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

[GR191511]

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.

 

[Orodruin]

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.

 

[GR191511]

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 $(\omega\bigwedge\tau)(X,Y,Z)=?$

 

[Orodruin]

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?

 

[GR191511]

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.

 

[Orodruin]

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.

 

[GR191511]

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.

 

[Orodruin]

But v1,v2,v3 is vector from

No, they are not.

 

[Orodruin]

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?

 

[GR191511]

Is your result fully anti-symmetric?

The question don't mention that. I'm not sure about it either.

 

[Orodruin]

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.

 

[GR191511]

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

 

[Orodruin]

Is that "alternating n-linear function on a vector space"? I have seen it...it confused me

Yes, that is just another way of putting it. See https://en.wikipedia.org/wiki/Alternating_multilinear_map

 

[GR191511]

Yes, that is just another way of putting it. See https://en.wikipedia.org/wiki/Alternating_multilinear_map

...I'm in China,I can't open wikipedia.Thank you all the same.