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Decomposable forms

  1. Dec 12, 2006 #1
    Sorry to keep bothering, but I am preparing an exam based on Spivak's book on forms (chapter 7 of tome 1).

    I need to prove that if [itex]\dim V \le 3[/itex], then every [itex]\omega \in \Lambda^2(V)[/itex] is decomposable, where an element [itex]\omega \in \Lambda^k(V)[/itex] is decomposable if [itex]\omega =\phi_1\wedge\dots\wedge\phi_k[/itex] for some [itex]\phi_i \in V^*=\Lambda^1(V)[/itex].

    I think I must use the inner product, but I am not sure. If [itex]\omega \in \Lambda^2(V)[/itex], then

    [tex]\omega=a_{12} \phi_1\wedge \phi_2+a_{13}\phi_1\wedge\phi_2+a_{23}\phi_2\wedge \phi_3[/tex]

    I know that if [itex]\{v_1,v_2,v_3\}[/itex] are a basis of [itex]V[/itex], then
    [tex]\begin{array}{l} i_{v_1}\phi_1\wedge\phi_2\wedge\phi_3=\phi_2\wedge\phi_3 \\
    i_{v_2}\phi_1\wedge\phi_2\wedge\phi_3=-\phi_1\wedge\phi_3 \\


    [tex]\omega=(a_{12}i_{v_3}-a_{13}i_{v_2}+a_{23}i_{v_1}) \phi_1\wedge\phi_2\wedge\phi_3[/tex]

    and given the linearity


    where [itex]v=a_{21}v_1-a_{13}v_2+a_{12}v_3[/itex].

    Does that prove the result?

    Other idea I had is to express [itex]\phi_i[/itex] in terms of the base of [itex]\Lambda^1(V)[/itex], but I seem to going nowhere.
  2. jcsd
  3. Dec 12, 2006 #2
    I think that I've got it.

    Let [itex]\phi_1,\phi_2\in \Lambda^1(V)[/itex], where
    \phi_1=a_1\varphi_1+a_2\varphi_2+a_3\varphi_3 \\


    [itex]\phi_1\wedge \phi_2=(a_1b_2-a_2b_1)\varphi_1\wedge\varphi_2+ (a_1b_3-a_3b_1)\varphi_1\wedge\varphi_3+ (a_2b_3-a_3b_2)\varphi_2\wedge\varphi_3[/tex]

    So, given [itex]\omega \in \Lambda^2(V)[/itex], there are (many?) [itex]\phi_1,\phi_2\in\Lambda^1(V)[/itex] such that [itex]\omega=\phi_1\wedge\phi_2[/itex].

    What do you guys think?
    Last edited: Dec 12, 2006
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