Prove that for a 2-form $\omega$ on $Tp\mathbb{R}^3$ there are linearly independent vectors $V_1, V_2$ s.t $\omega(V_1,V_2) = 0$.

Question

I need to better understand a proof for an exercise from Bachmans "A Geometric Approach to Differential Forms". It ask...

Exercise 2.16 - Prove that for a 2-form $\omega$ on $Tp\mathbb{R}^3$ there are linearly independent vectors $V_1, V_2$ s.t $\omega(V_1,V_2) = 0$.

When the problem was graded the grader let me know that part of my set-up was correct but I they didn't specify which part and aren't available for discussion between now and the final, and neither is the professor. Any tips or walk-thoughts would be greatly appreciated! Ok, so here's what I have.

$\omega = a(dx\wedge dy) + b(dy\wedge dz) + c(dx\wedge dz)$ we know that $a(dx\wedge dy) = a(<dx>\times<dy>)\cdot(V_1\times V_2)$

$b(dy\wedge dz) = b(<dy>\times<dz>)\cdot(V_1\times V_2)$

$c(dx\wedge dz) = c(<dx>\times<dz>)\cdot(V_1\times V_2)$
Without loss of generality let $a,b \neq 0$ and let $\lambda_1, \lambda_2$ be such that $V_1 \times V_2 = \lambda_1(<dx>\times <dy>) + \lambda_2(<dy>\times <dz>) = 0 $

Geometrically this means that we chose $V_1$ and $V_2$ which are perpendicular to the plane $\omega$ represents.

After that I got stuck and to be honest the grader seemed like he was in a hurry so I'm not sure how solid his notes on that problem are. Thanks again and good luck to everyone who has finals this and next month!

Answer 1

Here's a hint for a correct solution. If we evaluate $\omega=a\,dy\wedge dz + b\,dz\wedge dx + c\,dx\wedge dy$ (note that I've changed the orders of things) on the pair of vectors $V_1,V_2$, this will be the determinant of the $3\times 3$ matrix whose columns are $W,V_1,V_2$, where $W$ is the vector $(a,b,c)$. This in turn can be computed as $W\cdot (V_1\times V_2)$. Why? Can you solve the problem now? (This now sounds very close to what you were saying.)