# Independence of an integral of a differential form on a sphere

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 :)

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.

• There are really two steps here. First, show that the result is independent of the choice of $\beta$ and then, second, deduce that the result is independent of the choice of $\omega$. Commented May 10 at 16:40
• Okay, i showed that the result is independet of the choice of $\beta$, but i don't know what to do next to deduce the assumption... Commented May 10 at 16:53
• What you're missing is the observation that $f^*\eta\wedge d\beta_2 = f^*(\eta\wedge\omega_2) = f^*0=0$ and similarly with other terms. Commented May 10 at 21:01
• Oh yes, now i see why, thank you very much. Commented May 11 at 10:17
• I would recommend that you write/post a succinct answer to your question. :) Commented May 12 at 5:52

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 $$0$$.