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Bilinear forms and wedge products

  1. May 3, 2007 #1
    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 [tex]R^4[/tex] that cannot be written as a wedge product of two covectors.

    Here are my thoughts thus far: on [tex]R^2[/tex] 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 [tex]e_i[/tex]. I get something like:

    [tex]F(e_1,e_2)(v_1w_2-v_2w_1)[/tex] which easily maps into [tex]F(e_1,e_2)dx\wedge dy[/tex] 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 [tex]R^3[/tex] are a little more complicated and I have a question about that as well. If I look at [tex]F(v,w)[/tex] where [tex]v,w\in R^3[/tex] and expand it in a basis [tex]e_i[/tex] I get what you'd expect, a cross product looking expression. Now if I also take the wedge product of two 1-forms on [tex]R^3[/tex], say [tex]\alpha,\beta[/tex] where [tex]\alpha = \alpha_idx^i[/tex] and [tex]\beta=\beta_idx^i[/tex] 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 [tex]e_i[/tex]gives me that cross product looking thing, but with some additional constants which I denote as:

    [tex]F(e_i,e_j)=F_{ij}[/tex]. 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:

    If i suppose that I pick my [tex]\alpha_k[/tex] 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 [tex]R^3[/tex].

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

    Many thanks,

  2. jcsd
  3. May 3, 2007 #2


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    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.
  4. May 3, 2007 #3
    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,

  5. May 8, 2007 #4
    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 [tex]a\wedge b+c\wedge d[/tex]. 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.
  6. May 13, 2007 #5


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    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.
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