Is $H([f]) = \int_{S^{2n - 1}} \alpha \wedge d\alpha$ independent of all choices, defines a map $H: \pi_{2n - 1}(S^n) \to \mathbb{Z}$? Let $[f] \in \pi_{2n - 1}(S^n)$. Choose a smooth representative $f: S^{2n - 1} \to S^n$. Let $\omega$ be a smooth $n$-form on $S^n$ with$$\int_{S^n} \omega = 1.$$We know that$$f^*\omega = d\alpha$$for some $(n - 1)$-form $\alpha$ on $S^{2n - 1}$. With preceding notation, define$$H([f]) = \int_{S^{2n - 1}} \alpha \wedge d\alpha.$$Two questions.


*

*Is $H$ is independent of all choices?

*How do I see that $H$ defines a map $H: \pi_{2n - 1}(S^n) \to \mathbb{Z}$?

 A: One thing that you haven't mentioned is that $n$ must be at least $2$, otherwise the form $\alpha$ may not even exist: you need $H^n(S^{2n-1};\mathbb{R}) = 0$, but this isn't true for $n = 1$. I'll answer the first question, it's quite long already.
There are three choices made: first the choice of $f : S^{2n-1} \to S^n$ representing $[f] \in \pi_{2n-1}(S^n)$, then the choice of a volume form $\omega$ on $S^n$, and finally the choice of $\alpha \in \Omega^{n-1}(S^{2n-1})$ such that $f^* \omega = d\alpha$.

First fix $f$ and $\omega$. To see that the choice of $\alpha$ doesn't matter, let $\alpha'$ be such that $f^* \omega = d\alpha'$ too. Then $d(\alpha-\alpha') = 0$, so $\alpha-\alpha'$ is closed; but $H^{n-1}(S^{2n-1};\mathbb{R}) = 0$, thus $\alpha - \alpha' = d\beta$ for some $\beta \in \Omega^{n-2}(S^{2n-1})$. But then:
$$\int_{S^{2n-1}} \alpha \wedge d\alpha = \int_{S^{2n-1}} (\alpha'  + d\beta) \wedge d(\alpha' + d\beta) = \int_{S^{2n-1}} \alpha' \wedge d\alpha' + \int_{S^{2n-1}} d\beta \wedge d\alpha'.$$
But $d\beta \wedge d\alpha' = d(\beta \wedge d\alpha')$, and so by Stoke's theorem ($\partial S^{2n-1} = \varnothing$) the second integral is zero and the result is the same for $\alpha$ and $\alpha'$.

Now fix $f$ and suppose you have two volume forms $\omega$ and $\omega'$ on $S^n$, with $f^*\omega = d\alpha$ and $f^*\omega = d\alpha'$ (in both cases the choice of $\alpha$ and $\alpha$ will give the same result). Since $\int_{S^n} (\omega - \omega') = 0$, there is some form $\theta \in \Omega^{n-1}(S^n)$ such that $\omega - \omega' = d\theta$. This implies that
$$f^*(\omega - \omega' - d\theta) = 0 = d(\alpha - \alpha' - f^*\theta)$$
and since $H^{n-1}(S^{2n-1};\mathbb{R}) = 0$, this means that $\alpha - \alpha' - f^*\theta = d \tau$ for some $\tau \in \Omega^{n-1}(S^{2n-1})$. So now we compute:
\begin{align}
d\alpha \wedge \alpha
& = d(\alpha' + f^*\theta + d\tau) \wedge (\alpha' + f^*\theta + d\tau) \\
& = d\alpha' \wedge \alpha' + d\alpha' \wedge f^*\theta + d\alpha' \wedge d\tau + f^*d\theta \wedge \alpha' \\
& \quad + f^*(d\theta \wedge \theta) + f^* d\theta \wedge d \tau + d\tau \wedge \alpha + d\tau \wedge f^*\theta + d\tau \wedge d\tau \\
& = d\alpha' \wedge \alpha' + f^*(\omega \wedge \theta + d\theta \wedge \theta) + f^* d\theta \wedge \alpha' + d(\tau \wedge (d\alpha' + f^* d\theta))
\end{align}
Now for degree reasons, $\omega \wedge \theta + d\theta \wedge \theta \in \Omega^{2n-1}(S^n)$ is zero. The integral over $S^n$ of the last part is zero by Stoke's theorem, and $f^*d\theta \wedge \alpha' = d(f^* \theta \wedge \alpha')$ (again, by degree reasons the leftover term is zero), so by Stoke's theorem its integral over $S^n$ is zero. So finally $\int_{S^n} d\alpha \wedge \alpha = \int_{S^n} d\alpha \wedge \alpha$.

Finally, suppose you have two representatives $f,f' : S^{2n-1} \to S^n$. This means they are continuously homotopic. The homotopy between the two can be approximated by a smooth homotopy, say $F : S^{2n-1} \times [0,1] \to S^n$. Then $F^*\omega = dA$ for some $A \in \Omega^n(S^{2n-1} \times [0,1])$. Let $\alpha$ and $\alpha'$ be the restriction of $A$ to $S^{2n-1} \times 0$ and $S^{2n-1} \times 1$, respectively, then $f^* \omega = \alpha$ and $(f')^*\omega = \alpha'$ (because $F(-,0) = f$ and $F(-,1) = f'$). Applying Stoke's theorem to $\int_{S^{2n-1} \times [0,1]} dA \wedge A$ then shows $\int_{S^{2n-1}} d\alpha \wedge \alpha = \int_{S^{2n-1}} d\alpha' \wedge \alpha'$ (the leftover term is a pullback of a $(2n-2)$-form on $S^n$, hence zero), and we're done.
