Non isomorphic principal $G$-bundles

I think the "standard" argument would be that the Hopf fibration is classified by a nontrivial element of $H^1(S^2, S^1) \cong H^2(S^2, \mathbb{Z}) \cong \mathbb{Z}$ (I think more specifically by twice a generator but don't quote me on that) so the inverted bundle is presumably classified by the negative of this element, and since $\mathbb{Z}$ has no $2$-torsion no such bundle is equivalent to its inverse. Probably there's something to learn from trying to show that the diffeomorphism $\Phi$ doesn't exist more concretely though.

I apologize in advance for the long post, but, as Qiaochu mentioned, I wanted to learn something from the bare hands approach.

To set notation, I'm going to view $S^3\subseteq \mathbb{C}^2$ as the pairs $(z_1,z_2)$ of unit length vectors, so $|z_1|^2 + |z_2|^2 = 1$. I'm going to view $S^2\subseteq \mathbb{C}\oplus \mathbb{R}$ consisting of unit length vectors.

Then the Hopf map $\pi:S^3\rightarrow S^2$ is (according to Wikipedia), given by $\pi(z_1,z_2) = (2z_1\overline{z}_2, |z_1|^2 - |z_2|^2)$.

Let $U = \{(w,t)\in S^2: t > -\epsilon\}$ and $V = \{(w,t)\in S^2: t < \epsilon\}$ where I'm thinking of $\epsilon$ as some fixed very small positive number. Intuitively, $U$ is the northern hemisphere of $S^2$, except extended slightly below the equation, and $V$ is likewise essentially the southern hemisphere.

Proposition 1: The open sets $U$ and $V$ form a trivializing cover for the Hopf bundle.

Proof: Let's start with $\pi^{-1}(U)$. We'll start by finding a section $s_U:U\rightarrow \pi^{-1}(U)$. So, given $(w,t)\in U$, we want to associate to it $(z_1,z_2)\in \pi^{-1}(U)$. Let's try making the simplifying assumption that $z_1$ is real and positive.

So, we are solving $(2z_1 \overline{z}_2, z_1^2 - |z_2|^2) = (w, t)$ for $(z_1,z_2)$ under the assumption that $z_1$ is real. The first equation $2z_1 \overline{z}_2 = w$ can be solved for $z_2$, getting $z_2 = \frac{\overline{w}}{2z_1}$. Substituting this into the equation $z_1^2 - |z_2|^2 = t$, clearing denominators, we get a quadratic in $z_1^2$. Using the quadratic formula, together with the fact that $z_1 > 0$, we find $$z_1 = \sqrt{\frac{t+1}{2}}, \text{ and } z_2 =\frac{\overline{w}}{\sqrt{2t + 2}}.$$ Thus, our section $s_U$ is given by $s_U(w,t) = (z_1,z_2)$, with formulas for $z_1$, $z_2$ defined above. I will leave it to you to verify it's a section.

Armed with this section $s_U$, we define $f_U: U\times S^1\rightarrow \pi^{-1}(U)$ by $f_U(w,t,z) = s_U(w,t)z$. I'll leave it to you to verify that $f_U$ is an $S^1$-equivariant diffeomorphism with inverse $f^{-1}(z_1,z_2) = \left(\pi(z_1, z_2), \frac{z_1}{|z_1|}\right).$

In a similar fashion, we have a section $s_V:V\rightarrow \pi^{-1}(V)$ given by $s_V(w,t) = (z_1,z_2)$ with $$z_1 = \frac{w}{\sqrt{2-2t}} \text{ and } z_2 = \sqrt{\frac{1-t}{2}}.$$ This gives a trivialization $f_V:V\times S^1\rightarrow \pi^{-1}(V)$ given by $f_V(w,t,z) = s_V(w,t)z$ with inverse $f_V^{-1}(z_1,z_2) = \left(\pi(z_1,z_2), \frac{z_2}{|z_2|}\right).$ $\square$

Proposition 2: The composition $f_V^{-1}\circ f_U$ maps $(w,t,z)$ to $(w,t, \frac{\overline{w}}{|w|} z$.

Proof: We compute. \begin{align*} f_V^{-1}(f_U(w,t,z)) &= f_V^{-1}( s_U(w,t)z) \\ &= f_V^{-1}\left(\sqrt{\frac{t+1}{2}}z, \frac{\overline{w}}{\sqrt{2t+2}}z\right) \\ &= \left(\pi\left(\sqrt{\frac{t+1}{2}}z,\frac{\overline{w}}{\sqrt{2t+2}}z\right) , \frac{\frac{\overline{w}}{\sqrt{2t+2}}z}{\left|\frac{\overline{w}}{\sqrt{2t+2}}z\right|}\right) \\&= \left(w,t, \frac{\overline{w}}{|w|} z\right).\end{align*} $\square$

Using this, we can view $S^3$ as $(U\times S^1) \coprod (V\times S^1)/\sim$ where $(w,t,z)\in U\times S^1$ is identified with $(w,t, \frac{\overline{w}}{|w|} z)\in V\times S^1$ for any $t\in(-\epsilon,\epsilon)$. Using this description, the projection map $\pi$ is simply projection onto the $w$ and $t$ coordinates.

Now, let's show $\Phi$ cannot exist. To that end, let's assume $\Phi$ does exist. Note $(\ast)$ implies that $\Phi$ maps fibers to fibers. In particular, in our above description, $\Phi$ is given by a pair of maps $\Phi_U:U\times S^1\rightarrow U\times S^1$, $\Phi_V:V\times S^1\rightarrow V\times S^1$ which respect $\sim$.

Since $\Phi$ preserves each fiber, $\Phi_U(w,t,z) = (w,t,\phi_U(w,t,z))$ for some function $\phi_U$.

Proposition 3: The function $\phi_U$ has the property that $\phi_U(w,t,z) = \phi_U(w,t,1)z^{-1}$.

Proof: Using $(\ast\ast)$, we know that $$\Phi_U(w,t,z) = \Phi_U((w,t,1)z) = \Phi_U(w,t,1)z^{-1} = (w,t,\phi_U(w,t,1))z^{-1} = (w,t,\phi_U(w,t,1)z^{-1}).$$ On the other hand, $\Phi_U(w,t,z) = (w,t,\phi_U(w,t,z))$. Thus $\phi_U(w,t,1)z^{-1} = \phi_U(w,t,z)$ as claimed. $\square$

Of course, the above discussion applies equally well to $\Phi_V$. In particular, Proposition 3 is also true for $\phi_V$.

We'll now use the fact that $\Phi$ is well defined to find a relationship between $\phi_U$ and $\phi_V$.

Proposition 4: We have $\phi_U(w,0,1)\overline{w} = \phi_V(w,0,1)w$.

Proof: Since $(w,t,z)\in U\times S^1$ is identified with $(w,t,\frac{\overline{w}}{|w|}z)$ in $V\times S^1$, we must have $[\Phi_U(w,t,z)] = [\Phi_V(w,t,\frac{\overline{w}}{|w|} z]$ for all $(w,t,z)$ with $t\in (-\epsilon,\epsilon).$ Set $t=0$ (so $|w| = 1$) and set $z=1$.

Now, $\Phi_U(w,0,1) = (w,0,\phi_U(w,0,1))\in U\times S^1$, and so $(w,0,\phi_U(w,0,1))\sim (w,0, \overline{w} \phi_U(w,0,1))\in V\times S^1$. Since $\Phi_V(w,0, \overline{w} ) = (w,0, \phi_V(w,0, \overline{w}))$, the condition that $\Phi$ respect $\sim$ implies that $$(w,0,\overline{w}\phi_U(w,0,1)) = (w,0,\phi_V\left(w,0,\overline{w})\right).$$

Using Proposition 3 on the last coordinate, we conclude $\phi_U(w,0,1)\overline{w} = \phi_V (w,0,1)w$ as claimed. $\square$

We are now ready to reach a contradiction. Namely, we claim that $\phi_U(w,0,1)\overline{w} =\phi_V(w,0,1)w$ is contradictory. Viewing $\phi_U(\cdot,0,1):S^1\rightarrow S^1$, the degree of this map must be $0$ because $\phi_U$ extends to the disk $U$. Likewise, the degree of $\phi_V(\cdot,0,1)$ is $0$. Thus, the maps $\phi_U(\cdot, 0,1)$ and $\phi_V(\cdot,0,1)$ are homotopy to constants. It now follows from the equation $\phi_U(w,0,1)\overline{w} = \phi_V(w,0,1)w$ that the maps $w\mapsto \overline{w}$ and $w\mapsto w$ are homotopic. This is absurd since one has degree $1$ while the other has degree $-1$. This contradiction establishes that $\Phi$ cannot exist.