Wedge product of a 2-form with a 1-form

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
Let ##\omega## be 2-form and ##\tau## 1-form on ##R^3## If X,Y,Z are vector fields on a manifold,find a formula for ##(\omega\bigwedge\tau)(X,Y,Z)## in terms of the values of ##\omega## and ##\tau ## on the vector fields X,Y,Z.
I have known how to deal with only one vector field.But there are three vector fields,I have no idea.
 
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  • #2
GR191511 said:
I have known how to deal with only one vector field.
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.
 
  • #3
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.
Thank you.
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
GR191511 said:
Thank you.
I know ##\omega (X) = (a_idx^i)(b^j\frac{\partial }{\partial x^j})=a_ib^i##
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?
 
  • #5
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?
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.
 
  • #6
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.
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.
 
  • #7
Orodruin 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.
Thanks.But ##v_1,v_2,v_3## is vector from ##R^1## while X,Y,Z are three vector fields
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
GR191511 said:
But v1,v2,v3 is vector from
No, they are not.
 
  • #9
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.
Is your result fully anti-symmetric?
 
  • #10
Orodruin said:
Is your result fully anti-symmetric?
The question don't mention that. I'm not sure about it either.
 
  • #11
GR191511 said:
The question don't mention that. I'm not sure about it either.
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.
 
  • #12
Orodruin 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.
Is that "alternating n-linear function on a vector space"? I have seen it...it confused me
 
  • #14

1. What is the wedge product of a 2-form with a 1-form?

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.

2. How is the wedge product of a 2-form with a 1-form calculated?

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.

3. What is the purpose of the wedge product in mathematics?

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.

4. What are some properties of the wedge product?

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.

5. How is the wedge product related to the cross product?

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.

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