I've been working on the last part of problem 9., chapter 9 in Nakahara's Geometry, Topology and Physics all day, with no success, and am in need of some assistance. We are asked to compute the Hopf invariant of $\pi:S^3\to S^2$, where given $S^3 = \{\mathbf x = (x^1,x^2,x^3,x^3)\in\mathbb R \,|\, \|\mathbf x\|=1\}$, and similarly given $S^2$ (where we write the points as $(\xi^1,\xi^2,\xi^3)$), the map $\pi$ is defined by $$\xi^1=2(x^1x^3+x^2x^4)\\ \xi^2=2(x^2x^3-x^1x^4)\\ \xi^3=(x^1)^2+(x^2)^2-(x^3)^2-(x^4)^2.$$
But consider for now just some general $f:S^{2n-1}\to S^n$. Let $\Omega_n$ be a volume form on $S^n$ normalized to 1, and consider its pullback $f^*\Omega_n$. We verify it is closed and can be written as $f^*\Omega_n=d\omega_{n-1}$, $\omega_{n-1}\in \Omega^{n-1}(S^{2n-1})$. The Hopf invariant is defined as $$H(f)\equiv \int_{S^{2n-1}}\omega_{n-1}\wedge\mathrm d \omega_{n-1}.$$
After a few more questions, we are asked to compute $H(\pi)$, with $n=2$.
My approach was to take the god given volume element $vol_{\mathbb R^3}=\mathrm d\xi^1\wedge\mathrm d\xi^2\wedge\mathrm d\xi^3$ and take the interior product with a vector field everywhere normal to the 2-sphere, $N=\sum_i \xi^i\partial_{\xi^i}$, to obtain the volume form on $S^2$, $$vol_{S^2}=\xi^1 \mathrm d\xi^2\wedge\mathrm d\xi^3-\xi^2 \mathrm d\xi^1\wedge\mathrm d\xi^3+\xi^3 \mathrm d\xi^1\wedge\mathrm d\xi^2.$$
I don't care about the normalization yet so I take this to be $\Omega_2$. However, when I pull it back with $\pi$ the result is horribly long, and I won't even post it here. After that, I have, modulo my own errors in arithmetic, the result $\mathrm d \omega_1$, but I also need $\omega_1$. Everything just seems hopeless. I could get it by solving a system of PDEs but that seems also inaccessible.
I found a similar calculation in Topology for physicists by Schwarz and Levy, p.165:
The (7.4.1) they mention is just the integral definition of the invariant, whereas (6.3.1) is a volume form for the n-sphere $$\omega = \frac{1}{(1+\sum (x^i)^2)^n} \mathrm dx^1\wedge\dots\wedge\mathrm dx^n.$$
They haven't specified $f:S^3\rightarrow S^2$ anywhere, and their entire calculation is unclear to me. How did they get the one-form $\sigma$? In one of the questions, we are asked to prove that the invariant does not depend on the choice of $\omega_{n-1}$ - so did the authors simply take $d\omega_{n-1}$ to be the volume form (6.3.1) on the 2-sphere in spherical coordinates? Still, how did they get $\omega_1=\sigma$? Can anyone explain?
And, again, is the direct approach that Nakahara asks even doable? Or does it need to be a bit more indirect :)
Well, I will be answering my question for the benefit of people who may search for this exact thing in the future. I appreciate PVAL's valuable comments. In the end, it seems I had to simplify the volume form a bit - after that it all worked out.
First of all, I normalize the form $vol_{S^2}$ from my question $$\Omega_2=\frac{1}{4\pi}(\xi^1 \mathrm d\xi^2\wedge\mathrm d\xi^3-\xi^2 \mathrm d\xi^1\wedge\mathrm d\xi^3+\xi^3 \mathrm d\xi^1\wedge\mathrm d\xi^2)$$ and note that, since $(\xi^1)^2+(\xi^2)^2+ (\xi^3)^2=1$, we have $$\xi^1\mathrm d \xi^1+\xi^2\mathrm d \xi^2+\xi^3\mathrm d \xi^3=0$$ which we can use to eliminate for example $\mathrm d\xi^3$ on any open set $\subset S^2$ where $\xi^3\neq0$. We find that $$\Omega_2=\frac{1}{4\pi\xi^3}\,\mathrm d\xi^1\wedge\mathrm d\xi^2,$$ which is much easier to pullback. After lengthy calculation, and using the same thing as before, $$\mathrm d x^4=-\frac{1}{x^4}\sum_{i=1}^3 x^i \mathrm d x^i,$$ we get $$-\pi\cdot f^*\Omega_2=\mathrm dx^1\wedge\mathrm dx^2+\frac{x^1}{x^4}\cdot\mathrm dx^1\wedge\mathrm dx^3+\frac{x^2}{x^4}\cdot\mathrm dx^2\wedge\mathrm dx^3. $$
The first term is $\mathrm d ( x^1 \mathrm d x^2).$ To find the other two, we want to find some $\tilde\omega\in\Omega^1(S^3)$ such that $\mathrm d\tilde\omega=(\frac{x^1}{x^4}\cdot\mathrm dx^1+\frac{x^2}{x^4}\cdot\mathrm dx^2)\wedge\mathrm dx^3$, so we try $\tilde\omega=f(x^1,x^2,x^3)\mathrm dx^3$. This means we have to solve $$\partial_{x^1}f=\frac{x^1}{x^4}=\frac{x^1}{\sqrt{1-(x^1)^2-(x^2)^2-(x^3)^2}}$$ and the same but with $x^1\leftrightarrow x^2.$ The solution is obviously $f=-\sqrt{1-(x^1)^2-(x^2)^2-(x^3)^2}$. To get this, we had to say $x^4\mapsto +\sqrt{1-\dots}$, and this choice of the plus sign in front of the radical is valid only on open covers of $S^3$ disjoint from the ones on which we take the negative sign (since we are using a chart on which $x^4\neq0$). But, the opposite choice also reverses the sign of the solution, so we can write in general $\tilde\omega= -x^4 \mathrm d x^3.$
Therefore, $\omega_1=-\frac{1}{\pi}(x^1 \mathrm d x^2 - x^4 \mathrm d x^3)$, and $$\omega_1\wedge\mathrm d\omega_1=\frac{1}{\pi^2}(x^1 \mathrm dx^2\wedge\mathrm dx^3\wedge\mathrm dx^4-x^4 \mathrm dx^1\wedge\mathrm dx^2\wedge\mathrm dx^3).$$ Now from symmetry, $$H(\pi)=\frac{2}{\pi^2}\int_{S^3}x^1 \mathrm dx^2\wedge\mathrm dx^3\wedge\mathrm dx^4.$$ To calculate the integral we can use spherical coordinates:
$$x^1=\sin{\alpha}\sin{\beta}\sin{\gamma} \\ x^2=\sin{\alpha}\sin{\beta}\cos{\gamma} \\ x^3=\sin{\alpha}\cos{\beta} \\ x^4=\cos{\alpha} $$
where $\gamma\in(0,2\pi)$ and $\alpha,\beta\in(0,\pi)$. So, we calculate!
$$\int_{S^3}x^1 \mathrm dx^2\mathrm dx^3\mathrm dx^4=\int \sin^4{\alpha}\sin^3{\beta}\sin^2{\gamma}\:\mathrm d\alpha\mathrm d\beta \mathrm d\gamma=\frac{3\pi}{4}\cdot\frac{4}{3}\cdot\frac{\pi}{2}=\frac{\pi^2}{2}.$$
Thus, finally, $H(\pi)=1.$