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Let $J$ be the open interval $J = (0, 1) \subset \mathbb{R}$

Let the 1-differential form $\alpha =\Big( \dfrac{\log(x)}{1-y} + \dfrac{\log(1-x)}{y} \Big) \in \Omega^0(J^2)$

I want to compute $d\alpha$

I suppose I'd have to use the formula $d\alpha = \dfrac{\partial\alpha}{\partial x}dx + \dfrac{\partial\alpha}{\partial y}dy $

Heres's my attempt

$\dfrac{\partial\alpha}{\partial x}dx + \dfrac{\partial\alpha}{\partial y}dy = \Big(\dfrac{1}{x-xy} - \dfrac{1}{y-xy}\Big)dx + \Big(\log(x)(-\dfrac{1}{(1-y)^2}\log(x) - \dfrac{1}{y^2}\log(1-x)\Big)dy$

Of course, I feel like something is wrong.

Edit

Now I've almost entirely changed my post, but how would I compute $d\alpha$ if alpha was indeed a 1-form and not a 0-form. ($\alpha =\Big( \dfrac{\log(x)}{1-y} + \dfrac{\log(1-x)}{y} \Big)dy)$

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    As written, $\alpha$ is not a one-form, it is a function. – 2017-01-06
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    @user2520938 I haven't studied manifolds. And I don't get 0 – 2017-01-06
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    @MichaelAlbanese Yes I've edited my post and changed my computaion – 2017-01-06
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    I'm not going to verify your computation, you can have wolfram alpha do that. You should note though, that you start with a function, and the output is a $1$-form. – 2017-01-06
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    @user2520938 Thank you – 2017-01-06
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    @SylvesterStallone: I have changed $\Omega^1$ into $\Omega^0$, in order for the formulae to match the wording. – 2017-01-06
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    To answer your edit, you would use $d\alpha = \frac{\partial\alpha}{\partial x} dx \wedge dy$. Read up some more information [here](http://mathworld.wolfram.com/ExteriorDerivative.html). – 2017-01-06
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    @Zorawar No, this is not correct since $dx \wedge dy$ is a $2$-form and $d\alpha$ is just a $1$-form. EDIT: Ah I get it, $\alpha$ was changed to a $1$-form, didn't see that. – 2017-01-06
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    @MarvinF: In the edit they have asked what to do if $\alpha$ were a 1-form (although they actually say if $d\alpha$ were a 1-form, but I'm taking that to be a typo...) EDIT: No worries :) – 2017-01-06

1 Answers 1

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It is nearly correct. You have the $0$-form (i.e. a function) $\alpha =\Big( \dfrac{\log(x)}{1-y} + \dfrac{\log(1-x)}{y} \Big)$ and you want to compute the $1$-form $\text{d}\alpha$. Be clear that the exterior differential (one also calls this Cartan differential) of a $p$-form is given by a $(p+1)$-form. This is immediate from the representation theorem for $p$-forms.

Yes, $\text{d}\alpha = \dfrac{\partial\alpha}{\partial x}dx + \dfrac{\partial\alpha}{\partial y}dy$ is correct.

About your attempt

$$\dfrac{\partial\alpha}{\partial x}\text{d}x + \color{red}{\dfrac{\partial\alpha}{\partial y}}\text{d}y = \Big(\dfrac{1}{x-xy} - \dfrac{1}{y-xy}\Big)\text{d}x + \Big(\color{red}{\log(x)(-\dfrac{1}{(1-y)^2}\log(x) - \dfrac{1}{y^2}\log(1-x)}\Big)\text{d}y$$

The red thing is not correct since $\left(\frac{1}{1-y} \right)'=\frac{1}{(1-y)^2}$ because of the chain rule and therefore you get

$$\partial_y \alpha=\frac{\log(x)}{(1-y)^2}-\frac{\log(1-x)}{y^2}$$

instead of the red part above. Additionally you messed up something in the red part with the parenthesis.

EDIT: To answer the second part, if you have a $1$-form $$\beta =\Big( \dfrac{\log(x)}{1-y} + \dfrac{\log(1-x)}{y} \Big)dy=:B\ \text{d}y$$ then its exterior differential is given by the $2$-form

$$\text{d}\beta=\text{d}B \wedge \text{d}y=\partial_x B \ \text{d}x \wedge \text{d}y + \partial_y B \ \text{d}y \wedge \text{d}y=\partial_x B \ \text{d}x \wedge \text{d}y$$

where we used that $\text{d}{B}=\partial_x B \ \text{d}x+\partial_y B \ \text{d}y$ (as seen above for $\alpha$) and $\text{d}y \wedge \text{d}y=0$.