Proving Decomposability of Forms in Spivak's Book (Vol. 1, Chap. 7)

[AiRAVATA]

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 $\dim V \le 3$, then every $\omega \in \Lambda^2(V)$ is decomposable, where an element $\omega \in \Lambda^k(V)$ is decomposable if $\omega =\phi_1\wedge\dots\wedge\phi_k$ for some $\phi_i \in V^*=\Lambda^1(V)$.

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

$$\omega=a_{12} \phi_1\wedge \phi_2+a_{13}\phi_1\wedge\phi_2+a_{23}\phi_2\wedge \phi_3$$

I know that if $\{v_1,v_2,v_3\}$ are a basis of $V$, then
$$\begin{array}{l} i_{v_1}\phi_1\wedge\phi_2\wedge\phi_3=\phi_2\wedge\phi_3 \\ i_{v_2}\phi_1\wedge\phi_2\wedge\phi_3=-\phi_1\wedge\phi_3 \\ i_{v_3}\phi_1\wedge\phi_2\wedge\phi_3=\phi_1\wedge\phi_2 \end{array}$$

so

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

and given the linearity

$$\omega=i_v\phi_1\wedge\phi_2\wedge\phi_3$$

where $v=a_{21}v_1-a_{13}v_2+a_{12}v_3$.

Does that prove the result?

Other idea I had is to express $\phi_i$ in terms of the base of $\Lambda^1(V)$, but I seem to going nowhere.

 

[AiRAVATA]

I think that I've got it.

Let $\phi_1,\phi_2\in \Lambda^1(V)$, where
$$\begin{array}{l} \phi_1=a_1\varphi_1+a_2\varphi_2+a_3\varphi_3 \\ \phi_2=b_1\varphi_1+b_2\varphi_2+b_3\varphi_3 \end{array}$$

Then

[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 $\omega \in \Lambda^2(V)$, there are (many?) $\phi_1,\phi_2\in\Lambda^1(V)$ such that $\omega=\phi_1\wedge\phi_2$.

What do you guys think?

 

[tacman]

First of all, it is not a bother to ask questions and seek clarification. It shows that you are actively engaged in your studying and seeking to understand the material.

Now, to answer your question, your approach seems to be on the right track. The idea of using the inner product to prove decomposability is a common one, and it can be used in this case as well.

Your first step is correct in expressing \omega as a linear combination of \phi_i. Now, to prove decomposability, we need to show that there exists a \phi_i such that \omega=\phi_1\wedge\dots\wedge\phi_k.

To do this, we can use the Hodge star operator, denoted by * which maps k-forms to (n-k)-forms, where n is the dimension of the vector space V. In this case, since we are dealing with \Lambda^2(V), which is a 2-form, the Hodge star operator maps it to a 0-form (a scalar).

In general, for a 2-form \omega, the Hodge star operator is defined as *(\omega)=\frac{1}{2}\omega_{ij}\epsilon^{ij}, where \omega_{ij} are the components of \omega and \epsilon^{ij} is the Levi-Civita symbol.

Now, we can use the fact that for any \phi \in V^*, we have *(\phi\wedge\phi)=\phi\wedge*\phi=\phi\wedge\phi, since * is an involution (i.e., *^2=1).

Thus, we have *(\omega)=\frac{1}{2}(a_{12}\phi_1\wedge\phi_2+a_{13}\phi_1\wedge\phi_3+a_{23}\phi_2\wedge\phi_3)=\frac{1}{2}(\omega+\omega)=\omega.

This means that \omega is self-dual, i.e., *(\omega)=\omega.

Now, since \omega is self-dual, we can write it as \omega=\phi\wedge*\phi for some \phi \in V^*. But since we are dealing with a 2-form, we know that *(\phi)=\phi, so we can write \omega=\phi\