)dy(v_2)dz(X)+dx(v_2)dy(v_1)dz(v_2)-dx(X)dy(v_2)dz(v_1)]\
=[bdx(v_2)dz(v_1)-cdx(v_2)dy(v_1)]+[bdx(v_1)dz(v_2)-cdx(v_1)dy(v_2)+ady(v_1)dz(v_2)-ady(v_2)dz(v_1)]$

(The only term that gives me is $ady wedge dz$ the others do not fit me with the sign to form what I want.
How would it be? I do not handle this multiplication game well.

differential-forms

share|cite|improve this question

edited Jul 30 at 6:27

asked Jul 30 at 6:14

eraldcoil

386

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

I can not understand the practical way to calculate the following:

Let $X=afracpartial partial x+bfracpartial partial y+cfracpartial partial z$
show that $dxwedge dywedge dz(X,v_1,v_2)=adywedge dz-bdxwedge dz+cdxwedge dy.$

I have this:
$i_X(dxwedge dywedge dz)(v_1,v_2)=dxwedge dywedge dz(X,v_1,v_2)=dx(X,v_1,v_2)wedge (dywedge dz)(X,v_1,v_2)\
=dx(X,v_1,v_2)wedge (dy(X)dz(v_1)-dy(v_1)dz(X)+dy(X)dz(v_2)-dy(v_2)dz(X)+dy(v_1)dz(v_2)-dy(v_2)dz(v_1))\
=[dx(v_2)dy(X)dz(v_1)-dx(v_2)dy(v_1)dz(X)]+[dx(v_1)dy(X)dz(v_2)-dx(v_1)dy(v_2)dz(X)+dx(v_2)dy(v_1)dz(v_2)-dx(X)dy(v_2)dz(v_1)]\
=[bdx(v_2)dz(v_1)-cdx(v_2)dy(v_1)]+[bdx(v_1)dz(v_2)-cdx(v_1)dy(v_2)+ady(v_1)dz(v_2)-ady(v_2)dz(v_1)]$

(The only term that gives me is $ady wedge dz$ the others do not fit me with the sign to form what I want.
How would it be? I do not handle this multiplication game well.

differential-forms

share|cite|improve this question

edited Jul 30 at 6:27

asked Jul 30 at 6:14

eraldcoil

386

I can not understand the practical way to calculate the following:

Let $X=afracpartial partial x+bfracpartial partial y+cfracpartial partial z$
show that $dxwedge dywedge dz(X,v_1,v_2)=adywedge dz-bdxwedge dz+cdxwedge dy.$

I have this:
$i_X(dxwedge dywedge dz)(v_1,v_2)=dxwedge dywedge dz(X,v_1,v_2)=dx(X,v_1,v_2)wedge (dywedge dz)(X,v_1,v_2)\
=dx(X,v_1,v_2)wedge (dy(X)dz(v_1)-dy(v_1)dz(X)+dy(X)dz(v_2)-dy(v_2)dz(X)+dy(v_1)dz(v_2)-dy(v_2)dz(v_1))\
=[dx(v_2)dy(X)dz(v_1)-dx(v_2)dy(v_1)dz(X)]+[dx(v_1)dy(X)dz(v_2)-dx(v_1)dy(v_2)dz(X)+dx(v_2)dy(v_1)dz(v_2)-dx(X)dy(v_2)dz(v_1)]\
=[bdx(v_2)dz(v_1)-cdx(v_2)dy(v_1)]+[bdx(v_1)dz(v_2)-cdx(v_1)dy(v_2)+ady(v_1)dz(v_2)-ady(v_2)dz(v_1)]$

(The only term that gives me is $ady wedge dz$ the others do not fit me with the sign to form what I want.
How would it be? I do not handle this multiplication game well.

differential-forms

share|cite|improve this question

edited Jul 30 at 6:27

asked Jul 30 at 6:14

eraldcoil

386

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 6:27

edited Jul 30 at 6:27

edited Jul 30 at 6:27

asked Jul 30 at 6:14

eraldcoil

386

asked Jul 30 at 6:14

eraldcoil

386

asked Jul 30 at 6:14

eraldcoil

386

386

add a comment | 

add a comment | 

1 Answer
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accepted

Remember that
$$beginalign
dx wedge dy wedge dz
& = dx otimes dy otimes dz
+ dy otimes dz otimes dx
+ dz otimes dx otimes dy \
& - dx otimes dz otimes dy
- dy otimes dx otimes dz
- dz otimes dy otimes dx
endalign$$

Therefore,
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = dx(X) otimes dy(Y) otimes dz(Z)
+ dy(X) otimes dz(Y) otimes dx(Z) \
& + dz(X) otimes dx(Y) otimes dy(Z)
- dx(X) otimes dz(Y) otimes dy(Z) \
& - dy(X) otimes dx(Y) otimes dz(Z)
- dz(X) otimes dy(Y) otimes dx(Z)
endalign$$

In this case, $X = a partial_x + b partial_y + c partial_z,$ so
$dx(X) = a, dy(X) = b, dz(X) = c,$ which gives
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = a dy(Y) otimes dz(Z)
+ b dz(Y) otimes dx(Z) \
& + c dx(Y) otimes dy(Z)
- a dz(Y) otimes dy(Z) \
& - b dx(Y) otimes dz(Z)
- c dy(Y) otimes dx(Z) \
& = a (dy(Y) otimes dz(Z) - dz(Y) otimes dy(Z)) \
& + b (dz(Y) otimes dx(Z) - dx(Y) otimes dz(Z)) \
& + c (dx(Y) otimes dy(Z) - dy(Y) otimes dx(Z)) \
& = a (dy wedge dz)(Y, Z) + b (dz wedge dx)(Y, Z) + c (dx wedge dy)(Y, Z) \
& = (a , dy wedge dz + b , dz wedge dx + c , dx wedge dy)(Y, Z)
endalign$$

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answered Jul 30 at 7:15

md2perpe

5,7821922

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Remember that
$$beginalign
dx wedge dy wedge dz
& = dx otimes dy otimes dz
+ dy otimes dz otimes dx
+ dz otimes dx otimes dy \
& - dx otimes dz otimes dy
- dy otimes dx otimes dz
- dz otimes dy otimes dx
endalign$$

Therefore,
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = dx(X) otimes dy(Y) otimes dz(Z)
+ dy(X) otimes dz(Y) otimes dx(Z) \
& + dz(X) otimes dx(Y) otimes dy(Z)
- dx(X) otimes dz(Y) otimes dy(Z) \
& - dy(X) otimes dx(Y) otimes dz(Z)
- dz(X) otimes dy(Y) otimes dx(Z)
endalign$$

In this case, $X = a partial_x + b partial_y + c partial_z,$ so
$dx(X) = a, dy(X) = b, dz(X) = c,$ which gives
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = a dy(Y) otimes dz(Z)
+ b dz(Y) otimes dx(Z) \
& + c dx(Y) otimes dy(Z)
- a dz(Y) otimes dy(Z) \
& - b dx(Y) otimes dz(Z)
- c dy(Y) otimes dx(Z) \
& = a (dy(Y) otimes dz(Z) - dz(Y) otimes dy(Z)) \
& + b (dz(Y) otimes dx(Z) - dx(Y) otimes dz(Z)) \
& + c (dx(Y) otimes dy(Z) - dy(Y) otimes dx(Z)) \
& = a (dy wedge dz)(Y, Z) + b (dz wedge dx)(Y, Z) + c (dx wedge dy)(Y, Z) \
& = (a , dy wedge dz + b , dz wedge dx + c , dx wedge dy)(Y, Z)
endalign$$

share|cite|improve this answer

answered Jul 30 at 7:15

md2perpe

5,7821922

add a comment | 

up vote
1
down vote



accepted

Remember that
$$beginalign
dx wedge dy wedge dz
& = dx otimes dy otimes dz
+ dy otimes dz otimes dx
+ dz otimes dx otimes dy \
& - dx otimes dz otimes dy
- dy otimes dx otimes dz
- dz otimes dy otimes dx
endalign$$

Therefore,
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = dx(X) otimes dy(Y) otimes dz(Z)
+ dy(X) otimes dz(Y) otimes dx(Z) \
& + dz(X) otimes dx(Y) otimes dy(Z)
- dx(X) otimes dz(Y) otimes dy(Z) \
& - dy(X) otimes dx(Y) otimes dz(Z)
- dz(X) otimes dy(Y) otimes dx(Z)
endalign$$

In this case, $X = a partial_x + b partial_y + c partial_z,$ so
$dx(X) = a, dy(X) = b, dz(X) = c,$ which gives
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = a dy(Y) otimes dz(Z)
+ b dz(Y) otimes dx(Z) \
& + c dx(Y) otimes dy(Z)
- a dz(Y) otimes dy(Z) \
& - b dx(Y) otimes dz(Z)
- c dy(Y) otimes dx(Z) \
& = a (dy(Y) otimes dz(Z) - dz(Y) otimes dy(Z)) \
& + b (dz(Y) otimes dx(Z) - dx(Y) otimes dz(Z)) \
& + c (dx(Y) otimes dy(Z) - dy(Y) otimes dx(Z)) \
& = a (dy wedge dz)(Y, Z) + b (dz wedge dx)(Y, Z) + c (dx wedge dy)(Y, Z) \
& = (a , dy wedge dz + b , dz wedge dx + c , dx wedge dy)(Y, Z)
endalign$$

share|cite|improve this answer

answered Jul 30 at 7:15

md2perpe

5,7821922

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Remember that
$$beginalign
dx wedge dy wedge dz
& = dx otimes dy otimes dz
+ dy otimes dz otimes dx
+ dz otimes dx otimes dy \
& - dx otimes dz otimes dy
- dy otimes dx otimes dz
- dz otimes dy otimes dx
endalign$$

Therefore,
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = dx(X) otimes dy(Y) otimes dz(Z)
+ dy(X) otimes dz(Y) otimes dx(Z) \
& + dz(X) otimes dx(Y) otimes dy(Z)
- dx(X) otimes dz(Y) otimes dy(Z) \
& - dy(X) otimes dx(Y) otimes dz(Z)
- dz(X) otimes dy(Y) otimes dx(Z)
endalign$$

In this case, $X = a partial_x + b partial_y + c partial_z,$ so
$dx(X) = a, dy(X) = b, dz(X) = c,$ which gives
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = a dy(Y) otimes dz(Z)
+ b dz(Y) otimes dx(Z) \
& + c dx(Y) otimes dy(Z)
- a dz(Y) otimes dy(Z) \
& - b dx(Y) otimes dz(Z)
- c dy(Y) otimes dx(Z) \
& = a (dy(Y) otimes dz(Z) - dz(Y) otimes dy(Z)) \
& + b (dz(Y) otimes dx(Z) - dx(Y) otimes dz(Z)) \
& + c (dx(Y) otimes dy(Z) - dy(Y) otimes dx(Z)) \
& = a (dy wedge dz)(Y, Z) + b (dz wedge dx)(Y, Z) + c (dx wedge dy)(Y, Z) \
& = (a , dy wedge dz + b , dz wedge dx + c , dx wedge dy)(Y, Z)
endalign$$

share|cite|improve this answer

answered Jul 30 at 7:15

md2perpe

5,7821922

Remember that
$$beginalign
dx wedge dy wedge dz
& = dx otimes dy otimes dz
+ dy otimes dz otimes dx
+ dz otimes dx otimes dy \
& - dx otimes dz otimes dy
- dy otimes dx otimes dz
- dz otimes dy otimes dx
endalign$$

Therefore,
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = dx(X) otimes dy(Y) otimes dz(Z)
+ dy(X) otimes dz(Y) otimes dx(Z) \
& + dz(X) otimes dx(Y) otimes dy(Z)
- dx(X) otimes dz(Y) otimes dy(Z) \
& - dy(X) otimes dx(Y) otimes dz(Z)
- dz(X) otimes dy(Y) otimes dx(Z)
endalign$$

In this case, $X = a partial_x + b partial_y + c partial_z,$ so
$dx(X) = a, dy(X) = b, dz(X) = c,$ which gives
$$beginalign
(dx wedge dy wedge dz)(X,Y,Z)
& = a dy(Y) otimes dz(Z)
+ b dz(Y) otimes dx(Z) \
& + c dx(Y) otimes dy(Z)
- a dz(Y) otimes dy(Z) \
& - b dx(Y) otimes dz(Z)
- c dy(Y) otimes dx(Z) \
& = a (dy(Y) otimes dz(Z) - dz(Y) otimes dy(Z)) \
& + b (dz(Y) otimes dx(Z) - dx(Y) otimes dz(Z)) \
& + c (dx(Y) otimes dy(Z) - dy(Y) otimes dx(Z)) \
& = a (dy wedge dz)(Y, Z) + b (dz wedge dx)(Y, Z) + c (dx wedge dy)(Y, Z) \
& = (a , dy wedge dz + b , dz wedge dx + c , dx wedge dy)(Y, Z)
endalign$$

share|cite|improve this answer

answered Jul 30 at 7:15

md2perpe

5,7821922

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 7:15

md2perpe

5,7821922

answered Jul 30 at 7:15

md2perpe

5,7821922

answered Jul 30 at 7:15

md2perpe

5,7821922

5,7821922

add a comment | 

add a comment | 

 

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