How Do You Apply the Alt Operator to a Differential 2-Form in R^4?

[arthurhenry]

Suppose the standard coordinates in R^4 are x,y,x,w.
We have a differential 2-form a= z dx \wedge dy.
Trying to evaluate Alt(a)

I am trying to see this form as a bilinear form that acts on a pair of vectors so that I can apply the Alt operator formula. I am able understand the formula for the operator that makes a form alternating (i.e. The Alt operator), but I cannot apply it in this case.


I think I would have been, perhaps,be able to approach this problem one way or the other if I had found some source that can explain where these notions originate...originate in the sense: I see them in a math book and I would like to know why people needed this definition and to address what, etc. I tried reading different sources, but I am still here. (I have looked at David Bachmann's notes, they are very good, but I am still missing something)


Thank you for your time

 

[arkajad]

All you need is the precise mathematical definition of the wedge product and the the precise mathematical definition of the Alt operator. Without using these definitions you will not be able to prove anything.

You do not need for that to know where these notions originate. That is a separate subject.

 

[arthurhenry]

Thank you for responding Arkajad,

Yes, I agree. I believe -not much exaggration here-that almost all of mathematics can be done if one understood what a precise definition meant precisely what.
Having said that, I can only guess that you have never encountered the problem of not being able to apply a certain definition to the problem in hand. One does not have this problem because one does not appreciate the strenght of definitions or that he/she believes that there must be a way of solving a certain problem through some mysterious intuition or some insight.

If somebody asked me to prove that Cantor set is a perfect set, I realize that the problem is nothing more than understanding precisely what only 1 or 2 definitions involved mean and when these are satisfied. But,...would you not aggree that this might be challenging for a beginner?

Thank you

 

[Fredrik]

If \omega is a k-form and \eta is a l-form:

\omega\wedge\eta=\frac{(k+l)!}{k!l!}\operatorname{Alt}(\omega\otimes\eta)=\frac{1}{k!l!}\sum_\sigma (\operatorname{sgn}\sigma)\ ^\sigma(\omega\otimes\eta)

\omega\wedge\eta(X_1,X_2)=\frac{1}{k!l!}\sum_\sigma (\operatorname{sgn}\sigma)\omega\otimes\eta(X_{\sigma(1)},X_{\sigma(2)})=\frac{1}{k!l!}\sum_\sigma (\operatorname{sgn}\sigma)\omega(X_{\sigma(1)})\eta(X_{\sigma(2)})

You defined a=z dx \wedge dy. So...

\operatorname{Alt}(a)(X_1,X_2)=\frac{1}{2!}\sum_\sigma(\operatorname{sgn}\sigma)\ (^\sigma a)(X_1,X_2)=\frac{1}{2!}\sum_\sigma(\operatorname{sgn}\sigma)a(X_{\sigma(1)},X_{\sigma(2)})

=\frac{1}{2!}(a(X_1,X_2)-a(X_2,X_1))

a(X_1,X_2)=zdx\wedge dy(X_1,X_2)=z\frac{(1+1)!}{1!1!}\operatorname{Alt}(dx\otimes dy)(X_1,X_2)=2z\frac{1}{2!}\sum_\sigma(\operatorname{sgn}\sigma)\ ^\sigma(dx\otimes dy)(X_1,X_2)

=z\sum_\sigma(\operatorname{sgn}\sigma)dx\otimes dy(X_{\sigma(1)},X_{\sigma(2)})=z(dx\otimes dy(X_1,X_2)-dx\otimes dy(X_2,X_1))

a(X_2,X_1)=z(dx\otimes dy(X_2,X_1)-dx\otimes dy(X_1,X_2))=-a(X_1,X_2)

a(X_1,X_2)-a(X_2,X_1)=2a(X_1,X_2)

\operatorname{Alt}(a)(X_1,X_2)=a(X_1,X_2)

\operatorname{Alt}(a)=a

 

[arthurhenry]

Frederick,

I thank you for your clear (and patient) answer. This helps me, the definition is clear now.Thank you again