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Let $\alpha$ be a $1$-form on $\mathbb{R}^n$. Define the following which takes $k$-forms to $(k+1)$-forms.

$$D\omega := d\omega+\alpha \wedge \omega $$

Show that $D^2=0$ iff $D=e^{-f}de^{f}$ for some function $f$ and express $\alpha$ in terms of $f$.

I get that

$$\begin{align} D^2 \omega=D(D(\omega) &= D(d\omega + \alpha \wedge \omega) \\ &= d(\alpha \wedge \omega)+ \alpha \wedge d\omega + \alpha \wedge \alpha \wedge \omega \\ &= d\alpha \wedge d\omega + \alpha \wedge d\omega + \alpha \wedge \alpha \wedge \omega \end{align}$$

but I cannot see how to proceed. I also find it strange that first term is a $(k+3)$-form.

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  • $\begingroup$ The first term is a $(k+2)$-form, and is something you can compute further. Can you show that $D^2 = 0$ iff $\alpha$ is closed? $\endgroup$
    – Santiago Canez
    Commented Dec 28, 2014 at 21:16
  • $\begingroup$ Someone accidentally posted the following comment as an edit to the post: The following may be useful: $$d(\alpha \wedge \beta)= (d\alpha) \wedge \beta + (-1)^k \alpha \wedge (d\beta)$$ $\endgroup$
    – user14972
    Commented Dec 29, 2014 at 10:17
  • $\begingroup$ This arises in the context of the curvature of a connection on a line bundle. I wonder where you saw this exercise. Is it in a book? A set of lecture notes? I would be curious to see whatever resource you got it from. $\endgroup$
    – Steven Gubkin
    Commented Jan 2, 2015 at 14:07
  • $\begingroup$ Past exam paper. Why is this so interesting to you? $\endgroup$
    – Trajan
    Commented Jan 2, 2015 at 14:23
  • $\begingroup$ @1234 I am interested in teaching this stuff well, and this is a good problem. I treasure good problems. Where there is one, there is usually another. So I was curious if this resource was available somewhere. It looks like not. Thanks anyway! $\endgroup$
    – Steven Gubkin
    Commented Jan 2, 2015 at 18:39

1 Answer 1

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By definition of $D$, we get $$\begin{align} D^2 \omega=D(D(\omega) &= D(d\omega + \alpha \wedge \omega) \\ &= d(\alpha \wedge \omega)+ \alpha \wedge d\omega + \alpha \wedge \alpha \wedge \omega \\ &= d\alpha \wedge \omega-\alpha \wedge d\omega + \alpha \wedge d\omega + \alpha \wedge \alpha \wedge \omega\\ &= d\alpha \wedge \omega , \end{align}$$ where the second last equality follows from $d(\gamma \wedge \beta)= (d\gamma ) \wedge \beta + (-1)^k \gamma \wedge (d\beta)$ when $\gamma $ is a $k$-form, and the last equality follows from $\alpha \wedge \alpha =-\alpha \wedge \alpha =0$ since $\alpha$ is a $1$-form.

Then $D^2=0\iff d\alpha=0$ $\iff$ $\alpha$ is closed $1$-form in $\mathbb{R}^n$ $\iff$ $\alpha$ is exact (since $H^1_{DR}(\mathbb{R}^n)=0$) $\iff$ there exists a smooth function $f:\mathbb{R}^n\to\mathbb{R}$ such that $\alpha=df$.

Now, if $\alpha=df$, we can see that $D=e^{-f}de^f$, because $$e^{-f}d(e^f\omega)=e^{-f}e^fd\omega+e^{-f}d(e^f)\wedge\omega=d\omega+e^{-f}e^fdf\wedge\omega=d\omega+df\wedge\omega=d\omega+\alpha\wedge\omega=D\omega.$$

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  • $\begingroup$ It might be worth justifying that $d\alpha \wedge \omega = 0$ for all $\omega$ implies $d\alpha=0$. $\endgroup$
    – Steven Gubkin
    Commented Jan 2, 2015 at 18:44
  • $\begingroup$ Its briefly explained in $\mathbb{R}^n$ for my course $\endgroup$
    – Trajan
    Commented Jan 2, 2015 at 20:33

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