$ω = dx_1 ∧ dy_1 + dx_2 ∧ dy_2 + dx_3 ∧ dy_3$. Prove: $∧^1 ( \mathbb{R}^6 ) \ni η \to η ∧ ω ∧ ω \in ∧^5(\mathbb{R}^6)^{*}$ is a linear isomorphism.

Question

In $\mathbb{R}^6$ we take variables $x_1, x_2, x_3, y_1, y_2, y_3$ and a bilinear form $\omega \in \wedge^2(\mathbb{R}^6)^{*}$ given by: $$\omega = dx_1 \wedge dy_1 + dx_2 \wedge dy_2 + dx_3 \wedge dy_3$$

Prove that transformation: $\wedge^1 ( \mathbb{R}^6 )^{*} \ni \eta \longrightarrow \eta \wedge \omega \wedge \omega \in \wedge^5(\mathbb{R}^6)^{*}$

is a linear isomorphism.

I am just starting with a book on multivariable calculus and found such a question. From linear algebra I know that to prove that a transformation is a isomorphism it's enaugh to show that it's is one to one and it's onto. How would that tanslate to that case?

Answer 1

Hint Since $\dim \wedge^1 \Bbb R^6 = {6 \choose 1} = {6 \choose 5} = \dim \wedge^5 (\Bbb R^6)^*$, (taking for granted that the map is linear---it's straightforward to verify as much) it's enough to show that the kernel of the map is trivial. To do this, a straightforward option is to compute the image under the map of a general $1$-form $\eta = \eta_1 \,dx^1 + \cdots + \eta_6 \,dy^3 \in \wedge^1 (\Bbb R^6)^*$ in terms of the coordinate cobasis.