Independence of an integral of a differential form on a sphere

Question

Let $n \geq 2$ and $f: S^{2n-1} \rightarrow S^{n}$ be a smooth map. Let $\omega$ be an volume form on $S^n$ such that $\int_{S^n}w = 1$.

We know that there exists a (n-1) form $\beta \in \Lambda^{n-1}(S^{2n-1})$ such that

$d\beta = f^{*}w$ ($f^*$ is the pullback along $f$).

I want to show that the expression

$$\int_{S^{2n-1}}\beta \wedge d\beta$$

is independent on the choice of $w$.

---

What i've done so far:

Since ($\int_{S^n}\cdot$) is an isomorphism on $H^n(S^n)$, we can conclude that if we take two $\omega_1$, $\omega_2$ which fulfill the given conditions, they only differ by some $d\eta$. From this follows:

$d\beta_1-d\beta_2 = f^*d\eta=df^*\eta$.

And also since $H^n(S^{2n-1}) = 0$:

$\beta_1-\beta_2 = f^*\eta + d\alpha$.

Substituting this into the integral we get:

$\int_{S^{2n-1}}\beta_1 \wedge d\beta_1$ = $\int_{S^{2n-1}}(\beta_2 + f^*\eta + d\alpha) \wedge (d\beta_2 + df^*\eta)$ = $\int_{S^{2n-1}}\beta_2 \wedge d\beta_2 + \int_{S^{2n-1}} (\beta_2 \wedge df^*\eta + f^*\eta \wedge d\beta_2 + f^*\eta \wedge df^*\eta)$.

To finish the proof my idea is to show that $$\beta_2 \wedge df^*\eta + f^*\eta \wedge d\beta_2 + f^*\eta \wedge df^*\eta$$ is an exact form. But i didn't manage to do so and maybe someone has an idea how to do so or maybe comes up with a totally different idea...

Thank you for your help :)

Addendum:

If $n$ is uneven i've managed to do the proof, since then

$$d(\beta_2 \wedge f^*\eta + \frac{1}{2} f^*\eta \wedge f^*\eta)$$

equals the requiered form.

Answer 1

The Integral over the differential form $$\beta_2 \wedge df^*\eta + f^*\eta \wedge d\beta_2 + f^*\eta \wedge df^*\eta$$ is equal to $0$:

$d\beta_2 \wedge f^*\eta = f^*(\omega_2 \wedge \eta)$ and $\omega_2 \wedge \eta = 0$, because $\omega_2, \eta$ are n/n-1-forms on $S^n$, and the only $2n-1$ form on $S^n$ is $0$.

The same holds for $f^*\eta \wedge df^*\eta$.

For $\beta_2 \wedge df^*\eta$ we observe $d(\beta_2 \wedge f^*\eta) = d\beta_2 \wedge f^*\eta + (-1)^{n-1} \beta_2 \wedge df^*\eta$.

and thus $\beta_2 \wedge df^*\eta = \pm d(\beta_2 \wedge f^*\eta)$ and the corresponding integral is equal to $0$.