Consider the following Riemannian metric on $\mathbb{R}^2$: $$g=(1+y^2)dx\otimes dx+xy(dx\otimes dy+dy\otimes dx)+(1+x^2)dy\otimes dy $$
If $w^1$ and $w^2$ are 1-forms defined by $$w^1=\sqrt{1+y^2}dx+\frac{xy}{\sqrt{1+y^2}}dy, w^2=\sqrt{\frac{1+x^2+y^2}{1+y^2}}dy$$
then show that the Riemannian metric takes the form
$$g=w^1\otimes w^1+w^2\otimes w^2$$
Is there a way to convert $w^1$ and $w^2$ to matrices? And also can someone please guide me the way? Thanks!
As linear functionals, write $$w^1 = \begin{bmatrix} \sqrt{1+y^2} & \dfrac{xy}{\sqrt{1+y^2}}\end{bmatrix}\quad \mbox{and} \quad w^2 = \begin{bmatrix} 0 & \sqrt{\dfrac{1+x^2+y^2}{1+y^2}} \end{bmatrix}$$Then $w^1\otimes w^1$ corresponds on the matrix level to $(w^1)^\top w^1$ and similarly for $w^2$. Thus, is your goal is to show that $$(w^1)^\top w^1 + (w^2)^\top w^2 = \begin{bmatrix} 1+y^2 & xy \\ xy & 1+x^2\end{bmatrix}$$by directly computing the left side. You should be able to do this without running into any issues.