Writing a two-form as a wedge product

Question

Suppose a differential two-form $\Omega$ on $\mathbb{R}^2$ is defined by $\Omega_p(x, y)=p_2(x_1y_2-x_2y_1)$.

Then using coordinates $(p_1, p_2)$ for $\mathbb{R}^2$, this reads $$\Omega_p=\frac{1}{p_2}dp_1\wedge dp_2.$$

I don't understand why $p_2$ goes in the denominator when we use the wedge product. Can someone explain this to me please.

Answer 1

Hint: Write the tangent vectors $(r, s)_{\mathfrak{g}^*}(\mu)$ and $(u, v)_{\mathfrak{g}^*}(\mu)$ in terms of basis vectors for the tangent space of the orbit at $\mu$.

Apply $d\alpha\wedge d\beta$ to the pair of these tangent vectors and you will see that you get $\beta^2(rv-su)$.

(This is an exercise in your book at the end of the section.)

Answer 2

You sure that's it? Notice: $$(dp_1 \wedge dp_2)_{{\bf p}}({\bf x},{\bf y}) = \begin{vmatrix} (dp_1)_{{\bf p}}({\bf x}) & (dp_1)_{{\bf p}}({\bf y}) \\ (dp_2)_{{\bf p}}({\bf x}) & (dp_2)_{{\bf p}}({\bf y}) \end{vmatrix} = \begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} = \frac{1}{p_2}p_2(x_1y_2 - x_2y_1),$$hence $(dp_1 \wedge dp_2)_{{\bf p}}({\bf x},{\bf y}) = \frac{1}{p_2} \Omega_{{\bf p}}({\bf x},{\bf y})$.