- #1

GR191511

- 76

- 6

I have known how to deal with only one vector field.But there are three vector fields,I have no idea.

You should upgrade or use an alternative browser.

In summary, the conversation discusses the calculation of the wedge product of a 2-form and a 1-form on a manifold, when given three vector fields. The formula for this calculation is provided as well as a discussion on the anti-symmetry of n-forms. The conversation concludes with a suggestion to seek additional resources on this topic.

- #1

GR191511

- 76

- 6

I have known how to deal with only one vector field.But there are three vector fields,I have no idea.

Physics news on Phys.org

- #2

- 22,076

- 13,551

Please expand on what you mean by this. The wedge product of a 2-form and a 1-form is a 3-form and so must take 3 vector arguments.GR191511 said:I have known how to deal with only one vector field.

- #3

GR191511

- 76

- 6

Thank you.Orodruin said:Please expand on what you mean by this. The wedge product of a 2-form and a 1-form is a 3-form and so must take 3 vector arguments.

I know ##\omega (X) = (a_idx^i)(b^j\frac{\partial }{\partial x^j})=a_ib^i## but ##(\omega\bigwedge\tau)(X,Y,Z)=?##

- #4

- 22,076

- 13,551

In the case you have presented in the OP, ##\omega## is a 2-form so this is not true. What you have written here is true if it is a 1-form. It feels like you may have skipped some reading regarding how to go from 1-forms to higher order (0,n) tensors in general and n-forms in particular?GR191511 said:Thank you.

I know ##\omega (X) = (a_idx^i)(b^j\frac{\partial }{\partial x^j})=a_ib^i##

- #5

GR191511

- 76

- 6

Thank you!I'm learning from 《An introduction to manifolds》Loring W.Tu...All I know now is that:Orodruin said:In the case you have presented in the OP, ##\omega## is a 2-form so this is not true. What you have written here is true if it is a 1-form. It feels like you may have skipped some reading regarding how to go from 1-forms to higher order (0,n) tensors in general and n-forms in particular?

##\omega\bigwedge\tau = a_Ib_Jdx^I\bigwedge dx^J## and ##(f\bigwedge g)(v_1,v_2,v_3)=f(v_1,v_2)g(v_3)-f(v_1,v_3)g(v_2)+f(v_2,v_3)g(v_1)## ...I don't know what I should do next.

- #6

- 22,076

- 13,551

The first of those relations is the wedge between two 1-forms. The second describes a two-form ##f## with a one-form ##g## - which also happens to be the case you are asked about.GR191511 said:Thank you!I'm learning from 《An introduction to manifolds》Loring W.Tu...All I know now is that:

##\omega\bigwedge\tau = a_Ib_Jdx^I\bigwedge dx^J## and ##(f\bigwedge g)(v_1,v_2,v_3)=f(v_1,v_2)g(v_3)-f(v_1,v_3)g(v_2)+f(v_2,v_3)g(v_1)## ...I don't know what I should do next.

- #7

GR191511

- 76

- 6

Thanks.But ##v_1,v_2,v_3## is vector from ##R^1## while X,Y,Z are three vector fieldsOrodruin said:The first of those relations is the wedge between two 1-forms. The second describes a two-form ##f## with a one-form ##g## - which also happens to be the case you are asked about.

Maybe the answer is##(\omega\bigwedge\tau)(X,Y,Z)=\omega(X,Y)\tau(Z)-\omega(X,Z)\tau(Y)+\omega(Y,Z)\tau(X)##But It always feels like something is wrong.

- #8

- 22,076

- 13,551

No, they are not.GR191511 said:But v1,v2,v3 is vector from

- #9

- 22,076

- 13,551

Is your result fully anti-symmetric?GR191511 said:Maybe the answer is(ω⋀τ)(X,Y,Z)=ω(X,Y)τ(Z)−ω(X,Z)τ(Y)+ω(Y,Z)τ(X)But It always feels like something is wrong.

- #10

GR191511

- 76

- 6

The question don't mention that. I'm not sure about it either.Orodruin said:Is your result fully anti-symmetric?

- #11

- 22,076

- 13,551

Again, this seems to imply that you are missing information fundamental to n-forms (namely that they are fully anti-symmetric (0,n) tensors). If this is not covered by your textbook, I would suggest looking up another textbook that does.GR191511 said:The question don't mention that. I'm not sure about it either.

- #12

GR191511

- 76

- 6

Is that "alternating n-linear function on a vector space"? I have seen it...it confused meOrodruin said:Again, this seems to imply that you are missing information fundamental to n-forms (namely that they are fully anti-symmetric (0,n) tensors). If this is not covered by your textbook, I would suggest looking up another textbook that does.

- #13

- 22,076

- 13,551

Yes, that is just another way of putting it. See https://en.wikipedia.org/wiki/Alternating_multilinear_mapGR191511 said:Is that "alternating n-linear function on a vector space"? I have seen it...it confused me

- #14

GR191511

- 76

- 6

...I'm in China,I can't open wikipedia.Thank you all the same.Orodruin said:Yes, that is just another way of putting it. See https://en.wikipedia.org/wiki/Alternating_multilinear_map

The wedge product of a 2-form with a 1-form is a mathematical operation that combines two differential forms to create a new differential form. It is denoted by the symbol ∧ and is also known as the exterior product.

The wedge product of a 2-form with a 1-form is calculated by multiplying the two forms together and then taking the anti-symmetric part of the resulting product. This means that the order of the factors does not matter and any terms that are repeated with a different sign are cancelled out.

The wedge product is used in mathematics to define the exterior algebra, a mathematical structure that generalizes the concept of vectors and matrices to higher dimensions. It is also used in differential geometry to describe geometric objects such as curves and surfaces.

The wedge product has several important properties, including associativity, distributivity, and the fact that it is anti-commutative. It also satisfies the graded-commutative property, which means that the product of two forms of different degrees is equal to the product of their degrees in reverse order.

The wedge product is related to the cross product in three dimensions, but it is a more general operation that can be applied to any number of dimensions. In three dimensions, the cross product can be thought of as a special case of the wedge product, where the resulting form is a 3-form. However, in higher dimensions, the wedge product is a more powerful tool for describing geometric objects.

- Replies
- 13

- Views
- 2K

- Replies
- 13

- Views
- 1K

- Replies
- 73

- Views
- 3K

- Replies
- 1

- Views
- 402

- Replies
- 2

- Views
- 2K

- Replies
- 6

- Views
- 919

- Replies
- 3

- Views
- 306

- Replies
- 4

- Views
- 2K

- Replies
- 7

- Views
- 2K

- Replies
- 4

- Views
- 2K

Share: