# Showing a certain form is exact

I'm trying to solve the following:

Let $f: S^{2n - 1} \rightarrow S^n$ be a smooth map, and let $\omega$ be an n-form on $S^n$ such that $\int_{S^n} \omega = 1$. Show that $f^*\omega$ is exact, and if $f^*\omega = d\alpha$, then $\int_{S^{2n-1}} \alpha \wedge d\alpha$ is independent of the choice of $\omega$ and $\alpha$.

I don't really know how to go about answering this. If I knew $f^*\omega$ was closed, then it would be exact since $S^{2n - 1}$ has no nth cohomology. For the second part, if n was odd, then we would have $d(\alpha \wedge \alpha) = -2 \alpha \wedge d\alpha$, so $\alpha \wedge d\alpha$ is exact and hence 0 independent of $\omega$ and $\alpha$.

-
I don't think that you always get zero. I would guess that at least in the case $n=1$ you get a constant times the winding number. –  Jyrki Lahtonen Aug 16 '12 at 20:45

Note that $\omega$ is an $n$-form on $S^n$, an $n$-dimensional manifold. Therefore $d\omega = 0$ as it is a $(n + 1)$-form on $S^n$. So $\omega$ is closed. Since pullbacks commute with exterior differentiation, we have that $$d(f^\ast \omega) = f^\ast d\omega = 0,$$ so $f^\ast \omega$ is closed as well. Now apply what you said about $S^{2n-1}$ having trivial cohomology in degree $n$.

Now to show that $$\int_{S^{2n-1}} \alpha \wedge d\alpha$$ is independent of the choice of $\omega$ and $\alpha$, let $\omega' \in \Omega^n(S^n)$ be another $n$-form on $S^n$ such that $$\int_{S^n} \omega' = 1$$ and let $\alpha' \in \Omega^{n-1}(S^{2n-1})$ be such that $d\alpha' = f^\ast \omega'$. We will show that $$\int_{S^{2n-1}} \alpha \wedge d\alpha = \int_{S^{2n-1}} \alpha' \wedge d\alpha',$$ which implies the integral is independent of the choices.

Since $H^n(S^n; \mathbb{R}) \cong \mathbb{R}$, by deRham's theorem there is some $\tau \in \Omega^{n-1}(S^n)$ such that $$\omega' = \omega + d\tau.$$ Now $$d(\alpha' - \alpha - f^\ast \tau) = f^\ast(\omega' - \omega - d\tau) = 0$$ and $H^n(S^{2n-1}; \mathbb{R}) \cong 0$, so again by deRham's theorem there exists some $\eta \in \Omega^{n-2}(S^{2n-1})$ such that $$\alpha' = \alpha + f^\ast \tau + d\eta.$$ Hence we have that \begin{align*} \alpha' \wedge d\alpha' & = (\alpha + f^\ast \tau + d\eta) \wedge (d\alpha + f^\ast d\tau) \\ & = \alpha \wedge d\alpha + \alpha \wedge d(f^\ast \tau) + f^\ast(\tau \wedge (\omega + d\tau)) + d(\eta \wedge (d\alpha + f^\ast d\tau)) \\ & = \alpha \wedge d\alpha + \alpha \wedge d(f^\ast \tau) + d(\eta \wedge (d\alpha + f^\ast d\tau)), \end{align*} where in going to the third line we used the fact that $f^\ast(\tau \wedge (\omega + d\tau)) = 0$ since $\tau \wedge (\omega + d\tau) = 0$ as it is a $(n + 1)$ form on $S^n$. Now, since \begin{align*} \alpha \wedge d(f^\ast \tau) & = -d(\alpha \wedge f^\ast \tau) + d\alpha \wedge f^\ast \tau \\ & = -d(\alpha \wedge f^\ast \tau) + f^\ast (\omega \wedge \tau) \\ & = -d(\alpha \wedge f^\ast \tau), \end{align*} where once again $f^\ast (\omega \wedge \tau) = 0$ since $\omega \wedge \tau$ is an $(n+1)$-form on $S^n$, we get that $$\alpha' \wedge d\alpha' = \alpha \wedge d\alpha + d(-\alpha \wedge f^\ast \tau + \eta \wedge (d\alpha + f^\ast d\tau)).$$ So by Stokes' theorem, \begin{align*} \int_{S^{2n-1}} \alpha' \wedge d\alpha' & = \int_{S^{2n-1}} (\alpha \wedge d\alpha + d(-\alpha \wedge f^\ast \tau + \eta \wedge (d\alpha + f^\ast d\tau))) \\ & = \int_{S^{2n-1}} \alpha \wedge d\alpha, \end{align*} showing that the integral is independent of the choices.

Remark: The number $$H(f) = \int_{S^{2n-1}} \alpha \wedge d\alpha$$ is called the Hopf invariant of the map $f$. Here we showed that it only depends on $f$. You can also show that it only depends on the homotopy class of $f$, which isn't too difficult.

-
Thank you. Do you have any hints for the second part? Am I correct in the case n is odd? –  Joe Aug 16 '12 at 20:44
@Matt $\alpha \wedge d\alpha$ is indeed exact when $n$ is odd, so by Stokes' theorem the integral is zero, as you correctly said. Should it be $S^{2n+1}$ instead of $S^{2n-1}$? $\alpha \wedge d\alpha$ is a $(2n+1)$-form. The case for $n$ even is going to need to use the hypothesis $\int_{S^n} \omega = 1$. I will think about it and update my answer in a little while. –  Henry T. Horton Aug 16 '12 at 20:52
It is written 2n - 1. In that case it would be trivial. It's an old qual exam, so it must be a typo. –  Joe Aug 16 '12 at 20:56
@Matt Actually I was mistaken, $d\alpha = f^\ast \omega$ so $\alpha$ is a $(n-1)$ form, and hence $\alpha \wedge d\alpha$ is a $(2n-1)$-form. I rolled back your edit since the issue was actually nonexistent. In any case, I have a solution and will type it up momentarily. –  Henry T. Horton Aug 16 '12 at 21:25
I see. Thanks again for your answer, it is very clear. –  Joe Aug 16 '12 at 22:23