Find a $1$-form $ω$ on $\mathbb R^2 −\{(0,0)\}$ such that $ω(X) = 1$ and $ω(Y) = 0$.

Question

Please ı dont know what I need to do. thus, help me to solve.

Answer 1

For a 1-form $\newcommand{\pp}[2]{\frac{\partial #1}{\partial #2}}\mathrm{d}\xi$, acting on a vector field $\displaystyle\mathbf{v} = v_1 \pp{}{x} + v_2 \pp{}{y}$ means taking the derivative along the $\mathbf{v}$: $$ \mathrm{d}\xi(\mathbf{v}) = v_1 \pp{\xi}{x} + v_2 \pp{\xi}{y} $$ Hence let $\omega = f(x,y)\mathrm{d}x + g(x,y)\mathrm{d}y$, then $$ \omega(\mathbf{X}) = f\mathrm{d}x(\mathbf{X}) + g\mathrm{d}y(\mathbf{X}) =(-fy\pp{x}{x} + fx\pp{x}{y}) + (-gy\pp{y}{x} + gx\pp{y}{y}) = -fy + gx = 1 $$ Similarly we have: $$ \omega(\mathbf{Y}) = fx + gy = 0 $$ Then solving the following linear system for $f$ and $g$ will give you the answer: $$ \begin{cases} -yf(x,y) + xg(x,y) = 1 \\ xf(x,y) + yg(x,y) = 0 \end{cases} $$ and we have: $$ \omega = \frac{-y}{x^2+y^2} \mathrm{d}x + \frac{x}{x^2+y^2} \mathrm{d}y $$