Bilinear forms and wedge products

[homology]

Hey folks,

I'm reading "Symmetry in Mechanics" by S. Singer and I'm stuck on an exercise. It asks to find an antisymmetric bilinear form on R^4 that cannot be written as a wedge product of two covectors.

Here are my thoughts thus far: on R^2 its trivial to show that every antisymmetric bilinear form (ABF) is the product of two covectors. I just let F be an ABF and put in two vectors, v & w and expanded them in terms of a basis e_i. I get something like:

F(e_1,e_2)(v_1w_2-v_2w_1) which easily maps into F(e_1,e_2)dx\wedge dy and you can split up the wedge anyway you'd like to, to make two covectors whose wedge gives the same result as F.

Now on R^3 are a little more complicated and I have a question about that as well. If I look at F(v,w) where v,w\in R^3 and expand it in a basis e_i I get what you'd expect, a cross product looking expression. Now if I also take the wedge product of two 1-forms on R^3, say \alpha,\beta where \alpha = \alpha_idx^i and \beta=\beta_idx^i I get a 2-form with the coefficients you'd expect (they look like the components from a cross product).

Now the expansion of the ABF, F, using the basis e_igives me that cross product looking thing, but with some additional constants which I denote as:

F(e_i,e_j)=F_{ij}. If I wish to take that ABF and associate it to a product of covectors I thought I could just equate coeffcients and solve. If I do that I get the following system:

\alpha_1\beta_2-\alpha_2\beta_1=F_{12}
\alpha_2\beta_3-\alpha_3\beta_2=F_{23}
\alpha_1\beta_3-\alpha_3\beta_1=F_{13}
If i suppose that I pick my \alpha_k first I get a matrix that's inconsistent. But of course I'm doing something stupid here because you're supposed to be able to associate ABF with wedge products of covectors in R^3.

So as you can see, I'm lost. I'd appreciate some direction.

Many thanks,

Kevin

 

[mathwonk]

this is discussed in the l;ittle book once read communaly here by david bachman, possibly still available free or at least the running commentray here on all the exercieses. ask tom mattson.

 

[homology]

Thanks Math Wonk. I've read it and see now an example of a 2-form which is not a wedge of two covectors and I also see how I had gone awry in my method of associating 2-forms to wedges.

One last point, it seems as though 2-forms are "identical" to antisymmetric bilinear forms?

Thanks for your help,

Kevin

 

[explain]

Thanks Math Wonk. I've read it and see now an example of a 2-form which is not a wedge of two covectors and I also see how I had gone awry in my method of associating 2-forms to wedges.

One last point, it seems as though 2-forms are "identical" to antisymmetric bilinear forms?

Thanks for your help,

Kevin

Yes, 2-forms means the same as antisymmetric bilinear forms; 3-forms means antisymmetric trilinear forms, etc.

As for the question of representing a 2-form as a wedge product of two 1-forms. As far as I understand, one can make a general statement that it is sufficient to write n/2 wedge products to represent a 2-form in n-dimensional space. So in four dimensions any 2-form can be written as a\wedge b+c\wedge d. But I can't remember how this is proved. I seem to remember that this is called "Darboux theory" but I'm not sure any more.

 

[mathwonk]

the phrase "n-form" has more than one meaning depending on the author.

to me it means a field of antisymmetric multilinear functions, but as defined above and in bachman, the usage is for just one of them.