1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Dec 28 '14 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 Dec 29 '14 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 Jan 2 '15 at 14:07
  • $\begingroup$ Past exam paper. Why is this so interesting to you? $\endgroup$ – Permian Jan 2 '15 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 Jan 2 '15 at 18:39
2
+50
$\begingroup$

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 $$e^{-f}d(e^f\omega)=e^{-f}e^fd\omega+e^{-f}d(e^f)\wedge\omega=d\omega+e^{-f}e^fdf\wedge\omega=d\omega+df\wedge\omega=d\omega+\alpha\wedge\omega=D\omega.$$

$\endgroup$
  • $\begingroup$ It might be worth justifying that $d\alpha \wedge \omega = 0$ for all $\omega$ implies $d\alpha=0$. $\endgroup$ – Steven Gubkin Jan 2 '15 at 18:44
  • $\begingroup$ Its briefly explained in $\mathbb{R}^n$ for my course $\endgroup$ – Permian Jan 2 '15 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.