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Let $F = F(x,y)$, $G = G(x,y)$ be two functions on $E^2$ such that $F +iG = (x+iy)^3$. Calculate the values of $1$-forms $\omega = dF$ and $\sigma = dG$ on the vector field $A = r\partial r + \partial\phi$.

For $dF(A)$ in Cartesians I have got $3x^3 - 9xy^2$. I have been having issues calculating the value for the $1$-forms so if anyone could verify whether or not this answer is correct before I calculate $dG$ it would be greatly appreciated.

Alex
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user3450230
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  • I added latex to the question, but I'm not sure about that $E2$. Was that supposed to be $\mathbb{R}^2$? Or is there something we should know about $E$? – Alex Mar 30 '14 at 22:02
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    I guess it is $E^2$ that denotes euclidean space of dimension $2$, i.e. indeed $\Bbb R^2$. – Berci Mar 30 '14 at 22:03
  • sorry yes E2 = \mathbb{R}^2 @Alex – user3450230 Mar 30 '14 at 22:04
  • @Berci, OH! Actually, that seems pretty obvious now, making me think I've been looking at elliptic curves too much lately... :) – Alex Mar 31 '14 at 03:38

1 Answers1

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As I see it, you have a couple of options here when it comes to calculating the indicated quantities, both of which have their respective benefits. Before getting there, however, you will want to observe that $F + iG = (x + iy)^3$ implies that $F(x, y) = x^3 - 3xy^2$ and $G(x, y) = 3x^2y - y^3$.

The desired 1-forms expressed in Cartesian coordinates are thus \begin{align*} dF & = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy = \left(3x^2 - 3y^2\right) dx -6xy\,\, dy\\ &\\ dG & = \frac{\partial G}{\partial x} dx + \frac{\partial G}{\partial y} dy = 6xy\,\, dx + \left(3x^2 - 3y^2\right) dy\\ \end{align*}

Option 1: Observe that the vector fields $\partial r$ and $\partial \phi$ in cartesian coordinates as \begin{align*} \partial r &= \frac{x}{r}\frac{\partial}{\partial x} + \frac{y}{r} \frac{\partial}{\partial y}\\ \partial \phi &= -y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}, \end{align*} where $r$ is the radial distance from the origin and satisfies $r = \sqrt{x^2 + y^2}$.

Your vector field $A = r\partial r + \partial \phi$ can then be expressed in standard Cartesian coordinates as

$$ A = \left(x - y\right) \frac{\partial}{\partial x} + \left(x + y\right)\frac{\partial}{\partial y} $$

You can now evaluate $dF(A)$ and $dG(A)$ directly in cartesian coordinates using the linearity properties and the fact that the basis of vector fields $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ are dual to the basis of 1-forms $dx, dy$.

Option 2: Convert $F$ and $G$ to their representatives in polar coordinates by making the substitutions $x = r\cos \phi$ and $y = r \sin \phi$. You can then repeat the computation outlined above by first computing $dF$ and $dG$ in polar coordinates and then using using the linearity properties and the fact that the basis of vector fields $\partial r, \partial \phi $ are dual to the basis of 1-forms $dr , d\phi$.

The general observation here is that given a function in coordinates $F(x^1, x^2, \ldots ,x^n)$, the coordinate form representative of $dF$ is $$ dF = \sum\limits_{i = 1}^{n} \frac{\partial F}{\partial x^{i}} dx^i. $$

I hope that this helps address the issue.

THW
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