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Let $E_1$ and $E_2$ be vector fields on $U = \mathbb{R}^2\setminus (0,0)$ and let $\epsilon_1, \epsilon_2$ be the one-forms dual to $E_1$ and $E_2$ respectively. Suppose we have $$E_1(x,y) = \frac{x}{\sqrt{x^2+y^2}}\frac{\partial}{\partial x} + \frac{y}{\sqrt{x^2+y^2}}\frac{\partial}{\partial y}$$ and $$E_2(x,y) = \frac{-y}{\sqrt{x^2+y^2}}\frac{\partial}{\partial x} + \frac{x}{\sqrt{x^2+y^2}}\frac{\partial}{\partial y}$$

$\textbf{Question:}$ How do we compute $\epsilon_1$ and $\epsilon_2$?

Salazar_3854708
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  • Whats the metric? Euclidean? Then $\frac\partial{\partial x}= dx$ and $\frac\partial{\partial y}=dy$. – s.harp Jun 30 '20 at 17:14
  • I'm not sure what that means. Yes, $dx$ is dual to $\frac{\partial}{\partial x}$ and so forth, but how does that tell me what $\epsilon_1$ is? Remember, it has to be dual to $E_1$ we written above. – Salazar_3854708 Jun 30 '20 at 17:20
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    Use linearity at every point to get:$$\epsilon_1(x,y) = E_1(x,y)^* = \frac x{\sqrt{x^2+y^2}}(\frac{\partial}{\partial x})^+ \frac y{\sqrt{x^2+y^2}}(\frac{\partial}{\partial y})^ = \frac x{\sqrt{x^2+y^2}}dx+ \frac y{\sqrt{x^2+y^2}}dy$$ – s.harp Jun 30 '20 at 17:27
  • The matrix solution is elaborated upon here. – Antoni Parellada Jul 01 '22 at 01:25

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You just get a system of equations to solve. Set $\epsilon_1 = a\,dx + b\,dy$ and $\epsilon_2 = c\,dx + e\,dy$. Writing $\epsilon_i(E_j) = \delta_{ij}$ gives you a system of linear equations for $a,b,c,e$. (There are fancier ways to do this with change-of-basis matrices, but I won't bother.)

You should get \begin{align*} ax + by &= \sqrt{x^2+y^2} \\ cx + ey &= 0 \\ -ay + bx &= 0 \\ -cy + ex &= \sqrt{x^2+y^2}. \end{align*} I leave you to solve.

COMMENT: I usually come at this the other way. If I start with polar coordinates $(r,\theta)$, then $dr = \frac xr\,dx + \frac yr\,dy$ and $r\,d\theta = -\frac yr\,dx + \frac xr\,dy$. Can you get the dual vector fields from this?

Ted Shifrin
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