If $w^1$ and $w^2$ are 1-forms defined by $$w^1=\sqrt{1+y^2}dx+\frac{xy}{\sqrt{1+y^2}}dy, w^2=\sqrt{\frac{1+x^2+y^2}{1+y^2}}dy$$
Riemannian metric ($\mathbb{R^2}$) is given:
$$g=w^1\otimes w^1+w^2\otimes w^2$$
Find a $1$-form $\alpha$ s.t $$dw^1=-\alpha \wedge w^2$$ $$dw^2=\alpha \wedge w^1$$
First, I differentiate $w^1$,
$d(\sqrt{1+y^2}dx)=(0dx+\frac{y}{\sqrt{1+y^2}}dy+0dz)\wedge dx =\frac{y}{\sqrt{1+y^2}}dy \wedge dx$
$d(\frac{xy}{\sqrt{1+y^2}}dy)=\frac{y}{\sqrt{1+y^2}}dx \wedge dy$
So, $dw^1=0$. Same goes for $dw^2=\frac{x}{\sqrt{1+x^2+y^2}+\sqrt{1+y^2}}dx\wedge dy$. Can someone please guide me? Thanks!
Edit: $dw^2$ is not equal zero. Miscalculation.
First, I think you should get $dw^2=\displaystyle\frac{x}{\sqrt{1+x^2+y^2}\sqrt{1+y^2}}dx\wedge dy$ instead. To find the $1$-form $\alpha$, we let $\alpha= adx+bdy$. Then $dw^1=-\alpha \wedge w^2$ implies that $$0=-(adx+bdy)\wedge \left(\sqrt{\frac{1+x^2+y^2}{1+y^2}}dy\right) =-a\sqrt{\frac{1+x^2+y^2}{1+y^2}}dx\wedge dy,$$ which gives $a=0$. On the other hand, $dw^2=\alpha \wedge w^1$ implies that $$\displaystyle\frac{x}{\sqrt{1+x^2+y^2}\sqrt{1+y^2}}dx\wedge dy=(adx+bdy)\wedge \left(\sqrt{1+y^2}dx+\frac{xy}{\sqrt{1+y^2}}dy\right) =-b\sqrt{1+y^2}dx\wedge dy,$$ which gives $b=-\displaystyle\frac{x}{(1+y^2)\sqrt{1+x^2+y^2}}$. Therefore, we have $$\alpha=-\displaystyle\frac{x}{(1+y^2)\sqrt{1+x^2+y^2}}dy.$$