# Show $D^2=0$ iff $D=e^{-f}de^{f}$ for some function $f$ , where $D\omega := d\omega+\alpha \wedge \omega$

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.

• 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? Commented Dec 28, 2014 at 21:16
• Someone accidentally posted the following comment as an edit to the post: The following may be useful: $$d(\alpha \wedge \beta)= (d\alpha) \wedge \beta + (-1)^k \alpha \wedge (d\beta)$$
– user14972
Commented Dec 29, 2014 at 10:17
• 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. Commented Jan 2, 2015 at 14:07
• Past exam paper. Why is this so interesting to you? Commented Jan 2, 2015 at 14:23
• @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! Commented Jan 2, 2015 at 18:39

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 $$e^{-f}d(e^f\omega)=e^{-f}e^fd\omega+e^{-f}d(e^f)\wedge\omega=d\omega+e^{-f}e^fdf\wedge\omega=d\omega+df\wedge\omega=d\omega+\alpha\wedge\omega=D\omega.$$
• It might be worth justifying that $d\alpha \wedge \omega = 0$ for all $\omega$ implies $d\alpha=0$. Commented Jan 2, 2015 at 18:44
• Its briefly explained in $\mathbb{R}^n$ for my course Commented Jan 2, 2015 at 20:33